Designing a Pad foundation to Eurocode

Introduction

The simplest form of spread foundations used in transferring loads from a structure into the soil is the pad foundation. They are basically required to spread the concentrated actions from a vertical element into the founding soil provided that soil is not too deep to reach and is sufficiently stiff, in order words possesses good bearing capacity.

This post only covers the design of reinforced concrete pad foundation for single columns, it doesn’t discuss how the bearing capacity of the soil is obtained because this as already been covered in a previous post, also other forms of pad foundations such as combined, strip and strap foundations are not considered here as these would be the subject of a later post.

Principle of Pad Foundation

Pad foundations are designed to spread a concentrated force from vertical elements on the bearing stratum. Therefore the foundation must be sufficiently stiff such that the force is spread on the stratum without the applied pressure exceeding the permissible bearing capacity of the soil.

It is usually beneficial for the footing of a single column to be made square in shape but when it supports a bending moment in one direction, it becomes more economical to have a rectangular base with the longer side in the direction of the applied moment.

Bearing Pressure

The bearing pressure across a single pad foundations can take one of the following forms.

1. If the pad foundation is required to carry only axial action with no bending moment in both direction, the bearing pressure is uniform and is defined as

p=\frac { N }{ BL }

Where: N= Axial action; L= length of the footing; B=width of the footing

2. When bending moments are applied to a pad footing it is good practice to size the foundation so that the resultant force lies within the middle third of the base. To determine these the eccentricity must be less than one-sixth of the footing length. The bearing pressure is then defined as

p=\frac { N }{ BL } \quad \pm \frac { 6Ne }{ B{ L }^{ 2 } } \quad where\quad e=\frac { M }{ N }

Where: N= Axial action; M = Design Bending Moment L= length of the footing; B=width of the footing e = eccentricity

When the resultant force lies outside the middle third, the effective contact area between the pad footing and the soil is reduced. This is because the length of the pad is reduced as one side of the pad develops tension and so is lifted off the soil as the resultant force is applied to it due to its eccentricity.

In such conditions there is a possibility of uplift occurring as the pad rotates onto the soil. The value of the bearing stress being applied to the soil is calculated thus:

p=\frac { 2N }{ 3By } \quad where\quad y=\frac { L }{ 2 } -e

3. When moments are applied in both direction to the pad foundation, the bearing stress is given thus:

p=\frac { N }{ BD } \pm \frac { 6N{ e }_{ y } }{ B{ D }^{ 2 } } \pm \frac { 6N{ e }_{ x } }{ { B }^{ 2 }D }
where\quad { e }_{ y }=\frac { { M }_{ y } }{ N } \quad \& \quad { e }_{ x }=\frac { { M }_{ x } }{ N }

Where: N= Axial action; Mx = Design bending moment in the x-direction; My = Design bending moment in the y-direction L= length of the footing; B=width of the footing; ex & ey are the eccentricities in both directions

Partial Factors

When designing foundations of any kind to the Eurocodes, the partial factors to be applied to forces differ depending on what is being checked within the footing. In essence, when checking the applied stress against the overall strength of the soil, the partial factors are significantly reduced, whereas when the foundation itself is being designed, full partial factors apply.

BS EN 1990 defines the partial factors for sub-structures and splits them into geotechnical (GEO) and structural (STR). When checking the bearing stress that is being applied to the soil, the GEO partial factors apply.

{ E }_{ d }={ 1.0G }_{ k }+{ 1.3Q }_{ k,1 }+1.3{ \psi Q }_{ k,2 }

When carrying out the final design for the reinforcement the (STR) partial factors apply

{ E }_{ d }={ 1.35G }_{ k }+{ 1.5Q }_{ k,1 }+1.5{ \psi Q }_{ k,2 }

Reinforcement Design

The design of reinforcement in a pad foundation is similar to the design of a slab as discussed in a previous post. The only difference is that the slab is designed as an inverted cantilever slab subjected to the bearing pressure at the ultimate limit state (STR). The critical section in bending is basically taken as the column face as illustrated in Figure 1.

Fig 1: Pad foundation critical sections for design

Punching Shear Check

An additional check is required as regard to shear, as a matter of fact ,one of the governing criteria that determines the depth of the footing is its ability to resist ‘punching shear. This is a shear force that is developed around the perimeter of the vertical element the pad is supporting.

The punching shear is verified at two locations.

  • At the column face
  • At the first basic control perimeter (2d) from the face of column as shown in figure 1.

At the column face, the following expression must be verified.

