Stress Analysis: Completely Reversed Stresses, Modifying Factors, Stress Concentration (8 of 17)

Stress Analysis: Completely Reversed Stresses, Modifying Factors, Stress Concentration (8 of 17)


I just wanted to mention that. Alright, last time we talked about the SN
diagram, and how it, would be useless if we plot it on a rectangular, or even, semi-log
paper even though the semi-log is better it is a non-linear graph. So to make it all linear, we plot this on
a log-log graph, which you see on the screen. And we recognize 3 regions in this, this region,
called low cycle fatigue from 0 to 1,000 cycles. This region, called high cycle fatigue, from
1,000 to a million cycles. And the region beyond the million cycles which
is called region of infinite life. This diagram, in particular is drawn for steels. Not all materials have infinite life, or can
be designed for infinite life. The stress below which infinite life is, achievable,
is called the endurance limit or the endurance strength. Sometimes also called the fatigue limit. So all of these do apply. If we want to design for, an infinite life,
more than 10 to the 6th, it is very simple. We limit our stresses to this endurance limit. The variable stress. Remember we’re talking about a completely
reversed set of stresses. So in that case, whatever my maximum stress
is, I say it cannot go beyond this value, this Se or Se prime, there is a difference
between the two, I’ll talk about that. And if we want to design in this region, where
the number of cycles is less than 1,000, then, we use static design equation. In other words, we neglect the fatigue part. We say it doesn’t have that much effect on
it. It is in this middle region that we would
like to be able to do most of our analysis. As you see this is a straight line, in order
for me to be able to find the equation of that straight line, I need 2 points on that
line. And, those 2 points are relatively obvious,
one of them is right here, the other one is right there. So one of them is this, one of them is Se,
or Se prime, I’ll talk about the difference. As your stress, the other one is 10 to the
6. The difference between Se and Se prime is
that Se prime is the endurance limit of the rotating beam specimen. What you find in the lab, that specimen is
made according to ASTM standards, it has a polished surface, it has a specific dimension,
or dimensions, and its matter of testing should be exactly as they prescribe. Se, on the other hand, is the endurance limit
of the actual part that you will be using in practice. And we will see how to find Se from Se prime
in a few minutes. So, to get the equation of that line, here’s
1 point. The other point, of course is this. We assume the number of cycles to be about
1,000, so we will call the number of cycles 1,000. And the stress, as you can see from the graph,
is a little bit less than the tensile strength of the material. In many, books you find that they make life
easy for themselves. And they assume that value of the stress,
at this point, is 9/10 of Sut. Of course, they don’t just pick that off,
under dreams or whatever. But it is based on some experiments. But 9/10, your book however, uses a, variable
value. The variable there rather than .9 they just
use one called f, f, Sut. F can be found from, there are equations for
it, the easiest way to find f, is to go to this graph, which is in your book. It’s figure 618, and, the stressor is tensile
strength of the material, and the ordinate is the value of F. So if you have the material,
once again this is for steels, come in here, pick off the value of f, and use it in this
equation. To find an equation of the high cycle fatigue
part of the, SN diagram. So, for this line we have coordinates of that
point, we have coordinates of that point, we should be able to write the equation of
that line. The equation of that line is, Sf, called fatigue
strength, the ordinate, as you can see, that’s the value of stress for any number of cycle,
is equal to a and to the b. Because if you take the logarithm of both
sides of this. That’s what you get. And that’s the equation of a straight line. So this is like your y, and that’s like your
x, that’s the slope and that’s the intercept. You have two points on this, you substitute
Se or Se prime, whichever you’re interested in here. We substitute the number of cycles here, we
get two equations with two unknowns, the two unknowns are a and b. And you solve for the two unknowns, a and
b, and this is what you get. A equals, f Sut squared divided by Se, and
b equals, -1/3 log of f Sut divided by Se or Se prime, depending on which one you’re
calculating. You can find it for the rotating beam specimen
