This article presents a worked example on the structural design of pad foundation subjected to axial load and moment to EN 1992 -1-1-2004. The geotechnical aspects of the design shall be approached by using the prescriptive method described in Eurocode 7 pat 1 by adopting an allowable bearing pressure of 185KN/m2.
A pad foundation is required to support a column of 350 x 350mm which is bearing a permanent load of 1200KN (self-weight of column included) and a variable load of 400KN. The column is eccentrically placed such that permanent and variable moments between the column and the base are 80KNm and 65KNm respectively. Design the Pad foundation.
The steps in designing the pad foundation are as follows in the below worked example:
Determine the base area of the Pad Foundation
Let’s use a pad foundation of size of 3m x 3m as we had used in the worked example on the design of axially loaded pad foundation where the same column was located concentrically on the pad.
So, we will check the soil pressure beneath the foundation.
Total serviceability load on the base = 1.0Gk + 1.0Qk
= 1.0 x 1200 + 1.0 x 400
= 1600KN
Total serviceability Moment on the base = 1.0Gk + 1.0Qk
= 1.0 x 80 + 1.0 x 65
= 145KN
Base Pressure = N/BD ± 6M/BD²
Maximum Base Pressure = 1600 x 103 / (3 x 3) + 6 x 145 x 106/ (3 x 3²)
Maximum Base Pressure = 210KN/m²
Minimum Base Pressure = 1600 x 103/ (3 x 3) – 6 x 145 x 106/ (3 x 3²)
Minimum Base Pressure = 145.6 KN/m²
The maximum base pressure (210KN/m²) is greater than the allowable bearing pressure (185KN/m²), so we shall resize the footing. This also implies that locating the column eccentrically causes the pressure in the footing to be above the allowable bearing pressure which was not so when the column was concentrically located in this worked example
From inspection, the pad becomes adequate only when it has at least an area of 10.2m². Let’s maintain a square footing so that less moment is generated in the Pad. Let’s adopt a footing size of 3.25m x 3.25m. This gives a footing area of 10.56m² (still pretty close to 10.2m² and still economical!)
Let’s confirm the pressure below the newly adopted footing
Base Pressure = N/BD ± 6M/BD²
Maximum Base Pressure = 1600 x 103 / (3.25 x 3.25) + 6 x 145 x 106/ (3.25 x 3.25²)
Maximum Base Pressure = 176.8KN/m²
Minimum Base Pressure = 1600 x 103/ (3.25 x 3.25) – 6 x 145 x 106/(3.25 x 3.25²)
Minimum Base Pressure = 126.1 KN/m²
Since both the maximum base pressure is below the allowable pressure, then the footing size is adequate. The pressure diagram under the footing is shown below

Check Whether tension develops in the soil.
We shall check whether there is portion of ineffective bearing below the pad foundation by using the middle third rule
e = M/N
e = 145 x 10³/1600
e = 90.6mm
D/6 = 3250/6
D/6 = 541.7mm
Since e < D/6, then no tension in the soil.
Determine the Ultimate Earth Pressure
Ultimate axial load = 1.35(gk) + 1.5(qk)
1.35*1200 + 1.5*400 = 2220KN/m2
Ultimate Bending Moment = 1.35Gk + 1.5Qk
Ultimate Bending Moment = 1.35 x 80 + 1.5 x 65
= 205.5KN
Ultimate Earth Pressure = N/BD ± 6M/BD²
Maximum Ultimate Earth Pressure = (2220 x 10³)/ (3.25 x 3.25) + (6 x 205.5)/ (3.25 x 3.25²)
Maximum Ultimate Earth Pressure = 246KN/m²
Minimum Ultimate Earth Pressure = (2220 x 10³)/ (3.25 x 3.25) – (6 x 205.5)/(3.25 x 3.25²)
Minimum Ultimate Earth Pressure = 174KN/m²
Earth pressure under the ultimate loads is shown below

Eccentricity of the column base on Ultimate load
e = M/N
e = 205.5 x 10³/ 2220
e = 92.6mm
The eccentricity is still within the middle third (D/6) so no tension
Design the footing for Bending
Assume maximum rebar size = 12mm
Assume cover = 50mm
Assume thickness of foundation (h) = 600mm
Mean Effective depth (d) = h – c – θ
d = 600 – 50 -12 = 538
The bending is taken to the column face.
Length of the pad footing from the column face = ((3250/2) – (350/29)-2.6) = 1357.4
Pressure at column face = 174 + (246 – 174) x (3250 – 1357.4)/3250
Pressure at column face = 216KN/m²

