I want to continue our discussion of torsion and in this segment I want to look at cases of nonuniform torsion and also power transmission by shafts. So far we've only looked at cases of pure torsion, in other words, the bar was prismatic, constant cross sectional area and the torque is constant along the rod. Now I want to look at cases of nonuniform torsion, in other words, the bar may not be prismatic, the cross sectional area could be changing, and/or the torques can act anywhere along the rod, and can be varying along the rod. We'll look primarily at two cases. The first one here is where we have a situation where the torque is peacewise uniform, in other words, the torques are applied at fine art locations, for example, these gears here. But within each segment here, for example the segment CD the section is in pure torsion in other words its cross section is constant and the torque there is constant and this is called piecewise uniform. The second case which we wont look at in detail is where we have a continuously varying cross section and possibly a continuously variant torque along there. So, the method here will be for the first one, we will simply analyze each section separately by the equations of uniform torsion and add the deformations or the angular twists to get the total. And the second one is similar, except there the element of analysis is very small and again, we add up the results which becomes an integration. So let's first look at the case of piecewise uniform torsion. And let's suppose we have, for an example, this situation here, where we have two bars with different diameters and torques applied at four different locations A, B, C, D, as shown. So the analysis here is that each segment, for example, AB, BC, etc., is in pure torsion, therefore all of our previous torsion formula apply. For example, for the maximum shear stress and the angle of rotation V. So to analyze these problems, we'll assume a signed convention which is a normal right handed convention for talks. And for equilibrium, we simply put the summation of the talks acting on the element equal to zero and assume that any unknown torque is initially in the positive direction. This will give us the magnitude and direction of the torque in each segment and this we can obtain either by inspection or by drawing separate free body diagrams of each segment and putting the summation of the torques equal to zero, where the torques of course, are generally vector quantities. For example, for equilibrium of the whole rod in this case, the sum of the torques is equal to zero, in other words, T1 plus T2 minus T3, because it's in the other direction, plus T4, is equal to zero. Now to find the torque in each segment, let's suppose I first want to do the section, BC right here. So I can imagine that I take a hypothetical cut through here and just draw the freebody diagram of that segment to the left of that cut. And the freebody diagram looks like this. We have T1 acting And the torque in that section AB, alder note by TAB, which is in the direction as shown, which is positive according to our sign convention. So therefore, the summation of the torques is equal to zero, therefore we have TAB plus T1 is equal to zero so the torque in the segment AB is equal to minus T1. Next, we want to do the section BC, so again I can imagine a hypothetical cut somewhere through here at BC and I draw a free body diagram of the segment of the shaft to the left of that cut. And I get this diagram here. And now this segment is subject to two torques, T1 and T2 and the torque in that segment is TBC. So from equilibrium, the sum of the torques is equal to 0. In other words, TBC plus T1 plus T2 because those are both positive talks is equal to 0 so TBC is equal to minus T1 minus T2. Finally the last segment CD, I make my cut through here and there is my freebody diagram and this segment is now subject to three torques. T1, T1, and T3. And from equilibrium, some of the torques is equal to zero. So therefore TCD is equal to minus T1 plus T2 plus T3. And now we have the torques in each segment and we can apply our torsion formula to compute shear stresses and angle of twists. So the maximum stress in each segment is given by this equation and the total angle of twist is just the sum of the twists of each individual segment. In other words fee is equal to fee one plus fee two, etc, where fee the angle of twist is equal to TL/GIp for each segment. So therefore, we get phi is equal to summation of phi I, is equal to summation, T I, L I over G I L P I for each segment I. Let's do an example on that. So here we have a stepped steel shaft which is rigidly attached to a wall at the left hand end and it's twisted by two torques T1 and T2 which are rotating in opposite directions. And the diameters are given here and the magnitudes of the talks are given here. So the maximum shear stress is most nearly which of these alternatives? So, we'll be using our usual sheer formula of course, TMAX is equal to Tr over Ip, but initially, we don't know which segment will have the maximum sheer stress. So we have to calculate the sheer stress in both of those two segments. So to do that, first of all we'll compute the polar moments of inertia for each segment. Ip for a circle is Pi D to the 4th over 32. So Ip1 for segment one here Is Pi times it's diameter which is 56 millimeters or .056 meters over 32 which gives a polar moment of inertia as shown. Similarly the moment of inertia of segment two is Pi times it's diameter to the fourth over 32 which is this value. Next, we compute the torques. First of all, the torque in the segment TAB from equilibrium is equal to T2 minus T1, which is equal to 1.5 minus 3.5, which is equal to negative 2.0 kiloNewton meters. And similarly, the torque in the segment BC here, or segment two, is