Design of Shallow Foundation (Isolated Footings)

Procedure:-

Design of square footing :

Given:

Column size = 400*400 mm

Soil bearing capacity SBC = 150 KN/mm²

Compression axial load = 2000KN

Compressive strength fck = M₂₅

Steps involved in design of square footing :

• Determine the size of footing
• Upward pressure P₀
• Depth of footing on the basis of BM
• Area of R/F
• Check for one way shear
• Check for two way shear

Step 1 : Determine the size of footing

Assume self weight of footing equal to 10% of superimposed load

Wd = 10% * 2000 = 0.1 * 2000 = 200 KN

Now,

 Total load W = WL + WD = 2000 + 200 = 2200 KN

Now,

Area of footing (A) = W/SBC = 2200/150 = 14.66 m²

Now, let us take B = √A = √14.66 = 3.82 m

Hence, we provide the size of square footing is 4M * 4M

The area of footing is 16 m²

Step 2 : Net upward pressure

P₀ = 1.5 * W / size of footing = 1.5 * 2200 / 4*4 = 206.25 KN/m²

Step 3: Depth of the footing on the basis of BM

At the critical section of the footing face of the column from the edge of footing

        (4000/2) - (400/2) = 1800mm

   Therefore, load on critical section = σ x breadth of footing

                                                    = 206.25 x 4

                                      Load       = 825 kN/m

   So, the bending moment     = wl^2/2

                                            = (825 x 1.8^2)/2

                                             = 1336.5 kN-m

Mu = 0.138 * fck * bd2

                            = 0.138 x 25 x 4000 x d2

           1336.5*106 = 0.138 x 25 x 4000 x d2

                            d = 311.20 mm

 By the thumb rule, we can take 2 to 2.5 times higher value of d for shear check

     ∴ d= 778 mm

            D = d + effective cover = bar dia + half of bar dia 

                = 778 + 50+ 20 +10 ( Assume bar dia as 20mm)

           D = 860 mm

Step 4: Area of R/F

Bending moment = 1336.5 x 106

Percentage of reinforcement, Pt= (50fck/fy)(1-√4.6 x 13336.5*106/25x4000x7782)

                  Therefore, Pt = 0.1570 %

Area of steel =  (Pt/100) x bd

                   = (0.157/100) x 4000 x 778

           Ast= 4885 mm2

No of bars:

    Take 20mm bar

    No of bar = 4885/314.5

                   = 15.55 = 16 nos

Provide 16 nos of 20mm bar

  Astprovide = 16 x 314.5

                           = 5032 `mm^2

  ∴Ast > Astprovided

                      4885 > 5032

     Hence ok.

Check for clear spacing between bars:

As per IS 456 - table 15, for fefe415, the clear spacing between the bars should be less than 180mm

     c/c spacing = (4000-50-50-10-10)/(no of bars-1)= 258.66 mm

  clear spacing = 242.5- half of dia - half of dia  = 242.5-10-10= 238 mm

As per IS code value, our value is greater

so obtained clear spacing is greater than IS code value,

In order, the minimize the value reduce the bar size.

Take 16mm bar = 4885/200.96

          No. of bar = 24.30 = 24

  C/C spacing = (4000-50-50-8-8)/(No.of bars-1) = 168.89 mm

    Clear spacing = 168.86-8-8 = 152.86mm

     So, Clear spacing 152.86 < 180

        Hence Ok

Step 5: Check for one way shear

One way shear check at the critical section for one way shear at the distance, d from face of the column

        Vu= Upward pressure x Area of the highlighted section

                  = 206.25 x 4 x ?

Area of highlighted section = (4000/2)-(400/2)-778

                                        = 1.022

       Now,  

Vu = 206.25 x 4 x 1.022 = 843.15 kN

Nominal Stress:

τv = Vu/bd = (843.15 x 10³)/(4000 x 778) = 0.27 N/m          

Now compare  τv and  τc
For the τc value given by IS 456, Table 19

   Ptprovided = (100 Ast/bd = 100 x Ast/bd

       Ast = 24 nos of 16mm dia = 4823 mm²

 Pt= (100x 4823)/(4000x778) = 0.1549

   From IS 456, table 19

100A_(st)/bdvalueofgreaterthanorequal→0.15is0.28N mm^2`

 Therefore,  τv<  τc

               = 0.15<0.28

 Hence Ok

Step 6: Check for two way shear

At peripheral d/2 distance from face of the column

Area of highlighted section = 400+(778/2)+(778/2)

                                        = 1178 mm^2

  Therefore, area = 42-1.1782

                         = 14.6123 mm2

    Vu= Upward pressure x Area of highlighted section

              = 206.25 x 14.6123

         Vu = 3013.79 kN

         τv=  Vu/bd

Peripheral length = 4 x 1178

                    b= 4712 mm

      τv= (3013.79 x 103)/(4712 x 778)

          τv = 0.8221 N/mm2mm2

Compare τv and τc

For two way shear τc = Ksx τc

      For τc value taken from IS 456-2000 page number 58 and 58

 Ks= 1 (for square column)

  τc= 0.25o√fck0.25fck)= 1.25 N/mm2

 Therefore, τc= Ks x τc = 1 x 1.25= 1.25 N/mm2mm2

τv <  τc

                      0.8221 < 1.25

     Hence Ok

Reinforcement detail drawing:

RESULT:-

• Using the given details the square footing has been designed successfully.