Geotechnical Engineering

DHARM

BEARING CAPACITY 587

If Df = 1.5 m,

qult = γDf

1

1

2

+

−

F

HG

I

KJ

sin

sin

φ

φ

= 18 × 1.5

118

118

2

+°

−°

F

HG

I

KJ

sin

sin

= 96.8 kN/m^2.

Example 14.3: Calculate the ultimate bearing capacity of a strip footing, 1 m wide, in a soil for
γ = 18 kN/m^3 , c = 20 kN/m^2 , and φ = 20°, at a depth of 1 m. Use Rankine’s and Bell’s approaches.

φ = 20°

In Rankine’s approach, cohesion is not considered.

qult =

1

2

γ b Nγ + γ Df Nq

where Nγ =

1

2

NNφφ()^2 −^1 and Nq = Nφ^2

Nφ = tan^2 (45° + φ/2) = tan^2 55° = 2.04

Nγ =

1

2

× tan 55° (tan^4 55° – 1) = 2.256

Nq = tan^4 55° = 4.16

∴ qult =

1

2

× 18 × 1 × 2.256 + 18 × 1 × 4.16 = 95.2 kN/m^2

In Bell’s approach, cohesion is also considered.

qult = cNc +

1

2

γb Nγ + γDf Nq

where Nc = 2 NNφ()φ+^1 , Nγ =

1

2

NNφφ()^2 −^1 , and Nq = Nφ^2

∴ Nc = 2 tan 55° (tan^2 55° + 1) = 8.682

Nγ =

1

2

tan 55° (tan^4 55° – 1) = 2.256

Nq = tan^4 55° = 4.16

∴ qult = 20 × 8.682 +

1

2

× 18 × 1 × 2.256 + 18 × 1 × 4.16 = 268.8 kN/m^2.

Example 14.4: A strip footing, 1.5 m wide, rests on the surface of a dry cohesionless soil

having φ = 20° and γ = 19 kN/m^3. If the water table rises temporarily to the surface due to

flooding, calculate the percentage reduction in the ultimate bearing capacity of the soil. Assume

Nγ = 5.0. (S.V.U.—B.E., (Part-Time)—Apr., 1982)

φ = 20° Nγ = 5.0 b = 1.5 m Df = 0

Dry cohesionles soil, ∴ c = 0

qult = cNc +

1

2

γb Nγ + γDf Nq =

1

2

γ b Nγ in this case.

=

1

2

× 19 × 1.5 × 5.0 = 71.3 kN/m^2

If the water table rises temporarily to the surface due to flooding, reduction factors Rγ

and Rq shall be applied as the maximum values for the Nγ and Nq terms respectively.

In this case, Rγ = 0.5 is applied for Nγ-term.