Rigid Diaphragms 01, Centers of Mass and Rigidity (no concentrated masses)

This video runs through an example to compute centers of mass and rigidity for rigid diaphragms. To keep the example simple, only a uniform mass is considered. Please note that in practice concentrated sources of mass also need to be considered (e.g., the weights of concrete shear walls or permanent equipment). This video was recorded for CE 4650, Seismic Design, at Cal State LA.

[@drzlokita4life]

These lectures are very well done! Looking forward to seeing more of your videos.

[@tonatiuhrodrigueznikl7816]

Many thanks!

[@tonatiuhrodrigueznikl7816]

This simplified example only considers uniform masses. Please note that in practice concentrated sources of mass also need to be considered (e.g., the weights of concrete shear walls or permanent equipment).

[@rcsouljp]

Hello, thanks for your awesome courses. Is the course "previously in the course..." you mentioned at 2:30 still exist?

[@tonatiuhrodrigueznikl7816]

These videos are meant for a university course, so "previously in the course" may not always refer to a video on the channel. Sorry for the confusion.

[@krystiankwinta]

Hi. What happens if i have X-brace instead of shear walls? How to calculate center of rigidity? Can you provide example video for that case?

[@tonatiuhrodrigueznikl7816]

An example video probably won't come for a while, but the concepts here apply to any lateral system. For a braced frame, barring a more detailed analysis, look at this reference from the AISC: www.aisc.org/globalassets/modern-steel/steel-interchange/2013/012013_si.pdf.

[@krystiankwinta]

@@tonatiuhrodrigueznikl7816 Many Thanks for this article.

[@danielrocha7390]

Why was the mass of the walls not included in the Center of Mass calculation?

[@tonatiuhrodrigueznikl7816]

Excellent comment. I hope that other viewers see this comment and recognize that the video involves a simplifying assumption.
It is indeed more exact to include the weight of the walls in the calculations. People do this differently. It makes most sense to include the weight of the bottom half of the walls above and the top half of the walls below. These weights should be lumped at the centers of mass of each wall.

[@keyou1228]

@@tonatiuhrodrigueznikl7816I was gonna ask the same question.

[@emranangkal3521]

@@keyou1228 i was gonna ask this question too

[@brianchasecook20]

You just calculated the geometric center. Not the center of mass.

[@tonatiuhrodrigueznikl7816]

You are correct. The geometric center is the center of mass if the diaphragm is homogenous, which is the case if the weight of the walls is assumed to be smeared evenly over the diaphragms. Of course, it would be more precise (at the cost of complexity in an introductory video) to include the weights of the walls and their explicit location.

[@brianchasecook20]

​@@tonatiuhrodrigueznikl7816 Somewhat true. If you use, say, 250psf live load...a heavy storage load...you must consider a portion of that load in the seismic analysis, as you would with a snow load above 30psf. So regardless of the homogeneity of the diaphragm there can be different COMs. Also, the equations used in this example say COM = x bar times area, etc. When actually, the COM = the sum of the masses x distance over the sum of the masses. We had a young engineer in our office calculate the COM using your geometric center equations and when I was explaining it to them they said the picked it up on this video. We encourage our young engineers to do their own research so it is my responsibility to vet the resources. In the true scientific sense, COM never equals GC. They may happen to both equal the same value, but they themselves are not equal.

[@tonatiuhrodrigueznikl7816]

I'm embarrassed about the delay in my reply. I appreciate your attention to detail and to the education of your young engineers. I agree that this is not well explained for the reasons you describe. I intend to recut and repost the video; in the meantime, viewers, please make note of this important clarification.