---
title: "Bent-up bars, 600 mm effective depth"
subject: "Design of RCC Structure"
chapter: 5
key: "semester_7/design_of_rcc_structure:bent-up-bars-600-mm-effective"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/bent-up-bars-600-mm-effective
site: BCT Engineering
---

# Bent-up bars, 600 mm effective depth

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

- Asked in: 2075 Asoj, 14 marks
- Video: [Shear Reinforce Design | RCC Numerical | Bent up bar | Most important Questions](https://www.youtube.com/watch?v=z4alx3u28tw) — Sudip Dhakal
- 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-600-mm-effective
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-5

## Question

A reinforced concrete beam has an effective depth of 600 mm and a breadth of 400 mm. It contains 5 no of 25 mm dia bars out of which two bars are to be bent up at 45$^\circ$ near end of the support. Calculate shear resistance of bent up bars and additional stirrups needed if the factored shear force diagram is 250 kN at support and 0 kN at mid span of 6 m span beam. Use M20 grade concrete and Fe 415 steel.

## Short answer

Shear resistance of two bent-up 25 mm bars at 45° is 250.6 kN, but limited to 50% of shear reinforcement (38.0 kN) per IS 456. Additional 8 mm 2-legged stirrups at 225 mm c/c required. Shear reinforcement needed over 1.51 m from each support; nominal stirrups beyond.

## Answer

**Given:** $b=400$ mm, $d=600$ mm, 5-25 mm bars (2 bent up at 45$^\circ$, 3 straight), $V_u=250$ kN at the support reducing linearly to 0 at mid-span (span 6 m), M20, Fe415.

**Step 1: Shear force diagram.** Slope $w_u=\dfrac{250}{3}=83.33$ kN/m. At $d$ from the support:

$$
V_{u,d}=250-83.33\times0.6=200.0\ \text{kN},\qquad \tau_v=\frac{250\times10^3}{400\times600}=1.042\ \text{N/mm}^2\ (\text{at support})<2.8
$$

**Step 2: Concrete strength.** Straight bars continuing to the support: 3-25, $A_{st}=1472.6$ mm$^2$.

$$
p_t=\frac{100\times1472.6}{400\times600}=0.614\%,\quad\tau_c=0.516\ \text{N/mm}^2,\quad V_c=\tau_cbd=123.9\ \text{kN}
$$

$$
V_{us}=V_{u,d}-V_c=200.0-123.9=76.1\ \text{kN}
$$

**Step 3: Shear resistance of bent-up bars** ($A_{sb}=2\times490.87=981.7$ mm$^2$):

$$
V_{us,bent}=0.87f_yA_{sb}\sin45^\circ=0.87\times415\times981.7\times0.7071=250.6\ \text{kN}
$$

This is more than the whole requirement $V_{us}=76.1$ kN, but IS 456 cl. 40.4 limits bent-up bars to 50% of the shear reinforcement, i.e. $0.5\times76.1=38.0$ kN.

**Step 4: Additional stirrups** to carry $V_{us}-V_{bent}=38.0$ kN, with 8 mm 2-legged stirrups ($A_{sv}=100.5$ mm$^2$):

$$
s_v=\frac{0.87\times415\times100.5\times600}{38.0\times10^3}=573\ \text{mm}
$$

Limits: $\dfrac{0.87f_yA_{sv}}{0.4b}=227$ mm, $0.75d=450$ mm, 300 mm. Provide **8 mm 2-legged stirrups at 225 mm c/c**.

**Step 5: Extent of shear reinforcement.** Shear equals $V_c=123.9$ kN at $x=\dfrac{V_c}{w_u}=1.49$ m from mid-span, so shear reinforcement is needed over 1.51 m from each support; beyond it, nominal stirrups (minimum) are provided.

**Answer:** Shear resistance of the two bent-up bars $=250.6$ kN (usable 38.0 kN); 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. How is Vu,d found?
   A. 250 kN
   B. 250/2
   C. 83.33 kN
   D. Slope 250/3 = 83.33 kN/m; Vu,d = 250 - 83.33x0.6 = 200 kN
   Answer: D. Shear falls linearly to zero at mid-span. (Key point: Shear diagram)

2. What is Vus?
   A. Vu,d - Vc = 200 - 123.9 = 76.1 kN
   B. 200 kN
   C. 123.9 kN
   D. 250.6 kN
   Answer: A. Vc uses 3 straight bars: pt = 0.614%, tau_c = 0.516. (Key point: Vus)

3. What is the usable contribution of the bent bars?
   A. 250.6 kN
   B. 38.0 kN, half of Vus
   C. 76.1 kN
   D. 0
   Answer: B. 50% limit applies. (Key point: Usable bent value)

4. Over what length from each support is shear reinforcement needed?
   A. Whole 6 m
   B. 0.6 m
   C. About 1.51 m, up to where shear equals Vc (1.49 m from mid-span)
   D. 3 m
   Answer: C. x = Vc/wu = 123.9/83.33 = 1.49 m from mid-span. (Key point: Extent of shear reinforcement)

## 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 and stirrups, 5-25 mm bars](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Abent-up-bars-and-stirrups-5) — 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
- [Bent-up bars in shear strength](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Abent-up-bars-in-shear-strength) — Design of RCC Structure, chapter 5