Stress Analysis: Thick Walled Pressure Vessels, Press & Shrink Fits (4 of 17)

Stress Analysis: Thick Walled Pressure Vessels, Press & Shrink Fits (4 of 17)


No more questions? Okay. Then let’s quickly recap what we did last
week. We looked at theories of failure, both for
ductile materials and for brittle materials. For ductile materials, the two theories of
failure that you should be using are the maximum shear stress theory and the von Mises or the
maximum distortion energy . For the brittle materials, you should be using either brittle
Coulomb-Mohr or the modified Mohr. And all of them, of course, entail drawing
a locus of failure point or what we call strength lines. And then drawing the load line, finding where
the two cross. The results will give you the critical stresses
or the stresses that will define failure. Then you can simply find your factor of safety
by dividing those stresses by the applied stresses. And that will give you the factor of safety. We did some example problems. Let me do one more example problem for you. It is problem 523. It says for ASTM 30 cast iron — — A, find
the factors of safety using the brittle Coulomb-Mohr and modified Mohr theories. B, plot the failure diagram in the sigma A,
sigma B plane to scale and locate the coordinates of distress state. And C, estimate the factors of safety from
the two theories by graphical measurements along the load line. We rarely use graphical measurements. We use analytical ways of finding, in other
words, the equations in finding the results. However, it is important for you to understand
and be able to draw a graphical representation of both the strengths and the stress. And you should do this actually before you
start any calculations. Then you will have an idea, of course, provided
you do it correctly. You will have an idea of what the factor of
safety should be. So I’ll do this analytically. I will also sketch a plot of the strength
and load lines. And we’ll try to estimate what the factor
of safety is. So that’s for this problem and this is general
for a number of problems. This one in particular will have the following
stresses. Sigma X — — 15 KSI. Sigma Y is zero. XY is minus 10 KSI. This, of course, is given, the state of stress
at a point. In general, you will not be given these stresses
but you’ll be given a loading case from which you must identify critical points and for
each of those points draw or calculate stresses like this. So since we’ve done that before to share moment
diagrams and all of that, we start from here. But you should be able to take the problem
from the beginning to the end and this is somewhere in the middle. If we draw the stress element at this point
— — this is what you will be looking at. Now, a couple of points here. One, do you understand the meaning of that
number 30 ASTM — ASTM 30? By the way, do you know what ASTM stand for? No? This is important. I’ll put it out here or no, maybe I’ll put
it — — American Society of Testing and Materials. This is the official body that tells you how
to run tests. If we don’t have an official body to specifically
tell us how to run various tests, different people will do different things. For example, if I want to run a test on some
sort of steel weld, whatever I find downstairs, I’ll test that. Somebody else tests something else. There’s no way we can compare our results. So they tell you exactly what, for example,
the length should be, what the diameter should be, what the surface condition should be. All of these are set forth in ASTM. This is not just for, you know, tension and
compression tests or tests in mechanics. They have it for everything including soaps,
shampoos and anything else you can come up with. Next time you are in the shower, take a look
at the back of your shampoo bottle. It probably says complies with ASTM whatever
the number is. So that’s what ASTM stands for. ASTM 30 when applied to cast iron means what? Does anybody know what that 30 stand for? Percentage, the weight percentage of carbon. No, carbon — it’s a good thing you said that. Let’s talk a little bit about carbon. Is there a maximum amount of carbon that can
exist in steel? Of course, this is not steel. This is iron but that’s okay. Let’s do steel first. In other words, can I have a steel made of
50% iron and 50% carbon? No, so does anybody remember what the maximum
is? You guys who have taken 315 should know that. It’s 6.7. No, 6.7 is actually for cast iron. That’s the maximum amount of carbon that can
exist in an alloy of iron and carbon, 6.7%. Specifically for steel, the amount of carbon
in steel theoretically is limited to about 2.1% and that’s theoretical. If you get a steel like that, it is probably
— not probably, it’s definitely worthless. It is so brittle that you can’t use it for
anything. Practically no more than 1.2% to 1.3% carbon
can exist in steel. And those steel are very, very brittle as
well but there are some applications for those. So the number 30 in this case, so we can’t
have 30 for carbon in any type of a steel alloy or iron alloy, be it steel or cast iron. The number 30 gives you the approximate tensile
strength of the cast iron. So when you read ASTM 20, that means the tensile
strength is approximately 20. Thirty means it’s approximately 30 and here
they give you the tensile strength or you can find them. I don’t know whether they give them to you
here. You have to find them from the tables in the
back of the book. And if you do that, this is what you get. ST equals 31 KSI and SC equals 109 KSI. Tensile and compressive strengths of cast
iron. And you see that ST is approximately 30 KSI. That’s one point. The other point — oh, any questions on this
before I go to the next one? The way I do these stresses, specifically
this one, the one. What is — this set of stresses represents
stresses where? I don’t mean where in the structure but generally
when we have an element, that element is representing stresses where? At a point exactly, at a point, one point. Now you can pick the point wherever you like
and you’re going to come up with different stresses for that point. But this is stress at a point. And if that’s the case, then when we do Mohr
circle, we would consider this shear stress positive or negative? Okay, this one? Okay, we’re talking about shear stress at
one point. How can it both positive and negative? That doesn’t make sense. Shear stress at a point? Shear stress at one point? That’s it, it should have one sign, period. And there’s nothing wrong with the way we’ve
represented this point. It’s exactly right. We can actually draw Mohr circle with a different
type of definition but it’s very complicated if we do. Not very complicated, a bit complicated. But to signify or represent stresses and at
any point, there are nine components of stress. So let me draw those for you. I’ll leave that there. Stresses are among a group of entities called
