 Hi everyone. I want to walk through an example of torsional impact loading. So basically same as we've talked about, if we have a rotating system and it suddenly experiences a change in loading, how can we analyze what's going on? So I have an example on the screen here which basically shows two grinding wheels, one at either end of the shaft, a drive system in the middle, probably a pulley for like a belt drive and we have some dimensions given here and then we want to know what's happening if we're grinding on the right hand wheel and suddenly it becomes jammed. So we did something catastrophic and our material got jammed in there and it brought that that grinding wheel to a complete stop. What can we expect to happen to the shaft? So we've gotten some given information. The whole system is rotating at 2400 rpm when this happens. We have a bulk modulus of 79 gigapascals and when you think about what's going to happen, we've jammed the right hand side and then we have this mass spinning out on the left. And that mass all of a sudden is coming to a stop but it's got mass so it's got momentum which means it's going to try to continue turning and twist the shaft. So we've got the density of that grinding wheel material so that we can understand what's going on with that mass. So as we've done in our notes, a good place to start on this problem is using energy and the main source of energy in this system is our our grinding wheel that's rotating out on the left that suddenly comes to a stop and it's you can imagine that it's storing energy in that rotation you know high speed rotation of that that grinding disc. The shaft also would have some energy as well but being that it's relatively small diameter compared to the disc, it's probably not contributing nearly as much so we'll neglect that in this case. So we have our kinetic energy which for a rotating system has an equation that looks like one half i omega squared so i being mass moment of inertia and for a circular object mass moment of inertia is equal to one half mr squared so that's something we can look up and we don't know the mass but we do generally know how to solve for mass which is to take the volume so i'll start with area pi r squared and turn it into volume by multiplying by thickness and then multiply that multiply that by rho which is our density that we have given. So if I substitute these values m into the equation for i and then the resulting i into the equation for kinetic energy what I end up with is kinetic energy equal to one fourth pi r to the fourth t rho omega squared and I can start substituting quantities in here so one fourth pi r in this case we're talking about our rotating mass out here on the end so it's got a diameter of 120 millimeters so that would be 0.06 meters in radius taken to the fourth thickness that's to give us the volume of this grinding disc on the left so that's 20 millimeters so that's 0.02 meters I have 2000 kilograms per meter cubed and lastly I have 2400 rpm I'm going to convert that to radians per second by multiplying it by two two pi over 60 and that's all squared so now if I calculate all this out what I end up getting is 25.72 and if we worked through our units and carried that through carefully we'd find that we have units of newton meters which is what we'd expect for this kinetic energy scroll down a little bit I can find my mouse so from this kinetic energy we can go ahead and calculate our shear stress so this kind of you know gives us an example of why it's so useful in some cases to have these stress equations in terms of energy so rehashing the equation I had before and substituting in values I have two square root of 25.72 times 79 e to the ninth divided by volume and now just being careful we're talking about we're talking about volume of the shaft because the stress is being applied to the shaft the disc that is rotating that's our energy storage vessel and then that energy is going to get dissipated as stress through the shaft once this this jam happens so my volume is going to be pi r which in this case is 10 millimeters so 0.01 meters squared times length which is given as 250 millimeters so 0.25 close my square root and if I calculate all of these numbers I'm going to get 321.7 times 10 to the sixth pass scales which is of course 321.7 mega pass scales and this then represents the stress that I would expect to find now just out of curiosity we could actually calculate what we would expect the deflection to be for our part so if I rewrite deflection I have a standard equation for deflection which is tl over jg I can actually rewrite this in terms of tau which is tau l over rg if I make some substitutions so if I go ahead and calculate that I have 321.7 times 10 to the sixth length again is 0.25 radius 0.01 and g is 79 e to the ninth and all of that's going to come out to be approximately 0.1 radians which I could convert to degrees and I get about 5.7 degrees so summing all of that up if I go ahead and am grinding my material gets stuck and jams the whole thing one's going to be loud but that mass that's rotating out on the left hand side slams to a stop the whole shaft is going to experience a stress of 321.7 mega pass scales and it's going to rotate by 5.7 degrees you know and then probably sit there bouncing back and forth making a lot of noise so we could then compare that stress against our our yield criteria for the shaft see if we've permanently deformed our shaft as a result and kind of analyze the situation further as needed all right so I'll go ahead and stop there thanks