lecture_2_single-degree-of-freedom-system_formulation-of-equation-of-motion

CE-5113: DYNAMICS OF STRUCTURES By: Dr. Mohammad Ashraf (engineerashraf@yahoo.com) Office: CE: B109

Department of Civil Engineering, University of Engineering and Technology, Peshawar

Module-2 Single Degree of Freedom System: Formulation of Equation of Motion

Dynamics of Structures

2

Single Degree of Freedom System

A system whose dynamic response can be described by specifying displacement along only one coordinate.

True Engineering Systems are Multi-degree System; SDOF Systems are Idealization For many Engineering Systems SDOF Idealization is Satisfactory Dynamic Analysis of SDOF Systems is Simple and can be extended to more complex MDOF Systems

Dynamics of Structures

3

Components of Dynamic System

Components of a Dynamic System are: z z z z

Mass Resulting in Inertial Force (fI) Stiffness resulting in Spring Force (fS) Energy dissipating mechanism resulting in Damping Force (fD) External Exciting Force p(t)

From Direct Equilibrium (d’Alembert’s Principle):

f I + f D + f S = p (t ) mu&& + cu& + ku = p (t )

Dynamics of Structures

4

Inertia and Spring Force Â

Inertia Force: The force produced in a dynamic system due to: z z

Â

Translational Acceleration of mass (Inertial force, f I = mu&& ) Rotational Acceleration of mass (Inertial Moment, M I = Iθ&& )

Spring Force: Internal Forces induced in a body undergoing deformation. z z

z

Spring force exist both in static and dynamic system For elastic system the spring force is directly proportional to the deformation produced and is equal to stiffness times the deformation ( f S = ku ) For inelastic system the spring force is dependent on both displacement and velocity.

Dynamics of Structures

5

Damping Force Â

Damping Force: Damping is the process by which free vibration steadily diminishes in amplitude due to dissipation of energy and the corresponding force is called damping force.

There are many energy dissipating mechanisms: z z z z z

Â

Heating of the material due to internal friction Opening and closing of micro cracks Friction between structural and non-structural components Connections External devices

Equivalent Viscous Damping force f D = cu&

Dynamics of Structures

6

Formulation of Equation of Motion Â

A dynamic systems may be: 1.

2.

3.

4.

1.

System with Localized Mass and Localized Stiffness System with Distributed Mass and Localized Stiffness System with Localized Mass and Distributed Stiffness System with Distributed Mass and Distributed Stiffness

System with Localized Mass and Localized Stiffness:

f I + f D + f S = p (t ) mu&& + cu& + ku = p (t )

Dynamics of Structures

7

Formulation of Equation of Motion 1.

System with Localized Mass and Localized Stiffness

f S = kL2θ f = cdθ& D

f I = M (R + L1 )θ&& 2 MR M I = Iθ&& = θ&& 2 Taking moment about the hinged end

M I + f I (R + L1 ) + f D (d ) + f S (L2 ) = 0 MR 2 && θ + M (R + L1 )2 θ&& + cd 2θ& + kL22θ = 0 2 ⎛ R2 2⎞ M ⎜⎜ + (R + L1 ) ⎟⎟θ&& + cd 2θ& + kL22θ = 0 ⎠ ⎝ 2 ⎛ 3R 2 ⎞ + L12 + 2 RL1 ⎟⎟θ&& + cd 2θ& + kL22θ = 0 M ⎜⎜ ⎝ 2 ⎠ m*u&& + c*u& + k *u = 0 Dynamics of Structures

8

Formulation of Equation of Motion 2.

System with Distributed Mass and Localized Stiffness (m = Total Mass of Rod)

f S = ku ⎛b ⎞ f D = c⎜ u& ⎟ ⎝L ⎠ ⎛ u&& ⎞ f I = m⎜ ⎟ ⎝2⎠ 2 ⎛ u&& ⎞ mL ⎛ u&& ⎞ MI = I⎜ ⎟ = ⎜ ⎟ ⎝ L ⎠ 12 ⎝ L ⎠

Taking moment about the hinged end ⎛L⎞ M I + f I ⎜ ⎟ + f D (b ) + f S (L ) − p(t )(a ) = 0 ⎝2⎠ mL mL cb 2 u&& + u&& + u& + kLu = p(t )a 12 4 L mL cb 2 kL u&& + u& + u = p(t ) 3a La a * * * m u&& + c u& + k u = p(t ) Dynamics of Structures

9

Formulation of Equation of Motion 3.

System with Localized Mass and Distributed Stiffness Prob. 2.2

PL13 PL32 TL1 L2 F= + + 3EI1 3EI 2 GJ 1x363 1x123 12 x36 x12 F= + + 3x1472621 3x39062 1178097 F = 0.02971 _ in / kip k = 1 / F = 33.66 _ kips / in

The whole system can be shown as:

⎛ π (1)4 ⎞ ⎟ = 1,472,621 _ k − in 2 EI1 = 30 x10 ⎜⎜ ⎟ ⎝ 64 ⎠ 6

⎛ 1(0.25)3 ⎞ ⎟ = 39,062 _ k − in 2 EI 2 = 30 x10 ⎜⎜ ⎟ ⎝ 12 ⎠

mu&& + ku = 0

6

m = mass _ of _ flywheel u = Ver. _ displacement _ at _ flywheel

⎛ π (1)4 ⎞ ⎟ = 1,178,097 _ k − in 2 GJ = 12 x10 ⎜⎜ ⎟ ⎝ 32 ⎠ 6

Dynamics of Structures

10

Formulation of Equation of Motion 4.

