DHARM

STRESS DISTRIBUTION IN SOIL 383

(i) Square Area:
Imagine, as in Fig. 10.25, the area to be divided into four equal squares. The stress at A
will be four times the stress produced under the corner of the small square.

4m

4m

4m

A

Fig. 10.25 Uniform load on square area (Ex. 10.7)

m = 2/4 = 0.5, n = 2/4 = 0.5

Iσ =

1

4

21

1

2

1

21

1

22

22 22

22

22

1

22

π^2222

mn m n

mn mn

mn

mn

mn m n

mn mn

++

+++

++

++

+

++

++−

L

N

M

M

O

Q

P

P

.tan−

=

1

4

205 05025 0251

150 025 025

025 025 2

025 025 1

205050250251

150 025 025

1

π

×× + +

+×

++

++

+

×× + +

−×

L

N

M

M

O

Q

P

P

.... −

...

. ..
..

tan

....

...

=

1

4

05 15

1 5625

250

150

05 15

1 4375

1

π

..

.

.

.

tan ..

.

×+

L

N

M

M

O

Q

P

P

− = 0.0840

(The value may be obtained from Tables or Charts also.)

∴σz = 4 × 200 × 0.084 = 67.2 kN/m^2

(ii) Equivalent point load method:

Q = 200 × 16 = 3200 kN

σz = Q

zrz^2 2 52^52

32

1

3200

16

32

1

.

(/ )

[(/)]

(/ )

//

ππ

+

=× = 95.5 kN/m^2

(iii) Four equivalent point loads:

From Example 10.3, σz = 71.14 kN/m^2

Thus, percentage error in the equivalent point load method

=

(. .)

.

95 5 67 2

67 2

− × (^100) = 42.11
Percentage error in four equivalent point loads approach

(. .)
.
7114 67 20
67 20
− × (^100) = 5.86.
Example 10.8: A rectangular foundation, 2 m × 4 m, transmits a uniform pressure of 450 kN/m^2
to the underlying soil. Determine the vertical stress at a depth of 1 metre below the foundation
at a point within the loaded area, 1 metre away from a short edge and 0.5 metre away from a
long edge. Use Boussinesq’s theory.