# Torsion of Shafts

1. Sep 9, 2006

### Ascetic Anchorite

I am trying to calculate the diameter of a drive shaft. I would have thought that all I would need in order to determine the diameter of a shaft (concerning torsion) would be how much torque is going to be applied to the shaft and the shear modulus rating of the drive shaft material.

According to this web page http://www.engineeringtoolbox.com/torsion-shafts-d_947.html in order to determine shaft diameter, first I must know the diameter of the shaft! – Which is not at all helpful.

That site says:

And it also says:

So in order to determine the diameter of a shaft I need to know the Tmax of the shaft. But in order to know what the Tmax is I need to know the radius of the shaft!!

Am I missing something here?

2. Sep 9, 2006

### Hawknc

You've got two equations there and just the one variable to solve for (radius/diameter, whichever one takes your fancy). If anything, you have too much information. You can rearrange and substitute to get the variable you want if you know all the other values. All you should really need is Equation 2b on that site, though - if you know the maximum moment/torque applied, and the maximum allowable shear stress, you can directly solve for D.

3. Sep 9, 2006

### Ascetic Anchorite

What exactly do you mean by rearrange and substitute? It looks to me to be hopeless, unless I know the diameter of the shaft, which of course I do not as that is what I am trying to find and that is the whole purpose of the exercise!

As soon as I started this thread I realized that Tmax is just the maximum torque I want the shaft to cope with (is that right?). I do not know what is meant by σmax being the maximum shear stress? How do I find what the σmax is without knowing the diameter of the shaft? What exactly is maximum shear stress anyway (it sounds obvious, but that was before I realized I lived in a surreal twilight zone)?

Do you, or does anyone, know the formula for calculating a shaft diameter from knowing the maximum torque to be applied, the length of the shaft and the shear modulus rating of the shaft material?

This is an equation I thought would only take me 5 minutes to look up in a book, or online. But silly naive me, all of the engineering books I have contain pages of torsion formulae on shafts but every single one relies on already knowing the shaft diameter! This seems ridiculous to me. From a designers point of view torsion equations are supposed to tell the designer what the shaft diameter is meant to be, not the other way round!

4. Sep 9, 2006

### Q_Goest

Hi Ascetic,

The equation you're refering to:
Calculates the diameter of a solid shaft given that you know the torque being applied and the stress in the material. All this really says is that given some torque and some stress, the diameter of the shaft will be D.

Equation 4 is just equation 1, rearranged and with the polar moment of inertia for a solid shaft inserted. Note the polar moment of inertia is just a function of the diameter of the shaft.

Yes, that's correct. In fact, Tmax and σmax can be just torque and stress, you don't have to think of them as maximum torque or maximum stress. You can rearrange that equation any way you'd like to solve for the variable you'd like. And if you know two of the variables, you can solve for the third.

Note that you need to determine torque from some other evaluation of your system, and the stress you can allow is a function of the material properties.

The formula for calculating shaft diameter is a function of the torque you want to apply and the stress the material can handle. It is independent of shaft length or shear modulus. Those two variables would be needed to determine the amount of angular deflection in the shaft, but not the stress.

Equation 4 gives you the diameter of the shaft as a function of the torque you want to transmit and stress you can allow in your material.

5. Sep 9, 2006

### Hawknc

The shear modulus relates stress and strain, it's a measure of the stiffness of the material. It's really only relevant in this case if you want to know the deflection of the shaft. The maximum shear stress is probably the more relevant property for what you want, and it's a material property, i.e. independent of geometry. Equation 4 is the whole thing solved for D, so provided you know Tmax and σmax (one from your design conditions, the other based on the material you're using) you can find D easily enough.

Edit: beaten to the punch! :O

6. Sep 11, 2006

### Ascetic Anchorite

Thank you both very much for your help.

So the problem I have now is knowing how to identify this maximum stress figure, for a given material?

As you say it is a function of material properties, how am I to identify it? Is such a figure listed in a materials specification sheet? For example: I am looking at the spec’ sheet for grade 5 Titanium and it says ‘Ultimate Strength 896Mpa’, and also; ‘Yield Strength 827Mpa’ (maybe the Ti spec’ sheet is not ideal for an example, as Ti is not often used for drive shafts).

I have just found another couple of equations for determining shaft diameter, from a book. I have copied them out exactly as they appear in the book (click on link below, or see attachment):

Some of the nomenclature is guessed, as the book is not very clear in explaining its usage of symbols.

