Solution Manual Mechanical Vibration: Analysis, Uncertainties, and Control, Fourth Edition
Haym Benaroya
Mark L. Nagurka
Seon Han
Donald Zellman
Stephen Kuchnicki
Contents 2 Single Degree-of-Freedom Vibration: Undamped Vibration
1
3 SDOF VIBRATION: DAMPED VIBRATION
105
4 SDOF VIBRATION: GENERAL LOADING
185
5 ANALYTICAL DYNAMICS
261
6 Multi Degree-of-Freedom Vibration
357
7 Continuous Models for Vibration
523
8 Advanced Continuous Models for Vibration
669
9 Single Degree-of-Freedom Vibration: Probabilistic Loadings
723
10 Vibration Control
801
11 Nonlinear Vibration
829
v
Chapter 2 Single Degree-of-Freedom Vibration: Undamped Vibration Problems for Section 2.2 — Deterministic Modeling 1. The beam in Figure 2.30 vibrates as a result of loading not shown. State the necessary assumptions to reduce this problem to a one degree-of-freedom oscillator. Then derive the equation of motion.
Figure 2.30: Vibrating beam supported by springs.
Solution: In order for the system to be approximated by a one-DOF oscillator, one spatial coordinate is required to define its location. We need to assume that (i) the beam does not deform but moves as a rigid body, and (ii) the deflection at each end of the beam is the same, implying a symmetric loading (or only a moment is applied and the rotation at one end is equal and opposite to that at the other end). Under these limitations, a SDOF system can be used to model this behavior. Let () be the external force acting upward on the beam. Then, using Newton’s second law, X 2 = () − 2 = 2 2 2 + 2 = () 1
2
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
The system a natural frequency =
p 2
3
2. If a beam is supported continuously on a foundation, as shown in Figure 2.31, damping must be added to an idealized model to represent the viscous effects of the mat foundation. How would you idealize this system as a one degree-of-freedom oscillator? Derive the equation of motion.
Figure 2.31: Vibrating beam on mat foundation.
Solution: To be modeled as a one DOF system, it must be possible to describe the motion using one coordinate. We assume the beam is rigid and does not deform. We also assume that the mat behaves uniformly and has the same effect along the whole beam. Finally, the loading must be symmetrical so that the beam will translate only (although a pure rotation instead is also a possible single behavior). Given these assumptions, the equivalent system is the single mass oscillating in a vertical direction, where is the equivalent total stiffness of the mat, and is the total damping of the mat. Let () be the displacement, and () be the external force defined positive upward. The force must be distributed uniformly to avoid rotation. Then, ↑
X
=
() − − −
2 2
2 = 2
2 + + = () − 2 p The system has a natural frequency = It should be noted that we measure the displacement from the static equilibrium position, the weight term can be omitted. For example, the static displacement due to gravity is = Let () be the displacement measured from the static position (d for dynamic displacement). Then, the total displacement can be written as () = () + Upon substituting this into the equation of motion, we obtain
2 + = () + 2
4
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
3. An idealized one degree-of-freedom model is tested many times in order to estimate its natural frequency. It is relatively straightforward to measure its mass , but stiffness can only be measured approximately. How might one use the natural frequency data to estimate ? Solution: If we have test data on the natural frequency and we know we can then use this data to estimate the value of by using the average frequency in the equation = 2 How do we get the frequency data? By weighing the object, we can obtain the mass = where is the weight. Then we hang the weight on the spring, displace it and measure the period. This is actually the damped period. However, if damping is small, the undamped period, will be approximately equal to the damped period. Then = 2 From this we have again.
5
4. The cantilever beam in Figure 2.32 undergoes harmonic oscillation, being driven at its free end by a force of amplitude and nominal frequency . An examination of a long timehistory of the response shows slight fluctuations about an exact harmonic response. If we divide this long time-history into segments of one period (2) and superpose these, we obtain the set of curves shown in Figure 2.33. How serious are such fluctuations in the response of the beam? Is there a way to relate the magnitude of the fluctuation to the maximum response? What would be a reasonable way to specify the value of ?
Figure 2.32: Cantilever beam.
Figure 2.33: Overlapping time-histories.
Solution: These fluctuations do not seem significant based on a qualitative assessment of deviations, although there may be applications where the variations in loading might be important. For most applications, the beam motion is not heavily dependent on such minor deviations. We may quantify the magnitude of the fluctuations by measuring the amplitudes at one instant of time. For example, at one instant of time, the largest and smallest amplitudes might be and The fluctuation at this instant is defined as ∆ = ( − ) We can obtain a measure of the importance of the fluctuation by relating it to the amplitude of the response : ∆ Then, ∆ × 100 provides a percent fluctuation. A small percent signifies minor fluctuations. The frequency can be estimated from the curves by averaging the periods and then using the average period to calculate the frequency = 2
6
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
5. For each nonlinear equation of motion, linearize the equation about the indicated equilibrium position and discuss the range of validity of the linearized equation. Plot the relative error for the linearized equation of motion. Analytically solve the linearized equation of motion. (a) ̈ + 3 cos = 0 = −2 and zero initial conditions. (b) ̈ + 3 sin = 0 = 0
(0) = 05 and ̇ (0) = 0
(c) ̈ + 3 cos2 = 0 = −2 and zero initial conditions.
(d) ̈ + 33 = 3 √ (e) ̈ + = 1
= 1
and zero initial conditions.
= 1
and zero initial conditions.
Solution: The equation of motion should be linearized about the equilibrium position because the system oscillates about the equilibrium position. (a) ̈ + 3 cos = 0 (0) = 0 ̇(0) = 0 The equilibrium position is obtained by setting ̈ = 0 such that 3 cos = 0 Then, = · · · −2 2 32 · · · The numerical solution of this system for the initial conditions given shows that it oscillates about = −2. The nonlinear term, cos has a Taylor expansion cos ∼ cos ( ) − sin ( ) ( − ) + Near = −2 cos is approximated as ³ ´³ ³ ´ ´ − sin − + cos ' cos − 2 2 2 = + 2 Then, the linearized equation of motion is given by ³ ´ ̈ + 3 + = 0 (0) = 0 ̇(0) = 0 2
The percent relative error is defined by
¯ ¯ ¯ () − () ¯ ¯ ¯ relative error = ¯ ¯ ()
For this case, relative error is given by
³ ´¯ ¯ ¯ cos − + ¯ ¯ 2 ¯¯ relative error = ¯¯ ¯ cos ¯ ¯
7 and is plotted next.
The relative error is zero at = −2 where the nonlinear term is expanded about. The relative error is less than 10% approximately for −233 −081 rad Let us now compare the response obtained from the linearized and nonlinear equations of motion. Upon substituting cos with + 2 the linearized equation of motion becomes
̈ + 3 = −
3 2
whose analytical solution is possible. Solving the nonlinear equation of motion requires a numerical integration. Below shows a sample MATLAB program that solves and plots the response plots of the linear and nonlinear models. Note that MATLAB requires the EOMs to be in a state-space form. Figure below shows Matlab code and the results. The solution to the linear model oscillates at higher frequency.
8
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
(b) ̈ + 3 sin = 0 (0) = 05 ̇(0) = 0 The equilibrium position is at = for integer The initial conditions are such that the response oscillates about = 0 The nonlinear term, sin is linearized as sin ∼ = The relative error is ¯ ¯ ¯ sin − ¯ ¯ ¯ relative error = ¯ sin ¯ The relative error is plotted below, and we can estimate that the relative error is less than 10% for −075 075
9
The linearized EOM is ̈ + 3 = 0 (0) = 05 and ̇(0) = 0. The solution is () = 05 cos 1 7321 Let us plot these two solutions. Note that the solution to the linearized equation works very well.
(c) ̈ + 3 cos2 = 0 (0) = 05 ̇(0) = 0 The static equation is given by 3 cos2 = 0 The equilibrium positions are = · · · −2 2 32 · · · The next figure shows the Matlab code and the response of the original nonlinear system. The response of the system is unstable, and it does not oscillate about any of the possible stable equilibrium positions.
10
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
The nonlinear term, cos2 , is approximated as a linear function using a Taylor series, () ' ( ) + 0 ( ) ( − ) Since the response does not oscillate about a stable equilibrium position, the best option is to estimate the nonlinear term near the initial position (0) = 05 Then, the nonlinear term, cos2 expanded near = 05 is given by cos2 ' 0770 − 0842 ( − 05) = 119 − 0842 The relative error is given by
and plotted next.
¯ 2 ¯ ¯ cos − (119 − 0842) ¯ ¯ relative error = ¯¯ ¯ cos2
The relative error is never less than 10% for 0125 0873 The next figure shows the response of the nonlinear and the linearized model. The difference will grow to indefinitely.
11
(d) ̈ + 33 = 3 (0) = 05 ̇(0) = 0 The static equation is 33 = 3 and the equilibrium position is at = 1 The Taylor series up to the linear term is given by () ' ( ) + 0 ( ) ( − ) The nonlinear term in the equation is () = 3 which has a linear approximation of 3 ' 1 + 3 ( − 1) = 3 − 2 The relative error is given by ¯ ¯ 3 ¯ − (3 − 2) ¯ ¯ relative error = ¯¯ ¯ 3
Figure next shows the % relative error.
12
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
The relative error is zero at = 1 where the nonlinear term is expanded about. The relative error is less than 10% approximately for 085 124 rad The linearized equation of motion is obtained by substituting 3 with 3 − 2 or ̈ + 3 (3 − 2) = 3 ̈ + 9 = 9 The solution to the linearized EOM and the original nonlinear EOM are plotted below.
√ (e) ̈ + = 1 (0) = 05 ̇(0) = 0 The static equation is
p = 1
and the equilibrium position is at = 1 The Taylor series up to the linear term is given by () ' ( ) + 0 ( ) ( − ) √ The nonlinear term in the equation is () = which has a linear approximation of √ 1 ' 1 + ( − 1) 2 1 1 = + 2 2 The relative error is given by ¶¯ µ ¯√ ¯ 1 ¯ 1 ¯ − ¯ + ¯ ¯ 2 2 ¯ √ relative error = ¯¯ ¯ ¯ ¯ ¯ ¯
Figure next shows the % relative error.
13
The relative error is zero at = 1 where the nonlinear term is expanded about. The relative error is less than 10% approximately for 041 241 √ 1 1 The linearized equation is obtained by substituting with + Then, the linearized equation 2 2 of motion is given by 1 1 ̈ + = 2 2 Figure next shows the responses predicted by the linear and nonlinear models.
14
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
6. Consider the mass suspended by a nonlinear hardening spring as shown in Figure 2.34. The spring obeys = 3 and has an original length of The coordinate is the stretch of the spring, and is the displacement of the mass measured from its static equilibrium position. Derive the linearized equation of motion about its equilibrium position in terms of and also in terms of Find the natural frequency of this system.
Figure 2.34: A mass-spring system with a hardening spring.
Solution: The equation of motion is given by ̈ + 3 = The static deflection can be found by solving the static equation, 3 = r = 3 The Taylor expansion of 3 near = is given by () = 3
¯ ¯¯ ∼ ( ) + ( − ) ¯= = 3 + 32 ( − ) = 32 − 23 ³ ´23 = 3 −2
The linearized expression for 3 is substituted into the equation of motion so that µ ³ ¶ ´23 ̈ + 3 = −2 ³ ´23 = 3 ̈ + 3
(1)
15 The natural frequency of the system is v u ³ ´23 u µ ¶16 t 3 √ = = 3 13
The equation of motion can be rewritten in terms of the displacement measured from the static equilibrium position. The two coordinates, and are related by = + r
Substituting = + 3
into Equation (1), the equation of motion becomes r ¶ ³ ´23 µ 3 + = 3 ̈ + 3 ³ ´23 = 0 ̈ + 3
The equation of motion written in terms of the coordinate measured from the static equilibrium position is always homogeneous.
16
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
7. For each idealized model in Figures 2.35 to 2.38, draw a free-body diagram and derive the equation of motion using (a) Newton’s second law of motion and (b) the energy method. The block in Figure 2.36 slides on a frictionless surface. State whether the oscillations are linear or nonlinear. Determine the natural frequency of each model.
Figure 2.35: Mass oscillating in the vertical direction about equilibrium.
Figure 2.37: A simple pendulum.
Figure 2.36: Sliding oscillation of mass.
Figure 2.38: Torsional vibration.
Solution: Fig.2.35 For an oscillator vibrating in the gravitational direction , about static equilibrium, we have X = ̈ −1 ( + ) − 2 ( + ) + = ̈
where is the static deflection. With the dynamic displacement () = 0 the static deflection satisfies −1 − 2 + = 0 = 1 + 2
17 p Therefore, the equation of motion is ̈+(1 +2 ) = 0 with natural frequency = (1 + 2 ) Using an energy approach to the same problem, we first find the kinetic and potential energies, and then apply the principle of energy conservation + = constant. 1 = ̇2 2 1 = (1 + 2 ) ( + )2 − 2 1 1 ̇2 + (1 + 2 ) ( + )2 − = 2 2 Differentiating the last expression with respect to time, ̇̈ + (1 + 2 ) ( + ) ̇ − ̇ = 0 ̇(̈ + (1 + 2 ) ( + ) − ) = 0 Since ̇ 6= 0 we can divide it out of the expression, leaving the equation of motion. ̈ + (1 + 2 ) ( + ) − = 0 ̈ + (1 + 2 ) = 0 Fig.2.36 Sum the forces in the direction to obtain X = ̈
− = ̈
The equation of motion is ̈ + = 0 When the mass reaches the wall, it bounces back with a new velocity. The new velocity depends on the coefficient of restitution, The velocity after the bounce is = in the opposite direction. We re-solve the same equation of motion with this new initial velocity in the opposite direction. Using an energy approach to the same problem, we first find the kinetic and potential energies, and then apply the principle of energy conservation + = constant. 1 1 ̇2 = 2 2 2 1 1 2 2 = ̇ + = 2 2 =
Differentiating the total energy, and dividing the expression by ̇ we obtain the same equation of motion. Fig.2.37 For the oscillating pendulum, from the free-body diagram, we take the sum of the moments and set these equal to the mass moment of inertia about the base multiplied by the angular acceleration, X © = ̈ − sin = ̈
18
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
The negative sign is because the moment is in the opposite sense of the positive sign direction given to is the mass moment of inertia about the pivot point . For the point mass shown and massless p rod, = 2 Therefore, the EOM is ̈ + sin = 0 The natural frequency is given by = By the energy method, we substitute the kinetic and potential energies into the principle of the conservation of energy. 1 2 ̇ = − cos 2 1 2 + = ̇ − cos = constant 2 =
Differentiating the total energy with respect to time, ̇̈ + sin ̇ = 0 ̇( ̈ + sin ) = 0 Since ̇ 6= 0, we can divide it out of the expression, leaving us with the equation of motion. Fig.2.38 Take the moment about the vertical axis through the center of the disk to obtain X
= ̈
− = ̈
2 where is the mass polar moment of pinertia. For a solid disk = 2. The EOM is ̈+ = 0 and the natural frequency is =
Using an energy approach, we first find the kinetic and potential energies, and then apply the principle of energy conservation + = constant. 1 1 2 ̇ , = 2 2 2 1 2 1 2 + = ̇ + = 2 2 =
Differentiating the total energy and dividing the expression by ̇ we obtain the same EOM.
19
8. A disk of mass is mounted between two shafts with different properties, as shown in Figure 3.39. (a) What is the natural frequency of the system? (b) If the disk is rotated , where ¿ 1 rad, and then released, what will its angular position be at an arbitrary time ?
Figure 2.39: A disk mounted between two shafts. Solution: Note that 2 and 3 are in series. The equivalent stiffness is ¶−1 µ 1 2 3 1 0 + = = 2 3 2 + 3 Let us assume a rotation of then the springs exert moments in the opposite direction of the assumed motion. Summing the moment about the rotational axis, we have X = ̈ 0 −1 − = ̈
The equation of motion is then ¶ µ 2 3 = 0 ̈ + 1 + 2 + 3 {z } |
The natural frequency is
=
s
µ ¶ 2 3 1 + 2 + 3 1
Note that the polar mass moment of inertia of a solid disk is = 12 2 where is the radius of the disk. The response is purely sinusoidal, and () = 1 cos + 2 sin For a disk released from an initial position 0 with zero velocity, the response at an arbitrary time is given by () = 0 cos
20
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
9. Calculate the equivalent torsional spring constant for the stepped shaft shown in Figure 2.40. The two shafts have negligible mass and are made of a material with shear modulus 1 and 2 respectively. At the end of the shaft is a rigid disk of mass inertia Express your answers in terms of known quantities.
Figure 2.40: Stepped shaft. Solution: The formula (from Mechanics of Solid) for a torsional bar is where is the angle of rotation of the cross-section, is the torque, is the length of the bar, is the shear modulus, and is the polar area moment of inertia. Rewriting the equation, we have =
The torsional stiffness is The polar area moment of inertia of a circular bar is given by =
1 1 = 4 = 4 2 32 In this case, two torsional bars are in series so that the equivalent stiffness is ¶−1 µ 1 1 1 2 = + = 1 2 1 + 2 1 1 2 2 2 + 222 1
= 1 11 =
1 2 1 2 1 1 2 + 2 2 1
In terms of diameters, =
1 2 41 42 32 1 41 2 + 2 42 1
The equation of motion is given by ̈ + = 0
21
10. The valve mechanism in Figure 2.41 is in equilibrium when the rocker arm is horizontal. The system is assumed to be frictionless. Use an energy method to determine the natural angular frequency for small vibration about equilibrium in terms of and
Figure 2.41: A rocker-arm valve system. Solution: Let be the angle of rotation of the rocker arm in the counterclockwise direction measured from the horizontal position, which is also the static equilibrium position. The kinetic energy of the system is given by ³ ´2 1 ³ ´2 1 1 2 = ̇ + ̇ + ̇ cos |2 {z } |2 {z } |2 {z }
1 where = + 2 = 12 ( + )2 + 2 using the parallel axis theorem. The potential energy of the system is
= − sin () + sin () | {z } | {z }
1 + sin () + ( + sin ())2 | {z } |2 {z }
where is compression in the spring when the system is in static equilibrium. The total energy, is the sum of the kinetic and potential energies. The system is conservative, and the total energy is constant. Differentiating the total energy, we obtain 0 =
= 2 ̇̈ + ̇̈ ´ ³ 2 + 2 ̇ cos ̈ cos + ̇ sin
− cos () ̇ + cos () ̇ + cos () ̇ + ( + sin ()) cos ̇
22
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
Divide by ̇ to obtain 0 = 2 ̈ + ̈ ³ ´ 2 + 2 cos ̈ cos + ̇ sin
− cos () + cos () + cos () + ( + sin ()) cos ¢ ¡ Assume small rotations 2 1 and we obtain ¡ ¢ 0 = 2 + + 2 ̈ + ( + ) − + +
The static compression of the spring is obtained by setting = 0 and ̈ = 0 and 2 − + + = 0 − − = Then, the equation of motion is simplified to ¢ ¡ 0 = 2 + + 2 ̈ + 2
The natural frequency of this system is
=
s
2 2 + + 2
23
11. A rod is supported on two rotating grooved rollers, as depicted in Figure 2.42. The tubes rotate in opposite directions and the coefficient of friction between the tubes and the rod is Find the natural frequency of the system and describe the behavior of the rod if it is disturbed in the horizontal direction.
Figure 2.42: Rod supported by two rotating grooved rollers.
Solution: The rod oscillates because of the variation in the frictional force by each wheel. Let us consider a free-body diagram shown below.
The rod does not move up and down nor rotate. Therefore, we have X = 1 + 2 − = 0 µ µ ¶ ¶ X © = 1 + + 2 − = 0 2 2
We can find the normal forces as functions of : µ µ ¶ ¶ 1 1 1 = − and 2 = + 2 2 The force balance equation in the direction is given by X = 1 − 2 = ̈
µ
1 − 2 = ̈ ¶ 1 1 − − − = ̈ 2 2
24
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
The equation of motion is then
̈ + 2 = 0 The rod oscillates about the midpoint at the natural frequency of r = 2
25
12. A 25-kg block is suspended by two cables, as depicted in Figure 2.43. Assume small displacements. (a) What is the frequency of oscillation in Hz of the block in the direction if it is displaced slightly in this direction? (b) What is the period of oscillation in the direction if the block is displaced slightly in this direction?
Figure 2.43: A suspended block. Solution: (a) This is the case when the block oscillates on the plane. The corresponding free-body diagram is shown below, where the tension is assumed the same in both cables due to symmetry.
The block does not rotate, but translates. Summing the forces in the and direction, we obtain X (1) = −2 sin = ̈ X = 2 cos − = ̈ (2)
The accelerations in the and directions are written in terms of and derivatives of Note that the accelerations at the center of gravity are the same as those anywhere else on the block. We write the position of one of the points where the string is attached. Taking a second derivative with respect to time gives the accelerations. 2
= sin =⇒ ̈ = ̈ cos − ̇ sin
(3)
2
= (1 − cos ) =⇒ ̈ = ̈ sin + ̇ cos
(4)
26
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
Substituting Equation (3) into (1) and (4) into (2), the force balance equations are now ³ ´ 2 −2 sin = ̈ cos − ̇ sin ³ ´ 2 2 cos − = ̈ sin + ̇ cos
There are two unknowns, and both functions of time. We eliminate the tension to obtain a differential equation in terms of Multiplying the first by cos and the second equation by sin and adding them, we obtain ̈ + sin = 0 which is identical to a pendulum equation. The reason why the equation of motion is that of a pendulum is because the p block does not rotate. The natural frequency for small oscillations in units of Hz is = 2 = 2 (b) If the block oscillates in the plane, the free-body diagram is shown.
Taking a moment about the fixed point at the ceiling, we have X ¡ ¢ ª = + 2 ̈ ¡ ¢ − sin = + 2 ̈ 1 (2 + 2 ) The equation of motion is given by where = 12
¡ ¢ + 2 ̈ + sin = 0
For small oscillations, the natural frequency is s¡ s ¢ 2 +2 2 + (2 + 2 + 122 ) 1 12 = = 2 3 In Hz, 1 = = 2 4
s
(2 + 2 + 122 ) 3
27
13. Derive the equation of motion and natural frequency for a mass on the string that is under constant tension as shown in Figure 2.44. Assume small displacements and is much greater than the mass of the string.
Figure 2.44: Vibration of a mass on a string under tension.
Solution: When the string with mass is displaced from equilibrium, it is the vertical component of the string tension forces that restore the mass to its original position. The free-body diagram is shown below.
Summing the forces in the vertical direction, we obtain X = ̈
− sin 1 − sin 2 = ̈
For small angles, we can make the approximation that sin ≈ tan For the triangle with base tan = and for the triangle with base − tan ≈ ( − ) Therefore, the EOM becomes
The natural frequency is =
q
= ̈ − − − ¶ µ = 0 ̈ + ( − ) (−)
28
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
14. Continuing Problem 13, the string is stretched to the position shown in Figure 2.45. Calculate the natural frequency of the system using the following parameter values: = 2 lb, = 50 lb, and = 4 ft, where is the tension in the string for the configuration shown.
Figure 2.45: Mass on a stretched string.
Solution: From the previous problem, we found that the natural frequency is = Substitute the following values: = 50 lb = 2 ft = 4 ft 2 lb = = 322 ft/s2 The natural frequency is = =
s
s
( − ) 50 4 2322 2 (4 − 2)
= 284 rad/s
q
(−)
29
15. Derive the equation of motion for a uniform stiff rod restrained from vertical motion by a torsional spring of stiffness and two translational springs each of stiffness as shown in Figure 2.46. The torsional spring constant is determined by the application of a moment and the measurement of the angular displacement that is, = Calculate the natural frequency of oscillation. Let be the mass moment of inertia of the rod about the left restraint. State any assumptions.
Figure 2.46: Restrained rigid rod.
Solution: Assume that the springs are undeformed when the rod is horizontal. Let the point be the base of the rod or where the rod is attached to the torsional spring. Assume that the dimension of the torsional spring is negligible. Let be the mass of the rod. Assume is defined positive clockwise. The free-body diagram is shown below.
We assume small displacements of the end. In this way we can assume that essentially the springs remain vertical. Newton’s second law of motion, in moment form, can be applied for the sum of the external moments about the left end of the rod. It should be noted that the moment is taken positive in the positive direction. X y = ̈ − − ( sin )() − ( sin )() + 2 = ̈
where = 2 3 and sin is the deflection of the right end and is the moment arm of the spring force. Then, 2 ̈ + + 22 sin = 2 3 or ¡ ¢ 2 ̈ + 22 + = 3 2
30
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
where for small anglespsin ≈ gives the equation of linearized motion. The natural frequency for p 2 2 2 small motion is = 3 (2 + ) = (2 + ) Note that the gravity does not affect the natural frequency or the dynamic motion of the system. It only shifts the static equilibrium position.
31
16. A uniform rigid and massless rod is pinned at one end and connected to ground via a spring at the other end. At midpoint on the rod, a spring is connected to a mass which is connected to a fixed point via another spring, as shown in Figure 2.47. (a) Derive the equation of free vibration for the system. (b) Find the natural frequency. (c) Calculate the natural frequency given the following parameters: 1 = 20 lb/in, 2 = 30 lb/in, 3 = 40 lb/in, = 50 lb, and = 8 ft.
Figure 2.47: A coupled rod-mass system. Solution: (a) Equation of motion Assuming that the springs are undeformed when = 0 and = 0 the free-body diagrams corresponding to positive displacements are shown below. We will also assume that the system is on a horizontal plane to eliminate the effect of gravity.
For the mass, the force balance equation in the vertical direction is X ↑ = ̈ −1 − 2 ( − ) = ̈
(1)
32
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
The moment balance equation for the rod is ª
X
2 ( − )
= ̈
− 3 = 0 2
where = 0 because, from the problem statement, the rod is massless, and is twice (using similar triangles). Then, the moment balance equation for the rod is reduced to 2 ( − ) − 43 = 0 Solving for we find = (2 (2 + 43 )) Substituting this back into Equation (1), the equation of motion is given by µ ¶ 2 = ̈ −1 − 2 − 2 + 43 µ ¶ 22 ̈ + 1 + 2 + = 0 2 + 43 (b) The general expression for the natural frequency is sµ 1 + 2 + =
22 2 + 43
¶
1
(c) The numerical value for the natural frequency given 1 = 20 lb/in, 2 = 30 lb/in, 3 = 40 lb/in, = 50 lb, and = 8 ft is sµ ¶ 302 322 × 12 = 20 + 30 + 30 + 4 (40) 50 = 206 rad/s.
33
17. A component of uniform material and nonuniform cross-section is pinned at one end and supported by two springs as shown in the two configurations of Figure 2.48. The component is displaced slightly from equilibrium and released and observed to oscillate with a period of = 15 s. (a) For the system in the top figure, calculate the moment of inertia of the component with respect to the hinge axis of rotation if = 500 N/m and = 1 m. (b) For the system in the bottom figure, calculate the period of oscillation assuming the same parameter values.
Figure 2.48: A rigid rod supported by two springs.
Solution: (a) First Case - Assuming a positive rotation in the counterclockwise direction and positive displacement upward, we draw free-body diagram for the first case is shown next.
FBD for the first case. We sum the moment about and we find ª
X
= ̈
−22 sin = ̈ The linearized equation of motion is ̈ + 22 = 0
The natural frequency is =
p p 22 , and the natural period is = 2 = 2 22
34
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
Substituting = 500 N/m = 1 m, and = 15 s, we find =
µ
2
¶2
22 = 570 kg m2
(b) Second Case - The free-body diagram is shown next.
FBD for the second case. We sum the moment about to obtain ª
X
= ̈
( − sin ) = ̈
(1)
We sum the forces about to obtain X ↑ = 0 (massless point ) + − sin = 0
Solving for we obtain = sin 2 Substituting this into Equation (1), we obtain ¶ µ sin − sin = ̈ 2 ̈ + 2 sin = 0 2 The linearized equation of motion is ̈ + 2 = 0 2 Since the springs are attached in series, we could have used an equivalent stiffness of = (1 + 1)−1 = 2 to arrive at the identical equation of motion. Note that the stiffness of the second system is reduced √ by a factor of four compared to case (a). Then, the natural frequency is reduced by a factor of 4 or two. This will increase the period by a factor of two. The period of the second system is 3 s.
35
18. A body of mass is suspended by a spring of constant and attached to an elastic beam of length as shown in Figure 2.49. When the mass is attached to the spring measurements are taken of the spring extension and the end deflection of the beam Neglect the masses of the beam and the spring. (a) Estimate the natural frequency of the system. (b) Calculate the natural frequency if the mass is attached to the beam directly without a spring. (c) Calculate the natural frequency if the beam is assumed to be rigid. (d) Calculate the respective numerical values given = 12 mm, = 2 mm, and = 05 m.
Figure 2.49: Mass suspended by a spring from a beam. Solution: (a) In this system, a leaf spring is connected to a coil spring in series. The system can be simplified to
The beam can be modeled as a spring with stiffness This leaf spring has a stiffness of = 33 The combined stiffness is ¶−1 µ 1 1 = + The equation of motion is ̈ + = where is measured from the undeformed configuration. The natural frequency is s = ( + )
36
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
Note that we could have omitted from the free-body diagram if we measure the deflection from static equilibrium. (b) If the mass is attached to the beam directly without a spring, the natural frequency is obtained by letting = or let → ∞ in the natural frequency expression obtained in part (a): r = (c) If the beam is assumed rigid, we let → ∞ in the natural frequency expression obtained in part (a): r = (d) If a mass is placed, we find = 12 mm and = 2 mm. For a spring in series, the force in the springs are equal. That is = and = Then, =
and = 0002 0012
The answer to parts (a) (b) and (c) are (a)
(b) (b)
=
s
( + )
v u 1 1 u t ³ ´ = 265 rad/s = 1 1 + r r = = 700 rad/s = r r = = = 286 rad/s.
Note that and are not necessary to determine the natural frequencies.
37
19. A solid cylinder floating in equilibrium in a liquid of specific gravity is depressed slightly and released. A schematic is shown in Figure 2.50. Find the equilibrium position and solve for the natural frequency of oscillation assuming the cylinder remains upright at all times. Suppose the assumption that the cylinder remains upright is not reasonable. Then what difficulties do you foresee in the calculations, and how might they be resolved?
Figure 2.50: Solid body oscillating in a liquid.
Solution: We know from Archimedes’ principle that for a body immersed in a fluid, a buoyancy force acts in the opposite direction of gravity with a magnitude equal to the product of displaced volume and mass density of fluid. For water this mass density equals = 64 3 The displaced volume equals 2 where is the length of the cylinder that is immersed. We assume the cylinder remains vertical as it oscillates. The mass of the cylinder is 2 where is the length of the cylinder and is the mass density of the cylinder. The equilibrium position is given by the following calculations: we equate the buoyancy force to the weight of the cylinder:
= = (2 ) = 2
The forces in the vertical direction are those due to gravity: the weight of the cylinder and the buoyancy force. If the oscillation is taken about the equilibrium position, weight can be ignored. The buoyancy force acts like a spring force. Newton’s second law of motion in the vertical direction is given by X
=
−2 = 2 ̈ where is the displacement measured from the static position so that = + See the figure below for the definition of and
38
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
In standard form this EOM becomes ̈ + = 0 or ̈ + ( ) = 0 The specific gravity is the ratio of the density of a fluid to the density of water. If the cylinder does not remain vertical during oscillation, then in addition to vertical oscillations there will be rotational oscillations and the restoring force will be more complicated to evaluate.
39
20. The uniform rod is restrained by four translational springs and a torsional spring as shown in Figure 2.51. Determine the natural frequency of the system using the energy approach.
Figure 2.51: A rigid rod supported by four translational springs and a torsional spring. Solution: We will assume that the system is on a horizontal plane and the springs are undeformed when the bar is horizontal. The kinetic energy is given by 1 2 = ̇ 2 where is where the rod is attached to the hinge, and = + 2 In this case, = 2 12 and = 4 Then, 7 = 2 48 The potential energies are stored in the four translational springs and the torsional spring. The total potential energy is given by µ ¶2 µ ¶2 3 1 = + + 2 4 4 2 5 1 = 2 2 + 2 8 2 Note that gravitational potential energy can be omitted if is measured from the static equilibrium. The total energy is = + and it is conserved. Taking a derivative with respect to time, we have 7 10 = 0 = 2 ̇̈ + 2 ̇ + ̇ 48 8 Dividing by ̇ we have the equation of motion, µ ¶ 7 2 5 2 ̈ + + = 0 48 4 The natural frequency is =
s¡
5 2 + 4 7 2 48
¢
40
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
21. Derive Equation 2.26, () = (0) cos +
̇(0) sin
beginning with the equation of motion. Solution: Begin with Equation 2.20 with () set to zero: ̈ + 2 = 0 The solution is given in the form, () = 1 sin + 2 cos We can evaluate the arbitrary constants by satisfying initial conditions: (0) and ̇(0) : (0) = 2 ̇(0) = 1 Therefore, 2 = (0) and 1 = ̇(0) Equation 2.35 is found: () = (0) cos +
̇(0) sin
41
22. Derive the equation of motion and natural frequency for a disk in motion that is constrained to move vertically, but can rotate, as shown in Figure 2.52. The coordinate is the displacement of its center of mass, measured from the static equilibrium position, and the coordinate is the displacement of the center of mass measured from its location when the spring is unstretched. Let be the rotation of the disk measured from the static equilibrium position. The mass of the disk is , and the mass moment of inertia for the disk is 2 2 Assume that the cable is inextensible and the spring constant is
Figure 2.52: A disk suspended by an inextensible cable and a spring.
Solution: This is a single-degree-of-freedom system though two coordinates are given. Since the disk is not slipping, and are related by the kinematic relationship = In addition, the displacement of point is twice the displacement at Therefore, the stretch in the spring is 2 or 2 ( + )
42
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
From the free-body diagram, we have X ↓ = ̈
⇒ −2( + ) − + = ̈ X © = ̈
̈ ⇒ − 2( + ) =
Eliminating and combining above two equations, we have the equation of motion given by µ ¶ + 2 ̈ + 4 ( + ) = The natural frequency is v u 4 u =u t 1 2 + 2 + 2 2 r = 1633
v u = u t
4
The static deflection, is obtained by setting the dynamic displacement () to zero or =
4
It is also possible to derive the equation of motion in terms of The displacements and are off by the constant such that = + ̈ = ̈ From the free-body diagram, we have X ↓ = ̈ ⇒ −2 − + = ̈ X ̈ © = ̈ ⇒ − 2() = Eliminating we have
µ ¶ + 2 ̈ + 4 =
The static equilibrium is obtained by setting ̈ = 0 or =
4
43
23. Derive the equation of motion and natural frequency for mass shown in Figure 2.53. Assume the pulleys are massless, the cable is inextensible, and let the spring constant be The coordinate is the displacement of its center of mass, measured from the static equilibrium position, and the coordinate is the displacement of the center of mass measured from its location when the spring is unstretched.
Figure 2.53: A mass suspended in a pulley system. Solution: If the pulley’s mass is negligible, the tension can be assumed constant.
The displacements and are off by the constant such that = + The sum of the forces in the vertical direction gives X ↓ = ̈ ⇒ − 2 = ̈
The tension is same as the spring force, and the stretch in the spring is twice the displacement of the mass such that = (2) = (2) ( + )
44
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
Then, ̈ = − 4 ̈ + 4 = The natural frequency of the system is r
4 r = 2 =
The equilibrium position of the mass is obtained by setting the acceleration ̈ to zero or =
4
The equation of motion can be obtained in terms of the dynamic displacement () by replacing with + Then, we obtain ̈ + 4 = 0
45
24. For the two-mass system shown in Figure 2.54 derive the equation of motion in terms of 1 Assume the pulleys are massless, the cable is inextensible, and let the spring constant be The coordinates 1 and 2 are the displacements of the center of mass of 1 and 2 respectively, measured from the static equilibrium position. The coordinates 1 and 2 are the displacements of the center of mass of 1 and 2 respectively, measured from its location when the spring is unstretched.
Figure 2.54: Two masses suspended in a pulley system. Solution: The coordinates 1 and 2 measure the displacement of the masses from their original position when the spring is unstretched. The stretch in the spring is 2 and 1 and 2 are related by 21 = 2 The free-body diagrams are shown below.
Summing the forces in the direction defined by the positive coordinate directions, we obtain X ↓ = 1 ̈1 ⇒ 1 − 2 = 1 ̈1 X ↑ = 2 ̈2 ⇒ − 2 − 2 = 2 ̈2
Substituting ̈2 = 2̈1 and eliminating from the equation, the equation of motion becomes (1 + 42 ) ̈1 + 41 = 1 − 22
46
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
The natural frequency is =
r
4 (1 + 42 )
47
25. Derive the equation of motion and natural frequency for the disk shown in Figure 2.55. Assume the disk is rolling without slip.
Figure 2.55: Disk on an inclined plane connected to two springs. Since it is assumed that the disk rolls without slipping, there is no energy loss from friction. Solution:Assume that = 0 when the springs are unstretched. The free-body diagram is shown next.
For a disk rolling without slip, we can sum the moment about the contact point at so that P y & = ̈ −2 + sin & = ̈ where = + 2 For a solid disk, = 2 2 and = 152 In addition, for a nonslip disk, the rotation is related to kinematically by = Then, the equation of motion is given by ̈ + 22 = sin The natural frequency is = =
s r
22
r 2 = 115 15
48
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
26. Derive the equation of motion and natural frequency for the disk shown in Figure 2.56. Assume the disk is rolling without slip, and the cable is wrapped around the disk.
Figure 2.56: A rolling disk on an incline. Solution: The free-body diagram is
The sum of the moment about the instantaneous center (where the disk touches the ground) is X © = ̈ − 2 + sin = ̈
3 where = + 2 = 2 The stretch in the spring is twice the displacement of the cylinder 2 at its center = 2 In addition, the angle of rotation and the displacement are related by = Then, the EOM becomes − (2) 2 + sin =
3 2 ̈ 2
3 ̈ + 4 = sin 2
49 The natural frequency is =
r
p 4 = 1633 15
50
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
27. Derive the equation of motion and natural frequency for mass shown in Figure 2.57 Assume the pulleys are massless, the cable is inextensible, and let the spring constant be
Figure 2.57: A mass suspended in a pulley system.
Solution: The free-body diagrams for the moving pulley and the mass are shown below.
When the mass moves downward by the spring is stretched by 2 Therefore, 2 = 4
2 =
Summing the forces in the vertical direction for the mass, X ↓ = ̈ − = ̈ 4 ̈ + = 4 The natural frequency is =
r
p = 05 4
51
28. Derive the equation of motion and natural frequency for mass shown in Figure 2.58. Assume the pulleys are massless, and let the spring constants be 1 and 2 Let 1 and 2 be the displacements of the pulley and the mass measured from their locations when the springs are unstretched. Use 2 for the equation of motion.
Figure 2.58: A mass suspended by a pulley and springs.
Solution: The coordinates, 1 and are not related kinematically. For example it is possible to move the pulley up by the distance 1 while keeping it from rotating or setting to zero. Similarly, it is possible to rotate the pulley while keeping it from moving up and down. On the other hand, 2 is related to both 1 and We can find this relationship by obtaining the contribution toward 2 from 1 and individually. Letting = 0 we see that moving the pulley up by 1 upward will lower the mass also by 1 That is, 2 (while keeping = 0) = 1 Letting 1 = 0, we also see that rotating the pulley by will also lower the mass by That is, 2 (while keeping 1 = 0) = Then, 2 = 1 +
(1)
Next, we find the stretches in the springs. We find the individual contribution from 1 and separately. We find that the stretch in 2 is simply 1 and the stretch in 1 is − 1 2 = 1 1 = − 1 The next step is to draw free-body diagrams for the pulley and the mass, and write down Newton’s second law for the vertical direction. The free-diagrams are shown next.
52
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
It is also possible to use the following free-body diagram,
Summing the moment about the center of the pulley, we find that P ª = − 1 ( − 1 ) = ̈ where = 0 if we assume that the pulley has negligible mass and = 1 ( − 1 ) or the tensions on either side of the pulley are the same. Summing the forces in the vertical direction for the pulley, we obtain 21 ( − 1 ) − 2 1 = ̈1 = 0
Then, we find that 1 and are related by
1 =
21 21 + 2
Summing the forces in the vertical direction for the mass, we obtain P ↓ = ̈2 − 1 ( − 1 ) = ̈2
(2)
(3)
53 The equation of motion is obtained by replacing and 1 with 2 Using Equations (1) and (2) together, we find µ ¶ 21 − 21 + 2 ¶ 2 ( − 1 ) = µ 21 + 21 + 2 2 = 2 41 + 2 Finally, the equation of motion in (3) is rewritten as ̈2 +
1 2 2 = 41 + 2
The natural frequency of this system is =
r
1 2 1 41 + 2
54
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
Problems for Section 2.3 — Undamped Free Vibration 29. Derive Equation 2.30, () = 1 cos + 2 sin
(2.30)
beginning with the equation of motion. Solution: The equation of motion is ̈ + 2 = () Assume a general solution () = exp() Differentiate this function twice and substitute into the ODE to find: 2 + 2 = 0 The common factor exp() has been divided out of the expression. This provides us with two roots: 12 = ± Each of these is one solution to the governing ode. Therefore, () = 1 exp(1 )+2 exp(2 ) We substitute the two roots in this expression, and proceed as in the text, where the rest of the derivation is provided before our goal: Equation 2.30.
55
30. Show that the period of free vibration of a load weighing suspended from two parallel springs, as shown in Figure 2.59, is given by , s = 2 (1 + 2 ) and show that the equivalent stiffness is = 1 + 2 . Discuss the need to hang the weight asymmetrically, that is 1 6= 2 if 1 6= 2 , so that the extension of the springs is identical and that the ratio 1 2 = 2 1 .
Figure 2.59: Weight hanging from two parallel springs.
Solution: Let be the deflection of the mass defined positive downward, and let be the rotation of the rigid bar. For positive and we draw the free-body diagram shown below.
Free body diagram. The deflection of the left spring is approximately + 1 and of the right spring − 2 We have arbitrarily assumed that rotation of the bar is clockwise resulting that the beam deflects more at the left side than the right side. If for particular parameter values this is not true, then the signs will change in the resulting expressions to reflect this. By Newton’s second law of motion,
56
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
↓
X
=
̈
−1 ( + 1 ) − 2 ( − 2 ) =
̈
If the weight is hung asymmetrically such that 1 2 = 2 1 then the EOM becomes ̈ + (1 + 2 ) = 0 (1 + 2 ) ̈ + = 0 p where 2 = (1 + 2 ) and the period is then = 2 = 2 (1 + 2 ) If the above assumptions were not made, then we would need a second EOM that governs We study such problems in Chapters 8 and 9.
57
31. For the body suspended between two springs as in Figure 2.60, show that the period of oscillation is s = 2 (1 + 2 )
Figure 2.60: Body suspended between two springs. Solution: Let be the downward displacement of the body measured from the static position. The corresponding free-body diagram is shown below.
Applying Newton’s second law of motion in the vertical direction results in −(1 + 2 ) =
̈
Note that gravity is omitted because is measured from static equilibrium. Therefore, the EOM is ̈ +
(1 + 2 ) = 0
The natural frequency and the period are, respectively, r (1 + 2 ) 2 = =
58
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
32. A compound pendulum in the shape of a rectangle is supported at point and allowed to oscillate. The dimensions of the rectangle are given in Figure 2.61. Calculate the natural frequency for small oscillations.
Figure 2.61: A compound pendulum Solution: Draw a free-body diagram based on a positive rotation as shown below.
Taking a moment about , we have ª
X
= ̈ ¶ − − sin = ̈ 2 where is the mass moment of inertia of the plate about given by µ
2
= + ¶2 µ ¢ 1 ¡2 2 = + + − 12 2
59 where is the center of mass. The linearized equation of motion (for small oscillations) is given by ¶ µ ̈ + − sin = 0 2 and the natural frequency is =
s
v u u = t
¡
¢ − 2
¢ − 2 ¡ ¢2 1 2 2 ( + ) + − 12 2
¡
60
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
33. Two springs in series support weight as shown in Figure 2.62. If these springs are replaced by a single spring, find the equivalent stiffness as well as the period of oscillation. The solution is s 1 2 (1 + 2 ) = = 2 1 + 2 1 2
Figure 2.62: Two springs in series at left.
Solution: We need to find the equivalent stiffness of two springs in series. To do this attach a second coordinate to the point where the two springs are attached to each other. Call the deflection of this location, 2 and call the displacement of the weight 1 as shown below.
From a free-body of and the lower spring, = 1 (1 − 2 ) From a free-body of the upper spring, we have = 2 (2 ) Since 2 is an intermediate displacement, we can eliminate 2 using the second equation. Solve for 2 and substitute it into the first equation: = 1 (1 − 2 ) = 1 (1 − ); 2
61 or 1 ) = 1 1 2 1 + 2 ( ) = 1 1 2 1 2 = 1 1 + 2 (1 +
The equivalent stiffness is then 1 2 (1 + 2 ) The equation of motion is 1 2 1 = 0 ̈1 + 1 + 2 The natural frequency and period are s
1 2 (1 + 2 ) s (1 + 2 ) 2 = 2 = 1 2
=
62
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
34. A pendulum of mass and mass moment of inertia is suspended from a hinge, as shown in Figure 2.63. The center of gravity is located a distance from the hinge. For small oscillations, the period equals . Suppose the pendulum swings with an amplitude of angle from the vertical. Find the force exerted on the hinge when the pendulum is (a) at its maximum rotation and (b) at its vertical position.
Figure 2.63: Compound pendulum rotating about pinned connection. Solution: (a) A pendulum has an equation of motion given by ̈ + sin = 0 The linearized equation is ̈ + = 0 The problem says that for a small oscillations, the period is . That is, s 2 = 2 = This tells us of the compound pendulum if and are also known. At = the acceleration of the pendulum is in the tangential direction only and is ̇ or ̈. The normal acceleration toward is zero because = 0 at = ±
P = when =
63
From the equation of motion, we know ̈ ( = ) = − At = ± we write X = ( = ) = ̈ (− cos ̂ − sin ̂) ¶ µ − ̂ = (cos ̂ + sin ̂) µ 2 2 ¶ 2 2 cos ̂ + sin + ̂ ( = ) =
(Answer to part (a))
(b) When the pendulum passes through the vertical position, the tangential acceleration of ̇ is 2 zero, and the acceleration is ̇ in the normal direction only.
P = when = 0 Summing the forces at these locations, we have X
At = 0
2
= ̂
The normal velocity at = 0 is obtained from the conservation of total energy. At = 0 the total energy at = is converted to the kinetic energy. That is 1 ( − cos ) = 2 2 The velocity squared is then 2 ( = 0) = 2 (1 − cos ) The force balance equation then X
2
= ̂ 1 − ̂ = (2 (1 − cos )) ̂ The reaction force at = 0 is ( = 0) = (3 − 2 cos ) ̂
(Answer to part (b))
64
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
35. A bifilar pendulum of length 2 is suspended with two vertical strings, each of length , as shown in Figure 2.64. (Bifilar means fitted with or involving the use of two threads or wires.) Assuming small rotation of the strings, such that the bar is essentially horizontal with half its weight supported by each string, show that the period is s 2 = 3
Figure 2.64: A bifilar compound pendulum. Solution: Note that if we assume that the bar moves as a pendulum in plane, then the EOM will be the same as that for a simple pendulum, and therefore, the period is s 2 = 2 = Therefore, the problem refer to the period of the out-of-plane motion shown below.
Out-of-plane oscillation of a bifilar pendulum. Let us visualize the motion: the bar begins to rotate about its centerline, and as it rotates its ends also raise. However, we assume that this raising can be neglected if the angle of rotation out of the plane of the paper is small. By neglecting the motion in the direction, we can also approximate the tension in the each string as 2 For a small angle of rotation, the tension in each string
65 develops a restoring moment (like a spring). From the free-body diagram below
the restoring moment about the axis due to the tension in the each string is . The component of the tension perpendicular to the rod on − plane is approximately = = 2 and the restoring moment due to the two strings is 2 The restoring moment is set equal to the moment of inertia of beam about its center of mass (and rotation) times the angular acceleration ̈ : ̈ =
−2
2 2 ̈ + = 0 3 1 where we have used = 12 (2)2 Simplifying the EOM,
2 3 = 0 2 ̈ + 2 = 0
̈ +
where 2 = 32 2 The period is given by = 2 which equals the given expression. Note that using the vector notation may be more convenient. Taking the moment about the axis through the center of mass, we have, X
=
where the sum of the moment is given by
where
X
= × + ×
= cos + sin = − cos − sin ³ ´ ( − cos ) − sin + =
66
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
Considering the component only, we obtain the same EOM, 2 sin = ̈ 1 2 (2)2 ̈ + = 0 12 2 3 ̈ + 2 = 0 −2
The natural frequency and the natural period is given by r 3 = s 2 = 3
67
36. An inverted hinged pendulum with a mass at the top is suspended between two springs with constants , as shown in Figure 2.65. The rod can be assumed rigid and massless, and in the vertical position the springs are unstretched. For small motion, the springs can be assumed to remain horizontal. Show that the period of oscillation is =r
2 2 −
Figure 2.65: Stabilized inverted pendulum.
Solution: From the free-body diagram of the mass we see that a displacement to the right in the direction results in forces in the opposite sense due to each spring of − sin each.
Even though we can assume the springs to be horizontal, the motion of the rigid beam with the mass at the end is a rotational motion about the hinge.. Therefore, use Newton’s second law of motion in moment form: X y = ̈
where is the angle of the beam from the vertical and equals the moment of inertia about the hinge: = + 2 ≈ 2 since is very small when compared to 2 If you are wondering why the rod only transmits tension but not shear, it is because the mass is assumed to have no
68
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
inertia about its own center of gravity. If the pendulum has mass moment of inertia, shear must be included. Taking the moment about the hinge in the clockwise direction, we have: −(2 sin ) cos + () sin = 2 ̈; 22 sin = 0 ̈ + sin cos − 2 Even though we have not assumed small angles explicitly so far, we have assumed small angles in reality when we assumed that the spring forces act horizontally. Therefore, the above equations are not yet valid until we make the substitution: sin ≈ and cos ≈ 1 : 22 − = 0 2 ¸ ∙ 2 − = 0 ̈ + | {z }
̈ +
2
The term in the brackets is the natural frequency squared for small oscillations 2 , and the period is given by = 2
69
37. Consider the inverted simple pendulum shown in Figure 2.66. Initially the pendulum is in a perfectly vertical position. If it is displaced very slightly from the vertical, what are the stability characteristics of the system? What kind of motion is expected and what is the equation of motion? Discuss the long-term behavior.
Figure 2.66: Inverted pendulum.
Solution: We can start with the known pendulum equation given by ̈ +
sin = 0
For an inverted pendulum, we expand sin about = The Taylor series of sin about = is − sin Then, the equation becomes ̈ − = 0
The solution is given by () = 1 exp
³p ´ ³ p ´ + 1 exp −
The response is unstable about the equilibrium when it is in upright position. We cannot discuss long term behavior based on the linearized model because the model is valid only near the upright position. However, from energy conservation, we know that the pendulum will not be able to reach a height higher than initial height. If the pendulum is released from 5◦ from the vertical with zero initial velocity, it will oscillate between 5◦ and 355◦ measured from vertical. We can also derive the equation of motion measuring from the unstable equilibrium position as shown next.
70
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
Taking a moment about in the clockwise direction, we obtain X © = sin ̈ = sin ̈ − sin = 0
In order to investigate the behavior near = 0 we expand sin near = 0 The Taylor series of sin near = 0 is Then, the equation of motion becomes ̈ − = 0
71
38. For the system of Figure 2.67, what should the value of 3 be if 1 = 22 = 33 for a period of free vibration of 400 ms for = 25 kg?
Figure 2.67: Mass constrained by several springs.
Solution: Springs 1 2 and 3 are in parallel. This becomes more obvious when assume that the mass moves a distance of The corresponding spring force is 1 + 2 + 3 in the direction opposite to the assumed displacement p The equivalent stiffness is = 1 + 2 + 3 = 553 The natural period is = 2 = 2 For = 04 s and = 25 kg, = 61623 N/m. Then, 3 = 55 = 11204 N/m.
72
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
39. A slender bar of mass and length is supported at its base by a torsional spring of stiffness as per Figure 2.68. The bar rests in the vertical position when in equilibrium with the spring unstretched. Show that the differential equation of rotation from equilibrium is 2 ̈ + − sin = 0 3 2 For small vibration, that is, ¿ 1 show that the natural frequency is given by =
( − 2) 2 3
Figure 2.68: A slender bar of mass and length is supported at its base by a torsional spring of stiffness
Solution: Draw the free-body diagram first.
Take a moment about to obtain ©
X
= ̈
3 sin − = ̈ 2 3
73 We assume a small rotation (2 1) so that sin can be approximated as . The linearized equation of motion is µ ¶ 3 + − = 0 3 2 The natural frequency is
=
s
( − 2) 2 3
74
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
40. Solve Problem 39 for the initial conditions (0) = 0 and ̇(0) = ̇0 for two cases of the stiffness: and 2 2 What is the solution for = 2? Solution: If the equivalent stiffness of the linearized system is positive, and () will 2 oscillate about the static equilibrium position (vertical position). For the given initial conditions, the response is ̇0 () = sin q − 12 If the equivalent stiffness is negative, and the response will grow. where = 3 2 The solution near = 0 is given by ⎞ ⎛s 1 2 − () = 1 exp ⎝ ⎠ 3 ⎞ ⎛ s 12 − ⎠ +2 exp ⎝− 3 Upon substituting the initial conditions, we obtain s 3 ̇0 1 = 2 12 − s 3 ̇0 2 = − 2 12 −
The response is
⎛s
⎞ 1 − ̇0 3 2 () = exp ⎝ ⎠ 1 2 2 − 3 ⎛ s ⎞ s 1 2 − 3 ̇0 exp ⎝− ⎠ − 1 2 2 − 3 s
We must keep in mind that this solution is good for small only (2 1) If = 2 the linearized EOM is reduced to ̈ = 0 and has a solution () = 1 + 2 With given initial conditions, we find () = ̇0 This solution is unstable and is only valid for 2 1
75
41. A bar supported by a hinge at its base is held in place by a spring connected to a collar, as per Figure 2.69. The spring is unstretched when the bar is vertical. As the bar is displaced from equilibrium by an angle the collar slides on the frictionless bar so that the spring remains horizontal. Derive the equation of motion using an energy method, and then linearize the model for small amplitude motion. Identify the frequency of oscillation.
Figure 2.69: An inverted bar is held in place by a spring connected to a collar. Solution: The kinetic energy is given by 1 2 = ̇ 2 2 where = 3 The potential energy is given by 1 = cos + ( sin )2 2 2 The total energy is constant and is given by 1 1 2 = + = ̇ + cos + ( sin )2 2 2 2 Take a derivative to obtain 0 = ̈ − sin ̇ + ( sin ) cos ̇ 2 Divide by ̇ to obtain the equation of motion: ̈ − sin + 2 sin cos = 0 2 2 We assume small angles ( 1) so that sin ≈ and cos ≈ 1 The linearized equation of motion is ¶ µ 2 ̈ + − = 0 2 If the natural frequency is 2 s − 12 = 3
76
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
42. A simple pendulum of initial length 0 and initial angle 0 is released from rest. If the length is a function of time according to = 0 + , find the position ( ) of the pendulum at any time assuming small oscillations. The governing equation of motion will turn out to be (0 + )̈ + 2̇ + = 0 and in transformed Bessel form, 2 00 + 20 +
= 0 2
where = 0 + . (This problem requires the use of Bessel functions.) Solution: Let us first derive the equation of motion shown. We use Newton’s second law. The energy method using the total energy will not give the right answer because this is a non-natural system (a system with prescribed motion). Summing the moment about O, ( ) + (0 + ) sin = 0 where = ̇ and = (0 + )2 The moment balance equation becomes (0 + )2 ̈ + 2 (0 + ) ̇ + (0 + ) sin = 0 Dividing this equation by (0 + ) and assuming a small rotation will result in the equation of motion in the problem statement: (0 + )̈ + 2̇ + = 0 This equation can be transformed into a Bessel equation by change of variables. Let us start with (0 + )2 ̈ + 2 (0 + ) ̇ + (0 + ) = 0 Let = 0 + and the derivatives are given by = = µ ¶ 2 2 2 = = 2 2 Then, 2 2 00 + 22 0 + = 0 2 00 + 20 + 2 = 0 which is the second equation in the problem statement. From the theory of Bessel functions, the solution of differential equation 2 00 + (2 + 1) 0 + (2 2 + 2 ) = 0
77 where are constants, is the general equation ¤ £ = − ( ) + ( )
p where = 2 − 2 In the above problem, we have 2 + 1 = 2 = 0 = 12 2 = 2 = = 12 Making these substitutions, our solution for becomes ∙ ¸ √ √ 12 12 −12 = 1 (2 ) + 1 (2 ) ∙ √ −12 1 (2 (0 + )12 ) = (0 + ) ¸ √ 12 +1 (2 (0 + ) ) The first initial condition is that ( = 0) = 0 The second initial condition is that ̇( = 0) = 0 Therefore, √ √ ∙ ¸ 0 0 −12 1 (2 ) + 1 (2 ) 0 = (0 ) and ∙ √ ̇() = − 1 (2 (0 + )12 ) 32 2(0 + ) ¸ √ 12 +1 (2 (0 + ) ) √ √ ∙ 0 1 (2 (0 + )12 ) + 0 + ¸ √ 0 12 +1 (2 (0 + ) ) ̇(0)
√ √ ∙ ¸ 0 0 1 (2 ) + 1 (2 ) = − 2(0 )32 √ √ ¸ √ ∙ 0 0 0 0 1 (2 ) + 1 (2 ) + 0 √ √ ¸ √ ∙ 0 0 0 0 1 (2 ) + 1 (2 ) = − 0 + 20 0 = 0 We can use this last equation to find √ √ 0 0 0 0 0 ) + 1 (2 ) √ = 1 (2 2
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CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
Solve √ this equation simultaneously with the equation for 0 above for and (omit the argument 2 0 ) : √ 0 √ 0 1 − ((2 ))1 = 0 1 10 − 1 10 √ √ ((2 ))1 − 0 10 = 0 1 10 − 1 10 Use the identity ()0 () − 0 () () = 2() for = 1 1 10 − 10 1 = √ 0 and then √ √ 0 0 0 0 0 1 − 1 = 2 √ √ 0 0 0 0 0 = 1 − 1 2 From the identity 0 () = () − +1 () and a similar identity for 0 for = 1 we have √ √ 0 0 2 0 = − 2 ( ) √ √ 0 0 2 0 2 ( ) = The equation for becomes √ ∙ √ √ 0 0 2 p 2 0 = √ )1 ( 2 ( 0 + ) 0 + √ ¸ √ 2 p 2 0 −2 ( )1 ( 0 + ) Figure below shows the locations of the pendulum for 0 = 1 m, 0 = 15◦ and = 002 s−1 The snapshots are taken for the first 20 s at 0.02 s interval.
79
80
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
43. A homogeneous disk of weight and radius is supported by two identical cylindrical steel shafts of length as shown in Figure 2.70. From solid mechanics, for each shaft the relation between the moment on the disk and the angle of rotation of this disk is =
where is the shear modulus of the material, is the polar moment of inertia of the cross-section of each shaft, and is known as the torsional rigidity. Suppose this system is being designed for a particular application that requires the frequency of vibration to be . Find the value of in terms of the parameters to satisfy this requirement.
Figure 2.70: A homogeneous disk of weight is supported by two identical cylindrical steel shafts Solution: The mass moment of inertia of the disk is = 2 2 and the stiffness of the problem is 2 The equation of motion is given by ̈ + 2 The natural frequency is =
=0
p (2) 2 Writing in terms of we have s r 1 2 1 1 = = 2 2 2
The radius of the disk in terms of all the other parameters is r 1 1 =
81
44. When a manned craft is sent into space it is necessary to know the mass properties of the system, including the moment of inertia of astronauts on the flight, in order to accurately calculate its trajectory and fuel expenditures. The device in Figure 2.71 is one way to measure an astronaut’s moment of inertia. The horizontal platform is pinned at and supported a distance away by a spring with constant . When the astronaut is not present, the frequency of small vibration of the platform about is measured to be . When the astronaut is lying on the platform, the frequency of small vibration is measured to be . (a) Find the astronaut’s moment of inertia about the axis. (b) Assuming reasonable parameter values for the system and an the astronaut of height 2.0 m, calculate the moment of inertia.
Figure 2.71: System to measure astronaut’s moment of inertia.
Solution: (a) In absence of the astronaut, the platform oscillates at frequency The equation of motion is given by X ª = − − 2 = ̈ ̈ + 2 = −
where is the moment that the platform exerts due to its weight. This can be omitted if is measured from static equilibrium. The natural frequency is s 2 1 = ((1)) 2 If the astronaut lies on the platform, the moment equation becomes X ª = − − − 2 = ( + ) ̈
where is the moment that the astronaut exerts due to his weight. This can again be omitted if is measured from static equilibrium. The natural frequency of the combined system is s 1 2 = ((2)) 2 +
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CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
Let’s say that we know the spring constant, the length, and both frequencies, and but we do not know the moments of inertia. Then, we first find the moment of inertia of the platform, using Equation (1) and find using Equation (2). (b) Let us assume that = 10 000 N/m, = 700 Hz, and = 300 Hz. Then, = 207 kg·m2 and 919 kg·m2
83
45. A small pendulum is mounted in a rocket that is accelerating at a rate of 4g, as sketched in Figure 2.72. The pendulum is composed of a massless rod of length = 1 m that supports a block of mass = 05 kg. Assuming small oscillations, what is the rotational natural frequency of the pendulum?
Figure 2.72: An accelerating rocket with internal pendulum. Solution: Draw a free-body diagram.
Summing the forces in the transverse direction (̂ ) we have X = P where = − sin and = ̈ + 4 sin Then, the equation of motion becomes ³ ´ ̈ + 4 sin + sin = 0 ̈ + 5 sin = 0 2 For small p rotation ( 1, sin ≈ ), we have ̈ + (5) = 0 The natural frequency is = 5
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CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
46. For the undamped rocker arm sketched in Figure 2.73, determine the natural frequency of the system undergoing small amplitude oscillation using Newton’s second law of motion. Assume that the mass of the T-bar is negligible compared to the mass of the block.
Figure 2.73: Vibrating rocker arm. Solution: Draw a free-body diagram based on a rotation defined positive counterclockwise direction.±
The moment about is given by ª where = 2 and The equation of motion is
ª
X
X
= ̈
= −2 sin ( cos ) − sin
2 ̈ + 22 sin cos + sin = 0 ¡ ¢ Assuming small rotation 2 1 sin ≈ cos ≈ 1 we obtain the linearized equation of motion: ¡ ¢ 2 ̈ + 22 + = 0
The natural frequency is
=
p (22 + ) 2
85
47. For the system of Figure 2.74 with spring stiffness , bar mass and length , and end mass , derive the equation of motion for small oscillations about the horizontal equilibrium position assuming the support pin is frictionless. Based on the equation of motion, determine the effective mass and effective stiffness of the system.
Figure 2.74: Pinned bar-spring system. Solution: It is assume that the equilibrium position is when the bar is horizontal. We measure rotation from this static equilibrium and the gravity can be omitted from the problem. Draw a free-body diagram.
Summing the moment about O, we have ª where = 13 2 + 2 and ª
X
X
= ̈
µ ¶2 sin cos = − 2
Assuming small rotation, the linearized equation of motion is given by ¶ µ ¶2 µ 1 2 2 = 0 + ̈ + 3 2 Let us divide above equation by and rewrite the equation of motion in terms of the vertical displacement of the block, = Then µ ¶ 1 + ̈ + = 0 3 4 The equivalent stiffness is 4 and the equivalent mass is + 3
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CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
48. Consider a generalization of Example 2.10, where a mass moves to the left with speed on a platform, disconnected from two springs, as shown in Figure 2.75. Assuming that there is no friction, find the period of oscillation of the mass, given the data: 1 = 36 N/cm, 2 = 18 N/cm, = 25 kg, = 30 cm, = 100 cm, and = 6 m/s.
Figure 2.75: Mass disconnected from springs oscillating. Solution: The mass moves to the left with a constant velocity of Let = 0 when the mass comes into contact with 1 The equation of motion is given by ̈1 + 1 1 = 0 with 1 (0) = 0 and ̇1 (0) = where 1 is the position of the left face of the block defined positive to the left. The solution is valid until 1 () becomes zero again as the block leaves the spring p on the left. The block is in contact with the spring on the left for half the normal period or 1 The block travels the distance of − at velocity to the right until it comes in contact with the spring on the right. The time of p travel is ( − ) The block is in contact with the second spring for half the normal period or 2 After it loses contact, it travels the distance ( − ) at a constant velocity to reach where the block reaches 1 = 0 The period is − − = p +p + + 1 2 07 07 = p +p + + 6 6 3625 1825 = 655 s. Note that we can write a single equation of motion in terms of defined from the middle: ̈ + () = 0 where − − ≤≤ 2 2 = elsewhere.
() = 0 for −
87
49. A block of mass is suspended from the ceiling. The equation of motion is ̈ + = 0 where is measured from the static equilibrium position. Assume the initial conditions are (0) = 1 in and ̇(0) = 1 in/s. Solve for the response () for two cases: (a) the spring is assumed massless, and (b) the mass of the spring is included. Plot both solutions and comment on the importance of including the mass of the spring. Assume = 1 lb and a spring mass that is 10% of the block mass for = 01, 1, and 10 lb/in. What conclusions can be drawn?
Solution: Massless Spring: The EOM is given by ̈ + = 0, and its solution is given by = 1 cos + 2 sin where p = We use the initial conditions (0) = 1 in ̇(0) = 1 in/s to find the constants: 1 = 1 2 = 1 Thus, the solution to our differential equation is = cos + 1 sin with defined as q above. In the amplitude-phase form, we can write = 1 + (1)2 cos ( − )
Mass of Spring Included: The analysis including the mass of spring is identical to that for the massless spring, except we replace by = + 3 where is the spring mass. The natural frequency is then = p and the solution is given by = cos + 1 sin In the amplitude-phase form, we can write =
q ¡ ¢ 1 + (1 )2 cos −
Note that the units of the mass and stiffness must be converted to slug and lb/ft to obtain the natural frequency in terms of rad/s. (The lbm and slug are related by 1 slug = 32.2 lbm .) The figures below show the variation in the natural frequencies and the response amplitudes as functions of the spring mass and stiffness for = 1 lbm .
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CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
The natural frequency and amplitude as functions of spring mass. The mass is kept at 1/32.2 slug.
The natural frequency and amplitude as functions of stiffness. The mass is kept at 1/32.2 slug.
89
50. Two systems with discontinuities are shown in Figure 2.76. In each system the mass oscillates and comes into contact with springs at some distance ± . For each system, derive the equation of motion for a complete cycle.
Figure 2.76: (a) A mass with springs engaging at a distance. (b) A mass with multiple springs engaging at a distance. Solution: (a) The equation of motion is given by ̈ + () = 0 where () = 0 for − ≤ ≤ = elsewhere. (b) The equation of motion for this system is ̈ + () = 0 where () = 2 for − ≤ ≤ = 3 elsewhere. The easiest way to solve for the response and the period is to solve them in different segments.
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CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
Problems for Section 2.4 — Harmonically Forced Vibration 51. Derive Equation 2.37, () =
0 sin + 0 cos + [cos − cos ] 1 − ( )2
Solution: We begin with our known relationships, namely: ̈ + 2 = () cos () = The solution to this equation consists of two parts, the homogeneous and particular solutions, and they are given by () = 1 cos + 2 sin () = 1 cos The constant coefficients of integration, 1 and 2 are determined when the initial conditions are applied to the total solution. The coefficient for the particular solution must be such that the assumed solution, () satisfies the differential equation. Upon substituting the assumed solution, () into the differential equation, we obtain
Using the fact that =
¡ ¢ 1 2 − 2 cos = cos
p we find that: 1 =
1 − ( )2
Note that = the static response of the spring. Finally, we substitute 1 into the expression for (), and use the initial conditions to find: 1 = 0 − 2 =
0
1 − ( )2
Substituting 1 and 2 into our expression for () gives the solution: ¶ µ cos () = 0 − 1 − ( )2 0 + sin + cos 1 − ( )2
91
52. Solve the equation of motion,
1 () for: (a) () = cos 05, (b) () = cos 099, (c) () = cos , and (d) () = cos 2. Plot the responses and compare the results. Let = 1 with appropriate units. ̈ + =
Solution: Each of these loading cases has a homogeneous response and a particular response. The homogeneous solution is given by () = 1 cos + 2 sin Note that the natural frequency equals 1 rad/s We will only solve for the particular solution below. (a,b,d) Assume a particular solution () = cos + sin Differentiate this assumed solution twice and substitute it into the governing differential equation of motion to find ¡ ¢ ¡ ¢ 1 cos −2 + + sin −2 + = cos
¡ ¢ ¡ ¢ This identity is actually two equations: 1 − 2 = 1 and 1 − 2 = 0 Therefore, = 0 and ¡ ¢ ¢ ¡ = 1 1 − 2 The particular response is then () = 1 1 − 2 cos The total solution is given by () = 1 cos + 2 sin +
1 ¡ ¢ cos 1 − 2
(c) In this problem, the loading is at the same frequency as the natural frequency of the system. Since the system is undamped, we know to expect oscillation with unbounded growth. Assume a particular solution () = cos + sin Differentiate this assumed solution twice and substitute it into the governing differential equation of motion to find:
cos (2) + sin (−2) =
1 cos
This identity is actually two equations: 2 = 1 and −2 = 0 Then = 0 and = 12 The particular solution is () = (12) sin The total solution is
() = 1 cos + 2 sin + The following plot shows the unstable response for = 1.
1 sin 2
92
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION 15 10
x( t)
5 0 -5 -10 -15
0
5
10
15
20
25
t
Resonant response, () = 05 sin It should be noted that, in many of the vibration texts, the transient and homogeneous solution are used synonymously, and the steady-state and the particular solution are used synonymously. This is true for a damped system because the homogeneous solution decays with time. In this case, the homogeneous solution is the transient solution. However, in an undamped system, the homogeneous solution persists so that it is no longer the transient response (there is no transient response in an undamped system). This is not a gross error because there is no physical system that is truly undamped. Therefore, from an engineering perspective, the transient and homogeneous solution are the same, and likewise for the steady-state and the particular solutions.
93
53. For the oscillator that is beating according to the equation ̈ + 16 =
sin(4 − )
where is small but not equal to zero, what do you expect to happen as → 0? Relate the rate of growth of the response amplitude to the value of or to the value of 4 − . Solution: We know from our studies of resonance that a structure loaded harmonically by a force that oscillates at the natural frequency of the structure will oscillate with ever-increasing amplitudes. Therefore, even before analyzing the above governing differential equation, we know that for = 0 we have a resonance condition. Even for small we will have large amplitude motion. We can solve the differential equation for the forced response to find: 1 () = sin(4 − ) (4 − )2 − 16 +1 cos 4 + 2 sin 4 ´ ³ 4− Applying the initial conditions, we find 1 = 0 and 2 = − [4−]2 −16 4 The total solution is given by () ¶ µ 1 4− = sin 4 sin(4 − ) − (4 − )2 − 16 4 | {z } forced solution
1 + 0 cos 4 + sin 4 {z4 } | free solution
The free solution, the part of the solution due to the initial conditions, does not grow with time. In this current form, we can see the term −16 + (4 − )2 in the denominator of the forced response goes to zero as → 0 resulting in a response of unbounded amplitude given enough time. However, it does not show how the solution behaves in finite time. In order to investigate the behavior of the solution in finite time, look at the term, sin(4 − ) in the forced solution. This term is expanded as cos sin 4 − sin cos 4. The terms, cos sin sin 4 and cos 4 2 (4 − ) − 16 (4 − )2 − 16
describes sinusoidal functions that oscillates at 4 rad/s and whose magnitude varies slowly with frequency rad/s. The magnitude of the envelop function is 1/([4 − ]2 − 16) which approaches infinity as → 0 Therefore, resonance can be thought of beating with an infinite beat period and envelop that grows to infinity in time. In order to investigate how fast the response grows as → 0 we do the following. Using the method of partial fraction expansion, we write µ ¶ 1 1 1 1 − = 8 −8 [4 − ]2 − 16
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CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
The term, sin(4 − ) is expanded as cos sin 4 − sin cos 4 Then, ¶ µ 1 4− sin 4 sin(4 − ) − (4 − )2 − 16 4 () µ ¶ 1 1 1 − (cos sin 4 = 8 −8 ¶ 4− − sin cos 4 − sin 4 4 µ ¶ ∙µ ¶ 1 1 4− 1 − cos − sin 4 = 8 −8 4 − sin cos 4] µ 1 cos − 4− sin 4 = sin 4 − cos 4 8 −8 −8 ¶ cos − 4− sin 4 − sin 4 + cos 4 As → 0 the forced solution becomes
µ ¶ 1 1 0 − 0 − sin 4 + cos 4 lim () = →0 8 4
where the L’Hopital’s rule is used to evaluate the last two terms such that cos − 4− − sin + 14 1 4 lim = lim = →0 →0 1 4 sin cos lim = lim = →0 →0 1 Finally, the total solution is given by ¶ µ 1 1 1 − sin 4 + 0 cos 4 + cos 4 () = 4 32 8 As → 0 the solution grows linearly with time. This is a long way to find the resonant solution. The fast way will be set = 0 from the beginning in the equation of motion.
95
54. The block shown in Figure 2.77 is acted on by the force () = 100 + 25 sin 75 N and, after the transients have died out, it oscillates with an amplitude of 0.6 mm about a position 55 mm to the right of the static equilibrium position corresponding to the condition when no force is present. What is the mass of the block?
Figure 2.77: Block subjected to external force.
Solution: Let us assume that there is very small damping. It is small enough that the damping can be ignored when the oscillation amplitude is calculated, but the transient response can still die out given time. The new static equilibrium position is (static force) = 100 = 0055 m. Then, = 1000055 = 18182 N/m. The amplitude of the steady-state solution is given by || =
1 q (1 − 2 )2 + (2)2
Assuming that the damping is very small or zero, the amplitude is given by ¯ ¯ ¯¯ 1 ¯¯ || = ¯ 1 − 2 ¯
From the problem the amplitude of the dynamic portion of the force is 25 N, and the response amplitude is 0.0006 m. We just found that = 1818 N/m. Then, we find = 4891 The natural frequency is then, 75 = = = 1534 rad/s. 4891 The mass is 1818 = 7732 kg = 2 = (1534)2
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CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
55. For Example 2.12, what range of frequencies of the motion () must be excluded to keep the maximum force at less than 10 N? Solution: The equation of motion is 1 ̈ + = sin 2 The force at is + ( − 2) downward. Assuming that a very small damping is included in the model, the steady-state response is 1 1 () = sin 2 − 2 The force at is
¶ 1 sin = + − 2 1 − 2 2 2 ´ ´ ³ ³ 1 1 where oscillates between 1 = + 2 1−2 2 − 2 and 2 = − 2 1−2 2 − 2 We consider two cases. If |1 | is larger than |2 | Then, the frequency must be such that |1 | 10 N This is when the motion of the rod and the mass are in phase. The frequency must be such that 138 rad/s to have the force on less than 10 N If |2 | is larger than |1 | Then, the frequency must be such that 2 10 N This is when the motion of the rod and the mass are 180◦ out of phase. In this case, the frequency that satisfies this is 167 rad/s. To avoid force larger that 10 N, we must avoid 138 167 rad/s. It is much easier if we plot 1 and 2 as shown below. µ
97
56. A gantry crane system is shown schematically in Figure 2.78. The crane is a simple pendulum of length and mass . The gantry oscillates with the displacement () = sin If is very small, what gives the crane an amplitude of motion equal to 2? Assume = 1 m.
Figure 2.78: A model of a gantry for a crane. Solution: The amplitude of motion depends on the initial conditions especially if the system is undamped. Let us assume that the system has very small damping so that the transient effects die out. Then, we want to find the frequency of the base motion so that the steady-state response has an amplitude of 2 in the horizontal direction Draw a free-body diagram and sum the forces in the horizontal and vertical directions.
From Newton’s second law X
=
2 − sin = ( sin + sin ) X =
cos − =
2 (− cos )
98
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
Multiplying the first by cos and second by sin and adding the two equations, we have 2 ( sin + sin ) cos 2 + (− cos ) sin + sin = 0
and ̈ + sin = 2 sin cos Assuming small rotations, the equation of motion becomes 2 ̈ + = sin with a steady-state solution () =
2 1 sin − 2
The horizontal motion is = sin + sin ≈ + sin µ ¶ 1 2 + sin () = − 2 For the state-state amplitude of the bob to be 2, 1 + = 2 − 2
2
the base must have a frequency of = 221 rad/s (for = 1 m).
99
57. Solve for the response of the equation of motion, ̈ + 9 = 3 sin + cos 3 assuming zero initial conditions.
Solution: At first glance, we know that the system has the natural frequency of 3 rad/s, which coincides with the forcing frequency. Therefore, the response will consist of a harmonic response at the natural frequency of 3 rads/, another harmonic response at the forcing frequency of 1 rad/s, and a harmonic response that grows linearly with time at 3 rad/s. Exact solution is 1 3 1 1 () = − sin 3 + sin + sin 3 8 8 6 and is shown next. 2 1.5 1
x(t)
0.5 0 -0.5 -1 -1.5 -2
0
2
4
6
8
10
t
1 () = − 18 sin 3 + 38 sin + 16 sin 3 Let us also solve for the response due to each term on the right hand side separately. The exact solution for 00 + 9 = sin (0) = 0 0 (0) = 0 is () = − 18 sin 3 + 38 sin The harmonic loading is at a frequency of 1 while the natural frequency is 3 Therefore, we expect a periodic response as shown next.
100
CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
x(t)
0.5
0
-0.5
0
2
4
6
8
10
t
− 18 sin 3 + 38 sin The second loading case is given by 00 + 9 = cos 3 (0) = 0 0 (0) = 0 Here the loading is at the natural frequency, and thus, we expect unbounded growth. The exact solution is obtained by assuming a solution of the form () = cos 3 + sin 3 Differentiate and substitute into the differential equation of motion. We find 16 sin 3 1.5 1
x(t)
0.5 0 -0.5 -1 -1.5 -2 0
2
4
6
8
10
t 1 sin 3 6 1 1 cos + 23 sin cos2 − 16 sin − 36 cos 3 (Note that MAPLE evaluates the solution to be 19 cos3 − 12 which is identical to our solution. Plot the difference and you will obtain zero. This is just a word of warning on being too dependent on canned programs. We depend on them, but it is always a good idea to study the results and try to verify the results using back-of-the-envelope computations.)
101
58. Solve for the response of the governing equation of motion, ̈ + 2 =
sin
for arbitrary initial conditions. Solution: The free vibration has been solved previously, resulting in the general solution () = 1 cos + 2 sin where is the natural frequency We will only solve for the particular solution next. In this problem, the loading is at the same frequency as the natural frequency of the system. Since the system is undamped, we know to expect oscillation with unbounded growth. Assume a particular solution of the form, () = cos + sin Differentiate this assumed solution twice and substitute it into the governing differential equation of motion to find: 2 [ cos + sin ] 2 = −2 (sin ) − (cos ) 2 +2 (cos ) − (sin ) 2 The differential equation becomes −2 (sin ) − (cos ) 2 +2 (cos ) − (sin ) 2 +2 cos + sin = () sin Next we combine like terms: (−2 + (− 2 + 1)) sin +2 cos = () sin This equation is an identity of two equalities: −2 + (−2 + 1) = and 2 = 0 which must now be solved for and simultaneously: −2 − 2 + = 2 = 0 The solution is
1 2 Note that above simultaneous equations must be satisfied for all time The total solution is = 0 and = −
() = () + () = 1 cos + 2 sin −
1 cos 2
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CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
The constants of integration, 1 and 2 are determined from initial conditions. Let (0) = 0 and ̇ (0) = 0 and the total solution is given by ¶ µ 2 + sin () = cos + 22 1 cos − 2
103
59. Derive Equation 2.41, 2 () = [sin( + ) sin ] 2( + )
Solution: Begin with the equation () =
[cos − cos ] 1 − ( )2
Use the trig identities for sums and differences of angles to obtain ∙ ¸ ( + ) ( − ) 2 sin sin () = 1 − ( )2 2 2 and make the substitution = 2 + Then, 2 [sin( + ) sin ] 1 − ( )2 2 = ³ 2 2 ´ [sin( + ) sin ]
() =
− 2
2 2 [sin( + ) sin ] ( − )( + ) 2 2 = [sin( + ) sin ] 2(2 + 2)
=
Finally, we obtain () =
2 [sin( + ) sin ] 2( + )
Note that = ( − ) 2 and ( − ) 2 ( − ) ( + ) 2( + ) 2 = 2 2 ( − ) ( + ) = 2 µ ¶2 2 − 2 = =1− 2
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CHAPTER 2 SDOF VIBRATION: AN INTRODUCTION
60. A cylinder of mass is mounted in a water tunnel as shown in Figure 2.79 with the cylinder axis transverse to the flow direction. When there is no flow, a vertical force of on the cylinder results in a deflection of . With flow in the tunnel, shedding vortices impart alternating forces on the cylinder. The velocity of the water is in the direction, the distance between vortices is , and the magnitude of the lateral forces is The lateral forces can be modeled by the harmonic function () = sin . Find the amplitude of lateral steady-state response of the cylinder.
Figure 2.79: Cross-section of an elastically mounted cylinder excited by shedding vortices.
Solution: This problem can be modeled as a mass supported by a spring as shown below.
It is given that the force is required to deflect distance . This gives us the stiffness of the leaf spring as = It is said that the vortices are shed at every seconds. This gives us the forcing frequency of = 2 () The equation of motion of the simple model is ̈ + = sin where is the vertical deflection. The steady-state response (assuming very small damping so p that the transient response dies out) amplitude is ( − 2 ) where = and = 2 () It is assumed that and are known.
Chapter 3 Single Degree-of-Freedom Damped Vibration Problems for Section 3.4 — Free Vibration with Viscous Damping
1. Solve Equation 3.6, ̈ + 2 ̇ + 2 = 0 for 1
Solution: We assume the general solution () = exp() Differentiate twice and substitute into the differential equation to find: 2 + 2 ( ) + 2 exp() = 0 This simplifies to the quadratic characteristic equation 2 + 2 + 2 = 0 which is solved by the quadratic formula:
12 = (− ±
p 2 − 1)
For 1 the roots 12 are real and different, giving us pthe general solution: () = 1 exp(1 ) + 2 exp(2 ) Again, since 1 we know that − ± 2 − 1 0 and therefore, both roots are negative, as we expect, resulting in a decaying motion with no oscillation. √ As a numerical example, assume = 11 and = 1 rad/s Then, 12 = −11 ± 112 − 1 = −0 64174 −1 5583 Assuming 1 = 2 = 1 the solution is () = 1 exp(1 ) + 2 exp(2 ) = exp(−064174)+exp(−15583) which is plotted below. Note the initial conditions that correspond to 1 = 2 = 1 are (0) = 2 and (0) = −22 105
106
CHAPTER 3 SDOF VIBRATION: WITH DAMPING 2
x(t)
1.5
1
0.5
0
0
1
2
3
4
t
() = exp(−064174) + exp(−15583)
5
107
2. Derive Equation 3.9, () = − cos( − ) Solution: This equation is derived by rewriting the general solution for the case 1 (since this is an oscillatory solution) given by √ 2 √ 2 ´ ³ () = − 1 1− + 2 − 1−
where 1 and 2 must be complex conjugates to render () real. Using Euler’s formula, = cos + sin and letting 12 = ( ∓ ) 2 we obtain () = − ( cos + sin ) We write this in the amplitude-phase form as () = − cos ( − ) where cos = sin = The amplitude and the phase are then given by =
√ 2 + 2 and = tan−1
Care should be taken to find the phase angle in the correct quadrant. For instance, if 0 and 0 the phase angle is in the second quadrant.
108
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
3. (a) The response of an oscillator with mass = 1 kg is shown in Figure 3.51. What can you say about the properties of the oscillator and the response? Determine the equation of motion and the initial conditions. (b) One property is changed, with all others remaining the same. The resulting time history now looks like Figure 3.52. What can you say about the properties of the oscillator and the response? Determine the equation of motion and the initial conditions.
Figure 3.51: Free vibration for Problem 3(a).
Figure 3.52: Free vibration for Problem 3(b).
Solution: (a) The oscillator is undamped and vibrates with an amplitude of 1 m with a period of a little more than 6 seconds. It takes about 18.4 s to complete three cycles. Therefore, the natural frequency is = 2 ' 2(1843) = 102 rad/s Since there is no decay, the equation of motion is ̈ + 2 = 0 The initial displacement is zero. The initial velocity is about, by approximating the slope at = 0 s, 0405 = 08 m/s It is more accurate to calculate the initial velocity from the amplitude and phase. The free response is given by () = 0 cos + 0 sin With 0 = 0 we see that 0 is the amplitude of vibration. Therefore, the initial velocity is 0 = = 102 m/s We take the latter value of 102 m/s as the initial velocity. (b) Here, the one additional property is damping. We see the effect of damping in free vibration by the reduced peak amplitude of vibration with each cycle. The initial conditions are the same as above. The equation of free vibration is ̈ + 2 ̇ + 2 = 0 From the equation of the logarithmic decrement, we can estimate and therefore We know that µ
= ln +1
¶
2 =p 1 − 2
From the graph we can estimate ' ln(098089) = 0096. It is not necessary to use the two consecutive peaks. For example, we can use the first and the third peak. Write the logarithmic decrement as µ ¶ µ ¶ 2 1 = ln = ln 2 3
109 Then µ
¶ µ ¶ 1 2 2 = ln + ln 2 3 ¶ µ µ ¶ 1 1 1 2 1 = ln = ln 2 2 3 2 3 Using the first and the third peak, we have µ ¶ 1 098 = ln = 0101 2 08 q Let’s use this value. Therefore we can solve for = 2 (4 2 + 2 ) = 00162 Now, we can find the p damping constant to be = 2 = 00329 The damped natural frequency is = 1 − 2 = 102 The equation of motion is ̈ + 00329̇ + 104 = 0 In order to find the initial conditions, let us start from the solution () = −00165 ( cos 102 + sin 102) The initial displacement is (0) = = 0 The maximum response occurs when = 0 or () = −00165−00165 sin 102 + −00165 102 cos 102 = (−00165 sin 102 + 102 cos 102) −00165 = 0 102 tan 102 = 00165 = 152 467 · · · The maximum response is (152) = −00165(152) ( sin 102 (152)) = 0975 = 098 (read from plot). = 1005 The initial velocity is ¯ () ¯¯ = 102 ¯=0 = 1025 Let us keep only 2 significant figures, the equation of motion and initial conditions are ̈ + 0033̇ + 10 = 0 (0) = 0 and ̇ (0) = 10.
110
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
4. (a) Derive the equation of motion for a mass-spring-damper system in free vibration. Solve for the transient response () that is driven by the initial conditions (0) = 0 and ̇(0) = 0 . (b) Assume parameter values: = 1 lb/in, weight = 100 lb, 0 = 0, 0 = 10 in/s. Vary the damping constant so that both underdamped and overdamped responses can be demonstrated. Try values of such that the cases = 01 and = 09 are obtained. (c) For the case with = 01, vary the initial velocity 0 and study the variation of the first intercept (zero displacement) as a function of initial velocity. Solution: (a) The equation of motion for a free mass-spring-damper system is given by ̈ + ̇ + = 0
or
̈ + 2 ̇ + 2 = 0
The form of the solution changes depending on the value of 2 − 4 Overdamped Response: For 2 − 4 0 (equivalently 1) the system is overdamped, and the solution takes the form of √ √ −+ 2 −1 −− 2 −1 () = + Upon substituting the initial conditions, the constant coefficients of integration are found to be à à ! ! 1 + + 1 + p 2 − p 2 and = = 2 2 − 1 − 1 Critically Damped Response: For 2 −4 = 0 (equivalently = 1) the system is critically damped, and the solution takes the form of () = ( + ) exp (− ) where the constant coefficients of integration are = and = + Underdamped Response: For 2 − 4 0 (equivalently 0 1), the system is underdamped, and the solution takes the form of () = exp (− ) ( cos + sin ) p where = 1 − 2 . Upon substituting the initial conditions, the constant coefficients of integration are found to be + = and = An alternate form of the solution for the underdamped system is () = exp (− ) [ cos ( − )]
111 Upon substituting the initial conditions, we find that 0 = cos and 0 = −( cos + sin ) The amplitude and the phase, and can be solved for, and they are s ¶2 µ 0 + 0 0 + 0 2 = 0 + and = tan−1 0 where it should be checked that the correct phase angle is chosen. in/s. The mass is (b) It is given that = 1 lb/in, weight = 100 lb, 0 = 0, and 0 = 10 √ 2 = = 100(322 × 12) = p0259 lb-s in. The damping coefficient is = 2 The damped natural frequency is = 1 − 2 For = 01 and 0.9, we find = 01
= 09 √ √ = 2 × 01 × 0259 × 1 = 0102 lb s/in = 2 × 09 × 0259 × 1 = 0916 lb s/in p = 10259 = 1 97 rad/s = 197 rad/s √ √ = 197 1 − 012 = 1 96 rad/s = 197 1 − 092 = 0 857 rad/s
tan−1 ∞ = 2 rad
tan−1 ∞ = 2 rad
=01 () = 5 10−0197 sin 196
=09 () = 117−177 sin 0857
These two curves are plotted below. 5 4 3
x (t) (in.)
2 1 0 -1 -2 -3 -4
0
2
4
6
8
10
t (s)
=01 () and =09 () While the numbers 117 would lead us to expect that =09 () would have larger amplitudes than those of =01 () the very large value for quickly dissipates the oscillation. (c) We use the general solution and vary the initial velocity: () =
0 −0197 sin 196 196
112
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
Figure next shows the solution when the initial velocities are 1, 5, 10, 20 in/s. The amplitudes approach zero exponentially as → ∞ 10 8 6
x (t) (in.)
4 2 0 -2 -4 -6 -8
0
2
4
6
t (s)
() for various 0
8
10
113
5. A mass-spring-damper system is tested to determine the value of . Assume = 10 lb/in and = 2 slug. (a) If the vibrational amplitude is observed to decrease to 33% of its initial value after 2 consecutive cycles, what is the value of ? (b) If , rather than having the exact value given above, has a range of possible values of 8 to 12 lb/in with equal likelihood, then what is the range of possible values of ? Solution: (a) Vibration amplitude decreases to 33% of its initial value after 2 cycles: Let us keep ¡ three ¡ 100 ¢ figures. We can use the logarithmic decrement to evaluate We know ¢ 1 significant = 1 ln 100 ln = = 0554 We proceed by setting this value to: 2 33 2 0554 = p 1 − 2
q and solve for to find = 2 (4 2 + 2 ) = 00886 Note that 1 slug = 1 lb s2 /ft In order to calculate we first need to change the unit of stiffness to lb/ft (or p of mass to lb √ change the unit 2 s in) The stiffness is = 120 lb/ft. Now we can evaluate = 2 = 2(00886) 120(2) = 274 lb-s/ft. (b) Possible range for if has a range of possible values of 8 to 12 lb/in with equal likelihood: In this case we can substitute the upper and lower bounds into the equation for and thereby obtain the range of values for as follows. p p 2 min ≤ ≤ 2 max p p 2(00 886) 8(12)(2) ≤ ≤ 2(00 886) 12(12)(2) 2 45 lb-s/ft ≤ ≤ 301 lb-s/ft.
114
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
6. A periodic torque with a maximum value of 1.0 N-m at a frequency of 5.0 rad/s is applied to a flywheel suspended from a wire. The flywheel has a moment of inertia of 0.10 kg-m2 and the wire has a stiffness of 1 N-m/rad. A viscous component applies a damping moment of 0.5 N-m at an angular velocity of 1 rad/s. Calculate (a) the amplitude of the angular displacement from the rest position during the steady-state response, and (b) the amplitude of the moment applied to the viscous component during the steady-state response. Solution: The equation of motion is given by ̈ + ̇ + ̇ = 0 cos N-m = 05 N-m-s/rad, = 1 N-m/rad, 0 = 1 N-m and = 50 where = 010 kg-m2 = 05 1 rad/s ³ √ ´ p √ rad/s. Thus, = = 10 rad/s, and = 2 = 0791 (a) The angular displacement depends on the initial conditions that are not given. The problem must be asking for the maximum angular displacement during steady-state response. The steadystate response is given by
() =
1 0 q cos ( − ) (1 − 2 )2 + (2)2
where = For the values given, the amplitude is
1 1 1 0 q = q √ ¡ ¢2 = 0373 rad. 1 (1 − 2 )2 + (2)2 (1 − 225)2 + 2 (0791) 225
The maximum angular displacement is 0373 rad. (b) The maximum couple moment applied to the viscous component is ̇max or ̇max = max = 05 (0373) (5) = 0933 N-m.
115
7. A periodic force () = cos is applied to the damped system shown in Figure 3.53. Calculate the work done in one cycle during the steady-state response where = 50 N, = 10 rad/s, = 6 kN/m, = 200 N-s/m and = 25 kg.
Figure 3.53: Mass suspended between a spring and a damper.
Solution: The work done per cycle by the periodic force is given by Z · = Z 2 = () 0 We need to find the velocity. Although it is not mentioned in the problem statement explicitly, it is required to find the work done during one cycle after the transient effects have died out. Otherwise, the velocity depends on the initial conditions which are not given. The equation of motion is given by ̈ + ̇ + = cos which has a steady-state solution of cos ( − ) () = q 2 2 2 (1 − ) + (2)
p where = and tan = 2 (1 − 2 ) Then, the velocity is
() = −q sin ( − ) (1 − 2 )2 + (2)2
116
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
The work done per cycle is =
Z 2
()
0
Z 2
⎛
⎞
cos ⎝ q sin ( − )⎠ 2 2 0 (1 − 2 ) + (2) Z 2 2 cos sin ( − ) = −q (1 − 2 )2 + (2)2 0 ´ ³ 2 sin 2 − sin = q = −q 2 2 2 2 2 2 (1 − ) + (2) (1 − ) + (2) = −
For the values given, = 50 N, = 10 rad/s, = 6000 N/m, = 200 N-s/m and = 25 kg, we first find r = = 240 rad/s = 0646 = = = 0258 2 µ ¶ 2 −1 = tan = 0519 1 − 2 The work done per cycle is 2 sin = 0966 J =q 2 2 2 (1 − ) + (2)
117
8. This problem investigates the number of cycles required for a structural oscillation amplitude to decay to % of its maximum. (a) What is the expression for the logarithmic decrement in terms of and ? (b) Solve for in terms of and . (c) Plot as a function of for (i) = 70%, (ii) = 50%, (iii) = 20% using the exact formula, and then using the approximate formula for = 01 and = 05. Solution: (a) Expression for the logarithmic decrement in terms of and : The logarithmic decrement is defined as 1 2 = ln = = p 2 1 − 2 If instead of the decay over one cycle, we need the decay over cycles, we can write µ ¶ 1 100 = ln where is in percent. (b) Solve for in terms of and : In order to solve for in terms of and , we start with µ ¶ 1 100 2 = ln =p 1 − 2 The number of cycles, is then given by p µ ¶ 1 − 2 100 ln = 2
(c) Plot as a function of : Figure below show the plot of as a function of for = 70 50 and 20 The solid lines are the curves for the exact and dotted lines for the approximate ¡ solution ¢ 1 solution. The approximate solution is = 2 ln 100 Note that the approximate solution always overestimates
The number of cycles versus damping ratio.
118
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
9. Solve for in Equation 3.17 in general for arbitrary variables. Then evaluate numerically for the cases in Table 3.1. Plot each case and discuss trends.
Table 3.1: Cases for max Case (0) ( in) ̇ (0) (in/s)
(rad/s)
1
0
5
0.1
2
2
5
0
0.1
2
3
5
5
0.1
2
4
5
5
0.5
2
5
5
5
1.1
2
6
5
5
0.1
4
7
5
5
0.1
10
Solution: The responses are underdamped for all cases except for the fifth one, which is overdamped. Underdamped cases: Begin with the general solution for the underdamped system: () =
(0) exp(− ) cos( − ) cos
To find the maximum (), we solve for the time when is zero: − ( cos( − )) = − − cos ( − ) − − sin ( − ) = 0 We continue to solve for : − cos ( − ) − sin ( − ) = 0 tan ( − ) = − ∙ µ ¶ ¸ 1 −1 tan − + max = where = tan−1 ((̇(0) + (0) ) ((0) )) Overdamped cases: The overdamped response is given by () = 1 + 2
119 This response has a maximum when = 0 or 1 1 max + 2 2 max = 0 1 1 max = −2 2 max 1 (2 −1 )max = − 2 ¶ µ 1 1 max = ln − 2 − 1 2 For = 11 (fifth case), the system is overdamped. The EOM is given by ̈ + 44̇ + 2 = 0 The characteristic values, 12 and the constants of integration, and are 1 = −12835 2 = −31165 = 112287 and = −62287 The time when the maximum occurs is max = 01625 Case (0) (in) ̇ (0) (in/s)
(rad/s) (rad/s)
max (s)
1
0
5
0.1
2
1.99
1.570
0.7390
2
5
0
0.1
2
1.99
0.1002
0
3
5
5
0.1
2
1.99
0.5426
0.2224
4
5
5
0.5
2
1.73
0.8571
0.1925
5
5
5
1.1
2
N/A
N/A
01620
6
5
5
0.1
4
3.98
0.3382
0.0598
7
5
5
0.1
10
9.95
0.1984
0.0099
120
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
10. Coulomb used the following method to determine the viscosity of liquids. A thin plate of weight is suspended vertically and set into motion, first in air and then fully submerged in the particular liquid as shown in Figure 3.54. The time duration, 1 in air and 2 in the liquid, required for one oscillation is measured. The frictional force between the plate and fluid is estimated as 2, where 2 is the area of both sides of the plate, is the coefficient of viscosity, and is the plate velocity. The frictional force between the plate and air is negligible. Find the value of in terms of the above parameters.
Figure 3.54: Plate suspended in a liquid.
Solution: We have two regimes of behavior. In air, the plate moves without dissipation; there is no damping. In water, there is frictional resistance between the plate and the water. This experiment therefore consists of two parts, one a free vibration in air where there is no dissipation, and the other in water where there is dissipation. Let 1 and 2 be the period of undamped oscillation and the damped period of oscillation, respectively. For the undamped free vibration we have the following: ̈ + = 0 2 =
1 =
2
For the damped free vibration, we have ̈ + (2)̇ + = 0
q = 1 − 2
2 =
(2)
2 =
2
121 Here, it is assumed that the plate is fully submerged in the fluid at all times. We use the equation for and solve it for in the following way: 2 =
2 2 1 = p =p 2 1 − 1 − 2
22 (1 − 2 ) = 21 2 − 2 2 = 2 2 1 2 µ ¶2 2 − 2 = 2 = 2 2 1 2 µ ¶2 2 2 − 21 2 = 22 ¶2 2 µ 2 − 21 2 2 = 1 22 µ ¶p 2 2 − 21 2 = 1 2
122
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
11. A bell-crank mechanism depicted in Figure 3.55 is rotated slightly and released to oscillate in free vibration. Derive the damped frequency of oscillation and the critical damping constant The mass of the frame can be ignored compared to the mass attached to it at the end.
Figure 3.55: Bell-crank mechanism.
Solution: Draw a free body diagram based on positive rotation in the CCW direction.
Note that the force exerted by the damper is obtained by =
( sin ) = ̇ cos
123 Take a moment about to obtain ª
X
= ̈
− sin ( cos ) − ̇ cos ( cos ) − sin = 2 ̈ where the block is treated as a point mass so that = 2 and = 2 Note that if the dimensions of the block are not negligible compared to length we must include it in the expression for the mass moment of inertia of the block. Simplifying, we obtain the equation of motion: ¢ ¡ 2 ̈ + 2 cos2 ̇ + sin + 2 sin cos = 0 Assuming small angles, we obtain the linearized equation of motion given by ¡ ¢ 2 2 2 = 0 ̈ + ̇ + + |{z} |{z} | {z }
from which we find
r + 2 = = 2 2 2 2 √ = = q = p 2 2 2 +2 2 + 2 2 s r q + 2 2 4 2 = 1 − = 1 − 4 ( + 2 ) 2 2 1 p = 4 ( + 2 ) 2 − 2 4 22 √ 2 2p = 2 = = (2 ) ( + 2 ) 2 2 r
124
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
12. An automobile’s suspension system determines the vehicle’s response to inputs from uneven road surfaces. Figure 3.56 shows a drawing of a suspension system that includes a coil spring, a shock absorber, and connecting links to the car frame and the wheel. Next to this artist’s rendering is a simple model that can be used to estimate the vertical oscillations of the system. From a static state, the system is given an initial velocity of ̇(0) = 15 m/s. Calculate the response () of the simplified system for the following parameter values: = 40 kg, = 25 kN/m, and = 24 kN-s/m. There are no external forces acting on the wheel.
Figure 3.56: A cross-section of an automotive suspension system. Next to this sketch is a single degree-of-freedom simplified model.
Solution: The equation of motion is given by ̈ + ̇ + = 0 where = 40 kg, = 2400 N-s/m, and = 2500 N/m with initial conditions (0) = 0 m and ̇ (0) = 15 m/s. This system is over damped because 2 − 4 0 Assume a solution of the form, () = exp () and substitute into the equation of motion to obtain the characteristic equation, 2 + + = 0 402 + 2400 + 2500 = 0 The characteristic values are = −4658 and = −1342 The response is given by () = 1 exp (−4658) + 2 exp (−1342) Substitute the initial conditions to find 1 + 2 = 0 −46581 − 13422 = 15
125 Solving, the constants of integration are 1 = −04524 m and 2 = 04524 m The response is () = −04524 (exp (−4658) − exp (−1342)) m Note that exp (−1342) is the dominant term of the two.
126
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
Problems for Section 3.7 — Harmonic Excitation and Damped Response 13. For the system in Figure 3.57, the forcing per unit mass is () = 10 sin 15 ft/s2 . Solve for the transient response, the steady-state response, and the constants of integration for the following initial conditions. (a) (0) = 0 in, ̇(0) = 5 in/s (b) (0) = 5 in, ̇(0) = 0 in/s (c) (0) = 5 in, ̇(0) = 5 in/s. For the parameter values = 40 lb, = 085 lb-s/in, = 125 lb/in, write down the complete response. Plot the transient and steady-state responses separately, then plot both of these with the complete solution on the same graph. Compare the results and discuss the significance of the initial conditions on the total response as time evolves.
Figure 3.57: Harmonically driven oscillator.
Solution: The equation of motion is ̈ + () ̇ + (2) = 10 sin 15 where we are given and The mass is found from = = 40(322) = 1 242 slugs. If we use slugs, we must convert all the length units to ft. The force per unit mass, () is given in terms of s2 /ft. The damping and spring constants in proper units are = 1020 lb-s/ft and = 150 lb/ft. The initial conditions are converted to ft and ft/s. The specific EOM becomes ̈ + 821̇ + 2415 = 10 sin 15 where () is in ft. The natural frequency is = 1554 rad/s. First, we solve for the homogeneous solution. The first step is to establish whether the system is under, critically, or overdamped by evaluating The value of is given by =
1020 = 02642 = 2 2 × 1242 × 1554
Since 1 the system is underdamped and will oscillate. It has the solution form of () = exp(− )(1 cos + 2 sin ) p p where = 1 − 2 = 1554 1 − (02642)2 = 1499 rad/s
127 We next solve for the particular solution due to harmonic forcing. Assume the particular solution of the form () = 1 cos 15 + 2 sin 15 Differentiate this twice and substitute into the forced differential equation to find 1 = −007977 ft and 2 = 001069 ft. To this point, this solution is valid in general. The initial conditions are then applied to the sum of the homogenous and the particular solutions to find the constants 1 and 2 . The general solution is () = exp(−4105)(1 cos 1499 + 2 sin 1499) −007977 cos 15 + 001069 sin 15 The following equations must be satisfied in order to match the initial conditions: (0) = 1 − 007977 ̇(0) = − 1 + 2 + 15 (001069) The constants of integration, 1 and 2 are given by 1 = (0) + 007977 ̇(0) + ((0) + 007977) − 15 (001069) 2 = ̇(0) + (02642) (1554) ((0) + 007977) − 15 (001069) = 1499 For case (a), we have 1 = 007977 ft and 2 = 03447 ft. For case (b), we have 1 = 50798 ft and 2 = 1381 ft. For case (c), we have 1 = 50798 ft and 2 = 1714 ft. The responses are shown in the next figures.
128
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
14. For a mass-spring-damper system under harmonic loading per unit mass () = cos ft/s2 , solve for the magnification factor for the case where = 20 lb/in, = 40 lb, and then plot for a broad range of driving frequencies . If 0 ≤ ≤ 3 , discuss design considerations for the system as the force is varied from the rest state ( = 0) to its highest frequency ( = 3 ). Consider the damping cases: = 001 01 05. Discuss parameters of possible concern such as and whether material yielding could be of concern. Solution: The magnification factor, is given by the following relation: 1 =p 2 (1 − )2 + (2)2
where = The magnification ratios for = 001 0.1, and 0.5 are shown below for 0 ≤ ≤ 3
Magnification factor for = 001 0.1, and 0.5. As the damping pratio increases, the peak value for decreases. The maximum magnification ratio occurs max = 1 − 2 2 and the maximum is max =
1 p 2 1 − 2
The maximum magnification ratio for = 001 0.1, and 0.5 are max = 500 5025 and 11547 For the given parameters, = 40322 = 124 slugs, and = 20 lb/in = 240 lb/ft. The natural p frequency is = 240124 = 139 rad/s. The maximum displacement is ¶ µ ¡ 2 ¢ 124 slugs = max 1 ft/s max = max = max 240 lb/ft = 00052 max ft.
129 For = 001 0.1, and 0.5, max are max = 0260 00261 0.006 ft. For the designer, the interest lies with the frequency range where there is the greatest magnification. The concern with maximum displacement are two-fold. The first is that if a component vibrates with large amplitude, it may hit adjacent components. Secondly, if the vibration amplitude is too large, the system may start operating beyond the elastic and/or small displacement limit, for which the equations are no longer valid, fatiguing the materials.
130
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
15. Solve Problem 14 for = 40 lb/in and = 20 lb.
Solution: For the given parameters, = 20322 = 0621 slugs, and = 40 lb/in = 480 lb/ft. p The natural frequency is = 4800621 = 278 rad/s. The maximum displacement is ¶ µ ¡ 2 ¢ 0621 slugs max = max = max 1 ft/s 480 lb/ft = 00013 max ft. For = 001 0.1, and 0.5, max are max = 00650 00065 00015 ft.
131
16. Solve Problem 14 for = 20 lb/in and = 20 lb.
Solution: For the given parameters, = 20322 = 0621 slugs, and = 20 lb/in = 240 lb/ft. p The natural frequency is = 4800621 = 278 rad/s. The maximum displacement is ¶ µ ¡ 2 ¢ 0621 slugs max = max = max 1 ft/s 240 lb/ft = 00026 max ft. For = 001 0.1, and 0.5, max are max = 0130 00131 0003 ft.
132
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
17. For Problem 14 where = 20 lb/in and = 40 lb, solve for the steady-state response in ft/s2 to the following excitation forces per unit mass: (a) () = 100 sin (09 ) (b) () = 100 sin ( ) (c) () = 100 sin (11 ) (d) () = 100 sin (05 ) (e) () = 100 sin (2 ). Plot and discuss each case. Solution: For the given parameters, = 40322 = 1 24 slugs and = 20 lb/in = 240 lb/ft p The natural frequency is = 2401 24 = 139 rad/s Let us solve for a general case where () = sin where has units of ft/s2 The equation of motion is given by ̈ + 2 ̇ + 2 = sin ̈ + 2 (139) ̇ + 193 = sin The steady-state response (which is also equal to the particular solution for a damped system) is then given by () = 2 sin ( − ) where 2 is the static deflection if the system were to be forced by the static force per mass of is the magnification factor given by 1 =p (1 − 2 )2 + (2)2
where = and the phase angle is given by = tan−1
2 2 = tan−1 2 2 − 1 − 2
In all cases, the static deformation, 2 is 100 (240124) = 0518 ft. The table below shows and for each case. (a) 09
(b) 1
(c) 11
0.01
0.1
0.5
0.01 0.1
0.5
0.01 0.1
0.5
5.24
3.82
1.09
50.0 5.00
1.00
4.74 3.29
0.893
944 × 10−2
0758
1.363
2
2
304 2.33
176
2
133 (d) 05
(e) 2
0.01
0.1
0.5
0.01
0.1
0.5
1.33
1.32
1.11
0.333 0.330 0.277
133 × 10−2
0133
0588
313
3.00
255
As an example, the steady-state solution for part (c) with = 01 is plotted below. The steady-state solution is sin ( − ) = 0518 (329) sin (11(139) − 233) 2 = 170 sin (153 − 233) ft.
() =
2 1.5
x (t) (ft)
1 0.5 0 -0.5 -1 -1.5 -2
0
0.5
1
1.5
2
2.5
t (s)
Response for Part (c) and = 01
3
134
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
18. For what range of frequency ratios will the magnification factor be greater than 1 for the idealized model of Figure 3.58? For what range of frequency ratios will the magnification factor be greater than 86% of the peak response? Let = 1 slug, = 9 lb/in, and = 015. Show the cases on a plot of the magnification factor vs. frequency ratio.
Figure 3.58: Harmonically driven oscillator.
Solution: The magnification factor is given by the expression: 1 =p 2 2 (1 − )2 + (2 ) p p p For = 015 and = = (9 × 12) 1 = 104 rad/s, = 1 (1 − 2 )2 + (03)2
Magnification ratio for = 015
135 Range of frequency ratios for which 1 To find the range of frequency ratio for 1 we can either solve the inequality algebraically or read from the plot of Reading from the graph that 1, the frequency ratio is approximately less than 1.4. From the exact expression, 1 p 1 (1 − 2 )2 + (03)2
we find 1382 which is fairly close to the approximate value of 1.4. This means that if the loading frequency is in the range: 14 × = 14 × 104 = 146 rad/s, there will be a magnification of the response. Range of frequency for which 086 max The maximum magnification ratio occurs when = 0 The maximum magnification ratio p 2 occurs at max = 1 − 2 and the maximum value is max =
1 p 2 1 − 2
The plot shows that the peak response to be about 3.4, and 86% of the maximum value is 2.92. From the figure, we can estimate this occurs for 09 11 We can solve for thepfrequency ratio algebraically. The maximum magnification occurs when = 0 or max = 1 − 2 2 The maximum magnification is ¶ µ q 2 max = ( = max ) = 1 2 1 − For = 015 max = 338 which is very close to the approximate value of 3.4. It should be noted that the forcing frequency at which the maximum magnification occurs is not the natural frequency but slightly less than the natural frequency. The frequency ratio that satisfies 086 max is obtained by solving 1 p 086 (338) (1 − 2 )2 + (03)2 p 1 (1 − 2 )2 + (03)2 086 (338) µ ¶2 1 (1 − 2 )2 + (03)2 086 (338)
The range is found to be 0884 1063
136
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
19. For the system of Figure 3.58, what do the initial conditions have to be so that both integration constants and equal zero in Equation 3.37? Solution: The total solution consists of the homogeneous solution and the particular solution as given in Equation 3.37 or () = exp (− ) cos ( − ) + 1 cos + 2 sin
(3.37)
If the integration constants, and , are zeros, the particular solution alone must satisfy the initial conditions. That is, (0) = 1 ̇ (0) = 2
137
20. Derive and to verify Equations 3.38 and 3.39. Solution: Begin with Equation 3.37: () = exp(− ) cos( − ) + 1 cos + 2 sin The initial conditions are (0) and ̇(0) Applying the initial conditions to Equation 3.37, we obtain two simultaneous equations: (0) = cos + 1 ̇(0) = − cos + sin + 2 From the first equation, we derive Equation 3.38: =
(0) − 1 cos
(3.38)
From the first equation, we solve for cos : cos =
(0) − 1
From the second equation, we solve for sin : sin =
̇(0) − 2 + ((0) − 1 )
Dividing sin by cos we derive Equation 3.39: ∙ ¸ −1 ̇(0) − 2 + ((0) − 1 ) = tan ((0) − 1 )
(3.39)
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CHAPTER 3 SDOF VIBRATION: WITH DAMPING
21. For each case in Table 3.2, for the governing equation ̈+2 ̇+ 2 = cos , evaluate and , and 1 and 2 in Equation 3.37. Then plot the free and forced responses separately and then in sum. Let = 1 cm/s2 and = 1 rad/s. (a) = 05 rad/s = 01 (0) = 1 cm ̇(0) = 1 cm/s (b) = 1 rad/s
= 01 (0) = 1 cm ̇(0) = 1 cm/s
(c) = 10 rad/s = 01 (0) = 1 cm ̇(0) = 1 cm/s (d) = 05 rad/s = 01 (0) = 0 cm ̇(0) = 1 cm/s (e) = 05 rad/s = 01 (0) = 10 cm ̇(0) = 1 cm/s (f) = 05 rad/s = 01 (0) = 1 cm ̇(0) = 0 cm/s (g) = 05 rad/s = 01 (0) = 1 cm ̇(0) = 10 cm/s (h) = 05 rad/s = 001 (0) = 1 cm ̇(0) = 1 cm/s (i) = 05 rad/s = 095 (0) = 1 cm ̇(0) = 1 cm/s
Solution: Equation 3.37 is given by () = exp(− ) cos( − ) + 1 cos + 2 sin where 2 − 2 ( 2 − 2 )2 + (2 )2 2 2 = ( 2 − 2 )2 + (2 )2 (0) − 1 = cos ̇(0) − 2 + ((0) − 1 ) = tan−1 . ((0) − 1 ) 1 =
(3.1) (3.2) (3.38) (3.39)
Note that there are two sets of and that satisfy the last two equations. For example, if the argument of tan−1 is positive, can be in the first or third quadrant. If the argument of tan−1 is negative, can be in the second or fourth quadrant. Both sets of answers are correct. It is convenient to take between −2 and 2 rad The values for each case are tabulated next.
