WEBVTT 00:00:00.035 --> 00:00:03.135 Axial loading gave you a very comfortable answer. 00:00:03.239 --> 00:00:07.243 Pull a bar along its own axis and the stress is load over area, 00:00:07.326 --> 00:00:12.249 the same at every point of the section, and the strain is the same everywhere too. 00:00:12.353 --> 00:00:16.859 Torsion is the next load case, and almost none of that comfort survives. 00:00:18.255 --> 00:00:19.765 Here is the question. 00:00:19.962 --> 00:00:22.633 A round shaft carries a torque T. 00:00:22.737 --> 00:00:25.568 How is that torque shared out across the cross section, 00:00:25.651 --> 00:00:29.634 and how much can the shaft carry before the material gives way? 00:00:29.830 --> 00:00:32.185 We will answer it the way it should be answered, 00:00:32.268 --> 00:00:36.588 by looking at what the twist does to the material and letting the algebra follow. 00:00:37.892 --> 00:00:39.564 So here is the shaft. 00:00:39.668 --> 00:00:41.268 A solid circular bar, 00:00:41.351 --> 00:00:46.252 with its axis running along z, clamped at the left end and free at the right. 00:00:46.564 --> 00:00:48.794 Notice that the section is a circle. 00:00:48.898 --> 00:00:51.047 That matters more than you might think: 00:00:51.150 --> 00:00:55.132 a square shaft under torque warps out of plane, and everything we are about 00:00:55.132 --> 00:00:56.852 to do would fail for it. 00:00:57.648 --> 00:01:02.418 Now hang a torque T on the free end, and rule one straight line along the surface, 00:01:02.501 --> 00:01:06.588 running from a point A at the fixed end to a point B at the far end. 00:01:06.785 --> 00:01:09.758 Apply the torque, and the shaft twists. 00:01:09.861 --> 00:01:13.217 The far end rotates through an angle we will call phi. 00:01:13.828 --> 00:01:16.476 Watch the cross sections while it turns. 00:01:16.789 --> 00:01:18.889 Each one rotates in its own plane, 00:01:18.972 --> 00:01:22.525 and because it is a circle it lands straight back onto itself. 00:01:22.977 --> 00:01:26.261 Nothing warps out of plane, nothing changes diameter, 00:01:26.344 --> 00:01:29.166 and no section slides along the axis. 00:01:29.479 --> 00:01:31.720 Plane sections stay plane. 00:01:32.517 --> 00:01:35.594 Now paint four radii onto the end face. 00:01:35.791 --> 00:01:37.310 They swing round together, 00:01:37.393 --> 00:01:41.689 all through the same angle, and every one of them is still perfectly straight. 00:01:41.886 --> 00:01:44.847 The section turns like a rigid disc on a hub. 00:01:45.551 --> 00:01:49.256 The line along the surface is the one thing that does not survive. 00:01:49.452 --> 00:01:53.261 Here is where it started, in gray, and here is where it has gone. 00:01:53.713 --> 00:01:58.111 B has been carried round the rim while A stayed put, so the line has tilted over, 00:01:58.194 --> 00:02:00.401 and that tilt is a shear angle. 00:02:00.598 --> 00:02:03.197 Notice that the line has barely changed length. 00:02:03.280 --> 00:02:05.904 There is no stretching here, only shearing. 00:02:06.817 --> 00:02:10.811 So, three facts, and every one of them came out of the picture rather than out 00:02:10.811 --> 00:02:12.600 of a material property. 00:02:12.796 --> 00:02:15.386 Sections stay plane and circular. 00:02:15.490 --> 00:02:19.298 Each section rotates rigidly, so radii stay straight. 00:02:19.495 --> 00:02:23.873 And a line along the surface tilts through an angle, which we will call gamma. 00:02:24.785 --> 00:02:28.734 That last one is the one we can use, because gamma is a strain. 00:02:28.930 --> 00:02:32.356 Put a number on it, and the whole of torsion falls out. 00:02:33.491 --> 00:02:35.430 Now put a number on that tilt. 00:02:35.627 --> 00:02:36.998 Two pictures. 00:02:37.310 --> 00:02:41.735 On the left we are looking straight down the axis at the far face of the shaft. 00:02:41.931 --> 00:02:45.887 On the right I have taken the cylindrical surface at some radius rho, 00:02:45.971 --> 00:02:49.954 slit it along a line and unrolled it flat, which turns the whole question 00:02:49.954 --> 00:02:51.893 into plane trigonometry. 