--- title: "Bending stresses in a T-section beam" subject: "Strength of Materials" chapter: 3 key: "semester_4/strength_of_materials_mechanical:cast-iron-t-beam-bending-stresses" source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/cast-iron-t-beam-bending-stresses site: BCT Engineering --- # Bending stresses in a T-section beam Strength of Materials · Chapter 3: Pure Bending - Asked in: year not given, 8 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_4/strength_of_materials_mechanical/cast-iron-t-beam-bending-stresses - Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-3 ## Question A cast iron beam has a T-section: a flange 120 mm wide and 30 mm thick at the top, and a web 30 mm thick and 150 mm deep below it (overall depth 180 mm). It is simply supported over a span of 4 m and carries a uniformly distributed load of 12 kN/m, including self weight, over the whole span. (a) Locate the neutral axis and find I. (b) Find the maximum tensile and compressive bending stresses at mid-span. (c) If the allowable stresses are 40 MPa in tension and 150 MPa in compression, find the greatest UDL the beam can carry. ## Short answer Neutral axis 115 mm from bottom; I = 24.91×10^6 mm^4. At mid-span M = 24 kN m, tensile stress 110.8 MPa, compressive 62.6 MPa. Tension governs; allowable tensile stress 40 MPa limits safe UDL to 4.33 kN/m. ## Answer #### (a) Neutral axis and I Measure $\bar y$ from the bottom of the web. | Part | Area (mm$^2$) | $y$ from bottom (mm) | $Ay$ | |---|---|---|---| | Flange $120\times30$ | 3600 | 165 | 594 000 | | Web $30\times150$ | 4500 | 75 | 337 500 | | Total | 8100 | | 931 500 | $$ \bar y = \frac{931500}{8100} = 115\ \text{mm from bottom}, \quad 65\ \text{mm from top} $$ Using the parallel axis theorem: $$ I = \left[\frac{120\times30^3}{12} + 3600(165-115)^2\right] + \left[\frac{30\times150^3}{12} + 4500(115-75)^2\right] $$ $$ I = (0.27 + 9.0)\times10^6 + (8.4375 + 7.2)\times10^6 = 24.91\times10^6\ \text{mm}^4 $$ #### (b) Stresses at mid-span $$ M_{max} = \frac{wL^2}{8} = \frac{12\times4^2}{8} = 24\ \text{kN m} = 24\times10^6\ \text{N mm} $$ The load sags the beam, so the top (flange) is in compression and the bottom is in tension. $$ \sigma_c = \frac{M\,y_{top}}{I} = \frac{24\times10^6\times65}{24.91\times10^6} = 62.6\ \text{MPa} $$ $$ \sigma_t = \frac{M\,y_{bot}}{I} = \frac{24\times10^6\times115}{24.91\times10^6} = 110.8\ \text{MPa} $$ #### (c) Safe UDL Limit moment from tension: $M_t = \dfrac{\sigma_{t,all} I}{y_{bot}} = \dfrac{40\times24.91\times10^6}{115} = 8.66\times10^6$ N mm. Limit moment from compression: $M_c = \dfrac{150\times24.91\times10^6}{65} = 57.5\times10^6$ N mm. Tension governs (the bottom fibre is farther from the neutral axis and cast iron is weak in tension): $M = 8.66$ kN m. $$ w = \frac{8M}{L^2} = \frac{8\times8.66}{16} = 4.33\ \text{kN/m} $$ **Answer: NA 115 mm above the bottom; $I = 24.91\times10^6\ \text{mm}^4$; $\sigma_t = 110.8$ MPa and $\sigma_c = 62.6$ MPa (the 12 kN/m load is unsafe in tension); safe load $w = 4.33$ kN/m.** Placing the flange at the bottom (tension side) would be more suitable for cast iron. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the neutral axis of the T-section located? A. At mid-depth, 90 mm from the bottom B. At the centroid, from sum(A y)/sum(A) = 115 mm above the bottom C. At the centre of the flange D. At the top of the web Answer: B. For an unsymmetrical T the NA is at the centroid: 931500/8100 = 115 mm from the bottom. (Key point: Locating the neutral axis) 2. How is I computed for the T-section? A. Add bh^3/12 of each part only B. For each part add its own bh^3/12 plus A d^2, with d to the NA C. Use bh^3/12 of the overall rectangle 120 x 180 D. Use A y_bar^2 of the whole section Answer: B. The parallel axis theorem is applied to flange and web about the centroidal axis, giving 24.91 x 10^6 mm^4. (Key point: Parallel axis theorem) 3. At mid-span the 12 kN/m UDL sags the beam. Which result is correct? A. Top in tension 62.6 MPa, bottom in compression 110.8 MPa B. Top in compression 62.6 MPa, bottom in tension 110.8 MPa C. Both 86.7 MPa D. Top in compression 110.8 MPa, bottom in tension 62.6 MPa Answer: B. Sagging puts the top in compression (y = 65 mm, 62.6 MPa) and the bottom in tension (y = 115 mm, 110.8 MPa). (Key point: Which fibre has larger stress) 4. Why does tension govern the safe UDL? A. The bottom fibre is farther from the NA and cast iron is weak in tension B. Compression allowable stress is lower C. The flange is thin D. The span is long Answer: A. M_t = 40 I/115 = 8.66 kN m is far less than M_c = 57.5 kN m, so w = 8M/L^2 = 4.33 kN/m. 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