Isolated footing design as per IS 456-2000 code | structural design | civil engineering

Footing the structure which exists below the ground level; the strength of the building is depending upon the reinforcement details of the foundation. The reinforcement details depend upon the total load which is acts on building structures. The loads related to the gravity loads and gravity loads are considered in the foundation design. The load is taken as a point load which acts in the column section. In this article, I will explain to you the clear concept regarding the isolated footing design as per IS 456-2000 code by manual calculations. In the column design and assume the load is taken as 1200kN and dimensions of the columns are obtained as 400mmX400mm cross-section.

Basic formulas used in the design as per IS456-2000 code

For the design of RCC footing as per the IS456 code the following 8 basic formulas are used which are given as per IS 456 standards.

1. Area of footing = Total load/SBC
2. Soil pressure = qu = Total load/Area of footing
3. Factored shear force Vu1 = (quB/2) (B-C1-2d)
4. One way shear resistance = Vc1 = τcbd
5. Factored shear force / Two way shear = Vu2 = qu (B2-(C1+d)2)
6. Punching / Two-way shear resistance = Vc2 =Ksτcb0d
7. Ultimate moment Mu = (quB/8) (B-C)2
8. Mu = 0.87fyAstd(1-(Astfy/bdfck))

Steps used in Isolated footing design as per IS 456 code

The following six steps are used in the isolated footing design as per IS 456

1. Load calculations
2. Size of footings
3. Calculation of net upward pressure at ultimate load
4. Calculation of one-way shear to determine the depth
5. Check for punching shear / Two-way shear
6. Reinforcement design

Example of Isolated footing design as per IS 456-2000 code standards

Design RCC isolated footing for 400 mmX400 mm column size which carries a load of 1200kN on the column, take Soil bearing capacity of the soil (SBC) is 200kN/m2. Assume M20 grade concrete and Fe 415 grade steel.

Step 1: Load calculations

Given load P = 1200kN

Factor load (or) ultimate load = 1.5P = 1.5X1200 = 1800kN

Consider self-weight of footing and backfill is 10% column load

= (10%column load) = (10/100)1800 = 120kN

Now the total load on the column      = Factor load + 10% column load

= 1800+120 =1920 kN

Step 2: Size of footings calculations

As per step1, the total load is obtained as 1920 kN

Let us consider the square footing which is having length and width as B

So the area of footing is given by BXB = B2

As per the formula of area of footing is given by

B2 =  Total load/SBC = 1920/200 = 9.6 m2

By solving we can get the value of B = 3.1m

So final area of footing provided is = 3.1mX3.1m =9.61 m2

Step 3: Calculation of net upward pressure at ultimate load

The calculated ultimate load from step 1 is 1920kN

Soil pressure = qu = Total load/Area of footing  = 1920/9.61 = 199.80 kN/m2

So take soil pressure qu = 200 kN/m2

Upward pressure

Step 4: Calculation and check for one way shear to determine the depth

One way shear and moment

From formulae of factored shear force for one-way shear

Factored shear force Vu1 = (quB/2) (B-C1-2d)

By taking the values Vu1 = ((0.2X3100)/2)(3100-400-2d)

By solving the above equation we can get

Vu1 = 310(2700-2d) ————————– equation 1

By assuming the percentage of steel in the footing

Pt =0.15%

From table 19 of IS 456-2000 code

Design shear  = 0.36 N/mm2

And one way shear resistance is given by Vc1 = τcbd

By substituting the values Vc1    = 0.36X3100d

Vc1    =1116d ——————————- equation 2

By solving equation 1 and equation 2

310(2700-2d) = 1116d

We can get the value of d as 482.19mm

Let us consider the cover as 68mm

So overall depth is given by 482+68 = 550mm

Depth of isolated footing

Step 5: Check for punching shear / Two-way shear

Since we have formulae

Factored shear force / Two way shear = Vu2 = qu (B2-(C1+d)2)

Two-way shear and moment

By substituting the values in the above expression

Vu2    =       0.2 (31002 – (400+550)2)

=       1741.5 kN

Again we have punching / two-way shear resistance

Vc2     =       Ksb0d

Where b0     =       4(C1+d)

=       4(400+550)

=       3800

From clause 31.6.3 permissible shear stress ( IS 456-2000)

c = 0.25 fy1/2  = 0.25X201/2  = 1.118 N/mm2  and Ks =1

By substituting the values we can get

Vc2                      =       1X1.118X3800X550

=       2336.62 kN

So here Vc2  > Vu2, hence 550mm depth is safe

Step 6: Reinforcement design calculation

We have the ultimate moment expression from the formulae section

Mu = (quB/8) (B-C)2

By substituting the values we can get

Mu = ((0.2X3100)/8) (3100-400)2

Mu = 564.975 kN.m

By using the Mu expression we can easily calculate the Ast value

Mu = 0.87fyAstd(1-(Astfy/bdfck))

564.975 X 106 = 0.87X20XAstX550 (1-(AstX415/3100X400X20))

By solving the above equation we can get the value of Ast

So here Ast = 2951.4 mm2

Let us consider 16mm diameter bars

Area of 1 bar provided = µ/4(16)2 = 201 mm2

Number of bars required  = Ast/Area of 1 bar = 2951.4/201 = 14.68no’s

Take 15 numbers of 16 mm diameter bars

Spacing = (A/Ast)XB

= (201/2951.4)X3100

= 211.12 mm

So the final reinforcement for isolated footing as per IS 456-2000 code is obtained by using 16mm diameter bars at 210 C/C distance in both X direction and Y direction.

The final reinforcement view is shown in the below figure.

Final reinforcement details

Conclusion of complete isolated footing design as per IS 456-2000 code

Well, now the above-explained concepts are related to the complete design of isolated footing as per IS 456-2000 code. The detailed calculation is shown for a 1200kN load with a column size of 400mmX400mm dimensions. As per the final detailing the reinforcement details are obtained by using 16mm diameter bars at 210 C/C distance in both the X direction and Y direction.

Er. SP. ASWINPALANIAPPAN., M.E., (Strut/.,)., (Ph.D.,)

Structural Engineer

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