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More Than You Ever Wanted to Know About Mechanical Engineering, Part 4: Torsion
Torsion is actually simpler than you'd think it would be, but that doesn't mean it's necessarily intuitive. So let's consider a shaft with a torque, T', acting on it.
In order for the shaft not to rotate, there must be internal forces acting which amount to an opposite torque, T. (This is why I gave the external torque the T' rather than the T - mechanical engineers are primarily concerned with stresses, i.e. internal forces.)
Torque, you may recall, is force times perpendicular distance to the pivot. (Or, if you want to get all fancy and vector-y, T = r x F.) So if you pick a random point within the shaft at some distance from the center rho, there must be an incremental force dF acting perpendicular to rho. The incremental torque on that point will be rho x dF.
The sum of all these incremental torques must add up to the overall torque T. You see where this is going - this is a job for an integral.
You can take this a little farther. If you look at the picture above, you can see that there must be shear stress occurring on the element in the cross-section of the shaft, since the incremental torque is acting parallel across its face. If you recall, shear stress, tau, is the parallel force over the area. So we can solve that equation for force, substitute into our torque integral, and get a neat relationship between torque and stress.
Note that this tells you about overall stress in the shaft - it does not help you in figuring out what parts of the shaft are stressed the most. But we can figure it out with a little observation.
Remember back when I was talking about stress and strain and I mentioned that for most materials, there is a region below the yield point in a tension test where the relationship between stress and strain is linear? Well, if stress and strain are linearly related in that region, you can describe them mathematically as a straight line. For shear stress and strain, that relation is:
Gamma here is shear strain, and the constant G is the shear modulus, which is a property of whatever material you're using. The equivalent equation for normal stress is pretty much the same except it uses a different constant and different Greek letters. This relationship is called Hooke's Law, and we'll be seeing more of it.
But anyways, what this boils down to is that if you know the strain that occurs (i.e. the deformation an object experiences), you can figure out what kind of stress is going on. This is fucking fantastic - you can measure the deformation a part undergoes and calculate how much stress it experiences instead of doing a whole bunch of bullshit calculations. So let's go back to that shaft.
If you apply your torque and follow a point on the outside of the shaft, you can clearly imagine that it will twist some distance around the shaft.
Phi in the picture above is called the angle of twist. The shear strain, gamma, is the angle from BA to BA'. I'll leave the explanation of exactly why that is to another day - for now, just trust me on this. Knowing this, we can say get an expression relating shear strain to geometric parameters that can be directly measured.
The length of the shaft is constant, so gamma depends directly on rho and phi. This is intuitively obvious - naturally, you'll see the most deformation at the outer surface of the shaft, and the more twisting you do, the more deformation you'll get. You can do a little algebra with this and express the strain at some arbitrary point as a proportion of the maximum possible strain.
Stress, we've said, is directly proportional to strain (within that linear portion of the stress-strain curve, at least). So what this means is that we see the same proportional relationship between maximum stress and stress at any point within the shaft.
This is pretty cool - it tells us that stress in a twisting shaft is directly proportional to distance from the shaft center. But how do we relate this to the torque applied to the shaft?
Recall that integral we got above. We can substitute our expression for shear stress in there and get something pretty interesting.
That last integral may look a little familiar. It's the polar moment of inertia I talked about last time. Remember that moment of inertia tells you something about how likely an object is to experience bending or rotation around an axis - it's not too surprising to see it crop up here. It does make our equations a hell of a lot nicer.
We can rearrange this a little and generalize to any point within the shaft, not just the point of maximum torque.
And just for fun, here's a piece of steel I broke with a torsion tester. (You guessed it: it's a big machine that clamps both ends of a sample and twists the hell out of it.)
You can clearly see the twisting. We accidentally picked a steel rod instead of aluminum for this test, and it took fucking forever to break. Metals are way, way more tolerant of extreme deformation than you might expect.
(Originally posted to my personal blog, slightly revised for cross-posting.)
More Than You Ever Wanted to Know About Mechanical Engineering, Part 1: Stress and Strain
(Originally posted to my personal blog, slightly revised for cross-posting.)
All right, so let's go back to the very basics of mechanical engineering. Let's talk about stress and strain, baby.
Stress is usually denoted with the Greek letter sigma. The general equation for stress is given below.
"Force over area" may seem familiar - it's the same equation as for pressure, and stress is normally reported in the same units (Pascals, or psi if you're some sort of heathen US customary unit user). The difference is that pressure has to do with external forces, while stress deals with internal ones - you can think of it as the effect of the reaction force occurring in the body under load. At a microscopic level, it's the result of the complex interaction of intermolecular forces as molecules resist getting squashed into each other.
