Bearing resistance of the soil to SIA 267
Computes the bearing resistance of the soil under a shallow rectangular or strip footing to SIA 267:2013, drained and undrained, on the effective area, and checks overturning, sliding and the eccentricity in service. Every step shows its formula and the clause it comes from.
Basic SIA 267:2013
Part of the Basic plan
This module is included in the Basic plan and above, with its calculation note.
About this check
clauses, worked example and questions
Computes the bearing resistance of the soil under a shallow rectangular or strip footing to SIA 267:2013, drained and undrained, on the effective area, and checks overturning, sliding and the eccentricity in service. Every step shows its formula and the clause it comes from.
What it checks
- Design soil values: tan φ′d = tan φ′k/γφ, c′d = c′k/γc and cu,d = cu,k/γcu, clause 5.3.2.1, equation (5), with the factors of 5.3.2.2, table 1.
- Effective foundation: ea = Ma,d/EN,d, a′ = a − 2 · |ea|, the same in direction b, and A′d = a′ · b′, EN 1997-1 D.1.
- Overturning: (ea/a)² + (eb/b)² ≤ 1/9 with the favourable vertical force EN,d,inf, clause 8.5.3.6, equations (38) and (39).
- Drained bearing resistance: Nq, Nc and Nγ, the shape factors sq, sγ and sc, and the load inclination factors iq, iγ and ic, EN 1997-1 D.4; R/A′ = qc + qq + qγ, equation (D.2).
- Undrained bearing resistance: R/A′ = (π + 2) · cu,d · bc · sc · ic + q, EN 1997-1 D.3, equation (D.1).
- Both cases: qR,d = (R/A′)/γR and EN,d ≤ RN,d = qR,d · A′d, clause 8.5.2.1, equation (29), and clause 8.5.3.5, equation (37).
- Sliding, drained and undrained: ET,d ≤ RT,d + RE,d, clause 8.5.3.4, equation (36). Drained, RT,d = EN,d · tan δd/γR, clause 8.5.2.2, equation (31), with δk ≤ 35° (8.4.1.1, equation (27)). Undrained, RT,d = A′d · cu,d/γR, equation (32), at most 0.4 · EN,d where a gap can open at the base (EN 1997-1 §6.5.3(12)).
- Service state: the resultant in the central kern, |ea|/a + |eb|/b ≤ 1/6 (8.5.4.7), and the undrained capacity over the service load, Rk/EN, EN 1997-1 §6.6.2(16).
What you enter and what you get
You choose the model, partial factors or classical bearing capacity with a global safety factor (8.5.2.1), and the drainage condition, drained, undrained or both (8.5.1.6). You enter the shape, rectangular or strip, the sides a and b, the depth D and the base inclination α if any, the base roughness, then φ′k, c′k, cu,k, γe,1, γe,2 and the partial factors. For the actions at the base: EN,d, Ma,d, Mb,d, ET,a,d, ET,b,d and EN,d,inf; then γR, c′ in sliding, whether water can reach the base, RE,k, and the service actions.
You get qR,d for each case, ready to enter in the footing modules, and every check with its utilisation and clause. With the Basic plan, it prints as an A4 calculation note.
Assumptions and limits
- The pressure is taken as uniform on the effective area, centred on the resultant.
- A strip footing is computed per metre of length.
- c′ is left out of the sliding resistance unless its permanence is certain (8.4.1.2). RE,k counts only if the soil in front will stay.
- Without EN,d,inf, EN,d is used for sliding and overturning, which is not on the safe side when it contains variable loads.
- The undrained case uses the total unit weight for the overburden, the drained case the buoyant value below the groundwater table.
Worked example
A 2.4 × 3.0 m footing at D = 1.2 m, with φ′k = 30°, c′k = 5 kN/m² and cu,k = 120 kN/m², under EN,d = 2,400 kN, Ma,d = 240 kNm, ET,a,d = 150 kN and EN,d,inf = 1,900 kN, drained and undrained, and in service EN = 1,700 kN with Ma = 100 kNm. These are the module's own results for its worked example, computed when this page was built.