139
(rad/s) (a)
0.5
(0)
̇(0)
1
(cm) (cm/s) (rad/s) (cm)
2
(cm)
(cm)
(rad)
01
1
1
04975
−1310 01747 2.985 0.6860
(b) 1
01
1
1
0.9950
0
5000
(c)
10
01
1
1
99499
00101
00002 1.010 0.1993
(d) 05
01
0
1
04975
−1310 01747 2.219 0.9392
(e)
05
01
10
1
04975
−1310 01747 11.65 0.2423
(f)
05
01
1
0
04975
−1310 01747 2.313 -0.0514
(g)
05
01
1
10
04975
−1310 01747 20.11 1.456
(h) 05
0.01 1
1
05000
−1333 00178 3.065 0.7057
(i)
0.95 1
1
01561
−0512 06485 7.017 1.354
05
4.045 -1.321
(a) = 05, = 01 (0) = 1 ̇ (0) = 1 Using the tabulated values above, the solution for (a) is given by () = exp(−005) (213 cos 04975 + 189 sin 04975) {z } | transient
−1310 cos {z + 01747 sin } | steady-state
or
cos {z + 01747 sin } () = 2985 exp(−005) cos(04975 − 06860) + −1310 {z } | | transient
steady-state
The free and forced responses are similar but different from the transient and steady-state response. The free response is the part of the response due to the initial conditions only without the forcing term, and the forced response is the part of the response due to the forcing term only without the initial conditions. That is, the free and the forced responses satisfy the following equation of motion and initial conditions. ̈ + 2 ̇ + 2 = 0 (0) = 1 and ̇ (0) = 1
(1)
̈ + 2 ̇ + 2 = cos (0) = 0 and ̇ (0) = 0
(2)
Due to the principle of linear superposition, the total solution is the sum of the two or () = () + ()
140
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
It is not easy to separate out the free and forced responses from the transient and steady-state responses. It is better to solve the differential equations again. The solution to Equation (1) has the same form as the transient response and () = exp(−005) ( cos 04975 + sin 04975) Applying initial conditions, () = exp(−005) (cos 04975 + 211 sin 04975) The forced response has the form of () = exp(−005) ( cos 04975 + sin 04975) −1310 cos + 01747 sin Applying the zero initial conditions, we find = 131 and = −0219. The forced response is () = exp(−005) (131 cos 04975 − 0219 sin 04975) −1310 cos + 01747 sin Combining the free and the forced responses, we confirm that they add up to the total response. It should be noted that the steady-state response is not the same as the forced response. The next figures show the total, free, and forced responses.
Total response -, free response -., and forced response - -.
141
22. Derive Equations 3.44 and 3.45: 1 |()| = q (1 − ( )2 )2 + (2 )2 ∙ ¸ 2 −1 = tan 1 − 2 2 Solution: Let us start with a simple example. A complex number, + , can be drawn on the real-imaginary plane as shown below.
Another way of writing the complex number is to use its length and the angle such that + = cos + sin Using the Euler formula, we can write + = exp () where √ = 2 + 2 and = tan−1 Care must be taken to choose the correct quadrant for If the complex number is given by 1 ( + ) we can write 1 1 1 =√ exp (−) =√ + 2 + 2 exp () 2 + 2 where
= tan−1 Note that is in the same quadrant as + For the problem at hand, begin with Equation 3.44: () =
1 1 − 2 2 + 2
() can be written in the amplitude-phase form:
() = |()| exp (−) 1 1 |()| = q (1 − 2 2 )2 + (2 )2 = tan−1
2 1 − 2 2
142
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
The phase angle must be chosen so that it is in the same quadrant as the vector (1 − 2 2 ) + (2 ) on the complex plane.
143
23. Derive 1 and 2 in Equations 3.48 and 3.49: − 2 1 + 2 2 + 2 1 = 1 − 2 2 − 2 1 + 2 2 = 2
Solution: These are two simultaneous equations. The goal is to solve for 1 and 2 Let us put the equations in a matrix form, ⎡ ⎤⎡ ⎤ ⎡ ⎤ 2 − 2 2 1 1 ⎣ ⎦⎣ ⎦=⎣ ⎦ 2 2 −2 − 2 2 Inverting the coefficient matrix, ⎤ ⎡ ⎤⎡ ⎤ ⎡ 2 2 − −2 1 1 ⎦= ⎣ ⎦⎣ 1 ⎦ ⎣ 2 2 2 2 ( − ) + (2 ) 2 2 2 − 2 2 Note that 1 and 2 are
1 = 2 2 = 2 Then, ⎡ ⎣
1 2
⎤
⎦ = =
1 2 ( 2 − 2 ) + (2 )2
1 2 ( 2 − 2 ) + (2 )2
⎡
2 − 2 ⎣ ⎡ ⎣
−2 2 − 2
2
−2 3 2
(2 )
which is identical to the expression shown in the text.
⎤⎡ ⎦⎣
2 2
+ (2 − 2 ) 2
⎤
⎦
⎤ ⎦
144
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
24. Plot , Equation 3.52, vs. for = 005 010 and 025. Solution: It will be simpler to define = Then, ∙ ¸12 1 + (2)2 2 = (1 − 2 )2 + (2)2 For example, for = 025 : ( = 025) = 2
s
1 + (2(025))2 (1 − 2 )2 + (2(025)2
All three cases are plotted next.
for = 025
145
25. The base-excited system of Figure 3.59 is driven by a force with frequencies in the range 10 ≤ ≤ 20. For parameter values = 1 kg, = 9 N/cm, and = 10 cm, select a value of so that the average magnitude of the relative motion equals 1 cm over the frequency range.
Figure 3.59: A model of a base-excited system.
Solution: Let us start with the equation of motion, ̈ + ̇ + = ̇ + where () is the motion of the mass seen from a fixed frame. The relative motion of the mass, () is given by () = () − () The equation of motion in terms of () is given by ̈ + ̇ + ̇ = −̈ or ̈ + 2 ̇ + 2 ̇ = −̈ The amplitude ratio is then given by 2 1 =q =q 2 (2 − 2 ) + (2 )2 (1 − 2 )2 + (2)2 where is the amplitude of the response of the mass relative to the base motion, and is the amplitude of the base motion. We seek an average value of equal to 0.01 m or = 00101 = 01 over 1 ≤ ≤ 2. The average is given by R2 1
( ) = 01 R2 1
R2 The integral, 1 ( ) , is difficult to evaluate even for a given value of We can, however, integrate it numerically. A simple method is the trapezoidal rule. The figure below shows over R2 0 3 and the numerical value of the integral, 1 ( ) for 05 5.
146
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
Amplitude ratio vs. and
R2 1
( ) vs.
R2 From the figure, 1 ( ) is 0.1 when is about 3.5. The corresponding damping constant is = 2 = 21 N-s/m. Note that although () = () − () the amplitude does not obey the same rule. That is, 6= − . This is because of the phase differences among the base motion, total response, and the relative response. For example, the amplitude of 3 cos − 2 sin is not 5. Therefore, although tempting, we cannot find using − = = − 1
147
26. Solve for the response of the base-excited system governed by ̈+2 ̇+2 = 2 ̇+ 2 , where () = exp( ) for the response () = () exp(). Plot the response for parameter values: = 1 = 01 = 1 rad/s and = 05 rad/s.
Solution: The particular solution (which is also the steady-state response when damping is present) is () =
(2 + 2 ) exp () (2 − 2 ) + 2 | {z } ()
We can rewrite () in the amplitude-phase form such that q 4 + (2 )2
exp ( ( − )) () = q 2 ( 2 − 2 ) + (2 )2 where 2 2 2 = tan−1 2 − 2
= tan−1
where must be in the first quadrant (because both numerator and denominators are positive), and must be in either the first or second quadrant (because the numerator is always positive and the denominator can be either positive or negative). Then, the steady-state response is () = () exp () q 4 + (2 )2 = q exp ( ( + ( − ))) 2 2 2 2 ( − ) + (2 ) If the base excitation is a sine function, the response is also a sine function with a phase lag of − The solution is plotted for = 1 = 01 = 1 rad/s = 05 rad/s. We find that () = 0356 sin ( − 263)
148
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
0.4 0.3 0.2
x (t)
0.1 0 -0.1 -0.2 -0.3 -0.4
0
5
10
15
t
20
25
30
149
27. The response of a system excited at its base by () = sin is given by () = sin( − ) with
¸12 ∙ 1 + (2 )2 = (1 − 2 2 )2 + (2 )2
Use Figure 3.38 for displacement transmissibility in answering the following questions. (Do not solve any equations.) Assume negligible damping. (a) Assume that the maximum amplitude of the base is = 1 cm. A design for a rotating machine with this value of and with natural frequency = 3 rad/s is required so that ≤ 2 cm across all frequencies . Which value should be chosen?
(b) Suppose that instead of the criteria in (a), we require ≤ 1 cm and 10. The machine operates at = 9 rad/s. What are the options for satisfying this criterion? (c) What is the equation for the maximum force transmitted to the base in terms of what is the force transmissibility ( ) when the ( ) and ? For = 10, √ machine is running at ' 0, = 3 2 rad/s, and = 9 rad/s?
Solution: (a) Based on the displacement transmissibility plot in Figure 3.38, a possible value of that satisfies the criteria is = 050 A value of = 025 is too low, and is greater than 2 around = 1 (b) It is not possible to satisfy all the requirements for the whole frequency √ range. Therefore, one option is to control the vibration amplitudes for the range ≤ 2 A feedback control mechanism can be added that reduces the vibration amplitudes in this range. (c) The transmitted force is given by = −̈ = 2 sin( − ) where q = [1 + (2 )2 ][(1 − 2 2 )2 + (2 )2 ] We can write the magnitude of the transmitted force as ¸12 1 + (2 )2 | | (1 − 2 2 )2 + (2 )2 ¸12 µ ¶2 ∙ 1 + (2 )2 = (1 − 2 2 )2 + (2 )2 ¯ ¯ µ ¶2 ∙ ¸12 ¯ ¯ 1 + (2 )2 ¯ ¯ = ¯ ¯ (1 − 2 2 )2 + (2 )2 = 2
∙
150
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
If we multiply the displacement transmissibility by ( )2 we obtain the normalized force transmissibility. Therefore, we have the following equivalence for = 10 : ≈0
≈0 √ 3 2 = 1 4142 =⇒ =2 = 3 2 9 =⇒ = (3)2 = 60 = =3 3 3 √ Note that the normalized force transmissibility is always 2 when = 2 for any =⇒
151
28. A vibration sensor has a damping ratio = 055 and a natural frequency = 10 Hz. Estimate the lowest frequency that can be measured with 1% error.
Solution: Let us assume that the vibration sensor is a seismometer in for Problem 25. The relative motion, () is recorded. We want the properties of the seismometer such that () is close to the ground motion () and report it as the ground motion. The equation of motion is given by ̈ + 2 ̇ + 2 = −̈ where the ground motion is assumed harmonic, () = sin Let () = exp ( ) and after substitution, we find ( ) = 2
1 ( 2 − 2 ) + (2 )
= || exp ( ( − ))
The amplitude and the phase of the recorded motion are given by 2 || = q | | 2 2 2 (1 − ) + (2) µ ¶ 2 −1 = tan 1 − 2 where = As → ∞ the amplitude of the relative motion approaches that of the ground motion. It is also interesting to note that phase difference between the relative motion and the ground motion approaches 2. The absolute error is ⎞ ⎛ 2 ⎠ | | Absolute error = | | − || = ⎝1 − q 2 2 2 (1 − ) + (2) The percent relative error associated with reading the amplitude of the relative motion as the amplitude of the ground motion is | | − || × 100% | | ⎛ ⎞ 2 ⎠ 100% = ⎝1 − q 2 2 (1 − 2 ) + (2)
Percent relative error =
Clearly, the error depends on and Figure next shows the relative error as a function of for various values of
152
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
For the percent relative error to be within ±1% for = 055 the frequency ratio must be 62276 There is a small range of near 1 when the relative error falls below 1% However, as a practical matter, we would not design a seismometer hoping that falls in this small range. For a natural frequency of = 10 Hz, the seismometer can detect ground frequency greater than 62276 Hz.
153
29. An engine of mass 200 kg is to be supported on four helical springs. When the engine speed is 1000 rpm there is a primary vertical periodic force of maximum value 350 N due to the unbalanced reciprocating masses. Assume that the engine vibrates in the vertical direction with negligible damping, and there is no horizontal or angular movement. (a) Find the stiffness of each spring to limit the maximum total periodic force on the foundation to 25 N. (b) What will be the amplitude of vibration of the engine when its speed is 500 rpm? Use the minimum natural frequency found in part (a) Solution: The equation of motion for a system with a rotating unbalance is given by ̈ + 2 ̇ + 2 =
0 2 sin
From the problem statement, = 200 kg, = 1000 × 260 = 1047 rad/s, and 0 2 = 350 N. We assume that the damping is very small so that the transient solution dies out, and the steady-state response can be approximated using = 0. Then, the steady-state solution is given by () =
1 0 2 sin 2 ( − 2 )
(a) Find the stiffness so that | | 25 N. The force transmitted to the foundation is | | = | ()| ¯ ¯ ¯ 1 0 2 ¯¯ ¯ = ¯ 2 − 2 ¯
and we want this number to be less than 25 N Then, ¯ ¯ ¯ 0 2 ¯¯ 1 ¯ 25 ¯ 2 − 2 ¯ ¯ ¯ ¯ 1 ¯ 2¯ ¯ 25 0 ¯ 1 − 2 ¯ ¯ ¯ ¯ 1 ¯ ¯ ¯ 25 350 ¯ 1 − 2 ¯ ¯ ¯ ¯1 − 2 ¯ 350 25 Two ranges are possible,
1 − 2 1−
350 25
350 2 25 √ 13
or or or
¡ ¢ 350 − 1 − 2 25 350 2 +1 25 √ 15
154
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
√ where the first range is invalid. We find that 15. Equivalently, 270 rad/s and 146 × 105 N/m. There are four springs in parallel. Therefore, the stiffness of each spring must be less than 365 kN/m. (b) Amplitude of vibration at = 500 rpm. The amplitude of vibration is ¯ ¯ ¯ 1 0 2 ¯¯ ¯ ¯ 2 2 − ¯
where it was given that 0 2 was 350 N when = 1000 rpm. At = 500 rpm, 0 2 is reduced by a factor of 4. Let us use the natural frequency found in the previous section, = 270 rad/s. Then, the amplitude of vibration is ¯ ¯ ¯ ¯ ¯ ¯ ¯ 3504 1 1 0 2 ¯¯ ¯ = ¯ ¯ ¯ 2 − 2 ¯ 200 ¯ (27032 − 52332 ) ¯ = 2179 × 10−4 m, or 0.2179 mm.
155
30. A rotating machine component has an eccentricity of approximately = 01 mm in its center of mass resulting in a harmonic load on the structure. Assume, using the notation of the chapter, that 0 = 005, operating frequency = 100 rad/s, = 010, and = 10 rad/s. (a) Solve for the amplitude and phase of the response in the form of Equation 3.53. (b) Suppose amplitude must be reduced by 15%. What options exist if actively controlling the structure is not one? Discuss which parameters need to be changed. Solution: (a) The response is given by the equation () = sin( − ) 2 2 p = sin( − ) (1 − 2 2 )2 + (2 )2 2 ) = tan−1 ( 1 − 2 2 For the values given, solve for the ratio : 2 2 p = (1 − 2 2 )2 + (2 )2
1002 (10)2 = 005 × p = 00505 (1 − (100)2 (10)2 )2 + (2 × 010 × 10010)2
is a dimensionless quantity, and is given as 0.1 mm. Therefore, the response amplitude is 000505 mm. (b) Looking at the solution in part (a), we see that the ratio can be reduced by decreasing or increasing for example. It may improve the precision of the manufacturing process so that and/or are reduced. Note that as is increased, becomes smaller and the denominator in varies nonlinearly. Such parameter variations need to be examined carefully.
156
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
31. Derive the general equations of motion, Equations 3.55 and 3.56, (̈ − Ω̇2 cos Ω − Ω̈ sin Ω) + = 0
(̈ − Ω̇2 sin Ω + Ω̈ cos Ω) + = 0
Solution: Start with
2 ( + cos Ω) + = 0 2 Then differentiate with respect to time to find
(̈ − Ω̇2 cos Ω − Ω̈ sin Ω) + = 0 Note that the chain rule is used to differentiate cos Ω, and the chain rule and the product rule to differentiate ̇ sin : Ω cos Ω = cos Ω = −Ω̇ sin Ω Ω Ω ̇ sin Ω = Ω̈ sin Ω + Ω̇ sin Ω = Ω̈ sin Ω + Ω̇ sin Ω = Ω̈ sin Ω + Ω̇2 cos Ω Ω Similarly, starting with we obtain the second equation.
2 ( + cos Ω) + = 0 2
157
32. A simple model of a seismometer is shown in Figure 3.60. The top view shows how a pen attached to the seismometer mass traces out a pattern on a paper grid that is on a roller. The side view provides more details of the model. The absolute motion of the ground is defined by () the relative motion of the mass recorded by the seismometer is (), and parameter values are = 1 kg, = 3 N-s/m, and = 15 N/m. The seismometer is initially stationary until ground motion initiates motion, governed by the differential equation, ̈ + 2 ̇ + 2 = () where () = −̈ = 2 sin with = 15 mm and = 3 rad/s. Derive the response in general, and then for the specific parameter values. Discuss the possibilities of resonance.
Figure 3.60: A schematic of a seismograph.
Solution: From the equation of motion, we deduce that () is the motion of the mass measured on the seismometer or the relative motion of the mass with respect to the moving ground. From the parameters given, we find = 15 rad/s, q = 1 − 2 = 3571 rad/s,
2 =
3 = √ = √ = 0387 2 2 15 = 2 = 1 · 0015 · (3)2 = 0135 mm/s2
158
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
The equation of motion is given by ̈ + 3̇ + 15 = 0135 sin 3 The system is underdamped (because 1) and the solution takes the form of 2 () = exp (− ) (1 cos + 2 sin ) + q sin ( − ) | {z } 2 2 2 2 ( − ) + (2 ) homogeneous, transient | {z } particular, steady-state
where
−1
= tan
µ
2 2 − 2
¶
In this case, the homogeneous solution is identical to the transient solution, and the particular solution is identical to the transient solution. For the numerical values given, we have () = exp (−15) (1 cos 3571 + 2 sin 3571) + 00125 sin (3 − 0982) | {z } | {z } homogeneous, transient
particular, steady-state
Apply the initial conditions, (0) = 0 and ̇ (0) = 0 to find 1 and 2 :
(0) = 1 + 00125 sin (−0982) = 0 ̇ (0) = −151 + 35712 + 00125 (3) cos (−0982) = 0 The constants are 1 = 001038 mm and 2 = −0002036 mm. If the ground frequency is close to the natural frequency of the system, the response amplitude grows. From harmonically forced vibration, we found that the maximum magnification occurs p 2 when = 1 − 2 . As long as the system is damped, the response amplitude will not grow without a bound.
159 Problems for Section 3.8 — Periodic but Not Harmonic Excitation
33. A machine is loaded by a periodic “sawtooth” shaped force, as depicted in Figure 3.61. The load is assumed to have existed for a very long time. Model this force using: (a) a one-term Fourier series, (b) a three-term Fourier series, and (c) a five-term Fourier series. In each case solve for the response as a function of time. System properties are: = 1 kg, = 9 N/cm, = 015, = 1 s, and = 1 cm.
Figure 3.61: Sawtooth loading.
Solution: First of all, we realize that this problem can be solved exactly by solving the problem in discrete time intervals such that for 0 ≤ ≤ ̈2 + ̇2 + 2 = − for ≤ ≤ 2 .. .
̈1 + ̇1 + 1 =
The displacements and velocities are continuous at = for = 1 2 · · · Here, we use the Fourier expansion for the periodic forcing. The sawtooth function is given by 0 ≤ ≤ We first have to evaluate the Fourier coefficients as follows: Z 2 0 = = 0 Z 2 2 cos + (sin ) − 1 cos = 2 = 0 2 2 Z 2 2 sin = − 2 (− sin + (cos ) ) 2 2 = 0
160
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
Note that the integration by part was used to evaluate and For example, ¯ ¯ ¯ Z Z sin ¯¯ sin cos = − 1 ¯ |{z} | {z } |{z} |{z} 0 ¯ | {z }¯ | {z }
0
Since = 2 and is an integer, cos = cos 2 = 1 and sin = 0 Therefore, = 0 and = − The Fourier series representation of the force is then ∞ X 2 () = − sin 2 =1
The one-, three-, and five- term expansions are respectively, 2 1 () = − sin = 2 3 () = − sin − sin 2 − sin 3 2 2 3 5 () = − sin − sin 2 − sin 3 2 2 3 − sin 4 − sin 5 4 5 Similarly, the -term approximation is X 2 sin () = − 2 =1
Let us plot the three expansions in order, after substituting in our parameter values: 1 1 = 2 1 () = − sin 2 2 1 1 1 1 sin 4 − sin 6 3 () = − sin 2 − 2 2 3 1 1 1 1 1 1 sin 4 − sin 6 − sin 8 − sin 10 5 () = − sin 2 − 2 2 3 4 5
161
These now become the respective loading functions for our oscillator. We can solve for the response due to each term in the series and then add all solutions by linear superposition. We will solve the case for 5 () next. The equation of motion is the standard damped oscillator. Substitute the parameter values to find: 5 1 X 1 ̈ + 09̇ + 9 = − sin 2 2 =1
The homogenous solution is () = exp (− ) (1 cos + 2 sin ) = exp (−045) (1 cos 3 + 2 sin 3) The particular solution due to the constant term is 0 = 118. Recall that the particular solution to the harmonically forced system, ̈ + 2 ̇ + 2 = sin is given by 1 () = q¡ sin ( − ) ¢ 2 2 − 2 + (2 )2 = tan−1
2 0 ≤ ≤ 2 − 2
The particular solution to the forcing, − () = −
X 1
P
=1 (sin 2) is given by
¡ ¢ 1 q sin 2 − =1 (2 − (2))2 + (2 (2))2
= tan−1
2 (2) 0 ≤ ≤ 2 − (2)2
Combining the homogeneous and all the particular solutions, the total solution is
where
() = exp (−045) (1 cos 3 + 2 sin 3) X ¡ ¢ 1 1 1 q + − sin 2 − 18 =1 (9 − (2))2 + (18)2 = tan−1
18 9 − (2)2
The initial conditions are applied to the total solution to find the constants of integration, 1 and 2 The values of 1 and 2 will change depending on the number of forcing terms kept. The plots below show the exact solution, and the approximations using 1 () 3 () and 5 () It is interesting to note that the response when the one-term approximate forcing (1 ()) is used is just as accurate, although it is not a particularly good estimate for () itself.
162
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
Responses to () 1 () 3 () and 5 () The values used are = 1 kg, = 9 N/cm, = 015, = 1 s, and = 1 cm.
163
34. Solve Problem 33 with (a) = 2 s, and (b) = 05 s.
Solution: The form of the expansion is unchanged from the previous problem: X () = − sin 2 =1
=
2
(a) = 2 s: The expansions become: 1 1 − sin 2 1 1 1 3 () = − sin − sin 2 2 2 1 − sin 3 3 1 1 1 sin 2 5 () = − sin − 2 2 1 1 1 − sin 3 − sin 4 − sin 5 3 4 5 1 () =
So, then, we can plot the shapes of the three approximations for our forcing function:
Using the five-term expansion, we find that the differential equation of motion is: ̈ + 09̇ + 91 =
1 1 1 1 − sin − sin 2 − sin 3 2 2 3 1 1 − sin 4 − sin 5 4 5
The responses to the zero initial conditions are shown below.
164
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
Responses to () 1 () 3 () and 5 () The values used are = 1 kg, = 9 N/cm, = 015, = 2 s, and = 1 cm. (b) = 05 : The expansions for the forcing function become: 1 1 1 () = − sin 4 2 1 1 1 3 () = − sin 4 − sin 8 2 2 1 − sin 12 3 1 1 1 sin 8 5 () = − sin 4 − 2 2 1 1 1 − sin 12 − sin 16 − sin 20 3 4 5 The different approximations for the forcing are shown in the respective figures.
165
So, our differential equation is now: ̈ + 09̇ + 9 =
1 1 1 − sin 4 − sin 8 2 2 1 1 1 − sin 12 − sin 16 − sin 20 3 4 5
The responses to zero initial conditions are shown below.
Responses to () 1 () 3 () and 5 () The values used are = 1 kg, = 9 N/cm, = 015, = 05 s, and = 1 cm.
166
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
35. Solve Problem 33 with = 2 cm. Solution: For a linear system, if the forcing amplitude doubles so does the response. If you already solved Problem 33, simply multiply the response by 2
167
36. Solve Problem 33 with = 10. Solution: Recall that the Fourier expansion for the force is given by ∞ X 2 () = − sin 2 =1
Here, = 1 kg, = 9 N/cm, = 1, = 1 s, and = 1 cm The equation of motion is given by ̈ + 6̇ + 9 =
∞ 1 X 1 − sin 2 2 =1
The homogeneous solution is given by () = exp (−3) (1 + 2 ) Recall that the particular solution to the harmonically forced system, ̈ + 2 ̇ + 2 = sin is given by 1 sin ( − ) () = q¡ ¢ 2 2 2 2 − + (2 )
2 0 ≤ ≤ 2 − 2 P The particular solution to the forcing, − =1 (sin 2) is given by = tan−1
() = −
X 1
¡ ¢ 1 q sin 2 − =1 (2 − (2))2 + (2 (2))2
= tan−1
2 (2) 0 ≤ ≤ 2 − (2)2
Combining the homogeneous and all the particular solutions, the total solution is
where
() = exp (−3) (1 + 2 ) X ¡ ¢ 1 1 1 q + − sin 2 − 18 =1 (9 − (2))2 + (12)2 = tan−1
12 9 − (2)2
The figure below shows the responses when one-, three, and five-term expansions are used to approximate () Again, the responses to 1 () 5 () and 5 () are quite similar although the
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CHAPTER 3 SDOF VIBRATION: WITH DAMPING
forces themselves are not as similar.
Responses to () 1 () 3 () and 5 () The values used are = 1 kg, = 9 N/cm, = 1, = 1 s, and = 1 cm.
169
37. A body is subjected to the two harmonic motions, ³ ´ 1 () = 100 sin + 3´ ³ 2 () = 50 sin + 6 What additional harmonic motion should be applied to the body to bring it to static equilibrium? Solution: 3 () = − (1 () + 2 ()) ³ ³ ´ ´ = −100 sin + − 50 sin + 3 6 = −1455 sin ( + 08744) Another sinusoidal function with an identical frequency with amplitude −1455 and phase −08744 rad will cancel out 1 () and 2 ()
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CHAPTER 3 SDOF VIBRATION: WITH DAMPING
38. An undamped system is forced by the periodic function shown in Figure 3.62. Derive the Fourier series representation for this function, and solve for the response assuming zero initial conditions.
Figure 3.62: Triangle forcing function.
Solution: The Fourier series of a triangle function is given by ∞ ∞ X 0 X () = + cos + sin 2 =1 =1
where 2 =
Z 2
2 () cos () and = − 2
Z 2
() sin ()
− 2
We know right away that = 0 because the forcing function is an odd function. From the figure, we find that the forcing frequency is and the period is = 2 The coefficients, are found by Z 2 3 4 = () sin () − 4 µ µ ¶ ¶ Z Z 2 4 2 2 2 3 4 = 2− sin () + sin () − 4 4 Note that the limits of integration are changed to − 4 3 4 We can do so as long as we integrate over one period. We find 1 = The Fourier expansion is given by
8 8 8 3 = − 2 5 = ··· 2 9 252
µ ¶ 1 1 sin − 2 sin 3 + 2 sin 5 − · · · 3 5 ∞ +1 8 X (−1) = 2 sin =1 (2 − 1)2
8 () = 2
171 where = (2 − 1) The equation of an undamped system is given by ̈ + = () The solution is given by 1 8 X (−1)+1 1 sin () = 1 cos + 2 sin + 2 2 2 =1 (2 − 1) − 2 ∞
where the first two terms are the homogeneous solution. Applying the zero initial displacement, we find 1 = 0 Applying the zero initial velocity, we find 1 1 8 X (−1)+1 1 2 = − 2 2 2 =1 (2 − 1) − 2 ∞
which must approach a number. It is difficult to see what this number may be. One way around this problem is to solve for the response for 0 ≤ ≤ 2 (until the first discontinuity occurs). The equation of motion is given by ̈ + =
2
for 0 ≤ ≤ 2
The solution is given by
2 1 The constants of integration that satisfy the zero initial conditions are () = 1 cos + 2 sin +
1 = 0 and 2 = −
1 2 1
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CHAPTER 3 SDOF VIBRATION: WITH DAMPING
39. For Example 3.20, plot () and () for the case = 6 . Compare the results with those of Figure 3.50 and discuss. Assume = 1 s, 0 = 1 N, and = 01 Solution: The equation of motion is given by ̈ + 2 ̇ + 2 = () where () is a square wave with period and magnitude 0 that repeats every s, () = 0 for 0 2 = −0 for 2 It was found that () has a Fourier series: ∞ 40 X 1 () = sin( ), odd
where = 2 = 2 Recall a harmonically forced system, ̈ + 2 ̇ + 2 = sin has a particular solution given by 1 () = q¡ sin ( − ) ¢ 2 2 − 2 + (2 )2
2 0 ≤ ≤ 2 − 2 P 1 In our case, the system is forced by () = (40 ) ∞ odd sin( ) The corresponding particular solution is given by = tan−1
() =
∞ ¡ ¢ 1 40 X 1 r³ sin − ´2 ³ ´2 odd 2 2 1 − 2 + 2
= tan−1
2 ( ) 0 ≤ ≤ 1 − 2 ( )2
where = 2 is used. The homogeneous solution is given by
() = exp (− ) (1 cos + 2 sin ) The homogeneous solution is also the transient solution, which decays exponentially. Therefore, let us plot the particular solution only. Zero Damping: Let us first consider the case when = 0 and = 16 while 0 and are kept constant at 1 N and 1 s, respectively. It is mathematically much easier if there is no damping. The homogeneous solution and the particular solution are reduced to () = 1 cos + 2 sin ∞ 1 40 X 1 ³ ´ sin (2) () = odd 1 − 2 ¡ 1 ¢2 6
173 Applying the zero initial conditions to the total solution, we have (0) = 0 = (0) + (0) = 1 = 0 4 X 1 ³ ´ = 0 ̇ (0) = 0 = ̇ (0) + ̇ (0) = 2 + odd 1 − 2 ¡ 1 ¢2 ∞
6
giving
1 = 0 ∞ 1 4 X 1 ³ ´ = 0 2 = odd 6 1 − 2 ¡ 1 ¢2 6
Then, the solution is given by () = () The figure below shows the Fourier series of the force for = 5 and the corresponding response. The values used are = 1 s, 0 = 1 N, = 0 = 4 N/m, and = 1 kg. The solid line is for the exact solution and the dotted line is for the three term expansion solution. Note that the amplitude of the response is still at 05 but the small oscillations are more prominent.
Plots of () and () for the case = 6 We can compare these results to those of Figure 3.50. The results are qualitatively the same (that the system responds at the forcing frequency), but the exact shape differ because the parameter values differ. With Damping: For the problem with damping, it is given = 16 and = 01 while 0 and are kept constant at 1 N and 1 s, respectively. ∞ X ¡ ¢ 4 1 1 r³ () = sin 2 − ´ ¡ ¢2 2 ¡ ¢2 odd + 2 (01) 16 1 − 2 16 | {z }
= tan−1
2 (01) (16) 0 ≤ ≤ 1 − 2 (16)2
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CHAPTER 3 SDOF VIBRATION: WITH DAMPING
The amplitude and the phase of the particular solution are tabulated below:
1
3
5
7
(m) 13088 05608 07316 04232 (rad)
0.0343 0.1326 0.4993 2.5679
We will use the first 3 terms up to = 5 for the loading and response. The equations become: 4 X 1 sin(2) =135 µ ¶ 4 1 1 = sin(2) + sin(6) + sin(10) 3 5 X () = sin(2 − ) 3 () =
=135
=
1 [13088 sin(2 − 00343) + 05608 sin(6 − 01326) +07316 sin(10 − 04993)]
The figure below shows the Fourier series of the force for = 5 and the corresponding response. The values used are = 1 s, 0 = 1 N, = 01 = 4 N/m, and = 1 kg. The solid line is for the exact solution and the dotted line is for the three term expansion solution.
Plots of () and () for the case = 6
175
40. Solve Example 3.20 where the structure has viscous damping, that is, the governing equation of motion is ̈ + 2 ̇ + 2 = (), the forcing () is the same square wave, and = 2 rad/s.
Solution: We need to express the square wave in terms of a Fourier series. Following Example 3.20, we have the Fourier coefficients: = 0 ∙ ¸ 40 40 1 − cos( ) = = 2 Then, the forcing function is given by () = 4 0 ∞ X
() =
=135
=
odd.
P
1 odd sin The response is given by
¡ ¢ sin −
1 40 q 2 ( 2 − 2 2 ) + (2 )2
1 40 q 2 (1 − 2 2 2 ) + (2 )2 µ ¶ 2 −1 = tan 1 − 2 2 2 =
Assume = 1 s 0 = 1 N/kg = 4 N/m, and = 01 The forcing and natural frequencies are = 2 rad/s and = 4 rad/s. The amplitude multiplied by the spring constant (the spring constant is not given) and the phase are calculated below. Care should be taken so that is between 0 and rad.
1
3
(N) 16828 033
5
7
00484 00160
(rad) 0.1326 29060 30465 3.0795 Now the decision is how many terms in the series are needed. Usually between 5 and 10 terms are more than adequate. Suppose we keep the first 3 terms. () =
1 [16828 sin (2 − 01326) + 033 sin (6 − 29060) +00484 sin (10 − 30465)]
We would like to show each of the 3 terms above separately, in order to demonstrate how the majority of the response is in the first 1 or 2 terms of the series. Note the scale of the vertical axes.
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CHAPTER 3 SDOF VIBRATION: WITH DAMPING
The response to each Fourier components, = 1 3 and 5
177
41. Solve Example 3.20, as in Problem 40, with = 4 rad/s.
Solution: Following Example 3.20, we have: ∞ 40 X 1 sin = () = =135 4
This leads to: ¶ () = − sin 4 =135 ∞ X
µ
1 40 ( 2 − 2 2 ) 1 40 = (1 − 2 2 2 ) µ ¶ 2 −1 = tan 1 − 2 2 2
=
Let = 1 s 0 = 1 N/kg = 14 Then, 1 4 (1 − 2 42 ) µ ¶ 024 −1 = tan 1 − 2 42
=
Retaining the first three terms, we arrive at:
() =
1 (13581 sin 2 + 09701 sin 6 + 04527 sin 10)
The figure below shows the Fourier series of the force for = 5 and the corresponding response. The values used are = 1 s, 0 = 1 N, = 0 and = 4 N/m. The solid line is for the exact solution and the dotted line is for the three term expansion solution. (See Problem 37 for the justification for using = 4 N/m)
178
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
Plots of () and () for the case = 4
179
42. A base-excited structure is governed by the equation ̈ + 2 ̇ + 2 = 2 ̇ + 2 , where () is the base motion. Solve for the response if the base motion is given by the sawtooth function shown in Figure 3.61.
Figure 3.61: Sawtooth loading.
Solution: We need to express the sawtooth wave in terms of a Fourier series, knowing that: 0 ≤ ≤ ( + ) = () () =
Solving for the Fourier coefficients gives: Z 2 () = 0 = 0 ¶ µ Z 2 2 () sin = 0 = 0 ¶ µ Z 2 2 = () cos = − 0 The Fourier representation (up to = 3) for the force is given by ∞ X − sin () = + 2 =1
(In problem 33, = 1 s, and = 1 cm are used. For now, let us proceed without substituting any values.) The equation of motion is given by ̈ + 2 ̇ + 2 = 2 ̇ + 2 Ã∞ ! X cos = −2 =1 à ! ∞ X 1 1 + − sin +2 2 =1
180
CHAPTER 3 SDOF VIBRATION: WITH DAMPING
Recall that the particular solution of the problem, ̈ + 2 ̇ + 2 = 0 is = 0 2 Recall that the particular solution of the problem, ̈ + 2 ̇ + 2 = sin is 1 sin ( − ) () = q¡ ¢ 2 2 2 2 − + (2 ) = tan−1
2 0 ≤ ≤ 2 − 2
If the system is forced by cos , the response is simply cosine with the same amplitude and the same phase angle. The particular solution for the current problem is the summation of the particular solutions of the individual components and is given by
where
∞ ¡ ¢ X 1 q¡ () = − 2 cos − ¢2 2 =1 2 − ( )2 + (2 ( ))2 ∞ X ¡ ¢ 1 2 q¡ − sin − 2 ¢2 2 2 =1 − ( ) + (2 ( ))
= tan−1
2 2 − ( )2
181
43. Solve Problem 42 where the input is given by the square wave of Figure 3.63.
Figure 3.63: Square wave base motion.
Solution: Note that the square wave is an odd function. Thus, = 0 Solving for the remaining Fourier coefficients gives: ⎧ ⎨ 0 = ⎩ 4 So, the square wave is given by
∞ ∞ X X 4 4 () = sin = sin(2 − 1) (2 − 1) =135 =1
The equation of motion is given by ̈ + 2 ̇ + 2 = 2 ̇ + 2 ! µ ¶ ÃX ∞ 4 cos(2 − 1) = 2 =1 Ã∞ ! X 4 sin(2 − 1) + 2 (2 − 1) =1 The particular solution is then given by µ ¶ ∞ ¡ ¢ 1 4 X q¡ cos − () = 2 ¢2 =1 2 − 2 + (2 )2 −2
∞ X
¡ ¢ 4 1 q¡ sin − ¢2 (2 − 1) =1 2 − 2 + (2 )2
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CHAPTER 3 SDOF VIBRATION: WITH DAMPING
where = (2 − 1) and = tan−1
2 2 − 2
183
44. Solve Problem 42 where the input is given by the function sin 0 ≤ ≤ shown in Figure 3.64, repeated periodically.
Figure 3.64: Base motion driven by sin .
Solution: Note that the given wave can be written: () = sin 0 ≤ ≤ ( + ) = () Thus, = = 2 Also note that () is an even function; thus, = 0 and we can calculate as: 0 = = Thus, we can write () = The equation of motion is given by
2 ⎧ ⎨
⎩−
0 odd 4 (2 −1)
even
∞ 4 2 X − cos 4 =1 (42 − 1)
̈ + 2 ̇ + 2 = 2 ̇ + 2 Ã∞ ! X 4 −4 sin 4 = 2 2 − 1) (4 =1 Ã∞ ! X 4 cos 4 + 2 (42 − 1) =1
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CHAPTER 3 SDOF VIBRATION: WITH DAMPING
The particular solution is then given by () = 2 −2 where
∞ X 4
¡ ¢ −4 1 q¡ sin − ¢ 2 2 (4 − 1) =1 2 − 2 + (2 )2
∞ X =1
4
(42 − 1)
¡ ¢ 1 q¡ cos − ¢2 2 − 2 + (2 )2
= 4 and = tan−1
2 2 − 2
Chapter 4 Single Degree-of-Freedom Vibration: General Loading and Advanced Topics Problems for Section 4.1 — Arbitrary Loading: Laplace Transform 1. Solve the equations of motion using the Laplace transform approach, where () is the speed and () is the force: (a) ̇ + 2 = (), (0) = 1, and () = 0 (b) 2̇ + 2 = (), (0) = 0, and () = 10 (c) ̇ + 3 = (), (0) = 1, and () = 2 sin (3) Solution: (a) Take the Laplace transform to obtain () − (0) + 2 () = 0 Solve for () (0) 1 = +2 +2 Take the inverse Laplace transform to obtain () () =
() = −2 for ≥ 0 (b) Take the Laplace transform to obtain 2 () − 2 (0) + 2 () =
10
Solve for () () =
10 2 (0) + 10 = 2 + 2 2 ( + 1)
Take the inverse Laplace transform to obtain () ¡ ¢ () = 5 1 − − for ≥ 0 185
186
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
(c) Take the Laplace transform to obtain 3 () − (0) + 3 () = 2 2 +9 Solve for () 1 + 6 (2 + 9) 2 + 15 () = = +3 ( + 3) (2 + 9)
((1))
Observing the zeros of the denominator, = −2 ±3, we know in advance that the inverse Laplace transform will results in () = −3 + cos 3 + sin 3 where the constants, , and are to be determined. That is, () can be expanded such that () =
3 + 2 + 2 ( + 3) ( + 9) ( + 9)
The constants can be determined by combining the fractions again, (2 + 9) + ( + 3) ( + 3) ( + 3) (2 + 9) ( + ) 2 + (3 + 3) + 9 + 9 = ( + 3) (2 + 9)
() =
Comparing the previous expression for () with the original expression in Equation (1), we find + = 1 3 + 3 = 0 9 + 9 = 15 The constants are = 43 = −13 = 13 Then,
1 1 4 () = −3 − cos 3 + sin 3 for ≥ 0 3 3 3
187
2. Solve the equations of motion using the Laplace transform approach: (a) ̈ + 3̇ + 2 = 0 (0) = 1 ̇(0) = 0 (b) 2̈ + 5̇ + 3 = 0 (0) = 1 ̇(0) = 1 (c) ̈ + 2̇ + 2 = 0 (0) = 2 ̇(0) = 4 (d) ̈ + 2̇ + = 0 (0) = 1 ̇(0) = −1
(e) ̈ − 2̇ + 3 = 0 (0) = 0 ̇(0) = 2. Solution: (a) Take the Laplace transform,
2 () − (0) − ̇ (0) + 3 ( () − (0)) + 2 () = 0 ¢ ¡ 2 + 3 + 2 () = + 3 +3 () = 2 + 3 + 2 The roots of the denominator are = −1 and −2 Therefore, () = − + −2 whose Laplace transform is given by () =
+ +1 +2
where and are determined by comparing the two expressions for () () =
2 + 3 + 2
=
( + 2) + ( + 1) + = +1 +2 2 + 3 + 2
Equating coefficients of and 0 respectively, + = 1 2 + = 3 and solving them simultaneously gives = 2 and = −1 Then, () = 2− − −2 for ≥ 0 (b) Take the Laplace transform, ¢ ¡ 2 2 () − (0) − ̇ (0) + 5 ( () − (0)) + 3 () = 0 ¢ ¡ 2 2 + 5 + 3 () = 2 + 7 2 + 7 () = 2 2 + 5 + 3
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CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
The zeros of the denominator is = −1 and −32 and therefore, we expect () = − + −32 whose Laplace transform is given by () =
+ ( + 1) ( + 32)
The constants, and are determined by comparing the two expressions for () () =
( + 32) + ( + 1) (2 + 7) 2 + = 5 3 = 2 ( + 1) ( + 32) + 2 + 2 2 + 52 + 32
We find Coefficient of : + = 1 7 3 Coefficient of 0 : + = 2 2 Therefore, = 5 and = −4 Then, () = 5− − 4−32 for ≥ 0 (c) Take the Laplace transform, 2 () − (0) − ̇ (0) + 2 ( () − (0)) + 2 () = 0 ¢ ¡ 2 + 2 + 2 () = 2 + 8 2 + 8 () = 2 + 2 + 2 The zeros of the denominator is = −1 ± and therefore, we expect () = − ( cos + sin ) whose Laplace transform is given by () =
1 +1 + 2 ( + 1) + 1 ( + 1)2 + 1
The constants, and are determined by comparing the two expressions for () () =
+1 1 ( + 1) + 2 + 8 = + = 2 + 2 + 2 2 + 2 + 2 2 + 2 + 2 2 + 2 + 2
We find Coefficient of : = 2 Coefficient of 0 : + = 8
189 Therefore, = 2 and = 6 Then, () = − (2 cos + 6 sin ) for ≥ 0 (d) Take the Laplace transform, 2 () − (0) − ̇ (0) + 2 ( () − (0)) + () = 0 ¢ ¡ 2 + 2 + 1 () = + 1 +1 () = 2 + 2 + 1 The zeros of the denominator is = −1 and −1 and therefore, we expect () = − + − whose Laplace transform is given by () =
1 1 + +1 ( + 1)2
The constants, and are determined by comparing the two expressions for () () =
+1 2 + 2 + 1
=
1 1 ( + 1) + + 2 = +1 2 + 2 + 1 ( + 1)
We find Coefficient of : = 1 Coefficient of 0 : + = 1 Therefore, = 1 and = 0 Then, () = − (e) Take the Laplace transform, 2 () − (0) − ̇ (0) − 2 ( () − (0)) + 3 () = 0 ¢ ¡ 2 − 2 + 3 () = 2 () =
√ The zeros of the denominator is = 1 ± 2 and therefore, we expect √ √ () = sin 2 + cos 2
2
2 − 2 + 3
190
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
whose Laplace transform is given by √ 2 ( − 1) () = + 2 ( − 1) + 2 ( − 1)2 + 2 The constants, and are determined by comparing the two expressions for () √ √ 2 2 ( − 1) 2 + ( − 1) () = 2 = + = − 2 + 3 ( − 1)2 + 2 ( − 1)2 + 2 ( − 1)2 + 2 We find Coefficient of : = 0 √ Coefficient of 0 : 2 − = 2 Therefore, = Then, () =
√ 2 and = 0
√ √ 2 sin 2 for ≥ 0
191
3. Solve the equations of motion using the Laplace transform approach: (a) ̈ + 3̇ + 2 = () (0) = 1 ̇(0) = 0 (b) 2̈ + 5̇ + 3 = −3 (0) = 1 ̇(0) = 1 (c) ̈ + 2̇ + 2 = cos () (0) = 2 ̇(0) = 4 (d) ̈ + 2̇ + = −2 + sin (2) (0) = 1 ̇(0) = −1 (e) ̈ − 2̇ + 3 = (0) = 0 ̇(0) = 2.
Solution: (a) Take the Laplace transform, 1 2 () − (0) − ̇ (0) + 3 ( () − (0)) + 2 () = ¢ ¡ 2 + 3 + 2 () = + 3 + 1 2 + 3 + 1 () = 2 ( + 3 + 2) Using the partial fraction expansion method, () is expanded as 2 + 3 + 1 () = ( + 1) ( + 2) + + = +1 +2 Recombining the previous expression, we have () =
2 + 3 + 1 ( + 1) ( + 2) + ( + 2) + ( + 1) = ( + 1) ( + 2) ( + 1) ( + 2)
Equating the coefficients of Coefficient of 2 : + + = 1 Coefficient of : 3 + 2 + = 3 Coefficient of 0 : 2 = 1 Solving the three equations simultaneously, we find 1 1 = = 1 = − 2 2 Then, + + +1 +2 whose inverse Laplace transform is given by () =
() =
1 1 + − − −2 for ≥ 0 2 2
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CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
(b) Take the Laplace transform, ¢ ¡ 1 2 2 () − (0) − ̇ (0) + 5 ( () − (0)) + 3 () = +3 ¢ ¡ 2 2 + 5 + 3 () = 2 ( + 1) + 5 + 1 ( + 3) (2 + 7) ( + 3) + 1 () = ( + 1) (2 + 3) ( + 3) Using the partial fraction expansion method, () is expanded as 22 + 13 + 22 ( + 1) (2 + 3) ( + 3) + + = + 1 (2 + 3) + 3
() =
((1))
Recombining the previous expression, we have 22 + 13 + 22 ( + 1) (2 + 3) ( + 3) (2 + 3) ( + 3) + ( + 1) ( + 3) + ( + 1) (2 + 3) = ( + 1) (2 + 3) ( + 3)
() =
Equating the coefficients of Coefficient of 2 : 2 + + 2 = 2 Coefficient of : 9 + 4 + 5 = 13 Coefficient of 0 : 9 + 3 + 3 = 22 Solving the three equations simultaneously, we find 11 28 1 =− = 2 3 6 We can find the coefficients more efficiently if we use the residue method. Starting from Equation (1), multiply both sides by ( + 1) Then, =
22 + 13 + 22 =+ ( + 1) + ( + 1) (2 + 3) ( + 3) 2 + 3 +3 If we evaluate the previous expression at = −1 we are left with
2 − 13 + 22 = (1) (2) 11 = 2 Similarly, multiply Equation (1) by (2 + 3) and evaluate at = −32 to obtain 2 (94) − 13 (32) + 22 = (−12) (32) 28 = − 3
193 Taking the inverse Laplace transform of (1), we have −32 + −3 2 11 − 14 −32 1 −3 − = + for ≥ 0 2 3 6
() = − +
(c) Take the Laplace transform, 2 () − (0) − ̇ (0) + 2 ( () − (0)) + 2 () = 2 +1 ¢ ¡ 2 + 2 + 2 () = 2 + 8 +
2 + 1
(2 + 8) (2 + 1) + () = 2 ( + 2 + 2) (2 + 1)
The roots of denominator are = ± and −1 ± We expect the solution of the form, () = cos + sin + − cos + − sin which has a Laplace transform of 1 +1 1 () = 2 + 2 + + 2 +1 +1 ( + 1) + 1 ( + 1)2 + 1 Compare the previous expression with the original () we have 23 + 82 + 3 + 8 (2 + 8) (2 + 1) + = (2 + 2 + 2) (2 + 1) (2 + 2 + 2) (2 + 1) 1 +1 1 + 2 + + = 2 2 +1 +1 ( + 1) + 1 ( + 1)2 + 1 ( + ) (2 + 2 + 2) + ( ( + 1) + ) (2 + 1) = (2 + 2 + 2) (2 + 1)
() =
Equating the coefficients of Coefficient of 3 : + = 2 Coefficient of 2 : 2 + + + = 8 Coefficient of : 2 + 2 + = 3 Coefficient of 0 : 2 + + = 8 Solving the four equations simultaneously, we find 1 2 9 27 = = = ,= 5 5 5 5 Then, () = cos + sin + − cos + − sin 2 9 27 1 = cos + sin + − cos + − sin for ≥ 0 5 5 5 5
194
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
(d) Take the Laplace transform, 2 () − (0) − ̇ (0) + 2 ( () − (0)) + 1 () = ¡ 2 ¢ + 2 + 1 () = + 1 +
1 2 + 2 +2 +4
2 1 + 2 +2 +4 2 ( + 1) ( + 2) ( + 4) + (2 + 4) + 2 ( + 2) () = (2 + 2 + 1) ( + 2) (2 + 4) 4 + 33 + 72 + 14 + 16 = ( + 1)2 ( + 2) (2 + 4) The roots of denominator are = −1 −1 −2, and ±2 We expect the solution of the form, () = − + − + −2 + sin 2 + cos 2
which has a Laplace transform of 1 1 2 1 + + 2 + 2 () = 2 + +1 ( + 2) +4 +4 ( + 1) Compare the previous expression with the original () we have 4 + 33 + 72 + 14 + 16 ( + 1)2 ( + 2) (2 + 4) 1 1 1 2 = + + 2 + 2 2 + +1 ( + 2) +4 +4 ( + 1) ( + 1) ( + 2) (2 + 4) + ( + 2) (2 + 4) + ( + 1)2 (2 + 4) = ( + 1)2 ( + 2) (2 + 4) 2 ( + 1)2 ( + 2) + ( + 1)2 ( + 2) + ( + 1)2 ( + 2) (2 + 4)
() =
The coefficient can be found by multiply both sides by ( + 1)2 and evaluating the expressions at = −1 or 1 − 3 + 7 − 14 + 16 = 1·5 7 = 5 The coefficient can be found by multiplying both sides by ( + 2) and evaluating the expression at = −2 or = 1 Equating the coefficients of Coefficient of 4 : + + = 1 Coefficient of 3 : 3 + + 2 + 2 + 4 = 3 Coefficient of 2 : 6 + 2 + 5 + 8 + 5 = 7 Coefficient of : 12 + 4 + 8 + 10 + 2 = 14 Coefficient of 0 : 8 + 8 + 4 + 4 = 16
195 It is easier to solve these five simultaneous equations because we already found and The constant coefficients are 7 3 4 4 = = = 1, = − = − 25 5 25 25 Then, () = − + − + −2 + sin 2 + cos 2 4 3 4 − 7 − + + −2 − sin 2 − cos 2 for ≥ 0 = 25 5 25 25 (e) Take the Laplace transform, 2 () − (0) − ̇ (0) − 2 ( () − (0)) + 3 () =
1 2
¢ ¡ 2 1 − 2 + 3 () = 2 + 2
1 2 () = 2 − 2 + 3 22 + 1 = 2 ( − 2 + 3) 2 2+
√ The roots of denominator are = 1 ± 2 0 and 0 We expect the solution of the form, √ √ () = sin 2 + cos 2 + + which has a Laplace transform of () =
2 1 −1 1 + + + 2 2 2 ( − 1) + 2 ( − 1) + 2
Compare the previous expression with the original () we have 22 + 1 (2 − 2 + 3) 2 2 1 −1 1 = + + + 2 2 2 ( − 1) + 2 ( − 1) + 2 2 2 2 2 + ( − 1) + ( − 2 + 3) + (2 − 2 + 3) = (2 − 2 + 3) 2 ( + )3 + (2 − − 2 + )2 + (3 − 2) + 3 = (2 − 2 + 3) 2
() =
Equating the coefficients of
Coefficient of 3 : + = 0 Coefficient of 2 : 2 − − 2 + = 2 Coefficient of : 3 − 2 = 0 Coefficient of 0 : 3 = 1
196
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
The constant coefficients are = 1718 = −29 = 29, = 13 Then, () =
√ √ 17 2 2 1 sin 2 − cos 2 + + for ≥ 0 18 9 9 3
197
4. Solve the equations of motion using the Laplace transform approach: (a) ̈ + 2̇ + 3 = 5 cos 3 (0) = 3 ̇(0) = 4 (b) 4̈ + 5̇ + 5 = 4() (0) = 1 ̇(0) = 1, where () is the unit step function, (c) 3̈ + 3̇ + 6 = 3− + 2 cos 3 (0) = 2 ̇(0) = 4 (d) ̈ + ̇ + = () (0) = 0 ̇(0) = 0, where () is given by a square wave function with maximum amplitude 1 and period 1 (e) ̈ + 2̇ + 3 = cos 3 + cos 5 (0) = 0 ̇(0) = 0. Solution: The Laplace transform of the general equation ̈ + ̇ + = () is ¡ ¢ 2 () − (0) − ̇(0) + ( () − (0)) + () = ()
where () and () are the Laplace transforms of () and () respectively. Then, () =
() + ((0) + ̇(0)) + (0) 2 + +
The inverse Laplace transform of () gives () Let us now do the individual problems. (a) ̈ + 2̇ + 3 = 5 cos 3, (0) = 3, (0) = 4 In this case, = 1 = 2 = 3 and () = 5 (2 + 9) where () is obtained from the Laplace transform table (Table A.2). The Laplace transform of the response, () is given by ( 25+9 ) + 3 + 10 33 + 102 + 32 + 90 = () = 2 + 2 + 3 (2 + 9) (2 + 2 + 3) The inverse transform can be found using a symbolic program. Alternately, to perform the inversion by hand, () needs to be rewritten using the method of partial fraction expansion. () can be rewritten as 33 + 102 + 32 + 90 + + = 2 + 2 2 2 ( + 9) ( + 2 + 3) ( + 9) ( + 2 + 3) where the assumed numerators depend on highest order of the numerator of the original fraction. Equating, we find that ¡ ¢ ¡ ¢ 33 + 102 + 32 + 90 = ( + ) 2 + 2 + 3 + ( + ) 2 + 9 = ( + )3 + ( + + 2) 2 + (3 + 2 + 9) + 3 + 9 Matching coefficients and solving simultaneous equations, we obtain = −512 = 54 = 4112 = 11512 Then, () =
41 + 115 −512 + 54 + 2 ( + 9) 12 (2 + 2 + 3)
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CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
To transform () to () first we must rewrite in the following form 5 5 3 41 +1 ³ ´ + + 12 (2 + 9) 12 (2 + 9) 12 ( + 1)2 + ¡√2¢2 √ 2 74 1 ´ + √ ³ 12 2 ( + 1)2 + ¡√2¢2
() = −
The inverse transform is then given by
() = −1 [ ()] √ √ √ 5 41 − 37 2 − 5 sin 3 + cos 2 + sin 2 = − cos 3 + 12 12 12 12 = −0417 cos 3 + 0417 sin 3 + exp (−) (342 cos 141 + 4 36 sin 141) The response plot is shown at the end. (b) 4̈ + 5̇ + 5 = 4() (0) = 1 (0) = 1 In this case, = 4 = 5 = 5 and [ ()] = [4 ()] = 4 Then, () =
42 + 9 + 4 (42 + 5 + 5)
which is expanded (by partial fraction expansion) as () =
41 1 4 + 25 + 2 5 5 (4 + 5 + 5)
By completing the square for the denominator of the second term, we have () =
4 1 1 1 4 + 25 + ¡ ¢ 5 5 4 + 5 2 + 55 8
64
We express the second term in that form so that we can use ∙ ¸ + −1 = exp (−) cos ( + )2 + 2 ∙ ¸ −1 = exp (−) sin ( + )2 + 2 Then, 4 1 1 1 4( + 58 ) − 52 + 25 + () = ¡ ¢2 5 54 + 5 + 55 =
8 64 √ 5 + 8 558 41 1 9 +√ ¡ + ¡ ¢ ¢ 2 5 2 55 5 5 + 5 + 55 55 + + 8 64 8 64
199 The inverse Laplace transform of the first term is 4 () 5 and response is given by √ ¶Ã √ ! µ 55 55 4 9√ 1 5 − 58 () = + sin + exp 55 exp − cos 5 55 8 8 5 8 = 08 + exp (−0625) (121 sin 0927 + 02 cos 0927) (c) 3̈ + 3̇ + 6 = 3− + 2 cos 3 (0) = 2 ̇(0) = 4 In this case, = 3 = 3, = 6 [ ()] = [3− + 2 cos 3] = 3 ( + 1) + 2 (2 + 9) The Laplace transform of the response is given by () + ((0) + ̇(0)) + (0) 2 + + 3 + 22 + 3 (2 + 4) + 6 = +1 +92 3 + 3 + 6
() =
which can be rewritten as 1 7 − 9 5 55 + 208 1 1 − + 2 2 + 1 87 + 9 174 2 + + 2 + 12 1 1 7 1 3 275 = − + + 2 + 1 87 2 + 9 29 2 + 9 174 ( + 12)2 + 74 √ 72 1805 √ 7 + 1218 ( + 12)2 + 74
() =
We use the following relationships from Table A.2. ∙ ¸ 1 −1 = exp (−) + ∙ ¸ −1 = cos 2 + 2 ∙ ¸ −1 = sin 2 + 2 ∙ ¸ + −1 = exp (−) cos ( + )2 + 2 ∙ ¸ −1 = exp (−) sin ( + )2 + 2 Then, taking the inverse Laplace transform of () we obtain 1 7 1 exp (−) − cos 3 + sin 3 2 87 29 ¶ µ √ √ 1805 √ 275 7 sin 73 cos 73 + + exp (−05) 174 1218 = 05 exp (−) − 00805 cos 3 + 00345 sin 3 + exp (−05) (158 cos 0882 + 392 sin 0882)
() =
200
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
(d) ̈ + ̇ + = () (0) = 0 ̇(0) = 0 () is a square wave Method 1: This problem can be solved in two ways. The first way is to expand the force using the Fourier series: ∞ X
∞ X 2 2 sin = sin [(2 + 1) ] () = (2 + 1) =135 =0
Taking the Laplace transform of both sides of the differential equation gives: () =
( ()) 2 + + 1
The response is ¸ ∙ 1 8 8 8 () = 2 + + + ··· + + 1 2 + 4 2 2 + 36 2 2 + 100 2 This relationship must be rewritten to take the inverse Laplace transform manually. As an example, let us consider the inverse Laplace transform of 1/(2 + + 1) (2 + 2 ) The roots of denominator √ (poles) are = ± −12 ± 32 The inverse transform will have the form of √ √ 3 3 −05 −05 + + cos + sin cos sin 2 2 Therefore, we can write p 34 1 + 12 = + 2 2 2 2 ( + + 1) ( + ) ( + 12) + 34 ( + 12)2 + 34 + 2 + 2 2 + + 2 After some brute force algebra, we find = −
1 4 − 2 + 1
1 1 − 2 4 − 2 + 1 1 = 4 − 2 + 1 1 2 2 − 1 = √ 4 3 − 2 + 1 =
Therefore, the response is given by (∞ à √ √ ! X 1 3 3 1 − 2 () = 8 − exp (−05) cos + exp (−05) sin 4 2 − + 1 2 2 =1 ¶¾ µ 1 2 2 − 1 sin − 4 cos + √ − 2 + 1 3
201 where 1 = 2 2 = 6 3 = 10 · · · Method 2: The second method is to write the forcing using the unit step functions: () = () − 2 ( − 12) + 2 ( − 1) + · · · µ ¶ ∞ X 2 − (−1) = () + 2 =1 The Laplace transform of the force is µ ¶ 1 2 1 2 () = − exp − + exp (−) + · · · 2 µ ¶ ∞ X 1 2 = + exp − (−1) =1 2 Then, 1 1 () = 2 ++1
! µ ¶ exp − (−1) 1+2 2 =1
Ã
∞ X
which is rewritten as à ! !à √ ¶ µ ∞ X + 12 12 32 1 () = − −√ 1+2 exp − (−1) ( + 12)2 + 34 2 32 ( + 12)2 + 34 =1 We use (Appendix A) −1 [exp (−) [ ()]] = ( − ) ( − ) where
Then,
√ 1 32 + 12 12 [ ()] = − −√ 2 ( + 12) + 34 32 ( + 12)2 + 34 µ µ ¶ ¶ 2 − (−1) − () = () + 2 2 =1 ∞ X
where () is the inverse Laplace transform of [ ()] or √ √ ¶ ¶ µ µ 1 1 3 3 1 − √ exp − sin () = 1 − exp − cos 2 2 2 2 3 This is an exact solution, while the solution from the first approach is an approximation. (e) ̈ + 2̇ + 3 = cos 3 + cos 5 (0) = 0 ̇(0) = 0 The Laplace transform of () is [cos 3 + cos 5] =
2 + 9
+
2 + 25
202
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
Then, () + ((0) + ̇(0)) + (0) 2 + + + 2 2 = 2+9 +25 + 2 + 3
() =
Taking the inverse transform, we find "
# √ 53 47 2 () = exp (−) cos 141 − sin 141 438 438 +
1 5 11 1 sin 3 − cos 3 + sin 5 − cos 5 12 12 292 292
The response plots are shown next.
203
5. For the given transfer functions, find the response of the system due to the unit step input force () = () and zero initial conditions: (a) () F () = 1 ( + 2) (b) () F () = 1 (2 + 3 + 2) (c) () F () = (3 + 2) (2 + 3 + 2)
Solution: The Laplace transform of the unit step input is 1 F () = The response due to the input force with zero initial conditions is given by () 1 () × F () = F () F () −1 () = L [ ()]
() =
(a) The response is given by 1 1 +2 = + +2
() =
where =
1 1 and = − 2 2
Then, () = + −2 ¢ 1¡ 1 − −2 for ≥ 0 = 2
Check the initial conditions are zeros, (0) = 0 and ̇ (0) = 0 The response obtained from the transfer function always satisfy zero initial conditions.
(b) The response is given by () × F () F () 1 1 = 2 + 3 + 2 = + + +1 +2
() =
204
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
where
1 1 = , = −1 = 2 2
Then, () = + − + −2 1 1 = − − + −2 for ≥ 0 2 2 Again, the response satisfies the zero initial conditions, (0) = 0 and ̇ (0) = 0 (c) The response is given by () × F () F () (3 + 2) 1 = 2 ( + 3 + 2) + + = +1 +2
() =
where = 1, = 1 = −2 Then, () = + − + −2 = 1 + − − 2−2 for ≥ 0 Again, the response satisfies the zero initial conditions, (0) = 0 and ̇ (0) = 0
205
6. For the given transfer functions, find the response of the system due to the input force () = 2 cos (3) () and zero initial conditions: (a) () F () = 1 ( + 2) (b) () F () = 1 (2 + 3 + 2) (c) () F () = 1 (2 + 2 + 1) The Laplace transform of the unit step input is F () = 2 2 +9 The response due to the input force with zero initial conditions is given by () () × F () = 2 2 F () F () + 9 −1 () = L [ ()]
() =
(a) The response is given by 2 1 2 +2 +9 3 = + 2 + 2 +2 +9 +9
() =
(1) (2)
The expansion is determined by observing the roots of the denominator, = −2 and ±3 The response in the time domain will have a combination of −2 −3 3 or −2 cos 3 sin 3 The Laplace transforms of above time domain functions are 3 1 2 2 +2 +9 +9 The constant coefficients are found by combining Equation (2) using least common denominator and equating the result to Equation (1), and they are =− Then,
4 6 4 = and = 13 13 13
4 6 4 −2 + cos 3 + sin 3 for ≥ 0 13 13 13 Again, the response satisfies the zero initial conditions, (0) = 0 and ̇ (0) = 0 () = −
206
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
(b) The response is given by () × F () F () 1 = 2 2 2 + 3 + 2 + 9 3 = + + 2 + 2 +1 +2 +9 +9
() =
where
1 4 7 9 =− ,= =− = 5 13 65 65
Then, () = − + −2 + cos 3 + sin 3 9 4 7 1 cos 3 + sin 3 for ≥ 0 = − − + −2 − 5 13 65 65 Again, the response satisfies the zero initial conditions, (0) = 0 and ̇ (0) = 0 (c) The response is given by () × F () F () 1 2 2 = 2 ( + 2 + 1) + 9 3 + + 2 = 2 + 2 + 1 ( + 1) +9 +9
() =
where =
1 4 3 4 ,=− =− = 25 5 25 25
Then, () = − + −2 + cos 3 + sin 3 3 4 4 − 1 − − − cos 3 + sin 3 for ≥ 0 = 25 5 25 25 Again, the response satisfies the zero initial conditions, (0) = 0 and ̇ (0) = 0
207
7. Consider a system subjected to two input forces, as shown in Figure 4.30. Find the transfer functions, () F1 () and () F2 () Find the response () due to 1 () = and 2 () = () Use zero initial conditions, and let = 1 kg, 1 = 500 N/m, and 2 = 1000 N/m.
Figure 4.30: Two input system.
Let us use a intermediate coordinate where 2 () is applied and call it () Then, the free-body diagram for the mass and where 2 () is applied are shown next.
Summing the forces in the horizontal direction, we have 2 () + 1 ( () − ()) − 2 () = 0 1 () − 1 ( () − ()) = ̈ () Take the Laplace transform of both equations and they are F2 () + 1 () − (1 + 2 ) () = 0 F1 () − 1 () + 1 () = 2 () Solve for () using the first equation and substitute it into the second equation to obtain F1 () − 1 () + 1
F2 () + 1 () = 2 () 1 + 2
Solve for () : () =
1 + 2 F1 () 2 ( + 1 ) (1 + 2 ) − 12 +
1 F2 () 2 ( + 1 ) (1 + 2 ) − 12
208
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
The transfer functions are () 1 + 2 = 2 F1 () ( + 1 ) (1 + 2 ) − 12 () 1 = 2 F2 () ( + 1 ) (1 + 2 ) − 12 The response to respective inputs is () () F1 () + F1 () F1 () F2 () 1 1 1 + 2 1 = + 2 2 2 2 2 ( + 1 ) (1 + 2 ) − 1 ( + 1 ) (1 + 2 ) − 1 1 1 500 1500 + = 2 2 2 1500 + 500000 1500 + 500000
() =
The roots of the denominator are = 0 0 and 333 Then, the inverse Laplace transform has the form of p p () = cos 10003 + cos 10003 + + Then,
p 10003 () = 2 + 2 + + 2 ( + 10003) ( + 10003)
where
√ 3 30 1 3 1 =− = = =− 1000 10000 1000 1000
Then,
√ p p 3 3 30 1 1 cos 10003 − cos 10003 + + () = − 1000 10000 1000 1000
209 Problems for Section 4.2 — Step Loading 8. A mass of = 50 kg is suspended by a spring attached to a rope that is wound around a drum with radius = 04 m and moment of inertia about the point of rotation = 45 kg m2 , as shown in Figure 4.31. The spring of stiffness = 3 kN/m reduces the shock of an applied load. The system is initially at rest when a constant torque = 1 kN-m is applied to the system. (a) Derive an expression for the tension in the rope after a time (b) What is the maximum tension in the rope?
Figure 4.31: Mass connected to drum via spring.
Solution: This is a 2 DOF system, which is not covered yet in the text. The two coordinates are coupled ( () and ()), and it is difficult (if not possible) to find () without finding () simultaneously. However, it is still possible to find the time history of the tension in the rope using the method studied in this chapter. The figure below shows the free-body diagrams.
210
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
The free-body diagrams are drawn assuming that and are measured from static equilibrium. Summing the moments for the wheel, we obtain X ª = ̈ − ( − ) = ̈
where we denote tension as so that = ( − ) Summing the forces in the vertical direction for the mass, we obtain X ↑ = ̈ ( − ) = ̈
The tension in the rope is same as the spring force, ( − ) It should be noted that is the dynamic tension. The total tension is the summation of static tension () and the dynamic tension. Now, the goal is to write an equation of motion in terms of We multiply the moment equation by and the force equation by 1: ( − ( − )) = ̈ ( − ) = ̈ Subtracting the second equation from the first and replacing ( − ) with we obtain ̈ 1 ( − ) − = ¶ µ 2 ̈ + = 1+ 2
is 12 where is the radius of gyration of the drum. The dynamic tension is given Note that by 1 ¢ () = 1 sin + 2 cos + ¡ 2 + 1
where the natural frequency is
=
r
s
1+
2
Applying zero initial conditions (we assume that the system is initially at rest at its static equilibrium, not the undeformed configuration), we have (0) = 2 +
1 ¢ =0 ¡ 2 + 1
̇ (0) = 1 = 0 Then, the dynamic tension is given by () =
1 ¢ (− cos + 1) ¡ 2 + 1
211 The total tension is () =
1 ¢ (− cos + 1) + ¡ 2 + 1
(Answer to part (a))
The maximum tension in the rope is (when = s)
The numerical value is
1 ¢ + max = 2 ¡ 2 + 1 max = 2
1000 ³ 04
= 1245 N.
1
(Answer to part (b))
´ + 50 (98)
45 +1 50(04)2
212
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
9. Derive Equation 4.19: ∙ ¸ 1 − (cos + sin ) ≥ 0 () = 2 1 −
Solution: () here is the response to the step load applied at = 0 Let us start with the convolution integral, Z 1 () = ( ) − (− ) sin ( − ) 0
where ( ) = 1 for 0 We let = − and express the integral in terms of : Z 1 () = ( − ) − sin 0
The step force is turned on when − 0 or That is, ( − ) for the entire range of integration, 0 Then, Z 1 () = − sin 0 Integrate by parts once to obtain à ! ¯ Z ¯ 1 − 1 − − () = cos ¯¯ − cos 0 0 µ ¶ Z 1 − 1 1 − − = cos + − cos 0 We integrate by parts again to obtain à ! ¶2 Z µ 1 − 1 − 1 − cos + − 2 sin − − sin () = 0 ¶2 µ 1 1 − 1 − cos + − sin − () = − 2 2 2 R Note that we get back the integral, 0 − sin which is the same integral we are trying to integrate. Solve for () to obtain µ µ ¶¶ 1 − 1− cos + sin () = 2 ¡ ¢ where we used 2 = 1 − 2 2 Equation 4.19 is the response of a unit mass, = 1
213
10. Assuming 0 1 solve Equation 4.20, ∙ −0 ¸ 1 −1 () = L 2 + 2 + 2
Solution: Equation 4.20 is the inverse Laplace transform of the response to a unit step applied at = 0 . The inverse transform can be evaluated by using Borel’s theorem. To apply Borel’s theorem, we need to know the inverse transform of each component in the bracket above: exp(−0 )
and
1 2 + 2 + 2
We know that
and
½
¾ exp(−0 ) = ( − 0 ) L ½ ¾ 1 1 − −1 L = sin 2 2 + 2 + −1
where () is the unit step function, and by Borel’s theorem, the inverse transform is given by the convolution of these two functions: Z 1 () = ( − 0 ) exp(− ( − )) sin( ( − )) 0 Z 1 = exp(− ( − )) sin( ( − )) 0
214
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
11. Using the Laplace transform, solve ̈ + 2̇ + = () where () = ( − 1) − 2( − 2) + ( − 3) is in terms of unit step functions, and (0) = 0 and ̇(0) = 0. Solution: First, we write:
() =
which is shown below
⎧ ⎪ ⎪ 0 0 ≤ ≤ 1 ⎪ ⎪ ⎪ ⎪ ⎨ 1 1 ≤ ≤ 2
⎪ ⎪ −1 2 ≤ ≤ 3 ⎪ ⎪ ⎪ ⎪ ⎩ 0 ≥ 3
Taking the Laplace transform of the differential equation and solving for () gives ¢ ¡ 2 2 1 1 + 2 + 1 () = exp(−) − exp(−2) + exp(−3) 1 () = (exp(−) − 2 exp(−2) + exp(−3)) (2 + 2 + 1) µ ¶ 1 1 1 = − (exp(−) − 2 exp(−2) + exp(−3)) + ( + 1)2 + 1 So, we take the inverse transform to find () = ( − 1) ( − 1) − 2 ( − 2) ( − 2) + ( − 3) ( − 3) where () = 1 − − + −
215 Problems for Section 4.4 — Arbitrary Loading: Convolution Integral 12. Find the impulse response function of a system whose equation of motion is given by (a) ̇ + 4 = () (b) ̈ + 4 = () (c) ̈ + 5̇ + 4 = () (d) ̈ + 2̇ + = () (e) ̈ + 5̇ + 4 = 5˙ () + 4 () Solution: The impulse response function can be found in two ways. The first method is solving for () after setting () = () and initial conditions to zeros. The second method is finding the inverse Laplace of the transfer function. (a) Method 1: The impulse response () satisfies (a)
̇ + 4 = () Taking Laplace transform with zero initial conditions, we obtain () + 4 () = 1 () =
1 +4
Taking the inverse transform, we have () = −4 for 0 = 0 for ≤ 0 Method 2: The transfer function is given by 1 () = () +4 The inverse Laplace transform is () = −4 for 0 = 0 for ≤ 0 The method 2 is nearly identical to method 1. The transfer function when the force is the delta function is identical to () () () |() =1 = () Method 3: Integrating Equation (a) from 0− to 0+ we have Z 0+ 0−
(̇ + 4) =
Z 0+ 0−
= 1
()
216
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
where the integral of the delta function is 1. The left hand side becomes Z 0+ Z 0+ ¡ ¢ ¡ ¢ = 0+ − 0− = 0− 0− Z 0+ () = 0 4 0−
It was given that the initial condition is zero. That is, (0− ) = 0 and we have ¡ ¢ 0+ =
Z 0+ 0−
= 1
¡ ¢ () + 0−
Then, we can convert our original problem with the delta function into an initial value problem given by ̇ + 4 = 0 (0) = 1 whose solution is given by () = −4 (b) Methods 1 and 2: The impulse response function satisfies ̈ + 4 = () (0) = 0 and ̇ (0) = 0 Taking the Laplace transform, 2 () + 4 () = 1 () = The inverse transform gives () =
1 2 + 4
1 sin 2 2
Method 2: Integrate ̈ + 4 = () to obtain Z 0+ µ
¶ Z 0+ ̇ + 4 = () 0− 0− ¡ ¢ ¡ ¢ ̇ 0+ − ̇ 0− = 1
Since ̇ (0− ) = 0 ̇ (0+ ) = 1 Then, the corresponding initial value problem is ̈ + 4 = 1 ¡ ¢ ¡ ¢ 0+ = 0 and ̇ 0+ = 1
217 R 0+ You may ask what happened to 0− 4 This term must be zero because the limits of integral have zero width. (c) Methods 1 and 2: The impulse response function satisfies ̈ + 5̇ + 4 = () (0) = 0 and ̇ (0) = 0 Taking the Laplace transform, we have 1 1 = () = 2 + 5 + 4 3
µ
¶ 1 1 − +1 +4
Taking the inverse Laplace transform, the impulse response function is given by () =
¢ 1 ¡ − − −4 3
(d) Methods 1 and 2: The impulse response function satisfy
̈ + 2̇ + = 1 (0) = 0 and ̇ (0) = 0 Take the Laplace transform to obtain () =
1 1 = 2 + 2 + 1 ( + 1)2
The inverse Laplace transform is () = − (e) Methods 2: Take the Laplace transform (while applying zero initial conditions and (0) = 0) to obtain 2 () + 5 () + 4 () = 5 () + 4 () When () = () () = 1 and () = () by definition. Then () = The inverse Laplace transform is
5 + 4 163 −13 = + 2 + 5 + 4 +4 +1
1 16 () = − − + −4 3 3
218
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
13. Find the impulse response function of a system whose transfer function is given by (a) () = 1 (2 + 1) (b) () = 1 (22 + 3 + 1) (c) () = 1 (2 + 4 + 4) (d) () = 1 (2 + 9) (e) () = (5 + 4) (2 + 5 + 4)
Solution: (a) The impulse response function is also the inverse Laplace transform of the transfer function, which is given by 1 2 + 1 1 1 = 2 + 05
() =
The impulse response function is 1 () = −05 for 0 2 (b) The transfer function is given by 1 22 + 3 + 1 1 = (2 + 1) ( + 1) 1 2 − = 2 + 1 + 1 1 1 = − + 05 + 1
() =
The inverse Laplace transform is () = −05 − − for 0 (c) The transfer function is given by () = =
1 2 + 4 + 4 1 ( + 2)2
The inverse Laplace transform is () = −2 for 0
219 (d) The transfer function is given by 1 2 + 9 1 3 = 3 2 + 32
() =
The inverse Laplace transform is () =
1 sin (3) for 0 3
(e) The transfer function is given by 5 + 4 5 + 4 = 2 + 5 + 4 ( + 4) ( + 1) + = +4 +1
() =
where the inverse transform gives () = −4 + − The coefficients and are found by recombining the two fractions so that 5 + 4 ( + 1) + ( + 4) = ( + 4) ( + 1) ( + 4) ( + 1) The coefficients of must match or 5 = + 4 = + 4 Solving previous equations simultaneously, they are = 163 and = −13 Then, () =
16 −4 1 − − 3 3
220
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
14. Find the response of each of the systems given below to arbitrary inputs, (): (a) ̇ + 4 = () () = () (b) ̈ + 4 = () () = , ≥ 0
(c) ̈ + 5̇ + 4 = () () = 1 for 0 ≤ 2 and () = 0 for ≥ 2 (d) ̈ + 2̇ + = () () = ()
Solution: Using the convolution integral, the response is given by Z Z ( − ) ( ) or ( ) ( − ) () = 0
0
(a) The impulse response function is obtained from the transfer function. The transfer function is () =
1 +4
and the corresponding impulse response function is () = −4 The response is Z
−4(− ) ( ) 0 Z −4 4 =
() =
0
¢ 1¡ = −4 4 − 1 4 ¢ 1¡ = 1 − −4 for 0 4
(b) The transfer function is given by
() = and its inverse is () =
1 2 + 4
1 sin 2 2
The response is given by () =
Z 0
Z
( − ) ( )
1 ( − ) sin 2 2 0 Z Z 1 1 sin 2 − sin 2 = 2 0 2 0
=
221 The second integral requires the integration by parts. After integration, we obtain µ ¶ µ ¶ 1 1 1 1 () = − cos 2 − sin 2 − cos 2 4 4 8 4 1 1 = − sin 2 for 0 4 8 (c) The transfer function is given by µ ¶ 1 1 1 1 = − () = 2 + 5 + 4 3 +1 +4 and its inverse Laplace transform is
() = The response for 2 is given by
Z
() =
0
Z
¢ 1 ¡ − − −4 3
( − ) ( )
¢ 1 ¡ −(− ) − −4(− ) 0 3 1 1 − 1 = − + −4 4 3 12 =
The response for ≥ 2 is given by () =
Z 0
Z 2
( − ) ( )
¢ 1 ¡ −(− ) − −4(− ) 0 3 ¢ 1 1 −4 ¡ 8 ¢ ¡ 1 = − 2 − 1 − −1 3 34 − −4 = 213 − 2483
=
(d) The transfer function is given by () =
1 2 + 2 + 1
=
1 ( + 1)2
and its inverse Laplace transform is () = − The response is given by () =
Z 0
=
Z 0
( − ) ( ) −(− )
( − )
−
=
Z 0
−
−
Z
0
where the second integral requires integration by parts. After integration, we obtain () = 1 − ( + 1) −
222
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
15. Find the response of the undamped oscillator ̈ +4 = () to each of the following forces per unit mass using the convolution integral: (a) () = 1 − − ≥ 0
(b) () = cos 2 0 ≤ ≤
(c) () = cos 2 + 3 0 ≤ ≤
(d) () = cos 2 + cos 3 0 ≤ ≤ .
Solution: The impulse response function is given by 1 exp (− ) 1 = sin 2 2
() =
The convolution solution is () =
Z 0
( − ) ( ) or
Z 0
( − ) ( )
assuming that () is turned on at = 0 The second relationship is somewhat easier when () is discontinuous. (a) () = 1 − − ≥ 0 The convolution response is ¶ µ Z Z 1 −(− ) () = sin 2 ( − ) ( ) = (1 − ) 2 0 0 1 1 1 1 cos 2 − sin 2 = − − − 4 5 20 10 (b) () = cos 2 0 ≤ ≤ The convolution response is () =
Z 0
( − ) ( )
where () is turned off for A two part solution is needed. For 0 ≤ ≤ the response is given by Z Z 1 () = ( − ) ( ) = sin 2 ( − ) cos 2 0 0 2 1 = sin 2 for 0 ≤ ≤ 4 Note that the response grows linearly with time. This is because the forcing frequency is identical to the natural frequency.
223 For the response is given by Z Z Z () = ( − ) ( ) = ( − ) ( ) + ( − ) ( ) 0 0 µ ¶ Z 1 cos 2( − ) sin 2 + 0 = 2 0 1 for = sin 2 4 where we used ( ) = 0 for Note that that the amplitude of the response no longer grows with time. The response is plotted at the end. The two part solution is used. However, it should be noted that () and ̇ () are continuous at = . The displacement cannot have jumps, and the velocity is continuous because no impulsive force is applied. (c) () = cos 2 + 3 0 ≤ ≤ The convolution response is () =
Z 0
( − ) ( )
where the force is turned off for A two part solution is needed. For 0 ≤ ≤ the response is given by Z Z 1 sin 2 ( − ) (cos 2 + 3) ( − ) ( ) = () = 0 0 2 1 3 3 = sin 2 − cos 2 + for 0 ≤ ≤ 4 4 4 For the response is given by () =
Z 0
1 sin 2 ( − ) (cos 2 + 3) 2
1 = sin 2 4
for
To save time, what we could have done is to find the solution to () = 3 for 0 ≤ ≤ and simply add this solution to the solution of part (b). (d) () = cos 2 + cos 3 0 ≤ ≤ The convolution response is () =
Z 0
( − ) ( )
where the force is turned off for A two part solution is needed. Let us utilize the solution of part (b). We already obtained that the response to () = cos 2 for 0 ≤ ≤ is 1 sin 2 for 0 ≤ ≤ 4 1 = sin 2 for 4
() =
224
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
Let us find the response to () = cos 3 and add it to the solution of part (a). For 0 ≤ ≤ the response is given by Z Z 1 () = sin 2 ( − ) cos 3 ( − ) ( ) = 0 0 2 1 1 = cos 2 − cos 3 for 0 ≤ ≤ 5 5 and for () =
Z 0
1 sin 2 ( − ) cos 3 2
2 = cos 2 5
for
Combining with the solution of part (a), we have 1 1 1 sin 2 + cos 2 − cos 3 for 0 ≤ ≤ 4 5 5 2 1 = sin 2 + cos 2 for 4 5
() =
The response plots for part (a)-(d) are shown next.
225
16. Find the response of the damped oscillator ̈+2 ̇+ 2 = (), with 2 = 4 (rad/s)2 , to the forces per unit mass listed, and solve for the two underdamped cases of = 01 and = 09: (a) () = 1 − − ≥ 0
(b) () = cos 2 0 ≤ ≤
(c) () = cos 2 + 3 0 ≤ ≤
(d) () = cos 2 + cos 3 0 ≤ ≤ .
Solution: The impulse response function is given by () = where =
1 − sin
p 1 − 2 For = 01
() = 05025−02 sin 199
For = 09 () = 1147−18 sin 0872 The solution by the convolution integral is given by () =
Z 0
( − ) ( ) or
Z 0
( − ) ( )
The second expression is easier to use when the force is discontinuous. (a) () = 1 − − ≥ 0 The solution by the convolution integral is given by () = =
Z 0
( − ) ( )
0
(1 − −(− ) )
Z
1 − sin
The integral is not easy. The first part integrates to (Refer to part (b) for detailed procedure for integrating such integrals.) Z
1 − sin 0 Ã " #! 1 sin ( ) + cos ( ) = 2 1 − exp (− ) p 1 − 2
226
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
The second part integrates to Z 1 − −(− ) − sin 0 ¶ ∙ µ ¸¶ µ − 1 1 sin − exp (−) + exp (− ) cos + = 2 − 2 + 1 Together, the response is given by à " #! 1 () = 2 1 − exp (− ) p sin ( ) + cos ( ) 1 − 2 ¶ ∙ µ ¸¶ µ − 1 1 sin − exp (−) + exp (− ) cos + + 2 − 2 + 1 For = 0 it reduces to () =
1 1 1 1 − cos (2) − sin 2 − exp (−) 4 20 10 5
which agrees with the solution to part (a) of the previous problem. For = 01 the response is () = exp (−02) (−00326 cos (199) − 01125 sin (199)) + 025 − 02174 exp (−) For = 09 the response is () = exp (−18) (04643 cos (087) + 01393 sin (087)) + 025 − 07143 exp (−) The responses are plotted at the end. (b) () = cos 2 0 ≤ ≤ The solution by the convolution integral is given by Z () = ( − ) ( ) 0
Because the force is turned off at = a two-part solution is required, Z 1 − (− ) () = sin ( − ) cos 2 for 0 ≤ ≤ 0 Z 1 − (− ) = sin ( − ) cos 2 for 0 Again, this is not an easy integral to evaluate manually although it is a standard technique. The strategy is to combine the two sinusoidal functions first so that sin ( − ) cos 2 =
1 sin (cos ( − 2) + cos ( + 2) ) 2 1 − cos (sin ( − 2) + sin ( + 2) ) 2
227 Then, the response is given by Z 1 − () = sin [cos ( − 2) + cos ( + 2) ] 2 0 Z 1 − cos [sin ( − 2) + sin ( + 2) ] − 2 0 The integrals such as
Z
cos
0
can be evaluated by integrating the expression twice by parts twice. Let = and = cos ¯ Z ¯ 1 1 ¯ cos ( − 2) = sin ¯ − sin 0 0 =0
Z
We integrate the second term by parts again. This time, let = and = sin After R the second integration by parts, we can solve for 0 cos We find that Z 0
cos =
− exp (− ) ( cos − sin ) ( )2 + 2
Similarly, we find that Z − exp (− ) ( sin + cos ) sin = ( )2 + 2 0 Upon substituting the values, = 01 = 2 rad/s, and = 199 rad/s we find that the response is given by () = 125 sin 2 − 125 exp (−02) sin 199 for 0 ≤ ≤ = exp(−02( − )) (00396 cos 199 + 126 sin 199) − 126 exp(−02) sin 199 for For = 09 = 2 rad/s,and = 087 rad/s () = 0139 sin 2 − 0319 exp (−18) sin 0872 for 0 ≤ ≤ = exp(−18( − )) (−0125 cos 087 − 0293 sin 087) − 0319 exp(−18) sin (087) for (c) () = cos 2 + 3 0 ≤ ≤ Let us find the solution to () = 3 for 0 ≤ ≤ and combine the result with the solutions of part (b). The solution by the convolution integral is given by () =
Z 0
( − ) ( )
228
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
Because the force is turned off at = a two-part solution is required, Z 1 − (− ) () = sin ( − ) 3 for 0 ≤ ≤ Z0 1 − (− ) = sin ( − ) 3 for 0 For = 01 the integral evaluates to () = exp(−02) (−075 cos 199 − 0075 sin 199) + 075 for 0 ≤ ≤ = exp(−02 ( − )) (075 cos 199 + 0052 sin 199) − exp (−02) (075 cos 199 + 00754 sin 199) for ;For = 09 the integral evaluates to () = exp(−18) (−075 cos 087 − 155 sin 087) +075 for 0 ≤ ≤ = − exp(−18 ( − )) (130 cos 087 + 113 sin 087) − exp (−18) (075 cos 087 + 155 sin 087) for The response to () = cos 2 + 3 for 0 ≤ ≤ is the combination of the response to () = cos 2 for 0 ≤ ≤ and () = 3 for 0 ≤ ≤ For example, for = 01 the response to the current problem is () = exp(−02) (−075 cos 199 − 1325 sin 199) +075 + 125 sin 2 for 0 ≤ ≤ = exp(−02 ( − )) (07896 cos 199 + 1312 sin 199) − exp (−02) (075 cos 199 + 13354 sin 199) for (d) () = cos 2 + cos 3 0 ≤ ≤ Taking = 01 the convolution integral leads to: () = 125 sin(2) − 0189 cos(3) + 00454 sin(3) + exp(−02) (0189 cos(199) − 1306 sin(199)) for 0 ≤ ≤ = exp(−02( − )) [023 cos(199) + 12 sin(199)] + exp(−02) (0189 cos(199) − 131 sin(199)) for Taking = 09 the convolution integral leads to: () = 0139 sin(2) − 0035 cos(3) + 00762 sin(3) + exp(−18) (00353 cos(087) − 0508 sin(087)) for 0 ≤ ≤ = exp(−18( − )) (−0083 cos(087) − 0105 sin(087)) + exp(−18)(035 cos(087) − 0508 sin(087)) for
229 The responses for all cases are plotted next. The dotted lines are for = 09 and the solid line for = 01
230
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
17. Solve for the damped response beginning with the equation Z 1 sin ( − ) exp(− ) sin () = 0 In the same graph, plot the results for the following cases: = 10 15 20 25 30 rad/s where = 01, = 2 rad/s, and = 1 kg. Discuss.
Solution: The specific convolution integral to be solved is given by 1 () = √ 2 1 − 012
Z 0
√ sin ( − ) exp(−02 ) sin 2 1 − 012
which integrates to () = exp (−02) ( cos + sin ) + cos + sin p where the damped natural frequency is = 1 − 2 The coefficients, − depend on the natural frequency and the damping coefficient. Specifically: When = 10 the integral is reduced to () = exp (−02) (00437 cos 1 99 − 0 160 sin 1 99) − 00437 cos + 0 328 sin When = 15 rad/s () = exp (−02) (0175 cos 199 − 0368 sin 199) − 017531 cos 15 + 051132 sin 15 When = 2 rad/s () = exp (−02) (1 25 cos 1 99 + 0126 sin 199) − 1 25 cos 2 When = 25 rad/s () = exp (−02) (0165 cos 1 99 + 0483 sin 1 99) − 0165 cos 25 − 0371 sin 25 When = 3 rad/s () = exp (−02) (00454 cos 1 99 + 0290 sin 199) − 00454 cos 3 − 0189 sin 3 The responses are shown below, and the last plot is cross plots of all the responses.
231
We see that, as expected, the higher the driving frequency the higher frequency the response.
232
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
18. The stepped disk in Figure 4.32 is subject to the moment () = 1 − − ≥ 0 The spring is unstretched when = 0 Derive the equation of motion and solve for the response () assuming zero initial conditions and an underdamped system.
Figure 4.32: A stepped disk undergoing rotation in response to an impulsive torque.
Solution: Take the sum of the moments about the center of the disk, X © = ̈ − (2)2 − (2)2 ̇ = ̈
The equation of motion is then ̈ + (2)2 ̇ + (2)2 = = 1 − exp (−) We find
r
,=√ r s ¶2 µ 1− √ = 2
= 2
The homogeneous solution and the particular solution are (assuming 1) () = exp (− ) (1 cos + 2 sin ) () = + exp (−)
(1)
233 where and are found by substituting the particular solution back into the equation of motion. We find =
1 42
= −
1 − 42 + 42
The next step is to apply the initial conditions to the total solution. Although they are not explicitly given, we assume they are zero. Zero initial conditions give 1 1 − 42 − 42 + 42 1 ̇ (0) = 0 = − 1 + 2 + − 42 + 42
(0) = 0 = 1 +
and 1 1 + 2 4 − 42 + 42 µ ¶ 1 1 1 1 2 = + − − 42 + 42 − 42 + 42 42 1 = −
Therefore, the solution is given by () = exp (− ) (1 cos + 2 sin ) 1 1 exp (−) + 2 − 4 − 42 + 42 with and given in Equation (1) and 1 and 2 given in Equation (2).
(2)
234
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
19. Structures are sometimes subjected to very rapidly applied loads of extremely short duration. These types of loads are called blast or explosive loads. Consider how such a load time history may look. Figure 4.33(a) is a generic blast load. There is a rapid rise time along with an exponential-like decay, and by time 1 the load is effectively zero, where 1 ¿ 1 s. Such a time history is generally a very complicated function that may not be easily determined because of its sensitivity to many factors, such as the medium through which the load passes, be it the atmosphere or the Earth.
Figure 4.33: Measured (left) and approximate (right) blast time histories.
A common approach is to curve-fit the time history with an appropriate combination of functions. As a first approximation here, we replace the exact curve by the one shown in Figure 4.33(b). To draw such a straight-line approximation, estimate the area under the original curve (this is the impulse) and set it equal to the area under the triangle load, thus fixing the value of . The maximum value 0 is the same for both curves. Although not strictly true for such high strain-rate loading, we neglect the effects of damping for the early time response. Solve for the structural response for all time; 0 ≤ ≤ and ≥ . Show that: ∙ ¶ µ ¸ 1 0 1 − − cos + sin 0 ≤ ≤ () = 2 0 () = [sin (1 − cos ) 3 − cos ( − sin )]
Solution: The convolution integral can be applied to cases where the loading is discontinuous, but, as with any integral, the convolution integral must be split into continuous parts. Therefore, in this problem, since the idealized loading is continuous between 0 and we have two convolution integrals, one for 0 ≤ ≤ and another for ≥ The equation for the forcing is 0 (1 − )
235 1 for 0 ≤ ≤ The impulse response function is given by () = − sin in general, and for an undamped system, () = 1 sin Then, the response is
Z 1 () = 0 (1 − ) sin ( − ) 0 ≤ ≤ 0 0 − + + (cos ) − sin = − 3 For ≥
Z 1 0 (1 − ) sin ( − ) ≥ () = 0 0 sin (− + ) + (cos ) − sin = − 3
Assume the following parameter values: = 1 s, = 2 rad/s 0 = 1 N. Then, for 0 ≤ ≤ the response is given by () = − 213 (−2 + 2 + 2 cos 2 − sin 2) For ≥ the response is given by () = − 213 (sin (−2 + 2) + 2 cos 2 − sin 2) The response is plotted next. Note that the solid line is the solution for ≤ and the dotted line for ≥ 0.3 0.2
x (t) (m)
0.1 0 -0.1 -0.2 -0.3 -0.4
0
1
2
3
t (s)
4
5
236
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
20. Resolve Problem 19 including damping. Discuss the importance of damping by comparing the two results. Solution: The addition of damping greatly complicates the expressions of the previous problems. The convolution integral can be applied to cases where the loading is discontinuous, but, as with any integral, the convolution integral must be split into continuous parts. Therefore, in this problem, since the idealized loading is continuous between 0 and we have two convolution integrals, one for 0 ≤ ≤ and another for ≥ The equation for the forcing is 0 (1 − ) for 0 ≤ ≤ 1 The impulse response function is given by () = − sin Then, the response is 1 () =
Z 0
0 (1 −
− (− ) ) sin ( − )
0 ≤ ≤
This expression can be integrated by parts, or solved by using a symbolic solver such as MAPLE This would lead to a very long expression. We will show one example response below. For ≥ Z 1 0 (1 − )− (− ) sin ( − ) ≥ () = 0 Assume the following parameter values: = 1 kg, = 1 Rs = 2 s 0 = 1 N = 01 and = √ 1 2 1 − 012 = 199 rad/s. Then for 0 ≤ ≤ () = 199 (1 − )−001(2)(− ) sin (199( − )) 0 R 1 1 (1 − )−001(2)(− ) sin (199( − )) For ≥ () = 199 0 The integrals are not evaluated to plot the responses. MAPLE is capable of plotting functions in the present integral forms. The solid line is the response for , and the dashed line for
237
21. Solve for the response for all time of an underdamped oscillator that is driven by the forcing function shown in (a) Figure 4.34, and (b) Figure 4.35. Use the convolution integral.
Figure 4.34: Triangular loading function.
Figure 4.35: Sinusoidal loading function.
Solution: In this problem, forces are applied before = 0 In this case, we can reset the time so that the force are applied beginning at = 0 or redefine the convolution integral. (a) (Fig.4.34) After resetting the time so that it starts 2 seconds early, the force is shown below, and the force () is given by ⎧ ⎪ ⎪ 2 0 ≤ ≤ 05 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −2( − 1) 05 ≤ ≤ 10 ⎪ ⎪ ⎪ ⎪ ⎨ 0 10 ≤ ≤ 30 () = ⎪ ⎪ 2( − 3) 30 ≤ ≤ 35 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −2( − 4) 35 ≤ ≤ 40 ⎪ ⎪ ⎪ ⎪ ⎩ 0 ≥ 40
238
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
Now, within each time interval we apply the convolution integral as follows
where
R
2( − )( ) 0 ≤ ≤ 05 0 R 05 R () = 0 2( − )( ) + 05 (−2)( − 1 − )( ) R 10 R 05 () = 0 2( − )( ) + 05 (−2)( − 1 − )( ) R 05 R 10 () = 0 2( − )( ) + 05 (−2)( − 1 − )( ) R + 30 2( − 3 − )( ) R 05 R 10 () = 0 2( − )( ) + 05 (−2)( − 1 − )( ) R 35 R + 30 2( − 3 − )( ) + 35 (−2)( − 4 − )( ) R 05 R 10 () = 0 2( − )( ) + 05 (−2)( − 1 − )( ) R 35 R 40 + 30 2( − 3 − )( ) + 35 (−2)( − 4 − )( ) () =
() =
05 ≤ ≤ 10 10 ≤ ≤ 30 30 ≤ ≤ 35 35 ≤ ≤ 40 40 ≤
1 − sin
We have shown the full damped response. This can be significantly simplified if damping is eliminated. Note how the convolution integral is split into responses within continuous intervals of the loading. Just as we break apart any integral over discontinuities, we break apart the convolution integral over loading discontinuities. Note that we need to transform back to our original time by replacing with − 2 (b) (Fig.4.35) This loading is a sine function over the time period − ≤ ≤ We will solve this problem in two ways. The first method is the same method used in part (a). We will first shift the time, find the solution and shift the time back. The second method is to redefine the convolution integral so that it includes the effect of force that was applied before = 0 Method 1 We shift the time so that the loading begins at zero time, and then we can use the convolution integral. Let 1 = + Then at 1 = 0 = − and at 1 = = 0 The new loading is shown in the next figure.
239 The convolution response is given by ⎧R ⎨ 1 ( − ) sin ( − ) 0 ≤ ≤ 2 1 1 (1 ) = R 02 ⎩ ( − ) sin ( − ) ≥ 2 1 1 0
After the integral, the time needs to be shifted back by replacing 1 with + . We can shift the time even before the integrals are evaluated. We can rewrite the integral so that ⎧R ⎨ + ( + − ) sin ( − ) − ≤ ≤ 0 (1) () = ⎩ R 2 ( + − ) sin ( − ) ≥ 0
Method 2 We can also redefine the convolution integral for this problem. Recall that the convolution integral is defined as Z () = ( − ) ( ) 0
where ( − ) is the response at time due to the impulse ( ) applied at . In order to add the effect of the force applied starting from − s, the lower limit of the convolution integral is adjusted such that Z
() =
−
Then, the response is
( − ) ( )
⎧ R ⎨ ( − ) sin − ≤ ≤ . () = R − ⎩ ( − ) sin ≥ −
(2)
Note that the solutions given in Equations (1) and (2) are identical once the dummy variable in Equation (2) is replaced with another dummy variable, ̂ = + The impulse response function is given by () =
1 − sin
and the response is ⎧R ⎨ 1 − (− ) sin ( − ) sin − ≤ ≤ () = R− . ⎩ 1 − (− ) sin ( − ) sin ≥ −
Numerical Example To give a numerical solution, assume the following parameter values: = 2 rad/s = 01 Then, ⎧R ⎨ 1 ( − ) sin ( − ) 0 ≤ ≤ 2 1 1 (1 ) = R 02 ⎩ ( − ) sin ( − ) ≥ 2 0
1
1
240
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
where () = Then,
1 −02 sin 199 199
⎧R ⎨ 1 1 −02(1 − ) sin 199 ( − ) sin ( − ) 0 ≤ ≤ 2 1 1 (1 ) = R 02 199 1 ⎩ −02(1 − ) sin 199 (1 − ) sin ( − ) 1 ≥ 2 0 199
Figure next shows the responses for the first 20 seconds. The response decreases, although slowly, for 1 2 We can go back to our original time by shifting the time axis to the right by
Note how the convolution integral is split into responses within continuous intervals of the loading. Just as we break apart any integral over discontinuities, we break apart the convolution integral over loading discontinuities. Note that we need to transform back from 1 to to regain our original time scale.
241 Problems for Section 4.5 — Introduction to Lagrange’s Equation 22. Use Lagrange’s equation to derive the equation of motion for the simple pendulum of Example 4.15, except here assume that the mass is suspended on a rigid bar that is connected to the support via a torsional spring of stiffness . Solution: We will apply Lagrange’s equation for generalized coordinate Lagrange’s equation is µ ¶ − + = 0 ̇ where there is no dissipation or external forcing. is the kinetic energy and is the potential energy, and these are given by 1 1 = + 2 = (1 − cos ) + 2 2 2 1 2 1 2 = = (̇) 2 2 Next, we perform the necessary differentiations for the Lagrange equation: = 2 ̇ ̇ µ ¶ = 2 ̈ ̇ = 0 = sin + Substituting these into Lagrange’s equation yields the equation of motion: ̈ +
sin + 2 = 0
For small sin ≈ and the linearized equation of motion is ̈ + p frequency given by = ( + 2 )
¡
¢ + = 0 with the natural 2
242
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
23. For Example 4.16 derive the equation of motion using as the generalized coordinate. Then solve this equation with initial conditions 0 0 to find () = where =
0 sin − sin (cos − 1) + 0 2
p 43.
Solution: The disk on the inclined plane is shown next.
Disk on an inclined plane connected to two springs. Since it is assumed that the disk rolls without slipping, there is no energy loss from friction. Note that = ̇ and ̇ = ̇ where indicates the center of gravity. The kinetic and potential energies are 1 2 1 + 2 2 2µ ¶ 1 2 1 1 2 2 3 2 ̇ = ̇ = ̇ + 2 2 2 4 2 = + = ( − 0 ) − sin =
where is measured from the left wall, and 0 is the location of the disk when springs are unstretched. The potential energy stored in the spring is = 12 ∆2 where ∆ is the net stretch or compression. The net stretch is ∆ = −0 It should be noted that ∆ is not measured from the static equilibrium. We can now construct the Lagrangian, = − and use Lagrange’s equations to find: − = 0 ̇ 2 4 ( − 0 ) = sin ̈ + 3 3 Or, letting =
q
4 we get: 3
̈ + 2 ( − 0 ) =
2 sin 3
243 The solution to this differential equation is of the form: () = 1 cos + 2 sin +
µ
¶ 2 sin + 0 3 2
We use the given initial conditions to solve for the constants, finding: ¶ µ 2 sin + 0 (0) = 0 = 1 + 3 2 ̇(0) = 0 = 2 Then, the total solution is given by ¶ ¶ µ µ 0 2 2 sin + sin + 0 () = − 2 sin cos + 3 3 2 0 2 = sin − 2 sin (cos − 1) + 0 3 Replace 2 (3 2 ) with (2) and we obtain an identical expression given in the problem.
244
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
24. Derive the equation of motion of Example 4.16 using Newton’s second law of motion. Solution: We choose to use as the generalized coordinate, so it becomes convenient to approach the problem using the sum of moments instead of the sum of forces. Figure next shows the free-body diagram.
We can use one of the three versions of Newton’s second law for a rotational system. Note that we must define a clockwise moment to be positive here, since that is the direction of increasing . X = © X = + × © X = + × (Any moving point ). © is the angular momentum defined by where Z = ×
The first equation is easy to memorize but it does not eliminate the frictional force at , P and therefore, needs to be used in conjunction with the force balance equation along the incline ( = along the incline) The second equation can be reduced to a more familiar equation when is a fixed point on the body (so that is constant P with time) and also has a zero velocity. Typically such point is re-labeled as and we have © = ̈ Let us use the second relation, X = ̈ + ×
where for a disk that rolls without slip, = 0 The sum of moments about due to gravity and the spring forces is sin − 22 ( − 0 ) and, using the parallel axis theorem, the moment of inertia is = + 2 = 32 2 When the Substituting these into Newton’s second law gives: 3 2 2 4 ̈ + ( − 0 ) = sin 2 3 3
245 Problems for Section 4.6 — Notions of Control 25. For the problem of oscillator control, given by Equation 4.32, consider the specific governing equation ̈ + 2 ̇ + 2 = cos + () where , and () must be determined so that the maximum amplitude of the response is , where = 05. Since there is no single answer, describe how to proceed and what considerations must be made during the analysis. Provide one solution. Solution: Begin by assuming a specific structure, with known and Solve for the forced response to cos where () = 0 Compare the peaks of this response to If the peaks exceed this value, consider what would happen if is increased. If for the specific structure it is impossible to alter or then a control force becomes necessary. As a possible solution in general is the obvious one: () = − cos While this is excessive, it does suggest an approach to this problem. Other approaches are shown in Chapter 6.
246
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
26. How does randomness of excitation alter the analyst’s ability to evaluate structural response? Solution: If the loading has randomness, meaning that we are uncertain about its amplitude of frequency and oscillation, then the analyst has a more difficult time assessing structural response. Of particular concern is whether loading frequencies overlap structural natural frequencies, which result in response magnification. If the randomness is small, then it may be possible to use average loading. If the randomness is broader, then several possible loadings can be used to consider the various responses. The worst case of these is then taken in consideration in a design. Other approaches are shown in Chapter 5, where we will learn how to pick optimal, not extreme, values for parameters using their probability density.
247
27. What difficulties arise if a system parameter such as or is known only approximately? Solution: Problems can arise if uncertainties exist about system parameters. If the uncertainties are too large to allow the analyst to use the average values of or then it is necessary to consider the worst case scenario, that is, parameter values that result in the largest response amplitude or frequency. In Chapter 5 we will learn how to pick optimal values for parameters using their probability density functions.
248
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
Problems for Section 4.7 — Notions of Control 28. If the structure responds in a way that is unacceptable for a particular application, what options exist for the designer? Solution: Unacceptable response requires either a re-design of the structure or the addition of a control force. Sometimes both are necessary. If a re-design is not possible, then one requires the addition of a control force to counter the effects of environmental forces. In Chapter 6 the concepts of feedback control will be introduced.
249
29. If the analyst decides that a structure requires control in order to fulfill the needs of an application, describe the procedure by which a control force can be derived and then verified that it performs as expected. Solution: If a control force is required, then the benchmark against which a control force is compared is the uncontrolled response. Thus, the system response is evaluated without the addition of control. The difference between the uncontrolled response and the desired response is to be eliminated by the control force. As far as the system is concerned, the control force is an external force. In reality, the control force is generated as a function of system response. If the system response is large, the control force is large, and vice versa. Knowing the character of the environmental forces permits us to prescribe the control force. Of course, in many instances we cannot predict the specific nature of the external forces, only their general characteristics. In this case, feedback control is necessary, where the control force is governed by the real-time measurement of the response. Verification hinges on testing the system and control in a variety of expected loadings. There is significant discussion on control in the later chapter.
250
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
30. A simple system is represented by the governing equation ̈ + = () where = 1 kg, = 10 N/m, and () = cos N with zero initial conditions. Find the steady-state response () Now suppose that the system is constrained regarding its maximum allowable amplitude, |max | ≤ 005 m. Determine the needed () Solution: The system is undamped, and as a consequence the homogeneous solution does not die out with time. This is not realistic because in a real physical system there is always damping no matter how small it may be). Therefore, we assume that there is damping in the system so that the homogeneous solution decays with time, but small enough so that the steady-state response amplitude can be approximated using = 0Then, the steady-state response is given by () =
0 0 cos = cos − + 9
In order to reduce the amplitude to 0.05 m, the force magnitude must be between −005 × 9 N ≤ 0 ≤ 005 × 9 N. With the existing force magnitude of 1 N, we need the control force () = cos ( − ) where −145 N ≤ ≤ −055 N. Note that the control force is at the same frequency as the existing force. In addition, note that for above answer to be valid, the control force must be in phase with the existing force. For example, if the control force is 180◦ out of phase, the amplitude of sin + 0 cos is q 2 2 + | | = instead of +
251 Problems for Section 4.8 — The Inverse Problem 31. Consider the simple system, ̈+ = () where () is known but and are unknown. Suppose the initial conditions are both equal to zero. What can we determine about the values of and ? Solution: In this case where we don’t know the response (), we don’t have enough information to explicitly solve for and . If we are able to measure the response, then we can calculate its frequency assuming that () is harmonic. Also,if we can apply a constant () via an experiment we can measure the static displacement, which provides us with a means of relating and .
252
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
32. Consider the simple system, ̈ + = () where () and steady-state solution () are known and and are unknown. Suppose the initial conditions are both equal to zero. What can we determine about the values of and ? Solution: Here we know the steady-state response. If the force is harmonic then we can measure the frequency of oscillation. Also, if we can apply a constant force via an experiment we can measure the static displacement, which provides us with a means of relating and .
253
33. Consider the simple system, ̈ + = () where steady-state solution (), and are known and () is unknown. Suppose the initial conditions are both equal to zero. What can we determine about ()?
Solution: Here the unknown is the input force. If the steady-state solution is harmonic, then the input is also harmonic. From the steady-state solution we can measure the natural frequency of oscillation.
254
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
Problems for Section 4.9 — A Self-Excited System and Its Stability 34. Consider a damped oscillator where the force is a function of velocity, ̈ + ̇ + = 0 ̇ Discuss the stability of this system in terms of the parameters 0 . Solution: We can rewrite the equation of motion as ̈ + ( − 0 )̇ + = 0 This can be put into non-dimensional form as − 0 ̇ + = 0 ̈ + 2 ̇ + 2 = 0
̈ +
where 2 = ( − 0 ) and 2 = We know the general response to this system () = exp(− )(1 cos + 2 sin ) ( − 0 ) = exp(− )(1 cos + 2 sin ) 2 If ≥ 0 then the response decays as we know from before. If ≤ 0 then we have the exponential of a positive quantity, which grows with time in an unbounded way. Eventually, the system, will have amplitudes that are too large, rendering our linear governing differential equation invalid, and/or resulting in a system failure. Therefore, a velocity-dependent force can be stable or unstable, depending whether is greater or less than 0 Note that if = 0 the system is undamped.
255
35. Consider a damped oscillator where the force is a function of displacement, ̈ + ̇ + = 0 Discuss the stability of this system in terms of the parameters 0 . Solution: We can rewrite the equation of motion as ̈ + ̇ + ( − 0 ) = 0 Now, we have a possibility that the coefficient of can be negative. The characteristic values are p − ± 2 − 4 ( − 0 )
If − 0 0 the real part of the characteristic values are negative, and the response decays exponentially (as seen in the underdamped, overdamped, and critically damped cases) unless there is no damping, in which case, the response does not decay but persists over time with the same amplitude of vibration. The response is uniformly stable. If − 0 = 0 a solution of the type, ³ ´ () = 1 + 2 exp − is possible. At most, () has a constant solution, and the response is uniformly stable. If − 0 0 one of the characteristic value is positive and the other negative. The response grows exponentially, and the response is unstable.
256
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
Problems for Section 4.10 — Solution Analysis and Design Techniques 36. For an -- oscillator subjected to harmonic loading, cos , conceive a design so that the displacement response |()| for all time. Assume = 10 cm, = 1 kg, = 1 N/cm, and = 3 rad/s. In your design, specify and . To do this, find the upper limit for (). Here, we attempt to obtain quantitative values for and Solving the differential equation, we find that the response is given by (for an underdamped case) () = exp (− ) (1 cos + 2 sin ) +
1 sin 2
where is the magnification ratio is 1 =p (1 − 2 2 )2 + (2 )2
The constants of integration, 1 and 2 depend on the initial conditions. Upon applying zero initial conditions, the constants of integration, 1 and 2 are 1 = 0 and 2 = The total solution is ∙ µ ¶ ¸ 1 () = − exp (− ) sin + sin 2 The maximum occurs when = 0 or ∙ µ ¶ ¸ () 1 exp (− ) sin − exp (− ) cos + cos = 0 = 2 which means that the term in the bracket is zero, µ ¶ exp (− ) sin − exp (− ) cos + cos = 0
(2)
The idea is to find the time when the maximum occurs for a set of and from Equation (1) and substitute this time into Equation (1) to find the maximum value. This is not easy to do because you have to choose the time when the global maximum occurs. The problem is considerably more difficult because it is asking for the values of and for which |()| for all time instead of for the time after the transient effects died out. Instead of finding the exact maximum, what we can do is to find the upper limit for () We know that ¯ µ ¶ ¯ ¯ ¯ ¯− exp (− ) sin ¯ ≤ ¯ ¯ |sin | ≤ 1 ¯ µ ¶ ¯ ¯ ¯ ¯− exp (− ) sin + sin ¯ ≤ + 1 ¯ ¯
257 Note that if = , we can put tighter limits on the last expression so that √ √ |− exp (− ) sin + sin | ≤ 12 + 12 = 2 In general, the forcing frequency is not equal to the damped natural frequency. Then, the upper bound for the response () is given by ¯ ¯ µ ¶ ¯ () ¯ 1 1 ¯ ¯ ¯ ¯ ≤ 2 + 1
Let us rewrite above expression in terms of so that ¯ ¯ µ ¶ ¯ () ¯ ¯ ≤ 1 1 ¯ +1 ¯ ¯ 2 Ã ! ¯ ¯ ¯ () ¯ 1 1 1 ¯ ¯ +1 ¯ ¯ ≤ 2 p 1 − 2
Figure next shows the maximum values of | () | as a function of and We use the given values of = 1 kg and = 3 rad/s.
