Hi, I'm a mechanic and I need to settle a point with a fellow mechanic. I was describing how a 3/8"s extension twisted during use, and I said maybe I need one that has been hardened properly. As in *through hardened*. He said I am confusing hardness with rigidity. And to look into Young's Modulus. From my perspective, rigidity seems to be something that goes hand in hand with hardening. Although it is not clear if there is a direct correlation. What I am interested in is the angular deflection of a shaft from one end to the other when torque is applied. I have looked into transmission shaft design criteria just a bit, but over there they are not so concerned with angles of deflection as much as the ultimate yield strength and fatigue failures. Drill bits I believe are quite hard, and through hardened. Quite brittle too. An automotive drive shaft is not very hard at all, certainly. But has a great deal of torsional rigidity. If you can't increase a shaft's torsional rigidity by hardening it, well then how do you do it? Keeping the dimensions the same, is there a materials change or heat treatment change to be done? Thanks,
I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
The only thing I have really learned is that for a given amount of mass it seems it is better to make a shaft hollow than solid. This is because for a straight/solid shaft, the center does not move very much for each degree of angular deflection. When all the mass is concentrated at the outer edge, the circumference area, there is much more shearing action. This is because a circle's circumference increases non linearly(?). So there is an aspect of geometry to torsional rigidity. But I did not ever learn much about the materials aspect. It seems like the rigidity comes from the shear stress / strain properties. For hardened metals, I would think they do not deform elastically as much, but rather strain up to a point, and yield. About this I don't know very much. That is the extent of what I was able to find out.
From Wikipedia on Torsion, we have these relationships: [itex]T[/itex] is the torque applied; [itex]J_T[/itex] is the "geometry factor" (as you found out); [itex]r[/itex] is the distance between the center of rotation and the outer surface (the radius for a circular cross-sectional shape); [itex]l[/itex] is the length of the object the torque is being applied to; [itex]G[/itex] is the modulus of rigidity of the material (related to the Young's modulus your friend was talking about); [itex]\tau[/itex] is the shear stress; [itex]\theta[/itex] is the angle of deflection under the applied torque. The modulus of rigidity depends on the type of material and is not affected by hardening methods. It could be seen as the rotational spring stiffness of the material (How much deformation will happen under a given torque). But another material characteristic is the shear strength ([itex]\tau_{max}[/itex]) which is the maximum shear stress the material can support before shearing or permanent deformation occurs. Hardening will affect this characteristic. By rearranging the previous equations, you get the relationship between the maximum deflection and the maximum shear stress: [tex]\theta_{max} = \frac{l}{r}\frac{\tau_{max}}{G}[/tex] So if you harden the material, you will increase [itex]\tau_{max}[/itex], which mean you will be able to support a greater deflection before shearing occurs (and a greater torque as well). This would be the strength of the material, how strong it is. But, as your friend mentioned, the amount of deformation depends also on the modulus of rigidity: the smaller it is, the greater will be the maximum deflection. This would be the stiffness of the material, how rigid it is. Finally, as you found out on the Internet, there is a geometric factor as well. The longer is the object and the smaller is its outside radius, the greater can be the maximum deflection. But if you are only interested in the angular deflection vs torque, then it is only this relation that applies: [tex]\theta = \frac{l}{J_T}\frac{T}{G}[/tex] And - yes, your friend was right - it only depends on the modulus of rigidity and some geometric factors, and hardening doesn't affect it.
