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Pad footing design to SANS 10100

A complete worked example: an isolated pad footing under a 400 mm square column, from self-weight and bearing pressure to bending, shear, punching and minimum steel, with the SANS 10160-1 load combinations. Every number below comes from the free STATIX pad footing calculator and was checked by hand.

1. The problem

Input	Value
Column	400 × 400 mm, stub 0.5 m high
Characteristic loads at the top of the stub	permanent G: N = 550 kN, M_y = 20 kNm, H_x = 25 kN imposed Q: N = 250 kN, M_y = 10 kNm, H_x = 15 kN
Footing	2.5 × 2.5 m on plan, 600 mm thick, 0.5 m of soil on top
Allowable bearing pressure	q_all = 200 kPa
Materials	f_cu = 30 MPa, f_y = 450 MPa, cover 50 mm
Unit weights	concrete 24 kN/m³, soil 18 kN/m³
Load combinations	SANS 10160-1: serviceability G + Q; ultimate STR 1.2G + 1.6Q and STR-P 1.35G + 1.0Q (the worse governs)

Footing plan with Y12 at 125 mm both ways, elevation and soil bearing pressure diagram from the STATIX pad footing calculator — The free pad footing calculator with these inputs: plan with the bottom mat, elevation, and the bearing pressure from 123 to 180 kPa.

2. Self-weight

W_base = 2.5 × 2.5 × 0.6 × 24 = 90.0 kN W_stub = 0.4 × 0.4 × 0.5 × 24 = 1.9 kN W_soil = (2.5 × 2.5 − 0.4 × 0.4) × 0.5 × 18 = 54.8 kN ΣSW = 146.7 kN

The soil on top of the footing presses on the ground too, so it goes into the bearing check.

3. Bearing pressure (serviceability, G + Q)

The horizontal force acts at the top of the stub, 1.1 m above the underside of the footing, so it adds to the moment.

N = 550 + 250 + 146.7 = 946.7 kN M = (20 + 10) + (25 + 15) × (0.6 + 0.5) = 74.0 kNm e = M / N = 74.0 / 946.7 = 0.078 m (inside B/6 = 0.417 m: the whole base stays in contact) q_max = N/(B·L) × (1 + 6e/B) = 151.5 × 1.188 = 179.9 kPa ≤ 200 kPa OK q_min = 151.5 × (1 − 6e/B) = 123.1 kPa

Overturning (safety factor 16) and sliding (safety factor 11.8 with μ = 0.5) are far from critical here.

4. Design pressure for bending and shear (ultimate)

Bending and shear use the net pressure. The weight of the base and the soil on it pushes down as much as the ground pushes back under it, so it does not bend the base. Each combination of SANS 10160-1 is checked; STR governs here.

STR: N_u = 1.2 × 550 + 1.6 × 250 = 1 060 kN M_u = 1.2 × (20 + 25 × 1.1) + 1.6 × (10 + 15 × 1.1) = 57.0 + 42.4 = 99.4 kNm STR-P: N_u = 1.35 × 550 + 250 = 992.5 kN, M_u = 1.35 × 47.5 + 26.5 = 90.6 kNm (lower) mean pressure a = 1 060 / 6.25 = 169.6 kPa, e_u = 99.4 / 1 060 = 0.0938 m q_u = a (1 ± 6e_u/B) = 169.6 × (1 ± 0.225) = 207.8 kPa at one edge, 131.4 kPa at the other

5. Bending (SANS 10100-1 cl. 4.10.2.1 b and 4.3.3)

Each direction is a cantilever from the column face. The code says the reaction varies linearly across the base, so the moment at the face is that of the trapezoid of pressure on one side, over the full width.

