8.3 Boussinesq’s Problem
267
σ r = A 1
∞
z
r
2
−
1
2
z
2
r
2
+ z
2
−
5
2 dz
=
A 1
2
1
r 2 −
z
r 2
r
2
+ z
2
−
1
2
− z
r
2
+ z
2
−
3
2
σ z = A 1
∞
z
z
2
−
1
2
r
2
r
2
+ z
2
−
5
2 dz
=
A 1
2
z
r
2
+ z
2
−
3
2
(8.9)
τ r z =
3
2
A 1
∞
z
r z
r
2
+ z
2
−
5
2 dz =
A 1
2
r
r
2
+ z
2
−
3
2
σ θ = −
1
2
A 1
∞
z
r
2
+ z
2
−
3
2 dz
= −
A 1
2
1
r 2 −
z
r 2
r
2
+ z
2
−
1
2
On the plane z = 0, we find that the normal stress is zero and the shearing stress
is
τ r z =
1
2
A 1
r 2
(b)
From (a) and (b), it is seen that the shearing forces on the boundary plane are
eliminated if, −B(1 − 2v) +
A 1
2
= 0
Therefore, A 1 = 2B(1 − 2v)
Substituting the value of A 1 in Eq. (8.9) and adding together the stresses from
Eqs. (8.6) and (8.9), we get
σ r = B
(1 − 2v)
1
r 2 −
z
r 2
r
2
+ z
2
−
1
2
− 3r
2 z
r
2
+ z
2
−
5
2
σ z = −3Bz
3
r
2
+ z
2
−
5
2
σ θ = B(1 − 2v)
−
1
r 2 +
z
r 2
r
2
+ z
2
−
1
2
+ z
r
2
+ z
2
−
3
2
τ r z = −3Br z
2
r
2
+ z
2
−
5
2
(8.10)