[MUSIC] Hi and welcome back. Today we're going to talk about stress, strength and the factor of safety. This is the beginning of unit two, Static Failure. Last time we talked about CTE mismatch and that was the end of unit one, Material Properties in design. So, the learning objectives for today is one, to understand how to calculate the factor of safety, and two, to understand the considerations when you determine how large you want your factor of safety to be. Just quickly, we've talked about stress and strength before. We talked about how strength is a property of the material due to the structure, the processing, and the treatment it's been through. We typically have values of strength from test data and you can think of it as the capacity of an object, where stress is a state of an object at that point. It can be due to thermal or mechanical loading, sometimes you get residual stresses due to processing. You can calculate stress on an object if you know the load and the area, or the load and the geometry, and it's really a state that that object is in. And we talked about how if you have an isotropic homogeneous cylinder like we do here, and we're assuming it's behaving in a linear elastic manner, that if there's no load applied, it has a strength of 550 megapascals but it has zero stress. And then when we go ahead and we apply a load of say 500 newtons, our strength doesn't change, it's still 550 MPa, but now our stress has increased up to 500 Pa. So stress, state of an object strength capacity. I'm going to turn this over to you guys with a quick question. In mechanical component design, you're going to spend a lot of your time in this course calculating stress and determining strength based off of operating conditions. And then comparing the two and so why, why do we do that? Why is that so important what the stresses and strengths of components are? So and I give you a couple of minutes to think about that. Okay. Why do we care about what the stresses and strengths are in a mechanical component? Well the answer is that we're trying to avoid failure, right? We want this to design the component to withstand its operating environment to be able to function and to be able to function for a certain life. And when we say failure what we mean is that component has lost its functionality. So it's no longer meeting its performance criteria. It could have yielded which means it's undergone permanent deformation. It might have deflection, so if you have two shafts aligning gears and the shafts deflect and the gears can no longer mesh then you're going to have a lot of noise and you're going to have low efficiency so you might not allow a certain amount of deflection in your design. And clearly if you start to get fracture, that's a pretty bad sign. The performance criteria and the determination of what is failure and what is not is typically determined by both the design engineer and the program that they're working on. And so, what engineers do is they calculate something called a factor of safety and what a factor of safety does is it acts as a safeguard to failure and it gives you a certain amount of margin. In this course we'll call the factor of safety lower case n and what it's equal to is the loss of function strength. That's your strength where the part will no longer function. If a fracture means that your part can no longer function then your loss of function strength might be your fracture strength. If permanent deformation means your part can no longer function then your loss of function strength might your yield strength. In this class, we'll be primarily be looking at yield strength and that's divided by your allowable stress. What's the maximum stress you’re going to allow this part to see as a designer in operation? What is it going to see? Or maybe you'd be designing for failure. In failure, what will this part see. A couple of things to keep in mind when you use this equation, obviously you have to make sure your strength units and your stress units are the same, so don't have your strength in megapascals and your stress in KSI and you have to be looking at the same location at a part. You have to be comparing the stress at a certain point in that part to the strength at that point in the part. So there's some considerations to think about when you're determining what should your factor of safety be. Maybe you say on this program we're not going to have a factor of safety less than one point five. You might think is it flight critical which means on the aircraft something is flight critical if it's necessary for the aircraft to fly. You might think are there any redundant systems on this system that could back up and act if this component fails. How well do you know the materials that you're building this out of? Are you building this out of 6061 aluminum which has been pretty much studied for tens and tens of years. Or are you going to use a brand new alloy that there's not a lot of data on. How well do you know the loads and operating conditions? Are you absolutely positive, you know the loads that this part is going to see? Or is it possible that you guys could be underestimating or over estimating the loads? Is it possible that you have not foreseen one something that could happen, that could cause high loading in this part, and then is it a heritage design which just means have you utilized this design before and is there a lot of knowledge in testing already surrounding this design and you just modifying a few things to go forward. For heritage designs with really well known material properties, well known loads, well known operating conditions, where you're going to utilize a lot of testing on this design. That tends to happen in aerospace and you'll see fairly low factors of safety. In the one point two range or one point five range. Over on the pressure vessel side, they tend to go much higher, their factors of safety might be around eight. And so if we look at n, n is equal to our strength [SOUND] divided by stress, right? And as soon our stress becomes greater than our strength, that means that we've hit failure, right? We've either caused rupture or we've caused some sort of yield. And so if n is less than one, you're not in a safe design envelope. You should not be operating there or your analysis needs to be reevaluated so whenever n is less then one that means something is wrong. Something is wrong. If n is greater than one that's generally good but you need to see what did your program mandate. So what do the military standards or the automobile standards say that it's necessary for this type of design. With that, I'm going to refer you guys to worksheet two, which gives you this problem. Which is a 7075 T6 aluminum rod has a radius of two millimeters plus or minus point 02 millimeters. So there's it's tolerance. And the stress analyst has determined that this rod will see a maximum load ranging anywhere from 2,000 up to 2,700 newtons. Any plastic deformation of this rod will result in a complete loss of functionality. This rod is part of a class S component so in the US class S in aerospace means flight that means it's held to a little bit of a higher standard in both design and manufacturing. With the program mandated factor of safety of N equals three. PM and P found that lot AOO1927. Let me stop there, PM and P is that material processing group that most companies have. They'll test materials when they come in and they'll test them in lots so a lot of times you're getting material properties based off of a certain lot. This lot of aluminum had a strength a yield strength of 500 megapascals and an ultimate strength of 580 megapascals. And the question is, is the design currently within program spec? I'm going to let you guys attempt to work through this problem on your own. And then in the next module, we'll go through the problem together. But it's critical that you try to start it on your own so you can see where you struggle. Alright? So I will go ahead and see you guys next time. [MUSIC]