Design of Shallow Foundation (Isolated Footings)

Design of Shallow Foundation (isolated Footings)

Aim:

To design the square footing using the given details

Introduction:

• Footings which are provided under each column independently are called as isolated footings.
• They are usually square, rectangular, or circular in section.
• Footing is laid on pcc, before laying pcc, termite control liquid is sprayed on top face of pcc to restrict the termites to damage the footing.
• Isolated footings are provided where the soil bearing capacity is generally high and it comprises of a thick slab which may be flat or stepped or sloped.
• This type of footings are most economical when compared with the other kind of footings.

Procedure:

• Design of square footing:

Given.

Column size = 400x400 mm

Soil bearing capacity = 150 KN/mm²

Compression axial load = 2000 KN

Compressive strength fck = M25

Steps involved in depth of square footing

1. Determine the size of footing
2. Upward pressure
3. Depth of footing on the basis of B.M
4. Area of reinforcement
5. Check for one way shear
6. Check for two-way shear

Step:1 Size of footing:

Assume self-weight of footing equal to 105 of superimposed load.

Wd =10% x 2000

      = 0.1x2000 = 200 KN.

Now, Total load, W = WL + Wd

      = 200 + 200

  W = 2200 KN.

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

       = 2200 / 150 = 14.67 m³

Now, let us take, B = `sqrt A` = `sqrt 14.67`= 3.82

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

Therefore, Area of footing is 16 m².

Step:2 Net upward pressure (`rho`o)

 `rho`o = (Wx15) / (size of footing)

 `rho`o = (2200x1.5) / (16)

 `rho`o = 206.25 KN/m³

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

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

M = `rho` `frac{B}{8}`(B-b)²

    = 206.25x`frac{4}{8}`(4-0.4)²

    = 1336.5 KN/m

Now, Ultimate moment

(Mu) = 1.5 (M)

       = 1.5x1336.5

  Mu = 2004.75

To get the ultimate moment we have calculated that,

Mu = 1.5 FOS x Moment.

Now, Depth of footing (D),

Mu = 0.138 fck Bd²

2004.75 = 0.138 x 25 x 4000 x d²

d² = (2004.75 x`10^6`) / (0.138x25x4000)

d² = 145271.73

d = `sqrt 145271.73`

d = 381.14 

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

d = 381.14 x 2.5 = 952.85

D = d+50

D = 952.85+50

D =1003 mm

 

Step:4 Area of Reinforcement

Ast = 0.5 fck / fy [ 1- `sqrt 1-` `frac{4.6Mu}{fckBd²}`]  Bd

= 0.5x25 / 415 [ 1- `sqrt 1-` `frac{4.6x2004.75x10^6}{fckBd²}`] x 4000x953

Ast = 5985.3 mm²

Now, we have to find the % of the reinforcement

P = `frac{Ast}{Bd}`x 100

P = `frac{5985.3}{4000 X 953}`x 100

P = 0.157 %

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

 Ast = 5985.3 mm²

`pi/4`x 20² = 314.5

No:of bars = `frac{5985.3}{314.5}` = 20 bars.

Therefore, Spacing b/w the bars

= `frac{B}{no:of Bars}`

= `frac{4000}{20}`

= 200 mm c/c

Step:5 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) = `rho`oB {`frac{1}{2}`(B-b)-d}

= 206.25 x 4 {`frac{1}{2}`(4-0.4) – 0.952}

= 699.6 KN

Ultimate shear force (Vu) = 1.5xV

= 1.5x699.6 = 1049.4 KN.

Nominal shear stress = `frac{Vu}{Bd}` as per IS 456-2000

=`frac{1049.4X10³}{4000X952}` = 0.27 N/mm²

Now, design shear strength of concrete `tau`c

`tau`c = `tau`c K

K=1, because depth > mm

`tau`c=0.28 N/mm²

Then, `tau`v < `tau`c

Hence OK.

Now, `tau`v < `tau`c

Vu /Bd = K`tau`c

d=Vu / BK`tau`c

d=`frac{1049.4X10}{4000X1X0.28}`

d=937 mm

d provided = 953 > 937 mm

Therefore, OK for One-way shear.

 

Step:6 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 = 1353 mm

Now the Net shear force acting in the perimeter.

F  = `rho`o [ B² - bo²]

   = `rho`o [B² - (b+d)²]

   = 206.25 [4 –(1.35)²]

   = 449.10 KN.

Now ultimate shear force, Fu = 1.5 x F

= 1.5 x 449.10 = 673.65 KN

Now, from IS 456-2000

`tau`v= `frac{Fu}{4bd}`=  = 0.13 N/mm²

Now, permissible shear stress (kc `tau`c)

Where, Ks =1

`tau`c = 0.25 `sqrt fck` = 0.25 `sqrt 25` = 1.25

ks `tau`c = 1x1.25 = 1.25

Therefore, `tau`v < `tau`c

Hence, safe in the two-way shear

 

RESULT:

Hence the given details for the square footing has been designed successfully.