Design of Shallow Foundation (Isolated Footings)

Given:

Column size = 400x400mm

SBC = 150 kN/mm2</math>">mm2��2

Compression axial load = 2000 kN

fck</math>">fck��� = M25

Steps to design a square footing:

1) Size of Footing

2) Net upward Pressure

3) Bending moment

4) Depth of Footing

5) Reinforcement

6) One way and Two-way shear

Step 1:

Size of Footing:

 Load on column = 2000 kN

 Add 10% of column axial load for footing = 10% (2000)

                                                             = 2000 + 200

                                                             = 2200 kN

Area of Footing:

 Area of Footing = Load on footing/Safe bearing capacity

                        = 2200/150

                        = 14.66 m^2

            √14.66  = 3.82 

                         = 4m x 4m

 Therefore, provided area = 16m^2

Step 2:

Net upward pressure:

           σ = Force/Area

              = 1.5 x Total load on footing/Area

              = 1.5 x 2200/4 x 4

          σ  = 206.25 kN/m^2

Step 3:

Bending Moment:

 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

Step 4:

Depth of footing:

                 Mu</math>">Mu�� = 0.138 fck</math>">fck���bd2</math>">d2�2

                            = 0.138 x 25 x 4000 x d2</math>">d2�2

           1336.5 x 106</math>">106106 = 0.138 x 25 x 4000 x d2</math>">d2�2

                            d = 311.20 mm

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

     ∴</math>">∴∴ 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 5: 

Reinforcemet:

 Bending moment = 1336.5 x 106</math>">106106

Percentage of reinforcement, Pt</math>">Pt��= (50fck</math>">fck���/fy</math>">fy��)(1-</math>">√4.6 x 13336.5 106</math>">106106/25x4000x7782</math>">77827782)

                  Therefore, Pt</math>">Pt�� = 0.1570 %

Area of steel =  (Pt</math>">Pt��/100) x bd

                   = (0.157/100) x 4000 x 778

           Ast</math>">Ast���= 4885 mm2</math>">mm2��2

No of bars:

    Take 20mm bar

    No of bar = 4885/314.5

                   = 15.55 = 16 nos

Provide 16 nos of 20mm bar

  Ast</math>">Ast���provide = 16 x 314.5

                           = 5032 `mm^2

  ∴</math>">∴∴  Ast</math>">Ast��� > Ast</math>">Ast���provided

                      4885 > 5032

     Hence ok.

Step 6:

Check for clear spacing between bars:

As per IS 456 - table 15, for fe</math>">fe��415, 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

   ∴</math>">∴∴ clear spacing = 242.5- half of dia - half of dia bar 

                                         = 242.5-10-10

                 clear spacing   = 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

               ∴</math>">∴∴No. of bar = 24.30

                                               = 24

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

                     = 168.89

       ∴</math>">∴∴Clear spacing = 168.86-8-8

                                            = 152.86mm

     So, Clear spacing 152.86 < 180>

        Hence Ok

Step 7:

Check for Oneway Shear:

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

        Vu</math>">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</math>">Vu�� = 206.25 x 4 x 1.022

                   Vu</math>">Vu�� = 843.15 kN

Nominal Stress:

τv</math>">τv�� = Vu</math>">Vu��/bd

             = (843.15 x 103</math>">103103)/(4000 x 778)

    τv</math>">τv��= 0.27 N/m          

Now compare  τv</math>">τv�� and  τc</math>">τc��
      

For the τc</math>">τc�� value given by IS 456, Table 19

    Pt</math>">Pt��provided = (100 Ast</math>">Ast���/bd)

                          = 100 x Ast</math>">Ast���/bd

       Ast</math>">Ast��� = 24 nos of 16mm dia

                     = 4823 mm2</math>">mm2��2

  ∴</math>">∴∴ Pt</math>">Pt��= (100x 4823)/(4000x778)

                             = 0.1549

   From IS 456, table 19

100</math>">100100A_(st)/bdvalueofgreaterthanorequal→0.15is0.28N</math>">/bdvalueofgreaterthanorequal→0.15is0.28N/�����������������ℎ��or�����→0.15��0.28�mm^2`

 Therefore,  τv</math>">τv��<  τc</math>">τc��

               = 0.15<0>

 Hence Ok

Step 8:

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</math>">42−1.178242-1.1782

                         = 14.6123 mm2</math>">mm2��2

    Vu</math>">Vu��= Upward pressure x Area of highlighted section

              = 206.25 x 14.6123

         Vu</math>">Vu�� = 3013.79 kN

         τv</math>">τv��=  Vu</math>">Vu��/bd

Peripheral length = 4 x 1178

                    b= 4712 mm

      τv</math>">τv��= (3013.79 x 103</math>">103103)/(4712 x 778)

          τv</math>">τv�� = 0.8221 N/mm2</math>">mm2��2

Compare τv</math>">τv�� and τc</math>">τc��

For two way shear τc</math>">τc�� = Ks</math>">Ks��x τc</math>">τc��

      For τc</math>">τc�� value taken from IS 456-2000 page number 58 and 58

 Ks</math>">Ks��= 1 (for square column)

  τc</math>">τc��= 0.25fock</math>">0.25o√fck0.25����)

              = 1.25 N/mm2</math>">mm2��2

 Therefore, τc</math>">τc��= Ks</math>">Ks��x τc</math>">τc��

                             = 1 x 1.25

         τc</math>">τc��= 1.25 N/mm2</math>">mm2��2

∴</math>">∴∴  τv</math>">τv�� <  τc</math>">τc��

                      0.8221 < 1>

     Hence Ok

Reinforcement detail drawing: