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# Re: Fw: RIGID DIAPRAGM

• To: seaint(--nospam--at)seaint.org
• Subject: Re: Fw: RIGID DIAPRAGM
• From: Alex Gutierrez <alexgutz(--nospam--at)pworld.net.ph>
• Date: Tue, 03 Nov 1998 05:50:51 +0800

```
T. Eric Gillham wrote:

> Alexander:
>
> You will have two points per floor that you will need to find: center of
> mass and center of rigidity.  The diaphragm will rotate about the center of
> rigidity, and you can lump your mass for the level at the center of mass.
> If you are using a computer analysis program, you may need to calculate the
> center of mass as well as translational/rotational mass properties to be
> lumped there, which is just basic mass moments of inertia.  Your program
> will calculate the location of the center of rigidity.
>
> If you aren't using a computer program for the analysis, then you will need
> to calculate the (relative at least) rigidities of all the laterally stiff
> elements in both the X and Y directions.  Then just find the location of
> the resultant stiffness vectors in X and Y, which will give you the center
> of rigidity location.  I would suggest looking at a basic Masonry design
> handbook such as Gary Hart's manual, which will likely give you some basic
> examples of this procedure.
>
> T. Eric Gillham PE
>

In computing the center of rigidity, will I consider the floor below? the floor
above? or both? Say for a ten story building, say the column size , beam size
and shear wall thickness are all the same for the first five floors, and
another set members for the next five floors, does that mean that on the lower
five floors, it will have the same center of rotation and on the last five
floors, it will also rotate on another location center of rotation? Does it not
differ from floor to floor even if the member sizes are the same?

Alex Gutierrez

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