Design of Shallow Foundation (Isolated Footings)

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:- To design a square footing of column size 400x400mm and to provide sketch showing the size of the footing, maximum and minimum depth of the section, flexural reinforcement in both directions and anchorage of column steel.

INTRODUCTION:

The spread footing behaves like an inverted cantilever with loads applied in the upward direction. A spread footing is a quite rigid element therefore, the applied soil stresses are almost linear and in case of a symmetric (with respect to the pedestal) footing, they are orthogonal. These soil pressures constitutes the loads carried by the footing which behaves like a slab and is deformed. Soil pressures and deformation of an isolated spread footing.

• Footings and other foundation units transfer loads from the structure to the soil or rock supporting the structure.
• The distribution of soil pressure under a footing is a function of the type of soil and the relative rigidity of the soil and the foundation pad.

The most common type of footings used are

• Strip or wall footing
• Spread footing
• Stepped footing
• Tapered footing
• Pile cap
• Combined footing
• Mat or raft footing

PROCEDURE:

GIVEN:-

• Column Size = 400mm X 400mm
• Compression axial load = 2000 KN
• Safe soil bearing capacity = 150 KN/M^2
• Compressive Strength fck=M25
• depth = 2 metres below E.G.L
• Soil depth = 20 KN/M^3
• RCC density = 25 KN/M^3

Steps involved in Design of Footing:-

• Determine the size of footing
• Upward pressure
• Depth of footing on the basis of B.M
• Area of reinforcement
• Check for one way shear
• Check for two-way shear

SIZE OF FOOTING:-

Assume self-weight of footing equal to 1.5 of superimposed load.

Wd= 10% of 2000

= 0.1 X 2000 = 200 KN

Now, Total load,

W = WL + Wd

= 2000 + 200

W = 2200 KN

Now, Area of footing ;

A = W/SBC

= 2200/150 = 14.67 m^3

Now, let us take,

B = √AA = √14.67=3.82

Hence, we provide the size of square footing is = 4m X 4 m

Therefore, Area of footing is 16 m^2

NET UPWARD PRESSURE (ρo)

ρo = WX 1.5/size of footing

ρo = 2200 X1.5/16

ρo = 206.25 KN/m^3

DEPTH OF FOOTING ON THE BASIS OF B.M:-

The minimum B.M at the face of column and is given by,

M= ρ⋅B8(B-b)²

=206.25 X 4(4-0.4)^2/8

=1336.5KN/m

Now, Ultimate moment

Mu=1.5(M)

=1.5 X1336.5

Mu=2004.75

To get the ultimate moment we have calculated that,

Mu=1.5 x FOS X Moment

Now, Depth of footing (D),

Mu=0.138fckBd^2

2004.75=0.138X 25X 4000X d^2

d^2=2004.75X 1060X 138⋅25X 4000

d^2=145271.73

d= √145271.73

d=381.14

By the thumb rule, we take 2.5 times higher value of for shear check.

d=381.14⋅2.5=952.85

D=d+50

D=952.85+50

D=1003mm

AREA OF REINFORCEMENT:-

Ast=0.5fck/f(y)[1-√1−4.6Mu/fckBd²]Bd

=0.5X 25/415[1-√1−(4.6X 2004.75X 10^6/fckBd²)]X 4000X 953

Ast=5985.3mm2

Now, we have to find the % of the reinforcement

P= (Ast/Bd)100

P= (5985.3/4000X 953)100

P=0.157%

No of bars required for the reinforcement, we assume that 20mm dia bars,

Ast=5985.3mm^2

(π/4)X 20^2=314.5

No of bars = 5985.3/314.5 =20

Therefore, Spacing b/w the bars

= B/No.ofBars

= 4000/20

=200mm c/c

CHECK FOR ONE WAY SHEAR:-

For, One-way shear, critical section is located at a distance ‘d’ from the face of the column.

Now, shear force V= ρo⋅B{1/2(B-b)-d}

=206.25⋅4{12(4-0.4)–0.952}

=699.6KN

Ultimate shear force Vu=1.5 X V

=1.5 X 699.6=1049.4KN

Nominal shear stress = Vu/Bd as per IS 456-2000

=1049.4 X 10³/4000X 952 =0.27Nmm^2

Now, design shear strength of concrete τc

τc = τc X K

K=1 because depth > mm

τc=0.28N/mm²

Then,τv< τc...............Hence OK.

Now,τv< τc

Vu/Bd=K⋅τc

d=Vu/B⋅K⋅τc

d=1049.4 X10/4000X 1X 0.28

d=937mm

d provided =953>937mm

Therefore, OK for One-way shear.

CHECK FOR TWO WAY SHEAR:-

For, two-way shear, critical section is located at a distance ‘d/2’ from the face of the column all around.

The width bo=b+d=400+953=1353mm

Now the Net shear force acting in the perimeter.

F = ρo [B²-bo2]

= ρo [B²-(b+d)²]

=206.25[4–(1.35)²]

=449.10KN

Now ultimate shear force, Fu=1.5X F

=1.5X 449.10=673.65KN

Now, from IS 456-2000

τv= Fu/4bd= =0.13N/mm²

Now, permissible shear stress(ksτc)

Where,Ks=1

τc =0.25 √fck=0.25 X√25=1.25

ks ⋅τc =1X1.25=1.25

Therefore, τv< τc.........Hence, safe in the two-way shear

RESULT:-

Given square footing of size has been designed successfully.