{ N }_{ ED }\le { V }_{ RD,max }

Where: NED = is the design axial action; vRD,max= is the concrete maximum shear resistance defined as

{ V }_{ RD,max }=0.2\left[ 1-\frac { { f }_{ ck } }{ 250 } \right] { f }_{ ck }pd

At the first basic control perimeter (2d) from the face of the column, the following expression must be satisfied.

{ v }_{ ED }\le { v }_{ RD,c }

vRD,C is the design concrete resistance, vED is the applied punching stress while defined as

{ v }_{ ED }=\frac { { V }_{ ED } }{ { u }_{ 1 }d } \quad where\quad { V }_{ ED }={ N }_{ ED }-{ \triangle V }_{ ED }
\triangle { V }_{ ED }=pA
A={ c }_{ 1 }{ c }_{ 2 }+\pi { \left( 2d \right) }^{ 2 }+4({ c }_{ 1 }+{ c }_{ 2 })d
{ u }_{ 1 }=2({ c }_{ 1 }+{ c }_{ 2 })+4\pi d\\

Where

  • vED is the design punching force
  • ΔvED is the upward force within the control perimeter
  • NED is the design axial action
  • p=bearing stress at the ultimate limit state (STR)
  • A is the area within the control perimeter
  • u1 is the control perimeter
  • d is the effective depth.
  • c1 & c2 are the column dimensions

N:B vRD,C is as defined in the design of beams.

Procedure for Design

  • Determine the required size of the pad foundation using the permissible bearing stress and the critical loading at the serviceability limit state.
  • Determine the bearing pressures associated with the critical loading arrangement at the ultimate limit state.
  • Assume a suitable thickness for the footing, h and effective depth, d and verify punching shear at the column face
  • Design the bending reinforcement
  • Verify shear at the critical sections including punching shear at the basic control perimeter 2d from the column face.
  • Verify detailing requirements.

Worked Example

A pad foundation is required for a column 500mm x 500mm. The column carries 1750kN from permanent actions and 1050kN from variable actions. Design the pad foundation completely from C30/37 concrete with steel grade of 460mpa, Assuming the presumed bearing resistance of the soil is 150kN/m2.

Serviceability limit state
N={ 1.0G }_{ k }{ +1.0Q }_{ k }=1,0(1750)+1.0(1050)=2800kN

Assume 10% increase due to footing self weight =0.1×2800 =280kN

Total\quad load\quad =2800+280=3080kN
\quad Area=\frac { Total\quad load }{ Presumed\quad bearing\quad resistance } =\frac { 3080 }{ 150 } =20.53{ m }^{ 2 }
Assume\quad a\quad square\quad base:\quad length=width=\sqrt { 20.53 } =4.53m
Provide\quad a\quad square\quad base\quad of\quad 4.6m\times 4.6m
Ultimate Limit State
{ N }_{ ED }=1.35{ G }_{ k }+1.5{ Q }_{ k }=1.35(1750)+1.5(1050)=3937.5kN
soil\quad pressure;\quad p=\frac { { N }_{ ED } }{ Area } =\frac { 3937.5 }{ 4.6\times 4.6 } =186.08kN/{ m }^{ 2 }
Punching check @ column face

Assume h=800mm; Cover =50mm; and 20mm bars

d=h-({ c }_{ nom }+\phi /2)=800-(50+20/2)=740mm
Verify\quad that\quad { N }_{ ED }\le { V }_{ RD,max }
{ V }_{ RD,max }=0.2\left[ 1-\frac { { f }_{ ck } }{ 250 } \right] { f }_{ ck }pd
\\ 0.2\left[ 1-\frac { 30 }{ 250 } \right] 30\cdot (4\times 500)\times 740\times { 10 }^{ -3 }=7814.4kN
\\ Therefore\quad ({ N }_{ ED }=3937,5kN)<({ V }_{ Rd,max }=7814.4.kN
Flexural Design

As stated earlier the pad is designed as an inverted cantilever slab subjected to soil pressure. The lever arm. critical section is at the column face.

