# 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
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 by a specific percentage; 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 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;
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;
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, 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.
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 until you get the desired results. This is because the remaining moment should be redistributed to other elements.
Stiffness iteration cycles by using EXCEL.

1. Peter Manthorpe says:

Two different explanations of the reduction factor that contradict each other. I am confused…

2. Rana Waseem says:

Peter, there is only one definitaion of reduction factor in the article…tell me why are you confused?

3. Peter Manthorpe says:

Thank you so much for replying, I am desperate for your help!Basically I calculated the K value and got a similar result. Now surely as a test, If I put in the full K value into the "frame partial fixity springs" box then the resulting moments should be similar to that of a fully fixed connection?I did the check and the moments at the connection were 120KNm and when I applied partial fixity and entered the "full stiffness value" the result of bending moment was 80KNm.I have approached my lecturers with this problem and they dont know, you are saving my life thanks.Peter

4. Zeeshan says:

Nice work!!! Thanks.

5. fiasco22 says:

Could you please give few practical examples in which we would want to release moment, shear etc??

6. Rana Waseem says:

A practicle example would be…the design of baseplate connections lets say!You want to design your connections based on the forces you get from ETABS model on support…and you provided FIXED supports….(infinite stiffness)after realizing that the structural reactions are very high on fixed supports to design a reasonable connection you would definately want to reduce the support stiffness so it carry less reactions and forces..and redistribute the remaining forces back to structures for example in steel beams and columns in structure where they are capable to resist the additional forces resulting from redistribution…this way you save a lot in connection design….

7. shahidthanvi says:

Hi, My question is related to structural analysis and design as well. If an end connection of RC frame (beam-column joint) ; both beam and column rebars end up with conventional ACI hooks then what Partial fixity should be considered for this joint OR how to calculate it?

1. Rana Waseem says:

In practice, a connection between a concrete column and a concrete beam will be some what between a fully fixed and pinned connection. For example in ETABS, you dont need to specify the connection fixity of beam column connection. Joint or connection fixity between a column and beam will depend on stiffness of joining members. So in software this phenomenon is automatically calculated. You just need to provide enough development length for example by ACI hooks etc.

8. Hao Hu says:

Hi Rana, thanks for what you did in the example. The only problem also get many of us confused is the reduction factor. In the book “Non-linear static and cyclic analysis of steel frames with semi-rigid connections”.(Chan and Chui)
Chapter 5, it defines “n” as fixity factor, which is equal to zero for pinned joints and unity for rigid joints.
That might helps clarifying a little bit.
hhu4@ncsu.edu

9. Rana Waseem says:

You are talking about n…if n=0 then pin | if n=1 then rigid…right

10. engsam777 says:

Rana,
Why are u using 2EI/L instead of 4EI/L in calculating the fixed end moment.

11. engsam777 says:

Sorry the stiffness

12. baka kada says:

Exquisite explanation ….some lectures coud not make one understand for a whole semester…… i like your approach

1. Hiển Cò says:

in practicle, If i using 4EI/L instead of 2EI/L in calculating the fixed end stiffness, the results is right. Can U answer that?

13. Shams Ur REhman says:

i always get higher value of reactions at base when i analyse it through ETABS, how could i reduce these moments?

14. Tariq Zaid says:

Thank you very much for all these information.
My question is : Now in ETABS what is the value should I put ?
Should I calculate the stiffness for all members which I want to release ? Or there another way calculate the value?

15. Santiago Vallejo says:

For those confussed, On step 1 Rana means the stiffness factor K. Not the stiffness per-se.
http://classes.mst.edu/civeng217/concept/12/03/index.html

Nither 2EI/L OR 4EI/L is wrong, it deppends on the end conditions of the beam (or column)

1. Rana Waseem says:

16. Omer Anwaar says:

When i release moment 22 and 33 on both ends of a frame (beam) although i get the correct moments, but i am not able to design in etabs. Is it right to release moments on both ends?

17. nikunj thakor says:

dear sir
you say about iteration. but what should i change if my E,I n L value r fixed.
and if m taking n=10%(0.1)
then i got value 5195000 , by that vqlue my moment is getting higher. can you please give me some right direction

1. Rana Waseem says:

You have to keep iterate by guessing the next stiffness value (any number), until you get the desired results. There is no formula to calculate required stiffness in a 3d model. You have to keep on guessing and iterate for desired results.

18. Vijay Patil says:

Thank you Mr. Rana for the elaborate explanation. Just one simple question. As mentioned above in a comment I too used 2EI/L and put it as stiffness of the spring. However the Moments at the fixed end at that point reduce as compared to no spring specified. The closest the moment to the fixed end was when the beam stiffness at junctions was given as 10EI/L. So when I have to apply the reduction factor should I apply it to 10EI/L x Reduction factor ? Which would mean 2EI/L is just a guideline to start of iteration in 3D models.

19. praveena E says:

Thnks for description. It is helpfull to all structural Engineers

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