Partial Fixity in Etabs _ Waseem Rana

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Par ti al Fi xi ty i n ETABS | W ASEEM RAN A

WASEEM RANA cracking the structures

Partial Fixity in ETABS April April 16, 2013 ∙ by Rana Waseem ∙ in Stiffness. ∙ There There is no simple option to just release a specific percentage percent age of moments and shear at the supports. supports. The only way is to provide the reduced stiffness of   the members. Let’s have a look at different different options in ETABS for releases. You can access these options op tions by clicking: ASSIGN>FRAME clicking:  ASSIGN>FRAME LIN ES>Frame ES>Frame Releases/Partial Releases/Partial Fixity

(hps://waseemrana8.fil (hps://waseemrana8.files.wordpress.c es.wordpress.com/2013/04/839a om/2013/04/839af‑image001 f‑image001‑700322.png) ‑700322.png) There are many sets of combinations possible. You can get the details in ETABS help menu. For example it will not allow you to release torsion at both ends. The various check boxes you see in this form are for the release of a specific force; let’s say bending moment mo ment by a specific percentage; percen tage; one for start point of the section and the other for end point. Important Important point here is when when you select the option either START or END the boxes for spring values will be enabled. By default the values in these boxes is zero which means the stiffness (or the force) is reduced to ZERO To make partial frame releases, you need to put “FRAME PARTIAL FIXITY SPRINGS” values in the START and END boxes. First you need to calculate the stiffness of a FULLY FIXED support and this is calculated based on the support conditions. For example a fixed‑fixed end beam having uniformly distributed load will have support stiffness of 2EI/L. 1‑Here an important point is that L is the member length between supports in analysis direction (unsupported length). If the member is divided in let’ say 10 parts, you will not put the length of one part, rather the full unsupported length. 2‑After calculating the actual stiffness value of the connection, you need to multiply it by the reduction factor, by which you need to reduce the moment, shear etc. The reduction factor is; https://waseem r ana.com /2013/04/16/how- to- par ti al - fi xi ty- i n- etabs/

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Partial Fixity in ETABS | WASEEM RANA

REDUCTION FACTOR = (1‑n)/n

Where n is the percentage you want to reduce. For example if you want to reduce by 25% you will get REDUCTION FACTOR = (1‑0.25)/0.25 = 3 You need to multiply 2EI/L with 3 to get the final spring stiffness value and put it in ETABS. Also see the link shared by Santiago in comments below. Consider following two cases; Simple one frame analysis. Suppose I have a fixed end beam of 6m length. A load of 10kN/m is applied. The fixed end moments are wl²/12 = 30kN.m. E=2E+8 kPa, I=4.787E‑3 m^4 Now I want to make the ends partially fixed/pinned, reducing the moments by 70%; The REDUCTION FACTOR = (1‑0.7)/0.7 = 0.43 Full fixity stiffness = 2EI/L = 319133 kN.m Reduced stiffness = 0.43 x 319133 = 137227 kN.m If I put this value in frame releases option in ONE END only, I will get 30% of 30kN.m moment that’s 9kN.m. Now the rest of the 21kN.m will be distributed in the beam and at the support on the other end. Second case, when we have a full 3D integrated structure. We may not get the moment values reduced by that percentage by which we applied the reduction factor, meaning to say we wanted 50% reduction in moment values but after analysis we got only 35%. So this process is iterative. You have to change the stiffness values based on many iterations until you get the desired results. This is because the remaining moment should be redistributed to other elements.

(hps://waseemrana8.files.wordpress.com/2013/04/f78a9‑image002‑701660.jpg)  Stiffness iteration cycles by using EXCEL.

https://waseemrana.com/2013/04/16/how-to-partial-fixity-in-etabs/

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