Shear safety check for a simply supported RCC beam: A beam of width b = 25 cm, clear span = 5 m carries a uniformly distributed load w = 2000 kg/m (inclusive of self-weight). If the lever arm j·d = 45 cm, comment on shear safety near the supports based on nominal shear stress.

Introduction / Context:
Near beam supports, the shear force is maximum. Checking nominal shear stress against permissible values helps decide whether the section is adequate in shear or whether shear reinforcement or section revision is required.

Given Data / Assumptions:

• Width b = 25 cm.
• Clear span L = 5 m; simply supported; UDL w = 2000 kg/m (includes self-weight).
• Lever arm j·d = 45 cm (effective shear area taken as b * j * d).
• Typical permissible concrete shear stress τ_c for common RC percentages is higher than about 4.5 kg/cm^2 (varies with % steel).

Concept / Approach:
Support shear V = w * L / 2. Nominal shear stress τ_v = V / (b * j * d). Compare τ_v with τ_c (from code tables depending on percentage of steel). If τ_v ≤ τ_c, concrete alone is adequate in shear (nominal links still provided).

Step-by-Step Solution:
Compute shear: V = 2000 * 5 / 2 = 5000 kgf.Effective area: A_v = b * j * d = 25 * 45 = 1125 cm^2.Nominal shear stress: τ_v = 5000 / 1125 ≈ 4.44 kg/cm^2.Compare with typical τ_c (generally ≥ about 5 kg/cm^2 for usual reinforcement ratios): τ_v ≤ τ_c → section safe in shear.

Verification / Alternative check:
Using b * d instead of b * j * d gives an even lower τ_v; thus the conclusion remains conservative that the section is safe in shear.

Why Other Options Are Wrong:
is safe with stirrups / with inclined members: Unnecessary here since τ_v does not exceed τ_c.needs revision: Not warranted by the calculated τ_v.shear cannot be assessed: All required inputs are available.

Common Pitfalls:

• Forgetting that τ_c depends on percentage tension steel; always consult the relevant code table.
• Mixing units; keep kgf and cm consistent.

Final Answer:
safe in shear