second order tensors. And all they do is transfer according to the
equations that you know related to Mohr circle. So let me show you the nine components of
stress. This stress, what should be the subscripts
for that shear stress, Tom? YZ. Why YZ? This plane is the YZ plane. That is correct. But that’s not the YZ. That’s actually XY. Now you were told that it doesn’t matter if
you call them XY and YX and generally, in most problems, that’s true. It doesn’t. But each of these are represented with two
subscripts. The first subscript, X, in this case — —
subscript first — — direction of the outward normal to the plane on which the stress acts. And this is good not only for shear stresses
but also for normal stresses. And it is the correct way of representing
stresses. I’ll explain what this means exactly. Second — — direction of stress. So according to this definition, there are
two subscripts here. The first one indicates the direction of the
outward normal to the plane on which the stress acts. What do I mean by outward normal? If you have this plane, you can draw perpendicular
to it going in that direction or you can draw it perpendicular to it going
in that direction. This is called the inward normal. This is called the outward normal. Everybody okay with that? Going away from the plane, outward normal. The second subscript, so that’s X, direction
of outward normal to the plane. The second subscript is the direction of the
sheer stress itself. So this is tau XY. This says first subscript, direction of outward
normal so X. Second subscript, direction of the shear stress or sorry, direction of the
stress itself, sigma XX. Then we have another shear stress here. This and this is what, tau? Outward normal, X? Shear stress in the Z direction, XZ? Similarly, we will have this stress as sigma
YY. This stress as tau YX. Oh sorry, tau ZX, not YX. ZX — I’m sorry? Isn’t that sigma ZZ? Sigma? ZZ, yes, thank you. That’s why I made that mistake and then I
made that mistake. Thank you. Sigma ZZ. Tau ZX and then this one? Tau ZY. Same definition. Last plane, sigma YY and this one, tau YX
and that one, tau YZ. The two dimensional stress element that you
draw is the one that’s drawn when you’re looking along the Z axis. So this is what you have when you draw that. You’re looking in that direction, say you
get sigma XX, tau XY. Sigma XX here. Sigma YY here. Tau YX. Is everybody comfortable with this? That’s the nomenclature for stresses. Now, the beauty of this is that it doesn’t
just name the stress. If you look at the stress with its subscripts
and you have a set of axes, you can figure whether the stress is positive or negative
regardless of whether it’s normal or shear. So this is the way we do it. If the outward normal to the plane on which
this stress acts, in other words, the outward normal in this case, in the X direction, is
in the positive direction of the X axis then we call that subscript positive. That’s the case here. For the second subscript, if it is in the
positive direction of the axis, we call it positive. The product of those two signs give you the
sign of the stress. In this case, positive times positive gives
you positive. Therefore, this is a tensile stress. It tells you exactly what it is. On the other side, you may be thinking on
the other side of this phase, the outward normal is in the negative direction of the
X axis, negative. The normal stress itself is on the negative
direction of the X axis, another negative. That would be this one. That’s my X and Y. Negative direction of the X axis, outward
normal. Negative direction of the stress. Negative times negative is positive, a tensile
stress. Had this just been reversed, we’re a compressive
stress. This one would remain as positive. That one would be negative. And that would tell you that the stress is
compressive. The same thing is true for all of these. And if you do that, you see that whether or
not I draw a shear stress like this or like that, it’s not that this is negative, that’s
positive but they’re both positive. For example, this one, the outward normal
is in the positive direction of the X axis. Shear stress is in the positive direction
of the Y axis, positive shear stress. What about this one? Outward normal in the negative, shear stress
in the negative, positive shear stress. What about this one? Outward normal positive, shear stress positive,
positive. Outward normal negative, shear stress negative,
positive. All of these are positive shear stresses. Even though when we use Mohr circle, we plot
one of them as positive, the other one as negative. We just do that because it’s easier. There is a way to draw a Mohr circle using
this convention but we’re not going to talk about that here. We’ll just stick with the convention of Mohr
circle. But this needs a couple of points and I think
you should be aware of it. Any questions? Okay, so that’s how I do the stress. It’s just tau XY is minus 10 KSI. The stress that acts on this plane is negative. The outward normal is certainly not negative
therefore the shear stress must act in the negative direction of the Y axis and that’s
why it’s strong like it is. Does everybody understand that? Because given this set of numbers, you should
be able to draw that. So if there is anything murky here, please
let me know. Any question? All right, so going back to the problem, we
have this set of stresses. These are our strengths. And we would like to come up with some sort
of a factor of safety for this cast iron. To do that, we used failure theories. And in failure theories, we draw — two-dimensionally,
we draw the two principal stresses, sigma A and sigma B, for example. So we need to find the principal stresses,
sigma A and sigma B, for example. So we need to find the principal stresses. So I have sigma X as 15 and now I use Mohr
circle convention. And for that, that’s positive. So that’s 15 and 10, this is all in KSI. And then on the other surface, zero and 10,
zero and minus 10 that is. There’s the diameter of Mohr circle. And there is Mohr circle. If you calculate the values, this is a simple
Mohr circle case. So I’ll let, I’ll leave this up to you to
come up with these values. Sigma A equals 20 KSI and sigma B equals minus
5 KSI. So now, we have the principal stresses. We plot our failure locus. They asked for us to do this using brittle
Coulomb-Mohr and the modified Mohr. This is sigma A. That’s sigma B. Mark ST here
so that’s 10, 20, 30, and 31. And here — This is 31 KSI. That’s also 31 KSI. Then and notice we only use the first and
fourth quadrant. The second and third are very similar. And this being, of course, cast iron and a
brittle material, what’s important are tensile stresses. So this one will go 30, 60, 93 or 90 what? Oh, 109 — 90 okay, we’ll go down in here
