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Plain concrete footing or pedestal is a shallow foundation designed to take small loads from columns without using longitudinal tension steel. These are the most economical forms of shallow foundation. To sustain temperature and shrinkage stresses, the standard codes recommend to provide distribution reinforcement bars for plain concrete footings.

The design stipulations for plain concrete footings are provided in cls.34.1.2 and 34.1.3 of IS: 456-2000.

This article explains the theory behind the design of plain concrete footing with the help of an example.

Contents:

## Edge Thickness of Plain Concrete Footing For R.C.C Columns

As shown in the figure above, a plane passing through the bottom edge of the footing and the corresponding junction of the column makes an angle 'α' with the horizontal. The calculated maximum bearing pressure acting at the base of the pedestal is given by q_{a} in N/mm^{2}. This angle 'α' is given by the following equation as per IS:456-2000, cl.34.1.3.

f_{ck }is the characteristic strength of concrete at 28 days in N/mm^{2}

As per IS:456-2000, cl.34.1.2 and 34.1.3, the minimum thickness at the edge of the plain concrete footings shall be at least 150 mm for footings on soils. When footings are on piles, the minimum thickness at the edge of plain concrete footings shall be at least 300 mm.

## Design of Plain Concrete Footing for 400 x 400 mm Column

Consider an example to design a plain concrete footing for a column of 400 x 400 mm cross-section. It carries an axial load of 400 kN under service loads. Take safe bearing capacity of soil as 300 kN/m^{2} at a depth of 1m below the ground level. Use M20 and Fe 415 for the design purpose.

### Step 1: Transfer of Load at the Base of Column

Before starting the design procedure, it is necessary to check whether the total factored loads on the columns are safely transferred to the base of the column without the use of reinforcement.

Given Axial load P = 400 kN

Factored Axial Load Pu = 400 x 1.5 = 600 kN

As per cl.34.4, all the forces and moment acting at the base of the column must be safely transferred to the footing base by concrete compression and steel tension. The compression forces are transferred using direct bearing, and reinforcement help to transfer tension forces.

As per cl.34.4, the permissible bearing stresses on full area of concrete as per limit state method is given by:

The above bearing stress increases as the area of footing is large. To consider the effect of dispersion of concentrated load of column to footing as shown in Figure-2, the permissible bearing stress is given by:

In this problem, the bearing stress at the column-footing interface is governed by the column face. Hence A1 = A2 =400 x 400 =160000 mm^{2}. So, A1/A2 =1;

Hence, Equation-3 becomes

Permissible bearing stress = 0.45 x 20 x 1= 9 N/mm^{2}

Permissible bearing force = P_{br} = Permissible bearing stress x A1

P_{br} = (9 x 160000 )/1000 = 1440kN > P_{u} = 600 kN

As the bearing force, P_{br} at the column-footing interface is greater than the axial load P_{u}, it is possible to transfer the load P_{u} without any reinforcement.

### Step 2: Size of the Footing

Assume the weight of the footing and the back fill soil = 15% of Pu

Total vertical load = Pu + 15% of Pu = (1.15 x 400) kN

Base area required = Total vertical load / Bearing capacity of soil

= (1.15 x 400)/300 = 1.533 m^{2}

Provide a cross-section of 1250 x 1250 mm which gives an area of 1.5625 m^{2}

Bearing pressure q_{a} = (400 x 1.15) / (1.25 x 1.25) = 294.4 kN/m^{2 }

### Step 3: Thickness of Footing

From Equation-1,

From Figure-3,

**Hence, provide 1250 x 1250 x 670 mm Block of Plain Concrete**

### Step 4: Minimum Reinforcement

The plain concrete block of dimension 1250 x 1250 x 670 mm shall be provided with a minimum reinforcement of 0.12% of area of cross-section to account for temperature and shrinkage cracks.

Ast (min) = 0.12% x 1250 x 670 = 1005 mm^{2}

Assume 12 mm diameter bars;

Number of bars = Ast (min) / Area of one bar = 1005 / (113.04) = 8.89 = 9 nos

Spacing of bars = (1250 - 50 - 12 )/8 = 148.5 mm c/c [Where, cover = 50 mm]

**Hence, provide 9 nos. of 12 mm diameter bars @ 140 mm c/c**

### Step 5: Check for Gross Base Pressure

Service load = 400 kN

Weight of footing = Volume of footing x unit weight of plain concrete

= (0.67 x 1.25 x 1.25 ) x 24

= 25.125 kN

Weight of Soil above the footing = Volume of soil x unit weight of soil

Volume of soil = 0.330 x 1.25 x 1.25

Hence, weight of soil = 0.330 x 1.25 x 1.25 x 20 = 10.3125 kN

Total weight = 400 + 25.125 + 10.3125 = 435.4375 kN;

Gross base pressure qa = total weight / cross-sectional area

= 435.4375/ (1.25 x 1.25) = 278.68 kN/m^{2} < 300 kN/m^{2}

**Hence, the design is safe**.

## FAQs

**What is a plain concrete footing or pedestal?**

Plain concrete footing or pedestal is a shallow foundation designed to take small loads from column without using longitudinal or main reinforcement bars. These are the most economical forms of shallow foundation. To sustain temperature and shrinkage stresses, the standard codes recommend to provide distribution reinforcement bars for plain concrete footings.

**How much reinforcement is provided for plain concrete footing to sustain temperature and shrinkage stresses?**

To account for shrinkage and temperature stresses in plain concrete footing, a minimum distribution reinforcement of 0.12% of footing cross-section shall be provided as per IS:456-2000.

**Read More**

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Two Way Slab Design by Direct Design Method as per ACI 318-11

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