Recent content by Jony S

Extra question for clarification, the stress is the same (opposite signal) if the pressure is applied externally, right ? If the pressure is the same inside and outside the total stress has to be zero.

Yes it is :) Thanks for the help!

No. Searched a bit more and found this: https://nptel.ac.in/courses/Webcourse-contents/IIT-ROORKEE/strength of materials/lects & picts/image/lect16/lecture16.htm In the end of the first part there is a formula for displacement. This appears to be the combination of the two equations that we...

These concepts are fairly new to me so I'm not "familiar" with much. Are you talking about Poisson's ratio, i.e., the expansion/compression of the material in perpendicular directions ? so I would add this: σh - v(σr+σz)= E * strain ?

Homework Statement Hi all, this isn't exactly homework, but it is nonetheless a problem I would like to solve, so here goes. Consider a thin-walled ring/hoop with diameter "d" and thickness "t". I want to know the change of diameter "Δd" due to the stress caused by pressure "p" being applied...

True, the end solution is the same, but I wasn't sure if the moments at A/B were simply M+FB/Ay*L1, because it's wasn't clear to me if moment M is somehow distributed between A and B or if it actually is M in both points (plus the Fy*L1 of the other point) Thanks for the help :)

If I add forces F1 and F2 that produce moment M, but are symmetric so that it doesn't introduce translation to the system, I get the same solution (in the picture, sorry for bad handwriting :) )

Hey guys, sorry about the delay, yes forget about Fcy, that's silly, and yes Fxs will cancel, I got that :P Now, if I exclude my force "F" (the one that makes moment M) and just assume there's an applied moment M on point C /(with no "translational" force), I will ultimately get something...

I drew a new picture to explain the new approach. We have points A, B and C of a rigid structure. The force "F" is such that it's momentum on point C is M, so that M = F * L4. In order for the structure to be static, the reaction forces on points A, B and C will have to obey the system of...

of course i see what you mean now, yes the axis of rotation is on point E, but of course the forces from the moment applied there will distributed on points A and B, so it's the sum of the momentum from both those forces will equal M. Can't do the math now but I'll come back here tomorrow :)

I see your approach but I think this is a bit more complicated than I wanted, since there was no need to make this a "static" problem (last comment for more details :P) I simply need the forces on A or B that result from momentum M. edit: ok I mentioned the structure was pinned to the ground on...

Fx and Fy that I want are simply the forces from applied momentum M. If we approach it as a traditional "static" problem then I guess we can simply imagine that A and B are "screwed" to the ground, and whatever forces result from the momentum are the normal reaction forces on this screw, i.e...

Sorry about the picture, it's a png with transparency that's why it appears weird, something to do with the website's interface. If you click on the name of the file instead of the picture, a new tab opens with the correct display. What do you mean pick my axis conveniently ? is my proposed...

Homework Statement [/B] Hello, I would like to request some help in solving this mechanics problem. Consider the attached drawing of a solid structure which is pinned to the ground on points A and B. Distances AC=Xa, CB=Xb, CD=H1 and DE=H2 A torque with magnitude M is applied on point E. How do...