00:02:52.690 --> 00:02:54.154 Start on the left. 00:02:54.606 --> 00:02:57.764 The whole section has rotated through the angle phi. 00:02:57.961 --> 00:03:02.962 A material point P sitting at radius rho has been carried round to P prime, 00:03:03.046 --> 00:03:06.622 and the distance it has travelled is the arc rho phi. 00:03:06.819 --> 00:03:10.709 Radius times angle, with the angle measured in radians. 00:03:11.192 --> 00:03:13.387 And there is the key fact already. 00:03:13.584 --> 00:03:17.822 The distance a point moves is proportional to how far out it sits. 00:03:18.019 --> 00:03:21.433 Bring the point in toward the axis and it travels less. 00:03:21.630 --> 00:03:24.614 Push it out to the surface and it travels the most. 00:03:24.718 --> 00:03:26.840 Settle it back somewhere in between, 00:03:26.924 --> 00:03:31.255 and remember that a point sitting exactly on the axis does not move at all. 00:03:32.307 --> 00:03:34.061 Now the right hand picture. 00:03:34.374 --> 00:03:38.760 The line A B was ruled straight along the shaft, parallel to the axis, 00:03:38.844 --> 00:03:40.691 and its length is L. 00:03:40.887 --> 00:03:43.791 A sits at the fixed end and does not move. 00:03:44.405 --> 00:03:46.214 B sits on the far face, 00:03:46.298 --> 00:03:51.755 and B has just been carried sideways by exactly that same distance, rho phi. 00:03:52.551 --> 00:03:56.986 So the line has swung round to A B prime, and the angle at A between 00:03:56.986 --> 00:04:00.911 where it was and where it is now is the shear strain, gamma. 00:04:01.015 --> 00:04:04.754 Its tangent is that sideways movement divided by the length: 00:04:04.951 --> 00:04:06.832 rho phi over L. 00:04:07.315 --> 00:04:10.984 Torsional strains are small, well under a degree in any shaft 00:04:10.984 --> 00:04:16.209 you would actually use, so the tangent is the angle to any accuracy that matters. 00:04:16.661 --> 00:04:19.692 Shear strain is rho phi over L. 00:04:20.489 --> 00:04:23.287 Now look at what is fixed in that expression. 00:04:23.902 --> 00:04:27.769 Phi belongs to the whole shaft. So does L. 00:04:27.966 --> 00:04:32.169 Both of them are the same number for every single point in this cross section. 00:04:32.366 --> 00:04:35.803 The only thing that varies from point to point is rho. 00:04:36.344 --> 00:04:41.546 So shear strain is proportional to radius, and nothing but geometry went into that. 00:04:41.650 --> 00:04:43.125 Watch it happen. 00:04:43.438 --> 00:04:49.081 At the axis, where rho is nearly zero, the strain almost vanishes. 00:04:49.277 --> 00:04:52.250 Halfway out it is half of its largest value. 00:04:52.354 --> 00:04:59.182 And at the outer surface, at radius c, it reaches its maximum, c phi over L. 00:04:59.978 --> 00:05:03.566 Divide the two and the shaft's own dimensions cancel out. 00:05:03.879 --> 00:05:07.754 Gamma is rho over c times gamma max. 00:05:07.838 --> 00:05:11.620 A straight line, zero at the centre and largest at the surface. 00:05:11.704 --> 00:05:16.535 That is the strain distribution, and no material property has appeared yet. 00:05:17.332 --> 00:05:19.248 One last way to picture it. 00:05:19.445 --> 00:05:23.683 Think of the shaft as a nest of thin tubes, one inside the next. 00:05:23.880 --> 00:05:25.726 Each tube shears a little relative 00:05:25.726 --> 00:05:30.347 to its neighbour, and the further out you go the more of that sliding has piled up. 00:05:30.544 --> 00:05:32.074 The skin does the most work. 00:05:32.158 --> 00:05:33.958 The axis does none. 00:05:35.048 --> 00:05:36.800 Strain is geometry. 00:05:36.883 --> 00:05:41.261 Stress needs the material, and for that we need one constitutive law. 00:05:41.457 --> 00:05:45.716 Below the elastic limit, shear stress and shear strain are proportional. 00:05:45.799 --> 00:05:47.748 Tau equals G gamma, 00:05:47.831 --> 00:05:52.371 where G is the shear modulus, the shear counterpart of Young's modulus. 00:05:53.423 --> 00:05:55.920 Substitute the strain we just derived. 00:05:56.116 --> 00:05:59.496 Tau is G rho phi over L. 00:05:59.692 --> 00:06:04.683 G is a material constant, phi and L belong to the whole shaft, 00:06:04.766 --> 00:06:09.341 so once again the only quantity that varies across the section is rho. 00:06:09.445 --> 00:06:12.592 Shear stress grows linearly with radius. 