The upper limit of | () | is increased from 0.1 to 0.9 at the increment of 01 Note that the plot is the upper limit, not the actual maximum value. The case 1 is marked in the plot, and it shows that the actual maximum of 0.18 (estimated from the first plot) is well within the upper limit of 0.5. The figure above provides the tendency of the maximum response as and are varied. We can easily convert to and using = 2 and = 2
258
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
37. For an oscillator governed by ̈ + 2 ̇ + 2 = cos , conceive a design so that the displacement response |()| for duration 0 . Assume = 10 cm and = 1 kg. In your design, specify and . Solution: Given = 1 kg then rather than specify and we choose to specify and in satisfying the condition |()| 01 m We need to know the loading frequency before we can evaluate the response. We proceed in terms of the frequency ratio: This will be our second parameter. This suggests that we also study the magnification parameter : 1 =p (1 − 2 2 )2 + (2 )2
p Or = 1 (1 − 2 )2 + (02)2 Let us begin by assuming a reasonable value for = 010 Now we can plot as follows: beta
5
3.75
2.5
1.25
0
0
1
2
3
p = 1 (1 − 2 )2 + (02)2
4 r
We can write the equation of motion as ̈ + 02 ̇ + 2 = cos where = 1 kg. Then = 2 and = 02 At this point, for a given driving frequency we can examine whether the displacement criterion can be met. Suppose the driving frequency is low, say = 01 rad/s. We know that ≤ 01 since the inputp magnitude equals 1 Then, for the cutoff value = 01 the required is given by solving 01 = 1 (1 − 012 2 )2 + (02 × 01 )2 This results in a (real) value for This procedure can be repeated until a practical set of design parameters are found.
259
38. Solve Problem 37, but for the load shown in Figure 4.36.
Figure 4.36: Ramp loading.
Solution: Now the load is a ramp function with peak amplitude The simplest approach is to solve for the response, and then try various groups of parameter values. The general solution can be given by the convolution (Duhamel) integral: Z 1 () = exp(− [ − ]) sin ( − ) ( ) 0 where () is the equation of the forcing: () = 0 ≤ ≤ 1 Our goal is that |()| where 01 m
260
CHAPTER 4 SDOF VIBRATION: GENERAL LOADING
39. Solve Problem 37, but for the load as shown in Figure 4.37.
Figure 4.37: Simplified representation of a shock loading.
Solution: The load is a triangular idealization of a shock loading. The simplest approach is to solve for the general response and then try various groups of parameter values. The general solution can be given by the convolution (Duhamel) integral: Z 1 () = exp(− [ − ]) sin ( − ) ( ) 0 where () is the equation of the forcing: () = (1 − ) 0 ≤ ≤ 1 Our goal is that |()| where 01 m
Chapter 5 Variational Principles and Analytical Dynamics Problems for Section 5.3 — Virtual Work 1. Consider the system shown in Figure 5.21. Let 1 and be the displacement and angle of rotation of the pulley measured from their original locations when the spring is undeformed. Let 2 be the displacement of the cylinder, also measured from its original location when the spring is undeformed. The pulley has mass , radius and its mass moment of inertia about its centroidal axis is The cylinder has mass Find (a) the static equilibrium position of the system using the principle of virtual work, and (b) the equation of motion of the system for generalized coordinate 2 using d’Alembert’s principle.
Figure 5.21: Pulley-cylinder coupled system.
Solution: Static Equilibrium via Newton’s Law The first step is to identify that there is one degree of freedom and realize that there are two redundant coordinate. There are two constraint equations that relate the three coordinates, 1 2 261
262
CHAPTER 5 ANALYTICAL DYNAMICS
and Note that 1 is the vertical displacement of the center of the pulley. We can see that 1 = 2 = + 1 = 21 Let’s us use 1 as our generalized coordinate. The next step is to find the stretch. The stretch in the spring is 1 We draw the free body diagrams for the pulley and the mass.
The sum of the forces in the 1 direction for the pulley is 1 + 2 + − 1 = 0 The sum of the moment in the direction for the pulley is 2 − 1 = 0 The sum of the forces in the 2 direction for the cylinder is − 2 = 0 Eliminating tension from the equations, we obtain − 1 + 2 = 0 2 + 1 =
Static Equilibrium via Principle of Virtual Work Let us redraw the free-body diagram that contains both the pulley and the mass.
263
The virtual work done on the system is the dot product of the force vector and the virtual displacement where the force is applied. The displacements at locations and are = 0 = 1 downward = 1 + downward The corresponding virtual displacements are = 0 = 1 = 1 + The virtual work is = ( − 1 ) 1 + (1 + ) Now, the redundant coordinates must be eliminated. We eliminate by replacing it with 1 The virtual displacements are related by =
1 1
The virtual work becomes = ( − 1 ) 1 + () 21 = ( − 1 + 2) 1 Finally, the equilibrium equation is obtained by setting the coefficient of the virtual displacement zero, − 1 + 2 = 0 Then, 1 =
2 +
Using constraint equations, =
1 2 + 2 + and 2 = 2
264
CHAPTER 5 ANALYTICAL DYNAMICS
We should note that the virtual work done by the geometric constraint forces is always zero, and to prove this point, we write (due to 1 and 2 ) = 1 (1 − ) + 2 (1 + ) − 2 2 where (1 − ) is the displacement of the pulley where 1 is applied and (1 + ) is the displacement of the pulley where 2 is applied. The virtual work due to 1 and 2 is reduces to zero after we use the kinematic relations, = 1 and 2 = 21 Equation of Motion via d’Alembert’s Principle D’Alembert’s equation is given by ´ ³X ´ ´ ³X ³X − ̈1 1 + − ̈2 2 = 0 − ̈ +
where the first two terms are due to the pulley and the third term is due to the block. In this case, ´ ³ 0 − ̈ + ( − 1 − ̈1 ) 1 + ( − ̈2 ) 2 = 0
where 1 and 2 are related by
2 = 21 and = 1 We choose to use 1 as our generalized coordinate. Then, d’Alembert’s equation is given by ¶ µ ( − 1 − ̈1 ) 1 + 0 − 2 ̈1 1 + 2 ( − 2̈1 ) 1 = 0 µ ¶ + 2 − 1 − ̈1 − 2 ̈1 − 4̈1 1 = 0 The equation of motion is given by ¢ ¡ + 2 + 4 ̈1 + 1 = + 2
265
2. Consider the point mass supported by two linear springs shown in Figure 5.22. The springs are undeformed when the mass and the springs are along a straight line, where the coordinate defines the displacement of the mass. Do not assume small angles. Derive (a) the equilibrium equation using the principle of virtual work, and (b) the equation of motion of the system using d’Alembert’s principle.
Figure 5.22: Point mass supported by two identical springs.
Solution: The free body diagram for the mass is shown below. 0 is the new length of the spring or p 0 = 2 − 2
Newton’s Law The sum of the forces in the direction is P
= 0 − 2 (0 − ) 0 = 0 where 0 = The equilibrium equation is given by
p 2 + 2
´ ³p 2 2 + − p = 0 − 2 2 + 2
Static Equilibrium via Virtual Work
266
CHAPTER 5 ANALYTICAL DYNAMICS
The virtual work is given by =
Ã
³p ´ − 2 2 + 2 − p 2 + 2
!
Therefore, the equilibrium equation is given by ³p ´ 2 2 − 2 + − p = 0 2 + 2
which is identical to the ones obtained using Newton’s law. Equation of Motion via d’Alembert’s Principle D’Alembert’s equation is given by Ã
− 2
³X
´ − ̈ = 0 !
³p ´ 2 + 2 − p − ̈ = 0 2 + 2
Therefore, the equation of motion is ´ ³p 2 + 2 − p − ̈ = 0 − 2 2 + 2
267
3. Consider again the system shown in Figure 2.58, reproduced in Figure 5.23. The system consists of massless pulleys, a block of mass and two linear springs. The coordinate, , is the angle of rotation of the moving pulley, 1 is the displacement of the pulley center of mass, and 2 is the displacement of the mass. The displacements are measured from the positions where the springs are undeformed. (a) Find the static equilibrium position of the system using the principle of virtual work, and (b) derive the equation of motion of the system using d’Alembert’s principle for generalized coordinate 2
Figure 5.23: A mass suspended by springs.
Solution: Static Equilibrium via Principle of Virtual Work There are three coordinates used, and there is one kinematic relationship. Therefore, the system has two independent coordinates. Due to negligible mass of the moving pulley, the system model can be written in terms of a second order differential equation in terms of one coordinate. The displacement of the mass is kinematically related to the other two coordinates by
2 = 1 + The stretch in the spring 1 is − 1 while the stretch in the spring 2 is 1 2 = 1 1 = − 1 The free-body diagrams for the pulley and the mass are shown next.
268
CHAPTER 5 ANALYTICAL DYNAMICS
In order to find the virtual work, we may use either free-body diagrams. Let us use the last freebody diagram. The forces and are the reaction forces from the pin that are used to hold the pulley. These forces do not do work because the displacement at the pin is zero. The virtual work done is = 1 ( − 1 ) (1 − ) − 2 1 1 + () 2 where the term, 1 − is the displacement of the moving pulley where the spring force, 1 ( − 1 ) is applied. A care is taken so that displacement is in the same direction of the force. Similarly, we note that the downward force, 2 1 , is applied to the center of the pulley which has a downward displacement of −1 hence the negative sign in the virtual work expression. We eliminate 2 by replacing it with 1 + so that = (1 ( − 1 ) − 2 1 + ) 1 + (− ( − 1 ) + ) We find 1 ( − 1 ) − 2 1 + = 0 −1 ( − 1 ) + = 0 Solving them simultaneously, we find 2 1 and = 1 = 2
µ
2 + 2 1
Using kinematic equations, we have 2 = 1 + 4 = + 2 1 Equation of Motion via d’Alembert’s Principle
¶
269 D’Alembert’s equation is given by ³X ´ ³X ´ ³X ´ − ̈ + − ̈ + − ̈ = 0
where the first two terms are due to the pulley and the third term is due to the block. In this case, ³ ´ −1 ( − 1 ) − ̈ + (1 ( − 1 ) − 2 1 − ̈1 ) 1 + ( − ̈2 ) 2 = 0
where and are the mass moment and the mass of the pulley. They are assumed negligible in this problem. The coordinates are also related by 2 = 1 + Then, d’Alembert’s equation simplifies to ´ ³ 0 = −1 ( − 1 ) − ̈ + (1 ( − 1 ) − 2 1 − ̈1 ) 1 + ( − ̈2 ) 2 ³ ³ ´ ´ ³ ³ ´ ´ = −1 ( − 1 ) + − ̈1 + ̈ + − ̈1 + ̈ + 1 ( − 1 ) − 2 1 1 Then, we obtain
³ ´ −1 ( − 1 ) + − ̈1 + ̈ = 0 ³ ´ − ̈1 + ̈ + 1 ( − 1 ) − 2 1 = 0
(1) (2)
It is possible reduce this equation into a single equation in terms of 1 only. Divide the first equation by and subtract the second equation from it to obtain −21 ( − 1 ) + 2 1 = 0 21 + 2 = 1 21 Replacing from Equation (2), we finally obtain µ ¶ 41 + 2 2 ̈1 + 1 = 21 2 Equation of motion in terms of 2 can be obtained using the kinematic relationship 2 = 1 + where is replaced using Equation (3) so that ¶ µ 21 + 2 2 = 1 + 1 21 41 + 2 = 1 21 Then, the equation of motion is given by ¶µ ¶ µ 21 2 21 41 + 2 ̈2 + 2 = 21 41 + 2 2 41 + 2 1 2 ̈2 + 2 = 41 + 2
(3)
270
CHAPTER 5 ANALYTICAL DYNAMICS
4. Consider the system shown in Figure 5.24. Let be the displacement of the mass measured from its original location where the springs are undeformed. Assume that the pulleys are massless and the springs are linear. (a) Find the static equilibrium position of the system using the principle of virtual work, and (b) derive the equation of motion using d’Alembert’s principle.
Figure 5.24: A mass suspended by springs.
Solution: Newton’s Law The stretch in the springs are 1 = + and 2 = − If we assume that the pulleys are massless, the tension in the cable is constant such that = 1 ( + ) = 2 ( − )
((1))
Even if we do not realize this, we can eventually find that the tension is constant by taking a moment of the pulley as shown next. The free-body diagram for the pulley and the mass is shown next.
271
Summing the moment about the center of the pulley in the positive direction, we find that 2 ( − ) − 1 ( + ) = ̈
where the right hand side is zero because the pulley has a negligible mass. Therefore, 1 ( + ) = 2 ( − ) for both static and dynamic cases. The sum of the forces in the positive direction, we have − 1 ( + ) − 2 ( − ) = 0
((2))
Solving Equation (1) and (2) together, we have 1 + 2 = 41 2 1 2 − 1 = 41 2 Principle of Virtual Work The virtual work is
= ( − 1 ( + ) − 2 ( − )) + (2 ( − ) − 1 ( + ) )
Note that and are not kinematically related, and therefore, no further simplification is possible. We set the coefficients of virtual displacements to zero or − 1 ( + ) − 2 ( − ) = 0 2 ( − ) − 1 ( + ) = 0
which are identical to Equations (1) and (2). Therefore, we obtain the identical equilibrium positions. Equation of Motion via d’Alembert’s Principle D’Alembert’s equation is given by ´ ³X ´ ³X ´ ³X − ̈ + − ̈ = 0 − ̈ +
where the first two terms are due to the pulley and the third term is due to the block. In this case, (2 ( − ) − 1 ( + ) ) + (−2 ( − ) − 1 ( + )) + ( − ̈) = 0
which simplifies to
2 ( − ) − 1 ( + ) = 0 −2 ( − ) − 1 ( + ) + − ̈ = 0
This is a differential algebraic equation (DAE), where can be eliminated from the equation to reduce to the system model to a single differential equation in terms of . Solve for 1 2 − 1 = 1 + 2 Replacing in the second equation, and we obtain 41 2 = ̈ + 1 + 2
272
CHAPTER 5 ANALYTICAL DYNAMICS
5. Consider again the system shown in Figure 2.46, reproduced in Figure 5.25. The bar is horizontal when the massless springs are undeformed. Assume that the rod mass is , and the mass moment of inertia about the torsional spring is . (a) Find the static equilibrium position of the system using the principle of virtual work, and (b) derive the equation of motion of the system using d’Alembert’s principle.
Figure 5.25: Restrained rigid rod.
Solution: Principle of Virtual Work The virtual work is
µ
sin = − − 2 sin ( sin ) + 2 ¶ µ = − − 2 sin cos + cos 2
¶
The equilibrium equation is − − 2 sin cos + cos = 0 2 This equation must be solved numerically. For small angles, the equation is reduced to − − 2 + = 0 2 = 2 ( + 22 ) Equation of Motion via d’Alembert’s Principle D’Alembert’s equation is given by ³X ´ − ̈ = 0
where point is the pivot point where the torsional spring is attached. In this case, d’Alembert’s equation is µ ¶ − − 2 sin cos + cos − ̈ = 0 2
273 where = 2 3 The equation of motion is − − 22 sin cos + cos − ̈ = 0 2
274
CHAPTER 5 ANALYTICAL DYNAMICS
6. Consider again the system shown in Figure 2.56, reproduced in Figure 5.26. A disk is on an inclined plane supported by a spring. The coordinates and are zero when the spring is undeformed. The disk has mass moment of inertia about its mass center and has radius Assume that the disk rolls without slip. (a) Find the static equilibrium position of the system using the principle of virtual work, and (b) derive the equation of motion of the system using d’Alembert’s principle.
Figure 5.26: A rolling disk on an incline.
Solution: Principle of Virtual Work This is a one degree-of-freedom system because the disk rolls without slip. The contact point between the disk and the ground is the instantaneous center for the disk, and therefore, the coordinates, and , are related by = and the stretch in the spring is = 2 = 2 The free-body diagram of the disk is shown next.
275 The virtual work done on the disk by the weight, spring force, and moment about the center of gravity, and they are = sin − 2 (2) Another way to find the virtual work for a rigid body is to use P P = ( F) ·x + ( M ) ·θ P where F is sum of the all the forces on the body, is the displacement of the center of gravity, P M is the sum of the moment about the center of gravity and θ is the virtual angle of rotation. In this case, P ( F) ·x = ( sin − 2 − ) P ( M ) ·θ = ( − 2) = ( sin − 2 − ) + ( − 2) = ( sin − 2 − + − 2) = ( sin − 4) Note that the friction and the normal force do not do work because the displacement is zero where these forces are applied. The equilibrium equation is given by sin − 4 = 0 The equilibrium position is then sin 4 1 sin = 4
=
Equation of Motion via d’Alembert’s Principle For a rigid body that pivots about an instantaneous center or a stationary point, there are two versions of D’Alembert’ equations. They are ´ ³X ´ ³X − ̈ = 0 − ̈ +
or
³X
´ − ̈ = 0
For a ball that is rolling without slipping, the contact point qualifies as the pivot point. Using the second version, d’Alembert’s equation is given by ³ ´ − 2 + sin − ̈ = 0
where = 2 = 2 The equation of motion is given by ̈ + 42 = sin
276
CHAPTER 5 ANALYTICAL DYNAMICS
7. Consider the system comprised of the disk, pulley, and block shown in Figure 5.27. The spring is connected to the disk of mass at its mass center. The disk has radius and mass moment of inertia about its mass center. A cable is connected to and wrapped around the disk and connected to the block of mass . Assume that the pulley is massless and that the disk rolls without slip. (a) Find the static equilibrium position of the system using the principle of virtual work, and (b) derive the equation of motion of the system using d’Alembert’s principle.
Figure 5.27: Spring-disk-pulley mass system.
Solution: This is a one degree-of-freedom system. The coordinates and are related by = and the displacement of the block is 2 Figures below show the free-body diagrams of the disk and the mass.
Newton’s Law The sum of the forces in the vertical direction for the mass is = The sum of the moment about the instantaneous center, where the disk touches the ground, is P = 0 (2) − = 0
277 Substituting = we have =
2
Principle of Virtual Work The virtual work is done by the spring force and gravity only, and it is = (2) − = (2 − ) = 0 The virtual displacement for the force is (2) because the displacement of the system where is applied (at the mass) is 2 The equilibrium equation is given by 2 − = 0 which gives =
2
Equation of Motion via d’Alembert’s Principle D’Alembert’s equation is ´ ³X ´ ³X − ̈ = 0 − ̈ + ³ ´ − − ̈ + ( − (2̈)) (2) = 0
where = and the displacement of is 2 Then, d’Alembert’s equation becomes ¶ µ ̈ 1 + ( − (2̈)) (2) − − µ ¶ = − − 2 ̈ + − 2̈ = 0 The equation of motion is
µ ¶ 4 + 2 ̈ + = 2
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CHAPTER 5 ANALYTICAL DYNAMICS
8. Consider the two degree-of-freedom system shown in Figure 5.28. The springs are undeformed when the rods are horizontal. The mass of each rod is with length and mass moment of inertia about each center of gravity is (a) Find the static equilibrium position of the system using the principle of virtual work, and (b) derive the equation of motion of the system using d’Alembert’s principle.
Figure 5.28: Spring-constrained double pendulum.
Solution: This is a two degree-of-freedom system. The generalized coordinates are 1 and 2 The free-body diagram is shown next. Note that we may put the spring force at on either rod. The value of will change, but the total force on the each rod at will be the same.
Newton’s Law The sum of the force and the moment of the rod on the left is P → = + = 0 P ↑ = − − + sin 1 = 0 P ª = ( sin 1 − ) cos 1 − cos 1 + sin 1 = 0 2
279 The sum of the force and the moment of the rod on the right is P → = − = 0 P ↑ = − + ( sin 1 + sin 2 ) = 0 P ª = ( sin 1 + sin 2 ) ( cos 2 ) − cos 2 = 0 2
We have six equations and six unknowns. Let us first eliminate and Then, we have two equations in terms of 1 and 2 After some algebra, the two equations are ¶ µ 3 sin 1 − + (sin 1 + sin 2 ) cos 1 = 0 2 µ ¶ 1 2 (sin 1 + sin 2 ) − cos 2 = 0 2 The equilibrium positions, 1 and 2 , are obtained by solving these equations numerically. Principle of Virtual Work We can put the two rods together at so that the free-body diagram of the system is shown next.
The virtual work is given by ¶ µ ¶ µ sin 1 + sin 1 + sin 2 − sin 1 ( sin 1 ) = 2 2 − ( sin 1 + sin 2 ) ( sin 1 + sin 2 ) Operating on the displacements, we have cos 1 1 + cos 1 1 + cos 2 2 − sin 1 cos 1 1 2 2 − ( sin 1 + sin 2 ) ( cos 1 1 + cos 2 2 ) µ ¶ = cos 1 + cos 1 − sin 1 cos 1 − cos 1 ( sin 1 + sin 2 ) 1 2 µ ¶ + cos 2 − cos 2 ( sin 1 + sin 2 ) 2 2
=
280
CHAPTER 5 ANALYTICAL DYNAMICS
Setting the coefficients of 1 and 2 zeros, we have the equilibrium equations given by ¶ µ 3 2 2 − 2 sin 1 − sin 2 cos 1 = 0 2 µ ¶ 2 2 − sin 1 − sin 2 cos 2 = 0 2 These equations are identical to the ones obtained using Newton’s law. These equations must be solved numerically. We note that 1 = 2 = 90◦ is one of the solutions. This is when the bars are vertical. This is when the bar is in unstable equilibrium. The stable equilibrium position can be found by solving 3 − 22 sin 1 − 2 sin 2 = 0 2 − 2 sin 1 − 2 sin 2 = 0 2
If we assume small angles, the equations are reduced to 3 − 22 1 − 2 2 = 0 2 − 2 1 − 2 2 = 0 2
which have solutions 1 =
1 and 2 = − 2
Equation of Motion via d’Alembert’s Principle D’Alembert’s equation is ³X ´ 0 = − ̈1 1 1 ´ ´ ³X ³X + − ̈2 2 + − ̈2 2 2 2 ´ ³X + − ̈2 2 2
where the first term is the first rod, and the other two terms are for the second rod. The second rod does not have a pivot point that also has zero velocity. D’Alembert’s equation is µ ¶ 2 0 = − sin 1 cos 1 + cos 1 − ̈1 1 2 ¶ µ + − (sin 1 + sin 2 ) cos 2 − ̈2 2 2 + ( − (sin 1 + sin 2 ) − ̈2 ) 2 + (0 − ̈2 ) 2
281 where = 2 3, = 2 12 and 2 is the vertical displacement of the center of mass of the second rod and is related to 1 and 2 by sin 2 2 ̇2 = ̇1 cos 1 + ̇2 cos 2 2 2 = cos 1 1 + cos 2 2 2 2 = sin 1 +
2
̈2 = ̈1 cos 1 − ̇1 sin 1 +
2 ̈2 cos 2 − ̇2 sin 2 2 2
Similarly, cos 2 2 ̇2 = −̇1 sin 1 − ̇2 sin 2 2 2 = − sin 1 1 − sin 2 2 2 2 = cos 1 +
2
̈2 = −̈1 sin 1 − ̇1 cos 1 −
2 ̈2 sin 2 − ̇2 cos 2 2 2
Replacing 2 and ̈2 and setting the coefficients of 1 and 2 to zeros, we obtain two equations motion given by cos 1 − ̈1 2 + cos 1 ( − (sin 1 + sin 2 )) µ ¶ 1 1 2 2 2 − cos 1 ̈1 cos 1 − ̇1 sin 1 + ̈2 cos 2 − ̇2 sin 2 2 2 µ ¶ 1 1 2 2 2 − sin 1 ̈1 sin 1 + ̇1 cos 1 + ̈2 sin 2 + ̇2 cos 2 2 2 0 = − (sin 1 + sin 2 ) cos 2 − ̈2 2 + cos 2 ( − (sin 1 + sin 2 )) 2 µ ¶ 1 1 2 2 2 − cos 2 ̈1 cos 1 − ̇1 sin 1 + ̈2 cos 2 − ̇2 sin 2 2 2 2 µ ¶ 2 1 1 2 2 − sin 2 ̈1 sin 1 + ̇1 cos 1 + ̈2 sin 2 + ̇2 cos 2 2 2 2 0 = −2 sin 1 cos 1 +
They are simplified to ¶ µ 3 2 2 − 2 sin 1 − sin 2 cos 1 − ̈1 0 = 2 1 1 2 −2 ̈1 − 2 ̈2 cos (2 − 1 ) + 2 ̇2 sin (2 − 1 ) 2 2
282
CHAPTER 5 ANALYTICAL DYNAMICS ¶ µ 2 2 0 = − sin 1 − sin 2 cos 2 − ̈2 2 2 2 2 2 − ̈1 cos (2 − 1 ) − ̇1 sin (2 − 1 ) − ̈2 2 2 4
283 Problems for Sections 5.4 and 5.5 — Lagrange’s Equation and Hamilton’s Principle 9. Derive the equation of motion for the system in Figure 5.21 for generalized coordinate 2 using (a) Lagrange’s equation, and (b) Hamilton’s principle. Solution: The kinetic and potential energies of the system are given 1 1 1 2 ̇21 + ̇ + ̇22 2 2 2 1 = −1 − 2 + 21 2 and the three coordinates are related by the two kinematic relations by =
1 = 2 = 21 = 2 This system has one DOF, and we decided to use 2 as the generalized coordinate. The other two coordinates are 2 2 1 = and = 2 2 The kinetic and potential energies in terms of 2 are µ ¶ 1 2 1 2 1 2 1 ̇ + + ̇2 + ̇2 = + ̇22 = 2 4 2 2 42 2 2 4 42 µ ¶ 1 2 = − + 2 + 2 24 2 (a) Lagrange’s equations Lagrange’s equation is given by − + = 0 ̇2 2 2 µ ¶ + + ̈2 + 2 = + 2 4 4 4 2 (b) Hamilton’s Principle Hamilton’s principle states that
Z 2 1
( − ) = 0
Take the variation of the Lagrangian, and we obtain Z 2 ( − ) 1 ¶ ¶ ¶ µ Z 2 µ µ 1 2 1 2 = + + 2 − + ̇2 + 2 4 42 2 24 2 1 ¶ ¶ µ ¶ Z 2 µµ = + ̇2 ̇2 + + + 2 − 2 2 4 42 2 4 1
284
CHAPTER 5 ANALYTICAL DYNAMICS
Integrate the first term by parts to obtain Z 2 ( − ) 1 ¯2 ¶ µ ¯ ¯ + + ̇ = 2 2 ¯ 4 42 1 ¶ ¶ ¶ µ Z 2 µ µ − + + 2 − 2 2 + + ̈2 2 + 4 42 2 4 1 The terms evaluated at = 1 and 2 are zeros by definition, and we obtain ¶ ¶ ¶ µ Z 2 µ µ Z 2 − + + − 2 2 = 0 ( − ) = + ̈2 + 4 42 2 4 1 1 For any possible variation for 2 the integral must become zero, and therefore, the coefficient of 2 must be zero. The equation of motion is given by ¶ ¶ µ µ + + − 2 = 0 + ̈2 + − 2 4 4 2 4 or
µ
¶ ¶ µ + + + ̈2 + 2 = 4 42 4 2
285
10. Derive the equation of motion for the system in Figure 5.22 for generalized coordinate using (a) Lagrange’s equation, and (b) Hamilton’s principle. Solution: The kinetic and potential energies are given by =
(a) Lagrange’s equations Lagrange’s equation is given by
1 2 ̇ 2
´2 1 ³p 2 = − + + 2 − 2 − + = 0 ̇
The each term is = ̈ ̇ = 0 ! Ã ³p ´ p = 2 + 2 − 2 + 2
Then, the equation of motion is given by ³p ´ ̈ + 2 + 2 − p = 2 + 2 (b) Hamilton’s Principle Hamilton’s principle states that :
Z 2 1
( − ) = 0
In this case, the left hand side is given by Z 2 µ ´2 ¶ 1 ³p 2 1 2 ̇ + − + 2 − 2 2 1 Z 2 ³ ´ ³p ³p ´´ = 2 + 2 − 2 + 2 ̇ ̇ + − 1 ! Z 2 Ã ³p ´ ̇ ̇ + − 2 + 2 − p = 2 + 2 1
Integrate the first term by parts and we obtain ! Z 2 Z 2 Ã ³p ´ −̈ + − ( − ) = ̇|21 + 2 + 2 − p 2 + 2 1 1
286
CHAPTER 5 ANALYTICAL DYNAMICS
Then, the equation of motion is given by ³p ´ 2 2 −̈ + − + − p = 0 2 + 2
287
11. Derive the equation of motion for the system in Figure 5.23 for generalized coordinate 2 using (a) Lagrange’s equation, and (b) Hamilton’s principle. Solution: Three coordinates used and there is one kinematic relationship in this system. Therefore, there are two independent coordinates. The displacement of the mass is kinematically related to the other two coordinates by 2 = 1 + The stretch in the spring 1 is − 1 while the stretch in the spring 2 is 1 2 = 1 1 = − 1 The kinetic and potential energies are given by 1 2 ̇ 2 2 1 1 = 1 ( − 1 )2 + 2 21 − 2 2 2 =
Let us use 1 and 2 and eliminate Then, the potential energy is given by 1 1 = 1 (2 − 21 )2 + 2 21 − 2 2 2 (a) Lagrange’s equations Lagrange’s equations are − + = 0 ̇1 1 1 − + = 0 ̇2 2 2 Two Lagrange’s equations are necessary because there are two independent coordinates. They simplify to −21 (2 − 21 ) + 2 1 = 0 ̈2 + 1 (2 − 21 ) − = 0 Solve for 1 using the first equation to obtain 1 = and the equation of motion in terms of 2 is µ ̈2 + 1 2 −
21 2 2 + 41
¶ 41 2 − = 0 2 + 41 1 2 ̈2 + 2 = 2 + 41
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CHAPTER 5 ANALYTICAL DYNAMICS
(b) Hamilton’s Principle Hamilton’s principle states that
Z 2 1
In this case,
Z 2
( − ) = 0
( − )
Z 12 µ
¶ 1 2 1 1 2 2 ̇ − 1 (2 − 21 ) − 2 1 + 2 2 2 2 2
= Z 21 = (̇2 ̇2 − 1 (2 − 21 ) (2 − 21 ) − 2 1 1 + 2 ) 1
Integrate the first term by parts, Z 2 ( − ) 1
= ̇2 2 |21 Z 2 + (−̈2 2 − 1 (2 − 21 ) (2 − 21 ) − 2 1 1 + 2 ) 1
The variations evaluated at 1 and 2 are zeros (2 (1 ) = 2 (2 ) = 0) and the left hand side evaluates to Z 2 ( − ) 1 Z 2 = ((−̈2 − 1 (2 − 21 ) + ) 2 + (21 (2 − 21 ) − 2 1 ) 1 ) 1
The coefficients of 1 and 2 inside of the integral must be zeros at all times or −̈2 − 1 (2 − 21 ) + = 0 21 (2 − 21 ) − 2 1 = 0 The coordinate 1 may be eliminated, and the equation of motion becomes ̈2 +
1 2 2 = 2 + 41
289
12. Derive the equation of motion for the system in Figure 5.24 for generalized coordinate using (a) Lagrange’s equation, and (b) Hamilton’s principle.
Solution: The kinetic and potential energies are given by 1 2 ̇ 2 1 1 = 1 ( + )2 + 2 ( − )2 − 2 2 =
There are two independent coordinates, and (a) Lagrange’s equations Lagrange’s equations are − + = 0 ̇ + = 0 − ̇ The equations of motion are given by ̈ + 1 ( + ) + 2 ( − ) − = 0 1 ( + ) − 2 ( − ) = 0 The equation of motion can be written in terms of only by using the second equation to solve for Then, 1 − 2 1 =− 1 + 2 µ ¶ µ ¶ 1 − 2 1 − 2 ̈ + 1 1 − + 2 1 + = 1 + 2 1 + 2 The equation of motion is simplified to ̈ +
41 2 = 1 + 2
(b) Hamilton’s Principle Hamilton’s principle states that 0 = =
Z 2
( − )
Z 12 µ 1
¶ 1 2 1 1 2 2 ̇ − 1 ( + ) − 2 ( − ) + 2 2 2
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CHAPTER 5 ANALYTICAL DYNAMICS
Take the variation, Z 2 0 = (̇ ̇ − 1 ( + ) ( + ) − 2 ( − ) ( − ) + ) 1 Z 2 = (̇ ̇ − [1 ( + ) + 2 ( − ) + ] − [1 ( + ) + 2 ( − )] ) 1
Integrate the first term by parts and use (1 ) = (2 ) = 0 and we obtain Z 2 0= (− [̈ + 1 ( + ) + 2 ( − ) − ] − [1 ( + ) + 2 ( − )] ) 1
Setting the coefficients of and to zero, we obtain ̈ + 1 ( + ) + 2 ( − ) − = 0 1 ( + ) + 2 ( − ) = 0 These are identical to the ones we obtained using Lagrange’s equations. The equation of motion is given by 41 2 = ̈ + 1 + 2
291
13. Derive the equation of motion for the system in Figure 5.25 for generalized coordinate using (a) Lagrange’s equation, and (b) Hamilton’s principle. Solution: The kinetic and potential energies are given by 1 2 ̇ 2 1 1 = 2 + (2) ( sin )2 − sin 2 2 2 =
where of a rod is 2 3 (a) Lagrange’s equations Lagrange’s equation is + = 0 − ̇ ̈ + + 22 sin cos − cos = 0 2 (b) Hamilton’s Principle Hamilton’s principle states that Z 2 ( − ) 0 = 1 ¶ Z 2 µ 1 1 2 1 2 2 = ̇ − − (2) ( sin ) + sin 2 2 2 2 1 Take the variation of the integrand, and we obtain ¶ Z 2 µ ¡ 2¢ ̇ ̇ − − 2 sin cos + sin cos 0= 2 1 where we used the chain rule,
(sin ) = cos ¢ ¡ sin2 = 2 sin (sin ) = 2 sin cos
The first term of the integrand is integrated by parts, and we let (1 ) = (2 ) = 0 Then, ¶ Z 2 µ ¡ 2¢ 0= − ̈ − − 2 sin cos + sin cos 2 1 The coefficient of must be zero or
¡ ¢ − ̈ − − 22 sin cos + sin cos = 0 2
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CHAPTER 5 ANALYTICAL DYNAMICS
14. Derive the equation of motion for the system in Figure 5.26 for generalized coordinate using (a) Lagrange’s equation, and (b) Hamilton’s principle.
Solution: The kinetic and potential energies are given by 1 1 1 2 2 ̇ + ̇2 = ̇ 2 2 2 1 2 = ( + ) − sin 2 =
where is at the contact point and = + 2 Here, we used two coordinates and Since the disk rolls without slip and the two coordinates are related by = Then, the kinetic and potential energies in terms of are 1 1 1 2 2 ̇ ̇ + ̇2 = 2 2 2 2 1 = (2)2 − sin 2 =
(a) Lagrange’s equations Lagrange’s equation is − + = 0 ̇ ̈ + 4 − sin = 0 2 The equation of motion in terms of is ̈ + 42 − sin = 0 (b) Hamilton’s Principle Hamilton’s principle states that 0 = =
Z 2
( − )
Z 12 µ 1
¶ 1 2 1 2 ̇ − (2) + sin 2 2 2
Take the variation of the integrand, and we find ¶ Z 2 µ 0= ̇ ̇ − 4 + sin 2 1
293 Integrate the first term by parts and evaluate (1 ) = (2 ) = 0 Then ¶ Z 2 µ − 2 ̈ − 4 + sin 0= 1 The coefficient of must be zero so that the integral is zero for any choice for The equation of motion is given by − 2 ̈ − 4 + sin = 0
294
CHAPTER 5 ANALYTICAL DYNAMICS
15. Derive the equation of motion for the system in Figure 5.27 for generalized coordinate using (a) Lagrange’s equation, and (b) Hamilton’s principle.
Solution: The kinetic and potential energies are given by 1 1 2 ̇ + ̇2 2 2 1 2 = − ( + ) 2 =
where = + 2 , and the coordinates are related by = Let us use as our generalized coordinate. Then, the kinetic and potential energies are given by ¶ µ 1 + ̇2 = 2 2 1 = 2 − (2) 2 (a) The Lagrange’s equation is given by − + = 0 ̇ The each term is µ ¶ = + ̈ ̇ 2 = 0 = − 2 Then, the equation of motion is given by ¶ µ + ̈ + = 2 2 (b) Hamilton’s Principle Hamilton’s principle states that Z 2 ( − ) 0 = 1 ¶ ¶ Z 2 µ µ 1 1 2 2 = + ̇ − + (2) 2 2 2 1
295 Take the variation of the integrand, and we obtain Z 2 0 = ( − ) 1 ¶ ¶ Z 2 µµ = + ̇ ̇ − + 2) 2 1 Integrate the first term by parts and use (2 ) = (1 ) = 0 Hamilton’s equation becomes ¶ ¶ Z 2 µµ 0= + ̈ − + 2 2 1 The equation of motion is given by µ
¶ + ̈ − + 2 = 0 2
296
CHAPTER 5 ANALYTICAL DYNAMICS
16. Derive the equations of motion for the system in Figure 5.28 for generalized coordinates 1 and 2 using (a) Lagrange’s equations, and (b) Hamilton’s principle. The kinetic and potential energies are =
1 1 2 1 2 2 ̇1 + ̇2 + 2 |2 {z } |2 {z 2 } 1
2
1 1 = ( sin 1 )2 + ( sin 1 + sin 2 )2 2 2 µ ¶ − sin 1 − sin 1 + sin 2 2 2 where = + (2)2 and 2 is the velocity of the center of mass of the second rod. Note 2 that we are now allowed to use 12 ̇2 for the second rod because there is no such point in the 2
second rod that has zero velocity. The kinetic energy of the form, 12 ̇ can only be used if point exists on the body that has zero velocity. The velocity 2 is given by r2 µµ ¶ µ ¶ ¶ = cos 1 + cos 2 i+ sin 1 + sin 2 j 2 2 ¶ µ ¶ µ = − ̇1 sin 1 + ̇2 sin 2 i+ ̇1 cos 1 + ̇2 cos 2 j 2 2
v2 =
The velocity squared is then õ
¶2 ! µ 1 2 2 = 2 + ̇1 cos 1 + ̇2 cos 2 2 µ ¶ 1 2 2 = 2 ̇1 + ̇2 + ̇1 ̇2 cos (2 − 1 ) 4 1 ̇1 sin 1 + ̇2 sin 2 2
¶2
The kinetic energy is given by µ ¶ 1 1 1 1 2 2 2 2 2 = ̇1 + ̇2 + ̇1 + ̇2 + ̇1 ̇2 cos (2 − 1 ) 2 2 2 4 =
(a) Lagrange’s equation
1 1 2 (sin 1 )2 + 2 (sin 1 + sin 2 )2 2 2 µ ¶ 1 − sin 1 − sin 1 + sin 2 2 2
297 There are two independent coordinates, and therefore, there are two Lagrange’s equations. Lagrange’s equations are given by − + = 0 ̇1 1 1 − + = 0 ̇2 2 2 Each term in Lagrange’s equation is listed next: ³ ´ 1 1 = ̈1 + 2 ̈1 + 2 ̈2 cos (2 − 1 ) − 2 ̇2 ̇2 − ̇1 sin (2 − 1 ) ̇1 2 2 1 = 2 ̇1 ̇2 sin (2 − 1 ) 1 2 = 2 sin 1 cos 1 + 2 (sin 1 + sin 2 ) cos 1 − cos 1 − cos 1 1 2 ³ ´ 1 1 1 = ̈2 + 2 ̈2 + 2 ̈1 cos (2 − 1 ) − 2 ̇1 ̇2 − ̇1 sin (2 − 1 ) ̇2 4 2 2 1 = − 2 ̇1 ̇2 sin (2 − 1 ) 2 2 = 2 (sin 1 + sin 2 ) cos 2 − cos 2 2 2 The equations of motion are 1 1 2 ̈1 + 2 ̈1 + 2 ̈2 cos (2 − 1 ) − 2 ̇2 sin (2 − 1 ) + 2 2 3 2 sin 1 cos 1 + 2 (sin 1 + sin 2 ) cos 1 − cos 1 = 0 2 1 1 1 2 ̈2 + 2 ̈2 + 2 ̈1 cos (2 − 1 ) + 2 ̇1 sin (2 − 1 ) + 4 2 2 2 (sin 1 + sin 2 ) cos 2 − cos 2 = 0 2 (b) Hamilton’s Principle Hamilton’s principle states that Z 2 0 = ( − ) 1 µ ¶ Z 2 ½ 1 1 1 1 2 2 2 2 2 = ̇1 + ̇2 + ̇1 + ̇2 + ̇1 ̇2 cos (2 − 1 ) 2 2 2 4 1 1 1 − 2 (sin 1 )2 − 2 (sin 1 + sin 2 )2 2 2 µ ¶¾ 1 + sin 1 + sin 1 + sin 2 2 2
298
CHAPTER 5 ANALYTICAL DYNAMICS
Take the variation µ ¶ Z 2 n 1 1 2 0 = ̇1 ̇1 + ̇2 ̇2 + ̇1 ̇1 + ̇2 ̇2 + ̇2 cos (2 − 1 ) ̇1 4 2 1 µ ¶ 1 1 2 + ̇1 cos (2 − 1 ) ̇2 − ̇1 ̇2 sin (2 − 1 ) (2 − 1 ) 2 2 2 2 − sin 1 cos 1 1 − (sin 1 + sin 2 ) (cos 1 1 + cos 2 2 ) ¶¾ µ 1 + cos 1 1 + cos 1 1 + cos 2 2 2 2 Integrate the terms with ̇1 and ̇2 by parts to obtain ½ Z 2 n 1 1 2 − ̈1 1 − ̈2 2 − ̈1 1 + ̈2 2 + ̈2 cos (2 − 1 ) 1 0 = 4 2 1 ¾ ´ ³ 1 − ̇2 sin (2 − 1 ) ̇2 − ̇1 1 2 ½ ´ ³ 1 1 2 + − ̈1 cos (2 − 1 ) 2 + ̇1 sin (2 − 1 ) ̇2 − ̇1 2 2 2 ¾ 1 − ̇1 ̇2 sin (2 − 1 ) (2 − 1 ) 2 −2 sin 1 cos 1 1 − 2 (sin 1 + sin 2 ) (cos 1 1 + cos 2 2 ) ¶¾ µ 1 + cos 1 1 + cos 1 1 + cos 2 2 2 2 Collect terms 0 =
Z 2
1 1 {− ̈1 − 2 ̈1 − 2 ̈2 cos (2 − 1 ) + 2 ̇2 sin (2 − 1 ) ̇2 2 2 1 −2 sin 1 cos 1 − 2 (sin 1 + sin 2 ) cos 1 + cos 1 + cos 1 }1 2 1 +{− ̈2 − 2 ̈2 − 2 ̈1 cos (2 − 1 ) − 2 ̇1 sin (2 − 1 ) ̇1 4 1 −2 (sin 1 + sin 2 ) cos 2 + cos 2 }2 2
The two equations of motion are the coefficients of 1 and 2 and they are 1 1 2 0 = − ̈1 − 2 ̈1 − 2 ̈2 cos (2 − 1 ) + 2 ̇2 sin (2 − 1 ) 2 2 3 −22 sin 1 cos 1 − 2 sin 2 cos 1 + cos 1 2 1 1 1 2 0 = − ̈2 − 2 ̈2 − 2 ̈1 cos (2 − 1 ) − 2 ̇1 sin (2 − 1 ) 4 2 2 1 −2 (sin 1 + sin 2 ) cos 2 + cos 2 2
299
17. Consider again the system of Figure 2.52, reproduced in Figure 5.29. A disk of mass and mass moment of inertia is supported by a massless spring and a rope. Derive the equation of motion of the system using (a) Lagrange’s equation and (b) Hamilton’s principle, for generalized coordinate
Figure 5.29: A disk suspended by an inextensible cable and a spring.
Solution: This system has one independent coordinates. The coordinates are related by = The kinetic and potential energies are 1 2 1 2 ̇ + ̇ 2 2 1 = ( + )2 − 2 2 where for a disk is 2 Using the kinematic constraint equation, the kinetic and potential energies are rewritten as ¶ µ 1 + 2 ̇2 = 2 1 = (2)2 − 2 (a) Lagrange’s equation Lagrange’s equation is given by − + = 0 ̇ The equation of motion is given by µ ¶ + 2 ̈ + 4 − = 0 =
300
CHAPTER 5 ANALYTICAL DYNAMICS
(b) Hamilton’s Principle Hamilton’s principle states that Z 2 ( − ) 0 = 1 ¶ ¾ Z 2 ½ µ 1 1 2 2 = + 2 ̇ − (2) + 2 2 1 Take the variation of the integrand, and we obtain Z 2 ( − ) 1 ¶ ¾ Z 2 ½µ = + 2 ̇ ̇ − 4 + 1 Integrate the first term by parts Z 2 ( − ) 1 ¯2 Z 2 ½ µ ¶ ¶ ¾ µ ¯ ¯ − + 2 ̈ − 4 + = + 2 ̇¯ + 1 1
The variation of is zero at 1 and 2 ( (1 ) = (2 ) = 0) and we are left with Z 2 ( − ) 1 ¶ ¾ Z 2 ½ µ = − + 2 ̈ − 4 + 1
For the integration to be zero for any choice of (that satisfies (1 ) = (2 ) = 0), the term inside of curly braces must be zero or ¶ µ − + 2 ̈ − 4 + = 0
301
18. Consider again the system of Figure 2.53, reproduced in Figure 5.30. A mass is suspended by a pulley system with a spring. Derive the equation of motion of the system using (a) Lagrange’s equation and (b) Hamilton’s principle, for generalized coordinate
Figure 5.30
Solution: The kinetic and potential energies are 1 2 ̇ 2 1 = (2)2 − 2 =
where the stretch in the spring is 2 (a) Lagrange’s equation Lagrange’s equation is given by − + = 0 ̇ The equation of motion is given by ̈ + 4 − = 0 (b) Hamilton’s Principle Hamilton’s principle states that 0 = =
Z 2
( − )
Z 12 ½ 1
¾ 1 2 1 2 ̇ − (2) + 2 2
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CHAPTER 5 ANALYTICAL DYNAMICS
Take the variation of the integrand, and we obtain Z 2 ( − ) 1 Z 2 = {̇ ̇ − 4 + } 1
Integrate the first term by parts Z 2 ( − ) 1 Z 2 2 = ̇|1 + {−̈ − 4 + } 1
The variation of is zero at 1 and 2 ( (1 ) = (2 ) = 0) and we are left with Z 2 ( − ) 1 Z 2 = {−̈ − 4 + } 1
For the integration to be zero for any choice of (that satisfies (1 ) = (2 ) = 0), the term inside of curly braces must be zero or −̈ − 4 + = 0
303
19. Consider again the system of Figure 2.54, reproduced in Figure 5.31. Derive the equation of motion of the system using (a) Lagrange’s equation and (b) Hamilton’s principle, for generalized coordinate 1
Figure 5.31
Solution: The kinetic and potential energies are 1 1 1 ̇21 + 2 ̇22 2 2 1 2 = (2 ) − 1 1 + 2 2 2 =
where the coordinates 1 and 2 are related by 21 = 2 The kinetic and potential energies are rewritten as 1 1 1 1 ̇21 + 42 ̇21 = (1 + 42 ) ̇21 2 2 2 1 1 = (21 )2 − 1 1 + 22 1 = 421 + (−1 + 22 )1 2 2 =
(a) Lagrange’s equation Lagrange’s equation is given by − + = 0 ̇1 1 1 The equation of motion is given by (1 + 42 ) ̈1 + 41 − 1 + 22 = 0
304
CHAPTER 5 ANALYTICAL DYNAMICS
(b) Hamilton’s Principle Hamilton’s principle states that Z 2 ( − ) 0 = 1 ¾ Z 2 ½ 1 1 2 2 = (1 + 42 ) ̇1 − 41 + (1 − 22 )1 2 2 1 Take the variation of the integrand, and we obtain Z 2 ( − ) 1 Z 2 = {(1 + 42 ) ̇1 ̇1 − 41 1 + (1 − 22 ) 1 } 1
Integrate the first term by parts Z 2 ( − ) 1
= (1 + 42 ) ̇1 1 |21 Z 2 + {− (1 + 42 ) ̈1 1 − 41 1 + (1 − 22 ) 1 } 1
The variation of is zero at 1 and 2 ( (1 ) = (2 ) = 0) and we are left with Z 2 ( − ) 1 Z 2 = {− (1 + 42 ) ̈1 − 41 + 1 − 22 } 1 1
For the integration to be zero for any choice of (that satisfies (1 ) = (2 ) = 0), the term inside of curly braces must be zero or − (1 + 42 ) ̈1 − 41 + 1 − 22 = 0
305
20. Consider again the system of Figure 2.57, reproduced in Figure 5.32. Derive the equation of motion using (a) Lagrange’s equation, and (b) Hamilton’s principle, for generalized coordinate
Figure 5.32
Solution: The kinetic and potential energies are 1 2 ̇ 2 1 ³ ´2 = − 2 2 =
where the stretch in the spring is 2 (a) Lagrange’s equation Lagrange’s equation is given by − + = 0 ̇ The equation of motion is given by ̈ + − = 0 4 (b) Hamilton’s Principle Hamilton’s principle states that 0 = =
Z 2
( − )
Z 12 ½ 1
¾ 1 2 11 2 ̇ − + 2 24
306
CHAPTER 5 ANALYTICAL DYNAMICS
Take the variation of the integrand, and we obtain Z 2 ( − ) 1 ¾ Z 2 ½ 1 = ̇ ̇ − + 4 1 Integrate the first term by parts Z 2 ( − ) 1 ¾ Z 2 ½ 1 2 = ̇|1 + −̈ − + 4 1 The variation of is zero at 1 and 2 ( (1 ) = (2 ) = 0) and we are left with Z 2 ( − ) 1 ¾ Z 2 ½ 1 = −̈ − + 4 1 For the integration to be zero for any choice of (that satisfies (1 ) = (2 ) = 0), the term inside of curly braces must be zero or 1 −̈ − + = 0 4
307
21. For the simple pendulum of Figure 5.33, derive the governing equation of motion assuming (a) that is a point mass, and (b) that the mass is a sphere with small but finite mass moment of inertia. Given = 10 cm and = 5 kg, how significant is the effect of the mass moment of inertia if = 05 cm? Derive the governing equation of motion using (i) Newton’s second law, (ii) Lagrange’s equation, and (iii) Hamilton’s principle.
Figure 5.33: Simple pendulum.
Solution: (a,i) Newton’s second law of motion - Point Mass - To use Newton’s second law of motion requires a free-body diagram of the mass. The two forces in the FBD are the weight acting down and the string tension acting through Using Newton’s second law in moment form about point yields:
− sin = ̈ or 2 ̈ + sin = 0 since = 25 2 + 2 ≈ 2 Therefore, we see for inclusion of the dimension of the mass, we need to keep the term 25 2 in the equation of motion. See the next part for a comparison. (a,ii) Lagrange’s equation: Point Mass - The use of Lagrange’s equation requires the formulation of the kinetic and potential (gravitational and strain) energies. We need to define a datum from where the potential energy is set to zero. We choose the datum here to be at the low point of the pendulum swing, that is, the static equilibrium. Therefore, the gravitational potential and the kinetic energies are:
= (1 − cos ) 1 = (̇)2 2 (The expression for assumes that the mass has no dimension and therefore does not rotate about its own center of mass. Had we chosen the point as our datum, = − cos ) Then,
308
CHAPTER 5 ANALYTICAL DYNAMICS
= − = 12 (̇)2 − (1 − cos ) Lagrange’s equation is µ ¶ − = 0 ̇
2 ̈ + sin = 0 ̈ + sin = 0 (a,iii) Hamilton’s principle - Point Mass - Hamilton’s principle states that the variation of the integral of the Lagrangian equals zero: Z 2
( − ) = 0 ¶ Z 2 µ 1 2 (̇) − (1 − cos ) = 0 2 1 Z 2 ³ ´ 2 ̇ ̇ − sin = 0
1
1
The integral of the term ̇̇ is done by parts, resulting in ̇|21 − ̈ = −̈ since the variation is zero at the beginning and end of the time interval. Therefore, Z 2 ³ ´ −2 ̈ − sin = 0 1
and 2 ̈ + sin = 0
Divide by 2 and we obtain
sin = 0 (b,i) Newton’s second law of motion - Finite Inertia We assume that the connection to the bob is rigid so that the system is described by one-degreeof-freedom. In order to have a rigid connection, we must have a massless rod instead of a massless string so that moment and shear as well as tension can be transmitted. Consider the first free-body diagram shown below. ̈ +
309 Taking the moment about we have ª
X
= ̈
− sin = ̈ where = + The rod is massless, and therefore, = 0 The mass moment of inertia of the bob is obtained using parallel axis theorem, = + 2 = 22 5 + 2 It is also interesting to derive the equation of motion by breaking up the rod and the bob as shown in the second free-body diagram. Taking the moment about for the rod, X ª = ̈ = 0 ( − ) + = 0
(1)
For the bob, summing the moment about the center of mass and summing the forces in the transverse direction, Newton’s second law gives us X ª = ̈ + − = ̈ X =
− − sin = ̈
(2)
(3)
Adding Equations (1) and (2), we obtain 1 = ̈ and substituting the shear into Equation (3), we obtain 1 − ̈ − sin = ̈ ¢ ¡ 2 + ̈ + sin = 0
where 2 + = the mass moment of inertia of the bob about . For a spherical bob, = 22 5 The purpose of the second derivation is to show that the member that connects the bob with finite dimensions to must be able to transmit shear force and moment so that the system can be modeled with one degree of freedom. For instance, if a bob with finite dimensions is connected to by a string, this system must be modeled with minimum two degrees of freedom similar to a double pendulum. (b,ii) Lagrange’s equation: Finite Inertia - Including the effect of the mass geometry changes the kinetic energy because the mass (sphere) has a moment of inertia that has an effect when it rotates. The kinetic energy is given by 1 2 ̇ 2µ ¶ 1 2 2 2 2 = + ̇ 2 5
=
310
CHAPTER 5 ANALYTICAL DYNAMICS 2
Then the Lagrangian becomes: = − = 12 ̇ − (1 − cos ) The Lagrange’s equation becomes ̈ + sin = 0 sin = 0 ̈ + Substituting the numerical values, the mass moment of inertia about is given by ¶ µ 2 2 2 2 + = 5(0005)2 + 5(01)2 = 005005 kg-m2 = 5 5 and the EOM is given by 98 sin = ̈ + 98 sin = 0 01 5 × 98 × 01 sin = ̈ + 97 902 sin = 0 ̈ + 005005 ̈ +
where the second equation takes the inertia of the mass about its own center of mass We see then for these parameter values there is not a significant effect due to the rotational kinetic energy of the mass (98 versus 97902) (b,iii) Hamilton’s principle - Finite Inertia - In this case, the kinetic energy is given by 1 2 = ̇ 2 and the remaining calculations are the same. See part (a,ii) for the comparison.
311
22. The lightweight bar in Figure 5.34 is released from rest when the spring is undeformed at = 0. The two masses at and slide in frictionless guides that are horizontal at the left and vertical at the right. Formulate the equation of motion using (a) Newton’s second law, (b) Lagrange’s equation, and (c) Hamilton’s principle. Assume the bar is massless. Show that the equation of motion is ¡ ¢ 2 0 = (2 + 42 cos2 + 42 sin2 )̈ + 2 2 − 2 ̇ sin 2 1 2 +(5 + 4) cos + sin 2 2
Figure 5.34: Oscillation of constrained bar.
Solution: (a) Newton’s second law of motion - The center of mass is obtained using P3 ¯ = P=1 3 =1
Measuring from along the bar, we have ¯ = ( + 4 ( + )) 6 From kinematics, ¶ ´ − 4 ³ − 4 2 cos → ̈ = ̈ sin + ̇ cos = − 6 6 ´ 5 + 4 5 + 4 ³ 2 = sin → ̈ = ̈ cos − ̇ sin 6 6 µ
The free-body diagrams and the intermediate coordinates are shown in the figures next. When = 0 the horizontal location of mass is at = 0
312
CHAPTER 5 ANALYTICAL DYNAMICS
We have three equations to work with: the force balance equations in the and directions, and the moment balance equation about the center of mass. Summing the forces in the and directions, ´ − 4 ³ 2 ̈ sin + ̇ cos 6 ´ X 5 + 4 ³ 2 ̈ cos − ̇ sin ↓ = ̈ → −1 + 6 − ( sin ) = 6 µ ¶ X 5 + 4 − 4 cos + ( sin ) cos + 2 sin = ̈ © = ̈ → −1 6 6
−→
where
X
= ̈ → 2 =
µ
5 + 4 = 6
¶2
µ
− 4 + 6
¶2
µ
+ 2 + 4 6
¶2
Eliminating 1 and 2 from the equation, we are left with ¡ ¢ 2 (2 + 42 cos2 + 42 sin2 )̈ + 2 2 − 2 ̇ sin 2 1 +(5 + 4) cos + 2 sin 2 = 0 2 (b) Lagrange’s equation, Define the following coordinates: the position of mass that travels horizontally is from the left end; the position of mass that travels vertically is 1 from the top; the position of mass 4 is 2 from the top. Since there is one generalized coordinate we can relate all of these to and the geometry of the system: = (1 − cos ) 1 = sin 2 = + cos 2 = ( + ) sin
̇ = ̇ sin ̇1 = ̇ cos ̇2 = −̇ sin ̇2 = ( + )̇ cos
313 Now we can proceed to evaluate the kinetic and potential energies needed for the application of Lagrange’s equation: ¢ 1 ¡ 2 ̇ + ̇12 + 4̇22 + 4̇22 2 µ ¶ ³ ´2 ³ ´2 ³ ´2 1 2 2 2 = ̇ sin + ̇ cos + 4 ( + )̇ cos + 4 ̇ sin 2 ´ 1 ³ 2 2 2 2 = ̇ + 4( + )2 ̇ cos2 + 42 ̇ sin2 2 1 = −1 − 42 + 12 2 1 = −1 − 4(1 + sin ) + 12 2 1 = −5 sin − 4 sin + ( sin )2 2 =
These expressions can be re-written as the Lagrangian = − and then substituted into the respective Lagrange’s equation, as follows: ´ 1 ³ 2 2 2 2 2 2 2 2 = ̇ + 4( + ) ̇ cos + 4 ̇ sin 2 1 −5 sin − 4 sin − ( sin )2 ; 2 µ ¶ − = 0 ̇ = 2 ̇ + 4( + )2 ̇ cos2 + 42 ̇ sin2 ̇ µ ¶ = 2 ̈ + 4( + )2 ̈ cos2 + 42 ̈ sin2 ̇ 2
2
−8( + )2 ̇ cos sin + 82 ̇ sin cos
2 2 = −4( + )2 ̇ cos sin + 42 ̇ sin cos −5 cos − 4 cos − 2 sin cos The EOM becomes: 2
2
2 ̈ + 4( + )2 ̈ cos2 + 42 ̈ sin2 − 8( + )2 ̇ cos sin + +82 ̇ sin cos 2
2
+4( + )2 ̇ cos sin − +42 ̇ sin cos +5 cos + 4 cos + 2 sin cos = 0
or simplified to ¡ ¢ 2 (2 + 4( + )2 cos2 + 42 sin2 )̈ + 4 2 − 2 ̇ cos sin + (5 + 4) cos +2 sin cos = 0
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CHAPTER 5 ANALYTICAL DYNAMICS
and finally: ¡ ¢ 2 0 = (2 + 42 cos2 + 42 sin2 )̈ − 2 2 − 2 ̇ sin 2 1 +(5 + 4) cos + 2 sin 2 2 (c) Hamilton’s principle - Hamilton’s principle states that : Z 2 ( − ) = 0 1
Substitute the expressions for and from above: Z 2 Z 2 ∙ ´ 1 ³ 2 2 2 2 ̇ + 42 ̇ cos2 + 42 ̇ sin2 ( − ) = 2 1 1 ¸ 1 2 − (5 + 4) sin − ( sin ) 2 Take the variation to obtain
Z 2
( − ) 1 Z 2 h ³ ´ = 2 ̇ + 42 ̇ cos2 + 42 ̇ sin2 ̇ 1 ³ ´ 2 2 + −42 ̇ cos sin + 42 ̇ cos sin ¸ 1 2 − (5 + 4) cos − sin cos 2 Collect terms to obtain Z 2 ( − ) 1 Z 2 h ³ ´ = 2 ̇ + 42 ̇ cos2 + 42 ̇ sin2 ̇ 1 µ ³ ¶ ¸ ´ 1 2 2 2 2 2 + −4 ̇ cos sin + 4 ̇ cos sin − (5 + 4) cos − sin cos 2 Integrate the first integral by parts and find Z 2 ( − ) 1
´ i¯2 h ³ ¯ = 2 ̇ + 42 ̇ cos2 + 42 ̇ sin2 ¯ 1 Z 2 ∙ ³ ´ 2 2 2 2 2 − ̇ + 4 ̇ cos + 4 ̇ sin + 1 µ ¶¸ 1 2 2 2 2 2 − 4 ̇ cos sin − 4 ̇ cos sin + (5 + 4) cos + sin cos 2
315 where the terms evaluated at the two end times are zeros. The equation of motion is the expression multiplied by or ³ ´ 2 2 0 = 2 ̈ + 42 ̈ cos2 − 82 ̇ cos sin + 42 ̈ sin2 + 82 ̇ sin cos 1 2 2 +42 ̇ cos sin − 42 ̇ cos sin + (5 + 4) cos + 2 sin cos 2
The expression simplifies to ¡ ¢ 2 (2 + 42 cos2 + 42 sin2 )̈ + 2 2 − 2 ̇ sin 2 1 +(5 + 4) cos + 2 sin 2 = 0 2 which is the same expression obtained using Lagrange’s equation.
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CHAPTER 5 ANALYTICAL DYNAMICS
23. The spring supporting the rod-sphere system in Figure 5.35 is undeformed when the rod is horizontal. Assume the rod mass is negligible. The roller from which the spring is suspended permits the spring to maintain a vertical configuration. If the system is in this position when it is released from rest, derive the equation of motion using (a) the principle of virtual work along with d’Alembert’s principle, and (b) using Lagrange’s equation. Show that the equation of motion is µ ¶ 2 2 2 ( + ) + ̈ + 2 sin cos − ( + ) cos = 0 5
Figure 5.35: Oscillation of rod-sphere system.
Solution: (a) The principle of virtual work with d’Alembert’s principle - The external forces are gravity on the sphere and the spring on the rod. We can formulate the principle of virtual work in angular form and consider moments about (positive angle for increasing ):
³X
´
− ̈ = 0 µ µ ¶ ¶ 2 2 2 = − sin × cos + ( + ) cos − ( + ) + ̈ = 0 5 ¶ µ 2 =⇒ ( + )2 + 2 ̈ + 2 sin cos − ( + ) cos = 0 5 =
where sin cos = 12 sin 2 (b) using Lagrange’s equation - We need the kinetic and potential energies, and they are given by ¶ µ 1 2 2 2 1 2 2 + ̇ = ̇ = 2 2 5 1 = ( sin )2 − ( + ) sin 2
317 Lagrange’s equation is given by ¶ − = 0 ̇ ̈ + 2 sin cos − ( + ) cos = 0
µ
where = − µ ¶ = ̇ = ̈ ̇ ̇ = −2 sin cos + ( + ) cos Therefore, the Lagrange’s equation results in the same equation as in part (a).
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CHAPTER 5 ANALYTICAL DYNAMICS
24. Verify Equation 5.20, ¶ ¸µ ¶ ∙ µ 1 r − = ṙ · ṙ r̈ · ̇ 2
Solution: Equation 5.20 can be verified by expanding it and showing that the previous equation is a result: ∙ µ ¶ ¸µ ¶ ¶µ ¶ µ ¶ µ 1 1 1 − ṙ · ṙ = ṙ · ṙ − ṙ · ṙ ̇ 2 ̇ 2 2 ∙ µ ¶ ¸ (ṙ · ṙ ) (ṙ · ṙ ) 1 − = 2 ̇ µ ∙ ¸ ¶ ṙ ṙ 1 − 2ṙ 2ṙ = 2 ̇ Using the relation given by ṙ r = ̇ we obtain the desired relation: ∙ µ ¶ ¸µ ¶ ¶ µ 1 r ṙ r − − ṙ · = r̈ · ṙ · ṙ = ṙ · ̇ 2
319
25. For Example 5.10, use Newton’s second law of motion to derive the equations of motion.
Simple pendulum pivoted to a free block.
Solution: Let us draw free-body diagrams, and write the force balance equations based on them.
The force balance equations for the block is X
X
= 1 ̈ =⇒ sin = 1 ̈
(1)
= 0 =⇒ − 1 − cos = 0
where the second equation merely relates and The force balance for the bob is X
X
= 2 ̈2 =⇒ − sin = 2 ̈2
(2)
= 2 ̈2 =⇒ 2 − cos = 2 ̈2
(3)
The dynamic equations are first, third, and fourth equations. The system has two degrees-offreedom, and here we use and We need kinematic relations to express ̈2 and ̈2 in terms of
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CHAPTER 5 ANALYTICAL DYNAMICS
and and their derivatives. They are related by 2 = + sin
=⇒
2 = cos
=⇒
³ ´ 2 ̈2 = ̈ + ̈ cos − ̇ sin ³ ´ 2 ̈2 = −̈ sin − ̇ cos
(4) (5)
Substituting in the kinematic relations (Equations (4) and (5)) into the force balance equations (Equations 1,2, and 3), we have sin = 1 ̈ n ³ ´o 2 − sin = 2 ̈ + ̈ cos − ̇ sin ³ ´ 2 2 − cos = 2 −̈ sin − ̇ cos
(6) (7) (8)
The next step is to eliminate and from Equations (6)-(8). Adding Equations (6) and (7), we obtain ³ ´ 2 (2 + 1 ) ̈ + 2 ̈ cos − ̇ sin = 0
Multiplying Equation (7) by cos and multiplying (8) by sin and adding the two, we obtain 2 ̈ cos + 2 ̈ + 2 sin = 0
321
26. Derive Equation 5.28, assuming a linear model with = constant and = constant, []{̈} + []{} = {0}
Solution: The kinetic and potential energies are given by: 1 XX = ̇ ̇ 2 =1 =1
1 XX ≈ 2 =1 =1
Substituting these into Lagrange’s equation results in µ ¶ + = 0 ̇ " #! " # Ã XX XX + ̇ ̇ = 0 ̇ =1 =1 =1 =1 For example, for = 1 and = 2 (211 ̇1 + 12 ̇2 + 21 ̇2 ) + 211 1 + 12 2 + 21 2 = 0 and due to symmetry, 12 = 21 such that (211 ̇1 + 212 ̇2 ) + 211 1 + 212 2 = 0 211 ̈1 + 212 ̈2 + 211 1 + 212 2 = 0 This last equation can be seen to be the product for the first row of a matrix by a vector: ⎧ ⎫ ⎧ ⎫ i ⎨ ̈1 ⎬ h i ⎨ 1 ⎬ n o h + 11 12 = 0 11 12 ⎩ ̈ ⎭ ⎩ ⎭ 2
2
This can be generalized for all and any resulting in Equation 5.28.
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CHAPTER 5 ANALYTICAL DYNAMICS
27. Formulate the problem of Example 5.13 where the string is elastic with stiffness and can stretch. Derive the equations of motion. Solution: If the string is elastic, the system now has 2 DOFs. Let 1 be the position of 1 and 2 be the position of 2 as shown below.
Two-body single degree-of-freedom system shown with the virtual displacements 1 and 2 and the resisting friction forces F1 and F2 Then, the kinetic and the potential energies are given by =
1 1 1 ̇12 + 2 ̇22 2 2
1 = −1 1 sin 1 − 2 2 sin 2 + ( − 1 − 2 )2 2 where is the undeformed length of the elastic cord. Lagrange’s equations are ¶ µ − + = 0 ̇1 1 1 ¶ µ − + = 0 ̇2 2 2 where
and
¶ µ = 1 ̈1 = 1 ̇1 , ̇1 ̇1 = 0 = −1 sin 1 − ( − 1 − 2 ) 1 1 ¶ µ = 2 ̈2 = 2 ̇2 , ̇2 ̇2 = 0 = −2 sin 2 − ( − 1 − 2 ) 2 2
323 Then, the Lagrange’s equations reduce to 1 ̈1 + (1 + 2 − ) = 1 sin 1 2 ̈2 + (1 + 2 − ) = 2 sin 2
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CHAPTER 5 ANALYTICAL DYNAMICS
28. For the elastic pendulum discussed in Example 5.14, and shown in Figure 5.19, derive the governing equations of motion using Lagrange’s equation.
Figure 5.19: A spring pendulum.
Solution: This system has two generalized coordinates: and To use Lagrange’s equation, find the kinetic and potential energies: ¢ 1 ¡ 2 + 2 2 µ ³ ´2 ¶ 1 2 = ̇ + ̇ 2 1 = − cos + ( − 0 )2 2 =
where 0 is the undeformed length; when = 0 the spring is in equilibrium. The Lagrangian is = − The two equations of motion are then found as follows: µ ¶ − = 0 ̇ ³ 2 ´ ̇ + sin = 0 2 ̈ + 2̇̇ + sin = 0 and
µ
̇
¶
−
= 0
2 (̇) − ̇ − cos + ( − 0 ) = 0 2 ̈ − ̇ − cos + ( − 0 ) = 0
325 These results are the same as those obtained using Hamilton’s principle, as shown next. Hamilton’s principle, for a conservative system, says Z 2 = 0 1
Substitute the Lagrangian into the equation and take the variation to obtain ¸ µ Z 2 ∙ Z 2 ³ ´2 ¶ 1 1 2 2 + cos − ( − 0 ) = ̇ + ̇ 2 2 1 1 Z 2 h ³ i ´ 2 = ̇ ̇ + ̇ + 2 ̇ ̇ − sin + cos − ( − 0 ) 1
The terms with ̇ and ̇ are integrated by parts so that Z 2 Z 2 2 ̇ ̇ = ̇|1 − ̈ 1 1 Z 2 = − ̈ 1 Z 2 ¯2 Z 2 ³ ´ ¯ 2 2 2 ̇ ̇ = ̇¯ − ̇ 1 1 1 Z 2 ³ ´ = − 2 ̈ + 2̇̇ 1
where the variations evaluated at the two end points are zeros. Then, ¸ Z 2 ∙ Z 2 ³ 2 ´ 2 −̈ + ̇ − ̇ − sin − ( − 0 ) = 1 1 Z 2 nh ´ i o i h ³ 2 = −̈ + ̇ + cos − ( − 0 ) + − 2 ̈ + 2̇̇ − sin 1
R The coefficients of and must be zeros so that 12 is zero for any arbitrary variation in and Then, 2
̈ − ̇ − cos + ( − 0 ) = 0 2 ̈ + 2̇̇ + sin = 0 which are identical to the EOMs obtained using Lagrange’s equations.
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CHAPTER 5 ANALYTICAL DYNAMICS
29. In the pendulum shown in Figure 5.36 the length is variable. Derive the equations of motion. Discuss the possibility and implications if = ̇.
Figure 5.36: Variable length pendulum.
Solution: We derive the EOM using Lagrange’s equation. This is a single degree-of-freedom system in terms of Note that () is a given function. Since the pendulum string is inextensible, let be its total length and therefore = + or = − The position of the mass is = () sin () and = − () + () cos () The velocities are = ̇ sin + ̇ cos = −̇ + ̇ cos − ̇ sin =
=
The velocity squared is 2 = 2 + 2 2
= 2 ̇ + 2̇2 − 2̇2 cos + 2̇̇ sin where = − = − and ̇ = − cos The kinetic and potential energies as well as the Lagrangian are given by ´ 1 2 1 ³ 2 2 = ̇ + 2̇2 − 2̇2 cos + 2̇̇ sin 2 2 = − ( + cos ) = − ´ 1 ³ 2 = 2 ̇ + 2̇2 − 2̇2 cos + 2̇̇ sin + ( + cos ) 2 =
327 The Lagrange’s equation is given by µ ¶ − = 0 ̇ µ ¶ − + = 0 ̇ where ´ ³ = 2 ̇ + ̇ sin ̇ µ ¶ ³ ´ = 2 ̈ + ̇̇ + ̇2 sin + ̈ sin + ̇̇ cos ̇ ³ ´ = ̇2 sin + ̇̇ cos = sin The Lagrange’s equation is ´ ³ ´ ³ 2 2 2 ̈ + ̇̇ + ̇ sin + ̈ sin + ̇̇ cos − ̇ sin + ̇̇ cos + sin = 0
Simplifying and dividing above equation by 2 we have µ ¶ ̈ ̇ + sin = 0 ̈ + ̇ +
where = − sin and ̇ = − cos If = ̇ we cannot simply replace with ̇ in the equation of motion. We need to make the substitution the Lagrangian. The Lagrangian is same as before except that = − sin ̇ and ³ in ´ ³ ´ ̇ = − ̈ + ̇ cos ̇ The derivatives such as ̇ and ̇ will significantly change.
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CHAPTER 5 ANALYTICAL DYNAMICS
30. (a) For the elastically supported mass on a vertically moving base shown in Figure 5.37 derive the governing equation of motion first using d’Alembert’s principle, and then using Lagrange’s equation. Discuss the effect of base motion and show that it is equivalent to a force on mass equal to sin . (b) Repeat the problem where viscous damping exists due to the motion of the mass through the surrounding medium. Assume a damping constant denoted by .
Figure 5.37: Elastically supported pendulum on a moving base.
Solution: First let us derive the EOM using Newton’s second law. The stretch in the spring is sum of the static deflection and dynamic deflection, + ( − ) Take the sum of the forces in the vertical direction to obtain ↓
X
= ̈
−( − + ) + = ̈ ̈ + = = sin where the static equilibrium equation, = , is used. Let us use d’Alembert’s principle next, in conjunction with the principle of virtual work: P = ( − ̈) = 0 (−( − + ) + − ̈) = 0 =⇒ ̈ + = Finally let us use Lagrange’s equation to formulate this problem. The kinetic energy, potential energies, and Lagrangian are given by 1 2 ̇ 2 1 = ( − + )2 − 2 =
329 Substitute the Lagrangian into Lagrange’s equation to obtain the equation of motion, µ ¶ − + = 0 ̇ ̈ + ( − + ) − = 0 ̈ + = where the static equilibrium equation is used to simplify. If damping is included there is a dissipation term that must be added. The resulting equation of motion is ̈ + ̇ + = + ̇ = sin + cos
330
CHAPTER 5 ANALYTICAL DYNAMICS
31. Consider the motion of a pendulum that is supported by springs that are elastically restrained to horizontal motion, as depicted in Figure 5.38. Assume that the springs are massless and remain horizontal, that is small, and is a constant. Formulate the equations of motion using (a) Newton’s second law, (b) Lagrange’s equation, and (c) Hamilton’s principle. Show that the period is given by s + 2 = 2 2
Figure 5.38: Pendulum supported by horizontal springs.
Solution: (a) Newton’s Law: Let the point at which the pendulum is connected to the springs be the origin, with positive direction i to the right and positive direction j upwards. Begin with Lagrange’s equation. The position of the mass is given by r = ( + sin )i + (− cos )j The velocity and acceleration vectors are then,
³ ´ ³ ´ ̇ + ̇ cos i + ̇ sin j ³ ´ ³ ´ 2 2 a = ̈ + ̈ cos − ̇ sin i + ̈ sin + ̇ cos j
v =
The free body diagram is shown next.
331
Taking the moment about the pivot point we have X k = ̈k + ×
− sin = 0 + ( sin − cos ) n ³ ´ ³ ´ o 2 2 × ̈ + ̈ cos − ̇ sin i + ̈ sin + ̇ cos j ³ ´ ³ ´ 2 2 = sin ̈ sin + ̇ cos + cos ̈ + ̈ cos − ̇ sin = 2 ̈ + ̈ cos
Summing the forces in the direction, we have X = ³ ´ 2 −2 = ̈ + ̈ cos − ̇ sin The equations of motion are
´ ³ 2 ̈ + ̈ cos + sin = 0 ³ ´ 2 ̈ + ̈ cos − ̇ sin + 2 = 0 Just for demonstration purposes, let us use the moment balance equation, free-body that includes the rod, the moment balance equation is given by X ª = ̈ = 0
(1) (2) P
2 ( cos ) − sin = 0
where of the point mass is zero. The force balance in the direction gives X = ³ ´ 2 − = ̈ sin + ̇ cos
= ̈ For the
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CHAPTER 5 ANALYTICAL DYNAMICS
Solve for and substitute it into the moment equation. The moment equation is given by ³ ³ ´ ´ 2 2 ( cos ) − ̈ sin + ̇ cos + sin = 0 2
2 ̈ sin2 + 2 ̇ cos + sin − 2 cos = 0
(3)
The last equation is one of the equation of motion although it does not resemble Equation (1). This equation can be solved with Equation (2) to find the response. It is possible to make this equation look like Equation (1) by solving for 2 using Equation (2) and replace it in Equation (3). We find ³ ´ 2 −2 = ̈ + ̈ cos − ̇ sin
and Equation (3) becomes
³ ³ ´´ 2 2 2 ̈ sin2 + 2 ̇ cos + sin + cos ̈ + ̈ cos − ̇ sin
This simplifies to
2 ̈ + ̈ cos + sin = 0 which is identical to Equation (2). (b) Lagrange’s Equation: The kinetic and potential energies are
µ³ ´2 ¶ ´2 ³ 1 = ̇ + ̇ cos + ̇ sin 2 ´ 1 ³ 2 2 2 = ̇ + ̇ + 2̇̇ cos 2 1 = 2 2 − cos 2 = −
Therefore,
µ
¶ − = 0 ̇ µ ¶ ´´ ³ ³ = ̇ + ̇ cos ̇ ³ ´ 2 = ̈ + ̈ cos − ̇ sin
= −2; µ ¶ − = 0 ̇ µ ¶ ´´ ³ ³ 2 = ̇ + ̇ cos ̇ ³ ´ = 2 ̈ + ̈ cos − ̇̇ sin = −̇̇ sin − sin
333 The EOM are: ³ ´ 2 ̈ + ̈ cos − ̇ sin + 2 = 0 ³ ´ 2 ̈ + ̈ cos + sin = 0
(c) Equation of motion using Hamilton’s Principle:
Z 2 1
Z 2 µ
¶ ´ ¡ ¢ 1 ³ 2 2 2 2 ̇ + ̇ + 2̇̇ cos − − cos 2
( − ) = Z 21 ³ = ̇ ̇ + 2 ̇ ̇ + ̇ cos ̇ + ̇ cos ̇ 1 ´ −̇̇ sin − 2 − sin = 0
At this point we need to integrate by parts those terms that are variations of derivatives. Then, since the variations at theR beginning and R ending times are taken to be zero, we are left only with the integrals of the form () and () . Each of the terms in the parenthesis is now an EOM, as derived above. Period For small angles the following approximations holds: cos ≈ 1 sin ≈ The linearized EOMs are given by ³ ´ ̈ + ̈ + 2 = 0 ³ ´ 2 ̈ + ̈ + = 0 The period of the pendulum can be found as follows. The horizontal displacement of the pendulum is = + The two linearized EOMs are rewritten as ´ ³ ̈ + ̈ + = 0 2³ ´ ̈ + ̈ + = 0 Let us add the two equation to find µ
+ 2
The natural frequency is =
¶
s
̈ + = 0
2 + 2
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CHAPTER 5 ANALYTICAL DYNAMICS
and the period is = 2
s
+ 2 2
There is a better method that can reliably find the natural frequency and the period without relying on being able to reduce the two differential equations into one. We use the trial solution method. Assume () = and () = Θ where () and () share the same frequency, Substitute these into the linearized equations of motion to find ¡ ¢ − 2 ( + Θ) + 2 = 0 ¢ ¡ − 2 (Θ + ) + Θ = 0 from which, we have two algebraic equations, ¢ ¡ 2 − 2 − 2 Θ = 0 ¡ ¢ − 2 + − 2 Θ = 0
In order to avoid trivial solutions, the two equations must be linearly independent. That is, (2 − 2 ) −2 = − 2 ( − 2 )
Cross multiply and we have (2 + ) 2 − 2 = 0 s 2 = 2 +
335
32. For a pendulum supported from a horizontally restrained mass, as shown in Figure 5.39, derive the governing equations of motion using (a) Lagrange’s equation and (b) Hamilton’s principle. Identify the constraint equations. Discuss the effect of setting 1 = 0. Finally, simplify the governing equations if small motions are assumed.
Figure 5.39: Pendulum suspended from a horizontally restrained mass.
Solution: (a) Lagrange’s Equation: This is a two-degree-of-freedom system. Let us use 1 and The constraint equations are the following: 2 = 1 + sin 2 = cos The velocities of 2 are ̇2 = ̇1 + ̇ cos ̇2 = −̇ sin The kinetic energy is then ¡ ¢ 1 1 1 ̇21 + 2 ̇22 + ̇22 2 2 ³ ´ 1 1 2 = 1 ̇21 + 2 ̇21 + 2̇1 ̇ cos + 2 ̇ 2 2 1 1 2 = (1 + 2 )̇21 + 2 ̇1 ̇ cos + 2 2 ̇ 2 2
=
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CHAPTER 5 ANALYTICAL DYNAMICS
The change in gravitational potential energy is for mass 2 only, and the strain energy is due to the extension and compression of the two springs. This total potential equals 1 1 = 21 + 21 + 2 cos 2 2 Recall that the Lagrangian = − The virtual work done by the nonconservative force, () is = () i · r2 where the virtual displacement of 2 is r2 = 2 i + 2 j Then, = () 2 We decided to use 1 and as our generalized coordinates, and 2 is 2 = 1 + cos The virtual work done by () is = 1 + cos The generalized forces are 1 = and = cos These generalized forces will go on the right-hand sides of the respective EOMs generated by the Lagrange’s equations: ¶ µ − = () ̇1 1 µ ¶ − = () cos ̇ Performing the necessary differentiations, we find the two EOMs: 2
(1 + 2 )̈1 + 21 − 2 ̇ sin + 2 ̈ cos = () 2 ̈1 cos + 2 2 ̈ + 2 sin = () cos If 1 = 0 it will drop out of the inertia term of the first equation, and the equations of motion become identical to the previous problem with massless wheel. If we assume small angle motion, then sin ≈ cos ≈ 1 In general, if is small, it does not necessarily follow that its derivative 2 ̇ is also small. We make additional assumptions that the nonlinear terms such as ̇ are also small compared to the linear terms.
337
33. Fill in the missing steps in the derivation of the equations of motion for Example 5.8, that is, Equations 5.15 and 5.16.
A two-linked pendulum.
Solution: Let us start with the force and moment balances for the two links shown in the text. The equations are based on the free-body diagrams shown below. The external torque 2 may be applied to either joint . As long as there is a net torque of 2 when the two joints are combined.
The position vectors to the center of mass are r1 = 1 cos 1 i + 1 sin 1 j r2 = [2 cos + 1 cos 1 ]i +[2 sin + 1 sin 1 ]j The velocity vectors are v1 = −1 ̇1 sin 1 i + 1 ̇1 cos 1 j
v2 = [−2 ̇ sin − 1 ̇1 sin 1 ]i + [2 ̇ cos + 1 ̇1 cos 1 ]j
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CHAPTER 5 ANALYTICAL DYNAMICS
The accelerations of the center of mass are ³ ´ ³ ´ 2 2 a1 = −1 ̇1 cos 1 − 1 ̈1 sin 1 i+ −1 ̇1 sin 1 + 1 ̈1 sin 1 j ´ ³ 2 2 a2 = −2 ̇ cos − 2 ̈ sin − 1 ̇1 cos 1 − 1 ̈1 sin 1 i ³ ´ 2 2 + −2 ̇ sin + 2 ̈ cos − 1 ̇1 sin 1 + 1 ̈1 cos 1 j
where = 1 + 2 We can write three equations, two force balance and one moment balance equations, for each freebody. They are Link 1: X = 1 1 ³ ´ 2 (1) =⇒ + = 1 −1 ̇1 cos 1 − 1 ̈1 sin 1 X = 1 1 ³ ´ 2 =⇒ + − 1 = 1 −1 ̇1 sin 1 + 1 ̈1 sin 1 (2) X ¡ ¢ = 1 + 1 12 ̈1 ¡ ¢ =⇒ (1 cos 1 ) − (1 sin 1 ) + 1 − 1 1 cos 1 = 1 + 1 12 ̈1 (3) Link 2:
X
= 2 2
X
= 2 2
X
³ ´ 2 2 =⇒ − = 2 −2 ̇ cos − 2 ̈ sin − 1 ̇1 cos 1 − 1 ̈1 sin 1
(4)
³ ´ 2 2 =⇒ − − 2 = 2 −2 ̇ sin + 2 ̈ cos − 1 ̇1 sin 1 + 1 ̈1 cos 1
(5)
=⇒ (2 cos ) − (2 sin ) + 2 = 2 ̈
(6)
2 = 2 ̈
The unknowns in Equations (1)-(6) are 1 () and () The six equations can be algebraically manipulated to eliminate the reaction forces and to obtain two differential equations in terms of 1 () and () In order to simplify the math, we take the moment about instead of 2 for the second link. The equation is given by X ¡ ¢ = 2 ̈ + r2 × 2 a2 · k 2 − 2 2 cos = 2 ̈ + 2 2 cos 2 − 2 2 sin 2 ´ ³ 2 2 = 2 ̈ + 2 2 cos −2 ̇ sin + 2 ̈ cos − 1 ̇1 sin 1 + 1 ̈1 cos 1 ´ ³ 2 2 − 2 2 sin −2 ̇ cos − 2 ̈ sin − 1 ̇1 cos 1 − 1 ̈1 sin 1 ¡ ¢ 2 = 2 + 2 22 ̈ + 2 2 1 cos ( − 1 ) ̈1 + 2 2 1 ̇1 sin ( − 1 ) (EOM2)
339 This equation is identical to the first equation of motion given in the text after = 2 − 1 substitution. In order to obtain the second equation of motion, we solve for the reactions, and , using Equations (4) and (5), and they are given by ³ 2 ´ 2 = 2 2 ̇ cos + 2 ̈ sin + 1 ̇1 cos 1 + 1 ̈1 sin 1 ³ 2 ´ 2 = 2 2 ̇ sin − 2 ̈ cos + 1 ̇1 sin 1 − 1 ̈1 cos 1 − Substitute and into Equation (3) to obtain ´ ³ 2 2 2 2 ̇ sin − 2 ̈ cos + 1 ̇1 sin 1 − 1 ̈1 cos 1 − (1 cos 1 ) ³ 2 ´ 2 −2 2 ̇ cos + 2 ̈ sin + 1 ̇1 cos 1 + 1 ̈1 sin 1 (1 sin 1 ) ¡ ¢ +1 − 1 1 cos 1 = 1 + 1 12 ̈1
Simplify this equation to find
¡ ¢ 2 1 + 1 12 + 2 12 ̈1 + 2 1 2 cos ( − 1 ) ̈ − 2 1 2 ̇ sin 2 +2 1 cos 1 + 1 1 cos 1 = 1
(EOM1)
Let us use Lagrange’s equations to verify the equations of motion. The kinetic energy is given by ¢ 2 1 1¡ 1 2 1 + 1 12 ̇1 + 2 ̇ + 2 v2 · v2 2 2 2 ¢ 1¡ 1 2 2 1 + 1 12 ̇1 + 2 ̇ = 2 2 ½³ ´2 ³ ´2 ¾ 1 −2 ̇ sin − 1 ̇1 sin 1 + 2 ̇ cos + 1 ̇1 cos 1 + 2 2 ³ ´ ¢ 2 1 1 1¡ 2 2 2 1 + 1 12 ̇1 + 2 ̇ + 2 22 ̇ + 12 ̇1 + 22 ̇1 ̇1 cos ( − 1 ) = 2 2 2 ¡ ¢ ¢ 2 1¡ 1 2 = 1 + 1 12 + 2 12 ̇1 + 2 + 2 22 ̇ + 2 2 1 ̇̇1 cos ( − 1 ) 2 2
=
The potential energy is = 1 1 sin 1 + 2 (1 sin 1 + 2 sin ) Lagrange’s equation for 1 is − + = 1 ̇1 1 1
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CHAPTER 5 ANALYTICAL DYNAMICS
where ¡ ¢ = 1 + 1 12 + 2 12 ̇1 + 2 2 1 ̇ cos ( − 1 ) ̇1 ´ ³ ¡ ¢ = 1 + 1 12 + 2 12 ̈1 + 2 2 1 ̈ cos ( − 1 ) − 2 2 1 ̇ ̇ − ̇1 sin ( − 1 ) ̇1 = 2 2 1 ̇̇1 sin ( − 1 ) 1 = (1 1 + 2 1 ) cos 1 1 = 1 1 The equation of motion is ¡ ¢ 2 1 + 1 12 + 2 12 ̈1 + 2 2 1 ̈ cos ( − 1 ) − 2 2 1 ̇ sin ( − 1 ) + (1 1 + 2 1 ) cos 1 = 1
(EOM1)
Lagrange’s equation for is + = − ̇ 1 1 2 2 where (1 + 1 12 + 2 12 ) ̇1 + (2 + 2 22 ) ̇ + 2 2 1 ̇̇1 cos ( − 1 ) 2 2 ¡ ¢ = 2 + 2 22 ̇1 + 2 2 1 ̇1 cos ( − 1 ) ̇ ³ ´ ¢ ¡ = 2 + 2 22 ̈ + 2 2 1 ̈1 cos ( − 1 ) − 2 2 1 ̇1 ̇ − ̇1 sin ( − 1 ) ̇ = −2 2 1 ̇̇1 sin ( − 1 ) = 2 cos = 2 The second equation of motion is ¡ ¢ 2 2 + 2 22 ̈ + 2 2 1 ̈1 cos ( − 1 ) + 2 2 1 ̇1 sin ( − 1 ) +2 cos = 2
(EOM2)
In order to compare them with the equations of motion in the text book, replace with 1 + 2 to rewrite the equations of motion as ¡ ¢ 1 + 1 12 + 2 12 + 2 2 1 cos 2 ̈1 + 2 2 1 cos 2 ̈2 ³ ´2 −2 2 1 ̇1 + ̇2 sin 2 + (1 1 + 2 1 ) cos 1 = 1
341
¡ ¡ ¢ ¢ 2 + 2 22 + 2 2 1 cos 2 ̈1 + 2 + 2 22 ̈2 2
+2 2 1 sin 2 ̇1 + 2 cos (1 + 2 ) = 2
The latter equation is the second equation of motion. The sum of the first and the second equation is identical to the first equation of motion.
342
CHAPTER 5 ANALYTICAL DYNAMICS
34. Use Lagrange’s equation to derive the equations of motion governing the displacements 1 and 2 for the two degree-of-freedom system depicted in Figure 5.40.
Figure 5.40: Undamped two degree-of-freedom system.
Solution: The kinetic energy is given by 1 1 = 1 ̇21 + 2 ̇22 2 2 and the potential energy by 1 1 1 = 1 21 + 2 (2 − 1 )2 + 3 22 2 2 2 The Lagrangian is given by = − and Lagrange’s equations are − = 0 ̇1 1 − = 0 ̇2 2 Each term in the Lagrange’s equation is written below: = 1 ̇1 ̇1 = 2 ̇2 ̇2
= 1 ̈1 ̇1 = 2 ̈2 ̇2
= −1 1 + 2 (2 − 1 ) 1 = −2 (2 − 1 ) − 3 2 2
Then, Lagrange’s equations become 1 ̈1 + (1 + 2 ) 1 − 2 2 = 0 2 ̈2 − 2 1 + (2 + 3 ) 2 = 0
343
35. Use Lagrange’s equation to derive the equations governing the rotations 1 2 and 3 for the spring-connected triple pendulum system shown in Figure 5.41.
Figure 5.41: Triple pendulum coupled by springs.
Solution: The kinetic energy is given by 1 1 1 2 2 2 ̇1 + ̇2 + ̇3 2 2 2 ´ 1 2³ 2 2 2 = ̇1 + ̇2 + ̇3 2
=
where is the mass moment of inertia about the hinge or = 2 assuming that the mass has no dimensions (point mass assumption). The potential energy is given by =
1 1 ( sin 2 − sin 1 )2 + ( sin 3 − sin 2 )2 2 2 − (cos 1 + cos 2 + cos 3 )
The Lagrangian is = − Lagrange’s equations are − =0 ̇1 ̇1 =⇒ 2 ̈1 − 2 (sin 2 − sin 1 ) cos 1 + sin 1 = 0 − =0 ̇2 ̇2 =⇒ 2 ̈2 + 2 (sin 2 − sin 1 ) cos 2 − 2 (sin 3 − sin 2 ) cos 2 + sin 2 = 0 − =0 ̇3 ̇3 =⇒ 2 ̈3 + 2 (sin 3 − sin 2 ) cos 3 + sin 3 = 0
344
CHAPTER 5 ANALYTICAL DYNAMICS
For small angles, the equations become ¡ ¢ 2 ̈1 + + 2 1 − 2 2 = 0 ¡ ¢ 2 ̈2 + + 22 2 − 2 1 − 2 3 = 0 ¡ ¢ 2 ̈3 + + 2 3 − 2 2 = 0
345
36. A pendulum is suspended from a torsionally restrained disk, as in Figure 5.42, where is the angle of rotation of the disk, and is the relative angle of the pendulum measured from the disk. Derive the governing equation of motion using (a) Lagrange’s equation and (b) Hamilton’s principle. Compare the equation of motion to the equation of the simple pendulum.
Figure 5.42: Pendulum supported from a torsionally restrained disk.
Solution: (a) Lagrange’s Equation: We assume that the pendulum mass is at a point, relieving us from considering its mass moment of inertia about its center of mass. We develop the kinetic and potential energies needed for Lagrange’s equation: i2 1 1 h = (̇ + ̇) + ̇2 2 2 1 = − cos + 2 2 Therefore, the Lagrangian = − and its appropriate derivatives are given by: i2 1 1 1 h = (̇ + ̇) + ̇2 + cos − 2 2 2µ ¶ 2 = 2 (̈ + ̈) = 2 (̇ + ̇) ̇ ̇ = − sin µ ¶ 2 = (̇ + ̇) + ̇ = 2 (̈ + ̈) + ̈ ̇ ̇ = − From these we can formulate the Lagrange’s equation in each generalized coordinate: 2 ̈ + 2 ̈ + sin = 0 2 ̈ + ( + 2 )̈ + = 0
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CHAPTER 5 ANALYTICAL DYNAMICS
If we constrain the disk by setting and its derivatives to zero, the first equation becomes that of the pendulum: ̈ + sin = 0 R (b) Hamilton’s Principle: This principle says that the variation of Lagrangian must be zero, 12 = 0 ¸ Z 2 ∙ Z 2 i2 1 1 h 1 2 2 (̇ + ̇) + ̇ + cos − = 2 2 2 1 1 Z 2 h ³ ´ i 2 = (̇ + ̇) ̇ + ̇ + ̇ ̇ − sin − 1 Z 2 h i = 2 (̇ + ̇) ̇ + (2 ̇ + 2 ̇ + ̇) ̇ − sin − 1
Integrate the first two terms by parts to obtain Z 2 ³ ´ 2 2 2 2 = (̇ + ̇)|1 + ̇ + ̇ + ̇ |21 1 Z 2 nh i − 2 (̈ + ̈) + sin 1 o 2 2 +( ̈ + ̈ + ̈ + )
The variations evaluated at 1 and 2 are zeros, and the remaining integral must evaluate to zero for any variation in and Then, the equations of motion are 2 (̈ + ̈) + sin = 0 2 ̈ + 2 ̈ + ̈ + = 0
347
37. A rigid beam acts as a compound pendulum, suspended from an elastically restrained block that can undergo horizontal motion, as drawn in Figure 5.43. Derive the equations of motion using Lagrange’s equation. Consider the following cases: (a) no friction between the block and surface, and (b) viscous damping between the block and surface.
Figure 5.43: Compound pendulum pivoted to a block.
Solution: (a) Frictionless Case: We need to derive the kinetic and potential/strain energies. For the pendulum () we formulate the kinetic energy in the rectangular coordinate system: 1 2 − cos 2 2 1 1 2 2 2 + ) = (2) + ( 2 2" µ ¶2 µ ¶2 # 1 1 2 ̇ sin = (2)̇ + ̇ + ̇ cos + 2 2 2 2 Ã ! 2 2 ̇ 1 = 3̇2 + + ̇̇ cos 2 4
=
where we have used
µ ¶ µ ¶ + sin and = − cos = 2 2 ³ ´ 2 2 Thus, we have the Lagrangian = − = 12 3̇2 + 4̇ + ̇̇ cos − 12 2 + 2 cos from which we can derive the EOM using Lagrange’s equation for each generalized coordinate: µ ¶ − = 0 ̇ ´ ³ 3̇ + ̇ cos + = 0 2 ´ 1 ³ 2 3̈ + ̈ cos − ̇ sin + = 0 2
348
CHAPTER 5 ANALYTICAL DYNAMICS
and
µ
¶
= 0 Ã ! ´ 2 ̇ ³ + ̇ cos − −̇̇ sin − sin = 0 2 2 2 µ 2 ¶ ´ ³ ̈ + ̈ cos − ̇̇ sin − −̇̇ sin − sin = 0 2 2 2 2 ̈ + ̈ cos + sin = 0 2 ̇
−
(b) With Friction - Since friction only affects the motion of the block directly, only this equation is altered. Of course, since the motion of the block drives the motion of the pendulum, the friction affects this motion indirectly. The virtual work done by friction is = − (2) where the term multiplying is the generalized force associated with the virtual displacement = − (2) Lagrange’s equations including the work done by non-conservative forces are µ ¶ − = ̇ µ ¶ − = ̇ In our case, the left-hand sides remain the same. The right-hand side of the first Lagrange equation is changed to − (2) ´ 1 ³ 2 3̈ + ̈ cos − ̇ sin + = − 2 2
349
38. For the inverted simple pendulum of length and mass 2 supported on a cart of mass 1 as shown in Figure 5.44, derive the equations of motion if the cart is forced by ()
Figure 5.44: Inverted pendulum on a moving base.
Solution: Let us use Lagrange’s equations. The kinetic energy of the system is 1 1 = 1 ̇2 + 2 22 2 2 ³ ´2 ³ ´2 2 where 2 = ( + sin ) ̂+ ( cos ) ̂ and 22 = ̇ + ̇ cos + ̇ sin = ̇2 +2̇̇ cos +2 ̇ The kinetic energy is then ³ ´ 1 1 2 = 1 ̇2 + 2 ̇2 + 2̇̇ cos + 2 ̇ 2 2 ³ ´ 1 1 2 = (1 + 2 ) ̇2 + 2 2̇̇ cos + 2 ̇ 2 2 The potential energy is = 2 cos Lagrange’s equations are
µ ¶ − = ̇ µ ¶ − = ̇ where the virtual work done by () is = so that = () and = 0 Then, Lagrange’s equations are reduced to ³ ´ 2 (1 + 2 ) ̈ + 2 ̈ cos − ̇ sin = 2 2 ̈ + 2 ̈ cos − 2 sin = 0
350
CHAPTER 5 ANALYTICAL DYNAMICS
39. The system in Figure 5.45 oscillates about the equilibrium position denoted by the horizontal lines. The coil torsional springs are each undeformed when 1 = 2 = 0. If the link masses are neglected, derive the equations of motion to be (1 + 2 )12 ̈1 + 2 1 2 ̈2 cos (2 − 1 ) − 2 1 2 ̇2 ̇2 sin (2 − 1 ) + (1 + 2 ) 1 − 2 2 − (1 + 2 )1 cos 1 = 0 2
2 22 ̈2 + 2 1 2 ̈1 cos (2 − 1 ) + 2 1 2 ̇1 sin (2 − 1 ) +2 (2 − 1 ) − 2 2 cos 2 = 0
Figure 5.45: Torsionally restrained beams.
Solution: This is a 2 DOF system. We will use 1 and 2 Superimpose an − coordinate system at hinge with positive upwards and positive to the right. Let (1 1 ) be the coordinates of point and (2 2 ) be the coordinate of point In the following, we express the coordinates of points and in terms of 1 and 2 1 = 1 cos 1 1 = −1 sin 1 ´2 ³ ̇21 + ̇12 = 1 ̇1
̇1 = −1 ̇1 sin 1 ̇1 = −1 ̇1 cos 1
2 = 1 cos 1 + 2 cos 2 ̇2 = −1 ̇1 sin 1 − 2 ̇2 sin 2 2 = −1 sin 1 − 2 sin 2 ̇2 = −1 ̇1 cos 1 − 2 ̇2 cos 2 ´2 ³ ´2 ³ ̇22 + ̇22 = 1 ̇1 + 2 ̇2 + 21 2 ̇1 ̇2 cos (2 − 1 )
From these relations we can write down the kinetic energy of the system in terms of 1 , 2 , and their derivatives: ¡ ¢ 1 ¡ ¢ 1 = 1 ̇21 + ̇12 + 2 ̇22 + ̇22 2 2 µ ¶ ³ ³ ´2 1 ´2 ³ ´2 1 = 1 1 ̇1 + 2 1 ̇1 + 2 ̇2 + 21 2 ̇1 ̇2 cos (2 − 1 ) 2 2
351 The potential and strain energies are given by the following: 1 1 1 21 = 2 (2 − 1 )2 2 2 = −1 1 sin 1 − 2 (1 sin 1 + 2 sin 2 )
=
Now we can write the expression of the Lagrangian = − where equals the total potential: µ³ ¶ ³ ´2 1 ´2 ³ ´2 1 1 ̇1 + 2 ̇2 + 21 2 ̇1 ̇2 cos (2 − 1 ) = 1 1 ̇1 + 2 2 2 1 1 − 1 21 − 2 (2 − 1 )2 + 1 1 sin 1 + 2 (1 sin 1 + 2 sin 2 ) 2 2 We can use the Lagrangian to derive the two EOM for this system. Lagrange’s equations for 1 is derived from ¶ µ − = 0 ̇1 1 ¶ µ ´ ³ = 1 12 ̇1 + 2 12 ̇1 + 2 1 2 ̇2 cos (2 − 1 ) ̇1 = (1 + 2 )12 ̈1 + 2 1 2 ̈2 cos (2 − 1 ) ³ ´ −2 1 2 ̇2 ̇2 − ̇1 sin (2 − 1 ) = 2 1 2 ̇1 ̇2 sin (2 − 1 ) − 1 1 + 2 (2 − 1 ) 1 +(1 + 2 )1 cos 1
which yields the first EOM. 0 = (1 + 2 )12 ̈1 + 2 1 2 ̈2 cos (2 − 1 ) − 2 1 2 ̇2 ̇2 sin (2 − 1 ) + 1 1 − 2 (2 − 1 ) − (1 + 2 )1 cos 1 Lagrange’s for 2 is ¶ µ − = 0 ̇2 2 ¶ µ ´ ³ 2 = 2 2 ̇2 + 2 1 2 ̇1 cos (2 − 1 ) ̇2
³ ´ = 2 22 ̈2 + 2 1 2 ̈1 cos (2 − 1 ) − 2 1 2 ̇1 ̇2 − ̇1 sin (2 − 1 )
= −2 1 2 ̇1 ̇2 sin (2 − 1 ) − 2 (2 − 1 ) + 2 2 cos 2 2 which yields the second EOM given by 2
0 = 2 22 ̈2 + 2 1 2 ̈1 cos (2 − 1 ) + 2 1 2 ̇1 sin (2 − 1 ) +2 (2 − 1 ) − 2 2 cos 2
352
CHAPTER 5 ANALYTICAL DYNAMICS
40. Figure 5.46 shows a pulley system that is acted on by an applied moment (), resulting in oscillatory motion. Use Lagrange’s equation to derive the equations of motion. Identify the generalized forces and .
Figure 5.46: A pulley system subjected to a moment.
Solution: To use Lagrange’s equation derive the following energies: 1 2 1 2 ̇ + ̇ 2 2 1 1 = 1 ()2 + 2 ( − )2 2 2 =
Since the oscillation of mass is about equilibrium, the weight can be left out as being balanced by the static force in the spring. The Lagrangian is given by = − In addition, there is dissipation that can be included using Rayleigh’s dissipation function. This is added to the Lagrangian for the equation in where = ̇2 2 We also know there is an applied moment (). The equations of motion are then: 1 2 1 1 1 = ̇ + ̇2 − 1 ()2 − 2 ( − )2 2 2µ ¶ 2 2 − = () ̇ µ ¶ ³ ´ = ̇ = ̈ ̇ = −1 2 + 2 ( − )
353 and µ ¶ − + =0 ̇ ̇ µ ¶ = (̇) = ̈ ̇ = −2 ( − ) = ̇ ̇ If we had not included Rayleigh’s dissipation function or the applied moment above, we could have included these as non conservative forces/moments as follows. The virtual work done by the applied moment and by the dissipation are given by: = = () = = −̇ Therefore, = () and = −̇ The two EOM are: ̈ + 1 2 − 2 ( − ) = () ̈ + 2 ( − ) = −̇ Had there been an external force acting on the mass in the direction, this would have appeared on the right-hand side of the equation in
354
CHAPTER 5 ANALYTICAL DYNAMICS
41. The system sketched in Figure 5.47 depicts two shafts coupled through meshing gears. Assume that gears 2 and 4 have gear ratio = 2 4 , and that external moment () is applied to gear 1. Identify the constraint equation and the generalized coordinates, and derive the governing equation(s) of motion. Discuss how the value of affects the resulting behavior.
Figure 5.47: A gear-shaft system.
Solution: The constraint equation is derived from the gear ratio: = 2 4 = 4 2 Therefore, there are 3 degrees of freedom, and then 3 associated equations of motion. The kinetic and strain energies are given: 1 2 1 2 1 2 1 2 1 ̇ + 2 ̇ + 3 ̇ + 4 ̇ 2 1 2 2 2 3 2 4 ¢ 2 1 2 1 2 1¡ 2 + 2 4 ̇2 + 3 ̇3 = 1 ̇1 + 2 2 2 1 1 2 = 1 (1 − 2 ) + 2 (4 − 3 )2 2 2 1 1 2 = 1 (1 − 2 ) + 2 (2 − 3 )2 2 2 =
The external work is non-conservative and is given by 1 = 1 1 = ()1 Therefore, () will appear on the right-hand side of Lagrange’s equation in 1 The necessary derivatives are: ¶ µ = 1 ̈1 = 1 (1 − 2 ) ̇1 1 ¶ µ = (2 + 2 4 )̈2 = −1 (1 − 2 ) + 2 (2 − 3 ) ̇2 2 µ ¶ = 3 ̈3 = −2 (2 − 3 ) ̇3 3
355 Substituting these into the respective Lagrange’s equation yields the three EOM: 1 ̈1 + 1 (1 − 2 ) = () (2 + 4 )̈2 + (1 + 2 )2 − 1 1 − 2 3 = 0 3 ̈3 + 2 3 − 2 2 = 0 2
2
Chapter 6 Multi Degree-of-Freedom Vibration Problems for Section 6.2 — The Concepts of Stiffness and Flexibility
1. The two degree-of-freedom system in Figure 6.64 undergoes rectilinear motion. (a) Derive the flexibility influence coefficients. (b) Derive the stiffness influence coefficients. (c) Find the inverse of the flexibility matrix and show that this equals the stiffness matrix. (d) Write the matrix equation of motion.
Figure 6.64: Two degree-of-freedom system.
Solution: (a) Flexibility influence coefficients The flexibility matrix for this system has four elements. Since the flexibility influence coefficient matrix is symmetric, we need only derive three coefficients. First, we apply a unit force to mass 1 The resulting free-body diagram is shown below. Note that the forces in the vertical direction are the gravity and the normal force from the ground. They cancel each other, and they are excluded from the FBD.
357
358
CHAPTER 6 MULTI DOF VIBRATION
Since 2 is unstretched, 1 and 2 displace as a rigid body, 1 = 1 1 1 = 1 1 ; 1 1 = 1 The second mass translates to the right by the same distance, and therefore we find 11 = 21 = 11 Next, we apply a unit force to mass 2 The resulting free-body diagram is shown below.
This is a static problem, and the forces on each mass must sum zero. Then, we have 1 1 = 2 (2 − 1 ) 2 (2 − 1 ) = 2 = 1 We find 12 = 1 = 11 and 22 = 2 = 11 + 12 The flexibility influence coefficient matrix is ⎡ [] =
1 ⎣ 1 1 1
³
1 1 1 + 12 1
⎤
´⎦
(b) Stiffness influence coefficients Now, we apply a unit displacement to each degree-of-freedom and observe the resulting force. We draw a free-body diagram results from positive 1 and 2
First, we apply a unit displacement to the 1 coordinate and let 2 = 0. 1 and 2 are the forces necessary to maintain static equilibria. The result is 1 = (1 + 2 ) ≡ 11 and 2 = −2 ≡ 21
359 Applying a unit displacement to 2 and setting 1 = 0 gives us 1 = −2 ≡ 12 and 2 = 2 ≡ 22 Thus, the matrix of stiffness influence coefficients is: ⎡ ⎤ 1 + 2 −2 ⎦ [] = ⎣ −2 2
(c) Find the inverse of the flexibility matrix and show that this equals the stiffness matrix To see that [ ] and [] are inverses, we take the matrix product [ ][] : ⎤ ⎡ ⎡ ⎤ (1 +2 ) 2 −2 +2 − 1 1 0 1 ⎦=⎣ ⎦ [ ][] = ⎣ 1 −2 +2 −2 +2 2 + 2 0 1 1 1
Since the matrix product [][] is the identity matrix, [] is the inverse of [] or, []−1 = [] (d) Write the matrix equation of motion. We already have the [] matrix from part (b) of this problem. The [] matrix can be written as ⎡ ⎤ 1 0 ⎦ [] = ⎣ 0 2 Thus, we may write the matrix equations of motion as ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎬ ⎨ 1 + 2 −2 ⎨ 1 ⎬ ̈1 0 ⎣ ⎦ ⎦ ⎣ 1 + = {0} ⎭ ⎩ ⎩ ⎭ 0 ̈ − 2
2
2
2
2
360
CHAPTER 6 MULTI DOF VIBRATION
2. For the triple pendulum of Figure 6.65, determine the flexibility influence coefficients that relate the horizontal forces and the horizontal displacements. Find the inverse of the matrix of flexibility coefficients and then write the matrix equation of motion for this system.
Figure 6.65: Triple pendulum.
Solution: We will find the flexibility matrix that relate the horizontal forces to the horizontal displacements such that ⎡ ⎤ ⎡ ⎤ ⎢ 1⎥ ⎢ 1⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 2 ⎥ = [ ] ⎢ 2 ⎥ ⎣ ⎦ ⎣ ⎦ 3 3 where the influence matrix is given by
⎡
⎤
⎢ 11 12 13 ⎥ ⎢ ⎥ ⎢ 21 22 23 ⎥ ⎣ ⎦ 31 32 33 This matrix is symmetric, so we only have to derive six influence coefficients. First Column of the Flexibility Influence Matrix To find the first column of the flexibility matrix, we consider the figure below where a unit force is applied to mass 1 shown in the next figure.
361
We can then write the static force equations for mass 1. Begin by applying a unit force in the horizontal direction resulting in an offset from the vertical of 1 Define the distance that the mass is displaced in the horizontal direction to be 11 = 1 sin 1 Note that since the masses below are also displaced the same amount as mass 1 we have 21 = 31 = 11 Write the equations of static equilibrium for 1 : 1 sin 1 = 1 1 cos 1 = (1 + 2 + 3 ) From this we have tan 1 = 1 [(1 + 2 + 3 ) ] Therefore, we now have the first column of the flexibility matrix, 11 = 21 = 31 = 1 sin 1 1 = q 2 2 1 + (1 + 2 + 3 ) We also make use of the known symmetry properties: 12 = 21 and 13 = 31 If is small, then sin ' tan and the expression for 11 simplifies to 11 ' 1 (1 + 2 + 3 ) Second Column of the Flexibility Influence Matrix
Next, apply a unit force in the horizontal direction to mass 2 resulting in an offset from the vertical of 1 for the mass 1 and 2 for mass 2 as shown in the next figure.
362
CHAPTER 6 MULTI DOF VIBRATION
The distance that the mass is displaced in the horizontal direction is 2 or 12 = 1 sin 1 22 = 32 = 1 sin 1 + 2 sin 2 Newton’s second law for mass 1 and mass 2 yields: For mass 1: X −→ = 2 sin 2 − 1 sin 1 = 0 X ↑ = 1 cos 1 − 2 cos 2 − 1 = 0
(1) (2)
(3) (4)
For mass 2:
−→ ↑
X
X
= 2 sin 2 − 1 = 0
(5)
= 2 cos 2 − (2 + 3 ) = 0
(6)
Eliminating 1 and 2 from above equations, we find 1 (1 + 2 + 3 ) 1 tan 2 = (2 + 3 ) tan 1 =
The second equation is obtained by dividing Equation (5) by Equation (6). The first equation is obtained by dividing Equation (3) by Equation (4) after 2 is replaced by the expressions given in Equations (5) and (6). Then, we find 1 sin 1 = q 1 + (1 + 2 + 3 )2 2 1 sin 2 = q 1 + (2 + 3 )2 2
363 Then, we rewrite Equations (1) and (2) as 1 12 = 1 sin 1 = q 1 + (1 + 2 + 3 )2 2
1 2 22 = 32 = 1 sin 1 + 2 sin 2 = q +q 1 + (1 + 2 + 3 )2 2 1 + (2 + 3 )2 2
Now, we have the second column of the flexibility matrix. As we have expected, 12 = 21 So far, we have ⎡ ⎤ 1 1 √ √ 13 1+(1 +2 +3 )2 2 ⎢ 1+2 (1 +2 +3 )2 ⎥ ⎢ ⎥ 1 1 2 ⎥ √ √ √ + [] = ⎢ 23 2 2 2 ⎢ 1+(1 +2 +3 ) 2 ⎥ 1+(1 +2 +3 ) 2 1+(2 +3 ) 2 ⎣ ⎦ 1 2 1 √ √ √ + 33 2 2 2 2 2 2 1+(1 +2 +3 )
1+(1 +2 +3 )
1+(2 +3 )
Third Column of the Flexibility Influence Matrix Using the symmetry property of the [ ] matrix, we know the two elements of the [] matrix in advance. We know that 1 13 = 31 = q 1 + (1 + 2 + 3 )2 2
1 2 23 = 32 = q +q 2 2 2 2 1 + (1 + 2 + 3 ) 1 + (2 + 3 )
To find 33 we apply a unit force on mass 3 as shown in the next figure.
