--- title: "MoR of T-beam by limit state" subject: "Design of RCC Structure" chapter: 4 key: "semester_7/design_of_rcc_structure:mor-of-t-beam-by-limit" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/mor-of-t-beam-by-limit site: BCT Engineering --- # MoR of T-beam by limit state Design of RCC Structure · Chapter 4: Design of beams: Behavior in Flexure - Asked in: 2076 Asoj, 8 marks - Video: [# T- Beam # Moment of Resistance](https://www.youtube.com/watch?v=5F7Md2nsOVQ) — Vijay Srivastava - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/mor-of-t-beam-by-limit - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-4 ## Question A T-beam of effective flange width 1600 mm, depth of flange 110 mm, breadth of web 300 mm and overall depth 460 mm is reinforced with 4-20 mm bars in tension at an effective cover of 40 mm. Determine the moment of resistance of the section using M20 grade concrete and Fe415 grade steel in Limit state method of design. ## Short answer Neutral axis lies in flange, xu = (0.87fy Ast)/(0.36fck bf) = 39.38 mm < Df. Moment of resistance Mu = 0.87fy Ast (d - 0.42xu) = 183.05 kNm. ## Answer **Given:** $b_f=1600$ mm, $D_f=110$ mm, $b_w=300$ mm, $D=460$ mm, effective cover 40 mm so $d=420$ mm; 4-20 mm bars, $A_{st}=4\times\frac{\pi}{4}(20)^2=1256.6$ mm$^2$; M20, Fe415. **Step 1: Locate neutral axis.** Assume $x_u\le D_f$ (rectangular behaviour with width $b_f$). Equating compression and tension: $$ 0.36f_{ck}b_fx_u=0.87f_yA_{st}\ \Rightarrow\ x_u=\frac{0.87\times415\times1256.6}{0.36\times20\times1600}=39.38\ \text{mm} $$ $x_u=39.38x_u$, so the section is under-reinforced. **Step 3: Moment of resistance** $$ \begin{aligned} M_u&=0.87f_yA_{st}(d-0.42x_u)\\ &=0.87\times415\times1256.6\times(420-0.42\times39.38)\\ &=183.05\ \text{kN·m} \end{aligned} $$ **Answer:** $M_u=183.05$ kN·m (design moment of resistance of the T-section). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the neutral axis position first tested for the T-beam? A. Assume xu <= Df and equate 0.36 fck bf xu to 0.87 fy Ast B. Assume xu > Df and use the web width only C. Take xu = 0.48d directly D. Take xu = Df/2 Answer: A. Assume the NA lies in the flange (rectangular beam of width bf), solve xu from C = T and compare with Df. (Key point: Neutral axis location) 2. xu = 39.38 mm is less than Df = 110 mm. What does this show? A. The NA is in the web, so a T-beam formula is needed B. The NA is in the flange, so the section acts as a rectangular beam of width bf C. The section is over-reinforced D. The flange is in tension Answer: B. Since xu < Df the whole compression zone lies in the flange and the rectangular formulas with bf apply. (Key point: Result of NA check) 3. Why is xu compared with xu,max = 0.48d (Fe415)? A. To find the flange width B. To determine the self weight C. To confirm the steel yields first, i.e. the section is under-reinforced D. To check shear capacity Answer: C. If xu < xu,max the tension steel yields before concrete crushes, giving ductile failure and allowing 0.87fy to be used. (Key point: Under-reinforced check) 4. Which formula gives Mu for the under-reinforced section with NA in flange? A. Mu = 0.87 fy Ast (d - Df) B. Mu = 0.36 fck bw xu d C. Mu = 0.138 fck bf d^2 always D. Mu = 0.87 fy Ast (d - 0.42 xu) Answer: D. Tension force 0.87fyAst acts with lever arm d - 0.42xu (centroid of the stress block at 0.42xu from the top). 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