[MUSIC]

Hi and welcome back.

In today's module,

we're going to estimate the endurance limit working through an example problem.

So the learning outcome for

today is be able to estimate the endurance limit in an example problem.

And just a reminder,

we're working through this area to determine the endurance limit.

To solve a whole fatigue problem, we'll put all of these steps together and

we'll do that towards the end.

So here is the problem that we were given.

A rotating shaft – it's in fully reversed bending.

It's simply supported and there's a fillet radius of

0.1 inch between the two shafts, or the diameters.

The shaft is machined, and it gives us the surface conditions in the problem.

There's a requirement for the reliability to be 99.99%.

The shaft is made out of 5Cr-Mo-V Aircraft Steel with an ultimate

strength of 240 ksi and it's operating at 600 degrees Fahrenheit.

And the problem asks, what is the fully adjusted endurance limit?

And so, that's the endurance limit at operation.

Okay, so when we start working through the problem

we can see that we need to figure out ka, kb, kc,

kd, ke, kf and our endurance limit that would be in test data.

And so if you actually go into MIL Handbook 5J,

you're not going to find an SN curve for 5Cr-Mo-V Aircraft Steel.

So what we're going to do is we're going to estimate Se

prime based off of the equations off of Shigley,

which said that the endurance limit is 100 ksi for

steels with an ultimate strength greater than 200 ksi.

So since our steel is greater than 200 ksi,

we're going to say our endurance limit is 100 ksi.

The next step is to estimate our Marin factor, lowercase ka.

And the equation for that is a S ultimate to the b.

We're assuming here we don't have any in-house test data on the surface finish.

They said that the surface is machined.

And they gave us the surface conditions so we can go ahead and

plug directly in and say, 1.34 ksi times 240 ksi to the -0.085.

And our ka is equal to 0.84, so

we're already starting to reduce our endurance limit.

Our kb is our size factor.

Now, this shaft isn't bending so it is going

to have that uneven stress distribution through it.

So we definitely need to take into account the size of the shaft.

And so if we plug into the equation for

a rotating cylinder with a one inch diameter, so note,

I'm using a one inch diameter because I know I'm going to calculate the stress in

the smaller shaft because that's going to probably be my worst-case scenario.

I'm actually going to calculate the stress right

here at the stress concentration area.

And I'm going to use the smaller shaft diameter because it's going to be

the worst case scenario.

So I'm going to use a smaller shaft diameter in my kb calculations.

So, 0.879 times 1 inches to the negative 0.107.

And what I end up with is a kb of 0.879.

kc, because we're in bending, is 1.

And then kd we're going to estimate with the military standard.

So here's our tensile test data for

temperature at 600 degrees Fahrenheit.

We can see we have this reduction in strength.

So it says percentage of room temperature ultimate strength.

FTU means ultimate strength in the military standard.

And that's about 86%, so our kd is going to be 0.86.

So in this case, kd is going to be 0.86.

And then, the reliability is 99.99% and so in that case our ke,

if we look at the chart provided on the lecture slides or

in Shigley's or Fundamentals, it's going to be roughly 0.702.

So I can go ahead and solve for

the endurance strength by saying ka is

0.84 times kb which is 0.879

times kc which is 1 times kd is 0.86,

times ke which is 0.702.

I'm going to assume my kf is 1,

because I don't see anything in this operating condition that's alarming.

They are not talking about plating or residual stresses.