# Column footing design – with RCC column design pdf

Every RCC structure cab be divided into two parts:

1. The portion above the ground, called superstructure.
2. The portion which is below the ground level called sub-structure or foundation.

## Why Column footing design is necessary?

1. To safely transmit the loads and moments from the superstructure to the soil, so that the pressure on the soil does not exceed bearing capacity at any point.
2. To ensure safety with respect to permissible settlement, tilting in one direction, overturning, uplift pressure, etc.

## Classification of foundations

Depending upon the type of structure, distribution of loads, type and capacity of sub-soil, presence, and level of the water table, etc, different types of foundations are provided. Following are some common type of foundations:

1. Isolated footing
2. Square footing
3. Rectangular footing
4. Circular footing
5. Combined footing
6. Rectangular
7. Trapezoidal
8. Continuous or wall footing or strip footing
9. Strap footings
10. Raft or mat footing
11. Pile footing
12. Well foundation

## Isolated footing

The footings which are provided under single columns are called isolated footings. They are usually square or rectangular and rarely circular.

Even for columns of hexagonal, octagonal, circular, or any other shape, it is preferable to provide rectangular or square foundations. Isolated footings are provided when loads are small and the soil is not very poor, i.e. its load-bearing capacity is very sufficient.

## Steps of isolated Column footing design

Given:

Load on column, safe bearing capacity of soil, grade of concrete and steel.

### Calculate the area of footing.

(Note: If the depth of foundation is not given, the self-weight of the foundation is taken equal to 10% of service load. If the depth of the foundation is given. The self-weight of footing is taken tentatively equal to the weight of backfill soil.)

### Calculate the soil pressure due to factored column load only, as follows:

Alternatively, if the ratio of width to length of column and footing is assumed to be similar (say a/b), then bending moment is the same in both directions.

X = shorter dimension of footing

Y = Longer dimension of footing

### By one way shear criterion :

the critical section for one way shear is taken at a distance d (effective depth) from the column’s face.

If percentage of reinforcement to be provided is not yet known.

i.e. 0.2%. for M20, this value may be taken as 0.32N/mm2.

### By two-way shear criterion:

The critical section for two-way shear or punching hear as it is commonly called, is at a distance d/2 from the face of the column.

Depth of footing can be calculated by equating the two expressions, (3) and (4).

### Determine the area of reinforcement required by following equation;

The calculated reinforcement area should not be less than the minimum reinforcement and distribute as per IS code provision.

## Column footing design example

Q) Design a square footing of uniform thickness for an axially loaded column of 450 mm x 450 mm size. The load on column is 850 KN. Use M20 concrete and Fe 415 steel.

Solution:

wc = 850 KN

Self weight of footing, wf = 10% of wc = 85 KN

### Step-3Depth of footing by one-way shear criterion

From clause 34.2.4.1(a) of IS code 456:200, critical section for one way shear lies at a distance d from the face of column as shown in figure below;

### Depth of footing by two way shear

Critical section of two way shear lies at a distance d/2 from each face of column as shown in figure below:

Perimeter of critical section = 40(0.45 +d) = 1.80 + 4d

### Step-4 Shear force at critical section

= 251.85 x (2.25 x 2.25 – (0.45 + d)2)

= 1274.99 – 251.85 (0.2025 + d2 + 0.9d)   …. (3)

Shear force raised by the critical section

### Step-5Depth of footing by bending moment criterion

Mulim = Ru x bd2

= 2.76 x 2250 x d2 = 6210d2           …….(6)

Equating (5) and (6), we get

229.498 x 106 = 6210 d2

d = 192.24 mm

= 0.192 m      …………..(C)

From equations (A) , (B) and (C), the highest value of d obtained is 0.396 m.

Let us adopt       d = 400 mm

Overall depth = 400 + 4 + 50

= 458 say 460 mm (taking clear cover = 50 mm and 16 mm dia. Bars)

### Step-6Area of reinforcement – Footing reinforcement details

From ANNEX ‘G’ of code IS 456:2000;

### Step-7Check for development length

Happy Learning – Civil Concept

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