3D Rigid Body Equilibrium
HTML-код
- Опубликовано: 26 авг 2024
- Solution to a three dimensional rigid body equilibrium problem.
Topics/content included:
free body diagrams, equilibrium, equations of equilibrium, forces, moments, reactions, supports
This video screencast was created by Dr Terry Brown with Doceri on an iPad. Doceri is free in the iTunes app store. Learn more at www.doceri.com
Thank you! I just spent an age trying to understand the sign convention for moments in 3D and you just explained it perfectly. At 11:00 for anyone wondering.
thanks for the feedback. Glad to be of help. Good luck with your studies and career. Cheers
This is seriously the exactly same problem as the one that I happened to be looking at in that book right now when I clicked this video, haha.
Quite a coincidence.
Thank you very much! well explained😎
Thanks for the feedback. Glad you found it helpful.
You're a legend for sharing this Terry :)
This was a great refresher.
Glad you found it helpful.
In the sum of moments around the z-axis shouldn't it be: -Ray* √2 ? Cause it seems to be the perpendicular distance from the z-axis with origin in point D.
14:40
thanks for the question. Sorry for the delayed reply. √2 is the distance between points A and D, but it is not the distance from the z-axis perpendicular to the line of action of the force (RAy)
Very clear explanation
Incredibly helpful. Thank you
Hello, I am having difficulty to determine the moments. How do you the termine what forces create a moment about a given axis. I am kinda stuck visualizing it thanks
papa bless you , you're a good soul !
Thanks very much
You are a legend
Very well explained you are a very good teacher! I Must admit I wished you had solved it using the "clever" method. How would that method differ from this method? Would we simply use trig to find the distance of the AD axis to TBc and then calculate the moment to find TBc and then revert to the method used here to solve for the rest? Would you mind outlining this method a bit further please?
Hi Chris, thanks for the feedback. Glad you found the explanation helpful. The 'clever' way would be to use vector algebra. Probably won't have time to do solution this way any time soon as I do not currently teach the subject at my university in which this method is taught. Sorry.
why we not consider reaction moment at point A?
Sir I don't understand on how you are resolving the moments
It's a Ball and socket joint in D which allows rotations about all axes. So instead of taking the coordinate system at D, will it be technically correct to consider at A and take the moments about it?
Yes, you can take moments about any point you like. It will just lead to different equations that may be easier or more difficult to solve, but the answers should come out the same. In fact, it’s a good way to check your calculations and answers.
thanks man, love!
Thank you😊
thks for sharing the lesson i found it very usefull!
very helpful explanation but do have a question; why wasnt RAY included during the summation of moment taken at x-axis?
The line of action RAy is in the x-y plane and is perpendicular to the x-axis. Therefore the line of action of RAy would intersect with the x-axis in which case the perpendicular distance between line of action of RAy and the x-axis would be zero. Thus the moment of RAy about the x-axis is zero. Hope this helps. Thanks for the feedback.
Why isn't Ray included in the moment equation for y?
The line of action of RAy is parallel to the y-axis. Therefore it does not create a ‘turning effect’ about the y-axis. Thus the moment of RAy about the y-axis is zero. Also, your question is incomplete. When asking about a moment calculation you need to state the ‘point’ at which you want to do the calculation. On the y-axis you could be talking about B, D or E (or any other not labelled point). Hope this helps.
Shouldn't the sum of moment around the x axis be: Rdz (1) - W (.50) = 0
Tbc intersects both the Y and X Axis so there's no moment force for Tbc.
Tbc doesn't intersect X axis. Tbc is parallel to Z axis, and 1m away from D along Y axis
@@TerryBrownMechEng ooooh you're right. I thought point b was (0,0,0). I see it now... (0,0,0) is at point D. Brain fart haha. Just to confirm, if a force intersects an Axis - moment does not exist. Thanks for the quick response !
nice lesson thks !
Hi terry,
Why was Tbc was not included in the sum of moments at y axis?
Hi alexojha, thanks for your question. The line of action of the force Tbc passes through the y-axis. Therefore, the perpendicular distance from the line of action of the force Tbc to the y-axis is zero. Hence its moment effect about the y-axis is zero.
Once you got to 12:35 the chaos started. I don't think you have your forces labeled correctly. Also I don't believe you did all the moments correctly.
Thanks for the comment and feedback. When you say “chaos”, are you referring to the fact that lots of reaction forces turn out to be zero? This does seem counter-intuitive at first, but if you look at the original problem, you can see that the bearing at A is not actually needed for equilibrium for the loads and cable support specified. There is nothing to cause rotation around the y-axis or z-axis which leads to the result that the reaction forces at A are zero. If we included the weight of the bent rod, then we would have weight of the section AB causing a moment about the y-axis and you would get a non-zero answer for the vertical reaction at A. Which forces do you think I have labeled incorrectly? They all look ok to me but sometimes it’s easy to miss your own mistakes. Also, which moments do you think I did incorrectly? Again, they all look ok to me. What values did you get for all the reaction forces?
Thanks for the lesson, next time include the axis labels so we know what your on about
axis labels are on the original problem diagram at the beginning and added to my free body diagram at 6:15