 So I want to walk through a fatigue problem example which is actually from the textbook but I think provides a useful illustration of how to implement this process for understanding fatigue loading. So on the screen I have some setup for what we're talking about here. Basic ideas you have a part cylindrical part loaded axially with some sort of cyclic loading and we have some material properties that tell us the yield strength and the ultimate strength for this part. So the first thing that we need to start working on is is setting up our SN curve and setting up what we want to know or what we know about our SN curve. So to do that I mentioned in an earlier video that we need to know our 10 to the third limit and if we refer to the textbook figure 8.11 gives us that for axial loading we have a 10 to the third limit of 0.75 SU. So that's easy enough we can take 0.75 plug in our 150 KSI and we get 112 KSI. Great we also need to know our infinite life endurance limit where we say it's equal to SN prime CLCG CSCTCR and we can go through and kind of figure out what each of these is. Well SN prime and I've pulled in table 8.1 from the textbook because it provides useful insight to basically all of this information. SN prime gives us right down here in this small little bullet at the bottom which you can't probably read it says SN prime equals 0.5 SU for steel lacking better data. So it's kind of a generic answer when we don't have more specific information 0.5 SU. So coming back to my equation then I'm going to plug in 0.5 times 150 CL is my load factor for axial my table says that that's one CG for axial loading and let's see based on diameter we have a range of 0.7 to 0.9 let's assume that our diameter is less than two inches it isn't specified yet but let's assume that it is and we just need to pick something then so I'm going to say 0.9 let's see surface surface factor we would refer to figure 813 the setup for this problem says that it is a commercially polished surface so that's a pretty good pretty good surface and if we look at table 813 under these conditions I think we would get that or excuse me figure 813 0.9 would be our surface temperature the problem doesn't say anything about having high temperature so we can go ahead and use one and reliability doesn't specify anything that we want higher reliability so from our table over here we'll just start with the default of one great so we can calculate all that out and we end up getting 61 KSI for our endurance limit and that's pretty much all we need in order to fully define our SN chart so through the magic of television I have the SN chart down here already and basically what our two calculations just did is gave us this point here and this point here now on the chart it's showing points for 10 to the fourth cycles and 10 to the fifth cycles as well luckily we could just calculate those because we know everything else about it and I just want to talk about that for a second so say I wanted to know what this 10 to the fifth limit was so basically I would call that s 10 to the fifth there's something like that equals what and if you kind of remember any of your math courses you might remember that we can do something called linear interpolation so anytime we have a straight line and we know two points on that line we can interpolate between them to figure out a point that we don't know and effectively we do have a straight line now we do have to be careful because this straight line is only straight on log log axes so when we do our linear interpretation we have to take that into account so to start our linear interpretation we basically take two known quantities um and do like a rise over run type calculation so I need to remember this is log and my first point that I know is 112 over here that's my 10 to the third limit and my other point that I know is my endurance limit of one or of 61 and my x-axis change so this is the the run of my rise over run is 10 to the third to 10 to the sixth now I would also be using log scale here but if you recall log is base 10 so log of 10 to the something means that I just take that power and move it out front in the log and the 10 cancel so I can say 3 minus 6 and that's the same thing as using the log scale in this case and this has to be equal to one of my points and the point that I'm trying to interpolate so I'm going to take that 10 to the third point again one two minus log of the limit that I don't know which is s 10 to the fifth and now I'm talking about three compared to five as my run in this case and I've got everything in here all I have is this one unknown value that I can go ahead and calculate so you can obviously see the answer is over on my on my chart but effectively all I need to do is you know take this three minus five multiply that over divide that by the three minus six so negative two over negative six and multiply that by log 112 minus log 61 divide or excuse me subtract that from log 112 and then I would be end up with log of s 10 to the fifth is equal to something I can undo that log by taking 10 to the power of it so 10 to the power of the log gives me the s 10 and then 10 to the power of whatever number is on my right hand side gives me my s 10 to the fifth value in this case of 75 ksi so I can use this this linear interpretation or excuse me linear inter can't talk linear interpolation in order to find any value that I would want for any you know say particular life cycle on this chart the next