Question

A beam with a rectangular section of 120 mm × 40 mm is placed horizontally by mistake, (with width as 120 mm and depth as 40 mm) whereas it was designed to be placed vertically. (with width as 40 mm and depth as 120 mm). The ratio of section modulus will be 
1. \(\frac{1}{3}\)
2. \(\frac{1}{2}\)
3. \(\frac{1}{6}\)
4. \(\frac{1}{8}\)

Answer 1

Correct Answer - Option 1 : \(\frac{1}{3}\)

Concept:

Sectional Modulus (Z):

• It is the ratio of moment of inertia (I) of the beam cross-section about the neutral axis to the distance (ymax) of extreme fiber from the neutral axis,  \(Z = \frac{I}{y_{max}}\)
• The section modulus (Z) of the cross-sectional shape is significant in designing beams. It is a direct measure of the strength of the beam. A beam that has a larger section modulus than another will be stronger and capable of supporting greater loads.

• The unit of the Section modulus is mm3.

Shapes	Section Modulus
Circular	\(~Z_{cirular\;section}=\frac{\pi{d^3}}{32}\)
Rectangle	\(~Z_{Rectangular\;section}=\frac{b{d^2}}{6}\)

Calculation:

Given:

Case 1: When the beam is placed horizontally

b = 120 mm, d = 40 mm

\(~Z_{1}=\frac{b{d^2}}{6} = \frac{120\times{40^2}}{6} = 32000\; mm^3\)

Case 2: When the beam is placed vertically

b = 40 mm, d = 120 mm

\(~Z_{2}=\frac{b{d^2}}{6} = \frac{40\times{120^2}}{6} = 96000\; mm^3\)

The ratio of section modulus is, \(\frac{Z_1}{Z_2} = \frac{32000}{96000}=\frac{1}{3}\)