Thank you for your valuable information @Mr. Jeff Hanson! Hello everyone, I have a question about that, If we apply heat treatment to related shaft, will there be any improvement regarding the angle of twist or shaft safety? I mean, How can we calculate the effects of heat treatment in case of comparing the case hardened steel or induction hardened steel (20MnCr5 vs 4140)during calculation of torsion on the shaft?
so if we have hollow shaft and wanna know what is the shear stress at inside diameter which is Ri. Which J that we gonna use. I think we consider it like solidshaft and use the first one.
A bit late but the standard unit for pressure is the pascal (N/m^2), but since he did his calculations using mm, his final answer would have been in N/mm^2, which is the megapascal
Prolly too late but it's half of the outer diameter as we want to find the biggest shear stress(which is located at the further most point from the rotation center)
Is C = 20mm because it's half of 40mm? Why wouldn't it be C = 40mm since in the drawing he says that C is the length from the center of the shaft to the outer part of the shaft
Hello, not sure how correct I am on this, but since c is measured from the center of the shaft to the point of interest (which in this case is the outer part of the shaft) then c pretty much acts like the outer radius of the shaft which is 40mm/2 = 20mm. Hope this helped.
Love the Just a Moment Joke! ahaha
for some reason I laughed more than I should have hahahhaha
Loved the moment pun! Thank you
These videos have been more useful to me in 30 minutes than a week of lectures. You are the best.
absolutaley
What a master piece of lecture thumbs up ❤❤
That just a moment joke would have been perfect if there was an ad at that moment
Thank you for your valuable information @Mr. Jeff Hanson! Hello everyone, I have a question about that, If we apply heat treatment to related shaft, will there be any improvement regarding the angle of twist or shaft safety?
I mean, How can we calculate the effects of heat treatment in case of comparing the case hardened steel or induction hardened steel (20MnCr5 vs 4140)during calculation of torsion on the shaft?
thank you Doctor Hanson, you are a legend
so for the counterclockwise 8 kn-m.. if I were to block the left side of the cut, wouldnt it be going clockwise as theres only the 8 kn-m there?
best teacher
Thanks for the videos :)
Wonderful solids lessons!! really taking the time to explain the problems along with a good sense of humor :))
can someone please explain to me why he used C= 20mm as his furthest point of intersection? Shouldnt it be C = 40mm?
that is the outside diameter not the radius
you are awesome its so useful thanks
You're the best sir!
Professor, what if the element is a hollow disk? What is the formula we can use to calculate the torsional stress?
so if we have hollow shaft and wanna know what is the shear stress at inside diameter which is Ri. Which J that we gonna use. I think we consider it like solidshaft and use the first one.
Last minutes safe a man
RESPECT!!
Nice
Is it a procedure? I am not having alot of fun as I was in statics 🤔
Dr. Hanson, shouldn't CD be 7kN*m and not 8kN*m? 15-8=7
He did 15-7
“Just a moment” joke lost me a moment 😂
Can someone explain to me why he used MPa as the units for the final answer?
A bit late but the standard unit for pressure is the pascal (N/m^2), but since he did his calculations using mm, his final answer would have been in N/mm^2, which is the megapascal
PLEASE DOES SOME ONE HELP ME AND THE THE VERTION OF THE BOOK BECAUSE I WANT TO DOWNLOAD PLEASE HELP ME
please how can help me and tell me the name of this class book editor
@@abdirahmanalihajji8833 mechanics of materials by Hibbeler
might be a stupid question but why is it 20mm for C
Prolly too late but it's half of the outer diameter as we want to find the biggest shear stress(which is located at the further most point from the rotation center)
@@tonicakarun1731 isnt it suppose to be 40mm because it is from the centre of the circle no?
@@damhilmnNo, 40mm is the diameter. We need radius so 40 ÷ 2 = 20mm.
LOL 3:40
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Quality
Is C = 20mm because it's half of 40mm? Why wouldn't it be C = 40mm since in the drawing he says that C is the length from the center of the shaft to the outer part of the shaft
You just explained why C wouldn't be 40mm lmao. C has to be a radius (half of a diameter) which the 40mm would not be since it's given as a diameter.
You didn't treat internal forces on suspended frame
Hi, can someone help me determining the torque at the End-B of a shaft that is 40ft long and with an applied torque on one end-A of 30,000 ft-lb.
sorry but I'm confused why you used c= 20mm ? would appreciate if anyone can explain it to me. thank you...
Hello, not sure how correct I am on this, but since c is measured from the center of the shaft to the point of interest (which in this case is the outer part of the shaft) then c pretty much acts like the outer radius of the shaft which is 40mm/2 = 20mm. Hope this helped.
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Jai shree ram master sahab
THİS İS LİKE MATATABİ LOL