The best-ways to follow this learning from this video is not focusing on the formulas and calculations but listen the lectures most important facts,core information and concepts are stated even if it is fast explanations.
Thanks for your excellent way of delivering your lecture with nice illustration, hand writing and concise concept! I am a gear design engineer, and I would like to watch many lectures here to refresh my memory from university years. By the way, can you share with me how you create the electronic lectures? (device, software and on). I used my own note with ipad pro, but my hand writings in that device is not as expected, but when I look at yours, the hand writings are so natural!. It would be appreciate how you create the notes!
thanks for your effort, but i have a question ; when designing a shaft with gears you know torque changes at the same point of gear , can i take this as a torque fluctuation to get the Ta & Tm ?? or consider always torque is constant ??? also when can i consider Mm for a rotating shaft is not equal to zero???
I think there is a tiny mistake, r is not 3/16, it's 5/16 you can see this in minute 9:20 5/16=0.3125 thanks a lot for this video its very rich and useful
At 7:40 , the alternating shear should equal to transverse shear stress due to forces acting on gear. This is because the shaft is rotating hence causes the transverse shear stress to alternate.
Transverse shear is usually not considered for shaft design (unless the shaft is short and thick, which is usually not the case). The transverse shear values don't contribute much to the shearing stresses, and are therefore usually negligible.
Nice summary and explanation, but I have to ask how can I determine the stress concentration factors when designing for a new shaft, when I don't have any dimensions yet, like the fillet radius and the diameter of the shaft. How can I know kb ( size factor ), (r/d, D/d) to find Kt, Kts, Kf, Kfs. Thank you
Can we also use the same formula to determine "n" when Sut and Sy are given in MPa? (We would convert them to Pa and then substitute in the formula for n, correct?)
Hello! Any idea what would happen in the case where we have an alternating bending moment acting on a rotating shaft? What usually happens is that we have a static bending moment acting on a rotating shaft, and the rotation causes the bending load to fully alternate. But I have never found an example in which the bending moment itself is alternating (ex. Mb = 10 +/- 5 Nm) WHILE the shaft is rotating. What would we write in the fatigue failure criterion for the bending stress in such an example?
you really dedicate yourself into your videos, i can see that and i am thankfull for that. But you are reading? or you sound like you are reading. Tonning in your voice made me sleepy, your voice is nice no problem about that. Just tonning. i believe if you adjust this, it would be beter. thank you
The reason for the Marin Factors is purely experimental, but they match the values obtained using the von Mises stress for torsion only. Therefore, when using von Mises stresses, Kc should always be 1, as the von Mises stress is already accounting for it.
You’re right, thanks for pointing it out. However it’s only the 600 missing a zero in the sum of moments about A. Notice that the resultant 19000 and all the other values are still correct.
@@rustamali8309The problem gives you either the material or the properties. The properties are given in the back of some books, e.g. Shigleys Mechanical Engineering Design.
Great stuff man, really appreciated for those equations and methods. By the way is there any place where I can download or purchase your notes? If there is, then you do a huge favor to me.
![[WATCH THE NEW CORRECTED ONE] Shaft Design - Moments in 3D Orthogonal Planes - Example 1](http://i.ytimg.com/vi/CvPU9lUkJ_c/mqdefault.jpg)
![[WATCH THE NEW CORRECTED ONE] Shaft Design - Moments in 3D Orthogonal Planes - Example 1](/img/tr.png)
Correction: the fillet radius should be 5/16", not 3/16" for the example at 5:05. Thanks to John Finan for pointing this out.
Dude, this is insanely helpful for my final project! Thanks man!
This video said what my mech Eng teacher did not teached in 6 months of course. Thanks a lot :)
Didn't know there were such high quality videos on ME topics like this, soooo helpful as a student
Just discovered this channel. Absolutely insane!!
The best-ways to follow this learning from this video is not focusing on the formulas and calculations but listen the lectures most important facts,core information and concepts are stated even if it is fast explanations.
God bless you, you saved me a lo tof time
Amazing content!
superb
Thank you so much! A great refresher!!!!