or for the actual part that you have and we’ll see how to find one from the other. If you have this equation, notice if you require
a number of cycles, you can find out the maximum stress that you can put on. Once you have those. Notice that this is only a, both of those
are properties of the material. So you have a and b. You want 500,000 cycles, you put it in here,
it gives you the maximum, completely reverse stress you can put on the material. Alternatively, if you have the stress, you
put it over here, and solve for the number of cycles. So either of the two. So if you do that for example, if you have
the stress let’s say your stress, we’ll call it sigma max, you want to see how many cycles. That’s how many cycles. That’s the inverse of this equation. Okay. Now, we can always of course do a lot of tests
and come up with the value of Se, by running tests. Yes, that’s true. But in many instances that would be very cumbersome,
because every point that you see on that graph, is probably the average result of 100 tests. So it’s not just 25 or 30, or however many
there are, but multiplied by 100. So anytime you go to a new material, you can’t
be running that many tests, so that doesn’t make any sense. So, for steels we find that, when we run some
tests, that the, endurance limit of the material as tested in the rotating beam, in other words,
Se prime, is related to the tensile strength of the material via a graph, that looks like
this. This is for different types of steels. We notice that, up to some strength about
200 ksi, up to about 200 ksi of tensile strength. This is the relationship which, if you write
the equation for, turns out to be, half of Sut, approximately. This is approximate of course you can tell
by the number of points that are there. This is provided. And what is 200 ksi in, the metric system,
in terms of MPa? 1400, that’s right. Approximately 1400, the actual number is,
1 ksi, well it’s not really, that’s, approximate as well. It’s a little bit better approximation, but
just remember 1 in 7 ratio that would be fine. So that’s 1400 MPa. And then we see that once the stress goes,
the strength goes passed 200, ksi. Yeah, there’s a whole bunch of scatter over
here, but it appears as though, if I draw a line through all of this, it’s not changing. We got the same number of points on either
side of the line. So we say if the strength of the material
is more than 200 ksi, oh this is Se prime by the way. Please correct that. Se prime is equal to 100 ksi, which is 700
MPa. Now, in some of the, at least one of the example
problems that I’ll do for you, you will see that I’ve gone a little bit beyond that 700,
MPa mark, a little bit more than that. But that’s because of the previous addition
and how they were doing it. But that number is approximately correct. So that’s the value of Se prime. You remember I talked about the, hardness
test. This is a very simple test, very cheap, you
can do it very, very easily. In fact, there is portable equipment that
you can take to the field and run a hardness test right then and there, on a benign location,
a location that’s not carrying stress. And look what you can see, what you can get
from it. So if you have the Brinell hardness, you can
get Sut which is approximately half of that. And then if you have ductile material, Sy
is approximately, if you want a middle value, 75% of Sut. These are not exact number by the way, don’t
use these, you know to, design an automobile or send a, an airplane into the sky. But just numbers to check and have in the
back of your mind as, back of the envelope calculation if you will. This number, this is somewhere between .7
and .9, this is a pretty good value. And, what we get from Sut is also, Se prime. Provided it’s less than 200 ksi. So with one hardness test you can get all
of that information, that variable, the hardness test is, okay. Any questions? Now, not all materials, oh by the way, let
me show this too. Look at the top graph. That’s similar to the one we were looking
it except that it’s only drawn for 1,000 cycles, beyond. This is the high cycle fatigue, that’s the
infinite life part, and this is where all of the experimental points lie. Notice that this is not drawn through the
middle. Because if it is drawn through the middle
and you design with that, you’re asking for 50% of your design to fail, before they reach
the max or the required number of cycles. So, in here maybe, maybe only 5% of them might
fail, and that’s why we will take a look at a factor called reliability. How reliable are your calculations? Did you put it through the middle? In which case reliability’s is only 50%, 50%
of your samples will survive. Whatever your designing. So, that book uses the value of f as .9 rather
