Design of Shallow Foundation (Isolated Footings)

AIM -

1. To design a square footing for column of size 400 X 400 mm.

2. To do the following checks :

• If gross bearing pressure under the footing is within the 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 0.45times the maximum depth of the section, flexural reinforcement in both directions and anchorage of column steel.

GIVEN –

• Soil bearing capacity = 150 kN/m²
• Compression Axial load = 2000 KN
• Compressive Strength f_{ck =} 25
• Thickness of footing = 0.8 m (Assume)
• Depth 2m below E.G.L
• Soil depth = 20 KN/cu.m
• RCC density = 25 KN/cu.m

PROCEDURE-

1. Size of Footing :

Assume self weight of the footing equal to 10% of the superimposed load.

Wd = 10% of 2000

Hence, 0.1 X 2000

= 200 KN

Now , the total load = Wl + Wd

= 2000 + 200

W = 2200 KN

Now,

Area of footing (A) = W/SBC (SBC: soil bearing capacity)

= 2200/150

= 14.66 m²

Hence area of footing (A) = 14.66m²

Size of square footing = ˇ14.66

Hence, the size = 3.82 ~ 4m

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

And Area of footing (A) = 16m²

2. Net Upward Pressure:

P_o = W + 1.5 / size of footing

= 2200. 1.5 / 4. 4

Hence, P_o = 206.25 KN/m²

3. Depth of footing based on B.M:

The maximum BM at the face of column is given by ,

M = P_o (B/8) (B-b)²

M = 206.25 (4/8). (4-0.4)²

M = 1336.5 KN/m

Now, Ultimate Mu = 1.5 M

= 1.5 x 1336.5

Mu = 2004.75 KN/m

To get ultimate moment, we need

Mu = 1.5 (factor of safety) x moment.

Depth of footing :

Mu = 0.238 F_ck Bd²

2004.75 = 0.138 x 25x 4000 x d²

d² = 145271.73

d = 381.14

By using thumb rule, we take 2.5 times higher

Value of shear check,

d = 381.14 x 2.5

d = 952.85

D = d + 50 = 952.85 + 50

Hence D = 1003mm

4. Area of Reinforcement:

0.5 x 25/ 415 X [ 1- ˇ1-( 4.6 x 2004.75 x 10⁶)/( 25 x 4000 x 953²)] X 4000 X 953

Ast = 5985.3mm²

Now, to find the % reinforcement,

P = Ast/ Bd x 100

= ( 5985.3/ (4000x953)) x 100

P = 0.157 %

Number of bars required for reinforcement are assumed of 20mm dia.

Ast = 5985.3mm²

Pie / 4 x 20² = 314.5

No. of bars = 5985.3 / 314.5

= 20

Hence 20 bars are provided.

Spacing between the bars = B / no. of bars

= 4000/ 20

= 200

Hence, 200 mm c/c spacing is provided.

5. Check for One Way Shear –

For One Way Shear, Critical Section is located at distance d from the face of the column.

Now,

Shear force (U) = P_o. B { ½ (B-b) -d }

= 206.25 x 4 {2 (4-0.4) -0.952}

= 699.6 KN

Ultimate Shear force (V) = 1.5 V

= 1.5 x 699.6

= 1049.4 KN

Nominal Shear strength (Tu) = Vu/Bd

= (1049.4 x 10³⁾/(4000x951)

= 0.27 N/mm²

Now,

Design Shear Strength of Concrete T:

As per Is 456 2000, page 72

Tc’ = Tc K

K= 1

Tc for 0.157% of P is 0.28 N/mm²

[ Tc = 0.28 N/mm²]

Tv < Tc.

Hence OK.

Tv < Tc’

Vu/ Bd = K .Tc

d = Vu/B.K Tc

d = 937mm

d provided = 953mm > 937mm

Hence OK for One Way Shear.

Now,

Is 456 2000, Clause 31.6.2,

Tv = Fu/4bd

= 673.65 x 10³ (4x 1353x 953)

= 0.23 N/mm²

Now,

Is 246 2000, Page 58

Permissible shear stress (Ks Tc):

Ks = 1 max

Tc = 0.25 ˇf_ck

= 0.25 x ˇ25

= 1.25

Hence, Ks.Tc = 1x 1.25 = 1.25

Tv < Tc

Hence, safe in Two-way shear