--- title: "Shear reinforcement, 300x500 beam, 370 kN" subject: "Design of RCC Structure" chapter: 5 key: "semester_7/design_of_rcc_structure:shear-reinforcement-300x500-beam-370-kn" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/shear-reinforcement-300x500-beam-370-kn site: BCT Engineering --- # Shear reinforcement, 300x500 beam, 370 kN Design of RCC Structure · Chapter 5: Design for Shear and Torsion - Asked in: 2075 Chaitra, 8 marks - 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/shear-reinforcement-300x500-beam-370-kn - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-5 ## Question A reinforced concrete rectangular beam has an overall depth of 500 mm and breadth of 300 mm. It consists of 5-25 mm dia bars in tension and 3-16 mm dia bars in compression. Calculate the shear reinforcement needed for a factored shear force of 370 kN. Take M20 concrete and Fe415 grade (TOR) steel. Also check the spacing for minimum shear reinforcement. ## Short answer Nominal shear stress is 2.741 N/mm2, below 2.8; Vus=267.3 kN. Provide 10 mm 2-legged vertical stirrups at 95 mm c/c near the support, with maximum spacing 300 mm elsewhere. ## Answer **Given:** $b=300$ mm, $D=500$ mm, tension steel 5-25 mm ($A_{st}=2454.4$ mm$^2$), $V_u=370$ kN, M20, Fe415. Assume effective cover 50 mm: $d=450$ mm. (Compression steel does not affect shear design.) **Step 1: Nominal shear stress** $$ \tau_v=\frac{V_u}{bd}=\frac{370\times10^3}{300\times450}=2.741\ \text{N/mm}^2<\tau_{c,max}=2.8\ \text{N/mm}^2 $$ The section size is adequate. **Step 2: Concrete shear strength** $$ p_t=\frac{100\times2454.4}{300\times450}=1.818\%\ \Rightarrow\ \tau_c=0.761\ \text{N/mm}^2\ (\text{Table 19}) $$ $$ V_c=\tau_cbd=0.761\times300\times450=102.7\ \text{kN} $$ **Step 3: Shear reinforcement** $$ V_{us}=V_u-V_c=370-102.7=267.3\ \text{kN} $$ Using 10 mm 2-legged stirrups ($A_{sv}=157.1$ mm$^2$): $$ s_v=\frac{0.87f_yA_{sv}d}{V_{us}}=\frac{0.87\times415\times157.1\times450}{267.3\times10^3}=95\ \text{mm} $$ **Step 4: Check for minimum shear reinforcement and maximum spacing** $$ s_v\le\frac{0.87f_yA_{sv}}{0.4b}=\frac{0.87\times415\times157.1}{0.4\times300}=473\ \text{mm},\quad s_v\le0.75d=337.5\ \text{mm},\quad s_v\le300\ \text{mm} $$ The calculated spacing 95 mm is less than all these limits, so it governs. **Answer:** Provide 10 mm 2-legged vertical stirrups at 95 mm c/c near the support. The spacing may be increased towards mid-span up to a maximum of 300 mm (minimum shear reinforcement) where the shear decreases. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Is the section size adequate? A. No, tau_v exceeds tau_c,max B. Not required C. Yes, since tau_v < tau_c D. Yes, tau_v = 2.741 < tau_c,max = 2.8 N/mm2 Answer: D. Web crushing is avoided. (Key point: Check of size) 2. tau_c depends on which quantities? A. pt = 100Ast/bd (1.818%) and concrete grade B. Steel grade only C. Compression steel D. Overall depth only Answer: A. Compression steel does not affect shear design; tau_c = 0.761 for M20. (Key point: Tau_c basis) 3. How is sv computed? A. sv = 0.87fy Asv d/Vu B. sv = 0.87 fy Asv d/Vus with Vus = Vu - Vc = 267.3 kN C. sv = Vus/(0.87fy Asv) D. sv = 0.4b/0.87fy Answer: B. 10 mm 2L gives 95 mm. (Key point: Spacing formula) 4. What is the purpose of the 473 mm and 300 mm limits? A. They replace the calculated spacing B. They are the actual spacing C. They ensure minimum shear reinforcement and cap spacing; 95 mm is smaller so it governs D. They apply only at mid-span of T-beams Answer: C. The least spacing governs, and spacing can increase to 300 mm towards mid-span. 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