than f Sut. So that’s one of the books, this is I think
from Juvenile Books. Also, not all materials have endurance limits. This, slide is in your handout. This is, the SN diagrams for aluminum alloys
both rot and cast. Also, fatigue strength has a function of the
tensile strength of the material, this is for some aluminum alloy that’s 5 times 10
to the 8th. That’s what we usually use for aluminum. Notice I don’t, and this is also for aluminum
and magnesium, or is it just, yeah, it’s magnesium. Notice aluminum and magnesium both, they have
no endurance limit. In other words, it doesn’t matter how low
the stress is, there’s always a number of cycles for which the part will fail. There is no such thing as infinite life. So for those alloys, you must design for finite
life. Finite life would be the middle part of your
SN diagram. You have the number of cycles that you want. You have that, you find Sf, of course this
is not a, that equation for aluminum. But once you find the equation, you have the
number of cycles, you find SF and you design with that. So, that’s a little bit about some other materials. But in here it says, rot aluminum alloys,
what does the word rot mean? 315 people, no, no 315 people? You’ve heard of rot iron gates, right? Ever wonder why they call it rot iron? Now somebody in the previous class says it’s
spoiled fruit, which is probably right but it’s not spelled the same. Rot is an alloy or a metal, or a part, that
has been manufactured by plastic deformation. So you bend it into that shape. This is a type of manufacturing, and for example
you can have cold work material, you can have hot work material, forged materials, these
are all rot. They are formed by plastic deformation, as
opposed to cast, which obviously means you cast the molten metal in a mold or a form. Okay. [Background speakers] Let’s see if this is
a good place to stop. Yeah, it’s a good place to stop. Any questions? Okay, if you have any questions regarding
your exams, please see me now. Okay. So, here, we’ve just learned how to work with
this data that we get from fatigue analysis or tests of the rotating beam. And, it is the relationship between the applied
stress and the number of cycles. This is all good if you have a, test specimen
that you tested in the lab and if you want to use that some place, it’s all perfect. Nothing wrong with it. In almost no cases, is that true. We never, really make that little specimen,
according to ASTM standard with a polished surface and everything to actually use it
in a, an application. Therefore, since the parts that we do use,
in design, will have different surface conditions, they’re not polished. Will have different sizes, will be loaded
differently, maybe at a different temperature. All of these will have to be taken care of
before you can continue your analysis for the actual part that you’re designing. Or apply what we have just said to the actual
part that you’re designing. We do this by what are called, fatigue strength,
remember this is also called endurance limit. Modifying factors. Do you know what does your book call this? Does it call it fatigue limit or fatigue strength? I think your book calls fatigue strength a
general value, and then fatigue limit. I like, I myself like the word endurance limit,
because it tells you exactly what, you want to hear. Yeah, they use fatigue strength for a stress
so it must be using fatigue limit. So let me, rather than strength, make this
limit. Or endurance limit. These are factors that we multiply, by the
endurance limit of the rotating beam, Se prime, which we just found. It’s approximately half of the tensile strength. To get the endurance limit or the fatigue
limit of our part. The first one of these is called, surface
factor. The specimen that we use is highly polished,
yours may be machined, may be cold work, maybe hot work, maybe cast, what have you. The surface conditions are not the same. Again, and we will repeat these many times,
fatigue is entirely an experimental type of adventure. From which we get information to analyze and
design. So you will see me repeating this that everything
is found by experiment. There’s a graph of this, I’ll show you in
a just a bit. But there’s also an equation, ka equals a
Sut to the b. And, I want to warn you that the a and b here,
are not the same as the ones we just used for the equation of SN diagram. These are completely different, have nothing
to do with one another, do not confuse them with one another. These two constants, come from experiments
and they’re tabulated here, in your book. In table 6-2. For example, for ground, for ground you get