somewhat. That’s 109. That’s the brittle Coulomb-Mohr. Modified Mohr — you draw that too. There’s the modified Mohr. Actually, let me make this dash. That’s not a part of the failure line. Failure lines are the solid lines. Okay, now let’s plot our point. And try to figure out approximately what the
factor of safety should be. Our point is 20 and minus 5 so here’s 20. So that’s 10, that’s 5 so we’re over here. Now, let’s do other load line in a different
color. These two points identify the critical points
or limiting values of these stresses according to the two failure theories. So this is 20 and that’s minus 5. So all in KSI. Any questions? Everybody following what I’m doing? Estimate the factor of safety by looking at
what’s on the board. Remember this is an estimate. You don’t have to be exact. According to first, the modified Mohr because
it’s the easier of the two because it’s crossed up here, not down here. Had it crossed down here, it would be a little
bit more difficult. Graphically, how do you get the factor of
safety? That length over that length, right? Which is the same as this length over this
length. So estimate the factor of safety, 1.5 right? The exact factor of safety according to the
modified Mohr theory which is the easier of the two — 31 over 20. That comes out to be 1.55. Any questions? Now the brittle Coulomb-Mohr, of course, you
can use the equations if nothing else is set. But I am not a real fan of just simply plugging
numbers into equations. You need to know what you are doing and why
you are doing what you’re doing. So I will ask you to do the procedure that
I went over last time and I’m going to go over again today in order to find a factor
of safety. Not in all cases but in some simple cases
such as this, I will ask you to do that. So to do that, what we do is write an equation
for the failure line, brittle Coulomb-Mohr. And that is sigma A over 31 minus sigma B
over 109 equals one. You can check that equation by saying if sigma
B is zero, sigma A is 31. If sigma A is zero, sigma B is minus 109 here,
there is the line connecting the two. This is a straight line. Two points is all you need. The equation of the load line — this green
line is a load line. And clearly the equation for that is sigma
A equals negative five over 20 sigma B. You have one point and the line goes through the
origin. We now solve these two together. When I solve these two together, algebraically
what will I find? this must be equal to — Oh, you’re right. Yes, yes, yes. Yes, it is. Thank you. Thank you. Thank you. It is the opposite, absolutely. Sigma B equals that times sigma A. Yes, as
I said before, if you see anything that I write on the board that doesn’t make sense,
please say so. You’re either right which is very, very good
because I’ll correct my mistake very quickly. Or if you’re not, I’ll tell you why not and
then you won’t make that — you won’t make that mistake again. So it’s good all the way. So back to my question. If I solve these two equations together, I
will — what will I find? Of course, I’ll find answers for sigma A and
sigma B, right? This is two equations and two unknowns. But what do they represent? Sorry? I’m sorry, I didn’t hear. Intersection, the coordinates of the intersection
of the two straight lines, correct? That would be the coordinates of that point,
agreed? As the point of intersection, that is the
limiting value of these stresses. The limiting value of these stresses we called
strength. So when we solve these two together at the
same time, we’ll switch the names from sigma’s to S’s to signify strengths. So when we solve these two together, this
is what we’re going to get. And you only need one of them here. You don’t need both and I think I’ve only
found one. Prime SA equals 28.9 KSI. So all you do is substitute this for sigma
B and solve for sigma A and that’s what you get. The factor of safety according to this theory
is this length divided by that length which is the same as saying, that’s SA, by the way. That’s SA right there which is the same as
saying N equals SA over sigma A, 28.9 divided by 20 and that’ll give you N equals 1.45. And you expect to get a number similar to
that. It certainly has to be smaller than 1.5. And
you can estimate it. Maybe you look at it and say, “It’s 1.3,”
but that’s close enough. If you draw it to scale, you can actually
get good values up to two significant figures. But from your sketch which will be very helpful. So that’s the factor of safety. Any questions? Okay, so from now on, if given a problem and
ask for a factor of safety or a maximum value of a load or something like that, you should
no longer be saying that the factor of safety is equal to the new strength divided by the
maximum stress. Although sometimes that’s good like in this
case. But generally, you should be using a failure
theory. So in this case, you say, “I’m using brittle
Coulomb-Mohr or modified Mohr. Here is my answer.” Done, okay. So that concludes a look at various static
failure theories. And remember, these were all static failure
theories. No speed of loading involved. Now, we will take a look at — — pressure vessels. And what we say about pressure vessels — —
in many instances, it’s good for pipes as well. The only difference between a pressurized
pipe and a pressurized vessel is that in a pipe since the fluid flows through the pipe,
the pressure is really on the periphery of the pipe, not along the axis of the pipe. But in a vessel, you will have pressure on
the periphery and at the two . In other words, if you have a vessel due to an inside pressure,
for example. The diameter increases, so will the length. Everybody okay with that? In a pipe, the diameter increases. Of course, other stuff will happen too because
of Poisson’s ratio but we’re not going to go into that right now. You studied the case related to thin-walled
pressure vessels. So I’ll refresh your memory here. A vessel is considered to be thin walled if
its thickness is 1/20th or less of its diameter. So for that, this is what you must have. According to your book now, you may read numbers
a little bit different from that but generally, that’s pretty good. By doing this, we are — by saying that the
pressure vessels is thin walled, we are simplifying the calculations. But we want to make sure that what we have
is not too far off from the truth. So if we can keep our errors below about 5%
or so, then that’s good and it warrants the easiness of the problem. For such a pipe or vessel, there are two stresses
that you studied in ME 219 called the tangential — — or hoop stress. This is a stress that tends to increase the
circumference of the vessel, in other words, its diameter. You also had — oh, let me write the equation
for this. PD over 2T, we’ll talk about it, what you
do with the thickness in just a second. You also have sigma A called axial stress. This tends to increase the length of the vessel. So if I draw the vessel here to show these
stresses, sigma A acts like that and sigma T acts like that. The magnitude of sigma A is half of sigma
