Engineering Rock Mechanics

Question and answers: foundation and dope instability mechanisms 301

To obtain the stresses and displacements due to the complete loaded
area, we multiply the results for a single quadrant by 4. This is possible
because each quadrant of the loaded area has no structural influence on
any of the others.
For the single quadrant shown above, the displacement at a point on
the z-axis is given by

P (1 7.9)

4n G

UZ = - [ 1, (5,3) + 1, (0,O) - I, (5,O) - 1, (O,3)]

where

and

R = Jxm (17.10)

For a position on the ground surface, we have z = 0, and for this
particular foundation we have E = 800 MPa and w = 0.2 (giving
G = 333.3 MPa), with the applied pressure p being given by

p=-=-- -3MPa.

P

A 10x6

Substituting these values into Eq. (17.9) we obtain

u, = - r28.86 + 0.00 - 12.88 + 5.271

4rr G

x 10.71 = 7.672 x m

3

4n x 333.33

with the total displacement then being 424, = 4 x 7.672 x

is given by

= 0.031 m.

The stress at a point on the z-axis for the single quadrant shown above

P

4n

U, = - [I, (5,3) + 1, (0,O) - 10 (5,O) - 1n(033)1 (17.11)

where

and R is defined in Eq. (17.10).

with the appropriate values of x, y, and p into Eq. (17.11), we obtain

For a depth of 5 m we have z = -5, and substituting this, together

P

2n

0, = - [OB + 0.00 - 0.00 + 0.001

3

2n

= - x 0.85 = 0.41 MPa

with the total stress then being 4az = 4 x 0.41 = 1.63 MPa.
We can examine the distribution of both displacement and stress across
the extent of the loaded area by substituting the appropriate values of
x and y into Eq. (17.9) or Eq. (17.11), respectively. The asymmetry of
the problem under these circumstances means that we need to compute