Duhamel's Integral

NUMERICAL EVALUATION OF DUHAMEL’S INTEGRAL UNDAMPED SYSTEM In many practical cases the applied loading function is known only from experimental data as in the case of seismic motion and the response must be evaluated by a numerical method. For this purpose we use the trigonometric identity sin (t   )  sin wt cos w  cos wt sin w , in Duhamel’s integral. Assuming zero initial conditions, we obtain Duhamel’s integral, eq. (4.4), in the form

y(t )  sin t

1 t 1 t F (  ) cos  d   cos  t F ( ) sin  d m t 0 m t 0

(4.14a)

or

y(t ) 

1 A(t ) sin t  B(t ) cos t m

(4.14b)

where

A(t )   F ( ) cos d

(4.15a)

B(t )   F ( ) sin  d

(4.15b)

t

0 t

0

The calculation of Duhamel’s integral thus requires the evaluation of the integrals A(t) and B(t) numerically. Several numerical integration techniques have been used for this evaluation. The most popular of these methods are the TRAPEZOIDAL RULE and the SIMPSON’S RULE. Consider the integration of a general function I() t

A(t )   I ( )d 0

For trapezoidal rule

1 A(t )   I O  2 I1  2 I 2    2 I n 1  I n  2

(4.16)

1 A(t )   I O  4 I1  2 I 2    4 I n1  I n  3

(4.17)

and for Simpson’s rule

where n 

t must be an even number for Simpson’s rule. 

An alternative approach to the evaluation of Duhamel’s integral is based on obtaining the exact analytical solution of the integral for the loading function assumed to be given by a succession of linear

segments. This method does not introduce numerical approximations for the integration other than those inherent in the round off error, so in this sense it is an exact method. In using this method, it is assumed that F ( ) , the forcing function may be approximated by a segmentally linear function as shown. To provide a complete response history, it is more convenient to express the integrations in eq. (4.15) in incremental form, namely

A(ti )  A(ti1 )   F ( ) cos d

(4.18)

B(ti )  B(ti1 )   F ( ) sin  d

(4.19)

ti

ti 1 ti

ti 1

Where A(ti) and B(ti) represent the values of the integrals in eq. (4.15) at time ti. Assuming that the forcing function F() is approximated by a piecewise linear functions as shown in Fig. 4.6, we may write

F ( )  F (ti1 ) 

Fi (  ti1 ), ti

ti1    ti

(4.20)

Fi  F (ti )  F (ti1 )

ti  ti  ti1 The substitution of eq. (4.20) into eq. (4.18) and integration yield ti   F A(ti )  A(ti1 )    F (ti1 )  i   ti1  cos d ti 1 ti  

 A(ti1 )   F (ti1 ) cos d   ti

ti

ti 1

ti 1

(4.20a)

Fi   ti1 cos d ti

ti F Fi i  cos d   ti1 cos d ti 1 t ti 1 t i i

 A(ti1 )   F (ti1 ) cos d   ti

ti

ti 1

(4.20b)

(4.20c)

Consider the 1st and 3rd integrals and let u   and du  d



ti

ti 1

F (ti1 ) cos d F (ti1 )  cos u 

ti

du

ti 1



F (ti1 )





F (ti1 )



sin 

ti

(4.20d)

ti 1

sin ti  sin ti1 

ti ti Fi F du F ti1  cos d  i ti1  cosu    i ti1 sin t t i 1 i 1 ti ti  ti

(4.20e)

ti ti 1

(4.20f)



Fi t sin ti  sin ti1  ti i1

(4.20g)

To evaluate the 2nd integral we use “integration by parts”. dv  cos d Let u  

du  d Fi ti

v

sin  



t  ti sin   Fi   sin   i     cos  d    d  ti1   i 1 ti    ti 1  ti



Fi ti

Fi  ti 

(4.20h)

ti sin u         t sin  t  t sin  t  du  i i i  1 i  1   i 1  

(4.20i)

ti   1 ti sin ti   ti1 sin ti1   cos     ti 1  

(4.20j)

Fi ti sinti   ti1 sinti1  costi   costi1   2 ti

(4.20k)

Therefore

 F  sin ti  sin ti1  A(ti )  A(ti1 )   F (ti1 )  ti1 i   ti    Fi cos ti  cos ti1  ti sin ti  ti1 sin ti1  2 ti

(4.21)

By the same token

 F  cos ti1  cos ti  B(ti )  B(ti1 )   F (ti1 )  ti1 i   ti    Fi sin ti  sin ti1  ti cos ti  ti1 cos ti1  2 ti

(4.22)

Equations (4.21) and (4.22) are recurrent formulas for the evaluation of the integrals in eq. (4.15) at any time t  ti .

Determine the dynamic response of a tower subjected to a blast loading. The idealization of the structure and the blast loading are shown. Neglect damping.

NUMERICAL EVALUATION OF DUHAMEL’S INTEGRAL UNDAMPED SYSTEM

The response of a damped system expressed by the Duhamel’s integral is obtained in a manner entirely equivalent to the undamped analysis except that the impulse producing an initial velocity is substituted into the corresponding damped free-vibration equation.

dy(t )  e  t  

F ( )d sin D t    mD

(4.23)

Summing these differential response terms over the entire loading interval results in

y (t ) 

1 mD

t

 F ( )e

 t  

0

sin D t   d

(4.24)

Which is the response for a damped system in terms of the Duhamel’s integral. For numerical evaluation, we proceed as in the undamped case and obtain from eq. (4.24)

y (t )  AD (t ) sin Dt  BD (t ) cos Dt

e t mD

(4.25)

where ti

AD (ti )  AD (ti1 )   F ( )et cos Dtd ti 1

ti

BD (ti )  BD (ti1 )   F ( )et sin Dtd ti 1

(4.26)

(4.27)