Design of Square Isolated Footing

A comprehensive guide to RCC isolated footing design as per IS codes

Thumb Rules (As per IS Codes)

Parameter Specification Reference
Minimum Depth of Foundation 1.0 m IS 1080 (Clause 6.2)
Minimum Thickness of Footing 150 mm (on soil)
300 mm (on piles)		 IS 456 (Clause 34.1.2)
Minimum Grade of Concrete M20 (for Mild exposure) IS 456 (Table 4)
Clear Cover to Reinforcement 50 mm IS 456 (Clause 26.4.2.2)
Minimum Percentage of Steel 0.15% (Mild Steel)
0.12% (HYSD Steel)		 IS 456 (Clause 26.5.2.1)
Minimum Spacing Between Bars 150 mm or 6" -

Design Steps for Isolated Footing of Uniform Depth

Step 1: Determine the Size of the Footing

  1. Total Load:
    • Let ( Wu ) = Load on the column
    • Self-weight of footing, ( W1 = 0.1W ) (Assumed as 10% of W)
    • Total Load = W + W1
  2. Factored Load:
    Factored Load = 1.5 × (W + W₁)
  3. Ultimate Bearing Capacity:
    qult = 2 × Safe Bearing Capacity (SBC) of Soil
  4. Area of Footing:
    A = Factored Load / qult = 1.5 × (W + W₁) / (2 × SBC)
    For a square footing of side \( B \):
    B = √A

Step 2: Find Net Upward Ultimate Soil Pressure

The weight of the footing is not included for shear and moment calculations.

pu = (1.5 × W) / (B × B)

Step 3: Find Depth of Footing from Bending Moment

  1. Critical Section: At the face of the column
  2. Bending Moment (B.M.):
    Mu = pu × (B / 8) × (B - b)2
    where \( b \) is the width of the square column
  3. Calculate Effective Depth (d):
    • Use the limiting moment of resistance formula for a singly reinforced section:
      • For Fe 250: ( Mu = 0.148 fck Bd2 )
      • For Fe 415: ( Mu = 0.138 fck Bd2 )
      • For Fe 500: ( Mu = 0.133 fck Bd2 )
    • The calculated depth must be increased if required to satisfy shear checks
Note: Provide a minimum thickness of 150 mm.

Step 4: Calculate the Area of Steel Required

  1. Main Reinforcement:

    Solve for ( Ast ) using:

    Mu = 0.87 fy Ast [d - (fy Ast / (fck B)]
  2. Check Minimum Steel:
    Ast,min = 0.0012 × B × d (For HYSD bars)
  3. Distribution: Reinforcement is uniformly distributed over the entire width in both directions

Step 5: Check for One-Way Shear (Beam Shear)

  1. Critical Section: Located at a distance ( d ) from the face of the column
  2. Shear Force ( Vu ):
    Vu = pu × B × [ (B - b)/2 - d ]
  3. Nominal Shear Stress (tauv ):
    τv = Vu / (B × d)
  4. Shear Strength Check:
    • Find ( p = (100 × Ast) / (B × d) )
    • Obtain design shear strength ( τc ) from IS 456 (Table 11.1)
    • Check: ( τv < τc)
    • Also ensure ( τv < τv,max) (Table 11.3)
Note: Shear reinforcement is generally not provided in isolated footings.

Step 6: Check for Two-Way Shear (Punching Shear)

  1. Critical Section: Located at a distance ( d/2 ) from the periphery of the column
  2. Factored Shear Force (τc):
    Vu = pu [ B² - (b + d)² ]
  3. Perimeter of Critical Section (bo):
    bo = 4(b + d)
  4. Nominal Shear Stress ( τv ):
    τv = Vu / (bo × d)
  5. Permissible Shear Stress ( τc ):
    τc = ks × 0.25 √fck
    6. where ( ks = 0.5 + βc) and ( βc ) is the ratio of the shorter to longer side of the column
  • Check:
    ( τv < τc )

    • Step 7: Check for Development Length

    1. Critical Section: Along the face of the column
    2. Development Length ( Ld ):
      Ld = (87 fy φ) / (4 τbd)
      where ( τbd ) is the design bond stress from IS 456 (Table 11.5)
    3. Available Embedment Length:
      Lavailable = (B - b) / 2
    4. Check: ( Lavailable > Ld )

    Step 8: Summary and Sketch