Відео

Hola David. Cómo andas? La semana pasada me acordé de ti porque vi una película de una familia que tenía tu mismo apellido y me dije, dónde andará David?. Un abrazo.

Hi, you just need to change the values of the distances to each one of figures' centroids. For example, in the video, we have the triangle centroid locate at 1/3 of L. In your approach, this would change to 2/3 of L. The rectangle centroid position remains the same. Doing this substitution will give you the position of the trapezoid centroid from the pin support. I hope it helps.

Thanks for your comment. If you need more clarification, please, let me know. There is another video where I show how to deal with distributed loads in Macaulay's method. Check it out if you need --> ua-cam.com/video/5dovyJlb56Y/v-deo.html

Glad it helped!

Hi, you are correct in your assumptions. Now, with chain links, there are extra considerations when deriving the stresses from strain. The equation to calculate the stress outside of the ring when the section is perpendicular to the applied load is: \sigma_r = \frac{W}{A} \left[\frac{R^2}{\pi \left(R^2 + h^2 ight)} + \frac{R^2}{2h^2} \left( \frac{2R^2}{\pi \left(R^2 + h^2 ight )} -1 ight) \left(\frac{y_2}{R + y_2} ight ) ight ] + \frac{W}{2A} For the inner side, under same conditions, is: \sigma_r = \frac{W}{A} \left[\frac{R^2}{2h^2} \left( \frac{2R^2}{\pi \left(R^2 + h^2 ight )} -1 ight) \left(\frac{-y_1}{R - y_1} ight ) + \frac{R^2}{\pi \left(R^2 + h^2 ight)} + ight ] + \frac{W}{2A} These equations are written in Latex format. You can copy and paste them in latex.codecogs.com/eqneditor/editor.php to see them pretty written. Please, let me know if it helped or if you would like to see calculation examples and derivations of this part of the theory.

No that is really very helpful - thank you so much for taking the time@@johnbruzzo

Thanks. I hope it helped. Please, let me know if you would like to see something specific related to the topic. Bless you too!.