Isolated Footing Design — IS 456

Plan size, one-way shear, two-way (punching) shear, flexural steel, and development length — complete IS 456:2000 procedure with worked examples

Last Updated: March 2026

Key Takeaways

  • The plan area of a footing is fixed by the service load (not factored) and the Safe Bearing Capacity (SBC) of soil: A = (P + Wself) / SBC.
  • All structural checks (bending, shear) are performed with the factored net upward pressure: qu = Pu / A (self-weight already excluded from net pressure).
  • Two-way (punching) shear is checked at a perimeter d/2 from the column face; the permissible stress is ks τc where ks = 0.5 + βc ≤ 1.0 and τc = 0.25√fck.
  • One-way (beam) shear is checked at a section d from the column face; τv ≤ τc (from IS 456 Table 19, based on pt).
  • Bending moment for steel design is calculated at the column face; Mu = qu × B × (L′)²/2 per unit width where L′ is the cantilever projection.
  • Development length of footing bars must be available from the column face to the bar end; Ld = 47φ for Fe 415 in M20 concrete (deformed bars, tension).
  • Minimum depth from IS 456 Cl. 34.1.2: the depth at the edge of a footing must not be less than 150 mm for footings on soil and 300 mm for footings on piles.

1. Introduction to Isolated Footings

A footing is the lowest structural element of a building; it transfers load from the column (or wall) to the underlying soil. An isolated footing supports a single column and is the most common foundation type for columns of low-to-medium loaded buildings founded on soil with adequate bearing capacity.

The footing acts as an inverted cantilever: soil pressure pushes upward on the base and the column load pushes down at the centre. This creates bending and shear in the footing slab, requiring both flexural reinforcement (bottom mesh) and adequate depth to resist punching and one-way shear without extra shear reinforcement (IS 456 prefers no stirrups in footings; depth is instead increased).

IS 456:2000 covers isolated footing design primarily in Clause 34 (footings) and references Clause 31 (punching shear), Clause 26.5.2 (development length), and relevant tables for concrete shear capacity.

2. Types of Isolated Footings

TypeShapeTypical Use
Flat (Pad) FootingUniform thickness slab, square or rectangularMost common; columns with moderate load
Stepped FootingTwo or three steps reducing thickness toward edgeHeavily loaded columns; reduces concrete volume
Sloped (Tapered) FootingThickness tapers from column base to edgeEconomical for large footings; slope ≤ 1:2

In exam problems and standard practice, the flat (uniform depth) pad footing is used unless stated otherwise. All design equations in this article apply to flat pad footings.

3. Design Philosophy & Loading

Footing design uses a two-load system that often confuses students:

PurposeLoad UsedWhy
Sizing the plan area (L × B)Service (unfactored) load + footing self-weightSBC is a geotechnical (service) limit state value
Structural design (shear, bending, steel)Factored net upward pressure quStrength (Limit State) design of concrete section

The net upward pressure concept is key: soil pressure at the base includes the weight of the footing and soil overburden. These cancel with the self-weight of the footing in the structural design, leaving only the column load effect. Hence, the net factored pressure = Pu / A, where Pu is the factored column load and A is the footing plan area already determined at the service load stage.

4. Step 1 — Plan Size

For a column carrying a service axial load P (kN), assuming the footing self-weight and soil overburden = 10 % of column load (a common approximation):

Total load on soil = P + Wself ≈ 1.10 P

Required area A = 1.10 P / SBC

For a square footing: L = B = √A    (round up to next 50 mm)

For a rectangular footing: choose L/B ratio, then L = √(A × L/B), B = A/L

The 10 % self-weight allowance is an engineering approximation. In detailed design, the actual footing dimensions and depth are used to compute self-weight directly and the calculation is iterated once. For GATE and standard design problems, 10 % is the accepted shortcut.

