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Ultimate Bearing Capacity of a Strip Footing

A strip footing of width $B = 2.0\ \text{m}$ is founded at depth $D_f = 1.5\ \text{m}$ in a cohesionless sand.

Soil: $\gamma = 18\ \text{kN/m}^3$ , $c = 0$ , $\phi = 30^\circ$ . Bearing capacity factors: $N_q = 18.4$ , $N_\gamma = 22.4$ .

Using Terzaghi's equation for a strip footing, find the ultimate bearing capacity $q_u$ .

Given

B	2 m	Footing width
Df	1.5 m	Founding depth
γ	18 kN/m^3	Unit weight
Nq	18.4 -	Bearing capacity factor
Nγ	22.4 -	Bearing capacity factor

Hint 1

Terzaghi's equation has three terms — one drops out for clean sand.

Hint 2

q_u = cN_c + γD_f N_q + ½γBN_γ. Which term uses depth and which uses width?

Hint 3

The surcharge term uses D_f with no ½; the width term uses B with a ½.

Worked solution — try the problem first

Terzaghi's equation for a strip footing:

$q_u = cN_c + \gamma D_f N_q + \tfrac{1}{2}\gamma B N_\gamma$

The cohesion term vanishes since $c = 0$ for clean sand.

Surcharge term — the weight of soil beside the footing, at founding depth:

$\gamma D_f N_q = 18 \times 1.5 \times 18.4 = 27 \times 18.4 = 496.8\ \text{kPa}$

Width term — note the ½ and that it uses $B$ , not $D_f$ :

$\tfrac{1}{2}\gamma B N_\gamma = 0.5 \times 18 \times 2.0 \times 22.4 = 18 \times 22.4 = 403.2\ \text{kPa}$

Total: $q_u = 496.8 + 403.2 = 900\ \text{kPa}$

$\boxed{q_u = 900\ \text{kPa}}$

This is an ultimate value. Allowable bearing pressure applies a factor of safety, typically 3: $q_{all} = 300$ kPa. Designing to $q_u$ would put the footing at the point of collapse.

Note the shape factors here (1.0 and 0.5) are specific to strip footings; square and circular footings use different values.

Concepts:Bearing capacityShallow foundationsTerzaghi's equation

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