Advertisement

FAQ 003538 | How do I determine the effective lengths of frame columns in RFEM or RSTAB?

FAQ 003538 | How do I determine the effective lengths of frame columns in RFEM or RSTAB? Question:

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



Answer:

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 normal force state. 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 mode 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 shapes and, for each member, each buckling shape by a buckling 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 length result for the frame column results for each load situation. The effective length for the design of the respective load situation is the effective length for the buckling length whose buckling mode causes the column to buckle in the corresponding plane. Since this result may be different for each analysis due to the different load situations, the longest effective length of all calculated analyzes for a design is assumed for all load situations equally for the design - on the safe side. Example for Manual Calculation and RSBUCK / RF-STABILITY A 2D frame with a width of 12 m, a height of 7.5 m and pinned supports is provided. The column cross-sections correspond to I240 and the frame latch to an IPE 270. The columns are loaded with two different concentrated loads. l = 12 m h = 7.5 m E = 21,000 kN / cm² Iy, R = 5790 cm 4 Iy, S = 4250 cm 4 N L = 75 kN N R = 50 kN $EI_R=E\ast Iy_R=12159\;kNm^2$ $EI_S=E\ast Iy_S=8925\;kNm^2$ $\nu=\frac2{{\displaystyle\frac{l\ast EI_S}{h\ast EI_R}}+2}=0.63$ The result is 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 $\eta_{Ki}=4.408$ $sk_L=16.324\;m$ $sk_R=19.993\;m$



Go to FAQ page:





Did you find this FAQ helpful?

► If this video was helpful, we would appreciate your comments and likes.



Do you have another question? Take advantage of our free support.

► Frequently Asked Questions (FAQ):

► Ask Individual Question:



Follow us and stay updated on our current news:

► Instagram:

► Are you also on Facebook? Follow us here:



Keywords:

#Effective_length #Critical_load #Critical_load #Critical_load_factor #effective_length_coefficient #Buckling_mode

Effective length,Critical load,Critical load factor,effective length coefficient,Buckling mode,

Post a Comment

0 Comments