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

# Shear reinforcement of T-beam, 250 kN

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

- Asked in: 2072 Chaitra, 6 marks
- Video: [T-beam Shear Design Strength Example - Reinforced Concrete](https://www.youtube.com/watch?v=FKPbL5oqMZ4) — structurefree
- 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-of-t-beam-250
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-5

## Question

Find shear reinforcement required for a beam as shown in figure. Beam is subjected to design SF of 250 kN. Consider M25 and Fe500 grade of concrete and steel. [Figure: T-beam with flange width 1200 mm, flange depth 125 mm, web width 250 mm, overall depth 500 mm, reinforced with 4-20 mm dia bars]

## Short answer

For T-beam web (bw = 250 mm, d = 450 mm), τv = 2.22 N/mm² < τc,max = 3.1 N/mm². Concrete shear strength Vc = 75.2 kN (τc = 0.668 N/mm² for pt = 1.117%). Shear carried by stirrups Vus = 174.8 kN. Provide 10 mm 2-legged vertical stirrups at 175 mm c/c near support.

## Answer

**Given (from figure):** $b_f=1200$ mm, $D_f=125$ mm, $b_w=250$ mm, $D=500$ mm, 4-20 mm bars ($A_{st}=1256.6$ mm$^2$), $V_u=250$ kN, M25, Fe500. Assume effective cover 50 mm: $d=450$ mm. In a T-beam, shear is resisted by the web, so $b_w$ is used.

**Step 1: Nominal shear stress**

$$
\tau_v=\frac{V_u}{b_wd}=\frac{250\times10^3}{250\times450}=2.222\ \text{N/mm}^2<\tau_{c,max}=3.1\ \text{N/mm}^2\ (\text{M25})
$$

**Step 2: Concrete shear strength**

$$
p_t=\frac{100A_{st}}{b_wd}=\frac{100\times1256.6}{250\times450}=1.117\%\ \Rightarrow\ \tau_c=0.668\ \text{N/mm}^2
$$

$$
V_c=\tau_cb_wd=75.2\ \text{kN}
$$

**Step 3: Shear taken by stirrups**

$$
V_{us}=V_u-V_c=250-75.2=174.8\ \text{kN}
$$

Using 10 mm 2-legged stirrups ($A_{sv}=157.1$ mm$^2$), Fe500:

$$
s_v=\frac{0.87f_yA_{sv}d}{V_{us}}=\frac{0.87\times500\times157.1\times450}{174.8\times10^3}=176\ \text{mm}
$$

Limits: $\dfrac{0.87f_yA_{sv}}{0.4b_w}=683$ mm, $0.75d=337.5$ mm and 300 mm; the calculated value governs.

**Answer:** Provide 10 mm 2-legged vertical stirrups at 175 mm c/c near the support.

## Check yourself

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

1. Why use bw = 250 mm in tau_v?
   A. Flange resists shear
   B. bf gives conservative result
   C. Shear is resisted by the web of a T-beam
   D. It is given
   Answer: C. tau_v = 250e3/(250x450) = 2.222. (Key point: Width)

2. What is Vc?
   A. 250 kN
   B. 174.8 kN
   C. bf d tau_c
   D. tau_c bw d = 0.668x250x450 = 75.2 kN
   Answer: D. pt = 1.117%. (Key point: Vc)

3. Stirrup spacing result?
   A. 10 mm 2L at 175 mm c/c (from Vus = 174.8 kN)
   B. 10 mm at 300 mm
   C. 8 mm at 337 mm
   D. 10 mm at 683 mm
   Answer: A. sv = 0.87x500x157.1x450/174.8e3 = 176 mm. (Key point: Spacing)

## Related in Design of RCC Structure

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