| Input | Value | Ref. |
|---|---|---|
| Model for the bearing resistance | Partial factors on the soil and on the resistance | § 8.5.2.1 |
| Drainage condition | Both drained and undrained | § 8.5.1.6 |
| Shape of the footing | Rectangular | |
| Side of the footing in direction a a | 2.4 m | |
| Side of the footing in direction b b | 3.0 m | |
| Depth of the base below the ground D | 1.2 m | |
| Base of the footing, for sliding | Rough, failure in the soil: δ = φ′ | § 8.4.1.1 |
| Angle of internal friction φ′k | 30 ° | |
| Effective cohesion c′k | 5 kN/m² | |
| Undrained shear strength cu,k | 120 kN/m² | |
| Unit weight of the soil above the base γe,1 | 19 kN/m³ | |
| Unit weight of the soil below the base γe,2 | 10 kN/m³ | |
| Partial factor on tan φ′ γφ | 1.2 | § 5.3.2.2 t.1 |
| Partial factor on the effective cohesion γc | 1.5 | § 5.3.2.2 t.1 |
| Partial factor on the undrained shear strength γcu | 1.5 | § 5.3.2.2 t.1 |
| Vertical force EN,d | 2400 kN | § 8.5.3.7 |
| Moment in direction a Ma,d | 240 kNm | |
| Moment in direction b Mb,d | 0 kNm | |
| Horizontal force in direction a ET,a,d | 150 kN | |
| Horizontal force in direction b ET,b,d | 0 kN | |
| Vertical force with favourable weights (optional) EN,d,inf | 1900 kN | § 5.4.1.2 |
| Partial factor on the resistance γR | 1.0 | § 8.5.2.1 (30) |
| Effective cohesion in sliding | Not counted | § 8.4.1.2 |
| Can water or air reach the base? | Yes, water or air can reach the base | EN 1997-1 §6.5.3(12) |
| Vertical force EN | 1700 kN | |
| Moment in direction a Ma | 100 kNm | |
| Moment in direction b Mb | 0 kNm |
| Step | Value | Ref. |
|---|---|---|
| Area of the basea·b | A = 7.200 m² | |
| Overburden at the base levelγe,1·Dacts beside the footing and holds the soil down | q′ = 22.8 kN/m² | |
| Design angle of internal frictionatan(tan φ′k/γφ) | φ′d = 25.69 ° | § 5.3.2.1 (5) |
| Design effective cohesionc′k/γc | c′d = 3.3 kN/m² | § 5.3.2.1 (5) |
| Design undrained shear strengthcu,k/γcu | cu,d = 80.0 kN/m² | § 5.3.2.1 (5) |
| Friction angle of the baseφ′k ≤ 35° | δk = 30.00 ° | § 8.4.1.1 (27) |
| Design friction angle of the baseatan(tan δk/γφ) | δd = 25.69 ° | § 8.5.2.2 |
| Bearing factore^(π·tan φ′d)·tan²(45° + φ′d/2) | Nq = 11.47 | EN 1997-1 D.4 |
| Bearing factor(Nq − 1)·cot φ′d | Nc = 21.77 | EN 1997-1 D.4 |
| Bearing factor2·(Nq − 1)·tan φ′d, rough base (δ ≥ φ′/2) | Nγ = 10.08 | EN 1997-1 D.4 |
| Step | Value | Ref. |
|---|---|---|
| Eccentricity in direction aMa,d/EN,d | ea = 0.100 m | |
| Eccentricity in direction bMb,d/EN,d | eb = 0.000 m | |
| Reduced side in direction aa − 2·|ea| | a′ = 2.200 m | |
| Reduced side in direction bb − 2·|eb| | b′ = 3.000 m | |
| Effective areaa′·b′the area centred on the resultant, on which the pressure is taken uniform | A′d = 6.600 m² | EN 1997-1 D.1 |
| Effective width (the smaller side)min(a′ ; b′) | B′ = 2.200 m | EN 1997-1 D.1 |
| Effective length (the larger side)max(a′ ; b′) | L′ = 3.000 m | |
| Horizontal force√(ET,a,d² + ET,b,d²) | ET,d = 150.0 kN |
| Step | Value | Ref. |
|---|---|---|
| Eccentricity with the favourable vertical force, aMa,d/EN,d,inf | ed,a = 0.126 m | |
| Eccentricity with the favourable vertical force, bMb,d/EN,d,inf | ed,b = 0.000 m | |
| Overturning: resultant within the admissible eccentricityed/egr | η = 0.158 · Pass16 % of the capacity used | § 8.5.3.6 (38)(39) |
| Step | Value | Ref. |
|---|---|---|
| Shape factor1 + (B′/L′)·sin φ′d | sq = 1.318 | EN 1997-1 D.4 |
| Shape factor1 − 0.3·B′/L′ | sγ = 0.780 | EN 1997-1 D.4 |
