4.5 Time Domain Analysis: Non-proportional Damping
103
If the real parts of the eigenvalues {λ i }, i = 1, . . . , 2 n, are negative, the system
solution becomes
z ss (t)
2 n
i=1
u i
t
0
e
λ i (t−s)
v T b f(s)
i
v T
i u i
ds,
(4.91)
for large times, and is referred to as the steady-state solution.
The initial conditions for the modal coordinates {q i (t)} result from the initial
conditions
x 0 , ˙
x 0
in the physical space and the representation of z(t) in Eq. 4.87.
We have z 0 = u q 0 at time t = 0 from the initial conditions in the physical space.
The left multiplication of z 0 = u q 0 by v T gives v T z 0 = v T u q 0 = diag{v T
i u i q i,0 }
or
v T z 0
i
= v T
i u i q i,0 . This shows that the initial conditions for the modal
coordinates are
q i,0 =
v T z 0
i
v T
i u i
, i = 1, . . . , 2 n.
(4.92)
4.5.5 Free Vibration
The free vibration solution results from Eq. 4.90 by setting f(t) = 0. For completeness, we also present the solution by direct arguments for this special case. The
representation of z(t) in Eq. 4.87 and the equation of motion ˙
z(t) = a z(t) give
u ˙
q(t) = a u q(t) ( which becomes)
v
T u ˙
q(t) = v
T a u q(t) (by left multiplication with v
T )
v
T
i u i ˙
q i (t) = v
T
i a u i q i (t), i = 1, . . . , 2 n, ( by orthogonality).
The latter condition gives
˙
q i (t) = λ i q i (t) so that q i (t) = q i,0 e
λ i t
z(t) =
2 n
i=1
u i q i,0 e
λ i t ,
(4.93)
where {q i,0 } denotes initial conditions for {q i (t)}, which can be obtained by the
approach of the previous subsection.
Example 4.9 Consider a SDOF system with damping ratio ζ = 0.05 and natural
frequency ω = 6 rad/s, which is subjected to a unit force applied suddenly. The
oscillator is at rest at the initial time. Its displacement x(t) satisfies the differential
equation