6.5 Static Solutions
163
(J 2 − α J 4 )
∂ϕ
∂ x
− α J 4
∂
2 w
∂ x 2 − α
(J 4 − α J 6 )
∂ϕ
∂ x
− α J 6
∂
2 w
∂ x 2
+
+
1
4
(B 0 − 3α B 2 )
∂ϕ
∂ x
−
∂
2 w
∂ x 2
− 3α B 2
∂ϕ
∂ x
+
∂
2 w
∂ x 2
+
(6.114)
+9α
2 B 4
∂ϕ
∂ x
+
∂
2 w
∂ x 2
x=1
x=0
= ¯
M or ϕ|
x=1
x=0 = ¯
ϕ,
α
2 I 6
∂
3 w
∂ x∂t 2 − α (I 4 − α I 6 )
∂
2
ϕ
∂t 2 +
C 0 + 3αC 2
2
+
+ ( A 0 − 3α A 2 )
ϕ +
∂w
∂ x
+
1
4
∂
∂ x
B 0
∂ϕ
∂ x
−
∂
2 w
∂ x 2
− 3α B 2
∂ϕ
∂ x
+
∂
2 w
∂ x 2
+
+ J 0
∂u
∂ x
+
1
2
∂w
∂ x
2
∂w
∂ x
+ α
∂
∂ x
(J 4 − α J 6 )
∂ϕ
∂ x
− α J 6
∂
2 w
∂ x 2
− (6.115)
−3α (A 2 − 3α A 4 )
ϕ +
∂ w
∂ x
+ 9α
2 B 2
ϕ +
∂w
∂ x
+
+
3
4
α
∂
∂ x
B 2
∂ϕ
∂ x
−
∂
2 w
∂ x 2
− 3α B 4
∂ϕ
∂ x
+
∂
2 w
∂ x 2
x=1
x=0
= ¯
Q or w|
x=1
x=0 = ¯
w,
∂ϕ
∂ x
−
∂
2 w
∂ x 2
x=1
x=0
= 0
or
∂w
∂ x
x=1
x=0
=
____
∂w
∂ x
.
(6.116)
Here I 0 , I 2 , I 4 , I 6 , J 0 , J 2 , J 4 , J 6 and A 0 , A 2 , A 4 , B 0 , B 2 , B 4 are
defined through (6.88) and in the case of panel with rectangular cross section, we
have
J 0 = I 0 = 1 , J 2 = I 2 =
1
12
, J 4 = I 4 =
1
80
, J 6 = I 6 =
1
448
,
A 0 =
1
2 (1 + ν)
, B 0 =
1
2 (1 + ν)
l
h
2
,
A 2 =
1
24 (1 + ν)
, B 2 =
1
24 (1 + ν)
l
h
2
,
A 4 =
1
160 (1 + ν)
, B 4 =
1
160 (1 + ν)
l
h
2
, α =
4
3
.
(6.117)