Ultimate Moment (Med) = (216 x 3.250 x 1.3574 x 1.3574/2) + 0.5 x (246 – 216) x 1.3574 x 3250 x 2/3(1.3574) Ultimate Moment (Med) = 706.9KNm
The footing is to be designed for a bending moment of 706.9KNm
- Check whether section is to be designed as singly or doubly reinforced beam
$ K\,=\,\,\frac{M}{bd^2f_{ck}} $
$ K\,=\,\,\frac{706.9 x 10^6}{3250 x 538^2 x 30} $
= 0.025
Since K (0.025) < K’ (0.168); design as singly reinforced.
3) Calculate the lever arm (Z)
$ Z\,\,=\,\,d\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{K}{1.134}} \right) $
$ Z\,\,=\,\,538\left( 0.5+\sqrt{\text{0.25}-\,\,\frac{0.025}{1.134}} \right) $
Z = 525.8mm
Since 525.8 > 0.95d (511): use Z = 511mm
4. Calculate the area of steel
$ A_{st\,\,=\,\,\frac{M_{Ed}}{0.87f_{yk}Z}} $
$ A_{st\,\,=\,\,\frac{706.9 x 10^6}{0.87×500 x 511}} $
Ast = 3047mm2
Ast = 3047mm2/3m Ast = 1015mm2/m width Provide T16@150mm c/c (1407mm2 /m ) in both directions
5) Check Whether area of tensile steel provided satisfies minimum area requirement
Asmin = 0.26 (fctm/fyk )bt d ≥ 0.0013bh
fctm = 0.3 x fck 2/3 fctm fctm = 0.3 x 302/3 = 2.9 Asmin = 0.26 (2.9/500) 3250 x 538
Asmin = 2194mm2
Since 2194mm2 < 3047mm2, then minimum area of reinforcement requirement is satisfied.
Check for one-way shear
One way shear is checked at a distance of d from the column face
Footing length at d from column face = (3250/2) – (350/2) – 538 – 92.6
= 819.4mm
Area of shear influence = 819.4 x 3250 x 10^ (-6) = 2.7m²
Pressure at d from column = 174 + (246 – 174) x (3250 – 819.4)/3250
Pressure at d from column = 227.9KN/m²

Shear Force (VED) = ultimate earth pressure x shear area
VED = 0.5(246 + 227.9) x 3.250 x 1.0 x 2.7 = 631.3KN
vED = 631.3 x 10^3/ (3250 x 538) = 0.36N/mm²
Calculate the shear capacity without shear reinforcement
vRdc = {0.12K(100ρLfck)1/3} bwd ≥ vmin bwd
K = (1 +√200/d) ≤ 2.0
K = (1 +√200/538) = 1.6
ρL = Asl/bwd ≤ 0.02
ρl = (1407) / (1000 × 538) = 0.0026
vRdc = {0.12 x 1.6 (100 x 0.0026 x 30)1/3} ≥ vmin
vRdc = 0.38 N/mm²
vmin = {0.035K3/2fck1/2}
vmin = 0.035 x 1.6³/2 x 301/2
vmin = 0.39 N/mm²
Since vRdc (0.34N/mm²) < vmin (0.39 N/mm²); hence vRdc = 0.39 N/mm²
Since vED < vRdc then no punching shear reinforcement is required.
Check for Punching Shear
The punching shall be checked at the column perimeter and at perimeter 2d from the column face
Punching shear at column perimeter
Shear force acting on column = 2220KN
Shear stress at column perimeter (vEd ) = 2220 x 10³/ (1400 x 538)
vEd = 2.95N/mm²
Calculate the shear capacity of concrete
vRDmax = 0.3 (1 -fck/250)fcd
fcd = fck/1.5 = 30/1.5 = 20 N/mm²
vRDmax = 0.3 (1 -30/250)20 =5.28 N/mm²
vRDmax > vEd, the section of pad is sufficient.
Punching shear at 2d from the column face
Critical perimeter within 2d from the column surface = column perimeter + 4πd
= 4 x 350 + 4 x 3.142 x 538 = 8161.6mm
Area within the critical perimeter = (col length + 4d)² – (4-π) x 1040²
= (350 + 4 x 538)² – (4 – 3.142) x 1040²
= 5266632mm²
= 5.2 x 10^6 mm²
Area of the footing = 3.25 x 10³ x 3.25 x 10³ = 1.0 x 10^7
For punching shear at perimeter 2d from column face, the pressure under the footing will be assumed to be uniform ultimate earth pressure (246.7KN/m2) for simplicity.
Punching shear force (VEd) = 246.7 x (1 x 10^7 – 5.2 x 10^6) x 10^(-6)
VEd = 1303KN
Punching shear stress (vEd) = VEd/(u x d)
Punching shear stress (vEd) = 1303 x 10³/(8161 x 538)
vEd = 0.29N/mm²
(vRdc) is 0.39 N/mm2 as calculated earlier under one-way shear.
Since (vRdc) > vEd ; the section is adequate for punching shear and no reinforcement is required.