Shear safety check of an RCC beam carrying UDL: A simply supported RCC beam of span 6 m is 30 cm wide. The lever arm (j·d) is 55 cm. It carries a uniformly distributed load of 12 t/m (including self-weight). If the allowable shear stress in concrete is 5 kg/cm², assess the adequacy in shear and choose the correct action.

Introduction / Context:
Shear design in RCC beams requires checking the nominal shear stress against permissible limits. If the nominal shear exceeds the maximum allowable shear capacity of concrete (even with shear reinforcement), the section must be revised (e.g., increased depth or width) for safety.

Given Data / Assumptions:

• Beam span L = 6 m (simply supported).
• Uniformly distributed load w = 12 t/m = 12,000 kg/m (includes self-weight).
• Width b = 30 cm; lever arm j·d = 55 cm (given directly).
• Allowable shear stress in concrete τ_allow = 5 kg/cm² (working stress context).

Concept / Approach:
For a simply supported beam under UDL, the maximum shear force occurs at the supports and equals V = wL/2. The nominal shear stress is approximated as τ_v = V / (b * j·d) when the effective lever arm j·d is given. Compare τ_v with the allowable value. If τ_v far exceeds code limits (and likely the maximum permissible with reinforcement), the section must be revised.

Step-by-Step Solution:
Compute support shear: V = wL/2 = 12,000*6/2 = 36,000 kg.Nominal shear stress: τ_v = V / (b * j·d) = 36,000 / (30 * 55) = 36,000 / 1,650 = 21.82 kg/cm² (approx.).Compare with allowable: 21.82 kg/cm² » 5 kg/cm² → greatly exceeds limit.Conclusion: Even with shear reinforcement, such a high nominal shear typically surpasses maximum permissible limits; revise section.

Verification / Alternative check:
In classic working stress design tables, maximum shear stresses that concrete can sustain (even with stirrups) are much lower than ~22 kg/cm². Increasing depth (thus j·d and b·j·d) significantly reduces τ_v, offering a viable revision path.

Why Other Options Are Wrong:

• Safe in shear / with stirrups / with stirrups and inclined bars: The computed τ_v is far beyond permissible; these are inadequate remedies for such an exceedance.

Common Pitfalls:
Using d instead of j·d; forgetting to convert t/m to kg/m; checking shear at midspan instead of at supports; assuming stirrups can compensate unlimitedly for undersized sections.

Final Answer:
needs revision of section