--- title: "Design of member for bending, shear, torsion" subject: "Design of RCC Structure" chapter: 5 key: "semester_7/design_of_rcc_structure:design-of-member-for-bending-shear" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/design-of-member-for-bending-shear site: BCT Engineering --- # Design of member for bending, shear, torsion Design of RCC Structure · Chapter 5: Design for Shear and Torsion - Asked in: 2072 Chaitra, 6 marks; 2065 Kartik (old course), 8 marks; 2076 Asoj, 5 marks; 2075 Asoj, 6 marks; 2071 Chaitra, 4 marks; 2066 Bhadra (old course), 6 marks (asked 6 times) - Video: [Design of RC members for combined Bending, Shear and Torsion](https://www.youtube.com/watch?v=qCjNGbSdFWM) — RCET CIVIL - 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-of-member-for-bending-shear - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-5 ## Question Explain how a RC structural member subjected to bending moment, shear force and torsion is designed by IS code method (write down the step-by-step procedure). ## Short answer Convert torsion to equivalent shear Ve = Vu + 1.6Tu/b and equivalent moments Me1 = Mu + Mt, Me2 = Mt - Mu (if Mt > Mu). Design longitudinal steel for Me1 (and Me2 on compression face). Provide closed stirrups for combined shear and torsion: Asv/sv = Tu/(b1d1·0.87fy) + Vu/(2.5d1·0.87fy) with spacing limits. ## Answer A member under combined bending, shear and torsion is designed in IS 456:2000 (cl. 41) by converting the torsion into an **equivalent shear** and an **equivalent bending moment**, and then designing for flexure and shear in the usual way. #### Step-by-step procedure 1. **Analysis and loads:** find the factored $M_u$, $V_u$ and $T_u$ at the critical sections. 2. **Section size:** assume $b$, $D$ and cover; find $d$. 3. **Equivalent shear:** $$ V_e=V_u+1.6\frac{T_u}{b} $$ Equivalent nominal shear stress $\tau_{ve}=\dfrac{V_e}{bd}$. 4. **Check section size:** $\tau_{ve}\le\tau_{c,max}$ (Table 20: 2.8 for M20, 3.1 for M25). If not, increase the section. 5. **Equivalent moments:** $$ M_t=T_u\,\frac{1+D/b}{1.7},\qquad M_{e1}=M_u+M_t $$ $M_{e1}$ is resisted by longitudinal steel on the flexural tension face. 6. **Compression face steel:** if $M_t>M_u$, an extra moment $M_{e2}=M_t-M_u$ acts in the opposite sense and needs extra longitudinal steel on the compression face (designed as tension steel for $M_{e2}$). 7. **Longitudinal steel** from $M_{e1}$ (and $M_{e2}$) using the flexure formula: $$ M_e=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right) $$ If $M_{e1}>M_{u,lim}$, the section is doubly reinforced. 8. **Transverse reinforcement:** find $\tau_c$ from $p_t$ (Table 19). If $\tau_{ve}>\tau_c$, provide closed stirrups: $$ \frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_u}{2.5d_1(0.87f_y)}\ \ \ge\ \frac{(\tau_{ve}-\tau_c)b}{0.87f_y} $$ $b_1$, $d_1$ = centre-to-centre distances between the corner bars in the width and depth directions. 9. **Spacing limits:** $s_v\le x_1$, $\le(x_1+y_1)/4$ and $\le300$ mm, where $x_1$ and $y_1$ are the short and long dimensions of the stirrup. 10. **Detailing (cl. 26.5.1.7):** stirrups must be closed, with 135$^\circ$ hooks, and there must be at least one longitudinal bar in each corner of the stirrups. When a cross-sectional dimension exceeds 450 mm, provide additional longitudinal (side face) bars as in cl. 26.5.1.3: a total of 0.1% of the web area, divided equally between the two faces, at a spacing of not more than 300 mm or the web thickness. 11. **Check** deflection, cracking and development length. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is torsion converted to shear in IS 456? A. Ve = Vu + 1.6 Tu / b B. Ve = Vu - Tu/b C. Ve = Tu/b only D. Ve = Vu + Tu/D Answer: A. Equivalent shear adds 1.6 Tu/b to Vu, and tau_ve = Ve/(bd). (Key point: Equivalent shear) 2. How is the equivalent bending moment found? A. Me1 = Mu - Mt B. Me1 = Mu + Mt, Mt = Tu(1 + D/b)/1.7 C. Me1 = Tu only D. Me1 = Mu/Mt Answer: B. Torsion adds Mt to the flexural moment on the tension face. (Key point: Equivalent moment) 3. When is extra longitudinal steel needed on compression face? A. Always B. When Mt < Mu C. When Mt > Mu, for Me2 = Mt - Mu D. Never Answer: C. Torsion moment larger than bending moment reverses tension to the compression face. (Key point: Compression face steel) 4. What is checked first with tau_ve? A. tau_ve >= tau_c B. tau_ve = 0 C. tau_ve <= 0.1 fck D. tau_ve <= tau_c,max, else increase the section Answer: D. If tau_ve exceeds tau_c,max, the section is enlarged. 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