Geotechnical Engineering

DHARM

594 GEOTECHNICAL ENGINEERING

qsafe =

qnet ult

η

+ γDf =

756 1

012

3

18 0 6

FHG + IKJ

+×

L

N

M

M

M

M

O

Q

P

P

P

P

.

b. kN/m 2

Qsafe = qsafe × A =

5

3

756

1 012

3

10 8

b^2 + b

F

HG

I

KJ

+

L

N

M

M

M

M

O

Q

P

P

P

P

.

. kN

Equating Qsafe to 10,000, we have

420 b^2 1

012

FHG + IKJ

.

b + 18 b

(^2) = 10,000
Solving for b,
b = 4.72 m, say 4.80 m. (Df /b < 2.5 is satisfied)
L = 4.8/0.6 = 8.0 m
Hence, the size of the foundation required is 4.8 m × 8.0 m.
Example 14.19: Calculate the ultimate bearing capacity, according to the Brinch Hansen’s
method, of a rectangular footing 2 m × 3 m, at a depth of 1 m in a soil for which γ = 18 kN/m^2 ,
c = 20 kN/m^2 , and φ = 20°. The ground water table is lower than 3 m from the surface. The total
vertical load is 1350 kN and the total horizontal load is 75 kN at the base of the footing.
Hansen’s factors for φ = 20° are Nc = 14.83, Nq = 6.40, and Nγ = 3.54. Determine also the factor
safety.
Rectangular footing:
φ = 20°, Nc = 14.83, Nq = 6.40, Nγ = 3.54, Df = 1 m
(Hansen’s factors)
c = 20 kN/m^2 q = γDf = 18 × 1 = 18 kN/m^2 γ = 18 kN/m^3
b = 2 m L = 2 m A = 6 m^2 H = 75 kN V = 1350 kN
Hansen’s formula:
qult = c Ncscdcic + qNqsqdqiq +
1
2 γb Nγsγdγiγ
Shape factors:
sc = 1 + 0.2 b/L = 1 + 0.2 ×
2
3 = 1.133
sq = 1 + 0.2 b/L = 1.133
sγ = 1 – 0.4 b/L = 1 – 0.4 ×
2
3 = 0.733
Depth factors:
dc = 1 + 0.35 Df /b = 1 + 0.35 ×
1
2 = 1.175
dq = 1 + 0.35 Df /b = 1.175
dγ = 1.0