this is the, in the U.S. customary system, the metric system. 1.34 is the factor a, the factor b is the
same for both of them. Then you have machine, you have cold drawn,
hot drawn, forged what have you. You use these numbers in this equation along
with the tensile strength of the material, and you find the factor k. Simple as that. The next factor is called the size factor. This factor relates to the difference between
the size of the, rotating beam sample, which we test, and the actual part that you may
have. Assuming that actual part is also of circular
cross-section and it is going through rotating fatigue. Like a crankshaft, for example. Obviously it’s not the same size as the, specimen
that we use in the lab. So for these, we have a size factor, kb which
relates the actual endurance limit of the part to that of the, rotating beam specimen. Why should there be a size factor anyway? In fact we find that the larger the size,
the smaller this factor, and eventually we’re going to multiply it by Se prime. In other words, the larger the size, the smaller
the value of the endurance limit. The less stress the material will take before
it fails. Why should that be the case? Just because we have a large part. How about if we have a brittle material, should
part play any, or size play any part in that? In other words you have a tiny, a wire of
tiny diameter of brittle material versus one, 1 inch diameter. Should they have the same strength? Or different strengths? Those of you who have had 315 should be able
to answer that question. Brittle materials fail by crack propagation,
initiation and propagation. The larger the part, the more the probability
of a critical flaw or crack in the part. And therefore, the higher the probability
of its failure. So it is not an actual material property,
but is a statistical method. You may find that one of your, you test 100,
maybe in 100 you will find 1 that is larger but still has a larger strength too, that’s
possible, that’s statistics for you. It’s not all nice along the same line. But we find that in general, the larger the
size, the smaller the strength, so as long as the failure is in a brittle manner. And fatigue, is a brittle failure, and therefore,
the size plays a part. So we call this kb, and we use this equation,
in order to, convert Se prime to Se. So we have kb equals, and let me see if this
is exactly the same as what your book has, because they keep changing nomenclature on
us. So I want to check and make sure that’s right,
I’ll write it off of this one, kb equals, d over 0.3, to the -0.107 which is equal to
0.829d to the -0.107. Provided, d is less than or equal to 2 inches
greater than or equal to 0.11, inches. And kb is equal to 0.91 d to the -0.157, If
d is less than or equal to 10 inches, greater than or equal to, not equal to, just 2 inches. That’s for the English system. For the metric system, it’s the same, d divided
by 7.62, .3 and 7.62 respectively, are the diameter of the rotating beam specimen in
the English system and the metric system. So that’s why we divide everything by this. To the -0.107 and this is for d, less than
or equal to 51 millimeters greater than or equal to 2.79 millimeters. And finally, this is 1.15 d to the -0.157,
for d less than or equal to 254 millimeters, that’s 10 inches. And less than 51. Notice that it is possible for kb to be larger
than 1. For the same reasoning that I just talked
about, if we have a larger material, we end up with a smaller strength, it follows that
if I have a rotating beam, whose, everything else is the same as the rotating beam specimen,
except for the diameter. And the diameter is smaller than .3, then
that part will have a longer life, or it can take more stress. And therefore, the value of kb will turn out
to be larger than 1. But in almost all other conditions, these
factors, the modifying factors, are going to be less than 1, okay. Any questions? One thing that comes up when talking about,
the size factor, is that this is only good, if your part is rotating, is in rotating fatigue. Again, like a crankshaft. But what if it looks like this? It’s in bending, but this is not rotating
bending. The shaft doesn’t rotate. The force changes between plus and minus f. So in this case, this is what we do. We say, we will find a diameter called equivalent
diameter. For this type of, such that the cross-sect,
the part of the cross-sectional area that sees stresses, 95% of the maximum or higher,
will be the same in the two. Let me draw this and then I’ll repeat it again. In a rotating bending, and if any part of
this you don’t understand, please tell me. In rotating bending. And this shaft has a diameter, D. In rotating
