OpenSpecEngineering

Finds the ultimate and allowable bearing pressure under a shallow rectangular, square or strip footing on uniform soil. It uses Terzaghi’s general bearing capacity equation [1] with Vesić’s bearing capacity factors [2] and De Beer’s shape factors [3]. Engineers use it for a preliminary footing size before a geotechnical report gives the design value.

The bearing capacity equation

q_ult = c·N_c·s_c + q·N_q·s_q + 0.5·γ·B·N_γ·s_γ, with q = γ·D_f

N_q = e^π·tanφ·tan²(45° + φ/2)
N_c = (N_q − 1)/tanφ (5.14 at φ = 0)
N_γ = 2(N_q + 1)·tanφ

s_c = 1 + (B/L)(N_q/N_c), s_q = 1 + (B/L)·tanφ, s_γ = 1 − 0.4(B/L)

c is the soil cohesion, φ the friction angle, γ the soil unit weight, B the footing width (the shorter side), L its length and D_f the depth from the ground surface to the footing base. q is the overburden pressure at the base. The page takes the shorter side as B whichever way you enter them. For a strip footing, enter a long L. The net ultimate pressure is q_ult − q, the pressure beyond the weight of soil removed. Each is divided by the factor of safety, usually 3, for the allowable values.

The soil list fills c = 0 and typical φ and γ values of compacted granular fill by USCS group [4], [5]. Clays and silts are not listed: their bearing capacity depends on undrained strength, which has to come from tests. For a clay, enter its undrained strength as c with φ = 0.

Limits and common mistakes

• The page assumes a vertical, centered load and a dry, uniform soil. It has no depth, load-inclination or groundwater factors. If the water table is within about B below the base, use the buoyant unit weight.
• Bearing capacity is only half the check: settlement often limits the allowable pressure on sand, especially for wide footings.
• Use drained (c′, φ′) or undrained (s_u, φ = 0) parameters, not a mix.
• Small changes in φ change the result a lot: N_γ nearly doubles for every 4°.

Worked example

A column footing 2.0 m square will sit 1.2 m below grade. The soil report gives c′ = 0, φ′ = 32° and γ = 18.5 kN/m³. Find the allowable bearing pressure with a factor of safety of 3.

1. Keep Soil Custom and enter Cohesion 0 kPa, Soil Unit Weight 18.5 kN/m³ and Friction Angle 32°.
2. Enter Footing Width (B) 2 m, Footing Length (L) 2 m and Embedment Depth (D_f) 1.2 m.
3. Keep Factor of Safety 3.
4. Press Solve. The Solution shows: Nc: 35.4903
   Nq: 23.1768
   Nγ: 30.2147
   Surcharge q = γ·Df: 22.2 kPa
   Ultimate bearing capacity q_ult: 1,171.42 kPa
   Net ultimate q_net: 1,149.22 kPa
   Allowable bearing pressure q_all = q_ult / FS: 390.473 kPa
   Net allowable q_net / FS: 383.073 kPa

Check: the factors match Vesić’s published table for φ = 32° (35.49, 23.18, 30.22) [3]. With c = 0, q_ult = 22.2 × 23.177 × 1.625 + 0.5 × 18.5 × 2 × 30.215 × 0.6 = 836.0 + 335.4 = 1,171.4 kPa.