Shear Reinforcement (Stirrup) Design in RCC Beams as per IS 456

Beams carry shear force as well as bending moment. Concrete alone can resist some shear, but beyond that stirrups (shear reinforcement) are needed to stop diagonal tension cracks from splitting the beam. Here is the IS 456 limit state method, step by step.

Step 1: Nominal shear stress

τv = Vu / (b d)

where V_u is the factored shear force, b the width and d the effective depth.

Step 2: Compare with τ_c,max (Table 20)

If τ_v > τ_c,max, no amount of stirrups is enough — increase the section size.

Step 3: Design shear strength of concrete τ_c (Table 19)

τ_c depends on the tension steel percentage p_t = 100 A_st / (b d) and the grade of concrete. A few values:

Interpolate linearly for in-between values.

Step 4: Decide what to provide

• τ_v < τ_c: provide only minimum shear reinforcement: A_sv / (b s_v) ≥ 0.4 / (0.87 f_y).
• τ_c < τ_v < τ_c,max: design stirrups for the extra shear V_us = V_u − τ_c b d.

Spacing of vertical stirrups: sv = 0.87 fy Asv d / Vus

Maximum spacing: 0.75d or 300 mm, whichever is less.

Worked example

A beam 230 × 450 mm, effective depth 410 mm, M20 concrete, Fe 500 steel, tension steel 3 nos. 16 mm (603 mm²), factored shear V_u = 110 kN. Design 2-legged 8 mm stirrups.

IS 456 (clause 40.4) does not allow f_y greater than 415 N/mm² to be used for shear steel, so even with Fe 500 bars we use f_y = 415 in the stirrup formulas.

1. τ_v = 110 × 10³ / (230 × 410) = 1.17 N/mm², which is less than τ_c,max = 2.8 N/mm² — section is adequate.
2. p_t = 100 × 603 / (230 × 410) = 0.64%. From Table 19 (M20), τ_c = 0.48 + (0.64 − 0.50)/0.25 × 0.08 = 0.52 N/mm².
3. τ_v > τ_c, so design stirrups. V_c = 0.52 × 230 × 410 = 49.5 kN, so V_us = 110 − 49.5 = 60.5 kN.
4. A_sv for 2 legs of 8 mm = 2 × 50.3 = 100.5 mm².
5. s_v = 0.87 × 415 × 100.5 × 410 / (60.5 × 10³) = 246 mm.
6. Check limits: 0.75d = 307 mm, 300 mm, and the minimum-shear spacing 0.87 × 415 × 100.5 / (0.4 × 230) = 394 mm. The least is 246 mm.
7. Provide 2-legged 8 mm stirrups at 240 mm c/c near the supports, where the shear is highest.

Practical points

• Shear is highest near supports, so stirrups are often closer there and wider at mid-span (but never beyond the maximum spacing).
• The critical section for shear is usually taken at distance d from the face of the support.
• In earthquake zones, IS 13920 requires closer stirrups near beam ends (not more than d/4, 6 × the smallest bar diameter or 100 mm, over a length of 2d) with 135° hooks.