Define the distance that the mass is displaced in the horizontal direction to be 33 , and it is equal to 33 = 1 sin 1 + 2 sin 2 + 3 sin 3
364
CHAPTER 6 MULTI DOF VIBRATION
Summing the forces in the and directions for three masses, we obtain: For mass 1: 1 sin 1 = 2 sin 2 1 cos 1 = 2 cos 2 + 1 For mass 2: 2 sin 2 = 3 sin3 2 cos 2 = 3 cos 3 + 2 For mass 3: 3 sin 3 = 1 3 cos 3 = 3 Eliminating the tensions (1 2 and 3 ) from six equations above, we obtain three equations given by 1 (1 + 2 + 3 ) 1 tan 2 = (2 + 3 ) 1 tan 3 = 3 tan 1 =
Then, 33 = 1 sin 1 + 2 sin 2 + 3 sin 3 1 2 3 = q +q +p 1 + 23 2 1 + (1 + 2 + 3 )2 2 1 + (2 + 3 )2 2
Now substitute all of these expressions into the flexibility matrix to obtain ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ [] = ⎢ + + ⎥ ⎣ ⎦ + ++ where
1 = q 1 + 2 (1 + 2 + 3 )2
2 = q 1 + (2 + 3 )2 2 3 = p 1 + 23 2
365 If the angles are small, they are approximately 1 (1 + 2 + 3 ) 2 = (2 + 3 ) 3 = 3 =
and the flexibility matrix is given by ⎤ ⎡ ⎤ ⎡ 1 1 1 (1 +2 +3 ) ⎥ ⎢ 11 12 13 ⎥ ⎢ (1 +2 +3 ) (1 +2 +3 ) ⎥ ⎢ ⎥ ⎢ 1 1 2 1 2 ⎥ ⎢ 21 22 23 ⎥ = ⎢ (1 +2 +3 ) (1 +2 +3 ) + (2 +3 ) (1 +2 +3 ) + (2 +3 ) ⎦ ⎣ ⎦ ⎣ 2 2 3 1 1 1 31 32 33 + + + (1 +2 +3 ) (1 +2 +3 ) (2 +3 ) (1 +2 +3 ) (2 +3 ) 3 The stiffness matrix is the inverse of the flexibility matrix,
[] = []−1 ⎡ ⎤ 1 1 ( + ) − 0 ⎢ ⎥ ⎢ ⎥ 1 1 1 = ⎢ ( + ) − ⎥ − ⎣ ⎦ 1 0 − 1
The equation of motion is given by
[] {̈ ()} + [] { ()} = {0} where {} is the vector of horizontal displacements and [] is the mass matrix, given by ⎤ ⎡ 1 0 0 ⎥ ⎢ ⎥ ⎢ [] = ⎢ 0 2 0 ⎥ ⎦ ⎣ 0 0 3
It is possible to derive the flexibility or stiffness matrix that relate the angles to the torques.
366
CHAPTER 6 MULTI DOF VIBRATION
Problems for Section 6.3 — Equations of Motion 3. For the double pendulum of Figure 6.66, derive the equations of motion using (a) Newton’s second law, and (b) Lagrange’s equation.
Figure 6.66: Double pendulum.
Solution: We assume the pendula to be inextensible strings. (a) By Newton’s second law of motion, we need a free-body diagram for each mass.
Then since the motion of each mass is planar, we will have two EOM for each mass, even though there are only 2 degrees-of-freedom, 1 and 2 The additional equations are for the tensions in the strings. Therefore, the tensions can be removed from the equations resulting in the two EOM. Let 1 and 2 be the generalized coordinates. Consider mass 1. The and components of velocities are: ̇ = 1 ̇1 cos 1 ̇ = 1 ̇1 sin 1
367 Differentiate these expressions with respect to time and recalling the chain rule of differentiation, we find the respective accelerations in the and directions. Applying Newton’s second law to each mass, we obtain: For mass 1: ³ ´ X 2 = 1 =⇒ −1 sin 1 + 2 sin 2 = 1 1 ̈1 cos 1 − ̇1 sin 1 (1) ³ ´ X 2 = 1 =⇒ 1 cos 1 − 2 cos 2 = 1 + 1 1 ̈1 sin 1 + ̇1 cos 1 (2) Similarly for mass 2: X
= 1
X
= 1
=⇒
=⇒
³ ´ 2 −2 sin 2 = 2 2 ̈2 cos 2 − ̇2 sin 2 ³ ´ 2 +2 1 ̈1 cos 1 − ̇1 sin 1 ³ ´ 2 2 cos 2 = 2 + 2 2 ̈2 sin 2 + ̇2 cos 2 ³ ´ 2 +2 1 ̈1 sin 1 + ̇1 cos 1
(3)
(4)
We can eliminate 2 from Equations (3) and (4) by adding Equation (3) × cos 2 and Equation (2) × sin 2 so that (3) × cos 2 + (2) × sin 2 = 0 h i 2 2 2 ̈2 + 2 1 ̈1 cos (2 − 1 ) + ̇1 sin (2 − 1 ) + 2 sin 2 = 0
(EOM 1)
This is one of the equations of motion. The second equation of motion is obtained by eliminating 1 and 2 from Equations (1) and (2). Equations (3) and (4) are substituted into Equations (1) and (2) to eliminate 2 first. Then, Equations (3) and (4) become ³ ´ ³ ´ 2 2 −1 sin 1 = (1 + 2 ) 1 ̈1 cos 1 − ̇1 sin 1 + 2 2 ̈2 cos 2 − ̇2 sin 2 ³ ´ 2 1 cos 1 = (1 + 2 ) + (1 + 2 ) 1 ̈1 sin 1 + ̇1 cos 1 ³ ´ 2 +2 2 ̈2 sin 2 + ̇2 cos 2 Multiplying the first by cos 1 , multiplying the second by sin 1 and adding the two, we obtain (1 + 2 ) 1 ̈1 + 2 2 (cos 1 cos 2 + sin 1 sin 2 ) ̈2 2
+2 2 (cos 2 sin 1 − cos 1 sin 2 ) ̇2 + (1 + 2 ) sin 1 = 0 which simplifies to h i 2 (1 + 2 ) 1 ̈1 + 2 2 ̈2 cos (2 − 1 ) − ̇2 sin (2 − 1 ) + (1 + 2 ) sin 1 = 0
This is the second equation of motion.
(EOM 2)
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CHAPTER 6 MULTI DOF VIBRATION
(b) By Lagrange’s equation, we need to derive the kinetic and potential energies. For this we need the positions and velocities for each mass. The datum is taken to be the static equilibrium position. Take point as the origin with axis pointing to the right and axis pointing up. Then, r1 = (1 sin 1 ) i − (1 cos 1 ) j r2 = (1 sin 1 + 2 sin 2 ) i − (1 cos 1 + 2 cos 2 ) j ´ ³ ´ ³ v1 = 1 ̇1 cos 1 i + 1 ̇1 sin 1 j ´ ³ ´ ³ v2 = 1 ̇1 cos 1 + 2 ̇2 cos 2 i + 1 ̇1 sin 1 + 2 ̇2 sin 2 j
The kinetic, potential energies, and the Lagrangian are µ³ ³ ´2 1 ´2 ³ ´2 ¶ 1 = 1 1 ̇1 + 2 1 ̇1 cos 1 + 2 ̇2 cos 2 + 1 ̇1 sin 1 + 2 ̇2 sin 2 2 2 µ³ ¶ ³ ´2 1 ´2 ³ ´2 1 1 ̇1 + 2 ̇2 + 21 2 ̇1 ̇2 cos (2 − 1 ) = 1 1 ̇1 + 2 2 2 = 1 1 (1 − cos 1 ) + 2 (1 (1 − cos 1 ) + 2 (1 − cos 2 )) ; = − Now we are prepared to apply Lagrange’s equation for each generalized coordinate, 1 and 2 : ¶ µ − = 0 ̇1 1 ¶ µ − = 0 ̇2 2 where
µ
̇1
¶
´ ³ (1 + 2 )12 ̇1 + 2 1 2 ̇2 cos (2 − 1 ) ³ ³ ´ ´ = (1 + 2 )12 ̈1 + 2 1 2 ̈2 cos (2 − 1 ) − ̇2 ̇2 − ̇1 sin (2 − 1 )
=
= −2 1 2 ̇1 ̇2 sin (2 − 1 ) − (1 + 2 )1 sin 1 ; 1 and
µ
̇2
¶
=
´´ ³ ³2 2 2 ̇2 + 1 2 ̇1 cos (2 − 1 )
³ ´ = 2 22 ̈2 + 2 1 2 ̈1 cos (2 − 1 ) − 2 1 2 ̇1 ̇2 − ̇1 sin (2 − 1 )
= 2 1 2 ̇1 ̇2 sin (2 − 1 ) − 2 2 sin 2 2 The EOM are, therefore,
³ ´ 2 ̈2 cos (2 − 1 ) − ̇2 sin (2 − 1 ) + (1 + 2 )1 sin 1 = 0 ´ ³ 2 2 22 ̈2 + 2 1 2 ̈1 cos (2 − 1 ) + ̇1 sin (2 − 1 ) + 2 2 sin 2 = 0
(1 + 2 )12 ̈1 + 2 1 2
369 If we want to simplify these EOM by assuming small angles and small motions, then we can make the approximations:
2
cos (2 − 1 ) ' 1
2
2
2
̇2 sin (2 − 1 ) ' ̇2 (2 − 1 ) ' 0
̇1 sin (2 − 1 ) ' ̇1 (2 − 1 ) ' 0
sin 12 '
and then the linearized EOM become: (1 + 2 )12 ̈1 + 2 1 2 ̈2 + (1 + 2 )1 1 = 0 2 22 ̈2 + 2 1 2 ̈1 + 2 2 2 = 0 Small angles and small motions can be a bit vague, and we can quantify this statement by expanding the nonlinear EOMS near ≈ 0, ̇ ≈ 0 and ̈ ≈ 0 Let the first EOM be 1 and the second EOM be 2 They are functions of 1 2 ̇1 ̇2 ̈1 and ̈2 . The multi-variable Taylor series for 1 is given by ³ ´ 1 ∼ 1 1 = 0 2 = 0 ̇1 = 0 ̇2 = 0 ̈1 = 0 ̈2 = 0 ¯ ¯ ¯ ¯ 1 ¯¯ 1 ¯¯ 1 ¯¯ 1 ¯¯ + ̇1 + ̇2 1 + 2 + 1 ¯0 2 ¯0 ̇1 ¯0 ̇2 ¯0 ¯ ¯ 1 ¯¯ 1 ¯¯ + ̈1 + ̈2 + ̈1 ¯0 ̈2 ¯0 After taking the derivatives and evaluating them at 1 = 0 2 = 0 ̇1 = 0 ̇2 = 0 ̈1 = 0 and ̈2 = 0, we will find the linearized version of 1 . As an example, a few terms are evaluated next. ³ ´ 1 1 = 0 2 = 0 ̇1 = 0 ̇2 = 0 ̈1 = 0 ̈2 = 0 = 0 ¯ ³ ´¯ 1 ¯¯ 2 ¯ = sin ( − ) + cos ( − ) + ( + ) cos ̈ ̇ 2 1 2 2 2 1 2 1 2 2 2 1 1 2 1 1 ¯ 1 ¯0 0 = (1 + 2 )1 ¯ ¯ 1 ¯ = (1 + 2 )12 ¯ ̈1 0 ¯ 1 ¯¯ = 2 1 2 cos (2 − 1 )|0 = 2 1 2 ̈2 ¯0
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CHAPTER 6 MULTI DOF VIBRATION
4. Derive the characteristic equation and the modal ratios for the system shown in Figure 6.66. Solution: Let us start with the equations of motion obtained in Problem 3. (1 + 2 )12 ̈1 + 2 1 2 ̈2 + (1 + 2 )1 1 = 0 2 22 ̈2 + 2 1 2 ̈1 + 2 2 2 = 0 In matrix-vector form, we write ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎬ ⎨ ⎬ ⎨ (1 + 2 ) 0 ̈1 ( + 2 ) 1 2 2 1 ⎦ ⎦ ⎣ 1 +⎣ = {0} 0 2 ⎩ 2 ⎭ 2 1 2 2 ⎩ ̈2 ⎭ Substituting 1 () = exp() and 2 () = exp() we obtain ⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎨⎬ ⎨0⎬ (1 + 2 ) − 2 (1 + 2 ) 1 −2 2 ⎣ ⎦ = − − 2 ⎩ ⎭ ⎩ 0 ⎭ 2 1
2
2 2
The characteristic equation is obtained by setting the determinant of the matrix above equal to zero, or ⎡ ⎤ 2 (1 + 2 ) − (1 + 2 ) 1 −2 2 ⎦ = 0 det ⎣ −2 1 2 − 2 2 2 ¶ ¶ µ µ 2 2 2 + 1 2 2 4 2 + = 0 − − 1 2 2 1 2 (1 + 2 )
The modal ratio corresponding to the mode is given by µ ¶ 2 2 = (1 + 2 ) − 2 (1 + 2 ) 1 where is one of the roots of the characteristic equation.
371
5. The system shown in Figure 6.67 consists of two linear springs with stiffness 1 and 2 and a uniform bar of mass and length of 2 If the force is acting at a distance from the centerline as shown, derive Lagrange’s equation of motion. Assume small motions.
Figure 6.67: A bar restrained by two springs with asymmetric load.
Solution: Since the bar can translate and rotate, measure from the equilibrium of the center of mass of the bar and as the rotation in the counter-clock-wise direction about the center of mass as shown next. The stretch of spring 1 is − cos and the stretch of spring 2 is + cos
Assuming that the center of mass moves horizontally only, the potential energy is due to the springs only. The potential energy is then 1 1 = 1 ( − sin )2 + 2 ( + sin )2 2 2 The kinetic energy is due to both translation and rotation of the bar: = + where 1 2 ̇ 2 1 2 = ̇ 2 =
372
CHAPTER 6 MULTI DOF VIBRATION
where is the mass moment of inertia of the bar about its center of gravity and is = 2 12 The kinetic energy is given by 1 1 2 2 = ̇2 + ̇ 2 2 12 The force is a non-conservative force, so we need to determine the generalized forces due to The virtual work done by is = F·r = ( − sin ) = − cos The generalized forces are Lagrange’s equation is given by
= and = − cos − + = ̇ + = − ̇
1 1 The individual derivatives are = 1 ( − sin )2 + 2 ( + sin )2 2 2 = ̇; = ̈; ̇ ̇ = 0; = 1 ( − sin ) + 2 ( + sin ) = (1 + 2 ) + (−1 + 2 ) sin 1 1 2 ̇; = = 2 ̈ 12 ̇ 12 ̇ = 0 = −1 ( − sin ) cos + 2 ( + sin ) cos = (−1 + 2 ) cos + (1 + 2 ) 2 sin cos The equations of motion are ̈ + (1 + 2 ) + (2 − 1 ) sin =
1 2 ̈ + (2 − 1 ) cos + (1 + 2 ) 2 sin cos = − cos 12 For small rotation (2 1) the equations of motion are written as ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎬ ⎬ ⎨ ⎨ ⎨ 0 ⎬ ̈ + 2 (2 − 1 ) ⎣ 1 ⎣ ⎦ ⎦ + = 1 ⎭ ⎭ ⎩ ⎩ 2 2 ⎩ ̈ ⎭ 0 − (2 − 1 ) (1 + 2 ) 12
373
6. A simple lumped-parameter model of a building, shown as the inverted link in Figure 6.68, can be used for preliminary study of earthquake dynamics. Derive the equations of motion for this two degree-of-freedom system using (a) Newton’s second law, and (b) Lagrange’s equation. Let () be the prescribed ground motion, and () be the absolute position of the mass .
Figure 6.68: An earthquake loading model.
Solution: (a) Newton’s Second Law to derive the EOM by taking two free-body diagrams, one of the block and the other of the rod. There are two degrees-of-freedom. We use () and () where () is the horizontal coordinate of the block measured from inertial frame.
374
CHAPTER 6 MULTI DOF VIBRATION
Assume that the ground has displacement and velocity ̇ Then, the damper and the spring exert a force of − [ ( − ) + (̇ − ̇)] on the block. From Newton’s second law, summing the forces on the block in the horizontal direction gives − [ ( − ) + (̇ − ̇)] + = ̈
(1)
Similarly, summing the forces in the horizontal direction for the rod gives − = where is the acceleration of the rod’s center of mass in the horizontal direction obtained from 2 2 r = {( + sin ) i + cos j} 2 2 ´ ³ ´ ³ 2 2 = ̈ + ̈ cos − ̇ sin ̂ + −̈ sin − ̇ cos j {z } {z } | |
a =
Then, the force balance equation in the horizontal direction for the rod is ³ ´ 2 − = ̈ + ̈ cos − ̇ sin
(2)
Eliminating from Equations (1) and (2), we obtain ´ ³ 2 ( + ) ̈ + ̈ cos − ̇ sin − ( − ) − (̇ − ̇) = 0
This is one of the two equation of motion. Note that the spring and the damping force do not explicitly dependent on the position or velocity of the rod. Now, we examine the motion of the rod. To do so, we will look at the sum of the moments about the rod’s center of mass using X +© = ̈
To apply this equation requires knowledge of the reaction forces and Note that taking moments at any other point will result in an extra term in the equation. For instance, if we take a moment about a moving hinge (call it 0 ) that is not necessarily on the body we must use X +© 0 = ̈ + (0 × ) · k Using the latter moment balance equation, we have
− + sin = ̈ + { (sin i + cos j) ´ ³ ´ ´ ³³ 2 2 × ̈ + ̈ cos − ̇ sin ̂ + −̈ sin − ̇ cos j } · k ´ ³ ´´ ³ ³ 2 2 = ̈ + sin ̈ sin + ̇ cos + cos ̈ + ̈ cos − ̇ sin = ̈ + 2 ̈ + ̈ cos
We can rewrite the equation as sin − = 0 ̈ + cos ̈
375 where 0 = + 2 Assuming small angles and small motions, the linearized equations are ( + ) ̈ + ̈ − ( − ) − (̇ − ̇) = 0 ̈ + 0 ̈ + − = 0 Combining the two equations in matrix-vector form yields, ⎤⎧ ⎫ ⎤⎧ ⎫ ⎡ ⎡ ⎤⎧ ⎫ ⎡ ⎨⎬ 0 0 ⎨ ̇ ⎬ + ⎨ ̈ ⎬ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ + + 0 − ⎩ ⎭ 0 0 ⎩ ̇ ⎭ 0 ⎩ ̈ ⎭ ⎡ ⎤ ⎡ ⎤ = ⎣ ⎦ ̇ + ⎣ ⎦ 0 0
(b) Use Lagrange’s Equation - by first deriving the kinetic and potential energies. We implicitly assume that the rod is rigid, and that the block can be treated as a point mass. Because of the damping between the block and the environment, we have nonconservative work. The kinematic relationships of these two bodies are needed in order to derive the energies. The position of the rod from the axis is given by: r = ( + sin )i + cos j v = (̇ + ̇ cos )i − ̇ sin j
where i and j are unit vectors to the right and up, respectively. The energies are: µ ³ ´2 ¶ 1 1 2 1 2 2 = ̇ + (̇ + ̇ cos ) + ̇ sin + ̇ 2 2 2 ¶ µ ³ ´ 2 1 1 1 2 + ̇ = ̇2 + ̇2 + 2̇̇ cos + ̇ 2 2 2 ¶ µ 1 1 = 2 + 2 ( − )2 + cos 2 22 ¶ µ 1 2 (̇ − ̇) = 2 22
where is the Rayleigh dissipation function. We have two generalized coordinates: and We can now apply Lagrange’s equation, where = − : µ ¶ − + = 0 ̇ ̇ µ ¶ ´ ³ = ( + )̇ + ̇ cos ̇ ³ ´ 2 = ( + )̈ + ̈ cos − ̇ sin = − ( − ) = (̇ − ̇) ̇
376
CHAPTER 6 MULTI DOF VIBRATION
The first EOM is then ³ ´ 2 ( + )̈ + ̈ cos − ̇ sin + ( − ) + (̇ − ̇) = 0 ³ ´ 2 ( + )̈ + ̈ cos − ̇ sin + ̇ + = + ̇
2
which is a (nonlinear) base-excited system. If we can justify the following approximation: ̇ sin ≈ 2 ̇ ≈ 0 and cos ≈ 1 then the linearized base-excited motion is governed by ( + )̈ + ̈ + ̇ + = + ̇ We next proceed to derive the equation in : µ ¶ − = 0 ̇ µ ¶ ´ ´ ³ ³ 2 = ̇ cos + ̇ + ̇ ̇ ³ ´ = ̈ cos − ̇̇ sin + 2 ̈ + ̈ = −̇̇ sin − + sin
Therefore, this EOM becomes ´ ³ 2 ̈ cos + ̈ + ̈ + − sin = 0
Making the same smallness assumptions as before linearizes this equation: ¢ ¡ ̈ + 2 + ̈ + ( − ) = 0
377
7. Derive the equation of motion for the elastically restrained rigid beam shown in Figure 6.69 using (a) Newton’s second law, and (b) Lagrange’s equation.
Figure 6.69: Elastically restrained rigid beam suspended with a torsional spring.
Solution: We assume that the system is on a horizontal plane, and the torsional spring can only move horizontally. The springs, 3 and 4 may move freely vertically so that they only exert horizontal force. The free-body diagram for the rod is shown below.
The position vector of the center of the rod (positive to the right and up) is ¶ µ r = + sin i − cos j 2 2 The velocity vector is v = r =
µ ¶ ̇ + ̇ cos i + ̇ sin j 2 2
378
CHAPTER 6 MULTI DOF VIBRATION
The speed squared is 2
¶2 µ ¶2 µ ̇ sin = ̇ + ̇ cos + 2 2 µ ¶2 2 2 = ̇ + ̇̇ cos + ̇ 2
Differentiating the velocity vector with respect to time, the acceleration is a = =
v µ
¶ ´ ³ 2 2 ̈ + ̈ cos − ̇ sin i + ̈ sin + ̇ cos j 2 2 2
(a) Newton’s second law We need to sum forces in the direction and sum moments about the center of mass. Note that the only forces acting in the -direction are the spring forces; thus, we get: P = ∙ ´¸ ³ 2 − (1 + 2 ) − (3 + 4 ) ( + sin ) = ̈ + ̈ cos − ̇ sin 2
where we have substituted for ̈ on the left-hand side. Bringing all the terms to one side gives: ∙ ´¸ ³ 2 ̈ cos − ̇ sin + (1 + 2 + 3 + 4 ) + (3 + 4 ) sin = 0 ̈ + 2
Assuming small angles and small motions, the linearized equation of motion is ̈ + ̈ + (1 + 2 + 3 + 4 ) + (3 + 4 ) = 0 2
(EOM1)
To sum moments we use the attachment point, , of the torsional spring as a reference, and take the direction of increasing as positive. P × = + ¡ ¢ P ª = ̈ + × · k
where
(sin i − cos j) 2 ½µ ¶ ´ ¾ 2 ³ 2 ̈ sin + ̇ cos j × ̈ + ̈ cos − ̇ sin i + 2 2 2 ½ µ ¶¾ ³ ´ 2 2 sin ̈ sin + ̇ cos + cos ̈ + ̈ cos − ̇ sin k = 2 2 2 2 µ µ ¶ ¶ = ̈ cos + ̈ k 2 2
× =
379 Then, the moment balance equation become µ ¶ − − (3 + 4 ) ( + sin ) cos − sin = ̈ + ̈ cos + ̈ 2 2 2 ¶ µ 2 ̈ + ̈ cos = + 4 2 where the left hand side shows the moments due to the torsional springs and gravity. The second equation of motion is obtained by bringing all terms to one side, ¶ µ 2 ̈ + ̈ cos + + (3 + 4 ) ( + sin ) cos + sin = 0 + 4 2 2 The linearized equation of motion is µ ¶ ³ 2 ´ ̈ + (3 + 4 ) + (3 + 4 ) 2 + + = 0 ̈ + + 2 4 2
(EOM2)
Another version of moment equation For a point that is fixed to the body, we may use another formula given by P = ̈k + ×
In this case, the right-hand-side of this equation is ¶ µ ̈ + sin i − cos j × ̈i 2 2 = ̈ + ̈ cos 2 which is identical to ̈k + × (b) Lagrange’s equations of motion The energies are then given by: 1 2 1 2 + ̇ 2 Ã 2 µ ¶2 ! 1 1 2 2 = ̇2 + ̇̇ cos + ̇ + ̇ 2 2 2
=
1 1 1 (1 + 2 )2 + (3 + 4 )22 + (1 − cos ) + 2 2 2 2 2 1 1 1 = (1 + 2 )2 + (3 + 4 )( + sin )2 + (1 − cos ) + 2 2 2 2 2
=
where the gravitational potential 2 (1 − cos ) is retained since the beam rotates even though the springs are assumed to act only horizontally. These energies can be substituted into Lagrange’s
380
CHAPTER 6 MULTI DOF VIBRATION
equation, using = − ; for each generalized coordinate: and The two Lagrange’s equations are given by µ ¶ − = 0 ̇ µ ¶ − = 0 ̇ where µ
¶ µ ¶ 1 = ̇ + ̇ cos ̇ 2 µ ¶ µ ¶ 2 = ̈ + ̈ cos − ̇ sin ̇ 2 2 = (1 + 2 ) + (3 + 4 )( + sin ) − and ¶ 2 = ̇ cos + ̇ + ̇ 2 4 ̇ ¶ µ ¶ µ 1 2 = ̈ cos − ̇̇ sin + + ̈ ̇ 2 2 4 − = ̇̇ sin + (3 + 4 )( + sin ) cos + sin + 2 2 µ
Note that we can significantly simplify the equations. The resulting EOMs are given by ∙ ´¸ ³ 2 ̈ + ̈ cos − ̇ sin + (1 + 2 ) + (3 + 4 ) ( + sin ) = 0 2 µ ¶ 2 ̈ cos + + sin = 0 ̈ + + (3 + 4 ) ( + sin ) cos + 2 4 2 which are the same as those derived using Newton’s second law of motion. The linearized EOM are µ ¶ ̈ + ̈ + (1 + 2 ) + (3 + 4 )( + ) = 0 2 2 ̈ + ( + )̈ + (3 + 4 )( + ) + + = 0 2 4 2
381
8. Use Hamilton’s principle to derive the equations of motion for the structures shown in Figures (a) 6.66, (b) 6.68, and (c) 6.69.
Solution: For the system shown in Fig. 6.66 - We need to derive the kinetic and potential energies. For this we need the positions and velocities for each mass. The datum is taken to be the static equilibrium position. Take point as the origin with the axis pointing to the right and the axis pointing up. Then, r1 = (1 sin 1 ) i − (1 cos 1 ) j r2 = (1 sin 1 + 2 sin 2 ) i − (1 cos 1 + 2 cos 2 ) j ´ ³ ´ ³ v1 = 1 ̇1 cos 1 i + 1 ̇1 sin 1 j ´ ³ ´ ³ v2 = 1 ̇1 cos 1 + 2 ̇2 cos 2 i + 1 ̇1 sin 1 + 2 ̇2 sin 2 j where i and j are unit vectors to the right and up, respectively. The kinetic and potential energies are given by µ³ ³ ´2 1 ´2 ³ ´2 ¶ 1 = 1 1 ̇1 + 2 1 ̇1 cos 1 + 2 ̇2 cos 2 + 1 ̇1 sin 1 + 2 ̇2 sin 2 2 2 µ³ ¶ ³ ´2 1 ´2 ³ ´2 1 1 ̇1 + 2 ̇2 + 21 2 ̇1 ̇2 cos (2 − 1 ) = 1 1 ̇1 + 2 2 2 ´2 1 ³ ´2 ³ 1 = (1 + 2 ) 1 ̇1 + 2 2 ̇2 + 2 1 2 ̇1 ̇2 cos (2 − 1 ) 2 2 = 1 1 (1 − cos 1 ) + 2 (1 (1 − cos 1 ) + 2 (1 − cos 2 )) = (1 + 2 ) 1 (1 − cos 1 ) + 2 2 (1 − cos 2 ) To apply Hamilton’s principle, we substitute the Lagrangian, = − into Hamilton’s variational R 2 principle: 1 = 0 0=
Z 2 1
=
Z 2
⎡
´2 1 ³ ´2 ³ ⎢1 ( + ) + + 2 1 2 ̇1 ̇2 cos (2 − 1 ) ̇ ̇ ⎣ 1 2 1 1 2 2 2 {z } 2 2 1 | {z } | {z } | third term first term second term ⎤
− (1 + 2 ) 1 (1 − cos 1 ) − 2 2 (1 − cos 2 )⎦ | {z } | {z } fourth term
fifth term
382
CHAPTER 6 MULTI DOF VIBRATION
The variations R are with respect to 1 and 2 , and their derivatives. The variation and the integration can be interchanged. For example, the variation of the first term becomes
Z 2 1
³ ´2 1 (1 + 2 ) 1 ̇1 = 2 =
Z 2
Z 12
³ ´2 1 (1 + 2 ) 1 ̇1 2 (1 + 2 ) 12 ̇1 ̇1
1
This term includes a variation of a derivative, and it is replace by (integration by parts) Z 2
Z 2
Z 2
(̇1 1 ) − ̈1 1 1 ¯2 Z 2 ¯ = ̇1 1 ¯ − ̈1 1
̇1 ̇1 =
1
1
1
1
where the variation, 1 , evaluated at the initial and final times are zeros. Then,
Z 2 1
´2 ³ 1 (1 + 2 ) 1 ̇1 = − 2
The second term in the Lagrangian, 12 2
Z 2 1
Z 2
(1 + 2 ) 12 ̈1 1
(1)
1
³ ´2 2 ̇2 simplifies the same way so that
³ ´2 1 2 2 ̇2 = − 2
Z 2
2 22 ̈2 2
(2)
1
The product rule is applied to the third term, and the variation is given by Z 2
2 1 2 ̇1 ̇2 cos (2 − 1 ) 1 Z 2 Z 2 Z 2 = 2 1 2 ̇1 ̇2 cos (2 − 1 ) + 2 1 2 ̇1 ̇2 cos (2 − 1 ) + 2 1 2 ̇1 ̇2 cos (2 − 1 ) 1
1
1
383 where each term is simplified further by integrating by parts such that Z 2
2 1 2 ̇1 ̇2 cos (2 − 1 ) Z 2 h i 2 = 2 1 2 ̇2 cos (2 − 1 ) 1 |1 − 2 1 2 ̇2 cos (2 − 1 ) 1 1 Z 2 h i = − 2 1 2 ̇2 cos (2 − 1 ) 1 Z 2 1 2 1 2 ̇1 ̇2 cos (2 − 1 ) 1 Z 2 h i 2 = 2 1 2 ̇1 cos (2 − 1 ) 2 |1 − 2 1 2 ̇1 cos (2 − 1 ) 2 1 Z 2 h i = − 2 1 2 ̇1 cos (2 − 1 ) 2 Z 2 1 2 1 2 ̇1 ̇2 cos (2 − 1 ) 1 Z 2 = − 2 1 2 ̇1 ̇2 sin (2 − 1 ) (2 − 1 ) 1
(3)
(4)
(5)
1
The fourth and fifth terms are simplifies to Z 2
[− (1 + 2 ) 1 (1 − cos 1 ) − 2 2 (1 − cos 2 )] 1 Z 2 = [− (1 + 2 ) 1 sin 1 1 − 2 2 sin 1 2 ]
(6)
1
Finally, combining Equations (1)-(6), we obtain 0 =
Z 2
1
= − − −
Z 2
Z 12
Z 12
(1 + 2 ) 12 ̈1 1 −
Z 2 1
2 22 ̈2 2 −
Z 2 1
i h 2 1 2 ̇2 cos (2 − 1 ) 1
Z 2 i h 2 1 2 ̇1 ̇2 sin (2 − 1 ) (2 − 1 ) 2 1 2 ̇1 cos (2 − 1 ) 2 − 1
[(1 + 2 ) 1 sin 1 1 + 2 2 sin 1 2 ]
Z 12 ∙
2 1 2 ̇2 cos (2 − 1 )] + 2 1 2 ̇1 ̇2 sin (2 − 1 ) 1 + (1 + 2 ) 1 sin 1 ] 1 Z 2 ∙ i¸ h 2 − 2 2 ̈2 + 2 1 2 ̇1 cos (2 − 1 )] − 2 1 2 ̇1 ̇2 sin (2 − 1 ) + 2 2 sin 1 2 1
= −
(1 + 2 ) 12 ̈1 +
384
CHAPTER 6 MULTI DOF VIBRATION
The coefficients of 1 and 2 must be zeros so that the integral becomes zero for arbitrary variations, 1 and 2 Then, we obtain two equations given by i h 2 2 1 2 ̇2 cos (2 − 1 )] + 2 1 2 ̇1 ̇2 sin (2 − 1 ) + (1 + 2 ) 1 sin 1 = 0 (1 + 2 ) 1 ̈1 + i h 2 2 2 ̈2 + 2 1 2 ̇1 cos (2 − 1 ) − 2 1 2 ̇1 ̇2 sin (2 − 1 ) + 2 2 sin 1 = 0
Perform the differentiations, and we will obtain ´ ³ 2 2 (1 + 2 )1 ̈1 + 2 1 2 ̈2 cos (2 − 1 ) − ̇2 sin (2 − 1 ) + (1 + 2 )1 sin 1 = 0 ´ ³ 2 2 22 ̈2 + 2 1 2 ̈1 cos (2 − 1 ) + ̇1 sin (2 − 1 ) + 2 2 sin 2 = 0 which are identical to the equation of motion in Problem 4.
For the system shown in Fig. 6.68 - First derive the kinetic and potential energies. We implicitly assume that the beam is rigid, and that the block can be treated as a point mass. Because of the damping between the block and the environment, we have nonconservative work. The kinematics of these bodies are needed in order to derive the energies. The position and velocity of the beam with respect to the axis is given by: r = ( + sin )i + cos j v = (̇ + ̇ cos )i − ̇ sin j The respective energies are given by µ ³ ´2 ¶ 1 1 2 1 2 2 + ̇ = ̇ + (̇ + ̇ cos ) + ̇ sin 2 2 2 ¶ µ ³ ´ 2 1 1 1 2 + ̇ = ̇2 + ̇2 + 2̇̇ cos + ̇ 2 2 2 ¡ ¢ 1 1 2 = ( + ) ̇2 + ̇̇ cos + 2 + ̇ 2 2 ¶ µ 1 1 = 2 + 2 ( − )2 + cos 2 22 ¶ µ 1 2 (̇ − ̇) = 2 22 where is the Rayleigh dissipation function. We have two generalized coordinates: and To apply Hamilton’s principle, we substitute the energies into Hamilton’s variational principle. The difference here from the prior problem is that we must include the dissipation effects. We do this by using the extended Hamilton’s principle. The extended Hamilton’s principle says Z 2 ( + ) = 0 1 Z 2 ( − + ) = 0 1
385 where the virtual work done by the dissipative force is given by = − 2 (̇ − ̇) ( − ) where ( − ) = The variation in is zero because is prescribed. Now we interchange the variation and the integration, and apply the variation to each term in the integrand. The extended Hamilton’s principle says Z 2 0 = ( − + ) 1 Z 2 = ( + ) 1 Z 2 ∙ ¢ 2 1 1¡ 2 = ( + ) ̇2 + ̇̇ cos + + ̇ 2 2 1 ¸ 1 − 2 − ( − )2 − cos − (̇ − ̇) 2 2 2 Simplifying above equation, we obtain Z 2 h ¡ ¢ 0 = ( + ) ̇ ̇ + ̇ cos ̇ + ̇ cos ̇ − ̇̇ sin + 2 + ̇̇ 1 i − − ( − ) + sin − (̇ − ̇) 2
Let us simplify before we proceed with integration by parts by collecting terms such that Z 2 h i h ¡ 2 ¢ i 0 = ( + ) ̇ + ̇ cos ̇ + ̇ cos + + ̇ ̇ 1 i ³ ´ h + −̇̇ sin − + sin − ( − ) + (̇ − ̇) 2
We integrate the first two terms by parts to obtain Z 2 h i ( + ) ̇ + ̇ cos ̇ 1 h i ¯2 Z 2 h i ¯ = ( + ) ̇ + ̇ cos ¯ − ( + ) ̇ + ̇ cos 1 1 Z 2 h i = − ( + ) ̇ + ̇ cos 1 and
Z 2 h ¡ ¢ i ̇ cos + 2 + ̇ ̇ 1 Z 2 h h ¡ 2 ¡ ¢ i ¯¯2 ¢ i = ̇ cos + + ̇ ¯ − ̇ cos + 2 + ̇ 1 1
where the variations evaluated at the initial and final times are zeros.
(1)
386
CHAPTER 6 MULTI DOF VIBRATION
Then, Equation (1) is simplified to 0 =
Z 2
i ¡ ¢ i h h ( + ) ̇ + ̇ cos − ̇ cos + 2 + ̇ 1 i ³ ´ h + −̇̇ sin − + sin − ( − ) + (̇ − ̇) 2 −
We set the coefficients of and to zeros to obtain the two equations of motion given by i h ( + ) ̇ + ̇ cos + ( − ) + (̇ − ̇) = 0 2 ¡ 2 ¢ i h ̇ cos + + ̇ + ̇̇ sin + − sin = 0
After differentiation, we obtain
³ ´ 2 ( + )̈ + ̈ cos − ̇ sin + ̇ + = + ̇ ¡ ¢ ̈ cos + 2 + ̈ + − sin = 0
which are identical to the solutions given in Problem 5.
For the system shown in Fig. 6.69 - We substitute the kinetic and potential energies into Hamilton’s principle, and then apply the variation to each term under the integral given by
Z 2
= 0
1
where = − and
¶2 µ 1 2 2 1 2 2 = ̇ + ̇̇ cos + ̇ + ̇ 2 4 2 1 1 1 = (1 + 2 )2 + (3 + 4 )22 + (1 − cos ) + 2 2 2 2 2 1 1 1 = (1 + 2 )2 + (3 + 4 )( + sin )2 + (1 − cos ) + 2 2 2 2 2
and ¶2 µ 2 2 1 1 2 2 + ̇ = ̇ + ̇̇ cos + ̇ 2 4 2 1 1 − (1 + 2 )2 − (3 + 4 )( + sin )2 2 2 1 + (1 − cos ) + 2 2 2
387 The two coordinates that are used are and Let us take a variation of the Lagrangian. For example, the variation of the first term is ∙ ¶¸ µ 1 2 2 2 ̇ + ̇̇ cos + ̇ 2 4 ¶ µ 2 = ̇ ̇ + ̇ cos ̇ + ̇ cos ̇ − ̇̇ sin + ̇ ̇ 2 2 2 4 ¶ µ ¶ µ 2 ̇ cos + ̇ ̇ = ̇ + ̇ cos ̇ + 2 2 4 − ̇̇ sin 2 The variation of the Lagrangian is given by µ ¶ µ ¶ 2 = ̇ + ̇ cos ̇ + ̇ cos + ̇ ̇ − ̇̇ sin 2 2 4 2 + ̇ ̇ − (1 + 2 ) − (3 + 4 )( + sin )( + cos ) − sin − 2 = ̇ + ̇ + +
where ¶ µ = ̇ + ̇ cos 2 ¶ µ 2 ̇ cos + ̇ + ̇ = 2 4 = − ̇̇ sin − (3 + 4 )( + sin ) cos 2 + sin + 2 = −(1 + 2 ) − (3 + 4 )( + sin ) 0=
Z 2 1
Z 2 h i ̇ + ̇ + + = 1
The first two are integrated by parts so that 0=
Z 2 1
= |21 + |21 +
¸ Z 2 ∙ − − + + 1
where the terms evaluated at the initial and final times are zeros. Then, ¶ µ ¶ ¸ Z 2 ∙µ − 0= = + + − + 1 1 Z 2
388 For
CHAPTER 6 MULTI DOF VIBRATION R 2 1
to be zero for any arbitrary variations, and the system must satisfy + = 0 + = 0 −
−
or µ ¶ 2 − ̈ + ̈ cos − ̇ sin − (1 + 2 ) − (3 + 4 )( + sin ) = 0 2 2 2 − ̈ cos − ̈ − ̈ − (3 + 4 )( + sin ) cos − sin − = 0 2 4 2 The two equations of motion are identical to those obtained in Problem 6.
389
9. Derive the equations of motion for the problem of Example 6.5 using (a) Newton’s second law, (b) Lagrange’s equation, and (c) Hamilton’s principle, and then solve for the response.
Two degree-of-freedom idealization of a rigid beam.
Solution: (a) - Newton’s approach was done in Example 6.5. Here, we take a slightly different approach. For a rigid body in 2D motion, Newton’s law states that X
X
X
= =
= ̈
(1)
where is the center of mass or center of gravity. An alternative form of the moment equation may be used. If the moment is taken about an arbitrary point on the body (such as the point ) the moment equation is given by
where
X
k = ̈k + r × a =
1 ( + )2 12
(2)
390
CHAPTER 6 MULTI DOF VIBRATION
The acceleration of the center of mass is ³ ´ 2 a = 2 ( + ) j = ̈ + ̈ j
Summing the forces in the vertical directions, we obtain
³ ´ () − 1 ( − ) − 2 ( + ) = ̈ + ̈
(EOM 1)
1 ( − ) − 2 ( + ) = ̈ + ̈
(EOM 2)
Using the second version of the moment equation (Equation (2)) we obtain ³ ´ 1 ( − ) − 2 ( + ) = ̈ + ̈ + ̈ where we used the parallel axis theorem, = + 2
If we use the first version of the moment equation (Equation (1)), the moment equation is given by − + 1 ( − ) ( + ) − 2 ( + ) ( − ) = ̈
(3)
We can obtain the second equation of motion back by multiplying the first equation of motion by and adding it to Equation (3). Both moment equations are valid. (b) Lagrange’s equation - To apply Lagrange’s equation, first derive the energies of the system, assuming that the motion is only vertical, using the mass center at a distance of from point : 1 2 1 + 2 2 2 ´2 1 1 ³ 2 = ̇ + ̇ + ̇ 2 2 ´ 1 1 ³ 2 2 2 = ̇ + 2̇̇ + 2 ̇ + ̇ 2 2
=
1 1 1 ( − )2 + 2 ( + )2 2 2 ¢ 1 ¡ ¢ 1 ¡ 2 = 1 − 2 + 2 2 + 2 2 + 2 + 2 2 2 2
=
Substituting these into the Lagrangian = − we obtain the two EOM: µ ¶ = 0 − ̇ where
µ ¶ = ̇ + ̇ = ̈ + ̈ ̇ ̇ − = 1 − 1 + 2 + 2
391 The second Lagrange’s equation is
µ
¶
=0 µ ¶ 2 = ̈ + 2 ̈ + ̈ = ̇ + ̇ + ̇ ̇ ̇ − = −1 + 1 2 + 2 + 2 2 The (linear) EOM are then: ̇
−
̈ + ̈ + 1 − 1 + 2 + 2 = 0 ̈ + ̈ + ̈ − 1 + 2 1 + 2 + 2 2 = 0 2
Note that + 2 = In matrix-vector form, we have ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎨ ⎬ ⎨ () ⎬ 1 + 2 ⎨ ̈ ⎬ −1 + 2 ⎣ ⎦ ⎦ +⎣ = ⎭ ⎩ ⎭ 2 2 −1 + 2 1 + 2 ⎩ ⎭ ⎩ 0 ̈ (c) Hamilton’s principle - We need to use the extended Hamilton’s principle given by Z 2 Z 2 ( + ) = ( ( − ) + ) 1 1 Z 2 = ( + ) 1
where the Lagrangian was derived above. Then take the variation of each term as follows: Z 2 1 ⎤ ⎡ ´ 1 ³ 2 Z 2 1 2 2 2 + ̇ + 2 ̇ ̇ + ̇ ̇ 2 ⎦ 2 = ⎣ ¡ ¢ ¡ ¢ 2 2 1 1 2 2 2 2 1 − 2 1 − 2 + − 2 2 + 2 + The variation of work done by () is Z 2 1
=
Z 2
()
1
We can take the variation of each term in Lagrangian in turn: ´ 1 ³ 2 2 2 ̇ + 2̇̇ + ̇ = ̇ ̇ + (̇ ̇ + ̇ ̇) + 2 ̇ ̇ 2 1 2 ̇ = ̇ ̇ 2 ¸ ∙ ¢ 1 ¡ 2 − 1 − 2 + 2 2 = −1 + 1 ( + ) − 1 2 2 ∙ ¸ ¢ 1 ¡ 2 2 2 − 2 + 2 + = −2 − 2 ( + ) − 2 2 2
392
CHAPTER 6 MULTI DOF VIBRATION
The variations of the derivatives must be integrated by parts so that only variations of and remain: (̇) − ̈ ´ ³ ̇ − ̈ ̇ ̇ = ∙ ¸ (̇) − ̈ ̇ ̇ = ¸ ∙ ³ ´ ̇ − ̈ ̇ ̇ = ∙ ³ ¸ ´ 2 2 ̇ − ̈ ̇ ̇ = ̇ ̇ =
These can be substituted into Hamilton’s principle, recalling that, for example, ̇ ̇ = (̇) − ̈
R and 12 (̇) = ̇|2 − ̇|1 = 0 since the variation equals zero at the end times. After substituting all these terms and collecting terms according to their variations, and noting that the variations are arbitrary, means that the collection of terms multiplying them must equal zero: Z 2 {[ ] + [ ] } = 0; 1
Each group of terms in the square brackets must equal zero independently. This leads to the matrix EOM. Finally, we solve the matrix EOM: ⎤⎧ ⎫ ⎡ ⎡ ⎨ ̈ ⎬ 1 + 2 ⎦ ⎣ +⎣ ⎩ ̈ ⎭ − + 2 +
1
⎫ ⎤⎧ ⎫ ⎧ ⎨ ⎨ ⎬ () ⎬ −1 + 2 ⎦ = ⎭ 2 1 + 2 2 ⎩ ⎭ ⎩ 0
2
(You may have to read Section 7.4 before solving this problem.) To solve this undamped unforced system we assume a harmonic response () = cos , and () = Θ cos Differentiate these appropriately and substitute into the matrix EOM to find: ⎫ ⎡ ⎫ ⎧ ⎫ ⎡ ⎤⎧ ⎤⎧ 2 ⎨ ⎬ ⎨ 1 + 2 −1 + 2 − cos cos ⎬ ⎨ 0 ⎬ ⎣ ⎦ ⎦ +⎣ = ⎩ Θ cos ⎭ ⎩ 0 ⎭ 2 ⎩ −2 Θ cos ⎭ − + 2 + 2 1
2
1
2
We cancel the common cos factors and combine into one characteristic matrix equation: ⎫ ⎧ ⎫ ⎡ ⎤⎧ 2 2 ⎨ + 2 − −1 + 2 − ⎬ ⎨0⎬ ⎣ 1 ⎦ = − + − 2 2 + 2 − 2 2 ⎩ Θ ⎭ ⎩ 0 ⎭ 1
2
1
2
393 For the non-trivial solution, the determinant of the characteristic matrix must be set to zero: ¡ ¢¡ ¢ ¡ ¢2 1 + 2 − 2 2 1 + 2 2 − 2 2 − −1 + 2 − 2 = 0
which is a quadratic in 2 that can be easily evaluated for 21 and 22 where 12 are the respective natural frequencies. Each of these roots is now substituted into the characteristic matrix in order to obtain the modes of vibration, (Θ)12 For example: ¢ ¡ ¢ ¡ 1 + 2 − 21 + −1 + 2 − 21 Θ = 0 Therefore,
µ
Θ
¶
−1 + 2 − 21 = 1 + 2 − 21 1
and similarly the second modal ratio is given by µ ¶ −1 + 2 − 22 = Θ 2 1 + 2 − 22 These modal ratios are then combined as follows for the complete response: ⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎨ ⎬ ⎨1 ⎬ ⎨1 ⎬ = 1 ¡ ¢ cos ( 1 − 1 ) + 2 ¡ ¢ cos (2 − 2 ) ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ Θ 1
Θ 2
The arbitrary constants of integration 12 12 are obtained by satisfying the four initial conditions.
394
CHAPTER 6 MULTI DOF VIBRATION
Problems for Section 6.4 — Undamped Vibration
10. In Example 6.7 evaluate ̄1 ̄2 1 and 2 using the given p initial conditions: 1 (0) = 1 cm 2 (0) = −1 cm ̇1 (0) = 3 cm/s ̇2 (0) = −2 cm/s. Use = 10 rad/s. Solution: The solution found is ⎧ ⎧ ⎫ ⎫ Ã ! r ⎨ () ⎬ ⎨ 100 ⎬ 1 cos 114 = 1 − 1 ⎩ () ⎭ ⎩ 0845 ⎭ 2 ⎫ ⎧ Ã ! r ⎨ 100 ⎬ cos 205 + 2 − 2 ⎩ −0595 ⎭
with initial conditions: 1 (0) = 1 cm 2 (0) = −1 cm ̇1 (0) = 3cm/s and ̇2 (0) = −2 cm/s Substituting these into the matrix equation of motion and its derivative, we obtain the 4 equations to solve for the 4 unknowns ( 1 2 1 2 ) They are given by ⎫ ⎧ ⎨1 ⎬
⎫ ⎧ ⎨ 100 ⎬
⎧ ⎨ 100
⎫ ⎬
= 1 cos 1 + 2 cos 2 ⎩ −1 ⎭ ⎩ 0845 ⎭ ⎩ −0595 ⎭ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫ r r ⎨3 ⎬ ⎨ ⎬ ⎨ 100 100 ⎬ = 114 sin 1 + 205 sin 2 1 2 ⎩ −2 ⎭ ⎩ 0845 ⎭ ⎩ −0595 ⎭
The initial displacement conditions yield ⎡ ⎣
1 cos 1 2 cos 2
⎤
⎡
⎦ = ⎣
100
100
0845 −0595 ⎤ −0281 ⎦ = ⎣ 128 ⎡
⎤−1 ⎡ ⎦
⎣
1 −1
⎤ ⎦
The initial velocity conditions yield ⎡ ⎣
1 sin 1 2 sin 2
⎤
r
⎡
⎤−1 ⎡
⎣ 114 × 100 205 × 100 ⎦ 114 × 0845 205 × −0595 ⎡ ⎤ r −0131 ⎦ = ⎣ 154
⎦ =
⎣
1 −1
⎤ ⎦
395 Then, we find that r ¶ r ¶ µ µ 1 = arctan 047 2 = arctan 120 −0281 1281 1 = 2 = cos 1 cos 2 For
p = 10 rad/s, the numerical values are 1 = 0047 1 = −0281
2 = 0119 2 = 129
396
CHAPTER 6 MULTI DOF VIBRATION
11. Consider the problem of the “sliding pendulum” drawn in Figure 6.70, where a mass slides along a frictionless rod which is assumed to be uniform and thin. Use (a) Newton’s second law, and (b) Lagrange’s equation to derive the equations of motion. Can the governing equations be linearized? Under what conditions? What happens when = ?
Figure 6.70: Sliding pendulum.
Solution: This is a 2-DOF system. We will use and where is the distance between the mass and the pivot point. (a) Newton’s second law - Let us draw a free-body diagram as shown below. We assume that the rod is frictionless.
Summing the forces in the radial and the transverse directions for the mass , we obtain X = cos = X = − sin =
397 where is the normal force exerted by the rod. From kinematics, we know the accelerations in the radial and transverse directions are given by 2
= ̈ − ̇ = ̈ + 2̇̇ Then,
´ ³ 2 cos = ̈ − ̇ ³ ´ − sin = ̈ + 2̇̇
For the rod, sum the moment about the pivot point to obtain X = − − sin = ̈ 2 1 where = 2 . The constraint force, is eliminated and we have two equations of motion 3 given by 2
̈ − ̇ − cos = 0 ¶ µ ¡ ¢ 2 sin = 0 + ̈ + 2̇̇ + + 2 2
These are nonlinear equations where the terms, ̇ 2 ̈ 2̇̇ and sin are nonlinear. It should be noted that the equations are valid while the mass is sliding along the rod. They are not valid during collision. After the collision, a new set of initial conditions must be given. (b) Lagrange’s Equation - The system energies are µ ³ ´2 ¶ 1 1 2 2 = ̇ + ̇ + ̇ 2 2 ¶ µ µ ³ ´2 ¶ 1 1 2 2 1 2 ̇ + ̇ + ̇ = 2 3 2 µ ¶ = − + cos ; 2 = − where the datum for the potential energy is at Now we are ready to apply Lagrange’s equation for the two generalized coordinates and : µ ¶ − = 0 ̇ ¶ µ ¶ µµ ¶ 1 2 2 = ̇ + ̇ ̇ 3 ³ ´ 1 2 2 = ̈ + ̈ + 2̇̇ 3µ ¶ = − + sin ; 2
398
CHAPTER 6 MULTI DOF VIBRATION
µ
¶ − = 0 ̇ µ ¶ = (̇) = ̈ ̇ 2 = ̇ + cos
Therefore, the EOM are: ¶ µ ¶ µ 1 2 2 + ̈ + 2̇̇ + + sin = 0 3 2 2
̈ − ̇ − cos = 0 which is identical to the EOM obtained via Newton’s law. 2 The EOM can be linearized if physical arguments can be made that ̇̇ and ̇ can be ignored. In general this cannot be done unless ̇ is very small. If = at all times, ̇ = 0 and ̈ = 0 and the equations of motion become ¶ µ ¶ µ 1 2 2 + ̈ + + sin = 0 3 2 which is identical to the pendulum equation. Note that for = another normal constraint force is needed in the radial direction to enforce this condition.
399
12. Linearize and solve the equations of motion derived in Problem 11. State all assumptions necessary for the linearization and explain them physically. Solution: The equations of motion from the previous problem are: ¶ µ ¶ µ 1 2 2 + ̈ + 2̇̇ + + sin = 0 3 2 2
̈ − ̇ − cos = 0 2
These equations can be linearized if physical arguments can be made that ̇̇ and ̇ can be ignored. In general this cannot be done unless ̇ is very small. Some possible linearization justifications are 2 that the rod oscillates very slowly, and therefore ̇ is a small quantity. Then ̇ is much smaller. Similarly, if mass moves very slowly, then the product ̇̇ is an ignorable quantity. In addition we need to assume that is small. If so, this could lend credence to the possibilities of small velocities. The linearized EOM are then: µ ¶ µ ¶ 1 2 2 + 0 ̈ + + 0 = 0 3 2 ̈ − = 0 where 0 is the starting location for the mass. The second EOM describes a free falling mass. It is independent of and can be integrated twice to yield: = 2 2 + 1 + 2 This can be substituted into the first EOM. Physically, the solution for () grows without bounds, which is a problem. Further, when () is substituted into the equation for () we obtain an EOM with time-dependent coefficients. We have mentioned such equations in earlier chapters (when discussing stability) that such equations may have responses that grow without bounds, depending on the system parameter values.
400
CHAPTER 6 MULTI DOF VIBRATION
13. For the system shown in Figure 6.13, solve for the responses where 1 = 2 = 3 = , 1 = 2 = , and where the initial conditions are 1 (0) = 1 2 (0) = 1, and zero initial velocities, all in appropriate units. Plot the modes, identifying the nodes. Plot 12 vs. and . Solution: Response The matrix EOM become: ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎡ ⎬ ⎨ ⎨ ⎬ ⎨0⎬ 1 + 2 −2 ̈1 0 1 ⎦ ⎦ ⎣ 1 +⎣ = 0 2 ⎩ ̈2 ⎭ −2 2 + 3 ⎩ 2 ⎭ ⎩ 0 ⎭ becomes, after substitution of the above parameters: ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎨ ⎬ 0 ̈1 2 − ⎨ 1 ⎬ ⎨ 0 ⎬ ⎣ ⎦ ⎦ +⎣ = ⎩ ⎭ 0 − 2 ⎩ ⎭ ⎩ 0 ⎭ ̈ 2
2
Assume the following responses: 1 () = 1 exp() and 2 () = 2 exp() Differentiate and substitute appropriately to find the matrix-vector equation for the characteristic matrix: ⎫ ⎧ ⎫ ⎡ ⎤⎧ ⎨ ⎬ ⎨0⎬ 2 − 2 − 1 ⎣ ⎦ = − 2 − 2 ⎩ ⎭ ⎩ 0 ⎭ 2
2
The determinant of the matrix yields the characteristic equation: (2 − 2 ) −()2 = 0 which can be solved for the eigenvalues (frequencies squared): 212 = 3 Substituting these in turn into the last matrix equation allows us to find the eigenvectors (mode shapes): 21 = 100 11
22 = −100 12
Thus, the responses are given by: ⎫ ⎫ ⎧ ⎧ Ãr ! ⎨ 100 ⎬ ⎨ () ⎬ 1 cos = 1 − 1 ⎩ 100 ⎭ ⎩ () ⎭ 2 ⎧ ⎫ Ãr ! ⎨ 100 ⎬ 3 cos + 2 − 2 ⎩ −100 ⎭
(1)
The modes are {1 1} and {1 − 1} The first mode has no node, but the second mode has a node midway between mass one and two. We can plot the frequencies as functions of or
401 but it may more appropriate to plot frequency as a function of using appropriate units. To satisfy the given initial conditions, substitute the initial displacements and velocities in turn into the solution and its derivative to find the 4 equations needed to evaluate the four arbitrary constants of integration: 1 2 1 2 We find all constants equal zero except: 1 = 1 Only the first mode is present. This is because the initial position resembles the first mode. Mode shapes The modes are given by Equation (1) and plotted next.
Natural frequencies p p √ The natural frequencies are given by 3 They both are proportional to and 12 = √ inversely proportional to Figures next show the natural frequencies as functions of and while mass is kept constant at = 1 kg for the first plot and the stiffness is kept constant at = 1 N/m for the second plot
402
CHAPTER 6 MULTI DOF VIBRATION
14. Solve the general response of the system given in Problem 1: ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎬ ⎨ ̈1 0 (1 + 2 ) −2 ⎨ 1 ⎬ ⎣ 1 ⎦ ⎦ +⎣ = 0 ⎩ ⎭ 0 ⎩ ̈ ⎭ − 2
2
2
2
2
Solution: We will begin to solve this matrix EOM in general, and then we will assume the following numerical values for the parameters: 1 = 2 = 1 and 1 = 2 = 1 Of course, it would be an interesting exercise to try a variety of parameter values to gain a better feel for how the system is affected by different ratios of parameter values, that is 1 2 and 1 2 The matrix EOM become: ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎨ ̈ ⎬ ⎨ ⎬ ⎨0⎬ 1 0 + − 1 1 2 2 1 ⎣ ⎣ ⎦ ⎦ + = ⎩ ⎭ ⎩0⎭ 0 ⎩ ̈ ⎭ − 2
2
2
2
2
Assume that the responses are harmonic so that we can write 1 () = 1 exp() and 2 () = 2 exp() Upon substituting the assumed solutions, we obtain ⎧ ⎫ ⎫ ⎤⎧ ⎡ 2 ⎨0⎬ ⎬ ⎨ 1 + 2 − 1 −2 ⎦ ⎣ 1 exp() = ⎩0⎭ − − 2 ⎩ ⎭ 2
2
2
2
The characteristic equation (determinant set to zero) is given by ¡ ¢¡ ¢ 1 + 2 − 1 2 2 − 2 2 − 22 = 0 1 2 4 − [(1 + 2 ) 2 + 2 1 ] 2 + (1 + 2 ) 2 − 22 = 0
whose solutions are given by 21 = 22 =
q (1 + 2 ) 2 + 2 1 − [(1 + 2 ) 2 + 2 1 ]2 − 41 2 [(1 + 2 ) 2 − 22 ] 21 2 q (1 + 2 ) 2 + 2 1 + [(1 + 2 ) 2 + 2 1 ]2 − 41 2 [(1 + 2 ) 2 − 22 ] 21 2
(1)
(2)
If the mass and the stiffness matrices are positive definite, it is guaranteed that the natural frequencies are real. The natural frequencies are substituted into ([] − 2 []) {} = {0} to find {} Substituting 2 with 21 we obtain ⎫ ⎧ ⎫ ⎡ ⎤⎧ ⎨ ⎬ ⎨0⎬ 1 + 2 − 1 21 −2 11 ⎣ ⎦ = − − 2 ⎩ ⎭ ⎩ 0 ⎭ 2
2
2
1
21
so that the modal ratio is given by
2 11 = 21 1 + 2 − 1 21
403 where 21 is given by Equation (1). Similarly, the modal ratio that corresponds to the second natural frequency is given by 12 2 = 22 1 + 2 − 1 22
where 22 is given by Equation (2). Let us now solve a specific example. After substituting 1 = 2 = 1 and 1 = 2 = 1 the matrix EOM becomes ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎨ ⎬ ̈1 1 0 2 −1 ⎨ 1 ⎬ ⎨ 0 ⎬ ⎣ ⎦ ⎣ ⎦ + = ⎩ ⎭ ⎩ ⎭ ⎩0⎭ 0 1 ̈ −1 1 2
2
Given initial conditions, we can solve for the four arbitrary constants of integration: 1 2 1 2 Assume the following responses: 1 () = 1 exp() and 2 () = 2 exp() Differentiate and substitute appropriately to find the matrix equation for the characteristic matrix: ⎫ ⎧ ⎫ ⎡ ⎤⎧ 2 ⎨ 1 ⎬ ⎨ 0 ⎬ 2 − −1 ⎣ ⎦ = −1 1 − 2 ⎩ ⎭ ⎩ 0 ⎭ 2
The determinant of the matrix yields the characteristic equation: ¡ ¢¡ ¢ 2 − 2 1 − 2 − 1 = 0
which can be solved for the eigenvalues (frequencies squared): 212 = 0382 2617 Substituting these in turn into the last matrix equation allows us to find the eigenvectors (mode shapes): 21 = 1618 11
22 = −0617 12
Thus, the responses are given by: ⎫ ⎫ ⎧ ⎧ ⎨ 100 ⎬ ⎨ () ⎬ 1 cos (0618 − 1 ) = 1 ⎩ 1618 ⎭ ⎩ () ⎭ 2 ⎧ ⎫ ⎨ 100 ⎬ cos (1618 − 2 ) + 2 ⎩ −0618 ⎭
To satisfy the given initial conditions, substitute the initial displacements and velocities in turn into the solution and its derivative to find the 4 equations needed to evaluate the four arbitrary constants of integration: 1 2 1 2
404
CHAPTER 6 MULTI DOF VIBRATION
15. Solve the governing equation of Problem 2.
Solution: The linearized matrix EOM is given by [] {̈ ()} + [] { ()} = {0} where the mass matrix and the stiffness matrix is given by (from the solution of Problem 2) ⎡
⎡
⎤
1 ( + ) ⎢
0 0 ⎢ 1 ⎥ ⎢ ⎢ ⎥ [] = ⎢ 0 2 0 ⎥ [] = ⎢ ⎣ ⎣ ⎦ 0 0 3
− 1 0
− 1
0
− 1
1
⎤ ⎥
⎥ 1 ( + ) − 1 ⎥ ⎦
where 1 (1 + 2 + 3 ) 2 = (2 + 3 ) 3 = 3 =
After substituting { ()} = {} exp () we obtain an eigenvalue problem, ¡ ¢ [] − 2 [] {} = {0}
(1)
whose characteristic equation is given by
¯ ¯ ¯[] − 2 []¯ = 0
For simplicity, let us use to refer to a specific element in the stiffness matrix. For example, ⎡
11 12
0
⎤
⎢ ⎥ ⎢ ⎥ [] = ⎢ 12 22 23 ⎥ ⎣ ⎦ 0 23 −33 The characteristic equation is then, 0 = 1 2 3 6 + (2 3 11 + 1 3 22 − 1 2 23 ) 4 ¡ ¢ 2 2 + −3 11 22 + 1 22 23 + 2 11 23 + 1 23 2 2 −11 22 23 + 12 23 − 11 23
(2)
405 The three positive roots of the characteristic equation are the natural frequencies. Each natural frequencies are substituted back into the eigenvalue problem in Equation (1) to find the modal ratios. For instance, Equation (1) with 2 substituted with 2 is given by ¢ ¡ [] − 2 [] {} = {0} ⎫ ⎡ ⎤ ⎡ ⎤⎧ ⎪ ⎪ ⎪ ⎪ 0 11 − 2 1 12 0 ⎨ 1 ⎪ ⎬ ⎢ ⎥ ⎢ ⎥⎪ ⎢ ⎥ ⎢ ⎥ = ⎢ 0 ⎥ ⎢ ⎥ 2 12 22 − 2 2 23 ⎪ ⎪ ⎣ ⎦ ⎣ ⎦⎪ ⎪ ⎪ ⎪ 0 0 − − 2 ⎩ ⎭ 23
33
3
3
Using the first and the third equations, the modal ratios are given by
1 12 2 −33 − 2 3 =− and = − 2 11 − 2 1 3 23 If we arbitrarily let 1 = 1 the eigenvector is given by ⎫ ⎧ ⎫ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎪ ⎬ ⎨ 1 ⎪ 11 −2 1 − 12 2 = ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2 ⎪ 11 − 1 ⎪ 23 ⎩ ⎭ ⎪ ⎭ ⎩ ⎪ 2 3
−33 − 3
12
where 2 is the lowest root of Equation (2).
As a numerical example, we set all length and mass in the flexibility matrix of Problem 2 such that 123 = and 123 = , giving the flexibility matrix: ⎡ ⎤ 1 1 1
3 3 3 ⎥ ⎢ ⎢1 5 5 ⎥ [ ] = ⎢ ⎥ ⎣ 3 6 6 ⎦ 1 5 11 3 6 6
In order to obtain the equations of motion for the system, we need to invert the flexibility matrix, which gives the stiffness matrix. The result is: ⎤ ⎡ 5 −2 0 ⎥ ⎢ ⎥ ⎢ [] = ⎢ −2 3 −1 ⎥ ⎣ ⎦ 0 −1 1
The equations of motion are of the form []{̈} + []{} = 0 Since we have previously assumed all three mass is , our equations of motion are ⎧ ⎫ ⎤⎧ ⎫ ⎡ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ ⎪ ̈ 5 −2 0 ⎪ ⎨ ⎪ ⎨ 1⎪ ⎥⎪ ⎬ ⎢ ⎬ ⎥ ⎢ ̈2 + ⎢ −2 3 −1 ⎥ 2 = 0 ⎪ ⎪ ⎪ ⎦⎪ ⎣ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎩ ̈ ⎭ ⎭ 0 −1 1 3 3
406
CHAPTER 6 MULTI DOF VIBRATION
Now, we need only to solve these three coupled equations of motion. We follow the procedure of the direct method, assuming responses of the form exp () The resulting eigenvalue problem is given by ⎤ ⎛⎡ ⎞ 5 −2 0 ⎥ 2 ⎟ ⎜⎢ ⎥ ⎜⎢ ⎟ I⎟ {} = {0} ⎜⎢ −2 3 −1 ⎥ − ⎦ ⎝⎣ ⎠ 0 −1 1
Most graphic calculators can compute eigenvalue problems. In MATLAB, the command is [V D]=eig(k), where k is the given stiffness matrix, V is the matrix whose columns are eigenvectors, and D is the matrix whose diagonals are eigenvalues. p p An eigenvalue p analysis of the stiffness matrix leads to natural frequencies 1 = 0645 2 = 152 3 = 251 The corresponding eigenvectors are ⎫ ⎫ ⎫ ⎧ ⎧ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0215 ⎪ −0505 ⎪ −0836 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎬ ⎬ ⎨ ⎨ ⎨ {}1 = 0493 {}2 = −0683 {}3 = 0539 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎭ ⎭ ⎩ 0843 ⎪ ⎩ 0528 ⎪ ⎩ −0102 ⎪ After evaluating the eigenvectors, we find ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ () ⎪ 3 ⎨ 1 ⎪ ⎬ X = ̄ {} cos ( − ) 2 () ⎪ ⎪ ⎪ ⎪ =1 ⎪ ⎩ () ⎪ ⎭ 3 ⎫ ⎫ ⎧ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 02149 ⎪ −0505 ⎪ r r ⎪ ⎪ ⎪ µ µ ¶ ¶ ⎬ ⎬ ⎨ ⎨ − 1 + ̄2 −0683 cos 152 − 2 = ̄1 04927 cos 0645 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 08433 ⎭ ⎩ 0528 ⎭ ⎫ ⎧ ⎪ ⎪ ⎪ r ⎪ ⎪ −0836 ⎪ µ ¶ ⎬ ⎨ cos 251 +̄3 − 3 0539 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ −0102 ⎭
The constants ̄ and can be found from initial conditions, specifically, the initial displacement and velocity of each degree-of-freedom.
407
16. Linearize the governing equations of Problem 3 shown below and find the natural frequencies and mode shapes for 1 = 2 = and 1 = 2 = . State all assumptions in the linearization process: h i 2 (1 + 2 ) 1 ̈1 + 2 2 ̈2 cos (2 − 1 ) − ̇2 sin (2 − 1 ) + (1 + 2 ) sin 1 = 0 h i 2 2 2 ̈2 + 2 1 ̈1 cos (2 − 1 ) + ̇1 sin (2 − 1 ) + 2 sin 2 = 0
Solution: If we want to simplify these EOM by assuming small motions, then we can make the approximations: 2
cos (2 − 1 ) ' 1
2
2
2
̇2 sin (2 − 1 ) ' ̇2 (2 − 1 ) ≈ 0
̇1 sin (2 − 1 ) ' ̇1 (2 − 1 ) ≈ 0
sin 12 ' 12
The linearized EOM from Problem 3 are: (1 + 2 )12 ̈1 + 2 1 2 ̈2 + (1 + 2 )1 1 = 0 2 22 ̈2 + 2 1 2 ̈1 + 2 2 2 = 0 See the solution to Problem 3 for the nonlinear equations plus the assumptions made for linearization. The EOM in matrix-vector form is ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎫ 2 ⎬ ⎨ ⎨ ⎬ ⎨0⎬ ( + 2 )1 2 1 2 (1 + 2 )1 0 ̈1 1 ⎣ 1 ⎦ ⎦ +⎣ = 0 ⎩ ⎭ ⎩ 0 ⎭ 2 ⎩ ̈ ⎭ 2 1 2
2 2
2
2
2
2
We substitute { ()} = {} exp () to obtain the eigenvalue problem given by ⎫ ⎧ ⎫ ⎛⎡ ⎤ ⎡ ⎤⎞ ⎧ 2 ⎨ ( + 2 )1 0 1 ⎬ ⎨ 0 ⎬ ( + 2 )1 2 1 2 ⎝⎣ 1 ⎦ − 2 ⎣ 1 ⎦⎠ = ⎩ ⎭ ⎩0⎭ 0 2 2
2
2 1 2
2 2
(1)
2
The determinant of the matrix is given by ¡ ¢ 0 = 4 12 22 1 2 − 1 22 1 2 + 12 2 1 2 + 1 22 22 + 12 2 22 2 ¡ ¢ + 2 1 2 1 2 + 2 1 2 22
The roots are obtained using the quadratic equation. We obtain two positive natural frequencies. Next, we substitute each natural frequency into the eigenvalue problem in Equation (1) to find the mode shapes such that ⎫ ⎧ ⎫ ⎤⎧ ⎡ 2 2 2 ⎨ − 2 1 2 1 ⎬ ⎨ 0 ⎬ ( + 2 )1 − (1 + 2 )1 ⎦ ⎣ 1 = −2 − 2 2 ⎩ ⎭ ⎩ 0 ⎭
2 1 2
2
2
2 2
2
408
CHAPTER 6 MULTI DOF VIBRATION
The mode shape is given by ⎫ ⎧ ⎨ ⎬ 1
⎩ ⎭
=
2
⎧ ⎨1
⎫ ⎬
⎩ (1 +2 )21 −2 (1 +2 )12 ⎭ 2 1 2
Let us now make the following simplifying assumptions: 1 = 2 = and 1 = 2 = Then the governing equations become: ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎨ ⎬ 2 ̈1 2 0 ⎨ 1 ⎬ ⎨ 0 ⎬ ⎣ ⎦ ⎦ +⎣ = ⎩ ⎭ ̈ 0 ⎩ ⎭ ⎩0⎭ 2
2
Assume the following responses: 1 () = 1 exp() and 2 () = 2 exp() Differentiate and substitute appropriately to find the matrix-vector equation for the characteristic matrix: ⎫ ⎧ ⎫ ⎡ ⎤⎧ ⎨ ⎬ ⎨0⎬ 2 − 2 2 − 2 1 ⎣ ⎦ = − 2 − 2 ⎩ ⎭ ⎩ 0 ⎭ 2
The determinant of the matrix yields the characteristic equation: ¡ ¢¡ ¢ ¡ ¢2 2 − 2 2 − 2 − 2 = 0
which can be solved for the eigenvalues (frequencies squared): r r 1 = 1848 1 = 0765
Substituting these in turn into the last matrix equation allows us to find the eigenvectors (mode shapes): √ 22 21 √ = 2 = − 2 11 12 Thus, the responses are given by: ⎧ ⎧ ⎫ ⎫ r µ ¶ ⎨ () ⎬ ⎨ 100 ⎬ 1 = 1 √ cos 0765 − 1 ⎩ () ⎭ ⎩ 2 ⎭ 2 ⎧ ⎫ r µ ¶ ⎨ 100 ⎬ cos 1 848 − 2 + 2 ⎩ −√2 ⎭
To satisfy the given initial conditions, substitute the initial displacements and velocities in turn into the solution and its derivative to find the 4 equations needed to evaluate the four arbitrary constants of integration: 1 2 1 2
409
17. For the two degree-of-freedom system shown in Figure 6.71, determine the natural frequencies of oscillation about equilibrium and the respective modal ratios. Assume the following parameter values: 1 = 10 kg, 2 = 14 kg, 1 = 4 N/m, 2 = 5 N/m, and 3 = 6 N/m.
Figure 6.71: Two degree-of-freedom system in free vibration.
Solution: The equation of motion in matrix-vector form is given as ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎨ 1 ⎬ 1 0 ⎨ ̈1 ⎬ 1 + 2 −2 ⎣ ⎣ ⎦ ⎦ + = 0 0 ⎩ ̈ ⎭ − + ⎩ ⎭ 2
2
2
2
3
2
where 1 and 2 are measured from their respective static equilibrium positions. Assume ⎧ ⎫ ⎧ ⎫ ⎨ ⎬ ⎨ ⎬ 1 1 = exp() ⎩ ⎭ ⎩ ⎭ 2
2
and substitute this expression and its second derivative into the governing equations. Eliminating the exponential and combining expressions leads to the eigenvalue problem to be solved, ⎧ ⎫ ⎨ ¢ 1 ⎬ ¡ = {0} (1) [] − 2 [] ⎩ ⎭ 2
The characteristic equation is given by setting ¢ ¡ det [] − 2 [] = 0
410
CHAPTER 6 MULTI DOF VIBRATION
When the values are substituted in, the characteristic equation is given by ⎛⎡ ⎤ ⎡ ⎤⎞ 9 −5 10 0 ⎦ − 2 ⎣ ⎦⎠ = 0 det ⎝⎣ −5 11 0 14 140 4 − 2362 + 74 = 0
The eigenvalues are 21 = 04164 and 22 = 12693. The natural frequencies are 1 = 0645 rad/s and 2 = 113 rad/s The eigenvectors are obtained from Equation (1) with respective values of 2 substituted in. Equation (1) is given by ⎫ ⎛⎡ ⎤⎡ ⎤⎞ ⎧ 2 ⎨ 9 − 10 −5 10 0 1 ⎬ ⎝⎣ ⎦⎣ ⎦⎠ = {0} ⎩ ⎭ −5 11 − 14 2 0 14 2
and the eigenvectors are ⎧ ⎫ ⎫ ⎧ ⎫ ⎧ ⎫ ⎧ ⎨1 ⎬ ⎬ ⎨1 ⎬ ⎨1 ⎬ ⎨1 and = = ⎩ ⎭ ⎩ −07386 ⎭ ⎩ ⎭ ⎩ 09671 ⎭ 21 11 22 12
411
18. In Problem 6, using the governing equations of motion given below, assume a harmonic base motion and find the undamped response: ´ ³ 2 ( + ) ̈ + ̈ cos − ̇ sin − ( − ) − (̇ − ̇) = 0 ¡ ¢ + 2 ̈ + ̈ cos + − sin = 0
Solution: The linearized coupled EOM are: ( + )̈ + ̈ + ̇ + = + ̇ ¡ 2 ¢ ̈ + + ̈ + ( − ) = 0
In matrix-vector form this becomes: ⎤⎧ ⎫ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎡ ⎨⎬ ⎨ ̈ ⎬ 0 0 ⎨ ̇ ⎬ + ⎣ ⎦ ⎣ ⎦ ⎦ ⎣ + + 0 − ⎩ ⎭ 0 0 ⎩ ̇ ⎭ 2 + ⎩ ̈ ⎭ ⎡ ⎤ ⎡ ⎤ = ⎣ ⎦ ̇ + ⎣ ⎦ 0 0
For undamped system, we set = 0 The harmonic base motion is given by = sin The matrix EOM then becomes: ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎤⎧ ⎫ ⎡ ⎨ ⎬ ⎨ sin ⎬ ⎨ ̈ ⎬ + 0 ⎣ ⎦ ⎦ = +⎣ ⎭ ⎩ ⎭ 2 + 0 − ⎩ ⎭ ⎩ 0 ̈
Since this is a forced vibration problem, we initially find the homogeneous solution (by setting the force to zero) to find the two natural frequencies of vibration, and the two modes of vibration. Then we solve for the particular solution. The complete solution is the sum of the homogeneous and particular solutions. The initial conditions are satisfied using the complete solution. Homogeneous Solution
The homogeneous solution is of the form, ⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎨ ⎬ ⎨ ⎬ ⎨ () ⎬ 11 12 cos ( 1 − 1 ) + 2 cos ( 2 − 2 ) = 1 ⎩ ⎭ ⎩ ⎭ ⎩ () ⎭ 21
(1)
22
where 21 and 22 are the eigenvalues of the eigenvalue problem given by ⎫ ⎧ ⎫ ⎛⎡ ⎤ ⎡ ⎤⎞ ⎧ ⎨ 0 + 1 ⎬ ⎨ 0 ⎬ ⎝⎣ ⎦ − 2 ⎣ ⎦⎠ = ⎩ ⎭ ⎩0⎭ 0 − 2 +
2
(2)
412
CHAPTER 6 MULTI DOF VIBRATION
and {11 21 } and {12 22 } are the eigenvectors of the same eigenvalue problem. The natural frequencies are obtained by solving the characteristic equation given by ⎛⎡ ⎤ ⎡ ⎤⎞ 0 + ⎦ − 2 ⎣ ⎦⎠ = 0 det ⎝⎣ 2 0 − + ¡ ¢ ¢ ¡ 0 + 2 + 4 − + − + − 2 + 2 2 + ( − ) =(3) from which we obtain two positive natural frequencies. We order them in ascending order. After substituting the natural frequencies into the eigenvalue problem in Equation (2), we can find the eigenvectors given by ⎫ ⎧ ⎫ ⎧ ⎬ ⎨ ⎬ ⎨1 1 = −2 (+) for = 1 and 2. (4) ⎭ ⎩ ⎭ ⎩ 2 2
The four constants, 1 2 1 and 2 are obtained by applying the initial conditions to the total solution. Particular Solution For the particular solution, we specifically look for the solution that has the form of the input. Assume () = sin () = sin
(5)
Differentiating and substituting these into the EOM results in: ⎫ ⎤⎧ ⎫ ⎧ ⎡ 2 2 ⎨ ⎨ ⎬ ⎬ − − ( + ) ⎦ ⎣ = − 2 ( − ) − (2 + )2 ⎩ ⎭ ⎩ 0 ⎭
To solve for the non-trivial values of and we need to invert the matrix. We have called this matrix [] The inverse of [] equals the ratio of its adjoint and its determinant. Therefore, ⎧ ⎫ ⎫ ⎧ ⎨ ⎬ ⎨ adj[] ⎬ = ⎩⎭ det[] ⎩ 0 ⎭ ⎫ ⎤⎧ ⎡ 2 2 2 ⎨ ⎬ 1 ⎣ ( − ) − ( + ) ⎦ = det[] 2 − ( + ) 2 ⎩ 0 ⎭ Therefore,
¤ 1 £ ( − ) − (2 + ) 2 det[] ¤ 1 £ 2 = det[] =
(6)
413 where
£ ¤£ ¤ £ ¤2 det[] = − ( + )2 ( − ) − (2 + )2 − −2
Then substitute and into the following equation for the forced response: () = sin () = sin Total Solution
The total solution is given by (Equation (1) and (5)) ⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎨ () ⎬ ⎨ ⎬ ⎨ ⎬ 11 12 = 1 cos ( 1 − 1 ) + 2 cos ( 2 − 2 ) ⎩ () ⎭ ⎩ ⎭ ⎩ ⎭ 21 22 ⎧ ⎫ ⎨ ⎬ + sin ⎩⎭
where 1 and 2 are the roots of Equation (3), the eigenvectors are given in Equation (4), and and are given in Equation (6). Now, we apply the initial conditions to find the constants of integration, 1 2 1 and 2
414
CHAPTER 6 MULTI DOF VIBRATION
19. Find the response of the freely vibrating system of Problem 7, ¶ µ ̈ + ̈ + (1 + 2 ) + (3 + 4 )( + ) = 0 2 ̈ + ( + )̈ + (3 + 4 )( + ) + + = 0 2 2 2
Solution: The linearized EOM in matrix-vector form is: ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎡ ⎨ ⎬ ⎨⎬ ⎨0⎬ (3 + 4 ) ̈ 1 + 2 + 3 + 4 2 ⎦ ⎦ ⎣ +⎣ ¡ ¢ ⎩ ⎭ = ⎩ ⎭ ⎩ ⎭ 2 0 ( + ) ̈ ( + ) + + 2
3
4
3
4
2
Assume a free vibration response that is harmonic: () = 1 exp() and () = 2 exp() Differentiate and substitute appropriately to find the matrix equation for the characteristic matrix: ⎫ ⎧ ⎫ ⎡ ⎤⎧ 2 2 ⎨ (3 + 4 ) − 2 + 2 + 3 + 4 − 1 ⎬ ⎨ 0 ⎬ ⎣ 1 ⎦ = ¡ ¢ ( + ) − 2 2 ( + )2 + + − 2 ⎩ ⎭ ⎩ 0 ⎭ 3
4
3
4
2
2
The natural frequencies are obtained by setting the determinant of the above characteristic matrix to zero: ¶ ¶ µ µ ¡ ¢ 2 2 2 1 + 2 + 3 + 4 − (3 + 4 ) + + − 2 ¡ ¢2 2 − (3 + 4 ) − 2 = 0
which is a quadratic equation in 2 : 4 + 2 + = 0 where the constants and are functions of the system parameters resulting from carrying out the determinant. Solving for 212 these can be substituted into the characteristic matrix in order to obtain the modes: ¡ ¢ ¢ ¡ 1 + 2 + 3 + 4 − 21 1 + (3 + 4 ) − 21 2 2 = 0 This results in an expression for the first modal ratio (2 1 )1 Similarly for the second modal ratio: (2 1 )2 : ¡ ¢ ¢ ¡ 1 + 2 + 3 + 4 − 22 1 + (3 + 4 ) − 22 2 2 = 0 Once the modes and frequencies are known, the free vibration response is then: ⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎨⎬ ⎨1 ⎬ ⎨1 ⎬ = 1 cos( 1 − 1 ) + 2 cos(2 − 2 ) ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ 2
1
1
2
1
2
where the integration constants 1 2 1 and 2 are found using the initial conditions (2 initial displacements and 2 initial velocities).
415
20. Suppose the system of Equation 6.38, ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎬ ⎬ ⎨ ⎨ ⎨ 1 () ⎬ ̈1 1 0 11 12 ⎣ 1 ⎦ ⎦ +⎣ = ⎩ ⎭ ⎩ () ⎭ 0 ⎩ ̈ ⎭ 2
2
21
22
2
is forced by the vector {1 () 2 ()} = {1 cos 1 response using the direct method.
2
2 cos 2 } . Solve for the general
Solution: We solve for the forced response only. In general, the forced response is added to the free response, and the sum of the two are used to satisfy the initial conditions. Start with ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎨ ̈ ⎬ ⎨ ⎬ ⎨ cos ⎬ 1 0 1 11 12 1 1 1 ⎣ ⎦ ⎦ +⎣ = ⎩ ⎭ ⎩ cos ⎭ 0 ⎩ ̈ ⎭ 2
2
12
22
2
2
2
The easiest analytical approach is to solve for one forcing at a time, and then, by superposition, add the two solutions to obtain the complete response to both loads. Let us solve for the response to the first loading: 1 cos 1 Assume the following harmonic response: 1 () = cos 1 2 () = cos 1
Substitute these expressions and their second derivatives into the EOM and combine like terms, canceling the common cosine factors, to find: ⎡ ⎤⎧ ⎫ ⎧ ⎫ 2 ⎨⎬ ⎨ ⎬ 12 − 1 1 1 ⎣ 11 ⎦ = ⎩ ⎭ ⎩ ⎭ 2 − 0 12
22
2
1
To solve for the non-trivial values of and we need to invert the matrix. We have called this matrix [] The inverse of [] equals the ratio of its adjoint and its determinant. Therefore, ⎧ ⎫ ⎨⎬ ⎩⎭
=
⎡
22 − 2 21
adj[1 ] 1 ⎣ = det[1 ] det[1 ] −12
⎤⎧ ⎫ ⎨ ⎬ −12 1 ⎦ 11 − 1 21 ⎩ 0 ⎭
We can easily find the determinant of [] : det[] = (11 − 1 21 )(22 − 2 21 ) − (−12 )2 We can see from above that: ¢ 1 ¡ 22 − 2 21 1 det[1 ] 1 12 1 = − det[1 ] =
416
CHAPTER 6 MULTI DOF VIBRATION
Therefore, the response to the first force is ¢ 1 ¡ 22 − 2 21 1 cos 1 det[1 ] 1 12 1 cos 1 2 () = − det[1 ] 1 () =
Now we proceed to the second part of this problem: the response to the loading 2 cos 2 Assume responses: 1 () = cos 2 2 () = cos 2 Following the previous procedure, we obtain ⎤⎧ ⎫ ⎧ ⎫ ⎡ 2 ⎨ ⎬ ⎨0 ⎬ 12 − 1 2 ⎦ ⎣ 11 = 12 22 − 2 22 ⎩ ⎭ ⎩ 2 ⎭
To solve for the non-trivial values of and we need to invert the matrix. We have called this matrix [] The inverse of [] equals the ratio of its adjoint and its determinant. Therefore, ⎧ ⎫ ⎤⎧ ⎫ ⎡ 2 ⎨ ⎬ [ ] ⎨0 ⎬ 1 ⎣ 22 − 2 2 −12 2 ⎦ = = ⎩ ⎭ det[2 ] det[2 ] − − 2 ⎩ ⎭ 12
11
1
2
2
We can see from above that:
1 12 2 det[2 ] 1 (11 − 1 22 )2 = det[2 ] = −
Therefore, the response to the second force is: 1 12 2 cos 2 det[2 ] 1 (11 − 1 22 )2 cos 2 2 () = det[2 ] 1 () = −
The complete solution is the sum: ¢ 1 ¡ 1 22 − 2 21 1 cos 1 − 12 2 cos 2 det[2 ] det[2 ] 1 1 12 1 cos 1 + (11 − 1 22 )2 cos 2 2 () = − det[1 ] det[2 ]
1 () =
417
21. For the undamped vibration absorber problem governed by Equations 6.45 and 6.46, 1 ̈1 + (1 + 2 )1 − 2 2 = 1 sin 2 ̈2 − 2 1 + 2 2 = 0 assume that the absorber mass is also forced by 2 sin . Solve for the responses 1 () and 2 (). Obtain () and () as a function of . Solution: The solution procedure here is the same as the first part of the previous solution, except that we will work with the ratios =
2 1
2 =
1 1
2 =
2 2
The EOMs in matrix-vector form is ⎫ ⎤⎧ ⎫ ⎧ ⎡ ⎤⎧ ⎫ ⎡ 2 2 2 ⎨ ⎬ ⎬ ⎨ ⎨ + − 1 1 ⎬ 1 1 0 ̈1 ⎦ ⎣ ⎦ +⎣ = sin ⎩ ⎭ ⎩ ⎭ − 2 2 0 1 ⎩ ̈ ⎭ 2
2
2
2
Assume a solution of the form: 1 () = sin 2 () = sin Differentiate these expressions twice and substitute into the EOM to find ⎫ ⎤⎧ ⎫ ⎧ ⎡ 2 2 2 2 ⎨ ⎨ ⎬ 1 1 ⎬ ( + ) − − ⎦ ⎣ = − 2 2 − 2 ⎩ ⎭ ⎩ ⎭
2
2
To solve for the values of and we need to invert the matrix. We have called this matrix [] The inverse of [] equals the ratio of its adjoint and its determinant. Therefore, ⎧ ⎫ ⎫ ⎤⎧ ⎡ 2 2 2 ⎨ ⎬ adj[] ⎬ ⎨ − 1 ⎣ 1 1 ⎦ = = ⎩ ⎭ det[] det[] 2 ( 2 + 2 ) − 2 ⎩ ⎭
⎧ ⎫ ⎨⎬
sin
The forced response is then
⎧ ⎫ ⎨ ⎬ 1
⎩ ⎭ 2
The amplitudes, () and () are
=
⎩⎭
¾ ½ ¡ 2 ¢ 1 2 1 2 2 () = − + det[] 1 2 ¾ ½ ¡ 2 ¢ 1 2 1 2 2 2 + ( + ) − () = det[] 1 2 ¡ 2 ¢¡ 2 ¢ 2 2 2 4 det[] = ( + ) − − −
2
2
418
CHAPTER 6 MULTI DOF VIBRATION
22. A cargo ship containing many boxes is at sea and experiences rough motion. Consider two masses in the cargo hold that are in close proximity when mass 1 is subjected to a force = cos( + ) in line with the springs, as shown in Figure 6.72. The two masses are connected to each other via spring 2 and mass 1 is connected to the ship’s hull via a spring and a damper. Derive the equation for the relative motion of the two masses.
Figure 6.72: A two degree-of-freedom model of two masses on a cargo ship.
Solution: Although the system describes a system in a cargo ship experiencing rough motion, the problem is not posed as a base-excited problem, but instead a harmonically forced problem. We draw free-body diagrams for each mass and sum the forces in the horizontal direction. Note that 1 () and 2 () are total displacements.
The corresponding EOMs are 1 ̈1 + 1 ̇1 + (1 + 2 ) 1 − 2 2 = 2 ̈2 − 2 1 + 2 2 = 0 We are interested in the distance between the two masses. Let us define = 2 − 1 The actual clearance between the masses is + where is the undeformed length of the spring 2 Replace 2 with + 1 in the EOMs. Then, 1 ̈1 + 1 ̇1 + 1 1 − 2 = 2 ̈1 + 2 ̈ + 2 = 0
419 The relative motion of the two masses is given by (). The above equations are the equations of motion in terms of 1 () and () which must be solved simultaneously. Another approach would be to solve for the system response 1 2 and then evaluate = 2 − 1 It is possible to find a single differential equation in terms of only. For a linear system, take the Laplace transforms of the two differential equations, solve for () algebraically and convert back to the differential equation using inverse Laplace transform. Take the Laplace transform of the two differential equations to find 1 2 1 () + 1 1 () + 1 1 () − 2 () = () 2 2 1 () + 2 2 () + 2 () = 0 Solve for 1 () using the second equation such that 1 () = −
(2 2 + 2 ) 2 2
and substitute 1 () into the first equation to obtain ¡ ¢ (2 2 + 2 ) − 1 2 + 1 + 1 () − 2 () = () 2 2
Clean up by multiplying both sides by 2 to find ¢¡ ¢ ¡ 1 2 + 1 + 1 2 2 + 2 () + 2 2 2 () = − () 2 2 ¡ ¢ 1 2 4 + 2 1 3 + (1 2 + 2 1 + 2 2 ) 2 + 1 2 + 1 2 () = −2 2 () Take the inverse Laplace transform to find 1 2
4 3 2 () 2 + + ( + + ) + = − + 2 1 1 2 2 1 2 2 1 2 1 2 2 4 3 2 2
420
CHAPTER 6 MULTI DOF VIBRATION
23. The vibration characteristics of an airplane are very complex due to its intricate system of structures and substructures. Suppose we consider the much simplified model of Figure 6.73, which is a three degree-of-freedom representation of the fuselage and wings. Derive the equations of motion using both Newton’s second law and Lagrange’s equation. Initially assume arbitrary 1 and 2 . Then consider two simplifications: (a) symmetrical vibration of the wing-body combination, that is, 1 = 2 = , 1 = 2 = , and 1 = 2 = , and (b) small . Solve for the free vibration response after simplifying and linearizing the equations of motion. Discuss your assumptions.
Figure 6.73: A much simplified airplane model.
Solution: Assume that the mass of each wing is represented by a point mass, 1 that is a distance away from the body of the airplane. For the general vibration case, the system has three DOFs: 1 2 and We have the following position vectors if the origin is placed at = 0 directly beneath mass with unit vectors i and j defined as positive to the right and up, respectively. The position vectors are:
r = j r1 = − cos 1 i + ( + sin 1 )j r2 = cos 2 i + ( + sin 2 )j
The velocity vectors are:
v = ̇j v1 = ̇1 sin 1 i + (̇ + ̇1 cos 1 )j v2 = −̇2 sin 2 i + (̇ + ̇2 cos 2 )j
421 For Lagrange’s equation we need the following energies: µ³ ´2 ³ ´2 ¶ 1 1 2 = ̇ + 1 ̇1 sin 1 + ̇ + ̇1 cos 1 2 2 µ³ ´2 ³ ´2 ¶ 1 ̇2 sin 2 + ̇ + ̇2 cos 2 + 2 2 ³ ´2 1 ³ ´2 1 1 2 = ̇ + 1 ̇1 + 2 ̇2 2 2 2 ´ ³ 1 2 + (1 + 2 )̇ + ̇ 1 ̇1 cos 1 + 2 ̇2 cos 2 2
The potential energy comes from gravity and stiffness, and it is given by
1 1 = 1 21 + 2 22 − − 1 ( + 1 sin 1 ) − 2 ( + 1 sin 2 ) 2 2 If we assume that 1 and 2 are in static equilibrium positions when 12 = 0 (or their positions measured from the static equilibrium positions), we can omit the gravitational potential. Then, the potential energy due to stiffness is given by 1 1 = 1 21 + 2 22 2 2 Therefore, the Lagrangian is = − ³ ´2 1 ³ ´2 1 1 1 = ̇ 2 + 1 ̇1 + 2 ̇2 + (1 + 2 )̇ 2 2 2 2 2 ´ 1 ³ 1 +̇ 1 ̇1 cos 1 + 2 ̇2 cos 2 − 1 21 − 2 22 2 2
Lagrange’s equations are given by
¶ µ − = 0 ̇1 1 ¶ µ − = 0 ̇2 2 µ ¶ − = 0 ̇ where
µ
̇1
¶
=
´ ³ 1 2 ̇1 + ̇1 cos 1
= 1 2 ̈1 + 1 (̈ cos 1 − ̇ ̇1 sin 1 )
= −̇1 ̇1 sin 1 − 1 1 1
422
CHAPTER 6 MULTI DOF VIBRATION
and
µ
̇
¶
´´ ³ ³ ̇ + (1 + 2 )̇ + 1 ̇1 cos 1 + 2 ̇2 cos 2 = ̈ + (1 + 2 )̈ ´ ³ 2 2 + 1 ̈1 cos 1 − 1 ̇1 sin 1 + 2 ̈2 cos 2 − 2 ̇2 sin 2 =
= 0
Therefore, the EOMs are given by 1 2 ̈1 + 1 (̈ cos 1 − ̇ ̇1 sin 1 ) + ̇1 ̇1 sin 1 + 1 1 = 0 2 2 ̈2 + 2 (̈ cos 2 − ̇ ̇2 sin 2 ) + ̇2 ̇2 sin 2 + 2 2 = 0 ´ ³ 2 2 ( + 1 + 2 )̈ + 1 ̈1 cos 1 − 1 ̇1 sin 1 + 2 ̈2 cos 2 − 2 ̇2 sin 2 = 0
Note that 1 () = 2 () if the plane is symmetric (1 = 2 and 1 = 2 ) In this case, the number of degrees-of-freedom is two, and only one of the first two equations need to be solved simultaneously with the third. If these equations are to be linearized, we assume small angles so that cos 1 ' 1 and sin 1 ' 1 In addition, we need to assume that nonlinear terms such as the product ̇ ̇1 sin 1 ≈ ̇ ̇1 1 can be ignored. Then, the linearized EOMs are
or
1 2 ̈1 + 1 ̈ + 1 1 = 0 2 2 ̈2 + 2 ̈ + 2 2 = 0 ´ ³ ( + 1 + 2 )̈ + 1 ̈1 + 2 ̈2 = 0
⎤⎧ ⎫ ⎧ ⎫ ⎤⎧ ⎫ ⎡ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ̈1 ⎪ ⎪0⎪ ⎪ ⎪ 0 0 ⎪ 0 1 ⎬ ⎨ ⎪ ⎨ ⎪ ⎨ ⎪ ⎬ ⎢ 1 ⎬ ⎪ ⎢ 1 ⎥⎪ ⎥⎪ ⎢ ⎢ ⎥ ⎥ + = ⎢ ⎢ 0 2 2 ⎥ ⎥ 0 0 2 0 2 2 ̈2 ⎪ ⎪ ⎪ ⎪ ⎣ ⎣ ⎦⎪ ⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎩ ⎩ ⎭ ⎭ 0⎭ 0 0 0 ( + + ) ̈ ⎡
2
1
2
1
2
For a symmetric airplane, the two linearized EOMs are
2 ̈ + ̈ + = 0 ( + 2)̈ + 2̈ = 0 or
⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎨ ⎬ ̈ 0 ⎨⎬ ⎨0⎬ ⎦ ⎦ ⎣ +⎣ = 2 ( + 2) ⎩ ̈ ⎭ 0 0 ⎩⎭ ⎩0⎭ ⎡
2
At first glance, the non-symmetric mass matrix is worrisome. The symmetric mass matrix is recovered by dividing the second equation by 2. Then, the resulting equation is given by ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎫ 2 ⎨ ⎬ ̈ 0 ⎨⎬ ⎨0⎬ ⎣ ⎦ ⎦ +⎣ = (+2) ⎩ ̈ ⎭ 0 0 ⎩⎭ ⎩0⎭ 2
423 Free Vibration The associated eigenvalue problem is ⎛⎡ ⎤ ⎡ ⎤⎞ 2 0 ⎝⎣ ⎦ − 2 ⎣ ⎦⎠ {} = {0} (+2) 0 0 2 ⎤ ⎡ 2 2 2 − − ⎦ {} = {0} ⎣ − 2 − 2 (+2) 2
whose characteristic equation is
¢ ( + 2) ¡ − 4 ()2 = 0 − − 2 2 2 2
We find the eigenvalues are
( + ) 2 The first mode shape that corresponds to 1 = 0 is ⎧ ⎫ ⎨0⎬ {}1 = ⎩1⎭ 2 = 0 and
r
( + ) is 2 ⎫ ⎧ ⎬ ⎨ 1 {}2 = ⎩ − 2 ⎭ + 2
The second mode shape that corresponds to 2 =
The free response is given by ⎧ ⎧ ⎫ ⎧ ⎫ ⎨ ⎨ () ⎬ ⎨ 0 ⎬ ( + ) + = ⎩− ⎩ () ⎭ ⎩ 1 ⎭
⎫ ⎬ 1 2 ⎭ ( cos 2 + sin 2 ) + 2
where the integration constants , and are found using the initial conditions (2 initial displacements and 2 initial velocities). Substitute general initial conditions, (0) = 0 ̇ (0) = 0 (0) = 0 and ̇ (0) = 0 into the solution to find 0 =
r
( + ) 2 2 0 = − + 2 r ( + ) 2 2 0 = − 2 = − + 2 + 2 2
0 = 2 =
424
CHAPTER 6 MULTI DOF VIBRATION
The constants of integration are 2 0 + 2 2 = 0 + 0 + 2 = 0 r ( + ) = 0 2 = 0 +
Finally, the free response is ⎧ ⎫ ⎧ ⎫ ¶ ¶ µ ⎨ () ⎬ ⎨ 0 ⎬µ 2 2 + 0 + 0 = + ⎩ () ⎭ ⎩ 1 ⎭ 0 + 2 0 + 2 ⎞ ⎛ ⎧ ⎫ r ⎨ ⎬⎜ 1 ( + ) ⎟ ⎜0 cos 2 + r 0 sin ⎟ + ⎩ − 2 ⎭ ⎝ 2 ⎠ ( + ) + 2 2
425
24. What will change in our formulation and solution of Problem 23 if the symmetry assumption in the previous problem is removed? Discuss fully. Solution: If symmetry is applied, we are implicitly assuming that the responses 1 and 2 are identical. This can only occur if the loading is completely symmetrical as well. Of course, this would be a special case and not a general assumption. Such symmetry reduces the degree of the structure from 3 to 2. The three EOMs for the non-symmetric case are obtained in the previous problem.
426
CHAPTER 6 MULTI DOF VIBRATION
25. Two uniform rods and of linear density are suspended and connected by a spring as shown in Figure 6.74. The rods are in equilibrium when hanging in the vertical direction. (a) Derive the equations of motion of the system. (b) Find the natural frequencies of oscillation. (c) If bar is rotated due to a static force applied to rod , find the corresponding rotation of and the maximum force in the spring. (d) Given 1 = 05 m, 2 = 08 m, = 3 kN/m, = 4 kg/m, and = 5 degrees, evaluate the previous questions numerically.
Two pendula coupled by a spring.
Solution: (a) Equation of motion Let us use Lagrange’s equation to obtain the equations of motion The kinetic energy is given by 1 1 2 2 = ̇1 + ̇2 2 2 where = 1 12 3 and = 2 22 3 The potential energy is = −1
1 2 1 cos 1 − 2 cos 2 + ∆2 2 2 2
where ∆ is the stretch in the spring, ∆ = 2 sin 2 − 1 sin 1 where is the distance between and and also the length of the unstretched spring. Lagrange’s equations for this system are − + = 0 ̇1 1 1 − + = 0 2 2 2
427 which become ̈1 + 1
1 sin 1 + (2 sin 2 − 1 sin 1 ) (−1 cos 1 ) = 0 2
and
2 sin 2 + (2 sin 2 − 1 sin 1 ) (2 cos 2 ) = 0 2 If we assume small angles, the equations of motion are linearized and given by ̈2 + 2
1 ̈1 + 1 1 − 1 (2 2 − 1 1 ) = 0 2 2 ̈2 + 2 2 + 2 (2 2 − 1 1 ) = 0 2 Collecting terms, ¶ µ 1 2 ̈1 + 1 + 1 1 − 1 2 2 = 0 2 ¶ µ 2 2 ̈2 − 1 2 1 + 2 + 2 2 = 0 2 In matrix-vector form, ⎤⎧ ⎫ ⎧ ⎫ ⎤⎧ ⎫ ⎡ 1 2 ⎬ ⎨ −1 2 ⎥ ⎨ 1 ⎬ ⎨ 0 ⎬ 0 ̈1 ⎢ 1 2 + 1 ⎣ ⎦ +⎣ = ⎦ 2 0 ⎩ ̈2 ⎭ −1 2 2 + 22 ⎩ 2 ⎭ ⎩ 0 ⎭ 2 ⎡
(b) Natural frequencies The natural frequencies are found by setting the determinant of ([] − 2 []) equal to zero, ¯ ¯ ¯ ¯ 1 2 2 ¯ 1 + 1 − ¯ − 1 2 ¯ ¯ 2 ¯ ¯=0 2 ¯ 2 2 −1 2 2 + 2 − ¯¯ ¯ 2 ¶ µ 1 1 4 2 2 − 1 1 + 2 2 + 1 + 2 2 2 2 1 1 1 +( 1 22 1 + 12 2 2 + 2 1 2 1 2 ) = 0 2 2 4 (c) Rotation of and the maximum force in the spring The design of this mechanism requires the knowledge of the force in the spring. This is clearly a static case when a force is applied to to rotate it by If the bars are in dynamic motion, the quantities depend not only on the geometric configuration but also velocities and acceleration. Let’s use the Principle of Virtual Work. It helps to draw FBDs.
428
CHAPTER 6 MULTI DOF VIBRATION
The virtual work is done through gravity, and spring forces. For example, the virtual work done by is (1 sin 1 ) where (1 sin 1 ) is the virtual displacement though which the force did the virtual work. The total virtual work is given by ¶ ¶ µ µ 1 2 cos 1 + 2 cos 2 − ∆∆ 0 = = (1 sin 1 ) + 1 2 2 where ∆ is the stretch in the spring given by ∆ = 2 sin 2 − 1 sin 1 Then, the virtual work is given by ¶ µ 1 0 = = 1 cos 1 − 1 sin 1 + (2 sin 2 − 1 sin 1 ) 1 cos 1 1 2 ¶ µ 2 + −2 sin 2 − (2 sin 2 − 1 sin 1 ) 2 sin 2 2 2 from which we obtain two equations: 1 sin 1 + (2 sin 2 − 1 sin 1 ) 1 cos 1 = 0 2 2 −2 sin 2 − (2 sin 2 − 1 sin 1 ) 2 sin 2 = 0 2
1 cos 1 − 1
It is given 1 = then we can solve for the required force, and the angle of rotation, 2 The spring force is then calculated using = (2 sin 2 − 1 sin 1 ) If we assume small rotations, the static equilibrium equations are reduced to 1 1 − 1 1 + (2 2 − 1 1 ) 1 = 0 2 2 −2 2 − (2 2 − 1 1 ) 2 = 0 2
429 For 1 = 1 2 = 1 + 2 2 2 µ ¶ 1 2 1 2 = 1 + 1 − 1 2 + 2 2 2 ¶ µ 1 2 = 1 − 1 + 2 2 2 (d) Evaluate (a)-(c) given 1 = 05 m, 2 = 08 m, = 3 kN/m, = 4 kg/m, and = 5◦ The masses of the rods are 1 = 1 = 2 kg and 2 = 2 = 32 kg. The mass moments of inertia about the hinged points are = 1 12 3 = 0167 kg·m2 and = 2 22 3 = 0683 kg·m2 (a) The linearized equations of motion are ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎨ ̈ ⎬ ⎨ ⎬ ⎨0⎬ 0167 0 7549 −1200 1 1 ⎣ ⎦ ⎣ ⎦ + = 0 0683 ⎩ ̈ ⎭ −1200 1932 ⎩ ⎭ ⎩ 0 ⎭ 2
(b) The corresponding characteristic equation is
4 − 73632 + 166 × 105 = 0 The natural frequencies are = 476 rad/s 857 rad/s. (c) If the first rod is rotated by 5◦ (0.0873 rad), we find = 1706 N 2 = 00542 rad (3105◦ ) = −087 N (compressed)
2
430
CHAPTER 6 MULTI DOF VIBRATION
26. For the two degree-of-freedom system shown in Figure 6.75 derive the equations of motion. Assume that there is no slip between the cords that connect the masses to the pulley. The moment of inertia of the pulley about its point of rotation is How would the formulation be different if slippage does occur?
Figure 6.75: A system of two masses coupled through a pulley.
Solution: This is a one-DOF system if the cord does not slip. If the cord is allowed to slip, the motion of the wheel and the motion of the block are not related by a simple kinematic relationship, and it becomes a two-DOF system. We will derive the equations of motion for both cases. Let us draw the free-body diagrams for 1 2 and the wheels as shown below.
Summing the forces in the direction for 1 we obtain X → = − + 1 = 1 ̈
(1)
431 Summing the moment about for the wheels, we obtain X © = 2 2 − 1 1 = ̈ Summing the forces in the direction for 2 we obtain X ↓ = −2 + 2 = 2 ̈
(2)
(3)
No Slip If there is no slip, we have two kinematic relationships, = 1 = 2
(4) (5)
Let us use as our independent variable. This is a one-DOF system because we have three coordinates (, , and ) and two constraint equations (Equations (4) and (5)). We should always be in the habit of counting the number of unknowns and the number of available equations. We have 5 unknowns, 1 and 2 and 5 equations from Newton’s second law (Equations (1)-(3)) and two kinematic equations. The goal is to reduce them to a single equation in terms of () Solving for 1 and 2 using Equations (1) and (3), and substituting them into Equation (2), we obtain −2 ̈2 − (1 ̈ + ) 1 = ̈ We replace and ³using the kinematic relations (Equations (4) and (5)) and obtain the equation ´ 2 2 of motion, 2 ̈2 − 1 ̈ + 1 = ̈ Collecting terms, we have
With Slip
¡ ¢ + 2 22 + 1 12 ̈ + 12 = 0
If the slip is allowed, and if it is assumed that the friction between the cord and the pulley is negligible, the motion of the masses are independent of the motion of the pulley. The tension, which is constant along the cable, is not transferred to the pulley. In addition, we may assume that one cable connects both masses and the displacements and are, therefore, equal. (When no slip occurs, we have to assume that the cords that connect the pulley and the masses are cut) The system now has two degrees-of-freedom, () and () Equations (1)-(3) are given by − + = 1 ̈ (2 − 1 ) = ̈ − + 2 = 2 ̈ which give two EOMs, (1 + 2 ) ̈ + = 2 ̈ + (2 − 1 ) (2 ̈ − 2 ) = 0
432
CHAPTER 6 MULTI DOF VIBRATION
27. Complete the solution of Equation 6.55, ⎫ ⎫ ⎧ ⎧ ⎫ ⎧ ⎨ 1 () ⎬ ⎨1⎬ ⎨ 1 ⎬ cos(1 − 1 ) + 2 cos( 2 − 2 ) = 1 ⎩ () ⎭ ⎩1⎭ ⎩ −1 ⎭
(6.55)
2
if the initial conditions are given by
1 (0) = ̇1 (0) = Ω
2 (0) = ̇2 (0) = Ω
Solution: The solutions were given by ⎧ ⎫ ⎧ ⎫ ⎫ ⎧ ⎨1⎬ ⎨1 ⎬ ⎨ () ⎬ 1 cos ( 1 − 1 ) + 2 cos (2 − 2 ) = 1 ⎩1⎭ ⎩ −1 ⎭ ⎩ () ⎭ 2
Substitute the given initial conditions into the response equations and their derivatives to find the 4 equations needed to solve for the constants of integration given by ⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎨1⎬ ⎨1 ⎬ ⎨ ⎬ cos 1 + cos 2 = ⎩1⎭ 1 ⎩ −1 ⎭ 2 ⎩ ⎭ ⎧ ⎫ ⎧ ⎫ ⎫ ⎧ ⎨ Ω ⎬ ⎨ 1 ⎬ ⎨ 2 ⎬ = sin 1 + sin 2 ⎩Ω ⎭ ⎩ ⎭ 1 ⎩ − ⎭ 2
1
2
From the initial displacement conditions, we can solve for 1 cos 1 and 2 cos 2 , and they are given by ⎤ ⎡ ⎡ ⎤−1 ⎡ ⎤ cos 1 1 1 ⎦ = ⎣ ⎣ 1 ⎦ ⎣ ⎦ 2 cos 2 1 −1 ⎤ ⎡ 1 + ⎦ = ⎣ 2 − From the initial velocity conditions, we can solve for 1 sin 1 and 2 sin 2 and they are given by ⎡ ⎤ ⎤−1 ⎡ ⎤ ⎡ sin 1 Ω 2 ⎣ 1 ⎦ = ⎣ 1 ⎦ ⎣ ⎦ 2 sin 2 1 −2 Ω ⎤ ⎡ 1 (Ω + Ω ) 1 ⎦ = ⎣ 2 (Ω − Ω ) 2
433 We can now solve for 1 2 1 and 2 and they are ¶ ¶ µ µ Ω + Ω Ω − Ω 2 = arctan 1 = arctan 1 ( + ) 2 ( − ) + − 1 = 2 = 2 cos 1 2 cos 2
434
CHAPTER 6 MULTI DOF VIBRATION
Problems for Section 6.5 — Free Vibration with Damping: Direct Method 28. Derive Equations 6.69 and 6.70, 1 () = −1 (1 1 cos 1 + 10 2 sin 1 ) + −2 (2 3 cos 2 + 20 4 sin 2 ) 2 () = −1 (1 cos 1 + 2 sin 1 ) + −2 (3 cos 2 + 4 sin 2 )
(6.69) (6.70)
Solution: Begin with Equations 6.65 and 6.66. 1 () = 11 11 11 + 12 12 12 + 21 21 21 + 22 22 22 11
2 () = 11
12
+ 12
21
+ 21
22
+ 22
(6.65) (6.66)
where the roots of the characteristic equation are complex conjugates of each other: 11 = ̄12 and 21 = ̄22 The corresponding modal ratios are also complex conjugates of each other: 11 = ̄12 and 21 = ̄22 The constants of integration, also form complex conjugate pairs such that 11 = ̄12 and 21 = ̄22 Let us write the characteristic roots, modal ratios, and the constants of integration in terms of real numbers such that 11 = −1 + 1 21 = −2 + 2
12 = −1 − 1 22 = −1 − 2
11 = 1 + 1 21 = 2 + 2
12 = 1 − 1 22 = 2 − 2
1 (1 − 2 ) 2 1 21 = (3 − 4 ) 2 11 =
1 (1 + 2 ) 2 1 22 = (3 − 4 ) 2 12 =
where and are real. When these expressions are substituted into Equations 6.65 and 6.66., we obtain Equations 6.69 and 6.70. As an example, consider the first two terms of 2 (), 11 11 + 12 12 When 11 11 12 and 12 are replaced by the expressions above, we have 1 1 (1 − 2 ) (−1 +1 ) + (1 + 2 ) (−1 −1 ) 2 2 ½ ¾ 1 1 −1 1 − 1 = + (1 + 2 ) (1 − 2 ) 2 2
11 11 + 12 12 =
435 Using Euler’s formula, = cos + sin we can rewrite the above expression as ∙ ¸ 1 11 12 −1 1 (1 − 2 ) (cos + sin ) + (1 + 2 ) (cos − sin ) + 12 = 11 2 2 −1 (1 cos + 2 sin ) = which is identical to the first term of 2 () of Equation 6.69. Similarly, the first two terms of 1 () 11 11 11 + 12 12 12 are reduced to −1 {(1 1 + 2 1 ) cos 1 + (−1 1 + 2 1 ) sin 1 } In the text, we use 1 1 = 1 1 + 2 1 and 10 2 = −1 1 + 2 1
436
CHAPTER 6 MULTI DOF VIBRATION
29. Derive Equations 6.71 and 6.72, 1 () = 10 −1 cos( 1 − 01 ) + 20 −2 cos( 2 − 02 ) 2 () = 1 −1 cos( 1 − 1 ) + 2 −2 cos( 2 − 2 )
(6.71) (6.72)
Solution: Begin with Equations 6.69 and 6.70, and re-write terms such as exp (−1 ) 1 cos 1 + 2 sin 1 as 10 exp (−1 ) cos ( 1 − 01 ) Then expand the cosine term on the right-hand side in order to find the expression for 1 and 1 Do this for all the expressions in the response equations. As an example, −1 (1 cos + 2 sin ) = 10 −1 cos( 1 − 01 ) = 10 −1 (cos 1 cos 01 + sin 1 sin 01 ) Equating right-hand side with the left-hand side, we find 1 = 10 cos 01 and 2 = 10 sin 01 Then,
q 2 12 + 22 and 01 = arctan 1 Care should be taken so that 1 is chosen in the correct quadrant. 10 =
437
30. Use the direct method to solve the equations of motion for a system with the following property matrices, ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 5 −2 2 −1 ⎦ [] = ⎣ ⎦ [] = ⎣ ⎦ [] = ⎣ 0 1 −2 5 −1 1 Sketch this system.
