--- title: "Shear design for 230x450 beam, 5 m span" subject: "Design of RCC Structure" chapter: 5 key: "semester_7/design_of_rcc_structure:shear-design-for-230x450-beam-5" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/shear-design-for-230x450-beam-5 site: BCT Engineering --- # Shear design for 230x450 beam, 5 m span Design of RCC Structure · Chapter 5: Design for Shear and Torsion - Asked in: 2082 Baisakh, 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/shear-design-for-230x450-beam-5 - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-5 ## Question Design the shear reinforcement for a simply supported beam of span 5.0 m having size 230 x 450 mm. It carries a central point load of 30 kN. It is reinforced with 4 bars of 16 mm diameter out of which one bar is bent having the grade of Fe 415. Use two-legged vertical stirrups of 8 mm diameter. Take M20 grade concrete. ## Short answer At critical section d from support, τv=0.325 N/mm² < τc=0.525 N/mm², so minimum shear reinforcement governs. Provide 8 mm two-legged vertical stirrups at 300 mm c/c; the bent-up bar provides extra 51.3 kN capacity. ## Answer **Given:** $L=5$ m, $b=230$ mm, $D=450$ mm, central point load 30 kN (taken as service load), 4-16 mm bars with one bent up (3 bars, $A_{st}=603.2$ mm$^2$, continue to the support), M20, Fe415, 8 mm 2-legged stirrups. Assume effective cover 40 mm, $d=410$ mm. **Step 1: Design shear force** $$ \begin{aligned} \text{Self weight}&=0.23\times0.45\times25=2.587\ \text{kN/m},\quad w_u=1.5\times2.587=3.881\ \text{kN/m}\\ P_u&=1.5\times30=45\ \text{kN}\\ V_u&=\frac{w_uL}{2}+\frac{P_u}{2}=32.20\ \text{kN (at support)} \end{aligned} $$ At the critical section $d$ from the support: $V_{u,d}=V_u-w_ud=32.20-3.881\times0.41=30.61$ kN. **Step 2: Nominal shear stress** $$ \tau_v=\frac{V_{u,d}}{bd}=\frac{30.61\times10^3}{230\times410}=0.325\ \text{N/mm}^2<\tau_{c,max}=2.8\ \text{N/mm}^2 $$ **Step 3: Shear strength of concrete.** $p_t=\dfrac{100A_{st}}{bd}=\dfrac{100\times603.2}{230\times410}=0.640\%$, so $\tau_c=0.525$ N/mm$^2$ (Table 19, M20). Since $\tau_v<\tau_c$, no stirrups are required by calculation, but minimum shear reinforcement is provided (cl. 26.5.1.6): $$ \frac{A_{sv}}{b\,s_v}\ge\frac{0.4}{0.87f_y}\ \Rightarrow\ s_v\le\frac{0.87\times415\times100.5}{0.4\times230}=395\ \text{mm} $$ Other limits: $s_v\le0.75d=308$ mm and $\le300$ mm. **Step 4: Contribution of the bent-up bar** (available as extra capacity): $V_{us}=0.87f_yA_{sb}\sin45^\circ=0.87\times415\times201.1\times0.707=51.3$ kN. Total capacity $=\tau_cbd+V_{us,bent}=49.5+51.3$ kN $>V_u$. **Answer:** Provide 8 mm 2-legged vertical stirrups at 300 mm c/c throughout (minimum shear reinforcement governs); the bent-up bar adds an extra 51.3 kN. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At what section is the design shear for stirrups taken? A. At mid-span B. At the centre of the support C. At 2d from the support D. At distance d from the face of the support Answer: D. The critical section is d from the support, giving Vu,d = 32.20 - 3.881x0.41 = 30.61 kN. (Key point: Critical section) 2. tau_v = 0.325 and tau_c = 0.525 N/mm2. What does this mean? A. Concrete alone is enough, but minimum shear reinforcement must still be provided B. Stirrups must be designed for the excess shear C. The section is unsafe and must be enlarged D. No stirrups at all are needed Answer: A. When tau_v < tau_c only minimum stirrups per cl. 26.5.1.6 are provided. (Key point: Tau_v versus tau_c) 3. What governs the stirrup spacing in this design? A. The calculated 51 mm from shear demand B. The 300 mm maximum (with 0.75d = 308 mm and 395 mm from minimum steel) C. Bent-up bar spacing D. 0.75d = 308 mm alone as the minimum value Answer: B. The least of 395, 308 and 300 mm is 300 mm. (Key point: Spacing limits) 4. Why is Ast taken as 603.2 mm2 when finding tau_c? A. All four bars count since they are in the beam B. Only the bent bar counts C. Only the three straight bars continuing to the support count; the bent bar is excluded D. Ast is taken as the minimum steel Answer: C. Tau_c depends on the tension steel that continues to the support, so pt = 0.640%. 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