00:06:13.388 --> 00:06:18.312 So here is the stress at one point, a distance rho out from the axis. 00:06:18.508 --> 00:06:19.888 It acts tangentially, 00:06:19.971 --> 00:06:25.057 at right angles to the radius, and its size follows that straight line rule. 00:06:25.254 --> 00:06:29.167 Move the point outward and the arrow grows in exact proportion. 00:06:30.080 --> 00:06:31.775 Take it right out to the outer surface, 00:06:31.775 --> 00:06:36.211 at radius c, and the stress reaches its largest value there. 00:06:36.663 --> 00:06:39.308 Every point on that rim sits at the same radius, 00:06:39.391 --> 00:06:42.887 so the entire outer skin is at the maximum together. 00:06:44.102 --> 00:06:47.075 And here is the whole distribution across a radius. 00:06:47.271 --> 00:06:51.579 Nothing at the axis, rising straight out to tau max at the surface. 00:06:51.892 --> 00:06:57.326 Written down, that is tau equals rho over c times tau max. 00:06:58.123 --> 00:07:01.163 Now collect what all of that stress adds up to. 00:07:01.246 --> 00:07:05.819 Take a small patch of area, d A, sitting at radius rho. 00:07:05.902 --> 00:07:10.140 The force on that patch is stress times area, tau d A. 00:07:11.041 --> 00:07:15.579 That force acts tangentially, at a distance rho from the axis, 00:07:15.662 --> 00:07:19.900 so its moment about the axis is rho times tau d A. 00:07:20.201 --> 00:07:23.639 That is the torque this one little patch carries. 00:07:24.435 --> 00:07:27.847 The total torque is the sum over every patch in the section, 00:07:27.930 --> 00:07:32.470 which is the integral of rho tau d A taken over the area. 00:07:33.522 --> 00:07:36.297 Now substitute the stress distribution. 00:07:36.494 --> 00:07:40.091 Tau max over c is the same number everywhere in this section, 00:07:40.174 --> 00:07:42.286 so it comes straight outside the integral, 00:07:42.369 --> 00:07:46.561 and what is left behind is the integral of rho squared d A. 00:07:47.264 --> 00:07:51.352 Stop and look at that integral, because it is the whole point. 00:07:51.549 --> 00:07:55.044 There is no load in it and no material property in it. 00:07:55.148 --> 00:07:57.700 It contains nothing but the shape of the cross section, 00:07:57.783 --> 00:08:00.187 and where that shape puts its area. 00:08:00.802 --> 00:08:05.272 That integral is the polar second moment of area, J. 00:08:05.813 --> 00:08:08.345 So work it out for a solid circle. 00:08:08.542 --> 00:08:10.606 The natural element is a thin ring, 00:08:10.689 --> 00:08:14.579 because every point on a ring sits at the same radius rho. 00:08:14.683 --> 00:08:19.929 Give it a thickness d rho, and its area is its circumference times that thickness, 00:08:20.012 --> 00:08:22.381 two pi rho d rho. 00:08:23.178 --> 00:08:27.311 Substitute that in and the double integral collapses to an ordinary one 00:08:27.311 --> 00:08:32.188 in a single variable, running from zero at the axis out to c at the surface. 00:08:32.640 --> 00:08:36.763 Sweep the ring outward and you are sweeping through exactly that. 00:08:37.559 --> 00:08:42.318 Integrate rho cubed, and J comes out as pi c to the fourth over two. 00:08:42.401 --> 00:08:47.440 Notice the fourth power. Doubling the radius multiplies J by sixteen. 00:08:48.051 --> 00:08:51.558 Put J back into the torque relation and rearrange. 00:08:51.662 --> 00:08:58.662 The shear stress at radius rho is T rho over J. That is the torsion formula, 00:08:58.802 --> 00:09:02.866 and it is the exact counterpart of stress equals load over area. 00:09:03.779 --> 00:09:05.786 And phi came along for the ride. 00:09:05.869 --> 00:09:09.153 Thread the same substitution back through the strain we derived, 00:09:09.236 --> 00:09:12.392 and the angle of twist is T L over G J. 00:09:12.475 --> 00:09:16.098 Strength and stiffness both hang on the same integral. 00:09:16.895 --> 00:09:19.052 So the whole of torsion sits in J, 00:09:19.135 --> 00:09:22.213 and J is a statement about where the material is. 00:09:22.526 --> 00:09:25.580 Which raises exactly one design question. 00:09:25.684 --> 00:09:27.240 Where should you put it? 00:09:28.359 --> 00:09:33.106 So where should the material be? Here is the cross section of a solid steel shaft, 00:09:33.189 --> 00:09:37.776 thirty millimetres in radius, carrying a torque about its own axis. 00:09:38.828 --> 00:09:44.633 Go back to what J was actually measuring, the integral of rho squared d A. 00:09:44.737 --> 00:09:49.126 Every scrap of area is weighted by the square of its distance from the axis. 00:09:49.323 --> 00:09:52.876 So it is worth asking what the middle of this bar is earning. 