bending, all of the surface of the shaft sees the maximum stress, correct? Because it’s rotating. On all points of the surface. And if I want the area, which sees 95% of
the total stress or sorry, 95% of the maximum stress, or higher, then that would be this
cross area, agreed? Remember Mc over i. Either Mc over i, or M times .95c over i. That part sees 95% of the highest stress. We say the part that’s important really is
this part, because it’s near the surface, most flaws occur on the surface, therefore,
cracks will be developed on the surface a lot easier and will propagate a lot easier. Cracks are more likely to exist on the surface
than on the interior of the material. So, we use this area, we say this is the important
area. This part over here. Has those 2 areas, that see 95% of the maximum
stress or higher. The point over here, never sees any stress,
that’s on the neutral axis. So these are areas of high stress, we say,
this, which has this cross-section, is equivalent to a rotating shaft that has the same area,
we equate those 2 areas, and you can see those. Oh I forgot to put in the, sigma in there. See that here. This area that I’m talking about is this much,
.01046d squared. So that’s 0.01046d squared. And, this area is pi over 4, times d squared,
minus 0.95d, squared. We call this the equivalent rotating beam
diameter. And we solve for the diameter, which comes
out to be De equals 0.370d. So, this is equivalent to a rotating shaft,
whose diameter is this. Now we’re back there, and we put this equivalent
diameter here, here, wherever we want, and calculate kb. Any questions? Also, we may have some something like this,
this may be cross-sectional area. In this case, the parts that see 95% of the
maximum stress, are across part. Again, we said that area equal to this area,
and solve for the equivalent diameter, which is .808 square root of hp. And, we put that here to find the size factor. Any questions? For other, for other shapes you can find an
equivalent diameter doing exactly the same thing. And a couple more over here, an beam and a
channel, found for those, but you can find it for, any other type of, cross-section that
you want. Any questions? I’ll show you that a little bit later. So, where’s my, no I don’t want to talk about
temperature yet. That’s kd. So that’s kb, kc, load factor. The manner of loading in the rotating beam
is bending. So if the manner of loading in the part that
you have is also bending, there is no correction factor, there is no modifying factor. When I say there is no modifying factor, I
really mean that the modifying factor is 1. It doesn’t have any effect. So, that’s 1 for bending. 0.85 for axial load. And, 0.59, and I want to emphasize this for
pure, torsion. Only if you have reciprocating torsion, back
and forth, nothing else. If that torsion is a part of a combined loading,
in other words if it is accompanied by axial load or bending, you do not use that factor. So make sure that you make a note some place
and keep it in the back of your mind. When we do combined loading, we’ll take care
of that load factor in a different way, and I’ll talk about it when we come to it. This factor, 0.95, is theoretically 0.577. And it actually, if you take a look at the
von Mises equation, it comes from pure torsion case. Like this, this is the question in the, multiple
choice as well. So this is a pure torsion. That’s also exactly the same thing, same point. Notice that if I draw a, more circle, get
tau here. Get tau here. Those are your principle stresses. Pure torsion, at 45 degrees. Now, if you want to do, a von Mises on this,
sigma prime is equal to, the square root of sigma t squared, minus sigma t sigma c, plus
sigma c squared, this comes out to be root 3 tau, if you substitute for these values. Sigma c should be actually minus tau, so I
add tau here. And we say yielding occurs, when this von
Mises stress, sigma prime. Is equal to the yield strength of the material. So, Sy equals root 3 tau, therefore, tau equals,
1 over Sy, that is 0.577. That’s where that comes from, that relationship. But experimentally it’s .59, Any questions? The next factor, is the temperature factor,
kd. For KD, we find that when you increase the
temperature, initially, up to about, this is for steels again. Up to about 400 degrees, 3 to 400 degrees,
somewhere in here, there’s a tiny bit of increase in the tensile strength of the material. St is the tensile strength of the material,
at the temperature, is Rt, is the tensile strength of the material at room temperature. So this is the ratio of the, tensile strength
and the temperature that you have versus room temperature. But then after that, there’s a big drop, in
