Flexural strength comparison of two square beams: One beam has a square cross-section of side a oriented with sides horizontal; the other identical square is rotated so that a diagonal is vertical (diamond). What is the ratio of their flexural strengths (i.e., section modulus ratio Z_sides / Z_diagonal)?

Introduction / Context:Flexural strength under elastic bending is proportional to the section modulus Z = I / y_max. For the same material and allowable stress, comparing section moduli reveals which orientation is stronger in bending.

Given Data / Assumptions:

• Both sections are identical squares of side a.
• Comparison is about the horizontal neutral axis for vertical bending.
• Linear elasticity and small deflection; no stability effects.

Concept / Approach:For a square with sides horizontal: I = a^4 / 12 about the centroidal horizontal axis; y_max = a/2; thus Z_sides = (a^4 / 12) / (a/2) = a^3 / 6. For the same square rotated 45° (diagonal vertical): the second moment of area about any centroidal axis remains I = a^4 / 12 (shape is the same), but the half-depth becomes y_max = (a√2)/2 = a / √2, giving Z_diagonal = (a^4 / 12) / (a/√2) = (a^3 √2)/12.

Step-by-Step Solution:Z_sides = a^3 / 6.Z_diagonal = (a^3 √2) / 12.Ratio = Z_sides / Z_diagonal = (a^3 / 6) / ((a^3 √2)/12) = 12 / (6 √2) = 2 / √2 = √2 ≈ 1.414.

Verification / Alternative check:Numerical substitution, e.g., a = 100 mm, reproduces the same ratio independent of size.

Why Other Options Are Wrong:1 and 1/√2: would imply no change or the diamond being stronger; both are incorrect.2 and 3/2: overestimate the advantage; the exact factor is √2.

Common Pitfalls:Assuming I changes with rotation (it does not for the centroidal value of the square), or forgetting that y_max increases to a/√2 when the diagonal is vertical.

Final Answer:√2 (approximately 1.414).