5. Step 2 — Net Factored Upward Pressure

qu = Pu / A

Pu = 1.5 P (factored column load, DL+LL combination)

A = L × B (plan area from Step 1)

Units: qu in N/mm² (= MPa) when Pu in N and A in mm²

This uniform pressure acts over the entire base and produces bending and shear in the footing. For eccentric loading or combined axial + moment, the pressure varies across the base, but for purely axial load (common in isolated footing problems) it is uniform.

6. Step 3 — Depth from Two-Way (Punching) Shear

Punching shear failure occurs on a truncated pyramid surface just outside the column perimeter. IS 456 Cl. 31.6 specifies the critical perimeter at d/2 from the column face on all sides (where d is the effective depth of the footing).

6.1 Punching Shear Force

Vu,punch = Pu − qu × (a + d)(b + d)

where a × b = column dimensions, d = effective depth of footing

Critical perimeter bo = 2(a + d) + 2(b + d) = 2(a + b + 2d)

6.2 Permissible Punching Shear Stress (IS 456 Cl. 31.6.3)

τc,punch = ks × 0.25 √fck

ks = 0.5 + βc    but ≤ 1.0

βc = short side / long side of column

For square columns: βc = 1.0 → ks = 0.5 + 1.0 = 1.5 → limited to 1.0

So for square columns: τc,punch = 1.0 × 0.25 √fck

For M20: τc,punch = 0.25 √20 = 0.25 × 4.47 = 1.118 N/mm²

For M25: τc,punch = 0.25 √25 = 1.25 N/mm²

6.3 Design Condition

τv,punch = Vu,punch / (bo × d) ≤ τc,punch

Rearranging to find minimum d:

d ≥ Vu,punch / (bo × τc,punch)

Note: bo itself depends on d → solve iteratively or use the simplified form below

6.4 Simplified Approach for Square Column on Square Footing

For a square column (a × a) on a square footing (L × L):

Vu,punch = Pu − qu (a + d)²

bo = 4(a + d)

Setting τv = τc,punch:

Pu − qu(a+d)² = τc,punch × 4(a+d) × d

This is a quadratic in d — solve or trial-and-improve.

7. Step 4 — One-Way (Beam) Shear Check

One-way shear is checked across the full width B of the footing at a section located d from the column face (IS 456 Cl. 34.2.4).

Cantilever projection from column face to critical section: x1 = (L − a)/2 − d

Shear force per unit width: Vu,1way = qu × B × x1

Nominal shear stress: τv = Vu,1way / (B × d)

Permissible: τv ≤ τc (from IS 456 Table 19 based on pt = 100Ast/Bd)

If τv > τc, the depth d must be increased (shear reinforcement is generally not provided in footings). The one-way shear check is less critical than punching shear for most square footings but becomes governing for rectangular footings with a high L/B ratio.

IS 456 Table 19 — τc values for M20 concrete (excerpt)

pt = 100Ast/bd (%)τc (N/mm²) — M20τc (N/mm²) — M25
0.150.280.29
0.250.360.36
0.500.480.49
0.750.560.57
1.000.620.64
1.250.670.70

8. Step 5 — Bending Moment & Flexural Steel

The critical section for bending moment is at the column face (IS 456 Cl. 34.2.3). The footing acts as a cantilever of projection L′ from the column face to the footing edge.

8.1 Cantilever Projection

For a square column (a × a) on a square footing (L × L):

L′ = (L − a) / 2

8.2 Bending Moment at Column Face

Mu = qu × B × (L′)² / 2

For a square footing, B = L; the same moment applies in both directions (the footing is symmetric)

8.3 Required Area of Steel

Using the IS 456 Limit State design formula for a singly reinforced section (same as slab/beam):

Mu = 0.87 fy Ast d [ 1 − (Ast fy) / (B d fck) ]

Solve for Ast (quadratic in Ast)

Or use the design aid: Ast = Mu / (0.87 fy × 0.90 d) as a first estimate (assuming lever arm ≈ 0.90d)

8.4 Minimum Steel

Ast,min = 0.12 % of B × D (total depth)   for Fe 415 (same as slab: IS 456 Cl. 26.5.2.1)

Ast,min = 0.15 % of B × D   for Fe 250 (mild steel)

8.5 Bar Spacing

The footing reinforcement is provided as a mesh (two-way grid) in both directions. For square footings, the same Ast applies in both directions. For rectangular footings, IS 456 Cl. 34.3 requires that the closer-direction (shorter span B) receives a band of steel in the central B-width plus uniformly distributed steel in the outer portions.

