272
8 Elastic Solutions in Geomechanics
∂φ
∂θ
=
q
π
r [θ cos θ + sin θ ]
∂φ
∂θ
=
qrθ cos θ
π
+
qr sin θ
π
Similarly,
∂
2 φ
∂θ 2 =
qr
π
[θ(− sin θ) + cos θ ] +
qr
π
cos θ
∂
2
φ
∂θ 2 =
qr
π
cos θ +
qr
π
cos θ −
qrθ
π
sin θ
Therefore, Eq. (8.13) becomes
σ r =
1
r
q
π
θ sin θ
+
1
r 2
q
π
r cos θ +
q
π
r cos θ −
q
π
r θ sin θ
=
q
πr
θ sin θ +
2q
πr
cos θ −
q
πr
θ sin θ
Or σ r =
2q cos θ
πr
Now,
1
r
∂φ
∂θ
=
q
π
[θ cos θ + sin θ ]
∂
∂r
1
r
∂φ
∂θ
= 0
Hence, τ rθ = 0
The stress function assumed in Eq. (8.12) will satisfy the compatibility equation
∂
2
∂r 2 +
1
r
∂
∂r
+
1
r 2
∂
2
∂θ 2
∂
2
φ
∂r 2 +
1
r
∂φ
∂r
+
1
r 2
∂
2
φ
∂θ 2
= 0
Here, σ r and σ θ are the major and minor principal stresses at point P. Now, using
the above expressions for σ r , σ θ and τ rθ , the stresses in rectangular co-ordinate
system (Fig. 8.7) can be derived.
Therefore,
σ z = σ r cos
2
θ + σ θ sin
2
θ − 2τ r θ sin θ cos θ
Here, σ θ =0 and τ rθ =0
Hence, σ z = σ r cos
2
θ
2q
πr
cos θ
cos
2
θ
σ z =
2q
πr
cos
3
θ