--- title: "T-beam with overhang: max bending stress" subject: "Strength of Materials" chapter: 7 key: "semester_3/strength_of_materials:t-beam-4kn-m-10kn-overhang" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/t-beam-4kn-m-10kn-overhang site: BCT Engineering --- # T-beam with overhang: max bending stress Strength of Materials · Chapter 7: Theory of Flexure - Asked in: 2078 Bhadra, 10 marks - Video: [How to Find Maximum Bending Stress in a overhanging T-Beam | Strength of Materials | TU, PU, PoU](https://www.youtube.com/watch?v=rh4ZfpdRzew) — Mr. Structure: Building Minds with Neeraj Bhandari - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/t-beam-4kn-m-10kn-overhang - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7 ## Question Determine the maximum bending stress in the beam shown in figure below. [Figure: beam with a hinge at the left end and roller 6 m away, an overhang of 2 m beyond the roller; UDL 4 kN/m over the 6 m span; 10 kN downward load at the free end of the overhang. Cross-section: T-section with flange 10 + 10 + 10 = 30 cm wide and 10 cm thick, web 10 cm wide, stem total height 25 cm; all dimensions in cm.] ## Short answer Maximum bending stress is 15.12 N/mm² (compressive) occurring at the bottom fibre over support B; the maximum tensile stress is 8.76 N/mm² at the top fibre over B. ## Answer **Assumption:** flange 30 cm × 10 cm on top, web 10 cm wide, overall depth 25 cm (web depth 15 cm). Self-weight ignored. ```text 4 kN/m over A to B 10 kN vvvvvvvvvvvvvvvvvvvvvvvvvv | |========================|==========v A (hinge) B (roller) C |<-------- 6 m --------->|<- 2 m -->| ``` #### Reactions Taking moments about A: $$ R_B \times 6 = 4\times 6\times 3 + 10\times 8 \Rightarrow R_B = 25.33\ \text{kN} $$ $R_A = 24 + 10 - 25.33 = 8.67$ kN (check: $\Sigma F_y = 0$). #### Bending moments - At B (hogging): $M_B = -10\times 2 = -20$ kNm. - Zero shear at $x = R_A/4 = 2.167$ m from A, so the sagging moment is maximum there: $$ M_{max}^{+} = 8.67\times 2.167 - 4\times\frac{2.167^2}{2} = 9.39\ \text{kNm} $$ #### Section properties Measure from the base (cm): $$ \bar y = \frac{30\times10\times20 + 10\times15\times7.5}{300+150} = 15.83\ \text{cm} $$ $$ I = \frac{30\times10^3}{12} + 300(20-15.83)^2 + \frac{10\times15^3}{12} + 150(7.5-15.83)^2 = 20937.5\ \text{cm}^4 $$ Distance to top fibre $= 25 - 15.83 = 9.17$ cm; to bottom fibre $= 15.83$ cm. #### Stresses ($\sigma = My/I$) | Section | Top fibre | Bottom fibre | |---|---|---| | Max sagging, 9.39 kNm | 4.11 N/mm² (C) | 7.10 N/mm² (T) | | Support B, hogging 20 kNm | 8.76 N/mm² (T) | 15.12 N/mm² (C) | $$ \sigma = \frac{20\times10^6\times 158.3}{20937.5\times10^4} = 15.12\ \text{N/mm}^2 $$ **Answer: maximum bending stress = 15.12 N/mm² (compressive) at the bottom fibre over support B; the maximum tensile stress is 8.76 N/mm² at the top fibre over B.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is R_B found? A. Moments about B B. Moments about A: R_B x 6 = 4x6x3 + 10x8 C. R_B = 24 kN D. Equal to half the total load Answer: B. Gives R_B = 25.33 kN and R_A = 8.67 kN. (Key point: Reaction) 2. What is the bending moment at B and its nature? A. 20 kNm sagging B. zero C. 9.39 kNm sagging D. 20 kNm hogging from the overhang load Answer: D. M_B = -10 x 2 = -20 kNm, hogging. (Key point: Moment at support B) 3. Where is the maximum sagging moment? A. At mid-span (3 m) B. Where shear is zero, 2.167 m from A C. At A D. At B Answer: B. Zero shear at x = R_A/w = 2.167 m, giving 9.39 kNm. (Key point: Max sagging) 4. Why is the final maximum stress at the bottom fibre over B? A. The top fibre is farther B. Sagging moment is larger C. Tension always governs D. Largest moment (20 kNm) acts with the farthest fibre (15.83 cm), compression Answer: D. Hogging puts the bottom in compression: 15.12 N/mm^2 exceeds all other values. 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