9. Step 6 — Development Length Check (IS 456 Cl. 26.2.1)

The footing bars are in tension (bottom of the cantilever slab). The available length from the column face to the end of the bar is:

Lavailable = L′ − end cover

End cover = 50 mm (for footings in contact with soil, IS 456 Table 16)

Required: Lavailable ≥ Ld = 47φ (Fe 415, M20, deformed, tension)

Steel GradeConcrete GradeLd / φ
Fe 415M2047
Fe 415M2540
Fe 500M2057
Fe 500M2549

If the available length is less than Ld, either increase the footing projection (increase L) or reduce the bar diameter so that Ld = 47φ becomes smaller. Hooks may also be provided, though this is less common in footings.

Cover in Footings

IS 456 Table 16 specifies a nominal cover of 50 mm for reinforcement in footings cast against and permanently in contact with earth. If a plain concrete levelling course (PCC blinding layer) is provided beneath the footing, the cover may be reduced to 40 mm. Always use 50 mm unless PCC is explicitly stated.

10. Complete Design Procedure Summary

  1. Compute plan area: A = 1.10 P / SBC; choose L × B dimensions.
  2. Compute factored net pressure: qu = Pu / (L × B); Pu = 1.5 P.
  3. Determine depth from punching shear: Assume d, compute Vu,punch, check τv,punch ≤ τc,punch. Iterate until satisfied.
  4. Adopt total depth: D = d + cover + φ/2 (with cover = 50 mm for soil contact). Round D up to next 50 mm.
  5. Recompute effective depth: d = D − 50 − φ/2.
  6. Check one-way shear: τv = Vu,1way / (B × d) ≤ τc (Table 19). If not satisfied, increase D.
  7. Design flexural steel: Compute Mu at column face; solve for Ast. Check Ast ≥ Ast,min.
  8. Check development length: Lavailable = L′ − 50 ≥ Ld = 47φ (Fe 415, M20).
  9. Report and sketch: Plan and section views with all dimensions, bar spacing, and cover.

11. Worked Examples

Example 1 — Complete Design of Square Isolated Footing (M20 / Fe 415)

Problem: Design a square isolated footing for a 300 × 300 mm column carrying a service axial load of 900 kN. SBC of soil = 200 kN/m². Use M20 concrete and Fe 415 steel.

Step 1: Plan Size

Total load = 1.10 × 900 = 990 kN
Required area = 990 / 200 = 4.95 m²
Side = √4.95 = 2.225 m → Adopt L = B = 2.30 m (2300 × 2300 mm)

Step 2: Factored Net Pressure

Pu = 1.5 × 900 = 1350 kN = 1350 × 10³ N
A = 2300 × 2300 = 5.29 × 10⁶ mm²
qu = 1350 × 10³ / 5.29 × 10⁶ = 0.2552 N/mm²

Step 3: Depth from Punching Shear

τc,punch = 0.25 √20 = 1.118 N/mm²

Trial d = 400 mm:
Vu,punch = Pu − qu(a+d)² = 1350000 − 0.2552 × (300+400)² = 1350000 − 0.2552 × 490000 = 1350000 − 125 000 = 1 225 000 N
bo = 4(300+400) = 2800 mm
τv,punch = 1225000 / (2800 × 400) = 1225000 / 1120000 = 1.094 N/mm² < 1.118 ✓

d = 400 mm is adequate for punching shear.

