--- title: "Beam design for flexure, shear, torsion, 650x300" subject: "Design of RCC Structure" chapter: 5 key: "semester_7/design_of_rcc_structure:beam-design-for-flexure-shear-torsion" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/beam-design-for-flexure-shear-torsion site: BCT Engineering --- # Beam design for flexure, shear, torsion, 650x300 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/beam-design-for-flexure-shear-torsion - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-5 ## Question A rectangular RC beam of overall dimensions 650 mm by 300 mm is subjected to a factored bending moment of 85 kN-m, factored shear force of 110 kN and factored twisting moment of 25 kN-m. Design the beam for longitudinal and transverse reinforcements. Use M25 concrete and Fe415 grade steel. ## Short answer Longitudinal reinforcement: 4-16 mm bars at bottom (804 mm²), 2-12 mm at top, and 1-12 mm side face bar on each face. Transverse reinforcement: 8 mm 2-legged closed stirrups at 115 mm c/c. Equivalent shear Ve = 243.3 kN, equivalent moment Me1 = 131.57 kN·m; section is adequate as τve = 1.35 N/mm² < τc,max = 3.1 N/mm². ## Answer **Given:** $b=300$ mm, $D=650$ mm, $M_u=85$ kN·m, $V_u=110$ kN, $T_u=25$ kN·m, M25, Fe415. Assume effective cover 50 mm: $d=600$ mm; $b_1=300-100=200$ mm, $d_1=650-100=550$ mm. **Step 1: Equivalent shear and moment** (IS 456 cl. 41.3) $$ \begin{aligned} V_e&=V_u+1.6\frac{T_u}{b}=110+1.6\times\frac{25}{0.3}=243.3\ \text{kN}\\ M_t&=T_u\frac{1+D/b}{1.7}=25\times\frac{1+650/300}{1.7}=46.57\ \text{kN·m}\\ M_{e1}&=M_u+M_t=85+46.57=131.57\ \text{kN·m} \end{aligned} $$ $\tau_{ve}=\dfrac{243.3\times10^3}{300\times600}=1.352$ N/mm$^2<\tau_{c,max}=3.1$ N/mm$^2$ (M25), so the section is adequate. $M_tM_{e1}$, hence singly reinforced: $$ M_{e1}=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)\ \Rightarrow\ A_{st}=646\ \text{mm}^2 $$ Minimum steel $0.85bd/f_y=369$ mm$^2$. Provide 4-16 mm bars ($804$ mm$^2$) at the bottom and 2-12 mm at the top as hangers. **Side face bars:** $D=650$ mm $>450$ mm, so for torsion (cl. 26.5.1.7(d)) provide $0.1\%\times300\times650=195$ mm$^2$ (98 mm$^2$ per face): 1-12 mm bar on each face at mid-depth ($113$ mm$^2$), giving a spacing of about 275 mm $<300$ mm. **Step 3: Transverse reinforcement.** $p_t=0.45\%$, $\tau_c=0.46$ N/mm$^2$; $\tau_{ve}>\tau_c$. $$ \frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_u}{2.5d_1(0.87f_y)}=0.629+0.222=0.851\ \text{mm}^2/\text{mm} $$ Code minimum: $\dfrac{(\tau_{ve}-\tau_c)b}{0.87f_y}=0.739$ mm$^2$/mm, so 0.851 governs. For 8 mm 2-legged closed stirrups ($A_{sv}=100.5$ mm$^2$): $s_v=118$ mm. Limits: $x_1=225$ mm, $(x_1+y_1)/4=200$ mm, 300 mm. **Answer:** Longitudinal steel 4-16 mm at the bottom ($A_{st}=646$ mm$^2$ required for $M_{e1}=131.57$ kN·m), 2-12 mm at the top and 1-12 mm side face bar on each face; transverse reinforcement: closed 8 mm 2-legged stirrups at 115 mm c/c. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is Me1? A. 85 kN.m B. 85 - 46.57 C. 46.57 kN.m D. Mu + Mt = 85 + 46.57 = 131.57 kN.m Answer: D. Mt = 25x(1 + 650/300)/1.7. (Key point: Me1) 2. Why singly reinforced? A. Mu,lim = 372.5 kN.m exceeds Me1 B. Mt < Mu C. tau_ve is small D. Mild exposure Answer: A. No compression steel needed for flexure. (Key point: Singly reinforced) 3. Required 646 mm2 versus minimum 369 mm2. What is done? A. Provide 369 mm2 B. Provide at least the larger, 4-16 mm (804 mm2) C. Provide 2-12 mm only D. Ignore minimum Answer: B. Provided area must satisfy both. (Key point: Minimum steel) 4. Which value governs for Asv/sv? A. 0.739 B. 0.629 alone C. 0.851 mm2/mm (torsion plus shear) over 0.739 mm2/mm D. 0.222 alone Answer: C. 8 mm 2L closed at 115 mm c/c. 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