Bending strength for equal-area beams under the same bending moment A square beam and a circular beam have the same cross-sectional area and are subjected to the same bending moment. Based on section modulus and resulting extreme fiber stress, which statement about economy/strength is correct (assuming identical material and allowable stress)?

Introduction / Context:When two beams are made of the same material and carry the same bending moment, the bending stress at the extreme fiber is inversely proportional to the section modulus. For a fixed cross-sectional area, the shape that yields a larger section modulus produces lower bending stress and is therefore stronger and more economical in bending design.

Given Data / Assumptions:

• Same material and same allowable stress for both beams.
• Square beam side = a; circular beam diameter = d.
• Equal area condition: a^2 = (π d^2)/4.
• Pure bending; linear elastic behavior; no shear effects considered.

Concept / Approach:

Bending stress relation: σ = M / Z, where Z = I / c is the section modulus. For equal area, the beam with the larger Z has smaller σ for the same moment M, hence is stronger/more economical. Compute Z for square and circle under the equal-area constraint and compare magnitudes.

Step-by-Step Solution:

Verification / Alternative check:

Numerical example with a = 1 unit gives Z_square ≈ 0.167 and Z_circle ≈ 0.141. Since σ ∝ 1/Z, the square shows about 18% lower extreme fiber stress, confirming superior bending efficiency for equal area.

Why Other Options Are Wrong:

• Circular beam is more economical: False; its section modulus is smaller for equal area.
• Both equally strong/economical: False; Z differs noticeably.
• None of these: Incorrect because one shape is clearly better in bending for equal area.

Common Pitfalls:

• Comparing second moment I alone; the correct comparison uses section modulus Z = I/c.
• Forgetting to enforce equal area before comparing Z values.

Final Answer:

square beam is more economical.