Calculate Shear centre of asymmetrical section

[cookiem]

Please can someone tell me how to calculate the shear centre of a channel with a plate welded to the bottom? I'm trying to work out the torsion on a combined section but finding it difficult to get the shear centre calc, thanks

 

[cookiem]

I also need to work out the torsional constant and the torsional bending constant. thanks

 

[Lion06]

There is a procedure in Salmon and Johnson, and I think there is one in Timoshenko's Mechanics of Materials.

You'll need to fine the "product of inertia" to plug into the equations. I have a sample calc for the product of inertia. I can post it a little later - I'm at home right now.

 

[cookiem]

Thanks, I've calculated the second moments of area for the combined section, I'll see if i can find anything on Salmon and Johnson. Cheers

 

[Lion06]

Here is the product of inertia example calc.

 

[Lion06]

My apologies for teh previous attachment, that is used for finding Imax and Imin for unsymmetrical shapes.

The general procedure is this -

Determine the shear stress flow through the section. Armed with this information, you will determine the location at which the shear stresses have moments that sum to zero. This point is the shear center. For a channel it is typically outside the web, away from the direction the toes point.

 

[slickdeals]

This is a related question, but it pertains to finding the shear center for a building composed of a combination of box-walls, C-shaped walls and planar walls.

Are there any good examples/procedures to hand-calculate the center of rigidity (shear center) of the building?

We are Virginia Tech
Go HOKIES

 

[cookiem]

Thanks for all your help. I spoke to SCI who have said that they have carried out a calculation on the torsional resistance of a channel with a plate welded to the bottom flange and it provides minimal additional torsional resistance and therefore it should be ignored.

thanks again

 

[Lion06]

slick-

You raise an awesome point. I'm very interested in the same topic. I'm working on a 30 story building right now, and the lateral system is braced frames. Some frames in orthogonal directions share columns, so the result is some planar frames, some C-shaped frames, and some frames that resemble an angle in plan. The non-planar frames throw the center of rigidity off from what you expect based on stiffness. One other thing I noticed is that the center of rigidity is not constant. Our braced frames are continuous from ground to roof, but the center of rigidity starts off outside the building at the roof and slowly works it way to where you expect to see it for planar frames at the lowest elevated slab. When I think about the center of rigidity and compare it to a shear center, the shear center of a cantilevered channel is what it is. It doesn't change based on it's distance from the support (at least not from anything I've read or learned), but the center of rigidity of a lateral system seems to be changing based on the distance from ground (support). I struggle to understand why that is.


cookie-
I could have told you that the plate on the bottom wouldn't add much in the way of torsional resistance, you need to close the section to make that happen. It will change the location of the shear center, though, I can't say how much.

 

[slickdeals]

It doesn't change based on it's distance from the support (at least not from anything I've read or learned), but the center of rigidity of a lateral system seems to be changing based on the distance from ground (support).

That is precisely my question as well. I find that the point about which the diaphragm wants to twist actually changes based on how far up you are from the base. Intuitively, I guess it makes sense because your resistance to twist is decreasing, but I haven't found anything in the literature on this.

I think the problem becomes interesting when you have an unsymmetrical building.

Cookie:
Sorry to have hijacked your thread.

We are Virginia Tech
Go HOKIES

 

[slickdeals]

SEIT,
See attached screenshot from a C-shaped 30 story building with a rigid diaphragm constraint.

Take a look at the the CoR shift from top to bottom. I presume you are getting something along the same lines.

We are Virginia Tech
Go HOKIES

 

[Lion06]

Yes. My situation is similar. I can only imagine that it has something to so with the angle of twist, since that would change based on distance from the base. The location of the shear center (at least from everything I've learned and read) is not dependent on the location of the section from a support, only the proportions of the section.

 

[WillisV]

There are coupling effects between stories associated with determining C.O.R. for a multi-story building such that you do not get the same C.O.R. as you would just looking at the plan by itself. See below from ETABs documentation:

Calculate Diaphragm Centers of Rigidity

The Analyze menu > Calculate Diaphragm Centers of Rigidity command is a toggle. When a check precedes this command on the menu, ETABS will calculate the diaphragm centers of rigidity during the analysis. When no check precedes this command on the menu, this calculation is not performed.

The original concept of center of rigidity dates back to manual rigidity analysis techniques associated with the lateral analysis of single-story shear wall buildings. The center of rigidity was defined as the location of the centroid of the stiffnesses of single-story lateral resisting elements (typically planar) arbitrarily located in plan. For single-story structures the definition worked well because the stiffness for each wall frame was a 1 by 1 matrix with no interstory coupling or compatibility factors to complicate the problem. The analysis technique was extrapolated for multistory lateral analysis whereby multistory buildings were analyzed as a series of single-story buildings stacked over one another with no interstory displacement compatibility. Needless to say, for complex three dimensional structures this assumption was approximate at best.

Modern computer techniques do not require the explicit evaluation of the center of rigidity. However, the center of rigidity still needs to be evaluated because some building codes refer to it as a reference point to define design eccentricity requirements in multistory buildings.

In the general three-dimensional analysis of a building, where the behavior is coupled in plan as well as through the height of the structure, the center of rigidity requires a broader definition. In this broader definition when translational lateral loads are applied at the center of rigidity of a particular floor diaphragm, with no loads applied to any of the other floor diaphragms, the displacements of that diaphragm will have only translational components with no rotations. it should be noted that the resulting displacements of the diaphragms at other levels in general will contain translational as well as rotational components.

To evaluate the center of rigidity at a particular diaphragm, the structure is analyzed for three load cases. The loads are applied at the center of mass (or any arbitrary point). Load case 1 has a unit load applied in the global X direction and results in a diaphragm rotation of Rzx. Load case 2 has a unit load applied in the global Y direction and results in a diaphragm rotation of Rzy. Load case 3 has a unit moment applied about the global Z-axis, giving a diaphragm rotation of Rzz.

The center of rigidity relative to the center of mass (or the arbitrary point) is then given in by the coordinates (X, Y), where

 

[paddingtongreen]

I had the assignment, more than 20 years ago, to calculate the The actual masses and stiffnesses plus their centers for an existing reinforced concrete multi-floor building for input to a stick model. I started by manual calculation and persuaded the boss to let me use the spreadsheet, Lotus 123. Shortly afterwards one of the guys higher up the chain wrote a paper on using the new accounting programs for engineering analysis. I had discovered spreadsheets on my ZX81. I don't know of any engineers who used them before me (in truth, I'd rather not know the truth if I wasn't the first).

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[jberg]

My recommendation would be to download a free trial of ShapeBuilder by IES (

http://iesweb.com/products/shapebuilder/index.htm).

A single license is around $500, and I believe it will calculate the properties you listed.

JWB