Design of Shallow Foundation (Isolated Footings)
Aim: To design the footing using the given details
Procedure:
Calculation of Gross soil pressure:
- Unfactored load = 2000/1.5 = 1333 kN Assume thickness of footing (t) = 0.5 m
- Depth of footingas given in data (h) = 2 m, So soilsurcharge depth = 2 - 5 = 1.5 m Assume soildensity = 20 kN/cu.m
- Assume RCC density = 25 kN/cu.m
- Consider square shape footing,area of the Footing = b x b = b^2
Condition: The safe bearing capacity of soil should be greater than gross soil pressure.
150>(1333/b^2)+(1.5x20)+(0.5x25)
150>(1333/b^2)+42.5
b^2>(1333)/107.5
b>3.52m
So we can provide 3.6m size of square spread footing
Gross soil pressure=(1333/3.6^2)+(1.5x20)+(0.5X25)
=0.145N/sq.mm
Calculation of Net soil pressure:
Net soil pressure = Factoredload I Area of footing
=2000/3.6^2
=0.154N/sq.mm
Calculation of bearing pressure
=Net load/Column area
=2000000/400^2
=12.5N/sq.mm
Allowable bearing pressure as per clause 34.4 from IS456-2000,
= `sqrt ((A1)/(A2))`x0.45fck`
Where A1= Base area of Footing
A2= Base area of the column
`sqrt ((A1)/(A2))`but not greater than 2 in calculation
`sqrt ((A1)/(A2))`= 3
So Allowable bearing pressure = 2 x 0.45 x 25 = 22.5 N/sq.mm
Hence OK
Calculation of Oneway Shear check:
Considering concrete cover SOmm, the effective depth of footing shall be 500 - 50 =450 mm
Net Load on the shaded area = 1150 x 3600 x 0.154 = 637.56 kN
Design Footingarea at critical section = 3.6 x 0.45 = 1.62 sq.m
Oneway shear stress = 637560I 1620000 = 0.393 N/sq.mm
As per Table-20 from 15456-2000, the maximum concrete shear stress for M25 = 3.1 N/sq.mm.
Hence OK.
Calculation of Two way Shear check:
Considering first critical perimeter of two-way shear at "d/2" distance either side from column face. As shown below,
Net Load from the shaded area = (0.154x3.6x3.6) - (0.154x0.85x0.85) = 1884.5 kN
Design Footingarea at critical perimeter = 4 x 0.85 x 0.45 = 1.53 sq.m
Twoway shear stress = 1884500I1530000 = 1.23 N/sq.mm
As per Clause 31.6.3.1from 15456-2000, the permissible two-way shear stress for M25 concrete with 0.4 m square column,
ks= 0.4/0.4 = 1
ksTc= 1X 0.25`sqrt(fck)`
= 1.25 N/sq.mm
Hence OK.
Calculation of Flexural reinforcement:
Ultimate Footing Bending moment at face of column: `(wl^2)/2`
w = 0.154x1000 = 154 N/mm
l= (3600-400)/2 = 1600mm
BMD = 197.12 kNm/meter width of footing
As per Table-2 of SP16 guidelines, percentage of reinforcement for with Fe500 and M25,
pt= 0.2372%
So the flexure rebar per meter width = (0.2372/100)x1000x450= 1067 sq.mm/meter width
Calculation of One-way shear reinforcement:
One way shear stress=0.393N/sqmm
As per table -19 of IS456-2000,for concrete M25 with design shear stress 0.393N/sqmm,the required reinforcement is 100( As/bd) = 0.3077
= 1385 sq.mm/meter width
We will provide greater of the two reinforcement with 20mm dia with 225 mm spacing.
So, 1000/225 x 314 = 1395 sq. mm which is greater than the designed reinforcement.
Hence OK.
Calculation of Developmentlength:
As per cluase 26.2.1 from 15456-2000, the developmentlength of rebar should be minimum,
Ld = `(phisigmas)/(4tbd)`
`phi`= diameter of bar = 20mm
`sigmas`= al owable stress of rebar = 0.55 Fy (As per Note-1 from Table 22 of 15456-2000)
=0.55x500 = 275 N/sq.mm
Tbd= 1.4 (for M25 from clause 26.2.1.1 of 15456-2000)
Ld = 982 mm Minimum, rounding it off to Ld = 1000 mm
Calculation of stepped footing:
Critical distance of Oneway shear from column face = 450mm Critical distance of Twoway shear from column face= 225mm
We have to provide SOOmm footing thickness at least 450mm from column face, rounding it off to 5OOmm uniform around the column, as shown below.
Reinforcement detail drawing:
Conclusion:
Thus the required spread footing is designed and detailed as per the given inputs. All the required limite state criteria are verified and found to be ok.