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.

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

ANSWER:

AIM:

To design the sqaure footings using the given details of above content.

PROCEDURE:

DESIGN OF SQUARE FOOTING:

Column size = 400X400 mm

soil bearing capacity = 150 KN/mm2

Comprssion axial load = 2000KN

compressive strength fck = M25

Steps invloved in design of square footing:

1. Determine the size of footing
2. Upward pressure (`rho underset 0`)
3. Depth of footing on the basis of B.M.
4. Area of reinforcement.
5. Check for one way shear and two way shear.

STEP1:

Size of the footing.

Assume self-weight of footing equal to 10% of super imposed load.

`omega underset w`= 10% X 2000

= 0.1 X 2000

= 200KN

Now total load, w= `omega underset l ' + omega underset w`

= 2000+ 200

`omega`= 2200 KN

Now, Area of footing (A) = W/SBC

= 2200/150 = 14.66 `m^2`

Now Let us take B = `sqrt A`=`sqrt 14.66`

= 3.82

Hence, we provide the size of square footing is, 4m X4m

The area of footing is 16`m^2`

Step 2:

Net upward pressure (`rho underset 0`)

(`rho underset 0`) = w X1.5/ Size of footing.

=2200 X 1.5 / 4 X 4

= 206.25 KN/ m2

Step 3: depth of footing on the basis of B.M.

the maximum B.M at the face of the column and is given by

M = (`rho underset 0`) B/B (B-b)2

= 206.25 X 4/B (4-0.4)2

= 1336.5 KN/m

Now, the ultimate moment are calculated

Mu = 1.5 (FOS) x Moment.

Now, depth of footing (`omega`)

Mu = 0.138 fck bd2

2004.75 = 0.138 x25 x 4000 X d2

d2 = 145271.73

d = 381.14

check shear value by taken 2.5 times higher value.

d = 381.14 X 2.5

= 952.86

D = d+ 50

= 952.85 + 50

= 1003 mm

STEP 4 :

Are of reinforcement:

Ast = 0.5 fck/ fy [1-sqrt of 1-4.6Mu/fck Bd2]Bd

= 5985.3 mm2

Now we have to find the percent of the reinforcment.

P = Ast / bd X 100

= 5985.3/4000 X953 X100

= 0.157%

No of bars as w assume tha 20mm dia bars.

which means we have 5985.3/314.5 = 20

20 no. of bars.

spacing between bars = B/20 = 4000/20 = 200mm c.c.

step 5:

check for one way shear:

shear force (V) = (`rho underset 0`) B {1/2 (B-b)-d}

= 206.25 X 4{1/2(4-0.4) - 0.952}

= 699.6 KN

Ultimate shear force (Vu) = 1.5(v)

= 1.5 X 699.6

= 1049.4 KN.

Nominal shear stress = Vu /Bd

= 1049.4 X 10 3 / 4000 X 952

= 0.27N/mm2

Now design shear stength of concrete

Tc = Tc.K

K =1 , Because depth >mm

Tc for o.1575 of P is 0.28 N/mm2

Tv

hence ok.

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.

HENCE its ok.