step then is to take this data from my my s n curve and move it to my constant life fatigue diagram so just like in the example what we do is we take all of these data points and read them onto our chart and plot them extending the lines from wherever they cross on the the alternating axis down to where they cross the ultimate strength on the mean axis and that gives me the the chart that I need I add on my yield strength 120 and the lines connecting that onto my chart and I've I've constructed my my Goodman diagram now to give us something to solve for let's go ahead and say that we have an alternating load between 1000 pounds and 5000 pounds and we want to design for a safety factor of two and basically what we're talking about is what diameter of our round shaft that we're applying this load to would result in a safety factor of two so first we need to go ahead and calculate what our mean stress would be well we've got our alternating situation here right 1000 to 5000 pounds so let's say that we can take our mean force and divide that by an area area is unknown because we don't know the diameter that's what we're solving for so we get 3000 over a if we factor in a safety factor that means we're looking for 6000 over a so basically our stress we have to double our stress effectively if we want to take in that safe take into account that safety factor our alternating stress is basically what is our peak past that mean so if our mean is 3000 then our peak is 5000 that means our alternating load is 2000 because it's the difference between those two or with a safety factor I get 4000 over a now the way that that they've shown how to do this in in the book is to say well if we look at this graphically let's figure out what our slope of our our line over here is on our constant like fatigue diagram so that's 4000 over a over 6000 over a that gets 0.67 so that gives us the slope of this line here and if we charted this accurately enough we can kind of read some stuff from this chart then one of the things is if we were looking at the infinite life fatigue then we'd want to know what's going on right at this data point where my line my dash dotted line crosses my my infinite life infinite life chart and that's easy enough to find to do that I need to go ahead and set up an equation to solve for the intersection of these two points so I know what my two equations are right we can use the equation of a line give myself a little more space here so I know the equation of my two lines one is 0.67 again as I said what my slope of my dash dotted line is uh so that's uh slope times x which is sigma mean and it has a zero for a y intercept my other equation is my endurance limit line which goes from 61 down to 150 so I can set that equal to rise over one minus 61 over 150 sigma m plus y intercept of 61 so I have an equation if I add this uh component over to the left I have everything in terms of sigma m I can go ahead and solve that for sigma m and I would get 56.69 ksi and great now I have a a mean stress at which my endurance limit would be my where I would hit my endurance limit and I have a relationship from up above to area so I can say 6,000 over area now I have to be careful this is 56.69 ksi so 56,690 pounds per square inch so take that thousand into account and my area I can substitute in pi r squared if I do that and solve for r I get 0.1835 inches or a diameter of 0.367 inches great so what this is telling me is that if I want a safety factor of two my diameter for infinite life endurance should be 0.367 inches and in theory my part won't fail no matter how many cycles I would subject it to now suppose I wanted to save some money and this isn't a high use part so I only need it to last for a thousand cycles so if I come over here and follow this line up the one thing we should notice right away is well right here is the intersection with 10 to the third cycles however we have to be careful because I've already at that point crossed my yield stress limit so I'm going to get yielding due to static you know load exceeding the yield criteria before I make it to my 10 to the third line so in that case I need to calculate my limit based on that static yielding situation so I'm not going to write it all out but I could basically apply this same criteria set up two equations solve for sigma m plug in my 6000 a and maybe I should note that this is 10 to the third or excuse me 10 to the six cycles for 10 to the third cycles using the same process I get that my diameter would only need to be 0.326 inches so I could have a slightly smaller part if I don't have as high of a load or a cycle requirement um on on the thing so overall uh oops realize you can't even see what I'm writing there overall for fatigue I have a couple of things I need to take into count I need to figure out what my loading scenario is I need to determine what my mean and alternating stress are because they allow me to apply my situation to the constant life fatigue diagram and I also need to determine my limits so depending on how many cycles I want my part to last I need to figure out what those endurance limits or 10 to the third limit would be which is you know it's kind of like setting the yield criteria when we talk about when we talk about static loading and we use the failure theories when we're talking about fatigue we're comparing against whatever this endurance limit is which is a statistically determined quantity okay I'll go ahead and stop there