---
title: "Shear design with curtailed bars, 30x45 cm"
subject: "Design of RCC Structure"
chapter: 5
key: "semester_7/design_of_rcc_structure:shear-design-with-curtailed-bars-30x45"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/shear-design-with-curtailed-bars-30x45
site: BCT Engineering
---

# Shear design with curtailed bars, 30x45 cm

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

- Asked in: 2065 Shrawan (old course), 13 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-curtailed-bars-30x45
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-5

## Question

A RC beam has an effective depth of 45 cm and breadth of 30 cm. It contains 5-20 mm dia mild steel bars, out of which 2-20 mm dia bars are curtailed at a section where shear force at service load is 100 kN. Design the shear reinforcement if the concrete is M20.

## Short answer

Factored shear Vu = 150 kN. After curtailment (3-20 mm bars), Vc = 73.4 kN, Vus = 76.6 kN. Stirrups at 125 mm c/c for shear. At cut-off section, 2/3(Vc+Vus) = 101.4 kN < Vu, so excess stirrups required: 8 mm 2-legged at 75 mm c/c over 0.75d ≈ 340 mm from cut-off point.

## Answer

**Given:** $b=300$ mm, $d=450$ mm, 5-20 mm mild steel bars (Fe250), with 2 bars curtailed at a section where the service shear is 100 kN, M20. The factored shear is $V_u=1.5\times100=150.0$ kN. Stirrups: 8 mm 2-legged mild steel ($f_y=250$ N/mm$^2$, $A_{sv}=100.5$ mm$^2$).

**Step 1: Nominal shear stress**

$$
\tau_v=\frac{V_u}{bd}=\frac{150.0\times10^3}{300\times450}=1.111\ \text{N/mm}^2<\tau_{c,max}=2.8\ \text{N/mm}^2
$$

**Step 2: Concrete strength.** After curtailment, 3-20 mm bars remain: $A_{st}=942.5$ mm$^2$.

$$
p_t=\frac{100\times942.5}{300\times450}=0.698\%\ \Rightarrow\ \tau_c=0.543\ \text{N/mm}^2,\quad V_c=\tau_cbd=73.4\ \text{kN}
$$

**Step 3: Stirrups for shear**

$$
V_{us}=V_u-V_c=150.0-73.4=76.6\ \text{kN}
$$

$$
s_v=\frac{0.87f_yA_{sv}d}{V_{us}}=\frac{0.87\times250\times100.5\times450}{76.6\times10^3}=128\ \text{mm}
$$

Spacing limits: $\dfrac{0.87f_yA_{sv}}{0.4b}=182$ mm, $0.75d=337$ mm, 300 mm. Provide $s_v=125$ mm for shear.

**Step 4: Extra provisions where bars are curtailed** (IS 456 cl. 26.2.3.2). Flexural bars may be stopped in a tension zone only if (a) the shear at the cut-off point is not more than $\tfrac23$ of the shear capacity of the section (including the shear reinforcement provided), or (b) excess stirrup area is provided along each terminated bar over a distance $\tfrac34d$ from the cut-off point, such that

$$
A_{sv,excess}\ge\frac{0.4\,b\,s_v}{f_y},\qquad s_v\le\frac{d}{8\beta_b}
$$

where $\beta_b=\dfrac{\text{area of bars cut off}}{\text{total area of bars at the section}}=\dfrac{2}{5}=0.4$, so $s_v\le\dfrac{450}{8\times0.4}=141$ mm.

Check of option (a): with stirrups at 125 mm, $V_{us}=\dfrac{0.87\times250\times100.5\times450}{125}=78.7$ kN, so $\tfrac23(73.4+78.7)=101.4$ kN $<V_u=150.0$ kN. Option (a) is not satisfied, so excess stirrups (option b) are needed.

Option (b): the stirrups must carry the shear plus the excess:

$$
\frac{A_{sv}}{s_v}=\frac{V_{us}}{0.87f_yd}+\frac{0.4b}{f_y}=\frac{76.6\times10^3}{0.87\times250\times450}+\frac{0.4\times300}{250}=0.783+0.480=1.263\ \text{mm}^2/\text{mm}
$$

$$
s_v=\frac{100.5}{1.263}=80\ \text{mm}\ (<141\ \text{mm, OK})
$$

Provide 8 mm 2-legged stirrups at 75 mm c/c in this zone.

**Answer:** 8 mm 2-legged mild steel stirrups at 125 mm c/c for shear near the cut-off section, reduced to 75 mm c/c over a length $\tfrac34d\approx340$ mm from the cut-off point along the curtailed bars.

## Check yourself

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

1. What shear is used?
   A. 100 kN
   B. 1.5x100 = 150 kN (service shear factored)
   C. 66.7 kN
   D. 225 kN
   Answer: B. Design at ultimate. (Key point: Factored shear)

2. Which Ast gives tau_c after curtailment?
   A. 5-20
   B. 2-20
   C. 3-20 mm remaining (942.5 mm2)
   D. 1256 mm2
   Answer: C. pt = 0.698%, tau_c = 0.543. (Key point: Concrete strength)

3. What is the limit sv <= d/(8 beta_b) with beta_b = 0.4?
   A. 141 mm
   B. 337 mm
   C. 300 mm
   D. 182 mm
   Answer: A. beta_b = cut bars/total = 2/5; 450/3.2 = 141, and 125 mm is provided. (Key point: Spacing limit)

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