Mechanical Engineering: Particle Equilibrium (8 of 19) Attached Beam Under Compression
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- čas přidán 19. 06. 2015
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In this video I will calculate the compression force of a beam, attached to a wall, under compression.
Next video in the Particle Equilibrium series can be seen at:
• Mechanical Engineering...
What an incredible contribution to humanity.
Sir, you deserve Patreons.
Im majoring in electrical engineering but also watching all of your ME videos because of how great you teach. Thank youu!!
Cudos for EE. I have taught many ME courses at several universities and colleges as an adjunct professor. Some of my best students were EEs.
Could we check that the direction of T2 is exacttly the direction of the beam, by compting the angle teta tang-1(T2x/T2y)?
What if the load was placed was placed at a point further alongthe beam & away form the tie pivot as to create an Extra moment
I know that before applaying the equilibrium we have to define our system to know the exterior forces isn't it?
very good lecture Sir
I think this lecture is good to listen in light-mind.
MIT lectures need mind-preparation before listen...
Professor Biezen, thank for another fantastic analysis on Particles Equilibrium Attached Beam Under Compression, however some of the Sine and Cosine components are confusing.
Thank you. Yes in this example the angles are a bit of a brain twister.
Sir, Do you happen to know any playlists or websites which are related to math?
There are over a thousand math videos on this channel: here czcams.com/users/ilectureonlineplaylists?sort=dd&view=50&shelf_id=9 and here czcams.com/users/ilectureonlineplaylists?sort=dd&view=50&shelf_id=11
Hey professor Michel van Biezen, what if force is not given how could we approach to solve this problem? thank you .
If the force was not given you would have 4 unknowns. The equilibrium category for this 2-D problem would make the problem statically indeterminate.
Thanks. I'm trying to re-teach myself Newtonian mechanics lately. Wouldn't it have been much easier to assign the x-direction to the axis ALONG the beam itself (i.e., rotate the picture clockwise by 55°). The angle between the beam & T1 is 85° (i.e., 180° - 35° - 60°), and the angle between the beam and F is 55° (i.e., 20° + 35°). This would result in only 2 simple simultaneous equations that directly solve for both T1 & T2: [X-axis]: T2 = T1cos85° + Fcos55° & [Y-axis]: T1sin85° = Fsin55° (or T1 = F (sin55°/sin85°). Substituting second equation into the first gives: T2 = F ((sin55°/tan85°) + cos55°) = 645N and T1 = 822N.
There are many ways in which problems like this can be solved.
Thanks for making this video. Can I solve this problem with Lami's theorem?
Try it and see if you get the same answer. We learn a lot from trying......
Hello, thanks a lot for the amazing video, but I was just wondering why isn't there a force exerted by the wall on the beam.
There is a force, but it does not contribute to the torque on the beam.
@@MichelvanBiezen Thank you
it is not clear for my mind that this force T2 will be in the same direction of the beam, should we check this once T2x and T2y are calculated? and check that teta is 35°?
This a 2-force member (with the weight of the beam neglected). So the force is directed between the 2 support points of the beam. A FBD would clarify that, but this professor never utilizes it.
@@psilvakimo ;)
Great man
We appreciate the comment.
At 9:04 the equation was multiplied by 2... why may i ask?
To simplify it. (Solving for T1)
Hi sir, I don get why we have T2 isnt T1 enough to make the system at equilibrium. Also why did you put T2 in this specific direction. Is there something i should follow to know the direction of T2. last question when do we include T2
The goal here is to find the force T2 such that when applied there will be no compression or tension on the beam, which essentially equals the compression on the beam without T2
Thanks for your response but why in this direction is T2 did you follow a certain procedure
The direction of T2 should be on the same line as the bar.
sir, what is the application for this video?
An example like this is used to learn how to utilize the concept that if a system is in static equilibrium, then all the forced in each direction add up to zero, and all the moments (torques) add up to zero.
Can we say that T2 is beam's pin reaction?
Yes.
Sir what website should i search in order to get more exercises to practice?
Have you worked out all of the examples in the playlist?
Yes i did
There are many more examples in the physics playlists as well. I haven't looked for other sites on the internet for that but I am sure there are many.
Do you know any of them?
Excuse me, I don't understand why T2 is in the first quadrant and not in the third :/
T2 represents the beam pushing back against the 2 forces compressing it.
ahhhh. Thanks! How could we calculate the compression suffered by the beam caused by the applied force so? Do you have some classes where you treat the subject?
Thanks again, your videos are great :)
T2 has the same magnitude as the compression force on the beam.
thanks!
Why shouldn't you considered the support reaction at the hinge of the beam ?
The forces at the pivot point can be ignored, since they don't provide a torque.
@@MichelvanBiezen In which cases can we ignore it ?
When the action of the force goes through the point of rotation (pivot point) we can ignore them because they don't cause a torque to exist.
@@MichelvanBiezen you mean that, the pivot point has frictionless surfaces? Like, it's free to rotate.
What if the pivot is tight ?
i think that again you are looking for resultante and not equilibrium ? isn'i it?
He should mention that the beam is a 2-force member (neglecting the weight) and draw the free-body-diagram, which why the T2 force is directed the way it is. What I don't like about this guys lectures is that he never utilizes the FBD.
Your videos are not visible
Not sure what you mean by "not visible". We are able to see all the videos.
Allrieght
Glad you liked it. 🙂
In french please
We don't have the bandwidth to translate into different languages. Perhaps in the future if the channel keeps growing.
I would just solve this graphically.....
i enjoy all u r lectures-- they are excellent - amarjit
It took around 12 minutes to solve this problem. Is it logical to say the teacher should only give 10 problems for a 2 hours exam.
When i give a test, I usually give the students at leas 3 times as much time as it took me to solve the problems. (I believe that is the standard).
@@MichelvanBiezen Thank you for your reply. You are such a considerate person！
Thank you
nice bowtie
Thank you. 🙂