Design of Shallow Foundation (Isolated Footings)

I've attached the answer sheet for you.

Design a square footing for a column size of 400x400. The axial compression 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 2 meters below E.G.L. Participants can go for either a tapered or stepped footing. Besides the total axial load, it accounts for the self-weight of the ground and soil above it. Assume M25 concrete grade. Do the following checks:

If the gross bearing pressure under the footing is within the safe bearing capacity

Provide sufficient depth and longitudinal steel for resisting bending and shear.

Sufficient anchorage for column steel into the footing

If the bearing pressure at the column-footing joint is within 0.45 times the maximum concrete compressive strength

Could you provide a simple sketch showing the base size, maximum and minimum depth of the section, flexural reinforcement in both directions and anchorage of column steel?

Answer)

Column Size = 400 x400

The axial compression 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 the footing

= 2200 / ( 3.90 x 3.90)

 = 144.64 x 1.5 (factor)

= 216.96KN/m2

• Bending Moment

At critical section distance from the face of the column to the 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; hence shear stress is less than the permissible stress limit.

• 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 shear 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.