---
title: "Bent-up bars and stirrups, 5-25 mm bars"
subject: "Design of RCC Structure"
chapter: 5
key: "semester_7/design_of_rcc_structure:bent-up-bars-and-stirrups-5"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/bent-up-bars-and-stirrups-5
site: BCT Engineering
---

# Bent-up bars and stirrups, 5-25 mm bars

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

- Asked in: 2080 Baisakh, 10 marks
- Video: [Shear Design of Beam Explained | Vertical Stirrups, Bent-up Bars & Formula](https://www.youtube.com/watch?v=UjNSJgWMWWg) — Prikshit Civil Stories 
- 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/bent-up-bars-and-stirrups-5
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-5

## Question

An RC beam with effective depth of 550 mm and breadth of 400 mm contains 5#25 mm diameter bars, out of which two bars are to be bent-up at 45$^\circ$ near the end of the support. The beam carrying a uniformly distributed factored load of 100 kN/m over a 6 m clear span. Calculate the shear resistance of the bent-up bars and design the additional stirrups if needed. Use M20 grade concrete and Fe415 grade steel.

## Short answer

At d from support, Vud=245 kN and Vus=127.5 kN. Bent-up bars give 250.6 kN but are limited to 63.7 kN by the 50% rule. Provide 8 mm 2-legged stirrups at 225 mm c/c near the support.

## Answer

**Given:** $b=400$ mm, $d=550$ mm, 5-25 mm bars (two bent up at 45$^\circ$), $w_u=100$ kN/m on 6 m clear span, M20, Fe415.

**Step 1: Design shear force**

$$
V_u=\frac{w_uL}{2}=\frac{100\times6}{2}=300.0\ \text{kN (at support face)}
$$

Critical section at $d$ from the support face: $V_{u,d}=300.0-100\times0.55=245.0$ kN.

$\tau_v=\dfrac{245.0\times10^3}{400\times550}=1.114$ N/mm$^2<\tau_{c,max}=2.8$ N/mm$^2$, OK.

**Step 2: Shear resistance of concrete.** Bars continuing to the support: 3-25, $A_{st}=1472.6$ mm$^2$.

$$
p_t=\frac{100\times1472.6}{400\times550}=0.669\%\ \Rightarrow\ \tau_c=0.534\ \text{N/mm}^2\ (\text{Table 19})
$$

$$
V_c=\tau_cbd=0.534\times400\times550=117.5\ \text{kN}
$$

Shear to be carried by shear reinforcement: $V_{us}=245.0-117.5=127.5$ kN.

**Step 3: Shear resistance of the bent-up bars**

$$
V_{us,bent}=0.87f_yA_{sb}\sin\alpha=0.87\times415\times981.7\times\sin45^\circ=250.6\ \text{kN}
$$

IS 456 (cl. 40.4) allows bent-up bars to carry at most **half** of the shear reinforcement requirement, i.e. $0.5\times127.5=63.7$ kN. So the useful contribution of the bent bars is 63.7 kN.

**Step 4: Additional stirrups.** Balance shear $=127.5-63.7=63.7$ kN. For 8 mm 2-legged stirrups ($A_{sv}=100.5$ mm$^2$):

$$
s_v=\frac{0.87f_yA_{sv}d}{V_{us}}=\frac{0.87\times415\times100.5\times550}{63.7\times10^3}=313\ \text{mm}
$$

Limits: $\dfrac{0.87f_yA_{sv}}{0.4b}=227$ mm, $0.75d=412$ mm and 300 mm. Provide **8 mm 2-legged stirrups at 225 mm c/c** near the support.

**Answer:** Shear resistance of bent-up bars $=250.6$ kN (limited to 63.7 kN by the 50% rule); additional stirrups 8 mm 2L @ 225 mm c/c.

## Check yourself

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

1. Which bars are used to compute pt and tau_c?
   A. All five 25 mm bars
   B. Only the two bent-up bars
   C. Two bars only
   D. The 3-25 mm straight bars continuing to the support
   Answer: D. pt = 100x1472.6/(400x550) = 0.669%, tau_c = 0.534. (Key point: Bars counted for tau_c)

2. How is Vus (shear to be carried by reinforcement) found?
   A. Vu,d - Vc = 245.0 - 117.5 = 127.5 kN
   B. Vu at support face = 300 kN
   C. Vu,d + Vc
   D. Vc only
   Answer: A. Design shear at d from the support minus concrete shear. (Key point: Shear for reinforcement)

3. The bent bars can resist 250.6 kN. How much is credited?
   A. 250.6 kN in full
   B. 63.7 kN, limited to half of Vus
   C. 127.5 kN
   D. None
   Answer: B. IS 456 cl. 40.4 limits bent-up bars to 50% of the shear reinforcement shear. (Key point: Bent bar usable)

4. What stirrup spacing was provided?
   A. 8 mm 2L at 313 mm c/c
   B. 8 mm at 300 mm c/c
   C. 8 mm 2L at 225 mm c/c, limited by the minimum steel spacing (227 mm)
   D. 8 mm at 412 mm c/c
   Answer: C. Calculated 313 mm exceeds the 227 mm minimum-steel limit, so 225 mm governs. (Key point: Stirrup spacing)

## Close variants set in other exams

- [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×

## Related in Design of RCC Structure

- [Bent-up bars, 600 mm effective depth](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Abent-up-bars-600-mm-effective) — Design of RCC Structure, chapter 5, asked 2×
- [Shear at bent-up section, 8 mm stirrups](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Ashear-at-bent-up-section-8) — Design of RCC Structure, chapter 5