10.8 Beam Segments with Distributed Mass Having Shear Deformation…
421
SM ¯
ψ 2 =
L
α
β
2
− s
2
sinh bα − β
α
2
+ s
2
sin bβ
μAGbs 2
α 2 + β 2
D
(10.107)
SMY 1 =
L 2
αβ
2s 2 + α 2 − β 2
(1 − cosh bα cos bβ) −
2α 2 β 2 − α 2 s 2 + β 2 s 2
sinh bα sin bβ
E I b 2
α 2 + β 2
αβ
(10.108)
SMY 2 =
β
2
− s
2
α
2
+ s
2
[− cosh bα + cos bβ]
μAGbs 2
α 2 + β 2
D
(10.109)
SV Y 1 =
L
α
β
2
− s
2
cosh bα sin bβ + β
α
2
+ s
2
sinh bα cos bβ
E I bαβ
α 2 + β 2
D
(10.110)
SV Y 2 =
L
β
α
2
− s
2
sinh bα + α
β
2
+ s
2
sin bβ
E I bαβ
α 2 + β 2
D
(10.111)
Combining Eqs. (10.106)–(10.111) yields
D = L
2
2αβ
β
2
− s
2
α
2
+ s
2
− 2αβ
β
2
− s
2
α
2
+ s
2
cosh bα cos bβ
α
2
− β
2
α
2
β
2
− s
4
+ 4α
2
β
2 s
2
sinh bα sin bβ
/
μAG E I b
2
α
2
+ β
2
αβs
2
(10.112)
The dynamic stiffness coefficients mentioned above are based on Eqs. (10.82)
and (10.83), by assuming that b
2 r
2 s
2
< 1. However, when the condition b
2 r
2 s
2
> 1
occurs, α
= iα is to be used where i =
√ −1, and then the solutions of Timoshenko
beam equation will be
Y = C 1 cos bα
ξ + iC 2 sin bα
ξ + C 3 cos bβξ + C 4 sin bβξ
(10.113)
= iC
1 sin bα
ξ + C
2 cos bα
ξ + C
3 sin bβξ + C
4 cos bβξ
(10.114)
The integration constants are, however, same as given by Eqs. (10.88)–(10.91).
The dynamic stiffness coefficients for b
2 r
2 s
2
> 1 are
SM ¯
ψ 1 =
1
bD
α
β
2
− s
2
sin bα
cos bβ + β
α
2
+ s
2
sin bβ cos bα
E I
L
(10.115)
SM ¯
ψ 2 =
−1
bD
α
β
2
− s
2
sin bα
+ β
α
2
+ s
2
sin bβ
E I
L
(10.116)