Geotechnical Engineering

DHARM

588 GEOTECHNICAL ENGINEERING

∴ qult =

1

2

γ b Nγ. Rγ =

1

2

× 19 × 1.5 × 5.0 × 0.5 = 35.6 kN/m^2
The percentage reduction in the ultimate bearing capacity is thus 50 due to flooding and
consequent complete submergence.
FHGNote:γγγγsatis assumed to be itself here, and ′≈^1 satIKJ
2
Example 14.5: A continuous footing of width 2.5 m rests 1.5 m below the ground surface in
clay. The unconfined compressive strength of the clay is 150 kN/m^2. Calculate the ultimate
bearing capacity of the footing. Assume unit weight of soil is 16 kN/m^3.
(S.V.U.—B.E., (R.R.)—May, 1969)
Continuous footing b = 2.5 m Df = 1.5 m
Pure clay.
φ = 0° qu = 150 kN/m^2 γ = 16 kN/m^3

c =

qu

2

= 75 kN/m^2

For φ = 0°, Terzaghi’s factors are: Nγ = 0, Nq = 1, and Nc = 5.7.

qult = cNc +

1
2 γ^ b Nγ + γDf Nq = cNc + γDf Nq, in this case.
∴ qult = 5.7 × 75 + 16 + 1.5 × 1 = 451.5 kN/m^2 ≈ 450 kN/m^2.
Example 14.6: Compute the safe bearing capacity of a continuous footing 1.8 m wide,
and located at a depth of 1.2 m below ground level in a soil with unit weight γ = 20 kN/m^3 , c =
20 kN/m^2 , and φ = 20°. Assume a factor of safety of 2.5. Terzaghi’s bearing capacity factors for
φ = 20° are Nc = 17.7, Nq = 7.4, and Nγ = 5.0, what is the permissible load per metre run of the
footing?
b = 1.8 m continuous footing Df = 1.2 m
γ = 20 kN/m^3 c = 20 kN/m^2
φ = 20° Nc = 17.7
Nq = 7.4 Nγ = 5.0 η = 2.5
qult = cNc +

1

2

γ b Nγ + γ Df Nq

= 20 × 17.7 +

1

2 × 20 × 1.8 × 5.0 + 20 × 1.2 × 7.4

= 621.6 kN/m^2

qnet ult = qult – γDf = 621.6 – 20 × 1.2 = 597.6 kN/m^2

qnet safe =

qnet ult

η =

597 6

25

.

.

= 239 kN/m^2

qsafe = qnet safe + γDf = 239 + 20 × 1.2 = 263 kN/m^2

Permissible load per metre run of the wall = 263 × 1.8 kN = 473.5 kN.

Example 14.7: What is the ultimate bearing capacity of a square footing resting on the sur-

face of a saturated clay of unconfined compressive strength of 100 kN/m^2.

(S.V.U.—Four-year B. Tech.—Apr., 1983)

Square footing.

Saturated clay,

φ = 0° Df = 0.