👋 Introduction

Shear design is one of the essential check in reinforced concrete beam design, as shear force existed in most situation. This article aims to demonstrate the steps involved in calculating the shear resistance of a section with given shear links through worked numerical examples.

The concept behind and corresponding equations used can be found in other articles What is shear resistance of reinforced concrete section in Eurocode 2? and Fundamental Concept of Shear Reinforcement Design of Reinforced Concrete Section in Eurocode 2.

📐 Design Data

Below are the design information of the shear analysis of the section.

Geometry
Section
	Width	400mm
	Height	600mm
Reinforcement
	Shear links	R12@150c/c

Material
Concrete
	Strength	30 MPa
Rebar
	Strength	500 MPa

Assumption
$cover, cc$	=	$50mm$
$tension \ reinforcement \ dia., \phi_{tension}$	=	$25mm$

🔢 Calculation Steps

*Unless otherwise specified, all clauses and equations quoted herein are based on BS EN 1992-1-1:2004.

1. Determine compression strut angle

1.1Formula to be used

By equaling the below formulas, find the inclined angle of compression strut, θ
$V_{Rd,s}$	=	$V_{Rd,max}$
$\frac{A_{sw}}{s}zf_{ywd}cot\theta$	=	$\frac{\alpha b_{w}zv_{1}f_{cd}}{cot\theta + tan\theta}$
$\theta$	=	$sin^{-1}\sqrt{\frac{A_{sw}f_{ywd}}{\alpha b_{w}svf_{cd}}}$

1.2Compute required parameters

Effective Depth ^ⓘ
$d$	=	$h - cc - \phi_{shear} - \phi_{tension}/2$
	=	$600 mm - 50 mm - 12 mm - 25 mm /2$
	=	$525 mm$

Strength Reduction Factor of Concrete(Eq. 6.6N)
$v$	=	$0.6 \cdot (1 - \frac{f_{ck}}{250})$
	=	$0.6 \cdot (1 - \frac{30 MPa}{250})$
	=	$0.53$

1.3Calculate strut angle

Inclined angle of compression strut
$\theta$	=	$sin^{-1}\sqrt{\frac{A_{sw} \cdot f_{ywd}}{\alpha \cdot b_{w} \cdot s \cdot v \cdot f_{cd}}}$
	=	$sin^{-1}\sqrt{\frac{226mm^2 \cdot 500 MPa}{1 \cdot 400mm \cdot 150mm \cdot 0.53 \cdot 17MPa}}$
	=	$42.6°, OK, where 21.8° < \theta < 45° (Eq 6.7)$

2. Calculate shear resistance by the given section

Max Shear Resistance (From Concrete)(Eq. 6.9)
$V_{Rd,max}$	=	$\frac{\alpha \cdot b_{w} \cdot z \cdot v_{1} \cdot f_{cd}}{cot\theta + tan\theta}$
	=	$\frac{1 \cdot 400mm \cdot 150mm \cdot 0.53 \cdot 17MPa}{cot(42.6°) + tan(42.6°)}$
	=	$849kN$

Shear Resistance (From Shear Links)(Eq. 6.8)
$V_{Rd,s}$	=	$\frac{A_{w}}{s} \cdot z \cdot f_{ywd} \cdot cot\theta$
	=	$\frac{226mm^2}{150mm} \cdot 0.9 \cdot 525mm \cdot 500MPa \cdot cot(42.6°)$
	=	$388kN < 849kN, OK!$

∴ The Shear Capacity of the Section is 388kN.

📄 Need a Full Design Check Report?

If you want similiar calculation analysis with your own design data, feel free to try our application, what you need is just the design data value, and a ready-to-submit design PDF report will be generated in a minute.

The ready-to-submit PDF Report is not generated by AI but programmed by chartered engineer, which is accurate, no hallucination and same input same output.

#rc design #shear design #work example #structural

CivilSimple Team

The CivilSimple Team writes practical engineering guides for the profession and the curious. All articles are reviewed for technical accuracy before publication.