T. And of course, you finish all that by taking two sections and writing .
If I have an element, if wanted to draw these on an element like so — Yeah, I’ll just call
that sigma A. So I was with the picture on top. So I’ve drawn these as principal stresses. How do I know these are principal stresses? Yeah, but that’s the way I do it. If I put a shear stress there, there’ll be
shear stress there, right? How do I know not to draw a shear stress here? How do I know these are principal stresses? Saying — let me back track a little bit and
mention something so no confusion will take place here. When you say there’s no shear stress, you
have to qualify that. There’s no shear stress on these planes, right? There is shear stress on other planes, agreed? Because if I draw a Mohr circle for this very
quickly — — where is sigma T? There is shear stress. Just on these planes the way I have drawn
them. But how do I know to draw it like that? One more time please? The outside diameter gets larger, so does
the length. What does shear stress do to a material? How does it deform the material? What kind of deformation does it produce? It changes the shape? Change of shape or angular deformation, right? Think of it this way. We draw a little element here. And after we pressurize it, we take a look
at the angles of that element. And they will be 90 degrees, right? We can see that that’s what’s going to happen. Just like a balloon that you blow up. If you can think of a square with curved sides
on a balloon, it’s going to remain a square with curved sides when you blow it up. It’s just going to be larger. That’s all. Angles will not change. So no shear stress. That’s all we have, okay? So tangential and normal — oh sorry, tangential
and axial stresses and these are the equations. Your book brings in some maximum value of
this and let me show — yes, in order for me to be correct, I should do this too, right? This is a thin-walled vessel so it can’t be
. Your book brings in some maximum value of the stress and this is what they mean by that. If I look at this cross-section and let’s
put it down, I mean, let’s assume that this is thin walled. The way we find the stresses in reality is
that the tangential stress here is maximum and it varies like that. It goes to the outside and it becomes minimum. Not zero, you know. Anybody understands what I’ve drawn here? Everybody okay with that? If not, please tell me because once you tell
me that you understand these then it becomes closer for a quiz. The variation between this maximum and minimum,
if the wall thickness is small, it’s very, very small. So we essentially assume that there is no
variation and we use these equations and assume there’s a constant. In other words, we assume the stress distribution
is like this. A little bit smaller than the larger one,
a little bit larger than the smaller one, the average value of this stress. Your book actually calculates this value by
bringing in the T, the thickness and all of that. I don’t think that’s necessary because if
you want to treat it like that then you shouldn’t treat it as a thin-walled cylinder. Treat it as a thick-walled cylinder and as
we will see very shortly, you can get very good values from the equations of thick-walled
cylinders. So that’s thin-walled cylinders. Any questions? All right, let’s take a break and when you
come back, I will take a look at thick-walled cylinders. Okay. Now we take a look at thick-walled vessels
and pipes. Again, the only difference is that in one
case, you will have an axial stress in vessels. The length of the vessel will change. The other one, you will not. The length of the vessel does not change. Save for, of course, the effects of Poisson’s
ratio which for the time being, we’ll neglect it. A pressure vessel such you would see on the
screen in cross-section can be pressurized both from the inside with a pressure of PI
and from the outside with a pressure of PO. For example, if you have a pipe that has fluid
running through it and the pipe is housed in another vessel which is pressurized then
you have both internal pressure and external pressures. So this is the most — —
general type of pressurizing a vessel or a pipe. The stresses that are set up here, very similar
to the thin-walled pressure vessels and actually for thin-walled let me say something else
that, of course, we are neglecting that’s why I didn’t refer to it. But we’re not going to neglect it here. So the hoop stress is like so. The tensile stress goes into the board, right? Oh sorry, not the tensile. It is tensile but the axial. The axial stress goes into the board. Is there a stress in this direction? In thin-walled pressure vessels or thick-walled
for that matter, ? There is. What is it equal to? The pressure, that’s right. Remember pressure is force divided by the
area so it’s the stress. Yes, there is a pressure here that creates
what is called a radial stress. It’s in the direction of the radius. Everybody follows that? Same as what you see there, radial stress,
direction of the radius. But neglecting that stress introduces very,
very small errors in calculations for thin-walled cylinders. Not so for thick-walled cylinders, okay? So for these cylinders, what we are going
to do and what we’re looking at, let me redraw that here. We will take an element such as this. And if you take a look at it — — in the
plane that I’m showing the element in, there are two types of stresses. One acting on this plane, that’s the tangential
stress or the hoop stress. Very similar to thin walled. And the other one, sigma R in the direction
of the radius. Does everybody follow what I have on the board? The picture, does it make sense? The radial stress tends to change the thickness
of the vessel. The hoop stress tends to change the diameter
or the circumference of the vessel. There is, of course, if we have a vessel and
it’s capped, of course, there is a stress in the third direction. So there is a stress on this point as well
into the board so actually, it’s out of the board so I’ll show it like that. It’s a tensile stress. That’s like an arrow coming out. If it were a compressive stress, I would show
it like this, an arrow going in. That’s one way of showing the stress, three-dimensional
stress on a two-dimensional board. These are the two stresses that we will look
at extensively here. And to do that, you can do that, of course,
taking an element like this to finding out how much change there is in its circumference
and from that, to get to the stresses. But this is done using theory of elasticity. And since we are not going to do that here,
we’re just going to give you the results. And these are the results and let me put them
there. I’ll put them over here so I’m going to keep
them for a while. The tangential stress is equal to PI RI squared
minus PO RO squared minus RI squared RO squared times PO minus PI over R squared. Divided by RO squared minus RI squared. The radial stress, PI RI squared minus PO
RO squared — these equations are in your book — plus RI squared RO squared times PO
minus PI over R squared — — all over RO squared minus RI squared. In these equations, PI is the internal pressure. PO is the external pressure. RI is the inner radius. RO is the outer radius. What’s R? These are general equations for the tangential
and radial stress and they apply to all of the points between these two. So R is that. It’s a general radius. So if you want, for example, the stresses
right in the middle of the thickness for whatever reason, R is equal to RI plus RO over two,
average value of R. Is everybody following that? Like an X, it’s like a variable. So whatever point you’re interested in, R
will be radius at that point. If you would like to find the stress on the
inner circle — did it die? Boy, today everything dies. If you would like to find the stresses here,
R is equal to RI. You want it on the outer surface, R equals
RO and anywhere in between. General equations, there is of course, if
you’re talking about a vessel, an axial stress as well. But we generally don’t take that into consideration
unless it’s really specified. The axial stress, we assume it is uniform. In other words, it doesn’t change with R.