Again, my problem with those two equations is knowing what to put for the allowable shear and bending stresses and the bending moment strain (presumably these are functions of material properties too, and not dependant upon exact dimensions of the material to be used in a shaft?)?

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7. Sep 11, 2006

### Hawknc

The first equation in that image is just Equation 4 rewritten for D and is probably still the more useful one. Do you know the difference between yield strength and ultimate tensile strength?

8. Sep 11, 2006

### Ascetic Anchorite

No, not yet. I am to research this area (different types of material strengths) after I have learned how to calculate shaft sizes (that may seem an odd order in which to progress).

I only used 'yield strength' and 'ultimate tensile strength' as examples of the sort of information listed in material spec' sheets. Whether these two measures are what I need for shaft calculations, I would have no idea!

9. Sep 12, 2006

### Ascetic Anchorite

I think I understand this but if someone could confirm if I am correct, I would be very grateful.

For the diameter of a solid shaft:

D = 1.72 (Tmax/σmax)1/3

Example calculation:

I need a shaft that can cope with 10Nm. The shaft material is to be 6061-T6 aluminium (the spec’ sheet for which is partially posted below as an attachment). So the calculation would be:

Diameter = 1.72 * (10 / 207)1/3 = 0.027697262 = 27.7mm

So, basically the relevant section of a material’s spec’ sheet (when designing drive shafts for torsion) is the ‘shear strength’ figure, which for 6061-T6 aluminium is 207Mpa. Is that correct?

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10. Sep 12, 2006

### Q_Goest

Hi Ascetic. Couple problems. The equation you're using is correct, but the 1/3 is an exponent which means they are taking the quantity (Tmax/σmax) and raising it to the 1/3 power. Note that this is the same as taking the cube root.

Using your numbers, and correcting for the exponent, you should get 6.26 mm.

However, that's not all of the issue. A drive shaft is under cyclic loading, and even if it weren't, you shouldn't use a material right up to the point that it will start to deform (ie: yield strength). There are "factors of safety" which are applied depending on the accuracy with which you can predict stress. Also, because of the cyclic loading, there is a fatigue issue.

If you can be certain the maximum torque will not exceed 10 Nm, you can use that as the maximum torque. However, you may want to consider other operating conditions which could increase that value.

You also should examine how the stress may cycle in this material. There's too much to explain regarding fatigue, so rather than go through it, I'll simply throw out a value which I think would be reasonable for the maximum allowable shear stress in this particular material, which is around 60 mPa. Much more than that and cyclic stress could cause it to break.

Given the 10 Nm for torque and 60 mPa for allowable stress, you should come up with a diameter of 9.5 mm.

Note also that drive shafts are often hollow tubes. This puts more of the material under stress and reduces weight. The same reference you used for the solid shaft has an equation for the hollow shaft. You could use that one if you decide to use an aluminum tube instead of a solid shaft.

That's the short response. There's a lot to designing such things, and I really can't put it all in one post for you. Doing a thorough stress analysis on the drive shaft might also include other factors, and a Mohr's circle might be called for. A Goodman diagram for fatigue would help narrow down what maximum stress you should use, and there will inevitably be stress concentrations along the shaft which should be considered. Design often comes down to determining to the best of your ability, all of the factors you must consider, and then using your best judgment, determining what factor of safety you should add to that.

11. Sep 13, 2006

### Ascetic Anchorite

Thank you very much, QG.

Thank you for correcting me on the 1/3 exponent issue! That could have caused a few upsets! – Although at least my drive shaft would certainly not be under-engineered had I not realized the error of my calculation!

A new problem for me now seems to be a lack of material specification data on shear strength. The information I posted as an attachment on Aluminium was only used as it was the first spec’ sheet I found. I actually intend to use a much stronger material, as the shaft has to be as small as possible (which is why it will be solid). Although the Aluminium spec’ sheet contained specific shear strength data, it seems to be almost alone in doing so! I searched through the website it came from ( www.matweb.com ) and I clicked on the material data pages for other materials such as different types of steel. Unfortunately the steel spec’ sheets I looked at contained no listings for shear strength, only shear modulus (and modulus of elasticity, etc.). So I am not sure what I am to do now? I will search for other metal databases, but I sense I will not have a lot of luck.

