DC EC 08: cracking widths in beam under pure bending (SLS design)

Description

Geometry	Cross-section:	$b=200mm$ $h=400mm$
	Material:	C25/30 $E_c=31475\text{MPa}$ $E_s=200000\text{MPa}$
	Concrete cover:	$c=40mm$
	Practical reinforcement:	$A_{s1}=798mm^2$ divided over 2 bars $A_{s2}=107mm^2$ divided over 2 bars
	Creep factors:	$\phi_{deformation}=2$ $\phi_{stress}=1.36$
Loads		$M_{Ed.SLS.RC}=74\text{kNm}$ $M_{Ed.SLS.QP}=60\text{kNm}$
Standard		EN 1992-1-1 [- -]

Independent reference results

Open handcalculations

Determine mean bar diameter
$\phi_1=\sqrt{\frac{4 \cdot A_{s1.pr}}{\text{#}_{bars} \cdot \pi}}=\sqrt{\frac{4 \cdot 798mm^2}{2 \cdot \pi}}=22.54mm$

$\phi_2=\sqrt{\frac{4 \cdot A_{s2.pr}}{\text{#}_{bars} \cdot \pi}}=\sqrt{\frac{4 \cdot 107mm^2}{2 \cdot \pi}}=8.25mm$
Determine position of neutral line $x_{cr}$
$0.5\cdot b\cdot x_{cr}^{2}+\left ( \alpha _{cr}-1 \right )\cdot A_{s2}\cdot \left (x_{cr}-d_2 \right )=\alpha_{cr} \cdot A_{s1}\cdot \left ( d-x_{cr} \right )$

$\alpha_{cr}=\frac{E_s}{E_{cr.\infty}}=\frac{E_s}{\frac{E_c}{1+\phi}}=\frac{20 000 \text{MPa}}{\frac{31475\text{MPa}}{1+2}}=19.06$

$x_{cr}=165mm$
Determine maximum cracking distance $s_{r1.max}$
$h_{eff}=min\left( 2.5 \cdot \left( h-d \right),\frac{h-x_{cr}}{3},\frac{h}{2} \right)=78.32mm$

$A_{c.eff}=b \cdot h_{eff}= 15664mm^2$

$\rho_{pr.eff}=\frac{A_{s1.pr}}{A_{c.eff}}=0.051$

$k_1=0.8; k_2=0.5; k_3=3.4; k_4=0.425$

$s_{r1.max}=\left( d_1 - \frac{\phi_1}{2}\right) \cdot k_3 + \frac{k_1 \cdot k_2 \cdot k_4 \cdot \phi_1}{\rho_{pr.eff}} = 172.9mm$
Determine stress in tensile reinforcement
From

$M_{Ed.SLS.QP}=(b \cdot x_{cr}-A_{s2})\cdot 0.5 \cdot \varepsilon_c\cdot\frac{E_c}{1+\phi_{stress}}\cdot \left( d-\frac{x_{cr}}{3}\right) + A_{s2} \cdot \varepsilon_{s2} \cdot E_s \cdot \left( d-d_2 \right)$

And

$\varepsilon_c= \frac{\varepsilon_{s2} \cdot x_{cr}}{x_{cr}-d_2}$

The value of $\varepsilon_c$ can be deducted, namely 0.0008969.
Which can then be used to calculate the strain and stress in the tensile reinforcement

$\varepsilon_{s1}=\frac{\varepsilon_c \cdot \left( d-x_{cr} \right)}{x_{cr}}=0.001059$

$\sigma_{s1}=\varepsilon_{s1} \cdot E_s=211.89MPa$

$k_t=0.4$

$\varepsilon_{sm}-\varepsilon_{cm}=max\left( \frac{\sigma_{s1}-k_t \cdot \frac{f_{ctm}}{\rho_{pr.eff}}\cdot \left( 1-\alpha_{cr} \cdot \rho_{pr.eff} \right)}{E_s}, 0.6 \cdot \frac{\sigma_{s1}}{E_s} \right) = 0.00106$
Determine cracking width
$w_k=s_{r1.max} \cdot \left( \varepsilon_{sm}-\varepsilon_{cm} \right)= 172.9mm \cdot 0.00106=0.183mm$

Diamonds results and comparison

Cracking width in a SLS QP combination calculated by Diamonds (EN 1992-1-1 [- -])

Results	Independent reference	Diamonds	Difference
Cracking width SLS QP	0.183mm	0.185mm	1.27%

References

EN 1992-1-1: 2005 + AC: 2010
Bhatt, P., MacGinley, T., & Choo, B. S. (2014). Reinforced Concrete Design to Eurocodes: Design Theory and Examples, Fourth Edition. CRC Press.

Tested in Diamonds 2026.

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Description

Independent reference results

Diamonds results and comparison

References

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