Designing a Combined Footing-Worked Example

Spread foundations supporting two or more columns are classified as combined footings. Sometimes, it becomes necessary to combine the footing for two or more columns for two reasons:

If columns are so close to each other such that constructing single footings will result in an overlap.
When columns are very close to a property line such that a single footing cannot be constructed without projecting beyond the line.

In designing combined footings, it is always best to proportion the footing such that the resultant of the applied loads pass through the centroid of the footing area. This allows a uniform bearing pressure under the footing and helps to prevent differential settlement. However, sometimes, the footings are constraint from projecting in a certain direction due to the property line or site restrictions. Thus, it is not possible to proportion the footing for uniform pressure distribution, the eccentricities must, therefore, be calculated and the footing designed for varying earth pressure.

Alternatively, a strap foundation may be used where a property line or site constraint exists. However, a strap foundation might not be economical for closely spaced columns. The design of strap foundation has been the subject of a previous post, see: Designing a strap foundation to Eurocode

Design Procedures

The design procedure of a combined footing is almost the same with that of pad foundation. The only concern is proportioning the footing for uniform pressure distribution. Generally, the steps are as follows:

Determine the required size of the footing using the soil bearing resistance and the loads at serviceability limit state.
Proportion the footing for uniform pressure distribution by finding the location of the centroid and adjusting the footing appropriately.
Determine the bearing pressures at the ultimate limit state.

Assume a suitable thickness for the footing, calculate the effective depth and verify punching shear at the column face.
Analyze the footing in the longitudinal and transverse direction.
Design the bending reinforcement in the longitudinal and transverse direction.

Verify shear at the critical sections including punching shear at the basic control perimeter 2.0d from the column face.
Verify detailing requirements.

The analysis of the combined is idealized as an inverted overhanging beam subjected to the earth pressure at the ultimate limit state.

Worked Example

A combined footing is required in a proposed office building for four heavily loaded closely spaced columns. Design the footing completely assuming a presumed bearing resistance of 150kN/m2 using C30/37 concrete with 460mpa bars. The design data for the columns are presented in Table 1 and Figure 1 shows a part of the column layout.


Columns	Sizes	Gk (kN)	Qk (kN)
C1	350x350	900	600
C2	400x400	1200	800
C3	400x400	1200	800
C4	275x275	350	215

Serviceability limit state.

Let’s start by determining the actions on the footings.

{ N }_{ Ed,1 }=1.0{ G }_{ k }+1.0{ Q }_{ k }

=(1.0\times 900)+(1.0\times 600)=1500kN

{ N }_{ Ed,2 }=1.0{ G }_{ k }+1.0{ Q }_{ k }

=(1.0\times 1200)+(1.0\times 800)=2000kN

{ N }_{ Ed,3 }=1.0{ G }_{ k }+1.0{ Q }_{ k }

=(1.0\times 1200)+(1.0\times 800)=2000kN

{ N }_{ Ed,4 }=1.0{ G }_{ k }+1.0{ Q }_{ k }=(1.0\times 350)+(1.0\times 215)=565kN

Assume the self weight of the footing is 20% of the total axial loads @ serviceability.

0.2\times (1500+2000+2000+565)=1213kN

Total\quad loads=(1500+2000+2000+565+1213kN)=7278kN

Area\quad required=\frac { Total\quad loads }{ Bearing\quad resistance } =\frac { 7278 }{ 150 } =48.52{ m }^{ 2 }

Assume\quad a\quad square\quad base\quad \sqrt { 48.52 }=6.97m

Provide\quad a\quad 7m\times 7m\quad Footing\quad ({ Area }=49{ m }^{ 2 })

Proportioning the Footing

The base is unsymmetrically loaded, therefore we must proportion the base in the y-y and z-z direction inorder to have a uniform pressure distribution.

