DHARMBEARING CAPACITY 591Square b = 2.25 m Df = 1 m γ = 17.5 kN/m^3
qult = 1.3 c Nc + 0.4 γb Nγ + γDf Nq = 1.3 c Nc + γDf Nq, in this case since Nγ = 0
∴ qult = 1.3 × 30 × 5.7 + 17.5 × 1 × 1 = 239.8 kN/m^2
qnet ult = qult – γDf = 239.8 – 17.5 = 222.3 kN/m^2qsafe =qnet ult
η+ γDf =222 3
3.
+ 17.5 = 91.6 kN/m^2Safe load on the footing:
= qsafe × Area = 91.6 × 2.25 × 2.25 ≈ 463 kN.
Example 14.13: What is the ultimate bearing capacity of a circular footing of 1 m diameter
resting on the surface of a saturated clay of unconfined compression strength of 100 kN/m^2?
What is the safe value if the factor of safety is 3?
Diameter, D = 1 Df = 0
qu = 1.3 c Nc + 0.3 γ D Nγ + γ Df Nq
For saturated clay,
φ = 0°
∴ Nc = 5.7 Nq = 1 Nγ = 0Also c =1
2qu = 50 kN/m^2
∴ qult = 1.3 × 50 × 5.7 = 370.5 kN/m^2qsafe =qult
η , in this case,
since Nq = 1
∴ qsafe =370 5
3.
= 123.5 kN/m^2.
Example 14.14: A circular footing is resting on a stiff saturated clay with qu = 250 kN/m^2. The
depth of foundation is 2 m. Determine the diameter of the footing if the column load is 600 kN.
Assume a factor of safety as 2.5. The bulk unit weight of soil is 20 kN/m^3.
(S.V.U.—Four-year B. Tech.—Dec., 1982)
Circular footing: φ = 0°, Nc = 5.7, Nq = 1, Nγ = 0qu = 250 kN/m^2 c =1
2qu = 125 kN/m^2 Df = 2 m
Column load = 600 kN η = 2.5 γ = 20 kN/m^3
qult = 1.3 c Nc + 0.3 γ D Nγ + γ Df Nq = 1.3 c Nc + γ Df Nq, in this case.
∴ qult = 1.3 × 125 × 5.7 + 20 × 2 × 1 = 966 kN/m^2
qnet ult = qult – γDf = 966 – 20 × 2 = 926 kN/m^2qsafe =qnet ult
η + γDf =926
- 20 × 2 = 410 kN/m
2Safe load on the column
= qsafe × Area = 600 kN