It’s the same on the inside, the outside and every other point in between. In other words, when the vessel extends, it
remains a cylinder. It doesn’t go like this. That’s what we mean by that. So if you plot these stresses, actually we
don’t have a plot of this. We have a plot of a special case of this. So any questions on these before I go a step
further? I know I have not proven that this is the
case. I just told you to take that, take my word
for it. It’s actually not my word either. The people who did it. But you can use that and we’re going to use
it today as well in a special case anyway. A special case that is used often in many
of our applications is a case in which the external pressure is zero and there’s only
an internal pressure. Most vessels are like that. Most pipes are like that. So if that’s the case — — for PO equal to zero, we get these rather
simplified equations. Sigma T equals RI squared PI divided by RO
squared minus RI squared times 1 plus RO squared over RI squared. Sigma R equals RI squared PI divided by RO
squared. RO squared over RI squared. And for this special case where we only have
an internal pressure, the average axial stress — — is equal to PI RI squared divided by RO squared
minus RI squared. So these are the stresses. Sorry. This is not RI. This is R. Please correct that. And we need an R here. Can’t do without R. On the other one, there
is no R because that’s an average stress. It’s the same for all Rs. So R doesn’t appear there. Therefore — — if we have a situation like
this, we select an element at the most critical point. That means the point at which the stresses
are maximum. If we plot these stresses, and this is only
for internal pressure, if we plot these two equations, this is what we get. The tangent shows stress, which is a tensile
stress. It’s maximum at the inner surface and decreases
as it goes out. Still remains tension, of course. The radial stress — — is compressive. That’s why it’s plotted under here, under
the line. It’s compressive. But its magnitude is maximum here and it drops
off to zero at the end. Because, remember, sigma R is nothing but
the pressure on the inner and the outer surface. On the inner surface, it’s the internal pressure. On the outer surface, it’s the external pressure. That’s this stress in the radial direction. So if you only have an internal pressure,
that’s sigma R is equal to the internal pressure here and it’s equal to zero there. So if you want that, you can actually come
in here and say I’ll find the radial stress at R equals RI; that’s the internal surface. Put it over here, we’ll find minus PI. And if you want it at the outer surface, you
set it equal to RO and — sorry. In this one. You set it equal to RO and, of course, you
find zero. Because there is no pressure on the outer
surface. So that’s your sigma R. The existence of this
compressive sigma R, this compressive radial stress, does it add to the shear stress or
does it subtract from the shear stress? For example, if I compare this point to that
point, the inner point, the outer point. Which one do you think will give me the largest
shear stress even if there is no — even if that doesn’t decrease? Even if it stays constant, which it does not,
of course, but even if it stays constant, which one of those two — and assume it stays
constant, which one of those two would give me the maximum shear stress? The inside surface or the outside surface? Sorry? Well, let’s take a look. What do we have on the outer surface? Again, more circle. Remember, you got to do more circle in many
cases to get to these. What do we have on the outer surface? We have a sigma T and nothing else. Sigma RSU. So your maximum shear stress is that. Correct? Everybody with me on this? You understand the more circle and how I draw
it? Now, inner surface. And let’s assume this remains the same so
we can compare things correctly. So that’s sigma T again, and then I have something
over here. Right? Sigma R is compressive. Now, the maximum principal stress — sorry,
the minimum principal stress is not zero. It’s some negative number and therefore, I
get a higher shear stress. Therefore, the most critical location is on
the inner surface of this vessel for maximum shear stress, for yielding, that is. And remember that that’s what we calculate
for yielding. So this is the most critical point and most
of our calculations are done there. And these are the equations that govern the
variation of the stresses. Any questions? One application of thick walled pressure vessels
is in press and shrink fits. Do you know what press and shrink fits are? And — — what the difference is between the
two? No? Okay. Good. We’ll go over it. Doesn’t take very long. It’s very simple. You can understand it very easily. Press and shrink fits, or shrink fits is a
way of connecting two parts that are transmitting torque. For example, if you have a shaft and a gear
is sitting on that shaft, the gear load produces a torque. That torque needs to be transferred to the
shaft, which rotates the shaft. You somehow have to connect the gear to the
shaft. Of course, you can use your setscrew. You can locate it by a shoulder under setscrew. You can do a keyhole and a key way — a key
and a keyhole, rather. And you can also do it by press and shrink
fit. What we do in the two, and here’s the definition
of shrink fit. In order to put the two parts together, and