Good
Great stuff
Thank you very much!
Very good
Damn incredible
good job
Thanks for your excellent way of delivering your lecture with nice illustration, hand writing and concise concept! I am a gear design engineer, and I would like to watch many lectures here to refresh my memory from university years. By the way, can you share with me how you create the electronic lectures? (device, software and on). I used my own note with ipad pro, but my hand writings in that device is not as expected, but when I look at yours, the hand writings are so natural!. It would be appreciate how you create the notes!
thanks for your effort, but i have a question ; when designing a shaft with gears you know torque changes at the same point of gear , can i take this as a torque fluctuation to get the Ta & Tm ?? or consider always torque is constant ???
also when can i consider Mm for a rotating shaft is not equal to zero???
Thanks Sir
I think there is a tiny mistake, r is not 3/16, it's 5/16
you can see this in minute 9:20
5/16=0.3125
thanks a lot for this video
its very rich and useful
At 7:40 , the alternating shear should equal to transverse shear stress due to forces acting on gear. This is because the shaft is rotating hence causes the transverse shear stress to alternate.
Transverse shear is usually not considered for shaft design (unless the shaft is short and thick, which is usually not the case). The transverse shear values don't contribute much to the shearing stresses, and are therefore usually negligible.
Big like
How can I apply this to calculate shaft for shredder?
Nice summary and explanation, but I have to ask how can I determine the stress concentration factors when designing for a new shaft, when I don't have any dimensions yet, like the fillet radius and the diameter of the shaft. How can I know kb ( size factor ), (r/d, D/d) to find Kt, Kts, Kf, Kfs. Thank you
Can this information be used to design vertically suspended shafts?
only for high speed brain this can be understand consider slowing down
Can we also use the same formula to determine "n" when Sut and Sy are given in MPa? (We would convert them to Pa and then substitute in the formula for n, correct?)
if i have a solid shaftwithout notches the value for kf can be negleted right?
Hello! Any idea what would happen in the case where we have an alternating bending moment acting on a rotating shaft? What usually happens is that we have a static bending moment acting on a rotating shaft, and the rotation causes the bending load to fully alternate. But I have never found an example in which the bending moment itself is alternating (ex. Mb = 10 +/- 5 Nm) WHILE the shaft is rotating. What would we write in the fatigue failure criterion for the bending stress in such an example?
what is the app you demonstrating on?
you really dedicate yourself into your videos, i can see that and i am thankfull for that. But you are reading? or you sound like you are reading. Tonning in your voice made me sleepy, your voice is nice no problem about that. Just tonning. i believe if you adjust this, it would be beter. thank you
why is Kc equal to 1 when von misses stress is used, is it just a fact or is there a reason behind it
The reason for the Marin Factors is purely experimental, but they match the values obtained using the von Mises stress for torsion only. Therefore, when using von Mises stresses, Kc should always be 1, as the von Mises stress is already accounting for it.
at Ma you multiply 3.5 in with 600 instead of 6000
You’re right, thanks for pointing it out. However it’s only the 600 missing a zero in the sum of moments about A. Notice that the resultant 19000 and all the other values are still correct.
How you have take "' Se"' endurance stress ?
Can you rephrase the question?
I mean,
How did you get the value of Se=24ksi
Can you calculate at this comment platform PLEASE
@@rustamali8309 Se is given as part of the problem. No calculations were carried out to find it.
@@rustamali8309The problem gives you either the material or the properties. The properties are given in the back of some books, e.g. Shigleys Mechanical Engineering Design.
Great stuff man, really appreciated for those equations and methods. By the way is there any place where I can download or purchase your notes? If there is, then you do a huge favor to me.
Thanks! I'll find a way to share these soon, and let you know.
Very fast Sir. Want to understand but not able to cpture the speed even though it looks useful
3/16 is not 0.3125
thanks for catching that!
great lecture, very objective indeed. except, your monotonous way of speaking makes it very hard for me to focus.
👁 👁
👄
took my prof 7 weeks...
Tþ
Damn you lost me 1 minute in