Y-Y direction

Take moments about the centerlines of column C2 & C3

(1500+565)\times 4.5+1213{ e }_{ y }-R{ e }_{ y }=0

R=1500+2000+2000+565+1213=7278kN

{ e }_{ y }=\frac { (1500+565)\times 4.5 }{ (7278-1213) } =1.50m(approx)

{ y }_{ 1 }=\frac { l }{ 2 } -{ e }_{ y }=\frac { 7 }{ 2 } -1.5=2.0m

{ y }_{ 2 }=l-4.5-{ y }_{ 1 }=7-4.5-2=0.5m

Z-Z direction

Take moments about the centerlines of column C1 & C2

(565+2000)\times 3.5+1213{ e }_{ z }-R{ e }_{ z }=0

R=1500+2000+2000+565+1213=7278kN

{ e }_{ z }=\frac { (565+2000)\times 3.5 }{ (7278-1213) } =1.50m(approx)

{ z }_{ 1 }=\frac { b }{ 2 } -{ e }_{ y }=\frac { 7 }{ 2 } -1.5=2.0m

{ z }_{ 2 }=b-3.5-{ z }_{ 1 }=7-3.5-2=1.5m

Haven proportioned the footing, we can now proceed to the analysis and design at the ultimate limit state.

Ultimate limit state

{ N }_{ Ed,1 }={ 1.35G }_{ k }+1.5{ Q }_{ k }

(1.35\times 900+1.5\times 600)=2115kN

{ N }_{ Ed,2 }={ 1.35G }_{ k }+1.5{ Q }_{ k }

=(1.35\times 1200+1.5\times 800)=2820kN

{ N }_{ Ed,3 }={ 1.35G }_{ k }+1.5{ Q }_{ k }

=(1.35\times 1200+1.5\times 800)=2820kN

{ N }_{ Ed,4 }={ 1.35G }_{ k }+1.5{ Q }_{ k }

=(1.35\times 350+1.5\times 215)=795kN

Bearing\quad Pressure\quad { p }_{ u }=\frac { { N }_{ Ed,1 }+{ N }_{ Ed,2 }+{ N }_{ Ed,3 }+{ N }_{ Ed,4 } }{ Area }

=\frac { 2115+2820+2820+795 }{ 49 } =174.5kN/{ m }^{ 2 }

Punching verification at the column faces

verify\quad that\quad { N }_{ Ed }\le { V }_{ Rd,max }

Assume the over depth of the footing, h=975mm and the mean effective depth d=900mm

Column C1

{ N }_{ Ed }=2115kN\quad ;\quad assume\quad d=900m

{ V }_{ Rd,max }=0.2\left( 1-\frac { 30 }{ 250 } \right) 30(4\times 350)\times 900\times { 10 }^{ -3 }=6652.8kN

({ N }_{ Ed }=2115kN)\le ({ V }_{ Rd,max }=6652.8kN)

Column C2 & C3

{ N }_{ Ed }=2820kN\quad ;\quad assume\quad d=900mm

{ V }_{ Rd,max }=0.2\left( 1-\frac { 30 }{ 250 } \right) 30(4\times 400)\times 900\times { 10 }^{ -3 }=7603.2kN

({ N }_{ Ed }=2820kN)\le ({ V }_{ Rd,max }=7603.2kN)

Column C4

{ N }_{ Ed }=795kN\quad ;assume\quad d=900mm

{ V }_{ Rd,max }=0.2\left( 1-\frac { 30 }{ 250 } \right) 30(4\times 275)\times 900\times { 10 }^{ -3 }=5227.2kN

({ N }_{ Ed }=795kN)\le ({ V }_{ Rd,max }=5227.2kN)

Analysis

The footing is idealized as an inverted overhanging beam subjected to the earth pressure. The analysis will be carried out in the y-y and z-z direction for moment and shear. Since the beam is statically determinate simple rules of statics can be used to obtain the bending moment and shear diagram as presented in the following figures.

Analysis in the Y-Y Direction

Analysis in the Z-Z Direction

Flexural design

Bending in the Y-Y direction

Hogging in spans

{ M }_{ Ed }=271.4kN.m/m

assuming\quad mean\quad effective\quad depth=900mm

k=\frac { { M }_{ ed } }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 271.4\times { 10 }^{ 6 } }{ { 10 }^{ 3 }{ \times 900 }^{ 2 }\times 30 } =0.0112

z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d

z=0.95d=0.95\times 900=855mm

{ A }_{ s }\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 271.4\times { 10 }^{ 6 } }{ 0.87\times 460\times 855 } =793.18{ mm }^{ 2 }/m

Try\quad Y16-250mm\quad centeres(804{ mm }^{ 2 }/m)\quad subject\quad to\quad { A }_{ s,min }

