Dynamics cheat sheet

ζ > 1
⎧
√
⎨ −ω n ξ + ω n ξ 2 − 1
roots
⎩
√
−ω n ξ − ω n ξ 2 − 1
⎧ (
⎪⎨
−ξ+ √ (
ξ
Ae
2 −1
)ωt + Be
u (t) =
(
⎪⎩
−ξ+ √ )
ξ 2 −1
ω
u (t) = A 1 e
nt + B1 e
A = u′ (0)+u(0)ωξ+u(0)ω √ ((
ξ 2 −1+ F k β ξ+ √ )
ξ 2 −1
2ω √ ξ 2 −1
B = − u′ (0)+u(0)ωξ−u(0)ω √ ξ 2 −1+ F k β ((
ξ− √ ξ 2 −1
1
β = √
(1−r 2 ) 2 +(2ξr) 2
(
A 1 = − ˙u(t 1)+u(t 1 )ω n ξ− √ )
ξ 2 −1
√
2ω n ( ξ 2 −1
B 1 = ˙u(t 1)+u(t 1 )ξω n ξ+ √ )
ξ 2 −1
2ω n
√
ξ 2 −1
2ω √ ξ 2 −1
−ξ− √ )
ξ 2 −1 ωt +
F
π t
(
−ξ− √ )
ξ 2 −1 ω nt
k β sin (
)
ω sin θ− π cos θ t 1
)
)
ω sin θ− π cos θ t 1
)
t 1
− θ
0 ≤ t ≤ t 1
t > t 1
6 Tree view look at the different cases
This tree illustrates the different cases that needs to be considered for the solution of single degree of freedom
system with harmonic loading.
There are 12 cases to consider. Resonance needs to be handled as special case when damping is absent due to
the singularity in the standard solution when the forcing frequency is the same as the natural frequency. When
damping is present, there is no resonance, however, there is what is called practical response which occur when
the forcing frequency is almost the same as the natural frequency.
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