Solution: Responses This system is composed of two masses that oscillate horizontally. The first mass is connected on the left end to a wall by a spring and a damper. The second mass is connected on the right side to a wall by a damper only. The two masses are connected to each other by a spring and a damper. The variable notation for the masses, springs, and dampers is where = 1 2 for the springs and = 1 2 3 for the dampers. The EOM is then in general ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎬ ⎬ ⎨ ⎨ 1 + 2 −2 1 + 2 −2 ⎨ 1 ⎬ ̈1 ̇1 0 ⎦ ⎦ ⎦ ⎣ 1 +⎣ +⎣ ⎭ ⎩ 0 2 ⎩ ̈2 ⎭ −2 2 + 3 ̇2 −2 2 ⎩ 2 ⎭ ⎧ ⎫ ⎨0⎬ = ⎩0⎭
where 1 = 2 = 1 1 = 3 2 = 2 3 = 3 and 1 = 2 = 1 with appropriate units. The direct method requires that we assume responses of the form: 1 () = exp() and 2 () = exp() We differentiate these twice and substitute into the EOM to obtain ⎧ ⎫ ⎤ ⎤ ⎡ ⎤⎞ ⎛ ⎡ ⎡ ⎨0⎬ 0 + 2 −2 + 2 −2 ⎦ + ⎣ 1 ⎦+⎣ 1 ⎦⎠ exp() = ⎝2 ⎣ 1 ⎩0⎭ 0 2 −2 2 + 3 −2 2 ⎧ ⎫ ⎤⎧ ⎫ ⎡ 2 ⎨0⎬ ⎨ ⎬ −2 − 2 + (1 + 2 ) + (1 + 2 ) ⎦ ⎣ 1 exp() = ⎩0⎭ − − 2 + ( + ) + ⎩ ⎭ 2
2
2
2
3
2
There is no finite that will make exp() = 0 for all Therefore, ⎡ ⎤⎧ ⎫ ⎧ ⎫ 2 ⎨⎬ ⎨0⎬ + (1 + 2 ) + (1 + 2 ) −2 − 2 ⎣ 1 ⎦ = − − 2 + ( + ) + ⎩ ⎭ ⎩ 0 ⎭ 2
2
2
2
3
2
Using the given parameter values, we obtain ⎤⎧ ⎫ ⎧ ⎫ ⎡ 2 ⎨⎬ ⎨0⎬ + 5 + 2 −2 − 1 ⎦ ⎣ = −2 − 1 2 + 5 + 1 ⎩ ⎭ ⎩ 0 ⎭
438
CHAPTER 6 MULTI DOF VIBRATION
with characteristic matrix ¡ 2 ¢¡ ¢ + 5 + 2 2 + 5 + 1 − (−2 − 1)2 = 0
which needs to be solved numerically for the four values of Using MATLAB we find: 11 = −66241 12 = −28127 21 = −04417 22 = −01215 implying that all the = 0 Thus, there is no oscillation, only exponential decay. From these we can find the (generally) complex modes In this case, as in Example 7.10, the modes are not complex because of the nature of the property matrices. We find: 11 = −096 12 = 1114 21 = −771 22 = 0538 implying that all the = 0 Therefore, 1 () = −0961 exp(−66241) + 11142 exp(−28127) −7713 exp(−04417) + 05384 exp(−01215) 2 () = 1 exp(−66241) + 2 exp(−28127) +3 exp(−04417) + 4 exp(−01215) The constants of integration are evaluated using the 4 initial conditions. Sketch of the System There are many system configurations that may be represented by the given mass, damping, and stiffness matrices. Let us consider a system shown below.
The mass, damping, and stiffness matrices are given by ⎡ ⎤ ⎤ ⎡ 1 0 + 2 −2 ⎦ [] = ⎣ 1 ⎦ [] = ⎣ −2 2 + 3 0 2
⎡
[] = ⎣
1 + 2
−2
−2
2 + 3
⎤
⎦
We can choose 1 = 2 = 1 kg, 1 = 3 Ns/m, 2 = 2 Ns/m, 3 = 3 Ns/m, 1 = 1 N/m, 2 = 1 N/m, and 3 = 0 N/m to represent our current system.
439
31. Use the direct method to solve the equations of motion for a system with the following property matrices, ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 2 −1 3 −2 ⎦ [] = ⎣ ⎦ [] = ⎣ ⎦ [] = ⎣ 0 1 −1 1 −2 4 Sketch this system.
Solution: We can sketch the system by noting several characteristics of the system property matrices. The first is that the mass matrix is diagonal, generally implying that this is rectilinear motion rather than rotational motion. The damping and stiffness matrices are each fully coupled, but the second diagonal term is not the same multiple of the first diagonal term. A little thought will show that we have two masses, with mass 1 coupled to a fixed end by spring 1 and damper 1 on one side and coupled to mass 2 on the other side by spring 2 and damper 2 Mass 2 is attached to a fixed end by a spring 3 . Such system is shown below.
The general EOM for such system is given by ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎡ ⎬ ⎬ ⎨ ⎨ ⎨ ⎬ ⎨0⎬ 1 + 2 −2 1 + 2 −2 ̈1 ̇1 0 1 ⎦ ⎦ ⎦ ⎣ 1 +⎣ +⎣ = ⎩ ̇ ⎭ 0 ⎩ ̈ ⎭ − − + ⎩ ⎭ ⎩0⎭ 2
2
2
2
2
2
2
3
2
In this particular example, 1 = 2 = 1 1 = 2 = 1 and 1 = 1 and 2 = 3 = 2 with appropriate units. The direct method requires that we assume responses of the form: 1 () = exp() and 2 () = exp() We differentiate these assumed responses twice and substitute into the EOM to obtain ⎧ ⎫ ⎛ ⎡ ⎤ ⎤ ⎡ ⎤⎞ ⎡ ⎨0⎬ 0 + 2 −2 + 2 −2 ⎝2 ⎣ 1 ⎦ + ⎣ 1 ⎦+⎣ 1 ⎦⎠ exp() = ⎩0⎭ 0 2 −2 2 + 3 −2 2 ⎧ ⎫ ⎤⎧ ⎫ ⎡ 2 ⎬ ⎨0⎬ ⎨ −2 − 2 + (1 + 2 ) + (1 + 2 ) ⎦ ⎣ 1 exp() = ⎩0⎭ − − 2 + ( + ) + + ⎩ ⎭ 2
2
2
2
3
2
3
There is no finite that will make exp() = 0 for all Therefore, ⎡ ⎤⎧ ⎫ ⎧ ⎫ 2 ⎨⎬ ⎨0⎬ + (1 + 2 ) + (1 + 2 ) −2 − 2 ⎣ 1 ⎦ = − − 2 + + + ⎩ ⎭ ⎩ 0 ⎭ 2
2
2
2
2
3
440
CHAPTER 6 MULTI DOF VIBRATION
Using the given parameter values, we obtain ⎤⎧ ⎫ ⎧ ⎫ ⎡ ⎨⎬ ⎨0⎬ 2 + 2 + 3 − − 2 ⎦ ⎣ = − − 2 2 + + 4 ⎩ ⎭ ⎩ 0 ⎭ with characteristic matrix
or
¡ 2 ¢¡ ¢ + 2 + 3 2 + + 4 − (− − 2)2 = 0 4 + 33 + 82 + 7 + 8 = 0
which needs to be solved numerically for the four values of Using MATLAB we find 11 = −1180 + 1855 12 = −1180 − 1855 21 = −0320 + 1246 22 = −0320 − 1246 From these we can find the (generally) complex modes In this case, as in Example 7.11, the modes are complex because of the nature of the property matrices. We find 11 = −0986 − 0846 12 = −0984 + 0848 21 = 0984 − 0470 22 = 0982 − 0470 where 11 and 12 are complex conjugates as are 21 and 22 We have left them in their approximate form due to rounding. We then find 1 = 1180 2 = 0320 1 = −0984 2 = 0983
1 = 1855 2 = 1250 1 = 0846 2 = 0470
Now we can write the response equations, which turn out to be ⎡ ⎤ (−09841 − 08462 ) cos 1855 ⎦ 1 () = −1180 ⎣ + (08461 − 09842 ) sin 1855 ⎡ ⎤ − 0470 ) cos 1250 (−0983 3 4 ⎦ +−0320 ⎣ + (04703 − 09834 ) sin 1250 1 () = −1180 [1 cos 1855 + 2 sin 1855] +−0320 [3 cos 1250 + 4 sin 1250]
441
32. Use the direct method to solve the equations of motion for a system with the following property matrices, ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 5 −2 2 −1 ⎦ [] = ⎣ ⎦ [] = ⎣ ⎦ [] = ⎣ 0 1 −2 3 −1 1 Sketch this system.
Solution: Response This system is composed of two masses that oscillate horizontally. The first mass is connected on the left end to a wall by a spring and a damper. The second mass is connected on the right side to a wall by a damper only. The two masses are connected to each other by a spring and a damper. The variable notation for the masses, springs and dampers is where = 1 2 for the springs and = 1 2 3 for the dampers. The EOM is then in general ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎬ ⎬ ⎨ ⎨ 0 1 + 2 −2 1 + 2 −2 ⎨ 1 ⎬ ⎨ 0 ⎬ ̈1 ̇1 ⎣ 1 ⎣ ⎣ ⎦ ⎦ ⎦ + + = ⎭ ⎭ ⎩ ⎩ ⎩ ⎭ ⎩0⎭ 0 ̈ − + ̇ − 2
2
2
2
3
2
2
2
2
where 1 = 2 = 1 1 = 3 2 = 2 3 = 1 and 1 = 2 = 1 with appropriate units. The direct method requires that we assume responses of the form: 1 () = exp() and 2 () = exp() We differentiate these twice and substitute these expressions into the EOM. Using the given parameter values, we obtain ⎤⎧ ⎫ ⎧ ⎫ ⎡ 2 ⎨⎬ ⎨0⎬ + 5 + 2 −2 − 1 ⎦ ⎣ = −2 − 1 2 + 3 + 1 ⎩ ⎭ ⎩ 0 ⎭ with characteristic matrix
¡ 2 ¢¡ ¢ + 5 + 2 2 + 3 + 1 − (−2 − 1)2 = 0
which needs to be solved numerically for the four values of Using MATLAB we find: 11 = −5783 12 = −15112 21 = −04527 22 = −02528 implying that all the = 0 Thus, there is no oscillation, only exponential decay. From these we can find the (generally) complex modes In this case, as in Example 7.10, the modes are not complex because of the nature of the property matrices. We find: 11 = −1618 12 = 0618 21 = −16154 22 = 06181
442
CHAPTER 6 MULTI DOF VIBRATION
implying that all the = 0 Therefore, 1 () = −16181 exp(−5783) + 06182 exp(−15112) −161543 exp(−04527) + 061814 exp(−02528) 2 () = 1 exp(−5783) + 2 exp(−15112) +3 exp(−04527) + 4 exp(−02528) The constants of integration are evaluated using the 4 initial conditions. Sketch of the System There are many system configurations that may be represented by the given mass, damping, and stiffness matrices. Let us consider a system shown below.
The mass, damping, and stiffness matrices are given by ⎡ ⎤ ⎤ ⎡ 1 0 1 + 2 −2 ⎦ [] = ⎣ ⎦ [] = ⎣ 0 2 −2 2 + 3
⎡
[] = ⎣
1 + 2
−2
−2
2 + 3
⎤
⎦
We can choose 1 = 2 = 1 kg, 1 = 3 Ns/m, 2 = 2 Ns/m, 3 = 1 Ns/m, 1 = 1 N/m, 2 = 1 N/m, and 3 = 0 N/m to represent our current system.
443
33. Derive Equation 6.78 showing all steps. Solution: Let us start with the 1 = 1260
1 = 1430;
2 = 0115
2 = 0632
1 = −2178
1 = 1004;
2 = 0511
2 = 0131
and Then, the solution is given by 1 () = −1 ((1 1 + 2 1 ) cos 1 + (−1 1 + 2 1 ) sin 1 ) + −2 ((3 2 + 4 2 ) cos 2 + (−3 2 + 4 2 ) sin 2 ) 2 () = −1 (1 cos 1 + 2 sin 1 ) + −2 (3 cos 2 + 4 sin 2 ) or 1 () = −1260 ((−21781 + 10042 ) cos 1430 + (−10041 − 21782 ) sin 1430) + −0115 ((01153 + 01314 ) cos 0632 + (−01313 + 01154 ) sin 0632) 2 () = −1260 (1 cos 1430 + 2 sin 1430) + −0115 (3 cos 0632 + 4 sin 0632) It is clear that there are four constants to be found. The initial conditions are 1 (0) = 1 with all other initial conditions equal to zero. Applying the initial displacement conditions, we have 1 (0) = 1 = −21781 + 10042 + 01153 + 01314 2 (0) = 0 = 1 + 3 ̇1 (0) = 0 = −1260 (−21781 + 10042 ) + 1430 (−10041 − 21782 ) −0115 (01153 + 01314 ) + 0632 (−01313 + 01154 ) ̇2 (0) = 0 = −12601 + 14302 − 01153 + 06324 The four equations above are solved simultaneously to find 1 = −0460 2 = −0182 3 = 0460 and 4 = −0422 The responses are then ⎧ ⎫ ⎫ ⎫ ⎧ ⎤ ⎡⎧ ⎨ 0859 ⎬ ⎨ 0820 ⎬ ⎨ () ⎬ 1 sin 1430⎦ = −1260 ⎣ cos 1430 + ⎩ −0182 ⎭ ⎩ −0460 ⎭ ⎩ () ⎭ 2 ⎧ ⎫ ⎫ ⎤ ⎡⎧ ⎨ −0276 ⎬ ⎨ 0180 ⎬ sin 0632⎦ cos 0632 + + −0115 ⎣ ⎩ −0422 ⎭ ⎩ 0460 ⎭
444
CHAPTER 6 MULTI DOF VIBRATION
Problems for Section 6.6 — Modal Analysis 34. Derive Equation 6.79. Solution: Equation 6.79 is {}2 [] {}1 = 0
Begin with 21 [] {}1 = [] {}1 Pre-multiply it by{}2 so that 22 {}1 [] {}2 = {}1 [] {}2 where {}1 = 11
{}2 = 12
⎧ ⎪ ⎨
⎫ ⎪ ⎬
1 11 − 21 1 ⎪ ⎪ ⎩− ⎭ 12 ⎧ ⎫ ⎪ ⎪ ⎨ ⎬ 1 2
− 2 1 ⎪ ⎪ ⎩ − 11 ⎭ 12 ⎤ ⎡ 1 0 ⎦ [] = ⎣ 0 2 sµ ¶2 2 1 22 + 2 11 11 22 − 12 + 1 22 2 11 2 ∓ − 12 = 21 2 21 2 1 2
Substitute them into Equation 6.79 to find ½ ¾ ¢¡ ¢ 2 ¡ 2 2 {}1 [] {}2 = 11 12 1 + 2 11 − 1 1 11 − 2 2 12
We substitute 21 and 22 in, and we will be able to obtain 0 It is much
(6.79)
445
35. Show that the orthogonality relations hold when there is coupling in both mass as well as in stiffness matrices as long as they are symmetric. This requires a derivation of the characteristic matrix and equation, the eigenvalues and the eigenvectors. Solution: Start with the eigenvalue problem given by [] {} = 2 [] {} There are eigenvalues ( 2 ) and corresponding eigenvectors ({}) for DOF system. Let us confine ourselves to 2 DOF system for now. The two sets of eigenvalues and eigenvectors satisfy the eigenvalue problem so that [] {}1 = 21 [] {}1 [] {}2 = 22 [] {}2 Multiply the first equation by {}2 from the left and second by {}1 also from the left. Then, we have {}2 [] {}1 = 21 {}2 [] {}1
(1)
{}1 [] {}2 = 22 {}1 [] {}2 Transpose of the second equation is {}2 [] {}1 = 22 {}2 [] {}1
(2)
For a symmetric system, we have [] = [] and [] = [] Subtracting Equation (2) from Equation (1), we obtain ¡ ¢ 0 = 21 − 22 {}2 [] {}1
If the natural frequencies are distinct, the modes are orthogonal. That is {}2 [] {}1 = 0 for 21 6= 22
Note that the derivation of orthogonality is equally valid when the mass and stiffness matrices are fully populated.
446
CHAPTER 6 MULTI DOF VIBRATION
36. For the two degree-of-freedom system in Figure 6.76 undergoing longitudinal motion, derive the equations of motion utilizing each of the following approaches: (a) flexibility coefficients, (b) Newton’s second law of motion, (c) Lagrange’s equation, and (d) Hamilton’s principle. Then, with the equations of motion in matrix form, derive the natural frequencies and mode shapes, and solve for the responses. Assume the values 1 = 2 = 1 kg and all stiffnesses are 1 N/m. Solve via (i) the direct method and (ii) modal analysis. Find the response of each mass if the initial velocities equal zero and the initial displacements are in the ratio of the first mode. For a design application, it is necessary that |2 −1 | ≥ , where is some critical separation distance between the masses. Describe how to check that this condition is met, and what options exist for redesign of the system or feedback control if the condition is not met.
Figure 6.76: Two degree-of-freedom structure.
Solution: In all the following we assume linear behavior. (a) Flexibility coefficients - To derive the flexibility coefficients we will work with the following (static) set of displacement-force relations: ⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎨ 1 ⎬ 11 12 ⎨ 1 ⎬ ⎣ ⎦ = ⎩ ⎭ ⎩ ⎭ 2
21
22
2
The free-body diagram is shown below.
The static equations for 1 and 2 are given by 1 = 1 1 + 2 (1 − 2 ) 2 + 2 (1 − 2 ) = (3 + 4 ) 2
447 To find the first column of the flexibility matrix, we set 1 = 1 and 2 = 0 Then, the resulting displacements are the elements of the flexibility matrix, 1 = 11 and 2 = 21 The static equations are reduced to 1 = 1 1 + 2 (1 − 2 ) 2 (1 − 2 ) = (3 + 4 ) 2 Solve the second equation for 2 and substitute this into the first equation to obtain 2 1 ; 2 + 3 + 4 (1 + 2 ) (2 + 3 + 4 ) 1 − 22 1 1 = ; 2 + 3 + 4 2 + 3 + 4 1 = 1 2 + 1 3 + 1 4 + 2 3 + 2 4
second equation
2 =
first equation
Therefore, 2 + 3 + 4 1 2 + 1 3 + 1 4 + 2 3 + 2 4 2 21 = 1 2 + 1 3 + 1 4 + 2 3 + 2 4 11 =
To find the second column of the flexibility matrix, we set 2 = 1 and 1 = 0 The resulting static equations are 0 = 1 1 + 2 (1 − 2 ) 1 + 2 (1 − 2 ) = (3 + 4 ) 2 Solving for 1 and 2 we find 2 1 2 + 1 3 + 1 4 + 2 3 + 2 4 1 + 2 22 = 1 2 + 1 3 + 1 4 + 2 3 + 2 4 12 =
The flexibility is then given by ⎤ ⎡ ⎤ ⎡ 2 + 3 + 4 1 ⎣ 2 ⎦ ⎣ 11 12 ⎦ = 1 2 + 1 3 + 1 4 + 2 3 + 2 4 21 22 2 1 + 2 Let us also show the (easier) derivation by the stiffness coefficient approach: We will work with the following (static) set of force-displacement relations: ⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎨ ⎬ 11 12 ⎨ 1 ⎬ 1 ⎦ =⎣ ⎩ ⎭ ⎩ ⎭ 2
21
22
2
448
CHAPTER 6 MULTI DOF VIBRATION
First set 1 = 1 and 2 = 0 Then, from a free-body diagram of mass 1 with 1 acting on it while mass 2 is held fixed, and using the force displacement relations, we have: 1 = 11 1 = 1 + 2 Also, 2 = 21 1 = −2 Next, set 2 = 1 and 1 = 0 Then, from the free-body diagram of mass 2 and the force displacement relations, we have: 1 = 12 2 = −2 Also, 2 = 22 2 = 2 + 3 + 4 Therefore, the stiffness matrix is: ⎤ ⎡ −2 + 2 ⎦ ⎣ 1 −2 2 + 3 + 4
(b,ii) Newton’s second law of motion, and modal analysis - Draw two free-body diagrams, one for each mass as shown below.
Take the sum of the forces in the direction for each mass and apply Newton’s second law of motion, to find: −1 1 − 2 (1 − 2 ) = 1 ̈1 2 (1 − 2 ) − 4 2 − 3 2 = 2 ̈2 The EOM in matrix-vector form is ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎡ ⎬ ⎨ ⎨ ⎬ ⎨0⎬ −2 1 + 2 ̈1 0 1 ⎣ ⎦ ⎦ ⎣ 1 + = ⎭ ⎩ ⎩ 0 ̈ − + + ⎭ ⎩0⎭ 2
2
2
2
3
4
2
or []{̈} + []{} = {0} The eigenvalue problem is given by ([] − 2 []) {} = 0 or ¯ ¯ ¯ ¯ 2 ¯ ¯ 1 + 2 − 1 −2 ¯ = 0 ¯ ¯ ¯ 2 ¯ −2 2 + 3 + 4 − 2 ¯
In general, this is expanded and solved in sequence for the frequencies and modes. For the numerical values given above, this determinant becomes ¯ ¯ ¯ ¯ 2 ¯2− −1 ¯ ¯ ¯ = 0 ¯ ¯ 2 ¯ −1 3 − ¯
We find 212 = 138 362 The modes are then {}1 = {1 062} and {}2 = {1 normalize the modes as follows: 21 {}1 []{}1 = 1 =⇒ 1 = 085
− 162} We
449 We do the same for the second mode to normalize it, finding 2 = 0526 The normalized modes are then: ⎫ ⎫ ⎧ ⎧ ⎨ 0851 ⎬ ⎨ 0526 ⎬ {̂}2 = {̂}1 = ⎩ 0526 ⎭ ⎩ −0851 ⎭ Assume { ()} = {}1 1 () + {}2 2 () Then
[] ({}1 ̈1 + {}2 ̈2 ) + [] ({}1 1 + {}2 2 ) = {0} Pre-multiply this equation in sequence by {}1 and {}2 to find the decoupled EOM for the generalized (modal) coordinates: ̈1 + 21 1 = 0 ̈2 + 22 2 = 0 where () and 2 () are harmonic functions. The solution in physical space is then ⎫ ⎧ ⎨ 1 () ⎬ = {}1 1 () + {}2 2 () ⎩ () ⎭ 2 ⎧ ⎫ ⎧ ⎫ ⎨ 0851 ⎬ ⎨ 0526 ⎬ cos (1 − 1 ) + 2 cos ( 2 − 2 ) = 1 ⎩ 0526 ⎭ ⎩ −0851 ⎭ The initial conditions are satisfied using the following: ⎧ ⎫ ⎧ ⎧ ⎫ ⎧ ⎫ ⎫ ⎨1 ⎬ ⎨ (0) ⎬ ⎨ 0851 ⎬ ⎨ 0526 ⎬ 1 = 1 cos 1 + 2 cos 2 = ⎩ 062 ⎭ ⎩ (0) ⎭ ⎩ 0526 ⎭ ⎩ −0851 ⎭ 2 ⎫ ⎫ ⎫ ⎧ ⎫ ⎧ ⎧ ⎧ ⎨0⎬ ⎨ ̇ (0) ⎬ ⎨ 0851 ⎬ ⎨ 0526 ⎬ 1 = −1 sin 1 − 2 sin 2 = ⎩0⎭ ⎩ ̇ (0) ⎭ ⎩ 0526 ⎭ ⎩ −0851 ⎭ 2
In direct method, we let
{ ()} =
⎧ ⎫ ⎨⎬ ⎩⎭
After substituting it into the Equation of motion, we find ⎧ ⎫ ⎧ ⎫ ⎨⎬ ⎨⎬ = 0 + [] 2 [] ⎩⎭ ⎩⎭
Comparing this to the eigenvalue problem from the modal analysis, we find that 2 = −2 and { } is the eigenvector {} Therefore, the two solution methods become identical.
450
CHAPTER 6 MULTI DOF VIBRATION
For a design application, it is necessary that |2 − 1 | ≥ , where is some critical separation distance between the masses. Describe how you would check that this condition is met, and what options exist for a re—design of the structure. For the above free vibration problem, define a new variable = 2 − 1 and it is straightforward to track this difference for a particular set of initial conditions. All we have to do is to program () and check at each time instant that the responses are calculated whether ≤ If there are instances where the critical distance is exceeded, then it would be necessary to re-design the stiffness between the two masses. (c) Lagrange’s Equation - To use Lagrange’s equation, we need the kinetic and potential (strain) energies of the system: 1 1 1 ̇21 + 2 ̇22 2 2 1 1 1 1 = 1 21 + 2 (2 − 1 )2 + 3 22 + 4 22 2 2 2 2 ³ ´ Then, Lagrange’s equation for each generalized coordinate is ̇ + =0: =
¶ µ + = 0 ̇1 1 1 ̈1 + (1 + 2 ) 1 − 2 2 = 0 ¶ µ + = 0 ̇2 2 2 ̈2 + (2 + 3 + 4 ) 2 − 2 1 = 0
(d) Hamilton’s Principle - For the application of Hamilton’s principle, we substitute the Lagrangian (see part (c) above) into the principle: Z 2 = 0; 1
=
1 1 1 ̇21 + 2 ̇22 2µ 2 ¶ 1 1 1 1 2 2 2 2 1 1 + 2 (2 − 1 ) + 3 2 + 4 2 − 2 2 2 2
Interchanging the variation and the integral and then taking the variation of each term yields: 1 ̇1 ̇1 + 2 ̇2 ̇2 − 1 1 1 − 2 (2 − 1 ) (2 − 1 ) − (3 + 4 ) 2 2 We need to replace the variations of the derivative terms as follows: ( ̇ ) − 1 ̈1 1 1 1 1 1 ̇1 ̇1 = (1 ̇1 1 ) − 1 ̈1 1 ; 2 ̇2 ̇2 = ( ̇ ) − 2 ̈2 2 2 2 2 2 ̇2 ̇2 = (2 ̇2 2 ) − 2 ̈2 2 1 ̇1 ̇1 =
or
or
451 where the () terms vanish at the end times. Then, ⎛ ⎞ Z 2 (−1 ̈1 − 1 1 + 2 (2 − 1 )) 1 ⎝ ⎠ = 0 1 + (−2 ̈2 − 2 (2 − 1 ) − (3 + 4 ) 2 ) 2
from which, since the variations are independent and arbitrary, we obtain the respective EOM: −1 ̈1 − 1 1 + 2 (2 − 1 ) = 0 −2 ̈2 − 2 (2 − 1 ) − (3 + 4 ) 2 = 0
452
CHAPTER 6 MULTI DOF VIBRATION
37. In Problem 36, the system is forced with 1 () = 1 cos Ω1 and 2 () = 2 sin Ω2 acting on mass 1 and 2 respectively. Using modal analysis, solve for the forced response assuming initial conditions 1 (0) = 0 ̇1 (0) = 1 2 (0) = 2 and ̇2 (0) = 0.
Solution: Since arbitrary loading frequencies are given, there are two cases that should be considered in a design. The first is that one or both of the frequencies 1 and 2 are equal to the natural frequencies of vibration. The second case is that neither equals the natural frequencies. We will proceed only with the second case. Proceeding under this assumption, we can write the EOM as: ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎬ ⎬ ⎨ ⎨ ⎨ 1 cos 1 ⎬ 1 1 −2 0 1 + 2 ⎣ 1 ⎦ ⎦ +⎣ = 0 ⎩ ⎭ − + + ⎩ ⎭ ⎩ sin ⎭ 2
2
2
2
3
4
2
2
2
Now, we can proceed with an eigenvalue analysis of the system to find the natural frequencies and the mode shapes. Recall from Problem 36 that 1 = 2 = 1 kg and all stiffness are 1 N/m Eigenvalue Problem We first solve the homogeneous problem with forces set to zeros. Assume a harmonic solution, { ()} = {} exp () Upon substituting the assumed solution, we obtain the eigenvalue problem given by ⎧ ⎫ ⎛⎡ ⎤ ⎡ ⎤⎞ ⎧ ⎫ ⎨0⎬ ⎨ ⎬ 1 + 2 −2 0 1 ⎝⎣ ⎦ − 2 ⎣ 1 ⎦⎠ = ⎩0⎭ ⎩ ⎭ 0 2 −2 2 + 3 + 4 2 ⎫ ⎧ ⎫ ⎧ ⎤⎞ ⎛⎡ 2 ⎬ ⎨0⎬ ⎨ −1 1 2− ⎦⎠ ⎝⎣ = ⎩0⎭ ⎩ ⎭ −1 3 − 2 2
The natural frequencies are obtained by setting the determinant of the matrix, [] − 2 [] to zero, or ⎤⎞ ⎛⎡ 2 −1 2− ⎦⎠ = 0 det ⎝⎣ −1 3 − 2 4 − 52 + 5 = 0
Therefore, the natural frequencies are 1 = 1176 rad/s 2 = 1902 rad/s. We substitute each natural frequency into the eigenvalue problem to obtain the eigenvectors so that ⎧ ⎫ ⎛⎡ ⎤⎞ ⎧ ⎫ 2 ⎬ ⎨ ⎨0⎬ 2 − −1 1 ⎝⎣ ⎦⎠ = for = 1 2. ⎩ ⎭ ⎩0⎭ −1 3 − 2
2
453 We find
⎧ ⎫ ⎨ ⎬ 1
⎩ ⎭ 2
=
For given numerical values, we find ⎫ ⎧ ⎫ ⎧ ⎨ ⎬ ⎨ 1 ⎬ 1 = 1 ⎩ ⎭ ⎩ 0617 ⎭ 2
1
⎧ ⎨
⎫ ⎬
1
⎩ 2 − 2 ⎭
⎧ ⎫ ⎨ ⎬ 1
⎩ ⎭ 2
= 2
2
⎧ ⎨
1
⎫ ⎬
⎩ −1618 ⎭
We normalize the eigenvectors with respect to the mass matrix so that {̂} [] {̂} = 1 The normalized eigenvectors are {̂}1 =
⎧ ⎫ ⎨ 0851 ⎬ ⎩ 0526 ⎭
{̂}2 =
⎧ ⎨ 0526
⎫ ⎬
⎩ −0851 ⎭
Note that while these vectors have the same elements (except for a negative sign), they are linearly independent. Also, since the eigenvectors are normalized with respect to the mass matrix, which incidentally is an identity matrix, these {̂} vectors are of unit length. The modal transformation matrix [ ] can be written as the modal matrix, the matrix of normal modes, ⎡ ⎤ 0851 0526 ⎦ [ ] = ⎣ 0526 −0851 Modal Analysis
Following the procedure of modal analysis, we first transform our coordinates to modal coordinates using the relationship given by { ()} = [ ] { ()}
Then, the equation of motion becomes
[] [ ] {̈ ()} + [] [ ] { ()} = { ()}
Multiply on the left by [ ] and use the orthogonality conditions [ ] [][ ] = [] [ ] [][ ] = [Ω2 ] to obtain
⎡ ⎣
1 0
⎤
⎡
⎦ {̈ ()} + ⎣
21
0 22
⎤
⎦ { ()} = [ ] { ()}
0 0 1 ⎫ ⎧ ⎤ ⎡ ⎤ ⎨ 1 0 138 0 08511 cos 1 + 05262 sin 2 ⎬ ⎣ ⎦ {̈ ()} + ⎣ ⎦ { ()} = ⎩ 0526 cos − 0851 sin ⎭ 0 1 0 362 1 1 2 2 ⎡
454
CHAPTER 6 MULTI DOF VIBRATION
where the two modal coordinates, 1 () and 2 () are decoupled. Note that the initial conditions in coordinates are ⎫ ⎧ ⎫ ⎧ ⎨ 0851 ⎬ ⎨ 0526 ⎬ 2 1 {̇ (0)} = [ ] {̇ (0)} = (1) { (0)} = [ ] { (0)} = ⎩ 0526 ⎭ ⎩ −0851 ⎭ 2
1
The solutions, 1 () and 2 () are given by
08511 05262 cos 1 + sin 2 2 138 − 1 138 − 22 05261 08512 cos 1 − sin 2 2 () = 2 cos 190 + 2 sin 190 + 2 3618 − 1 3618 − 22
1 () = 1 cos 118 + 1 sin 118 +
(2)
where the first two terms on the right-hand sides of each equation are the homogeneous solution and the last two terms are the particular solutions. The homogeneous solutions can be written in the amplitude-phase form such that 08511 05262 cos 1 + sin 2 2 138 − 1 138 − 22 05261 08512 cos 1 − sin 2 2 () = 2 cos (190 − 2 ) + 2 3618 − 1 3618 − 22 1 () = 1 cos (118 − 1 ) +
It is easier however to apply the initial conditions to the first set of equations (Equation (2)) to find 1 1 2 and 2 Let us apply the initial displacement in Equation (1) to find 08511 138 − 21 05261 −08512 = 2 + 3618 − 21 05262 = 1 +
Solving for 1 and 2 we obtain
08511 138 − 21 05261 2 = −08512 − 3618 − 21
1 = 05262 −
Similarly, applying the initial velocity conditions, we obtain 05262 2 138 − 22 08512 05261 = 1902 − 2 3618 − 22 08511 = 1181 +
=⇒ =⇒
04372 2 138 − 22 04482 2 = 02771 + 2 3618 − 22
1 = 07211 −
The constants of integration, 1 1 2 and 2 are substituted into Equation (2). Finally, the response is given by { ()} = [ ] { ()} = {̂}1 1 () + {̂}2 2 ()
455
38. If Problem 36 is modified to include damping between the two masses as shown in Figure 6.77, solve for the case with no forcing using (a) modal analysis, and (b) the direct method.
Figure 6.77: Two degree-of-freedom system with damping.
Solution: The EOM for the damped system can be written in matrix-vector form, ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎬ ⎨ ⎬ ⎨ ⎨ ⎬ ⎨0⎬ −2 0 − ̇1 1 + 2 ̈1 1 ⎣ 1 ⎦ ⎦ ⎦ +⎣ +⎣ = 0 ⎩ ̈ ⎭ − ⎩ ̇ ⎭ − + + ⎩ ⎭ ⎩0⎭ 2
2
2
2
2
3
4
2
The question is now whether the damping is proportional or not. What we require for proportional damping is that the following relation exists between the property matrices: [] = [] + []; ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ −2 − 0 + 2 ⎣ ⎦ = ⎣ 1 ⎦ + ⎣ 1 ⎦ 0 2 − −2 2 + 3 + 4
In order for this to be true, we have to equate the respective elements in the matrix equation: = 1 + (1 + 2 ) − = −2 = 2 + (2 + 3 + 4 )
Due to the symmetry in the matrices, there are in reality 3 and not 4 equations. The second equation defines = 2 Then, substituting into the first and third equations, we can solve each for Since these two expressions must be equal to each other, we equate them to find: =
1 1 + 2 1 2 + 3 + 4 − = − 1 1 2 2 2 2
(1)
This is the general relation that must be satisfied by the system parameters in order that the damping be proportional. We can immediately see the special case: if 1 = 2 then 1 = 3 + 4 In Problem 36, all the mass and stiffness are set to 1 with appropriate units. This violates Equation (1), and therefore, the damping of the current system with the parameters of Problem 36 is not proportional. Only the direct method can be used. However, we will proceed with both methods without substituting the numerical values.
456
CHAPTER 6 MULTI DOF VIBRATION
(a) Modal Analysis - First, we rewrite the EOM as [] {̈} + ( [] + []) {̇} + [] {} = 0 where and are given in Equation (1). The first step is to find the natural frequencies and the mode shapes of the undamped problem. The eigenvalue problem to be solved is ¢ ¡ 2 − [] + [] {} = {0}
The eigenvectors, {} must be normalized with respect to [] MATLAB command, [P, D] = eig(k,m) will produce the modal matrix, P, and the diagonal matrix of eigenvalues, The columns of the modal matrix are the normalized eigenvectors. The second step is to transform the current system using {} = [ ] {} We then multiply the EOM by [ ] from the left to find the transformed EOMs: ⎛ ⎡ ⎤⎞ ⎤ ⎡ 21 0 21 0 ⎦⎠ {̇} + ⎣ ⎦ {} = {0} {̈} + ⎝ + ⎣ 2 2 0 2 0 2 or
¡ ¢ ̈1 + + 21 ̇1 + 21 1 = 0 ¡ ¢ ̈2 + + 22 ̇2 + 22 2 = 0
The solution forms depend on whether the system is over, under, or critically damped. The damping ratios are given by = ( + 2 ) 2 The third step is to apply the initial conditions to find the constants of integration that appear in 1 () and 2 () We know {} and {} are related by {} = [ ] {} and {} = [ ] [] {} The initial conditions for { ()} are given by { (0)} = [ ] [] { (0)} {̇ (0)} = [ ] [] {̇ (0)}
Finally, we transform back to our original coordinate using {} = [ ] {} Modal Analysis - Numerical Example Let us consider a system given by ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎬ ⎨ ⎬ ⎨ 1 −1 3 −1 ⎨ 1 ⎬ ̇1 1 0 ̈1 ⎦ ⎦ ⎣ ⎦ +⎣ +⎣ = 0 ⎭ ⎩ ⎭ ⎩ −1 1 −1 3 ⎩ ⎭ ̈ ̇ 0 1 2
2
2
where = −2 and = 1 Solving the eigenvalue problem, we find 1 = 1414 = 2 = 2 rad/s, and ⎫ ⎫ ⎧ ⎧ ⎨ − √1 ⎬ ⎨ − √1 ⎬ 2 2 {̂}1 = {̂}2 = ⎩ − √1 ⎭ ⎩ √1 ⎭ 2 2 ⎤ ⎡ − √12 − √12 ⎦ [ ] = ⎣ − √12 √12
√ 2 rad/s and
457 The EOM in terms of modal coordinates is given by ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎡ 0 0 ⎨ ̇1 ⎬ 2 0 ⎨ 1 ⎬ 1 0 ⎨ ̈1 ⎬ ⎦ ⎦ ⎣ ⎦ +⎣ +⎣ =0 ⎩ ⎩ ⎭ ⎭ 0 2 0 4 ⎩ ⎭ 0 1 ̈ ̇ 2
2
2
An interesting feature of this choice of coordinates is that all the damping is now in the second mode; the first mode will exhibit an undamped free response. To be specific, note that 1 = 0 2 = 05 Thus, the equations for the modal response are: ⎧ ⎫ ⎧ ⎫ ¡√ ¢ ⎨ 1 ⎬ ⎨ ⎬ 1 cos 2 − 1 = ⎩ ⎭ ⎩ exp (−) cos ¡√3 − ¢ ⎭ 2
2
2
where and can be found from the given initial conditions. We then use the transformation {()} = [ ]{()} to arrive at the response in physical coordinates: ⎧ ⎫ ⎫ ⎧ √ ¡√ ¢ ⎨ ⎬ ⎨ −1 cos 2 − 1 − 2 exp(−) cos( 3 − 2 ) ⎬ 1 1 √ = √ ¡√ ¢ ⎩ ⎭ 2 ⎩ −1 cos 2 − 1 + 2 exp(−) cos( 3 − 2 ) ⎭ 2 Note that we may evaluate the initial conditions in either the modal coordinates or the physical coordinates to solve for the constants of integration. (b) Direct Method - Consider a system whose matrix EOM is given by [] {̈} + [] {̇} + [] {} = 0
We substitute the assumed solution of the form, {} = {} exp () into the EOM to obtain ¡ 2 ¢ [] + [] + [] {} exp () = 0
The determinant of the matrix, 2 [] + [] + [] must be zero in order to avoid trivial solution for {} This equation is called characteristic equation whose solution renders eigenvalue The eigenvalues may be real or complex, but the real part of is never positive if the damping coefficients are positive. ¡ ¢ We substitute back into the eigenvalue problem, 2 [] + [] + [] {} = {0} to find the eigenvectors, {} It should be noted that if a pair of eigenvalues are complex conjugate of each other, so are the corresponding eigenvectors. There will be pairs of eigenvalues for DOF system. In the following, we consider two cases: A pair of eigenvalues are real and negative, and a pair of complex with negative real part. Case I: When a pair of eigenvalue is real and negative. Then the solution is given by {} = 1 {}1 exp (1 ) + 2 {}2 exp (2 ) Case II: When a pair of eigenvalues is complex and given by − ± the solution is given by { ()} = {} exp ((− + ) ) + ̄ {̄} exp ((− − ) ) ª © = exp (−) {} exp ( ) + ̄ {̄} exp (− )
458
CHAPTER 6 MULTI DOF VIBRATION
Split {} and into real and imaginary parts such that {} = { } + { } and = + Euler’s formula is used to expand the exponential functions. The constant coefficients must be complex conjugate pairs so that { ()} is real. Then, { ()} = exp (−) [( + ) ({ } + { }) (cos + sin ) + ( − ) ({ } − { }) (cos − sin )] = exp (−) [2 ( − ) cos − 2 ( + ) sin ]
(1)
where and are unknown constants of integration. They will be found using initial conditions.
Direct Method - Numerical Example
Again, we begin with the equations of motion in physical coordinates, namely: ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎤⎧ ⎫ ⎡ ⎨ ̇ ⎬ ⎨ ⎬ ⎨ ̈ ⎬ 1 −1 3 −1 1 1 1 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ + + = 0 −1 1 ⎩ ̇2 ⎭ −1 3 ⎩ 2 ⎭ 0 1 ⎩ ̈2 ⎭ ⎡
1 0
Now, we find the eigenvalues and eigenvectors of the system. The resulting characteristic polynomial is: 4 + 23 + 62 + 4 + 8 = 0 Solving this polynomial gives us: √ 12 = −1 ± 3 √ 34 = ± 2 There are two pairs of eigenvalues, and therefore, the total solution will be the combination of two of Equation (1). ¡ ¢ The eigenvalues are substituted into 2 [] + [] + [] {} = {0} to find the corresponding eigenvalues. For example, ⎛⎡ ⎝⎣
2
0
0 2
⎤
⎡
⎦+⎣ ⎛⎡
− −
⎤
⎦+⎣
2 + + 3 ⎝⎣ − − 1
⎡
3 −1 −1 3
− − 1
2 + + 3
⎤⎞
⎦⎠ {} = {0} ⎤⎞
⎦⎠ {} = {0}
459 The eigenvectors are ⎡ {} = ⎣ ⎡
+ 1 2 + + 3
⎤ ⎦
⎤ ⎡ √ ⎤ √ −1 + 3 + 1 3 ⎦=⎣¡ ⎦=⎣ √ ⎦ {}1 = ⎣ √ √ ¢ ¡ ¢ 2 21 + 1 + 3 −1 + 3 + −1 + 3 + 3 − 3 ⎤ ⎡ ⎤ ⎡ ⎡ √ √ ⎤ 2 + 1 −1 − 3 + 1 − 3 ⎦=⎣¡ ⎦=⎣ √ ⎦ {}2 = ⎣ √ √ ¢ ¡ ¢ 2 22 + 2 + 3 −1 − 3 + −1 − 3 + 3 3 ⎤ ⎡ ⎤ ⎡ ⎡ √ √ ⎤ 3 + 1 2+1 1+ 2 ⎦=⎣¡ √ ¢ ⎦=⎣ {}3 = ⎣ √ √ ⎦ 2 23 + 3 + 3 2 + 2+3 1+ 2 ⎤ ⎡ ⎤ ⎡ ⎡ √ √ ⎤ 4 + 1 − 2 + 1 1− 2 ⎦=⎣¡ √ ¢ ⎦=⎣ {}4 = ⎣ √ ⎦ √ 2 24 + 4 + 3 − 2 − 2 + 3 1− 2 1 + 1
⎤
⎡
Then, the part of the solution that corresponds to 1 and 2 is given by
{ ()}1 = exp (−) [2 ( { } − { }) cos − 2 ( { } + { }) sin ] ⎡ √ ⎤ ⎡ √ ⎤ ⎞ ⎛ √ √ 3 3 = exp (−) ⎝21 ⎣ √ ⎦ cos 3 − 21 ⎣ √ ⎦ sin 3⎠ − 3 − 3 ⎤ ⎡ ⎞ ⎛⎡ √ √ ⎤ √ √ 21 3 21 3 = exp (−) ⎝⎣ √ ⎦ cos 3 − ⎣ √ ⎦ sin 3⎠ −21 3 −21 3
where we identified
⎡ √ ⎤ √ 3 = −1 = 3 { } = {0} { } = ⎣ √ ⎦ − 3
Note that we can rewrite { ()}1 as
⎡ ⎤ √ 1 ⎦ { ()}1 = 1 exp(−) cos( 3 − 1 ) ⎣ −1
The part of the solution that corresponds to 3 and 4 is given by { ()}2 = exp (−) [2 (2 { } − 2 { }) cos − 2 (2 { } + 2 { }) sin ] ⎧ ⎫ ⎧ √ ⎫⎞ ⎧√ ⎫ ⎧ ⎫⎞ ⎛ ⎛ ⎬ ⎨1⎬ ⎨ 2⎬ ⎨ ⎨1⎬ √ √ 2 ⎠ sin 2 ⎠ cos 2 − 2 ⎝2 √ − 2 √ + 2 = 2 ⎝2 ⎩1⎭ ⎩ 2⎭ ⎩ 2⎭ ⎩1⎭ ⎫ ⎫ ⎧ √ ⎧ ⎨ ⎨ 2 − 2√2 ⎬ √ √ 2 22 + 22 ⎬ 2 2 cos sin 2 2 − = √ √ ⎩ 2 2 + 2 ⎭ ⎩ 2 − 2 2 ⎭ 2 2 2 2
(2)
460
CHAPTER 6 MULTI DOF VIBRATION
where we identified
⎧ ⎫ ⎧√ ⎫ ⎨1⎬ ⎨ 2⎬ √ { } = √ = 0 = 2 { } = ⎩1⎭ ⎩ 2⎭
Note that we can rewrite { ()}2 as
⎡ ⎤ √ 1 { ()}2 = 2 cos( 2 − 2 ) ⎣ ⎦ 1
(3)
with the same phase angle used for both displacements. Such simplification is not possible if the damping is not proportional. For non-proportional damping, the responses, 1 () and 2 () will have different phase angles. The total solution is (Equation (2) and (3)) { ()} = { ()}1 + { ()}2
⎡
⎡ ⎤ √ √ 1 1 ⎦ + 2 cos( 2 − 2 ) ⎣ ⎦ = 1 exp(−) cos( 3 − 1 ) ⎣ −1 1 ⎤
which are identical to the ones obtained via modal analysis.
461
39. The two degree-of-freedom damped and forced system of Figure 6.78 oscillates about its equilibrium position. Use Lagrange’s equation to formulate the problem and then solve the equations of motion for () = cos . Check whether proportional damping is a valid model here. If not, how can the system be modified so that such a model is valid? Make this modification, and then solve via modal analysis.
Two degree-of-freedom general system.
Solution: Let us first formulate the EOM using Lagrange’s equation. We assume small displacements and rotations. In this way we can assume that the spring and damping forces act essentially in the vertical direction. The kinetic, potential, and Rayleigh dissipation energies are given next: 1 1 1 ̇21 + 2 ̇22 2 2 1 1 = 1 21 + 2 (22 − 1 )2 2 2 1 2 = ̇2 2 =
The strain energy 12 2 (22 − 1 )2 was obtained using similar triangles. The generalized force on mass 1 is 1 = (). We can now apply Lagrange’s equation for each generalized coordinate: ¶ µ + + = ̇ ̇ 1 ̈1 + (1 + 2 ) 1 − 22 2 = cos 2 ̈2 + ̇2 + 42 2 − 22 1 = 0 We can write this in matrix-vector form: ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎬ ⎨ ⎬ ⎬ ⎨ ⎨ ⎨ 0 0 0 1 + 2 −22 ̇1 cos ⎬ ̈1 1 ⎣ 1 ⎦ ⎦ ⎦ +⎣ +⎣ = ⎭ ⎩ ⎭ ⎩ −2 4 0 ⎩ ̈ ⎭ 0 ⎩ ̇ ⎭ 0 2
2
2
2
2
2
462
CHAPTER 6 MULTI DOF VIBRATION
We see that the property matrices are symmetrical, as expected. Because the damping matrix has a zero entry in row 1 column 1 above, the damping is non-proportional. There are many ways to modify our current problem so that damping becomes proportional. One possible modification is to add another damper at mass 1 so that [] = [] Then = 12 1 ̇21 + 1 ̇2 The EOM then becomes 2 2 2 ⎫ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎡ ⎬ ⎬ ⎬ ⎨ ⎨ ⎨ ⎨ 1 0 1 + 2 −22 cos ⎬ ̈1 ̇1 1 0 ⎦ ⎦ ⎦ ⎣ 1 +⎣ +⎣ = ⎭ ⎩ ⎭ ⎩ 0 ⎩ ̈ ⎭ 0 ⎩ ̇ ⎭ 0 −2 4 2
2
2
2
2
2
2
where = 2 2 and 1 = 2 1 2 In this instance, modal analysis for proportional damping can be applied. We can solve these equations for forced vibration, assuming zero initial conditions. Response The first step is to solve the undamped, free vibration problem for the eigenvalues and eigenvectors. The eigenvalue problem is given by ¢ ¡ 2 − [] + [] {} = {0}
The second step is to let { ()} = [ ] { ()} where { ()} is the modal coordinate: [] [ ] {̈ ()} + [] [ ] {̇ ()} + [] [ ] { ()} = { ()} The third step is to multiply the matrix EOM by [ ] from the left. We now have
[ ] [] [ ] {̈ ()} + [ ] [] [ ] {̇ ()} + [] [ ] { ()} = { ()} ⎫ ⎫ ⎡ ⎫ ⎧ ⎧ ⎤⎧ 2 ⎨ ̇ () ⎬ ⎨ () ⎬ ⎨ ̈ () ⎬ 0 1 1 1 1 ⎦ + +⎣ = [ ] { ()} ⎩ ̇ () ⎭ ⎩ ̈ () ⎭ 0 22 ⎩ 2 () ⎭ 2 2
The damping ratios are = (2 ) Depending on () can be over, under or critically damped. The initial conditions are transformed using { ()} = [ ] [] { ()} Once we obtain { ()} we transform it back to the original coordinate using { ()} = [ ] { ()} to find the final solution.
463
40. Solve Example 6.14 with the equation of motion ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎨ ⎬ ⎬ ⎨ ⎨ ̈1 1 () ⎬ 1 0 3 −2 ⎣ ⎦ ⎣ ⎦ + = 0 ⎩ ̈ ⎭ −2 2 ⎩ ⎭ ⎩ () ⎭ 2
2
2
p for 1 () = cos(066 ) and 2 () = 0. Plot the displacement time histories.
Solution: In Example 6.14 we performed a free vibration analysis leading to frequencies 12 = p p 0662 214 The normalized modes were found to be: ⎧ ⎫ ⎨ ⎬ 0615 1 {}1 = √ ⎩ 0788 ⎭
⎧ ⎫ ⎨ ⎬ 0788 1 {}2 = √ ⎩ −0615 ⎭
To account for the forcing, we need to calculate the following product:
{()} = [ ] { ()} ⎫ ⎡ ⎤⎧ p ⎬ ⎨ ) 0615 0788 cos(0660 1 ⎣ ⎦ = √ ⎭ 0788 −0615 ⎩ 0 ⎫ ⎧ p 1 ⎨ 0615 cos(0660 ) ⎬ = √ p ⎩ 0788 cos(0660 ) ⎭
These are now the forcing functions for the modal EOM. Assuming zero initial conditions, we have: 1 1 () = 1
Z
1 ( ) sin 1 ( − ) Z p p 1 1 p √ = 0615 cos(0660 ) sin 0662 ( − ) 0662 0 √ ³ ´ p p = 233 cos 0660 − cos 0662 Z 1 2 ( ) sin 2 ( − ) 2 () = 2 0 Z p p 1 1 √ p = 079 cos(0660 ) sin 214 ( − ) 2136 0 √ ³ ´ p p = 01914 cos 0660 − cos 214 0
464
CHAPTER 6 MULTI DOF VIBRATION
Once these are integrated, the physical responses are given by {()} = [ ]{()} : ⎧ ⎫ ⎫ ⎡ ⎤⎧ ⎨ () ⎬ ⎨ 1 ⎣ 0615 0788 ⎦ 1 () ⎬ 1 √ = ⎩ () ⎭ 0788 −0615 ⎩ 2 () ⎭ 2 ⎫ ⎧ ⎨ ³ ´ p p 1 0615 ⎬ 233 cos 0660 − cos 0662 = ⎩ 0788 ⎭ ⎧ ⎫ ⎨ ´ ³ p p 0788 ⎬ 1 + 01914 cos 0660 − cos 214 ⎩ −0615 ⎭
Figure below shows 12 () as functions of
p
465 p 41. In Problem 40,solve the same equation of motion, except with 1 () = cos(2 ) and 2 () = 0. Plot the displacement time histories. Solution: In Example 6.14 we performed a free vibration analysis leading to frequencies 12 = p p 0662 214 The normalized modes were found to be: ⎫ ⎫ ⎧ ⎧ ⎬ ⎨ ⎨ 0615 0788 ⎬ 1 1 {}2 = √ {}1 = √ ⎩ 0788 ⎭ ⎩ −0615 ⎭
To account for the forcing, we need to calculate the following product:
{()} = [ ] { ()} ⎫ ⎡ ⎤⎧ p ⎨ 0615 0788 cos(200 ) ⎬ 1 ⎦ = √ ⎣ ⎭ 0788 −0615 ⎩ 0 ⎫ ⎧ p ⎬ ⎨ ) 0615 cos(200 1 = √ p ⎩ 0788 cos(200 ) ⎭
These are now the forcing functions for the modal EOM. Assuming zero initial conditions, we have: Z 1 1 () = 1 ( ) sin 1 ( − ) 1 0 Z p p 1 1 p √ 0615 cos(200 ) sin 0662 ( − ) = 0662 0 √ p p = 0173(cos 0662 − cos 200 ) Z 1 2 ( ) sin 2 ( − ) 2 () = 2 0 Z p p 1 1 p √ 0788 cos(200 ) sin 214 ( − ) = 214 0 √ ´ ³ p p = 136 cos 200 − cos 214 Once these are integrated, the physical responses are given by {()} = [ ]{()} : ⎧ ⎫ ⎫ ⎡ ⎤⎧ ⎨ () ⎬ ⎨ 1 ⎣ 0615 0799 ⎦ 1 () ⎬ 1 = √ ⎩ () ⎭ 0799 −0615 ⎩ 2 () ⎭ 2 ⎫ ⎧ ⎨ p p 1 0615 ⎬ 0173(cos 0662 − cos 200 ) = ⎩ 0788 ⎭ ⎧ ⎫ ⎨ ´ ³ p p 0788 ⎬ 1 136 cos 200 − cos 214 + ⎩ −0615 ⎭
466 Figure below shows 12 () as functions of
CHAPTER 6 MULTI DOF VIBRATION p
467
42. In Problem 40,solve for the response where the system damping is (a) [] = 2[], (b) [] = 3[], (c) [] = 2[] + 3[]. In each discuss the results, especially regarding how the p damping for each mode depends on the proportional damping model. Let be 1 rad/s if necessary. Solution: The difference between this problem with damping and the previous problem without damping is that the convolution (Duhamel) responses must now include the more general impulse response function. The modal analysis is appropriate because damping is proportional. Modal Analysis The EOM from Problem 40 is given by ⎧ ⎫ ⎡ ⎫ ⎤⎧ ⎫ ⎧ ⎡ ⎤⎧ ⎫ ⎨ ̇ ⎬ ⎨ ⎬ ⎨ () ⎬ 3 −2 0 ⎨ ̈1 ⎬ 1 1 1 ⎦ ⎣ ⎦ + [] +⎣ = ⎩ ̇ ⎭ −2 2 ⎩ ⎭ ⎩ () ⎭ 0 ⎩ ̈ ⎭ 2
2
2
2
p for 1 () = cos(0660 ) and p Example 6.14 we performed a free vibration analysis 2 () = 0. In p leading to frequencies 12 = 0662 214 The normalized modes were found to be: ⎫ ⎫ ⎧ ⎧ ⎬ ⎬ ⎨ ⎨ 0615 0788 1 1 {}2 = √ {}1 = √ ⎩ 0788 ⎭ ⎩ −0615 ⎭ ⎡ ⎤ 0615 0788 1 ⎦ [ ] = √ ⎣ 0788 −0615
We let { ()} = [ ] { ()} and multiply the EOM by [ ] from the left to obtain the EOM in terms of modal coordinate: ⎫ ⎧ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎨ 1 () ⎬ ⎨ ̈1 ⎬ ⎨ ̇1 ⎬ ⎨ 1 ⎬ [ ] [] [ ] + [ ] [] [ ] + [ ] [] [ ] = [ ] ⎩ () ⎭ ⎩ ̈ ⎭ ⎩ ̇ ⎭ ⎩ ⎭ 2 2 2 2 ⎧ ⎫ ⎡ ⎧ ⎫ ⎧ ⎫ ⎤⎧ ⎫ 2 ⎨ ̇ ⎬ ⎬ ⎨ ⎨ ⎨ ̈ ⎬ 1 0 1 () ⎬ 1 1 1 ⎦ + [ ] [] [ ] +⎣ = [ ] (1) ⎩ ̇ ⎭ ⎩ () ⎭ ⎩ ̈ ⎭ 0 2 ⎩ ⎭ 2
2
2
2
where we let
{()} = [ ] { ()} ⎫ ⎡ ⎤⎧ p ⎨ cos(0660 ) ⎬ 0615 0788 1 ⎦ = √ ⎣ ⎭ 0788 −0615 ⎩ 0 ⎫ ⎧ p ⎨ 0615 cos(0660 ) ⎬ 1 = √ p ⎩ 0718 cos(0660 ) ⎭
2
468
CHAPTER 6 MULTI DOF VIBRATION
These are now the forcing functions for the modal EOM. Assuming zero initial conditions, we have: Z 1 () = 1 ( )1 ( − ) 0 Z 2 ( )2 ( − ) (2) 2 () = 0
The response is given by {()} = [ ] {()} Impulse Response Functions The general response for each mode is given by Z () = ( ) ( − ) 0
where () is the impulse response function, which is the solution to ̈ () + 2 () + 2 () = () ¡ ¢ (0) = 0 and ̇ 0− = 0
Above problem is posed as an initial value problem such that
̈ () + 2 () + 2 () = 0 ¡ ¢ (0) = 0 and ̇ 0+ = 1
For a underdamped system, the impulse response function is given by q 1 () = p exp (− ) sin 1 − 2 2 1 − 1 exp (− ) sin =
(3)
For a critically damped system, the impulse response function is given by () = exp (− ) For an overdamped system, the impulse response function is given by µ ∙µ ¶ ¸ ∙µ ¶ ¸¶ q q 1 2 2 () = p 2 exp − + − 1 − exp − − − 1 2 − 1
(4)
(5)
It should be noted that if the forcing function is simple, it may be easier to use the trial solution method than to use the convolution integral. Responses for Part (a), (b) and (c)
469 (a) [] = 2[] : For this case, = = 2 s−1 and = 0 The damping matrix is ⎡ ⎤ ⎡ ⎤ 0 2 0 ⎦=⎣ ⎦ [] = 2 ⎣ 0 0 2 and Equation (1) is ⎫ ⎧ ⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ 2 ⎨ ⎬ ⎬ ⎨ ⎨ ⎨ ̈ ⎬ 0 2 0 ̇1 1 1 () ⎬ 1 1 ⎦ ⎦ +⎣ +⎣ = [ ] ⎩ () ⎭ ⎩ ̈ ⎭ 0 2 ⎩ ⎭ 0 2 ⎩ ̇ ⎭ 2
2
2
2
(6)
2
We need to evaluate and for each DOF: ¶ µ 1 1 = + 1 2 1 Ã ! p 2 1 p = = 151 2 0662 ¶ µ 1 + 2 2 = 2 2 Ã ! p 2 1 p = = 0468 2 214
We can immediately see that the first modal response decays with exponential constant of 1 1 = 151 × 0662 = 100, and the second response at 2 2 = 0468 × 214 = 100. Both modal responses decay at the same rate as expected since the term equals cannot p 0 and are identical. Wep proceed with the response unless the numerical value for is given. Depending on , the response can be overdamped, critically damped, or underdamped. As a numerical example, we obtain the responses using = 1 s−1 Then, 1 () is overdamped with 1 = 151, and 2 () is underdamped with 2 = 0467. The impulse response functions are given in Equations (5) and (3) or µ ∙µ ¶ ¸ ∙µ ¶ ¸¶ q q 1 2 2 exp − 1 + 1 − 1 1 − exp − 1 − 1 − 1 1 1 () = p 2 2 1 − 1 ³ h³ ´ i h³ ´ i´ √ √ 1 2 2 exp −151 + 151 − 1 0662 − exp −151 − 151 − 1 0662 = √ 2 1512 − 10662 = 0668 {exp (−0250) − exp (−175)} 1 exp (− 2 2 ) sin 2 2 √ 1 √ exp (−0468 × 214) sin 214 1 − 04682 = 214 1 − 04682 = 0529 (sin 189) exp (−100)
2 () =
470
CHAPTER 6 MULTI DOF VIBRATION
The impulse response functions and the modal forces are substituted into Equation (2) to find 1 () and 2 () such that
1 () =
Z 0
Z
1 ( )1 ( − )
1 √ 0615 cos(0660 )0668 {exp (−0250 ( − )) − exp (−175 ( − ))} 0 1 = √ {0000672 cos(0660) + 0467 sin(0660) −0206 exp(−0250) + 0206 exp(−175)} =
and
2 () =
Z 0
Z
2 ( )2 ( − )
1 √ 0788 cos(0660 ) {0529 (sin 189 ( − )) exp (−100 ( − ))} 0 1 = √ {0173 cos(0660) + 00551 sin(0660) −0173 exp(−100) cos(189) − 0111 exp(−100) sin(189)} =
Note that you may find it easier to solve Equation (6) directly without using Duhamel’s integral. The response is then given by ⎧ ⎫ ⎨ () ⎬
⎫ ⎤⎧ ⎬ ⎨ () 1 ⎣ 1 1 ⎦ = √ ⎩ () ⎭ 0788 −0615 ⎩ 2 () ⎭ 2 ⎤ ⎡ () + 0788 () 0615 1 ⎣ 1 2 ⎦ = 07881 () − 06152 () ⎡
0615 0788
Figure next shows the responses 1 () and 2 (). Note that since we assumed = 1 we can write 1 () as 1 ()
471
Responses for Probem 8.42, Part (a). It is assumed = 1 s−1 (b) [] = 3[] : For this case, = 0 and = 3 s: ⎡ ⎤ ⎡ ⎤ 3 −2 9 −6 ⎦=⎣ ⎦ [] = 3 ⎣ −2 2 −6 6
We need to evaluate and for each DOF: ¶ µ 1 1 = + 1 2 1 ´ p p 1³ 3 × 0662 = 0993 = 2µ ¶ 1 + 2 2 = 2 2 ´ p p 1³ 3 × 214 = 321 = 2 We can immediately see that p the first modal response decays with an exponential decay p constant p 1 = 0993 × 0662 = 0657, and the second response at 2 2 = 321 × of 1p 214 = 687 The second modal response decays much more rapidly than the first. As a numerical example, let = 1 Then, the impulse response functions are (Equations (3) and (5)) q 1 1 () = p exp (− 1 1 ) sin 1 − 21 1 2 1 − 1 1 √ 1 exp (−0657) sin 1 − 09932 0662 = √ 1 − 09932 0662 = 128 (sin 00782) exp (−0657)
472
CHAPTER 6 MULTI DOF VIBRATION
µ ∙µ ¶ ¸ ∙µ ¶ ¸¶ q q 2 2 exp − 2 + 2 − 1 2 − exp − 2 − 2 − 1 2
1
p 2 22 − 12 h³ ´ i √ 1 2 exp −321 + 321 − 1 214 = √ 2 3212 − 1214 h³ ´ i √ 1 2 √ exp −321 − 321 − 1 214 − 2 3212 − 1214 = 00766 {exp (−0342) − exp (−134)}
2 () =
The modal coordinates are given by (Equation (2)) 1 () =
Z 0
Z
1 ( )1 ( − )
1 √ 0615 cos(0660 )128 (sin 00782 ( − )) exp (−0657 ( − )) 0 1 = √ {000177 cos(0660) + 0710 sin(0660) −000177 exp(−0657) cos(00782) − 601 exp(−0657) sin(00782)}
=
and 2 () =
Z 0
Z
2 ( )2 ( − )
1 √ 0788 cos(0660 )00766 {exp (−0342 ( − )) − exp (−134 ( − ))} 0 = 00257 cos(0660) + 00561 sin(0660) − 00292 exp(−0342) + 000351 exp(−134)
=
The response is then given by ⎫ ⎫ ⎧ ⎡ ⎤⎧ ⎬ ⎨ () ⎬ ⎨ () 0615 0788 1 ⎣ 1 1 ⎦ = √ ⎩ () ⎭ 0788 −0615 ⎩ 2 () ⎭ 2 ⎤ ⎡ () + 0788 () 0615 1 ⎣ 1 2 ⎦ = 07881 () − 06152 () Figure next shows the responses 1 () and 2 (). (c) [] = 2[] + 3[] : For this case, = 2 s−1 and = 3 s: ⎡
[] = 2 ⎣
0 0
⎤
⎡
⎦ + 3⎣
3 −2 −2 2
⎤
⎡
⎦=⎣
2 + 9
−6
−6
2 + 6
⎤
⎦
473
Figure 6.1: Responses for Probem 8.42, Part (b). It is assumed = 1 s−1
We need to evaluate and for each DOF: 1 1 = 2
µ
Ã
+ 1 1
¶
! p 2 p + 3 × 0662 0662 ¶ µ 1 + 2 2 = 2 2 Ã ! p 2 1 p + 3 × 214 = 2 214 1 = 2
Again, assume = 1 s−1 for numerical purposes. Then, both 1 () and 2 () are overdamped with 1 = 250 and 2 = 367 The impulse response functions are (Equation (5)) 1
µ ∙µ ¶ ¸ ∙µ ¶ ¸¶ q q 2 2 exp − 1 + 1 − 1 1 − exp − 1 − 1 − 1 1
p 2 21 − 11 = 0330 {exp (−0138) − exp(−31718)} µ ∙µ ¶ ¸ ∙µ ¶ ¸¶ q q 1 2 2 2 () = p 2 exp − 2 + 2 − 1 2 − exp − 2 − 2 − 1 2 2 2 − 12 = 00660 {exp (−0296) − exp(−155)} 1 () =
474
CHAPTER 6 MULTI DOF VIBRATION
The modal coordinates are given by (Equation (2)) Z 1 () = 1 ( )1 ( − ) 0 Z 1 √ 0615 cos(0660 )0330 {exp (−0138 ( − )) − exp(−317 ( − ))} = 0 1 = √ {000272 cos(0660) + 0282 sin(0660) −00616 exp(−0138) + 00613 exp(−317)} and
2 () =
Z 0
Z
2 ( )2 ( − )
1 √ 0788 cos(066 )00660 {exp (−0296 ( − )) − exp(−155 ( − ))} 0 1 = √ {00261 cos(0660) + 00655 sin(0660) −00294 exp(−0296) + 000335 exp(−155)}
=
The response is then given by ⎧ ⎫ ⎫ ⎡ ⎤⎧ ⎨ () ⎬ ⎨ 0615 0788 1 () ⎬ 1 1 ⎦ = √ ⎣ ⎩ () ⎭ 0788 −0615 ⎩ 2 () ⎭ 2 ⎤ ⎡ 1 ⎣ 06151 () + 07882 () ⎦ = 07881 () − 06152 ()
Figure next shows the responses 1 () and 2 ().
Responses for Probem 8.42, Part (c). It is assumed = 1 s−1
475
476
CHAPTER 6 MULTI DOF VIBRATION
43. A vibrating mass is found to be oscillating with an amplitude that is too large. To reduce this amplitude, an auxiliary system is added as shown in Figure 6.79. This problem generally occurs when the forcing frequency is too close to the natural frequency of the primary system.
Original system and system with vibration absorber.
The auxiliary system acts as a damped vibration absorber. (a) Derive the equation of motion of the system plus absorber and then solve to determine what values of absorber mass , stiffness 2 , and damping must be selected in order to minimize the vibration amplitude of primary mass . Assume that = 025 and () = 0 cos . (b) Suppose that we can accept a design where 2 1 is selected so that neither natural frequency of the combined system, 1 and 2 , is closer than 5% to the driving frequency . Devise the system that achieves this criterion. Solution: This problem of the damped vibration absorber falls into the category of a 2 DOF damped system with the governing EOM given by ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎨ ⎬ ⎨ ⎬ ⎬ ⎨ ⎨ 0 ̈1 − ̇1 1 + 2 −2 () ⎬ 1 ⎣ ⎦ ⎦ ⎦ +⎣ +⎣ = ⎩ ⎭ ⎩ 0 ⎭ 0 ⎩ ̈ ⎭ − ⎩ ̇ ⎭ − 2
2
2
2
2
where 1 = 3 = 0 2 = 2 = 0 1 = 1 = 2 = and 3 = 0 We also note that the damping matrix can be written as a proportion of the mass and stiffness matrices using the concept of proportional damping. Therefore, the solution to this system is well-known to us. This is a well-known problem with a well-known solution. It is known that, in the absence of damping, that placing a secondary mass and spring that has the same ratio as the ratio of the primary mass and the spring (1 p = 2 with = 0 in our case) will eliminate the response at the original natural frequency of 1 Therefore, the simple answer for part (a) is to let = 0 and 2 = 1 () = 0251 In the next part, we will verify this claim in addition to obtaining the response amplitude at other values of (1 ) (2 ) and in the presence of damping. (a) Response amplitude
477 It is important to note that the response amplitude refers to the amplitude of the response that specifically responds at the forcing frequency The total response would not make sense because it depends on the initial conditions. In order to avoid confusion, let us denote the driving frequency as Let () = 0 exp ( ) and the response is given by { ()} = {} exp ( ). We are selectively looking for the solution at the forcing frequency (particular solution). The EOM is then ⎫ ⎧ ⎫ ⎛⎡ ⎤ ⎡ ⎤⎞ ⎧ ⎤ ⎡ 2 ⎨ − 0 1 ⎬ ⎨ 0 ⎬ − + 2 −2 ⎝⎣ ⎦+⎣ ⎦⎠ ⎦+⎣ 1 = ⎩ ⎭ ⎩0 ⎭ 0 − 2 − −
2
2
2
The response {} is then given by ⎧ ⎫ ⎤−1 ⎧ ⎫ ⎡¡ ¢ 2 ⎨ ⎬ ⎨ ⎬ −2 − 1 + 2 − + 1 0 ⎦ = ⎣ ¡ ¢ ⎩ ⎭ ⎩0 ⎭ −2 − 2 − 2 + 2 ⎫ ⎧ 2 ⎨ 2 − + ⎬ 0 = ⎭ det () ⎩ + 2
where refers to the matrix in the square brackets, and
¡ ¢ det () = 4 − (2 + 1 + 2 ) 2 + 1 2 + 1 − 3 − 3
Note that the imaginary terms indicate that there is a phase shift in the response from the forcing function. The response amplitude of the primary mass is then ¯ ¯ ¯ ¯ 2 − 2 + ¯ ¯ ¡ ¢ |1 | = 0 ¯ ¯ 4 2 3 3 ¯ − (2 + 1 + 2 ) + 1 2 + 1 − − ¯
Also, |1 | = 0 when = 0 2 = 1 and 2 = 1
Use of dimensionless parameters will simplify the expression. Let 2 1 = ̂ = √ ̂ = ̂ 2 = 1 2 1 2
Note that = 1 when the forcing frequency is at the natural frequency of the original system. Then, dividing the top and bottom by 1 2 , we obtain ¯ ¯ 2 −2 + ¯ ¯ ¯ ¯ 1 2 ¯ ¯ ´ ´ ³ |1 | = 0 ¯ 4 ³ 3 3 ¯ 1 2 2 1 2 1 2 ¯ − + + + + − − ¯ 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 ¯ ¯ ¯ 1 − ̂ 2 2 + ̂ 0 ¯¯ ¯ ¢ ¡ ¢ = ¯ 2 4 ¡ ¯ 1 ¯ ̂ − 1 + ̂2 + ̂ 2 + 1 + ̂̂ 2 1 − 2 − ̂2 2 ¯ v ¡ ¢ u 2 2 2 + (̂)2 1 − ̂ 0 u t = ¡ ¢ ¢2 ¡ ¡ ¡ 2 ¢¢2 1 ̂ 4 − 1 + ̂2 + ̂ 2 + 1 + ̂̂ 2 1 − 2 − ̂2 2
478
CHAPTER 6 MULTI DOF VIBRATION
We can plot the dimensionless response amplitude of the primary mass, |1 | (0 ) as a function of the dimensionless forcing frequency, for various values of ̂, ̂ and ̂ The figures that follows shows the dimensionless response amplitude (also called magnification ratio) for ̂ = 025 and ̂ 2 = 1. It is reasonable to assume that the forcing frequency is at the natural frequency of the original system. Therefore, we are interested in the response amplitude when = 1 We expect to see zero response at = 1 and ̂ = 0 as we claimed earlier.
Magnification ratio as a function of dimensionless forcing frequency. ̂ is varied from 0 to 1.4 at 0.2 interval.
The MATLAB code that was used to create the figure above is given: clear mr=0.25; r=[0:0.002:1.5]; cr=[0:0.2:1.4]’; wr=1; W=1; [R Cr]=meshgrid(r,cr); Amp=sqrt(((1-R.^2./W.^2).^2+(Cr.*R).^2)./((R.^4./W.^2 ... -(1+1./W.^2+mr).*R.^2+1).^2+(Cr./W.^2.*R.*(1-R.^2-R.^2./W.^2)).^2)); plot(r, Amp) axis([0 1.5 0 12])
479 Since we are interested in the response amplitude when = 1 only, we can plot |1 ( = 1)| (0 1 ) as a function of ̂ 2 (recall ̂2 = (1 ) (2 )) and ̂ while keeping ̂ = 025
Magnification ratio when the system is forced at the natural frequency of the original system ( = 1) as a function of (1 ) (2 ). ̂ is varied from 0 to 1.4 at 0.2 interval.
The MATLAB code that was used to create the figure above is given: clear mr=0.25; R=1; cr=[0:0.2:1.4]’; wr=[0:0.02:5]; [W Cr]=meshgrid(wr,cr); Amp=sqrt(((1-R.^2.*W.^2).^2+(Cr.*R).^2)./((R.^4.*W.^2 ... -(1+W.^2+mr).*R.^2+1).^2+(Cr.*W.^2.*R.*(1-R.^2-R.^2.*W.^2)).^2)); plot(wr, Amp) Note that ̂ 2 = 0 when the secondary mass is zero. As we have known all along, we can completely eliminate the response if we set ̂ = 0 and select the secondary system such that 2 = 1 When 2 = 1 the damping increases the response amplitude. Another way to reduce
480
CHAPTER 6 MULTI DOF VIBRATION
the response amplitude is select the system with large (1 ) (2 ) or simply a more flexible secondary system (small 2 ) than the original system. This may be impractical because the displacement of the secondary mass may become too large. It is interesting to note that increasing damping does not always reduce the response amplitude. For low (1 ) (2 ) it is beneficial to have low damping while the opposite is true for high (1 ) (2 ) The value of (1 ) (2 ) when the effect of damping is reversed is a function of In this case where = 025 (1 ) (2 ) = 1086 is when the effect of damping is reversed. Small damping is desirable for (1 ) (2 ) 1086, and large damping is desirable for (1 ) (2 ) 1086 When (1 ) (2 ) = 1086 the damping has no effect on the response amplitude. (b) Selection of 2 so that natural frequencies are not close to the forcing frequency The natural frequency squared are the eigenvalue of the eigenvalue problem given by ¢ ¡ [] − 2 [] {} = {0}
Note that damping does not enter into the natural frequency calculations. In our case, we can write ⎡ ⎤ 1 + 2 − 2 −2 ⎣ ⎦ {} = {0} 2 −2 2 − The characteristic equation is the determinant of the matrix and is given by 4 − (2 + 1 + 2 ) 2 + 1 2 = 0 Re-write as 4
−
µ
1 2 2 + +
¶
2 +
1 2 = 0
Divide by (1 )2 so that the characteristic equation becomes ¶ µ 2 2 2 4 ̂ 2 + +1+ = 0 ̂ − 1 1 1 ¶ µ 2 2 2 4 ̂ − ̂ 2 + +1+ = 0 1 1 1 p where ̂ = 1 There are two dimensionless parameters that determine the frequency ratio (the ratio of the natural frequency to the forcing frequency), and they are 2 and 1 where it is given that = 025 Note that we can also use 1 and 2
481 which are the dimensionless parameters used in part (a). The criteria that the natural frequencies are not within 5% of the forcing frequency can be written as ̂ 105 and ̂ 095 The following figure shows the two natural frequencies (normalized by the natural frequency of the original system) for various values of 2 1 with vertical bars at ̂ = 095 and 105
The MATLAB code that generated the figure is: clear mratio=0.25; Rratio=[0:0.05:1.5] for i=1:length(Rratio); rr=Rratio(i); rsquare=roots([1 -(rr+1+rr*mratio) rr]); rsquare=sort(rsquare); r1(i)=sqrt(rsquare(1)); r2(i)=sqrt(rsquare(2)); end plot(Rratio, r1,’k’, Rratio, r2,’k’)
482
CHAPTER 6 MULTI DOF VIBRATION hold on plot(0.95*ones(16), [0:0.1:1.5],’k’) plot(1.05*ones(16), [0:0.1:1.5],’k’) ylabel(’\it omega_n/omega_f’) xlabel(’\it k_2/m\omega_f^2’)
As it turns out, it is not too difficult to meet the condition that the natural frequencies are not within 5% of the forcing frequency. We need to select 2 1 such that it is greater than 0.075. The problem asks for the value of 2 that will satisfy this criteria. We cannot give an explicit value for 2 but as a ratio (2 ) (1 ) such that 2 2 1 = 1 2 2 = 1 1 For = 025, we have 2 1 03 or 33 1 2 That is, the natural frequency of a system composed of the secondary mass and the spring must be √ greater than 03 times the natural frequency of the original system. This answers the questions posed in this problem. We have concluded in part (a) that the best secondary system is zero or light damping with (1 ) (2 ) = 1 The additional criteria of (1 ) (2 ) 33 does not alter our previous conclusions.
483 Problems for Section 6.9.1 – Unrestrained Systems 44. Figure 6.80 may be used as a simple model of a mobile robotic arm undergoing forced motion. The forcing is at the base of the arm. Derive the matrix equation of motion, and solve it in general for the response if (a) () = (), and (b) () = cos 2. Solve this problem in two ways: (i) using a constraint matrix, and (ii) directly without a constraint matrix.
Simple model of a robotic arm.
Solution: The equations of motion are given by 1 ̈1 = + (2 − 1 ) + (̇2 − ̇1 ) 2 ̈2 = −(2 − 1 ) − (̇2 − ̇1 ) In matrix-vector form, we have ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎤⎧ ⎫ ⎡ ⎬ ⎨ ⎬ ⎨ 0 − ̇1 − ⎨ 1 ⎬ ̈1 ⎣ 1 ⎦ ⎦ ⎦ +⎣ +⎣ ⎭ ⎭ ⎩ ⎩ ̈2 ̇2 0 2 − − ⎩ 2 ⎭ ⎧ ⎫ ⎨ ⎬ = ⎩0 ⎭
Let us call the mass matrix [] damping matrix [] and stiffness matrix [] In our case, we have a proportional damping so that [] = [] We use modal analysis to find the solution to the current problem. The steps that we take to solve this problem are given as follows: 1. Solve the eigenvalue problem given by − 2 [] {} + [] {} = 0 2. Expand the solution to the current problem in terms of the eigenvectors found in the previous step such that { ()} = {}1 1 () + {}2 2 ()
484
CHAPTER 6 MULTI DOF VIBRATION
where () are modal coordinates to be determined. 3. Substitute into the EOM and use the orthogonality property to find the equations for the modal coordinates, ̈ () + 2 ̇ () + 2 () = () where () is called the modal force and is given by () = {} { ()} 4. Solve for () We are given zero initial conditions for this problem, and the solution can be written using the convolution integral, Z () = ( − ) ( ) 0
where () is the impulse response function or the solution to ̈ () + 2 ̇ () + 2 () = () and zero initial conditions. When the solution () is obvious, the convolution integral is not necessary. 5. Write the solution, { ()} = {}1 1 () + {}2 2 () It should be noted that the constraint matrix method is applicable for the first step only. There is no reason why the momentum of the system be conserved when there is an external force. Step 1: Eigenvalue problem Eigenvalue Problem Using the Constraint Matrix The eigenvalue problem is given by ⎧ ⎫ ⎡ ⎤⎞ ⎧ ⎫ ⎛⎡ ⎤ ⎬ ⎨0⎬ ⎨ 0 1 − ⎦⎠ ⎝⎣ ⎦ − 2 ⎣ 1 = ⎩0⎭ ⎩ ⎭ 0 2 − 2 ⎧ ⎫ ⎤⎧ ⎫ ⎡ ⎨0⎬ ⎨ ⎬ − − 2 1 1 ⎦ ⎣ = ⎩0⎭ − − 2 ⎩ ⎭ 2
2
The frequency equation is given by ¡ ¢¡ ¢ − 2 1 − 2 2 − 2 = 0 1 2 4 − (1 + 2 ) 2 = 0 The natural frequencies are 0 = 0 1 = The corresponding mode shape is {}0 =
r
1 + 2 1 2
⎧ ⎫ ⎨1⎬ ⎩1⎭
485 The second mode shape is {}1 =
⎧ ⎨
1
⎫ ⎬
⎩ − ⎭ 1
2
The second mode shape can also be obtained from conservation of linear momentum. The mode shapes of the homogeneous problem are related such that 1 1 + 2 2 = 0 or 2 = −1 1 2 The homogeneous problem can be written entirely in terms of 1 . The constraint matrix relates {1 2 } to 1 such that ⎫ ⎧ ⎫ ⎧ ⎬ ⎨ ⎬ ⎨ 1 1 = ⎩ ⎭ ⎩ − ⎭ 1 2 1 2 {z } | [{]
We can get the second natural frequency back from the equation of motion written in terms of 1 only such that [{] [][{]̈1 + [{] [][{]1 = 0 µ ¶ ¶2 µ 1 1 ̈1 + 1 + 1 = 0 1 1 + 2 2
where we find the natural frequency associated with the second mode is ´ ³ 1 1+ 2 21 = 1 (1 + 2 ) = 1 2 The mode shape that corresponds to this frequency is {}1 = [{] ⎫ ⎧ ⎬ ⎨ 1 = ⎩ − ⎭ 1
2
The mode shapes can be normalized with respect to the mass matrix so that {} [] {} = 1 The normalized mode shapes are ⎧ ⎫ ⎨1⎬ 1 {}0 = √ 1 + 2 ⎩ 1 ⎭ ⎧ ⎫ r ⎨ ⎬ 2 1 1 {}1 = √ 1 + 2 1 ⎩ − ⎭ 1
2
486
CHAPTER 6 MULTI DOF VIBRATION
Eigenvalue Problem Without Using Constraint Matrix First we solve the free vibration problem. Assume the following harmonic response in free undamped vibration: ⎧ ⎫ ⎧ ⎫ ⎨ ⎬ ⎨⎬ 1 = ⎩ ⎭ ⎩⎭ 2
Differentiating this and substituting into the matrix EOM, combining terms and canceling the exponential that is common to all terms, we find: ⎡ ⎤⎧ ⎫ ⎧ ⎫ 2 ⎨⎬ ⎨0⎬ − 1 − ⎣ ⎦ (1) = − − 2 ⎩ ⎭ ⎩ 0 ⎭ 2
The determinant of the characteristic matrix yields a quadratic in 2 that is solved for the two natural frequencies. The characteristic equation is then ¢¡ ¢ ¡ − 1 2 − 2 2 − 2 = 0 1 2 4 − (1 + 2 ) 2 + 2 − 2 = 0 4 − (1 + 2 ) 2 = 0 ¤ £ 1 22 1 2 − (1 + 2 ) 2 = 0
The eigenvalues (natural frequency squared) are
20 = 0 and 21 =
(1 + 2 ) 1 2
The mode shape associated with each natural frequency is obtained by substituting the natural frequencies into the eigenvalue problem (Equation (1)). However, for 2-DOF unrestrained system, it is easy to obtain them using our physical intuition. The mode shape that is associated with 0 = 0 must be { 1 1 } because it must correspond to the rigid body motion. There must be no relative motion and no deformation in the spring. The mode shape associated with 1 must be such that there is no translation. The center of mass does not translate, and this can only happen if the ratio of the displacements of 1 to that of 2 is −2 1 Therefore, the second mode shape must be { 2 −1 } If we assume that the initial conditions are zero, then we can proceed directly to the forced response. We first need to normalize the above modes in order to use the modal analysis approach. Let us normalize the modes now using the relation {} []{} = 1 Then, let the normalized modes be given by {} : ⎤⎧ ⎫ ⎡ ⎨1⎬ 0 1 1 2 ⎦ ⎣ = 1 =⇒ 0 = √ 0 {1 1} 0 2 ⎩ 1 ⎭ where = 1 + 2
487 The normalized {}1 is found by ⎡ 21 {2
⎫ ⎤⎧ ⎨ 2 ⎬ 1 0 ⎦ ⎣ = 1 − 1 } 0 2 ⎩ −1 ⎭
=⇒
1 1 = √ 1 2
where = 1 + 2
Then,
⎫ ⎧q 2 ⎬ ⎨ 1 1 q {}1 = √ ⎩ − 1 ⎭ 2
Step 2-3: Equation of Motion for Modal Coordinates via Modal Analysis The free response of an undamped and unforced system will be given by: ⎧ ⎫ ⎨ ⎬ 1 = (1 + 2 ) {}0 + {}1 cos (1 − 1 ) ⎩ ⎭ 2
Our goal is to find the forced response of a damped system. The damping is proportional in our case, and we will perform modal analysis. We assume that the solution to our current problem can be expanded in terms of modes of the undamped and unforced system. That is, { ()} = {}0 0 () + {}1 1 () where the modal coordinates, 0 () and 1 () are to be found. In a matrix notation, we write { ()} = [] { ()} where the column of the modal matrix [ ] is the mode shape, {} Substituting this and its second derivative into the forced matrix-vector EOM, and left-multiplying by {}0 and {}1 respectively, yields the decoupled EOM in terms of modal coordinates 0 () and 1 () Use the orthogonality given by {} [] {} = {} [] {} = 2 to simplify the equation. Note that [] = [] and another orthogonality property holds. {}0 [] {}0 = 2 Simplification yields the ordinary differential equations for the modal coordinates given by ⎫ ⎧ ⎨ () ⎬ 1 = √ () ̈0 + 20 ̇0 + 20 0 = {}0 ⎭ ⎩0 ⎧ ⎫ r ⎨ () ⎬ 2 1 2 2 =√ () ̈1 + 1 ̇1 + 1 1 = {}1 ⎩0 ⎭ 1
488
CHAPTER 6 MULTI DOF VIBRATION
We call the right-hand sides modal forces or () In our case, 1 0 () = √ () r 1 2 () 1 () = √ 1 Using the convolution integral, we have () =
Z 0
( − ) ( )
The impulse response function, 0 () is the solution to ̈0 = () with zero initial conditions. The solution can be readily found if we realize that impulse force changes velocity instantaneously. The change in velocity is equivalent to the impulse/mass. That is, ⎧ ⎧ ⎫ ⎫ ⎨ ⎨ ⎬ ⎬ ̈0 = 0 ̈0 = () =⇒ ⎩ (0) = 0 and (0+ ) = R 0+ () = 1 ⎭ ⎩ (0) = 0 and (0− ) = 0 ⎭ 0 0 0 0 0−
The impulse response function is
0 () =
The impulse response function, 1 () is the solution to ̈1 + 21 ̇1 + 21 1 = () with zero initial conditions. The form of the impulse response function depends on the value of 1 ( 1 = 1 2) If the system is underdamped, the impulse response function is given by 1 () =
1 exp (− 1 1 ) sin ( 1 ) 1
If the system is overdamped, the impulse response function is given by 1 () = 1 exp (1 ) + 2 exp (2 ) where 1 and 2 are roots of
2 + 21 + 21 = 0 If the system is critically damped, the impulse response function is given by 1 () = (1 + 2 ) exp () where is the repeated root.
489 We change our forced problem to an initial value problem with zero initial displacement and initial velocity of 1 such that ⎧ ⎫ ⎧ ⎫ 2 2 ⎨ ̈ + 2 ̇ + 2 = () ⎬ ⎨ ⎬ ̈1 + 1 ̇1 + 1 1 = 0 1 1 1 1 1 =⇒ ⎩ (0) = 0 and (0− ) = 0 ⎭ ⎩ (0) = 0 and (0+ ) = R 0+ () = 1 ⎭ 1
1
1
1
0−
Step 4 Forced Response
Part (a) () = () When () = cos 2 the modal coordinate, 0 () is given by 1 0 () = √ (1 − cos 2) 4 We can now find the explicit expression for the modal coordinate that corresponds to the rigid mode, 1 () but not for 2 () The equations of motion were given by 1 ̈1 = √ () r 2 2 1 2 ̈2 + 2 ̇2 + 2 2 = √ () 1 The modal force 1 () is simple so that we do not need to use the convolution integral. When () is a unit step function, the modal coordinate, 1 () is given by 2 1 () = 1 + 2 + √ | {z } 2 | {z } homogeneous particular
where the constant coefficients of integration must be zeros to satisfy the zero initial conditions. Then, 2 1 () = √ (1) 2 We write the solution for 2 () in terms of Duhamel integral (mainly because of its simplicity) r Z 1 2 √ 2 () = ( − ) 2 ( ) 1 0 r Z 1 2 ( ) 2 ( − ) = √ 1 0 r Z 1 2 = √ 2 ( − ) (2) 1 0
490
CHAPTER 6 MULTI DOF VIBRATION
Part (b) () = cos 2 When () = cos 2 the modal coordinate, 1 () is given by 1 1 () = √ (1 − cos 2) 4
(3)
The modal coordinate, 2 () is given by r Z 1 2 √ 2 () = ( − ) 2 ( ) 1 0 r Z 2 1 cos 2 ( − ) 2 ( ) = √ 1 0 Step 5 Final Solution Now we can transform back to the physical domain 12 : ⎧ ⎫ ⎫ ⎧ ⎫ ⎧q 2 ⎨ () ⎬ ⎬ ⎨ ⎬ ⎨ 1 1 1 1 1 q () =√ 1 () + √ ⎩ () ⎭ ⎩1⎭ ⎩ − 1 ⎭ 2 2 2
where 1 () and 2 () are given in Equations (1)-(2) for case (a) and (3)-(4) for case (b).
(4)
491
45. The system shown in Figure 6.81 undergoes rotational motion. Derive the equations of motion and solve for the response in terms of the initial conditions, given by 1 (0) ̇1 (0) 2 (0), and ̇2 (0). Solve this problem in two ways: (a) using a constraint matrix, and (b) directly without a constraint matrix.
Two disk shaft system.
Solution: It is assumed that the ends are free to rotate so that this is an unrestrained problem. Example 6.1 and Problem 1 show how to solve such a problem using the constraint matrix for a three-DOF system and a two-DOF system, respectively. We should note that we use the constraint matrix to somewhat simplify the eigenvalue problem. The equation of motion is given by n o [] ̈ () + [] {} = 0
where
⎡
[] = ⎣
1 0 0 2
⎤
⎡
⎦ and [] = ⎣
2 −2 −2 2
⎤
⎦
Before we continue, it is worthwhile to remind ourselves what are and In this problem, is the polar mass moment of inertia of the disk. For a solid disk with mass and radius it is Z 2 =
1 2 2 where is the distance from the rotational axis (also called the beam axis). It should be noted that it is sometimes denoted as to distinguish it from the polar area moment of inertia, which is also often denoted as The polar area moment of inertia is Z = 2 =
=
1 2 2
492
CHAPTER 6 MULTI DOF VIBRATION
where = 2 Note that the area moment inertia can be obtained by replacing with From a course in mechanics of materials, we also know that the end torque, and the torsional angle of deformation, are related by = where is the polar area moment of inertia in this case. Then, the stiffness in this problem is . For a rod with radius of 1 = 4 2 Now that we are clear what and stand for in this problem, let us proceed with the eigenvalue problem. (a) Eigenvalue Problem using the Constraint Matrix We know right away the first natural frequency and the mode shape are ⎧ ⎫ ⎨1⎬ 1 0 = 0 and {}0 = √ 1 + 2 ⎩ 1 ⎭
√ The constant coefficient, 1 1 + 2 was determined by normalizing the mode shape such that {}0 [] {}0 = 1 The subsequent eigenvalues and eigenvectors are found using a reduced eigenvalue problem, which is based on the fact that the momentum must be conserved. That is 1 1 +2 2 = 0 or 2 = −1 1 2 Then, ⎫ ⎧ ⎫ ⎧ ⎬ ⎨ 1 ⎬ ⎨ 1 = ⎩ ⎭ ⎩ − ⎭ 1 2 1 2 | {z } [{]
The equation of motion is written in terms of 1 only such that
[{] [][{]̈1 + [{] [][{]1 = 0 µ ¶ µ ¶2 1 1 ̈1 + 2 1 + 1 1 + 1 = 0 2 2 where we find the natural frequency associated with the second mode is 2 (1 + 1 2 ) 1 2 (1 + 2 ) = 1 2
21 =
The mode shape that corresponds to this frequency is {}1 = [{]1 ⎫ ⎧ ⎬ ⎨ 1 = ⎩ − ⎭ 1 2
493 The mode shape is normalized with respect to the [] matrix so that ⎧ ⎫ r ⎨ ⎬ 2 1 1 √ {}1 = 1 + 2 1 ⎩ − ⎭ 1
2
(b) Eigenvalue Problem without using the Constraint Matrix The eigenvalue problem to be solved is given by
−2 [] {} + [] {} = 0 We first find 2 that will make the determinant of ([] − 2 []) zero. We will find two values, one of which is zero. We then substitute each 2 to find the corresponding eigenvector {} Initial Value Problem The problem asks for the free solution when the system is subject to initial conditions. The free solution is given by ⎧ ⎫ ⎨ () ⎬ 1 = {}0 0 () + {}1 1 () ⎩ () ⎭ 2
= {}0 (1 + 2 ) + {}1 (1 cos 1 + 2 sin 1 )
where the constants of integration, 1 2 1 and 2 are obtained from initial conditions. We have two choices. We can transform the initial conditions and apply them to { ()} or apply the initial conditions to the total solution above. The first method is recommended when we have more than 2 DOFs.
494
CHAPTER 6 MULTI DOF VIBRATION
46. Two identical disks are connected by an elastic shaft of stiffness . The moment of inertia of each disk with respect to the axis of the shaft is . The system is at rest when a constant moment is applied instantaneously to one of the disks. Determine the motion of the system assuming that the mass of the disks is much greater than that of the shaft. Solve this problem in two ways: (a) using a constraint matrix, and (b) directly without a constraint matrix.
Solution: The system is subject to a constant torque Assume that the constant torque is applied to the first disk. The equation of motion is given by ⎫ ⎤⎧ ⎫ ⎧ ⎤⎧ ⎫ ⎡ ⎨ ⎨ ⎬ ⎬ ⎨ − 1 ⎬ 0 ̈1 ⎦ ⎣ ⎦ +⎣ = − ⎩ 2 ⎭ ⎩ 0 ⎭ 0 ⎩ ̈2 ⎭ ⎡
From Problem 2, we found that (with 2 replaced with , 1 and 2 replaced with ) ⎧ ⎫ 1 ⎨1⎬ 0 = 0 and {}0 = √ 2 ⎩ 1 ⎭
⎧ ⎫ ⎨ 1 ⎬ 2 1 and {}1 = √ 21 = 2 ⎩ −1 ⎭
Now, we perform modal analysis. We let { ()} = [] { ()} where [ ] is the modal matrix to find the equations of motion for the modal coordinates. ̈0 = √ 2 ̈1 + 21 1 = √ 2 It is given that the initial conditions are zeros. Then, 2 0 () = √ 2 2 Ã
√ 1 () = 2 2
1 − cos
r
! 2
495 Now we can transform back to the physical domain 12 : ⎫ ⎧ ⎨ () ⎬ 1 = {}0 0 () + {}1 1 () ⎩ () ⎭ 2 ⎫ ⎧ ⎫ ⎧ ⎬ ⎨ ⎨ 2 1 1 ⎬ 1 1 √ √ √ √ (1 − cos 1 ) + = 2 ⎩ 1 ⎭ 2 2 2 ⎩ −1 ⎭ 2 2 ⎧ ⎫ ⎧ ⎫ Ã ! r ⎨ 1 ⎬ ⎨ 1 ⎬ 2 2 + 1 − cos = ⎩ −1 ⎭ 4 ⎩ 1 ⎭ 4
496
CHAPTER 6 MULTI DOF VIBRATION
47. In Example 6.20, derive [] Solution: The element of the stiffness matrix is the torque at disk resulting from unit displacement at the disk. For example, torques needed at disk 1, 2, and 3 in order to have a unit angle of twist at disk 1 (while the angles of twist at other disks are kept at zero) are 1 1 −1 1 and zero, respectively. That is, 11 = 1 1 12 = −1 1 13 = 0 The torques needed at disk 1, 2, and 3 in order to have a unit angle of twist at disk 2 are −1 1 1 1 + 2 2 and −2 2 respectively. That is, 21 = −1 1 22 = 1 1 + 2 2 23 = −2 2 Lastly, the torques needed at disk 1, 2, and 3 in order to have a unit angle of twist at disk 3 are 0 −2 2 and 2 2 respectively. That is, 11 = 0 12 = −2 2 13 = 2 2 The stiffness matrix is then, ⎡
1 1
−1 1
0
⎤
⎥ ⎢ ⎥ ⎢ [] = ⎢ −1 1 1 1 + 2 2 −2 2 ⎥ ⎦ ⎣ 0 −2 2 2 2
497
48. In Example 6.20, derive the eigenvalue problem for the constrained system and show that the frequency and modes are given by Equations 6.135 and 6.136.
Solution: Equations 6.135 and 6.136 are given by ⎫ ⎧ ⎨ 10 ⎬
r
{Θ}1{ = ⎩ 013 ⎭ ⎧ ⎫ r ⎨ 077 ⎬ 2 = 152 {Θ}2{ = ⎩ −10 ⎭
1 = 093
The inertia and the stiffness matrices are given by ⎡ ⎡ ⎤ ⎤ 0 0 −1 0 ⎢ 1 ⎢ 1 ⎥ ⎥ ⎢ ⎢ ⎥ ⎥ [] = ⎢ 0 2 0 ⎥ and [] = ⎢ −1 1 + 2 −2 ⎥ ⎣ ⎣ ⎦ ⎦ 0 0 3 0 −2 2
(6.135)
(6.136)
It is given that 1 = N-m/rad, 2 = 2 N-m/rad, and the inertias are 1 = kg-m2 /rad, 2 = 2 kg-m2 /rad, 3 = 3 kg-m2 /rad. Then ⎤ ⎡ ⎤ ⎡ − 0 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ [] = ⎢ 0 2 0 ⎥ and [] = ⎢ − 3 −2 ⎥ ⎦ ⎣ ⎦ ⎣ 0 −2 2 0 0 3
The momentum is conserved or 1 + 22 + 33 = 0 Let us eliminate 3 Then, ⎧ ⎫ ⎡ ⎤ ⎪ ⎪ ⎧ ⎫ ⎪ 0 ⎪ ⎢ 1 ⎪ 1 ⎪ ⎥⎨ ⎬ ⎬ ⎨ 1 ⎥ ⎢ 1 ⎥ 2 = ⎢ 0 ⎪ ⎪ ⎦ ⎩ 2 ⎭ ⎣ ⎪ ⎪ ⎪ ⎭ ⎩ ⎪ −13 −23 3 | {z } [{]
Then, the inertia and the stiffness matrices of the modified problem are given by ⎡ ⎤ 2 ⎣ 2 1 ⎦ [] = [{] [] [{] = 3 1 5 ⎡ ⎤ 11 1 ⎦ [] = [{] [] [{] = ⎣ 9 1 59
498
CHAPTER 6 MULTI DOF VIBRATION
The eigenvalue problem is given by
The eigenvalues are the roots of
¢ ¡ [] − 2 [] {}{ = {0}
(1)
¡ ¢ det [] − 2 [] = 0 ⎡ ⎤ ⎡ ⎤⎞ ⎛ 2 2 1 ⎦⎠ 1 ⎣ 11 1 ⎦ det ⎝ − 2 ⎣ = 0 9 1 59 3 1 5 µ ¶2 19 2 4 +2 = 0 − 6 The natural frequencies are
r
1 = 093
r and 2 = 152
We find the eigenvector that corresponds to these natural frequencies by substituting 2 into Equation (1). For 2 = 087 Equation (1) becomes
Then,
⎛ ⎡ ⎤ ⎡ ⎤⎞ ¶ µ 11 1 2 1 ⎝ ⎣ ⎦ − 0871 2 ⎣ ⎦⎠ {Θ}1{ = {0} 9 1 59 3 1 5 ⎡ ⎤ 609 × 10−2 −0470 ⎣ ⎦ {Θ}1{ = {0} −0470 365 {Θ}1{ = 1
Similarly, for 22 = 230 Equation (1) becomes
Then,
⎧ ⎫ ⎨ 10 ⎬ ⎩ 013 ⎭
⎛ ⎡ ⎡ ⎤ ⎤⎞ ¶ µ 11 1 2 1 ⎝ ⎣ ⎦ − 230 2 ⎣ ⎦⎠ {Θ}2{ = {0} 9 1 59 3 1 5 ⎡ ⎤ −184 −142 ⎣ ⎦ {Θ}2{ = {0} −142 −109 {Θ}2{ = 2
⎧ ⎫ ⎨ 077 ⎬ ⎩ −1 ⎭
499 The mode shapes are then, ⎡
1
0
⎤
⎡
0
⎤
⎡
⎤
10 ⎥ ⎥ ⎢ ⎥ ⎥ ⎢ 0 1 ⎥ {Θ}1{ = 1 ⎢ 013 ⎥ ⎦ ⎦ ⎣ −13 −23 −042 | {z }
⎢ ⎢ {Θ}1 = ⎢ ⎣
[{]
⎡
1
⎤
077 ⎥ ⎥ ⎢ ⎥ ⎥ ⎢ 0 1 ⎥ {Θ}2{ = 2 ⎢ −1 ⎥ ⎦ ⎦ ⎣ −13 −23 041 | {z }
⎢ ⎢ {Θ}2 = ⎢ ⎣
[{]
The natural frequencies and the mode shapes are
0 = 0 rad/s
r
rad/s
r
rad/s
1 = 093
2 = 152
⎡ ⎤ 1 ⎢ ⎥ ⎢ ⎥ {Θ}0 = ⎢ 1 ⎥ ⎣ ⎦ 1 ⎤ ⎡ 10 ⎥ ⎢ ⎥ ⎢ {Θ}1 = ⎢ 013 ⎥ ⎦ ⎣ −042 ⎤ ⎡ 077 ⎥ ⎢ ⎥ ⎢ {Θ}2 = ⎢ −1 ⎥ ⎦ ⎣ 041
500
CHAPTER 6 MULTI DOF VIBRATION
49. Solve Example 6.20 , where 1 = 3 N-m/rad, 2 = 10 N-m/rad and 1 = 5 kg-m2 /rad, 2 = 21 , 3 = 2 2. Solution: Start with Example 6.20 and substitute the given parameter values to obtain [] and [] given by ⎡ ⎤ ⎡ ⎤ [] = ⎣
10 2
2 30
⎦
[] = ⎣
13 27
27 113
⎦
The matrix EOM is given by ⎤⎧ ⎫ ⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎡ ⎨ ⎬ ⎨0⎬ ⎬ ⎨ 13 27 10 2 ̈1 1 ⎦ ⎣ ⎦ +⎣ = 27 113 ⎩ 2 ⎭ ⎩ 0 ⎭ 2 30 ⎩ ̈2 ⎭
Assume the harmonic response: 1 = Θ1 cos 2 = Θ2 cos Substitute these and their derivatives, respectively, into the EOM to find the characteristic matrix, ⎫ ⎧ ⎫ ⎡ ⎤⎧ 2 2 ⎨ −10 + 13 −2 + 27 Θ1 ⎬ ⎨ 0 ⎬ ⎣ ⎦ = −2 2 + 27 −30 2 + 113 ⎩ Θ ⎭ ⎩ 0 ⎭ 2
The characteristic equation is then ¡ ¢¡ ¢ ¡ ¢2 −10 2 + 13 −30 2 + 113 − −2 2 + 27 = 0
from which we find: 212 = 060 417 These can be substituted into the characteristic matrix in order to obtain the constrained modes: ½ ¾ Θ1 −221 + 27 = Θ2 1{ −10 21 + 13 −2(060) + 27 = 3 69 = −10(060) + 13 ⎫ ⎫ ⎧ ⎫ ⎧ ⎧ ⎬ ⎨ 100 ⎬ ⎨ 100 ⎨ Θ1 ⎬ = = or ⎩ (136857) ⎭ ⎩ 0 271 ⎭ ⎩Θ ⎭ 2 1{
The second constrained modal ratio is given by: ½ ¾ Θ1 −222 + 27 = Θ2 2{ −10 22 + 13 −2(417) + 27 = −0 65017 = −10(417) + 13 ⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎨Θ ⎬ ⎨ 100 ⎬ ⎨ 100 ⎬ 1 = = ⎩Θ ⎭ ⎩ −(10650) ⎭ ⎩ −1 54 ⎭ 2 2{
501 The constraint matrix was found in Example 6.20 to be ⎤ ⎡ 1 0 ⎥ £ ¤ ⎢ ⎥ ⎢ { = ⎢ 0 1 ⎥ ⎦ ⎣ −1 −2
We can recover the elastic modes using the constraint matrix as follows: ⎤ ⎡ ⎫ ⎧ ⎧ ⎫ 1 0 ⎨ Θ1 ⎬ ⎥ ⎨ 100 ⎬ ⎢ ⎥ ⎢ = ⎢0 1 ⎥ ⎩Θ ⎭ ⎦ ⎩ 0 271 ⎭ ⎣ 2 1 −1 −2 ⎤ ⎡ ⎧ ⎧ ⎫ ⎫ 1 0 ⎨Θ ⎬ ⎥ ⎨ 100 ⎬ ⎢ 1 ⎥ ⎢ = ⎢0 1 ⎥ ⎩Θ ⎭ ⎦ ⎩ −154 ⎭ ⎣ 2 2 −1 −2 The complete response is then:
{()} = (1 + 2 ){Θ}0 +
2 X =1
{Θ}1 cos( − )
502
CHAPTER 6 MULTI DOF VIBRATION
50. In Example 6.21, explain how Equation 6.137 arises. Solution: The equation of motion is given by ⎤⎧ ⎫ ⎧ ⎫ ⎤⎧ ⎫ ⎡ ⎡ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪0⎪ ⎪ ⎪ ̈ ⎪ − 0 ⎪ 0 0 ⎪ ⎨ ⎪ ⎬ ⎨ 1⎪ ⎥⎪ ⎨ ⎪ ⎬ ⎢ ⎬ ⎪ ⎥⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ 0 0 ⎥ ̈2 + ⎢ − 2 − ⎥ 2 = 0 ⎪ ⎪ ⎪ ⎪ ⎦⎪ ⎣ ⎦⎪ ⎣ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎩ ⎩ ⎭ ⎭ 0⎭ 0 − 0 0 ̈ 3
3
The corresponding eigenvalue problem is ⎛⎡ ⎤ ⎡ ⎤⎞ − 0 0 0 ⎜⎢ ⎥ ⎢ ⎥⎟ ⎜⎢ ⎥ ⎥⎟ 2⎢ ⎜⎢ − 2 − ⎥ − ⎢ 0 0 ⎥⎟ {} = {0} ⎝⎣ ⎦ ⎣ ⎦⎠ 0 − 0 0
Divide by and let = 2 to obtain ⎤ ⎤⎞ ⎡ ⎛⎡ 1 −1 0 1 0 0 ⎥ ⎥⎟ ⎜⎢ ⎢ ⎥ ⎥⎟ ⎢ ⎜⎢ ⎜⎢ −1 2 −1 ⎥ − ⎢ 0 1 0 ⎥⎟ {} = {0} ⎦ ⎦⎠ ⎝⎣ ⎣ 0 −1 1 0 0 1
The characteristic equation is the determinant of the matrix multiplying {} and is given by ⎤ ⎛⎡ ⎤⎞ ⎡ 1 −1 0 1 0 0 ⎥ ⎜⎢ ⎥⎟ ⎢ ⎥ ⎜⎢ ⎥⎟ ⎢ det ⎜⎢ −1 2 −1 ⎥ − ⎢ 0 1 0 ⎥⎟ = 0 ⎣ ⎦ ⎝⎣ ⎦⎠ 0 −1 1 0 0 1 ³ ´2 ³ ´ ³ ´3 4 −3 = 0 − which factors to
´³ ´ ³ −1 −3 =0 µ ¶µ ¶ − −3 = 0
or
(6.137)
503
51. For Example 6.21, show the steps in the derivation of the three modes of vibration. Solution: In Example 6.21, we have the three frequencies of vibration squared: 0 3 These are then substituted into the characteristic matrix: ⎫ ⎧ ⎫ ⎤⎧ ⎡ ⎪ ⎪ ⎪ ⎪0⎪ ⎪ ⎪ ⎪ − 2 − 0 ⎬ ⎨ ⎪ ⎨ 1 ⎪ ⎬ ⎪ ⎥⎪ ⎢ ⎥ ⎢ 2 = ⎥ ⎢ − 0 2 − − 2 ⎪ ⎪ ⎪ ⎦⎪ ⎣ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎩ ⎭ 2 0⎭ 3 0 − − For 20 = 0 :
⎫ ⎧ ⎫ ⎤⎧ ⎪ ⎪ ⎪ ⎪0⎪ ⎪ 10 ⎪ ⎪ − 0 ⎪ ⎬ ⎨ ⎪ ⎨ ⎬ ⎪ ⎥⎪ ⎢ ⎥ ⎢ = 0 ⎢ − 2 − ⎥ 20 ⎪ ⎪ ⎪ ⎦⎪ ⎣ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎩ ⎭ 0⎭ 0 − ⎡
30
we find: 10 = 20 = 30 or {}0 = {1 1 1} Similarly, for 21 = we have ⎫ ⎧ ⎫ ⎡ ⎤⎧ ⎪ ⎪ ⎪0⎪ ⎪ ⎪ ⎪ ⎪ 0 − 0 ⎬ ⎨ ⎪ ⎨ 11 ⎪ ⎬ ⎪ ⎢ ⎥⎪ ⎢ ⎥ = 0 ⎢ − − ⎥ 21 ⎪ ⎪ ⎣ ⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩0⎪ ⎩ ⎭ ⎪ 0 − 0 31
from which we can immediately find 21 = 0 and 11 = −31 = 1 This mode is then {}1 = {1 0 −1} Finally, for 22 = 3 we have ⎫ ⎧ ⎫ ⎡ ⎤⎧ ⎪ ⎪ ⎪ ⎪0⎪ ⎪ 12 ⎪ ⎪ −2 − 0 ⎪ ⎬ ⎨ ⎪ ⎨ ⎬ ⎪ ⎢ ⎥⎪ ⎢ ⎥ = 0 ⎢ − − − ⎥ 22 ⎪ ⎪ ⎪ ⎣ ⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎩ ⎭ 0⎭ 0 − −2 32
from which we find 12 = −22 2 = 32 Then, {}2 = {1 −2 1} The complete solution is then given by 2 X {()} = (1 + 2 ){}0 + {}1 cos( − ) =1
504
CHAPTER 6 MULTI DOF VIBRATION
Problems for Section 6.11 — Periodic Structures
52. (a) For the three degree-of-freedom system shown in Figure 6.82, derive the equations of motion and solve for the frequencies and modes of vibration for the arbitrary stiffnesses 1 2 3 4 and masses 1 2 3 . Then simplify these results for the case where all the stiffnesses have the value = 1 N/m and all the masses have the value = 1 kg. Plot the modes and responses for 1 (0) = 05 cm and assume that all other initial conditions are equal to zero.
Figure 6.82: Three degree-of-freedom periodic system.
(b) For the previous system, suppose that 1 = 3 = 4 = = 1 N/m, 1 = 2 = 3 = = 1 kg, and 2 = (1 + ) = (1 + 005), where signifies a stiffness imperfection. Solve for the frequencies and modes of vibration and plot the modes and responses for 1 (0) = 05 cm and all other initial conditions equal zero. Compare the results with and without imperfections. How do these results compare with those discussed in the section. Discuss.