Step 4: Total Depth

Assuming 16 mm dia bars, cover = 50 mm:
D = 400 + 50 + 8 = 458 mm → Adopt D = 460 mm
Revised d = 460 − 50 − 8 = 402 mm

Step 5: One-Way Shear Check

Critical section at d = 402 mm from column face:
x1 = (2300−300)/2 − 402 = 1000 − 402 = 598 mm
Vu,1way = 0.2552 × 2300 × 598 = 0.2552 × 1375400 = 351 200 N
τv = 351200 / (2300 × 402) = 351200 / 924600 = 0.380 N/mm²

Need τc ≥ 0.380 N/mm². From IS 456 Table 19 for M20, pt = 0.25 % gives τc = 0.36 N/mm² (slightly less). At pt = 0.30 % (interpolated) τc ≈ 0.39 N/mm² > 0.380 ✓. Verify after steel design.

Step 6: Bending Moment & Steel

L′ = (2300 − 300)/2 = 1000 mm
Mu = qu × B × (L′)² / 2 = 0.2552 × 2300 × 1000² / 2 = 0.2552 × 2300 × 500 000 = 293 480 000 N·mm = 293.5 kN·m
Ast (estimate) = Mu / (0.87 fy × 0.90 d) = 293.5 × 10⁶ / (0.87 × 415 × 0.90 × 402)
= 293.5 × 10⁶ / 130 217 = 2254 mm²
Ast,min = 0.12 % × 2300 × 460 / 100 = 1270 mm² < 2254 ✓

Provide 12 – 16 mm dia bars: Ast = 12 × 201.1 = 2413 mm² > 2254 ✓
Spacing = (2300 − 2 × 50) / (12 − 1) = 2200/11 = 200 mm c/c

pt Verification for One-Way Shear

pt = 100 × 2413 / (2300 × 402) = 241300 / 924600 = 0.261 %
From Table 19 (M20), at pt = 0.261 %: τc ≈ 0.37 N/mm² > τv = 0.380 N/mm²

Marginally close. Increase to 13 – 16 mm dia: Ast = 2614 mm²; pt = 0.283 %; τc ≈ 0.38 N/mm² ≈ τv. Alternatively increase d by 10 mm (adopt D = 470 mm, d = 412 mm), which reduces τv further. Adopt D = 470 mm, d = 412 mm, 12 – 16 mm @ 200 mm c/c (both ways).

Step 7: Development Length

Ld = 47 × 16 = 752 mm
Lavailable = L′ − 50 = 1000 − 50 = 950 mm > 752 mm ✓

Result Summary

Footing size2300 × 2300 mm (plan)
Total depth D470 mm
Effective depth d412 mm
Bottom reinforcement (both ways)12 – 16 mm dia @ 200 mm c/c
Cover (to main bar)50 mm
Net upward pressure qu0.2552 N/mm²

Example 2 — Footing for a Circular Column (M25 / Fe 415)

Problem: A 400 mm diameter circular column carries a service load of 1200 kN. SBC = 250 kN/m². Design a square footing using M25 and Fe 415.

Equivalent Square Column

For punching shear, convert the circular column to an equivalent square with the same area:

Area of circular column = π/4 × 400² = 125 664 mm²
Equivalent side a = √125664 = 354 mm → Use a = 354 mm

Plan Size

A = 1.10 × 1200 / 250 = 5.28 m²; L = √5.28 = 2.30 m → Adopt 2300 × 2300 mm

Factored Pressure

qu = 1.5 × 1200 × 10³ / (2300²) = 1800000 / 5290000 = 0.340 N/mm²

Depth from Punching Shear

τc,punch = 0.25√25 = 1.25 N/mm²

Trial d = 400 mm:
Vu,punch = 1800000 − 0.340 × (354+400)² = 1800000 − 0.340 × 568516 = 1800000 − 193 295 = 1 606 705 N
bo = 4(354+400) = 3016 mm
τv = 1606705 / (3016 × 400) = 1606705 / 1206400 = 1.332 N/mm² > 1.25 ✗
Trial d = 450 mm:
Vu,punch = 1800000 − 0.340 × (354+450)² = 1800000 − 0.340 × 646416 = 1800000 − 219 781 = 1 580 219 N
bo = 4(354+450) = 3216 mm
τv = 1580219 / (3216 × 450) = 1580219 / 1447200 = 1.092 N/mm² < 1.25 ✓

