Section Modulus Comparison – Square vs Circle In strength of materials and structural analysis, find the ratio of the section modulus of a square (side B) to that of a circle (diameter D), both about their centroidal axes.
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A(B^3 / 6) : (π * D^3 / 32)
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B(B^3 / 4) : (π * D^3 / 32)
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C(B^3 / 6) : (π * D^3 / 64)
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D(B^3 / 12) : (π * D^3 / 32)
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ENone of the above
Answer
Correct Answer: (B^3 / 6) : (π * D^3 / 32)
Explanation
Introduction / Context:Section modulus Z measures flexural strength: Z = I / y_max, where I is the second moment of area and y_max is the distance to the extreme fiber. We compare Z for square and circular sections about centroidal axes.
Given Data / Assumptions:
- Square: side B.
- Circle: diameter D.
- Both about centroidal axes with maximum stress at the outermost fiber.
Concept / Approach:Use standard formulas: I_square = B^4 / 12, y_max = B / 2; I_circle = π D^4 / 64, y_max = D / 2. Then compute Z for each and take the ratio.
Step-by-Step Solution:Z_square = (B^4 / 12) / (B / 2) = B^3 / 6Z_circle = (π D^4 / 64) / (D / 2) = π D^3 / 32Required ratio = (B^3 / 6) : (π D^3 / 32)
Verification / Alternative check:Dimensional check shows Z scales with length^3; both expressions have B^3 or D^3, confirming consistency.
Why Other Options Are Wrong:
- (B^3 / 4) or (B^3 / 12): arise from mixing up I or y_max.
- (π D^3 / 64) is incorrect for the circle's Z; it belongs to I's denominator before dividing by D/2.
Common Pitfalls:Confusing I with Z or using radius instead of diameter in circle formulas causes factor-of-2 errors.
Final Answer:(B^3 / 6) : (π * D^3 / 32)