now we call these collar and shaft, although they may be hubs of gears and whatnot. We call this the collar and that’s the shaft. In this case, a hollow shaft. Why would you like to have a hollow shaft
anyway? Sorry? It is lighter and, at the same time, you’re
taking materials out of locations of minimal stress. Right? Center of the shaft minimal stress. And that’s true, that’s one of the reasons
why. But most likely, you’re going to have to spend
a lot more money making it a hollow shaft than you’re going to gain by not paying for
something as cheap as steel. Now, if you’re talking titanium, that’s a
little bit different. So what would be another reason? Most of these shafts heat up in places where
they work. So if you now are able to pass a fluid through
this, to cool the shaft, that would be one reason you would want a hollow shaft. They’re not very common. You don’t see them very often, but those are
some of the reason. But the reason for the stress, perfectly valid,
and you do save weight. No question about that. In other words, if you put the economy aside,
that’s a perfect answer. So in order to connect the two, here’s what
you do. You make the outside diameter of the shaft
a little bit larger than the inside diameter of the collar. Then, you heat the collar or cool the shaft,
or both, until the shaft fits inside of the collar relatively easily, like a running fit. You’ve had fits in 233. Right? Fits and tolerances? No? Yes. All of you have had it. A running fit is basically what it means. You put the shaft inside of the collar and
it easily moves within the collar. So running fit. So you heated up the collar, cooled the shaft,
or both, and put the shaft inside of the collar. Now, you let them come back to room temperature. The shaft would like to grow to its original
size. The collar would like to shrink to its original
size. Neither will be able to do that and they end
up somewhere in the middle. And by doing so, a pressure is set up between
the two parts, which connects the two parts. That’s called a shrink fit. A press fit is very similar. The result is very similar, but the procedure
is like this. Put a chamfer on the shaft. Your collar remains the same. So the tip of the shaft fits inside the collar. You press it down. You have to overcome the friction between
the two parts. That’s called a press fit. But the result is the same. The two parts are connected, pressure is set
up between the two parts, the shaft will be in compression on the outside. The collar will be pressurized from the inside
like this and that is one of the applications of thick-walled pressure vessels. If we can find, knowing, for example, what
the difference is here, that difference — I’ll draw another picture, much better than this,
hopefully. We know what that difference is. That’s a radial — what is called radial interference. We know what that is and using that, we can
come up with the pressure that’s set up inside or at the interface between the shaft and
the collar. The, we can go in here and analyze the collar
as though it just had a pressure inside of it using these equations. So the task, therefore, is to find a relationship
between the interference, in this case the radial interference, and the pressure that’s
set up between the two parts if we shrink or press fit. Everybody understands what we’re supposed
to do? Right? To come up with this — Now, your book just
outright gives you these equations and it says, here, use them. You may like that. I don’t. So this is in your handout. So let me draw another picture here, or a
bigger picture. I should have done in the first place anyway. That’s the center line of the shaft and the
shaft could be hollow or solid, it doesn’t matter. That’s called delta. The radial interference, the difference between
the outer radius of the inner member, the shaft, and the inner radius of the outer member,
the collar. Okay? So once press or shrink fitted, the shaft
would like to come back to here. The collar would like to come back to here
because it’s now been pressed up somewhat or moved somewhat because of, say, temperature. Neither will be able to do that. They end up somewhere in the middle. This is very similar to the indeterminate
problems that you had in strength. So let us assume that eventually, when equilibrium
comes about, this is where they end up. Does that make sense? Shaft cannot go back to its original size. Collar cannot go back to its original size. So they compromise. Then, the collar will have grown by that amount;
we call it delta O. That’s the radial change in the collar. And the shaft will have decreased its — —
radius by delta I. Anything you don’t understand, please tell
me. Therefore —
We design — when we design the interference, we actually come up with this delta. What’s should delta B? I’ll do an example problem. As you can see, the magnitude of delta is
equal to the magnitude of delta I plus the magnitude of delta O. Never mind that one
is an increase and the other one is decrease, magnitude. Therefore, that top equation. Okay? Now, we say the strain in the tangential direction
in the outer member — the outer member is the collar. The strain in the tangential member in the
collar at the interface, because all of this is happening at the interface, the interface
is called R. The radius of the interface, where they end up at the end, is capital R.