Adopt d = 450 mm → D = 450 + 50 + 8 = 508 mm → D = 510 mm, d = 452 mm

Bending Moment & Steel

L′ = (2300 − 354)/2 = 973 mm
Mu = 0.340 × 2300 × 973² / 2 = 0.340 × 2300 × 473 264.5 / 2 = 370 000 000 N·mm = 370 kN·m
Ast = 370 × 10⁶ / (0.87 × 415 × 0.90 × 452) = 370 × 10⁶ / 146 500 = 2527 mm²

Provide 13 – 16 mm dia @ 180 mm c/c: Ast = 2614 mm² ✓ (both ways)

Development Length

Ld = 40φ = 40 × 16 = 640 mm (M25, Fe 415)
Lavailable = 973 − 50 = 923 mm > 640 ✓

Example 3 — GATE-Style: Find Safe Load Given Footing Data

Problem (GATE CE type): A 2.5 × 2.5 m square isolated footing has a total depth of 500 mm and carries a 350 × 350 mm column. Concrete M20, Fe 415. The footing has 14 – 16 mm dia bars each way. Check whether the footing can safely carry a service load of 1000 kN from the column. SBC = 180 kN/m².

Check 1: SBC

Area provided = 2500 × 2500 = 6.25 m²
Soil pressure = 1.10 × 1000 / 6.25 = 176 kN/m² < 180 kN/m² ✓

Check 2: Punching Shear

d = 500 − 50 − 8 = 442 mm
qu = 1.5 × 1000 × 10³ / 2500² = 1500000 / 6250000 = 0.240 N/mm²
Vpunch = 1500000 − 0.240 × (350+442)² = 1500000 − 0.240 × 627 264 = 1500000 − 150 543 = 1 349 457 N
bo = 4(350+442) = 3168 mm
τv = 1349457 / (3168 × 442) = 1349457 / 1400256 = 0.964 N/mm² < 1.118 ✓

Check 3: One-Way Shear

x1 = (2500−350)/2 − 442 = 1075 − 442 = 633 mm
Vu,1way = 0.240 × 2500 × 633 = 379 800 N
τv = 379800 / (2500 × 442) = 379800 / 1105000 = 0.344 N/mm²
Ast = 14 × 201.1 = 2815 mm²; pt = 100 × 2815/(2500 × 442) = 0.254 %
τc (M20, 0.25 %) = 0.36 N/mm² > 0.344 ✓

Conclusion

The footing is adequate for all checks at 1000 kN service load. ✓

12. Common Mistakes

Mistake 1: Using Factored Load to Size the Plan Area

What happens: Students multiply P by 1.5 and then divide by SBC to get the plan area. This overestimates the required size by 50 %, producing an unnecessarily large and expensive footing.

Root cause: Confusion between geotechnical (service) and structural (factored) limit states. SBC is a soil property determined from service-level settlements and shear failure — it is not a factored value. Only structural checks use Pu.

Fix: Plan area = (service load + self-weight allowance) / SBC. Factored load Pu = 1.5 P is used only for qu, shear, and bending calculations.

Mistake 2: Locating the Critical Punching Shear Perimeter Incorrectly

What happens: Some students place the punching perimeter at d from the column face (instead of d/2), significantly underestimating the punching shear force and producing an unsafe footing.

Root cause: Mixing up the one-way shear critical section (d from face) with the two-way shear critical section (d/2 from face). IS 456 Cl. 31.6.1 and 34.2.4 are different clauses with different rules.

Fix: Two-way (punching) shear perimeter is at d/2 from column face; one-way shear critical section is at d from column face.

Mistake 3: Computing Bending Moment at the Column Centreline Instead of the Face

What happens: Using the full half-span (L/2) as the cantilever instead of L′ = (L − a)/2. This overestimates Mu by including the moment inside the column footprint, which does not exist as a bending effect in the footing slab.

