RCC Beam: Flexural Capacity and Shear Reinforcement Dispute
A 300 mm wide × 500 mm deep simply supported RCC beam of clear span 6.0 m carries a factored uniformly distributed load. Analysis gives a factored design moment of 175 kN·m at mid-span and a factored shear of 220 kN at the support face. Concrete is M25 and steel is Fe415. Tension reinforcement is 4 bars of 20 mm diameter (1256 mm²) and compression reinforcement is 2 bars of 16 mm diameter (402 mm²), both with 40 mm effective cover, giving an effective depth of 460 mm. The site engineer has scheduled two-legged 8 mm Fe415 stirrups at 150 mm centre-to-centre throughout the span. The contractor now proposes increasing the stirrup spacing to 250 mm, arguing that the beam is lightly loaded and that stirrup steel is a major cost item. The checking engineer must decide whether the bar schedule as issued is safe, and whether the contractor's proposal can be entertained at all.
1. Establish whether the section is under-reinforced and compute its ultimate moment of resistance. Is the beam safe in flexure against the factored moment?
Read the worked answer
Force equilibrium (first assume compression steel yields):
T = 0.87 f_y A_st = 0.87 × 415 × 1256 = 453,455 N
C_sc = 0.87 × 415 × 402 = 145,142 N
C_c = 453,455 − 145,142 = 308,313 N → x_u = 308,313 / (0.36 × 25 × 300) = 114.2 mm
x_u/d = 114.2/460 = 0.248, against x_u,max/d = 0.48 for Fe415. The section is under-reinforced; steel yields first and failure will be ductile.
Verify the assumption: ε_sc = 0.0035 (x_u − d′)/x_u = 0.0035 (114.2 − 40)/114.2 = 0.00227, which exceeds ε_y = 0.87 × 415/(2 × 10⁵) = 0.00180. Compression steel does yield, so the assumption is valid.
Moment capacity:
M_u = 0.36 f_ck b x_u (d − 0.42 x_u) + f_sc A_sc (d − d′)
= 308,313 × (460 − 47.96) + 145,142 × 420
= 127.0 × 10⁶ + 61.0 × 10⁶ = 188.0 × 10⁶ N·mm = 188 kN·m
Since 188 kN·m > 175 kN·m, the beam is safe in flexure, but the reserve is only about 7%. That margin gives no room for a lower cover, a higher grade of steel substitution, or any future loading revision - worth recording in the check sheet.
2. Check the shear capacity of the beam with the scheduled 8 mm two-legged stirrups at 150 mm c/c. Is the schedule adequate against the 220 kN factored shear?
Read the worked answer
Nominal shear stress: τ_v = 220,000/(300 × 460) = 1.59 N/mm², well below τ_c,max = 3.1 N/mm² for M25, so the web will not crush. The deficiency, if any, is in stirrup steel.
Concrete contribution: p_t = 100 × 1256/(300 × 460) = 0.91%. Interpolating IS 456 Table 19 for M25 between p_t = 0.75% (0.64) and 1.00% (0.70) gives τ_c ≈ 0.68 N/mm².
V_c = 0.68 × 300 × 460 = 93,840 N
Stirrup contribution required: V_us = 220,000 − 93,840 = 126,160 N
For 2-legged 8 mm stirrups, A_sv = 100.5 mm²:
s_v = 0.87 f_y A_sv d / V_us = (0.87 × 415 × 100.5 × 460)/126,160 = 132 mm
At 150 mm c/c the stirrups supply only 16.69 × 10⁶/150 = 111,270 N, giving a total capacity of 93,840 + 111,270 = 205 kN < 220 kN. The schedule is inadequate by about 7%.
Adopt 8 mm two-legged at 125 mm c/c over the end regions (capacity 93,840 + 133,520 = 227 kN), and step to 150 mm beyond the section where V_u falls below 205 kN. Maximum spacing limits (0.75d = 345 mm and 300 mm) and minimum shear steel (s_v ≤ 0.87 × 415 × 100.5/(0.4 × 300) = 302 mm) are both satisfied, so spacing, not minimum steel, is the governing criterion.
3. The contractor argues that 250 mm spacing still satisfies the 'minimum stirrup' rule and should be accepted to save steel. Give your ruling with reasons.
Read the worked answer
The argument confuses two different requirements. The minimum shear reinforcement clause exists only to guard against sudden brittle shear failure in members where computed shear is low; it does not grant permission to fall below the shear strength required by analysis.
At 250 mm c/c: V_us = 16.69 × 10⁶/250 = 66,760 N; total capacity = 93,840 + 66,760 = 160.6 kN, against a demand of 220 kN. The member would be under-strength by 27% - a genuine collapse risk, not a code technicality. The minimum-steel rule (s_v ≤ 302 mm) and maximum-spacing rule (≤ 0.75d = 345 mm) are both met, which is exactly why they cannot be used as a proxy for adequacy.
Ruling: reject the proposal. Permitted alternatives, in order of preference: (i) keep 250 mm spacing only in the middle third where V_u is below 160 kN, with 125 mm c/c near supports for at least 1.5d; (ii) widen the web, which raises V_c and reduces τ_v; (iii) use 10 mm stirrups at wider spacing if handling is the real driver. Any of these should be issued as a revised bar bending schedule with a shear envelope drawn along the span, not as a uniform spacing.