00:09:53.580 --> 00:09:56.492 Shade in everything inside half the radius, 00:09:56.575 --> 00:09:59.525 everything within fifteen millimetres of the axis. 00:09:59.629 --> 00:10:03.238 By area, that core is a quarter of the whole bar. 00:10:03.321 --> 00:10:07.942 A quarter of the weight you carry around, and a quarter of the steel you paid for. 00:10:08.646 --> 00:10:13.044 But J for that core alone is pi times fifteen to the fourth over two, 00:10:13.127 --> 00:10:16.866 and fifteen to the fourth is one sixteenth of thirty to the fourth. 00:10:17.063 --> 00:10:17.852 So a quarter 00:10:17.852 --> 00:10:21.858 of the material is contributing about six percent of the torsional stiffness. 00:10:22.055 --> 00:10:25.063 The middle of a shaft is very close to dead weight. 00:10:25.604 --> 00:10:29.517 So take it out, and put it where the rho squared can get at it. 00:10:29.714 --> 00:10:34.440 Here is a tube, outer radius fifty millimetres, inner radius forty. 00:10:34.636 --> 00:10:38.431 Its area is pi times fifty squared minus forty squared, 00:10:38.514 --> 00:10:43.833 which is pi times nine hundred, and that is exactly the area of the solid bar. 00:10:44.029 --> 00:10:46.770 Same steel per metre, same weight. 00:10:47.567 --> 00:10:49.388 Now put numbers against them. 00:10:49.471 --> 00:10:52.338 Both sections have an area of about two thousand eight hundred 00:10:52.338 --> 00:10:56.101 and thirty square millimetres, so a metre of each weighs the same. 00:10:56.897 --> 00:11:00.483 J for the solid bar is pi over two times thirty to the fourth, 00:11:00.566 --> 00:11:03.968 about one point two seven million millimetres to the fourth. 00:11:04.165 --> 00:11:09.167 For the tube it is pi over two times fifty to the fourth minus forty to the fourth, 00:11:09.250 --> 00:11:11.863 and that comes to five point eight million. 00:11:12.060 --> 00:11:15.799 Four and a half times as much, for the same weight of steel. 00:11:16.712 --> 00:11:18.280 That is stiffness. 00:11:18.476 --> 00:11:22.807 Strength is a slightly different sum, because the torsion formula puts the peak 00:11:22.807 --> 00:11:27.138 stress right at the outer surface, where rho equals c. 00:11:27.335 --> 00:11:31.966 The torque you can carry at a given allowable stress is tau times J over c, 00:11:32.049 --> 00:11:34.116 and the tube has the bigger c. 00:11:34.599 --> 00:11:35.749 Work it out. 00:11:35.946 --> 00:11:38.754 Forty two thousand cubic millimetres for the solid bar, 00:11:38.837 --> 00:11:41.334 one hundred and sixteen thousand for the tube. 00:11:41.530 --> 00:11:45.363 Two point seven times the torque, at exactly the same weight. 00:11:45.676 --> 00:11:48.022 That is why a drive shaft is a tube. 00:11:48.505 --> 00:11:53.205 There are limits, of course. Make the wall too thin and the tube buckles, or dents, 00:11:53.288 --> 00:11:57.351 long before the material anywhere near yields, and a hollow shaft costs more 00:11:57.351 --> 00:11:59.210 to make and more to join. 00:11:59.313 --> 00:12:03.189 But the trend is not subtle, and it is why almost every drive shaft, 00:12:03.272 --> 00:12:08.242 every bicycle frame and every aircraft control tube you will ever meet is hollow. 00:12:09.039 --> 00:12:11.884 So, the whole argument in four lines. 00:12:11.987 --> 00:12:15.773 Geometry alone gave us shear strain proportional to radius. 00:12:15.877 --> 00:12:20.162 An elastic material turned that into shear stress proportional to radius. 00:12:20.358 --> 00:12:24.188 Integrating the moment of that stress over the area gave the torque, 00:12:24.271 --> 00:12:28.335 and dragged the polar second moment out into the open as an integral. 00:12:28.532 --> 00:12:31.816 And because that integral weights area by rho squared, 00:12:31.899 --> 00:12:34.895 metal near the axis earns almost nothing. 00:12:35.807 --> 00:12:40.264 And all of it came out of one picture. A line ruled along the surface of a bar, 00:12:40.347 --> 00:12:42.937 and what happened to that line when you twisted it.