Geotechnical Engineering

DHARM

592 GEOTECHNICAL ENGINEERING

∴ 600 =^410

4

×πd^2

∴ d =

4600

410

×

m = 1.365 m

A diameter of 1.5 m may be adopted in this case.

Example 14.15: A column carries a load of 1000 kN. The soil is a dry sand weighing 19 kN/m^3

and having an angle of internal friction of 40°. A minimum factor of safety of 2.5 is required

and Terzaghi factors are required to be used. (Nγ = 42 and Nq = 21).

(i) Find the size of a square footing, if placed at the ground surface; and,

(ii) Find the size of a square footing required if it is placed at 1 m below ground surface

with water table at ground surface. Assume γsat = 21 kN/m^3.

(i) At ground surface:

φ = 40° Dry sand, N = 42 Nq = 21

Let the size of the footing be b m.

qult = 0.4 b × 19 × 42

Since Df = 0, qnet ult = qult

∴ qsafe = qnet ult =

qult

η =

0 4 19 42

25

.

.

b××

= 128 b kN/m^2

Qsafe = 128 b × b^2 = 1000

∴ b =

1000

128

(^3) m = 2 m
(ii) At 1 m below ground surface with water table at ground surface:
γ′ = γsat – γw = (21 – 10) = 11 kN/m^3
N = 42, Nγ = 21
qult = 0.4 × γbNγ + γDf Nq
qult = 0.4 b × 11 × 42 + 11 × 1 × 21 = (185 b + 231) kN/m^2
qnet ult = qult – γDf = 185 b + 231 – 11 × 1 = (185 b + 220) kN/m^2
qsafe =
qnet ult
η

• γDf =
  ()
  .
  185 200
  25
  L b+ + 11
  NM
  O
  QP
  kN/m^2
  ∴ Qsafe =
  185 220
  25
  L b+ + 11
  NM
  O

. QP

b^2 = 1000

Solving by trial and error, b = 2 m.

b = 2 m.

Example 14.16: What is the ultimate bearing capacity of a rectangular footing, 1 m × 2 m, on
the surface of a saturated clay of unconfined compression strength of 100 kN/m^2?

Rectangular footing:

b = 1 m L = 2 m Df = 0 qu = 100 kN/m^2

∴ c =

1

2

qu = 50 kN/m^2