So let me go back to this, to clarify all of this, and then we’ll come back. RI — — is the inner radius of the shaft. RO is the outer radius of the collar. So if you have a solid shaft, RI would be
zero. It’s solid. R is the radius of the interface, where they
end up at the end. So an easy question that I’m about to write,
this is the meaning of the various terms. Well, actually, I won’t write them. I’ll go through them here. So the strain — the tangential strain in
the outer member — and remember, we have the state of stress such as this, tangential
stress, that which changes the circumference. It’s sigma T over E minus nu sigma R over
E. Agreed? That’s what you have here. Well, actually, no. I’m one step ahead of myself. Let’s first do the geometry. The tangential strain at the interface for
the outer member is the final circumference minus the initial circumference divided by
the initial circumference. Agreed? Meaning of strain. The final circumference is calculated on the
basis of R plus delta O. Actually, I’ve shown this — I mean, show R to here at the original
interface. I should show it to here. R plus delta O. The collar has grown that much from its original
dimension, whatever it was. So the circumference is 2Pi-r. The original circumference was just 2Pi-R
divided by the original circumference gives you delta over R. That’s the strength. Therefore, the change in that radius delta
O for the outer member is equal to R times epsilon 2 from that equation. Now we know epsilon 2 is equal to sigma 2
divided by U minus nu times sigma R divided by U. This is your stress-frame relationship in
two dimensions. And we substitute here for — in this equation
for epsilon 2 to come up with this equation. So you’re wondering where all of this comes
from, the same time when we substituted that for this, you also substitute these stresses
for sigma T and sigma R. This is from — for the outer member, outer member is under an
internal pressure, and those are the stresses for an internal pressure. So you put the stresses here from there. You put all of this over here, and you solve
for delta O. And that’s what you get. Any questions? Is there anything you don’t understand? There’s two or three steps here, but this
is something you should understand. So we figured out what the strain was, from
which we calculated the change of radius. Then substituted for the strain in terms of
the stresses, and substituted for the stresses in terms of the pressure, because that’s what
we wanted to do. We wanted to find a relationship between the
change of radius and the pressure. There’s the pressure. But we’ve only done that for the outer member. This is not all of the interference, only
part of the interference. We’ll go through a similar process, exactly
the same thing this time for the shaft. We say what’s the strain in the shaft, do
exactly the same thing, and come up with this equation. And notice that in here we are using different
— or different nomenclature EO and nu O for the outer member, EI and nu I for the inner
member, just in case they’re not made of the same material, two different ones. And then according to that equation, we add
the two magnitudes to come up with delta. So there is the equation of delta as a function
of pressure, or vice versa. If you have delta you can find the pressure. If you have the pressure you can find delta,
whichever you like. And notice that there is nothing in there
that we don’t know. Material property, interface radius, outer
radius of the outer member, interface radius, Poisson’s ratio. Same thing over here. If this — if the two materials are the same,
steel collar, steel shaft, then the nu O, EO, and all of those will be the same. Will be E and nu, and then those actually
will cancel out, and you get a simplified equation such as that. That’s if the collar or the shaft are made
of the same material. Any questions? So given some sort of a torque, you should
be able to come up with the required interference. Or given the interference, you should be able
to come up with the pressure, and therefore, the stresses, and check for factor of safety,
and so on and so forth, okay. Any questions? One more thing here briefly, that’s rotating
rings or discs as the case may be. If this is a rotating ring for a disk, due
to the rotation, there will be changes in the dimensions of this, right? Just think of a very, very soft material that’s
rotating. Dimensions are going to change. Will there be any stresses in this if it rotates? Okay, think of it this way, you have a bungee
cord in your hand. You put a weight at the end of the bungee
cord, and you rotate the bungee cord. What happens to the length of the bungee cord? It increases. Same thing here. It’s just not a bungee cord. Due to normal component of acceleration, forces
are set up that will change the circumference of the disk as well as its thickness. If I take an element such as this here and
want to draw the stresses on this element, what would be the difference between the stresses
here and the stresses there? That for — vasopressurized on the inside. This for the rotating ring. What happens to this thickness? Will it increase or will it decrease? So you think it won’t change? Okay, let me ask you a different question. Maybe you have heard this. It doesn’t have to be a circle, either. There is a hole here. Could be a square. We heat this. What happens to it? What happens to the hole? Will it get larger or will it get smaller? It will get larger. All dimensions increase. Now you may be thinking, “Well, this thing
grows in all directions, so it’s going to grow into this hole.” Well, no, that’s not the way it is. The hole actually grows. Same thing in here, except that it’s not pressure. What you are doing is as they — you’re stretching
this all over, meaning there will be stretching of the circumference as well as the stretching
of the thickness. So if you draw the element, that’s the difference. Sigma R is tensile. It is not compressive. In that case, sigma R is compressive. That reduces the thickness. This increases the thickness. The equations for the stresses are in your
book. I believe they’re on page 129, long and drawn
out. I’m not going to write them. And as you would expect, they are a function
of the normal component of acceleration, rho omega squared. Yes, page 129, equations 355. And you should just read those and make sure
you understand what it says. Of course, you should not even try to memorize
these equations. Okay, any questions before we do example problems? All right, let’s do problem 573. Even though this is chapter three, the applications
I’m going to do a problem from five — chapter five so that we apply failure theories as
well, like I said . So 573. This is a solid steel shaft, has a with ASTM
20 cast iron — — whose E equals 14.5 MSI, shrink fitted to
it. The shaft diameter is — — 2.001 plus or
minus 0.0004 inches. And the hub — —
is 2.000 plus 0.0004, and minus 0.0000. With ID the specification for the gear hub
R, 2-point — inches internal diameter. This is ID. OD for the hub, 4 plus or minus 1/32. Yeah, 4.00. All in inches. Using the midrange values and the modified
Mohr theory, estimate the factor of safety guarding against fracture in the gear hub
due to the shrinkage. One point regarding what was just said is
