Section modulus of a rectangular section – proportional dependence For a rectangular cross-section of width b and overall depth d, the elastic section modulus Z about the centroidal horizontal axis is proportional to which of the following?

Mechanical Engineering Strength of Materials Difficulty: Easy
Choose an option
  • A
    Area of the section only
  • B
    Square of the area of the section
  • C
    Product of the area and the depth (b d × d)
  • D
    Product of the area and the width (b d × b)
  • E
    Half of the moment of inertia only

Answer

Correct Answer: Product of the area and the depth (b d × d)

Explanation

Introduction / Context:The elastic section modulus Z connects bending moment to extreme fibre stress via sigma_max = M / Z. For rectangular sections used in beams, knowing how Z scales helps with rapid sizing.

Given Data / Assumptions:

  • Rectangular section with width b and depth d.
  • Bending about the centroidal horizontal axis (through mid-depth).
  • Linear elastic behavior.

Concept / Approach:Moment of inertia about the centroidal horizontal axis is I = b d^3 / 12. Distance to the extreme fibre is y_max = d / 2. Hence section modulus Z = I / y_max = (b d^3 / 12) / (d/2) = b d^2 / 6. Since area A = b d, it follows Z ∝ A * d.

Step-by-Step Solution:I = b d^3 / 12.y_max = d / 2.Z = I / y_max = (b d^3 / 12) / (d/2) = b d^2 / 6.Express in terms of area: A = b d → Z = (A d) / 6 → proportional to A * d.

Verification / Alternative check:Dimensional check: Z has units of length^3; A * d has units of length^3, confirming consistency.

Why Other Options Are Wrong:Area only or area squared give incorrect scaling. Area × width becomes A * b = b^2 d, not proportional to b d^2. Half of I is not Z; Z depends on I divided by y_max.

Common Pitfalls:Using I directly instead of Z; forgetting to divide by the extreme fibre distance.

Final Answer:Product of the area and the depth (b d × d)

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