Total and Effective Stress - Elements of Soil Mechanics - page 84

Fig. 3.11 Example 3.5.

(ii) This example illustrates how the method can be used for points outside the founda-
tion area (Fig. 3.11b ). The foundation is assumed to extend to the point K (Fig. 3.11c )
and is now split into two rectangles, AEKH and HKFD.

For both rectangles:

B
z

2 25
3

. L

z

6 25
3

.

m = = = 0 75 . ; n = = = 2 08 .

From Fig. 3.10, I σ = 0.176, therefore σ z = 0.176 × 2 × 200 = 70.4 kPa.
The effect of rectangles BEGK and KGCF must now be subtracted.
For both rectangles:

2 25
3

. 1 75

3

.

m = = 0 75 . ; n = = 0 58 .

From Fig. 3.10, I σ = 0.122 (strictly speaking m is 0.58 and n is 0.75, but m and n are
interchangeable in Fig. 3.10) . Hence:

σ z = 0 122 2 200 . × × = 48 8 . kPa

Therefore the vertical stress increment due to the foundation

= 70 4 48 8 . − . = 21 6 . kPa

Circular foundations can also be solved by this method. The stress effects from such a foundation may
be found approximately by assuming that they are the same as for a square foundation of the same area.

Example 3.6: Vertical stress increments beneath circular
foundation

A circular foundation of diameter 100 m exerts a uniform pressure on the soil of 450 kPa.
Determine the vertical stress increments for depths up to 200 m below its centre.

Solution:

π 100
4

×

2

Area of foundation = = 7850 m

2