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-6 Numerical Evaluation of Dynamic Response

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Introduction 

Analytical Solution of Equation of Motion is usually not possible when: z z



There are two possible alternatives: z z

 



The exciting force varies arbitrarily or The system is nonlinear Evaluation of Duhamel’s Integral by Numerical Methods Direct Numerical Integration of Equation of Motion

Duhamel’s integral is applicable to only linear system Direct Integration is more general and applicable to both linear and nonlinear systems Also the direct integration methods can be extended to multi-degree of freedom system

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Numerical Evaluation of Duhamel’s Integral 

Duhamel’s Integral for a linear damped system is given by: u (t ) =



1 mωd

ωζ ( ∫ p(τ )e t

t −τ )

0

sin ωd (t − τ )dτ − − − − − (a )

Expansion of Equation (a) is: u (t ) =

1 −ωζt t e ∫ p(τ )eωζτ (sin ωd t cos ωdτ − cos ωd t sin ωdτ )dτ 0 mω d

u (t ) = Ae −ωζt sin ωd t − Be −ωζt cos ωd t − − − − − − − − − − − −(b) 



Where: A=

1 mω d

∫e

ωζτ

p (τ ) cos ωdτdτ

B=

1 mω d

∫e

t

ωζτ

p (τ )sin ωdτdτ

t

0

0

The value of A and B are evaluated using time step numerical methods Dynamics of Structures

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Numerical Evaluation of Duhamel’s Integral (Cont..) 

Numerical Integration is basically the determination of area under the curve drawn between time as abscissa and integrand as ordinate. Several methods are available to determinate the approximate area: z z z

Rectangular Summation [Equation (1)] Trapezoidal Method [Equation (2)] Simpson’s Method [Equation (3)]

t

0

ydt = ∆t ( yo + y1 + y2 + ..... + y N −1 ) − − − − − − − − − − − − − − − − − − − (1)

∆t ( yo + 2 y1 + 2 y2 + ..... + 2 y N −1 + y N ) − − − − − − − − − − − − − (2) 2 t ∆t ∫0 ydt = 6 ( yo + 4 y1 + 2 y2 + 4 y3 + 2 y4 + ..... + 2 y N −2 + 4 y N −1 + y N ) − − − (3)

t

0

ydt =

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Example 8.9 and 8.10 

A water tower idealized as SDOF system is subjected to half sine wave loading. Calculate the displacement history for the first 1.0 sec using numerical evaluation of Duhamel’s integral for h = 0.1 sec. 8.9: Neglect Damping 8.10: Damping = 10%

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Example 8.9 (2nd Ed.) u (t ) = Ae

−ωζt

sin ωd t − Be

− ωζt

cos ωd t

A=

1 mω d

t

B =

1 mω d

t

0

0

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e ωζτ p (τ ) cos ω d τ d τ e ωζτ p (τ ) sin ω d τ d τ

7

Example 8.10 (2nd Ed.) u (t ) = Ae

−ωζt

sin ωd t − Be

− ωζt

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cos ωd t

A=

1 mω d

t

B =

1 mω d

t

0

0

e ωζτ p (τ ) cos ω d τ d τ e ωζτ p (τ ) sin ω d τ d τ

8

4

Example 8.10 (2nd Ed.)

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Direct Integration Methods 





These methods are based on appropriately selected expressions that relate response parameters (displacement, velocity and acceleration) at a given interval of time (ti+1) to their values at one or more time points (ti-1, ti) In general two expressions must be specified. The equation of motion provides the 3rd expression to determine the three unknown parameters. The three important requirements for a numerical method are: z

z

z



Convergence: As the time step decreases the numerical solution should approach the exact solution Stability: The numerical solution should be stable in the presence of numerical round-off errors Accuracy: The numerical solution should be close enough to the exact one

These requirements depend upon: z z

The expressions selected and The magnitude of time interval Dynamics of Structures

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Central Difference Method 

The central difference expressions for velocity and acceleration are: u − 2ui + ui −1 u −u u&&i = i +1 u&i = i +1 i −1 (∆t )2 2∆t



The 3rd expression is obtained by substituting the above two expressions in the equation of motion: m



u −u ui +1 − 2ui + ui −1 + c i +1 i −1 + kui = pi 2 2 ∆t (∆t )

Re-arranging the above expression ⎡ m ⎡ m ⎡ 2m ⎤ c ⎤ c ⎤ + − u ⎢ ⎥ui +1 = pi − ⎢ ⎥ui −1 − ⎢k − 2 2 2⎥ i ⎣ (∆t ) 2∆t ⎦ ⎣ (∆t ) 2∆t ⎦ ⎣ (∆t ) ⎦



For i = 0, the expression becomes ⎡ m ⎡ m ⎡ c ⎤ c ⎤ 2m ⎤ + − u ⎢ ⎥u1 = p0 − ⎢ ⎥u −1 − ⎢k − 2 2 2⎥ 0 ∆ ∆ 2 t 2 t ⎣ (∆t ) ⎦ ⎣ (∆t ) ⎦ ⎣ (∆t ) ⎦ Dynamics of Structures

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Central Difference Method (Cont..) 

