Design of Shallow Foundation (Isolated Footings)
Aim : Design of Shallow Foundation (Isolated Footings)
Introduction :
Foundation is a structure which transfers its gravity loads to earth from superstructure.
Factors on which footing types depends are,
- Bearing Capacity of the soil.
- The load applied from structure to the foundation.
- The type of Superstructure is going to construct.
- The Condition of the soil.
- The depth of the soil at which safe bearing strata exists.
Procedure :
Given Data :
P = 2000 KN
SBC = 150 KN/m^2
Fck = 25 N/mm^2
Column size = 400 x 400 (0.4 x 0.4)m
Step 1
Total load on soil = load on column + self weight of footing
2000 + 10 % of 2000
= 2200 KN
Find The area of Footing:
Area of Footing = `frac{Total load of soil}{SBC}`
= `frac{2200}{150}`
= 14.66 m^2
Size of footing = B x B = 14.66 m^2
B = 3.83 m = 4m
Hence area of footing = 4mx4m.
Step 2
Upward soil pressure = Total Load / Area of footing
= 2200 / 4x4
137.5 KN/m^2 > 150 KN / m^2
Net upward soil pressure = Load / Area
= 2000/4x4
= 125KN/m^2
Factored net soil pressure = 1.5 x Net soil upward pressure
= 1.5 x 125
= 187.5 KN/m^2
Step 3
Calculation of depth of footing
Case 1
For 1-way shear (beam shear)
Shear force (V) = P x (1.8-d) x B
= 187.5 x (1.8-d) x 4
Nominal shear stress (`tau`v) = `frac{Vu}{Bd}`
`frac{187.5 xx (1.8-d) xx4}{4d}`
Design strength of concrete (Tc) = ?
Assume 0.25% of M25 (IS code 875 Table -19)
Tc = 0.36 n/mm^2 = 360 KN/m^2
Nominal shear < K. Design shear
`tau`v < K.`tau`c
(K = 1 from IS code 875 for depth >300 mm.)
By putting the values-
`frac{187.5xx(1.8-d)xx4}{4d}<1 xx0.36xx10^3`
337.5-187.5d < 360d
d > 0.616m
d > 616 mm or d= 620 mm
Minimum Reinforcement for 1-way shear = `frac{0.12xx4000xx620}{100}`
Ast min = 2976 mm^2
Provided Reinforcement = `frac{0.25xx4000xx620}{100}`
Astprovided = 6200 mm^2 > Astmin hence Ok.
Case
For 2-way shear
Net Punching force = Fp
Fp = P [ `B^2xx(b-d)^2`]
= 187.5 [`4^2-(0.4-d)^2`]
Now Punching stress (tv) = `frac{Fp}{Perimeterxxd}`
=`frac{187.5[16-(0.4+d)^2]}{4xx(0.4+d)xxd}`
Punching stress`< Kc xx tau`
`tauc = 0.25`sqrtFck`
1.25 N/mm^2 = 1250 KN/m^2
Now, Ks = 0.5`beta` `le`1
`beta = 0.4/0.4` = 1
Hence : Ks = 0.5x1
=1.5 `ge` 1
Ks = 1
Ks.tc = 1 x 1.25 = 1.25 = 1250 KN
Punching stress < Kc x tc (Putting these values)
`frac{187.5[16-(0.4+d)^2]}{4xx(0.4+d)xxd}` < 1250
calculating 'd'
d > 780 mm
Minimum reinforcement = 3456 mm^2
Ast provided = 7200 mm^2
Calcultion of B.M
B.M = wl^2 x 4 /2
= 911 KN
For under Reinforces section :
mu < = Mulim
911 x 10^6 = 0.36fck . xulim .b [d-0.42 x 0.48d]
911 x 10^6 = 0.35 x 25 x 0.48 x 4000 x [4-0.42x0.48d]
hence; d = 257 mm
Area of steel :
`span>`frac{0.5fck}{fy}``[frac{1-sqrt(1-4.6Mu)}{FckBd^2}]Bxxd` `
Hence by putting the values
Ast = 3686 mm^2
% os steel = `frac{Ast xx100){Bd}`
= 0.13%
No of bar for mm dia
=`frac{3686}{113}`
= 32
Spacing of bar = 4000- (2x50)/32
= 122 mm
Checking for bearing pressure = 0.12% of BD
Minimum Distribution of steel Ast min = 0.12 x 1/100 x 4000x750
= 3600 mm^2
Transfer of load,
`sigmabr` = 0.45fck`sqrt(A1)/(A2)` (IS code 875 clause 34.4)
Area of footing (A1)= 8.5 m^2 ac
Area of column (A2) = 0.16 m^2
`sqrt((A1)/(A2)) = 7.2 >2`
Hence; `sqrtfrac{A1}{A2}` = 2
`sigmabr`=0.4x25x2 = 20 N/mm^2
*Actual bearing pressure = factored load/Area
= 206 KN/m^2 = 0.206 N/mm^2\
Hence ; `sigmabr`> Actual bearing pressure.
For Development Length;
`frac{phixxsigmas}{4taubd}`(IS code 875 Clause 26.2.1)
Ld = 657 mm
Result : Hence we have calculated and drawing all the necessary data for columns.
