C5: DESIGN OF FOOTING AS PER IS CODE

COLUMN FOOTING DESIGN

DATA:

Size of column                                  =    230*230mm

Spacing of column                            =    3m

Total factored load on each column =   210 KN/m²

Ultimate bearing capacity of soil      =  200 KN/m²

                      Fck = 20 N/mm² & fy  = 415 N/mm²

LOADS ON FOOTING:

Total load on column             = 210 KN

Self weight of footing (10%) = 210*(10/100) =  21 KN

Total ultimate load ,          Pu = 210 + 21        =  231 KN

SIZE OF FOOTING:

Area of footing = (231/200) = 1.155m²

Adopt a footing of size 1.1m*1.1m

Adopt width of strap beam, b = 200mm

DESIGN OF FOOTING:

Soil pressure, Pu = 210/(1.1*1.1) = 173.55 KN/m² < 225 KN/m²

Cantilever projection of footing   = 0.5(1.1*0.4) = 0.22m

Ultimate design moment        Mu = 0.5 Pu L²

                                                      = 0.5*173.55*0.22²

                                                                                = 4.199 KNm

Effective depth of footing       d = √(Mu / 0.138 fck b)

                                                    = √ [ (4.199*10⁶) / (0.138*20*10³)]

                                                    = 39mm

Hence adopt effective depth    d = 250mm

Overall depth                           D = 300mm

    

Mu = 0.87*fy*Ast*d*[(1)-((Ast*fy) / (fck*b*d))]

             4.199*10⁶ = 0.87*415*Ast*250[(1)-((Ast*415) / (20*1000*250))]

             4.199*10⁶ = 90,262.5*Ast*[(1)-((8.3*10^-5)*(Ast))]

              7.4917875Ast²-90,262.5Ast+(4.199*10⁶) = 0

                           Ast = 467mm².

Minimum reinforcement = (0.012 * 1000 * 300) = 360mm²

Adopt 10mm dia bars at 200mm c/c (Ast = 393mm²) as main reinforcement.

DESIGN OF STRAP BEAM:

Factored load on beam Wu = 1.5*173.55 = 260.325 KN/m

Neglecting the small cantilever portion of the beam:

                  Mu = 0.125*Wu*L²

                           = 0.125*260.32*3²

                         = 292.86 KNm

                   Vu = 0.5*Wu*L      

                         = 0.5*260.32*3      

                         = 390.48 KN

Depth of strap beam computed based on moment.

                Assuming,Ʈ_c = 1.2 N/mm²

                                   d = (Vu) / (b*_Ʈc)

                                       = (292.86*10³) / (200*1.2)

                                       = 1.22 N/mm²

Adopt effective depth d = 1150mm

Overall depth              D = 1200mm

    

  Mu = 0.87*fy*Ast*d*[(1)-((Ast*fy) / (fck*b*d))]

             292.86*10⁶ = 0.87*415*Ast*1150[(1)-((Ast*415) / (200*1150*20))]

             292.86*10⁶ = 415,207.5*Ast*[(1)-((9.6217*10^-5)*(Ast))]

              37.4589375Ast²-415,207.5Ast+(292.86*10⁶) = 0

                           Ast = 757mm².

Provide 4 bars of 16mm dia ( Ast = 804mm²)

Shear stress _Ʈv = (Vu / bd)

                        = (390.38*10³) / (200*1150)

                        = 1.69 N/mm²

    100 Ast / bd = (100*804) / (200*1150) = 0.35

Refer table 19 of IS:456-2000 and read out the permissible shear stress as:

                           Ʈ _c = 0.4 N/mm² < Ʈ_v

Hence shear reinforcement are required to resist the balanced shear force computed as:

                        Vus = [ 390.48 – (0.4*200*1150)*10^-3]

                                = 298 KN

Using 8mm diameter 4 legged stirrups.

The spacing is:

                        Sv = (0.87*415*4*50*1150) / (298*10³)

                             = 278mm

Adopt 8mm diameter 4 legged stirrups at 250mm c/c in the strap beam.