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Home›IS Codes›IS 456:2000›Clauses›Cl. 40.1
IS 456:2000 — Plain and Reinforced Concrete — Code of Practice
IS 456:2000 — Clause 40.1

Nominal Shear Stress

Clause 40.1 defines the nominal shear stress τv as Vu/(b·d) for beams. This is compared against the design shear strength of concrete τc (Table 19), which depends on the tension steel percentage and concrete grade. If τv > τc, shear reinforcement is required. If τv > τc,max (Table 20), the section must be redesigned.

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Key Requirements

  • •Nominal shear stress τv = Vu / (b × d) — must not exceed τc,max from Table 20
  • •If τv ≤ τc/2: no shear reinforcement needed (but minimum stirrups still required)
  • •If τc/2 < τv ≤ τc: provide minimum shear reinforcement (Clause 26.5.1.6)
  • •If τc < τv ≤ τc,max: design shear reinforcement for excess shear (τv − τc)
  • •If τv > τc,max: section is inadequate — increase depth or concrete grade
  • •Minimum shear reinforcement: Asv/(b·sv) ≥ 0.4/0.87fy (Clause 26.5.1.6)

Reference Tables

Table 19 — Design Shear Strength of Concrete τc (N/mm²)
100Ast/(b·d)M20M25M30M35M40+
≤ 0.150.280.290.290.290.3
0.250.360.360.370.370.38
0.500.480.490.50.50.51
0.750.560.570.590.590.6
1.000.620.640.660.670.68
1.250.670.70.710.730.74
1.500.720.740.760.780.79
1.750.750.780.80.820.84
2.000.790.820.840.860.88
2.250.810.850.880.90.92
2.500.820.880.910.930.95
2.750.820.90.940.960.98
≥ 3.000.820.920.960.991.01
For pt% between tabulated values, interpolate linearly. For pt > 3%, use τc at 3%.
Table 20 — Maximum Shear Stress τc,max (N/mm²)
GradeM20M25M30M35M40
τc,max2.83.13.53.74

Formulas

τv = Vu / (b × d)
Nominal shear stress
τv = Nominal shear stress (N/mm²)Vu = Factored shear force at the section (N)b = Breadth of member (mm)d = Effective depth (mm)

Practical Notes

✓For a typical 230 × 450 beam (d=410) with 2 nos 16 mm bars (Ast=402 mm²): pt = 0.43%. In M25: τc ≈ 0.46 N/mm². Maximum Vu without shear steel = 0.46 × 230 × 410 = 43.4 kN.
✓Critical section for shear is at distance 'd' from the face of support (Clause 22.6.2.1). Shear force at the face of support is not used for design.
✓IS 13920 (seismic) overrides: shear design must account for plastic hinge capacity at beam ends, not just analysis forces.

Common Mistakes

⚠Using τv at the face of support instead of at 'd' from face — the critical section for shear is at distance 'd', reducing the design shear force.
⚠Forgetting minimum shear reinforcement — even when τv < τc, minimum stirrups are always required in beams (Clause 26.5.1.6).
⚠Not interpolating Table 19 for intermediate pt% values — jumping to the next higher row overestimates τc.

Frequently Asked Questions

Related Resources

Cl. 40.4Cl. 38.1Cl. 26.5Punching Shear Check (Two-Way Shear)Stirrup & Tie Spacing RulesRcc Design
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Cl. 39.3
Limit State of Collapse — Short Column Under Axial Load
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Cl. 40.4
Design of Shear Reinforcement
View all 15 clauses of IS 456:2000 →