This article presents a worked example on the structural design of axially loaded pad foundation to EN 1992 -1-1-2004. The geotechnical aspects of the design shall be approached 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. 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

Total serviceability load on the base = 1.0Gk + 1.0Qk

                    = 1.0 x 1200 + 1.0 x 400 = 1600KN

Base Area = (Service Load)/ (allowable bearing pressure)

                    = 1600/185 = 8.6m²

Provide a pad footing of dimension 3 x 3m which gives an area of 9m2  

Determine the Ultimate Earth Pressure

Ultimate axial load = 1.35(gk) + 1.5(qk)

                   1.35*1200 + 1.5*400 = 2220KN/m2

Ultimate Earth Pressure = 2220/9 = 246.7KN/m2  

 

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 = (3000/2-350/2) = 1325

Ultimate Moment (Med) = 246.7 x 3 x 1325 x 1325/2 = 649.6KNm

The footing is to be designed for a bending moment of 649.6KNm  

1. Check whether section is to be designed as singly or doubly reinforced beam

$ K\,=\,\,\frac{M}{bd^2f_{ck}} $

$ K\,=\,\,\frac{649.6 x 10^6}{3000 x 538^2 x 30} $

 

= 0.02

Since K (0.02) < 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{0.02}{1.134}} \right) $

Z = 525.9mm

Since 525.9 > 0.95d (511): use Z = 511mm

 

4. Calculate the area of steel

$ A_{st\,\,=\,\,\frac{M_{Ed}}{0.87f_{yk}Z}} $

 

$ A_{st\,\,=\,\,\frac{649.6 x 10^6}{0.87×500 x 511}} $

 

Ast = 2800mm²

Ast = 2800mm²/3m

Ast = 933mm²/m width

Provide T16@200mmc/c (1005mm²) in both directions  

5) Check Whether area of tensile steel provided satisfies minimum area requirement

 A_smin = 0.26 (f_ctm/f_yk )b_t d ≥ 0.0013bh

fctm = 0.3 x f_ck ^2/3 fctm

fctm = 0.3 x 30^2/3 = 2.9

 A_smin  = 0.26 (2.9/500) 3000 x 538

  A_smin      =   2025.7mm²

     Since 2025.7mm² < 2800mm², 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 = 3000/2-350/2 – 538 = 787

Area of shear influence = 787 x 3000 x 10-6 = 2.36m2

Shear Force (V_ED) = ultimate earth pressure x shear area

V_ED  = 246.7 x 2.36 

V_ED = 582.38KN  

 

Calculate the shear resistance without shear reinforcement

v_{Rdc  x bwd    = {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   

ρ_l = (1005) / (1000 × 538) = 0.002

v_{Rdc  = {0.12 x 1.6 (100 x 0.002 x 30)^1/3} ≥ vmin} 

v_Rdc = 0.34 N/mm²

 

_{vmin = {0.035K3/2fck^1/2}}

_{vmin = 0.035 x 1.6³/2 x 30^1/2}

_vmin = 0.39 N/mm²

Since v_Rdc (0.34N/mm²) < v_min (0.39 N/mm²); hence v_Rdc = 0.39 N/mm2

Since v_ED < v_Rdc  then no punching shear reinforcement is required.

Shear resistance (V_Rdc )=  v_Rdc b_wd = 0.39 x 3000 x 538 = 631.9KN

Since V_ED < V_Rdc then no 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

 

Calculate the shear capacity of concrete

v_RDmax = 0.3 (1 -f_ck/250)f_cd

f_cd = fck/1.5 = 30/1.5 = 20 N/mm2

v_RDmax = 0.3 (1 -30_/250)20

=5.28 N/mm²

Shear resistance (v_RDmax) = 5.28 x 4(350) x 538 x 10-6 

v_{RDmax = 3976.8KN}

v_RDmax > V_Ed, the section of pad is sufficient.

 

Note: The shear force (V_Ed) for one-way shear is checked against the shear resistance of concrete without reinforcement (V_Rdc). This could also have been achieved by checking the shear stress (v_Ed) for one-way shear against shear capacity of concrete without reinforcement (v_Rdc). This is also applicable to punching shear at the column perimeter. The axial load on the column could have been converted to shear stress at the perimeter of the column and checked against the concrete shear capacity.

 

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 x 10³  x 3 x 10³ = 9 x 10^6

Punching shear force (V_Ed) = 246.7 x (9 x 10^6- 5.2 x 10^6) x 10^(-6)

V_Ed = 920KN

Punching shear stress (v_Ed) = V_Ed/(u x d)

Punching shear stress (v_Ed) = 920 x 10³/(8161 x 538) 

v_Ed  = 0.2N/mm²

(v_Rdc) is 0.39 N/mm2 as calculated earlier under one-way shear.

Since (v_Rdc) > v_Ed ;  the section is adequate for punching shear and no reinforcement is required.

 

Detailing