Design of Shallow Foundation (Isolated Footings)

Design of Shallow Foundation (Isolated Footings)

AIM : 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.

Please do go through the solved examples in the document provided with the challenge.

PROCEDURE :

Given:

Column size = 400x400mm

SBC = 150 kN/mm²

Compression axial load = 2000 kN

fck = M25

Steps to design a square footing:

1) Size of Footing

2) Net upward Pressure

3) Bending moment

4) Depth of Footing

5) Reinforcement

6) One way and Two-way shear

Step 1:

Size of Footing:

Load on column = 2000 kN

Add 10% of column axial load for footing = 10% (2000)

= 2000 + 200

= 2200 kN

Area of Footing:

Area of Footing = Load on footing/Safe bearing capacity

= 2200/150

= 14.66 m^2

√14.66 = 3.82 = 4m x 4m

Therefore, provided area = 16m^2

Step 2:

Net upward pressure: σ = Force/Area

= 1.5 x Total load on footing/Area

= 1.5 x 2200/4 x 4

σ = 206.25 kN/m^2

Step 3:

Bending Moment:

At the critical section of the footing face of the column from the edge of footing

(4000/2) - (400/2) = 1800mm

Therefore, load on critical section = σ x breadth of footing

= 206.25 x 4

Load = 825 kN/m

So, the bending moment = wl^2/2

= (825 x 1.8^2)/2

= 1336.5 kN-m

Step 4: Depth of footing:

Mu = 0.138 fckbd² = 0.138 x 25 x 4000 x d² 1336.5 x 10⁶ = 0.138 x 25 x 4000 x d²

d = 311.20 mm

By the rule, we can take 2 to 2.5 times higher value of d for shear check

Hence take ; d= 778 mm

D = d + effective cover = bar dia + half of bar dia

= 778 + 50+ 20 +10 ( Assume bar dia as 20mm)

D = 860 mm

Step 5: Reinforcement:

Bending moment = 1336.5 x 10⁶

Percentage of reinforcement, Pt= (50fck/fy)(1-√4.6 x 13336.5 10⁶/25x4000x778²)

Therefore, Pt = 0.1570 %

Area of steel = (Pt/100) x bd

= (0.157/100) x 4000 x 778

Ast= 4885 mm²

No of bars: (Take 20mm bar)

No of bar = 4885/314.5

= 15.55 = 16 nos

Provide 16 nos of 20mm bar

Ast provided = 16 x 314.5

= 5032 `mm^2

Therefore,

Ast > Ast provided ; 4885 > 5032 ; Hence ok.

Step 6:

Check for clear spacing between bars:

As per IS 456 - table 15, for fe 415, the clear spacing between the bars should be less than 180mm

c/c spacing = (4000-50-50-10-10)/(no of bars-1)

= 258.66 mm

Therefore ,

clear spacing = 242.5- half of dia - half of dia bar

= 242.5-10-10

clear spacing = 238 mm

As per IS code value, our value is greater so obtained clear spacing is greater than IS code value,

In order, the minimize the value reduce the bar size.

Take 16mm bar = 4885/200.96

Therefore,

No. of bar = 24

C/C spacing = (4000-50-50-8-8)/(No.of bars-1)

= 168.89

Clear spacing = 168.86-8-8

= 152.86mm

So, Clear spacing 152.86 < 180

Hence Ok

Step 7: Check for Oneway Shear:

One way shear check at the critical section for one way shear at the distance, d from face of the column

Vu = Upward pressure x Area of the highlighted section

= 206.25 x 4 x ?

Area of highlighted section = (4000/2)-(400/2)-778

= 1.022

Now, Vu = 206.25 x 4 x 1.022

Vu = 843.15 kN

Nominal Stress:

τv = Vu/bd = (843.15 x 10³)/(4000 x 778)

τv= 0.27 N/m

Now compare τv and τc

For the τc value given by IS 456, Table 19

Pt provided = (100 Ast/bd) = 100 x Ast/bd

Ast = 24 nos of 16mm dia

= 4823 mm²

Therefore , Pt= (100x 4823)/(4000x778) = 0.1549

From IS 456, table 19

100A_(st)/bd

Value of greater than or equal→0.15is0.28Nmm^2`

Therefore, τv< τc

= 0.15<0.28

Hence Ok

Step 8: Check for Two way shear:

At peripheral d/2 distance from face of the column

Area of highlighted section = 400+(778/2)+(778/2)

= 1178 mm^2

Therefore, area = 42−1.1782

= 14.6123 mm²

Vu= Upward pressure x Area of highlighted section = 206.25 x 14.6123

Vu = 3013.79 kN

τv= Vu/bd

Peripheral length = 4 x 1178

b= 4712 mm

τv= (3013.79 x 10³)/(4712 x 778)

τv = 0.8221 N/mm²

Compare τv and τc

For two way shear τc = Ksx τc

For τc value taken from IS 456-2000 page number 58 and 58

Ks= 1 (for square column)

τc= 0.25o√fck) = 1.25 N/mm²

Therefore, τc= Ks x τc = 1 x 1.25

τc= 1.25 N/mm²

Therefore ,

τv < τc 0.8221 < 1.25

Hence Ok

RESULT :

Reinforcement detail drawing: