8.5 Cerrutti’s Problem
277
Fig. 8.10 Cerrutti’s problem
R
z
z
P(x,y,z)
y
y
x
x
P
.
u y =
P
4π G R
x y
R 2 − (1 − 2v)
x y
(R + z) 2
(8.19a)
u z =
P
4π G R
xz
R 2 + (1 − 2v)
x
R + z
(8.19b)
and the stresses are
σ x = −
Px
2π R 3
−
3x
2
R 2 +
(1 − 2v)
(R + z) 2
R
2
− y
2
−
2Ry
2
(R + z)
(8.20)
σ y = −
Px
2π R 3
−
3y
2
R 2 +
(1 − 2v)
(R + z) 2
3R
2
− x
2
−
2Rx
2
(R + z)
(8.20a)
σ z =
3Pxz
2
2π R 5
(8.20b)
τ xy = −
Py
2π R 3
−
3x
2
R 2 +
(1 − 2v)
(R + z) 2
−R
2
+ x
2
+
2Rx
2
(R + z)
(8.20c)
τ yz =
3Pxyz
2π R 5
(8.20d)
τ zx =
3Px
2 z
2π R 5
(8.20e)
Here, R
2
= x
2
+ y
2
+ z
2
From the above, it is clear that the stresses approach to zero for large value of R. As
seen from the x-component of the displacement field, it is observed that the particles
are displaced in the direction of the point load. The y-component of displacement
moves particles away from the x-axis for positive values of x and towards the xaxis for negative x. The plot of horizontal displacement vectors at the surface z =0
is shown in Fig. 8.11 for the special case of an incompressible material. Vertical