l = (2.5 − 0.4) / 2 = 1.05 m pressure at the column face q₀ = 169.6 × (1 + 12 × 0.0938 × 0.2 / 2.5²) = 175.7 kPa; at the edge q₁ = 207.8 kPa M_x = q₀ l²/2 + (q₁ − q₀) l²/3 = 175.7 × 0.551 + 32.1 × 0.368 = 96.9 + 11.8 = 108.6 kNm per metre M_y = 169.6 × 1.05² / 2 = 93.5 kNm per metre (no eccentricity that way) d = 600 − 50 − 12 = 538 mm (mean depth of the two-layer mat, with the 12 mm bars chosen) K = M / (b d² f_cu) = 108.6×10⁶ / (1000 × 538² × 30) = 0.013 ≤ K′ = 0.156 z = d [0.5 + √(0.25 − K/0.9)] = 530 mm, limited to 0.95d = 511 mm A_s,req = M / (0.87 f_y z) = 108.6×10⁶ / (0.87 × 450 × 511) = 543 mm²/m A_s,min = 0.13% × 1000 × 600 = 780 mm²/m ← governs Provide Y12 at 125 mm both ways = 905 mm²/m OK

This base is thick for its load, so minimum steel governs, not bending. Taking the peak edge pressure over the whole cantilever (207.8 × 1.05²/2 = 114.5 kNm/m) is on the safe side, but it is not what cl. 4.10.2.1 b) asks for.

6. Shear and punching (SANS 10100-1 cl. 4.10.3, 4.3.4 and 4.6.2)

100A_s/(bd) = 100 × 905 / (1000 × 538) = 0.17 v_c = (0.75/1.4) × (f_cu/25)^1/3 × (0.17)^1/3 × (400/d)^1/4 = 0.29 MPa One-way shear at 1.5d from the column face (cl. 4.10.3.3.2 a): strip a_v = 1.05 − 1.5 × 0.538 = 0.243 m; mean pressure on it 207.8 − (207.8 − 131.4) × 0.243 / (2 × 2.5) = 204.1 kPa V = 204.1 × 0.243 × 2.5 = 124.0 kN; v = V/(B d) = 124.0×10³ / (2500 × 538) = 0.09 MPa ≤ 0.29 MPa OK Punching at 1.5d: u = 4 × (400 + 2 × 1.5 × 538) = 8 056 mm; V = 169.6 × (6.25 − 2.014²) = 372.1 kN β = 1 + 1.5M/(Vx) = 1 + 1.5 × 99.4 / (372.1 × 2.014) = 1.20; v = βV/(ud) = 0.10 MPa ≤ 0.29 MPa OK Column face: u₀ = 1 600 mm; β = 1 + 1.5 × 99.4/(1 060 × 0.4) = 1.35 v₀ = βV/(u₀d) = 1.35 × 1 060×10³ / (1600 × 538) = 1.66 MPa ≤ v_max = 4.11 MPa OK

SANS 10100-1 lets the one-way section of a pad footing sit at 1.5d from the face (cl. 4.10.3.3.2 a); BS 8110 puts it at d. The shear on the strip uses the linear pressure, not the mean. The depth factor (400/d)^1/4 = 0.93 is kept as calculated: SANS has no floor of 1.0 on it.

β allows for the column moment passing into the base (cl. 4.6.2.2.2 eq. 8). The limit at the column face is v_max = 0.75√f_cu = 0.75√30 = 4.11 MPa.

7. Result

Check	Result
Bearing pressure (G + Q)	179.9 kPa ≤ 200 kPa
Bending steel	Y12 @ 125 both ways (minimum steel governs; M = 108.6 kNm/m)
One-way shear at 1.5d	0.09 ≤ 0.29 MPa
Punching at 1.5d	0.10 ≤ 0.29 MPa
Column face	1.66 ≤ 4.11 MPa
Quantities	about 85 kg of steel and 3.8 m³ of concrete

Run this example yourselfThe free STATIX pad footing calculator opens with exactly these inputs. Change any value and every step updates, with a calc sheet. It can also check the bearing capacity of the soil.
Open the free pad footing calculator

A worked example for learning, not a design. Check every value against the current code and your own project, and have the design checked and signed by a registered professional engineer.

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