Solution: (a) We can derive the EOM using Newton’s second law from the free-body diagram for each mass, but rather, we derive these via Lagrange’s equation. The kinetic and strain energies are given by ¢ 1¡ 1 ̇21 + 2 ̇22 + 2 ̇22 2 ¤ 1£ 2 1 1 + 2 (2 − 1 )2 + 3 (3 − 2 )2 + 4 23 = 2 =
The Lagrangian is given by = − and Lagrange’s equation is
µ
̇
¶
−
=0
505 for each degree-of-freedom. Applying Lagrange’s equation for each degree-of-freedom and then putting the EOM into matrix-vector form yields ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪0⎪ ⎪ ⎪ ⎪ ̈1 ⎪ ⎪ 1 ⎪ + 2 −2 0 0 0 ⎬ ⎨ ⎪ ⎬ ⎢ 1 ⎬ ⎪ ⎨ ⎪ ⎨ ⎪ ⎢ 1 ⎥⎪ ⎥⎪ ⎢ ⎢ ⎥ ⎥ = ⎢ 0 2 0 ⎥ ̈2 + ⎢ −2 ⎥ 0 2 + 3 −3 2 ⎪ ⎪ ⎪ ⎪ ⎣ ⎣ ⎦⎪ ⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎭ ⎩ ⎩ 0⎭ 0 0 3 ̈3 0 −3 3 + 4 3
Assume a harmonic response for each mass: 1 () = cos 2 () = cos 3 () = cos Substitute these and derivatives into the matrix EOM to obtain the characteristic matrix equation: ⎤⎧ ⎫ ⎧ ⎫ ⎡ 2 ⎪ ⎪ ⎪0⎪ ⎪ ⎪ ⎪⎪ + 2 − 1 −2 0 ⎬ ⎨ ⎪ ⎬ ⎪ ⎨ ⎪ ⎥⎪ ⎢ 1 ⎥ ⎢ 2 = ⎥ ⎢ −2 0 2 + 3 − 2 −3 ⎪ ⎪ ⎪ ⎦⎪ ⎣ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ 2 0⎭ 0 −3 3 + 4 − 3
Setting the determinant of the characteristic matrix to zero yields the characteristic equation: 6 − 64 + 102 − 4 = 0 The roots are equal to 21 = 0586 22 = 200 23 = 341 Now substitute these as well as the given parameter values into the above characteristic matrix equation to find the eigenvectors (modes): ⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪0⎪ ⎪ ⎪ ⎪⎪ 2 − 2 −1 0 ⎬ ⎨ ⎪ ⎬ ⎪ ⎨ ⎪ ⎢ ⎥⎪ ⎢ ⎥ 2 ⎢ −1 ⎥ = 0 2 − −1 ⎪ ⎪ ⎪ ⎣ ⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ 2 0⎭ 0 −1 2− Substituting each 2 in turn yields the respective modal ratio. These are ⎧ ⎫ ⎧ ⎫ ⎫ ⎫ ⎫ ⎧ ⎧ ⎧ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 100 ⎪ 100 ⎪ 100 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎬ ⎨ ⎪ ⎨ ⎪ ⎬ ⎬ ⎬ ⎬ ⎨ ⎨ ⎨ ⎨ ⎪ = = = 1414 0 −1414 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎭ ⎩ ⎪ ⎩ ⎪ ⎭ ⎭ ⎭ ⎭ ⎩ 100 ⎪ ⎩ −100 ⎪ ⎩ 100 ⎪ ⎩ ⎪ 1
2
3
The normalized modal matrix ( modes are the column of the modal matrix) is given by ⎤ ⎡ 05 −0707 −05 ⎥ ⎢ ⎥ ⎢ [ ] = ⎢ 0707 0 0707 ⎥ ⎦ ⎣ 05 0707 −05 We now can write the general solution as ⎧ ⎧ ⎫ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ () cos 077 + sin 077 1 ⎪ ⎪ ⎪ ⎨ 1 ⎪ ⎨ 1 ⎬ ⎬ 2 () = [ ] 2 cos 141 + 2 sin 141 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ () ⎪ ⎩ cos 184 + sin 184 ⎪ ⎭ ⎭ 3 3 3
506
CHAPTER 6 MULTI DOF VIBRATION
or ⎧ ⎪ ⎪ 05 ⎪ ⎨
⎧ ⎫ ⎪ ⎪ ⎪ () ⎪ ⎪ ⎨ 1 ⎪ ⎬
⎫ ⎪ ⎪ ⎪ ⎬
⎫ ⎧ ⎪ ⎪ ⎪ −0707 ⎪ ⎪ ⎪ ⎬ ⎨
= 0707 (1 cos 077 + 1 sin 077) + 0 2 () ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ 05 ⎪ ⎩ () ⎪ ⎩ 0707 ⎭ 3 ⎫ ⎧ ⎪ ⎪ ⎪ ⎪ −05 ⎪ ⎪ ⎬ ⎨ + 0707 (3 cos 184 + 3 sin 184) ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ −05 ⎪
⎪ ⎪ ⎪ ⎭
(2 cos 141 + 2 sin 141)
At this point the arbitrary constants, and can be evaluated by satisfying the given initial conditions. Apply the initial displacement conditions first to obtain ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 1⎪ ⎬
⎫ ⎧ ⎪ ⎪ ⎪ 05 ⎪ ⎪ ⎪ ⎬ ⎨
= [ ] 2 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎭ ⎩0 ⎪ ⎭ 3
Use the orthogonality condition, [ ] [] [ ] to find
⎫ ⎧ ⎪ ⎪ ⎪ ⎪ 05 ⎪ ⎪ ⎬ ⎨
⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ 1⎪
2 = [ ] [] 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩0 ⎭ ⎩ ⎪ 3
⎪ ⎪ ⎪ ⎭
=
⎧ ⎪ ⎪ 025 ⎪ ⎨
⎫ ⎪ ⎪ ⎪ ⎬
−03536 ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ −025 ⎪
Apply the zero initial velocity conditions to obtain 1 = 2 = 3 = 0
(b) For the previous system suppose that 1 = 3 = 4 = = 1 N/m, 1 = 2 = 3 = = 1 kg, and 2 = (1 + ) = (1 + 005) : we need to begin with the general characteristic matrix except that we have to substitute the new parameter values given above: ⎤⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪0⎪ ⎪ ⎪ ⎪⎪ 0 + 2 − 1 −2 ⎬ ⎨ ⎪ ⎬ ⎪ ⎨ ⎪ ⎥⎪ ⎢ 1 ⎥ ⎢ 2 = ⎥ ⎢ −2 0 2 + 3 − 2 −3 ⎪ ⎪ ⎪ ⎦⎪ ⎣ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ 2 0⎭ 0 − + − ⎡
2
3
becomes:
⎡
3
3
⎤⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪0⎪ ⎪ ⎪ ⎪⎪ ⎬ ⎨ ⎪ ⎬ ⎪ ⎨ ⎪ ⎥⎪ ⎥ 2 ⎥ = 0 205 − −1 ⎪ ⎪ ⎪ ⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ 2 0⎭ −1 2−
205 − 2 −105
⎢ ⎢ ⎢ −105 ⎣ 0
4
0
507 The natural frequencies and the normalized modal matrix are given by 21 = 05878 22 = 20238 23 = 34884 ⎤ ⎡ −05056 06898 −05182 ⎥ ⎢ ⎥ ⎢ [ ] = ⎢ −07041 00172 07099 ⎥ ⎦ ⎣ −04986 −07238 −04770
The solution is written as ⎧ ⎧ ⎫ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ () cos 07667 + 1 sin 07667 ⎪ ⎪ ⎪ ⎪ ⎨ 1 ⎪ ⎨ 1 ⎬ ⎬ 2 () = [ ] 2 cos 14226 + 2 sin 14226 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ () ⎪ ⎩ cos 18677 + sin 18677 ⎪ ⎭ ⎭ 3
3
3
Applying the initial displacements, we find ⎫ ⎫ ⎧ ⎧ ⎫ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −02528 ⎪ ⎪ ⎪ ⎪ 05 ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ ⎬ ⎪ ⎨ 1⎪ ⎨ ⎬ = 03449 2 = [ ] [] 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎪ ⎪ ⎩ ⎩ 0 ⎭ ⎩ −02591 ⎭ 3
The figures below show the responses. They show minimal differences. The effects discussed in the text do not show because there are only three masses.
The response of the three masses. Solid line is when 2 = 1, and dotted line is when 2 = 105
508
CHAPTER 6 MULTI DOF VIBRATION
509
53. In Problem 52, part (b), solve assuming 1 = 2 = 3 = 4 = and 1 = 3 = and 2 = (1 + ). Use the same parameter values. Discuss.
Solution: We begin with the characteristic matrix equation: ⎤⎧ ⎫ ⎧ ⎫ ⎡ ⎪ ⎪ ⎪0⎪ ⎪ ⎪ ⎪⎪ 1 + 2 − 1 2 −2 0 ⎬ ⎨ ⎪ ⎬ ⎪ ⎨ ⎪ ⎥⎪ ⎢ ⎥ ⎢ 2 = ⎥ ⎢ −2 0 2 + 3 − 2 −3 ⎪ ⎪ ⎪ ⎦⎪ ⎣ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ 2 0⎭ 0 −3 3 + 4 − 3 and substitute the given parameter values to find ⎡ ⎤⎧ ⎫ ⎧ ⎫ 2 ⎪ ⎪ ⎪0⎪ ⎪ ⎪ ⎪⎪ 2 − −1 0 ⎬ ⎢ ⎨ ⎪ ⎬ ⎪ ⎨ ⎪ ⎥⎪ ⎢ ⎥ 2 ⎢ −1 ⎥ = 0 2 − 105 −1 ⎪ ⎪ ⎪ ⎣ ⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ 2 0⎭ 0 −1 2− The natural frequencies and the normalized modal matrix are given by 21 = 05785 22 = 19513 23 = 33750 ⎤ ⎡ −04930 06989 −05182 ⎥ ⎢ ⎥ ⎢ [ ] = ⎢ −07008 00341 07126 ⎥ ⎦ ⎣ −05032 −06972 −04616
The solution is written as ⎧ ⎫ ⎪ ⎪ ⎪ () ⎪ ⎪ ⎨ 1 ⎪ ⎬
⎧ ⎫ ⎪ ⎪ ⎪ cos 07606 + 1 sin 07606 ⎪ ⎪ ⎪ ⎨ 1 ⎬
2 () = [ ] 2 cos 13969 + 2 sin 13969 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ () ⎪ ⎩ cos 18371 + sin 18371 ⎪ ⎭ ⎭ 3
3
Applying the initial displacements, we find ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 1⎪ ⎬
3
⎫ ⎧ ⎪ ⎪ ⎪ 05 ⎪ ⎪ ⎪ ⎬ ⎨
= [ ] [] 0 2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩0 ⎪ ⎩ ⎪ ⎭ 3 ⎫ ⎧ ⎪ ⎪ ⎪ ⎪ −02465 ⎪ ⎪ ⎬ ⎨ = 03494 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ −02591 ⎭
510
CHAPTER 6 MULTI DOF VIBRATION
The figures below show the responses. They show minimal differences. The effects discussed in the text do not show because there are only three masses.
The response of the three masses. Solid line is when 3 = 1, and dotted line is when 3 = 105
511
54. Derive Equation 6.144. Solution: Let us start with
A two-degree-of-freedom system. The equation of motion is given by ⎡ ⎤⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎨ ̈ ⎬ ⎨ ⎬ ⎨0⎬ 1 0 + − 1 1 2 2 1 ⎣ ⎣ ⎦ ⎦ + = ⎩ ⎭ ⎩0⎭ 0 ⎩ ̈ ⎭ − 2
2
2
2
2
The eigenvalue problem is given by ⎛⎡ ⎤ ⎤⎞ ⎡ 1 + 2 −2 1 0 ⎝⎣ ⎦ − ⎣ ⎦⎠ {} = {0} 0 2 −2 2 where = 2 The characteristic equation is given by ⎤ ⎡ −2 1 + 2 − 1 ⎦ = 0 det ⎣ −2 2 − 2
1 2 − 1 2 − 2 1 − 2 2 + 2 1 2 = 0
Divide by 1 2 to obtain ¸ 1 2 1 + 2 2 + + = 0 − 1 2 1 2 2
∙
We know that there are two roots to this quadratic equation, and we denote them as 1 and 2 The quadratic equation is written as ( − 1 ) ( − 2 ) = 0 Expanding it, we have 2 − (1 + 2 ) + 1 2 = 0
Matching the coefficients, we find that
1 + 2 =
1 + 2 2 1 2 + 1 2 = 1 2 1 2
(6.144)
512
CHAPTER 6 MULTI DOF VIBRATION
Problems for Section 6.12 — Inverse Vibration Problem 55. Derive Equations 6.150 to 6.153. Solution: Begin with Equations 6.146-6.148 given by 1 =
2 2
2 =
1 1
3 =
Assume = 1 + 2 We can now solve for 1 2 and 12
2 1
513
56. Suppose that for the two degree-of-freedom system of Figure 6.54 we obtain the data: 1 = 4 Hz 2 = 10 Hz 3 = 8 Hz, and 1 + 2 = 20 kg. Solve for the system parameters 1 2 1 2 . Solution: We have the following parameters: p p1 2 = 2 × 10 = 2 p 3 = 2 × 8 = 3
1 = 2 × 4 =
Use Equations 6.146-6.148 to evaluate such that
2 = (1 + 2 − 3 ) 2 ¡ ¢ = (16 + 100 − 64) 4 2 = 52 4 2 1 1 2 = 2 = 1 1 + 2 − 3 ¡ ¢ 16 × 100 ¡ 2 ¢ 4 = 308 4 2 = 52 2 (3 − 1 ) (2 − 3 ) = 3 = 1 1 + 2 − 3 ¡ ¢ (64 − 16) (100 − 64) ¡ 2 ¢ = 4 = 332 4 2 52
1 =
From Equations 6.150-6.153 we are able to evaluate the system properties: 1 2 1 + 3 52 × 308 ¡ 2 ¢ 4 20 = 1500 2 = 52 + 332 1 3 2 = 1 + 3 52 × 332 ¡ 2 ¢ = 4 20 = 1620 2 52 + 332 1 1 = 1 + 3 52 20 = 12 2 = 52 + 332 3 2 = 1 + 3 332 = 20 = 779 52 + 332 1 =
514
CHAPTER 6 MULTI DOF VIBRATION
Problems for Section 6.13 — Fluid Sloshing in Containers 57. Since liquids have damping characteristics, suggest possible equivalent mechanical models for a sloshing liquid that includes damping. Solution: Assuming that the sloshing can be approximated as linear, then we can assume a viscous damping oscillator model for the sloshing behavior.
515
58. Suppose a container with interior fluid is in an aircraft. The aircraft leaves on its flight with full tanks and arrives at its destination with almost empty tanks. How could this fact be incorporated in the equivalent mechanical model? Solution: As the tanks empty the sloshing behavior generally increases in amplitude. We can imagine that the sloshing results in an assumed harmonic/periodic set of impacts on the interior of the container. This could be modeled as a harmonic/periodic forcing with increasing amplitudes in time.
516
CHAPTER 6 MULTI DOF VIBRATION
59. Suppose a container with interior fluid is in a spacecraft in orbit around the Earth. What additional considerations need to be incorporated in the equivalent mechanical model? Solution: In space there is microgravity. Any interior fluid evolves toward equilibrium, which is a spherical volume in microgravity. We can decide to block such formations by including screens and baffles in the tank. Sloshing will be generally slow dynamically.
517 Problems for Section 6.14 — Stability of Motion 60. In Example 6.23, solve for () analytically for the case 2
Solution: The equation of motion is given by ̈ +
¢ 1¡ − 2 cos = 0
This is still a linear equation. You have to turn it into Mathieu’s equation.
518
CHAPTER 6 MULTI DOF VIBRATION
61. Derive Equations 6.154 and 6.155 using Newton’s second law of motion. Solution: Equations 2 ̈ + 2 ̈ sin + sin = 0 2
̈ + ̈ sin + ̇ cos + =
(6.154) (6.155)
Let us draw a free-body diagram first.
For the mass, sum the forces in the and directions to obtain, respectively, X −→ = − sin = X ↑ = cos − =
(1) (2)
where and are the accelerations in the and directions. This is a single degree-of-freedom system, and we choose as our independent variable. From kinematics, we can write ³ ´ 2 2 = 2 ( sin ) = ̈ cos − ̇ sin ³ ´ 2 2 = ̈ () − 2 ( cos ) = ̈ () + ̈ sin + ̇ cos Substitute these expression into force balance equations (Equations (1) and (2)), and we obtain ´ ³ 2 (3) − sin = ̈ cos − ̇ sin ³ ³ ´´ 2 cos − = ̈ () + ̈ sin + ̇ cos (4) Multiply Equation (3) by cos multiply Equation (4) by sin and add the two equations to eliminate We then obtain − sin = ̈ + ̈ () sin
Rearrange the terms to obtain Equation 6.154 given by
̈ + (̈ () + ) sin = 0
519 In order to obtain Equation 6.155, sum the forces in the vertical directions such that X ↑ = − cos = 0
(5)
Use Equation (4) and (5) to eliminate cos and to obtain ´´ ³ ³ 2 = ̈ () + ̈ sin + ̇ cos +
Rearrange the terms to obtain Equation 6.155 given by 2
̈ + ̈ sin + ̇ cos + =
(6.1)
520
CHAPTER 6 MULTI DOF VIBRATION
Problems for Section 6.15 — Rayleigh’s Quotient 62. For a two disk torsional system (see Figure 6.81) with its two ends fixed to walls, follow the procedure of Example 6.24 to evaluate the two approximate frequencies of oscillation. Explain the procedure used to guess the respective mode shapes. Check that the approximate fundamental frequency is above the exact frequency and that the second frequency is below the exact frequency.
Solution: Here we take a restrained system undergoing torsional motion. Before solving it approximately, we solve it exactly for free vibratory motion. We need to derive the EOMP using free-body diagrams for each disk and apply Newton’s second law of motion in torque form: = ̈ The stiffness of the shafts connecting the rotors are = () From the free-body diagrams, we find 1 ̈1 + (1 + 2 ) 1 − 2 2 = 0 2 ̈2 + (2 + 3 ) 2 − 2 1 = 0 Assuming harmonic motion of the form cos ( + ) for each disk results in the equation given by ⎫ ⎧ ⎫ ⎡ ⎤⎧ ⎨ ⎬ ⎨0⎬ 1 + 2 − 1 2 −2 1 ⎣ ⎦ = − + − 2 ⎩ ⎭ ⎩ 0 ⎭ 2
2
3
2
2
Taking the determinant of the characteristic matrix leads to the characteristic equation given by ¢¡ ¢ ¡ 1 + 2 − 1 2 2 + 3 − 2 2 − 22 = 0
from which we find the two natural frequencies of this system by solving the quadratic equation in 2 or ¸ ∙ 1 + 2 2 + 3 2 1 2 + 2 3 + 3 1 4 + + = 0 − 1 2 1 2 In order to simplify our comparisons, we assume 1 = 2 = 3 = 1 and 1 = 2 = 1, with appropriate units. Then the quadratic equation becomes 4 − 42 + 3 = 0
√ from which, 1 = 1 rad/s and 2 = 3 = 1 7321 rad/s With these frequencies, the modes (1 2 )12 can be found from the characteristic matrix equation, and then the free vibration is the superposition of the two harmonics, as we have seen before. The modal ratios are µ ¶ 1 = 1 2 1 µ ¶ 1 = −1 2 2
521 Now we can approximate the solution of the frequency problem using the energy procedure. Assume here that 1 () = 1 cos and 2 () = 2 cos The kinetic energy, and its maximum, are given by 1 1 2 2 1 ̇1 + 2 ̇2 2 2 1 1 max = 1 (1 )2 + 2 (2 )2 2 2 =
The corresponding potential energy expressions are given by 1 2 1 1 1 1 + 2 (2 − 1 )2 + 3 22 2 2 2 1 1 2 max = (1 + 2 ) 1 + (2 + 3 ) 22 + 2 1 2 2 2 =
Equate max to max and solve for 2 to obtain 2 =
(1 + 2 ) 21 + (2 + 3 ) 22 + 22 1 2 1 21 + 2 22
Assuming the same parameter values as above, we have 2 =
221 + 222 + 21 2 21 + 22
We now assume ratios 1 2 for each mode and then evaluate the respective frequency. Assume for the first mode 1 2 = 1 Then, 621 = 3 2 = 221 For the second mode assume 1 2 = −1 Then, 2 =
2 21 = 1 221
We see that in this instance we obtain the same frequencies as the exact case. This is because we used the exact modal ratios. Try this problem with different numerical values. You will see that Rayleigh coefficients are highly sensitive to the assumed modal ratios. The reason for this is that the mode shapes are not sensitive to the exact values of the mass and the stiffness.
522
CHAPTER 6 MULTI DOF VIBRATION
63. How would Rayleigh’s quotient need to be altered for an unrestrained system? Apply the method to a general two degree-of-freedom undamped system in unrestrained rectilinear motion. Solution: It is straightforward to show that Rayleigh’s quotient will work for an unrestrained system. For example, begin with Example 6.25. Take this system and reduce it to a unrestrained 2 DOF system by setting the stiffness to zero at the wall. Doing this, the stiffness and the mass matrices become ⎡ ⎤ ⎡ ⎤ 1 −1 1 0 ⎦ and [] = ⎣ ⎦ [] = ⎣ −1 1 0 1
Note that we cannot use the procedure of Example 6.25 to estimate the first mode shape because [] has no inverse and we cannot derive the static displacement. But, we know the first mode shape here because this is an unrestrained system with first mode: {}1 = {1 1} Rayleigh’s quotient for the first natural frequency is {}1 []{}1 21 = {}1 []{}1 Here, we only need to evaluate the numerator given by ⎡ ⎤⎧ ⎫ 1 −1 ⎨ 1 ⎬ ⎦ = 0 {}1 []{}1 = {1 1} ⎣ −1 1 ⎩ 1 ⎭
Thus, we see that the first natural frequency estimate is 0 as it should be. To estimate the second natural frequency we could assume ⎧ ⎫ ⎨1 ⎬ {}2 = ⎩ −1 ⎭ and Rayleigh’s quotient for the second natural frequency is s {}2 []{}2 2 = {}2 []{}2 r r √ 4 = 2 = 2
Chapter 7 Continuous Models for Vibration Problems for Section 7.2 — Vibration of Strings 1. Derive Equation 7.4
∙ ¸ 2 () + ( ) = () 2
(7.4)
using Newton’s second law of motion.
Solution: Figure 7.4 shows the forces that act on the incremental string element The free body diagram is shown next. If we make the small angle assumption, we can write ≈ tan and tan ≈ sin We recognize tan as The partial derivative is used because is a function of both and We can assume that the tension is also a function of both and but we can eliminate the functional dependence on time by assuming that the dynamic range of tension is much smaller than the static tension. Then, we can write ( ) as () and as or Summing the forces in the vertical direction, we obtain X 2 ( ) = × 2 µ ¶ µ ¶ 2 ( ) ( ) () + sin ( ) + − () sin ( ) + ( ) = × 2 Replacing sin with and then replacing with we obtain µ ¶µ µ ¶ ¶ ( ) ( ) 2 ( ) ( ) () + + − () + ( ) = × 2 Replacing with where is the mass per unit length, and some simplification, we find µ ¶ µ ¶ ( ) ( ) ( ) 2 ( ) () + + = 2 Dividing by µ ¶ µ ¶ ( ) ( ) ( ) 2 ( ) () + + = 2 µ ¶ µ ¶ ( ) ( ) 2 ( ) () + = 2 523
524
CHAPTER 7 CONTINUOUS MODELS
Taking the limit as → 0 terms with drop out. Then, the equation of motion is given by µ ¶ ( ) 2 ( ) () = 2
525
2. Solve for the string response, Equation 7.27, given the following initial conditions: (a) ( 0) = 0, ̇( 0) = () (b) ( 0) = sin , ̇( 0) = 0 (c) ( 0) = 0, ̇( 0) = sin (d) ( 0) = sin , ̇( 0) = sin . In all cases, sketch the initial conditions and plot the response ( ). Solution: Equation 7.27 is ( ) =
∞ X =1
r
2 ³ ´ sin sin + cos
(7.27)
Given initial conditions, ( 0) and ̇ ( 0) the constant coefficients are obtained using the orR thogonality condition, 0 = where () is the normalized modal function given by r 2 () = sin Applying the initial displacement, ∞ r X 2 sin ( 0) = =1 Multiplying the both sides by () and integrating over 0 to ! Ãr ! r Z Z Ãr ∞ X 2 2 2 sin = sin sin ( 0) 0 0 =1 The only non-zero term on the right-hand side is when = so that r Z 2 = ( 0) sin 0 Similarly, we find =
Z 0
r
̇( 0)
2 sin
(1)
(2)
(a) ( 0) = 0, ̇( 0) = () Zero initial displacement gives = 0 for all The initial velocity is used to find Substituting ̇ ( 0) with () in Equation (2), we have r Z 2 = () sin 0
526
CHAPTER 7 CONTINUOUS MODELS
(b) ( 0) = sin , ̇( 0) = 0 Zero initial velocity results in = 0 Before we proceed with initial displacement, the initial conditions must satisfy the boundary conditions. Otherwise, it will not make physical sense. In this case, the initial displacement satisfy the boundary conditions at = 0 and Applying the initial displacement, we find r Z 2 = sin ( 0) 0 r . 1 = 2 All the other coefficients except for 1 are zero. This makes sense because the initial displacement is proportional to the first mode shape, and therefore, the first mode should be the only mode present. Applying the second initial condition results in = 0 (c) ( 0) = 0, ̇( 0) = sin For ̇( 0) = sin we will find that 1 = (d) ( 0) = sin , ̇( 0) = sin We will find 1 =
r
q
and all the other coefficients are zero. 2
and 1 = 2
r
2
527
3. Derive the expression for the tension required in a simply supported transmission line modeled as a string of length and linear density such that its fundamental frequency for transverse vibration is 1 What is the value of the tension where = 20 m, = 5 kg/m, and 1 = 15 Hz? Solution: The fundamental frequency of a fixed-fixed string is r 1 = The tension is
µ µ ¶2 ¶2 = 21 = 1
Substitute = 20 m, = 5 kg/m, 1 = 15 Hz to find = 1 800 000 N.
528
CHAPTER 7 CONTINUOUS MODELS
4. A string is stretched between = 0 and = and has a variable density = 0 + , where 0 and are constants. The initial displacement is (), and the string is released from rest. (a) If the tension is constant, then show that the governing equation of motion is
2 2 = 2 2
for 0 , and 0, with the boundary conditions (0 ) = 0, ( ) = 0, and the initial conditions ( 0) = (), ( 0) = 0. (b) Show that the frequencies of normal mode vibration are given by = 2, where are the positive roots of 13 ()−13 () = 13 ()−13 () where 2 = 0 3
r
0
2(0 + ) = 3
r
0 +
Solution: Let ( ) = () () Separate variables 00 ̈ = = − 2 which results in two differential equations, one for and the other for ̈ + 2 = 0 2 00 + = 0 We are given = 0 + Define = = 0 + Then, = = =⇒ ≡ µ ¶ 2 00 = = 2 2 0 =
(1)
529 Equation (1) becomes 2 2 = 0 + 2 2 2 = 0 + 2 2 2 2 = 2
2
A form of Bessel’s equation is in general: ( 0 )0 + ( + −2 ) = 0 with the complete solution = [1 ( ) + 2 ( )] where 1− 2 2−+ = p2 2 || = 2q− +
=
=
(1 − )2 − 4 2−+
In this problem, = 0 = 1 = 0 = 2 Therefore, = 12 = 32 = 2 || 3 and = 13 Then, ∙ µ ¶ µ ¶¸ 2 2 32 32 1 13 + 2 −13 || || = 3 3 " à à ! !# r r + + 2 2 0 0 (0 + ) + 2 −13 (0 + ) = (0 + )12 1 13 3 3 12
Then, ¤ £ (0) = 0 1 13 () + 2 −13 () r 2 0 0 = 3 And, ¤ £ () = (0 + )12 1 13 () + 2 −13 () r 2 0 + 0 + = 3
530
CHAPTER 7 CONTINUOUS MODELS
Then, the characteristic equation is given by ⎡ ⎤⎧ ⎫ ⎧ ⎫ () −13 () ⎨ 1 ⎬ ⎨ 0 ⎬ ⎣ 13 ⎦ = 13 () −13 () ⎩ 2 ⎭ ⎩ 0 ⎭ or
13 () −13 () − −13 () 13 () = 0
531
5. Consider the uniform string with constant mass per unit length constant tension , and length shown in Figure 7.36. One end of the string is fixed, and the other end is constrained to move in the vertical direction. Find the frequency equation, the first three natural frequencies, and plot the first three mode shapes.
Figure 7.36 Solution: The equation of motion and the boundary conditions are given by 2 2 2 = 2
¯ ¯¯ (0 ) = 0 and = 0 ¯
where the second boundary condition says that the vertical component of the tension must be zero. Write ( ) = () () then ̈ = 00 (0) () = 0 and 0 () () = 0 We divide previous equations by () (which is possible because above equations must be satisfied for any choice of ) we have 00 ̈ = (0) = 0 and 0 () = 0 We let ̈ = − 2 for harmonic solutions, and therefore, 2 =0 (0) = 0 and 0 () = 0 00 +
For convenience, we let 2 = 2 where is the wave number. () has a general solution of the form () = cos + sin which must satisfy the boundary conditions. Apply (0) = 0 and 0 () = 0 we have (0) = = 0 0 () = cos = 0
532
CHAPTER 7 CONTINUOUS MODELS
The frequency equation is cos = 0 whose first three solutions are
1 1 3 1 5 ··· 2 2 2 The corresponding natural frequencies are r µ ¶ 1 1 3 1 5 ··· = 2 2 2 =
The first three mode shapes are 1 2 1 3 2 () = 2 sin 2 = 2 sin 2 1 5 3 () = 3 sin 3 = 3 sin 2
1 () = 1 sin 1 = 1 sin
The next figure shows the first three mode shapes as a function of
533
6. Consider the uniform string with constant mass per unit length constant tension , and length shown in Figure 7.37. One end of the string is fixed, and the other end is attached to a spring with stiffness that is constrained to move in the vertical direction. Find the frequency equation, the first three natural frequencies, and plot the first three mode shapes. Use the ratio = 3
Figure 7.37 Solution: The equation of motion and the boundary conditions are given by
2 2 = 2 2
¯ ¯¯ − (0 ) = 0 and ( ) = 0 ¯0
where the first boundary condition is obtained from the balancing the forces on the massless ring in the vertical direction. The corresponding free-body diagram is shown next. The forces must be drawn based on the displacement and angle of rotation in the positive direction.
Figure 7.1: After separating variables, the spatial coordinate () satisfy 00 + 2 = 0 0 (0) − (0) = 0 and () = 0 where we let 2 = 2 for convenience. () has a general solution of the form () = cos + sin
((1))
534
CHAPTER 7 CONTINUOUS MODELS
which must satisfy the boundary conditions. Apply 0 (0) − (0) = 0 and () = 0 we have 0 (0) − (0) = − = 0 () = cos + sin = 0
We can solve for using the first equation so that the mode shape is given by ¶ µ () = cos + sin
((1))
Substituting in the second boundary condition, we obtain the frequency equation given by cos + sin = 0 which is rewritten as cos + sin = 0 We are given = 3. Then, we can solve for It is better to plot the left hand side first to examine where the roots occur and estimate the roots or find the roots numerically. Most graphing calculators are capable of tracing the plot so that we can estimate the roots. Matlab function fzero is used to find the roots numerically.
The first three wave numbers are = 24556 52329 82045 · · · The corresponding natural frequencies are r r r = 24556 52329 82045 · · · The first three mode shapes (Equation (1)) are µ ¶ 246 cos 246 + sin 246 1 () = 1 3 µ ¶ 523 cos 523 + sin 523 2 () = 2 3 µ ¶ 820 cos 820 + sin 820 3 () = 3 3
535 Figure next shows the first three mode shapes as a function of
536
CHAPTER 7 CONTINUOUS MODELS
7. Consider the uniform string with constant mass per unit length constant tension , and length shown in Figure 7.38. One end of the string is fixed, and the other end is attached to a mass that is constrained to move in the vertical direction. Find the frequency equation, the first three natural frequencies, and plot the first three mode shapes. Use the ratio = 2
Figure 7.38 Solution: The equation of motion and the boundary conditions are given by
2 2 = 2 2
¯ ¯¯ 2 ( ) (0 ) = 0 and − = ¯ 2
where the second boundary condition says that the sum of the forces in the direction (upward) is equal to the product of and its upward acceleration. The free-body diagram must be drawn so that the forces are due to displacement and the angle of rotation in the positive directions.
Write ( ) = () () then ̈ = 00 (0) () = 0 and − 0 () () = () ̈ () We divide previous equations by () (which is possible because above equations must be satisfied for any choice of ) we have 00 ̈ = (0) = 0 and − 0 () = ()
̈ () ()
537 We let ̈ = − 2 for harmonic solutions, and therefore, 2 =0 (0) = 0 and − 0 () = − () 2 00 +
For convenience, we let 2 = 2 where is the wave number. () has a general solution of the form () = cos + sin which must satisfy the boundary conditions. The second boundary condition is rewritten in terms of − 0 () = − () 2 − 0 () = − () 2 0 () −
2 () = 0
Apply (0) = 0, and we have = 0. Apply 0 () −
2 () = 0 and we have
2 ( cos + sin ) = 0 ¶ µ 2 cos − sin = 0
− sin + cos −
The equation is satisfied if = 0, but it will result in a trivial solution of () = 0 Therefore, the terms in the parenthesis must be zero. The frequency equation is cos −
2 sin = 0
The frequency equation can be rewritten as ¶ µ sin = 0 cos − We check whether = 0 is possible solution. If = 0 the corresponding natural frequency is also zero and the form of the mode shape changes to 0 () = + If we apply the boundary conditions, we obtain = = 0 which is also trivial. Therefore, = 0 is not a valid wavenumber. We are given = 2 then, the frequency equation is 1 cos − sin = 0 2 which has solutions = 108, 364 658 · · ·
538
CHAPTER 7 CONTINUOUS MODELS
The corresponding natural frequencies are r = (108, 364 658 · · · ) The first three mode shapes are 108 364 2 () = 2 sin 2 = 2 sin 658 3 () = 3 sin 3 = 3 sin 1 () = 1 sin 1 = 1 sin
The next figure shows the first three mode shapes as a function of The end mass is not heavy enough such that the it behaves as a fixed-free string.
539
8. Consider the uniform string with constant mass per unit length constant tension , and length shown in Figure 7.39. One end of the string is attached to a mass and the other end of the string is constrained by a spring with stiffness Both ends are constrained to move in the vertical direction. Find the frequency equation, the first three natural frequencies, and plot the first three mode shapes. Use the ratios = 2 and = 100
Figure 7.39
Solution: The equation of motion and the boundary conditions are given by 2 2 = 2 2 ¯ ¯ ¯¯ ¯¯ = ̈ (0 ) and − ¯ ¯
0
− ( ) = 0
where the boundary conditions are obtained from the free-body diagrams shown next. The freebody diagrams must be drawn so that the forces are due to displacement and the angle of rotation in the positive directions.
Write ( ) = () () then ̈ = 00 0 (0) () − (0) ̈ () = 0 and 0 () () + () () = 0 We divide the equation of motion by () we have 00 ̈ =
540
CHAPTER 7 CONTINUOUS MODELS
which we may let ̈ = − 2 for harmonic solutions, and therefore, 2 =0 0 (0) + 2 (0) = 0 and 0 () + () = 0 00 +
For convenience, we let 2 = 2 where is the wave number. () has a general solution of the form () = cos + sin 0 () = − sin + cos which must satisfy the boundary conditions given by 2 (0) + (0) = 0 and 0 () + () = 0 0
Apply the first boundary condition, and we obtain +
2 = 0
((1))
Apply the second boundary condition, and we obtain (− sin + cos ) + ( cos + sin ) = 0 (− sin + cos ) + ( cos + sin ) = 0
((2))
Solve for in Equation (1) and substitute it into Equation (2) to obtain 2 = 0 (− sin + cos ) − ( cos + sin ) ¶ µ 2 2 cos − sin = 0 − sin + cos − The equation is satisfied if = 0, but it will result in a trivial solution of () = 0 Therefore, the terms in the parenthesis must be zero. The frequency equation is − sin + cos −
2 2 cos − sin = 0
The frequency equation can be rewritten as − sin +
cos − ()2 cos − () sin = 0
We are given = 2 and = 100 Then, the frequency equation is − sin + 100 cos −
1 1 ()2 cos − 100 () sin = 0 2 2
541 which has solutions = 107, 361 652 · · ·
The corresponding natural frequencies are r = (107, 361 652 · · · ) The mode shapes are obtained by eliminating or from () = cos + sin using Equation (1). The mode shape is given by µ ¶ 1 () = cos + sin The first three mode shapes are
¶ 2 cos 107 + sin 107 1 () = 107 µ ¶ 2 cos 361 + sin 361 2 () = 361 ¶ µ 2 cos 652 + sin 3 () = 652 µ
The next figure shows the first three mode shapes as a function of
Note that if as we increase , the boundary will act as a fixed support, and the mode shapes will look identical mirror image of the fixed-mass string of the previous problem. Take the frequency equation of the current problem and take a limit as → ∞ or → 0 lim − sin +
→0
cos − ()2 cos − () sin
() sin = 0 which is identical to the frequency equation of the previous problem. = cos −
542
CHAPTER 7 CONTINUOUS MODELS
9. Consider the fixed-fixed uniform string made of two sections shown in Figure 7.40. The left section of the string has a constant mass per unit length and the right section of the string has the constant mass per unit length 2 (a) From a free-body diagram and Newton’s second law of motion, under what assumptions can the tension be considered constant? (b) Assuming constant tension in both strings, find the frequency equation, the first three natural frequencies, and plot the first three mode shapes.
Figure 7.40
Solution: If we assume that the string moves in the vertical direction only, the sum of the forces in the direction must be zero at any point on the string cos − cos = 0 If we assume small vibration, cos ≈ 1 and therefore, ≈ The equation of motion, the boundary conditions, and the transition conditions are given by 2 1 2 1 = for 0 ≤ 1 ≤ 2 21 2 2 2 2 for 0 ≤ 2 ≤ 2 : 2 2 = 22 1 : 1 (0 ) = 0 2 : 2 (0 ) = 0 1 : 1 ( ) = 2 ( ) 2 : 10 ( ) + 20 ( ) = 0
1 :
where the last transition condition may seem wrong. The negative sign is due to the fact that the coordinate 2 goes from right to left. The free body diagram at the transition point is shown next.
The tension must be drawn so that the strings have positive slopes. For the heavier string, 2 is
543 defined positive to the left, and therefore, the positive slope means that the function increases as we go to the left. Summing the forces in the vertical direction, we obtain − 10 ( ) − 20 ( ) = 0 Separating ( ) into () () and letting ̈ = 2 the equations motion and boundary/transition conditions in terms of 1 () and 2 () are 1 : 100 + 2 1 = 0 for 0 ≤ 1 ≤ 2 : 200 + 22 2 = 0 for 0 ≤ 2 ≤ 1 : 1 (0) = 0 2 : 2 (0) = 0 1 : 1 () = 2 () 2 : 10 () + 20 () = 0 Let 2 = 2 for convenience, and from the equations of motion, we obtain 1 (1 ) = cos 1 + sin 1 √ √ 2 (2 ) = cos 21 + sin 22 Applying the boundary conditions, and we find = 0 and = 0 Applying the transition conditions, we find √ sin = sin 2 √ √ cos = − 2 cos 2 from which we obtain the frequency equation ³√ ´ √ √ 2 sin cos 2 + cos sin 2 = 0 We first need to test whether = 0 (rigid body motion) is possible. A fixed-fixed string cannot have a rigid body motion, and therefore, = 0 will not give a valid solution. The possible wave numbers are = 124 318 56236 .... √ The mode shapes are (after substituting = 0 = 0 and = sin sin 2) 1 (1 ) = sin 1 √ sin 2 (2 ) = √ sin 22 sin 2 The first three mode shapes are plotted.
544
CHAPTER 7 CONTINUOUS MODELS
545
10. Plot the displacement response of the string in Figure 7.36 at = as a function of time if the initial displacement is ( 0) = 005 and the initial velocity is zero. Solution: This is an initial value problem. We start from the equation of motion and the boundary conditions.
2 2 = 2 2
¯ ¯¯ (0 ) = 0 and = 0 ¯
We expand the solution ( ) in terms of eigenfunctions such that ( ) =
∞ P
() ()
((1))
=1
where the eigenfunctions () satisfy (see solution Problem 5) 00 () + 2 () = 0 2 = 2 1 1 3 1 5 () = sin where = ··· 2 2 2 The eigenfunctions are orthogonal such that Z () () = 0 for 6= 0
Let us normalize the eigenfunctions such that Z () () = 1 for = 0 Z 2 sin2 = 1 0 r 2 = The normalize eigenfunctions are
r
2 sin Getting back to Equation (1), upon substituting Equation (1) into the equation of motion, we have () =
∞ P
=1
̈ () () = =
∞ P
=1 ∞ P =1
00 () () ¡ ¢ () − 2 ()
546
CHAPTER 7 CONTINUOUS MODELS
We multiply the previous equation by (), integrate from 0 to and use the orthogonality condition to find ̈ () = − 2 () ̈ () + 2 () = 0 which has a solution of r r + sin () = cos We also expand the initial condition in terms of eigenfunctions such that ∞ P (0) () ( 0) = 005 = =1
By multiplying both sides by () integrating from 0 to and use the orthogonality condition, we find Z 005 () = (0) 0 For the first three eigenfunctions, we find Ãr ! Z 2 00287 1 (0) = sin = √ 005 2 0 Ã ! r Z 2 3 000318 2 (0) = sin = − √ 005 2 0 Ãr ! Z 5 000115 2 3 (0) = sin = √ 005 2 0 We apply these initial conditions, and we fin
r 00287 cos 1 () = √ 2 r 000318 3 2 () = − √ cos 2 r 000115 5 3 () = √ cos 2
Finally, the solution in Equation (1) is Ãr !Ã r ! √ 2 ( ) = sin 00287 cos 2 2 !Ã Ãr r ! √ 2 3 3 sin −000318 cos + 2 2 !Ã Ãr r ! √ 2 5 5 sin 000115 cos + 2 2 +···
547 At = the response is à r ! ³√ ´ ( ) = 2 sin 00287 cos 2 2 r ! ¶Ã µ √ 3 3 −000318 cos 2 sin + 2 2 r ! ¶Ã µ √ 5 5 000115 cos 2 sin + 2 2 +··· p Figure next shows the response ( ) as a function of Note that the initial condition ( ) = 005 is not quite satisfied. This is because only the first few terms in the expansion is used.
548
CHAPTER 7 CONTINUOUS MODELS
11. Plot the displacement response of the string in Figure 7.36 at = if the string is initially horizontal and is subject to the sinusoidal input force at = () = 0 sin Use = 1 m, = 100 N, = 02 kg/m, = 10 rad/s, and 0 = 5 N. Solution: We start from the equation of motion and the boundary conditions.
2 2 = 2 2
¯ ¯¯ (0 ) = 0 and − + () = 0 ¯
We can move the force () from the boundary condition to the equation of motion such that 2 2 = + () ( − ) 2 2 ¯ ¯¯ = 0 (0 ) = 0 and − ¯
This way, the boundary conditions remain homogenous, and we already have a set of eigenfunctions from the solutions to the Problem 5. We found the eigenfunctions for the homogenous version of this problem are given by 00 () + 2 () = 0 2 = 2 r 2 1 1 3 1 5 () = sin where = ··· 2 2 2 The eigenfunctions are orthonormal such that Z () () = . 0
We expand the solution ( ) in terms of eigenfunctions such that ( ) =
∞ P
() ()
((1))
=1
We substitute Equation (1) into the equation of motion to obtain ∞ P
=1
̈ () () = =
∞ P
=1 ∞ P =1
00 () () + () ( − ) ¡ ¢ () − 2 () + () ( − )
We multiply the previous equation by (), integrate from 0 to and use the orthogonality condition to find ̈ () = − 2 () + ()
549 where () =
Z 0
() ( − ) ()
= () () The first three terms are 1 () = ()
r
2 r
2 2 () = − () r 2 3 () = () The time varying function () has a solution of r r 0 1 + sin + sin () = cos 2 2 − Applying zero initial conditions, we find ⎞ ⎛ r ⎟ ⎜ 1 0 ⎟ ⎜− r () = + sin sin ⎠ ⎝ 2 2 − Finally, the complete solution is ∞ P ( ) = () () =1 ⎞ ⎛ Ãr ! r ∞ ⎟ ⎜ P 1 2 0 ⎟ ⎜− r = sin sin + sin ⎠ ⎝ 2 =1 − 2 Figure next shows the response at = using the first three terms.
550
CHAPTER 7 CONTINUOUS MODELS
12. Plot the displacement response of the string in Figure 7.38 at = as a function of time if the initial displacement is ( 0) = 005 and the initial velocity is zero. The orthogonality condition for the eigenfunctions must be obtained first. Solution: Orthogonality We start from the solutions for Problem 7. The eigenvalue problem is given by 00 + 2 = 0 (0) = 0 and 0 () − 2 () = 0 These equations must be satisfied for every eigenfunctions, and therefore, we can write for the and eigenfunctions as 00 + 2 = 0 00 + 2 = 0
((1)) ((2))
(0) = 0 and (0) = 0 0 () − 2 () = 0 and 0 () − 2 () = 0
((3))
and also for the boundary conditions,
((4))
Multiply Equation (1) by () and multiply Equation (2) by () and take the difference to find ¡ ¢ 00 − 00 = 2 − 2
Integrate it over the length of the string to find Z Z ¡ ¢ ¡ 2 ¢ 2 00 00 − = − 0
Integrate the left hand side by parts to obtain Z Z ³ ´ ¡ ¢ ¡ ¢¯ 00 00 0 0 ¯ 0 0 0 0 − − = − 0 − 0
((5))
0
0
where the term in the integral cancels. We substitute the boundary conditions given in Equation (3) and (4) to find Z ¡ ¢ 00 − 00 0
= () 0 () − () 0 () − (0) 0 (0) − (0) 0 (0) = () 2 () − () 2 () ¢ ¡ 2 = − 2 () ()
551 We rewrite Equation (5) by replacing the left hand side with the previous equation, and we have ¶ µZ ¡ 2 ¢ 2 − + () () = 0 0
For 6= we have the orthogonality condition of Z
+
0
or more commonly,
Z 0
() () = 0 for 6=
+ () () = 0 for 6=
For strings with varying mass per unit length, we need to replace 2 = 2 () and () cannot come out of the integral in Equation (5). Therefore, the second orthogonality condition with inside the integral is the correct expression. Let us also normalize the eigenfunctions so that Z + () () = 1 for = 0
We found the eigenfunctions in Problem 7, and they are 108 364 2 () = 2 sin 2 = 2 sin 658 3 () = 3 sin 3 = 3 sin 1 () = 1 sin 1 = 1 sin
If we normalized them, we find (also using = 2) 108 1198 sin 1 () = √ 364 1340 sin 2 () = √ 658 1384 sin 3 () = √ Eigenfunction Expansion Method We assume that the solution to the current initial value problem can be expanded in terms of these eigenfunctions so that ∞ P () () ( ) = =1
We substitute this expression into the original partial differential equation to obtain
∞ P
=1
() ̈ () =
∞ P
=1
00 () ()
552
CHAPTER 7 CONTINUOUS MODELS
The next step is to use the orthogonality condition to reduce above equation in terms of () only. Replace 00 () with − 2 () using the eigenvalue problem, and we obtain ∞ P
=1
() ̈ () =
At = we evaluate the previous equation to
∞ P
() ̈ () =
=1
∞ P
− 2 () ()
((6))
∞ P
− 2 () ()
((7))
=1
=1
We multiply Equation (6) by () and integrate and multiply Equation (7) by () to obtain Z Z ∞ ∞ P P () () ̈ () = () − 2 () () =1
0
0
=1
∞ ∞ P P () () ̈ () = () − 2 () () =1 =1
Adding the two expressions, we see a term such as Z ∞ ∞ P P + () () =1
0
=1
which we recognize as Then, the sum is written as
̈ ()
∞ P
=1
= − 2 ()
µZ 0
µZ 0
¶ + () ()
¶ () () + () ()
̈ () + 2
() = 0
which has the solution of the form () = cos
r
+ sin
r
where the wave numbers are already obtained in Problem 7. The initial velocity is zero, and therefore, we have = 0The initial displacement must be expanded the same way. We are given ( 0) =
∞ P 005 () (0) = =1
((8))
We evaluate the initial displacement at = to obtain ( 0) =
∞ P 005 = () (0) =1
((9))
553 We are now ready to apply the orthogonality condition. Multiply Equation (8) by () and integrate over 0 and to obtain Z 0
∞ P 005 () = =1
µZ 0
¶ () () (0)
Multiply Equation (9) () given by () 005 =
∞ P
=1
() () (0)
Adding the two previous equations, and we obtain Z 0
005 () + () 005 = (0)
where we use the eigenfunctions found in Problem 7 and normalized here. They were 108 1198 sin 1 () = √ 364 1340 sin 2 () = √ 658 1384 sin 3 () = √ We find the first three (0), and they are √ 1 (0) = 00456 √ 2 (0) = −000226 √ 3 (0) = 000053 Applying initial conditions, we find r √ 108 1 () = 00456 cos r √ 364 2 () = −000226 cos r √ 658 3 () = 000053 cos
554
CHAPTER 7 CONTINUOUS MODELS
Finally, the solution is given by ( ) =
∞ P
=1
() ()
r ! ¶Ã √ 1198 108 108 = √ 00456 cos sin r ! ¶Ã µ √ 364 364 1340 sin −000226 cos + √ r ! ¶Ã µ √ 658 658 1384 sin 000053 cos + √ +··· µ
The figure next shows above three-term solution.
555 Problems for Section 7.3 — Axial Vibration of Beams 13. Consider a system consisting of discrete masses connected by springs in series as shown in Figure 7.41. Derive the equation of motion for mass using Newton’s second law, arrange the equation in incremental form, take the limit, and derive the differential equation of motion for the longitudinal vibration of a thin beam.
Figure 7.41: Discrete mass and spring system constrained to move horizontally.
Solution: Using Newton’s equation of motion for mass the free-body diagram and the equation of motion are given by
̈ = (+1 − ) − −1 ( − −1 ) = (∆ ) − −1 (∆−1 ) where we defined the incremental change as ∆ = +1 − and ∆−1 = −−1 The incremental change in the net spring force acting on the mass can be written as (∆ ) − −1 (∆−1 ) = ∆ ( ∆ ) Thus, 2 = ∆ ( ∆ ) 2 Divide both sides by the length between masses, ∆ to obtain
∆ ( ∆ ) 2 = 2 ∆ ∆ From Strength of Materials, the stiffness of a uniform rod in tension is or = = ∆
556
CHAPTER 7 CONTINUOUS MODELS
As ∆ becomes small, each beam element becomes “uniform” over ∆ Inserting this into the equation of motion: µ ¶ ∆ ∆ 2 = ∆ 2 ∆ ∆
Now, let → ∞ and ∆ → 0 and we see that
= () ∆ →0 ∆ lim
where () is the mass per unit length. In addition, ∆ = ∆ →0 ∆ ( ) ∆ lim = ∆ →0 ∆ lim
Note that ( ) is a function of the spatial variable, as well as time so that the total derivative in 2 2 must be replaced with a partial derivative. Thus, µ ¶ 2 ( ) ( ) = 2
557
14. Derive Equation 7.32. Solution: Equation 7.32 is Z 2 ½Z ∙ 1
0
¸ ∙µ µ ¶ ¶ ¸ ) 2 = 0 () () − () 2 − 0
The equation that proceeds Equation 7.29 is µ ¶ µ ¶ ¸ Z 2 ∙Z Z − = 0 () () 1 0 0
(7.32)
(7.29)
We can rewrite the first integral as µ ¶ ¶ Z 2 Z Z µZ 2 () () () = 0 0 1 1
R R and = () ) to obtain Integrate by parts ( = − with = () ! ¯ µ ¶ µ ¶ Z 2 Z à Z 2 Z ¯¯ 2 () ¯ − () = () 0 0 1 1 1
Hamilton’s principle assumes that the ( ) is prescribed at the beginning and the end points such that ( 1 ) = ( 2 ) = 0 Then, we have µ ¶ µ ¶ Z Z 2 Z 2 Z = − () (1) () 0 0 1 1 We do the same thing for the second integral to obtain µ ¶ Z 2 Z () − 0 1 ¶ ¯ Z 2 Z µ ¶ µ ¯ ¯¯ + () = − () 0
1
0
Combining Equations (1) and (2), we obtain Equation 7.32.
(2)
558
CHAPTER 7 CONTINUOUS MODELS
15. For the uniform fixed-fixed beam in longitudinal vibration shown in Figure 7.42 derive the expressions for the natural frequencies and the mode shapes. Sketch the first three modes.
Figure 7:42: Fixed beam in longitudinal vibration.
Solution: The eigenvalue problem is given by ∙ ¸ − = 2 (0) = 0 () = 0 For this uniform beam, the eigenvalue problem can be written as 2 + 2 = 0 2 2 2 = The solution to this harmonic equation is given by () = 1 sin + 2 cos Applying the two boundary conditions yields the two equations for the constants of integration: (0) = 2 = 0 () = 1 sin = 0 from which sin = 0 since 1 6= 0 Thus,
= 1 2 3 2 and 2 = r r = or = 2 =
Therefore, the modes are given by () = sin = sin
= 1 2 3
559 R This expression can be normalized in order to define the coefficients using the relation 0 2 = 1 For the graphs of the modes we assume that the coefficients equal 1, and that = 10 :
The MATLAB code used for plotting the mode shapes are given below. clear L=10; for r=1:3 x=[0:0.02:L]’; b=r*pi/L; plot(x, sin(b*x)) hold on end
560
CHAPTER 7 CONTINUOUS MODELS
16. For Equation 7.37, solve for the constants and by satisfying the initial conditions. Solution: Refer to Equation 7.37 and its time derivative. Set = 0 in order to create two equations that must be solved for the two unknowns, and These two equations are the following: ∞ r X 2 (2 − 1) sin ( 0) = 2 =1 r ∞ r X 2 (2 − 1) (2 − 1) sin ̇( 0) = 2 2 =1 Solve for the two constants by multiplying both sides of each equation by the orthogonal function sin
(2 − 1) 2
and integrate all expressions over the length of the beam: Z Z ∞ r X 2 (2 − 1) (2 − 1) (2 − 1) = sin ( 0) sin sin 2 2 2 0 0 =1 Z 0
r ∞ r X 2 (2 − 1) (2 − 1) ̇( 0) sin = 2 2 =1 Z (2 − 1) (2 − 1) · sin sin 2 2 0
Due to the orthogonality condition, the integral Z (2 − 1) (2 − 1) sin = 0 except when = sin 2 2 0 Then, Z 0
(2 − 1) = ( 0) sin 2
0
(2 − 1) = ̇( 0) sin 2
Z
r r
2 2 2 (2 − 1) 2
r
2
These equations can be readily solved for and and explicitly so when the initial value functions of are given, ( 0) and ̇( 0) and therefore the integrals can be evaluated, r Z (2 − 1) = ( 0) sin 2 0 2 r r Z 2 (2 − 1) = ̇( 0) sin (2 − 1) 2 0 2
561
17. Complete the solution of Example 7.2 for the modes and the complete solution. Solution: Let us start with the spatial solution, () = sin
+ cos
which must satisfy the boundary conditions, (0) = 0 and 0 ()+ () = 0 These conditions result in =0 and the frequency equation that must satisfy, tan = −
The dimensionless wave number, can be found once the nondimensional number is given. We can generalized the result if we use dimensionless parameters, and We will use the following notations for the dimensionless parameters, = ̂ and = ̂ The mode shape is given by () = sin ̂ ̂ where is included so that the mode shapes are normalized. That is, Z 2 = 1 0
1 = √ qR 1 2 0 sin ̂ ̂̂
The normalizationpconstant varies depending on the value of ̂ The corresponding natural frequency is = The complete solution is
( ) =
∞ X
() ( cos + sin )
=1
The coefficients of integration, and , are found by applying the initial conditions. For example, if ( 0) and ̇ ( 0) are given, we find that Z = ( 0) () 0 Z 1 ( 0) () = 0
562
CHAPTER 7 CONTINUOUS MODELS
18. Consider the free axial motion of a beam with the elastic boundary conditions of Example 7.2, and solve for the first four natural frequencies, modes, and the complete solution, for the following cases: (a) = N/m, = 10 m (b) = 10 N/m, = 1 m (c) = 10 N/m, = 1 m (d) = 10 N/m, = 10 m.
Solution: From Example 7.2, we have the general expressions for the frequencies and modes: tan = − () = sin 2 = 2
(a) = N/m, = 10 m We need to solve for the roots of tan 10 = − or tan 10 + = 0
From this we can determine the roots as approximately 0 028 057 087 Note that 0 is not a valid answer because rigid-body motion is not possible.
(b) = 10 N/m, = 1 m We need to solve for the roots of tan = −10or tan + 10 = 0
563
From this we can read the roots as 0 28 57 and 87 Again, 0 is not a valid wave number.
(c) = 10 N/m, = 1 m We need to solve for the roots of tan = −10or tan + 10 = 0
From these graphs, we determine the roots to be approximately 0 16 47 and 79 These numbers are approximately, 0, 2 32 52 and so on. Again, 0 is not a valid wave number.
(d) = 10 , = 10 We need to solve for the roots of tan 10 = −10or tan 10 + 10 = 0
564
CHAPTER 7 CONTINUOUS MODELS
The roots are 0 020 049 080 Again, 0 is excluded.
565
19. An unrestrained beam lies on a horizontal and smooth surface and at = 0 is forced at one end by (), as shown in Figure 7.43. Derive the general response for any forcing function using the known eigenvalues and eigenfunctions. Then evaluate the specific steady-state response for the following specific forcing functions: (a) () = 0 (), where () is the unit step function (b) () = sin Ω for the case where Ω = 2 (c) () = 1 − 0 , 0 ≤ ≤ 0
(d) () = (), where () is the Dirac delta function.
Figure 7.43: Beam on smooth (frictionless) surface.
Solution: The equation of motion for this problem is given by
2 2 − = () (0 ) 2 2
with boundary conditions given by ¯ ¯¯ = 0 ¯ =0 ¯ ¯¯ ¯ = 0 =
Note that (0 ) is the Dirac delta function and is needed because the force is applied at = 0 instead of the whole domain. We have solved parts of the eigenvalue problem before, that is, we have solved for a free end. Here, both ends are free and therefore we need to evaluate the constants again. We can see that the motion of this beam is an axial vibration superposed on a rigid-body translation. The eigenvalue problem results in the general modal function: () = 1 sin + 2 cos 0 () = 0 0 (0) = 0 Applying the boundary conditions results in the following: 0 (0) = 0 = 1 0 () = 0 = −2 sin
566
CHAPTER 7 CONTINUOUS MODELS
Since we impose 2 6= 0 for otherwise we have a trivial solution, the following must hold: sin = 0 = =
= 1 2
The natural frequencies are then, 2 = 2 ³ ´2 = The modal function is () = 2 cos() for = 1 2 · · ·
(1)
There is a rigid-body mode: 0 () = 1 +2 that corresponds to 0 = 0 1 and 2 are the constants of integration. Applying the boundary conditions, we find that 0 () = 2 The constants, 2 and 2 can be determined so that the modal functions are normalized such that Z 0
2 () = 1 for = 0 1 2 · · ·
Then, we find that 1 ̂0 () = √ r 2 cos for = 1 2 · · · ̂ () =
(2)
For the general solution we have ( ) =
∞ X
()̂ ()
=0
where () satisfies ̈ () + 2 () =
1 ()
where () =
Z
( )̂ ()
0
=
Z
() (0 ) ̂ ()
0
= () ̂ (0)
(3)
567 Note the difference between the generalized force, () and the end force, () The general solution is then given by 1 () = cos + sin + | {z } | free solution
Z
( ) sin ( − ) 0 {z } forced solution Z (0) = cos + sin + ( ) sin ( − ) | {z } 0 {z } | free solution forced solution
for = 1 2 · · · For = 0 we the modal coordinate is the solution to ̈0 = 0 ( ) The modal functions are given above. The initial conditions can be satisfied by applying them to the general solution and solving for and The total solution is then given by Equation 3 where ̂ () is given by Equation 2. For an undamped case, the steady-state response is the total response because the homogeneous solution does not decay with time. The total response consists of the homogeneous solution that requires the knowledge of initial conditions and the particular solution. Since the initial conditions are not given, we can assume an arbitrary initial conditions. If we assume zero initial displacements and zero initial velocities, the forced solution (convolution solution) is the steady-state solution. Another possibility is to assume that there is a small damping in the system to eliminate the need for the initial conditions. From an engineering purposes, all systems have damping. Therefore, the homogeneous solution always decays with time, and we can take the particular solution to be the steady-state solution. We will take this approach. The modal coordinates, () satisfy ̈ () + 2 () = () ̂ (0)
(4)
In the following, we will find the particular solutions for the given end forcing. The steady-state response is given by Equation 1 where () is the particular solution of Equation 4 and ̂ () is the normalized modal function given in Equation 2. (a) () = 0 () The modal coordinates satisfy ̈ () + 2 () = 0 ()̂ (0) where 1 ̂0 (0) = √
and
̂12··· (0) =
For = 0 we have 1 ̈0 = 0 () √
r
2
568
CHAPTER 7 CONTINUOUS MODELS
The particular solutions are 0 2 0 () = √ 2 20 √ () = for = 1 2 · · · 2 for ≥ 0 The steady-state solution is ( ) =
∞ X
()̂ ()
=0
where ̂ () is given in Equation 2. (b) () = sin Ω The modal coordinates satisfy ̈ () + 2 () = sin Ω̂ (0) where 1 ̂0 (0) = √ The particular solutions are
and
̂12··· (0) =
r
2
1 1 sin Ω 0 () = − √ Ω2 r 1 2 () = sin Ω for = 1 2 · · · 2 − Ω2 (c) () = 1 − 0 N, 0 ≤ ≤ 0 The modal coordinates satisfy ¶ µ 1− ̂ (0) for 0 ≤ ≤ 0 0 = 0 for ≥ 0
̈ () + 2 () =
where 1 ̂0 (0) = √ The particular solutions are
and
̂12··· (0) =
r
2
µ ¶ 1 2 1 3 1 0 () = √ − for 0 ≤ ≤ 0 6 0 2 = 0 for ≥ 0
569 and r
2 0 − for 0 ≤ ≤ 0 2 0 = 0 for ≥ 0
() =
for = 1 2 · · · (d) () = () N The modal coordinates satisfy ̈ () + 2 () = ()̂ (0) where 1 ̂0 (0) = √ The particular solutions are
and
̂12··· (0) =
r
2
1 0 () = √ r 2 1 () = sin for = 1 2 · · · Note that these are the impulse response functions multiplied by ̂ (0)
570
CHAPTER 7 CONTINUOUS MODELS
20. Find the response of a uniform beam clamped at = 0 and free at = subjected to the longitudinal force, ( ) = [ − ( − )] () Solution: The eigenvalue problem for the longitudinal beam is given by µ ¶ − () = 2 () () and the boundary conditions of clamped free beam are (0) = 0 ¯ () ¯¯ () = 0 ¯=
The natural frequencies and normal modes are r = (2 − 1) 2 2 r 2 () = sin (2 − 1) for = 1 2 · · · 2 The forced response is ∞ P () () ( ) = =1
where () are the natural modes, which are already found, and Z 1 ( ) sin ( − ) () = 0 sin + (0) cos + ̇ (0) for = 1 2 · · · where Z () = () ( ) 0 Z () () ( 0) (0) = 0 Z () () ̇ ( 0) ̇ (0) = 0
Assuming zero initial conditions, ( 0) = 0 and ̇ ( 0) = 0 we find (0) = 0 and ̇ (0) The generalized force is Z r 2 sin (2 − 1) [ − ( − )] () () = 2 0 r 2 ( − ) sin (2 − 1) () = 2
571 and 1 () = = = 2
Z 0
r
r
( ) sin ( − )
2 ( − ) sin (2 − 1) 2
Z 0
sin ( − )
( − ) 2 sin (2 − 1) [1 − cos ] 2
Replace with their values and r
µ ¶ 2 ( − ) sin (2 − 1) 2 !# r × 1 − cos (2 − 1) 2 2
42 () = (2 − 1)2 2 " Ã
where the sine term can be simplified to ¶ µ ( − ) sin (2 − 1) 2 ³ ³ ³ ³ ´ ´ ´ ´ = sin (2 − 1) cos (2 − 1) − cos (2 − 1) sin (2 − 1) 2 2 2 ³2 ´ −1 = (−1) cos (2 − 1) 2 Then,
( ) =
∞ P
() ()
=1
³ ³ ´ ´ sin (2 − 1) (2 − 1) 8 2 2 = 2 2 =1 (2 − 1) " Ã !# r × 1 − cos (2 − 1) 2 2 −1 cos ∞ (−1) P
572
CHAPTER 7 CONTINUOUS MODELS
21. Derive the natural frequencies and mode shapes of the longitudinally vibrating uniform beam of Figure 7.44. Sketch the first three mode shapes. What will change in the analysis if the beam is vertical?
Figure 7.44: Transverse vibration of a uniform cantilever beam.
Solution: If the orientation of the beam changes, only the static deflection will be changed. The natural frequencies and the mode shapes will be the same for both horizontal and vertical beams. This is a clamped-free beam. The general expression for the mode shape is given by () = cos + sin + cosh + sinh The boundary conditions are (0) = 0 0 (0) = 0 00 () = 0 and 000 (0) = 0 Apply the boundary conditions to obtain + = 0 + = 0 − cos − sin + cosh + sinh = 0 sin − cos + sinh + cosh = 0 − cos − sin − cosh − sinh = 0 We can solve for three coefficients in terms of one coefficient. Let us express and in terms of so that = −
(cos + cosh ) (sin + sinh ) (cos + cosh ) = (sin + sinh )
= −
We can proceed with substitution to the final boundary condition (that we have not used yet to solve for and ) to find the characteristic equation. Another option is to write the boundary
573 conditions in a matrix form such that ⎤⎡ ⎤ ⎡ ⎤ ⎡ 0 1 0 1 0 ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ 0 ⎥ ⎢ 0 1 0 1 ⎥⎢ ⎥ = ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢ − cos − sin cosh sinh ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎦⎣ ⎦ ⎣ ⎦ ⎣ 0 sin − cos sinh cosh
The characteristic equation is the determinant of the 4×4 matrix and is given by cos cosh + 1 = 0 The normalized wave numbers that satisfy the characteristic equation are = 18751 46941 78548 The mode shape is given by µ (cos + cosh ) sin () = cos − (sin + sinh ) ¶ (cos + cosh ) sinh − cosh + (sin + sinh ) The first three mode shapes are plotted next..
574
CHAPTER 7 CONTINUOUS MODELS
22. A cantilever beam has its free end stretched uniformly so that the original length becomes 0 , and then it is released at = 0 as shown in Figure 7.44. Begin with the general solution for the axial response of a beam, Ã r ∞ X (2 − 1) (2 − 1) sin ( ) = sin 2 2 =1 ! r (2 − 1) + cos 2 then satisfy the boundary conditions to fix the arbitrary constants, and finally derive the particular response.
Figure 7.44: Stretched cantilever beam with original length
Solution: Let us derive the general solution. We start with the equation of motion for a longitudinal beam given by 2 2 2 − 2 = 0 on 0 ≤ ≤ and the boundary conditions are
The initial conditions are given by
¯ ¯¯ = 0 (0 ) = 0 and ¯=
0 − ( 0) =
and
¯ ¯¯ = 0 ¯=0
Writing ( ) as () () we obtain two ordinary differential equations given by
2 () + 2 () = 0 2 2 () + 2 () = 0 2
575 The eigenvalue problem is the spatial equation. The complete eigenvalue problem is given by 2 () + 2 () = 0 2 ¯ ¯¯ = 0 (0) = 0 and ¯=
from which we find the eigenfunctions, wave numbers, and the natural frequencies given by r 2 sin ̂ () = (2 − 1) = r 2 = for R = 1 2 · · · Note that ^ is used to indicate that the eigenfunction is normalized according to ̂ () = 1 0
Then, the general solution to the initial value problem is given by ( ) = =
∞ X =1 ∞ X
̂ () () ̂ () ( sin + cos )
=1
After substituting ̂ () and , we obtain à r ∞ r X 2 (2 − 1) (2 − 1) ( ) = sin sin 2 2 =1 ! r (2 − 1) + cos 2
(1)
Note that there is a typographical error in the problem statement. should not appear in () p The factor 2 in the eigenfunction can be omitted. Let us now find the constants of integration by applying the initial conditions. Applying the initial velocity, we have ̇( 0) = 0 =
∞ X
̂ ()
=1
which gives = 0 for all Applying the initial displacement, we obtain ∞ X 0 − ̂ () ( 0) = = =1
576
CHAPTER 7 CONTINUOUS MODELS
Multiplying the both sides by ̂ () and integrating over the length of the beam, we have Z
Z X ∞
0
0 − ̂ () =
0
0 − ̂ () =
Z
0
̂ () ̂ ()
=1
where the only nonzero term in the series on the right-hand side occurs for = due to the orthogonality condition. Substituting the explicit expression for ̂ () we have r Z 0 − (2 − 1) 2 for = 1 2 · · · sin = 0 2 We can replace with at this point. The integral has an analytical expression, and are given by r 42 0 − 2 = − (−1) (2) 2 2 (2 − 1) The first three coefficients are
r 0 − 2 42 1 = 2 r ¶ µ 0 − 2 −42 2 = 9 2 r µ 2¶ 0 − 2 4 3 = 25 2 The response to the initial conditions is given by Equation 1 with coefficients given by = 0 and Equation 2.
577
23. Consider the stepped shaft made of the same material of density and modulus shown in Figure 7.45. Determine the characteristic equation for the natural frequencies of the system as it undergoes axial motion. Let 1 ( ) denote the displacement in the shaft with the fixed end, and 2 ( ) be the displacement in the shaft with free end.
Figure 7.45: Stepped shaft undergoing axial motion.
Solution: The equations of motion are 2 2 = for 0 ≤ 1 ≤ 1 1 2 21 2 2 2 2 = 2 2 for 0 ≤ 2 ≤ 2 2 1
Note that the areas 1 and 2 can be canceled from the equations, so that we can write a single partial differential equation that applies for the whole range. For now, let us keep the two equations. In addition, we let 2 = 0 at the free end. The boundary conditions are 1 (0 ) = 0 ¯ 2 ¯¯ = 0 ¯2
In addition to the two conditions at the ends, there are two more conditions where the transition occurs. The displacement and the transmitted force must be continuous. We write 1 (1 ) = −2 (2 ) ¯ ¯ 1 ¯¯ 2 ¯¯ 1 = 2 1 ¯1 2 ¯2
578
CHAPTER 7 CONTINUOUS MODELS
The reason why the displacements at the joint are equal and opposite is that 1 is defined positive to the right and 2 is defined positive to the left. We are now ready solve the eigenvalue problem. Let 1 (1 ) = (1 ) () and 2 () = (2 ) () Note that 1 (1 ) and 2 (2 ) share the same time function. We can write () = exp () because the motion is sinusoidal in time. The eigenvalue problem is now given by 2 2 = 0 for 0 ≤ 1 ≤ 1 + 21 2 2 = 0 for 0 ≤ 2 ≤ 2 + 22 and (0) = 0 0 (0) = 0 (1 ) = − (2 ) 1 0 (1 ) = 2 0 (2 )
(7.1)
From the spatial ordinary differential equations, we have the forms of (1 ) and (2 ) given by (1 ) = cos 1 + sin 1 (2 ) = cos 2 + sin 2 where =
p Applying the boundary and transition conditions (Equation 1), we obtain = 0 = 0 sin 1 = − cos 2 1 cos 1 = −2 sin 2
Solving for in terms of using the third relation, and substituting into the fourth relation we find the characteristic equation, 1 cos 2 cos 1 − 2 sin 2 sin 1 = 0 The mode shapes are given by (1 ) = sin (1 ) sin 1 cos (2 ) (2 ) = − cos 2 This completes the solution to the eigenvalue problem. For demonstration purposes, let us find the mode shapes for the case 1 2 = 4 and 1 2 = 12. Then, the characteristic equation is given by 4 cos 2̂ cos ̂ − sin 2̂ sin ̂ = 0
579 where ̂ = 1 The roots, found numerically, are ̂ = 06847, 1.5708, 2.4569. The mode shapes are ¶ µ 1 (1 ) = sin ̂ = sin (1 ) 1 ¶ µ sin 1 sin ̂ 2 (2 ) = − = − cos ̂ cos (2 ) 1 cos 2 cos ̂2 1 For convenience, we use one independent variable, and the mode shape that are valid for the entire domain. Let = 1 for 0 ≤ ≤ 1 = 1 + 2 − 2 for 0 ≤ 2 ≤ 2 and () = () for 0 ≤ ≤ 1 = − () for 1 ≤ ≤ 1 + 2
The figure below shows the first mode shape with = 1 Note that there is an abrupt slope change at the transition. The slope becomes four times steeper in order for the tension to be continuous.
MATLAB code used for the plot is given below.
580 clear x1=[0:0.05:1]’; U1=sin(0.6847*x1); x2=[0:0.05:2]; U2=sin(0.6847)/cos(0.6847*2)*cos(0.6847*x2); plot(x1,U1,3-x2, U2)
CHAPTER 7 CONTINUOUS MODELS
581
24. Consider the uniform beam shown in Figure 7.46. Solve for the lowest natural frequency of axial motion of the steel beam given the following parameter values: = 210 GN/m2 = 7850 kg/m3 = 12 m, = 100 kN/m, and = 3 cm2
Figure 7.46: Axially vibrating beam with spring boundary.
Solution: The equation of motion is given by‘
2 2 = 2 2
and the boundary conditions are given by (0 ) = 0 ¯ ¯ ¯¯ + ( ) = 0
Let ( ) = () exp () The eigenvalue problem is given by 00 +
2 =0
and (0 ) = 0 0 () + ( ) = 0 The mode shape, () takes the form of () = cos + sin p where = Upon applying the boundary conditions, we obtain () = sin
and cos + sin = 0 We can substitute the values at this point to find or we can delay substitution a bit longer. We can rewrite the characteristic equation as cos + sin = 0
582
CHAPTER 7 CONTINUOUS MODELS
Now, the nondimensional wave number, can be found for various values of In our case, we are given = 210×109 N/m2 = 7850 kg/m3 = 12 m, = 100 000 N/m, and = 3×10−4 m2 . One single number that determines the nondimensional wave number is = 525 which has no units. The characteristic equation, 525 cos + sin = 0 can be solved for graphically or using root finding programs. In MATLAB, one would write fzero(@(b) 525*b*cos(b)+sin(b), 1) where 1 is the initial guess. It finds the root = 1572 The first four roots are = 0 1.572, 4.713, 7.854. Note that = 0 is not a valid wave number because it results in zero natural frequency which is not possible for this problem. Zero is an acceptable answer only when the geometry permits rigid-body motion. The first three wave numbers are = 1572 4713 7854 = 1886 5656 9425 The first three natural frequencies are s = = 0952 × 104 2854 × 104 4757 × 104 The mode shape is given by () = sin where is usually normalized according R to 0 2 () = 1 The figures below show the first three mode shapes as functions of . Note that we need to know only one nondimensional number, to plot the mode shapes. Plotting mode shapes is usually a good check whether we have the correct wave numbers and natural frequencies because the mode shapes are intuitive.
583
MATLAB code used for the plot is given below. clear x=[0:0.05:1]’; U1=sin(1.572*x); U2=sin(4.713*x); U3=sin(7.854*x); plot(x,U1,x, U2,x,U3)
584
CHAPTER 7 CONTINUOUS MODELS
25. Solve for the first three modes and frequencies of the nonuniform longitudinally vibrating beam of Figure 7.47 beginning with the known equation of motion. The following properties hold: () = 2(1 − ) and () = 2(1 − ) (a) Solve this problem for arbitrary , and (b) then reduce the solution for the case where = 0. How far can this problem be solved analytically?
Figure 7.47: Elastic constraint of a longitudinally vibrating beam. At position the beam elongates an amount ( ) At the right end there is a spring of stiffness constant
Solution: For this particular problem, we can find the mode shapes and natural frequencies analytically. (a) The eigenvalue problem for the axially vibrating beam (with non uniform properties) is given by ∙ ¸ − () = 2 ()() The boundary conditions for this problem are (0) = 0 () () + () = 0 0
where the first boundary condition is the no displacement condition at the left end, and the second boundary condition is the force balance at the right end connected to the spring. We are given the area and mass distributions. Substituting these into the eigenvalue problem yields: ∙ ¸ ³ ³ ´ ´ 1 − = 2 1 − () − Expanding this equation, we obtain ³ ³ ´ 2 1 ´ 2 − 1 − = () + 1 − 2
585 Rewriting this equation in standard form, and performing the transformation = − noting that = − = − and 2 2 = 2 2 we have µ µ ¶ ¶ − − 2 2 − + () = 0 2 2 2 + ( − )() = 0 ( − ) 2 − 2 + 2 = 0 (1) 2 + where 2 = 2 and is now a function of The boundary conditions are: ¯ ¯¯ (0) ¯ − (0) = 0 0 () = 0 The extensional rigidity at = 0 is zero. Therefore, we have (0) = 0 We can rewrite Equation 1 as 2
¤ 2 £ + + ()2 − 2 = 0 2
for the case = 0 This is a Bessel ODE of order 0 with general solution () = 1 0 () + 2 0 ()
(2)
Applying the boundary conditions, we obtain (0) = 1 0 (0) + 2 0 (0) () = 1 0 () + 2 0 () Since 0 () approaches −∞ as → 0 we set 2 = 0 The condition that () = 0 leads to 0 () q = 0 The first three roots of 0 () are = 2.4048, 5.5201, and 8.6537, where =
The roots of Bessel functions are readily available in literature. Therefore,
r
r 24048 = 1 = 1 r r 55201 = 2 = 2 r r 86537 = =
The mode shape is () = 0 ( ( − ))
586
CHAPTER 7 CONTINUOUS MODELS
(b) If the end is free, the boundary conditions are ( = 0) = 0 0 ( = ) = 0 The derivations are the same up to Equation 2 of part (a). The boundary conditions in term of are ( = ) = 0 0 ( = 0) = 0 Apply these boundary conditions to the general solution given in Equation 2, and we obtain 1 0 () + 2 0 () = 0 −1 1 (0) − 2 1 (0) = 0 Since 0 () approaches −∞ as → 0 we set 2 = 0 Then, the eigenfunction is given by () = 1 0 () The condition 0 () = 0 gives the same series of as part (a).
587
26. Consider the longitudinally vibrating uniform beam shown in Figure 7.48. One end of the beam is fixed and the other end is free with an attached concentrated mass Derive the frequency equation, the first three natural frequencies, and plot the first three mode shapes for axial vibration. Use = 1
Figure 7.48: Axially vibrating cantilever beam fixed at one end and with an end point-mass at the other end. Solution: The equation of motion and the boundary conditions are given by
2 2 = 2 2
¯ ¯¯ 2 ( ) (0 ) = 0 and − = ¯ 2
where the second boundary condition says that the sum of the forces to the right is equal to the product of and its acceleration. The free-body diagram must be drawn so that the forces are due to displacement and the angle of rotation in the positive directions.
Write ( ) = () () then ̈ = 00 (0) () = 0 and − 0 () () = () ̈ () We divide previous equations by () we have ̈ 00 = (0) = 0 and − 0 () = ()
̈ () ()
588
CHAPTER 7 CONTINUOUS MODELS
We let ̈ = − 2 for harmonic solutions, and therefore, 2 =0 + (0) = 0 and − 0 () = − () 2 00
For convenience, we let 2 = 2 where is the wave number. () has a general solution of the form () = cos + sin which must satisfy the boundary conditions. The second boundary condition is rewritten in terms of − 0 () = − () 2 − 0 () = − () 2 0 () −
2 () = 0
Apply (0) = 0, and we have = 0. Apply 0 () −
2 () = 0 and we have
2 ( cos + sin ) = 0 ¶ µ 2 cos − sin = 0
− sin + cos −
The equation is satisfied if = 0, but it will result in a trivial solution of () = 0 Therefore, the terms in the parenthesis must be zero. The frequency equation is cos −
2 sin = 0
The frequency equation can be rewritten as ¶ µ sin = 0 cos − We check whether = 0 is possible solution. If = 0 the corresponding natural frequency is also zero and the form of the mode shape changes to 0 () = + If we apply the boundary conditions, we obtain = = 0 which is also trivial. Therefore, = 0 is not a valid wavenumber. We are given = 1 then, the frequency equation is cos − sin = 0 which has solutions = 0860, 343 644 · · ·
589 The corresponding natural frequencies are r = (0860, 343 644 · · · ) The first three mode shapes are 0860 343 2 () = 2 sin 2 = 2 sin 644 3 () = 3 sin 3 = 3 sin 1 () = 1 sin 1 = 1 sin
The next figure shows the first three mode shapes as a function of The end mass is not heavy enough such that the it behaves as a fixed-free string.
590
CHAPTER 7 CONTINUOUS MODELS
27. Derive the orthogonality condition for the eigenfunctions of the beam shown in Figure 7.48.
Solution: We start from the solutions for Problem 26. The eigenvalue problem is given by 0
( 0 ) + 2 = 0 (0) = 0 and 0 () − 2 () = 0 These equations must be satisfied for every eigenfunctions, and therefore, we can write for the and eigenfunctions as 0
(0 ) + 2 = 0 ¢0 ¡ 0 + 2 = 0
(1) (2)
(0) = 0 and (0) = 0 0 2 () − () = 0 and 0 () − 2 () = 0
(3) (4)
and also for the boundary conditions,
Multiply Equation (1) by () and multiply Equation (2) by () and take the difference to find ¡ ¡ ¢0 ¢ 0 (0 ) − 0 = 2 − 2
Integrate it over the length of the string to find Z 0
0 (0 ) −
¡ ¡ ¢0 ¢ 0 = 2 − 2
Z
(5)
0
Integrate the left hand side by parts to obtain Z
¡ ¢0 0 (0 ) − 0 0 Z ³ ´ ¢¯ ¡ 0 0 0 ¯ 0 0 − 0 = − 0 − 0
where the term in the integral cancels. We substitute the boundary conditions given in Equation (3) and (4) to find Z 0
¡ ¢0 0 (0 ) − 0
= () 0 () − () 0 () − (0) 0 (0) − (0) 0 (0)
= () 2 () − () 2 () ¢ ¡ = 2 − 2 () ()
591 We rewrite Equation (5) by replacing the left hand side with the previous equation, and we have ¶ µZ ¡ 2 ¢ 2 − + () () = 0 0
For 6= we have the orthogonality condition of Z 0
+ () () = 0 for 6=
592
CHAPTER 7 CONTINUOUS MODELS
28. Consider the longitudinally vibrating uniform beam shown in Figure 7.49. The ends of the beam are supported by linear springs with stiffness 1 and 2 Derive the frequency equation, the first three natural frequencies, and plot the first three mode shapes for axial vibration. Use 1 = 05 and 2 = 2
Figure 7.49: Axially vibrating beam supported by springs at both ends.
Solution: The equation of motion and the boundary conditions are given by 2 2 2 = 2
¯ ¯ ¯¯ ¯¯ − 1 (0 ) = 0 and − − 2 ( ) = 0 ¯0 ¯
The boundary conditions are obtained from the free-body diagrams of fake masses between the springs and the two ends of rod as shown in the figure next. The tension in the rod is The axial displacement ( ) is defined positive to the right, and the spring at = 0 is stretched and the spring at = is compressed.
Write ( ) = () () then ̈ = 00 0 (0) () − 1 (0) () = 0 and 0 () () + 2 () () = 0 We divide previous equations by () we have ̈ 00 = 0 (0) − 1 (0) = 0 and 0 () + 2 () = 0
593 We let ̈ = − 2 for harmonic solutions, and therefore,
2 =0 0 (0) − 1 (0) = 0 and 0 () + 2 () = 0 00 +
For convenience, we let 2 = 2 where is the wave number. () has a general solution of the form () = cos + sin which must satisfy the boundary conditions so that 0 (0) − 1 (0) = − 1 = 0 0 () + 2 () = (− sin + cos ) +2 ( cos + sin ) = 0
(1) (2)
Using the first equation, we have = 1 = (1 ) () and we can write the mode shape as µ ¶ 1 1 () = cos + sin (3) The two equations (1) and (2) are rewritten as ⎤⎡ ⎤ ⎡ ⎤ ⎡ −1 0 ⎦⎣ ⎦ = ⎣ ⎦ ⎣ − sin + 2 cos cos + 2 sin 0 For non-trivial solutions, the two simultaneous equations must be linearly dependent or the square matrix must have zero determinant. That is, −1 ( cos + 2 sin ) − (− sin + 2 cos ) = 0 We multiply by 2 and collect terms, we have µ ¶ µ ¶µ ¶¶ µ 1 2 2 1 2 − − cos + () − sin = 0 We are given 1 = 05 and 2 = 2 then ¢ ¡ −25 cos + ()2 − 1 sin = 0 The roots are found numerically, and they are
= 134 376 665 · · · The first three mode shapes in Equation (3) are ¶ µ 1 sin 134 1 () = cos 134 + 05 134 ¶ µ 1 sin 376 2 () = cos 376 + 05 376 µ ¶ 1 sin 665 3 () = cos 665 + 05 665
594 and they are plotted next.
CHAPTER 7 CONTINUOUS MODELS
595
29. Consider the two identical longitudinally vibrating uniform beams shown in Figure 7.50 connected by a linear spring. Derive the frequency equation, the first three natural frequencies, and plot the first three mode shapes for axial vibration. Use = 01
Figure 7.50: Two cantilever beams connected by spring undergoing axial vibration.
Solution: The equations of motion and the boundary conditions are 2 2 = for 0 ≤ ≤ 2 2 2 2 2 = 2 for 0 ≤ ≤ ¯ ¯¯ (0 ) = 0 and − − ( ( ) + ( )) = 0 ¯ ¯ ¯¯ + ( ( ) + ( )) = 0 (0 ) = 0 and ¯
Note that and ( ) are defined positive to the left. The total compression in the spring is ( ) + ( ) We draw the free-body diagrams for the fake masses inserted between the ends of the bar and the spring as shown in the figure next. Summing the forces for the fake masses, we obtain the last two boundary conditions.
Write ( ) = ( ) () and ( ) = ( ) () and then the equations of motion and boundary conditions become ̈ = 00 for 0 ≤ ≤
̈ = 00 for 0 ≤ ≤ (0) = 0 0 − () () − ( () + ()) () = 0 (0) = 0 0 () () + ( () + ()) () = 0
596
CHAPTER 7 CONTINUOUS MODELS
We divide previous equations by () and let ̈ = − 2 for harmonic solutions. Then, the equations of motion and boundary conditions become 2 = 0 for 0 ≤ ≤ 2 = 0 for 0 ≤ ≤ 00 + 00 +
(0) = 0 0 − () − ( () + ()) = 0 (0) = 0 0 () + ( () + ()) = 0
For convenience, we let 2 = 2 where is the wave number. () has a general solution of the form ( ) = cos + sin ( ) = cos + sin Apply the first and the third boundary conditions, and we obtain = 0 and = 0 Apply the other two conditions, and we obtain − ( cos ) − ( sin + sin ) = 0 ( cos ) + ( sin + sin ) = 0 We find that = and therefore, the mode shapes are given by ( ) = sin ( ) = sin Getting back to the previous equations, we can have non-trivial solution only if ( cos ) + 2 sin = 0 cos + 2 sin = 0 cos + 2 (1) sin = 0 The solutions to this frequency equation are = 229 509 810 · · · The first three modes are plotted. The gap at = 1 is the spring extension/compression at each mode.
597
598
CHAPTER 7 CONTINUOUS MODELS
30. Plot the response of the concentrated mass in Figure 7.48 due to the initial displacement ( 0) = 002 and zero initial velocity. Solution: The equation of motion and the boundary conditions are given by 2 2 2 = 2
¯ ¯¯ 2 ( ) (0 ) = 0 and − = ¯ 2
where = 1 The initial condition is given by ( 0) = 002 We found previously that the eigenfunctions and eigenvalues for this problem satisfy the eigenvalue problem 2 =0 (0) = 0 and − 0 () = − () 2 00 +
and satisfy the orthogonal condition given by Z + () () = 0 for 6= 0
For = 1 we also found that the first three wave numbers, natural frequencies, and the eigenfunctions are = 0860, 343 644 · · ·
The corresponding natural frequencies are r = (0860, 343 644 · · · ) The first three mode shapes are 1 () = 1 sin 1 = 1 sin 0860 2 () = 2 sin 2 = 2 sin 343 3 () = 3 sin 3 = 3 sin 644
(7.2)
We use the method of eigenfunction expansion to find the response by letting ( ) =
∞ P
() ()
=1
The goal is to find () Substitute the expansion into the original equation of motion to obtain
∞ P
=1
() ̈ () −
∞ P
=1
00 () () = 0
599 Use 00 + ( 2 ) = 0 to replace 00 with − ( 2 ) so that
∞ P
() ̈ () +
=1
∞ P
=1
2 () () = 0
((1))
Next, we eliminate () using the orthogonality condition. Multiply Equation (1) by () and integrate over the length of the beam to obtain Z ∞ Z ∞ P P () () ̈ () + 2 () () () = 0 =1 0 0 =1 Z ∞ ¡ ¢ P 2 () () = 0 ((2)) ̈ () + () =1
0
Multiply Equation (1) by () and evaluate it at = to obtain
∞ P
() () ̈ () +
=1
∞ P
=1
2 () () () = 0
∞ ¡ ¢ P ̈ () + 2 () () () = 0
((3))
=1
Add Equations (2) and (3) to obtain ½Z ¾ ∞ ¡ ¢ P 2 () () + () () = 0 ̈ () + () =1
0
̈ () + 2 () = 0
Then, () has a solution of the form, () = cos + sin The initial conditions (0) and ̇ (0) are obtained from ( 0) and ̇ ( 0) The initial displacement ( 0) is expanded such that ( 0) =
∞ P
() (0) = 002 =1
((4))
Multiply Equation (4) by () and integrate over the length of the beam to obtain Z ∞ Z P ( 0) () = () () (0) 0
0
=1
Multiply Equation (4) by () and evaluate at = to obtain ( 0) () =
∞ P
() () (0)
=1
Add Equations (4) and (5) and use the orthogonality condition to find R ( 0) () + ( 0) () (0) = 0 R 2 2 () + () 0
((5))
600
CHAPTER 7 CONTINUOUS MODELS
Similarly,
R 0
̇ (0) =
̇ ( 0) () + ̇ ( 0) () R 2 2 () + () 0
If we normalized the eigenfunctions, the denominator would be 1. The eigenfunctions may be normalized now or we can proceed as it is. After substituting = and 1 () = 1 sin 0860 2 () = 2 sin 343 3 () = 3 sin 644 we obtain 00260 1 00010 2 (0) = − 2 0000256 3 (0) = 3 1 (0) =
We then find r 00260 cos 0860 1 () = 1 r 00010 2 () = − cos 343 2 r 0000256 3 () = cos 644 3 Finally, the solution is given by ( ) =
∞ P
=1
() ()
à ! r ³ ´ = 00260 sin 0860 cos 0860 à ! r ³ ´ cos 343 −00010 sin 343 à ! r ³ ´ cos 644 +0000256 sin 644 +···
601 The position of the concentrated mass is Ã
! ( ) = 00260 sin (0860) cos 0860 ! Ã r −00010 sin (343) cos 343 ! Ã r +0000256 sin (644) cos 644 +··· r
602
CHAPTER 7 CONTINUOUS MODELS
Problems for Section 7.4 — Torsional Vibration of Shafts 31. Consider a fixed-fixed uniform torsional beam with length torsional rigidity , and mass per unit length . Derive the frequency equation, the first three natural frequencies, and plot the first three mode shapes. Solution: The equation of motion and the boundary conditions are 2 2 = 2 2 (0 ) = 0 and ( ) = 0
We let ( ) = Θ () () and the eigenvalue problem is given by −2 Θ () = Θ00 () Θ (0) = 0 and Θ () = 0 The eigenfunction has the form of Θ () = cos
r
2 + sin
r
Apply the boundary conditions, and we find = 0 and r 2 = 0 sin Then,
The first three eigenfunctions are
r
2 = 2 3 · · · 2 Θ2 () = sin 3 Θ3 () = sin Θ1 () = sin
2
603
32. The uniform shaft of Figure 7.51 is subjected to the torque 0 sin at its free end. (a) Find the steady-state vibration response by solving the free vibration problem, including the forcing function in a boundary condition. (b) Solve this problem as a forced vibration problem where the free end is left unforced and the forcing function is an external loading in the equation of motion.
Figure 7.51: Torsional vibration of a shaft.
Solution: Both (a) and (b) are forced vibration problems whether the harmonic torque is included as a boundary condition or a distributed load that appears on the boundary condition. For (a), the problem can be posed as a set of partial differential equation and boundary conditions given by ∙ ¸ 2 () = () 2 for 0 ≤ ≤ (0 ) = 0 ¯ ¯¯ ¯ = 0 sin (1)
where the external torque term appears in the boundary condition.
For (b), the problem can be posed as a set of partial differential equation and boundary conditions given by ∙ ¸ 2 () + ( − ) 0 sin = () 2 for 0 ≤ ≤ (0 ) = 0 ¯ ¯¯ ¯ = 0 (2)
where the external torque term appears in the equation of motion. Let us use constant cross-section so that () = and () =
604
CHAPTER 7 CONTINUOUS MODELS
(a) When the external torque is included in the boundary condition. When the system is forced sinusoidally, it is reasonable to assume that the system also partly responds at the forcing frequency. This is the particular solution, which is also the steady-state solution if there is even small damping. The particular solution is given by the product, ( ) = Θ() () = Θ() sin where the spatial dependency is not yet known. Substituting this assumed solution into the equation of motion and boundary conditions (Equation 1), we obtain Θ00 sin + 2 Θ sin = 0 Θ (0) = 0 Θ () sin = 0 sin 0
2 Θ=0
⇒
Θ00 +
⇒
Θ0 () = 0
where we can define 2 = 2 The solution to the harmonic space equation is given by Θ() = cos + sin It should be noted that and are known. is the forcing frequency and is the corresponding wave number. The particular solution also must satisfy the boundary conditions. At the fixed end Θ(0) = 0 and at the free end the restoring torque equals the applied (external) torque, Θ0 () = 0 From the first boundary condition, we have = 0 The second boundary condition yields the equation ( cos ) = 0 =
0 cos
Therefore, ( ) = where =
p
0 sin sin cos
(b) When the external torque is included in the equation of motion. Let us again assume ( ) = Θ()() = Θ() sin Substituting this expression into the equation of motion and boundary conditions (Equation 2), we obtain ¡ ¢ Θ00 () + − − 0 = −2 Θ() for 0 ≤ ≤ Θ (0) = 0 Θ0 () = 0 where − is used so that the torque is applied slightly left of = From the ordinary differential equation, we have Θ () = cos + sin +
0 ( − − ) sin ( − )
605 where () is the unit step function. Applying the boundary conditions, we have Θ (0) = 0 = Θ0 () = − sin + cos +
0 ( − − ) sin ( − )
¢ 0 ¡ − − cos ( − ) ¢ ¡ ¢ 0 ¡ − − cos − − = 0 Θ0 () = cos − 1 0 = cos −
where we used
¢ ¡ ¢ ¡ − − = 0 and − − = 1
Then, Θ () =
1 0 ( − − ) 0 sin + sin ( − ) cos
0 1 sin which coincides with the answer from part (a). For 0 ≤ Θ () = cos
606
CHAPTER 7 CONTINUOUS MODELS
33. For the torsion problem of Section 7.4.1, complete the general solution for ( ) assuming the arbitrary initial conditions ( 0) and ̇( 0) Solution: In the text, we have solved the eigenvalue problem that resulted in the eigenfunctions and frequency equation, and the natural frequencies given by Θ̂ () = sin tan = r = where Θ̂ () is the normalized eigenfunction. The actual normalization process will be shown in a separate section labeled as "Orthogonalization." In the text, the first three dimensionless wave numbers are obtained for = 1 They are 1 = 0869 2 = 3426 and 3 = 6437 Given p we can find the natural frequencies as well. The general solution to the initial value problem is given by ( ) =
∞ X
Θ̂ () ( sin + cos )
(1)
=1
Applying the initial conditions, we obtain ( 0) = ̇( 0) =
∞ X =1 ∞ X
Θ̂ ()
(2)
Θ̂ ()
(3)
=1
We now need an orthogonality condition. Let us write the orthogonality condition as Θ̂ () · Θ̂ () = where we have not defined the inner product · yet. For all the other cases we have see in this book where there is no discrete inertia attached directly to the beam or rod, the inner product is defined as Z ̂ () · ̂ () ≡ ̂ () ̂ () = 0
In our case, let us take the inner product, Equation 2 · Θ̂ () to find such that ( 0) · Θ̂ () =
∞ X =1
=
Θ̂ () · Θ̂ ()
607 where the orthogonality is used to simplify the right-hand side. Let us take the inner product, Equation 3 · Θ̂ () to find such that ̇( 0) · Θ̂ () =
∞ X =1
Θ̂ () · Θ̂ ()
= In summary, the constant coefficients are given by =
1 ̇( 0) · Θ̂ ()
= ( 0) · Θ̂ ()
(4) (5)
The solution to the initial value problem is then given by Equation 1 with constant coefficients given by Equation 4. Now, let us find how the eigenfunctions are orthogonal and how the inner products are defined. Orthogonality For problems with discrete inertial attached directly to a rod or a beam, the orthogonality condition is slightly different from the problems without discrete inertia. We start with the eigenvalue problem given by () Θ00 () + 2 () Θ () = 0 Θ (0) = 0 0 2 () Θ () − Θ () = 0
(6)
for = 1 2 · · · Note that we need not assume a constant cross-section. We can write the eigenvalue problem in terms of eigenfunction such that () Θ00 () + 2 () Θ () = 0
(7)
Multiplying Equation 6 by Θ (), multiplying Equation 7 by Θ (), and subtracting the second from the first, we have ¢ ¡ ¢ ¡ () Θ00 Θ − Θ00 Θ = 2 − 2 () Θ Θ
Integrating from = 0 to and integrating by parts, we obtain Z ¡ ¢ () Θ00 Θ − Θ00 Θ 0 Z ¢ ¡ 2 2 () Θ Θ = − 0 Z ¢¯ ¡ 0 ¢ ¡ 0 0 ¯ () Θ0 Θ0 − 0 () Θ0 Θ0 () Θ Θ − Θ Θ 0 − 0 Z ¢ ¡ 2 () Θ Θ = − 2 0
608
CHAPTER 7 CONTINUOUS MODELS
Applying the boundary conditions, Θ (0) = 0 and () Θ0 () = 2 Θ () we obtain 2 Θ () Θ () − 2 Θ () Θ () =
¡ 2 ¢ − 2
Z
() Θ () Θ ()
0
Above expression simplifies to ¸ ∙Z ¡ 2 ¢ 2 − () Θ () Θ () + Θ () Θ () = 0 0
For distinct natural frequencies, 6= we have the orthogonality condition given by Z 0
() Θ () Θ () + Θ () Θ () = 0 for 6=
We can normalize the eigenfunctions such that Z () Θ̂2 () + Θ̂2 () = 1 0
Going back to the eigenfunction at the very top of the solution, Θ̂ () = sin the coefficient must be determined according to the normalization rule above. Then, we can define the inner product as Z
Θ̂ () · Θ̂ () ≡
() Θ̂ () Θ̂ () + Θ̂ () Θ̂ ()
0
The constant coefficients in Equations 4 and 5 are given by
1 ̇( 0) · Θ̂ () Z 1 () ̇( 0)Θ̂ () + ̇( 0)Θ̂ () = 0
=
= ( 0) · Θ̂ () Z () ( 0)Θ̂ () + ( 0)Θ̂ () = 0
This concludes the solution to this problem.
609
34. For the torsion problem of Section 7.4.1, solve for the frequencies, the modes and the total response ( ) of the system for the cases: (a) = (b) = 10 (c) = 10. Solution: Here, we will find the natural frequencies and the mode shapes. Let us continue from the eigenvalue problem that resulted in the eigenfunctions and frequency equation, and the natural frequencies given by Θ () = sin tan = r = for = 1 2 · · · Now, let us consider the three cases given in the problem. (a) tan = 1 or tan = 1 () The figure below shows the two curves, tan and 1 Graphically, we can see that the roots are near 2 32 and 52 · · · We estimate them to be = 086 34 64
More exact values can be found using a root solver such as the fzero.m function in MATLAB. The suggested command is fzero(@(b) b*sin(b)-cos(b), pi/2) for the first root. The wave numbers found this way are = 08603 34256 64373 Note that fzero(@(b) b*tan(b)-1, pi/2) does not give the correct root at = 08603 although sin() − cos() = 0 and tan() − 1 = 0
610
CHAPTER 7 CONTINUOUS MODELS
Figure 7.2: Instead, it gives the value of when the function tan () approaches infinity. Avoid using functions that has vertical asymptotes. The first three mode shapes are plotted at the end. Note that they are not normalized but are set to 1. (b) tan = 10 (or equivalently, sin − 10 cos = 0) The first three roots are = 14289 43058 72281 (c) tan = 01 (or equivalently, sin − 01 cos = 0) The first three roots are = 03111 31731 62991 The figures below shows the mode shapes. We should note how the slope where the disk is attached varies with varying We should not be looking at the relative amplitudes.
611
35. Consider the stepped shaft consisting of two uniform segments of lengths 1 and 2 torsional rigidities 1 and 2 , and mass moments of inertia per unit length 1 and 2 shown in Figure 7.52. Set up the eigenvalue problem for the torsional vibration of the system and derive the frequency equation. Show that as 2 → 0, lim 2 2 =
2 →0
where is the mass moment of inertia of a disk. Find the frequency equation and the orthogonality condition.
Figure 7.52: Two torsional bars in series.
Solution: We treat each section separately and match the boundary conditions where they join. The eigenvalue problem is given by 2 1 (1 ) + 1 2 1 (1 ) = 0 for 0 ≤ 1 ≤ 1 21 2 2 () 2 + 2 2 2 () = 0 for 0 ≤ 2 ≤ 2 2 1 (0) = 0 0 2 (2 ) = 0 1 (1 ) = 2 (0) 1 01 (1 ) = 2 02 (0)
1
where is the same for both sections. Note that the coordinates 2 starts from 0 and ends at 2 Restarting the coordinate system for the second beam can greatly simply the algebra. The solutions are of the form 1 () = 1 sin 1 + 2 cos 1 for 0 ≤ 1 ≤ 1 2 () = 3 sin 2 + 4 cos 2 for 0 ≤ 2 ≤ 2 where 2 =
1 2 2 2 and 2 = 1 2
(1)
612
CHAPTER 7 CONTINUOUS MODELS
Applying the boundary conditions, we obtain 1 (0) = 0 =⇒ 2 = 0 Apply the second boundary condition, and we obtain 02 (2 ) = 0 =⇒ 3 cos (2 ) − 4 sin (2 ) = 0 Solve for 3 : 3 = 4
sin 2 cos 2
(2)
Apply the third condition, and 1 (1 ) = 2 (0) =⇒ 1 sin 1 = 4 Solve for 4 : 4 = sin 1 1
(3)
Apply the fourth condition, and 1 01 (1 ) = 2 02 (0) =⇒ 1 1 cos 1 = 2 3 Replace 3 with Equation (2) and replace 4 again with Equation (2) to obtain ¶ µ sin 2 1 1 cos 1 = 2 4 cos 2 ¶ µ sin 2 = 1 2 sin 1 cos 2 Finally, let us rewrite the previous equation as µ ¶¶ µ sin 2 = 0 1 1 cos 1 − 2 sin 1 cos 2 In order to avoid trivial solution, the term in the parenthesis must be zero. This is the frequency equation and is given by ¶ µ sin 2 = 0 1 cos 1 − 2 sin 1 cos 2 By dividing it by sin 1 , it can be simplified even more, 1 cot 1 − 2 tan 2 = 0 As 2 → 0 tan 2 → 2 and the frequency equation becomes 1 cot 1 − 2 2 2 = 0
613 Recall the definitions of and given in Equation (1) or 1 2 2 2 and 2 = 1 2 2 2 2 2 2 1 2 2 = = 1 2 1 2 1
2 =
Replace 2 and divided by 1 cot 1 and we obtain. 2 1 2 2 = 0 1 2 2 cot 1 − 2 = 0 1 2 2 1 tan 1 = 0 1− 1 1
1 cot 1 − 2
Also, lim 2 2 =
2 →0
We can let 1 = and 1 = Then, the frequency equation becomes tan =
This is the frequency equation of Problem 38. The orthogonality condition is given by Z 1 Z 1 1 1 (1 ) 1 (1 ) 1 + 2 2 (2 ) 2 (2 ) 2 = 0 for 6= 0
0
For a discrete disk with mass moment of inertia of the orthogonality condition is Z () () + () () = 0 for 6= 0
614
CHAPTER 7 CONTINUOUS MODELS
36. In the oil drilling industry, a drill-tube is used for well-boring. It is a tube of outer diameter and inner diameter is made of steel of density and modulus of rigidity At the end of the drill-tube there is a drill bit that does the actual drilling. The bit rotates, enabling it to cut through the toughest of material. The bits are generally made of manufactured diamonds. Assuming a tube length and a bit moment of inertia derive an expression for the first two natural frequencies of torsional vibration, assuming that the tube is fixed at the upper end. Solution: A tube is subject to a given torque on one end and has a lumped inertia (representing the drill bit) on the other as shown in the figure below.
The equation of motion is given by 00 − ̈ = 0 for 0 ≤ ≤ and the boundary conditions are given by () + 0 (0 ) = 0 −0 ( ) = ̈| where ¢ ¡ 2 − 2 4 ¢ ¡ 4 = − 4 32
=
The boundary conditions are obtained using Newton’s second law. The internal torque is defined positive in the CCW direction on the positive face and negative in the CW direction on the negative face as shown in the first figure below.
615
The boundary condition at = 0 is obtained by inserting a fake mass. Summing the moments (positive in the positive direction), we obtain X = () + 0 (0 ) = 0
The boundary condition at = is obtained by summing the moments for the drill bit, and we obtain X = −0 ( ) = ̈|
The eigenvalue problem (the spatial equation of the homogeneous problem) is given by Θ00 − 2 Θ = 0 for 0 ≤ ≤ and the boundary conditions are given by Θ0 (0) = 0 −Θ0 () + 2 Θ () = 0
(1) (2)
Θ () = 1 cos + 2 sin
(3)
The modal function is given by
where
r
(4) The wave numbers, , must be such that the modal function must satisfy the boundary conditions. Substituting the assumed solution (Equation 3) into the boundary conditions (Equations 1 and 2), we obtain =
2 = 0 − (1 sin ) + 1 cos = 0 2
The second equation can be rewritten as 2 cos = 0 2 − sin + cos = 0 − sin + cos = 0
− sin +
Finally, the wave numbers are obtained from the characteristic equation given by the last expression. Once the ratio, is given, the dimensionless wave number, can be obtained. The natural frequencies are obtained using Equation 4.
616
CHAPTER 7 CONTINUOUS MODELS
37. Derive Equations 7.53 and 7.54. Solution: Equations 7.53 and 7.54 are [()Θ0 ()] = − 2 ()Θ() ()Θ0 () = 2 ()Θ()
(7.53) (7.54)
With Θ(0) = 0 they constitute an eigenvalue problem for a torsional beam with a disk at one end. They come from the equation of motion, 2 ( ) ( ) = () () 2
(7.50)
and the boundary conditions (0 ) = 0 ( ) 2 ( ) () = − () 2
(7.52)
The solution, ( ), is written as Θ () () The spatial part of the problem is called the eigenvalue problem.
617
38. A circular elastic shaft of length has one end clamped and a rigid disk attached at the other end, as shown in Figure 7.53. If denotes the modulus of rigidity, the area polar moment of inertia of the shaft, the mass polar moment of inertia of the shaft per unit length, and the mass moment of inertia of the disk about its center axis, calculate the potential and kinetic energy expressions associated with the rotational motion, and derive the equation of motion using Lagrange’s equation.
Figure: 7.53: Torsional vibration of a shaft with a rigid disk.
Solution: Let ( ) be the internal torque and ( ) be the angle of rotation of the crosssection. From Mechanics of Materials, we have the relationship ( ) = ()
( )
where () is the torsional stiffness, and ( ) is the shear strain ( ) The energy stored in an incremental element is µ ¶2 ( ) 1 1 ( ) = ( ) ( ) = () 2 2 The energy stored in the bar is = =
Z
Z 0
µ ¶2 1 ( ) () 2
The kinetic energy of an incremental element is µ ¶2 ( ) 1 = () 2 The total kinetic energy of the bar is ¯ ¶2 µ µ ¶2 Z 1 ( ) ¯¯ ( ) 1 = () + 2 ¯= 0 2
618
CHAPTER 7 CONTINUOUS MODELS
Note that the potential and kinetic energies are function of only. The Lagrange’s equation is given by ¡ ¢ ¡ ¢ ¡ ¢ ̄ ̄ ̄ − = 0 + ̇ 0 where ̄ is the integrand of or µ ¶2 µ ¶2 ( ) ( ) 1 1 − () ̄ = () 2 2 The equation of motion is given by ∙ µ ¶¸ ∙ µ ¶¸ ( ) ( ) () + () = 0 Note that the disk, is not included in the equation of motion. It would be included in the boundary conditions.
619 Problems for Section 7.5 — Transverse Vibration of Beams 39. Derive the frequency equation, the first three natural frequencies, and plot the first three mode shapes of a clamped-clamped uniform beam undergoing transverse vibration. Solution: The equation of motion and the boundary conditions are given by
4 2 + = 0 4 2 (0 ) = 0 ¯ ¯¯ = 0 ¯0
( ) = 0 ¯ ¯¯ = 0 ¯
Let ( ) = () () and ̈ () () = − 2 Then the eigenvalue problem is given by 0000 () −
We let 4 =
2 () = 0 (0) = 0 0 (0) = 0 () = 0 0 () = 0
2 and the eigenfunction is in the form () = cos + sin + cosh + sinh
Apply the boundary conditions to obtain + = 0 + = 0 cos + sin + cosh + sinh = 0 − sin + cos + sinh + cosh = 0 or in the matrix form ⎡ ⎤ 0 1 0 1 0 ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ 0 ⎥ ⎢ 0 1 0 1 ⎥⎢ ⎥ = ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢ cos sin cosh sinh ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎦⎣ ⎦ ⎣ ⎦ ⎣ 0 − sin cos sinh cosh ⎡
⎤⎡
⎤
620
CHAPTER 7 CONTINUOUS MODELS
Note that second and the fourth equation were simplified by diving them by This is usually not a good idea because = 0 can be a valid eigenvalue if a rigid body motion is permitted (pinned-free, free-free, sliding-sliding boundary conditions). The clamped-clamped beam cannot have a rigid body motion, and therefore, the coefficient was canceled earlier on. The frequency equation is the determinant of the square matrix or −2 cos cosh + 2 = 0 The numerical solutions are = 474 785, 110 · · · The mode shapes are found by expressing and in terms of Using the boundary conditions, we find = − = −
cos − cosh sin − sinh
= − =
cos − cosh sin − sinh
Then, the mode shape is cos − cosh sin sin − sinh cos − cosh sinh ) − cosh + sin − sinh
() = (cos −
The first three mode shapes are − 09825 sin 474 − cosh 474 + 09825 sinh 474 ) 2 () = 2 (cos 785 − 10008 sin 785 − cosh 785 + 10008 sinh 785 ) 3 () = 3 (cos 110 − 10000 sin 110 − cosh 110 + 10000 sinh 110 )
1 () = 1 (cos 474
and they are plotted next.
621
622
CHAPTER 7 CONTINUOUS MODELS
40. Derive the frequency equation, the first three natural frequencies, and plot the first three mode shapes of a clamped-pinned uniform beam undergoing transverse vibration.
Solution: The equation of motion and the boundary conditions are given by
4 2 + = 0 4 2 (0 ) = 0 ¯ ¯¯ = 0 ¯ 0
( ) = 0 ¯ 2 ¯¯ = 0 2 ¯
Let ( ) = () () and ̈ () () = − 2 Then the eigenvalue problem is given by 0000 () −
We let 4 =
2 () = 0 (0) = 0 0 (0) = 0 () = 0 00 () = 0
2 and the eigenfunction is in the form () = cos + sin + cosh + sinh
Apply the boundary conditions to obtain + = 0 + = 0 cos + sin + cosh + sinh = 0 2 − cos − 2 sin + 2 cosh + 2 sinh = 0 or in the matrix form ⎡
⎡ ⎤ 0 1 0 1 0 ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ 0 ⎥ ⎢ 0 1 0 1 ⎥⎢ ⎥ = ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢ cos sin cosh sinh ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎦⎣ ⎦ ⎣ ⎦ ⎣ 0 − cos − sin cosh sinh ⎤⎡
⎤
623 The frequency equation is the determinant of the square matrix or 2 sin cosh − 2 cos sinh = 0 The numerical solutions are = 393 707, 102 · · ·
The mode shapes are found by expressing and in terms of Using the boundary conditions, we find cos = − sin = − cos = sin Then, the mode shape is
¶ µ cos cos sin − cosh + sinh () = cos − sin sin
The first three mode shapes are
− 10008 sin 393 − cosh 393 + 10008 sinh 393 ) 2 () = 2 (cos 707 − 10000014 sin 707 − cosh 707 + 10000014 sinh 707 ) 3 () = 3 (cos 102 − 1000000008 sin 102 − cosh 102 + 1000000008 sinh 102 ) and they are plotted next. 1 () = 1 (cos 392
624
CHAPTER 7 CONTINUOUS MODELS
41. Derive the frequency equation, the first three natural frequencies, and plot the first three mode shapes of a clamped-uniform beam with a concentrated mass at the free end undergoing transverse oscillation. Use = 1
Solution: The equation of motion and the boundary conditions are given by
2 4 + = 0 4 2 (0 ) = 0 ¯ ¯¯ = 0 ¯0 ¯ 2 ¯¯ = 0 2 ¯ ¯ ¯ 2 ¯¯ 3 ¯¯ = 2¯ 3 ¯
Let ( ) = () () and ̈ () () = − 2 Then the eigenvalue problem is given by
We let 4 =
0000
2 () − () = 0 (0) = 0 0 (0) = 0 00 () = 0 000 () = − 2 ()
2 and the eigenfunction is in the form () = cos + sin + cosh + sinh
Apply the boundary conditions to obtain + =0 + = 0 − 2 cos − 2 sin + 2 cosh + 2 sinh = 0 3 sin − 3 cos + 3 sinh + 3 cosh = − 4 ( cos + sin + cosh + sinh )
625 or in the matrix form ⎡ ⎤ 1 0 1 0 ⎢ ⎥ ⎢ ⎥ 0 1 0 1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ − cos − sin cosh sinh ⎢ ⎥ ⎣ ⎦ sin + cos − cos + sin sinh + cosh cosh + sinh ⎡ ⎤ ⎡ ⎤ 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢0⎥ ⎥ ⎢ ⎥ ×⎢ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢0⎥ ⎣ ⎦ ⎣ ⎦ 0
The frequency equation is the determinant of the square matrix or cos sinh − sin cosh + cosh cos + 1 = 0 The numerical solutions are = 125 403, 713 · · · The mode shapes are found by expressing and in terms of Using the boundary conditions, we find = −
cos + cosh sin + sinh
= − cos + cosh = sin + sinh Then, the mode shape is µ ¶ cos + cosh cos + cosh () = cos − sin − cosh + sinh sin + sinh sin + sinh The first three mode shapes are ³ ´ 1 () = 1 cos 125 − 08649 sin 125 − cosh 125 + 08649 sinh 125 ³ ´ 2 () = 2 cos 403 − 1006 sin 403 − cosh 707 + 1006 sinh 403 ´ ³ 3 () = 3 cos 713 − 09999 sin 713 − cosh 102 + 09999 sinh 713 and they are plotted next.
