Pad foundation

Design of Eccentrically Loaded Pad Foundation – Worked Example

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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

Earth pressure under service load
Earth pressure under service load

  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

Earth pressure under ultimate load
Earth pressure under ultimate load

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

Pressure at the face of column
Pressure at the face of column

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

  1. 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

Earth pressure at d from the column surface
Earth pressure at d from the column surface

 

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.  

Author: Amuletola Rasheed

You can reach Amuletola Rasheed via amuletola@fppengineering.com

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