they say use the midrange values. In other words, I would take this at 2.001
plus 0.004 minus 0.004, so it would just be 2.001. That’s the midrange value. In reality you should not be using midrange
values, but that’s what the problem says. For this one would be 2.000 plus .0004 and
minus .0000. And it says using midrange values, find what
the — Sorry. Let’s read that over here. Ah, modified Mohr theory, estimate the factor
of safety, very good. So we have this. Midrange value is 2.001. That’s for the diameter of the shaft. That is somewhat larger than the diameter
of the collar, so this one would be minus — actually I should write this as interference
is equal to minus 2.0002 divided by 2, which will give you 0.0004 inches. Midrange value of the diameter of the shaft,
midrange value of the collar. The difference between the two divided by
2 gives you the radial interference. They have everything in terms of radial, so
we do this in terms of radial as well, radial interference. Then we have delta. Actually, before we go there, is it good to
take midrange values? If you really want to look for failure you
should look at the worst case, not the middle case. You should look at the case where the shaft
is the largest it can be and the collar is the smallest that it can be, because that
will give you the maximum interference. But they say to use these, so we use these. If we now use that in the equation for delta,
we can come up with P. So I’ll just leave it up to you to show that this is equal to
2613 PSI from that equation. The — this is for the collar. The shaft is steel. Shaft, steel. And for it use E equals 30 MSI. And nu equals 0.292. For the collar, 14.5 MSI, nu equals 0.211. Put them into that second equation from the
bottom, solve for P. You should actually program these on your calculators. You don’t want to do this every time you want
to solve a problem. Program it and put the numbers in. So sigma T, over there, at R equal to RI,
most critical point. We’ll get equal to 4355 PSI. Sigma R, of course, equals minus P. Once you
have those two you can easily solve for the factor of safety. Again, I’ll let you do that. Factor of safety comes out to be 5.05, according
to the modified — do they want modified Mohr? Modified Mohr, yes. Any questions? Okay, one more example problem. Everybody is clear on what you should do to
get — I mean, you know what to do — how to get to this final answer? It’s exactly the same as the previous problems
we did. Once you find these two, don’t forget that
this is compression. And I should say something about the value
of R, capital R. What do I substitute for capital R? Remember I had a sketch there, R was the initial
radius plus delta O, or some such thing, or minus delta I and all of that? But remember that these are only what, .0004? So for R what do I substitute? I’ll just substitute the radius of the shaft,
period. Of course, you can go ahead and substitute
radius plus .0044. So if the radius is 2 or 1, you get 1.00004. And figure out how much difference that makes
in your calculations. And the answer is none, because we generally
don’t go more than three significant figures anyway. So don’t try to figure out should I use this
plus plus .004 minus .0 — it’s the interface radius. It’s the nominal radius of the shaft or the
collar before you do any shrink of . That’s the interface radius, okay? All right, let’s take a quick look at this
problem and call it a day. This is in your handout again. And this is the type of problem that you should
be able to do. Nobody gives you the stresses outright or
the interference. So this problem says — and you can read it
in your handouts — the collar one inch thick and two inches long is to be designed for
a 1018 cold drawn steel shaft to transmit a torque of 10,000 foot pounds. The collar and the four-inch shaft are made
of the same material, and the coefficient of friction between them is .4. Calculate the precise dimensions of the collar. Make all necessary checks. Minimum factor of safety, 1.5. And they give you E, nu, and the yield strength
of the material, and RO, R, and RI are also given. And if they were not given you could probably
decipher it from the statement of the problem. First thing you have to do is convert this
business of torque being transmitted into something that we can work with. In other words, the pressure. The torque comes from stresses or forces called
tangential. So let me draw the shaft here. This is what I mean by FT, friction forces
between the shaft and the collar generate the torque. Where do friction forces come from? They come from normal forces. So that’s what I mean by F. And all of these
combined, so the tangential force is the torque multiplied by 12 to get it in inches, divided
by the radius, FT times R, 60,000 pounds. The normal force is that divided by the coefficient
of friction. FT equals mu FN. So 150,000 pounds. Pressure, therefore, is 150,000 divided by
the area on which it acts. The area is the area of the shaft. It’s this area [tapping]. Agreed? That’s where the pressure acts. Divided by that area, that you get the actual
pressure, divide the force by . Then use the equation for delta that I just had up on the
board. Everything is given here. You’ve just found P. You find the radial interference. Multiply it by 2. You get the diameter interference. And therefore, the interference should be
approximately 3,000ths of an inch. Now you have — this — the problem would
be over if you didn’t have to check anything else. But you’ve got to make sure that the collar
can take all of this pressure and not yield. So sigma T, sigma R — these are exactly the
same equations we just used. And they’re all calculated at the interface. That’s where the stresses are maximum. So this is all for the interface. When we have those two, then we can find them
— I’ve given you maximum normal stress theory to see how far off it is. Maximum normal stress yields strength divided
by the maximum stress, 3 1/2. Maximum shear stress requires the value of
tau max, so there’s tau max. Sigma X minus sigma min divided by 2, 10.7. And the maximum shear stress theory gives
you 2 1/2, so this is okay, because we needed 1 1/2. Maximum distortion energy gives you 2.8, and
that we’ve done before. So that’s a little bit larger than that, and
that’s what we would expect. You would expect it to be a little bit larger
or the same, a maximum the same. A couple of things we have neglected here,
one is the actual shear stresses that transferred the torque. We said there are no shear stresses, so we
have neglected that. and at the edge of the collar and the shaft
there will be some stress concentration because of the collar pressing on the shaft. We have neglected that as well. So take a look at this. If you have any issues with it — this is
the type of problem that you should be very comfortable with. So if you have any issues with this, please
let me know first thing next time. I’ll see you on Thursday.

8 comments on “Stress Analysis: Thick Walled Pressure Vessels, Press & Shrink Fits (4 of 17)

  1. change your Mood Post author

    Sir , What it means to Positive and negative shear stress , plz provide all possible cases , both in words and symbolic pattern . thank you for this amazing lecture sharing .

    Reply

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