626
CHAPTER 7 CONTINUOUS MODELS
627
42. Perform the necessary variations to derive Equation 7.62. Solution: Equation 7.62 is derived by beginning with the equation just before Equation 7.62 and integrating by parts and combining like terms. As an example of one term, begin with the first term under the integral µ ¶ Z 2 Z () 0 1
Our goal is to rewrite this in a form where we have the variation of the variable instead of the variation of the derivative of the variable. In order to do this, we interchange the time and space integrals, and interchange the variation with the partial derivative, and integrate by parts (in time) as follows: Z 2 () () = () |21 1 Z 2 2 − () 2 1 The first term on the right-hand side equals zero since the variation is zero at 1 and 2 The second term is in the form we need. There will be other terms with the factor emerging when we integrate the other terms by parts. All these terms are combined and lead to the first line of Equation 7.62. Since the variation is arbitrary and independent of the variation of the second generalized coordinate the factor of each variation is identically zero. This is how the equations of motion are derived. Additional terms result, i.e. the third line in Equation 7.62, which specify possible boundary conditions.
628
CHAPTER 7 CONTINUOUS MODELS
43. Equations 7.63 and 7.64 are solved using the assumed solutions ( ) = () exp() and ( ) = Ψ() exp(). Find the characteristic equation.
Solution: We must first consider the free problem where ( ) = 0 The equations of motion for a Timoshenko beam are given by ∙ µ ¶¸ 2 () − − () 2 = 0 (7.63) µ ¶ µ ¶ 2 () + () − − 2 ()() 2 = 0 (7.64) After we substitute ( ) = () exp() and ( ) = Ψ() exp() we can separate the spatial equations out such that 0
(() ( 0 − Ψ)) + () 2 = 0
(()Ψ)0 + () ( 0 − Ψ) + 2 ()() 2 Ψ = 0 These equations are part of the eigenvalue problem. We need boundary conditions in terms of () and Ψ () in order to complete the eigenvalue problem.
629
44. In Equations 7.67 and 7.68, eliminate ( ) and derive the governing equation for ( ). What physical motion does this equation govern? Solution: For constant cross-section, the equations of motion for the Timoshenko beam become ¶ µ 2 2 − − + ( ) = 0 (7.67) () 2 2 µ ¶ 2 2 2 + (7.68) − − 2 2 = 0 We first differentiate Equation 7.67 with respect to , µ 3 ¶ 2 3 − + =0 − 3 2 2
(1)
We solve for using Equation 7.68 and differentiate it to find
2 2 = − 2 + 2 2 3 2 4 4 3 = 2 − 4 + 2 2 2 4 2 4 3 2 = − + 2 2 2 2 4
3
(2) (3)
3
Replacing 3 and 2 in Equation 1 using Equations 2 and 3, we find an equation of motion in terms of only, ¶ µ 2 4 2 4 2 − + − 2 2 4 2 2 ¶ µ 2 4 2 4 + − + − 2 2 2 4 = 0
Simplifying we obtain µ ¶ 4 2 ()2 4 4 2 + 2 + = 4 − + 2 2 4 Note that the homogeneous part of the equation is identical to that for ( ) (see Equation 7.74). This equation governs the angle of rotation, ( ) only. It should be noted that the equation requires four boundary conditions, and they all need to be in terms of
630
CHAPTER 7 CONTINUOUS MODELS
45. Derive the Bernoulli-Euler beam equation, Equation 7.74. State all assumptions. Solution: In this text, we start with the Timoshenko model to derive the Bernoulli-Euler beam model. The commonly used assumptions in deriving the Bernoulli-Euler beam equation are 1. There is no deformation due to shear, 2. There is no warping, and plane remains plane. 3. The radius of curvature is approximated as 00 , an approximation to the exact formula = 32 00 (1 + 02 ) It is consistent with the small angle assumption. 4. The rotational kinetic energy is negligible compared to the translational kinetic energy.
631
46. For the simply supported uniform beam of Figure 7.54, undergoing transverse motion, derive the natural frequencies and the natural modes of vibration. Sketch the first three mode shapes. What will change in the analysis if the beam is on an inclined surface?
Figure 7.54: Simply supported uniform beam.
Solution: The equation of motion and the boundary conditions are given by
2 4 − = 0 for 0 ≤ ≤ 2 4 ¯ 2 ¯¯ =0 (0 ) = 0, 2 ¯(0) ¯ 2 ¯¯ ( ) = 0, = 0 2 ¯ ()
If the beam is now on a slope, gravity has a component in the axial direction of the beam. It add constant external force along the longitudinal direction. It will affect the static equilibrium. The modes of vibration are an intrinsic quality of the beam (its mass and geometry) and its boundary conditions. The natural frequencies and the mode shapes do not change. The equation of motion and boundary conditions are given by
2 4 − = − cos for 0 ≤ ≤ 2 4 ¯ 2 ¯¯ =0 (0 ) = 0, 2 ¯(0) ¯ 2 ¯¯ ( ) = 0, =0 2 ¯ ()
where cos is the component of the gravity in the transverse direction. It is a constant once the inclination angle is given. When the inclination angle is 0 the gravity acts only in the transverse direction. That is, ( ) and ( ) are related by ( ) = ( ) + ()
632
CHAPTER 7 CONTINUOUS MODELS
where () is the solution to 4 = − cos 4 ¯ 2 ¯¯ = 0 (0) = 0 2 ¯0 ¯ 2 ¯¯ () = 0, = 0 2 ¯
−
It is easy to show that
cos () =
µ
¶ 1 4 1 3 3 − + 24 12 24
Note that () is the static solution to the static distributed load of cos in the axial direction. Let us now find the natural frequencies. The equation of motion for a simply supported horizontal beam is given by 2 4 2 − 4 = 0 for 0 ≤ ≤ We let ( ) = () exp () to obtain the eigenvalue problem
4 + 2 = 0 for 0 ≤ ≤ 4
with ¯ 2 ¯¯ (0) = 0 =0 2 ¯0 ¯ 2 ¯¯ = 0 () = 0 2 ¯
The mode shape is of the form,
() = cos + sin + cosh + sinh where 4 =
2
Applying the boundary conditions, we find + = 0 − + = 0 sin + sinh = 0 2 − sin + 2 sinh = 0 Then, = 0 = 0 = − sin sinh
633 and the frequency equation is sin sinh = 0 The wave numbers are = 0 2 3 · · · where 0 is not a valid wave number because rigid-body motion is not possible for the given boundary conditions. For these wave numbers, we find that = 0 Then, the mode shape is given by () = sin where = The first three natural frequencies are r
2 r ³ ´ 2 =
=
The mode shapes look exactly same as the mode shapes of a fixed-fixed string. The figure below shows the first three mode shapes with = 1
634
CHAPTER 7 CONTINUOUS MODELS
47. For the transverse vibration of a simply supported uniform beam, solve for the transient response if the initial conditions are given by ¶ µ 2 3 ̇( 0) = 0 −3 2 +2 3 ( 0) =
Solution: A simply supportedpbeam has mode shapes of the form, () = sin , where = We find that = 2 so that the mode shapes are normalized with respect to R or 0 2 () = 1 Beginning with the displacement: r ∞ ³ ´ X 2 ( ) = ( cos + sin ) sin =1 and apply the initial conditions: r
³ ´ 2 sin r ∞ ³ ´ X 2 sin 0 = =1
∞ X 2 3 ( − 3 2 + 2 3 ) = =1
From the second equation, we see that the coefficient must equal zero. To evaluate the constants multiply the first equation by the function (to take advantage of orthogonality) q we ¡ ¢ 2 sin and integrate over the beam length: r ³ ´ 2 3 2 ( − 3 2 + 2 3 ) sin 0 Z ∞ ³ ´ ³ ´ X 2 sin sin = 0 =1 Z
where the right-hand side of the first equation equals zero except when = Then, we find r Z ³ ´ 2 2 3 = ( − 3 2 + 2 3 ) sin 0 r 2 6 = (1 + (−1) ) 4 3
635
48. A two-span uniform beam may be used as a simple model of a bridge. For this model, sketched in Figure 7.55, derive the frequency equations.
Figure 7.55: Two-span continuous uniform beam.
Solution: We need to have one modal function for each span. In addition we have the usual boundary conditions at each fixed end, plus compatibility at the center support. That is, the displacement, slope, moment, and shear must be continuous at the middle support. The modal functions are given by 1 (1 ) = 1 sin 1 + 1 cos 1 + 1 sinh 1 + 1 cosh 1 2 (2 ) = 2 sin 2 + 2 cos 2 + 2 sinh 2 + 2 cosh 2 The boundary conditions are 1 (0) = 0 1 (0) = 0 1
2 (0) = 0 2 (0) = 0 2
Note that 1 = 0 at the left end and 2 = 0 at the right end. At the middle support, the transition conditions are 1 () = 0 2 () = 0 1 () 2 () = − 1 2 2 22 () 1 () = − 21 22 Applying these boundary conditions, we can find the relations between the 8 constants: 0 = 1 + 1 0 = 2 + 2 0 = 1 + 1 0 = 2 + 2 0 = 1 (sin − sinh ) + 1 (cos − cosh ) 0 = 2 (sin − sinh ) + 2 (cos − cosh )
636
CHAPTER 7 CONTINUOUS MODELS 1 (cos − cosh ) − 1 (sin + sinh ) = −1 (cos − cosh ) +1 (sin + sinh ) −1 (sin + sinh ) − 1 (cos + cosh ) = 1 (sin + sinh ) +1 (cos + cosh )
We reduce the number of constant coefficients to one. After substitution, we find the characteristic equation given by (1 − cos cosh )(sin cosh − cos sinh ) = 0 This completes our solution. We can find the dimensionless wave numbers, in the following way. The frequency equation is satisfied when 1 − cos cosh = 0 or sin cosh − cos sinh = 0 where ‘or’ stands for the ‘logical or’ or the union of the two solutions. We obtain one set of from the first equation, and the second set from the second equation. The wave numbers, are then sorted in ascending order. The first four wave numbers are = 393 473 707 110
637
49. The beam of Figure 7.56 is released and rotates as a rigid body and impacts at an angular speed of Ω rad/s. Assume that upon impact the beam latches onto the support and there is no rebound and no loss in energy. At impact the beam transfers all of its rotational kinetic and potential energies into transverse vibrational motion. (a) Derive the response after impact. (b) How would the analysis proceed if there is a rebound?
Figure 7.56: A freely falling rotating beam.
Solution: (a) There are two aspects of this problem. The first is the beam rotation (assumed as a rigid-body) before impact. The second aspect is after impact (assumed with no rebound) when the beam undergoes transverse vibration. Let us consider the rotational aspect first, which allows us to calculate the angular velocity Ω just before the impact. We assume that the center of gravity of the beam is at the center. Since there is no dissipation of energy during rotation, the total energy remains constant: ∆ Kinetic Energy = ∆ Potential Energy 1 2 Ω = 2 2 1 1 ( 2 )Ω2 = 2 3 2 3 Ω2 = 2 where the moment of inertia of the beam is about the base. Now we can consider the motion after impact. The initial conditions are zero displacement and initial velocity given by Ω : ( 0) = 0 ( 0) = Ω The general solution for transverse displacement is given by ( ) =
∞ X =1
sin
( sin + cos )
638
CHAPTER 7 CONTINUOUS MODELS
The initial displacement condition leads to velocity condition results in
P∞
=1 sin = 0 implying that = 0 The initial
∞ X ( 0) = Ω = sin =1
Multiplying both sides by sin() and integrating over the domain we obtain Z 0
Z ∞ X = sin Ω sin sin 0 =1 cos = −2 Ω 2 Ω 2 Ω = − cos = − (−1)
Therefore, the response is given by ( ) = −
∞ X 2 Ω
(−1) sin
sin
√ ∞ 3 X (−1) sin sin = − =1 =1
(b) If there is a rebound, this becomes an impact problem. We need to know the coefficient of restitution to find the velocity after the impact. This velocity is then used as initial velocity.
639
50. Determine the characteristic equation for the transverse vibration of a beam that is pinned at one end and free at the other end. Solution: This is a hinged-free beam. The mode shape is given by () = cos + sin + cosh + sinh The boundary conditions are (0) = 0 00 (0) = 0 00 () = 0 and 000 (0) = 0 Applying the boundary conditions, we obtain + = 0 − = 0 − cos − sin + cosh + sinh = 0 sin − cos + sinh + cosh = 0 We find that = 0 = 0 sin = sinh − sin + sinh = 0 We can proceed with substitution to the final boundary condition (that we have not used yet to solve for and ) to find the characteristic equation, and it is given by − cos sinh + sin cosh = 0 This completes our solution. We can proceed a little further by obtaining the wave numbers and the mode shapes. The first four normalized wave numbers that satisfy the characteristic equation are = 39266 70686 102102 The mode shape is given by µ ¶ sin sinh () = sin + sinh
640
CHAPTER 7 CONTINUOUS MODELS
51. List the boundary conditions for the transversely vibrating beams of Figures 7.57 and 7.58.
Figure 7.57: Transversely vibrating beam with end mass.
Figure 7.58: Transversely vibrating beam with elastic support.
Solution: For the beam of Figure 7.57, the left end is fixed and the right end has a mass attached. Therefore, at the fixed end there is no displacement (0 ) = 0 and no slope 0 (0 ) = 0 At the right end the shear in the beam is equal and opposite to the force due to the inertia of the mass: 000 ( ) = ̈( ) and there is no moment resistance: 00 ( ) = 0 The shear condition can be found from the free-body diagram. With the displacement defined positive pointing downward, the shear, given by 000 ( ) on the positive face (whose normal is in the positive direction) of the beam is positive upward. Due to Newton’s third law, we have an equal and opposite force on the mass, Summing the forces in the vertical direction, we have ↓
X
= ̈( )
000 ( ) = ̈( )
641
For the beam of Figure 7.58, the left end is fixed and the right end has a translational and torsional spring attached. At the fixed end the boundary conditions are as above. At the right end we have the satisfaction of the shear boundary condition: 000 ( ) − ( ) = 0 and the moment boundary condition: 00 ( ) − 0 ( ) = 0 It helps to see the boundary conditions if we place a fake mass (with zero mass) between the beam and the spring elements. The sign convention says the moment given by 00 is defined positive clockwise on the positive face of the beam.
642
CHAPTER 7 CONTINUOUS MODELS
52. For the transversely vibrating beam of Figure 7.59, find the modes and the first three natural frequencies of vibration.
Figure 7.59: Transversely vibrating beam with end roller.
Solution: The equation for the general modal function for an Euler beam is () = sin + cos + sinh + cosh where 4 = 2 The boundary conditions are given by (0) = 0 0 (0) = 0 0 () = 0 000 () +
1 4 () = 0
The last boundary conditions comes from the fact that the shear force (due to the body at the right end of the beam) equals the mass of the body times its acceleration: 000 ( ) = ̈( ) Using the product solution ( ) = ()(), we have 000 () = ()¨ = − () 2 or 000 () = − 2 () Replacing 2 with 4 we obtain the last boundary condition. The four boundary conditions are used to obtain three of the four constants of integration From the first three boundary conditions, we find = − = − (sin + sinh ) = (cos − cosh )
643 Applying the last boundary condition, we have 3 [− cos + sin − cosh − sinh ] 4 ( sin + cos − sinh − cosh ) = Replacing with (sin + sinh ) (cos − cosh ) we obtain 0 = − cos sinh − sin cosh + (− cos cosh + 1) Once the mass ratio, is known we can find the normalized wave number,
644
CHAPTER 7 CONTINUOUS MODELS
53. Complete the solution of Equation 7.97 for the initial conditions ( 0) = () and ̇( 0) = ().
Solution: Equation 7.97 is ( ) =
∞ X
( )
=1
= =
∞ X =1 ∞ X
() cos( − ) () (cos cos + sin sin )
=1
Now apply the initial conditions: () =
∞ X
() cos
=1
() =
∞ X
() sin
=1
We will solve for cos and sin in order to obtain the expression for ( ) Multiplying the initial condition expressions by () and integrating over the domain, we obtain Z
() () =
0
Z 0
() () =
∞ X =1 ∞ X =1
cos
Z
() ()
0
sin
Z
() ()
0
The integrals on the right-hand sides equal zero except if = Therefore, Z () () cos = 0 Z 1 () () sin = 0 These can now be evaluated, given the initial conditions and upon substituting the mode shapes.
645
54. Derive Equation 7.102 beginning with Equation 7.101. Solution: Equation 7.102 is Z
()
0
2 [()00 ()] = 2 2
In order to derive Equation 7.102, first derive Equation 7.101 given by Z ()00 ()00 () = 2 0
Now we can integrate left-hand side of this equation by parts twice such that Z ()00 ()00 () 0 Z 0 00 0 ( () 00 ()) 0 () = () () ()|0 − 0 00 0 = () () ()|0 − () 00 () ()|0 Z 00 + ( () 00 ()) () 0
The terms evaluated at the boundaries are zeros. Therefore, we have Equation 7.102.
(7.102)
(7.101)
646
CHAPTER 7 CONTINUOUS MODELS
55. Derive Equation 7.106 starting with Equation 7.105. Solution: Equation 7.106 is the second orthogonality condition given by Z
()
0
2 ̂ 2 ̂ = 2 = 1 2 · · · 2 2
(7.106)
Let us start with Equation 7.105 given by " # Z 2 2 ̂ ̂ 2 () 2 = 2 = 1 2 · · · 0 We first integrate the h left-hand i side of Equation 7.105 by parts. We use
2 ̂ () = and 2
2 () ̂2
= Then, we have
R
(7.105)
= | −
" # 2 ̂ 2 ̂ 2 () 2 0 #¯ Z # " " 2 ̂ ¯¯ 2 ̂ = ̂ ̂ () 2 ¯ − () 2 ¯ 0
R
Let
Z
0
Integrating the second terms by parts one more time by letting ̂ = and
we obtain # " ̂ 2 ̂ () 2 0 #¯ Z " # " 2 ̂ 2 ̂ 2 ̂ ¯¯ ̂ = () 2 () 2 ¯ − ¯ 2 0
h i 2 ̂ () 2 = ,
Z
0
Note that the terms evaluated at the boundary are zeros (they are the same boundary conditions obtained from Hamilton’s principle.) Then, the left-hand side of Equation 7.105 becomes " # Z Z 2 ̂ 2 2 ̂ 2 ̂ () 2 ̂ () 2 () 2 = 2 0 0 Equating this with the right-hand side of Equation 7.105, we obtain Equation 7.106.
647
56. Fill in the steps in the derivation of Equation 7.110. Solution: Equation 7.110 is given by " # ∞ ∞ X X 4 ̂ 2 () 4 + ()̂ () ̂ + 2 ̂ = =1 =1
(7.110)
Start with the equation of motion
4 2 + + = ( ) 4 2
(7.107)
We expand the solution and the force in terms of modal functions such that (method of eigenfunction expansion) ∞ ∞ X X () ̂ () and ( ) = () ̂ () (1) ( ) = =1
=1
The modal functions, ̂ satisfy the eigenvalue problem given by (the undamped homogeneous spatial equation) 4 ̂ 4 − 2 ̂ = 0 (2) with homogeneous boundary conditions, and they are normalized such that Z ̂ ̂ = 0
Let us now get back to the expansion in Equation (1). Substitute (1) into Equation 7.107, and we obtain Equation 7.110.
648
CHAPTER 7 CONTINUOUS MODELS
57. Solve Equation 7.113 for zero initial conditions and for each of the following forcing functions: (a) ( ) = (), where () is the unit step function (b) ( ) = sin Ω (c) ( ) = − , for ≥ 0
(d) ( ) = sin Ω for 0 ≤ Ω ≤ 2.
Solution: Equation 7.113 is ( ) =
∞ X
̂ ()
Z ∙Z 0
=1
0
¸
( )̂ () ( − )
(7.113)
for a transversely vibrating damped beam. The impulse response function () is the solution to ̈ () + ̇ () + 2 () = () where and are, respectively, the damping coefficient and the mass per length of the original partial differential equation (Equation 7.107), and is the natural frequency of the undamped and unforced problem (related eigenvalue problem). Its explicit expression for 0 ≤ 1 is () = where =
¡ ¢ 1 exp − sin
1 and = 2
q 1 − 2
The damping ratio decreases as the natural frequency increases. Therefore, the impulse response function that corresponds to higher modes is more likely to be underdamped. The mode shape, () is the mode shape of the related eigenvalue problem. The related eigenvalue problem is that of the system with no damping and no external forcing. For the forces that have no dependence on the spatial coordinate (parts (a)-(d)), the first integral is easy. The forcing function, ( ) will come out of the integral, Z
( )̂ () = ( )
0
Z
̂ ()
0
The second integral, Z ∙Z
¸
0
0
( )̂ () ( − ) Z Z = ( ) ( − ) × ̂ () 0
0
649 R is not so easy to perform because the integral, 0 ( ) ( − ) may be difficult to perform by hand. It is advisable to utilize a symbolic code such as MAPLE or MATHEMATICA when working out the integrations. The general solution is then given by ( ) =
∞ X
̂ ()
0
=1
=
∞ X
̂ ()
R 0
0
∙Z 0
=1
The integral
Z ∙Z
¸
( )̂ () ( − )
( ) ( − )
¸ µZ 0
̂ ()
(7.113) ¶
̂ () can be found once we know the specific boundary conditions of the problem.
In the following, we will evaluate (a) ( ) = () N
³R 0
´
( ) ( − ) for each part. MAPLE is used for integration.
We have Z
=
0
( ) ( − )
0
( ) ( − )
Z Z
¡ ¢ 1 exp − ( − ) sin ( − ) 0 ¡ ¢¡ ¢ exp − cos + sin − ´ ³ = − 2 2 + 2 =
(b) ( ) = sin Ω N We have Z 0
¡ ¢ 1 exp − ( − ) sin ( − ) 0 ¡ ¢¡ £ ¤¢ 1 = {exp − 2Ω cos ( ) + sin ( ) Ω 2 2 + Ω3 − Ω 2 £ ¤ −2 Ω cos (Ω ) − sin (Ω ) Ω2 − 2 2 − 3 } =
where
Z
( ) ( − ) sin Ω
¡ ¢¡ ¢ = 2 2 + Ω2 − 2 Ω + 2 2 2 + Ω2 + 2 Ω + 2
650
CHAPTER 7 CONTINUOUS MODELS
(c) ( ) = − N, for ≥ 0 Z 0
Z
( ) ( − )
¡ ¢ 1 exp (− ) exp − ( − ) sin ( − ) 0 ¡ ¢¡ ¢ − exp − cos − sin + sin + exp (−) ¡ ¢ = 2 − 2 + 2 2 + 2
=
(d) ( ) = sin Ω N for 0 ≤ Ω ≤ 2 rad/s The integral,
R 0
( ) ( − ) is the same as part (b).
651
58. Derive the equation of motion and boundary conditions for a transversely vibrating beam using Hamilton’s principle. Assume small deflections. Let equal the mass moment of inertia per unit length of the beam about its neutral axis. Figure 7.60 provides a sketch of the beam along with positive sign conventions.
Figure 7.60: Transversely vibrating beam subjected to an arbitrary continuous load ( ) along with boundary moments and shear forces.
Solution: Timoshenko beam was derived using Hamilton’s principle in the text. In the text, there was no consideration of rotary inertia due to rotation. Here, we will do the same for a beam undergoing translation but include rotation (but no shear deformation as in the Timoshenko beam), where there is an arbitrary applied load, and given boundary shears and moments. The resulting governing equation of motion that includes the effect of rotary inertia is known as Rayleigh’s equation. The strain energy for small deflections is 1 = 2
Z 0
µ 2 ¶2 2
The kinetic energy is due to translation and to rotation. The translation is in the direction of the deflection leading to the kinetic energy due to translation, for a beam of mass per unit length µ ¶2 Z 1 2 0 The rotation is about an axis perpendicular to the neutral axis and the plane of vibration. The kinetic energy due to this rotation is equal to the product of the moment of inertia times one half of the square of the angular velocity 2 µ 2 ¶2 Z 1 2 0 The total kinetic energy due to these two effects is then µ 2 ¶2 µ ¶2 Z Z 1 1 + = 2 0 2 0
652
CHAPTER 7 CONTINUOUS MODELS
We now have to account for the effects of the external distributed forcing and the boundary shear forces and moments. The work done by these actions is equal to the negative of the respective potential energies µ ¶ µ ¶ Z = − ( )() − + 0 + − 0 0 0 0 The equation of motion is given by Hamilton’s principle: Z 2 ( − − ) = 0 1
Therefore:
Z 2 1
⎡
R
⎢ 0 ⎣
½
1 2
¾ ⎤ + ( )() ⎥ ⎦ = 0 ¡ ¢ ¡ ¢ + − 0 0 − + 0 0
¡ ¢2
+ 12
³ 2 ´2
− 12
³ 2 ´2 2
Following the procedures of the variational calculus, and noting that the virtual displacement must be specified such that = 0 at 1 and 2 and therefore, () = () = 0 at these times, then the above variation becomes (using the chain rule of differentiation) ⎡ o ⎤ n Z 2 R 2 2 2 2 + − 2 2 + ⎣ 0 ⎦ = 0 ¡ ¢ ¡ ¢ 1 + − 0 0 − + 0 0
Now, our goal is to obtain all variations in the generalized variable Therefore, we need to integrate by parts. To do this, we interchange the integration in time with the integration in space if necessary. For example, Z Z 2 Z 2 Z = 0 0 1 1 ¯ ¯ Z 2 Z 2 ¯ ¯2 2 ¯ ¯ = ¯ ¯ − and 2 1 1 1
Doing this for the other terms, we find the need to integrate by parts twice: ¯ ¯ ¶ µ Z Z 2 ¯ 2 ¯ 2 2 ¯ + 2 = − ¯¯ 2 2 − ¯ 2 0 0 0 ¯ ¶ ¯ µ ¶¯ ¯ µ ¯ ¯ ¯ 2 ¯¯ 2 ¯ ¯ +¯ 2 ¯¯ = − ¯ 2 ¯ 0 0 ¶ Z 2 µ 2 2 − 2 0
Similarly for the cross-derivative term (interchange order of integrations and integrate with respect to time first): ∙ ¸ µ ¶ Z 2 Z 2 2 2 2 2 2 = − 1 1 1
653 Then integrate these with respect to to obtain ¶ ¸ Z 2 ∙ µ 2 2 2 = − 1 1 0 µ ¶ Z 2 Z 2 2 + 1 0
Z 2
Performing all the integrations and combining terms leads to the following: µ ¶ ¸ µ ¶ Z 2 Z ∙ 2 2 2 2 2 − 2 + 2 + − 2 1 0 ¶¸ ¸ µ Z 2 ∙∙ 2 2 − − 1 0 ¶ µ ¶ ¸ ¾ Z 2 ½∙ µ 2 2 2 − − + 1 0 = 0 Since the variations are arbitrary and independent, then the expression in the first brackets, when set equal to zero, becomes the governing equation of motion. By the same reasoning, ¶ µ 2 =0 either 2 = or ¶ µ µ ¶ 2 2 2 − = or = 0 either
654
CHAPTER 7 CONTINUOUS MODELS
59. Solve for the response of the cantilever beam sketched in Figure 7.61.
Figure 7.61: Beam with harmonic end load.
Solution: This is a cantilever beam of length with harmonic end load equal to cos Ω The beam is subject to a harmonic forcing, and the goal is to find the response at the harmonic forcing (particular solution). When there is damping, this corresponds to the steady-state response. In the absence of damping, the total response is a summation of the particular solution (that responds at frequency Ω) and the homogeneous solution (that responds at the damped natural frequency). We assume that ( ) = () cos Ω where () is not the modal function in this case. It simply describes the spatial variation of the response when it responds at Ω Then, we find that () satisfies 0000 () − Ω2 () = 0 Then, () must be of the form, () = 1 sin + 2 cos + 3 sinh + 4 cosh where 4 =
³
´ Ω2
where the wave number, takes on a specific value given the beam properties ( and ) and the forcing frequency, Ω This is not to be confused with the wave number at the natural frequencies. The spatial function, () must still satisfy the boundary conditions of a cantilever beam forced at = (0) = 0 0 (0) = 0 00 () = 0 000 () = where ( ) is defined positive upward and positive to the right. Applying the four boundary conditions, we find the values of all the coefficients (unlike the eigenvalue problem where we can
655 only find one fewer coefficient) in terms of and Ω That is, 2 + 4 = 0 1 + 3 = 0 −1 sin − 2 cos + 3 sinh + 4 cosh = 0 −1 cos + 2 sin + 3 cosh + 4 sinh = The coefficients are ⎤−1 ⎡ ⎤ ⎡ ⎤ ⎡ 0 1 0 1 0 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⎢ 2 ⎥ 1 0 1 0 0 ⎥ ⎥ ⎢ ⎥ ⎥ = ⎢ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ − sin − cos sinh cosh ⎥ ⎢ 0 ⎥ ⎢ 3 ⎥ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ − cos sin cosh sinh 4 ⎤ ⎡ − cos − cosh ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ sin + sinh 1 ⎥ ⎢ = ⎥ 2 cos cosh + 2 ⎢ ⎢ − cos − cosh ⎥ ⎦ ⎣ − sin − sinh
where the inversion is done using a symbolic code. The part of the solution that responds at the forcing frequency (particular solution) is given by ( ) = () cos Ω = (1 sin + 2 cos + 3 sinh + 4 cosh ) cos Ω
The total response is ( ) =
∞ X
() ( cos + sin ) + () cos Ω
=1
where () is the modal function and is the natural frequency of the unforced cantilever beam. The constant coefficients of integration, and are obtained from initial conditions.
656
CHAPTER 7 CONTINUOUS MODELS
60. In Example 7.7, solve for the case where = 1 4 s. Solution: We have = 2 1 2 1 = 1 Therefore, at = 1 4 = 2
odd
2 1 = 2 2 1 4
Then 2 odd, 2 9 = cos cos 2 2 = 0
cos = cos
Therefore, ( 2 41 ) = 0
657
61. Find the response of a simply supported beam subjected to the external moment at = ( ) = () 00 ( − ) Solution: The equation of motion and the boundary conditions are
4 2 + = ( ) 4 2 (0 ) = 0 ¯ 2 ¯¯ 2 ¯ = 0 0
where
( ) = 0 ¯ 2 ¯¯ 2 ¯ = 0
2 ( − ) 2 where ( − ) is the unit step function. The first derivative of the unit step is Dirac delta function, ( ) = ()
( − ) where Dirac delta function now has the same unit as 1 A concentrated force of () applied at can be thought of as a distributed load of infinite magnitude applied at = or ( ) = () 0 ( − ) = () ( − ) In this case, a concentrated moment applied at can be thought of as a distributed load ( ) = () 0 ( − ) The unit of 0 ( − ) has the same as unit as 12 The fundamental equation that defines derivatives of the delta function is Z Z () (−1) () () () () ≡ − The eigenvalue problem is given by ( − ) =
0000 − 2 = 0 (0) = 0 00 (0) = 0 () = 0 00 () = 0 Simply supported transverse beam has natural frequencies and normalized eigenfunctions given by ³ ´2 r = r 2 ̂ () = sin
658
CHAPTER 7 CONTINUOUS MODELS
and satisfies the orthonormal relationship Z
̂ () ̂ () =
0
Using Eigenfunction expansion method, assume ( ) =
∞ P
̂ () ()
=1
where () is to be determined. Substitute this expanded form of the solution into the equation of motion to obtain ∞ P
=1
̂0000 () () + ̂ () ̈ () = () 0 ( − )
Replace ̂0000 () with 2 ̂ () and ∞ ¡ ¢ P ̈ () + 2 () ̂ () = () 0 ( − )
=1
Multiply both sides by ̂ () and integrate over the length of the beam. Use the orthogonality condition to find Z
̈ () + 2 () =
0
() 0 ( − ) ̂ ()
Using the fundamental property of the delta function given by Z 0
0
Z
̂ () ( − ) 0 ¯ ̂ () ¯¯ = − ¯ ¯
̂ () ( − ) ≡ −
=
we rewrite the differential equation as
¯ ¯ () ̂ ¯ ̈ () + 2 () = − () ¯ ¯ = r 2 = − () cos r 2 (−1)−1 = () The general convolution solution is r () =
2 (−1)−1
Z 0
( − ) ( )
659 where () =
1 sin
The final solution is ( ) =
∞ P
=1 ∞ P
̂ () ()
2 (−1)−1 sin = =1
Z
1 sin ( − ) ( ) 0
660
CHAPTER 7 CONTINUOUS MODELS
Problems for Section 7.6 — Other Transverse Beam Vibration Cases 62. Derive the equation of motion for the transversely vibrating beam with axial force by applying Hamilton’s principle.
Solution: We need to derive ∙ ¸ 2 2 2 () − + ( ) = −() 2 2 2 where the transversely vibrating beam with axial force was derived using earlier results. To apply Hamilton’s principle, we need the kinetic and potential energies. We have the following: µ ¶2 Z 1 = 2 0 Z 1 2 = 2 2 0 where we have assumed that the bending strain results only in a deflection, and no rotation or shear distortion. The total moment is given by + =
2 2
where, in addition to the bending moment there is a moment due to the axial force resulting in Therefore, ¸ 2 Z ∙ 2 2 − 2 0 # Z " µ 2 ¶2 2 1 − 2 = 2 2 0
1 = 2
However, we will find that this approach does not work. We must rethink this problem. The moment due to 2 2 depends on the angle of rotation itself. Therefore, the potential energy must have a factor of 12 However, the moment due to does not depend on the angle of rotation. The potential energy is given by =
Z " 0
1 2
# µ 2 ¶2 2 − 2 2
A better method of obtaining the potential energy is through the strain energy, but it is beyond the scope of this text.
661 Hamilton’s R 2 principle states that the variation of the integral in time of the Lagrangian is equal to zero: 1 ( − ) = 0 Therefore, first evaluate: µ ¶ ¾ Z ½Z 2 Z 2 = 0 1 1 µ 2 ¶¸ ¾ µ 2 ¶ Z 2 ½Z ∙ Z 2 2 2 − 2 = 2 1 1 0 ½ µ 2 ¶ µ 2 ¶¾¸ ¾ Z 2 ½Z ∙ 2 2 = 2 − + 2 2 2 0 1 ¸ ¾ ¶ µ 2 ¶ Z 2 ½Z ∙µ 2 1 2 2 − − 2 = 2 2 0 1 The goal is to express in terms of the instead of 0 , ̇ or 00 This is achieved by integrating by parts. Let us start with the kinetic energy. ¯ µ ¶ µ ¶ Z 2 Z 2 ¯¯ 2 = ¯ − 1 1 1 Z 2 2 = − 2 1 where the terms evaluated at 1 and 2 vanish. The potential energy requires integration by parts twice. We have ¶ µ 2 ¶ Z µ 2 1 2 − 2 2 0 µ ¶ µ ¶ ¶ µ ¶¯ Z µ 2 1 ¯¯ 2 = 2 − − 2 − ¯ 2 0 0 ¶ ¸¯ ¶ µ ¶ µ ∙µ ¯ 2 2 1 − 2 − ¯¯ = 2 − 2 0 ¶ Z 2 µ 2 2 − + 2 0 Combining terms, we obtain Z 2 ( − ) 1 µ ¶¸ Z 2 Z ∙ 2 2 2 = − 2 + 2 2 − 0 1 ¶ µ ¶ µ ¶ ¸¯ Z 2 ∙µ ¯ 2 2 2 − − 2 − ¯¯ + 1 0
where we obtain the equation of motion µ ¶ 2 2 2 − 2 + 2 2 − = 0 for 0 ≤ ≤
662
CHAPTER 7 CONTINUOUS MODELS
and the possible boundary conditions, µ ¶ µ ¶¯ 2 ¯¯ 2 − = 0 ¯0 µ ¶ ¯ ¯ 2 2 − ¯¯ = 0 0
663
63. Solve Equation 7.119 by setting the determinant of the coefficients equal to zero, resulting in a polynomial that is solved for the frequencies. Use the following parameter values: = 30 × 106 lb/in2 , = 10 × 104 in4 , 1 = 100 lb/in, 2 = 1000 lb/in, 3 = 500 lb/in, 4 = 2000 lb/in, and = 1000 in. Solution: Equation 7.119 is given by ¯ ¯ ¯ ¯ 3 3 ¯ −1 − −1 ¯ ¯ ¯ ¯ ¯ ¯ 21 22 23 24 ¯ ¯ ¯= 0 ¯ ¯ ¯ −2 − −2 ¯ ¯ ¯ ¯ ¯ ¯ 41 42 43 44 ¯
where
21 = − 3 cos − 3 sin
22 = 3 sin − 3 cos
23 = 3 cosh − 3 sinh
24 = 3 sinh − 3 cosh and 41 = − sin + 4 cos 42 = − cos − 4 sin 43 = sinh + 4 cosh 44 = cosh + 4 sinh Let us rewrite Equation 7.119 in terms of non-dimensional quantities so that ¯ ¯ ¯ ¯ 3 ¯ 1 3 ()3 ¯ −1 − () −1 2 1 ¯ ¯ ¯ ¯ ¯ 21 3 22 3 23 3 24 3 ¯ ¯ ¯ = 0 ¯ ¯ ¯ ¯ −1 − 2 −1 2 ¯ ¯ ¯ ¯ ¯ 41 4 42 4 43 4 44 4 ¯
where
21 = − 3 ()3 cos − sin 3 3 22 = 3 ()3 sin − cos 3 3 23 = 3 ()3 cosh − sinh 3 3 24 = 3 ()3 sinh − cosh 3 3
(7.119)
664
CHAPTER 7 CONTINUOUS MODELS
and 41 () sin + cos = − 4 4 42 () cos − sin = − 4 4 43 () sinh + cosh = 4 4 44 () cosh + sinh = 4 4 Then, we can write the characteristic equation in terms of five non-dimensional parameters, (1 3 ) (3 3 ) (2 ) (4 ), and At this point, substituting the values for the first four non-dimensional parameters will simplify the algebra considerably. Let us use = 30 × 106 lb/in2 , = 10 × 104 in4 , 1 = 100 lb/in, 2 = 1000 lb-in, 3 = 500 lb/in, 4 = 2000 lb-in, and = 1000 in. Then, = 30 = 6 = 3 × 106 = 15 × 106 3 3 1 3 2 4 Even with these values substituted in, the characteristic equation is still a complicated expression. At this point, we can utilize MATLAB to find The code is given next. function main10_42 fzero(@deter, 0.5) fzero(@deter, 1)
function out=deter(b) EI1=30; EI2=6; EI3=3e6; EI4=1.5e6; A=[EI1*b^3, -1, -EI1*b^3, -1; -EI3*b^3*cos(b)-sin(b), EI3*b^3*sin(b)-cos(b), EI3*b^3*cosh(b)-sinh(b), ... EI3*b^3*sinh(b)-cosh(b); -1, -EI2*b, -1, EI2*b; -EI4*b*sin(b)+cos(b), -EI4*b*cos(b)-sin(b), EI4*b*sinh(b)+cosh(b), ... EI4*b*cosh(b)+sinh(b)]; out=det(A); end
end
665 The nested function deter(b) calculates the determinant of the matrix above for a given value of This way, we avoid obtaining an explicit expression for the characteristic equation. The line, fzero(@deter, 0.5), uses a built-in function fzero.m to calculate the roots of the function defined in deter(b) near 0.5. The program will give 04219 as a root near = 05 and 11865 as a root near = 1 Then, the first two values of the wave numbers are (keeping three significant figures) 1 = 1 = 422 m 2 = 2 = 1190 m.
666
CHAPTER 7 CONTINUOUS MODELS
64. Derive Equation 7.120. Solution: Equation 7.120 is given by 4 + 4
µ
− 2 +
¶
= 0
(7.120)
which is the equation that governs the modal function of an Euler beam on an elastic foundation. Let us start with the equation of motion for a uniform beam,
2 4 + = − 4 2
We let ( ) = () () then 0000 + = − Dividing by and rearranging, we obtain ̈ 0000 + =− The right-hand side is at most a function of time and the left-hand side is at most a function of For them to be equal to each other, they must be a constant number, ̈ = 2 0000 + = 2 −
where the constant number is positive and the frequency squared of () The spatial equation is rewritten such that ¡ ¢ 0000 − 2 − = 0 (7.120)
667
65. Derive Equation 7.122. Solution: Equation 7.122 is the expression for the natural frequency of an Euler beam with elastic support. The natural frequencies are given by s ³ ´2 µ ¶4 = + (7.122) This is a continuation of the previous problem. Starting with Equation 7.122, we let 4 =
2 −
For a hinged-hinged beam, the wave numbers are = Then, = = which is Equation 7.122 with =
s
4 + s ³ ´4
+
668
CHAPTER 7 CONTINUOUS MODELS
66. Discuss the modal participation factor for a vibrating beam that is subjected to loads as shown in Figure 7.62. (Refer to Section 6.6.2 for a discussion of the modal participation factor.)
Figure 7.62: Modal participation factor.
Solution: The modal participation factor is a measure of the importance of a particular mode to the complete response on an DOF structure. Looking at the figure, we observe two loading cases and the first two modes of a transversely vibrating beam. We see that the first mode is qualitatively similar to the first loading shape, and the same is true for the second mode and load. The implication is that for loading case 1, the first mode is a major contributor to the complete response. For loading case 2, the second mode is a major contributor to the complete response. Therefore, given either loading case, it is likely that the first few (3-5) modes will be adequate to represent the complete response. In general, if a particular mode has positive and negative parts in the same physical locations as the load, then the load activates that mode most highly.
Chapter 8 Advanced Continuous Models for Vibration Problems for Section 8.1 — Vibration of Membranes 1. Identify applications where a membrane model could be useful. Solution: Examples include: inflatable structures, very thin and flexible panels, and drum heads.
669
670
CHAPTER 8 ADVANCED CONTINUOUS MODELS
2. Derive Equation 8.8. Solution: Equation 8.8 is ( ) =
∞ X ∞ X
2 √ sin 1 sin =1 =1
Begin with (two equations above Equation 8.8) ( ) = =
∞ ∞ X X =1 =1 ∞ ∞ X X
( ) ( ) ()
=1 =1
where () = 1 cos + 2 sin Letting = 0 and ( 0) = ( ) results in Equation 8.8.
(8.8)
671
3. Derive Equation 8.11. Solution: The detailed derivation can be found in the text (Equations 8.8 to 8.11). Here, we present a slightly varied version. Equation 8.11 is r Z Z 1 = 2 sin = 1 2 · · · (8.11) ( ) sin 0 0 Begin with Equation 8.8 given by ∞ ∞ X X 2 √ sin ( ) = 1 sin =1 =1 sin and integrating over the domain, we obtain Multiplying Equation 8.8 by sin Z Z sin ( ) sin 0 0 Z Z X ∞ X ∞ 2 √ sin 1 sin = 0 0 =1 =1
The right-hand side of the equation is rewritten as µZ ¶ µZ ¶ ∞ X ∞ X 2 √ = sin sin sin sin 1 0 0 =1 =1
where we recognize that
Z
sin
Z
sin
0
and
0
Then,
sin = for = 2 = 0 for 6= sin = for = 2 = 0 for 6=
r 2 1 = √ 1 1 = 4 2 Equating the right and left-sides of the equations, we have r Z Z 1 ( ) sin sin = 1 2 0 0 Solving for 1 we have r Z Z 1 = 2 sin = 1 2 · · · ( ) sin 0 0 Replacing with and with , we obtain Equation 8.11.
672
CHAPTER 8 ADVANCED CONTINUOUS MODELS
4. Derive Equation 8.12. Solution: Equation 8.12 is r Z Z 2 sin ( ) sin 2 = 0 0 Begin with ∞ ∞ X X ( ) ( ) =
= 1 2
(8.12)
=1 =1
for the membrane displacement. To satisfy the initial velocity, take the derivative with respect to time of Equation 8.7 given by 2 ( ) = √ sin × (1 cos + 2 sin ) (8.7) sin and set = 0 Doing this we obtain the following equation: ( 0) = ( ) ∞ X ∞ X 2 √ sin = ( ) 2 sin =1 =1
We can follow the derivation succeeding Equations 8.7 for the initial displacement or do the following. Multiplying the both sides by sin sin and integrate over the domain, we obtain Z Z ( ) sin sin 0 0 µZ ¶ µZ ¶ ∞ ∞ X X 2 √ sin sin sin sin 2 = 0 0 =1 =1 Using the orthogonality properties, we recognize that Z sin = for = sin 2 0 = 0 for 6=
and
Z 0
Then,
Z Z
sin
sin = for = 2 = 0 for 6=
2 ( ) sin sin = √ 1 4 0 r 0 Z Z 2 sin for = 1 2 1 = ( ) sin 0 0 Replacing with and with we obtain Equation 8.12.
673
5. Derive Equation 8.15. Explain the need for = 0 + 2. Solution: Equation 8.15 is 2 00 1 0 Θ00 ( + + 2 ) = − = 2 Θ
(8.15)
Begin with Equation 8.13 given by ∇2 ( ) + 2 ( ) = 0 where ∇2 ≡
plus boundary conditions,
(8.13)
1 2 2 1 + + 2 2 2
Assuming a solution of the form ( ) = ()Θ() Equation 8.13 can be rewritten as ∇2 (Θ) + 2 Θ = 0 2 1 2 Θ 1 Θ + Θ + 2 + 2 Θ = 0 2 2 Multiplying the last equation by 2 Θ and rearranging the equation, we obtain 1 2 2 1 2 Θ + + 2 2 = 0 + 2 Θ 2 2 2 1 1 2 Θ ( 2 + + 2 ) = − Θ 2 The left-hand side is a function of at most while the right-hand side is a function of at most. For the two sides to be equal, they must be equal to a constant number. We call this constant number 2 The boundary conditions that Θ () = Θ ( + 2) and Θ0 () = Θ0 ( + 2) requires that is an integer. Therefore, we have Equation 8.15 given by 2 2 1 1 2 Θ = 2 ( 2 + + 2 ) = − Θ 2
674
CHAPTER 8 ADVANCED CONTINUOUS MODELS
6. Derive Bessel’s equation and solve it in general. (Information about Bessel’s equation can be found in differential equations textbooks.)
Solution: Bessel’s equation arises frequently when the domain is circular. Consider the differential equation given below. µ 2 ¶ 2 1 2 = + 2 2
which is an one-dimensional wave equation in the radial direction. The equation describes a system where the wave propagates at speed radially. This is the simplest wave equation that result in Bessel’s equation. Assume the product solution: ( ) = () () Substituting appropriately results in the equation: µ 2 ¶ 2 1 2 2 = + 2 Dividing the previous equation by () () and separating the functions of and we obtain 1 2 1 1 2 = − 2 = + 2 2 2 where 2 is a constant number. Therefore, we have one (harmonic) equation in time and one equation in the radial coordinate given by ̈ + ()2 = 0 1 00 + 0 + 2 = 0 The second equation is of the Bessel form. The general solution is given by () = 1 0 () + 2 0 () where 0 is the Bessel function of the first kind, and 0 is the Bessel function of the second kind.
675
7. Derive Equations 8.23, 8.24, and 8.25. (Symbolic manipulation code such as Mathematica may be useful for the derivations.) Solution: Equations 8.23, 8.24, and are given by 2 = 0 2 = 1 2 = 2
1
(8.23)
2 12 [ 0 ] 2
(8.24)
2 2 +1 [ ]
2 2 2 +1 [ ]
(8.25)
Start with Equations 8.21 and 8.22 given by 0 = 0 0 ( 0 ) ¢ ¡ 1 = 1 sin ¢ ¡ 2 = 2 cos
They are normalized according to
Z 2 Z 0
0
For the mode shape, 0 we have Z 2 Z 0
2 ( ) = 1
2 2 0 0 ( 0 ) = 1
0
R
02 ( 0 ) is (from an integral table) Z ¢ 2 ¡ 2 0 ( 0 ) + 12 ( 0 ) 02 ( 0 ) = 2 0 ¢ ¡ For our case where the ends are fixed, = 0 for all and Then, we have Z 2 02 () = 12 ( 0 ) 2 0 The integral,
0
Therefore, the constant coefficient is given by Equation 8.23 or 2 = 0
For the mode shape, 1 we have Z 2 Z 0
0
1 2 12 ( 0 )
¢ ¡ 2 1 2 sin = 1
(8.21) (8.22)
676
CHAPTER 8 ADVANCED CONTINUOUS MODELS
¢ R ¡ The integral, 0 2 is Z ¢ ¢ ¢ ¢ ¢¢ ¡ ¡ ¡ ¡ ¡ ¡ 2 2 = 2 − 2 +1 + 1 +1 2 0
Using the boundary conditions, the integral is reduced to Z ¢ ¡ 2 2 2 = +1 ( 1 ) 2 0 R 2 R 2 The integral, 0 cos2 = 0 sin2 = 12 Then, we obtain Equations 8.24 and 8.25 or 2 2 1 = 2 =
2
2 2 +1 ( )
677
8. Derive by means of Hamilton’s principle the differential equation of motion and the associated boundary conditions for a thin rectangular membrane. Give the physical interpretation of the boundary conditions. Solution: The surface area of a rectangular membrane in its deformed state is s µ ¶2 µ ¶2 Z Z 1+ + = 0 0 where the membrane is defined on domains 0 ≤ ≤ and 0 ≤ ≤ and ( ) is the height of the membrane. The area of the undeformed membrane is Z Z = 0
µ
Expanding the integrand assuming small as s µ ¶2 µ 1+ +
0
¶2
¶2
¶2 and , we can approximate the integrand µ ¶2 µ ¶2 1 1 ≈1+ + 2 2 µ
The potential energy of the deformed membrane is the tension times the change in area from the undeformed reference: ⎡s ⎤ µ ¶2 µ ¶2 Z Z ⎣ 1 + + − 1⎦ = 0 0 " # µ ¶2 µ ¶2 Z Z + = 2 0 0 where is the tension per unit length. The kinetic energy of an infinitesimal area is 1 2 = 2 2 where is the density, mass per unit volume. Integrating over the surface area, Z Z 1 2 = 2 0 0 2 Hamilton’s extended principle is
Z 2
( + ) = 0
1
where
Z Z µ ¶ µ ¶ = 0 0
678
CHAPTER 8 ADVANCED CONTINUOUS MODELS
and = − + Z Z ∙µ ¶ µ ¶ µ ¶ µ ¶¸ + = 0 0 Z Z = 0
0
where is the upward force per unit surface area or negative pressure. Switching the variation and differentiation such that µ ¶ () = the variation of the kinetic energy is =
Z Z 0
0
()
Integrate by parts: Z 2
() 1 ¯ µ ¶ Z 2 ¯¯ 2 = ¯ − 1 1 µ ¶ Z 2 = − 1 Then,
Z 2 1
= −
µ ¶ 0
Z 2 Z Z 1
0
Similarly, the variation of the potential energy is Z 2 1
=
Z 2 Z Z ∙ 1
0
0
¸ () +
Integrate by parts: Z 2
Z 2 Z Z
Z 2 Z Z () = + 0 0 0 0 1 1 1 ¶¯ Z 2 Z Z 2 Z 2 Z µ ¯ ¯ ¯ − = 2 0 0 0 1 1 0 ¶¯ Z 2 Z µ Z 2 Z Z 2 ¯ ¯ + ¯ − 2 0 0 0 1 1 0
679 The variation of the non-conservative work remains the same. Assemble terms, and then the Hamilton’s principle becomes Z 2 ( + ) 1 ¶ ¶ µ ¶ µ 2 Z 2 Z Z µ 2 − + 2 + + = 2 0 0 1 ¶¯ ¶¯ Z 2 Z µ Z 2 Z µ ¯ ¯ ¯ − ¯ − ¯¯ 1
0
1
0
0
0
= 0
Since Hamilton’s principle is satisfied for any arbitrary or independent of choice of ( ) we obtain the equation of motion given by ¶ µ ¶ µ 2 2 + + 2 + = 0 for 0 ≤ ≤ and 0 ≤ ≤ − 2 and the boundary conditions must satisfy ¶¯= µ ¯ ¯¯ = 0 =0 ¶¯= µ ¯ ¯¯ = 0 =0
or must vanish Similarly, For the boundary conditions at = 0 and = either the boundary conditions at = 0 and = either or must vanish. For example, (0 or ) = 0 means that the variation is zero and is explicitly known or prescribed at that boundary. = 0 means that the horizontal (transverse) component of the tension is zero and the membrane is free to deflect transversely.
680
CHAPTER 8 ADVANCED CONTINUOUS MODELS
9. Derive the frequency equation for a uniform annular membrane defined over the domain ≤ ≤ and fixed at the boundaries at = and =
Solution: The eigenvalue problem of circular membrane is given by ∇2 ( ) + 2 ( ) = 0 where 2 = ()2 and the Laplacian operator is defined by ∇2 =
2 1 1 2 + + 2 2 2
The boundary conditions are ( ) = 0 ( ) = 0 ( ) = ( + 2) ( + 2) ( ) = Assume a solution of the form, ( ) = () Θ () to obtain: 2 Θ 2 2 + Θ = 0 ¶ µ 2 1 2 2 + + − 2 = 0 2
681 The boundary conditions also separate such that ( ) = 0 ⇒ () Θ () = 0 ⇒ () = 0 ( ) = 0 ⇒ () Θ () = 0 ⇒ () = 0 ( ) = ( + 2) ⇒ () Θ () = () Θ ( + 2) ⇒ Θ () = Θ ( + 2) ( + 2) Θ ( + 2) ( ) = ⇒ () Θ () = () Θ ( + 2) ⇒ Θ () = The solution Θ () has the form of Θ () = 1 sin + 2 cos Due to periodic boundary condition in the direction: Θ () = Θ ( + 2) Θ ( + 2) Θ () = After the periodic boundary conditions are applied, we find Θ () = Θ ( + 2) ⇒ 2 = 1 sin 2 + 2 cos 2 Θ ( + 2) Θ () = ⇒ 1 = 1 cos 2 − 2 sin 2 We rewrite the previous equations in a matrix form: ⎡ ⎤⎡ ⎤ ⎡ ⎤ sin 2 cos 2 − 1 0 ⎣ ⎦⎣ 1 ⎦ = ⎣ ⎦ cos 2 − 1 − sin 2 2 0
In order to avoid trivial solution, the two equations must be linearly dependent or the determinant of the matrix of the coefficient must be zero: − sin2 2 − (cos 2 − 1)2 = 0 sin2 2 + cos2 2 − 2 cos 2 + 1 = 0 1 − cos 2 = 0 Therefore, must be an integer. Note that 2 = 1 (cos 2 − 1) = 0 When and we can rewrite Θ () as Θ () = 1 sin The differential equation in () is a Bessel equation, with the solution: () = 3 () + 4 ()
682
CHAPTER 8 ADVANCED CONTINUOUS MODELS
Apply the boundary conditions, () = 0 and () = 0 : () = 3 () + 4 () = 0 () = 3 () + 4 () = 0 Solve for 3 using the first equation: 3 = −4
() ()
and substitute it into the second equation: µ ¶ () − () + () 4 = 0 () The frequency equation is −
() () + () = 0 ()
or () () − () () = 0
(1)
There will be a series of that satisfy the frequency equation, and we will use the index The eigenfunction () with 3 replaced is µ ¶ ( ) ( ) + ( ) () = 4 − ( ) The eigenfunction ( ) is then µ ¶ ( ) ( ) = sin − ( ) + ( ) () where is the product of two constant coefficients 3 and 1 For demonstration purposes, let us assume = 3 When = 0 Θ () = 0 and the eigenfunction is trivial. We will start from = 1 The frequency equation is ³ ´ ³ ´ − () = 0 () () (3) − (3) () = 0 The solutions to the frequency equation are = 1 : = 1636 3179 4738 = 2 : = 1868 3324 4840 Figure next shows the eigenfunctions for the first six cases.
683
684
CHAPTER 8 ADVANCED CONTINUOUS MODELS
10. For the simply supported rectangular membrane of Section 8.1.1, suggest possible initial displacements ( ). Select one and solve for the complete solution. Assume zero initial velocity. Solution: Since the membrane is simply supported, any expression for ( ) must equal zero at the boundaries of the membrane. Possible initial displacements include a double sine of particular amplitude, or a “pyramid” type shape that almost comes to a point. However, the “point” can be problematic because of derivative discontinuities. Other possibilities include any arbitrary displaced membrane volume. The difficulty is in obtaining an analytical expression for this displacement function. The case we pick is the simplest case given by ( ) = sin
sin
which satisfies the boundary conditions. Of course, a more general version would be ( ) = sin
sin
for any integer and The constant coefficient of integration in Equation 8.11 is r Z Z 1 = 2 sin = 1 2 ( ) sin 0 0 r Z Z = 2 sin sin sin sin 0 0 We find r 1 = 2 for = 1 and = 1 4 = 0 for all the other and Then, the total solution is given by (below Equation 8.7) 2 sin cos 11 sin ( ) = √ q p where 11 = 11 and 11 = (1)2 + (1)2 . The membrane response at the first natural frequency only.
685
11. Find the response of a uniform rectangular membrane to the initial displacement ( ) = ( − ) ( − ) with zero initial velocity.
Solution: Let us assume that the ends are fixed and the rectangle has a dimension of × The eigenvalue problem for a uniform rectangular membrane is given by ∇2 ( ) + 2 ( ) = 0 where 2 = ()2 and 2 = We assume the four sides are fixed so that (0 ) = 0 ( ) = 0 ( 0) = 0 ( ) = 0 We obtained the normal modes and natural frequencies: 2 sin sin ( ) = √ r³ ´ ´ ³ 2 2 + for = 1 2 · · · = Since we have only an initial displacement, we assume that the initial velocity is zero and the distributed force ( ) is also zero. Thus, the response is ( ) =
∞ P ∞ P
=1 =1
where
( ) ()
() = (0) cos Z Z 2 sin ( 0) sin √ (0) = 0 0 Z Z 2 sin ( − ) ( − ) = √ sin 0 0 Integrate over first:
Z
¡ ¢ − 2 sin 0
686
CHAPTER 8 ADVANCED CONTINUOUS MODELS
which requires integration by parts twice: Z
¡ ¢ − 2 sin 0 ¯ Z ¡ ¢ ¯¯ 2 cos cos = − − + ( − 2) ¯0 0 ¯ ¯ µ ¶2 ¯ ¯ ¢ ¡ ¯ ¯ + ( − 2) cos sin = − − 2 ¯ ¯0 ¯ 0 µ ¶2 Z − (−2) sin 0 ¯ ¯ µ ¶2 ¯ ¢ ¡ ¯ ¯¯ 2 cos + ( − 2) sin = − − ¯ ¯0 ¯ 0 ¯ µ ¶3 ¯¯ −2 cos ¯ ¯ 0 µ ¶3 = 2 (1 − cos ) ⎧ µ ¶3 ⎪ ⎨4 , odd = ⎪ ⎩ 0 even.
Similarly, the integration over is
Z
¡ ¢ − 2 sin 0 ⎧ ³ ´3 ⎨4 , odd = ⎩ 0 even.
µ ¶3 ³ 2 ´3 (0) = √ 4 , odd and 4 We can rewrite it as p µ ¶3 µ ¶3 5 5 1 1 (0) = 32 , = 1 2 · · · 6 2 − 1 2 − 1 Then,
p µ ¶3 µ ¶3 5 5 1 1 cos () = 32 6 2 − 1 2 − 1
687 The total response is then ( ) =
∞ P ∞ P
=1 =1
( ) ()
∞ P ∞ sin 64 P = 6 =1 =1 × cos 2 2
where =
sµ
2 − 1
(2 − 1) (2 − 1) sin 3 3 (2 − 1) (2 − 1)
¶2
+
µ
2 − 1
¶2
The only the odd modes that contribute to the response such as 11 13 31 , 33 · · · This is expected because the initial displacement is symmetric about the line = 2 and = 2 so the contributing modes must be as well.
688
CHAPTER 8 ADVANCED CONTINUOUS MODELS
12. Repeat Problem 10 for an assumed initial velocity ( ), with zero initial displacement.
Solution: Since the membrane is simply supported, any expression for ( ) must equal zero at the boundaries of the membrane. Possible initial velocities include a double sine of particular amplitude, or a “pyramid” type shape that almost comes to a point. However, the “point” can be problematic because of derivative discontinuities. Other possibilities include any arbitrary velocity contour. The difficulty is in obtaining an analytical expression for this velocity function. For example, it is possible that for some reason one part of the membrane has a significant initial velocity and another has negligible velocity, perhaps due to an impact. Such an initial velocity distribution can be established experimentally. The case we pick here is the simplest one given by ( ) = sin
sin
which satisfies the boundary conditions. Of course, a more general version would be ( ) = sin
sin
for any Then, we can proceed with Equation 8.12 to evaluate 2 given by r Z Z 2 ( ) sin 2 = sin = 1 2 0 0 r Z Z 2 = sin sin sin sin 0 0 We find r 2 = 2 for = 1 and = 1 4 = 0 for all the other and Then, the total solution is given by (below Equation 8.7) 2 sin sin 11 sin ( ) = √ q p where 11 = 11 and 11 = (1)2 + (1)2 . The membrane response at the first natural frequency only.
689
13. Convert the wave equation from Cartesian coordinates, µ 2 ¶ 2 2 2 = + 2 2 2 to polar coordinates, 2 = 2 2
µ 2 ¶ 1 1 2 + + 2 2 2
using the transformation = cos and = sin .
Solution: Begin with the Cartesian coordinates and make the following transformations given by = cos = sin Then, = + = cos + sin = + = (− sin ) + ( cos )
(1) (2)
Solve the above for and in order to find the differential operator. We first multiply Equation (1) by cos and second by sin and subtract the second from the first to obtain cos
¢ sin ¡ 2 − = cos + sin2 =
We multiply Equation (1) by sin and second by cos and add the two to obtain sin
cos + =
Then, the differential operators are related by sin = cos − cos = sin + The second derivative with respect to is given by µ ¶ µ ¶µ ¶ sin sin = cos − cos − µ ¶ µ ¶ sin cos − cos = cos µ ¶ µ ¶ sin sin sin cos + −
690
CHAPTER 8 ADVANCED CONTINUOUS MODELS
where µ ¶ 2 cos cos = cos2 2 µ ¶ µ ¶ 1 1 2 sin = cos sin − 2 + cos µ ¶ µ ¶ sin 2 sin cos = − sin + cos µ ¶ µ ¶ 2 sin sin sin cos = + sin 2 2 Then,
µ
¶
µ
¶
µ
¶
2 2 cos sin 2 cos sin 2 − + 2 2 2 2 2 sin sin + 2 + 2
= cos2
Similarly, we find that
µ ¶µ ¶ cos cos = sin + sin + µ ¶ µ ¶ cos sin + sin = sin µ ¶ µ ¶ cos cos cos sin + +
which simplifies to
2 2 cos sin 2 cos sin 2 − + 2 2 2 2 2 sin sin + 2 + 2
= sin2
Combining these and using the trigonometry identity cos2 + sin2 = 1, we obtain 1 2 2 2 2 1 + + = + 2 2 2 2 2
691 Problems for Section 8.2 — Vibration of Plates 14. Identify applications where a plate model would apply. Solution: Pistons, cylinder heads, pumping devices, optical lenses, floors in structures.
692
CHAPTER 8 ADVANCED CONTINUOUS MODELS
15. Derive by means of Hamilton’s principle the equation of motion of a square plate. Solution: We first need to form the kinetic and potential energies. It is assumed that the displacements in the and directions are zeros (in-plane displacements). The only non-zero displacement is ( ) in the direction perpendicular to the plate. The kinetic energy is given by Z 1 = (1) ̇2 2 where is the area. Note that is the incremental mass, The non-conservative work done by ( ) acting in the direction is (2)
= ( ) The potential energy is given by =
Z 2 Z ÃZ X 3 X 3 −2
R P3 P3
=1 =1
!
where =1 =1 is the volumetric strain energy. The expressions for the strain and stress are given in in the text on page 789 and reproduced next. = − = − = −2 ( + ) = ( + ) 2 1− 1 − 2 ( + ) = ( + ) = − 2 1− 1 − 2 = − = 1+ 1+
= −
The volumetric strain energy is given by Z X 3 3 X =1 =1
=
Z Z
+
Z
+
Z
+
Z
Z Z 2 = ( + ) + ( + ) + 2 2 1− 1− 1+ µ ¶ 1 2 2 1 2 + + + = 2 1− 2 2 1 + ¢ ¤ 1 2 £¡ 2 2 2 = + 2 + + 2 (1 − ) 1 − 2 2
693 Then, integrating over we obtain Z Z 2 ¢ ¤ 1 £¡ 2 2 2 = 2 + 2 + + 2 (1 − ) 2 1 − −2 2 Z ¢ ¤ 1 £¡ 2 2 2 + 2 + + 2 (1 − ) = 2
where
=
(3)
3 1 − 2 12
Lagrange’s Equation Just to confirm that we are on the right track, we can use Lagrange’s equation to find the equation of motion given by ( ) = −
2 2 2 − − + 2 2 ̇
where = − , where the kinetic and potential energies are given in Equations (1) and (3), and ( ) is the generalized non-conservative force associated with given in Equation (2). Performing the differentiations, we obtain ½ 2 ¾ 2 2 ( ) = ( + ) + 2 ( + ) + 2 (1 − ) ( ) + ̈ 2 = { + + 2 + 2 (1 − ) } + ̈ = ( + + 2 ) + ̈ = ∇4 + ̈
Hamilton’s Principle Hamilton’s principle (extended principle) says Z ( + ) = 0 or
Z
( − + ) = 0
R Let us evaluate the second integral term by term. Begin with
Z
Z µZ
¶ 1 2 ̇ = 2 ¶ Z µZ 1 2 ̇ = 2 ¶ Z µZ ̇ ( ̇) =
694
CHAPTER 8 ADVANCED CONTINUOUS MODELS
Integrating by parts, we obtain ¶ Z Z µ Z ̇ ()| − = ̈ ()
where the first term becomes zero because Hamilton’s principle assumes no variation at and Then, ¶ Z µZ Z
R Let us now evaluate
−̈ ()
=
Z
=
Z
Z µZ
¶ ( ) = ¶ Z µZ ( ) =
where ( ) is the force per unit area and ( ) is the force. R The evaluation of is long. The variation of the potential energy is given by
Z
¾ Z ½ Z ¢ ¤ 1 £¡ 2 2 2 = + 2 + + 2 (1 − ) 2
Let us evaluate the first term to demonstrate the procedure: ¾ ¾ Z ½ Z Z ½ Z Z 1 2 1 2 = 2 2 Z Z Z = ( )
0
0
where the rectangular plate is defined on 0 ≤ ≤ and 0 ≤ ≤ Integrating the right-hand side by parts twice, we obtain µZ ¶ ¾ Z ½Z = ( )|0 − ( ) 0 µZ ¶ ¾ Z ½Z = ( )|0 − ()|0 + () =
0
( ( )|0 − ()|0 )
Z Z
+
0
Z Z
()
695 R We evaluate all the other terms in this way, and we will obtain Z = ( + + 2 ) () Z + ( ( )|0 − ()|0 ) 0 Z ´ ³ + |0 − |0 0 Z ¡ ¢ + |0 − |0 Z 0 ³ ´ + ( )|0 − ()|0 0 Z Z + (1 − ) |0 − (1 − ) |0 0 0 Z Z + (1 − ) |0 − (1 − ) |0 0
Combining the integrand of
R R
0
() we obtain the equation of motion given by
−̈ − ( + + 2 ) + ( ) = 0 The boundary conditions are algebraically complicated. The boundary terms of (second line to sixth line of ) are simplified to Z 0 = {− ( + + (1 − ) ) 0
+ ( + ) + (1 − ) }|0 Z {− ( + + (1 − ) ) + 0
+ (1 − ) + ( + ) }|0
(4)
The coefficients of ( = 0 ) ( = ) ( = 0 ) ( = ) ( = 0 ) ( = ) ( = 0) ( = ) ( = 0) ( = ) ( = 0) and ( = ) must R be zeros so that ( + ) = 0 Let us look at the coefficient of at the boundary = 0 given by − ( (1 − ) + + ) |=0 = 0 Note that the coefficient simplifies to the shear force in the direction such that − (1 − ) − − =
+ =
The term, |=0 = 0 means that on the boundary = 0 either the shear force in the direction (acting on the face whose normal is in the direction) is zero or the displacement must be known
696
CHAPTER 8 ADVANCED CONTINUOUS MODELS
The boundary terms in Equation (4) can be rewritten as Z
{ − − }|0 0 Z + { − − }|0 0
from which we obtain possible boundary conditions |0 = 0 |0 = 0 |0 = 0 |0 = 0 |0 = 0 |0 = 0
(4)
697
16. Derive Equations 8.27 and 8.28. Solution: Equations 8.27 and 8.28 are given by − + = 0 + − = 0
(8.27) (8.28)
Most of the derivation for Equation 8.27 is done in the text. Summing the moments about the axis, we obtain ¶ ¶ µ µ − + + − + µ ¶ + + + = 0 2 2 which becomes
+ + + 2 = 0 Dividing by and taking the limit as → 0 and → 0 the last term disappears, and we obtain Equation 8.27. For Equation 8.28, we sum the moments about the axis to obtain µ µ ¶ ¶ − + + − + + µ ¶ − + = 0 + 2 2 −
which simplifies to + − = 0 2 Dividing by and taking the limit as → 0 and → 0 the last term disappears, and we obtain Equation 8.28.
698
CHAPTER 8 ADVANCED CONTINUOUS MODELS
17. Derive Equation 8.29. Solution: Equation 8.29 is given by 2 2 2 2 = −( ) + + − 2 2 2 2
(8.29)
We differentiate Equation 8.27 with respect to and Equation 8.28 with respect to Then add the two equations to obtain Equation 8.29. Recall that = −
699
18. Derive Equation 8.33. Solution: Equation 8.33 is given by ¶ µ 4 2 4 4 + + 2 + = ( ) 4 2 2 4 2 We start with Equations 8.30-8.32 given by ¶ µ 2 Z 2 2 = − + 2 = 2 −2 ¶ µ 2 Z 2 2 = = − + 2 2 −2 Z 2 2 = − = − = (1 − ) −2
(8.33)
(8.30) (8.31) (8.32)
where the stresses are written in terms of displacements. Taking the appropriate derivatives and substituting into Equation 8.29 yield Equation 8.33.
700
CHAPTER 8 ADVANCED CONTINUOUS MODELS
19. Derive Equation 8.38. Solution: Equation 8.38 is the expression for 2 ( ) and is given by 2 ( ) = 5 sinh 1 sinh 1 + 6 sinh 1 cosh 1 + 7 cosh 1 sinh 1 + 8 cosh 1 cosh 1
(8.38)
Start with Equation 8.36 given by ¢ ¡ 2 ∇ − 2 2 ( ) = 0
We let 2 ( ) = () () and substitute in to Equation 8.36. Then, we obtain 2 2 2 2 + − 2 = 0 2
2
2
2
If we let = 21 and = 21 the constants are related by 21 + 21 = 2 Then, the ordinary differential equations for () and () are given by 2 − 21 = 0 2 2 − 21 = 0 2 The solutions are () = sinh (1 ) + cosh (1 ) () = sinh ( 1 ) + cosh ( 1 ) where and are constant coefficients of integration. Then, 2 ( ) = () () = 5 sinh 1 sinh 1 + 6 sinh 1 cosh 1 +7 cosh 1 sinh 1 + 8 cosh 1 cosh 1
(8.36)
701
20. Show that Equation 8.39 must be true for the given boundary conditions. Solution: Equation 8.39 is ( ) = 1 sin sin
(8.39)
The boundary conditions are simply supported all around the plate boundary. This means there is no resistance to moment, and there is no deflection at the boundary. The displacement conditions are ( = 0 and ) = 0 ( = 0 and ) = 0
⇒ ⇒
( = 0 and ) = 0 ( = 0 and ) = 0
and the moment conditions are ( = 0 and ) = 0
⇒
( = 0 and ) = 0
⇒
( = 0 and ) = 0
⇒
( = 0 and ) = 0
⇒
¯ 2 2 ¯¯ + =0 2 2 ¯=0 and ¯ 2 ¯¯ =0 ¯=0 and ¯ 2 ¯¯ =0 ¯=0 and ¯ 2 2 ¯¯ + = 0 2 2 ¯=0 and
where the moments are expressed in terms of displacements (Equation 8.30-8.32). Substituting Equation 8.39 into boundary conditions, we find that it satisfies all 12 conditions. Looking at the general solution, Equation 8.38, we have terms that include sines and cosines, and hyperbolic sines and cosines. The cosine terms cannot satisfy the zero displacement conditions at the = 0 boundaries. The hyperbolic sine and cosine terms grow without bound for larger or and therefore cannot satisfy the boundary conditions. Only the sin sin term can satisfy the boundary conditions.
702
CHAPTER 8 ADVANCED CONTINUOUS MODELS
21. Derive coefficients
2 = √
using the normalization procedure given for by Equation 8.40. Solution: Equation 8.40 is given by ( ) =
∞ ∞ X X
̂ ( ) ()
=1 =1
The normalization procedure requires the evaluation of the double integral given by Z Z 2 2 sin = 1 sin2 0 0 where Z
= 2 Z 0 = sin2 2 0 sin2
2 Then, = 1 Therefore, we obtain the desired value. 4
(8.40)
703
22. Derive the differential equation of motion for a nonuniform plate with varying thickness. Assume that the material properties such as Young’s modulus and Poisson’s ratio remain constant. Solution: The derivation of equation of motion for a nonuniform plate is nearly identical to that for a uniform plate. From the text, Newton’s law is simplified to 2 2 2 2 = − + + − 2 2 2 2
(1)
where is the moment per unit length, ( ) is the distributed load per unit area, is the density, is the thickness of the plate, and ( ) is the displacement of the plate in the direction. In terms of displacements, we have ¶ µ 2 Z 2 2 = − + 2 = 2 −2 ¶ µ 2 Z 2 2 = = − + 2 2 −2 Z 2 = − = − −2
= (1 − )
2
where = 3 12(1 − 2 ) Let us assume that the thickness may vary such that ( ) while the material properties such as and are constant. Therefore, the flexural rigidity, , varies with and Differentiate the moment expressions and substitute them into Equation (1). The necessary derivatives are = − ( + ) − ( + ) 2 = − ( + ) − 2 ( + ) − ( + ) 2 = − ( + ) − ( + ) 2 = − ( + ) − 2 ( + ) − ( + ) 2 = (1 − ) + (1 − ) 2 = (1 − ) + (1 − ) + (1 − ) + (1 − )
704
CHAPTER 8 ADVANCED CONTINUOUS MODELS
After substitution, the equation of motion becomes − ( + ) − 2 ( + ) − ( + ) − ( + ) − 2 ( + ) − ( + ) −2 ( (1 − ) + (1 − ) + (1 − ) + (1 − ) ) 2 = − + 2 The left hand side is rewritten as − ( + 2 + ) − 2 ( + ) − 2 ( + ) + (1 − ) (−2 ) − − − − The terms,− − are rewritten as −2 + (1 − ) Then, the left hand side is reduced to − ( + 2 + ) − 2 ( + ) − 2 ( + ) + (1 − ) ( − 2 + ) − − − − Note that ∇4 = + 2 + ∇2 = + ∇2 ∇2 = ( + ) ( + ) = + + + Then, the equation of motion becomes ¡ 2 ¢ ¡ 2 ¢ ∇ +2 ∇ + ∇2 ∇2 − (1 − ) ( − 2 + ) =
̈ + ∇4 + 2
705 Problems for Section 8.3 — Approximate Methods 23. Derive Equations 8.54. Solution: Equation 8.54 is given by ⎫ ⎪ 11 = ⎪ ⎪ ⎪ 5 ⎪ ⎪ R ⎪ ⎪ 12 = 0 Y1 Y2 = = 21 ⎪ ⎪ ⎪ 6 ⎪ ⎪ ⎪ R ⎪ ⎪ 2 ⎬ 22 = 0 Y2 = 7 R 4 ⎪ ⎪ 11 = 0 [Y100 ]2 = 3 ⎪ ⎪ ⎪ ⎪ ⎪ R ⎪ 6 00 00 ⎪ 12 = 0 Y1 Y2 = 3 = 21 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ R ⎪ 12 ⎭ 00 2 22 = 0 [Y2 ] = 3 R
Y12 = 0
(8.54)
Start with Y1 = ()2 and Y2 = ()3 as implied on page 804. Substitute these functions in, and we will find and For example, 11 =
Z 0
Y12 =
Z 0
³ ´2
=
5
706
CHAPTER 8 ADVANCED CONTINUOUS MODELS
24. In Section 8.3.2, show how the parameters and are derived. Solution: In Example 8.2, we used Y1 = 2 and Y2 = 3 We can find the parameters and by using Z = ()Y00 Y00 0 Z ()Y Y = 0
Then,
11 =
Z
4 = 4 Z 12 = 62 12 = 21 = 0 Z 22 = 362 = 123 0
0
and
Z
1 2 2 = 5 5 0 Z 1 2 3 = 6 12 = 21 = 6 0 Z 7 3 3 = 22 = 7 0
11 =
The stiffness and mass matrices can also be found by matching the coefficients. In Example 8.2, we have 2 X 2 = (41 + 62 ) = 21 1 =1
2 X 2 3 = (61 + 122 ) = 22 2 =1
and 2 ∗ 5 1 6 2 X + = = 21 1 5 6 =1
2 ∗ 6 1 7 2 X + = = 22 2 6 7 =1
707 from which we find 11 = 2 12 = 21 = 32 22 = 63 1 5 11 = 10 1 12 = 21 = 6 12 1 7 22 = 14 The values are half of what we found previously, but they will result in the same eigenvalues and eigenvectors.
708
CHAPTER 8 ADVANCED CONTINUOUS MODELS
25. Derive Rayleigh’s quotient for a longitudinally vibrating beam. Solution: We begin with the statement of the eigenvalue problem given by ∙ ¸ − () = 2 ()() Proceeding as we did for the application of Rayleigh’s quotient to the transversely vibrating beam, for a particular mode () we multiply both sides by a different mode () and integrate over the length of the beam such that ∙ ¸ Z Z 2 () = () () () − 0 0 (This approach can be used to derive the orthogonality conditions for the axially vibrating beam as was done for the transversely vibrating beam.) Here, for Rayleigh’s quotient, we integrate by parts. Since we are considering a single mode, set = to find ∙ ¸ Z Z 2 − () 2 () () = 0 0 Then,
Rayleigh’s quotient is given by
¤ R £ () − 0 2 = R 2 () () 0
Ω21 =
−
∙ ¸ Y1 Y1 () () 0 Z ()[Y1 ()]2
Z
0
where Y1 () is a guess for the eigenfunction 1 () Note that Y1 () must satisfy the geometric boundary condition of the system. Note that the same Rayleigh’s quotient can be obtained by setting the maximum kinetic energy equal to the maximum potential energy given by µ ¶2 Z 1 () = 2 0 µ ¶2 Z 1 = () 2 0 Now assume ( ) = () () where () = cos Then, substitute into and set = and solve for 2 noting that arbitrary constant cancels out as do the harmonic functions we we assume max and max
709
26. Estimate the fundamental frequency for the tapered beam of Figure 8.17 where () = (1 − ) (a) and () = (1 − ) p Compare your result to the exact value of 1 = 240 2 .
Tapered beam in longitudinal vibration.
Solution: Let us use the Rayleigh quotient given by ∙ ¸ Z Y1 − Y1 () () 0 2 Ω1 = Z 2 ()[Y1 ()] 0
where Y1 () must at least satisfy the geometric boundary condition. This expression was obtained in Problem 25. The subscript 1 denotes that it looks similar to the first mode shape. Let us use Y1 () = 2 Then,
Ω21 =
−
Z
2
Z0 0
[(1 − )2]
(1 − ) (2 )2
= 10 2 The Rayleigh quotient is
r
Ω1 = 316
1
710
CHAPTER 8 ADVANCED CONTINUOUS MODELS
If a better guess is used, the Rayleigh quotient will be closer to the fundamental frequency (still greater than the fundamental frequency). For example, if we use Y1 () = sin
2
which is the first mode shape of a uniform rod, we obtain v Z u £ ¡ ¢¤ u sin 2 (1 − ) sin 2 u− u 0 Ω1 = u Z t ¡ ¢2 (1 − ) sin 2 0 r 1 = 241 p which is much closer to the exact answer of 2.40 2
711
27. For Problem 26, estimate the first two natural frequencies using the Rayleigh-Ritz procedure. Assume a trial function of the form Y() = 1 2 +2 3 . Compare the fundamental frequencies estimated by the Rayleigh-Ritz and the Rayleigh quotient methods. P Solution: The trial function can be written as () = =1 Y () where, here, = 2 and 2 3 Y1 () = and Y2 () = From Section 8.3.2, begin with the axial vibration version of Equation 8.41 (see previous problem), where () = (1 − ) and () = (1 − ) R () 2 0 = R = () () () 0
Into this ratio we substitute the two term approximation for () and its derivative. Then, we have P P =1 =1 2 = P P =1 =1 Z = ()Y Y 0 Z ()Y0 Y0 = 0
Following the procedure in Section. 8.3.2, we differentiate the above ratio with respect to the constants and find − 2 = 0 We have from Section 8.3.2 general expressions (for the two term approximations) In order to complete the derivation of the two term approximation (and the resulting two degreeof-freedom system), we need to evaluate the explicit expressions for and such that Z ()Y12 11 = 0 Z ()Y1 Y2 = 21 12 = 0 Z 22 = ()Y22 0 Z 2 () (Y10 ) 11 = 0 Z ()Y10 Y20 = 21 12 = 0 Z 2 22 = () (Y20 ) 0
712
CHAPTER 8 ADVANCED CONTINUOUS MODELS
where () = (1 − ), () = (1 − ) and where Y1 () = 2 and Y2 () = 3 Now we evaluate the integrals given by Z 1 11 = (1 − )4 = 5 30 0 Z 1 12 = (1 − ) 5 = 6 42 0 Z 1 22 = (1 − ) 6 = 7 56 0 Z 1 11 = (1 − )(2)2 = 3 3 0 Z 3 12 = (1 − )(63 ) = 4 10 0 Z ¡ ¢ 3 22 = (1 − ) 32 2 = 5 10 0
Therefore, we have the 2 DOF approximate system characteristic equation given by ⎤⎧ ⎫ ⎧ ⎫ ⎡ 1 3 3 2 1 5 4 2 1 6 − 30 − 42 ⎨ 1 ⎬ ⎨ 0 ⎬ 10 ⎦ ⎣ 3 = 3 4 − 2 1 6 3 5 − 2 1 7 ⎩ ⎭ ⎩ 0 ⎭ 10
42
10
2
56
The characteristic equation is obtained by setting the determinant of the matrix to zero or 1 1 1 1 3 3 ( 3 − 2 5 )( 5 − 2 7 ) − ( 4 − 2 6 )2 = 0 3 30 10 56 10 42 1 2 2 8 1 1 − 10 2 + 4 2 12 = 0 100 600 35280 35280 2 35280 ()2 + = 0 4 − 600 2 100 2 4
Using the quadratic formula, we obtain = 29 4 ∓ 2 22 = (29 4 ∓ 226) 2 = (68 520) 2 r 12 = (26 72) 2
212
sµ
35280 600
¶2
µ
35280 −4 100
¶
If we compare the fundamental frequency here (26) to that found in the previous problem using Rayleigh’s quotient, we see that this approximate result is not as accurate. This has to do with
713 the approximate mode shape we chose in this approximation. Note that the trial function satisfied the left-hand boundary condition, but did satisfy the right-hand boundary condition ( (0) = 0 and 0 () = 0). Even so, we have a reasonable approximate frequency. The second frequency is not as likely to be a good approximation. As discussed in the section, we generally need twice as many terms in the trial function as the number of accurate frequencies.
714
CHAPTER 8 ADVANCED CONTINUOUS MODELS
28. For the tapered beam sketched in Figure 8.17 undergoing transverse vibration, estimate the first two natural frequencies using the Rayleigh-Ritz procedure. Assume a trial function of the form Y() = 1 2 + 2 3 and beam properties, () = () = (1 − ) and () = (12)[(1 − )]3
Tapered beam in longitudinal vibration.
Solution: We assume a mode shape that is of the order 3: () = 1 2 + 2 3 We can follow Example 8.2. max is given by Z 1 3 ³ ´3 00 2 max = ( ) 1− 2 12 0 Z ´3 1 3 ³ 1− (21 + 62 )2 = 2 12 0 ¶ µ 6 3 3 2 3 2 2 1 + 1 2 + 2 = 24 5 5 Then ¶ X µ 2 6 2 21 + 2 = 1 5 =1 ¶ X µ 2 3 6 6 3 max 2 1 + 2 = = 2 2 24 5 5 =1
max 3 = 1 24
where 3 3 2 12 = 12 20 3 2 3 3 22 = 21 = 20 20 11 =
715 The maximum kinetic energy is given by Z ³ ´ 1 ( )2 1− = 2 0 Z ³ ¢2 ´¡ 2 1 1 + 2 3 1− = 2 µ0 6 ¶ 8 7 2 2 1 + 21 2 + 2 ; = 2 30 42 56 ∗ max 5 6 = 1 + 2 1 30 42 ∗ 6 7 max 1 + 2 = 2 42 56 ∗ max
Therefore, 5 30 6 21 = 42
11 =
6 42 7 22 = 56 12 =
We can place these terms above into a characteristic matrix given by ⎡ ⎤⎧ ⎫ ⎧ ⎫ 3 2 5 3 2 2 6 12 − 30 20 − 42 ⎨ 1 ⎬ ⎨ 0 ⎬ ⎣ ⎦ = 3 2 20 − 2 6 42 3 3 20 − 2 7 56 ⎩ ⎭ ⎩ 0 ⎭ 2
Defining 2 = 2 4 the characteristic determinant is µ
2 2 − 12 30
¶µ
¶ µ 2 ¶2 2 2 2 − − − = 0 20 56 20 42 104 − 2732 2 + 5884 = 0
and the natural frequencies are p (2732 )2 − 4(10)5884 212 = 2(20) 2 = 6 825 ± 5 6462 = (1 179 12 471)2 p 1 = 1 0858 2 4 p 2 = 3 5314 2 4 2732 ±
716
CHAPTER 8 ADVANCED CONTINUOUS MODELS
Problems for Section 8.4 — Variables Not Separating 29. In Example 8.3, derive the general solution for 0 () = exp(−). Solution: The complete solution is given in Section 8.3. We need to substitute 0 () and integrate to obtain ( ) = 0 () 0 () + ( ) ¶ µ 3 3 = 0 () 1 − + 2 23 Z ∞ X 1 2 () − ¨0 ( ) sin ( − ) 0 (sinh − sin ) 2 0 =1 where () is given by Equation 8.89, and the remainder of the parameters are defined within the Example.
717
30. Solve the problem of Example 8.3 where the only nonzero boundary condition is 1 () = 2 (0 )2 . Solution: This is a problem where a cantilever beam is subject to a harmonic moment 1 () at the free end. The nonhomogeneous term occurs in the boundary condition, and the goal is to transform this problem so that the nonhomogeneous term occurs in the equation of motion. Boundary Conditions Given the above boundary condition, we proceed with the general transformation given by ( ) = ( ) +
3 X
() ()
=1
Only one function, 1 () is non-zero so that ( ) = ( ) + 1 ()1 () where 1 () = + + 2 + 3 The goal is to find the constant coefficients, , and such that the boundary conditions are homogeneous when the problem is posed in terms of ( ) The boundary conditions in terms of ( ) are given by (0 ) = (0 ) + 1 ()1 (0) = 0 ( ) = ( ) + 1 ()1 () = 1 () ¯ ¯ ¯ ¯¯ 1 ¯¯ ¯¯ = + 1 () =0 ¯= ¯= ¯= ¯ ¯ ¯ 2 ¯¯ 2 ¯¯ 2 1 ¯¯ = + 1 () = 0 2 ¯ 2 ¯ 2 ¯ =0
=0
=0
With (0 ) = ( ) = |= = 2 2 |=0 = 0 we have
We find that
1 (0) = 0 1 () = 1 ¯ 1 ¯¯ = 0 ¯= ¯ 2 1 ¯¯ = 0 2 ¯=0 = 0 + + + 3 = 1 + 2 + 32 = 0 = 0 2
718
CHAPTER 8 ADVANCED CONTINUOUS MODELS
from which we find = 0 3 = 2 = 0 = −
1 23
Equation of Motion So far, we have found that if we use ( ) instead of ( ) the boundary conditions become homogeneous. The two variables are related by ( ) = ( ) + 1 ()1 () where 1 () = (32 − 3 23 ) The next step is to write the equation of motion in terms of ( ) The equation of motion for a beam is given by + ̈ = 0 where we adapted the notations, =
4 ( ) 2 ( ) and ̈ = 4 2
Now, we write µ
3 3 − 3 ̈ = ̈ + ¨1 ()1 () = ̈ + ¨1 () 2 2 3 = + 1 ()1 () = + 1 () 3
¶
so that the equation of motion is now given by ¶ µ 3 3 3 − + ̈ = −1 () 3 − ¨1 () 2 23
(1)
We now have successfully moved the nonhomogeneous term to the equation of motion using ( ) instead of ( ) Instead of solving a problem with a time varying moment on the boundary, we are now solving a problem with a time varying distributed load (right-hand side of the equation above). Let the time varying distributed load be ( ) such that ¶ µ 3 3 3 − ( ) = −1 () 3 − ¨1 () 2 23 where 1 () is the moment applied at the boundary. We proceed using the method of eigenfunction expansion as done for Example 8.4. We expand both the solution ( ) in terms of the eigenfunctions of the related homogeneous problem, () such that ( ) =
∞ X =1
() ()
(2)
719 where () satisfies the eigenvalue problem given by () − 2 () = 0 for 0 ≤ ≤ (0) = () = 0 () = 00 (0) = 0
(3)
for = 1 2 · · · In addition, we have an orthogonality condition given by Z () () = 0 for 6= 0
Note that using the orthogonality condition, we can solve for () and () such that (multiply both sides by () and integrate over the domain) R ( ) () (4) () = 0 R 2 () 0 In Example 8.3, we already found that the eigenfunction is given by 2 ³ ´ () = − sin sinh sinh sin sinh
(8.89)
where the wave number satisfies
tan = − tanh and it is related to the natural frequency via µ 2 ¶14 ≡ Note that the wave number is denoted as . Going back to ( ) in Equation 1 with the right-hand side written as ( ), we replace ( ) in terms of the expansion expressions. Then, ∞ X
()̈ () + () () = ( )
=1
Replacing () with 2 () (Equation 3), we have ∞ X
()̈ () + 2 () () = ( )
=1
∞ X =1
h i () ̈ () + 2 () = ( )
We multiply the both sides by () and integrate over the domain. Using the orthogonality property, we obtain R ( ) () 2 ̈ () + () = 0 R 2 () 0
720
CHAPTER 8 ADVANCED CONTINUOUS MODELS
We call the right-hand side generalized force () The general solution is given by Z 1 () = cos + sin + ( ) sin ( − ) 0
(5)
where and are obtained from the initial conditions such that (with the use of Equation (4)) R
( 0) () R 2 () 0 R ̇ (0) 1 0 ̇ ( 0) () = = R 2 () = (0) =
0
(6) (7)
0
where
( 0) = ( 0) − 1 (0)1 () ̇( 0) = ̇( 0) − ˙1 (0)1 () Without knowing 1 () we cannot obtain the generalized force () nor the convolution solution. With the expressions for the constant coefficients of integrations, and given in Equations (6) and (7), ( ) is given by ∞ X ( ) = () () (2) =1
where () is given in Equation (5) and () is given in Equation 8.89. Now we have all the parameters, and the solution to our original problem is given by ( ) = ( ) + 1 ()1 ()
721
31. Derive Equation 8.96. Solution: Equation 8.96 is given by − ( − U 2 ) + 2 U + ( + ) = 0
(8.96)
The equation is derived using Lagrange’s equation. The potential energy is due to bending strain energy and due to work done by tension, and is given by Z Z 1 1 2 2 + = 0 2 0 2 The kinetic energy is due to the vibrating pipe and to the motion of the fluid, and it is given by (page 834) Z ¤ 1 £ = [U 2 + ( + U )2 ] + 2 2 0
The Lagrangian is given by Z ¤ 1 £ 2 2 − = − = [U 2 + ( + U )2 ] + 2 − 2 0
Lagrange’s equation is
2 − 2
µ
L
¶
+
µ
L
¶
+
µ
L
¶
= 0
where L is the integrand of the Lagrangian. Let us evaluate Lagrange’s equation term by term such that µ ¶ L L 2 = − = − ⇒ 2 ¶ µ L L = U( + U ) − = U( + U )− ⇒ ¶ µ L L = ( + U ) + = ( + U ) + ⇒ Combining all the terms, we obtain the equation of motion. − ( − U 2 ) + 2 U + ( + ) = 0
722
CHAPTER 8 ADVANCED CONTINUOUS MODELS
32. Derive Equation 8.97. Solution: Equation 8.97 is given by ¡ ¢ + 2U + U 2 − 2 = 0
(8.97)
This is the equation of a traveling string reduced from a pipe flow equation. The original equation is given in Equation 8.96 or − ( − U 2 ) + 2 U + ( + ) = 0 where U is the velocity of the fluid, is the density of the fluid, is the density of the pipe, is the cross-sectional area for the fluid and is for the pipe. We assume that pipe is no longer present such that = 0 and = 0 while keeping the tension. In addition, let be the mass of the string per unit length, Then,
where 2 =
−( − U 2 ) + 2U + = 0 ¡ ¢ 2U + U 2 − 2 + = 0
Chapter 9 Single Degree-of-Freedom Vibration: Probabilistic Loadings Problems for Section 9.1 — Introduction 1. Twenty five samples of a steel beam were chosen and tested for the Young’s modulus. Eight had a modulus of = 30 × 106 psi. Two had a modulus of = 29 × 106 psi. Fifteen had a modulus of = 305 × 106 psi. Estimate the following probabilities: (a) Pr{ 295 × 106 psi}, (b) Pr{ 30 × 106 psi}, (c) Pr{ 28 × 106 psi}.
Solution: (a) Pr{ 295 × 106 psi} = (8 + 15)(8 + 2 + 15) = 23 = 092 25 (b) Pr{ 30 × 106 psi} = 1525 = 35 = 06 (c) Pr{ 28 × 106 psi} = 1
723
724
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
Problems for Section 9.2 — Motivation 2. For the following applications, make a list of parameters and forces that are needed to analyze the problem and distinguish between those that can be assumed deterministic and those that must be described probabilistically: (a) airplane design, (b) ship design, (c) turbomachinery design, (d) mechanical watch movements, (e) computer hard drive, (f) automobile internal combustion engine, (g) automobile body.
Solution: In all these applications, most of the geometric dimensions can be assumed deterministic. The forcing is generally random and requires a probabilistic model. (a) for airplane design we need the mass distribution, stiffness distribution, effect of fuel mass and motion inside wing tanks (probabilistic), input forces from engines, atmospheric forces (probabilistic), movement by people inside cabin (probabilistic) (b) for ship design we need the mass distribution, stiffness distribution, effect of fuel mass and motion inside fuel tanks (probabilistic), input forces from engines, ocean and wind forces (probabilistic), movement by people inside the ship (probabilistic) (c) for turbomachinery design we need the mass distribution of the components, stiffness distribution as well, effect of motion of linkages and other components (probabilistic), as well as torques applied to the system (d) mechanical watch movements are primarily kinematic, forces are small, here due to the high precision, it will be necessary to make sure that all the motions are congruent and that the many moving parts do not have intersecting paths (e) computer hard drives are also high precision items, the high speed of drive rotation and the very close read/write heads are in extremely close proximity and very precise knowledge of the mass and stiffness properties are needed, as well as the effects of the air inside the container and how it is compressed as the components move with respect to each other in high speed (f) auto IC engines are very complex machines with many moving parts operating in high precision at fast speeds, at high temperatures, so we need material thermal properties, mass-stiffness-damping properties in order to consider the dynamics, as well as the internal driving forces with every ignition (g) automobile bodies are large with complicated distributed masses, with also complicated distributed stiffness and damping properties, input forces come from aerodynamics forces, from people inside the cabin, and significant forces input from the wheels in all directions, and of course very large impact forces in case of an accident.
725
3. Discuss the possible shortcomings of Miner’s rule. Solution: Miner’s rule assumes that the order in which the various stress amplitudes occurs is irrelevant. It assumes failure occurs at a damage total equal to one. These are both only approximate and have to be generalized to include broader definitions.
726
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
4. How would wind forces be different than ocean wave forces? Solution: Wind forces are different than ocean forces since wind is of a much lower density, and a vibrating structure in the ocean experiences large added masses.
727
5. How can the differences in material properties at different temperatures be considered for applications where the system operates across a large temperature differential? Solution: If a system operates in large ranges of temperatures, then the thermal expansion/contraction properties need to be understood and it becomes necessary to match these properties across diverse materials.
728
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
Problems for Section 9.3 — Random Variables 6. Write the meaning of each of the following mathematical expressions in words only: R∞ (a) () (b) (−∞) (c) (+∞) (d) −∞ () = 1. Solution: (a) This is the cumulative distribution function that quantitatively describes the likely values of the random variable , telling us the probability that the random variable is less than or equal to a particular value (b) This equals zero because the probability that the random variable is less than −∞ is zero. (c) This equals one because all values of the random variable are less than +∞ (d) This integral equals the area under the probability density function. The area under the density function between two values represents the probability that the random variable will be in that range. Therefore, integrating over all values of means that it is certain that the random variable will be in the range −∞ to +∞ This expression is also used to find the unknown constant coefficient in the probability density function.
729
7. What does the fact that (1 ) ≤ (2 ) imply about the values of 1 and 2 ? Why? Solution: It means that 2 ≥ 1 since () is a monotonically increasing function. Therefore, we must be moving to the right on the abscissa, going from 1 to 2
730
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
8. Can the schematics in Figure 9.53 be cumulative distribution functions? Why?
Figure 9.53: Possible cumulative distribution functions.
Solution: Only Figure (a) can be a cumulative distribution function since it satisfies the conditions of being positive and non-decreasing.
731
9. Can the schematics in Figure 9.54 be probability density functions? Why?
Figure 9.54: Possible probability density functions.
Solution: All can be probability density functions except Figure (b). A density function must be zero or greater. It can be a spike or discontinuous.
732
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
10. Discuss the differences in the ways the cumulative distribution and the density depict random variability. Are there situations where one has the advantage over the other? Provide an example for the advantage of each. Solution: The cumulative distribution adds increments of probability when going from smaller parameter values to larger parameter values. The probability density is a normalized histogram that depicts higher probabilities by higher peaks in density; density functions are more commonly used to visualize how the probability is spread, and for calculations, but in some applications, distribution functions are used. For example, the term fragility is used in reliability theory to denote probability of damage.
733
11. If Pr{ ≤ 1 } = 01 and Pr{ ≤ 2 } = 02, sketch the probability density function for random variable . What is Pr{1 ≤ 2 }? Solution: We know that the total area under the density function must equal 1. Therefore, since Pr{ ≥ 2 } = 08 and Pr{ ≤ 1 } = 01 that only leaves a probability area equal to 0.1. This is the probability Pr{1 ≤ ≤ 2 } Or, Pr{1 ≤ ≤ 2 } is simply Pr {1 ≤ ≤ 2 } = Pr{ ≤ 2 } − Pr{ ≤ 1 } = 01
734
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
12. For the following density functions, evaluate the normalization constant and the respective probability, where and are constants: (a) −|| −∞ +∞; Pr{ ≤ 0}
(b) − 4 10; Pr{ ≤ 5}
(c) (2 − 1 ) 0 2; Pr{ ≤ 0}.
Solution: We will use the normalization property of the density function to evaluate the constants. R +∞ −|| R +∞ −|| R +∞ − (a) −∞ = 1 Since −∞ = 2 0 = 1 then = 12 Therefore, Pr{ ≤ 0} = R0 (12) −∞ − = 12
R 10 R 10 (b) 4 −| = 1 Since 4 − = −−10 + −4 = 00 1827 = 1001827 = 54 735 Therefore, R5 Pr{ ≤ 5} = 54 735 4 − = 6337 R2 R2 (c) 0 (2 − 1 ) = 1 Since 1(2 − 0) 0 = 1 = 1 Therefore, Pr{ ≤ 0} = 0
735
13. Can the following be probability densities? (a) 3(2 − 1 ) 0 ≤ ≤ 1 (b) exp 0 ≤ ≤ ∞
(c) ln 1 ≤ ≤ 2 (d) 2 ≤ ≤ 5, where , , and are constants. Solution: We will use the normalization property of the density function to evaluate the constants. R1 R1 (a) 0 3(2 − 1 ) must equal 1, where 2 = 1 and 1 = 0 Therefore, since 3 0 = 3 this cannot be a density function. (b) This cannot be a density function because the normalization integral is unbounded and has a value of infinity. R2 R2 (c) We evaluate to make this a density function. 1 ln must equal 1. Since 1 ln = 0 38629 = 1038629 = 2 5887 The density function is then 25887 ln (d) A constant can be aR density function, a uniform R 5 density function, if the area beneath it equals 5 one. Therefore, since 2 must equal 1, and 2 = 3 then = 13
736
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
14. Let have the probability density () = 2 9, 0 ≤ ≤ 3. Is this a legitimate density function? Find the following probabilities: (a) Pr{1 ≤ ≤ 2} (b) Pr{ ≤ 1} (c) Pr{ ≥ 3}. Solution: To show that this is a legitimate density function, () must be positive for the range of given.RIn addition, the area under () must be 1 Performing the normalization requirement, 3 we obtain 0 (2 9) = 1 Thus, this is a density function. (a) Pr{1 ≤ ≤ 2} =
(b) Pr{ ≤ 1} =
R1 0
R2 1
7 2 9 = 27 = 0 25926
1 2 9 = 27 = 3 7037 × 10−2 (c) Pr{ ≥ 3} = 0
737
15. What is the probability that a continuous random variable takes on a particular value, that is, Pr{ = }?
Solution: For a continuous random variable, the probability that it takes on any one value is zero since that probability is equivalent to the area under the density function.
738
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
16. For the density function () = , 0 ≤ ≤ 1, evaluate so that () becomes a probability density function. Then find Pr{ 03}.
Solution: First normalize the density in order to evaluate Then, since R 03 Pr{ 03} = 2 0 = 009
R1 0
= 12 = 2 Then
739
17. For the density function () = 1 − ||, −1 ≤ ≤ 1, sketch (), show that it is a density, and find Pr{−12 ≤ 13}. Solution: The plot of 1 − || shown below is an isosceles triangle with vertices at = −1 1 and () = 1 The total area equals 1 1
0.8
f (x)
0.6
0.4
0.2
0 -1
-0.5
0
0.5
1
x
Next, Pr{−12 ≤ ≤ 13} =
R0
R 13 (1 − ) + (1 − ) = 65 = 0 90278 72 −12 0
740
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
18. The direction at which the wind strikes a tower is a random variable with density function () = cos , −4 ≤ ≤ 4. The angle is measured from due east. (a) Evaluate . (b) Sketch the density as a function of . (c) Calculate the probability that |4| (d) Calculate the probability that −10 ≤ 10 (e) Sketch the cumulative distribution function.
Solution: Let us first evaluate Since √
R 4
cos = −4
√ √ √ 2 = 1 2 = 22 We sketch in the
range ±4 Now we can sketch 22 cos as shown next.
1.45 1.4 1.35
f ()
1.3 1.25 1.2 1.15 1.1 1.05 1 -0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
The probability that |4| is zero. The probability that −10 ≤ ≤ 10 is Z 10 √ 2 cos = 0 4370 Pr{− ≤ Θ ≤ } = 10 10 −10 2 √
The cumulative distribution function is given by the integral of the density function, 22 sin , for values of in the range ±4
741 Problems for Section 9.4 — Mathematical Expectation 19. Find the mathematical expectation of the random variable with density function () = (2 − 1 ), 1 ≤ ≤ 2 . Also evaluate the mean square value, the variance, and the coefficient of variation. Solution: First we evaluate the normalization constant in the following way. Since 1 ) = 1 we find that = 1 The mean value or mathematical expectation is given by Z 2 Z 2 1 1 () = = {} = = 1 + 2 2 2 1 1 (2 − 1 ) The mean square value is given by Z 2 Z 2 © 2ª 2 () = = 1
1
R 2 1
1(2 −
1 1 1 2 = 21 + 2 1 + 22 (2 − 1 ) 3 3 3
The variance is given by
© ª 1 1 1 1 1 2 = 2 − {}2 = 21 + 2 1 + 22 − ( 1 + 2 )2 3 3 3 2 2 1 2 1 1 2 1 = − 2 1 + 2 = (2 − 1 )2 12 1 6 12 12 The coefficient of variation equals the ratio of the standard deviation to mean, and it is given by √1 (2 − 1 ) 1 2 − 1 = = 112 =√ 1 + 2 2 3 1 + 2 2 1
742
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
20. Find the mathematical expectation of the random variable with density function () = ln , 1 ≤ ≤ 3, where is a constant to be determined. Also evaluate the mean square value, the variance, and the coefficient of variation. R3 R3 Solution: The normalization constant is found by setting 1 ln to 1. Since 1 ln = 3 ln 3 − 2 = 1 (3 ln 3 − 2) Therefore, the probability density function is given by () =
1 ln for 1 ≤ ≤ 3 (3 ln 3 − 2)
The expectation, mean square, variance, and the coefficient of variation are given by Z 3 Z 3 ln () = = 2272 = (3 ln 3 − 2) 1 1 Z 3 Z 3 © 2ª ln 2 () = 2 = 5401 = (3 ln 3 − 2) 1 1 © ª 2 = 2 − 2 = 5401 − (2272)2 = 02399 √ 02399 = = 02156 = 2272
743
21. The following values of a variable are known: 2.3, 3.5, 3.5, 4.9, 3.7, 0.7, 4.1. Find the average value, the mean square value, the variance, and the coefficient of variation. Solution: The expectation, mean square, variance, and the coefficient of variation are given by 23 + 2 × 35 + 49 + 37 + 07 + 41 = 3 2429 7 © ª 232 + 2 × 352 + 492 + 372 + 072 + 412 = 121130 2 = 7 © ª 2 = 2 − 2 = 12113 − (32429)2 = 15966 √ 15966 = 038969 = = 32429 =
744
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
22. Compare the mean values of the following two random variables. For the first random variable we have the data: 3.0, 3.5, 4.0, 4.5, 5.0, 5.5, and 6.0. For the second random variable we have the probability density () =
1 3 ≤ ≤ 6 (2 − 1 )
Do we expect them to have the same mean values? Solution: For the discrete random variable, the mean value is (3 + 35 + 4 + R45 + 5 + 55 + 6)7 = 6 4 5 For the uniformly distributed continuous variable, the mean is given by 3 (6 − 3) = 4 5 Both models of the random variable have the same mean. This is only the case because the discrete case has evenly spaced values along the same range as does the continuous variable. Any slight change will result in different means.
745 Problems for Section 9.5 — Useful Probability Densities 23. (a) The stiffness property of a new material is established by initial testing. The preliminary results show a uniform scatter of data in the range 9 lb/in ≤ ≤ 11 lb/in. There is no additional information. Calculate the mean value, standard deviation, coefficient of variation, and the probability that the stiffness is greater than 95 lb/in. (b) The above spread of data is found to be too large, and so additional testing is carried out. A plot of the data shows the density in Figure 9.55. Calculate the same values as before and draw conclusions.
Figure 9.55: Density of stiffness data.
Solution: The uniform density is a constant with value that is the inverse of the range. In this case, the density is 12 over the range, 9 lb/in ≤ ≤ 11 lb/in, such that () =
1 for 9 ≤ ≤ 11 2
For the preliminary data, we calculate the mean, standard deviation, and the coefficient of variation to be Z 11 Z 11 1 = () = = 10 lb/in 2 9 9 sZ q 11 2 2 = { } − = 2 () − 102 = 0577 lb/in 9
=
0577 = 00577 = 10
The probability that the stiffness is greater than 9.5 lb/in is Pr { ≥ 95} =
Z 11
1 = 0750 95 2
746
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
For the new data, we have a modified density function that is no longer uniform. We need the equation of the new density, which is just a straight line between = 9 and = 11, and is given by 1 7 () = − 6 6 The mean value, standard deviation, coefficient of variation, and the probability that the stiffness is greater than 95 lb that correspond to the new data are ¶ Z 11 Z 11 µ 7 1 = − = 10 1 lb/in () = 6 6 9 9 s ¶2 Z 11 µ q 1 7 2 2 2 − − 101112 = 0568 lb/in = { } − = 6 6 9 0568 = 00562 = = 101 ¶ Z 11 µ 1 7 − = 0 813 Pr { ≥ 95} = 6 6 95
747
24. Suppose a random variable has a probability density function with a very broad range of values. Even though the density function is well defined, what options does a designer have when dealing with such a broad parameter variability? Solution: For cases with very broad density functions the designer can determine whether the operating range is in a narrow range and in that case the designer can still work with that variable. Otherwise, it may be necessary to perform tests in order to revise and update the density to a narrower range to get more data on the parameter.
748
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
25. Several steel beams delivered to the factory are to be used as columns. The test data for the lot from which these beams were selected are known to have a Young’s modulus in the range 29 × 106 psi ≤ ≤ 31 × 106 psi, with all values equally likely. If the columns have a cross sectional area of 144 in2 and are loaded with 100 000 lb, what are the possible stress and strain ranges? Solution: Since all values are equally likely, the density is uniform. The stress equals the force per unit area; both parameters are deterministic, and therefore so is the stress, = 100000144 = 694 44 psi The strain, on the other hand, depends on a random variable. = and 69444 69444 ≤≤ 6 31 × 10 29 × 106 or 2 2401 × 10−5 ≤ ≤ 2 3946 × 10−5
749
26. Show that the discrete Poisson-distributed random variable with discrete density Pr{ = } =
− = 0 1 2 !
has an expectation {} = .
Solution: Use the definition of the mathematical expectation to show {} =
∞ X
−
! =
=0 ∞ X
=
=1
∞ X =1
−
( − 1)! =
− −1 ( − 1)! =
where we used the expansion for exponential: =
∞ X
∞ X =1
− −1 ( − 1)!
− ! = − =
=0
P∞
=1 !
750
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
27. If a mechanical component fails according to the exponential distribution with a mean value of 5000 h, what is the probability that the component will fail by 1000 h? How many additional hours before this probability is doubled? Sketch the density function. Solution: For the exponential, = R1 mean. Then,R = 15000 The probability that the compo1000 1000 nent will fail by 1000 h is given by 0 − = 0 (15000)−5000 = 018127 = 18127% To find the additional hours before this probability is doubled, it is easier to use the cumulative distribution. () = 1 − − = 2(018127) Solve this for = ln(1 − 2(018127))(−15000) = 2251 3 h, or an additional 22513 − 1000 = 1251 3 h
751
28. Derive Equations 9.13 and 9.14 showing that {} = and { 2 } = 2 + 2 Solution: We would like to derive the mean and variance of the Gaussian density. The mean is defined by ¸ ∙ Z ∞ 1 − 2 √ exp − ( {} = ) 2 −∞ 2 Z ∞ 1 √ ( + ) exp(− 2 2) = = 2 −∞ This expression wasR evaluated using MAPLE, but it is a standard integral. It is fairly easy to see ∞ that the first part, −∞ exp(− 2 2) integrates to zero, Z ∞ ¯∞ exp(− 2 2) = − exp(− 2 2)¯−∞ = 0 −∞
R∞ The second part, −∞ exp(− 2 2) is integrated in the polar domain. Here is the procedure for integrating such an integral. First, we recognize that we can write the integral as sµZ ¶ Z Z =
∞
exp(− 2 2) =
−∞
∞
−∞
exp(− 2 2) ×
∞
exp(−2 2)
−∞
where and are dummy variables. We then write 2 + 2 = 2 and = so that sZ Z ∙ ¸ ∞ ∞ 1 2 2 exp − ( + ) = 2 −∞ −∞ sZ Z ¸ ∙ 2 ∞ 1 2 = exp − 2 0 0 s ¸¯∞ ∙ 1 2 ¯¯ = − (2) exp − ¯ 2 0 √ = 2 Finally, we find that {} =
Z ∞
1 √ ( + ) exp(− 2 2) = 0 + = 2 −∞
752
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
Similarly for the root mean square, we have Z ∞ 2 1 − 2 2 √ exp(− ( { } = ) ) 2 −∞ 2 Z ∞ 1 √ ( + )2 exp(− 2 2) = 2 + 2 = 2 −∞
753
29. For the Gaussian random variable , numerically estimate the probability Pr{ 0} for (a) mean value = 10 and standard deviation = 10, and (b) mean value = 10 and standard deviation = 1. Can either of these be used as models for a positive definite variable? Explain. Solution: The Gaussian probability density function is given by ¸ ∙ 1 1 − 2 ) () = √ exp − ( 2 2 (a) For the given values of = 10 and = 10 the probability, Pr{ 0}, is given by ¸ ∙ Z 0 Z 0 1 1 − 10 2 √ exp − ( Pr{ 0} = ) = 0 15866 () = 2 10 −∞ −∞ 10 2 Either numerical integration or normal table is needed. As a check (again either numerically integrating or using a normal table), the remaining probability is found to be ¸ ∙ Z ∞ 1 1 − 10 2 √ exp − ( Pr{ 0} = ) = 0 84134 2 10 10 2 0
and the sum of the two equals one, as it should since the complete area under the density is evaluated. These integrals were computed numerically using MAPLE (b) For this part the standard deviation is much smaller. Therefore, we expect that the spread of values is much smaller, and much less area is in the negative region. ¸ ∙ Z 0 1 1 − 10 2 √ exp − ( Pr{ 0} = ) = 7 6199 × 10−24 2 1 −∞ 1 2 This number is effectively zero. Again as a check, let’s find the remaining probability. Z ∞ 1 1 − 10 2 √ exp(− ( Pr{ 0} = ) ) = 10 2 1 1 2 0
The numerical scheme returns an integral value 1.0. Round-off error effectively ignores values for ≤ 0 Therefore, in the second case, this model can be used for a positive definite variable. In the first case, almost 16% of the possible values are negative, and, therefore, this is an inappropriate model for a positive definite variable.
