6.11 The Effect of Circular Holes on Stress Distributions in Plates
185
σ
r =
pb
2
2
b 2 − a 2
1 −
a
2
r 2
σ
θ =
pb
2
2
b 2 − a 2
1 +
a
2
r 2
(b) The second part of the stress σ
r and σ
θ are functions of θ . The boundary
conditions for this are:
σ
r =
p
2
cos 2θ for r = b
τ
r θ = −
p
2
sin 2θ for r = b
These stress components may be derived from a stress function of the form,
φ = f (r ) cos 2θ
because with
σ
r =
1
r 2
∂
2
φ
∂θ 2 +
1
r
∂φ
∂r
and σ
θ =
1
r 2
∂φ
∂θ
−
1
r
∂
2
φ
∂r ∂θ
Now, the compatibility equation is given by,
∂
2
∂r 2 +
1
r
∂
∂r
+
1
r 2
∂
2
∂θ 2
f (r ) cos 2θ = 0
But
∂
2
∂r 2 f (r ) cos 2θ +
1
r
∂
∂r
f (r ) cos 2θ +
1
r 2
∂
2
∂θ 2 f (r ) cos 2θ
= cos 2θ
∂
2
∂r 2 f (r ) +
1
r
∂
∂r
f (r ) −
4
r 2 f (r )
Therefore, the compatibility condition reduces to
cos 2θ
∂
∂r 2 +
1
r
∂
∂r
−
4
r 2
2
f (r ) = 0
As cos 2θ is not in general zero, we have
∂
∂r 2 +
1
r
.
∂
∂r
−
4
r 2
2
f (r ) = 0