Geotechnical Engineering

DHARM

596 GEOTECHNICAL ENGINEERING

dγ = 1.0

Inclination factors:

iq = iγ = 1 for purely vertical loading.

(i) G.W.L. at 8 m below natural ground level:

γ = γd in both terms.

∴ qult = 15.3 × 1.5 × 18.40 × 1.133 × 1.263 × 1

+

1

2

× 15.3 × 2 × 18.08 × 0.733 × 1 × 1

= 604.27 + 202.76 = 807.03 kN/m^2

(ii) G.W.L. at 1.5 m below natural ground level:

γ = γd in the first term and γ = γ′ in the second term.

∴ qult = 15.3 × 1.5 × 18.40 × 1.133 × 1.263 × 1

+

1

2

× 9.40 × 2 × 18.08 × 0.733 × 1 × 1
= 604.27 + 124.58 = 728.85 kN/m^2
Thus, there is about 10% decrease in bearing capacity as the water table rises to the
level of the base of the footing.

Example 14.21: A foundation 2 m × 3 m is resting at a depth of 1 m below the ground surface.
The soil has a unit cohesion of 10 kN/m^2 , angle of shearing resistance of 30° and unit weight of
20 kN/m^3. Find the ultimate bearing capacity using Balla’s method.

Balla’s equation:

qult = cNc + qNq +

1

2

γb Nγ

Df /b =

1

2

= 0.5

c

γb

=

×

10

20 2 = 0.25

For φ = 30° and

c

γb = 0.25

ρ = 1.9 from the charts for Df /b = 0.5

For φ = 30° and ρ = 1.9, from the relevant charts, Balla’s factors are:

Nc = 37

Nq = 25

Nγ = 64

Hence, qult = 10 × 37 + 20 × 1 × 25 +

1

2

× 20 × 2 × 64

= 370 + 500 + 1280 = 2,150 kN/m^2.

(Note. Strictly speaking, Balla’s method is applicable only for continuous footings).

Example 14.22: A plate load test was conducted on a uniform deposit of sand and the follow-
ing data were obtained: