Mechanics of Materials: Lesson 23 - Shear Stress Due to Torsion, Polar Moment of Inertia

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[@haiiqbal]

Love the Just a Moment Joke! ahaha

[@javierhidalgo1465]

for some reason I laughed more than I should have hahahhaha

[@reubendsouza6465]

Loved the moment pun! Thank you

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absolutaley

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What a master piece of lecture thumbs up ❤❤

[@iversondescartin3948]

That just a moment joke would have been perfect if there was an ad at that moment

[@selimkarakullukcu3184]

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?

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[@LarryLovestein23]

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?

[@helenma9154]

can someone please explain to me why he used C= 20mm as his furthest point of intersection? Shouldnt it be C = 40mm?

[@cortlandjonas4069]

that is the outside diameter not the radius

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[@imranbu-khamseen8273]

Professor, what if the element is a hollow disk? What is the formula we can use to calculate the torsional stress?

[@dehagunay3390]

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.

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RESPECT!!

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Nice

[@HashemAljifri515]

Is it a procedure? I am not having alot of fun as I was in statics 🤔

[@jonathansanchez2854]

might be a stupid question but why is it 20mm for C

[@tonicakarun1731]

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)

[@damhilmn]

@@tonicakarun1731 isnt it suppose to be 40mm because it is from the centre of the circle no?

[@Lezyz]

​@@damhilmnNo, 40mm is the diameter. We need radius so 40 ÷ 2 = 20mm.

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[@michaelhawkins565]

Dr. Hanson, shouldn't CD be 7kN*m and not 8kN*m? 15-8=7

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He did 15-7

[@udeogaranyacollins3514]

You didn't treat internal forces on suspended frame

[@Natuaralised_Can_Fin]

“Just a moment” joke lost me a moment 😂

[@ma.lonavelfrancisco427]

Can someone explain to me why he used MPa as the units for the final answer?

[@roshannaidu13]

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

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@@abdirahmanalihajji8833 mechanics of materials by Hibbeler

[@slater202]

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

[@s7vxn]

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.

[@prashantjha6292]

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[@puteridania00]

sorry but I'm confused why you used c= 20mm ? would appreciate if anyone can explain it to me. thank you...

[@HarounSh]

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.

[@michaelmohan3272]

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.

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