# Quick Torsion question

1. May 18, 2005

### morry

Hi guys,

Just a quick question. I have been attempting some problems to do with torsion.

Most are pretty standard, you know, find the torque, angular rotation etc.

However Im stuck on this one. I have calculated the torque and rotation for a shaft thats at its proportional limit. But the next part asks me to find the same things, but this time the beam is fully plastic. How can I do this?? I have no idea.

Thanks guys.

2. May 18, 2005

### Gokul43201

Staff Emeritus
Morry, it is most efficient to post the exact question.

3. May 18, 2005

### morry

Ok here it is:

What torque and angular rotation are produced if the bar is deformed until it is fully plastic? Assume perfectly elasto-plastic material behaviour.

I am given the yield stress and elastic modulus. Dont worry about giving me specific answers, just a quick understanding is all Im after.

Cheers.

Last edited: May 18, 2005
4. May 18, 2005

### FredGarvin

At first thought, they may be asking you to use the ultimate yield stress as your failure criteria in stead of the yield stress. I am not quite sure what "perfectly elasto-plastic material behaviour" really means.

5. May 18, 2005

### morry

The trouble is, they dont give us anymore information about the Uts. I think its a weird question. Its good to see Im not the only one who thinks this way.

6. May 18, 2005

### PerennialII

Are you familiar with analyses using rigid or ideal plastic plasticity descriptions? The first thing coming to mind in a question such as this one is that they'd want you to do a limit load / plastic collapse analysis. Usage of solely yield strength in this case implies probably that hardening in all forms is neglected.

7. May 18, 2005

### Gokul43201

Staff Emeritus
Perennial, I suspect this is much simpler still, and yes, strain hardening must be neglected.

The stress-strain curve, I believe, looks as shown in the attachment. This is what I recall is refered to as "perfectly elastoplastic".

That said, I believe there is insufficient data to answer the second question : the angular rotation at failure. Clearly, the torque will be the same as in the elastic case.

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8. May 18, 2005

### PerennialII

yeah, that looks familiar ... if I was to answer the question would do a collapse analysis with the material model ... which can be done with "ease" for a simple geometry such as a shaft - however, can be that they're after something much simpler (or complex) ... depending on the "degree" of answer they're after ...

9. May 18, 2005

### morry

Thanks for the replies guys. So you think that the ultimate tensile stress will be the same as the yield stress?

Its only for 2nd year solid mechanics so the answer shouldnt be too complicated. We didnt cover torsion in great detail, only the basics.

Thanks again.

10. May 19, 2005

### PerennialII

In this case the UTS is the same as YS ... since you don't have info on the former (or course noting the obvious about the relationships of yield properties in shear versus tension). In these sorts of plasticity simplifications you often see people use either YS, UTS or mean of them .... and since you only got one it simplifies.

11. May 19, 2005

### morry

Well, i finally sorted it out. Turns out for shafts, that when they go beyond their proportional limit, the twisting moment turns out to be (4/3)*Ty, where Ty is the max moment at proportional limit.

So it was pretty simple. And of course, once you have this new torque, finding the angular rotation is easy.

Thanks for the help guys!