Geotechnical Engineering

DHARM

BEARING CAPACITY 595

Inclination factors (Revised):

ic = iq =^1

2

−

+

F

HG

I

KJ

H

VAccotφ

=^1

75

1350 6 20 20

2

−

+× × °

F

HG

I

cot KJ

= 0.913

iγ = (iq)^2 = 0.913^2 = 0.833

∴ qult = 20 × 14.83 × 1.133 × 1.175 × 0.913 + 18 × 6.40 × 1.133 × 1.175 × 0.913

+

1

2 × 18 × 2 × 3.54 × 0.733 × 0.83

= 360.5 + 140.0 + 38.9 = 539.5 kN/m^2

Vertical load that can be borne,

Qult = qult × Area = 539.4 × 6 = 3236 kN

Factor of safety = 3236/1360 ≈ 2.4.

Example 14.20: What is the ultimate bearing capacity of a footing resting on a uniform sand
of porosity 40% and specific gravity 2.6, if φ = 30° (Hansen’s factors: Nc = 30.4, Nq = 18.4, Nγ =
18.08 at a depth of 1.5 m under the following conditions:

(i) Size 2 m × 3 m, G.W.L. at 8 m below natural ground level; and

(ii) Size 2 m × 3 m, G.W.L. at 1.5 m below natural ground level.

(S.V.U.—Four-year B.Tech.—Dec., 1982)

Hansen’s formula:

φ = 30°; Nc = 30.14; Nq = 18.4; Nγ = 18.08

qult = c Ncscdcic + qNqsqdqiq +

1

2 γb Nγsγiγ

Since the soil is sand, c = 0, and the first term vanishes.

n = 40%, G = 2.60 Df = 1.5 m

e =

n

()n

.

.1

04

− 06

= = 0.67

γd =

G

e

γw

()

..

1.

260 981

+ 167

= × = 15.30 kN/m^3

γ′ =

()

()

..

.

G

e

− w

+

(^1) = ×
1
160 9 81
167
γ
= 9.40 kN/m^3
Since q = γDf ,
qult = γDf Nqsqdqiq +
1
2
γb Nγsγiγ
Shape factors:
sq = 1 + 0.2 b/L = 1 + 0.2 × 2/3 = 1.33
sγ = 1 – 0.4 b/L = 1 – 0.4 × 2/3 = 0.733
Depth factors:
dq = 1 + 0.35 Df /b = 1 + 0.35 ×
15
2
.
= 1.263