| Shape factor(sq·Nq − 1)/(Nq − 1) | sc = 1.348 | EN 1997-1 D.4 |
| Exponent of the inclination factorsmL·cos²θ + mB·sin²θ; mB = (2 + B′/L′)/(1 + B′/L′); mL = (2 + L′/B′)/(1 + L′/B′); θ: angle of ET,d to L′ | m = 1.577 | EN 1997-1 D.4 |
| Load inclination factor[1 − ET,d/(EN,d + A′·c′d·cot φ′d)]^m | iq = 0.905 | EN 1997-1 D.4 |
| Load inclination factor[1 − ET,d/(EN,d + A′·c′d·cot φ′d)]^(m + 1) | iγ = 0.850 | EN 1997-1 D.4 |
| Load inclination factoriq − (1 − iq)/(Nc·tan φ′d) | ic = 0.896 | EN 1997-1 D.4 |
| Cohesion termc′d·Nc·bc·sc·ic | qc = 87.7 kN/m² | |
| Overburden termq′·Nq·bq·sq·iq | qq = 312.0 kN/m² | |
| Self-weight term0.5·γe,2·B′·Nγ·bγ·sγ·iγ | qγ = 73.5 kN/m² | |
| Bearing resistance per unit of effective areaqc + qq + qγ | R/A′ = 473.1 kN/m² | EN 1997-1 D.4 (D.2) |
| Design bearing pressure(R/A′)/γRto be entered in the footing module | qR,d = 473.1 kN/m² | |
| Design bearing resistanceqR,d·A′d | RN,d = 3,122.6 kN | § 8.5.2.1 (29) |
| Drained bearing resistanceEN,d/RN,d | η = 0.769 · Pass77 % of the capacity used | § 8.5.3.5 (37) |
| Step | Value | Ref. |
|---|---|---|
| Total overburden at the base levelγe,1·Dwith the total unit weight as γe,1 for this case | q = 22.8 kN/m² | |
| Shape factor1 + 0.2·B′/L′ | sc = 1.147 | EN 1997-1 D.3 |
| Load inclination factor0.5·(1 + √(1 − ET,d/(A′·cu,d))) | ic = 0.923 | EN 1997-1 D.3 |
| Bearing resistance per unit of effective area(π + 2)·cu,d·bc·sc·ic + q | R/A′ = 458.2 kN/m² | EN 1997-1 D.3 (D.1) |
| Design bearing pressure(R/A′)/γRto be entered in the footing module | qR,d = 458.2 kN/m² | |
| Design bearing resistanceqR,d·A′d | RN,d = 3,023.9 kN | § 8.5.2.1 (29) |
| Undrained bearing resistanceEN,d/RN,d | η = 0.794 · Pass79 % of the capacity used | § 8.5.3.5 (37) |
| Step | Value | Ref. |
|---|---|---|
| Vertical force acting with the horizontal force | EN,d,inf = 1,900.0 kN | |
| Design sliding resistance, drainedEN,d·tan δd/γR, c′ not counted | RT,d = 914.1 kN | § 8.5.2.2 (31) |
| Sliding, drainedET,d/(RT,d + RE,d) | η = 0.164 · Pass16 % of the capacity used | § 8.5.3.4 (36) |
| Design sliding resistance, undrainedA′d·cu,d/γR ≤ 0.4·EN,d (gap at the base possible) | RT,d = 528.0 kN | § 8.5.2.2 (32) |
| Sliding, undrainedET,d/(RT,d + RE,d) | η = 0.284 · Pass28 % of the capacity used | § 8.5.3.4 (36) |
| Step | Value | Ref. |
|---|---|---|
| Eccentricity in direction aMa/EN | ea = 0.059 m | |
| Eccentricity in direction bMb/EN | eb = 0.000 m | |
| Resultant in the central kerne/ekern | η = 0.147 · Pass15 % of the capacity used | § 8.5.4.7 |
| Step | Value | Ref. |
|---|---|---|
| Undrained bearing capacity over the service load((π + 2)·cu,k·sc + q)·A′/EN, characteristic cu,k, vertical load | Rk/EN = 2.95 | EN 1997-1 §6.6.2(16) |
How this module is verified
Verified clause by clause against SIA 267:2013 and against independent calculations used on real projects, before release.
Questions
What does the warning on R_k/E_N mean?
In the worked example the ratio is 2.95. Below 3, settlements must be calculated, EN 1997-1 §6.6.2(16). The pad footing module computes them from the soil layers.
Which depth D should I enter?
From the lowest ground surface next to the footing that will remain, allowing for future excavations and trenches.
Which plan includes this module?
The Basic plan, with its calculation note.
Related
- Isolated pad footing
- Raft foundation under columns
- Foundations: all modules
- SIA 267: modules and clauses
Norm: SIA 267:2013 · Page updated: 29 October 2026