---
title: "Shear at bent-up section, 8 mm stirrups"
subject: "Design of RCC Structure"
chapter: 5
key: "semester_7/design_of_rcc_structure:shear-at-bent-up-section-8"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/shear-at-bent-up-section-8
site: BCT Engineering
---

# Shear at bent-up section, 8 mm stirrups

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

- Asked in: 2069 Chaitra, 10 marks
- Video: [Design of Shear Reinforce | RCC Numerical | Bent up Bar | Stirrups Design](https://www.youtube.com/watch?v=y5VzBZ9NvPo) — 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/shear-at-bent-up-section-8
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-5

## Question

Find the shear resisting capacity of a rectangular beam of 300 mm x 500 mm at the section of bent up bar. Angle of inclination of bent up bar is 45$^\circ$. Consider M20 and Fe415 grade of concrete and steel. [Figure: 8 mm dia @ 150 mm c/c two legged vertical stirrups; 4-25 mm dia tension bars of which 2-25 mm dia are bent up]

## Short answer

At bent-up section (d = 450 mm), concrete shear Vc = 74.6 kN, stirrup shear Vus = 108.9 kN, bent-up bar shear limited to 108.9 kN (50% rule). Total factored shear capacity = 292.4 kN. Web crushing limit Vmax = 378 kN > 292.4 kN, so section is adequate.

## Answer

**Given:** $b=300$ mm, $D=500$ mm (assume overall; effective cover 50 mm, $d=450$ mm), 4-25 mm bars of which 2 are bent up at 45$^\circ$, 8 mm 2-legged stirrups @ 150 mm c/c, M20, Fe415.

**Step 1: Shear capacity of concrete.** At the section where the bent-up bars act, the straight bars are 2-25, $A_{st}=981.7$ mm$^2$:

$$
p_t=\frac{100\times981.7}{300\times450}=0.727\%\ \Rightarrow\ \tau_c=0.553\ \text{N/mm}^2
$$

$$
V_c=\tau_cbd=0.553\times300\times450=74.6\ \text{kN}
$$

**Step 2: Stirrups** ($A_{sv}=2\times50.27=100.5$ mm$^2$):

$$
V_{us}=\frac{0.87f_yA_{sv}d}{s_v}=\frac{0.87\times415\times100.5\times450}{150}=108.9\ \text{kN}
$$

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

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

IS 456 cl. 40.4: bent-up bars can carry at most half of the shear reinforcement, i.e. not more than the stirrups. So the usable bent-bar contribution is $\min(250.6,\ 108.9)=108.9$ kN.

**Step 4: Total shear capacity**

$$
V_u=V_c+V_{us}+V_{us,bent}=74.6+108.9+108.9=292.4\ \text{kN}
$$

Upper limit from $\tau_{c,max}=2.8$ N/mm$^2$: $V_{max}=2.8\times300\times450=378$ kN, which is more than the above, so the section is not limited by web crushing.

**Answer:** Shear resisting capacity of the section at the bent-up bar $=292.4$ kN (factored). Without the 50% limit, the full bent-bar value would be $250.6$ kN.

## Check yourself

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

1. Which bars give Vc at the bent-up section?
   A. All 4 bars
   B. The 2 straight bars (981.7 mm2), Vc = 74.6 kN
   C. Bent bars only
   D. None
   Answer: B. pt = 0.727%. (Key point: Concrete part)

2. How is the stirrup contribution found?
   A. tau_c b d
   B. 0.87fy Asv sin45
   C. 0.87 fy Asv d/sv = 108.9 kN
   D. Asv/sv
   Answer: C. 8 mm 2L at 150 mm. (Key point: Stirrup part)

3. Bent bars give 250.6 kN. What is counted?
   A. 250.6
   B. 74.6
   C. 0
   D. 108.9 kN, limited to the stirrup value (half of total reinforcement shear)
   Answer: D. The 50% rule means bent bars can carry no more than the stirrups. (Key point: Bent bar credit)

4. What is the total capacity?
   A. Vc + Vus + Vus,bent = 74.6 + 108.9 + 108.9 = 292.4 kN
   B. 74.6 + 108.9
   C. 378 kN
   D. 74.6 + 250.6
   Answer: A. It is below Vmax = 378 kN, so no web crushing. (Key point: Total)

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