DC EC 06: 2nd order effect in a column

Description

Geometry	Cross-section:	$b=250mm$ $h=400mm$
	Column height: Buckling length:	$2m$ $4m$
	Material:	C25/30
	Concrete cover:	$c=40mm$
	Creep factor $\varphi _{t.\infty}$ :	2
	Reinforcement distribution:	Identical Only upper and lower
Load	Design forces:	$N_{Ed.ULS}=300kN$ $M_{Ed.ULS}=120kNm$ $M_{Ed.SLS QP}=88.9kNm$
Standard		EN 1992-1-1 [- -]

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

Diamonds looks for the most economical reinforcement ratio between Az and Ay by default. This is does not always results in a calculation along the local y’- axis or one along the local z’-axis. When loads (and/or eccentricities) are present along the y’- and z’ axis, the calculation is done around the resulting axes of the loads/eccentricities. Because this calculation is no longer in one specific plane, it becames hard to recalculate by hand. In this example, we’ll do some adjustments to the model, so that the calculation is in one particular plane parallel to one the local axes and can be reproduced by hand calculations.

choose only top and bottom reinforcement and set the reinforcement distribution to ‘identical’.
This way, you ask Diamonds to calculate equal reinforcements amounts for the top and bottom reinforcement. This is a reinforcement distribution for which graphs/tables are available in literature so we can do a hand calculation.
In this example we want to focus on the 2nd order effects. Therefor we set the accidental eccentricity to zero.
By default, Diamonds will apply the 2nd order eccentricity in different directions. Since the cross section is not a square, and the loads are applied around the strong axis, we can already predict that the worst direction for the 2nd order effect, will be around the weak axis. When that happens, the loads no longer act in one particular plane.
So to force Diamonds into applying the 2nd order effect into the same direction as the already present bending moment, we deselect the buckling verification around the weak axis.
According to Eurocode 2, reinforcement is required for the forces in ULS, possibly increased with additional reinforcement to limit the stresses in SLS. We don’t know 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

Eurocode 2 offers simplified methods (EN 1992-1-1 §5.8.7 and §5.8.8) to include 2nd order effects and imperfections into the results. The method based on nominal curvature (EN 1992-1-1 §5.8.8) is implemented in Diamonds. This method results in an increased bending moment.

Is a 2nd order calculation required? (EN 1992-1-1 §5.8.3.1)

This column has the same ULS loads has validation example DC EC 05. So we just reuse the reinforcement amounts:

$A_s=2 \cdot 514mm^2=1028mm^2$

$A_c=A - As = 250mm \cdot 400 mm - 1028mm^2= 98972mm^2$

$I_{0.c}=\frac{b \cdot h^3}{12}=\frac{250mm \cdot (400m)^3}{12}=133333cm^4$

(5.14) $\begin{align*} \lambda = \frac{l_0}{\sqrt{\frac{I_{0.c}}{A}}}= \frac{4m}{\sqrt{\frac{133333cm^4}{250mm \cdot 400 mm}}}=34.64 \end{align*}$

$n=\frac{N_{Ed.ULS}}{A \cdot f_{cd}} = \frac{300\text{kN}}{100000mm^2 \cdot 16.67\text{MPa}}=0.18$

$A=1.7; B=1.1; C=0.7$

(5.13N) $\begin{align*} \lambda_{lim} = \frac{ 20 \cdot A \cdot B \cdot C}{\sqrt{n}}= \frac{ 20 \cdot 0.7 \cdot 1.1 \cdot 0.7}{\sqrt{0.182}}=25.41 \end{align*}$

Because $\lambda$ is bigger than $\lambda_{lim}$ , second order effects need to be taken into account.

