406
10 Dynamic Direct Stiffness Method
Fig. 10.6 Example 10.2
where
γ 1 =
sin λL 1 cosh λL 1 − cos λL 1 sinh λL 1
1 − cos λL 1 cosh λL 1
λL 1
γ 1 =
sinh λL 1 − sin λL 1
1 − cos λL 1 cosh λL 1
λL 1
L 1 is the span of member 1.
Assuming p = 133 rad/s
and
λ
4
=
mp
2
E I
=
95.8 × 133
2
8.28 × 10 7 = 0.3782
Therefore,
b 1 = λL 1 = 0.3782 × 6.1 = 2.31
From Fig. 10.5, or by direct computation of γ and γ from the relations given by
Eq. (10.47), we obtain γ = 3.7147 and γ = 2.2170. Substituting these values in Eq.
(a) for the first member, we obtain
[K (λ)] 1
θ 1
θ 2
=
8.28 × 10
7
6.1
3.7174 2.2170
2.2170 3.7174
(b)
Similarly, for the second member
b 2 = λL 2 = 0.3782 × 9.15 = 3.465
For this value of b 2 , γ = 2.1535 and γ = 3.4911
Therefore, for the second member
[K (λ)] 2
θ 2
θ 3
=
8.28 × 10
7
9.15
2.1535 3.4911
3.4911 2.1535
(c)