---
title: "Prove shear stirrup spacing or Ld formula"
subject: "Design of RCC Structure"
chapter: 6
key: "semester_7/design_of_rcc_structure:prove-shear-stirrup-spacing-or-ld"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/prove-shear-stirrup-spacing-or-ld
site: BCT Engineering
---

# Prove shear stirrup spacing or Ld formula

Design of RCC Structure · Chapter 6: Design for bond and development length

- Asked in: 2068 Baisakh (old course), 5 marks
- Video: [Derivation for development length ! RCC !](https://www.youtube.com/watch?v=vYC_vbV7fYo) — RS EDUCATION
- 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/prove-shear-stirrup-spacing-or-ld
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-6

## Question

Prove that $S_v = \frac{0.87 f_y A_{sv} d}{V_s}$ or $L_d = \frac{0.87 f_y \phi}{4\tau_{bd}}$. The symbols have their usual meanings.

## Short answer

For L_d: T = 0.87 f_y πφ²/4 equals bond force τ_bd πφ L_d, giving L_d = 0.87 f_y φ/(4 τ_bd). For stirrup spacing: a 45° crack crosses d/S_v stirrups, each carrying 0.87 f_y A_sv, so V_s = 0.87 f_y A_sv d/S_v, hence S_v = 0.87 f_y A_sv d/V_s.

## Answer

#### Proof of $L_d=\dfrac{0.87f_y\,\phi}{4\,\tau_{bd}}$

Tension in a bar of diameter $\phi$ at design stress $0.87f_y$:

$$
T=0.87f_y\times\frac{\pi\phi^2}{4}
$$

Resisting bond force along embedded length $L_d$ at average design bond stress $\tau_{bd}$ on the surface area $\pi\phi L_d$:

$$
F=\tau_{bd}\,\pi\phi\,L_d
$$

Equating $T=F$:

$$
0.87f_y\frac{\pi\phi^2}{4}=\tau_{bd}\pi\phi L_d\ \Rightarrow\ L_d=\frac{0.87f_y\,\phi}{4\,\tau_{bd}}
$$

#### Proof of $S_v=\dfrac{0.87f_yA_{sv}d}{V_s}$

Consider a diagonal crack at 45$^\circ$ in a beam of effective depth $d$. The horizontal length of the crack is about $d$ (for a vertical stirrup the crack crosses $d/s_v$ stirrups).

```text
        crack at 45 deg
   ____________/_________
  |           /  |  |  |
  |          /   |  |  |  stirrups, spacing Sv
  |_________/____|__|__|
  |<--- d --->|
```

Number of stirrups crossing the crack $=\dfrac{d}{S_v}$. Each stirrup (two legs, total area $A_{sv}$) carries design tension $0.87f_yA_{sv}$ at yield. The vertical component of the force carried by the stirrups equals the shear $V_s$ assigned to them (vertical stirrups):

$$
V_s=0.87f_yA_{sv}\times\frac{d}{S_v}\ \Rightarrow\ \boxed{S_v=\frac{0.87f_yA_{sv}d}{V_s}}
$$

where $V_s=V_u-\tau_cbd$ is the shear to be carried by the stirrups.

## Check yourself

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

1. In the proof of Ld, what is the surface area over which bond acts?
   A. π φ Ld
   B. 2 φ Ld
   C. π φ² / 4
   D. π φ² Ld
   Answer: A. Bond acts on the perimeter πφ times the embedded length Ld, giving F = τbd πφ Ld. (Key point: Proof of Ld)

2. In the proof of Sv, how many vertical stirrups cross a 45° crack?
   A. d × Sv
   B. Sv / d
   C. d / Sv
   D. 2d / Sv
   Answer: C. The horizontal crack projection is about d, so the number of stirrups at spacing Sv is d/Sv. (Key point: Stirrup crack model)

3. What force does each stirrup (total area Asv) carry at yield?
   A. 0.87 fy Asv / d
   B. 0.87 fy d
   C. 0.87 fy Asv
   D. fck Asv
   Answer: C. At design yield each stirrup resists 0.87 fy Asv, and Vs = 0.87 fy Asv (d/Sv). (Key point: Stirrup force)

4. How is Vs defined in Sv = 0.87 fy Asv d / Vs?
   A. Vs = Vu - τc b d, the shear left after concrete's share
   B. Vs = τc b d
   C. Vs = Vu / 2
   D. Vs = Vu + τc b d
   Answer: A. Concrete carries τc b d; the remaining Vu - τc b d must be carried by the stirrups. (Key point: Shear for stirrups)

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