Bearing Capacity Factor Nc

Nc=(Nq−1)cot⁡ϕN_c = (N_q - 1)\cot\phi

Worked example: Nc 46.12 with Nq 33.30 → φ = 35.005° — press Try an example to run it live, then adjust anything.

Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Try different units for next level excitement!

Learning zone

Bearing Capacity Factor Nc explained

NqNcφ

NcN_c is not an independent result but a consequence: Prandtl's plasticity solution ties the cohesion term to the surcharge term through Nc=(Nq−1)cot⁡φN_c = (N_q - 1)\cot\varphi. With NqN_q = 18.40 at φ = 30°, NcN_c = 17.40 ÷ tan 30° = 30.14 — again the tabulated value. At φ = 35°, NqN_q = 33.30 gives NcN_c = 46.12.

The interesting case is φ = 0, where cot φ blows up and the formula is useless — but the limit is finite and famous: NcN_c → π + 2 = 5.14. That is the undrained bearing capacity of saturated clay, qu=5.14 cuq_u = 5.14\,c_u for a strip footing and about 6.2 cu6.2\,c_u for a square or circular one, and it is the single most-used number in shallow foundation design on clay. The trap is applying the φ = 0 analysis with a drained strength, or vice versa: undrained (total stress, cuc_u, φ = 0) governs immediately after construction, drained (effective stress, c′, φ′) governs decades later, and a clay foundation must be checked for both. Skempton's 1951 paper set out that two-case discipline and it has not been improved on.

Bearing Capacity Factor Nc formula

Nc=(Nq−1)cot⁡ϕN_c = (N_q - 1)\cot\phi
Where
  • NcN_c= Cohesion bearing capacity factor
  • NqN_q= Surcharge bearing capacity factor
  • ϕ\phi= Angle of internal friction (°)

Missing one of these? Work it out first, then come back