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Center of rigidity

15 REPLIES 15
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Message 1 of 16
peijingx
5419 Views, 15 Replies

Center of rigidity

I have a simple model with unsymmetrical braces on Y direction, the center of rigidity seems not what supposed to be. Please see attached pictures, one is from robot, the other one is from spredsheet. The spreadsheet show the center of rigidity is at the middile of two braces, bur Robot shows very close to center of gravity. What is the definition of the center of rigidity in Robot?  We want to see the center of rididity for resisting horizontal force? I have attached my model too.

 

Thanks,

 

peijing

15 REPLIES 15
Message 2 of 16
tony.ridley
in reply to: peijingx

How can centre of rigidity be in between the two rows of braces? 

 

Considering bracing only (no columns) the centre of rigidity MUST be slightly offset, as the bracing on the top and bottom (in the Y axis) is offset to the centroid of the two braces running in the X axis.  And that is ignoring all columns. 

 

Include columns and the centre shifts further close to the centre of gravity.  Check your spreadsheet.

Message 3 of 16
Rafal.Gaweda
in reply to: tony.ridley

Tony,

 

You are right 🙂

 

Full symmetry :

fullsym.jpg



Rafal Gaweda
Message 4 of 16
tony.ridley
in reply to: Rafal.Gaweda

well, his screen shot from spreadsheet shows only 4 items, not all the columns etc. surprise?
Message 5 of 16
Tuctas
in reply to: tony.ridley

  Robot doesn't take into account the lateral stiffness that braces provide to the structure (as concern the calculation of the center of rigidity). In help topic it shows that only the bending moment of inertia of the vertical elements are taken into account.

  If someone wishes to take into account the contribution of braces at the calculation of the center of rigidity he should manually try to find an equivalent value of moment of inertia for the 2 columns beside each brace.  

Message 6 of 16
peijingx
in reply to: Tuctas

We want to see the center of rigidity of lateral force resisting elements.  In this model the lateral force resisting elements are 4 braces as show on the Plan file. The braced frame are much stiffer than column frame, see attached Defection file, 50 times deflection value different. To calculate the center of rigidity, we should consider only brace frames. Refer to the OBC 2006 clause 4.1.8.3.(3) and (4), SFRS taking 100% lateral forces, SFRS include braced frame, shear wall, moment frame. Also we should consider torsional effect refer to OBC clause 4.1.8.11.(8), we need correct loaction of center of resistance (i.e.rigidity) of SFRS to calculate torsional  moments..

Message 7 of 16
Pawel.Pulak
in reply to: peijingx

As written by Tuctas Robot does not consider bracings in the calulations of  the center of rigidity - see these Help topics:

http://docs.autodesk.com/RSAPRO/2014/ENU/filesROBOT/GUID-BF3E6582-6B9C-40DF-BE12-21E33F151F4E.htm

and

http://docs.autodesk.com/RSAPRO/2014/ENU/filesROBOT/GUID-890C3B44-F7EE-4BB1-8C39-19F445B5B94E.htm

 

The improvement request to consider them is registered for some time but not implemented yet - Tuctas knows about it.

 

Regards,


Pawel Pulak
Technical Account Specialist
Message 8 of 16
tony.ridley
in reply to: Pawel.Pulak

OK, I didn't understand that.  But, how then is the centre of rigidity shifting if the bracing does not alter it?

 

 

Message 9 of 16
Tuctas
in reply to: tony.ridley

  That’s a good question Tony! (although the differences are small they should't exist..).

 

  I am attaching a screen capture from an approach that shows in which way you can calculate the “equivalent stiffness” of columns that are attached to braces. A displacement equal to 1.00m is imposed to the top nodes and then the results of base reaction in the same direction are obtained. You should isolate the 2D frame that consists of columns, beam and braces of each panel and constrain all d.o.f and then impose the displacement. The physical meaning of stiffness is: K=F/D for D=1.00, so in this way you find the “equivalent” value of Keff for each column. Then you calculate the equivalent value of bending moment of inertia Ieff : Keff=12*E*Ieff/(h^3) => Ieff=Keff*h^3/(12*E) and you replace this number to the properties of column’s section. Of course the above is just a simplified approach (for different kind of reasons).

 

  I should mention something else that is more important than my previous approach. Robot (as well as other programs too and that’s because many codes usually allow such an approach) calculates the “Center of Rigidity” of a storey that has no precise physical meaning because it uses a simplified approach that gives realistic results only for one storey buildings. In any other case (i.e in the general case of asymmetric multistory buildings) the calculated “Center of Rigidity” isn’t the point in which when a horizontal load is applied there is no rotation but only translation... If you are interested you can take a look at a related paper in the link below where the calculation of the fictitious elastic axis (optimum torsion axis) is proposed that is much closer to reality than the calculated “Center of Rigidity” (this method is already implemented in the Greek Seismic Code -EAK- some years ago). In the past, I requested for this method to be implemented in Robot as it has a general purpose. I hope sometime to become a tool for the Robot user!  

http://www.iitk.ac.in/nicee/wcee/article/13_833.pdf

Message 10 of 16
Tuctas
in reply to: Tuctas

So, was it investigated why did those (small) differences occur in this model?
Message 11 of 16
Pawel.Pulak
in reply to: Tuctas

Hi Tuctas,

yes, it was investigated.

The following descriptions in help are incomplete:

 

a/ "The rigidity center R (X, Y, Z) of the section at the level of the gravity center of a story is calculated considering the columns, the walls and the core walls."

b/ "R - Coordinates of the center of rigidity of the section at the height of the story's center of gravity, considering the columns, walls and core walls." 

When the virtual section/cut is made at the level of the gravity center of a story not only vertical columns are considered but also sloped members. So the bracing are considered too. But it is not a full stiffness of braced frame ! - only the moment of inertia and position of the bracing itself (treated as if the column).

 

It is the reason why R is not reported precisely for X=Y= 12.00


Pawel Pulak
Technical Account Specialist
Message 12 of 16
Tuctas
in reply to: Pawel.Pulak

  Thank you for replying Pawel,

 

  I suspected for such an explanation.. I am not sure if it wotrhs for any kind of correction about that... I hope that some day my suggestion about this issue will be implemented instead!

 

  Regards

  

 

  

Message 13 of 16
cdmatheus
in reply to: Rafal.Gaweda

Hello Everyone

 

I am working on a research thesis, that consist on the contribution of rigidity of the staircases to a steel building and I want to know if I model the slab of the stair (inclined slab), does RSAP consider the rigidity of the slab for calculations of center of rigidity?

 

And if I model the stair with inclined beam/column, does RSAP consider the rigidity of the inclined member for the center of rigidity?

 

I am new in the forum and I'd appreciate your help.

 

Thank you Mr.

 

Message 14 of 16

An inclined bar element will be considered in the same way as a vertical column.

 

inclined bar.PNG

 

If you find your post answered press the Accept as Solution button please. This will help other users to find solutions much faster. Thank you.



Artur Kosakowski
Message 15 of 16

But that is wrong

 

Regards

Message 16 of 16

Thank for answer Mr. Artur

 

I really appreciate that information.

 

But I was wondering what about landing slab, is it considerer for center of rigidity?

 

In the attached image, is rigidity and mass of landing slab 1 (green circle) considered to Story 1? and what about landing slab 2 (red circle), Story two maybe?

 

To finish my questions, In what Story is considered rigidity and mass of highlighted blue element?

 

Regards.

 

 

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