---
title: "Shear reinforcement, 250x500 beam, Fe500"
subject: "Design of RCC Structure"
chapter: 5
key: "semester_7/design_of_rcc_structure:shear-reinforcement-250x500-beam-fe500"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/shear-reinforcement-250x500-beam-fe500
site: BCT Engineering
---

# Shear reinforcement, 250x500 beam, Fe500

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

- Asked in: 2074 Asoj, 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-reinforcement-250x500-beam-fe500
- 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 depth) is subjected to a factored shear force of 110 kN. The beam is reinforced with 3-22 mm dia. bars in tension. Design the shear reinforcement. Consider M20 concrete and Fe500 steel.

## Short answer

Nominal shear stress τv = 0.88 N/mm² < τc,max = 2.8 N/mm². Concrete shear strength Vc = 74.9 kN (τc = 0.599 N/mm² for pt = 0.912%). Shear carried by stirrups Vus = 35.1 kN. Provide 8 mm 2-legged vertical stirrups at 300 mm c/c (maximum spacing governs).

## Answer

**Given:** $b=250$ mm, $d=500$ mm, $V_u=110$ kN, 3-22 mm bars ($A_{st}=1140.4$ mm$^2$), M20, Fe500.

**Step 1: Nominal shear stress**

$$
\tau_v=\frac{V_u}{bd}=\frac{110\times10^3}{250\times500}=0.880\ \text{N/mm}^2<\tau_{c,max}=2.8\ \text{N/mm}^2
$$

**Step 2: Concrete shear strength**

$$
p_t=\frac{100\times1140.4}{250\times500}=0.912\%\ \Rightarrow\ \tau_c=0.599\ \text{N/mm}^2\ (\text{Table 19, M20})
$$

$$
V_c=\tau_cbd=0.599\times250\times500=74.9\ \text{kN}
$$

$\tau_v>\tau_c$, so shear reinforcement is required.

**Step 3: Stirrups**

$$
V_{us}=110-74.9=35.1\ \text{kN}
$$

Using 8 mm 2-legged stirrups ($A_{sv}=100.5$ mm$^2$), Fe500:

$$
s_v=\frac{0.87f_yA_{sv}d}{V_{us}}=\frac{0.87\times500\times100.5\times500}{35.1\times10^3}=622\ \text{mm}
$$

**Step 4: Spacing limits**

- Minimum shear reinforcement: $s_v\le\dfrac{0.87f_yA_{sv}}{0.4b}=437$ mm
- $s_v\le0.75d=375$ mm
- $s_v\le300$ mm

The calculated spacing is larger than 300 mm, so the maximum spacing governs.

**Answer:** Provide 8 mm 2-legged vertical stirrups at 300 mm c/c.

## Check yourself

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

1. What is Vc?
   A. 110 kN
   B. 35.1 kN
   C. tau_v bd
   D. tau_c b d = 0.599x250x500 = 74.9 kN
   Answer: D. pt = 0.912% gives tau_c = 0.599. (Key point: Concrete shear)

2. What shear do stirrups carry?
   A. Vus = 110 - 74.9 = 35.1 kN
   B. 110 kN
   C. 74.9 kN
   D. 0
   Answer: A. tau_v = 0.88 > tau_c so reinforcement is needed. (Key point: Stirrup shear)

3. Calculated spacing is 622 mm. What is provided?
   A. 622 mm
   B. 300 mm c/c, the maximum spacing limit
   C. 437 mm
   D. 375 mm
   Answer: B. Limits are 437, 375 and 300 mm; the least governs. (Key point: Final spacing)

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