Design of Shallow Foundation (Isolated Footings)
Given:
Column size = 400x400mm
SBC = 150 kN/`mm^2`
Compression axial load = 2000 kN
`f_(ck)` = 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:
`M_u` = 0.138 `f_ck`b`d^2`
= 0.138 x 25 x 4000 x `d^2`
1336.5 x `10^6` = 0.138 x 25 x 4000 x `d^2`
d = 311.20 mm
By the thumb rule, we can take 2 to 2.5 times higher value of d for shear check
`therefore` 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:
Reinforcemet:
Bending moment = 1336.5 x `10^6`
Percentage of reinforcement, `P_t`= (50`f_ck`/`f_y`)(1-`root`4.6 x 13336.5 `10^6`/25x4000x`778^2`)
Therefore, `P_t` = 0.1570 %
Area of steel = (`P_t`/100) x bd
= (0.157/100) x 4000 x 778
`A_(st)`= 4885 `mm^2`
No of bars:
Take 20mm bar
No of bar = 4885/314.5
= 15.55 = 16 nos
Provide 16 nos of 20mm bar
`A_(st)`provide = 16 x 314.5
= 5032 `mm^2
`therefore` `A_(st)` > `A_(st)`provided
4885 > 5032
Hence ok.
Step 6:
Check for clear spacing between bars:
As per IS 456 - table 15, for `f_e`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 = 24.30
= 24
C/C spacing = (4000-50-50-8-8)/(No.of bars-1)
= 168.89
`therefore`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
`V_u`= 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, `V_u` = 206.25 x 4 x 1.022
`V_u` = 843.15 kN
Nominal Stress:
`tau_v` = `V_u`/bd
= (843.15 x `10^3`)/(4000 x 778)
`tau_v`= 0.27 N/m
Now compare `tau_v` and `tau_c`
For the `tau_c` value given by IS 456, Table 19
`P_t`provided = (100 `A_(st)`/bd)
= 100 x `A_(st)`/bd
`A_(st)` = 24 nos of 16mm dia
= 4823 `mm^2`
`therefore` `P_t`= (100x 4823)/(4000x778)
= 0.1549
From IS 456, table 19
` 100 `A_(st)`/bd value of greater than or equal to 0.15 is 0.28 N/`mm^2`
Therefore, `tau_v`< `tau_c`
= 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 = `4^2-1.178^2`
= 14.6123 `mm^2`
`V_u`= Upward pressure x Area of highlighted section
= 206.25 x 14.6123
`V_u` = 3013.79 kN
`tau_v`= `V_u`/bd
Peripheral length = 4 x 1178
b= 4712 mm
`tau_v`= (3013.79 x `10^3`)/(4712 x 778)
`tau_v` = 0.8221 N/`mm^2`
Compare `tau_v` and `tau_c`
For two way shear `tau_c` = `K_s`x `tau_c`
For `tau_c` value taken from IS 456-2000 page number 58 and 58
`K_s`= 1 (for square column)
`tau_c`= `0.25 root of_(ck`)
= 1.25 N/`mm^2`
Therefore, `tau_c`= `K_s`x `tau_c`
= 1 x 1.25
`tau_c`= 1.25 N/`mm^2`
`therefore` `tau_v` < `tau_c`
0.8221 < 1.25
Hence Ok
Reinforcement detail drawing:
