A combined footing is a type of shallow foundation where a common base is provided for two closely spaced columns. When a common base in form of a strip is provided for three or more rows of columns, it is better described as a continuous footing. 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 can also be used where property boundaries would not permit the design of separate bases.

The design of a combined footing involves the geotechnical and structural design, where the proper proportioning of the dimensions, thickness of the base, and proper reinforcement is provided. The foundation of a combined footing must not undergo excessive settlement or shear failure, and the footing itself must be strong enough to resist the bending moment and shear forces as a result of the superstructure load. The thickness of combined footings is usually governed by shear force considerations.

In the design of combined footings, the centre of gravity of the two column loads should as practically as possible coincide with the centre of area of the base. The base should preferably be rectangular in plan and symmetrically disposed about the line of loads. If it is not practicable to proportion the base as described above, the load will be eccentric, and the centre of pressure of the higher ground pressure will have the same eccentricity relative to the centre of the base.

If the base is thick enough, the pressure distribution diagram will be trapezoidal for eccentric loads, but uniform for concentric loads. If the base is relatively thin, the pressure distribution will be variable, with the maximum pressure occurring directly under each column.

The most common method of design of combined footings is the rigid approach, which makes the following assumptions.

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

However, flexible method can also be used for the design of combined footings, taking into account soil-structure interaction.

**Solved example on the design of combined footing**

Two (300 x 300)mm square columns spaced at a distance of 2.45 m c/c are loaded as shown below. The foundation is founded on soil of a bearing capacity of 125 kN/m^{2}. It is desired to design the footing to satisfy all requirements using the concrete grade of 30 N/mm^{2} and steel of yield strength 500 N/mm^{2}. The concrete cover is 50mm.

At serviceability limit state;

P_{Ed} = 1.0G_{k} + 1.0Q_{k}

Total service load on column P_{1} = 665 + 122 = 787 kN

Total service load on column P_{2} = 825 + 145 = 970 kN

Total load on both columns = 787 + 970 = 1757 kN

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

S_{w }= 0.1 × 1757 = 175.7 kN

Area of footing required = Total service load/Allowable bearing capacity = (1757 + 175.7)/125 = 15.46 m^{2}

Adopt a rectangular base = 6.5m x 2.5m (Area provided = 16.25 m^{2})

To locate the centroid of the footing, let us take moment about column P_{1};

(970 × 2.45) – 1757*x* = 0

On solving, *x* = 1.352m from column P_{1}

Let the projection of the footing from the lighter column be L_{1}

Therefore;

L_{1} + 1.352m = (Total length of footing)/2 = 6.5/2 = 3.25m

Therefore, L_{1} = 3.25 – 1.352 = 1.898 m (say 1.9m)

Hence L_{3} = 6.5 – (1.9 + 2.45) = 2.15m

Alternatively, the footing can be proportioned using a straight-forward formula based on the calculations done above;

L_{3} = 0.5L – P_{1}L_{2}/(P_{1} + P_{2}) = (6.5/2) – (787 × 2.45)/(787 + 970) = 2.15m

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

We can check the preliminary thickness of 600 mm 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;

P_{1} = 1.35G_{k} + 1.5Q_{k }= 1.35(665) + 1.5(122) = 1080.75 kN

P_{2} = 1.35G_{k} + 1.5Q_{k }= 1.35(825) + 1.5(145) = 1331.25 kN

Total ultimate load = 1080.75 + 1331.25 = 2412 kN

V_{Ed }< V_{Rd,max}

Where;

V_{Ed} = βV_{Ed} /u_{0}d

V_{Rd,max} = 0.5*v*f_{cd}

Considering Column P_{2} bearing the maximum axial load, V_{Ed} = 1331.25 kN

β = 1.5 (an approximate value from *clause 6.4.3(6)* of EN 1992-1-1:2004)

u_{0 }= column perimeter = 2(300) + 2(300) = 1200 mm*v *= 0.6[1 – f_{ck}/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;

The minimum thickness of footing d_{min} = 315.263 mm

Hence the depth provided exceeds the minimum required.

