👋 Introduction

The calculation of crack width with a given section is not straight-forward, which consist of determine the neutral axis of the section under elastic stage and also finding the stress and strain at tension reinforcement. For quick steps guide, below is an example, hope you find it helpful.

📐 Design Data

Below are the design information of the torsion design of the section.

Geometry
Section
	Width	500mm
	Height	700mm
Reinforcement
	Tension Area	2412 mm²
	Shear links	R12@150c/c

Design Action
	Moment My	270kNm

Material
Concrete
	Strength	30 MPa
Rebar
	Strength	500 MPa

Assumption
$cover, cc$	=	$50mm$

🔢 Calculation Steps

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

0. Material factor to be used

$E_{s}$	=	$200 GPa$
$E_{cm}$	=	$33GPa$
$\alpha_{e}$	=	$\frac{E_{s}}{E_{cm}} = 6.09$

1. Find strain difference for eq 7.9

1.1 Find stress at reinforcement level

Effective Depth ^ⓘ
$d$	=	$h - cc - \phi_{shear} - \phi_{tension}/2$
	=	$700 mm - 50 mm - 12 mm - 32 mm /2$
	=	$429 mm$

Neutral Axis
$x$	=	$\frac{-(\alpha(A_{st}))+ ((\alpha \cdot (A_{st})^2 + 4((0.5 \cdot b)\alpha(d A_{st}))^{0.5}}{2 \cdot 0.5 \cdot b} = 164mm$

$A_{st}$	=	$2413 mm^2, Area of rebars in tensile side$
$\alpha$	=	$6.09, ratio \frac{E_{s}}{E_{cm}}$
$b$	=	$500mm , Width of section$
$d$	=	$622mm , Effective depth of section$

Moment of inertia
$I_{cr}$	=	$\frac{b \cdot x^{3}}{3} + \alpha \cdot A_{st} \cdot (d - x)^2$
	=	$\frac{500mm \cdot (164mm)^{3}}{3} + 6.09 \cdot 2413mm^2 \cdot (622mm - 164mm)^2$
	=	$3.82 x 10^{9} mm^{4}$

Strain at concrete top
$ε_{c}$	=	$\frac{M_{sls} \cdot x}{I_{cr} \cdot E_{c}}$
	=	$\frac{270kNm \cdot 164mm}{3.84 x 10^{9} mm^{4} \cdot 3.3 x 10^{4} MPa}$
	=	$0.00035$

Strain at tens. rebar
$ε_{st}$	=	$\frac{d - x}{x} \cdot ε_{c}$
	=	$\frac{622mm - 164mm}{164 mm} \cdot 0.00035$
	=	$0.00098$

Stress of tens. rebar
$\sigma_{st}$	=	$ε_{st} \cdot E_{s}$
	=	$0.00098 \cdot 2 x 10^{5} MPa$
	=	$196MPa$

1.2Find Coefficent for Eq 7.9

Effective Height at Concrete(Fig. 7.1 & Cl 7.3.2(3))
$h_{c,eff}$	=	$min( 2.5 \cdot (h - d), \frac{h - x}{3})$
	=	$min( 2.5 \cdot (700mm - 622mm), \frac{700mm - 164mm}{3})$
	=	$179mm$

Effective Depth
$d$	=	$h - cc - \phi_{shear} - \phi_{tension}/2$
	=	$600 mm - 50 mm - 8 mm - 25 mm /2$
	=	$530 mm$

Steel Ratio(Eq. 7.10)
$\rho_{p,eff}$	=	$\frac{A_{s}}{b \cdot h_{c,eff}}$
	=	$\frac{2413mm^{2}}{500mm \cdot 179mm}$
	=	$0.027$

1.3Find the strain difference

$\epsilon_{sm} - \epsilon_{cm}$	=	$\frac{\sigma_{s} - k_{t} \cdot (f_{ct,eff}/\rho_{p,eff}) \cdot (1 + \alpha \cdot \rho_{p,eff})}{E_{s}} ≥ 0.6 \frac{\sigma_{s}}{E_{s}}$
$None$		$None$
$k_{t}$	=	$0.6, Cl. 7.3.4, for short term$
$f_{ct,eff}$	=	$2.9MPa$

Min. strain diff(Eq. 7.9)
$min. \epsilon_{sm} - \epsilon_{cm}$	=	$0.6 \frac{\sigma_{s}}{E_{s}}$
	=	$0.6 \frac{196MPa}{2 x 10^{5} MPa}$
	=	$0.00059$

Strain diff. at A<sub>st</sub> level(Eq. 7.9)
$\epsilon_{sm} - \epsilon_{cm}$	=	$\frac{\sigma_{s} - k_{t} \cdot (f_{ct,eff}/\rho_{p,eff}) \cdot (1 + \alpha \cdot \rho_{p,eff})}{E_{s}}$
	=	$\frac{196MPa - 0.6 \cdot (2.9MPa/0.027) \cdot (1 + 6.09 \cdot 0.027)}{2 x 10^{5}MPa}$
	=	$0.00061 > 0.00059, take 0.00061$

2. Find maximum spacing

Spacing(Fig. 7.2)
	=	$5(c + \phi / 2)$
	=	$5(50mm + 32mm / 2)$
	=	$330mm > 160mm, take s_{cr,max} = k_{3} \cdot c + k_{1} \cdot k_{2} \cdot k_{4} \cdot \frac{\phi}{\rho_{p,eff}}$

Maximum Crack Spacing(Eq. 7.11)
$s_{cr,max}$	=	$k_{3} \cdot c + k_{1} \cdot k_{2} \cdot k_{4} \cdot \phi_{tension} / \rho_{p,eff}$
	=	$3.4 \cdot 50mm + 0.8 \cdot 0.5 \cdot 0.425 \cdot 32mm / 0.027$
	=	$371 mm$

$None$		$None$
$k_{1}$	=	$0.8$
$k_{2}$	=	$0.5$
$k_{3}$	=	$3.4$
$k_{4}$	=	$0.425$

Crack Width(Eq. 7.8)
$w_{k}$	=	$s_{cr,max} \cdot (\epsilon_{sm} - \epsilon_{cm})$
	=	$371mm \cdot (0.00061)$
	=	$0.23 mm$

∴ The Max. Crack Spacing of the section is 0.23mm

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CivilSimple Team

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