A simply supported rectangular beam 10 cm × 18 cm with span 400 cm carries a uniformly distributed load that induces a maximum bending stress of 100 kg/cm². What is the intensity of the load per metre length?

Introduction / Context:
This problem links bending stress limits to allowable loading on a simply supported beam under a uniformly distributed load (UDL). It tests the use of bending stress formula with section properties and the maximum moment under UDL.

Given Data / Assumptions:

• Section: b = 10 cm, h = 18 cm.
• Span: L = 400 cm (4 m).
• Max bending stress σ_max = 100 kg/cm².
• Loading: UDL of intensity w (kg/cm) along the span; simply supported.

Concept / Approach:
UseM_max(UDL) = w * L^2 / 8and the flexure formulaσ_max = M_max * c / Iwith c = h/2 and I = b * h^3 / 12.

Step-by-Step Solution:
I = b h^3 / 12 = 10 * 18^3 / 12 = 4860 cm^4.c = h/2 = 9 cm.M_max = σ_max * I / c = 100 * 4860 / 9 = 54000 kg·cm.Equate to UDL moment: w L^2 / 8 = 54000 → w = 54000 * 8 / 400^2 = 432000 / 160000 = 2.7 kg/cm.Convert to per metre: w_per_m = 2.7 * 100 = 270 kg/m.

Verification / Alternative check:
Back-calculate σ using w = 270 kg/m to confirm M_max and σ_max match 100 kg/cm².

Why Other Options Are Wrong:

• 240, 250, 260, 280 kg/m do not reproduce σ_max = 100 kg/cm² for the given geometry and span.

Common Pitfalls:

• Forgetting to convert w between kg/cm and kg/m consistently.
• Using c = h instead of h/2 in the flexure formula.

Final Answer:
270 kg/m