# Partial Fixity in ETABS

August 15, 2017 | Author: Hemal Mistry | Category: Stiffness, Mechanical Engineering, Mathematics

#### Short Description

Partial Fixity in ETABS Let’s take the case PARTIAL FIXITY in ETABS. There is no simple option to just release a spec...

#### Description

PARTIAL FIXITY IN ETABS  Let’s take the case PARTIAL FIXITY in ETABS. There is no simple option to just release a specific percentage of moments and shear at the supports. The only way is to provide the reduced stiffness of the members. Let’s have a look at different options in ETABS for releases. You can access these options by clicking: ASSIGN>FRAME LINES>Frame Releases/Partial Fixity

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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 checkboxes you see in this form are for releasing (making 100% pin connection). One for start point of the section and the other for end point. Important point here is when you select the option either START or END the boxes for spring values will be enabled. By default the values in those boxes is zero which means the stiffness is reduced to ZERO so making it PIN connection. To make partial frame releases (say only 50% of the moment), you need to put the “FRAME  PARTIAL FIXITY SPRINGS” values in the START and END boxes. So, WHAT value I put for Fixity Springs

1- First you need to calculate the stiffness of the FULLY FIXED support and this calculated as k=4EI/L Where k = Fully fixed stiffness of the connection, E=Modulus of elasticity of the member, I=Moment of inertia in the direction of analysis, L=length of the member between supports.

Here the important point is that L is the member length between supports in that particular 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:REDUCTION FACTOR = n/(1‐n)  Where n is the percentage you want to reduce. For example if you want to reduce by 25% you will get REDUCTION FACTOR = 0.25/(1-0.25) = 0.33 You need to multiply 0.33 with 4EI/L to get the final spring stiffness value and put it in ETABS. There are following 2 cases involved:I)

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, so that I get only 30% of the moment I’m getting now. (30% of 30kN.m is 9kN.m.). The REDUCTION FACTOR = 0.3/(1-0.3) = 0.43 Full fixity stiffness = 4EI/L = 638266.67 kN.m Reduced stiffness = 0.43 x 638266.67 = 274454.67 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.

II)

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 unless you get the desired results. This is because the remaining moment should be redistributed to the other elements of the structure. Please see the picture below for the iteration process.