Weight comparison: square vs circular beam with equal bending capacity\n\nA square beam and a circular beam have the same length, the same allowable bending stress, and resist the same bending moment (i.e., same required section modulus). What is the ratio of their weights (square : circular)?

Introduction / Context:
Choosing a cross-section to minimize weight for a target bending capacity is central to efficient design. Here we compare square and circular solid sections sized to deliver the same section modulus (hence same allowable moment at the same allowable stress).

Given Data / Assumptions:

• Equal length and material (same density), so weight ∝ cross-sectional area.
• Allowable normal bending stress is identical for both sections.
• Required section modulus Z is the same for both beams.

Concept / Approach:
For a square of side a: I = a^4 / 12 and c = a/2, hence Z_square = I / c = a^3 / 6. For a circle of diameter D: I = π D^4 / 64 and c = D/2, hence Z_circle = π D^3 / 32. Setting Z_square = Z_circle gives the size ratio, then weight ratio equals area ratio.

Step-by-Step Solution:
Equate section moduli: a^3 / 6 = π D^3 / 32.Solve for a/D: (a/D)^3 = (3 π) / 16.Therefore a/D = [(3 π) / 16]^(1/3) ≈ 0.838.Areas: A_square = a^2; A_circle = π D^2 / 4.Weight ratio (square : circular) = A_square / A_circle = (a^2) / (π D^2 / 4) = (4 / π) * (a/D)^2 ≈ (4 / π) * 0.838^2 ≈ 0.894.Hence square : circular ≈ 0.894 : 1 ≈ 1 : 1.12, i.e., 1/1.12.

Verification / Alternative check:
Dimensional and numerical checks confirm the ratio. Using more digits for π barely changes the result; the qualitative conclusion is that the circular section is slightly heavier for the same bending capacity.

Why Other Options Are Wrong:
1/2 and 1/1.5 significantly understate the square’s weight. “1” would imply equal weight, which is not the case at equal Z. “2” is far off.

Common Pitfalls:
Comparing at equal area instead of equal Z; forgetting that weight scales with area when length and density are fixed; using wrong expressions for Z.

Final Answer:
1/1.12