150
5 Two-Dimensional Problems in Cartesian Co-ordinate System
φ =
a 2
2
x
2
+
b 3
2
x
2 y +
d 3
6
y
3
+
d 5
6
x
2 y
3
−
d 5
30
y
5
(5.46a)
Now, by definition,
σ x =
∂
2
φ
∂ y 2 = d 3 y + d 5
x
2 y −
2
3
y
3
(5.46b)
σ y =
∂
2
φ
∂ x 2 = a 2 + b 3 y +
d 5
3
y
3
(5.46c)
τ xy = −b 3 x − d 5 x y
2
(5.46d)
The following boundary conditions must be satisfied.
(i)
τ xy
y=±h
= 0
(ii)
σ y
y=+h
= 0
(iii)
σ y
y=−h
= −q
(iv)
+h
−h
(σ x ) x=±L dy = 0
(v)
+h
−h
τ xy
x=±L
dy = ±q L
(vi)
+h
−h
(σ x ) x=±L y dy = 0
The first three conditions when substituted in Eqs. (5.46c) and (5.46d) give
− b 3 − d 5 h
2
= 0
a 2 + b 3 h +
d 5
3
h
3
= 0
a 2 − b 3 h −
d 5
3
h
3
= −q
which gives on solving
a 2 = −
q
2
, b 3 =
3q
4h
, d 5 = −
3q
4h 3
Now, from condition (vi), we have
+h
−h
d 3 y + d 5
x
2 y −
2
3
y
3
y dy = 0
Simplifying,