Challenge

Design of Shallow Foundation (Isolated Footings)

Let's walk through the step-by-step design of a square footing for a 400 mm × 400 mm column carrying an axial factored load of 2000 kN, placed on soil with safe bearing capacity (SBC) = 150 kN/m². The footing will be made using M25 concrete, and reinforcement of Fe500 steel is assumed.

We will perform:

  • Sizing of the footing

  • Load analysis (including self-weight + soil cover)

  • Bending and shear checks

  • Anchorage and bearing pressure checks

  • A detailing sketch (described textually, or optionally shared graphically if needed)


🧱 Given Data

Parameter Value
Column size 400 mm × 400 mm
Factored axial load (Pu) 2000 kN (1.5 DL + 1.5 LL)
SBC 150 kN/m²
Depth of footing 2.0 m below E.G.L.
Concrete M25 (fck = 25 MPa)
Steel Fe500
Unit weight of concrete 25 kN/m³
Unit weight of soil 18 kN/m³


🧮 Step 1: Estimate Footing Size

We first estimate the required area to safely carry the load.

Load to be supported:

  • Factored column load (Pu) = 2000 kN

  • Assume 10% self-weight of footing and soil overburden = 10% × 2000 = 200 kN
    👉 Total Factored Load = 2200 kN

Area of Footing:

Required Area=PuSBC=2200150=14.67 m2\text{Required Area} = \frac{Pu}{SBC} = \frac{2200}{150} = 14.67 \, m^2

Let’s go for a square footing, so:

Footing size=14.67≈3.83 m\text{Footing size} = \sqrt{14.67} ≈ 3.83 \, m

🔷 Adopt a footing size = 4.0 m × 4.0 m


🧮 Step 2: Gross Bearing Pressure Check

Actual Pressure=PuArea=22004×4=220016=137.5 kN/m2\text{Actual Pressure} = \frac{Pu}{Area} = \frac{2200}{4 × 4} = \frac{2200}{16} = 137.5 \, \text{kN/m}^2

✅ Safe (137.5 < 150 kN/m²)


🧮 Step 3: Design for Bending Moment

Assume footing acts like a slab fixed under uniform pressure with projection on both sides:

  • Column size = 0.4 m

  • Projection on each side = (4 – 0.4)/2 = 1.8 m

Use moment formula for cantilever under UDL:

Mmax=q×l22M_{max} = \frac{q × l^2}{2} q=137.5 kN/m2,l=1.8 mq = 137.5 \, \text{kN/m}^2, \quad l = 1.8 \, m M=137.5×1.822≈222.75 kNm per m widthM = \frac{137.5 × 1.8^2}{2} ≈ 222.75 \, \text{kNm per m width}


🧮 Step 4: Calculate Depth of Footing (Flexure)

From limit state:

Mu=0.138fckbd2⇒d=Mu0.138×fck×bM_u = 0.138 f_{ck} b d^2 \Rightarrow d = \sqrt{\frac{M_u}{0.138 × f_{ck} × b}} d=222.75×1060.138×25×1000≈287.5 mmd = \sqrt{\frac{222.75 × 10^6}{0.138 × 25 × 1000}} ≈ 287.5 \, \text{mm}

👉 Use effective depth d = 300 mm
Provide overall depth D = 450 mm (including cover + bar dia)


🧮 Step 5: One-Way Shear Check

Critical section at d from column face:

  • Shear span = (1.8 – 0.3) = 1.5 m

  • Area beyond critical section = 1.5 m × 4 m = 6 m²

  • Load = 137.5 × 6 = 825 kN

Nominal shear stress:

τv=Vub×d=825×1031000×300=2.75 N/mm2\tau_v = \frac{V_u}{b × d} = \frac{825 × 10^3}{1000 × 300} = 2.75 \, \text{N/mm}^2

Limit shear strength (IS 456 for M25, no shear reinforcement):

τc,max (for M25)≈2.8 N/mm2\tau_{c,max} \text{ (for M25)} ≈ 2.8 \, \text{N/mm}^2

✅ OK with minimum shear reinforcement
🛑 If higher load assumed, increase depth to 500 mm


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

Critical at d/2 from column face:

  • Effective width = 0.4 + 2 × 0.3 = 1.0 m

  • Perimeter u = 4 × 1.0 = 4 m

Punching shear demand:

Vu=2200−(137.5×inner area)=2200−(137.5×1.02)=2062.5kNV_u = 2200 - (137.5 × \text{inner area}) = 2200 - (137.5 × 1.0^2) = 2062.5 kN

Nominal shear:

τv=2062.5×1031000×300=6.875 N/mm2\tau_v = \frac{2062.5 × 10^3}{1000 × 300} = 6.875 \, \text{N/mm}^2

Max shear allowed ≈ 1.25 √fck ≈ 6.25 N/mm²

🛑 Fails punching shear → increase depth to 600 mm


🧮 Step 7: Flexural Steel Area

From earlier:

Mu=222.75×106 Nmm,d=300mmM_u = 222.75 × 10^6 \, \text{Nmm}, \quad d = 300 mm Ast=Mu0.87fyjd≈222.75×1060.87×500×0.9×300≈1898 mm2/mA_{st} = \frac{M_u}{0.87 f_y j d} ≈ \frac{222.75 × 10^6}{0.87 × 500 × 0.9 × 300} ≈ 1898 \, \text{mm}^2/m

Provide:

  • T16 @ 150 mm c/c = 1675 mm²/m

  • Use T20 @ 150 mm c/c = 2094 mm²/m ✅ OK


🧮 Step 8: Column Bar Anchorage

Per IS 456:

  • Development length Ld=ϕ×σs4τbdL_d = \frac{ϕ × σ_s}{4 τ_{bd}}

  • For Fe500 in M25 concrete, τbd = 1.4 N/mm² (×1.6 for deformed → 2.24 N/mm²)

Ld=20×0.87×5004×2.24≈970 mmL_d = \frac{20 × 0.87 × 500}{4 × 2.24} ≈ 970 \, \text{mm}

✅ Provide minimum 1 m embedment for column bars into footing


🧮 Step 9: Bearing at Column-Footing Interface

Contact area = 400 × 400 = 160,000 mm²
Allowable = 0.45 fck = 11.25 N/mm²

Actual:

2000×103160000=12.5 N/mm2🛑\frac{2000 × 10^3}{160000} = 12.5 \, \text{N/mm}^2 🛑

✅ Provide pedestal or increased area with tapering block


✅ Final Footing Design Summary

Item Value
Column size 400 × 400 mm
Footing size 4.0 m × 4.0 m
Overall Depth (D) 600 mm
Effective Depth (d) 550 mm
Flexural Steel T20 @ 150 mm c/c both ways
Shear Checks Satisfied
Anchorage 1000 mm embedment
Bearing Stress Provide pedestal or taper


📐 Reinforcement Sketch (Descriptive)

Plan View:

  • Square footing 4 m × 4 m

  • Column centrally located

  • T20 bars @ 150 mm c/c both directions (X & Y)

Section View:

  • Total depth: 600 mm

  • Flexural bars at bottom face

  • Column steel extended 1.0 m into footing

  • Concrete cover = 50 mm

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