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## Robot Structural Analysis

Contributor
Posts: 12
Registered: ‎08-13-2013

# Center of rigidity

958 Views, 11 Replies
09-12-2013 08:51 AM

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

Valued Mentor
Posts: 652
Registered: ‎09-07-2011

# Re: Center of rigidity

09-12-2013 06:15 PM 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.

Product Support
Posts: 4,935
Registered: ‎04-26-2010

# Re: Center of rigidity

09-13-2013 02:20 AM in reply to: tony.ridley

Tony,

You are right :-)

Full symmetry :

Rafal Gaweda
Valued Mentor
Posts: 652
Registered: ‎09-07-2011

# Re: Center of rigidity

09-13-2013 04:44 AM in reply to: Rafal.Gaweda
well, his screen shot from spreadsheet shows only 4 items, not all the columns etc. surprise?
Distinguished Contributor
Posts: 163
Registered: ‎10-10-2012

# Re: Center of rigidity

09-13-2013 08:17 AM 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.

Contributor
Posts: 12
Registered: ‎08-13-2013

# Re: Center of rigidity

09-16-2013 11:40 AM 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..

Product Support
Posts: 472
Registered: ‎06-23-2008

# Re: Center of rigidity

09-17-2013 02:38 AM 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
Valued Mentor
Posts: 652
Registered: ‎09-07-2011

# Re: Center of rigidity

09-17-2013 06:09 AM in reply to: pp2008

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

Distinguished Contributor
Posts: 163
Registered: ‎10-10-2012

# Re: Center of rigidity

09-18-2013 01:46 AM 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