Design of Shallow Foundation (Isolated Footings)
AIM -
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 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 the 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.
ANS -
Aim -
of this challenge is manually to calculate the square footing reinforcement detail with respect tolimit state of design.
THE Following steps are use to design footing.
1. calculation for gross soil pressure for footing area.
2. net soil pressure
3. initial depth of footing calculation.
4. Flexural reinforcement calcultaion.
5. one way and two-way shear check.
Given That -
1. factored load = 2000 kn
2. column size = 400x400
3. Bearing capacity of soil = 150 kn/m2
4. Given footing depth/ height = 2 m.
Assume,
Density of soil = 20 kn/cum
Density of Rcc = 25 kn
Assume thickness of footing = 500mm = .5m
1. calculation for gross soil pressure for footing area -
surcharge pressure of soil = height of the footing - thickness of the footing
= 2 - 0.5 = 1.5 m.
Total ( unfactored load ) Load = 2000/1.5 = 1333 kn
square shape of the footing = bxb
As per codal reccomendation.
the safe bearing capacity of the soil should be greater than gross soil pressure.
150> (1333/bxb)+(1.5*20)+(0.5*25)
150>(1333/bxb)+42.5
`b^2`>1333/107.5
b>3.52m
provide square footing 3.6
Gross soil pressure = `(1333/3.6^2)+(1.5*20)+(.5*25)`
= 0.145 N/Sq MM
2. Net soil pressure and initial footing depth
= factored load / area of footing
= 2000/ (`3.6^2)`
`P_n` = 0.154 N/mm2
As per is code 456-2000 clause 34.4
Allowable bearing pressure = `root``A1/A2.0.45.Fck`
where A1 = Area of footing at base
A2 = Arae of column at base
`root``A1/A2`Not greater than 2
`root`A1/A2 = 3
So allowable bearing pressure = 2*0.45*0.25 = 22.5 kn/mm
hence ok.
permissible bearing stress 0.45*fck = 11.25 N/mm2
permissible bearing stress multiplied with area factor 11.25*2 = 22.5N/mm2
One way shear check
for one way shear assuming
concrete cover = 50mm
effective depth of footing = 500-50 = 450 mm
consider one side of the footing as critical section is shown
one way critical shear area
= footing area /2
= 3.6/2 = 1.8
= 1.8 - .2(half area of coluimn)+.45( effective depth of the footing)
= 1.15 m .
which shown in above plan.
load at the hatched (one way shear)
= 1150*3600*0.154 = 638 kn
footing area at critical section = 3.6*0.45 = 1.62 m2
one way shear stress = 637560/1620000 = 0.393 n/mm2
Using codal reference is456-2000 table 20
Allowable maximum shear stress for m25 =3.1 kn/mm2
hence ok.
Two way shear calculation
considering,
critical perimeter at d/2 distance from the column face
two way shear area = (0.154*3.6*3.6) - (0.154*0.85*0.85) = 1884.5kn
Area of footing at critical perimeter 4*0.85*0.45 = 1.53m2
two way shear stress = 1884500/1530000 = 1.213n/m3
Refering the is code 456-2000 clause 31.6.3.1
for m25 grade of concrete allowable two way shear stress = `k_st_c`
`k_s = 0.4/0.4 = 1`
`k_st_c = 1*0.25rootfck`
= 1.25 n/mm2
Flexural Reinforcement calculation.
At face of column,
Ultimate footing bending moment = `wl^2/2`
w = 0.154*1000 = 154 N/mm
l = (3600 * 400 )/2
= 1600mm
ultimate bending moment = 197 kn/m of footing width.
Referring to sp -16 Table -2
% of reinforcement = `m_u/bd^2` for fe 500 and m25
`p_t = 23%`
flexural reinforcement /m width = (0.2372/1000)*1000*450 = 1067mm2/m.
one way shear Reinforcement calculation.
one way shear stress calculation = 0.393 N/mm2
Referring to the code is 456 - 2000
For M25 Grade Concrete Required Shear reinforcement for the shear stress = 0.393 n/mm2
100(as/bd)
= 0.30
= 1385.2 mm2/m.
providing reinforcement of 20mm diameter at 225 mm spacing which greater then the required reinforcement
hence ok.
Development length calculation
41 x dia = development length is providing 41times into dia
= 41*20 = .820m
Actual bearing stress at the column base
= (2000*1000)/(400*400)
= 12.5 N/mm2
= 12.5 N/mm2 Actual stress is less than permissible 22.5 N/mm2 hence ok.
we can proceed with a 500mm depth footing for further design check.
if we assume 50mm cover of footing.
the effective depth(d) is 450mm
hence design shear stress is less than shear capacity
Using Autocad software to show the rebars.
