30 August 2019

FAQ 003538 EN

# How do I determine the effective lengths of frame columns in RFEM or RSTAB?

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

RSBUCK and RF-STABILITY perform an eigenvalue analysis for the entire model with a certain state of normal 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 from this. The effective lengths and effective length factors are then determined for this lowest buckling load.

The result shows, depending on the required number of eigenvalues, the critical load factors with the corresponding buckling curves and for each member - according to its mode shape - effective length about the strong and the minor axis.

Since usually, every load case has a different normal force state in the elements, a separate corresponding 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 designing the respective load situation.

Since this result may be different for each analysis due to the different load situations, the longest effective length of all calculated analyzes - equal for all load situations - is assumed for designing on the safe side.

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

l = 12 m
h = 7.5 m
E = 21000 kN/cm²
Iy,R = 5790 cm4
Iy,S = 4250 cm4

NL = 75 kN
NR = 50 kN

$EI_R=E\ast Iy_R=12159\;kNm^2$
$EI_S=E\ast Iy_S=8925\;kNm^2$

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

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 RSBUCK or RF-STABILITY.

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

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