Reinforced concrete beam under UDL: An RCC beam (width = 25 cm, effective depth = 50 cm) spans 6 m simply supported and carries a uniformly distributed load of 3000 kg/m inclusive of self-weight. If the lever arm constant j = 0.865, what is the maximum shear stress at the support?

Introduction / Context:
Design for shear in reinforced concrete relies on the shear force near supports and the effective resisting depth. Some textbooks compute nominal shear stress using the concrete's effective shear area b * j * d, where j is the lever-arm constant. This problem reinforces shear calculation and consistent unit handling for a simply supported beam under UDL.

Given Data / Assumptions:

• Span L = 6 m; UDL w = 3000 kgf/m (includes self-weight).
• Section: b = 25 cm, effective depth d = 50 cm.
• Lever arm constant j = 0.865, hence j * d = 43.25 cm.
• Simply supported beam; maximum shear occurs at supports.

Concept / Approach:
Support shear V for a simply supported beam with UDL is V = w * L / 2. Many RCC design approaches take nominal shear stress as tau_v = V / (b * j * d). Using j allows a slightly smaller effective shear area than b * d, giving a higher tau_v and a conservative check.

Step-by-Step Solution:
Compute support shear: V = w * L / 2 = 3000 * 6 / 2 = 9000 kgf.Compute effective area: b * j * d = 25 * (0.865 * 50) = 25 * 43.25 = 1081.25 cm^2.Nominal shear stress: tau_v = V / (b * j * d) = 9000 / 1081.25 ≈ 8.325 kg/cm^2.Round to two significant figures: ≈ 8.3 kg/cm^2.

Verification / Alternative check:
If one used b * d (i.e., 25 * 50 = 1250 cm^2), tau_v = 9000 / 1250 = 7.2 kg/cm^2. The j-factor approach is more conservative and aligns with the provided key option of 8.3 kg/cm^2.

Why Other Options Are Wrong:
7.6 or 11.4 kg/cm^2: Do not match consistent calculations using either b * d or b * j * d.21.5 kg/cm^2: Far too high for the given section and load.6.0 kg/cm^2: Underestimates shear stress.

Common Pitfalls:

• Mixing units (kgf, N) or forgetting to use j in the shear area when that convention is adopted.
• Computing midspan shear instead of support shear.

Final Answer:
8.3 kg/cm^2