Bearing Capacity and Shallow Foundations - Elements of Soil Mechanics - page 263

Solution:

(i) It may be assumed that immediately after construction the clay will be in an und-
rained state. The relevant soil parameter is therefore c u = 24 kPa.
From Fig. 9.6: N c = 5.7, N q = 1.0, N γ = 0.0.

q = cN 1 0 3

24 5 7 1 0 3 2 4

( + . B L / ) + γ zN u c q

= × . ( + . × / ) + × × 20 1 1

= 177 3 . kPa

(ii) It can be assumed that, after some years, the clay will be fully drained so that the
relevant soil parameters are φ′ = 25° and c ′ = 0.
From Fig. 9.6: N c = 25.1, N q = 12.7. N γ = 9.7.

q = γ γ 0 0

20 1 12 7 0 5 20 2 9 7 1 0 2 2 4

zN + . 5 BN 1 ( − . 2B L / )

u q

= × × . + . × × × . ( − . × / )

= 428 6 kPa .

Example 9.2: Ultimate bearing capacity (Terzaghi); effect of φ ′

A continuous foundation is 1.5 m wide and is founded at a depth of 1.5 m in a deep
layer of sand of unit weight 18.5 kN/m 3 .
Determine the ultimate bearing capacity of the foundation if the soil strength param-
eters are c ′ = 0 and φ ′ = (i) 35°, (ii) 30°.

Solution:

(ii) From Fig. 9.6: for φ′ = 35°, N c = 57.8, N q = 41.4, N γ = 42.4. For a continuous footing:
q = ′ + 0 5

18 5 1 5 41 4

c N γ zN + . γ BN

u c q γ

= . × × . . + 0 5 18 5 1 5 42 4 . × . × × . .

= 1737 kPa

(ii) From Fig. 9.6: for φ′ = 30°, N c = 37.2, N q = 22.5, N γ = 19.7.

q

u =

18 5 1 5 22 5 . × × . . + 0 5 18 5 1 5 19 7 . × . × × . .

= 898 kPa

The ultimate bearing capacity is reduced by about 48% when the value of φ′ is
reduced by about 15%.

9.4 Determination of the safe bearing capacity

Lumped factor of safety approach

The value of the safe bearing capacity is simply the value of the net ultimate bearing capacity divided by
a suitable factor of safety, F. The value of F is usually not less than 3.0, except for a relatively unimportant
structure, and sometimes can be as much as 5.0. At first glance these values for F appear high but the