--- title: "Design steps of doubly reinforced section" subject: "Design of RCC Structure" chapter: 4 key: "semester_7/design_of_rcc_structure:design-steps-of-doubly-reinforced-section" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/design-steps-of-doubly-reinforced-section site: BCT Engineering --- # Design steps of doubly reinforced section Design of RCC Structure · Chapter 4: Design of beams: Behavior in Flexure - Asked in: 2080 Baisakh, 6 marks - Video: [19. Design Steps for Doubly Reinforced Beam Section #RCC #GTU #DRCS #DOUBLYbeam](https://www.youtube.com/watch?v=ARdycscfr3I) — THE CIVILOCONCEPT - 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/design-steps-of-doubly-reinforced-section - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-4 ## Question Explain about design steps of doubly reinforced section with neat sketches. ## Short answer When M_u exceeds M_u,lim = 0.138 fck b d² (Fe415), provide compression steel. Compute extra moment M_u2 = M_u - M_u,lim. Find compression steel stress f_sc from strain at limiting NA, then A_sc = M_u2 / [(f_sc - f_cc)(d-d')]. Tension steel: A_st = A_st1 + A_st2 with A_st1 for limiting moment and A_st2 balancing compression steel. Check limits and provide closely spaced stirrups to prevent buckling. ## Answer A doubly reinforced beam is used when the factored moment $M_u$ exceeds the limiting moment $M_{u,lim}=0.138f_{ck}bd^2$ (Fe415) of the section whose size is restricted. ```text b strain stress forces +---------+ d' 0.0035 0.446fck | o Asc | |\ fcc|==| Cc = 0.36fck.b.xu | | | \xu | | Cs = Asc(fsc-fcc) | | -+--NA | | | | | \ | o o Ast| | \es T = 0.87fy.Ast +---------+ ``` **Steps** 1. Assume $b$, $D$, cover $d'$; find $d$ and factored moment $M_u$. 2. Compute $M_{u,lim}=0.36f_{ck}b\,x_{u,max}(d-0.42x_{u,max})$ using $x_{u,max}=0.53d$, $0.48d$ or $0.46d$ for Fe250, Fe415, Fe500. 3. If $M_u>M_{u,lim}$, find the extra moment $M_{u2}=M_u-M_{u,lim}$, to be carried by a steel couple. 4. Find strain in compression steel at the limiting NA: $\varepsilon_{sc}=0.0035\left(1-\dfrac{d'}{x_{u,max}}\right)$. Get the stress $f_{sc}$ from the stress-strain curve (IS 456 Fig. 23 / SP16 Table F). Concrete stress displaced by steel $f_{cc}=0.446f_{ck}$. 5. Compression steel: $$ A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')} $$ 6. Tension steel: $$ A_{st}=A_{st1}+A_{st2},\quad A_{st1}=\frac{0.36f_{ck}b\,x_{u,max}}{0.87f_y},\quad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y} $$ 7. Check limits: $A_{st}\ge0.85bd/f_y$; $A_{st}$ and $A_{sc}$ not more than 4% each of $bD$; $d'/d\approx0.1$-$0.2$. 8. Detailing: provide stirrups spacing $\le$ the least of $16\phi$ of compression bar, 300 mm and $b$, so that compression bars do not buckle. Check shear and deflection (with compression steel factor $k_c$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the moment carried by the steel couple found? A. Mu2 = Mu - Mu,lim B. Mu2 = Mu + Mu,lim C. Mu2 = Mu,lim D. Mu2 = Mu/2 Answer: A. The limiting section takes Mu,lim and the compression/tension steel pair takes the rest. (Key point: Extra moment) 2. How is the strain in compression steel found? A. 0.0035 (1 - d'/xu,max) B. 0.0035 (1 + d'/xu,max) C. 0.002 (d'/xu,max) D. 0.87 fy / Es Answer: A. By similar triangles on the strain diagram at the limiting NA. (Key point: Compression steel strain) 3. What is the area of compression steel? A. Mu2 / ((fsc - fcc)(d - d')) B. Mu2 / (0.87 fy d) C. Mu / (fsc d') D. Mu2 / (fcc d) Answer: A. The couple uses the net compression stress fsc - fcc and the arm d - d'. (Key point: Asc formula) 4. Why are closely spaced stirrups provided around compression bars? A. To resist buckling of the compression bars B. To increase flexure capacity C. To reduce cover D. To reduce deflection Answer: A. Spacing is limited to the least of 16 phi, 300 mm and b so the bars do not buckle. 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