Module 12: Elastic torsion formula

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来自 Georgia Institute of Technology 的课程

Mechanics of Materials II: Thin-Walled Pressure Vessels and Torsion

228 个评分

Georgia Institute of Technology

Mechanics of Materials II: Thin-Walled Pressure Vessels and Torsion

228 个评分

This course explores the analysis and design of thin-walled pressure vessels and engineering structures subjected to torsion. ------------------------------------------------ The copyright of all content and materials in this course are owned by either the Georgia Tech Research Corporation or Dr. Wayne Whiteman. By participating in the course or using the content or materials, whether in whole or in part, you agree that you may download and use any content and/or material in this course for your own personal, non-commercial use only in a manner consistent with a student of any academic course. Any other use of the content and materials, including use by other academic universities or entities, is prohibited without express written permission of the Georgia Tech Research Corporation. Interested parties may contact Dr. Wayne Whiteman directly for information regarding the procedure to obtain a non-exclusive license.

从本节课中

Elastic Torsion of Straight Cylindrical Shafts

In this section, we will learn how to analyze and design for elastic torsion of straight cylindrical shafts.

Module 9: Definition and examples of real world torsional engineering applications3:50

Module 10: Torsional shearing strain5:37

Module 11: Torsional shearing stress4:42

Module 12: Elastic torsion formula5:56

Module 13: Polar moment of inertia3:57

Module 14: Angle of twist2:32

与讲师见面

Dr. Wayne Whiteman, PE
Senior Academic Professional
Woodruff School of Mechanical Engineering

[SOUND] Hi, this is module 12

of Mechanics of Materials II.

We've now completed looking at torsional shearing stress and strain.

In this module we're going to develop an important relationship called the elastic

torsion formula.

And so today's learning outcome is to derive that elastic torsion formula.

So, here's where we, what we did in the last couple of modules.

First of all, we looked at strain.

We found that the torsional shear strain was maximum at the outer surface, and that

it varied linearly, the shear strain did, with the radial distance from the center.

And then we also looked at stresses and we found that shear stresses also vary

linearly with the distance rho from the center of our circular bar.

And so now we want to in this module, relate

the shear stress to the applied torque that we have on our engineering member.

And so I've shown my member here.

At a distance rho from the center I've shown a little

differential element of area.

And on that little differential of area we have a shearing stress.

So the stress times that little dA is equal to some dF or

differential piece of force.

And so now I can say that the differential torque for that little

element is equal to the force which is dF times it's moment arm which is rho.

And so I can substitute now in dF is tau dA.

And when I do that I can say okay, but we know that tau is

varies linearly with a distance from the center.

So it's rho over r times tau max or the maximum shear stress.

And when I substitute that in I have a rho and a rho so I get a rho squared.

And I have tau max over r, so I've got tau max over r rho squared dA.

Now if I want to find the total torque, how would I do that?

And what should you say is,

okay I should integrate over the cross section, the entire cross section.

So T, the total torque is equal to the integral of the entire cross section

of DT and I substitute in DT.

I've pulled out t max over r from the integral because those are constant,

they do not vary.

And this quantity, the integral over the area of rho squared dA we're going

to define as the polar moment on inertia, and we're going to give it the symbol J.

And so there's J, the integral over the area of rho squared dA.

And here again that's result.

Now we can say that T is equal to tau max over r and this whole quantity here is J.

When I substitute in, I rearrange that now so

that's tau max is equal to Tr over J or

the torque applied times the radius over the polar moment of inertia.

Recall also, that again,

the shear stress varies as we move out from the center a distance rho,

varies linearly, and it's proportional to the tau max.

And so if I rearrange that, I have tau max is equal to r times the shear stress over

rho or the radius over the distance from the center times the shearing stress.

And so tau max also equals Tr over J, which we developed up here, so

I can put that in.

And I can finally solve for

the shearing stress at any point on my circular bar across section.

It's equal to T rho over J and that's what we've called the Elastic Torsion Formula.

And so the shearing stress on a transverse plane, any transverse plane

is a at a distance rho from the center of the axis is equal to this value here.

And so what we've done is we've related shear stress to applied torque.

Here is our applied torque, here's the resulting shear stress and again a very

important relationship called the Elastic Torsion Formula Torque is the resisting

torque generally obtained from a free body diagram or equilibrium equation.

So you take an engineering member, you draw a free body diagram, and

you can solve for the torque, the torque being applied at any particular point.

And we'll do that in examples later in this course.

You'll note that the torque T is greater for larger J.

The resisting torque is greater for larger J, or polar moment of inertia.

If T's going to go up,

J goes up proportionately to keep the same shear stress.

And so J, the polar moment of inertia,

is larger when we have more area further from the axis of rotation.

And you can see that here.

The more area we have at a further distance rho, the greater our value of

J is and therefore we're going to have a greater resistance to torque.

So J is a measure of this resistance to twisting or torsion.

And let's look at our example here with our pool noodle.

So that more area I have further from the center, or the more

bigger J polar moment of inertia, the more resistance to torque I'm going to have.

And so that's where we'll leave off and we'll pick it up again next time.

[MUSIC]

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