4.4 Time Domain Analysis: Proportional Damping
87
which gives
T m ¨
q(t) +
T k q(t) = 0,
by left multiplication with T . The orthogonality condition of Eq. 4.10 implies
diag{ ˜
m i } ¨
q(t) + diag{ ˜
k i } q(t) = 0 or, equivalently, ˜
m i ¨
q i (t) + ˜
k i q i (t) = 0,
i = 1, . . . , n
so that
¨
q i (t) + ω
2
i q i (t) = 0, i = 1, . . . , n,
(4.53)
by using Eq. 4.9.
The differential equations of Eq. 4.53 describe the motion of n undamped single
degree of freedom (SDOF) systems with natural frequencies {ω i } in free vibration
so that (see results for SDOF systems)
q i (t) = q i,0 cos(ω i t) +
˙
q i,0
ω i
sin(ω i t),
(4.54)
where (q i,0 , ˙
q i,0 ) are the ICs for these oscillators, which can be obtained, e.g., from
Eq. 4.31 or Eq. 4.32. The free vibration solution of the MDOF system is
x(t) =
n
i=1
i q i (t) =
n
i=1
i
q i,0 cos(ω i t) +
˙
q i,0
ω i
sin(ω i t)
.
(4.55)
Example 4.5 Consider the cantilever in Fig. 4.8 with the masses m 1 = 2 and m 2 =
1 at the cantilever mid height and its free end. The mass and stiffness matrices of
this two degree of freedom system are
m =
2 0
0 1
and k =
16 −5
−5 2
.
Our objective is to find the system displacement x(t) for the initial conditions x 0 = 0
and ˙
x 0 = 0. The following three steps deliver the displacement vector.
Step1: Modal analysis:
– Modal frequencies: det
− ω 2 m + k
= 0, i.e.,
−2ω 2 + 16 −5
−5
−ω 2 + 2
= 0 ⇒
ω 2
1 = 0.3632
ω 2
2 = 9.6968.
87
which gives
T m ¨
q(t) +
T k q(t) = 0,
by left multiplication with T . The orthogonality condition of Eq. 4.10 implies
diag{ ˜
m i } ¨
q(t) + diag{ ˜
k i } q(t) = 0 or, equivalently, ˜
m i ¨
q i (t) + ˜
k i q i (t) = 0,
i = 1, . . . , n
so that
¨
q i (t) + ω
2
i q i (t) = 0, i = 1, . . . , n,
(4.53)
by using Eq. 4.9.
The differential equations of Eq. 4.53 describe the motion of n undamped single
degree of freedom (SDOF) systems with natural frequencies {ω i } in free vibration
so that (see results for SDOF systems)
q i (t) = q i,0 cos(ω i t) +
˙
q i,0
ω i
sin(ω i t),
(4.54)
where (q i,0 , ˙
q i,0 ) are the ICs for these oscillators, which can be obtained, e.g., from
Eq. 4.31 or Eq. 4.32. The free vibration solution of the MDOF system is
x(t) =
n
i=1
i q i (t) =
n
i=1
i
q i,0 cos(ω i t) +
˙
q i,0
ω i
sin(ω i t)
.
(4.55)
Example 4.5 Consider the cantilever in Fig. 4.8 with the masses m 1 = 2 and m 2 =
1 at the cantilever mid height and its free end. The mass and stiffness matrices of
this two degree of freedom system are
m =
2 0
0 1
and k =
16 −5
−5 2
.
Our objective is to find the system displacement x(t) for the initial conditions x 0 = 0
and ˙
x 0 = 0. The following three steps deliver the displacement vector.
Step1: Modal analysis:
– Modal frequencies: det
− ω 2 m + k
= 0, i.e.,
−2ω 2 + 16 −5
−5
−ω 2 + 2
= 0 ⇒
ω 2
1 = 0.3632
ω 2
2 = 9.6968.