[MUSIC] Hi and welcome back. In today's module, we're going to continue covering the factor of safety in Unit Two, Static Failure. So the learning outcome today, again is to calculate the factor of safety. Hopefully, you guys have gone through worksheet two and completed it to the best of your ability. And then today, we're going to go through the solution. So again, Factor of Safety is the loss-of-function strength divided by your allowable stress and here's the problem that we had. So, we saw this aluminum rod had a certain radius and there was a range of load that could be applied to it. It was a static load. It said that, any deformation could cause complete loss of component functionality. And it gave us an ultimate strength and a yield strength and a program mandated factor of safety that the design needs to hit, which was n=3. So, what we start out knowing is that our factor of safety is equal to strength divided by stress. Now in this case, it says that any plastic deformation results in a loss of component functionality. So, our loss-of-function strength is going to be our yield strength. And most of the time in this class, you'll be looking at yield strength. So, the next thing we need to figure out is our stress. So, we know our yield strength. Our yield strength is equal to 500 megapascals that was given down here in the problem and we know that since this is a tensile axial load, that our stress is equal to force divided by area. So now, we need to think about which force do we use and which area do we use? Let's start with area. First off, it's a rod. So, it's going to look something like this and the load applied is going to look something like that. So, our cross-sectional area is going to be pi r squared. So the problem is we've been given a radius of 2 millimeters plus or minus 0.2 millimeters and here's the trick in design, you always want to start with your worst case scenario. You can possibly back out of that worst case scenario if you need to, but you want to start at worst case and worst case scenario is your highest stress. Your highest stress is going to give you your lowest factor of safety. So, we need to calculate what would be the highest stress and our highest stress will be caused by the lowest cross-sectional areas. So we're going to assume that since it's acceptable to have a rod that could be 2 minus 0.22 millimeters that, that's what's happening. So our area is going to be equal to pi times 2-0.22, these are both in millimeters squared and what's going to happen is I convert this to meters just so I can keep all my calculations straightforward. So pi r squared and what I get is 9.95 times 10 to the -6 meters squared. So, that's the smallest possible area that we would allow for this rod. So, let's look at our force. Our maximum stress is going to occur at the largest possible force and they said, it could see a static maximum tensile load ranging anywhere from 2,000 up to 2,700. So, we want to use the 2,700 number. So now we can say, sigma equals 2,700. Largest force causes largest stress divided by smallest area and what we get is that our stress is equal to 271.2 megapascals. So then, now I have my stress, I have my strength, I'm going to go ahead and calculate my factor of safety. So n is equal to 500 megapascals, which is the yield strength divided by 271.2 megapascals and I get an n equal to roughly 1.84, which is not 3. It's not as big as 3. So currently, you're not meeting the program mandated factor of safety. So, there's a couple options that engineers could do here. What you would see happen is engineers could go back to the program and say, you know what? In this case, we really know this material. We really understand the loads on this design. We feel like a factor of safety of 1.8 is appropriate and maybe the program would lower the factor of safety. If there was any question though, you could just increase your cross-sectional area of the rod. You could change out the material of the rod, so you had a higher strength or you could see if there's anyway to lower the operating load on the rod and that would help you meet that factor of safety of three. So, that's worksheet two and that's it for this module. I look forward to seeing you guys next time. [MUSIC]