Thank you Jack, This is a bit beyond my education, particularly using math to describe the relationships. But I am trying to wrap my head around it nonetheless. It seems to me that Hardening a steel shaft means that it will break with less applied torsional/shear stress than one that isn't hardened. So in other words, it will not affect the angular deflection / applied torque relationship. It would seem that that relationship is mainly decided by the materials and the geometric properties of the specific shaft, and you can see as much from the equations, if I understood right. To me that is really interesting. What it says to me is that the torsional rigidity, G, is based off of some kind of frictional properties that go with the mechanical shear forces of the material. I forget what the nature of hardness is derived from. I think it may have been from the crystalline structure of the material. I forget but I think in Steel it is related to Martensite and Austenite which if I remember right, are different crystal structures of iron & carbon characteristic of different heat treatment levels. I think the thing for me to do would be to go back and study what Hardness really is, and what it is derived from, and also, what are the (molecular level?) forces that influence Shear strength. I am glad I took the time to look into this. Thanks for your help
Hardened steel will break with less deformation but at a higher shear strength. It's like the reed and the oak: The reed can bend (deform) a lot before breaking and the oak breaks suddenly (no deformation). But you can break the reed with a lot less force than the oak. It is often desirable to have deformation as a safety factor. Even if it is less strong (you need a bigger part), you can have some indication that the part will break before it happens and then choose to replace it before an accident happens.
Hardness has to do with wear resistance and resistance to indentation. If you try to put a punch mark in a hardened steel file, you will not leave much of a mark at all. Stiffness has to do with the relation between stress and strain (Young's Modulus, E, which is also related to the Shear Modulus, G) and the geometry of the situation. If you had a drill bit made out of aluminum, not only would it be too soft (lacking hardness), but it would be way to flexible because Young's modulus for aluminum is about 1/3 that for steel.
First of all hardening is a surface property, it doesn't have direct relationship with torsional rigidity. i say it direct because when you harden it, during the process the whole component goes through it and that might affect torsional rigidity, but i assume it independent of hardening process,because torsional rigidity is not a function of your hardened surface. now as you said I would like you to picturize it when drill bits come into action they come in contact with the surface of plate and the whole action begins with surface of the drill bit it is the first point of contact and from that point stress is induced to internal structure of the drill bit, hence more the hardness matters a lot. but this is not the case of shafts. the shafts have to transfer torque that is independent of surface and depends on inertia. Now you said automotive shafts are not hardened but they have good torsional rigidity its because, torsional rigidity is function of polar moment of inertia. Now θ=T*L/(Jt*G) or T/θ= Jt*G/L (torsional rigidity) T is the torque applied; JT is the "geometry factor" (as you found out); r is the distance between the center of rotation and the outer surface (the radius for a circular cross-sectional shape); l is the length of the object the torque is being applied to; G is the modulus of rigidity of the material (related to the Young's modulus your friend was talking about); τ is the shear stress; θ is the angle of deflection under the applied torque now in equation for torsional rigidity you can see that more the Jt better is the torsional rigidity and similarly it is affected by G. this is the reason why hollow shafts are preferred because they offer far more polar moment of inertia for same mass for solid shaft J=(π / 32) (D4 - d4) for hollow shaft so for increasing torsional rigidity either increase Jt or G.
Being an old racer and not one to re-visit my engineering books I can tell you a few facts about axles. Typical production cars have surface or case hardened axles. This means the axle goes through a heat treat process that leaves a very hard surface to a depth of 1/32” or about 0.030” about .762 mm. The core is soft and this is intentional. Street cars constantly undergo flex and bending due to road surface faults like pot holes , bumps etc.. The axel must be able to flex and bend. Race car axels are heat treated to 3/32” and purpose built drag racing axels are Through Hardened . This leaves a very brittle axle but one that is ultimate in handling shear. Inn drag racing you are not so concerned with pot holes and flexing but you want maximum shear handling. These axels would snap under various road conditions. We in the road racing series go one step further. We use gun drilling. This is drilling a hole thru the length of the axel. We hollow out the inside diameter. This lightens the rotating weight. The the axel is case or surface hardened. Now have two hard surface planes to handle the shear and the meat left after gun drilling will provide the flex needed for bending forces. As with all things in life it is all about compromise. In this case.. it is all about AREA.
Hardening does not have to be limited to surface (case) hardening. Some parts are through hardened, but not often because it usually just makes them brittle. As you said though, hardening simply does not affect torsional stiffness.