Magnitude of 2nd order eccentricity (EN 1992-1-1 §5.8.3.2)

$n_u=1+\frac{A_s \cdot f_{yd}}{A_c \cdot f_{cd}}=1+\frac{1028mm^2 \cdot 434.7\text{MPa}}{98972mm^2 \cdot 16.6\text{MPa}}=1.27$

$n_{bal}=0.4$

Correction factor $K_r$ depending on the axial force (this factor is called $K_2$ in ENV):

(5.36) $\begin{align*} K_r=min\left ( \frac{n_u-n}{n_u-n_{bal}},1 \right )=1 \end{align*}$

$\beta =0.35+\frac{f_{ck}}{200\text{MPa}}-\frac{\lambda }{150} = 0.35+\frac{25\text{MPa}}{200\text{MPa}}-\frac{34.64}{150} = 0.244$

(5.19) $\begin{align*} \varphi _{ef}=\frac{M_{Ed.SLS QP}}{M_{Ed.ULS}} \cdot \varphi _{t.\infty} = \frac{88.9\text{kNm}}{120\text{kNm}} \cdot 2 = 1.48 \end{align*}$

Correction factor $K_{\varphi}$ depending on the creep coefficient (this factor is called $K_1$ in ENV):

(5.37) $\begin{align*} K_{\varphi}=max\left (1+ \beta \cdot \varphi _{ef},1 \right )=1.362 \end{align*}$

$c=8 \qquad \qquad \text{because constant bending moment, see EN 1992-1-1 §5.8.8.2 (4)}$

The radius of curvature:

(5.34) $\begin{align*} r_0 &= \frac{0.45\cdot d \cdot E_s }{f_{yd} } \\ &=\frac{0.45\cdot 360mm \cdot 200000\text{MPa}}{434.7\text{MPa}}=74.52m\\ r&=\frac{r_0}{K_r \cdot K_{\varphi}}\\ &= \frac{74.52m}{1 \cdot 1.362} = 54.73m\\ \end{align*}$

The second order excentricity:

$\begin{align*} e_2 &=\frac{l_{0}^{2}}{c} \cdot \frac{1}{r}\\ &=\frac{(4m)^{2}}{8} \cdot \frac{1}{54.73m} =36.54mm\\ \end{align*}$

Magnitude of design bending moment (2nd order effect included)

The additional moment for 2nd order effects equals:

(5.33) $\begin{align*} M_{Ed.2nd.order}=N_{Ed.ULS} \cdot e_2 = 300\text{kN} \cdot 36.54mm = 11.0 \text{kNm} \end{align*}$

The total design moment becomes:

$M_{Ed}= M_{Ed.ULS} + M_{Ed.2nd.order} = 120\text{kN} \cdot 11 \text{kNm} = 131.0 \text{kNm}$

Reinforcement calculation

The principle for calculating the longitudinal reinfordement is entirly the same as in validation example DC EC 05. Only $\mu_d$ differs:

$\mu_d=\frac{M_{Ed}}{b\cdot d^2 \cdot f_{cd}}=\frac{131\text{kNm}}{250mm\cdot (360mm)^2\cdot 16.6\text{MPa}}=0.243$

Graph 10.2a from Gewapend beton: numeri (click here to open the graph) shows $\omega _1=\omega _2$ as a function of $\mu_d, v_d$ and $\delta_1$ . Using this graph and the results above, we can deduct:

$\omega _1=\omega _2=0.172$

$A_{s1}=A_{s2}=\omega_1 \cdot b \cdot d\cdot \frac{f_{cd}}{f_{yd}}= 0.172 \cdot 250mm \cdot 360mm \cdot \frac{16.6\text{MPa}}{434.8\text{MPa}} = 593mm^2$

Diamonds results and comparison

Longitudinal reinforcement calculated by Diamonds (EN 1992-1-1 [- -])
The reinforcement due to ULS is shown using a thin line, the reinforcement due to ULS + SLS + buckling with a thick line. Because the SLS verification were turned off, the thick line represents the reinforcement for ULS + buckling.

Results	Independent reference	Diamonds	Difference
Longitudinal reinforcement	593mm²	593mm²	0%

Remarks and notes:

Don’t forget the calculate the buckling lengths or impose them manually.
In the calculation of $n$ , Diamonds uses $A$ instead of $A_c$ . This is because at the moment he starts his calculations, the longitudinal reinforcement $A_s$ has not yet been unambiguously determined.

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 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.
Gruyaert, E., & Minne, P. (2019). Gewapend beton: numeri.
This reference is a summary of Gewapend beton (2002) but with updated formula and principles according to EN 1992-1-1. This document contains multiple graphs and tables helping the design of reinforced concrete.

Tested in Diamonds 2025.

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