Uniform-strength beam (constant depth d) under a central load – required width as a function of distance x A simply supported beam of span L carries a central point load W. If the beam is to be of uniform strength while keeping depth d constant, what should the width b(x) be at a distance x from a support (0 ≤ x ≤ L/2)?

Introduction / Context:
Shaping a beam for uniform strength keeps the extreme-fibre bending stress constant along the span. For a rectangular beam with constant depth, the width must vary to match the bending moment diagram.

Given Data / Assumptions:

• S.S. beam, span L, central point load W.
• Rectangular section, constant depth d.
• Allowable bending stress = σ_allow.
• Elastic behavior; small deflection theory.

Concept / Approach:
For 0 ≤ x ≤ L/2, bending moment is M(x) = R × x = (W/2) × x. For a rectangular section, Z = b d^2 / 6. Uniform strength requires σ_allow = M / Z = 6M / (b d^2) → b(x) = 6M / (σ_allow d^2).

Step-by-Step Solution:
M(x) = (W/2) x.b(x) = 6 × [(W/2) x] / (σ_allow d^2) = (3 W x) / (σ_allow d^2).At midspan (x = L/2), b_max = (3 W L) / (2 σ_allow d^2).

Verification / Alternative check:
The required b(x) is linear in x (triangular distribution), matching the triangular moment diagram for a central point load.

Why Other Options Are Wrong:

• (b) Double counts; would imply σ varying with x.
• (c) Wrong dimensions and not a function of x.
• (d) Incorrectly suggests constant width.
• (e) Quadratic in x, inconsistent with linear M(x).

Common Pitfalls:
Mixing up allowable stress with material strength; always use design (allowable) stress for sizing.

Final Answer:
b(x) = (3 W x) / (σ_allow d^2)