Root cause: IS 456 Cl. 34.2.3 specifically states the critical section for bending is at the face of the column. The soil pressure under the column itself transfers directly as compression into the column and does not create bending in the slab.

Fix: Always use L′ = (L − a)/2, not L/2, as the cantilever span for the moment calculation.

Mistake 4: Neglecting to Check Development Length

What happens: Steel is designed and sized correctly but the check that Lavailable ≥ Ld is skipped. If the footing projection is short (e.g., close-spaced columns) or large bars are used, the bars may not develop their full yield strength, leading to anchorage failure.

Root cause: Development length is checked as the final step and is sometimes treated as an afterthought. In GATE problems it is a common trap question.

Fix: After selecting bar diameter, always verify Lavailable = L′ − 50 ≥ 47φ (Fe 415/M20). If insufficient, reduce φ (use more, smaller bars) or increase L.

Mistake 5: Applying the Wrong Cover Value

What happens: Using 25 mm or 40 mm cover in footing calculations instead of the 50 mm required for members cast directly against earth (IS 456 Table 16). This reduces the effective depth d and may cause the shear or bending capacity to be underestimated.

Root cause: Cover rules for beams (25–30 mm) are habitually applied to footings. Footings have higher cover requirements because the reinforcement is permanently in contact with moist soil.

Fix: For footings cast against earth without a PCC blinding layer: cover = 50 mm. With a PCC blinding layer below: cover may be reduced to 40 mm (verify with IS 456 Table 16 footnote).

13. Frequently Asked Questions

Q1. Why is the net upward pressure calculated from the factored column load, while the plan area is from the service load?

The plan area is sized to satisfy a geotechnical limit — the Safe Bearing Capacity of soil — which is derived from service-level tests (plate load tests, SPT/CPT correlations) and represents the pressure at which settlements become excessive or shear failure is imminent under working conditions. It is not a factored value, so the service load is used. Once the area is fixed, the structural design of the footing slab (steel, depth) is done using the Limit State Method, which requires factored loads. The net pressure qu = Pu/A is therefore a factored stress used purely for the structural design of the concrete section, not for soil bearing. The “net” qualifier means the self-weight of the footing and overburden are excluded — they cause equal and opposite stresses in the soil and the slab, and cancel out.

Q2. What is the difference between punching shear and one-way shear failure, and which is usually more critical?

Punching shear is a localised failure on a truncated pyramid around the column; the slab punches through the footing like a cookie cutter. It is checked on the perimeter at d/2 from the column face. One-way shear is a wide-beam failure across the full width of the footing, similar to a beam shear crack; checked at d from the column face. For square footings on square columns, punching shear is almost always the more critical (governing) check because the failure perimeter is small relative to the large shear area in one-way shear. For rectangular footings with a high aspect ratio (L/B > 2), one-way shear in the short direction can sometimes govern.

Q3. Is it necessary to provide shear reinforcement (stirrups) in isolated footings?

IS 456 does not prohibit shear reinforcement in footings, but the standard practice (and the preferred approach in IS 456 design) is to increase the depth d until τv ≤ τc without any shear steel. This avoids the practical difficulty of bending and placing stirrups in a heavily congested bottom cage and ensures that the shear resistance comes from the concrete section rather than from steel that could corrode near the soil face. Shear reinforcement may become necessary in very shallow footings or when headroom constraints prevent increasing depth, but it requires careful detailing of anchorage within the limited slab depth.

Q4. How is an isolated footing designed differently when the column also carries a moment (eccentric loading)?

When the column transfers both axial load P and moment M to the footing, the soil pressure is no longer uniform. The pressure varies linearly: q = P/A ± M×y/I, where y is the distance from the centroid and I is the second moment of the plan area. For the footing to remain in full contact with the soil (no tension), the resultant must lie within the middle third of the base (eccentricity e = M/P ≤ L/6). If e > L/6, the footing area is increased, typically by shifting the footing centroid toward the more heavily loaded side, or a combined footing is used. The design for shear and bending must then use the maximum pressure (qmax) at the critical sections rather than the uniform qu, making calculations more involved.

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