8.3 Boussinesq’s Problem
265
Fig. 8.4 Boussinesq’s
problem
σ r =
∂
∂z
v∇
2
φ −
∂
2
φ
∂r 2
σ θ =
∂
∂z
v∇
2
φ −
1
r
∂φ
∂r
(8.5)
σ z =
∂
∂z
(2 − v)∇
2
φ −
∂
2
φ
∂z 2
τ r z =
∂
∂r
(1 − v)∇
2
φ −
∂
2
φ
∂z 2
Therefore, by substitution, we get
σ r = B
(1 − 2v)z
r
2
+ z
2
−
3
2
− 3r
2 z
r
2
+ z
2
−
5
2
σ θ = B[1 − 2v]z
r
2
+ z
2
−
3
2
(8.6)
σ z = B
(1 − 2v)z
r
2
+ z
2
−
3
2
+ 3z
3
r
2
+ z
2
−
5
2
τ r z = −B
(1 − 2v)r
r
2
+ z
2
−
3
2
+ 3r z
2
r
2
+ z
2
−
5
2
Now, the shearing forces on the boundary plane z =0 are given by
τ r z =
−B(1 − 2v)
r 2
(a)