Shear & Torsion in RCC Members Notes

Last reviewed 16 Sept 2026 · 7 min read

How RCC beams fail in shear

Near supports, shear force is high. Shear and bending combine to produce principal tensile stresses inclined at about 45°. When these exceed concrete's tensile strength, diagonal tension cracks form. Without web reinforcement, a beam can fail suddenly by:

diagonal tension (inclined crack running to the compression face),
shear compression (crushing of the compression zone above the crack),
shear bond / splitting along the tension bars.

Shear failures are brittle, so IS 456 always provides a minimum of shear reinforcement in beams.

Shear is resisted by: concrete in the compression zone, aggregate interlock across cracks, dowel action of longitudinal bars, and stirrups or bent-up bars crossing the cracks.

Nominal shear stress

τ_{v} = \frac{V _{u}}{b d}

For members of varying depth: $τ_{v} = \frac{V _{u} \pm \frac{M _{u}}{d} tan β}{b d}$ — the negative sign when the bending moment increases numerically in the same direction as the depth increases.

Design shear strength of concrete, τc

$τ_{c}$ depends on concrete grade and the tension steel percentage $p_{t} = 100 A_{s t} / (b d)$ at the section (bars continuing at least $d$ beyond it).

Code ProvisionIS 456 Table 19 — τc (N/mm²), selected values

$p_{t}$ (%)	M20	M25	M30
≤ 0.15	0.28	0.29	0.29
0.25	0.36	0.36	0.37
0.50	0.48	0.49	0.50
0.75	0.56	0.57	0.59
1.00	0.62	0.64	0.66
1.25	0.67	0.70	0.71
1.50	0.72	0.74	0.76
2.00	0.79	0.82	0.84
≥ 3.00	0.82	0.92	0.96

Interpolate linearly between values.

Code ProvisionIS 456 Table 20 — maximum shear stress τc,max (N/mm²)

M15	M20	M25	M30	M35	M40 and above
2.5	2.8	3.1	3.5	3.7	4.0

If $τ_{v} > τ_{c, ma x}$ , the section must be enlarged — shear reinforcement cannot help.

Design procedure for shear

$τ_{v} = V_{u} / b d$ at the critical section.
If $τ_{v} > τ_{c, ma x}$ → revise section.
If $τ_{v} \leq τ_{c}$ → provide minimum shear reinforcement (except in minor members such as lintels, where $τ_{v} < 0.5 τ_{c}$ ).
If $τ_{c} < τ_{v} \leq τ_{c, ma x}$ → design shear reinforcement for $V_{u s} = V_{u} - τ_{c} b d$ .

FormulaShear reinforcement (IS 456 cl. 40.4)

Vertical stirrups:

V_{u s} = \frac{0.87 f _{y} A _{s v} d}{s _{v}} \Rightarrow s_{v} = \frac{0.87 f _{y} A _{s v} d}{V _{u s}}

Inclined stirrups at angle $α$ : $V_{u s} = \frac{0.87 f _{y} A _{s v} d}{s _{v}} (sin α + cos α)$

Single bent-up bar (or group at one section): $V_{u s} = 0.87 f_{y} A_{s v} sin α$ — bent-up bars may resist not more than half the total shear reinforcement demand.

$A_{s v}$ = total area of stirrup legs (a 2-legged 8 mm stirrup gives $2 \times 50.3 = 100.5$ mm²).

Code ProvisionMinimum shear reinforcement and spacing

Minimum: $\frac{A _{s v}}{b s _{v}} \geq \frac{0.4}{0.87 f _{y}}$ → $s_{v} \leq \frac{0.87 f _{y} A _{s v}}{0.4 b}$
Maximum spacing of vertical stirrups: 0.75d or 300 mm, whichever is less.
For inclined stirrups at 45°: spacing not more than $d$ (and 300 mm).
Where $τ_{v} > τ_{c}$ , stirrups must also satisfy the design spacing above.

Enhanced shear strength near supports

Where a support reaction causes compression in the region and the section is within $2 d$ of the support face, IS 456 allows the design shear strength to be increased to

τ_{c}^{'} = \frac{2 d τ _{c}}{a _{v}} (not exceeding τ_{c, ma x})

where $a_{v}$ is the distance of the section from the support face. The critical section for shear in beams with such supports is commonly taken at distance $d$ from the support face.

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Shear & Torsion in RCC Members

How RCC beams fail in shear

Nominal shear stress

Design shear strength of concrete, τc

Design procedure for shear

Enhanced shear strength near supports

This chapter is in the syllabus of

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