What I am wondering is this: is it possible to determine a material’s shear strength, or strength to resist torsion, by looking at other figures on the spec’ sheet? I have attached a copy of the main spec’s of a particular grade of steel (also found here: http://www.matweb.com/search/SpecificMaterial.asp?bassnum=M1015G ). Is it possible to determine a metal’s torsion resistance strength by examining the information given in that spec’ sheet alone?

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12. Sep 13, 2006

### Q_Goest

6061-T6 is a good choice if you want to use aluminum. There are other aluminum alloys that are slightly stronger and have better fatigue strength, but 6061-T6 is nicely weldable, readily available, and overall is a decent choice for a high strength aluminum.

Regarding the relationship between shear and normal stress (yield and ultimate tensile stress are normal stress values) there are rules of thumb, but nothing you can truly calculate one from the other. There's a table that gives those values here:
http://www.roymech.co.uk/Useful_Tables/Matter/shear_tensile.htm
and a discussion on the engineering forums here:

Suggestion: Set up an Excel spread sheet so you only need to input the variables and the spread sheet calculates the value you're looking for. No sense in calculating everything through by hand every time, that gets monotonous and can lead to errors. I have hundreds of spreadsheets set up into different catagories on my computer so I can do all the common calculations very quickly and accurately. For this spreadsheet, I'd suggest incorporating different cross sections as well, such as a hollow tube or a square or rectangular tube, etc... Do some calcs using the hollow tube especially. You'll find the OD doesn't increase too much, but the weight can be reduced significantly. If you're looking for help with Excel, you can post questions here or at engineering forums here:

13. Sep 18, 2006

### Ascetic Anchorite

Thanks once again, QC, and sorry for my delay in responding!

Thank you very much for those links. I registered an eng-tips quite a while ago, but never posted. When I went to post recently I realized I have forgotten my password. I requested a reminder, but none was forthcoming. I tried to re-register, but was unable to! Maybe this is because my ISP uses proxies?

The Excel idea is a good one too. I have just recently learned how to program basic information into it. I tend to use OpenOffice Calc, as it seems to work better on my PC, particularly the graphing function. Nasty learning curve for a first timer, and not a lot of readily available information on how to work with that program, but it was a lot easier than many dedicated graphing programs, that’s for sure.

14. Oct 16, 2008

Reviving an old post but I have a similar question.

In my case I do not know σmax or at least how to calculate it from material properties.
Material is 1045
UTS 84800
YTS 74300
Tmax = 591.55 in-lbs

D = 1.72 (Tmax/σmax)^1/3

Any help would be appreciated.

Regards

15. May 12, 2009

### Benson

Hi guys,

I have a general question about torsion in shafts.

I know that the shear stress due to pure torsion of a solid round shaft is a maximum over the entire surface, and is of magnitude Tr/J. However, how does one account for the direction of the shear stress when performing a Mohr's circle analysis, or generally combining it with other stresses that may be present?

Take the example of a shaft with an attached pulley, which experiences combined shear, bending and torsion:

At the top of the shaft cross section (element A), where the bending stress is maximum and positive, and direct shear stress is zero, it is clear that the bending stress is in the axial direction, and that the shear stress due to torsion is as depicted. The Mohr's circle can then be constructed with the points (Mc/I, -Tr/J) and (0, +Tr/J) forming the diameter of the circle.

But what if I wanted to represent the stress state at an element on the side of the shaft? That is, at a point on the surface, but 90 degrees clockwise from element A, where the direct shear would be maximum (4V/3A). Is the torsional shear stress not then on a different plane to that at point A - a plane oriented 90 degrees from the one at point A? How does one then combine the direct shear and the torsional shear, and construct the Mohr's circle for that point?

Benson

16. May 12, 2009

### Benson

Hi again,

I think I solved my problem...

Using the same coordinate system they are using, the stress state on an element 90 degrees clockwise from element A would be pure shear (all direct stresses are zero), but this time in the x-z plane, and the magnitude would be Tr/J + 4V/3A. This gives a Mohr's circle of a radius equal Tr/J + 4V/3A, with the principal stresses being +/-(Tr/J + 4V/3A) and at an angle of 45 degrees.

However, on an element 90 degrees counterclockwise from element A, the direct shear and torsional shear are again in the x-z plane, but in opposite directions, and hence the magnitude of the shear, and hence also the principal stresses at 45 degrees, would therefore be Tr/J - 4V/3A.

Am I right?

Benson