188
6 Two-Dimensional Problems in Elasticity …
2 A + 6Bb
2
−
6C
b 4 −
2D
b 2 = −
p
2
Solving the above, we get
B = −
pa
2 b
2
2
a 2 − b 2
3
If “a” is very small in comparison to b, we may write B ∼ = 0.
Now, taking approximately,
D =
a
2 p
2
C = −
a
4 p
4
A = −
p
4
Therefore, the total stress can be obtained by adding part (a) and part (b). Hence,
we have
σ r = σ
r + σ
r =
p
2
1 −
a
2
r 2
+
p
2
1 +
3a
4
r 4 − 4
a
2
r 2
cos 2θ
(6.58)
σ θ = σ
θ + σ
θ =
p
2
1 +
a
2
r 2
−
p
2
1 +
3a
4
r 4
cos 2θ
(6.59)
and
τ r θ = τ
r θ = −
p
2
1 −
3a
4
r 4 +
2a
2
r 2
sin 2θ
(6.60)
Now, At r = a, σ r = 0
∴ σ r = p − 2 p cos 2θ
When θ =
π
2
or
3π
2
σ θ = 3 p
When θ = 0 or θ = π
σ θ = −p