duhamel's integral
TRANSCRIPT
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 wwtwwtt sincoscossin)(sin , in Duhamel’s integral.
Assuming zero initial conditions, we obtain Duhamel’s integral, eq. (4.4), in the form
dFm
tdFm
ttyt
t
t
t 00
sin)(1
coscos)(1
sin)( (4.14a)
or
ttBttAm
ty
cos)(sin)(1
)(
(4.14b)
where
dFtAt
0cos)()(
(4.15a)
t
dFtB0
sin)()(
(4.15b)
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()
dItAt
0)()(
For trapezoidal rule
nnO IIIIItA 121 2222
1)(
(4.16)
and for Simpson’s rule
nnO IIIIItA 121 4243
1)(
(4.17)
where
t
n 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
dFtAtAi
i
t
tii
1
cos)()()( 1 (4.18)
i
i
t
tii dFtBtB
1
sin)()()( 1 (4.19)
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
iii
i
ii ttt
t
FtFF
),()()( 111 (4.20)
)()( 1 iii tFtFF
1 iii ttt
The substitution of eq. (4.20) into eq. (4.18) and integration yield
dtt
FtFtAtA
i
i
t
ti
i
iiii
1
cos)()()( 111 (4.20a)
i
i
i
i
t
ti
i
it
tii dt
t
FdtFtA
11
coscos)()( 111 (4.20b)
i
i
i
i
i
i
t
ti
i
it
ti
it
tii dt
t
Fd
t
FdtFtA
111
coscoscos)()( 111 (4.20c)
Consider the 1st and 3rd integrals and let u and ddu
i
i
i
i
i
i
t
t
it
ti
t
ti
tFduutFdtF
111
sin)(
cos)(cos)( 111
(4.20d)
11 sinsin)(
ii
i tttF
(4.20e)
i
i
i
i
i
i
t
ti
i
it
ti
i
it
ti
i
i tt
Fduut
t
Fdt
t
F
111
sincoscos 111
(4.20f)
11 sinsin
iii
i
i tttt
F
(4.20g)
To evaluate the 2nd integral we use “integration by parts”.
Let u ddv cos
ddu
sinv
i
i
i
i
i
i
tt
ti
it
ti
i dt
Fd
t
F
11
1
sinsincos
(4.20h)
i
i
t
iiii
i
i duu
ttttt
F
1
sinsinsin 11
(4.20i)
i
i
t
t
iiii
i
i ttttt
F
1
cos1
sinsin 11
(4.20j)
1112coscossinsin
iiiiii
i
i ttttttt
F
(4.20k)
Therefore
1111
sinsin)()()( ii
i
iiiii
tt
t
FttFtAtA
1112sinsincoscos
iiiiii
i
i ttttttt
F
(4.21)
By the same token
ii
i
iiiii
tt
t
FttFtBtB
coscos)()()( 1
111
1112coscossinsin
iiiiii
i
i ttttttt
F
(4.22)
Equations (4.21) and (4.22) are recurrent formulas for the evaluation of the integrals in eq. (4.15) at any
time itt .
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.
tm
dFetdy D
D
t sin)(
)( (4.23)
Summing these differential response terms over the entire loading interval results in
t
D
t
D
dteFm
ty0
sin)(1
)(
(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)
D
t
DDDDm
ettBttAty
cos)(sin)()( (4.25)
where
i
i
t
tD
t
iDiD tdeFtAtA1
cos)()()( 1 (4.26)
i
i
t
tD
t
iDiD tdeFtBtB1
sin)()()( 1 (4.27)