the tensile strength of the material. This ratio St over Srt can be used as your
temperature factor. If you know sigma prime, you can use it, St
over Srt. And most of the time it’s less than 1, sometimes
it’s a little bit more. If you don’t have it, use this ratio, find
the tensile strength at the temperature that you desire, and then use the equation Se prime
equals .5 Sut, to find your endurance limit. In other words, use that tensile strength,
rather than the room temperature tensile strength. So that’s, a little bit about the temperature
effect. Reliability effect ke, it depends on how you,
design for example. Like I said before, if you design with a line
through the middle of all of the experimental points, you’re asking for a reliability of
50%. Half of your samples will survive, half of
them won’t. If you, anything more than that, you have
to bring down this curve. So remember this curve here, looks like this,
and your experimental points we had them like this. Now that’s your 50% curve, you want something
less than that, so that maybe 1 or 2 experimental points fall below that. There’s no such thing as 100% certainty, this
is all probability. So whatever probability you would like, take
a look in here, here’s, there is the variance we’re not going to talk about statistics over
here very much, but this is your reliability factor. Take off the reliability factor from here,
and you use it as ke. And then you have, so let’s write that too. And then you have a factor called kf. This is called miscellaneous effects. Anything and everything else we haven’t talked
about you can put in here. As an example, if the environment is prone
to corrosion, you might want to have a factor less than 1 here, to account for that. If the part has been plated, it’s also found
that its fatigue strength decreases with plating. If it has been, sharpened, fatigue strength
increases. Do you know what, you all know what plating
is right? Yes? Everybody knows what plating is? Okay. And does everybody know what shot-peening
is? Everybody knows what shot-peening, so these
are fair questions for the multiple choice for the next exam alright. But if you don’t, you better ask, so don’t
be bashful. But shot-peening I don’t think all of you
know, it’s very seldom that everybody knows what that is. It’s a process in which you bombard the surface
of the material with very tiny hard pellets, traveling at high velocity. And they, plastically deform the surface and
produce something similar to pock marks. At the same time, they leave residual compressive
stresses on the surface. So that when your tensile stressors are applied
to the surface, before they can propagate any cracks, they have to overcome those compressor
stressors first. So your, you’ll be allowed to put in more
stress. So when you have that, these affects are included
in the miscellaneous effects factor. Let me show some of the stuff that I talked
about, and this graph, these graphs are in your handout by the way. At the top graph here, this is from another
book, they call their factors c knot k. So that’s the surface factor, rather than
using that equation they give some graphs. These are just as good. What is, what is interesting to see is this,
that plating, reduces the fatigue strength of the material. This is the unplated part, plated part. It reduces the fatigue strength, plating does. Combined with shot-peening. Something very interesting happens. Again these are SN diagrams, This is nickel
plated, this is unplated. So nickel plated has a much lower SN curve. Even after we peen it, after we plate it,
we peen it and then nickel plate in. Then, peen first, nickel plate later. Still is not as good as the unplated plate. However, if you plated and then peen, then
its strength will increase. Because now you have those pock marked for
residual stresses on the actual outer surface of the material. Whereas in the previous peening, we did under
the plating material. Also when you’re plating, you have to take
into consideration, the fact that, not plating sorry, this is similar to plating, but it’s
not the same thing. Case hardening, this is a process in which
you, strengthen the outer surface of a material. It can be done in different ways, it can be
done via temperature, a process called quenching, it can also be done by infusing carbon into
the surface of steel. What you do is increase the carbon content,
the outer part of the steel, increasing the carbon content increases the tensile strength
and therefore, Se as well. So if this is your stress, and this your strength
for the case, that little part is the case. Then you say well my strength is less than,
my stress is less than strength so I’m good. No big deal. However, you should be careful here. Just when the case ends, and it’s not a black
