We have had two twisted shafts in a row so I want to do some math to get the max shear stress before deformation.
I found that a square shaft formula is:
Tmax = (2/9)* Shearmax bb*b
I went over to matweb and guessed low carbon steel that has been extruded to get that nice square bar and then zinc plated. So I am going with the average of 1.13*10^7 psi for the shear modulus. That would be 4900 in lb of force which seems too high for this kind of activity. What is the stressor causing this?
Last time we had a shaft twist, we calculated the maximum load it could have been under to be 118 in lbs (we were gearing down the output from four motors 1:3 and putting it all through one axle). So the correct answer should probably be less than this.
However, I’ll ask the original question slightly differently: what is G and J for the shaft material we’re working with?
Tempering, annealing, casting, rolling and all sorts of stuff you do to the shaft metallurgically as well as physically affects the shaft characteristics.
Should I assume low carbon steel that has no real treatment to make it better? The zinc plating I believe adds relatively little added strength.
The minimum torque to cause defleciton is from the yield strength. That starts the twisty motion. By having 1.5 inches from the drive gear to the arm, that created a length allowing for some twist. Enough twists, that made it look like a drill bit causing the shear stress to kick in and break the shaft.
How to solve twisty shafts? Reduce the length allowing torsion (thus setting L to 0).
How to do that in Vex? Put the arm on the gear and not allow the shaft to be in such high torsional stress. But the math is so much more interesting and a great teaching tool beyond driving killer robots and feeling bad when they break. Tell them why they broke so they can build a better robot next time. That’s the bridge to real engineering… You know, the “E” in STEM.](http://www.flickr.com/photos/66331260@N08/8511003123/)
I haven’t taken statics/dynamics yet, but from looking at the formula, it doesn’t seem that delta L can ever be 0. Rather, we can give the equation a maximum value for the amount the length of the object changes, and compute the force from that…
You can’t set L or ∆L to 0 in that equation. If the shaft has 0 length then it doesn’t exist, so the equation is meaningless. Also, there is no such thing as a material that doesn’t deform at all under a load. ∆L=0 means the change in length of the shaft is 0, i.e. the shaft doesn’t deform. If the load is nonzero but the shaft doesn’t deform then you have described an impossible situation, which is why the equation gives you an impossible solution.
For most engineering materials, including steel, a tensile force causes a proportional change in length (∆L ∝ F) until the force gets too large, at which point the material will deform permanently. Before this point, however, the object will behave a lot like a spring. What Young’s modulus does is let you calculate the stiffness of that spring. For shear, you use the shear modulus instead.
Unless someone out there has access to a materials testing machine, the best way to get answers to these questions would probably be to ask VEX directly. They seem to usually be pretty keen to help teams out by providing this sort of information and they should be able to tell you the carbon content of the shafts and how they’ve been processed.
I’m pretty sure the length of a shaft shouldn’t affect the torque at which it fails, provided the shaft isn’t under any other load and if you ignore the fact that longer shafts are more likely to contain imperfections.
It’s not the length of the shaft itself, it’s the length from the part you put the torque/force on the shaft - e.g. the grear - to where you want to drive it - e.g. the arm. The old design did not have the arm attached to the gear. There was 1.5-2 inches from the driving gear to the attached lifting arm. So it gave a bit of length for the shaft to twist.
If the arm is attached to the gear, you don’t give any real length on the shaft to twist it. (L goes to 0 in the twisty angle formula)
Bending to breaking is a fun thing in metal. This is not shear stress but it still holds true. You have a certain range where the metal basically does not do much movement at all. At the yield strength is where it starts to give and move about (twist the shaft). After taking so much it will break.
If you go on to study mechanical engineering (or a few other types) you get plenty of labs to test this yield/break stuff and measure the physical properties.
We’ll leave it there and not talk about repeated stresses and weakening over time from abuse.