--- title: "Shear and torsion design, 300x400 beam" subject: "Design of RCC Structure" chapter: 5 key: "semester_7/design_of_rcc_structure:shear-and-torsion-design-300x400-beam" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/shear-and-torsion-design-300x400-beam site: BCT Engineering --- # Shear and torsion design, 300x400 beam Design of RCC Structure · Chapter 5: Design for Shear and Torsion - Asked in: 2079 Bhadra, 10 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-and-torsion-design-300x400-beam - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-5 ## Question A simply supported RCC beam 300 mm wide and 400 mm deep (effective) is reinforced with a 4-20 mm diameter bars. Design the shear reinforcement if M 25 grade of concrete and TOR steel bars is used and beam is subjected to a shear force of 130 kN and torsional moment 45 kN-m at service state. ## Short answer With service loads factored, the 300 mm section fails tau_cmax (4.62>3.1 N/mm2). Increase width to 400 mm and provide 12 mm 2-legged closed stirrups at 85 mm c/c; longitudinal steel must also be designed for Me1. ## Answer **Given:** $b=300$ mm, $d=400$ mm, 4-20 mm bars ($A_{st}=1256.6$ mm$^2$), M25, Fe415. Loads at service state: $V=130$ kN, $T=45$ kN·m. Assume $D=450$ mm and effective cover 50 mm. **Step 1: Factored loads (service $\times$ 1.5)** $$ V_u=1.5\times130=195.0\ \text{kN},\qquad T_u=1.5\times45=67.5\ \text{kN·m} $$ **Step 2: Equivalent shear and check of section** (IS 456 cl. 41.3) $$ V_e=V_u+1.6\frac{T_u}{b}=195.0+1.6\times\frac{67.5}{0.3}=555.0\ \text{kN},\qquad \tau_{ve}=\frac{555.0\times10^3}{300\times400}=4.62\ \text{N/mm}^2 $$ $\tau_{ve}>\tau_{c,max}=3.1$ N/mm$^2$ (M25). The 300 mm wide section is therefore **not adequate**; the size must be increased (IS 456 cl. 41.3.1). Increase the width to $b=400$ mm: $$ V_e=195.0+1.6\times\frac{67.5}{0.4}=465.0\ \text{kN},\qquad \tau_{ve}=\frac{465.0\times10^3}{400\times400}=2.91\ \text{N/mm}^2<3.1\ \text{(OK)} $$ **Step 3: Stirrups for the revised section.** $b_1=400-100=300$ mm, $d_1=450-100=350$ mm. $p_t=\dfrac{100\times1256.6}{400\times400}=0.79\%$, $\tau_c=0.58$ N/mm$^2$ (Table 19, M25); $\tau_{ve}>\tau_c$. $$ \frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_u}{2.5d_1(0.87f_y)}=1.781+0.617=2.398\ \text{mm}^2/\text{mm} $$ The code minimum $\dfrac{(\tau_{ve}-\tau_c)\,b}{0.87f_y}=2.577$ mm$^2$/mm is larger and governs, so $A_{sv}/s_v=2.577$ mm$^2$/mm. For 12 mm 2-legged closed stirrups ($A_{sv}=226.2$ mm$^2$): $s_v=88$ mm. Limits: $x_1=325$ mm, $(x_1+y_1)/4=175$ mm, 300 mm. **Provide 12 mm 2-legged closed stirrups at 85 mm c/c.** **If the given loads are taken as already factored** (so $V_u=130$ kN, $T_u=45$ kN·m on the original 300 mm section): $V_e=370.0$ kN, $\tau_{ve}=3.083$ N/mm$^2$ (just below 3.1, OK), $p_t=1.05\%$, $\tau_c=0.65$ N/mm$^2$, $b_1=200$ mm, $d_1=350$ mm: $$ \frac{A_{sv}}{s_v}=1.781+0.411=2.192\ \text{mm}^2/\text{mm}\ \Rightarrow\ s_v=103\ \text{mm for 12 mm 2L} $$ giving 12 mm 2-legged stirrups at 100 mm c/c. **Answer:** With the loads treated as service loads, the 300 mm width fails the $\tau_{c,max}$ check; with $b=400$ mm provide 12 mm 2L closed stirrups at 85 mm c/c (longitudinal steel to be provided for $M_{e1}$ in addition). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The given V and T are service values. What is done first? A. Use them directly as factored B. Divide by 1.5 C. Multiply by 1.15 D. Multiply by 1.5 to get Vu = 195 kN and Tu = 67.5 kN.m Answer: D. Design is done at ultimate loads. (Key point: Factoring) 2. Why is the 300 mm wide section inadequate? A. tau_ve = 4.62 exceeds tau_c,max = 3.1 N/mm2 B. tau_ve is below tau_c C. pt is too high D. Stirrups would be too close Answer: A. IS 456 cl. 41.3.1 requires enlarging the section; width is raised to 400 mm. (Key point: Section adequacy) 3. Which governs for the revised section? A. 2.398, the larger value B. (tau_ve - tau_c)b/0.87fy = 2.577 mm2/mm, larger than 2.398 C. 1.781 alone D. 0.617 alone Answer: B. The larger of the two expressions is used, giving 12 mm 2L at 85 mm c/c. (Key point: Governing Asv/sv) 4. If the loads were already factored, what changes? A. The section still fails B. No stirrups are needed C. Vu = 130 and Tu = 45 on 300 mm give a smaller requirement, 12 mm 2L at 100 mm c/c D. Spacing would become 85 mm Answer: C. Here tau_ve = 3.083 < 3.1, so the original section passes. 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