Design of Shallow Foundation (Isolated Footings)

Q.Design a square footing for a column size of 400x400. 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 a depth of 2 metres below E.G.L. Participants are free to go for either a tapered or stepped footing. Besides the total axial load, also account for the self-weight of the footing and soil above it. Assume M25 concrete grade. Do the following checks:

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

Provide a simple sketch showing the size of the footing, maximum and minimum depth of the section, flexural reinforcement in both directions and anchorage of column steel.

AIM: Design of shallow foundation (isolated footing )

PROCEDURE :

steps involved in footing design :

1. calculations of gross soil pressure for footing area.

2. calculations of net soil pressure for flexure and shear design

3. initial depth of footing calculations

4. flexural reinforcement calculations

5. one way shear check

6. two way shear check

1. calculations of gross soil pressure for footing area

unfactored load = 2000/1.5 = 1333kn

assume thickness of footing (t)= 0.5m

depth of footing given in data(h) = 2m

so, soil surcharge depth = 2-0.5=1.5

assume soil density = 20kn/cu.m

assume RCC density = 25kn/cu.m

conside square shape footing, area of footing = b x b= b^2

unfactored load with surcharge X self weight of footing/allowable soil pressure

= (2000/1.5 x 1.1)/150

= 9.78m^2

assume we have square footing

9.78/2 = 3.12

consider 3.2 meter sided isolated footing

gross soil pressure= (1333/3.2^2) + (1.5 x 20) + (0.5 + 25)

= 0.145 N/sq.mm

2. net soil pressure and initial footing depth

net soil pressure = factored load / actual area of footing

= 2000/(3.2^2)

= 195.31 kn/m.sq, 0.195n/mm.sq

assume depth of footing 500mm and checking it for bearing stress as per IS 456-2000, clause 34.4

calculation of bearing pressure :

net load /column area = 2000000/400^2

= 12.5 N/sq.mm

allowable bearing pressure as per IS 456-2000,clause 34.4 from IS456-2000

√= consider this sign as underroot

√A1/A2 X 0.45Fck

A1= Base area of footing

A2= Base area of column

√A1/A2 but not greater than 2 in calculation

so allowable bearing pressure = 2 x 0.45 x 25 N/sq.mm Hence Ok

3. bending moment and flexural reinforcement

Bending moment due to UDL = wL²/ 2

Where,

w= net pressure into UDL.

w = 195.3 x 3.2 = 652 kN/m

= wL² / 2

= 625 x 1.6² / 2

M_u = 800 kNm

Area of steel for 1m

A_st = 0.5x 25/ 500 x (1- √1- (4.6x 800 x 10⁶)/ 25 x 1000 x 500²) x 1000 x500

A^st = 1450 mm²

Total A_st required = 4350mm²

No of bars

= M_u / Π/4 x d^2.

= 4350 / Π/4 x 16²

= 22 nos

Spacing,

S = L / Nos of bars

S = 3200 / 22

S = 145.45 mm

Therefore,

Provide 22 bars of 16mm dia at 140mm spacing.

4. check for oneway shear stress

Shear force in the one way area = 0.95 x 3.2 x 195.3 kN

= 593.7 kN

Shear stress at one way cut face = t_v = 593.7 x 1000 / (3200 x 450)

= 0.412 N/mm²

Shear capacity check as per IS 456-2000 table 19.

= 0.425 N/mm2

Design shear stress is less than capacity shear stress.

Hence okay.

5. check for two way shear or punching shear

Shear force in two way

= (3.2 x 3.2) – (0.85 x 0.85) x 195.3

= 1859 kN

Design shear stress at two way cut face

T_{v =} 1859 x 1000 / 3450 x 450

= 1.25 N/mm2

Check for shear capacity as per IS 456-2000 cl 31.6.3.1

= 0.25x √fck

= 0.25x √25

= 1.25 N/mm2

Design shear stress is less than shear capacity hence okay.

RESULT-

Provide a 3.2x3.2m footing with a depth of 500 mm.

Provide reinforcement of 20mm dia bars in both directions.

Provide anchorage for column and footing at 750mm.