Consider a rectangular hollow section having overall width B and overall depth D, and let the width and depth of the symmetrically placed hole be b and d respectively. Then the section modulus (Z) is
- A(BD³ − bd³) / 6d
- B(BD³ − bd³) / 12D
- C(BD³ − bd³) / 12d
- D(BD³ − bd³) / 6DCorrect
Solution
Concept
Section modulus of a hollow section
Formula
Z = I / ymax = [(BD³ − bd³)/12] / (D/2) = (BD³ − bd³) / 6D
Step-by-step solution
1. I = (BD³ − bd³) / 12. 2. ymax = D/2. 3. Z = I / ymax = (BD³ − bd³) / 6D.
Explanation
The moment of inertia of the hollow section is the difference of the outer and inner rectangles. Dividing by the extreme fibre distance D/2 gives Z.
Why the other options are wrong
Dividing by d or by 12D/12d uses the wrong distance to the extreme fibre.
Shortcut
Z = I / (D/2).
Common mistake
Dividing by the depth of the hole.
Correct answer: D) (BD³ − bd³) / 6D