To start the procedure we require the value of u-1. which is calculated from the central difference expressions as: u −u u − 2u0 + u−1 u&0 = 1 −1 u&&0 = 1 2∆t (∆t )2



From the simultaneous solution of the above two equations: (∆t )2 u&& u−1 = u0 − ∆tu&0 + 0 2



Initial displacement (u0) and velocity (ù0) are given and initial acceleration (ü0) is determined from equation of motion as: p0 − cu&0 − ku0 m ∆t 1 < The method is stable when: T π u&&0 =



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Example 8.10 ⎡ m ⎡ ⎡ m 2m ⎤ c ⎤ c ⎤ u − + ⎥ui −1 − ⎢k − ⎢ ⎥ui +1 = pi − ⎢ 2 2 2⎥ i ⎣ (∆t ) 2∆t ⎦ ⎣ (∆t ) ⎦ ⎣ (∆t ) 2∆t ⎦

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Example 8.10

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Constant Acceleration Method 

The expressions for displacement and velocity are: (∆t )2 u&& − − − (1) u& = u& + ∆tu&& − − − −(2) u = u + ∆tu& + i +1



i

i

2

i +1

i

i

i

The equation of motion is: mu&&i +1 + cu&i +1 + kui +1 = pi +1



Solving the above three equations for ün+1: u&&i +1 =



⎛ (∆t )2 1 ⎧⎪ ⎨ pi +1 − kui − (c + ∆tk )u&i − ⎜⎜ ∆tc + m ⎪⎩ 2 ⎝

⎞ ⎫⎪ k ⎟⎟u&&i ⎬ − − − −(3) ⎠ ⎪⎭

Starting from the initial conditions, the response parameters can be calculated at any time instant using equations (1), (2) and (3) Dynamics of Structures

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Newmark’s Method 

In 1959, Newmark devised a series of numerical integration formula of he form:

u&i +1 = u&i + ∆t (1 − γ )u&&i + (∆tγ )u&&i +1 − − − − − − − − − −(1a)

(

)

ui +1 = ui + (∆t )u&i + ∆t 2 (0.5 − β )u&&i + ∆t 2 β u&&i +1 − − − −(1b) 

The parameters γ and β define: z z

 

For γ = β = 0, the method reduces to constant acceleration method Other Newmark methods are: z z



Variation of acceleration over the time step and Determine the stability and accuracy characteristics of the method

Average Acceleration Method (γ = 1/2 and β = 1/4) Linear Acceleration Method (γ = 1/2 and β = 1/6)

Newmark methods are stable when:

∆t 1 ≤ T π 2

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γ − 2β 16

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Average Acceleration Method 

The expressions for displacement and velocity are: ui +1 = ui + ∆tu&i +

u&i +1 = u&i +

(∆t )2 u&& + (∆t )2 u&& 4

i

4

i +1

− − − (1)

∆t ∆t u&&i + u&&i +1 − − − − − − − (2) 2 2



From the equation of motion: ⎧⎪ ⎛ ∆t (∆t )2 1 ⎜ c+ & ( ) p − ku − c + ∆ tk u − u&&i +1 = ⎨ 1 i + i i ⎜ 2 4 ⎛ ∆t ∆t 2 ⎞ ⎪⎩ ⎝ ⎜⎜ m + c + k ⎟⎟ 2 4 ⎝ ⎠



Starting from the initial conditions, the response parameters can be calculated at any time instant using equations (1), (2) and (3)

⎞ ⎫⎪ k ⎟⎟u&&i ⎬ − − − −(3) ⎠ ⎪⎭

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Linear Acceleration Method 

The expressions for displacement and velocity are: ui +1 = ui + ∆tu&i +

u&i +1 = u&i + 



(∆t )2 u&& + (∆t )2 u&& 3

i

6

i +1

− − − (1)

∆t ∆t u&&i + u&&i +1 − − − − − − − (2) 2 2

From the equation of motion: ⎧⎪ ⎛ ∆t (∆t )2 1 ⎜ c+ & ( ) − − + ∆ − p ku c tk u u&&i +1 = ⎨ 1 i + i i ⎜ 2 3 ⎛ ∆t ∆t 2 ⎞ ⎪⎩ ⎝ ⎜⎜ m + c + k ⎟⎟ 2 6 ⎝ ⎠

⎞ ⎫⎪ k ⎟⎟u&&i ⎬ − − − −(3) ⎠ ⎪⎭

Starting from the initial conditions, the response parameters can be calculated at any time instant using equations (1), (2) and (3)

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lecture_10_numerical-evaluation-of-dynamic

By: Dr. Mohammad Ashraf (engineerashraf@yahoo.com) Office: CE: B109 Department of Civil Engineering, University of Engineering and Technolog...