System with Distributed Mass and Distributed Stiffness 1.

2.

Lumped Mass Procedure Generalized Coordinates

Suppose the virtual displacement at the left end is: δ z ψ ( x ). The virtual displacement at the right is: dψ ⎧ ⎫ δ z ⎨ψ ( x ) + ⎩

dx

dx ⎬ ⎭

The total internal virtual work by the spring force is:

dψ ⎧ ⎫ d δ W i = − f S δ z ψ ( x ) + ( f S + ∆ f S )δ z ⎨ψ ( x ) + dx ⎬ dx ⎩ ⎭

δWi =

L

∫ 0

u ( x, t ) = z (t )ψ ( x) ∂ψ ∂u = EA( x) z (t ) ∂x ∂x f I = m dxu&& = m dx&z&(t )ψ ( x) f s = EA( x)

dψ ⎛ dψ ⎞ f S δz dx = δ zz (t ) ∫ EA ( x ) ⎜ ⎟ dx dx dx ⎠ ⎝ 0 L

2

The external virtual work done by the inertial force fI and the tip force p(t) is: L

δWe = −δz.&z&(t ) ∫ m ( x)ψ 2 ( x)dx + p(t )δzψ ( L) 0

Dynamics of Structures

11

Formulation of Equation of Motion From principle of virtual displacement, the external virtual work must be equal to internal virtual work. ⎛ dψ ⎞ 2 δzz (t ) ∫ EA( x)⎜ ⎟ dx = −δz.&z&(t ) ∫ m ( x)ψ ( x)dx + p (t )δzψ ( L) ⎝ dx ⎠ 0 0 2

L

L

⎛ dψ ⎞ &z&(t ) ∫ m ( x)ψ 2 ( x)dx + z (t ) ∫ EA( x)⎜ ⎟ dx = p(t )ψ ( L) dx ⎝ ⎠ 0 0 L

L

2

m* &z&(t ) + k * z (t ) = p * where L

m = ∫ m ( x)ψ 2 ( x)dx, *

0

⎛ dψ ⎞ k = ∫ EA( x)⎜ ⎟ dx, ⎝ dx ⎠ 0 L

2

*

p * = p(t )ψ ( L) Dynamics of Structures

12

Effect of Gravity Load Let ut defines the position of spring from un-stretched and u defines its position from the static equilibrium

f I + f D + f S = p (t ) + mg mu&&t + cu&t + kut = p (t ) + mg mu&& + cu& + k (u + ∆ st ) = p (t ) + mg mu&& + cu& + ku = p (t ) + mg − k∆ st As __ mg = k∆ st Therefore mu&& + cu& + ku = p (t ) Thus when the displacement coordinate is selected with reference to position of static equilibrium, the force of gravity does not appear in the equation of Motion Dynamics of Structures

13

Effect of Gravity Load (Cont..)

However the force due to gravity must be considered when it acts as: z z

Restoring force (as in case of simple pendulum) Destabilizing force (as in the case of inverted pendulum)

(mLθ&&)L + (mg )L sin θ = 0 g θ&& + θ = 0 _____ sin θ = θ L

(mLθ&&)L + kLθ (L cosθ ) − (mg )L sin θ = 0 mg ⎞ ⎛ mθ&& + ⎜ k − ⎟θ = 0 _____ sin θ = θ , _ cos θ = 1 L ⎠ ⎝ Dynamics of Structures

14

Axial Force Effect L + f S b − SLθ = 0 2 mL2 && mLθ L θ+ + kbθb − SLθ = 0 12 2 2 mL2 && θ + kb 2θ − SLθ = 0 3 m && ⎛ kb 2 S ⎞ θ + ⎜⎜ 2 − ⎟⎟θ = 0 L⎠ 3 ⎝ L

M I + fI

b

m * u&& + (k * − kG )u = 0

kb 2 S Where _ u = θL, _ k * = 2 , _ kG = L L

kG is known as the Geometric Stiffness

Dynamics of Structures

15

Effect of Support Motion ut = u g + u mu&&t + cu& + ku = 0 m(u&&g + u&&) + cu& + ku = 0 mu&& + cu& + ku = 竏知u&&g

Dynamics of Structures

16

Problem 2.1 f I = (m + M )u&& 2 mR u&& mRu&& M I = Iθ&& = = 2 R 2 f S = k (2u ) = 2ku

Now taking moment about point of rotation of the pulley i.e. the right end: M I + f I R + f S (2 R ) = 0 mRu&& + (m + M )u&&R + 2ku (2 R ) = 0 2 ⎛ 3m ⎞ + M ⎟u&& + 4ku = 0 ⎜ ⎝ 2 ⎠ m * u&& + k * u = 0 3m + M , _ k * = 4k Where _ m* = 2 Dynamics of Structures

17