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.

jouheyna hedhliPost authorthe name of the book used please

Stevo BoePost author47:00 is where the analysis of the thick walled pressure vessels start.

The Rabbit Was Always Beneath My HatPost authorWhat is the R

Amanda BralickPost authorThis is awesome; it helps so much

ahmed attiaPost authorCan someone please mention the name of the book.

jobless indianPost authorI want derivation of lames theorem

Marwan ٍBa HurmuzPost authorThin Walled at 37:00

Thick Walled at 47:22

change your MoodPost authorSir , 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 .