156
5 Two-Dimensional Problems in Cartesian Co-ordinate System
∂φ
∂ x
=
H
π
z
2
x 2 + z 2
∂
2
φ
∂ x 2 = −
2H
π
xz
2
x 2 + z 2
2
∂
3
φ
∂ x 3 =
2H
π
z
2
3x
2
− z
2
x 2 + z 2
3
∂
4
φ
∂ x 4 =
H
π
24xz
4
− 24x
3 z
2
x 2 − z 2
4
Substituting the above values in (i), we get
4
π
1
x 2 + z 2
4
24xz
4
− 24x
3 z
2
+ 64x
3 z
2
− 24xz
4
− 8x
5
+ 8x
5
− 40x
3 z
2
= 0
Hence, the given stress function is admissible.
Therefore, the stresses are
σ x =
∂
2
φ
∂z 2 = −
24
π
x
3
x 2 + z 2
2
σ y =
∂
2
φ
∂ x 2 = −
24
π
x
2
x 2 + z 2
2
and
τ xy =
∂
2
φ
∂ x∂z
= −
24
π
x
2 z
x 2 + z 2
2
Example 5.3
Given the stress function φ = −
F
h 3
x y
2
(3h − 2y). Determine the stress components
and sketch their variations in a region included in y = 0, y = h, x = 0, on the side x
positive.
Solution The given stress function may be written as
φ = −
3F
h 2
x y
2
+
2F
h 3
x y
3
∴
∂
2
φ
∂ y 2 = −
6F x
h 2
+
12F
h 3
xz