shaft torque-twist question

[lfw618]

I have what may be a simple question, I don’t know if I’m just overthinking this. I am looking through the test data of Torque twist curves on a couple shafts. Yield and rupture torque seem more straightforward. However I would like to compare toughness between various shafts. I understand for a stress strain curve that toughness will be the area under the curve. Toughness determination does not seem as obvious to me, for shafts with varying cross section.

Using the area under the curve of the torque-twist diagram seems like it would be flawed because the sections of the shaft that have different cross sections will be straining at different rates, and the overall twist is heavily dependent on the length of each cross section.

I’ll give a hypothetical example to help illustrate my point. I’ve sketched up two shafts. For discussion purposes ignoring any stress concentrations, and assume the shaft is in pure torsion. The shafts are simpler than the ones I’m dealing with but the principle is the same. On both shafts the ends are the same diameter. The first shaft has a relatively large diameter in the middle of the shaft, the second shaft slopes down to a diameter just slightly larger than on the ends. Both shafts should rupture at the same torque since the smallest diameter on the ends is the same. The Second shaft will overall have much more twist, since the smaller diameter in the center will twist much more than the larger diameter on the first shaft.

If I take the area under the curves for the torque twist diagrams, it would seem that the second shaft is much tougher than the first. Would this actually be the case? Is having a smaller diameter in the center actually making the shaft effectively tougher, as there is more overall twist available before rupture? Or would they both be similarly tough in function because the strains at the the ends are still the same—since they’re still the same area cross section and under the same torque? Or am I off base on all of this?

Ideally I’d like to convert the curves from Torque-twist to stress-strain, but I’m not 100% sure how to do this, especially for the plastic region.

Thanks in advance for the help!

 

[handleman]

As I understand it, toughness has to do with absorbing energy from shock. Making your shaft thinner in the middle makes it more "springy" and it can absorb more displacement/shock energy without that energy causing the shaft to reach its yield stress. The more energy it can absorb (which is the area under the torque/twist curve) before reaching failure, the tougher the shaft.


I'm not sure what you're after with stress/strain and plastic region...

 

[Compositepro]

Your definition of toughness has nothing to do with toughness. It is the energy stored per unit of torsional stress. You do not want the shaft to be storing energy for no reason. The sudden change in diameter causes an increase in stiffness, and this results in an increase in stress at that point. This is generally where things break.

 

[lfw618]

handleman -- OK so you would say that the area under the torque-twist curve would relate to actual toughness, and specifically in this case the smaller diameter results in more allowable deformation, allowing for more energy to be absorbed before fracture?

I guess I was trying to find a way to convert the torque twist curve to a stress-strain curve. But thinking more on that I'm not sure it would be possible to get a meaningful graph given the variations in x-section.

Compositepro -- You do not think area under torque-twist would relate to toughness? And yes I agree the sudden change in diameter would be a stress riser. The examples I gave were just hypothetical to provide a simplified example of what I was trying to ask. Sometimes those stress risers are not avoidable where seal grooves, bearing journal, lube holes or splines are required.

 

[Compositepro]

You are looking at elastic energy stored in the shaft, not the elastic energy stored in the material of the shaft. Reducing diameter increases the strain in the material, it does not make the material capable of more strain.

 

[lfw618]

Right, this is more or less what I was going back and forth with. Because yes, the material itself is not actually any tougher, but if you look at each shaft at rupture, I believe the one with the reduced diameter will have more strain energy. Would that not make that shaft tougher in function? It has the ability to absorb more energy before fracture (so long as that reduced diameter stays larger than the end diameters).

 

[EdStainless]

By that logic you should keep the shaft constant diameter full length since that will distort the most and store the most energy. But do you want the system 'softer'?
But as the shaft twists it is trying to deflect, resulting in side load on bearings and possible whirling issues.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy

 

[handleman]

You always have to take into account the desired function of the component.

The less torsionally rigid the shaft is, the more shock it can absorb, but that stored energy may have undesirable effects as it's released. Or it might be good.

Consider the difference between pulling a stump out of the ground with a truck vs. fall protection in rock climbing. I've seen lots of footage of stumps going through the back windows of pickup trucks due to too much stored energy in a rope. The rope may be plenty strong, but it stores WAY too much energy as it stretches. A chain, much less so. On the other hand, if you tried to use a chain for climbing protection you'd snap your back in half because, even if it has the same ultimate tensile strength it absorbs much less energy.

 

[lfw618]

Thanks for the responses, this has helped me understand better. My thinking now is that there may be issues with making a shaft with too little torsional stiffness (energy release, critical speed, increased side load) but in some instances it may be desirable to increase the impact resistance of the part, and can be accomplished just through geometry changes.