Design of Shallow Foundation (Isolated Footings)

Column Size = 400 x400

The compression axial load for the load combination of (1.5 DL + 1.5 LL) is 2000 KN

The safe soil bearing capacity is 150 KN/m2 at Foundation depth below E.G.L = 2metres

Grade of concrete for footing = M25

• Footing Size

column load 2000KN

10% of footing self weight = 2000x10/100 = 200KN

Axial load on footing = 2000+ 200 = 2200KN

 

• Area of footing

Af = Load on footing / safe bearing capacity = 2200/150 = 14.66m2

 

B = L = Ѵ14.66 = 3.83m Say B = L = 3.90m  = 3900mm

 

• Net Upward pressure

Factored net soil pressure = load on footing /area of footing

= 2200 / ( 3.90 x 3.90)

 = 144.64 x 1.5 (factor)

= 216.96KN/m2

 

• Bending Moment

At critical section distance from face of column to edge of footing = l = (3.9 -0.4) /2 = 1.75m

Load on critical section = 4 x 216.96 = 867.84KN/m

 

BM = w l ^2 /2 = 867.84 x 1.75 x 1.75 / 2 =1328.88KNm

 

 

• Depth of footing

                Mu= 0.133 fck b d2    for Fe 500   

                     d =  Ѵ ( Mu /( 0.133 fck b) ) 

                  =  Ѵ 1328.88x10^6 / 0.133 x 25 x 4000 

                      =  315.99mm2 Say 650mm

                        Cover = 50mm      

                        Overall Depth of footing = d + C + (bar dia ( say 16mm)/2)= 650 + 50 + 16/2 = 708mm   Say 750mm

                          D = 750mm

• Reinforcement Requirement

Mu/bd2 = 1328.88 x10^6 / 4000 x 700^2 = 0.678  Say 0.7  for 0.7 Fy =500  Pt = 0.167%  from  SP-16 table 3

Ast = 0.167 x 4000 x 700 /100 = 4676 mm2

Numbers of bar = 4676/ 201 = 23.26   Say 24Nos      

 24X201 = 4824mm2 > 4676mm2        Spacing = 201 x 4000/ 4824 = 166.66mm c / c   

Say 16mm bar @ 150mm C/C

 

• Check for one way shear

The critical section for shear at the distance d from the face of the column

 

Shear force = Upward soil pressure x (B - b) / 2 -d = 216.96 x (4-0.4)/2 -0.7 = 238.66KN

 

100 Ast / b d = 100 X 4824 / 4000 x 700 = 0.172

 

Tv = V/bd = 238.66/4000x700 = 0.085n/mm2

  

Tc = 0.30 N/mm2  as per table 19

 

Tv < Tc>

 

• Check for Two way shear

The critical section for two way shear is located at a distance of d/2 from the face of the column

 

Shear force on critical section

V = 238.66 ( 4x4 – 1.1 x 1.1) =3155.8 KN

 

Periphery of the section b0 = 2 ( 1.1 + 1.1) = 4.4m = 4400mm

 

Tv  = Vu / b0 d    =  3155.8 / 4400 x 700 = 1754.44 KN = 1.025N/mm2

 

 The permissible shear stress = ks Tc                         

                                   

ks =(0.5+Bc) > 1                                  

                                   

Bc= b/b =         400/400 =  1.00                      

                                   

ks =      0.5+ 1  = 1.50             

                                   

Permissible shesr stress (Tcxk) (N/mm2) =    1.50

                                   

0.25 x SQRT (fck) =      1.25                

                                   

Tv        <          Tc       

1.25                 1.50    

 

Hence shear stresses are within safe permissible limits

                           

                      

• Reinforcement detail for footing with drawing as follow below

Please find the attached answer sheet