**Introduction**

Combined footings are normally employed in cases where two or more columns are closely spaced such that their individual pad footings would overlap each other. They are also used where property boundaries would not permit design of separate bases. The most common method of design of combined footings is the rigid approach, which makes the following assumptions;

(1)The footing is infinitely rigid, and therefore, the deflection of the footing does not influence the pressure distribution.

(2) The soil pressure is distributed in a straight line or a plane surface such that the centroid of the soil pressure coincides with the line of action of the resultant force of all the loads acting on the foundation.

**Solved example**

Two (300 x 300)mm square columns spaced at a distance 2.45m is loaded as shown below. The foundation is founded on a soil of bearing capacity 150 kN/m

^{2}. It is desired to design the footing to satisfy all requirements using concrete grade of 30 N/mm

^{2 }and steel of yield strength 500 N/mm

^{2}. The concrete cover is 50mm.

**Solution**

P

_{Ed}= 1.0Gk + 1.0Qk

Total service load on column P1 = 665 + 122 = 787 kN

Total service load on column P2 = 825 + 145 = 970 kN

Total load on column = 787 + 970 = 1757 KN

Assuming 10% of the service load to account for the self weight of the footing;

Sw = 0.1 × 1757 = 175.7 KN

Area of footing required = Total service load/Allowable bearing capacity = (1757 + 175.7)/200 = 9.6635 m

^{2}

Adopt a rectangular base = 5m x 2m (Area provided = 10 m

^{2})

To locate the centroid of the footing, let us take moment about column P1;

(970 × 2.45) - 1757(

*x*) = 0

On solving,

*x*= 1.352m from column P1

Let the projection of the footing from the lighter column be L

_{1}

Therefore;

L

_{1}+ 1.352m = (Total length of footing)/2 = 5/2 = 2.5

Therefore, L

_{1}= 2.5 - 1.352 = 1.148m (say 1.15m)

The final disposition of the footing is given below assuming a trial footing depth of 600mm.

^{}

^{We can check the preliminary thickness of 600mm by considering the punching shear at the column perimeter. At the column perimeter, the maximum punching shear stress should not be exceeded.}

^{}

^{At ultimate limit state; }

^{P1 = 1.35Gk + 1.5Qk = 1.35(665) + 1.5(122) = 1080.75 kN}

P2 = 1.35Gk + 1.5Qk = 1.35(825) + 1.5(145) = 1331.25 kN

^{Total ultimate load = 1080.75 + 1331.25 = 2412 KN}

^{VEd < VRd,max}

^{}

^{Where;}

V

_{Ed}= βV

_{Ed}/u

_{0}d

V

_{Rd,max}= 0.5

*v*f

_{cd}

Considering Column B3 bearing the maximum axial load, V

_{Ed}= 1850 kN

β = 1.5 (an approximate value from clause 6.4.3(6))

u

_{0 }= column perimeter = 2(300) + 2(300) = 1200 mm

*v*= 0.6[1 - fck/250] (strength reduction factor for concrete cracked in shear)

*v*= 0.6[1 - 30/250] = 0.528

f

_{cd}= α

_{cc}f

_{ck}/γ

_{c}f

_{cd}= (1.0 × 30)/1.5 = 20 N/mm

^{2}

_{}V

_{Rd,max}= 0.5 × 0.528 × 20 = 5.28 N/mm

^{2}

_{}

V

_{Ed}= (1.5 × 1331.25 × 1000) / (1200mm × d)

Therefore;

1996874/1200d = 5.28

On solving; d

_{min}= 315.263mm

Hence the depth is ok.

Earth pressure intensity = 2412/10 = 241.2 KN/m

^{2}

For a width of 2m, q = 241.2 × 2 = 482.4 kN/m

**Structural Design**

**Bottom Reinforcement**

We are supposed to take our maximum moment from the column face which can be easily determined as shown below.

M

_{Ed}= [482.4 × (1.4 - 0.15)

^{2}]/2 = 376.875 kNm

Effective depth (d) = 600 - 50 - 10 = 540 mm (assuming B20mm bars).

M

_{Ed}= 376.875 KNm/m

b = 2000mm

k = M

_{Ed}/(f

_{ck}bd

^{2})

k = (376.875 × 10

^{6})/(30 × 2000 × 540

^{2}) = 0.0215

Since k < 0.167 No compression reinforcement required

z = d[0.5+ √(0.25 - 0.882k)] = z = d[0.5+ √(0.25 - (0.882 × 0.0215)] = 0.95d

A

_{s1}= M

_{Ed}/(0.87f

_{yk}z)

A

_{s1 }= (376.875 × 10

^{6})/(0.87 × 500 × 0.95 × 540) = 1688 mm

^{2}

Provide 10B16mm @ 225mm c/c BOT (A

_{Sprov}= 2010 mm

^{2}/m)

Read Also....

