8.3 Boussinesq’s Problem
269
T z = 3Bz
2
r
2
+ z
2
−2
Integrating the above, we get the applied load P.
Therefore,
P =
π/2
z
T z 2πr
r
2
+ z
2
1
2 dψ
=
π/2
z
3Bz
2
r
2
+ z
2
−2
2πr
r
2
+ z
2
1
2
dψ
=
π/2
z
(6π B) cos
2
ψ sin ψdψ
= 6π B
π/2
0
cos
2
ψ sin ψdψ
Now, solving for
π/2
0
cos
2
ψ sin ψdψ, we proceed as below
Put cos ψ = t
i.e. −sinψ, dψ = dt
If cos0=1, then t =1
If cos
π
2
= 0, then t =0
Hence,
1
0 −t
2 dt = −
t
3
3
0
1
=
t
3
3
1
0
=
1
3
Therefore, P =2π B
Or B =
P
2π
Substituting the value of B in Eq. (8.10), we get
σ z = −
3P
2π
z
3
r
2
+ z
2
−
5
2
σ r =
P
2π
(1 − 2v)
1
r 2 −
z
r 2
r
2
+ z
2
−
1
2
− 3r
2 z
r
2
+ z
2
−
5
2
σ θ =
P
2π
(1 − 2v)
−
1
r 2 +
z
r 2
r
2
+ z
2
−
1
2
+ z
r
2
+ z
2
−
3
2
τ r z = −
3P
2π
r z
2
r
2
+ z
2
−
5
2
Putting R =
√
r 2 + z 2 and simplifying, we can write