--- title: "Cantilever 5 m: maximum and 25 mm bending stress" subject: "Strength of Materials" chapter: 7 key: "semester_3/strength_of_materials:cantilever-5m-8kn-15kn-bending-stresses" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/cantilever-5m-8kn-15kn-bending-stresses site: BCT Engineering --- # Cantilever 5 m: maximum and 25 mm bending stress Strength of Materials · Chapter 7: Theory of Flexure - Asked in: 2071 Chaitra, 8 marks - Video: [How to find Bending stresses in a Cantilever Beam || Example solved](https://www.youtube.com/watch?v=kQdWNgKd1WE) — Civil 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/cantilever-5m-8kn-15kn-bending-stresses - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7 ## Question A cantilever beam 5m in length is subjected to the loads as shown in figure. Determine the maximum bending stresses in the beam. Also, determine the value of bending stress 25 mm below from the top surface of the beam. [Figure: cantilever fixed at A, length 5 m, carrying an 8 kN/m UDL over its length and a 15 kN point load at the free end B; cross-section: rectangular web 50 mm wide and 120 mm high standing on a flange 200 mm wide and 50 mm thick (T-section).] ## Short answer Max hogging moment 175 kNm. Neutral axis 56.875 mm from base. Maximum bending stress 544.2 N/mm² (tension, top). Stress 25 mm below top surface = 423.9 N/mm² (tension). ## Answer **Section (as drawn):** web 50 mm wide × 120 mm high standing on a flange 200 mm × 50 mm. So the flange is at the bottom, the web top is the top surface, and overall depth is 170 mm. ```text 8 kN/m 15 kN vvvvvvvvvvvvvvvvvvvvvvv v |######################### A 5 m B ``` #### Maximum bending moment (at fixed end A, hogging) $$ M_{max} = \frac{8\times5^2}{2} + 15\times5 = 100 + 75 = 175\ \text{kNm} $$ Top fibres are in tension, bottom fibres in compression. #### Section properties (from base, mm) $$ \bar y = \frac{200\times50\times25 + 50\times120\times110}{10000+6000} = 56.875\ \text{mm} $$ $$ I = \frac{200\times50^3}{12} + 10000(25-56.875)^2 + \frac{50\times120^3}{12} + 6000(110-56.875)^2 = 36.377\times10^6\ \text{mm}^4 $$ Distance to top $= 170 - 56.875 = 113.125$ mm; to bottom $= 56.875$ mm. #### Maximum stresses $$ \sigma_{top} = \frac{175\times10^6\times113.125}{36.377\times10^6} = 544.2\ \text{N/mm}^2\ \text{(T)} $$ $$ \sigma_{bottom} = \frac{175\times10^6\times56.875}{36.377\times10^6} = 273.6\ \text{N/mm}^2\ \text{(C)} $$ #### Stress 25 mm below the top surface $y = 113.125 - 25 = 88.125$ mm above the NA (tension): $$ \sigma = \frac{175\times10^6\times88.125}{36.377\times10^6} = 423.9\ \text{N/mm}^2\ \text{(T)} $$ **Answer: maximum bending stress = 544.2 N/mm² (tension, top) at the fixed end; stress 25 mm below top = 423.9 N/mm² (tension).** These stresses are far above the yield stress of mild steel (about 250 MPa), so the section as given is not adequate for these loads. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Where is M_max and what is its value? A. Fixed end, 100 kNm B. Free end, 175 kNm C. Fixed end, 175 kNm hogging D. Mid-span, 87.5 kNm Answer: C. wL^2/2 + PL = 100 + 75 = 175 kNm. (Key point: Max moment) 2. Which fibre is in tension? A. Neither B. Bottom C. Top D. Web only Answer: C. Hogging cantilever has the top in tension. (Key point: Tension fibre) 3. Why is the top stress (544 N/mm^2) larger than the bottom? A. Because of tension B. Load acts on top C. Top fibre is farther from the NA (113 vs 57 mm) D. Top has less area Answer: C. The NA is nearer to the wide flange at the bottom. (Key point: Governing distance) 4. What y is used for a point 25 mm below the top? A. 113.125 mm B. 25 mm C. 31.9 mm D. 88.125 mm above NA Answer: D. 113.125 - 25 = 88.125 mm giving 423.9 N/mm^2 tension. 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