lever\quad arm\quad l=\left( \frac { 4.6-0.5 }{ 2 } \right) =2.05m
{ M }_{ ED }=\frac { p{ l }^{ 2 } }{ 2 } =\frac { 186.08\times { 2.05 }^{ 2 } }{ 2 } =391kN.m/m
k=\frac { { M }_{ ED } }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 391\times { 10 }^{ 6} }{ 1000\times { 740 }^{ 2 }\times 30 } =0.024<0.168
z=d\left[ 0.5+\sqrt { 0.25-\frac { k }{ 1.134 } } \right] \le 0.95d
0.95d\quad =0.95\times 740=703mm
{ A }_{ s }=\frac { { M }_{ ED } }{ 0.87{ f }_{ yk }z } =\frac { 391\times { 10 }^{ 6 } }{ 0.87\times 460\times 703} =1390{ mm }^{ 2 }/m
Try\quad T20\quad @\quad 200mm\quad c/c\quad ({ A }_{ sprov }=1570{ mm }^{ 2 }/m)
Verify Punching at Basic Control Perimeter (2.0d)
{ V }_{ ED }=N_{ ED }-\triangle { V }_{ ED }
\triangle { V }_{ ED }=pA
A={ 500 }^{ 2 }+4(500+500)740+\pi \left( 2\times 740 \right) ^{ 2 }=10.09m^{ 2 }
{ p }_{ 1 }=2({ c }_{ 1 }+{ c }_{ 2 })+4\pi { d }=2(500+500)+4\pi ({ 740 })=11,303mm
\triangle { V }_{ ED }=186.08\times 10.09=1877.55kN
{ V }_{ ED }=3937.5-1877.55=2060kN
{ v }_{ ED }=\frac { { V }_{ ED } }{ { p }_{ 1 }d }
{ v }_{ ED }=\frac { 2060\times { 10 }^{ 3 } }{ 11303\times 740 } =0.25mpa
{ v }_{ RD.c }=0.12k(100\rho { f }_{ ck })^{ 1/3 }\ge 0.035{ k }^{ 3/2 }\sqrt { { f }_{ ck } }
k=1+\sqrt { \frac { 200 }{ d } } =1+\sqrt { \frac { 200 }{ 740 } } =1.52<2
\rho =\frac { { A }_{ s } }{ bd } =\frac { 1570 }{ { 10 }^{ 3 }\times 740 } =2.21\times { 10 }^{ -3 }
{ v }_{ RD,c }=0.12\cdot 1.52(100\cdot 2.21\times 10^{ -3 }\times 30)^{ 1/3 }\ge 0.035\times { 1.52 }^{ 3/2 }\sqrt { 30 } =0.36mpa
{ (v }_{ ED }=0.25mpa)<({ v }_{ RD,c }=0.36mpa)
Transverse Shear

The critical section for shear is taken at 1.0d from the column face as shown in figure 1

{ V }_{ ED }=p\left( 2.05-1.0d \right) =186.08\times (2.05-0.740)=243.76kN/m
{ v }_{ ED }=\frac { { V }_{ ED } }{ bd } =\frac { 243.76\times { 10 }^{ 3 } }{ 1000\times 740 } =0.32mpa
{ v }_{ RD,c }\quad as\quad before\quad =0.4mpa
{ (v }_{ ED }=0.32mpa)<{ (v }_{ RD,c }=0.36mpa)
Verify Minimum Area of Steel
{ A }_{ s,min }=0.26\frac { { f }_{ ctm } }{ { f }_{ yk } } bd\ge 0.0013bd
{ f }_{ ctm }=0.3{ { f }_{ ck } }^{ 2/3 }=0.3({ 30}^{ 3/2 })=2.9mpa
=0.26\frac { 2.9}{ 460 } \cdot { 10 }^{ 3 }\cdot 740\ge 0.0013\cdot { 10 }^{ 3 }\cdot 740
{ A }_{ s.min }=1213{ mm }^{ 2 }/m<(A_{ s,prov }=1570mm^{ 2 }/m
USE\quad T20-200mm\quad BOTH\quad WAYS

Further Reading and References

  • Tomlinson M. J. (2001) Foundation Design and Construction (7th ed.) Zug, Switzerland: Prentice Hall
  • Mosley W., Bungey J. and Hulse R. (2007) Reinforced Concrete Design to Eurocode 2 (6th ed.) Basingstoke, UK: Palgrave Macmillan
  • Institution of Structural Engineers (2013). Designing a Concrete Pad Foundation- Technical Guidance Notes (Level 2).

THANKYOU!!!

28 Replies to “Designing a Pad foundation to Eurocode”

  1. hello victor,
    in your calculation under punching at basic control perimeter(2.0d) for pad footing.
    Area,A = 6.08m? how
    P1 = 10670.8mm? how
    Also, VRd,c = 0.35mpa which is less than 0.37mpa but you prefer to round it up to 0.4mpa why?

    1. Thank you for your observations, the expression for calculating the critical perimeter and the loaded area is defined under punching shear check. Secondly, whenever the applied shear stress is greater than the concrete shear resistance. You can either increase the depth of the footing, increase the reinforcements, or increase the concrete class. This post is just a guide to the design of pad footings. However, the post has been updated. You might want to take a second look at it.

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