 30 August 2019

003538

# How are the effective lengths of frame columns determined in RFEM or RSTAB?

The easiest way to do this is to use the RF‑STABILITY (RFEM) or RSBUCK (RSTAB) add-on modules.

RF‑STABILITY and RSBUCK perform an eigenvalue analysis for the entire model with a certain state of the axial force. The axial forces are increased iteratively until the critical load case is reached. This stability load is characterized in the numerical calculation by the determinant of the stiffness matrix becoming zero.

If the critical load factor is known, the buckling load and the buckling curve are determined by using this. The effective lengths and the effective length factors are then determined for this lowest buckling load.

Depending on the required number of eigenvalues, the results show the critical load factors with the corresponding buckling curves, and the effective length about the major and the minor axis for each member, depending on the mode shape.

Since every load case has usually a different state of the axial force in the elements, a separate belonging effective length result for the frame column arise for each load situation. The effective length whose buckling mode causes the column to buckle in the corresponding plane is the correct length for the design of the respective load situation.

Since this result may be different for each design due to the different load situations, the longest effective length of all calculated analyses is assumed as equal for all load situations.

###### Example for Manual Calculation and RF-STABILITY/RSBUCK
There is a 2D frame with a width of 12 m, a height of 7.5 m and simple supports. The column cross-sections correspond to I240 and the frame beam to IPE 270. The columns are subjected to two different concentrated loads.

l = 12 m
h = 7.5 m
E = 21,000 kN/cm²
Iy,R = 5,790 cm4
Iy,S = 4,250 cm4

NL = 75 kN
NR = 50 kN

$EI_R=E\ast Iy_R=12,159\;kNm^2$
$EI_S=E\ast Iy_S=8,925\;kNm^2$

$\nu=\frac2{{\displaystyle\frac{l\ast EI_S}{h\ast EI_R}}+2}=0.63$

This results in the following critical load factor:

$\eta_{Ki}=\frac{6\ast\nu}{(0.216\ast\nu^2+1)\ast(N_L+N_R)}\ast\frac{EI_S}{h^2}=4.4194$

The effective lengths of the frame columns can be determined as follows:

$sk_L=\pi\ast\sqrt{\frac{EI_S}{\eta_{Ki}\ast N_L}}=16.302\;m$

$sk_R=\pi\ast\sqrt{\frac{EI_S}{\eta_{Ki}\ast N_R}}=19.966\;m$

The results from the manual calculation correspond very well with those from RF‑STABILITY and RSBUCK.

###### RSBUCK
$\eta_{Ki}=4.408$
$sk_L=16.322\;m$
$sk_R=19.991\;m$

###### RF-STABILITY
$\eta_{Ki}=4.408$
$sk_L=16.324\;m$
$sk_R=19.993\;m$

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• Updated 30 November 2020   