Відео

To properly resist lateral loads, you need at least two lines of resistance, so I can't answer your question.

@@tonatiuhrodrigueznikl7816 on other sides I have retaining wall but those are not upto roof height but I have steel column and beam in another side.

The book we use in my class is Seismic Principles by Richards (www.amazon.com/Seismic-Principles-Paul-W-Richards-dp-B08KSLPPWL/dp/B08KSLPPWL/ref=dp_ob_title_bk). The equation for rigidity in this video applies to cantilevered walls, which is a safe assumption for wood shear walls. Note that there is a typo in the video. The factor in the second term in the denominator should be "0.3", not "3".

Design of diaphragms and collectors will depend on the specific material and structural type and is outside the scope of the video.

You are correct. Thank you for your comment. Thankfully, the resulting calculations (1.4286 and 2.2382) are correct; they use the 0.3 factor.

You're allowed to omit rotational DOFs at moment-free ends provided that you use the right stiffness factors and force recovery process. We do this to reduce the problem to a 3x3 instead of the 5x5 it would otherwise be.

You scared me there for a second, but the video is indeed correct. For loading in the x-direction, you move the CM in the y-direction, thus making the moment arm (perpendicular to the direction of loading) larger. Thus, the values given in the video are indeed correct.

Can you provide more details? Around 1:25 in the video there is a division by d that results in an Rd term in the denominator. Maybe that is the reason for your question.

@@tonatiuhrodrigueznikl7816 thanks for clearing it up

Seismic Force Resisting System

you have been such a help dont mind my grammer 😀

Yes, the curve will be continuous.

@@tonatiuhrodrigueznikl7816 won't the change in cross section affect the curvature?

@@aswinkumar731 Yes. There will be a discontinuity in the curvature but slope and displacement will be continuous.

We use an appendix from a published structural analysis book. The title doesn't come to mind, but any intermediate or advanced structural analysis text will have similar appendices.

You are indeed correct. Thank you for pointing that out. It was a typographical error - the subsequent shear calculation *is* still correct. I'll need to recut an repost the video; in the meantime, viewers, please make note of the error.

Yes. You can see an example in this video: ua-cam.com/video/cvGvipX3brE/v-deo.html.

Collectors are certainly a difficult topic to visualize and I see the value in your suggestion. Thank you for making it; I'll keep it in mind.

b will always be the same because we are considering two loadings *on the same member*

Thank you for the suggestion. At the moment, sharing files is beyond the scope of work for this UA-cam channel. I'll keep it in mind for the future.

You are correct. The R for these is 2.2443. However, this is apparently just at typographical error and the subsequent calculations are correct. I'll get around to recutting and reposting the view sometime, but for now, viewers, please make note of the error.

The exact design checks will depend on the material use (e.g., concrete or steel). An answer to that will require taking the relevant design course.

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.

Thanks. These were done simply with PowerPoint and a bit of patience.

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.

​@@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.

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.

Thanks for the feedback. Glad you like them.

Hi Dr. Tonatiuh, I watched all the diaphragm topics which you presented and I really enjoyed it. Thanks

Thank you. Good suggestion. Strength eccentricity is an advanced topic, so no promises about when that topic might appear.

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.

@@tonatiuhrodrigueznikl7816 Many Thanks for this article.