The way in which a force is applied affects the way stress occurs in a body. Take a look at the image below.
This is pretty straightforward. Force F is pushing down on the area of a x b, so the resulting stress is given as sigma = F/ab. This is a case of axial loading, in which force is applied straight into the area of interest. The stress generated is called a normal stress ('normal' in the sense of 'perpendicular'). It'll be oriented along the same axis as the force - if you put too much load on that column, the stress will become too large, and it'll crumple.
But what about the situation below?
If we just consider normal stress, there doesn't seem to be a lot going on. You wouldn't ever expect to see this thing buckle along the a axis, and looking at that stress equation, you'd think that the surface bc is too large for there to be significant stress occurring. (Assuming a distributed load, not a concentrated point load.) But you know from intuitively that a beam loaded in this way is way likelier to fail than the column above. So what's going on?
It turns out that just considering the normal stress isn't enough. You have to consider shear stress as well. Here's how shear stress is occurring in the beam above:
(There's more going on here than just shearing stress, but we'll get to that.)
Shear occurs with a force acting parallel to the surface of interest. It's usually written as tau.
Shear stress is just as important (often more important) than normal force in engineering. Unfortunately, it's frequently less obvious. It should be noted that these equations are for average stress - there are situations where averages aren't sufficient for design and analysis, but we'll stick with them for now.
So let's take a look at an example. Suppose you have a bolt joining two plates together. For simplicity's sake, we'll disregard the effects of threading.
Suppose the nut is rusted fast and some frustrated dude takes a pair of pliers and straight up tries to pull the bolt out, exerting a force along the axis of the bolt.
The surface that will experience the load in this case is the part of the nut contacting the bottom plate. This will be experience a normal stress, as the force is exerted perpendicularly to the surface of interest. If the guy pulls with a force of 150 N and the part of the nut contacting Plate B has a surface area of about 3 x 10^-4 square meters, a normal stress of 500,000 Pa, or 0.5 MPa will occur. (That sounds like a lot, but Pascals are pretty small. It's about 72.5 psi.)
How about if the plates are being pulled in opposite directions?
In this case, a shear stress will occur in the bolt - the force is occurring parallel to the cross-sectional area of the bolt. If it's the same 150 N force, the shear stress will be larger than the normal stress in the previous example, since the cross-section of the bolt is smaller than the surface of the nut that contacts the plate.
It's also worth noting that stresses are occurring in the plates themselves when this is going on. When the plates are being pulled apart, they'll experience a stress where the bolt presses against them - the area of interest here is that entire interior surface in contact with the length of the bolt. This kind of stress that fasteners exert on things they hold together is called bearing stress.
All right, cool. That's stress. So what's strain?
Strain is the deformation a body experiences under an applied load. It's usually denoted as epsilon, and it's just the ratio of change in length to original length.
Intuitively, it's obvious that there's a relationship between stress and strain. You put enough force on something and it bends. The exact nature of that relationship is complicated and varies considerably between materials.
The graph above is a stress-strain curve for an aluminum sample undergoing a tension test (image from Statics and Mechanics of Materials, by Beer, Johnston, DeWolf, and Mazurek). A tension test is exactly what it sounds like - you stick a piece of metal in a big machine that grips it on both ends and pulls until it breaks, continuously recording the force (from which stress can be calculated) and elongation of the specimen (from which strain can be calculated). The end result of a tension test looks like this:
It's pretty rad.
Anyway, you can see that there's a portion of the stress-strain curve (from 0 to 0.004 in the example above) that's very nearly linear. If the applied force is removed in this region, the sample will return to its original dimensions like nothing happened. The end of that linear section is the yield stress. This is the point at which the sample will start to experience permanent deformation. The peak of the curve is the ultimate stress - after this is reached, you'll see that the stress appears to decrease before rupture. The stress isn't actually decreasing here - this is an artifact of the way the test is measuring stress. What's going on is that the material is necking where it's going to break - that is, there's a smaller cross-sectional area (and a correspondingly greater stress) than the test accounts for. In reality, the stress occurring in the sample will continue to increase steadily until rupture.
This curve is for aluminum, which is a ductile material, i.e. a material which will bend before it breaks. Brittle materials behave quite differently, breaking abruptly without extensive deformation beforehand, and even the curve for another ductile material may look quite different in the particulars.
In engineering design, you always want to ensure that the stresses a part experiences will not approach the yield stress of the material. It's a simple idea with a lot of complicated applications.
