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

Footing the structure which is exists below the ground level; the strength of the building is depending upon the reinforcement details of foundation. The reinforcement details are depends 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 is acts in the column section. In this article I will explain you the clear concept regarding the isolated footing design as per IS 456-2000 code by manual calculations. In the column design as per previous article the load is taken as 1200kN and dimension of the columns are obtained as 400mmX400mm cross section.

Follow our previous articles for the complete column, beam, one way slab and two way slab design by using manual method

RCC column design as per IS 456-2000 code.

RCC beam design as per IS 456-2000 code.

one way slab design as per IS 456-2000 code.

RCC two way design by using IS 456 code.

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 = q_u = Total load/Area of footing
3. Factored shear force Vu₁ = (q_uB/2) (B-C₁-2d)
4. One way shear resistance = Vc₁ = τcbd
5. Factored shear force / Two way shear = Vu₂ = q_u (B²-(C₁+d)²)
6. Punching / Two way shear resistance = Vc₂ =Ksτcb₀d
7. Ultimate moment Mu = (q_uB/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 400mmX400mm column size which carries load of 1200kN on the column, take Soil bearing capacity of 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 back fill 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 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

Calculated ultimate load from step 1 is 1920kN

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

So take soil pressure q_u = 200 kN/m²

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

From formulae of factored shear force for one way shear

Factored shear force Vu₁ = (q_uB/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 assume percentage of steel in 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 cover as 68mm

So over all depth is given by 482+68 = 550mm

Step 5: Check for punching shear / Two way shear

Since we have formulae

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

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 fy^1/2  = 0.25X20^1/2  = 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 formulae section

Mu = (q_uB/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 Mu expression we can easily calculate Ast value

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

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

By solving 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 15numbers 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 as use 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.

The complete concepts of isolated footing design as per IS 456-2000code is explained in my YouTube channel Civil engineering by shravan. See below for complete 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 1200kN load with the column size of 400mmX400mm dimensions. As per the final detailing the reinforcement details are obtained as use 16mm diameter bars at 210 C/C distance in both X direction and Y direction.

Please follow our previous posts here

What is load combination ? different load combinations used as per IS 456-2000 code

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How to apply complete G+1 building gravity loads in staad pro software

What is slab? Types of slabs? Difference between one way slab and two way slab ? click here to read

What is minimum and maximum center to center distance between RCC columns click here to read

Please watch interesting concepts in my YouTube channel Civil engineering by shravan. Please feel free to text us at contact us page for any quarries.

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