Dynamics cheat sheet

8.4 Response to impulsive loading
8.4.1 impulse input
Undamped system with impulse
mü + ku = F 0 δ(t)
with initial conditions u (0) = 0 and u ′ (0) = 0.Assuming the impulse acts for a very short time period from
0 to t 1 seconds, where t 1 is small amount. Integrating the above differential equation gives
∫ t1
0
müdt +
∫ t1
0
kudt =
∫ t1
0
F 0 δ(t)
Since t 1 is very small, it can be assumed that u changes is negligible, hence the above reduces to
since we assumed u ′ (0) = 0 and since ∫ t 1
0
∫ t1
0
∫ t1
müdt =
0
( ) d ˙u
m dt =
dt
∫ ˙u(t1 )
˙u(0)
∫ t1
0
∫ t1
0
d ˙u = F 0
m
˙u (t 1 ) − ˙u (0) = F 0
m
˙u (t 1 ) = F 0
m
F 0 δ(t)
F 0 δ(t)
∫ t1
0
∫ t1
0
∫ t1
0
δ(t)
δ(t)
δ(t)
δ(t) = 1 then the above reduces to
˙u (t 1 ) = F 0
m
Therefore, the effect of the impulse is the same as if the system was a free system but with initial velocity given
by F 0
m
and zero initial position. Hence the system is now solved as follows
With u (0) = 0 and u ′ (0) = F 0
m
. The solution is
mü + ku = 0
u impulse (t) = F 0
sin ωt
mω
If the initial conditions were not zero, then the solution for these are added to the above. From earlier, it was
found that the solution is u (t) = u(0) cos ωt + u′ (0)
ω
sin ωt, therefore, the full solution is
u (t) =
under-damped with impulse c < c r , ξ < 1
due to IC only
due to impulse
{ }} { { }} {
u(0) cos ωt + u′ (0)
ω
sin ωt + F 0 sin ωt
mω
mü + c ˙u + ku = δ(t)
ü + 2ξω ˙u + ω 2 u = δ(t)
with initial conditions u (0) = 0 and u ′ (0) = 0.Integrating gives
∫ t1
0
müdt +
∫ t1
0
c ˙udt +
41
∫ t1
0
kudt =
∫ t1
0
F 0 δ(t)