DC EC 03: longitudonal reinforcement in column under pure compression

Description

Geometry	Cross-section:	$b=h=250mm$
Geometry	Material:	C35/40
Load	Design axial force:	$N_{Ed.ULS}=1700kN$
Standard		EN 1992-1-1 [- -] and [BE]

How to model a column in Diamonds that can be compared with handcalculations

In this example we don’t want 2nd order effects. Therefor we deselecty everything.
According to Eurocode 2, reinforcement is required for the forces in ULS, possibly increased with additional reinforcement to limit the stresses in SLS. It is not known in advance whether additional reinforcement will be required for SLS. Since we are only doing a ULS calculation in this example, we turn off the stress limits in SLS in Diamonds (by making a copy of C25/30 and editing the copy).

Independent reference results

Open handcalculations

According to EN 1992-1-1 [- -]

$N_{Ed,ULS}=\left ( A-A_s \right ) \cdot f_{cd} + A_s\cdot f_{yd}$

Deduct $A_s$ :

$A_s=\frac{N_{Ed,ULS}-A \cdot f_{cd}}{-f_{cd}+f_{yd}}$

Diamonds neglects $-f_{cd}$ in the denominator:

$A_s=\frac{N_{Ed,ULS}-A \cdot f_{cd}}{f_{yd}}$

Because we have pure compression the compression strain is limited to $\varepsilon=0.2$ % (EN 1992-1-1 Figure 6.1). Resulting in $f_{yd}=\varepsilon \cdot E_s=0.2$ % $\cdot 200000MPa = 400MPa$ .

$A_s=\frac{1700kN-(250mm)^{2} \cdot 23.3MPa}{400MPa}=604mm^2$

Divide reinforcement over 4 sides:

$A_{s, on each side}=\frac{604mm^2}{4}=151mm^2$

According to EN 1992-1-1 [BE]

$N_{Ed,ULS}=\left ( A-A_s \right ) \cdot \alpha_{cc} \cdot f_{cd} + A_s\cdot f_{yd}$

Deduct $A_s$ :

$A_s=\frac{N_{Ed,ULS}-A\cdot \alpha_{cc} \cdot f_{cd}}{-\alpha_{cc} \cdot f_{cd}+f_{yd}}$

Diamonds neglects $-\alpha_{cc} \cdot f_{cd}$ in the denominator:

$A_s=\frac{N_{Ed,ULS}-A\cdot \alpha_{cc} \cdot f_{cd}}{f_{yd}}$

Because we have pure compression the compression strain is limited to $\varepsilon=0.2$ % (EN 1992-1-1 Figure 6.1). Resulting in $f_{yd}=\varepsilon \cdot E_s=0.2$ % $\cdot 200000MPa = 400MPa$ .

$A_s=\frac{1700kN-(250mm)^{2} \cdot 0.85 \cdot 23.3MPa}{400MPa}=1151mm^2$

Divide reinforcement over 4 sides:

$A_{s, on each side}=\frac{1151mm^2}{4}=288mm^2$

Diamonds results and comparison

According to EN 1992-1-1 [–]

Longitudonal reinforcement calculated by Diamonds (EN 1992-1-1 [- -])

Results	Independent reference	Diamonds	Difference
Longitudonal reinforcement	151mm²	151mm²	0,00%

According to EN 1992-1-1 [BE]

Longitudonal reinforcement calculated by Diamonds (EN 1992-1-1 [BE])

Results	Independent reference	Diamonds	Difference
Longitudonal reinforcement	288mm²	288mm²	0,00%

References

EN 1992-1-1: 2005 + AC: 2010
Van Hooymissen, L., Spegelaere, M., Van Gysel, A., & De Vylder, W. (2002). Gewapend beton. Academia Press
Keep in mind that the calculations in this book are made done using the NBN B15-002/ ENV 1992-1. Both old standards. However, this book still remains a good reference if you want to understand how reinforcement calculations work.

Tested in Diamonds 2024r01.

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Description

Independent reference results

According to EN 1992-1-1 [- -]

According to EN 1992-1-1 [BE]

Diamonds results and comparison

According to EN 1992-1-1 [–]

According to EN 1992-1-1 [BE]

References

Article Attachments

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