Earth pressure intensity at ultimate limit state = (1080.75 + 1331.25)/16.25 = 148.43 kN/m^{2}

For a width of 2.5m, q = 148.43 × 2.5 = 371.075 kN/m

Bending moment at column P_{1} = (371 × 1.9^{2})/2 = 669.65 kNm

Bending moment at column P_{2} = (371 × 2.15^{2})/2 = 857.47 kNm

The maximum bending moment in the span will occur at the point of zero shear;

M_{x} = 371*x*^{2}/2 – 1080.75(x – 1.9) = 185.5*x*^{2} – 1080.75*x* + 2055.425

∂M_{x}/∂*x* = 371*x* – 1080.75 = 0

On solving; *x* = 1080.75/371 = 2.91 m

M_{max }= 185.5(2.91)^{2} – (1080.75 × 2.91) + 2055.425 = 479.275 kNm

The shear force just to the left of column P_{1} = (371 × 1.9) = 704.9 kN

The shear force just to the right of column P_{1} = 704.9 – 1080.75 = -375.85 kN

The shear force just to the left of column P_{2} = -375.85 + (371 × 2.45) = 533.1 kN

The shear force just to the right of column P_{2} = 533.1 – 1331.25 = -798.15 kN

As an be seen, there is no hogging moment in the combined footing.

**Bottom Reinforcement**

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

M_{Ed} = 857.47 kNm

Effective depth (d) = 600 – 50 – 10 = 540 mm (assuming 20mm bars).

b = 2000mm

k = M_{Ed}/(f_{ck}bd^{2} )

k = (857.47 × 10^{6})/(30 × 2500 × 540^{2} ) = 0.039

Since k < 0.167 No compression reinforcement required

z = d[0.5 + √(0.25 – 0.882k)] = z = d[0.5 + √(0.25 – (0.882 × 0.039)] = 0.95d

A_{s1} = M_{Ed}/(0.87f_{yk} z)

A_{s1 }= (857.47 × 10^{6})/(0.87 × 500 × 0.95 × 540) = 3842 mm^{2}

Provide 13H20mm @ 200mm c/c BOT (A_{Sprov} = 4082 mm^{2})

Top reinforcement

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

A_{smin} = 0.0013bd = (0.0013 × 1000 × 540) = 702 mm^{2}

Provide H12mm @ 150mm c/c BOT (A_{Sprov} = 753 mm^{2}/m)

**Transverse Reinforcement**

Cantilever arm = (2.5 – 0.3)/2 = 1.1 m

Designing per unit strip (per 1000 mm)

M_{Ed} = [148.43 × 1.1^{2}]/2 = 89.8 kNm

Effective depth (d) = 600 – 50 – 10 = 540 mm (assuming H20mm bars).

k = M_{Ed}/(f_{ck}bd^{2} )

k = (89.8 × 10^{6})/(30 × 1000 × 540^{2} ) = 0.010

Since k < 0.167 No compression reinforcement required

z = d[0.5+ √(0.25 – 0.882k)] = z = d[0.5+ √(0.25 – (0.882 × 0.010)] = 0.95d

A_{s1} = M_{Ed}/(0.87f_{yk} z)

A_{s1 }= (89.8 × 10^{6})/(0.87 × 500 × 0.95 × 540) = 402 mm^{2}

Provide H12mm @ 200mm c/c BOT (A_{Sprov} = 565 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 the 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} = 798.15 kN

V_{Ed,red} = V_{Ed} – ∆V_{Ed}

Where ∆VEd 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 = 148.43 kN/m^{2}

Unit weight of concrete = 25 kN/m^{3}

Hence, ∆V_{Ed} = (148.43 × 1.9044) – (25 × 0.6 × 1.9044) = 254.1 kN

V_{Ed,red} = 1331.25 – 254.1 = 1077.15 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} = (1077.15 × 1000) /(5520 × 540) = 0.3061 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) = A_{s}/bd = 565/(1000 × 540) = 0.001046

ρ_{z} (in the longitudinal direction) = A_{s}/bd = 4082/(2500 × 540) = 0.003023

ρ_{1} = √(ρ_{x }× ρ_{z}) = 0.001778 < 0.02

V_{Rd,c} = [0.12 × 1.608 (100 × 0.001778 × 30 )^{(1/3)}] = 0.337 N/mm^{2}

0.3061 N/mm^{2} < 0.337 N/mm^{2 }

The section is ok for punching shear.

With this, the reinforcement detailing can be drawn.

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As per EN-7,Design of Foundations shall be performed in one of the 3 Design approaches, Direct Design case.Which Design approach to follow depends on the country of the project.

each approaches seriously affect the size and reinforcement detail. So, for a complete analysis and design of foundations as per Eurocode, applying one or all of the design approaches may seem reasonable.

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