Learn: Classical bearing capacity — basic form
A simple shallow-foundation bearing model separates resistance into cohesion, surcharge and soil-weight contributions. This lesson evaluates that three-term expression using supplied bearing-capacity factors.
What the formula is saying
The terms are cNc, qNq and 0.5γB Nγ. Each is a pressure contribution; summing them estimates the stated idealized ultimate bearing pressure, not an allowable service pressure.
Read the symbols in plain language
- c′
- Effective cohesion
Intercept of the effective-stress shear-strength relation; it is not an undrained strength substituted without changing the model.
kPaOne kilopascal equals one kN/m² and 1000 Pa.
- Nc
- Bearing factor Nc
Bearing factor Nc. Multiply the chosen cohesion parameter by its compatible bearing-capacity factor.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- q
- Surcharge at base
Surcharge at base. Use the pressure at foundation level with the corresponding surcharge factor.
kPaOne kilopascal equals one kN/m² and 1000 Pa.
- Nq
- Bearing factor Nq
Bearing factor Nq. Use the pressure at foundation level with the corresponding surcharge factor.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- γ
- Soil unit weight
Weight force per unit volume, including gravity. Mass density in kg/m³ cannot be entered directly in a kN/m³ field.
kN/m³Use kN/m³ as the base unit shown here. c and q are kPa, γ is kN/m³ and B is m. Nc, Nq and Nγ are dimensionless factors from one consistent model. The final pressure is kPa.
- B
- Footing width
Footing width. Unit weight times width forms a pressure scale modified by the chosen Nγ factor.
mMetres measure length; 1 m = 1000 mm.
- Nγ
- Bearing factor Nγ
Bearing factor Nγ. Unit weight times width forms a pressure scale modified by the chosen Nγ factor.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- qult
- Result to find
Classical bearing capacity — basic form. The result is the sum for this stated simplified model, before omitted corrections and design checks.
kPa
Sort out the units first
c and q are kPa, γ is kN/m³ and B is m. Nc, Nq and Nγ are dimensionless factors from one consistent model. The final pressure is kPa.
Assumptions before calculating
This is an idealized study model with supplied soil parameters and loading. Ground investigation, drainage condition, groundwater, geometry and the governing design approach must be established by the responsible geotechnical design process.
This is a first-generation Eurocode teaching relationship or an explicitly simplified coefficient calculation. The numbers supplied here are exercise data, not a recommendation for any country. Check the adopted edition, relevant clause, National Annex, applicability conditions and all other limit states before any real design.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find qult and explain the result in the stated output unit.
- c′ · Effective cohesion
- 0 kPa
- Nc · Bearing factor Nc
- 30
- q · Surcharge at base
- 36 kPa
- Nq · Bearing factor Nq
- 18.4
- γ · Soil unit weight
- 18 kN/m³
- B · Footing width
- 2 m
- Nγ · Bearing factor Nγ
- 15.7
Calculate the cohesion contribution
Multiply the chosen cohesion parameter by its compatible bearing-capacity factor.
(0) × (30) = 0 kPaCalculate the surcharge contribution
Use the pressure at foundation level with the corresponding surcharge factor.
(36) × (18.4) = 662.4 kPaCalculate the soil-weight contribution
Unit weight times width forms a pressure scale modified by the chosen Nγ factor.
0.5 × (18) × (2) × (15.7) = 282.6 kPaAdd the three ultimate-pressure terms
The result is the sum for this stated simplified model, before omitted corrections and design checks.
(0) + (662.4) + (282.6) = 945 kPa
Does this worked answer make sense?
If c = 0, the cohesion term disappears but other terms can remain. Changing B affects only the soil-weight term in this isolated expression; a full design can have additional width effects.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- c′ · Effective cohesion
- 10 kPa
- Nc · Bearing factor Nc
- 20
- q · Surcharge at base
- 30 kPa
- Nq · Bearing factor Nq
- 10
- γ · Soil unit weight
- 18 kN/m³
- B · Footing width
- 1.5 m
- Nγ · Bearing factor Nγ
- 8
Calculate the cohesion contribution
Multiply the chosen cohesion parameter by its compatible bearing-capacity factor.
(10) × (20) = 200 kPaCalculate the surcharge contribution
Use the pressure at foundation level with the corresponding surcharge factor.
(30) × (10) = 300 kPaCalculate the soil-weight contribution
Unit weight times width forms a pressure scale modified by the chosen Nγ factor.
0.5 × (18) × (1.5) × (8) = 108 kPaAdd the three ultimate-pressure terms
The result is the sum for this stated simplified model, before omitted corrections and design checks.
(200) + (300) + (108) = 608 kPa
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- c′ · Effective cohesion
- 0 kPa
- Nc · Bearing factor Nc
- 30
- q · Surcharge at base
- 40 kPa
- Nq · Bearing factor Nq
- 15
- γ · Soil unit weight
- 19 kN/m³
- B · Footing width
- 2.5 m
- Nγ · Bearing factor Nγ
- 12
Find: Learn: Classical bearing capacity — basic form
A hint, not the answer
The terms are cNc, qNq and 0.5γB Nγ. Each is a pressure contribution; summing them estimates the stated idealized ultimate bearing pressure, not an allowable service pressure.
c and q are kPa, γ is kN/m³ and B is m. Nc, Nq and Nγ are dimensionless factors from one consistent model. The final pressure is kPa.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Calculate the cohesion contribution
Multiply the chosen cohesion parameter by its compatible bearing-capacity factor.
(0) × (30) = 0 kPaCalculate the surcharge contribution
Use the pressure at foundation level with the corresponding surcharge factor.
(40) × (15) = 600 kPaCalculate the soil-weight contribution
Unit weight times width forms a pressure scale modified by the chosen Nγ factor.
0.5 × (19) × (2.5) × (12) = 285 kPaAdd the three ultimate-pressure terms
The result is the sum for this stated simplified model, before omitted corrections and design checks.
(0) + (600) + (285) = 885 kPa
Avoid the common trap
Do not mix Nc, Nq and Nγ from incompatible bearing theories. Do not confuse surcharge q at founding level with the applied footing pressure, or report ultimate pressure as allowable pressure.
When this method applies — and when it does not
Assume a strip-foundation-style, level, centrally and vertically loaded, homogeneous-soil model without the omitted correction factors. Shape, depth, inclination, groundwater, effective width, local shear, settlement and design partial factors are not automatically included.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
One idea understood. Keep going.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Classical bearing capacity — basic form. Read the relevant soil phase, seepage, earth-pressure, settlement or foundation topic. Effective stress, drainage and idealized geometry determine whether the relationship applies.
- University of the West of England — GeotechniCAL: soil mechanics and foundations
- University of the West of England — GeotechniCAL: bearing capacity
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.