754
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
30. Compare the lognormal and the Rayleigh probability density functions, both of which govern positive-definite variables. Compare the shapes of the plotted functions and how they shift as parameters are varied. Discuss the differences, if any, due to the fact that one is a function of two parameters and the other a function of one parameter. Solution: Comparing the lognormal and Rayleigh probability density functions: (i) the Rayleigh has most of its probability density centered about its lower values while the lognormal shifts to larger values, (ii) the lognormal can be fit using two parameters while the Rayleigh depends on one parameter value, perhaps in some instances making it “stiffer” and more difficult to fit to data,and (iii) the lognormal has “Gaussian” properties allowing us to use Gaussian tables, but the Rayleigh is a simpler function.
755 Problems for Section 9.6 — Two Random Variables 31. Given the joint density function 1 (2 − 1 )(2 − 1 ) 2 ≤ ≤ 4 and 1 ≤ ≤ 3
( ) =
evaluate {}, { }, ( ), and . Note that ( ) = ()(). Solution: Denote 1 and 2 as the lower and upper limit of the range for We can easily evaluate the necessary values since this is a two-dimensional uniform density. Z 4Z 3 1 {} = = 30 (4 − 2)(3 − 1) 2 1 Z 4Z 3 1 { } = = 20 (4 − 2)(3 − 1) 2 1 Z 4Z 3 1 { } = = 60 (4 − 2)(3 − 1) 2 1 The covariance equals ( ) = { } − {}{ } = 6 − 2 × 3 = 0 The correlation coefficient is given by = ( ) and since the covariance equals zero, then so does the correlation coefficient.
756
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
32. and have a joint probability density given by ( ) = −(+) ≥ 0 ≥ 0 Find Pr{ ≥ ≥ 2} and sketch the region in the plane that defines the region of integration. Solution: This is a double integration. Pr{ ≥ ≥ 2} =
Z ∞Z ∞ 2
−(+) =
1 exp (−14) 2
The area of integration is a triangular region in the plane bounded by the lines = = 2 that goes to infinity as shown below.
Region of integration.
757
33. In Example 9.11 verify Equation 9.23, Var{} = Var{} + Var( } − 2
Solution: Begin with Var{} = { 2 } − 2 It is given that = − Then, Var{ − } = {( − )2 } − ( − )2 Expanding the squares, we obtain ¡ ¢ Var{} = { 2 − 2 + 2 } − 2 − 2 + 2 ¢ ¡ © ª ¢ ¡ © ª = 2 − 2 + 2 − 2 − 2 { } + 2
Using the relations, Var{} = { 2 }−2 Var{ } = { 2 }−2 , and { } = + , we obtain Var{} = Var{} + Var{ } − 2
758
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
Problems for Section 9.7 — Random Processes 34. Strain gages are placed at two locations on a wing that is being tested in a wind tunnel. The first gage is near the fixed base of the wing and the other gage is near the free wing tip. Two tests are performed: (a) the wing vibration is recorded for low velocity laminar flow and (b) for high velocity turbulent flow. For each of these tests, discuss the following: (i) Describe and compare the time histories you would expect to see recorded by each strain gage; (ii) Discuss the type of cross correlation 1 2 ( ) you would expect between 1 and 2 , the locations of the gages. Solution: For laminar flow, we would expect very slow variations in flow velocities at both the tip and the base. If the flow profile is essentially the same along the length of the wing, meaning that the flow speed is about constant, then we may expect a high degree of correlation along the beam. Where there will be less correlation will be at the boundaries. The free tip will vibrate with larger motion than points near the fixed end. A narrow band process may be a good model. For turbulent flow past the wing, we expect large variations in flow velocity (speed and direction) between two points in the flow, even if these points are close to each other. This means that the wing will experience a significant variety of forces, even at points on the beam that are near to each other. Therefore, we expect low correlation between points more than a “turbulence scale distance” apart. A broad band model is necessary.
759
35. In Figure 9.56, discuss whether the processes are stationary.
Figure 9.56: Random processes.
Solution: In the three time histories shown, the first is oscillatory, perhaps in several frequencies. The general character of the motion seems to be regular and steady state, and likely stationary. The periodic large fluctuations appear in the second figure. Still, this process or time history appears regular and steady state, and likely stationary. The third time history is nonstationary. Its motion begins at = 0 and appears transient.
760
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
36. From the expressions for wind and ocean wave power spectra, which parameters appear important for a proper characterization of the energy distribution? Solution: In both wind and ocean wave power spectra, wind speed is a critical parameter in establishing the energy distribution. For the ocean wave, if there is also a current, then its speed and direction is also a critical parameter. Other important parameters that are more difficult to discern are the medium properties, e.g., density of fluid.
761
37. Referring to the plot of the Pierson-Moskowitz ocean wave height spectrum (Figure 9.28), discuss the physical relationship between wind speed and wave height. Solution: From Figure 9.28, we see that the relationship between wind speed and wave height is on a general level, intuitive. The higher the wind speed the more energy is in the system, as shown by the area under the spectral density. Further, there is a larger density area with higher wave amplitudes, but the frequency content shifts to lower frequencies in all wind speed regimes.
762
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
38. For the Pierson-Moskowitz power spectrum, evaluate numerically the areas under the first 20 frequency bands of width ∆ = 01 rad/s for (a) = 10 m/s, and (b) = 20 m/s. What conclusions can be drawn regarding the effect of wind speed on the frequency distribution of wind energy for this spectrum? Could similar conclusions have been determined by directly studying the equation for the spectrum?
Solution: The integrations will be done numerically. The Pierson-Moskowitz spectrum is given by ¸ ∙ 81 × 10−3 × 982 98 4 () = ) m2 s. exp −073( 5 Figure next shows the spectra when = 10 m/s and 20 m/s.
16 V = 10 m/s V = 20 m/s
14 12
S ( )
10 8 6 4 2 0
0
0.5
1
1.5
2
Pierson-Moskowitz Spectra for = 10 and 20 m/s.
The integral for the first frequency band for = 10 m/s is: Z 01 0
81 × 10−3 × 982 98 4 exp(−073( ) ) = 1 7022 × 10−2925 5 10
The comparable integral for = 20 m/s is Z 01 81 × 10−3 × 982 98 4 ) ) = 7 9509 × 10−183 exp(−073( 5 20 0 While both are extremely small numbers and effectively zeros, the higher velocity increases the energy (area under power spectrum) by an order of magnitude. The table that follows provides
763 the other areas. Note the frequency bands that contain the most energy. These are going to be of concern to structural designers as possible sources of resonant behavior. Also note that the peak frequency ranges are different.
1 − 2 = 10 m/s
= 20 m/s
1 − 2 = 10 m/s
= 20 m/s
0.1-0.2
6 5736 × 10−184
1 7458 × 10−11
1.0-1.1
3 5048 × 10−2
5 9503 × 10−2
0.2-0.3
2 2822 × 10−37
2 5611 × 10−2
1.1-1.2
2 6394 × 10−2
3 8099 × 10−2
0.3-0.4
1 0911 × 10−12
0 86738
1.2-1.3
1 9422 × 10−2
25250 × 10−2
0.4-0.5
6 0518 × 10−6
1 4639 MAX
1.3-1.4
1 4225 × 10−2
1 7245 × 10−2
0.5-0.6
1 5946 × 10−3
0 98308
1.4-1.5
1 0466 × 10−2
1 2092 × 10−2
0.6-0.7
1 5887 × 10−2
053836
1.5-1.6
7 769 × 10−3
8 6764 × 10−3
0.7-0.8
3 8324 × 10−2
0 29176
1.6-1.7
5 8305 × 10−3
6 3537 × 10−3
0.8-0.9
4 7691 × 10−2 MAX 0 16412
1.7-1.8
4 4268 × 10−3
4 7376 × 10−3
0.9-1.0
4 3806 × 10−2
1.8-1.9
3 4003 × 10−3
35900 × 10−3
9 6668 × 10−2
764
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
39. Convert the power spectra of Figure 9.57 to one-sided equivalent spectra that are functions of cyclic frequency (Hz).
Figure 5.34: Broad-band and narrow-band spectra.
Solution: The relation between one-sided spectrum ( ) and the two-sided spectrum () is given by the relation () = 4() Then, the solutions for part (a) and (b) are (a) () = 40 for 0 ≤ ≤ (2) (b) () = 40 for 1 (2) ≤ ≤ 2 (2)
765
40. Compare Equations 9.33 and 9.34. In particular, discuss their respective units and how the energy distribution differs. Solution: In comparing equations 9.33 for the wave height spectrum and 9.34 for the wind velocity spectrum, the units reflect the particular physical process: [units2 -s], for displacement we have m2 s and for velocity we have (ft/s)2 s or ft2 /s, both spectra are empirical and have built-in assumptions and limitations, the wave height spectrum is a function of velocity, as expected.
766
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
41. Look up examples of the El Centro earthquake spectrum online. Is there a spectral density? How is the information presented? Solution: The website https://sites.google.com/site /kolukulasivasrinivas /matlab-fea/seismicanalysis /response-spectra(5 February 2017) provides a set of spectra (displacement, velocity and acceleration response spectra) as well as a discussion.
767 Problems for Section 9.8 — Random Vibration 42. Beginning with Equation 9.46, (1 ) =
Z ∞
−∞
( 1 ) (1 − 1 ) 1
derive Equation 9.49, ( ) =
Z ∞Z ∞ −∞
−∞
()() ( + − )
Solution: The procedure is given in the text.
768
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
43. Following Example 9.12, we would like to examine the sensitivity of the output spectral density () and the mean-square response {()2 } to various combinations of parameter values, where 0 is the input spectrum. For the following cases, plot |()|2 and (), and evaluate {()2 } for unit mass, 0 30 rad/s and 0 ft2 s: (a) = 01 rad/s and = 01 (b) = 05 rad/s and = 01 (c) = 10 rad/s and = 01 (d) = 20 rad/s and = 01 (e) = 01 rad/s and = 05 (g) = 05 rad/s and = 05 (h) = 10 rad/s and = 05 (i) = 20 rad/s and = 05 Compare the results and draw general conclusions. Solution: The relevant equations are the following: 1 ( 2 − 2 )2 + (2 )2 0 () = 2 2 2 ( − ) + (2 )2 0 {()2 } = |()|2 =
Since |()|2 is valid for positive we only consider this range of frequencies. The mean square response is written as 0 1 1 0 {()2 } = 2 = 2 () () 2 3 For a unit mass, the mean square response is {()2 } = 0 23 The expressions for |()|2 and ()0 are identical, and they are plotted for various values of and next. The mean square response for each case is shown below. = 01 rad/s = 05 rad/s = 10 rad/s = 20 rad/s = 01 1571 × 104 = 05 7854 × 103
125.7
15.71
1.964
62.83
7.854
0.9817
769 x 10
5
600 2
n= 0.1
2
|H (i )|
|H (i )|
2
3
1 0
0
0.05
0.1
0.15
Frequency,
0
0.5
1
Frequency,
2 n= 1
n= 2
1.5 2
20
|H (i )|
|H (i )|
2
30
10 0
200 0
0.2
n= 0.5
400
1 0.5
0
0.5
1
1.5
Frequency,
2
0
0
1
2
Frequency,
3
4
Figure 9.1: |()|2 and ()0
44. Using the fundamental relation between input and output spectra, find the response spectrum for the oscillator governed by ̈ + 2 ̇ + 2 =
1 ()
with = 10 kg, = 015, = 5 N/m, where the spectrum for () is (a) the PiersonMoskowitz (Equation 9.33), with wind speed = 25 km/h, (b) the wind spectrum (Equation 9.34) where ̄10 = 20 mi/h, and = 001. For each, numerically evaluate the mean square response for the following parameters. Plot each response spectrum.
Solution: There is a typographical error. The unit of mass is supposed to be kg. For a SDOF oscillator whose equation of motion is given by ³ ´ ̈ + 2 ̇ + 2 = () the fundamental input-output relation is given by () = |()|2 ()
770
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES 0.5
0.3
2
|H (i )| (m/N)
0.4
0.2
0.1
0
0
0.5
1
Frequency, (rad/s)
1.5
2
1 1 Figure 9.2: |()|2 = 100 0 25− 955 2 +4
where 1 1 2 (2 − 2 )2 + (2 )2 1 1 1 1 p = = 100 (12 − 2 )2 + (2 (015) 12)2 100 0 25 − 0955 2 + 4
|()|2 =
The frequency response function is plotted below. The mean square response is given by 2
{ } = (a) The Pierson-Moskowitz spectrum is
Z ∞
()
−∞
81 × 10−3 × 2 4 ) ) ≥ 0 exp(−073( 5 Let the force of the wave be proportional to the wave elevation with a proportionality constant of N/m so that = 2 - - () =
which now has the proper units of N2 s. For = 25 km/h = 6 9444 m/s, the force spectrum becomes ¶ µ −3 × 982 98 2 81 × 10 4 () = ) exp −073( 5 6 9444 ¶ µ 2 8953 2 0 7779 exp − = 5 4
771 The response spectrum, () is obtained by multiplying the input spectrum by the frequency response function, |()| () = |()|2 () µ ¶µ ¶¶ µ 1 1 2 8953 2 0 7779 = exp − 100 0 25 − 09552 + 4 5 4 The input and the response spectra are shown below.
1.5
x 10
-3
SXX ( ) /k (m /s) (N/m)
2
0.1
0.04 0.02 0
1
2
0.06
2
2 SP-M ( ) (m /s)
0.08
0
1
2
3
Frequency, (rad/s)
4
0.5
0
0
1
2
3
Frequency, (rad/s)
4
(a) Pierson-Moskowitz spectrum, (b) The response spectrum.
The mean square response is then Z ∞ 2 { } = () 0 ¶¶ µ Z ∞µ 1 1 2 8953 2 0 77792 × exp − = 100 0 25 − 0 955 2 + 4 5 4 0 = 000063952 m2 (b) A wind spectrum for horizontal velocity is () =
4̄102
̄10
µ
ft2 s
¶
¡ ¢2 (2 + ̄10 )56 40002933 = 4 (001) (2933)2 (2 + ( · 40002933)2 )56 46928 2 = 5 ft s. (18600 2 + 2) 6 43598 2 = 5 m 2 (18600 + 2) 6
772
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
where = 001, = 4000 ft, and ̄10 = 2933 ft/s are substituted in. In terms of angular velocity, () = 2 ( ) Let the force of the wind be proportional to the horizontal velocity with the proportionality constant N/m so that () = 2 2 () The response spectrum is () = |()|2 () ⎞ ⎛ ! Ã 1 43598 1 ⎟ ⎜ 2 p 2 ³ = ⎝ 5 ⎠ ´ 2 2 2 ¡ ¢ 100 (12 − ) + (2 (015) 12) 6 2 18600 2 +2
The wind horizontal velocity and the response spectra are plotted below.
(a)
500
0
0.5
1
Frequency, (rad/s)
1.5
50
2
1000
60
40
2
SXX ( ) /k (m /s) (N/m)
2
S ( ) (m /s)
1500
0
(b)
70 2
2000
30 20 10 0
0
0.5
1
Frequency, (rad/s)
(a) The wind spectrum, (b) The response spectrum.
1.5
773
45. Consider Figure 9.37 and describe in your own words the fundamental result depicted there graphically. Solution: Figure 9.37 demonstrates two concepts. First is the fundamental principle for stochastic dynamics of linear systems that the output spectrum equals the input spectrum times the system function. Second is the computational ease and physical justification for the white noise assumption. The product of curves shows that the system function for a lightly damped system filters the input spectrum about the system natural frequency, thus making irrelevant, mostly, the parts of the white noise input far above and far below that natural frequency, thus yielding accurate practical results.
774
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
Problems for Section 9.9 — Stochastic Response of a Linear MDOF System 46. Reduce the methodology of this section to the specific case of a two degree-of-freedom system. Solution: For a specific two DOF system, we begin with the following: 1 () = 11 1 () + 12 2 () 2 () = 21 1 () + 22 2 () Then we use these equations in all the subsequent derivations. For example, we will have expressions such as 1 1 ( ) and 2 2 ( ); 1 1 () and 2 2 () In this way, the expressions are specified for a two DOF system only.
775
47. Following the methodology of this section, study a forced vibrating system with the following property matrices, ⎡ ⎤ ⎡ ⎤ [] = ⎣
1 0 0 1
⎦
[] = ⎣
2 −1
−1 1
⎦
where each component force vector {1 () 2 ()} is white noise. Transform the coupled equations of motion into modal coordinates and proceed with an analysis of this two degreeof-freedom system. Discuss any assumptions needed in progressing through the steps of the analysis. Solution: This is an undamped system. The methodology developed in Section 9.9 for the modal analysis of a damped structure was based on the general modal equation given by ̈ + 2 ̇ + 2 = ()
= 1 2
In our case = 2 In addition, we need to relate the modal forces 12 to the physical forces that are given here as white noise. We need to begin with the physical equations, utilizing [] and [] along with the force vector. The modal analysis procedure requires finding the modes and frequencies, then normalizing the modes. This leads to the above equation in the modal coordinates. Our modal forces are () = {} { ()} where { () = {1 () 2 ()} . For a two-DOF system, then = 1 1 + 2 2 , = 1 2 The procedure then follows essentially that given in Section 9.9, except that our system is undamped. Therefore, where simplifications were made assuming “lightly damped” system, here, this is exact. This will lead to X 2 | ()|2 () () =
776
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
Problems for Section 9.10 — Lunar Seismic Structural Analysis 48. Refer to Figure 9.45 and the equation that it represents to discuss the physical nature of that description. How do variations in the parameters that define the power spectrum affect the velocity spectrum? Solution: Figure 9.45 shows a power spectrum in three dimensions. The spectrum is a function of time and frequency. Thus, it is a nonstationary response, and the equation that defines this curve is a function of frequency and time.
777
49. Compare the seismic risk for a surface lunar structure as compared to that for a buried lunar structure. Solution: A buried lunar structure can be more susceptible to seismic activity because the complete structure will experience seismic input, whereas a surface structure experiences only input from its base.
778
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
Problems for Section 9.11 — Random Vibration of Continuous Structures 50. Describe three applications where a continuous structure is loaded by a random force. Solution: Wind on a skyscraper; turbulent wind on an aircraft; ocean waves on a ship; earthquake loading on a structure.
779
51. Suppose a simply supported beam is forced by a random load with mean value and autocorrelation ( ) = exp(− ), where is a constant. Solve for the mean, autocorrelation, and spectral density of the response.
Solution: We make the following assumptions: no damping, the modal approach will be used, and given the above expressions that stationarity is acceptable. Let us write a general expression for the solution () () =
∞ X
̂ ()
µZ µZ 0
=1
0
¶
( − ) ̂ () ( )
¶
where ̂ () is the normalized mode shape and ( ) is the impulse response function given by r 2 ̂ = sin 1 sin (1) () = where the natural frequencies are = Mean of the Total Response
µ
¶2 r
(2)
The mean value of the response given by {( )} =
∞ X
̂ ()
=1
Z Z 0
If { ( − )} = we have {( )} =
∞ X =1
̂ ()
0
{ ( − )} ̂ () ( )
µZ µZ 0
0
¶
̂ () ( )
¶
where can come out of the integral if it is constant in space as well as in time. Then, ! ÃZ ÃZ r ! ∞ r X 2 2 1 {( )} = sin sin sin 0 0 =1 ∙µ ³ ¸ ∞ ´¶ 1 X 2 1 − (−1) = sin (1 − cos ) =1 2 where is given in Equation (2). It is expected that is a function of time even though the forcing has a constant mean because we took the total solution (to random forcing and zero initial
780
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
conditions) rather than the particular solution. The part of the response responding at the natural frequency remains. The particular solution, of course, will have a constant mean. Autocorrelation of the Total Response The autocorrelation is given by (page 798) (1 2 ; ) ≡ { (1 ) (2 + )} (∞ Z Z X ( 1 − ) ̂ () 1 () = ̂ () ×
=
0
=1 ∞ X
̂ ()
=1 ∞ ∞ XX
0
Z + Z 0
0
̂ (1 ) ̂ (2 )
( 2 + − ) ̂ ( 2 ) 2 ()
Z + Z ∙Z Z 0
=1 =1
)
0
0
0
{ ( 1 − ) ( 2 + − )}
×̂ ( 1 ) ̂ ( 2 ) 1 2 ] () () ¸ ∙Z Z ∞ ∞ X X ̂ (1 ) ̂ (2 ) ̂ ( 1 ) ̂ ( 2 ) 1 2 = =1 =1 Z + Z
×
0
0
0
0
(1 2 ; − + ) () ()
It is given that (1 2 ; ) = exp(− ) Then, (1 2 ; − + ) = exp(− ( − + )) Substitute this into the expression above, and we obtain ¸ ∙Z Z ∞ X ∞ X (1 2 ; ) = ̂ (1 ) ̂ (2 ) ̂ ( 1 ) ̂ ( 2 ) 1 2 =1 =1 Z + Z
×
0
0
0
0
exp(− ( − + ) () ()
Substituting Equations (1)-(2), we obtain ∙ ³ Z Z ´¸2 2 1 − (−1) ̂ ( 1 ) ̂ ( 2 ) 1 2 = 0 0 Z + Z exp(− ( − + ) () () = Long Expression 0
0
(1 2 ; ) =
∞ X ∞ X
̂ (1 ) ̂ (2 ) =1 =1 Z + Z
× Response Spectral Density
0
0
∙Z Z 0
̂ ( 1 ) ̂ ( 2 ) 1 2
0
exp(− ( − + ) () ()
¸
781 To evaluate the spectral density of the response, we need first the spectral density of the loading () For this expression, take the Fourier transform of the correlation function ( ) This is then substituted in () which is then substituted into the response spectral density (1 2 ) Another option is to take the Fourier transform of (1 2 ; ) to obtain (1 2 ) The solution is omitted here because of the difficulty in obtaining the explicit expression for (1 2 ; )
782
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
52. For Problem 51, suppose calculations need to be simplified. Can the autocorrelation of the force be replaced by a triangular spike around = 0? Try this replacement, justify the simplification, and solve. Compare with the results of Problem 51. Solution: In principle, this substitution for ( ) can be made in the calculations of the previous problem. This would simplify the calculations of integrals such as Z Z 2 ( ) (1 ) (2 )1 2 ( ) = 0
0
With such an approximation, we need to be certain that we have captured the essence of the correlation structure. But also, we need to be careful that when we transform to the frequency domain of the spectral density, that is () we have captured here the frequency content of the original random process.
783
53. What complications occur in the derivations of this section if the force is not assumed to be stationary? Solution: If the force is not stationary, then the mean value is no longer a constant, but a function of time, and the autocorrelations are not functions of time difference ( ) but rather functions of specific times (1 2 )
784
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
54. Derive Equation 9.74. Solution: Equation 9.74 is given by 1 XX ∗ (1 2 ; ) = 2 () () ()̂ (1 )̂ (2 ) =1 =1 ∞
∞
(11.47)
Start with the equation below given by
(1 2 ; ) =
∞ ∞ X X
( )̂ (1 )̂ (2 )
=1 =1
The spectral density is a Fourier transform of (1 2 ; ) Take a Fourier transform to obtain Z ∞ () = (1 2 ; )− −∞
=
∞ ∙Z ∞ ∞ X X =1 =1
−∞
−
( )
¸
̂ (1 )̂ (2 )
Let us now evaluate the Fourier transform of the autocorrelation, ( ) The autocorrelation is given by Z ∞Z ∞ 1 ( ) = 2 ( + − ) () () −∞ −∞ The Fourier transform of the autocorrelation is Z ∞ () ≡ ( )− ∙Z ∞ Z ∞ ¸ Z−∞ ∞ −(−+) 1 = () () () (11.43) 2 −∞ −∞ −∞ where the integration is performed over . Recall that the Fourier transform of the impulse response function is the frequency response function such that Z ∞ () ≡ ∗ () Z ∞−∞ − () ≡ () −∞
Then,
() = We recognize
Finally,
µZ ∞
−
−∞
¶ µZ ∞ ¶ µZ ∞ ¶ 1 () × () × () 2 −∞ −∞
Z ∞
−
−∞
(1 2 ; ) =
1 () = () 2
1 XX ∗ () () ()̂ (1 )̂ (2 ) 2 =1 =1 ∞
∞
785 Problems for Section 9.12 — Monte Carlo Simulation 55. Generate 100 random numbers that are governed by an exponential density function. Solution: We first need 100 random numbers that are distributed uniformly between 0 and 1. This is done using the linear congruential generator such as +1 = ( + ) (mod ) For example, let = 421, = 1663 and = 7875 Let us start with 1 = 100 Then, 2 = 4388 3 = 6261 The random numbers are normalized by to give uniformly distributed random numbers between 0 and 1 such that the first three numbers are 100 4388 6261 ··· 7875 7875 7875 We will call these uniform random numbers The 100 uniform random numbers between 0 and 1 are shown in the table below. 0.0127 0.6578 0.6362 0.9479 0.5930 0.5714 0.8832 0.5283 0.5067 0.8184 0.5572 0.1356 0.0474 0.2924 0.8709 0.7826 0.0277 0.6061 0.5178 0.7629 0.7950 0.3068 0.1519 0.3303 0.8420 0.6871 0.8655 0.3773 0.2223 0.4008 0.9262 0.3713 0.1497 0.2615 0.7065 0.4850 0.5967 0.0418 0.8202 0.9319 0.1507 0.5291 0.2409 0.2860 0.6644 0.3761 0.4212 0.7996 0.5114 0.5564 0.6686 0.9803 0.6254 0.6038 0.9156 0.5606 0.5390 0.8508 0.4959 0.4743 0.6797 0.9248 0.5032 0.4150 0.6601 0.2385 0.1502 0.3953 0.9737 0.8855 0.3843 0.5627 0.0744 0.9195 0.0979 0.6097 0.4547 0.6331 0.1449 0.9900 0.9821 0.0938 0.5389 0.3173 0.4291 0.8742 0.6526 0.7643 0.2094 0.9878 0.6733 0.7183 0.0968 0.8085 0.8536 0.2320 0.9437 0.9888 0.3672 0.0790 Once we have the uniform random numbers between 0 and 1, we can now find the random numbers distributed according to other density function. We will use the inverse transform method for this problem.. The probability density function for the exponential density is given by () = exp (− ( − )) The cumulative distribution, and its inverse are given by Z () = () = 1 − exp (− ( − )) 1 = − (ln (1 − ) + )
(1)
786
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
We substitute the uniform random numbers in the place of and the resulting will be distributed according to the exponential distribution in Equation (1) so that 1 = − (ln (1 − ) + ) The 100 random numbers that corresponds to exponential distribution (with = 1 and = 0) are shown in the table below. 0.0128 1.0723 1.0111 2.9553 0.8990 0.8473 0.8473 0.7513 0.7066 1.7060 0.8147 0.1457 0.0485 0.3459 2.0468 1.5260 0.0281 0.9316 0.7295 1.4394 1.5850 0.3664 0.1647 0.4009 1.8454 1.1619 2.0064 0.4736 0.2515 0.5121 2.6067 0.4641 0.1622 0.3031 1.2260 0.6635 0.9081 0.0427 1.7159 2.6873 0.1634 0.7532 0.2756 0.3368 1.0918 0.4718 0.5468 1.6075 0.7161 0.8129 1.1043 3.9280 0.9819 0.9259 2.4717 0.8224 0.7745 1.9024 0.6849 0.6430 1.1386 2.5879 0.6996 0.5361 1.0790 0.2724 0.1628 0.5030 3.6387 2.1668 0.4849 0.8271 0.0773 2.5194 0.1030 0.9407 0.6065 1.0028 0.1565 4.6020 4.0227 0.0985 0.7742 0.3817 0.5605 2.0727 1.0572 1.4453 0.2350 4.4071 1.1186 1.2671 0.1018 1.6529 1.9213 0.2640 2.8779 4.4941 0.4577 0.0823
787
56. Generate 100 random numbers that are governed by a Gaussian density function.
Solution: We first generate 100 random numbers that are uniformly distributed between 0 and 1 using the linear congruential generator such as +1 = ( + ) (mod ) with = 421, = 1663 and = 7875 The uniform random numbers are shown in the table below. 0.0127 0.6578 0.6362 0.9479 0.5930 0.5714 0.8832 0.5283 0.5067 0.8184 0.5572 0.1356 0.0474 0.2924 0.8709 0.7826 0.0277 0.6061 0.5178 0.7629 0.7950 0.3068 0.1519 0.3303 0.8420 0.6871 0.8655 0.3773 0.2223 0.4008 0.9262 0.3713 0.1497 0.2615 0.7065 0.4850 0.5967 0.0418 0.8202 0.9319 0.1507 0.5291 0.2409 0.2860 0.6644 0.3761 0.4212 0.7996 0.5114 0.5564 0.6686 0.9803 0.6254 0.6038 0.9156 0.5606 0.5390 0.8508 0.4959 0.4743 0.6797 0.9248 0.5032 0.4150 0.6601 0.2385 0.1502 0.3953 0.9737 0.8855 0.3843 0.5627 0.0744 0.9195 0.0979 0.6097 0.4547 0.6331 0.1449 0.9900 0.9821 0.0938 0.5389 0.3173 0.4291 0.8742 0.6526 0.7643 0.2094 0.9878 0.6733 0.7183 0.0968 0.8085 0.8536 0.2320 0.9437 0.9888 0.3672 0.0790 Then, we use the inverse transform method. The Gaussian probability density function is given by ¸ ∙ 1 ( − )2 () = √ exp − − ∞ ≤ ∞ 2 2 2 A closed form expression for the inverse of the cumulative distribution does not exist. We will use numerical inverse transform method. The cumulative distribution is given by Z ( ) = () −∞
where ( ) is replaced with the uniform random numbers between 0 and 1. The inverse transform can be achieved by looking up a Gaussian table (from probability or statistics book) or using the relation ¶¶ µ µ − 1 for ≤ 1 − erf − √ () = 2 2 ¶¶ µ µ 1 − √ = for ≥ 1 + erf 2 2
788
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
Note that the error function is an odd function so that we can write ¶¶ µ µ 1 − () = for all 1 − erf − √ 2 2 Most, if not all, mathematics programs have built-in error function and inverse error function commands. To find that corresponds to () we first solve for such that √ = + 2 erf −1 (2 () − 1) For = 0 and = 1 the MATLAB command to find the value of that corresponds to () = 06 is sqrt(2)∗erfinv(2*0.6-1). This will give us 0.2533. The 100 normally distributed random numbers are shown in the table below. −22353 04064
03483
16252
02353
01800
11910
00709
00167
09093
01439
−11002 −16710 −05463 11305
07810
−19160 02692
00447
07157
08241
−05050 −10284 −04391 10028
04877
11055
14482
−03284 −10377 −06388 05433
−00377 02448
−03127 −07643 −02514 −17304 09161
14904
−10333 00731
−07034 −05652 04244
−03157 −01988 08403
00285
04360
20604
03197
02632
01526
−00103 −00645
04670
14383
00081
−02147 04126
−02943 01577
−14437 14017
20991
−13175 00977
04490
05779
13758
−12936 02784 10519
10398
−07112 −10355 −02655 19384
12027
−01137 03402
−10586 23252
03923
07203
−08085 22511
−07323 15870
22844
−03392 −14119
−04752 −01787 11463
−13002 08724
00980
01420
789
57. Generate a time history for the Pierson-Moskowitz spectral density where = 40 m/s, using the parameter values of Equation 9.33. Compare these results with those of Figure 9.50.
Solution: Equation 9.33 is µ ³ ´4 ¶ 81 × 10−3 2 () = , 0 ≤ exp −074 5 Instead of using Borgman’s method where the frequencies are chosen so that they represent an equal area under the spectral density, we choose the frequencies according to the probability density given by () , 0 ≤ () 0
() = R ∞
where the denominator ensures that the area under () is unity. With = 98 m/s2 and = 40 m/s, the probability density is given by () = 001066
¡ ¢ 1 exp −000267 4 5
The cumulative distribution is given by Z
(̃) ̃ ¢ ¡ = exp −00026674
() =
0
The closed-form expression for the inverse exists, and it is
¶14 µ 0002667 = − ln Random frequencies distributed according to the spectral density are obtained by substituting the uniform random numbers in place of so that ¶14 µ 0002667 = − ln
((1))
For example, for = 0144 the corresponding frequency is 0193The table below shows 16 sets of uniform numbers and random frequencies generated according to Equation (1).
790
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
0.0127 0.1572 0.9821 0.6198 0.5572 0.2599 0.6733 0.2865 0.7950 0.3284 0.6578 0.2825 0.9262 0.4319 0.1356 0.1911 0.1507 0.1938 0.3068 0.2180 0.6686 0.2853 0.3713 0.2278 0.6797 0.2883 0.5291 0.2544 0.3843 0.2298 0.9803 0.6052 The time series is given by ̄( ) =
r
0 X sin(̄ − ̄ + ) 4 =1
where = 81 × 10−3 2 , = 074 ( )4 , = 25 m/s, and is uniform random number between 0 and 2. In addition, the wave number, ̄ is related to the sample frequency, ̄ through the dispersion relationship given by 2 = tanh where is the water depth. The time history of the wave profile at = 0 is s 16 81 × 10−3 2 X ̄() = sin(−̄ + ) 4 (074 4 25) 16 =1
and is plotted next. The random phases used for the figure are 3.5693, 2.3274, 4.4154, 3.4342, 2.7953, 4.3641, 3.9038, 4.9940, 6.0120, 3.2835, 5.5301, 1.0867, 6.1559, 1.7056, 1.5854, and 5.5024.
791
792
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
58. Describe the Borgman method for the generation of a random time history in physical terms. Why does it work?
793 Problems for Section 9.13 – Inverse Vibration with Uncertain Data 59. For Example 9.13, derive the expressions for 1 and 21
Solution: In Example 9.13, we determine 1 2 and 3 from experiment to be 1 = 0382 Hz2 2 = 2618 Hz2 and 3 = 1000 Hz2 with coefficient of variation = 001 The total mass, , is exactly 20 kg. The first two are measured for the two-DOF system shown in Figure 9.51, and the last eigenvalue is measured for the one-DOF system shown in Figure 9.52. These values are taken to be the means. The eigenvalues and the system parameters, and are related in two steps. They are first related through 1 2 and 3 Equations 6.146-6.148 relate the eigenvalues to these intermediate variables (reproduced here). 1 = 1 + 2 − 3 1 2 2 = 1 + 2 − 3 (3 − 1 )(2 − 3 ) 3 = 1 + 2 − 3 and the 1 2 and 3 are related to the system parameters through Equations 6.150-6.153 (reproduced here). 1 2 1 + 3 1 3 2 = 1 + 3 1 1 = 1 + 3 3 2 = 1 + 3 1 =
To find the mean and the variance of 2 , we first find the means and variances of 1 2 and 3 Using first term approximations, we have ' (1 2 3 ) ¶2 ¶2 ¶2 µ µ µ 2 2 2 1 + 2 + 23 ' 1 2 3 for = 1 2 and 3 Means
794
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
Since the means are much easier to find, we will find them first. The numerical values for the means of the intermediate variables, 12 3 are 1 = 1 + 2 − 3 = 0382 + 2618 − 1000 = 2000 Hz2 1 2 0382 · 2618 = 0500 Hz2 = 2 = 1 + 2 − 3 0382 + 2618 − 1000 (3 − 1 )(2 − 3 ) (1000 − 0382)(2618 − 1000) = 0500 Hz2 3 = = 1 + 2 − 3 0382 + 2618 − 1000 We continue to relate the means of to the means of the system parameters by 1 2 2000 · 0500 2 = 0800 = 1 + 3 2000 + 0500 1 3 2000 · 0500 2 = 0800 = 2 = 1 + 3 2000 + 0500 1 2000 2 = 160 = 1 = 1 + 3 2000 + 0500 3 0500 = 2 = 0400 2 = 1 + 3 2000 + 0500 1 =
Variances For the variances, let us write them in a matrix-vector form for the sake of simplicity. In a matrix form such that ⎧ ⎫ ⎧ ⎫ 2 ⎪ 2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 1 ⎪ ⎬ ⎨ 1 ⎪ ⎬ 2 2 2 = 1 |= 2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 2 ⎪ ⎭ ⎩ 2 ⎪ ⎭ 3 3 where 1 is the Jacobian given by
⎡
1 ⎢ 1
⎢ 2 1 = ⎢ ⎣ 1 3 1
Note that 1 is evaluated at the mean values.
1 2 2 2 3 2
⎤
1 3 ⎥
2 ⎥ ⎥ 3 ⎦ 3 3
The variances of the system parameters are related to the variances of by ⎧ ⎫ 2 ⎪ ⎧ ⎫ ⎪ ⎪ 1 ⎪ ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 2 ⎪ ⎬ ⎨ 1 ⎪ ⎬ 2 2 = 2 |= 2 ⎪ ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ 3 ⎭ ⎪ ⎩ 2 ⎪ ⎭ 2
795 where
⎡
1 ⎢ 1
⎢ 2 ⎢ 1 2 = ⎢ ⎢ 1 ⎢ ⎣ 1 2 1
1 2 2 2 1 2 2 2
⎤
1 3 ⎥ 2 ⎥ ⎥ 3 ⎥
1 ⎥ ⎥ 3 ⎦
2 3
Then, the variances of the system parameters are related to the variances of the eigenvalues by ⎧ ⎫ 2 ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 2 ⎪ ⎬ 2
It is given that
⎪ ⎪ ⎪ 21 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 2 ⎪ ⎭ 2
⎧ ⎫ 2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 1 ⎪ ⎬
= 2 |= 1 |= 22 ⎪ | {z }| {z } ⎪ ⎪ ⎪ ⎪ ⎩ 2 ⎪ ⎭ 4×3 3×3
(1)
3
21 = 2 21 = 0012 × 03822 = 146 × 10−5 22 = 2 22 = 0012 × 26182 = 685 × 10−4 23 = 2 23 = 0012 × 03822 = 146 × 10−5 The first Jacobian is ⎡
1 ⎢ ⎢ 2 (1 +2 −3 )−1 2 1 = ⎢ (1 +2 −3 )2 ⎣ (3 −2 )2 (1 +2 −3 )2
−1
1 1 (1 +2 −3 )−1 2 (1 +2 −3 )2 (3 −1 )1 (1 +2 −3 )2
(2)
⎤ ⎥
⎥ 1 2 ⎥ 2 (1 +2 −3 ) ⎦ 2 1 +22 −22 3 −23 1 +23 +21 (1 +2 −3 )2
The first Jacobian evaluated at = is ⎡
1
−1
1
⎤
⎥ ⎢ ⎥ ⎢ 1 |= = ⎢ 1059 −0059 0250 ⎥ ⎦ ⎣ −1059 0059 0750 The second Jacobian is
⎡
⎢ ⎢ ⎢ 2 = ⎢ ⎢ ⎢ ⎣
2 3 (1 +3 )2 23 (1 +3 )2
1 1 +3
3 (1 +3 )2
0
3 − ( + )2
0
1
3
0
2 1 − (+ )2
⎤
⎥ ⎥ ⎥ ⎥ ⎥ 1 − ( + )2 ⎥ 1 3 ⎦ 1
3
12 (1 +3 )2
1 (1 +3 )2
(3)
796
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
The second Jacobian evaluated at = is ⎡
0040 0800 −0160
⎢ ⎢ ⎢ 0040 2 |= = 20 ⎢ ⎢ | {z } ⎢ 0080 ⎣ 4×3 −0080
0 0 0
⎤
⎥ ⎥ 0640 ⎥ ⎥ ⎥ −0320 ⎥ ⎦ 0320
(4)
Finally, the variances of the system parameters are (Equation (1) with Equations (2)-(4) substituted in) ⎧ ⎫ 2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 2 ⎪ ⎬ 2
⎪ ⎪ ⎪ 21 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 2 ⎪ ⎭ 2 ⎤ ⎡ ⎤⎡ ⎤ 0040 0800 −0160 ⎡ −5 ⎥ ⎢ 146 × 10 1 1 −1 ⎥ ⎢ ⎥⎢ ⎥ ⎢ 0040 0 0640 ⎥ ⎢ ⎥⎢ ⎥ ⎥⎢ −4 = 20 ⎢ 1059 −0059 0250 ⎥ ⎢ 685 × 10 ⎥ ⎥⎢ ⎢ ⎣ ⎦ ⎣ ⎦ ⎢ 0080 0 −0320 ⎥ −5 ⎦ −1059 0059 0750 ⎣ 146 × 10 −0080 0 0320 ⎡ ⎤ 923 × 10−5 ⎢ ⎥ ⎢ ⎥ ⎢ 101 × 10−3 ⎥ ⎢ ⎥ = ⎢ ⎥ −4 ⎢ 766 × 10 ⎥ ⎣ ⎦ −766 × 10−4
797
60. In Example 9.13, formulate expressions for the mean value and variance of 3 .
Solution: Use Equation 6.148 for 3 given by 3 = from which we have
(3 − 1 ) (2 − 3 ) 1 + 3 − 2
¢¡ ¢ ¡ 3 − 1 2 − 3 [3 ] ' 1 + 3 − 2
Then use Equation 9.85 to estimate the variance such that ¶2 ¶2 ¶2 µ µ µ 3 3 3 2 2 2 3 = 1 + 2 + 23 1 2 3 Now we need to go through the derivatives of 3 with respect to each We find 3 (3 − 2 ) 2 = 1 (1 + 2 − 3 )2 3 (3 − 1 ) 1 = 2 (1 + 2 − 3 )2 3 2 1 + 22 − 22 3 − 23 1 + 23 + 21 = 3 (1 + 2 − 3 )2 which must be evaluated at the mean value before being substituted back into Equation (1).
(1)
798
CHAPTER 9 SDOF VIBRATION: PROBABILISTIC FORCES
61. Derive Equation 9.85. Solution: To obtain the variance 2 find the expression for [2 ] and then [ ]2 The difference of these two expressions is the desired variance. Terms of higher order than 2 have been set to zero.
799
62. In the inverse problem with uncertainty, suppose that instead of a fixed constraint, we take an elastic constraint in order to generate an additional equation (as shown in Figure 9.58). The benefit of using a spring as a constraint, rather than just fixing mass 2 as in Figure 9.52, is that 3 can be varied to represent a free end using 3 = 0, or a fixed end using 3 = ∞. Derive the ratios needed to estimate the mean values and variances of the structural parameters. Assume that each mass is exactly 20 kg.
Two degree-of-freedom system.
Solution: Let us assume that 12 are known from measurements, and the total mass is known exactly. The individual stiffnesses and the masses are unknown. Therefore, we look to creating a second but related problem that will generate an additional equation without creating an additional unknown. To proceed with the inverse problem, assume that mass 2 is attached via a spring of constant 3 to a fixed wall. 3 must now be also evaluated, but we will be generating two additional equations. We can obtain the relations for 34 by using the figure for the equation of motion. We have already obtained the general equation for the squares of the frequencies, and it is 1 22 + 2 11 2 sµ 1 2 ¶2 2 1 22 + 2 11 11 22 − 12 ∓ − 21 2 1 2
212 =
(9.1)
which is an algebraically complicated equation. However, we now have 4 equations needed to solve for the stiffinesses and masses, where the total mass is assumed known. The frequency equation for the second system, which is the equation above equation with 11 = 1 + 2 22 = 2 + 3 12 = 21 = −2 substituted in, is 1 2 2 − (1 [2 + 3 ] + 2 [1 + 2 ]) + (1 + 2 ) (2 + 3 ) − 22 = 0 Define the ratios:
2 1 2 3 2 = 3 = 4 = 2 1 1 2 These must be solved for 12 and 123 as functions of and 1 =
Chapter 10 Vibration Control Problems for Section 10.1 — Motivation 1. In each of the following systems, identify the control component(s), if possible. Describe qualitatively how the control system affects system behavior: (a) airplane wing, (b) elevator, (c) standard transmission in car, (d) automatic transmission in car, (e) hand holding a pen and writing, (f) computer..
Solution: (a) The airplane wing is engineered; by extending and articulating the flaps the lift and drag are altered. If on autopilot the system is purely engineered, but if human controlled, then it may be considered to be hybrid. (b) An elevator has engineered control systems to ensure that the car smoothly reaches the desired floor. The human initiates the command to call the elevator car, and then provides the input to indicate the desired floor. (c) The standard transmission is hybrid. (d) The automatic transmission is engineered with the only human dependence being the implicit one of speed. (e) The hand holding a pen is human controlled. (f) the computer has many engineered controls, for example, control of the position of the head above a spinning disk in a hard drive.
801
802
CHAPTER 10 SDOF VIBRATION: CONTROL
2. Identify advantages and disadvantages of passive and active control. Solution: Advantage Passive Control • Simple design changes can be made
Disadvantages • Limited: if operating conditions
(mass, springs, dampers).
change. Performance not guaranteed.
• Does not require external power.
• May introduce unwanted resonant frequencies.
Active Control
• Effective: desired performance can
• Requires sensors, actuators,
be achieved even if system parameters controllers, and external power. or operating conditions change.
• Generally more costly and complex. • May introduce unwanted instabilities.
803 Problems for Section 10.3 — Feedback Control 3. Derive the transfer functions for the following governing equations of motion: (a) ̈ + 2̇ + 4 = cos (b) ̈ + ̇ + = cos (c) ̈ + ̇ + = 1 cos + 2 sin (d) ̈ + = (), () = 1 ≥ 0
(e) ̈ + ̇ + = − cos Ω.
Solution: The transfer function is the ratio of the Laplace transform of the output to Laplace transform of the input, assuming zero initial conditions. Suppose the general equation is the following: 2 ̈ + 1 ̇ + 0 = 2 ̈ + 1 ̇ + 0 Then the transfer function is () L() = where zero initial conditions are taken, () L() 2 2 + 1 + 0 = 2 2 + 1 + 0
() =
For all cases, we let the forces (the right-hand side) to be the input, () Then, (2 + + )() = () 1 () = () = 2 () + + The response is given by () = () () To obtain the Laplace transform of the response, we only need to multiply the transfer function by the Laplace transform of the input. In the following, the transfer function and the Laplace transform of the response are obtained. We assume that the inputs start on at = 0 (a) If we let cos be the input () the transfer function, () is given by () =
1 2 + +
=
1 2 + 2 + 4
and the Laplace transform of the input, () is given by () = L (cos · ()) =
2 + 2
804
CHAPTER 10 SDOF VIBRATION: CONTROL
Finally, the Laplace transform of the output, () is given by µ ¶ ¶µ 1 () = () () = 2 + 2 + 4 2 + 2 1 = 4 3 + 2 + (4 + 2 ) 2 + 2 2 + 4 2 (b) (), () and () are given by () =
1 2 + +
() = L ( cos · ()) = () =
1
2 + + 2 + 2
2 + 2 =
4 + 3 + ( 2 + ) 2 + (2 + 2 ) + 2
(c) (), () and () are given by () =
1 2 + +
1 2 () = L ((1 cos + 2 sin ) · ()) = 2 + 2 2 + 2 ¶ + µ 1 1 2 () = + 2 2 2 2 + + + + 2 (d) (), () and () are given by () =
1 2 +
() = L ( ()) =
1
() =
=
1
1
2 +
1 3 +
(e) (), () and () are given by () =
1 2 + +
¢ ¡ () = L − cos Ω · () =
() = =
1
2 + +
+ ( + )2 + Ω2
+ ( + )2 + Ω2 +
4 + ( + 2) 3 + ( ( 2 + Ω2 ) + + 2) 2 + (2 + (Ω2 + 2 )) + (Ω2 + 2 )
805
4. For each of the equations in the previous problem, discuss how appropriate choices of and can be made to either maximize or minimize the effect of the transfer function. Can you think of applications where either maximum or minimum effects are desirable? Solution: If we have the option of changing the system parameters then we can maximize or minimize the transfer function, but not necessarily over all Thus, the first question is over what frequency range do we need to optimize the transfer function.
806
CHAPTER 10 SDOF VIBRATION: CONTROL
Problems for Section 10.4 — Performance of Feedback Control Systems 5. For the step response () governed by the equation ̈ + 2 ̇ + 2 = () where () is the unit step function, evaluate rise time , peak time , overshoot, and settling time where = 002 for the following cases: (a) = 001, = 1 rad/s, (b) = 001, = 2 rad/s, (c) = 001, = 05 rad/s, (d) = 01, = 1 rad/s, (e) = 01, = 2 rad/s, (f) = 01, = 05 rad/s, (g) = 05, = 1 rad/s, (h) = 05, = 2 rad/s, (i) = 05, = 05 rad/s. Solution: The step response is ∙ µ ¶¸ 1 (− ) cos + sin () = 2 1 −
≥ 0
The rise time is the time till the first crossing of the static displacement, the peak time is the time to the first peak, the overshoot is the maximum displacement minus the static displacement, and the settling time is the time after which the amplitudes are within ± They are given by Equations 10.4, 10.5 and 10.7 and reproduced below. Ã p ! 2 1 1 − p = tan−1 − 2 1 − p = 1 − 2 ´ ³ p − ln 1 − 2 () = The table next shows the rise time, peak time, and settling time for each case (in units of seconds). (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
1.58 0.790 3.16 1.68 0.840 3.36 2.42 1.21 4.84
3.14 1.57
(002) 391
196
6.28 3.16 1.58
6.31 3.63 1.81 7.26
782
78.3 8.11 4.06 16.2
39.2 19.6
807 0.35 0.3
x (t) (m)
0.25 0.2 0.15 0.1 0.05 0
0
5
10
15
20
25
30
t (s)
Figure 10.1:
6. For the equation of motion ̈ + ̇ + = (), with = 9 kg, = 4 N-sm, = 4 N/m and () is the unit step load, find the response. If the response is oscillatory, determine the possible modifications that need to be implemented to make the system critically damped. Plot both original and modified step response. Solution: The unit step response for the current system is ∙ µ ¶¸ 1 (− ) 1− cos + sin () = 2 p p √ where = = 23 = 2 = 0333 and = 1 − 2 = 0629 To make√the system critically damped we need to set = 10 That is, we need 2 − 4 = 0 or = 2 Currently, = 0333 and the system is underdamped. The damping needs to be increased, mass needs to be decreased, and/or the stiffness needs to be decreased. For example, if the mass and stiffness cannot be varied, the damping coefficient must be increased to 12 N-s/m The response is shown below. The solid line is for the original system, and the dotted line is when is increased to 12 N-s/m.
808
CHAPTER 10 SDOF VIBRATION: CONTROL
7. What applications exist where speed of response is more important than accuracy of response? Are there applications where the opposite is true? For what applications are both of critical importance? Solution: With this question we see one of the key trade-offs, speed versus accuracy. In some applications speed is more important, and in other applications accuracy is more important. Sometimes both are important and some sort of optimal solution becomes necessary. An example of the need for a swift response where modest accuracy is acceptable is the steering of a car. Of course, the level of accuracy must be sufficient so that we can guide the car to within several inches of a location. An example of the need for accuracy over speed is in medical apparatus. In surgery it is extremely important to have high accuracy, but speed may be a less important quality. Applications where both are crucial are, for example, military targeting applications. Of course, in situations where high accuracy is necessary, costs increase dramatically.
809
8. What is the maximum value of the step response as a function of . Solution: The step response of the system given by ̈ + 2 ̇ + 2 = 2 () is µ ¶ (− ) cos + sin ≥ 0 () = 1 − The peak time is given by =
p 1 − 2
The peak response value is when ( ) or !Ã ! Ã cos p + sin p ( ) = 1 − exp − p 1 − 2 1 − 2 1 − 2 Ã !Ã ! = 1 − exp − p cos + p sin 1 − 2 1 − 2 ! Ã = 1 + exp − p 1 − 2
(6.5)
We can deduce what the peak response look like as a function of When there is no damping, the peak response is ( = 0) = 2, twice the steady-state value. When the system is critically damped, lim→1 ( ) = 1 The peak response, ( ) is a monotonically decreasing function of The peak response as a function of is plotted next.
810
CHAPTER 10 SDOF VIBRATION: CONTROL
2
Peak Response
1.8
1.6
1.4
1.2
1
0
0.2
0.4
0.6
0.8
1
Figure 10.2:
9. Identify the poles and zeros for the following second order systems, and discuss system stability: (a) ̈ + 2̇ + = cos 3 (b) ̈ + ̇ + 01 = cos 3 (c) ̈ + 300̇ + 10 = cos 3 (d) ̈ + 2̇ + = 1 + 2 ̇ (e) ̈ + = cos (f) ̈ + = (() + ̇).
Solution: The Laplace transform of the response, (), is obtained by taking the Laplace transform of the equation of motion and solving for () (a) ̈ + 2̇ + = cos 3 Taking the Laplace transform, we obtain (2 + 2 + 1) () = () =
(2 + 9)
(2 + 2 + 1)(2 + 9)
811 The zeros occur for values where the numerator is zero: = 0 The poles occur for values where the denominator is zero: = ±3 and a multiple pole at = −1 None of the poles have positive real parts, and therefore the response is stable. (b) ̈ + ̇ + 01 = cos 3 Taking the Laplace transform and solving for () we obtain () = 2 ( + + 01)(2 + 9) The zeros occur at = 0 and the poles occur at = ±3 and = −05 ± have positive real parts, and therefore the response is stable.
√ 015 None of the poles
(c) ̈ + 300̇ + 10 = cos 3 Taking the Laplace transform and solving for () we obtain () = 2 ( + 300 + 10)(2 + 9)
√ The zeros occur at = 0 and the poles occur at = ±3 and = −150± 1502 − 10 = −333×10−2 and −300 × 102 None of the poles have positive real parts, and therefore the response is stable. (d) ̈ + 2̇ + = 1 + 2 ̇ Taking the Laplace transform and solving for () we obtain 1 + 2 () () = 2 ( + 2 + 1) The zero of the transfer function is at −1 2 , and the two identical poles are at = −1 The output will be stable if the poles of () do not have positive real parts. (e) ̈ + = cos Taking the Laplace transform and solving for () we obtain () = 2 ( + 1)(2 + 2 ) The zero occurs at = 0 and the poles occur at = ± and ± The response is marginally stable. (f) ̈ + = [() + ̇] Taking the Laplace transform and solving for () we obtain () () = 2 ( − + 1)
The zeros of the response occur at the zeros of the input, p () The poles of the response occurs at the poles of the input, () in addition to at = 2 ± 2 4 − 1 Both poles of the transfer function are positive if is positive, and both poles of the transfer function are negative if is negative. If is positive, the response is unstable. If is negative and the poles of the input, () do not have positive real parts, the response is stable.
812
CHAPTER 10 SDOF VIBRATION: CONTROL
10. For the given transfer functions, determine the governing differential equation of motion: (a) () = (2 + 2)(2 + 2 + 5) (b) () = ( − 1)(32 + 4)
(c) () = ( + 1)(2 + + 1). Solution: (a) () = (2 + 2)(2 + 2 + 5) The differential equation is ¶ 2 + 2 () () = 2 + 2 + 5 (2 + 2 + 5) = (2 + 2) ̈ + 2̇ + 5 = 2̇ + 2 µ
where is the differential operator (b) () = ( − 1)(32 + 4) The differential equation is ¶ −1 () () = 32 + 4 (32 + 4) = ( − 1) 3̈ + 4 = ̇ − µ
(c) () = ( + 1)(2 + + 1) The differential equation is ¶ +1 () () = 2 + + 1 (2 + + 1) = ( + 1) ̈ + ̇ + = ̇ + µ
813
11. For a gain factor = 2, use the pole-zero maps in Figure 10.29 to find each transfer function.
Figure 10.29: Pole-zero locations in the complex plane.
Solution: For the first figure, the transfer function, () is given by () =
2( + 2) ( + 1)( + 4)
where we see the gain 2 in the numerator, as well as the zero at = −2 in the numerator and the poles at = −1 and = −4 in the denominator. For the second figure, the poles are complex conjugates. Then, 2( + 2) ( + [−1 + ])( − [−1 − ]) 2( + 2) = (2 + 2 + 2)
() =
814
CHAPTER 10 SDOF VIBRATION: CONTROL
12. For responses (a) to (d) in Figure 10.30, explain why the response appears stable or unstable.
Figures 10.30: Four possible time-histories.
Solution: The responses shown in Figures (a), (c), and (d) appear stable since they are bounded. The response shown in Figure (b) unstable since it is unbounded.
815 Problems for Section 10.5 — Control of Response 13. For the transfer function
Θ () = 2 Θ () +
solve for () in terms of and (). How do variations in the values of parameters and affect the behavior of ()? Discuss and show if a bounded () can be selected to destabilize response (). Solution: From the transfer function, we have (2 + )Θ () = Θ () Transforming to the time domain, we have ̈ + = This is a standard equation of motion, with no damping. It can be solved for () in terms of () using, for example, the convolution equation once () is known. Parameter values can be substituted into the equation to determine how these affect response, for example, whether the output, () is stable if is positive (assuming inertia, is always positive).
816
CHAPTER 10 SDOF VIBRATION: CONTROL
14. For a rotating element with controller, the transfer function is given by Equation 10.14, repeated here, Θ () (1 + ) = 2 Θ () + + Explain how to select control parameters and that meet performance specifications, for example, the shortest settling time for a step input. What is the effect of parameter on the response? Solution: Rewrite the transfer function in differential equation form, ̈ + ̇ + = + ̇ where we assume the input is the step function, and we assume parameter values for and First solve the “free vibration” problem. Then solve for the “forced response.” Then vary parameter values for and to determine whether such a maximum decay can be accomplished in this trial and error approach. For such problems, optimization methods are ideal. If () is a step function, the equation of motion is given by ̈ + ̇ + = () + () The response will have the shortest settling time if the system is critically damped for the systems with the same . That is, ( )2 − 4 = 0
817 Problems for Section 10.6 — Parameter Sensitivity 15. A system transfer function is given by () =
2 + 1
Derive the sensitivity function and determine the value(s) of that minimize sensitivity. Plot () for = 1 10 100 on one set of axes and draw conclusions from the comparison. Plot the sensitivity function for each of the values, also on one set of axes. Which value of reduces sensitivity? Solution: Let us first plot the magnitude of the transfer function as a function of as shown below.
To derive the sensitivity function we need the derivative : µ ¶ 3 = = − 2 + 1 2 + 1 (2 + 1)2 The sensitivity function is then ¸ 3 − 2 2 + 1 (2 + 1) 2 +1 ¸ ∙ (2 + 1) 3 − = 2 + 1 (2 + 1)2 1 = 2 ( + 1)
=
∙
818
CHAPTER 10 SDOF VIBRATION: CONTROL
Now we can plot the sensitivity for various values as shown next. We see from both the sensitivity equation and from the graphs that for larger , the sensitivity goes to zero very rapidly whereas for smaller the sensitivity goes to zero much less rapidly.
Larger values of ensure that sensitivity reduces with but magnitude of sensitivity is unity at ?
819
16. Repeat Problem 15 for the transfer function 2 () = 2 + + 1
Solution: The sensitivity is given by The derivative is =
µ
2 2 + + 1
Then, the sensitivity function is µ 1−
¶
=
4 2 − 2 + + 1 (2 + + 1)2
2 (2 + + 1)
which is plotted next for = 1 10 and 100.
¶
=
+1 2 + + 1
820
CHAPTER 10 SDOF VIBRATION: CONTROL
17. For the general transfer function () =
1 () + 2 () 3 () + 4 () + 2 5 ()
where () are polynomials in , find the general sensitivity as a function of parameter . Solution: The sensitivity with respect to is given by which is given by 2 (3 + 4 + 2 5 ) − (1 + 2 ) (4 + 25 ) 3 + 4 + 2 5 = 1 + 2 (3 + 4 + 2 5 )2 =
2 3 − 1 4 − 21 5 − 2 2 5 1 2 3 + 4 + 5 1 + 2
=
2 3 − 1 4 − 21 5 − 2 2 5 (3 + 4 + 2 5 ) (1 + 2 )
821 Problems for Section 10.7 — State Variable Models 18. For the mechanical system shown in Figure 10.31, determine the differential equations of motion. Then, determine the state-space model where the input is the force and the output is the displacement of mass acted upon by the force.
Figure 10.31: Mechanical system with two masses, two dampers, and one spring acted on by an external force.
Solution: Equation of Motion Draw free-body diagrams for each mass as shown below.
Applying Newton’s second law to each mass, we obtain −1 + 2 (̇2 − ̇1 ) = 1 ̈1 −1 ̇2 − 2 (̇2 − ̇1 ) + () = 2 ̈2 They are rearranged so that 1 ̈1 + 2 ̇1 − 2 ̇2 + 1 = 0 2 ̈2 + (1 + 2 ) ̇2 − 2 ̇1 = ()
822
CHAPTER 10 SDOF VIBRATION: CONTROL
State-Space Form Model The equations of motion are coupled second-order ODEs. They can be recast into an equivalent set of four coupled first-order ODEs. One way is to define variables which are derivatives of the displacements such that (1) (2)
3 = ̇1 4 = ̇2 Substituting these into the equations of motion, we obtain 1 ̇3 + 2 3 − 2 4 + 1 = 0 2 ̇4 + (1 + 2 ) 4 − 2 3 = ()
(3) (4)
Solving for the highest order derivatives, we can rewrite Equations (1)-(4) as ̇1 = 3 ̇2 = 4 2 2 3 + 4 − 1 1 1 1 (1 + 2 ) 2 ̇4 = − 4 + 3 + () 2 2 ̇3 = −
The four coupled first-order are written in vector-matrix form as ⎫ ⎧ ⎫ ⎡ ⎤⎧ ⎫ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ 0 0 1 0 0 ⎪ ̇1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎪ ⎪ ⎪ ⎪ ⎪ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎪ ⎪ ⎪ ⎪ ⎪ ⎥ ⎨ ⎨ ⎬ ⎨ ̇ ⎬ ⎢ 0 0 0 ⎥ 2 1 0 ⎬ 2 ⎥ =⎢ () ⎢ ⎥⎪ ⎪ + ⎪ 2 2 ⎪ ⎪ ⎪ ⎢ ⎪ ⎪ ⎪ ⎪ ⎪ ⎥ ⎪ − 0 ̇ 0 − 3 ⎪ 3 ⎪ ⎪ ⎪ ⎪ 1 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎣ 1 ⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ( + ) 2 1 ⎪ 1 2 ⎩ ⎩ ⎭ ⎭ ⎩ ̇ ⎭ 0 0 2 − 2 4 4 2
The four state variables are the two displacements and the two velocities. To solve for the response, the force, () as well as the four initial conditions, (0) for = 1 · · · 4 must be specified. Note that if we have second-order ODEs written in the form, ⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎨ ̇ ⎬ ⎨ ⎬ ⎨ ̈ ⎬ 1 1 1 +C +K = F M ⎩ ̇ ⎭ ⎩ ⎭ ⎩ ̈ ⎭ 2
the state-space model is written as ⎡ {̇} = ⎣
2
0
I
−M−1 K −M−1 C
2
⎤
⎡
⎦ {} + ⎣
[0] [M]−1 F
⎤
⎦
823 where
{} =
⎧ ⎫ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ ⎪ 2
⎪ ⎪ ⎪ 3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ ⎪ 4
⎡
0=⎣
0 0 0 0
⎤
⎡
⎦ and I = ⎣
1 0 0 1
⎤
⎦
824
CHAPTER 10 SDOF VIBRATION: CONTROL
19. Calculate the transfer function of the linear system described by the following state and output equations, ⎧ ⎫ ⎡ ⎫ ⎧ ⎫ ⎤⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪0⎪ ⎪ ⎪ ⎪ ̇ () −2 −5 −4 ⎪ 1 () ⎪ ⎪ ⎪ ⎬ ⎨ ⎪ ⎨ 1 ⎪ ⎬ ⎢ ⎥⎨ ⎬ ⎪ ⎢ ⎥ ̇2 () = ⎢ 1 −1 0 ⎥ 2 () + 1 () ⎪ ⎪ ⎪ ⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎣ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ () ⎪ ⎩1⎪ ⎩ ̇ () ⎪ ⎭ ⎭ ⎪ 0 1 0 3 3 ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ 1 () ⎪ ⎪ ⎬ n o⎨ () = 3 1 0 2 () ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ () ⎪ ⎭ 3 Is the transfer function proper, strictly proper, or not proper?
Solution: For this third-order system, we have ⎡
⎤
⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪0⎪ ⎬ ⎨
−2 −5 −4 ⎥ ⎢ n o ⎥ ⎢ A = ⎢ 1 −1 0 ⎥ b = 1 c = 3 1 0 , and = 0 ⎪ ⎪ ⎦ ⎣ ⎪ ⎪ ⎪ ⎪ ⎩ 1⎭ 0 1 0
The transfer function denominator polynomial is the determinant of (I − A) called the characteristic polynomial, () Then, ⎡
+2
5
4
⎤
⎥ ⎢ ⎥ ⎢ () = det (I − A) = det ⎢ −1 + 1 0 ⎥ ⎦ ⎣ 0 −1 = ( + 2) [ ( + 1)] − 5 [−] + 4 [1] ¢ ¡ = 2 + 3 + 2 + 5 + 4 = 3 + 32 + 7 + 4
Next, we need the adjoint of (I − A) to find the numerator polynomial given by ⎡
+2
5
4
⎤
⎥ ⎢ ⎥ ⎢ adj (I − A) = adj ⎢ −1 + 1 0 ⎥ ⎦ ⎣ 0 −1
825 This is the transpose of the cofactor matrix. The cofactor elements are the signed minors: ⎡ ⎤ +1 0 ⎦ = ( + 1) (I − A)11 = det ⎣ −1 ⎡ ⎤ −1 0 ⎦= − (I − A)12 = − det ⎣ 0 ⎡ ⎤ −1 + 1 ⎦=1 (I − A)13 = det ⎣ 0 −1 ⎡
− (I − A)21 = − det ⎣ ⎡
(I − A)22 = det ⎣
(I − A)31 = det ⎣
5
−1
+2 4 ⎡
− (I − A)23 = − det ⎣ ⎡
5 4
0
⎤
⎦ = − (5 + 4)
⎤
⎦ = ( + 2)
+2 5
4
⎤
⎡
( + 1)
0
−1
⎤
⎦=+2
⎦ = −4 ( + 1) +1 0 ⎡ ⎤ +2 4 ⎦ = −4 − (I − A)32 = − det ⎣ −1 0 ⎡ ⎤ +2 5 ⎦ = ( + 2) ( + 1) + 5 = 2 + 3 + 7 (I − A)33 = det ⎣ −1 + 1
Then, the cofactor matrix is given by
1
⎤
⎢ ⎥ ⎢ ⎥ co factor (I − A) = ⎢ − (5 + 4) ( + 2) + 2 ⎥ ⎣ ⎦ 2 −4 ( + 1) −4 + 3 + 7
The adjoint matrix (transpose of the cofactor matrix) is given by ⎡ ⎤ ( + 1) − (5 + 4) −4 ( + 1) ⎢ ⎥ ⎢ ⎥ adj (I − A) = ⎢ ⎥ ( + 2) −4 ⎣ ⎦ 1 +2 2 + 3 + 7
826
CHAPTER 10 SDOF VIBRATION: CONTROL
Since = 0 the numerator polynomial is () = c adj (I − A) b ⎡ ⎤⎧ ⎫ ⎪ ⎪0⎪ ( + 1) − (5 + 4) −4 ( + 1) ⎪ ⎬ ⎨ ⎪ ⎥⎪ n o⎢ ⎢ ⎥ = 3 1 0 ⎢ ⎥ 1 ( + 2) −4 ⎪ ⎣ ⎦⎪ ⎪ ⎪ ⎪ ⎪ 1 +2 2 + 3 + 7 ⎩ 1 ⎭ =
n
3 ( + 1) + −3 (5 + 4) + ( + 2) −12 ( + 1) − 4
= −3 (5 + 4) + ( + 2) − 12 ( + 1) − 4 = 2 − 25 − 28 The transfer function is () =
2 − 25 − 28 () = 3 () + 32 + 7 + 4
Since = 0 the transfer function is strictly proper.
⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪0⎪ ⎬ o⎨ 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩1⎪
827
20. Determine the transfer function of state and output equations in Example Problem 10.10 if A b and c are unchanged but = 1 Is the transfer function proper, strictly proper, or not proper? Solution: The denominator polynomial is the characteristic polynomial given by det (I − A) as shown in the solution of the example problem, ⎤ ⎡ +2 0 −1 ⎥ ⎢ ⎥ ⎢ () = det (I − A) = det ⎢ −1 + 2 01 ⎥ ⎦ ⎣ −1 −1 + 1 = 3 + 52 + 7 + 1
The adjoint of (I − A) is
⎡
⎢ ⎢ adj (I − A) = ⎢ ⎣
2 + 3 + 2
1
+2
+1
+ 3 + 1
1
+3
+2
2 + 4 + 2
2
The numerator polynomial is
⎤
⎥ ⎥ ⎥ ⎦
() = c adj (I − A) b + () ⎡ ⎤⎧ ⎫ ⎪ ⎪ ⎪1⎪ 2 + 3 + 2 1 +2 ⎬ ⎨ ⎪ ⎥⎪ n o⎢ ⎢ ⎥ 2 = 2 1 −1 ⎢ + 1 ⎥ 0 + 3 + 1 1 ⎪ ⎣ ⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 2 1⎭ +3 +2 + 4 + 2 ¡ ¢ +1 3 + 52 + 7 + 1 ¢ ¡ ¢ ¡ = 2 + 4 + 3 + 3 + 52 + 7 + 1 = 3 + 62 + 11 + 4 Thus, the transfer function is () =
3 + 62 + 11 + 4 () = 3 () + 52 + 7 + 1
Since the degree of the numerator does not exceed the degree of the denominator, the transfer function is proper.
828
CHAPTER 10 SDOF VIBRATION: CONTROL
21. Determine the controllability and observability of the system ⎫ ⎡ ⎫ ⎧ ⎫ ⎧ ⎤⎧ ⎨ ⎨ ̇1 () ⎬ 1 () ⎬ ⎨ 0 ⎬ 0 1 ⎦ () =⎣ + ⎩ ̇ () ⎭ −2 −3 ⎩ () ⎭ ⎩ 1 ⎭ 2
2
⎧ ⎫ i ⎨ 1 () ⎬
h
() = 1 1
⎩ () ⎭
2
If the system is uncontrollable or unobservable, explain why. Solution: The controllability matrix is M = [b Ab] where
Therefore,
⎡
A =⎣
0
1
−2 −3
⎤
⎦ b =
⎡
M = ⎣
0 1 1 −3
⎤
⎧ ⎫ ⎨0⎬ ⎩1⎭
⎦
M is not singular (its determinant is −1). Therefore, the system is controllable. The observability matrix is ⎡ ⎤ where Therefore,
M = ⎣
c
cA
⎦
h i c= 1 1 . ⎡
M = ⎣
1
1
−2 −2
⎤
⎦
Since M is singular, the system is not observable. This can be explained by considering its transfer function, () +1 +1 1 = 2 = = () + 3 + 2 ( + 1) ( + 2) +2 The factor ( + 1) appears in the numerator and denominator, and hence can be cancelled. It represents a mode at = −1 which in the time-domain corresponds to a term of the form exp (−) It does not appear in the output, although it is present as one of the states within the system, and thus the system is not fully observable.
Chapter 11 Nonlinear Vibration Problems for Section 11.3 — The Phase Plane 1. Construct the phase diagram for the simple harmonic oscillator ̈ + 2 = 0
Solution: We can rewrite the equation as ̇ = ̇ = − 2 There is one equilibrium point at = 0 = 0 The phase paths are given by ̇ 2 = =− ̇ This equation is separable and easily integrated: = − 2 2 = − 2 2 + Therefore the phase diagram is a family of ellipses concentric with the origin. All solutions are periodic and harmonic. The equilibrium point is stable and the origin is a center, which is stable.
829
830
CHAPTER 11 NONLINEAR VIBRATION
2. Construct the phase diagram for the equation ̈ − 2 = 0
Solution In this problem ̇ = ̇ = 2 There is one equilibrium point at = 0 = 0 The phase paths are given by 2 = This equation is separable and easily integrated to give = 2 2 = 2 2 + These paths are hyperbolic with asymptotes given by the equation = ± ( = 0) . This type of equilibrium point is a saddle point, which is unstable since any small displacement from the equilibrium point involves solutions that keep growing.
831
3. Plot the Equations in Example 11.4. Solution: At the end of the example, we have 2 =− 3 which indicates that the phase paths have a constant slope of −23 Any line with slope of −23 is a phase path, = −23 + as shown below.
832
CHAPTER 11 NONLINEAR VIBRATION
4. Write the first-order Equations in Example 11.4 as a second-order differential equation. Solve this differential equation for (), then find the equation for the slope and relate this result to the final result in the example problem.
Solution: Time Domain Method The two first-order equations are given by ̇ = 3 + 2 ̇ = −2 − 2 We can stay in the time domain. Take the first equation and solve for : =
1 (̇ − 3) 2
Differentiate once:
1 (̈ − 3̇) 2 Replace and ̇ in the second equation with Equations 1 and 2: ̇ =
1 1 (̈ − 3̇) = −2 − 2 (̇ − 3) 2 2 1 (̈ − 3̇) = −2 − ̇ + 3 2 ̈ − ̇ − 2 = 0 The solution is given by () = 1 2 + 2 − Then,
The slope on the phase plane is
() = 1 22 − 2 − 1 () = (̇ − 3) 2 ¢ 1¡ −1 2 − 42 − = 2 ¢ () 1¡ = −21 2 + 42 − 2
1 (−21 2 + 42 − ) () () 2 = = 1 22 − 2 −
833 The phase plot crosses the axis when = 0 or at ¢ 1¡ −1 2 − 42 − = 0 2 1 2 = −42 − Then, the slope is given by 1 (82 − + 42 − ) 2 1 12 2 = = − = −82 − − 2 − 2 −9 3 Laplace Transform Method The two first-order equations are given by ̇ = 3 + 2 ̇ = −2 − 2 The easiest to way to eliminate one variable is to use the Laplace transform method. Taking Laplace transforms, we have () = 3 () + 2 () () = −2 () − 2 () Solve for () using the second equation: () =
−2 () +2
Substitute () into the first equation: µ () = 3 −
¶ 4 () +2
Multiply both sides by + 2 : ¢ ¡ 2 + 2 () = (3 + 6 − 4) () 2 () − () − 2 () = 0 Take inverse Laplace transform, and ̈ − ̇ − 2 = 0
834
CHAPTER 11 NONLINEAR VIBRATION
Problems for Section 11.4 — Perturbation or Expansion Methods 5. Using ̇ = investigate the equation ̈ + (2 + ̇2 − 1)̇ + = 0 for how damping adds and removes energy from the system depending on the values of and ̇ Solution: Let ̇ = ̇ = −(2 + 2 − 1) − and
−(2 + 2 − 1) − = A straightforward way to imagine the possible trajectories is to envision a circle of radius 1 such that 2 + 2 = 1 and then select key locations in the phase plane to figure out whether the trajectory is positive or negative. Here is how one can do this. Consider = 0 Then for positive 1 is negative. For positive 1 is positive. This tells us that these trajectories approach the circle. Next, again for = 0 for negative such that || 1 the slope is negative. And for negative such that || 1 the slope is positive. Again both trajectories at these locations approach the circle. A repetition of this method for = 0 leads to similar conclusions. In summary, trajectories outside the circle decay toward the circle, and trajectories inside the circle grow toward the circle as → ∞ The circle is therefore a limit cycle. Outside the limit cycle the system is damped and inside the limit cycle there is energy added to the system (negative damping). Another way to see this is to look at the coefficient of ̇ in the equation of motion, 2 + ̇2 − 1 The damping coefficient is positive outside of the circle, indicating energy loss, while it is negative inside of the circle, indicating energy gain.
835
6. For the equation ̈ + 2 0 ̇ + 20 = 0 (0) = ̄0 and ̇ (0) = ̄0 compare the standard expansion solution with the exact solution.
Solution: The exact solution is () = −0 ( cos + sin ) where the damped natural frequency is √ = 0 1 − 2 Apply initial conditions, and (0) = ̄0 = ̇ (0) = ̄0 = − 0 + Then, the exact solution is −0
() =
µ ¶ √ √ ̄0 + 0 ̄0 2 2 ̄0 cos 0 1 − + √ sin 0 1 − 0 1 − 2
Expand this exact solution about = 0 That is, ¯ ¯ ¯¯ 1 2 ¯¯ () ∼ ( = 0) + + 2 + · · · ¯ ¯ 2 =0 2 =0 The expansion is
̄0 sin 0 0 + (−0 ̄0 cos 0 + (̄0 − ̄0 ) sin 0 ) ½ ¶ µ ¾ ¢ 1¡ 1 ̄0 2 2 ̄0 2 + sin 0 −̄0 + ( 0 ) ̄0 cos 0 + − ̄0 0 + ( 0 ) 2 2 0 0 +···
() ∼ ̄0 cos 0 +
Now, let us find the perturbation solution by expanding such that = 0 + 1 + 2 2 + · · · Substituting above expression into the governing equation, we have ¡ ¢ ¢ ¡ ̈0 + ̈1 + 2 ̈2 + · · · + 2 0 ̇0 + ̇1 + 2 ̇2 + · · · ¡ ¢ + 20 0 + 1 + 2 2 + · · · = 0
(1)
836
CHAPTER 11 NONLINEAR VIBRATION
Collect terms of the powers of and equate them to zeros: 0 : ̈0 + 20 0 = 0 with 0 (0) = ̄0 and ̇0 (0) = ̄0 1 : ̈1 + 20 1 = −2 0 ̇0 with 1 (0) = 0 and ̇1 (0) = 0 2 : ̈2 + 20 2 = −2 0 ̇1 with 2 (0) = 0 and ̇2 (0) = 0 The solution for 0 () is 0 () = 0 cos 0 + 0 sin 0
(2)
where the constants of integrations, 0 and 0 are found by applying the initial conditions or 0 = ̄0 and 0 =
̄0 0
The equation for 1 () is ̈1 + 20 1 = −20 ̇0 ̈1 + 20 1 = −220 (−̄0 0 sin 0 + ̄0 cos 0 ) The solution is 1 () = 1 cos 0 + 1 sin 0 µ ¶ ̄0 + − − ̄0 0 cos 0 + (−̄0 ) sin 0 0 Apply the zero initial conditions and we find ̄0 0 1 = ̄0 1 =
Then, The equation for 2 () is
1 () = −̄0 0 cos 0 + (̄0 − ̄0 ) sin 0
(3)
̈2 + 20 2 = −20 ̇1
The solution obtained after applying zero initial conditions is ¶ µ ¢ 1¡ 1 ̄0 2 2 2 2 () = −̄0 + ̄0 0 cos 0 + − 0 ̄0 + ̄0 0 sin 0 2 2 0 The expansion solution (combining Equations 2,3, and 4) is
() ' 0 + 1 + 2 2 ̄0 = ̄0 cos 0 + sin 0 0 + (−̄0 0 cos 0 + (̄0 − ̄0 ) sin 0 ) µ ¶ ¶ µ ¢ 1¡ 1 ̄0 2 2 2 2 + − 0 ̄0 + ̄0 0 sin 0 −̄0 + ̄0 0 cos 0 + 2 2 0 +··· This expansion solution agrees with the exact solution in the limit → 0
(4)
837
7. Obtain an approximation to the forced response of period 2 for the equation 1 00 + + 013 = cos 4
Solution: Here we have = 01 Expand according to ( ) = 0 ( ) + 1 ( ) + 2 2 ( ) + · · · Then, ¡ ¢3 3 = 0 ( ) + 1 ( ) + 2 2 ( ) + · · · ¡ ¢ ¡ ¢ = 30 + 320 1 + 2 30 21 + 320 2 + · · · ¡ ¢ ¡ ¢ 3 ' 30 + 320 1 + 2
and the equation of motion becomes ³ 00 ¡ 2 ¢´ 1 ¡ ¡ ¢¢ ¡ ¡ ¢ ¡ ¢¢ 00 0 + 1 + 2 + 30 + 320 1 + 2 = cos + 0 + 1 + 4
Now group according to order of the terms in such that
1 00 0 : 0 + 0 = cos 4 1 00 1 : 1 + 1 = −30 4 .. . The solution to the 0 equation is
4 0 ( ) = − cos 3
Then, the 1 equation becomes µ ¶3 1 4 1 + 1 = − − cos 4 3 64 cos3 = 27 µ ¶ 64 3 1 = cos + cos 3 27 4 4 16 16 cos + cos 3 = 9 27 00
We specifically look for a solution that has a forcing period (or a fraction of the forcing period), and it is given by 64 64 1 ( ) = − cos − cos 3 27 945
838 Then,
and for = 01
CHAPTER 11 NONLINEAR VIBRATION µ ¶ ¡ ¢ 4 64 64 ( ) = − cos + − cos − cos 3 + 2 3 27 945 µ ¶ 4 64 64 (01 ) = − cos + 01 − cos − cos 3 3 27 945 = −1 570 cos − 6 × 10−3 cos 3
Note that the solution when = 0 is (0 ) = −1333 cos which is also the zeroth order solution. The figures below show the zeroth and first order solutions. Note the discrepancy, and we can see that the zeroth order solution is underestimates the amplitude of the response by 15% or so
We can see from the two graphs that the 1 order correction term is several orders of magnitude smaller than the first order term. The application will determine the importance of the correction. Note that it is impossible to compare the analytical solutions to a numerical solution because the analytical solutions are only the part of the total solution. These analytical solutions are only the part of the solution that oscillates at the forcing period (or a fraction of the forcing period).
839
8. Obtain an approximation to the forced response of period 2 for the equation 1 00 + + 013 = cos 2
Solution: We follow the same procedure used in the solution of problem 5. Again, = 01 Expand according to ( ) = 0 ( ) + 1 ( ) + · · · The cubic term, 3 expands to ¡ ¢ 3 ' 30 + 320 1 + 2
and the equation of motion becomes 00
00
0 + 1 +
¡ ¢ 1 (0 + 1 ) + 30 + 2 = cos 2
Now group according to order of the terms in such that
1 00 0 : 0 + 0 = cos 2 1 00 1 : 1 + 1 = −30 2 .. . The solution to the 0 equation is 0 ( ) = −2 cos Then, the 1 equation becomes 1 00 1 + 1 = − (−2 cos )3 2 3 = 8 cos µ ¶ 3 1 = 8 cos + cos 3 4 4 = 6 cos + 2 cos 3 We specifically look for a solution that has a forcing period (or a fraction of the forcing period), and it is given by 4 cos 3 1 ( ) = −12 cos − 17 Then, ¶ µ ¡ ¢ 4 cos 3 + 2 ( ) = −2 cos + −12 cos − 17
840
CHAPTER 11 NONLINEAR VIBRATION
and for = 01 µ ¶ 4 (01 ) = −2 cos + 01 −12 cos − cos 3 17 = −32 cos − 00235 cos 3 Note that the solution when = 0 is (0 ) = −2 cos which is also the zeroth order solution. The figure below shows the zeroth order and first order solutions. Note that the first order solution underestimates the amplitude of the response by 375% or so.
841
9. For the forced, quasi-harmonic system, (a) derive Equation 11.36, (b) derive Equation 11.37. Solution: Equations 11.36 and 11.37 are µ ¶ 3 2 1 + 0 − 4 0 µ ¶−1 µ ¶ 3 3 2 2 1 − 0 ' 1 + 0 4 4 2 = 20
(11.36) (11.37)
Start with Equation 11.35 given by 20 ≡ (1 + ) 2 Solve for to obtain
1 =
Substitute into Equation 11.35 given by
µ
(11.35)
¶ 20 −1 2
3 0 + 03 − 2 = 0 4 to obtain
1
Solve for 2 . The steps are
µ
¶ 20 3 − 1 0 + 03 − 2 = 0 2 4
¢ 1¡ 2 3 0 − 2 0 + 03 2 − = 0 4 ¶ µ 0 20 3 3 0 0 − 2 = − 4 Solving for 2 we obtain − 0 2 2 = 3 3 00 0 − 4
0 − 20
= 3 4
02 − 1
=
20 − 0
1 − 34 02
We can write the last expression in a slightly different form if the term, 34 02 is small compared to 1 using a Taylor series. The Taylor series of 1 (1 − ) near = 0 is µ µ ¶¯ µ ¶¯ ¶¯ 2 ¯ ¯ ¯ ¡ 3¢ 1 2 1 1 1 ¯ ¯ + ¯ ' + + ¯ ¯ ¯ 1− 1 − =0 1 − =0 2 1 − =0 2 ¡ ¢ = 1 + − 2 + 3
Keeping the first two terms, we obtain Equation 11.37 or
1 3 ' 1 + 02 3 2 4 1 − 4 0
842
CHAPTER 11 NONLINEAR VIBRATION
Then, 2
¶µ ¶ µ 3 2 2 1 + 0 = 0 − 0 4 µ ¶ ¡ ¢ 3 + 2 = 20 1 + 02 − 4 0
Keeping the terms up to the first order in we obtain Equation 11.36 or µ ¶ 3 2 2 2 = 0 1 + 0 − 4 0
843
10. Provide a detailed discussion of the amplitude curves shown in Figure 11.12. Discuss the variation of the amplitude as a sweep is made of the frequency from small values to larger values. Solution: Figure 11.12 demonstrates the nature of the nonlinear response of the Duffing oscillator for a variety of parameter values. An important point is that even with zero damping there is no blow-up of amplitudes at the “natural frequency” because of the slant, either to the right or left. (See Figure 11.13 and the next problem for the case where damping exists.)
844
CHAPTER 11 NONLINEAR VIBRATION
11. Provide a detailed discussion of the amplitude curves shown in Figure 11.13. Discuss the variation of the amplitude as a sweep is made of the frequency from small values to larger values. Solution: Figure 11.13 corresponds to the case of the Duffing oscillator with damping, in which case the curves are “closed” and do not go off to ever-increasing amplitudes. Here, we see paths along the curves as the frequency is increased or decreased, and because of the curved nature of the response amplitudes, jumps are possible, and there are regions along the response where certain amplitudes are not stable. This does not happen for undamped Duffing oscillators.
845
12. Solve the following equation to order 2 ̈ + = 2 1 where (0) = 0 and ̇ (0) = 0
Solution: Using Lindstedt’s method, () and are expanded as () = 0 () + 1 () + 2 2 () + · · · = = 0 + 1 + 2 2 + · · · where we let 0 be the natural frequency of the linear system or 1. The derivatives with respect to is transformed such that = = µ ¶ 2 2 2 = = 2 2 The equation of motion becomes 2 2 + = 2 2
Substituting the expansion expressions, the equation of motion becomes ¡ ¢2 ¡ 00 ¢ 1 + 1 + 2 2 + · · · 0 () + 001 () + 2 02 () + · · · ¢ ¡ + 0 () + 1 () + 2 2 () + · · · ¢2 ¡ = 0 () + 1 () + 2 2 () + · · ·
(1)
The initial displacement conditions are such that 0 (0) = 0 and (0) = 0 for 1. For velocities, ¯ ¯¯ ̇ (0) = ¯ =0 ¡ ¢¡ ¢ = 1 + 1 + 2 2 + · · · 00 () + 01 () + 2 02 () + · · · = 00 (0) + (1 00 (0) + 01 (0)) + 2 (02 (0) + 2 00 (0) + 1 01 (0)) + · · · We let the coefficients of be zeros. That is, 00 (0) = 0 1 00 (0) + 01 (0) = 0 02 (0) + 2 00 (0) + 1 01 (0) = 0
(2)
846
CHAPTER 11 NONLINEAR VIBRATION
Collect terms in Equation 1 with the same powers of 0 : 000 + 0 = 0 0 (0) = 0 and 00 (0) = 0 1 : 001 + 1 = 20 − 2 1 000 1 (0) = 0 and 01 (0) = − 1 00 (0) = 0 00 2 : 002 + 2 = 20 1 − 2 1 001 − ( 21 + 2 2 )0 − 1 1 2 (0) = 0 and 02 (0) = − 2 00 (0) − 1 01 (0) = 0 .. . The 0 order solution is 0 (0) = 0 cos The 1 order equation is 001 + 1 = 20 − 2 1 000 001 + 1 = 20 cos2 + 21 0 cos To avoid secular terms in the solution, the coefficient of cos must be zero or 1 = 0 1 ( ) = 1 cos + 1 sin 1 − 20 (−3 + cos 2 ) 6 Apply zero initial conditions, and we find 1 1 2 1 ( ) = 20 − 0 cos − 20 cos 2 2 3 6 The 2 order equation is given by (with 1 = 0) 00
002 + 2 = 20 1 − 2 2 0 ¶ µ 1 1 3 5 3 1 = − 0 + 2 2 0 + 0 cos − 30 cos (2 ) − 30 cos (3 ) 3 6 3 6 To avoid secular terms in the solution, the coefficient of cos must be zero or 5 2 2 0 + 30 = 0 6 2 = −
5 2 12 0
The 2 order solutions is 4 29 3 1 1 2 ( ) = − 30 + 0 cos + 30 cos 2 + 30 cos 3 9 144 9 48
847 Assembling terms, the expansion solution is () = 0 () + 1 () + 2 2 () + · · · µ ¶ 1 2 1 2 20 − cos − 0 cos 2 = 0 cos + 2 0 3 6 µ ¶ 1 3 29 3 1 3 1 3 2 cos + 0 cos 2 + 0 cos 3 + − 0 + 3 144 0 9 48 where =1−
5 2 2 + · · · 12 0
848
CHAPTER 11 NONLINEAR VIBRATION
13. Obtain approximately the solutions of period 2 of the equation1 ̈ + Ω2 − 2 = Γ cos
0
Solution: We need to consider two cases when the natural frequency is near the forcing frequency of 1 rad/s, and when it is far away from the forcing frequency. Excitation Far From Resonance Replace Ω2 by 2 the notation we use elsewhere for natural frequency of the linear system. Expand according to ( ) = 0 () + 1 () + · · · Then, 0 : ̈0 + 2 0 = Γ cos 1 : ̈1 + 2 1 = 20 .. . The periodic solution for 0 () is 0 () =
Γ 2 − 1
cos
The free response does not have a period of 2 and therefore those coefficients are set to zero. The input to the second order equation is 20 ¶2 µ Γ 2 cos2 0 = 2 − 1 µ ¶2 1 Γ = (1 + cos 2) 2 − 1 2 Substitute this into 1 equation. The free response is again set to zero, and then 1 () =
Γ2 Γ2 cos 2 + 22 ( 2 − 1)2 2 ( 2 − 1)2 ( 2 − 4)
Summing the zeroth and first order solutions, we obtain ( ) = 0 () + 1 () ½ ¾ Γ Γ2 Γ2 cos 2 = 2 cos + + − 1 2 2 (2 − 1)2 2 ( 2 − 1)2 (2 − 4) It should be noted that solution above is valid when the natural frequency is far away from the forcing frequency of 1 rad/s (and the multiple of forcing frequency such as 2 rad/s).
849 Weak Forcing Near Resonance One needs to consider the near-resonance case ( ≈ 1 rad/s). A stable (bounded) solution exists if the forcing magnitude is small (order of ). Let 2 = 1 + Γ = The governing equation becomes
¢ ¡ ̈ + = cos + 2 −
(Note that if we took Γ = then the expansion leads to only secular solutions.) The expansion results in the following, 0 : ̈0 + 0 = 0 1 : ̈1 + 1 = cos + 20 − 0 2 : ̈2 + 2 = 21 − 1 .. . The first order solution is 0 = 0 cos + 0 sin After substituting 0 into the first order equation, we obtain ¢ 1¡ 2 ̈1 + 1 = 0 + 20 + ( − 0 ) cos − 0 sin 2 ¢ 1¡ + 20 − 20 cos 2 + 0 0 sin 2 2 Secular terms vanish if 0 = 0 and 0 = Then 0 = () cos and µ ¶2 µ ¶2 1 1 + cos 2 ̈1 + 1 = 2 2
and
µ ¶2 µ ¶2 1 1 1 () = − cos 2 2 6 +1 cos + 1 sin
where 1 ad 2 are found by eliminating the secular terms in 2 () The equation for 2 () is given by ̈2 + 2 = 21 − 1 It is not important what 2 () is if we are only keeping the first order term. However, 2 () must be such that it does not have secular terms. The right-hand side must be free of cos and sin Then, µ ¶2 5 1 = 0 −1 + 6 µ ¶2 7 −1 + 1 = 0 6
850
CHAPTER 11 NONLINEAR VIBRATION
We find that 1 = 1 = 0 to eliminate the secular terms in 2 () At this point we can add the two term approximate solution and can compare with non-resonant case such that () = 0 + 1 = sin +
à µ ¶ ! µ ¶2 2 1 1 − cos 2 2 6
Suppose ' 1 rad/s Then 2 = 1 + Γ = Γ rather than Γ above, and ¢ ¡ ̈ + = 2 − Γ cos 0
After expanding we obtain
0 : ̈0 + 0 = Γ cos 1 : ̈1 + 1 = 20 − 0 .. . 0 only has the secular solution, and the method falls apart. That is why we look for a solution when Γ = .
851
14. Derive the forced periodic response of the equation 00 + (9 + ) − 3 = Γ cos
Solution: Here it is assumed that is small and that and Γ are not too large. This leads to a reasonable perturbation approximation with few terms. Expand according to ( ) = 0 ( ) + 1 ( ) + · · · and look for solutions having the 2 period of the forcing. Combine the terms from the governing equation, ¡ ¢ 00 + 9 + − 3 = Γ cos which becomes
Then,
¡ ¢ (0 + 1 )00 + 9 (0 + 1 ) + [0 + 1 ] − [0 + 1 ]3 = Γ cos 00
0 : 0 + 90 = Γ cos 00 1 : 1 + 91 = 30 − 0 .. . The solution to the 0 order equation is (homogeneous plus particular solution, assume = cos substitute into the 0 order equation and find = Γ8) 0 ( ) = 0 cos 3 + 0 sin 3 +
Γ cos 8
This is substituted into the 1 order equation to obtain 00
1 + 91 = 30 − 0 where the right-hand side of the equation is given by ¶ µ 3 2 3 3 1 3 3 2 1 = 0 − 0 + 0 + Γ + 0 Γ cos 3 4 0 4 2048 128 µ ¶ 3 3 3 2 3 2 + −0 + 0 + 0 0 + 0 Γ sin 3 + · · · 4 4 128 Now, the reason why the homogeneous solution is kept is evident. The coefficients of cos 3 and sin 3 comes not only from the forcing but also from the homogeneous solution. In order to eliminate secular terms in 1 () the coefficients of cos 3 and sin 3 must be zeros, or 3 1 3 0 Γ2 + Γ3 − 0 + 0 (20 + 20 ) = 0 128 2048 4 3 −0 + 3430 + 34 20 0 + 0 Γ2 = 0 128
852
CHAPTER 11 NONLINEAR VIBRATION
There are two equations and two unknowns, but it is not guaranteed that we will have one solution because the equations are nonlinear. It is somewhat easier to solve for and if we replace 2 + 2 with 2 Then, ¶ µ 3 2 3 2 1 Γ − + 0 0 = − Γ3 128 4 2048 3 2 3 2 0 = − Γ 4 128 We find that 0 = 0 This is expected because 0 = 0 means that the forcing and response is in phase. This is expected because there is no damping) and 0 is a root of µ ¶ 3 3 3 2 1 0 + Γ − 0 + Γ3 = 0 4 128 2048 One root is readily found, and it is
sµ ¶ 3 2 4 = − Γ 128 3
Now, let us compare the numerical results with the approximate solution (zeroth order solution). Let = 04 = 1 and Γ = 2 The zeroth order solution is 1 0 ( ) = 0 cos 3 + cos 4 where 0 = 00043 −11014 or 10971 The figure on the left shows the three approximate solutions (0_1 () 0_2 () and 0_3 ()), and the figure on the right shows the numerical solution calculated based on zero initial conditions. The first approximate solution is 1 0_1 () = 00043 cos 3 + cos 4
853
The result is not reassuring. The reason why even the best approximate solution does not fit the numerical solution well is because the approximate solution is the approximate solution to the particular solution only (recall we specifically looked for a solution that responds at the forcing period). We can make a better comparison if we add an arbitrary damping in the numerical solution so that the homogeneous solution (transient solution) will eventually die out. The figure next shows the numerical solution based on zero initial conditions with an arbitrary damping coefficient of = 01 (0 in the equation of motion).
The figure shows a very good agreement between the approximate solution and the numerical solution after the transient response has died out. This may come as a surprise because of two reasons: 1. The approximate solution is for the particular solution of an undamped system, and the numerical solution is the total solution with damping. Recall that the amplitude of the particular solution depends on damping. In our case, the damping coefficient is small such that the two solutions agree well after the transient effects have died out. With increasing we expect the results will deviate. 2. The value of used here (0.4) is not so small. The results shown in the first plots make it seem that the approximate solution is not good at all, but in reality, it is such a good approximation that it works well with a relative large even when only one term is used. A better approximate solution can be obtained if the damping is included in the derivation for the approximate solutions. However, it is much more algebraically challenging to include damping, and the added complexity results in minor improvement. You may ask what are the other two approximate solutions for? The other two solutions may be valid when a different set of initial conditions are used.
854
CHAPTER 11 NONLINEAR VIBRATION
15. Solve Equation 11.38 for Ω ' 1 using the approach suggested in the footnoted Equation 11.43. Solution: Equation 11.38 is given by 00 + Ω2 + 3 = Γ cos
(11.38)
We investigate the response near resonance. A stable solution (unlike in the linear case) is possible when the forcing is weak. We let Ω2 = 1 + Γ =
(11.43)
where and are not too large or too small. We expand the solution in terms of such that = 0 + 1 + 2 2 + · · · Substitute into the equation of motion to obtain 00 + (1 + ) + 3 = cos and 0 : 000 + 0 = 0 + 3 = Γ cos ¡ ¢ 1 : 001 + 1 = −30 − 0 + cos
From the zeroth order equation, we obtain
0 ( ) = 0 cos + 0 sin Substitute 0 ( ) into the first order equation to obtain 001 + 1 = 1 The coefficients of cos and sin on the right-hand side of the equation must be zeros to avoid secular terms in 1 ( ) The right-hand side is given by 3 3 3 3 1 = ( 2 + + 3 + ) cos + ( 2 + 3 + ) sin + Higher harmonics. 4 4 4 4 We let the coefficients of cos and sin zeros so that 3 2 3 0 0 + 0 + 30 + = 0 4 4 3 2 3 3 0 + 0 + 0 = 0 4 0 4
855 We find that 0 = 0. This is not surprising because the response must be in phase with the forcing for an undamped system. We find that 0 satisfies a cubic equation given by 3 0 + 30 + = 0 4 Therefore, there are three possible values of 0 Which one should be chosen depends on the initial condition of the problem. For a numerical example, see the second part of Problem 9. The zeroth order solution is then 0 ( ) = 0 cos The first order solution is found using the same procedure. Determine the coefficients 1 and 1 so that no secular term is present in 2 ()
856
CHAPTER 11 NONLINEAR VIBRATION
16. For the subharmonic response, derive (a) Equations 11.45 to 11.47, (b) Equation 11.48, (c) Equation 11.49, and (d) Equation 11.50. Solution: Equations 11.45 to 11.47 are µ ¶2 Ω ̈0 + 0 = cos Ω 3 µ ¶2 µ ¶2 ¢ Ω Ω ¡ ̈1 + 1 = − 0 + 30 3 3 µ ¶2 µ ¶2 ¢ Ω Ω ¡ ̈2 + 2 = − 1 + 320 1 3 3
(11.45) (11.46) (11.47)
and Equations 11.48 to 11.50 are
Ω 9 cos Ω − 3 8Ω2 ¶2 µ 9 9 4 02 − 0 + 2 + =0 2 2 8Ω 8Ω 3 Ω2 20 ≡ (1 + ) 2 = (1 + ) 9 Start with Equation 11.26 given by ¡ ¢ ̈ + 2 = −2 + 3 + cos Ω ¿ 1 0 () = 0 cos
Let = Ω3 and assume the following expansion in
( ) = 0 () + 1 () + 2 2 () + · · · Substitute the expanded ( ) into the equation of motion, and we obtain µ ¶2 ¢ ¢ ¡ Ω ¡ 2 0 + 1 + 2 2 + · · · ̈0 + ̈1 + ̈2 + · · · + 3 µ ¶2 ³ ¡ ¢ ¡ ¢3 ´ Ω = − 0 + 1 + 2 2 + · · · + 0 + 1 + 2 2 + · · · + cos Ω 3
where
¢3 ¡ ¢ ¡ ¢ ¡ 0 + 1 + 2 2 + · · · = 30 + 320 1 + 2 30 21 + 320 2 + 3
Then, we obtain Equations 11.45-11.48 given by µ ¶2 Ω 0 : ̈0 + 0 = cos Ω 3 µ ¶2 µ ¶2 ¢ Ω Ω ¡ 1 : ̈1 + 1 = − 0 + 30 3 3 µ ¶2 µ ¶2 ¢ Ω Ω ¡ 2 : ̈2 + 2 = − 0 + 320 1 3 3
(11.48) (11.49) (11.50)
(11.26)
857 The solution to the zeroth order equation is 0 () = 0 cos
Ω Ω + 0 sin + ¡ ¢2 cos Ω Ω 3 3 − Ω2 3
Substitute 0 () into the right-hand side of the first order equation to obtain µ ¶ 3 3 3 3 2 3 Ω 3 2 2 2 0 + 0 = 0 Γ + 0 + 0 + 0 Γ + 34 0 0 − 0 Γ cos 2 4 4 4 3 ¶ µ Ω 3 3 3 3 + − 0 0 Γ + 0 + 0 Γ2 + 03 + 0 2 0 sin 2 2 4 4 3 +higher harmonics, where we let
9 =− 2 Γ = ¡ ¢2 Ω 8Ω − Ω2 3
for simplicity. To avoid secular solution, the coefficients of cos Ω3 and sin Ω3 must be zeros. We right away see that 0 = 0 in order for the coefficient of sin Ω3 to be zero. Then, the zeroth order solution is given by 9 Ω 0 () = 0 cos − 2 cos Ω 3 8Ω 0 must satisfy ¶ µ 3 3 3 3 2 2 + Γ0 + + Γ 0 = 0 4 0 4 2 Then, 0 = 0 or must satisfy the quadratic equation given by Equation 11.49 or ¶2 µ 9 9 4 2 = 0 0 + 2 + 0 − 2 2 8Ω 8Ω 3
In the text, we define 2
20 ≡ 2 (1 + )
Replacing 2 with Ω9 we obtain Equation 11.50 or
20 = (1 + )
Ω2 9
858
CHAPTER 11 NONLINEAR VIBRATION 0.2 0.15 0.1
x (t)
0.05 0 -0.05 -0.1 -0.15 -0.2
0
20
40
60
80
100
t
¢ ¡ Figure 11.1: () = 18 cos − cos 13 17. Find the subharmonic response of order 1 for the linear equation, ̈ +
1 = Γ cos 2
Solution: Subharmonic resonance refers to the case when the system responds at the natural period instead of the forcing period. This occurs in the linear system as well as the nonlinear system when the forcing frequency is a multiple of the natural frequency. For the given differential equation, the solutions are of the form, 1 1 Γ () = cos + sin − cos 1 − −2 If is an integer, the period is 2 instead of the forcing period of 2 The response is said to be a subharmonic of order 1 Any damping will cause the disappearance of this motion, whereas for the nonlinear equation, the subharmonic is stable for a range of parameters, even with damping. The figure below shows the response when = 3 and Γ = 1 Zero initial conditions are used. It shows that the response period is 6
859
18. Consider the following Duffing equation, ̈ + + 3 = Γ cos with Γ 0 The values of for which the subharmonics occur are unknown. Expand and in a perturbation solution. Solution: Let = Then the governing equation becomes 2 00 + + 3 = Γ cos Because of the cubic nonlinearity, we expect subharmonics of order 13 We are specifically looking for the response with a period that is three times the forcing period. For the transformed equation, it would be 6. Use the following expansions ( ) = 0 ( ) + 1 ( ) + · · · = 0 + 1 + · · · Make the above substitutions to find ( 0 + 1 )2 (0 + 1 )00 + (0 + 1 ) + (0 + 1 )3 = Γ cos Then, 0 : 20 000 + 0 = Γ cos 1 : 20 001 + 1 = −2 0 1 000 − 30 .. . The zero order equation can be written as 000 +
Γ = 2 cos 2 0 0 0
We know that the period is 6 therefore,
and
√ 2 6 = p =⇒ 0 = 3 20 1 = 2 0 3
Thus, the 13 subharmonic is stimulated by the applied frequency of three times the natural frequency of the linearized equation ( = 0) The solution to the zeroth order equation is 1 Γ 1 cos 0 ( ) = cos + sin − 3 3 8
860
CHAPTER 11 NONLINEAR VIBRATION
The constants and are found from conditions in the solution for 1 ( ) We require 000 and 30 as input to the first order equation, 1 1 1 1 Γ 000 ( ) = − cos − sin + cos 9 3 9 3 8 This equation does not set the values of or The 30 term, however, does, ( Ã ) µ ¶2 ! ¡ ¢ Γ 3 Γ 1 30 = − 2 + 2 + 2 2 − 2 cos 4 8 8 3 ! ( Ã ) µ ¶2 Γ 1 Γ 3 2 2 + + +2 sin + 4 8 4 3 + cos sin terms + higher harmonics.
To avoid resonance, the coefficients of cos 13 and sin 13 must be set to zero. However, in addition to and 1 is also unknown, so we need an additional equation. But since further equations will not provide additional conditions, 1 is taken as arbitrary. We can see also that one possible solution is that = 0 This considerably simplifies further analysis.
861
19. For the combination harmonic response, derive Equation 11.54.
Solution: Equation 11.54 is given by ̈1 + 2 1 = 1 cos Ω1 + 2 cos Ω2 + 3 [cos (2Ω1 + Ω2 ) + cos (2Ω1 − Ω2 ) ] + 4 [cos (Ω1 + 2Ω2 ) + cos (Ω1 − 2Ω2 ) ] + 5 cos 3Ω1 + 6 cos 3Ω2
(11.54)
Let us start with 0 = 1 cos Ω1 + 2 cos Ω2 which is substituted into the right-hand side of equation ̈1 + 2 1 = − 2 30 We can write 30 = (1 cos Ω1 + 2 cos Ω2 )3 = (1 cos Ω1 )3 + 3 (1 cos Ω1 )2 (2 cos Ω2 ) + 3 (1 cos Ω1 ) (2 cos Ω2 )2 + (2 cos Ω2 )3 Let us tackle one term at a time as shown next: µ ¶ 1 3 3 3 3 1 cos Ω1 = 1 cos 3Ω1 + cos Ω1 4 4 µ ¶ 1 1 1 2 2 cos (2Ω1 − Ω2 ) + cos (2Ω1 + Ω2 ) + cos Ω2 3 (1 cos Ω1 ) (2 cos Ω2 ) = 31 2 4 4 2 µ ¶ 1 1 1 2 2 3 (1 cos Ω1 ) (2 cos Ω2 ) = 31 2 cos (2Ω2 − Ω1 ) + cos (2Ω2 + Ω1 ) + cos Ω1 4 4 2 µ ¶ 1 3 cos 3Ω2 + cos Ω2 (2 cos Ω2 )3 = 32 4 4 where we used cos cos = 12 [cos( + ) + cos( − )] repeatedly. For example, cos3 Ω1 = cos Ω1 × (cos Ω1 cos Ω1 ) 1 = cos Ω1 × [cos(2Ω1 ) + 1] 2 1 1 = [cos Ω1 × cos(2Ω1 )] + cos Ω1 2 2 1 1 = [cos(3Ω1 ) + cos(Ω1 )] + cos Ω1 4 2 1 3 = cos 3Ω1 + cos Ω1 4 4
862
CHAPTER 11 NONLINEAR VIBRATION
Combine all the terms, we have ¶ ¶ µ µ 3 3 3 3 3 3 2 3 2 0 = + 1 2 cos Ω1 + + 2 cos Ω2 4 1 2 4 2 2 1 3 3 + 21 2 cos (2Ω1 − Ω2 ) + 21 2 cos (2Ω1 + Ω2 ) 4 4 3 3 + 1 22 cos (2Ω2 − Ω1 ) + 1 22 cos (2Ω2 + Ω1 ) 4 4 1 3 1 3 + 1 cos 3Ω1 + 2 cos 3Ω2 4 4 Multiply 30 by −2 and we obtain Equation 11.54, where the constants are related by µ ¶ 3 3 3 2 2 1 = − + 1 2 4 1 2 µ ¶ 3 3 3 2 2 + 2 2 = − 4 2 2 1 µ ¶ 3 2 2 2 3 = − 4 1 µ ¶ 3 2 2 1 2 4 = − 4 µ ¶ 1 3 2 5 = − 4 1 µ ¶ 1 3 2 6 = − 4 2
863 Problems for Section 11.5— Mathieu Equation 20. Referring to Figure 11.14, describe physically the exchange of energy between the pendulum, the moving container, and the external force. Discuss in terms of energies and motion. Solution: Figure 11.14 shows a container with an interior pendulum. The container is forced, causing it to move and also resulting in motion of the pendulum. The exact governing equations 11.55 and 11.56 are coupled equations of motion. Under the assumption of harmonic support motion it is possible to combine the two governing equations into one where the “stiffness” term is time dependent and can have both positive and negative values, leading to the possibility of unstable motion in the case when the coefficient is negative. (See the next problem.)
864
CHAPTER 11 NONLINEAR VIBRATION
21. Suppose in Equation 11.60 that = 1, = 1, and , representing the harmonic motion of the support, is variable. What limitations, if any, are there on stable motion? Solution: For the values = 1, = 1, if 2 cos , the term can be negative and the solutions have coefficients with positive exponential terms, with amplitudes growing without bounds.
865
22. Derive Equation 11.67. Solution: Equation 11.67 is given by =4+
5 2 + ··· 12
(11.67)
and it is the transition curve for 0 = cos 2 Let us start with Equation 11.62-11.64 with = 2 They are given by ̈0 + 40 = 0
(11.62)
̈1 + 41 = −( 1 + 2 cos 2)0
(11.63)
̈2 + 42 = −( 1 + 2 cos 2)1 − 2 0
(11.64)
The first equation provides the zeroth order solutions ⎧ ⎨ cos 2 0 () = ⎩ sin 2
For 0 () = cos 2 the first order equation becomes
̈1 + 41 = −( 1 + 2 cos 2) cos 2 = − 1 cos 2 − cos 4 − 1 Then, 1 = 0 to prevent secular terms. With 1 = 0 we obtain 1 1 cos 4 1 () = − + 4 12 Substitute this into the second order equation to obtain ̈2 + 42 = −( 1 + 2 cos 2)1 − 2 0 ¶ µ 1 1 − cos 4 − 2 cos 2 = 2 cos 2 4 12 µ ¶ 1 1 = − 2 cos 2 − cos 2 cos 4 2 6 ¶ µ 1 1 1 − 2 − cos 2 − cos 6 = 2 12 12 5 We must set 2 = 12 to avoid secular terms. Then,
= 2 + 1 + 2 2 + · · · 5 = 4 + 2 + · · · 12
(11.67)
866
CHAPTER 11 NONLINEAR VIBRATION
23. Derive Equation 11.68. Solution: Equation 11.68 is given by =4−
1 2 + ··· 12
(11.68)
and it is the equation for the transition curve for 0 = sin 2 Let us start with Equation 11.62-11.64 with = 2 They are given by ̈0 + 40 = 0 ̈1 + 41 = −( 1 + 2 cos 2)0
̈2 + 42 = −( 1 + 2 cos 2)1 − 2 0
The first equation provides the zeroth order solutions ⎧ ⎨ cos 2 0 () = ⎩ sin 2
For 0 () = sin 2 the first order equation becomes
̈1 + 41 = −( 1 + 2 cos 2) sin 2 = − 1 sin 2 − sin 4 Then, 1 = 0 to prevent secular terms. With 1 = 0 we obtain 1 () =
1 sin 4 12
Substitute this into the second order equation to obtain ̈2 + 42 = −( 1 + 2 cos 2)1 − 2 0 µ ¶ 1 = −2 cos 2 sin 4 − 2 sin 2 12 1 = − cos 2 sin 4 − 2 sin 2 ¶ µ6 1 1 sin 6 = − − 2 sin 2 − 12 12 1 to avoid secular terms. Then, We must set 2 = − 12
= 2 + 1 + 2 2 + · · · 1 = 4 − 2 + · · · 12
867 Problems for Section 11.6 — Van der Pol Equation 24. Considering Equations 11.69 and 11.70, describe the physical coupling between the two equations. Describe the energy exchange between the systems, with reference to individual terms in the equations. Solution: Equations 11.69 and 11.70 are given by 00
0
+ 2 + = 2 0 00 0 0 − + ( )3 + 2 =
(11.69) (11.70)
Considering equations 11.69 and 11.70, we see two coupled equations used to model the interaction between a simple structure and the vortices shed. The structure is a linear oscillator and the fluid is modeled as a van der Pol equation, which has a property that allows it to cycle between negative and positive damping, that is, between energy being added to the system and energy being removed form the system. It also has a limit cycle, and behaves in a way that it never comes to rest. This is a classical set of equations used for modeling vortex-induced oscillation.
868
CHAPTER 11 NONLINEAR VIBRATION
25. Derive Equation 11.75. Solution: This is the case when the van der Pol equation is subject to weak forcing near resonance. This is done because we know that there is a stable solution even if the forcing frequency is near the natural frequency. Equation 11.75 is given by 000 + 0 = 0
¡ ¢ 001 + 1 = −2 1 000 − 20 − 1 00 + cos
(11.75)
Start with Equation 11.73 given by
¢ ¡ ̈ + 2 − 1 ̇ + = cos
(11.73)
We let = for convenience, and we obtain ¡ ¢ 2 00 + 2 − 1 0 + = cos
If the excitation is soft and near resonance, we can let = and = 1 + 1 We expand such that ( ) = 0 ( ) + 1 ( ) + · · · Substitute the expanded solution into the equation of motion to find ¡ ¢ (1 + 1 )2 (000 + 001 + · · · ) + (1 + 1 ) (0 + 1 + · · · )2 − 1 (00 + 01 + · · · ) + (0 + 1 + · · · ) = cos Collecting terms, we finally obtain 0 : 000 + 0 = 0 ¡ ¢ 1 : 001 + 1 = −2 1 000 − 20 − 1 00 + cos
The third order equation is
¡ ¢ ¡ ¡ ¢¢ 2 : 002 + 2 = −21 001 − 21 000 − 20 − 1 01 − 20 1 + 20 − 1 00
(11.75)
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