Sagging at Supports

{ M }_{ Ed }=349.0kN.m/m

assuming\quad mean\quad effective\quad depth=900mm

k=\frac { { M }_{ ed } }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 349.0\times { 10 }^{ 6 } }{ { 10 }^{ 3 }{ \times 900 }^{ 2 }\times 30 } =0.0144

z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d

z=0.95d=0.95\times 900=855mm

{ A }_{ s }\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 349.00\times { 10 }^{ 6 } }{ 0.87\times 460\times 855 } =1019.6{ mm }^{ 2 }/m

Try\quad Y16-175mm\quad centeres(1148{ mm }^{ 2 }/m)\quad subject\quad to\quad { A }_{ s,min }

Bending in the z-z direction

Hogging in span

As we can see, from the moment diagram, the hogging moment is small and insignificant, therefore we can provide the minimum area of steel.

Sagging at supports

{ M }_{ Ed }=349.0kN.m/m

assuming\quad mean\quad effective\quad depth=900mm

k=\frac { { M }_{ ed } }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 349.0\times { 10 }^{ 6 } }{ { 10 }^{ 3 }{ \times 900 }^{ 2 }\times 30 } =0.0144

z=d\left[ 0.5+\sqrt { 0.25-0.882k } \right] \le 0.95d

z=0.95d=0.95\times 900=855mm

{ A }_{ s }\frac { { M }_{ Ed } }{ 0.87{ f }_{ yk }z } =\frac { 349.00\times { 10 }^{ 6 } }{ 0.87\times 460\times 855 } =1019.6{ mm }^{ 2 }/m

Try\quad Y16-175mm\quad centeres(1148{ mm }^{ 2 })\quad subject\quad to\quad { A }_{ s,min }

Verify Minimum Area of Reinforcement

{ A }_{ s,min }=0.26\frac { { f }_{ ctm } }{ { f }_{ yk } } bd\ge 0.0013bd

{ f }_{ ctm }=0.3({ 30) }^{ 2/3 }=2.9mpa

=0.26\times \frac { 2.9 }{ 460 } \times { 10 }^{ 3 }\times 900\ge 0.0013\times { 10 }^{ 3 }\times 900

{ A }_{ s,min }=1475.2{ mm }^{ 2 }/m

As we can see, the minimum area of steel is critical in every section, designed for previously therefore we can provide the minimum area of steel and proceed with the design.

USE\quad Y16-125mm\quad Centres\quad (1608{ mm }^{ 2 }/m)\quad \\ Top\quad \& \quad Btm\quad in\quad Both\quad Directions

Punching verification at basic control perimeter

By inspection, the control perimeter for columns C1, C2 & C4 are outside the footing, therefore punching check cannot be verified. Also since the footing is considerably deep, punching will not be critical in C3. However, you may choose to verify, the procedure is the same as for a pad foundation.

Transverse Shear

The critical section for shear is taken at 1.0d from the support. Shear is more critical in the y-y direction and will be used for the verification.

{ v }_{ Ed }\le { v }_{ Rd,c }

The expression written above must be satisfied, else we might have to increase the concrete strength, section depth or the reinforcements.

{ V }_{ max }=457.6kN/m

{ V }_{ Ed }=457.6-(0.9\times 174.5)=300.45kN/m

{ v }_{ Ed }=\frac { { V }_{ Ed } }{ bd } =\frac { 300.45\times { 10 }^{ 3 } }{ { 10 }^{ 3 }\times 900 } =0.33mpa

Concrete shear resistance

{ 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 }{ 900 } } =1.47

\rho =\frac { { A }_{ s } }{ bd } =\frac { 1608 }{ { 10 }^{ 3 }\times 900 } =0.00179

{ v }_{ Rd,c }=0.12\times 1.47(100\times 0.00179\times 25)^{ 1/3 }\ge

0.035\times { 1.47 }^{ 3/2 }\sqrt { 25 }

{ v }_{ Rd,c }=0.34mpa

{ (v }_{ Ed }=0.33mpa)<({ v }_{ Rd,c }=0.34mpa) o.k

This concludes the design of the combined footing, by inspection, transverse shear was more critical, and was what governed the thickness of the footing. Figure 7 shows the detail drawing for the footing.