Find the maximum permissible torque for a shaft of known dimensions
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- čas přidán 13. 09. 2024
- This mechanics of materials tutorial goes over how to find the maximum permissible torque for a shaft of know dimensions.
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Thanks for watching, I hope it helps!
Perfect! Exactly what I needed and very well explained. Thank you!
Awesome! Glad to hear it :)
@@Engineer4Free What is you have three sleeves of material shafts?
Cracking video, cheers!
Isn't the equation for polar moment of inertia for this problem should be J = pi/4 * (Ro^4-Ri^4) ?
Fantastic, thank you so much :)
Thanks for the explanation. What surprised me that the length of the rod doesn't have any effect for this one. But now when I think about it more, it makes sense. The length only changes how much twist can be in the rod before it brakes. Thank you for opening my eyes :D
Yeahhhh cool eh. The longer the member, the greater the angle of twist will be, when all else is the same. You might be interested in checking the other torsion videos I did, they are numbers 14-21 here: engineer4free.com/mechanics-of-materials 👌👌
Just watched your vid on c++ spaces and tabs! Can you make an update version maybe? Btw great vid
+ShadowPlayz haha nice what program are you if you're watching both courses? For the next little while I'm just focussing on making new mechanics of materials tutorials!
No bro, i just watch your vids because they are educational and very fun to watch :) I hope you start with some more programming tho
That's awesome, good for you!! Keep it up and keep checking back :)
why did he do 'to the power of 4' for J?
is 2,893Nm = 2,89KNm ?
Sorry for the late comment, where does one find the allowable shear stress?
Fallen down in the same problem . did u find it?
Did either one of you find this?
@@adilfarooq9009 mine is given in the question
It depends from material
what if the shaft is from 2 irons
Hey Anas, then it would be similar to this example, www.engineer4free.com/4/angle-of-twist-due-to-torsion-in-a-shaft-with-varying-cross-section but rather than having J1 and J2, we'd have G1 and G2. Assuming the cross section is the same size and that the 2 sections are not overlapping. Also worth checking out the pure bending of composite materials videos (#29-24) here: engineer4free.com/mechanics-of-materials They're about bending, not torsion, but some concepts are transferrable
@@Engineer4Free thank you sir 🙏
U saved my life 😂
Are you sure you want to use the inner radius to earn the Maximum Torque out of this?
Because when we rearrange the equation for Torque it'll be: T = ( Shear stress * I polar ) / ( distance from center ) while we have already fixed values for each of: Shear stress = 85 Mpa & I polar = 1021017.612 mm^4 . And what we have left is the value of: Distance from the center, and since it is in the denominator we need to decrease it's value as much as possible so the Torque's value rise to it's Max.. .
and in order to do that is to choose the inner radius of the shaft ( 20 mm ) to obtain our MAXIMUM Torque.. am i right ?
i think you are wrong...we are essentially looking for the torque which caused the max shear force...so we put in the c at which the shear force is at it's maximum, and that is the outer dimension (30 mm)
you may want to refer to the equation before it is rearranged for T to visualise this
take this with a grain of salt though, I'm just a student, but I'm open for discussion :)
Why did you not use J=(π×d4)÷32
dont think you need the answer anymore😂 but its because it has an opening
Because instead of D^4 he used r^4. In the video C1 & C2 are radius. d=2*r. Therefore (2*r)^4= 16r. 16/32 gives the 1/2 you see in the denominator. I was seriously confused too. Had to look that one up so I will post here if others were scratching there head the same way I was.