--- title: "Section modulus: rectangle versus circle" subject: "Strength of Materials" chapter: 3 key: "semester_4/strength_of_materials_mechanical:section-modulus-rectangle-versus-circle" source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/section-modulus-rectangle-versus-circle site: BCT Engineering --- # Section modulus: rectangle versus circle Strength of Materials · Chapter 3: Pure Bending - Asked in: year not given, 5 marks - Video: [Section Modulus Of Rectangular And Circular Sections || Strength Of Material - 1 || Civil](https://www.youtube.com/watch?v=NzyYFMeEexo) — VeTeach - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/section-modulus-rectangle-versus-circle - Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-3 ## Question Define section modulus. For the same allowable bending stress and the same cross-sectional area, compare the moment of resistance of a rectangular section whose depth is twice its width with that of a solid circular section. Why are I-sections preferred in practice? ## Short answer Section modulus Z = I/y_max. For equal area, rectangle (depth = 2×width) has Z = 0.667b³, solid circle Z = 0.399b³, giving 67% more moment capacity. I-sections put material in flanges far from neutral axis, maximizing Z per weight. ## Answer #### Section modulus The **section modulus** is $Z = I/y_{max}$, where $I$ is the second moment of area about the neutral axis and $y_{max}$ the distance to the extreme fibre. From the flexure formula the moment of resistance is $$ M_r = \sigma_{all}\,Z $$ so for a given stress, the larger $Z$, the larger the moment the section can carry. Unit: mm$^3$ or m$^3$. #### Comparison for equal area Rectangle of width $b$ and depth $h = 2b$: $$ A = 2b^2, \qquad Z_r = \frac{bh^2}{6} = \frac{b(2b)^2}{6} = 0.667\,b^3 $$ Circle with the same area: $$ \frac{\pi d^2}{4} = 2b^2 \;\Rightarrow\; d = 1.596\,b $$ $$ Z_c = \frac{\pi d^3}{32} = \frac{\pi (1.596b)^3}{32} = 0.399\,b^3 $$ $$ \frac{M_{r,\text{rect}}}{M_{r,\text{circle}}} = \frac{Z_r}{Z_c} = \frac{0.667}{0.399} = 1.67 $$ For example, with $b = 100$ mm: $Z_r = 666.7\times10^3$ mm$^3$ and $Z_c = 398.9\times10^3$ mm$^3$. With $\sigma_{all} = 10$ MPa they carry 6.67 kN m and 3.99 kN m respectively. The rectangle (on edge) is about 67 % stronger in bending for the same weight of material. #### Why I-sections In bending the material near the neutral axis carries little stress. An I-section puts most of the area in the flanges, far from the NA, so $I$ and $Z$ are large for a small area. Its web resists shear. It therefore gives the highest strength-to-weight ratio, which is why rolled I-sections are used for beams. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does section modulus Z = I / y_max determine? A. The moment of resistance M_r = sigma_all Z B. The shear capacity C. The deflection only D. The weight of the beam Answer: A. From the flexure formula, M_r = sigma_all x Z; larger Z means more moment capacity. (Key point: Section modulus meaning) 2. For equal area, with a rectangle h = 2b against a circle, what is the ratio of moments of resistance? A. About 1.67 in favour of the rectangle B. About 0.6 in favour of the rectangle C. About 1.0 (equal) D. About 1.67 in favour of the circle Answer: A. Z_r = 0.667 b^3 and Z_c = 0.399 b^3, so the ratio is 1.67. (Key point: Rectangle vs circle comparison) 3. Why are I-sections preferred for beams? A. Most area is in the flanges far from the NA, giving large Z for little area B. Most area is near the NA C. They have zero shear stress D. They are easy to bend Answer: A. Material near the NA carries little stress; flanges far away give high I and Z, and the web resists shear, so strength-to-weight is highest. 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