--- title: "Bending stress diagram; moment in flanges" subject: "Strength of Materials" chapter: 7 key: "semester_3/strength_of_materials:l-section-bending-stress-diagram-flange" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/l-section-bending-stress-diagram-flange site: BCT Engineering --- # Bending stress diagram; moment in flanges Strength of Materials · Chapter 7: Theory of Flexure - Asked in: 2066 Jestha (old course), 8 marks - Video: [Problem 4.3 | Strength of Materials | Beer & Johnston | Bending Stress | Max Moment in I-Beam](https://www.youtube.com/watch?v=voWBmaqIGWg) — Inside Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/l-section-bending-stress-diagram-flange - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7 ## Question A beam of I-section (printed "L section") 20cm deep and 10cm wide has flanges 3cm thick and web 2cm thick. It carries a bending moment of 10 kNm at a section. Draw bending stress diagram showing values of maximum stresses in the section of the beam. Calculate the moment carried by the flanges. ## Short answer Maximum bending stress is 20.67 N/mm² (tension and compression). The flanges carry 9.05 kNm, which is 90.5% of the total moment. ## Answer **Section (I-section):** overall depth 20 cm, flange width 10 cm, flange thickness 3 cm, web 2 cm thick and 14 cm deep. $M = 10$ kNm. #### Moment of inertia $$ I = \frac{100\times200^3 - 80\times140^3}{12} = 48.373\times10^6\ \text{mm}^4 $$ #### Bending stresses ($\sigma = My/I$) - Extreme fibre ($y = 100$ mm): $$ \sigma_{max} = \frac{10\times10^6\times100}{48.373\times10^6} = 20.67\ \text{N/mm}^2 $$ - At the flange-web junction ($y = 70$ mm): $$ \sigma_j = \frac{10\times10^6\times70}{48.373\times10^6} = 14.47\ \text{N/mm}^2 $$ #### Bending stress diagram ```text y (mm) stress (N/mm^2) +100 20.67 (C) *-----------+ top fibre \ | +70 14.47 (C) *---------+ flange-web junction \ | 0 0 (NA) *-----+ neutral axis / | -70 14.47 (T) *---------+ flange-web junction / | -100 20.67 (T) *-----------+ bottom fibre ``` Stress is zero at the neutral axis, varies linearly with distance, and is compressive on one side and tensile on the other (for a sagging moment, top compressive). #### Moment carried by the flanges Flanges (both, from $y = 70$ to $100$ mm): $$ M_f = 2\times\frac{\sigma_{max}}{100}\times100\times\frac{100^3 - 70^3}{3} = 9.05\ \text{kNm} $$ Web (from $-70$ to $70$): $$ M_w = \frac{\sigma_{max}}{100}\times20\times\frac{2\times70^3}{3} = 0.95\ \text{kNm} $$ Check: $9.05 + 0.95 = 10$ kNm. **Answer: maximum stress = 20.67 N/mm² (tension and compression); moment carried by the flanges = 9.05 kNm (90.5% of the total).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does bending stress vary with depth? A. Linearly, zero at the NA, tension on one side, compression on the other B. Uniform across the depth C. Parabolic with maximum at NA D. Maximum at the NA Answer: A. sigma = My/I is linear in y. (Key point: Stress variation) 2. What is the maximum stress and where? A. 14.47 N/mm^2 at the extreme fibres B. 20.67 N/mm^2 at the extreme fibres (y = 100 mm) C. 20.67 N/mm^2 at the NA D. 14.47 N/mm^2 at the NA Answer: B. sigma = 10x10^6 x 100/48.373x10^6 = 20.67 N/mm^2. (Key point: Maximum stress) 3. What is the stress at the flange-web junction? A. 20.67 N/mm^2 B. 0 C. 14.47 N/mm^2 at y = 70 mm D. 10 N/mm^2 Answer: C. sigma = My/I at y = 70 mm gives 14.47 N/mm^2. (Key point: Stress at junction) 4. How is the flange moment found? A. Take half the total moment B. Use only the web area C. Use sigma_max x flange area D. Integrate stress x width x y over y from 70 to 100 mm for both flanges, giving 9.05 kNm Answer: D. Flange carries 9.05 kNm (90.5%) and the web the remaining 0.95 kNm. 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