In theory of vibrations, Duhamel's integral is a way of calculating the response of linear systems and structures to arbitrary time-varying external perturbation.
The response of a linear, viscously damped single-degree of freedom (SDOF) system to a time-varying mechanical excitation p(t) is given by the following second-order ordinary differential equation
m | {d2x(t) |
If a system initially rests at its equilibrium position, from where it is acted upon by a unit-impulse at the instance t=0, i.e., p(t) in the equation above is a Dirac delta function δ(t), , then by solving the differential equation one can get a fundamental solution (known as a unit-impulse response function)
h(t)= \begin{cases} | 1 |
{m\omegad |
c\ne0
p(t)=\delta(t-\tau)
h(t-\tau)=
1 | |
{m\omegad |
t\ge\tau
Regarding the arbitrarily varying excitation p(t) as a superposition of a series of impulses:
p(t) ≈ \sum\tau{p(\tau) ⋅ \Delta\tau ⋅ \delta}(t-\tau)
x(t) ≈ \sum\tau{p(\tau) ⋅ \Delta\tau ⋅ h}(t-\tau)
\Delta\tau\to0
x(t)=
t | |
\int | |
0 |
{p(\tau)h(t-\tau)d\tau}
x(t)=
1 | |
{m\omegad |
The above SDOF dynamic equilibrium equation in the case p(t)=0 is the homogeneous equation:
{d2x(t) | |
\bar{c}= | c | ,\bar{k}= |
m |
k | |
m |
xh(t)=C1 ⋅
| ||||||
e |
+\sqrt{\bar{c}2-4 ⋅ \bar{k}}\right)t}+C2 ⋅
| ||||||
e |
+\sqrt{\bar{c}2-4 ⋅ \bar{k}}\right)t}
A=
1 | |
2 |
\left(\bar{c}-\sqrt{\bar{c}2-4\bar{k}}\right), B=
1 | |
2 |
\left(\bar{c}+\sqrt{\bar{c}2-4\bar{k}}\right), P=\sqrt{\bar{c}2-4\bar{k}}, P=B-A
xh(t)=C1e + C2e
This solution has the form:
xp(t)=
\int{\bar{p(t) | |
⋅ |
eAtdt} ⋅ e-At-\int{\bar{p(t)} ⋅ eBtdt} ⋅ e-Bt
xp(t)=
Qt ⋅ e-At-Rt ⋅ e-Bt | |
P |
x(t)=xh(t)+xp(t)=C1 ⋅ e+C2 ⋅ e+
Qt ⋅ e-At-Rt ⋅ e-Bt | |
P |
dx | |
dt |
=-Ae-At ⋅ C2-Be-Bt ⋅ C1+
1 | \left[ | |
P |
Qt |
⋅ e-At-AQt ⋅ e-At-
Rt |
⋅ e-Bt+BRt ⋅ e-Bt\right]
Qt |
=p(t) ⋅ eAt,
Rt |
=p(t) ⋅ eBt
C1,C2
x(t)|t=0=0:C1+C
|
=0
C1+C
|
\left.
dx | |
dt |
\right|t=0=0:-A ⋅ C2-B ⋅ C1+
1 | |
P |
⋅ [-A ⋅ Q0+B ⋅ R0]=0
A ⋅ C2+B ⋅
C | ||||
|
⋅ [B ⋅ R0-A ⋅ Q0]
\left.{\begin{alignat}{5} C1&& +&& C2&& =&&
R0-Q0 | |
P |
&\\ B ⋅ C1&& +&& A ⋅ C2&& =&&
1 | |
P |
⋅ [B ⋅ R0-A ⋅ Q0]\end{alignat}}\right|{\begin{alignat}{5} C1&& =&&
R0 | |
P |
&\\ C2&& =&& -
Q0 | |
P |
\end{alignat}}
C1
C2
x(t)= | Qt-Q0 |
P |
⋅ e-
Rt-R0 | |
P |
⋅ e
Qt-Q0
Rt-R0
x(t)=
1 | |
P |
⋅ \left[
t{\bar{p}(\tau) | |
\int | |
0 |
⋅ eA\taud\tau} ⋅ e-At
t{\bar | |
-\int | |
0 |
p(\tau) ⋅ eB\taud\tau} ⋅ e-Bt\right]
c=2\xi\omegam, k=\omega2m
\bar{c}=2\xi\omega,\bar{k}=\omega2
P=2\omegaDi, A=\xi\omega-\omegaDi, B=\xi\omega+\omegaDi
\omegaD=\omega ⋅ \sqrt{1-\xi2}
x(t)= | 1 |
\omegaD |
t | |
\int | |
0 |
{\bar{p}(\tau)e-\xi\omega(t-\tau)\sin(\omegaD(t-\tau))d\tau}