--- title: "L-beam reinforcement for given moment" subject: "Design of RCC Structure" chapter: 4 key: "semester_7/design_of_rcc_structure:l-beam-reinforcement-for-given-moment" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/l-beam-reinforcement-for-given-moment site: BCT Engineering --- # L-beam reinforcement for given moment Design of RCC Structure · Chapter 4: Design of beams: Behavior in Flexure - Asked in: 2072 Kartik, 14 marks - Video: [Design Moment Strength Calculation of a RC T-beam - Reinforced Concrete](https://www.youtube.com/watch?v=HSjMacgA3Bc) — structurefree - 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/l-beam-reinforcement-for-given-moment - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-4 ## Question A L-beam has a flange of effective width 900 mm and depth of 100 mm. The web below is 250 mm x 500 mm. Determine the amount of reinforcement required for the cross-section if it has to carry a factored bending moment of 615 kN-m and SF of 50 kN. Adopt M20 concrete mix and Fe 500 grade steel. ## Short answer Limiting moment 495.04 kN·m < 615 kN·m, so doubly reinforced. Provide tension steel 2943 mm² (6-25 mm) and compression 592 mm² (3-16 mm); 8 mm two-legged stirrups at 200 mm c/c satisfy shear and minimum requirements. ## Answer **Given:** $b_f=900$ mm, $D_f=100$ mm, $b_w=250$ mm, web depth below flange 500 mm; $M_u=615$ kN·m, $V_u=50$ kN (factored); M20, Fe500. **Assumptions:** overall depth $D=500+100=600$ mm; effective cover 50 mm, so $d=550$ mm and $d'=50$ mm. **Step 1: Limiting moment of the L-section.** $D_f/d=100/550=0.182\le0.2$, so the flange thickness $D_f$ is used directly. For Fe500, $x_{u,max}=0.46d=253.0$ mm $>D_f$ (neutral axis in the web). $$ \begin{aligned} M_{u,lim}&=0.36f_{ck}b_wx_{u,max}(d-0.416x_{u,max})+0.45f_{ck}(b_f-b_w)D_f\left(d-\frac{D_f}{2}\right)\\ &=202.54+292.50=495.04\ \text{kN·m} \end{aligned} $$ Since $M_u=615>M_{u,lim}=495.04$ kN·m, compression steel is required (doubly reinforced L-beam). **Step 2: Compression steel** $$ M_{u2}=615-495.04=119.96\ \text{kN·m} $$ $$ \varepsilon_{sc}=0.0035\left(1-\frac{50}{253.0}\right)=0.00281\ \Rightarrow\ f_{sc}=414.2\ \text{N/mm}^2\ (\text{Fe500 curve}),\quad f_{cc}=0.446\times20=8.92 $$ $$ A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{119.96\times10^6}{(414.2-8.92)(550-50)}=592\ \text{mm}^2 $$ **Step 3: Tension steel** $$ A_{st1}=\frac{0.36f_{ck}b_wx_{u,max}+0.45f_{ck}(b_f-b_w)D_f}{0.87f_y}=2392\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=552\ \text{mm}^2 $$ $$ A_{st}=A_{st1}+A_{st2}=2943\ \text{mm}^2 $$ Provide 6-25 mm dia ($2945$ mm$^2$, in layers) and 3-16 mm dia compression bars ($603$ mm$^2$). **Step 4: Shear (V$_u$ = 50 kN)** $$ \tau_v=\frac{V_u}{b_wd}=\frac{50\times10^3}{250\times550}=0.364\ \text{N/mm}^2 $$ $p_t=100A_{st}/(b_wd)=2.14\%$, $\tau_c=0.80$ N/mm$^2$ (IS 456 Table 19, M20). Since $\tau_v<\tau_c$, only minimum shear reinforcement is needed: $\dfrac{A_{sv}}{b\,s_v}\ge\dfrac{0.4}{0.87f_y}$. For 8 mm 2-legged stirrups ($A_{sv}=100.5$ mm$^2$): $s_v\le\dfrac{0.87\times500\times100.5}{0.4\times250}=437$ mm; also $\le0.75d=412$ mm and 300 mm. Provide 8 mm 2-legged stirrups at 200 mm c/c (closer than 300 mm to support the compression bars). **Answer:** $A_{st}=2943$ mm$^2$ (6-25 mm), $A_{sc}=592$ mm$^2$ (3-16 mm); shear: 8 mm 2-legged stirrups at 200 mm c/c. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is Df/d = 0.182 checked in the limiting moment of an L-section? A. To decide the stirrup spacing B. To find the number of bars C. To check deflection D. If Df/d <= 0.2 the flange thickness Df is used directly in the formula Answer: D. For Df/d <= 0.2, Df is used as the flange depth; otherwise the reduced yf is used. (Key point: Df/d check) 2. What are the two parts of Mu,lim for the L-beam with NA in web? A. Web rectangle 0.36fck bw xu,max(d-0.416xu,max) plus flange overhang 0.45fck(bf-bw)Df(d-Df/2) B. Web moment only C. Flange moment with width bw D. 0.138 fck bf d^2 only Answer: A. The web block and the uniformly stressed flange overhang are added, giving 202.54 + 292.50 = 495.04 kN·m. (Key point: Limiting moment parts) 3. Mu = 615 kN·m versus Mu,lim = 495.04 kN·m. What follows? A. Increase the flange width only B. Compression steel is needed to carry Mu2 = 119.96 kN·m C. No extra steel; just add more tension bars D. Section is over-safe Answer: B. Mu > Mu,lim means the section would be over-reinforced if only tension steel were added. (Key point: Compression steel need) 4. tau_v = 0.364 is less than tau_c = 0.80. What shear steel is provided? A. Full design for Vus = Vu - tau_c b d B. No stirrups at all C. Minimum shear reinforcement, Asv/(b sv) >= 0.4/(0.87 fy) D. Bent-up bars only Answer: C. When tau_v < tau_c only minimum stirrups are needed; spacing is also limited to 0.75d and 300 mm. 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