and white line, but it’s shown that way here. Where the case is gone here, and this is the
real material, core material now there your stress may be higher than the strength of
the coring material. And you may get failure. So you should check not only where the stress
is maximum, but just under the, case as well. And finally, you may have stress concentration,
in the material. We incorporate stress concentration, your
book may apply it to strength, in other words Se. Or it may be they apply it, or may apply it
to the strength. My suggestion to you, is always apply it to
the stress. In other words, increase the stress, don’t
decrease the strength. Because you will always be right. If you increase, if you decrease the strength,
then in combined loading, you will have different stress concentration factors in which one
you’ll use. You don’t know. But I’ll emphasize this again when I come
to, stress concentration in combined loading. But, taking a look at all of the, modifying
factors, we find Se from Se prime via this equation. This is now the endurance limit of your part. Multiply all of those factors, multiply them
by the endurance limit of the rotating beam and you’ll have the endurance limit of your
part. Not including stress concentration. And so as far stress concentration is concerned,
we see that, when a material is loaded in fatigue, they, the effect of stress concentration
is somewhat reduced. This is mainly because in fatigue, failure
occurs by a propagation of a crack and at the tip of the crack. Let’s say it’s propagating in that direction. At the tip of the crack you have an area of
plastic deformation. Remember the stress there is extremely high,
it passes the tensile strength of the material, only locally. That’s the only way a crack can propagate. And since it is plastically deforming, as
we talked about during the first day, it cuts down on the stress concentration factor. And it also increases strength of the material
due to strain hardening. So, the effect of the, stress concentration
factor is reduced and it is reduced via, using a factor called notch, sensitivity. Sensitivity. The equation, is kf equals 1 plus q times
kt minus 1. This notch sensitivity, as the name implies,
tells you how sensitive the material is to exist into the notch. For example if you have cast iron, which already
in it has a whole lot of flakes, of carbon, with a high stress concentration factor, if
you have something like this. It’s probably not going to affect it. The stress raises inside of the material,
those carbon flakes. So this won’t have any effect. The value of q, can be found from a couple
of graphs, in your book, 620 and 621. 620 for a normal stresses, and 621 for shift
[phonetic] stressors. They are related to the tensile strength of
the material, as well as, the radius of the notch. The radius of the notch is very small, you
can see that notch sensitivity would be very, very low. And if the tensile strength of the material
is high, notch sensitivity will be high as well. So, what is kt in this case? KT is, geometric, stress concentration factor. In other words what you have already used
and you find from the tables. You go to the tables where you have these,
discontinuities, you find kt, you put it into this equation, you get the notch sensitivity,
which is never greater than 1, in here. Notice that for a notch sensitivity, of 1,
that means the material is fully notch sensitive, then the value of kf and the value of kt,
they are equal. Other than that, the value of kf is always
less than kt. So the effect is minimal. So this for normal stresses and this is for
stresses. Any questions? Okay, let’s take a look at a couple of, examples
now, these are in your handout. Says develop the expression for fatigue strength
as a function of number of cycles for a 32 millimeter diameter machine rod under non-rotating
bending. The steel has a tensile strength of 710 MPa. Oh I thought this was 1400, no it’s 700, so
that’s good. We, we are trying to develop, this equation. This of course is for the middle part of the
graph. That’s the equation we’re looking for. Sut, tensile strength, 710, Se prime, half
of that, 355. Surface factor, a Sut to the b we pick off
these two constants from the table. I think it was 6-2 but I’m not sure I don’t
have the number. Oh yeah, it says right here 6-2, .781. So it’s just straight plug into this equation. Since the rod is non-rotating, an equivalent
diameter is found. This is what I was talking about here. And for a non-rotating ground the equivalent
diameter is 37% of the actual diameter. This means 37% of the area of that rod sees
