--- title: "Design reinforcement under torsion, 400x750" subject: "Design of RCC Structure" chapter: 5 key: "semester_7/design_of_rcc_structure:design-reinforcement-under-torsion-400x750" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/design-reinforcement-under-torsion-400x750 site: BCT Engineering --- # Design reinforcement under torsion, 400x750 Design of RCC Structure · Chapter 5: Design for Shear and Torsion - Asked in: 2082 Bhadra, 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/design-reinforcement-under-torsion-400x750 - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-5 ## Question Design the reinforcement required for a rectangular section, 400 mm wide and 750 mm deep, subjected to an ultimate twisting moment of 150 kNm, combined with an ultimate hogging bending moment of 400 kNm and an ultimate shear force of 110 kN. Assume M25 concrete, Fe500 TMT steel and mild exposure conditions. Check for deflection and bond is not required. ## Short answer Equivalent shear Ve = Vu+1.6Tu/b = 710 kN; torsional moment Mt = Tu(1+D/b)/1.7 = 253.7 kN·m; Me1 = Mu+Mt = 653.7 kN·m. Top reinforcement 6–25 mm, bottom 2–16 mm, side face 2–10 mm per side, closed stirrups 12 mm @ 115 mm c/c. ## Answer **Given:** $b=400$ mm, $D=750$ mm, $T_u=150$ kN·m, $M_u=400$ kN·m (hogging), $V_u=110$ kN, M25, Fe500, mild exposure. Assume effective cover 50 mm: $d=700$ mm. Centre-to-centre distances of the corner bars: $b_1=400-2\times50=300$ mm, $d_1=750-2\times50=650$ mm. **Step 1: Equivalent shear and moment** (IS 456 cl. 41.3) $$ \begin{aligned} V_e&=V_u+1.6\frac{T_u}{b}=110+1.6\times\frac{150}{0.4}=710.0\ \text{kN}\\ M_t&=T_u\,\frac{1+D/b}{1.7}=150\times\frac{1+750/400}{1.7}=253.7\ \text{kN·m}\\ M_{e1}&=M_u+M_t=400+253.7=653.7\ \text{kN·m (on the tension face, top)} \end{aligned} $$ Since $M_t=253.7$ kN·m is less than $M_u=400$ kN·m, $M_{e2}=M_t-M_u$ is negative, so no extra steel is needed on the compression face (provide nominal bars there). **Step 2: Check the section.** $\tau_{ve}=\dfrac{V_e}{bd}=\dfrac{710.0\times10^3}{400\times700}=2.54$ N/mm$^2$ $<\tau_{c,max}=3.1$ N/mm$^2$ (M25). The section size is adequate. For Fe500, $x_{u,max}/d=0.46$: $M_{u,lim}=0.36\times0.46(1-0.42\times0.46)f_{ck}bd^2=0.1336f_{ck}bd^2=654.67$ kN·m $\ge M_{e1}=653.7$ kN·m, so the section can (just) be singly reinforced for $M_{e1}$. **Step 3: Longitudinal steel (tension face, top)** $$ M_{e1}=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)\ \Rightarrow\ A_{st}=2647\ \text{mm}^2 $$ Provide 6-25 mm dia bars ($2945$ mm$^2$) at the top, $p_t=1.05\%$. **Side face bars:** since $D=750$ mm $>450$ mm, cl. 26.5.1.7(d) requires extra longitudinal bars: $0.1\%\times400\times750=300$ mm$^2$, i.e. 150 mm$^2$ per face. Provide 2-10 mm bars on each face ($157$ mm$^2$), so the bar spacing up the side is less than 300 mm. **Step 4: Transverse (stirrup) reinforcement** (IS 456 cl. 41.4.3). $\tau_c=0.65$ N/mm$^2$ for $p_t=1.05\%$ (Table 19). Since $\tau_{ve}>\tau_c$, closed stirrups are needed: $$ \frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_u}{2.5d_1(0.87f_y)}=1.768+0.156=1.924\ \text{mm}^2/\text{mm} $$ Also $\dfrac{A_{sv}}{s_v}\ge\dfrac{(\tau_{ve}-\tau_c)\,b}{0.87f_y}=1.732$ mm$^2$/mm. The first value governs. Using 12 mm 2-legged closed stirrups ($A_{sv}=226.2$ mm$^2$): $s_v=118$ mm. Maximum spacing: $\le x_1=325$ mm, $\le\dfrac{x_1+y_1}{4}=250$ mm and $\le300$ mm. **Provide 12 mm closed 2-legged stirrups at 115 mm c/c.** **Answer:** Top steel 6-25 mm dia ($A_{st}=2647$ mm$^2$ required for $M_{e1}=653.7$ kN·m); bottom 2-16 mm bars (nominal, at the corners); side face bars 2-10 mm on each face; closed stirrups 12 mm @ 115 mm c/c. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the equivalent shear Ve obtained for combined shear and torsion? A. Ve = Vu + Tu/b B. Ve = Vu + 1.6 Tu/D C. Ve = Vu - 1.6 Tu/b D. Ve = Vu + 1.6 Tu/b Answer: D. IS 456 cl. 41.3.1 adds 1.6Tu/b to the shear; here Ve = 110 + 1.6x150/0.4 = 710 kN. (Key point: Equivalent shear) 2. What is the equivalent moment Me1 on the tension face? A. Me1 = Mu + Mt, where Mt = Tu(1 + D/b)/1.7 B. Me1 = Mu - Mt C. Me1 = Mu + Tu D. Me1 = Mt only, torsion replaces bending Answer: A. Torsion is converted to an extra bending moment Mt added to Mu, giving 400 + 253.7 = 653.7 kN.m. (Key point: Equivalent moment) 3. Why is no extra compression-face steel (Me2) needed here? A. Because the beam is hogging, compression steel is never needed B. Mt (253.7) is less than Mu (400), so Me2 = Mt - Mu is negative C. Because Me1 exceeds Mu,lim D. Because shear is small Answer: B. Me2 acts on the opposite face only when Mt exceeds Mu; here it does not. (Key point: Compression face steel) 4. Which expression gives the governing Asv/sv for torsion plus shear? A. Tu/(b d 0.87fy) only B. (Ve - Vc)/(0.87fy d) with no torsion term C. Tu/(b1 d1 0.87fy) + Vu/(2.5 d1 0.87fy) D. Vu/(b1 d1 0.87fy) + Tu/(2.5 d1 0.87fy) Answer: C. Cl. 41.4.3 uses the centre-to-centre of corner bars b1, d1; it gave 1.768 + 0.156 = 1.924 mm2/mm. 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