Skip to content
CivilSolve

Geotechnical

Terzaghi Bearing Capacity

Calculate ultimate and gross allowable bearing capacity for shallow footings using supplied Terzaghi factors, footing shape, and factor of safety.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Load a sample problem:

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

square footing in dense sand (the motivating exam problem)

Given

footing shape
square
width
2 m
depth
1.5 m
unit weight
18 kN/m^3
nc
57.8
nq
41.4
ngamma
42.4
factor of safety
3

Assumptions

  • Terzaghi general shear failure; homogeneous soil; vertical centric load; water table deep (use effective unit weights otherwise); N-factors as provided by the problem.

Solution steps

  1. Assemble the three capacity terms

    Cohesion, surcharge from the founding depth, and soil self-weight below the footing each contribute.

    q_ult = sc·c·Nc + γ·Df·Nq + sγ·0.5·γ·B·Nγ

    cohesion term = 0 kPa; surcharge term = 1118 kPa; width term = 610.6 kPa

  2. Ultimate capacity

    Sum the three contributions.

    q_ult = sum of the terms

    q_ult = 1728.4 kPa

    = 1728.4 kPa

  3. Gross allowable capacity

    Divide the ultimate capacity by the factor of safety. (Some texts subtract the surcharge first for a NET allowable — check which your course uses.)

    q_allow = q_ult / FS

    q_allow = 1728.4 kPa / 3 = 576.12 kPa

    = 576.12 kPa

Results

Ultimate bearing capacity q_ult

1728.4kPa

Gross allowable bearing capacity

576.12kPa

Where this answer was checked
source
Terzaghi q_ult = sc·c·Nc + γ·Df·Nq + sγ·0.5·γ·B·Nγ2×2 m, Df = 1.5 m, γ = 18, φ = 35° factors given, FS = 3
verified by
hand-recomputed
derivation
c = 0. Surcharge: 18·1.5·41.4 = 1117.8. Width: 0.8·0.5·18·2·42.4 = 610.56. q_ult = 1728.4 kPa; q_allow = 576.1 kPa.

Example 2

strip footing in c-φ soil

Given

footing shape
strip
width
1.5 m
depth
1 m
unit weight
17 kN/m^3
cohesion
25 kPa
nc
17.7
nq
7.4
ngamma
5

Assumptions

  • Terzaghi general shear failure; homogeneous soil; vertical centric load; water table deep (use effective unit weights otherwise); N-factors as provided by the problem.

Solution steps

  1. Assemble the three capacity terms

    Cohesion, surcharge from the founding depth, and soil self-weight below the footing each contribute.

    q_ult = sc·c·Nc + γ·Df·Nq + sγ·0.5·γ·B·Nγ

    cohesion term = 442.5 kPa; surcharge term = 125.8 kPa; width term = 63.75 kPa

  2. Ultimate capacity

    Sum the three contributions.

    q_ult = sum of the terms

    q_ult = 632.05 kPa

    = 632.05 kPa

Results

Ultimate bearing capacity q_ult

632.05kPa

Where this answer was checked
source
Terzaghi with cohesion, strip shape factors 1.0B = 1.5, Df = 1, γ = 17, c = 25; Nc = 17.7, Nq = 7.4, Nγ = 5.0
verified by
hand-recomputed
derivation
1·25·17.7 = 442.5; 17·1·7.4 = 125.8; 1·0.5·17·1.5·5 = 63.75. q_ult = 632.05 kPa.

Example 3

circular footing shape factors (1.3 / 0.6)

Given

footing shape
circular
width
2 m
depth
1 m
unit weight
18.5 kN/m^3
cohesion
10 kPa
nc
37.2
nq
22.5
ngamma
19.7

Assumptions

  • Terzaghi general shear failure; homogeneous soil; vertical centric load; water table deep (use effective unit weights otherwise); N-factors as provided by the problem.

Solution steps

  1. Assemble the three capacity terms

    Cohesion, surcharge from the founding depth, and soil self-weight below the footing each contribute.

    q_ult = sc·c·Nc + γ·Df·Nq + sγ·0.5·γ·B·Nγ

    cohesion term = 483.6 kPa; surcharge term = 416.3 kPa; width term = 218.7 kPa

  2. Ultimate capacity

    Sum the three contributions.

    q_ult = sum of the terms

    q_ult = 1118.5 kPa

    = 1118.5 kPa

Results

Ultimate bearing capacity q_ult

1118.5kPa

Where this answer was checked
source
Terzaghi circular sc = 1.3, sγ = 0.6D = 2 m, Df = 1, γ = 18.5, c = 10; Nc = 37.2, Nq = 22.5, Nγ = 19.7
verified by
hand-recomputed
derivation
1.3·10·37.2 = 483.6; 18.5·1·22.5 = 416.25; 0.6·0.5·18.5·2·19.7 = 218.67. q_ult = 1118.5 kPa.