the maximum stress down to 95%, of the maximum stress. Then put it in the equation for kb, and you
find the factor kb. If nothing is said, in this case, we assume
that it’s bending, so kc equals 1. Nothing is said, kd and ke we always assume
equal to 1, although the reliability factor, ke is not equal to 1, it is not a very good
idea. But generally, if nothing is said, just take
it equal to 1. Se therefore, is the product of all of these
factors, ka, kb, kc, and all of those, multiplied by Se prime. 355, so that gives you 265. In order to find the value of f in the equation
for. For a and b in this equation, remember a was
equal to, 0.9 Sut, or not 0.9, I’m used to 0.9. F Sut, squared, divided by Se. The value of f we get from the graph. However, that graph is, in the English system,
not in the metric system. So what I’m doing in this step here, just
in case you’re wondering, is converting 710 MPa to 103 ksi. And from the graph, I read f equals .84. Then, calculate the values of a and b according,
a is on the board I just re-wrote it, b you have it as -1/3 log of f Sut divided by Se. So there’s the value of d. There’s the value of a, and that your equation. That’s the this equation. For the values of a and b, substitute the
two. Any questions? Next, we’re going to use the same rod, so
now assume that the rod is used as a lever beam of length 300 millimeters. And loaded with the force f that is tipped
that fluctuates between plus f and minus f. Using a factor of safety of 2 respective fatigue
failure, find the maximum possible value of fa for infinite life and b for a life of 10
to the 5 cycles. So in this case, you, have to find the value
of this f, the applied load. The maximum stress in this rod, occurs over
here, never mind the stress concentration at this time, we’ll get to that later. And it’s equal to 32 M divided by pi d cubed,
in terms of f is that. M is equal to 300f in millimeters, and newtons. For infinite life, this maximum stress, must
be equal to Se, we also require a factor of safety of 2. So divide that by 2. And, we solve for the force f, 1422, for infinite
life, that’s the maximum force you can have, if infinite life is required. For a life of 10 to the 5, we can no longer
use Se. So if I may draw that graph again here. So life of 10 to the 6 we use Se. Life of 10 to the 5, we need that value. That’s our maximum. That we can find from the equation for Sf,
the equation we just found. Sf equals 1348, 10 to the 5 to the power of
-.1172, 350, so this is 350. It must of course, be, larger than the value
of Se. So there is 350, and now we say sigma max
at, this is covered with a sheet. So this is Sf over 2, please put a 2 under
that Sf, right there. So that should read like this, it should read
sigma max, equals Sf over 2, last time it was Se over 2, 2 lines up, Se over 2, now
Sf over 2. This is our strength now. And we find the value of f, 1876. And finally, assuming f equals 2,000 newtons,
will the part have infinite life? If yes find the factor of safety. If no, find the life of the rod. So in this case, we would like to take stress
concentration into account as well. So this is this business of kf, as a function
of kt. These are the dimensions from the picture
that you have, from appendix A59, we get kt equals 1.8, kt is the geometric stress concentration
factor, what you have already used. We find that equal to 1.8. The notch sensitivity from figure 620 and
the tensile strength of the material, we find is .8. KF, this is this equation here. KF equals q times kt minus 1, plus 1, 1.66. And this, kf must always be less than, or
at best, equal to kt, it can never be larger than kt. For infinite life, 265 is the maximum stress
we can have, the endurance limit. So 265 is equal to the stress concentration
factor, times this stress, 32 times 300f divided by pi times 32 cubed. For infinite life the maximum stress should
be 1712. You can’t have anymore than that. The force that’s given however, is 2,000 newtons. So obviously there is no infinite life, there’s
finite life. We find the life of the part by calculating
the maximum stress with the force of 2,000. 309 MPa, and then again, using this equation,
and solving for M, the number of cycles that will take, that comes out to 287,000 cycles. Okay. Any questions? Okay that will be all. If there is a quiz on Tuesday, it will cover
fatigue. If I collect homework problems, it’ll be all
of the homework problems pertaining to fatigue. I think it goes back to, time before the last
exam when I started fatigue. So 2 sessions. We have done fatigue up to, but not including
fluctuating loads. So everything is for, that part of fatigue.

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