**Design of Cantilever Retaining Wall Supporting Lateritic Earthfill**

**Design of Continuous Beam and Slab footing to BS 8110-1:1997**

**Top reinforcement**

There is no hogging moment on the footing as shown in the bending moment diagram. However, provide minimum reinforcement at the top.

A

_{smin}= 0.0013bd = (0.0013 × 1000 × 540) = 702 mm

^{2}

Provide B12mm @ 150mm c/c BOT (A

_{Sprov}= 753 mm

^{2}/m)

**Transverse Reinforcement**

Cantilever arm = (2.0 - 0.3)/2 = 0.85m

M

_{Ed}= [482.4 × 0.85

^{2}]/2 = 174.267 kNm

Effective depth (d) = 600 - 50 - 10 = 540 mm (assuming B20mm bars).

M

_{Ed}= 174.267 kNm

b = 1000mm (designing per unit strip)

k = M

_{Ed}/(f

_{ck}bd

^{2})

k = (174.267 × 10

^{6})/(30 × 1000 × 540

^{2}) = 0.0199

Since k < 0.167 No compression reinforcement required

z = d[0.5+ √(0.25 - 0.882k)] = z = d[0.5+ √(0.25 - (0.882 × 0.0199)] = 0.95d

A

_{s1}= M

_{Ed}/(0.87f

_{yk}z)

A

_{s1 }= (174.267 × 10

^{6})/(0.87 × 500 × 0.95 × 540) = 781 mm

^{2}

Provide B16mm @ 250mm c/c BOT (A

_{Sprov}= 804 mm

^{2}/m)

**Check for shear**

Clause 6.4.2(2) of EN 1992-1-1:2004 states that control perimeters at a distance less than 2d should be considered where the concentrated force is opposed by a high pressure (e.g. soil pressure on a base), or by the effects of a load or reaction within a distance 2d of the periphery of area of application of the force.

Let us investigate this at a distance d from the face of the column according to clause 6.4.4(2) of Eurocode 2.

Using the maximum shear force V

_{Ed}= 675.36 kN

V

_{Ed,red}= V

_{Ed}- ∆V

_{Ed}

Where ∆V

_{Ed}is the net upward force within the control perimeter considered i.e. upward pressure from soil minus self weight of base.

Control area = 1.38m × 1.38m = 1.9044 m

^{2}

Soil pressure = 241.2 KN/m

^{2}

Unit weight of concrete = 25 kN/m

^{3}

Hence, ∆V

_{Ed}= (241.2 × 1.9044) - (25 × 0.6 × 1.9044) = 430.775 kN

V

_{Ed,red}= 1331.25 - 430.775 = 900.475 kN

v

_{Ed,red}= V

_{Ed,red}/ ud

where u is the control perimeter = 4(2d + 300) = 4(2 × 540 + 300) = 5520mm

v

_{Ed,red}= (900.475× 1000) / (5520 × 540) = 0.3021 N/mm

^{2}

**Shear Resistance of section**

V

_{Rd,c}= [C

_{Rd,c}.k.(100ρ

_{1}f

_{ck})

^{(1/3)}] × 2d/a ≥ (V

_{min}× 2d/a)

C

_{Rd,c}= 0.18/γ

_{c}= 0.18/1.5 = 0.12

k = 1 + √(200/d) = 1 + √(200/540) = 1.606 > 2.0, therefore, k = 1.608

V

_{min}= 0.035k

^{(3/2)}f

_{ck}

^{0.5}

V

_{min}= 0.035 × (1.608)

^{1.5}× 30

^{0.5}= 0.391 N/mm

^{2}

^{}ρ

_{x}(in the transverse direction) = As/bd = 804/(1000 × 540) = 0.00148

ρ

_{z}(in the longitudinal direction) = As/bd = 2010/(2000 × 540) = 0.00186

ρ

_{1}= √(ρ

_{x }× ρ

_{z}) = 0.00165 < 0.02

V

_{Rd,c}= [0.12 × 1.608 (100 × 0.00165 × 30 )

^{(1/3)}] × 2 = 0.6577 N/mm

^{2}

0.3021 N/mm

^{2}< 0.6577 N/mm

^{2 }

We can now verify that the section is ok for punching shear.

With this, the reinforcement detailing can be drawn.

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