--- title: "Symmetrical I-beam: moment, flange and web share" subject: "Strength of Materials" chapter: 7 key: "semester_3/strength_of_materials:symmetrical-i-beam-max-moment-flange" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/symmetrical-i-beam-max-moment-flange site: BCT Engineering --- # Symmetrical I-beam: moment, flange and web share Strength of Materials · Chapter 7: Theory of Flexure - Asked in: 2066 Bhadra (old course), 10 marks - Video: [Shape Factor Problem No 1 ( Symmetrical I Section )](https://www.youtube.com/watch?v=Yemto1mozKU) — Stan Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/symmetrical-i-beam-max-moment-flange - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7 ## Question An I section beam (symmetrical) has 200mm wide flanges and overall depth 500mm. Each flange is 25mm thick and the web is 20mm thick. Determine (i) the maximum bending moment that should be imposed in the section if the tensile or the compressive stress is not to exceed $40\ \text{N/mm}^2$ (ii) What percentage of the moment is resisted by flanges and web? ## Short answer Maximum bending moment is 114.63 kNm. The flanges resist 78.8% of the moment and the web 21.2%. ## Answer **Given:** flanges 200 mm × 25 mm, overall depth 500 mm, web 20 mm thick (web depth $500 - 50 = 450$ mm). $\sigma_{max} = 40\ \text{N/mm}^2$. ```text +-----------+ 25 | | | | 450 +-----------+ 25 200 wide, web 20 ``` #### Moment of inertia $$ I = \frac{200\times500^3 - 180\times450^3}{12} = 7.1646\times10^8\ \text{mm}^4 $$ #### (i) Maximum bending moment Extreme fibre distance $y = 250$ mm: $$ M = \frac{\sigma I}{y} = \frac{40\times7.1646\times10^8}{250} = 114.63\times10^6\ \text{Nmm} = 114.63\ \text{kNm} $$ #### (ii) Moment shared by flanges and web Stress at distance $y$ is $\sigma = 40y/250$. Flanges (both, from $y = 225$ to $250$): $$ M_f = 2\int_{225}^{250}\frac{40y}{250}\,200\,y\,dy = 2\times\frac{40\times200}{250}\times\frac{250^3 - 225^3}{3} = 90.33\ \text{kNm} $$ Web (from $-225$ to $225$): $$ M_w = \frac{40\times20}{250}\times\frac{2\times225^3}{3} = 24.30\ \text{kNm} $$ Check: $90.33 + 24.30 = 114.63$ kNm. | Part | Moment (kNm) | Share | |---|---|---| | Flanges | 90.33 | 78.8% | | Web | 24.30 | 21.2% | **Answer: (i) $M_{max} = 114.63$ kNm; (ii) flanges carry 78.8% and the web 21.2% of the moment.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is I of the I section found? A. I = 200x500^3/12 B. I = [200x500^3 - 200x450^3]/12 C. I = [200x500^3 - 180x450^3]/12 (outer rectangle minus voids) D. I = 20x450^3/12 only Answer: C. The voids on each side of the web are 90 mm wide each, 180 mm total, 450 mm deep. (Key point: Moment of inertia) 2. Which relation gives the allowable moment? A. M = sigma I/y with y = 225 mm B. M = sigma A y C. M = sigma y/I D. M = sigma I/y with y = 250 mm, giving 114.63 kNm Answer: D. The limiting stress occurs at the extreme fibre, 250 mm from the NA. (Key point: Maximum moment) 3. How does bending stress vary through the section? A. Linearly with distance from the NA, sigma = 40y/250 B. Uniformly over the flange C. Parabolically D. Maximum at the NA Answer: A. Bending stress is proportional to y. (Key point: Stress distribution) 4. What share of the moment do the flanges carry? A. About 21.2% B. About 78.8% (90.33 kNm) C. About 50% D. About 95% Answer: B. Flanges are far from the NA, so they carry the greater part; the web carries 24.30 kNm, 21.2%. 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