equal to T2, which is equal to 1.5 kiloNewton meters. So notice that the direction of the torque T2 here is positive and the direction of the torque T1 is negative according to our sign convention. Now we can compute the stresses the maximum shear stresses in each segment. So the maximum shear stress in segment 1 is the torque in that segment times diameter of a tube, divided by IP1. And substituting in the numbers, the torque is minus 2. But here I don't need to worry about the sign because I just wanted to get the magnitude of the stresses. So computing the numbers that is equal to 58 mega pascals. Now I do the same calculation for segment two so TauMAX two is equal to it's torque times it's diameter of the two divided by it's polar moment of inertia and we know all those values, substitute them in and we find that the maximum shear stress in the smaller tube is 54.3 megapascals. So comparing these two numbers, we see that the maximum shear stress occurs in segment 1, the larger tube, so the answer is 58 megapascals, or B. Now, continuing with that same problem, now we're given that the length of these segments are 1 meter and 0.75 meters, and the shear modulus is 27 gigapascals and we want to compute the twist at the end C here. It's most nearly which of these alternatives. So, from the previous problem, here are the torques in each of the different segments and here are the polar moments of inertia of the two segments and here is our basic relationship to the angle of twist Phi. Under uniform torsion fee is equal to TL over GIP so therefore the total twist is just the summation of the total twists in the two segments which in this case then is equal to TAB L1 divided by G1 IP1 plus similar expression for the segment BC. So that is equal to this and now of course the signs of the torque are important because we have to get the directions of the twist here. So here the talk is negative 2.0 and computing those out the twist in the individual segment are there and the final answer is minus .019 radians or converting to degrees Is equal to minus 1.1 degrees. So the answer in this case, we are only asked about the magnitude of the twist so the answer is D. We note however the direction of the twist here is a negative number. Minus 1.1 degrees. In other words, this and here is rotating in this direction. In the counterclockwise direction if you look at this end from this direction, in the clockwise direction. Now some other topics that might come up are gear shafts, for example the situation shown here, and these are just applications of the same formula, the procedure is the same, compute the torque in each segment, and then apply our formula of uniform torsion to each segment and add up the twists to get the total amount of twist. If was have nonuniform torsion and/or nonuniform cross section, for example a situation like this where the diameter could be changing and also potentially, we could have a continuously varying torque along there, then the procedure is the same, we do our analysis on a small element of the tube of length Dx. Find the twist of that element and then apply our uniform torsion formula to that element and finally, we add up all the twists of the individual elements. In other words, we integrate over the length. So our previous summation formula here, becomes, in this case, replaced by an integration. Now, another topic that I won't cover is stresses and strains in pure shear. And if you remember that when we look to the uniactual forces on rods we have normal stresses, however uniactual force can also give rise to shear stresses on plains which are oriented at some arbitrary angle to the axis of the member. Similarly just as uniaxial force gives rise to shear stresses, so pure shear can give rise to normal stresses, as shown here. And this would be a topic in a more detailed course, but I won't cover this here. Finally, I want to mention power transmission because this is one of the most important uses of shafts to transmit power from motors or electric motors or etc. So, typical questions are what would be the amount of defamation of twist in a rod which is transmitting power at some known torque and some speed. So the relevant equations here of that power depends on torque and also the angular speed of the shaft. One of the relationships between power, sheer stress and strains etc. Well firstly remember that power is the rate of doing work, and work in other words work divided by time, so if we are pushing something along in a linear fashion at a steady speed, let's suppose I was pushing this with some force add a velocity v then work is force times distance. So therefore power which is work over time is force times distance over time, however, distance over time is just velocity. So therefore the power hen I'm moving something, or pushing something linearly, this simply force sence of velocity, F times v. So, the exactly analysis equation to that for shafts is power is equal to torque times angular speed, so replace force by toque and linear velocity by angular velocity and that is the equation for the power of a rotating shaft. And the fundamental units with be either foot pounds per second or Newton meters per second and again, in this equation, remember that Omega must be in radians per second for those units to be correct. And some conversion formulas that we'll often use are that 550 foot pounds per second is a horsepower, a Newton meter per second is a Watt and 1,000 watts is a kilowatt. And one final equation which is useful is that often, the angular speed will be given in revolutions per minute, rpm, and the conversion to Omega in radians per second is 2 Pi n over 60. And once we know the torque then we can apply all of our previous equations for twist and stresses, etc. And this concludes our discussion of torsion.