--- title: "Beam design for bending, torsion, shear, 425x550" subject: "Design of RCC Structure" chapter: 5 key: "semester_7/design_of_rcc_structure:beam-design-for-bending-torsion-shear" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/beam-design-for-bending-torsion-shear site: BCT Engineering --- # Beam design for bending, torsion, shear, 425x550 Design of RCC Structure · Chapter 5: Design for Shear and Torsion - Asked in: 2081 Baisakh, 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/beam-design-for-bending-torsion-shear - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-5 ## Question Design a beam 425 mm x 550 mm subjected to factored bending moment of 100 kN-m, twisting moment of 28 kN-m and shear force at critical section of 100 kN. The beam is under mild exposure condition. Use M20 grade concrete and Fe415 grade steel. ## Short answer Ve = Vu+1.6Tu/b = 205.4 kN, Me1 = Mu+Mt = 137.8 kN·m. Longitudinal steel 3–20 mm (942 mm²), top 2–12 mm, side face 1–16 mm each side. Closed 8 mm stirrups at 125 mm c/c. ## Answer **Given:** $b=425$ mm, $D=550$ mm, $M_u=100$ kN·m, $T_u=28$ kN·m, $V_u=100$ kN, M20, Fe415, mild exposure. Assume effective cover 50 mm: $d=500$ mm; $b_1=425-100=325$ mm, $d_1=550-100=450$ mm. **Step 1: Equivalent forces** (IS 456 cl. 41.3) $$ \begin{aligned} V_e&=V_u+1.6\frac{T_u}{b}=100+1.6\times\frac{28}{0.425}=205.4\ \text{kN}\\ M_t&=T_u\frac{1+D/b}{1.7}=28\times\frac{1+550/425}{1.7}=37.8\ \text{kN·m}\\ M_{e1}&=M_u+M_t=100+37.8=137.8\ \text{kN·m} \end{aligned} $$ $M_tM_{e1}$, so the section is singly reinforced. **Step 3: Longitudinal steel** $$ M_{e1}=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)\ \Rightarrow\ A_{st}=831\ \text{mm}^2 $$ Provide 3-20 mm bars ($942$ mm$^2$, $p_t=0.44\%$). Minimum steel $0.85bd/f_y=435$ mm$^2$, OK. Provide 2-12 mm bars at the top corners as hangers. **Side face bars:** $D=550$ mm $>450$ mm, so cl. 26.5.1.7(d) requires extra longitudinal bars of $0.1\%\times425\times550=234$ mm$^2$ (117 mm$^2$ per face). Provide 1-16 mm bar on each face at mid-depth ($201$ mm$^2$), giving a vertical spacing of about 225 mm $<300$ mm. **Step 4: Stirrups.** $\tau_c=0.45$ N/mm$^2$; since $\tau_{ve}>\tau_c$: $$ \frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_u}{2.5d_1(0.87f_y)}=0.530+0.246=0.776\ \text{mm}^2/\text{mm} $$ This is more than the code minimum $\dfrac{(\tau_{ve}-\tau_c)b}{0.87f_y}=0.609$ mm$^2$/mm. For 8 mm 2-legged closed stirrups ($A_{sv}=100.5$ mm$^2$): $s_v=129$ mm. Limits: $x_1=350$ mm, $(x_1+y_1)/4=206$ mm, 300 mm. **Answer:** Longitudinal steel $A_{st}=831$ mm$^2$ (3-20 mm bars at the bottom), 2-12 mm at the top, 1-16 mm side face bar on each face; closed 8 mm 2-legged stirrups at 125 mm c/c. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is Mt computed? A. Mt = Tu(1 + b/D)/1.7 B. Mt = 1.6 Tu/b C. Mt = Tu/1.7 D. Mt = Tu(1 + D/b)/1.7 = 37.8 kN.m Answer: D. Tu = 28, D/b = 550/425, so Mt = 28x2.294/1.7 = 37.8 kN.m. (Key point: Mt value) 2. Why is the section singly reinforced? A. Mu,lim (293.17 kN.m) exceeds Me1 (137.8 kN.m) B. Because Mt < Mu C. Because torsion is small D. Because Fe415 is used Answer: A. If Me1 were beyond Mu,lim, compression steel would be required. (Key point: Singly or doubly) 3. What minimum tension steel check is used? A. 0.12% of bD B. 0.85 b d / fy = 435 mm2 C. 0.4 b s/0.87fy D. Equal to Asv Answer: B. A_st = 831 mm2 provided as 3-20 mm (942 mm2) exceeds this minimum. (Key point: Minimum steel) 4. What data does the stirrup design use? A. Gross b and D: Asv/sv = 0.2 B. Only shear Vu with b d C. b1 = 325, d1 = 450 mm: Asv/sv = 0.530 + 0.246 = 0.776 mm2/mm D. Ve with b1 only Answer: C. Tu term 0.530 plus Vu term 0.246 gives 0.776, so 8 mm 2L spaced at 125 mm c/c. 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