---
title: "Shear design with torsion, 250x500"
subject: "Design of RCC Structure"
chapter: 5
key: "semester_7/design_of_rcc_structure:shear-design-with-torsion-250x500"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/shear-design-with-torsion-250x500
site: BCT Engineering
---

# Shear design with torsion, 250x500

Design of RCC Structure · Chapter 5: Design for Shear and Torsion

- Asked in: 2081 Bhadra, 8 marks
- Video: [Design for Shear Reinforcement in Beams - Problems](https://www.youtube.com/watch?v=9esi2xeKCNM) — Stan Academy
- 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-with-torsion-250x500
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-5

## Question

A rectangular RC beam of size 250 mm x 500 mm (effective size) is subjected to factored shear force of 110 kN and torsional moment of 20 kNm. The beam is reinforced with 3-20 mm diameter bars in tension side. Design the shear reinforcement. Consider M20 concrete and Fe 500 steel.

## Short answer

Equivalent shear Ve = Vu+1.6Tu/b = 238 kN, τve = 1.904 N/mm² > τc. Stirrup requirement: Asv/sv = 0.906 mm²/mm. Provide 8 mm closed two-legged stirrups at 110 mm c/c.

## Answer

**Given:** $b=250$ mm, $d=500$ mm, $V_u=110$ kN, $T_u=20$ kN·m, 3-20 mm bars ($A_{st}=942.5$ mm$^2$), M20, Fe500. Assume overall depth $D=550$ mm and effective cover 50 mm, so $b_1=250-100=150$ mm, $d_1=550-100=450$ mm (centre to centre of corner bars).

**Step 1: Equivalent shear** (IS 456 cl. 41.3.1)

$$
V_e=V_u+1.6\frac{T_u}{b}=110+1.6\times\frac{20}{0.25}=238\ \text{kN}
$$

$$
\tau_{ve}=\frac{V_e}{bd}=\frac{238\times10^3}{250\times500}=1.904\ \text{N/mm}^2<\tau_{c,max}=2.8\ \text{N/mm}^2\ \text{(OK)}
$$

**Step 2: Concrete strength.** $p_t=\dfrac{100\times942.5}{250\times500}=0.754\%$, $\tau_c=0.561$ N/mm$^2$ (Table 19). $\tau_{ve}>\tau_c$, so transverse reinforcement is needed.

**Step 3: Stirrup area** (cl. 41.4.3)

$$
\frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_u}{2.5d_1(0.87f_y)}=\frac{20\times10^6}{150\times450\times435}+\frac{110\times10^3}{2.5\times450\times435}=0.681+0.225=0.906\ \text{mm}^2/\text{mm}
$$

Check with $\dfrac{(\tau_{ve}-\tau_c)b}{0.87f_y}=0.772$ mm$^2$/mm (smaller, so the above governs).

For 8 mm 2-legged closed stirrups ($A_{sv}=100.5$ mm$^2$): $s_v=\dfrac{100.5}{0.906}=111$ mm.

**Step 4: Spacing limits:** $s_v\le x_1=175$ mm, $\le\dfrac{x_1+y_1}{4}=162$ mm, $\le300$ mm.

**Answer:** Provide 8 mm dia closed 2-legged stirrups at 110 mm c/c.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is Ve for Vu = 110 kN, Tu = 20 kN.m, b = 250 mm?
   A. 130 kN
   B. 110 kN
   C. 190 kN
   D. 238 kN
   Answer: D. Ve = 110 + 1.6x20/0.25 = 238 kN. (Key point: Equivalent shear)

2. What is the check on tau_ve = 1.904 N/mm2?
   A. It must be below tau_c,max = 2.8 N/mm2 for M20, and it is
   B. It must be below tau_c, and it fails
   C. It must exceed tau_c,max
   D. It need not be checked with torsion
   Answer: A. Tau_ve < tau_c,max means section size is adequate; tau_ve > tau_c means stirrups are needed. (Key point: Section check)

3. Which value of Asv/sv governs: 0.906 or 0.772 mm2/mm?
   A. 0.772, the smaller value
   B. 0.906, the larger value
   C. Their average
   D. Neither; minimum steel governs
   Answer: B. The larger of the torsion-shear formula and (tau_ve - tau_c)b/0.87fy is used. (Key point: Governing Asv/sv)

4. Why are closed stirrups used?
   A. They save steel compared to open ones
   B. They are needed only for deformed bars
   C. Torsion needs closed ties to resist the shear flow around the section
   D. They increase tau_c
   Answer: C. Closed stirrups at 110 mm c/c resist torsional shear around the perimeter. (Key point: Closed stirrups)

## Related in Design of RCC Structure

- [Shear reinforcement, 250x500 beam, Fe500](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Ashear-reinforcement-250x500-beam-fe500) — Design of RCC Structure, chapter 5
- [Shear and torsion design, 300x400 beam](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Ashear-and-torsion-design-300x400-beam) — Design of RCC Structure, chapter 5
- [Bent-up bars shear, 550 mm effective depth](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Abent-up-bars-shear-550-mm) — Design of RCC Structure, chapter 5, asked 3×
- [Beam design for flexure, shear, torsion, 650x300](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Abeam-design-for-flexure-shear-torsion) — Design of RCC Structure, chapter 5
- [Design 200x400 beam for bending, shear, torsion](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Adesign-200x400-beam-for-bending-shear) — Design of RCC Structure, chapter 5
- [Shear reinforcement of T-beam, 250 kN](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Ashear-reinforcement-of-t-beam-250) — Design of RCC Structure, chapter 5

## Related in other subjects

- [Rectangular Sal column, 150 kN + 15 kNm](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atimber-column-sal-4m-150kn-15knm) — Design of Steel and Timber Structure, chapter 11
- [Greatest bolt force, 100 kN at 500 mm](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abolt-force-shaft-100kn-500mm) — Design of Steel and Timber Structure, chapter 4
- [Unsupported beam, M=470 kNm, V=180 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-unsupported-3-5m-470knm) — Design of Steel and Timber Structure, chapter 7
- [Frame AFD, SFD, BMD (20 kNm, 15 kN)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-20knm-15kn) — Strength of Materials, chapter 1