In this blog you will learn Manual Design Of Footing of G+5 Building Using I.S. 456:2000 with the help of example.

Step by step guid to clear concept

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## Steps:

## Given:

- Grade of Concrete =
**M20** - Grade of Steel =
**Fe500** - Clear Cover to Reinforcement (c) =
**50 mm** - Column Dimension =
**300 X 600 mm** - Net Soil Bearing Capacity (S.B.C.) =
**300 kN/m**^{2} **Assume Axial Load (P) from Column to Footing = 1500 kN**

## Load Calculation (Step 1):

Load From Column = 1500 kN

Self Weight of Footing = Assume 10% of the column load

= (10/100) X 1500

= 150 kN

———————————————————–

Total Weight (W) = 1650 kN

## Area of Footing (A) (Step 2):

Area of Footing (A) = Total Weight (W) / Net Soil Bearing Capcity (S.B.C.)

= 1650 / 300

= 5.5 m2

Take Equal Projections From Both Sides,

5.5 = (0.6 + 2x) . (0.3 + 2x)

5.5 = 0.18 +1.2x + 0.6x + 4×2

5.5 = 0.18 +1.8x + 4×2

Therefore, by Solving Above Equation,**x = 0.95 m**

Take Projection of 1 m on Both Side.

Therfore, Size of Footing = (0.3+ 2 X 1 ) & (0.6 + 2 X 1) = 2.3 X 2.6 m

Therefore Area of Footing Provided = 2.3 X 2.6 = 5.98 m2

## Upward Soil Pressure (P_{au}) (Step 3):

P_{au }= (Factored Load of Column / Actual Area of Footing Provided)

= 1.5 X 1500 / 5.98**P _{au}= 376.25 kN/m^{2}**

## Bending Moment (B.M.) (Step 4):

The Critical Section for B.M. Along the Column Face as Shown in the Figure 2,

### From X-axis:

= (2.6 – 0.6) / 2

= 1 m

Consider a Strip of 1m.

Therefore, M_{ux} = [P_{au} X 1 X (C_{x}^{2}/2)]

= [376.25 X 1 X (1^{2 }/2)]

**M _{ux} = 188.125 kN.m**

### From Y-axis:

= (2.3 – 0.3) / 2

= 1 m

Therefore, M_{uy} = [P_{au} X 1 X (C_{y}^{2}/2)]

= [376.25 X 1 X (1^{2 }/2)]

**M _{uy} = 188.125 kN.m**

## Depth of Footing Required (d_{req}) (Step 5):

Equate The Maximum Bending Moment (Max. of M_{ux} & M_{uy}) With Limiting Moment of Resistance ,

Consider a Strip of 1m.

M_{umax}= M_{ulim}

188.125 X 10^{6} = 0.133 f_{ck}.b.d_{req}^{2}

188.125 X 10^{6 }= 0.133 X 20 X 1000 X d_{req}^{2}**d _{req} = 265.94 mm**

Provide 12 mm Diameter Bars at a Clear Cover of 50 mm.

Therefore, Overall Depth (D) = 265.94 + 50 + (12/2)

= 321.94 mm ≈ 600 mm …(rounded -off on higher side)

**Overall Depth (D) = 450 mm**

Therefore, Effective Depth Provided (d_{provided}) = 600 – 50 – (12/2) = 544 mm

## Area of Steel Calculation (A_{st}) (Step 6):

### A_{stx} Along X-Direction:

A_{stx }= 826.8 mm^{2}

Provide 12 mm Diameter Bars.

Spacing = (π/4 × 12² ÷ 826.8) ×100

= 136.8 mm ≈ 125 mm

Provide 12 mm Diameter Bars @ 125 mm c/c

### A_{sty} Along Y-Direction:

A_{sty }= 826.8 mm^{2}

Provide 12 mm Diameter Bars.

Spacing = (π/4 × 12² ÷ 826.8) ×100

= 136.8 mm ≈ 100 mm

Provide 12 mm Diameter Bars @ 100 mm c/c

## Check For One Way Shear (Step 7):

The Critical Section for One Way Shear is Taken at a Distance of ‘d’ From Column Periphery.

The Offset x = C_{x} – d = 1 – 0.544 = 0.456 m

Similarly, y = C_{y} – d = 1 – 0.544 = 0.456 m**Depth of Footing at Critical Section (d’) = 544 mm**

**Shear Force Along X-Direction :-**V

_{ux}= x . L . P

_{au}= 0.456 X 1 X 376.25 = 171.57 kN

**Shear Force Along Y-Direction :-**

V_{uy} = y . B . P_{au} = 0.456 X 1 X 376.25 = 171.57 kN

**V _{umax} = 209.2 kN**

### Nominal Shear Stress (τ_{v}):

τ_{v}_{ }= V_{u}_{max}/ b’.d’ = [(171.57 X 10^{3}) / (1000 X 544)] = 0.32 N/mm^{2}

### % of Steel (P_{t}):

P_{t }= 100 A_{stmax} / b’d’ = [(100 X 1130) / (1000 X 544)] = 0.21 %

### Design Shear Stress (τ_{c}):

τ_{c} = 0.32 N/mm^{2 } > 0.32 N/mm^{2}**As, τ _{c} > τ_{v} … Safe in One Way Shear**

P_{t} | τ_{c }( For M20) |

<0.15 | 0.28 |

0.21 | ? |

0.25 | 0.36 |

## Check For Two Way Shear (Step 8):

The Critical Section for Two Way Shear is Taken at a Distance of ‘d/2’ From Column Periphery.

Depth of Footing at Critical Section (D’) = 750 mm

Effective Depth at Critical Section = d’ = 600 – 50 – (12/2 ) = 544 mm

b’ = 2 . Perimeter of Critical Section

= 2 X (1144 + 844)

= 3976 mm

Shear Force on Shaded Area is Given by,

V_{u} = (Footing Area – Critical Section Area) . Upward Soil Pressure

V_{u} = [(2.6 X 2.3) – (1.144 X 0.844)] X 376.25**V _{u} = 1886.7 kN**

### Nominal Shear Stress (τ_{v}):

τ_{v}_{ }= V_{u}_{max}/ b’. d’ = [(1886.7 X 10^{3}) / (3976 X 544)] = 0.87 N/mm^{2}

### Design Shear Stress (τ_{c}’):

τ_{c}‘= k_{s}. (0.25√f_{ck} )

Where, k_{s}= Minimum of, i) 1

ii) 0.5 + β_{c }= 0.5 + 0.5 = 1

Where, β_{c }= (Breadth / Depth)_{column} = 300 / 600 = 0.5

Therefore, k_{s}= 1

Therefore, τ_{c}‘= 1 X (0.25√20) = 1.118 N/mm^{2 }> 0.87 N/mm^{2}**As, τ _{c} > τ_{v} … Safe in Two Way Shear**

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