Strength of Materials II: Strain Transformation, Mohr’s Circle for Strain (14 of 19)


Still I have problem with ma. Ma is as important
or more important than ra. Ra is a reaction, ma is a reaction here. You have also here,
I don’t know what I call there b, so you have an e also rb and mb, that’s four unknown and
therefore, you have two extra unknown here. Not one extra unknown. It was in your homework,
previous homework in single directive method and in superposition method. I have given
you this problem which was a fixed support at both and then there are lots of funny things
that some of you are going to do and my guess is, many of you understand the whole scenario
correctly. You make occasional error. You put– you lose one or two or three points.
But there are some also student which makes static error which is not supposed to be at
this level. So I explain as we go along. So first of all, you start writing your ei. First
of all, this is a differential equation. So when you write ei– y over dx squared equal
to m as a function of x. Any function can be given here. However, we have discussed
that. You have ra and you have ma at the beginning, you should start from there. Now, sign, many
of you in this class, especially not the other class, you have difficulty with this sign.
Everybody sees the moment of that part is negative, they put a negative here. That I
do not buy it, that is wrong, that is absolutely wrong. We are not talking about the static
moment. We are talking about the internal moment, the reaction of the beam to the action.
Everything that you learn in MA218 for the– I don’t know this is the 10th time I’m repeating
there, some of you are paying attention, this is only for benefit of those that still have
a little problem with MA218. Everything that we did in MA218 when you write m sigma equal
to mc over i, that m is not external moment, it is internal moment. Everybody understand
that that you have to go to the body. I don’t care what you have here, you have– they have
r here, m here, load here et cetera. You go to that section. There is only one n. There
is only one shear and there is only one moment. Everybody understand what I’ve said? That
defines all of your stresses, is that correct or not? Therefore, we are talking about this
m, not that m, is that correct? Therefore if this goes this way, this must be going
that way, yes or no? We go through balancing. Therefore, this system is a positive system
or negative system? Negative, or else, you can do it like that which I have done it many,
many times, the action is negative. The reaction would be, how many times have you seen me
doing that? This has not apparently registered for some people, at least about 10 or 15 people
put this wrong. So it is ra y is positive, the action is negative. The reaction is the
reaction of the beam to that is a positive. So this become ra x minus zero to the power
of 1, 1 because the moment is linear. Is that correct or not? With the same reasoning, ma
is going positive. The reaction to it is negative. If you draw it that way. Some of you draw
it on the other direction. So that gives you a positive mode, everybody– but that’s OK.
If you draw it that way, this should be minus ma. If you draw it opposite to that, then
it should be plus. Everybody understand it, yes? This system, action and reaction. This
is negative, we talked about it many times. This action and that’s reaction. Now, this
is this blue one. This is inside the beam. Is that correct or not? Yes? Anyhow, I think
that everybody by now should recognize that, but I saw that and that was surprising. X
minus zero to the power of zero and nothing else happening until at 1, which is a load
constant load. Shear is linear, moment is a parabola. Then you put again, the action
is? Look at the action, if I put the loads here, the action is positive, the reaction
of beam to it is, negative. So minus 12 x minus 1 to the power– that’s 1b there is
not– to the power of 2 over 2, is that correct or not? Yes, since this goes all the way down,
this is what I don’t like. If you miss that one, you miss probably 6 point or 5 point
or 6 point or 7 point, I don’t know because you cannot miss that, I gave you the 10, 12
homework. Six of them has this in it. Is that– how can you miss that, is that correct or
not? Then what? Then this goes all the way to the end then I have, this is because when
I wrote here, it goes all the way to the end, is that correct or not? Then I have to subtract
this much of it which is plus 12x minus y. Where does it start? X minus 3 to the power
of 2 over 2, is that correct or not? That is your moment equation for entire beam, correct?
Then the rest is automatic, I don’t care how you do that because this is very simple, then
you said ei, dy with the x or theta equal to ra x minus zero to the power of 2 over
2 minus mx minus zero. By the way, you can drop this zero because zero has no effect
on your quantity. So minus 12– but here, you cannot, but these are not parenthesis.
These are– this is another cardinal mistake that you or you make in this class. These
are not parenthesis, these are? Bracket, bracket has a special format which we discussed in
class with you guys. X minus 3 to the power of 3 over 6. If you miss that one, then the
whole system collapses. So that’s why I don’t understand how you did your homework, some
of you because this was very much simpler than your homework. Ra x minus zero to the
power of 3 over 6, minus ma x minus zero to the power of 2 over 2 minus 12x minus 2 to
the power of 4 over 24 plus 12x minus 3 to the power 4 over 24 plus c1x plus c2. Now
here come another point which is very interesting, many of you have seen in the homework that
I did on the class on the board. Many times I put c1 and c2 equal to zero, is that always
the case? Of course, not. This is the reason for it and that reason are these two, is that
correct or not? Since I have a moment there, therefore that means it’s not rotating, it’s
like that. Therefore add x equal to zero, theta equal to zero or you can say theta a
equal to– because this is for the theta, for a equal to zero. Therefore, that gives
you zero, zero. This is– some of you calculate that minus 1 to the power of 3, I warned you
about. If you do that, you haven’t heard anything about the singularity method, everybody understand
that. So this is negative, therefore negative must be equal to zero, this is negative zero,
therefore c1 become equal to zero. And similarly, add x equal to zero, y equal to zero but fortunately,
many of you knew what you are doing. This is zero, this is zero. Again, this is negative,
therefore zero, this is negative therefore zero, this is zero c2 equal to zero. And that
is– that’s the– what my objection come to your static system. You use this and you get
c1 or c2 equal to zero, yes or no? What is this? No, what is that? Is it the fixed support?
If you understand what you did at a you should do it for b as well, yes or no? What kind
of class don’t know how to do it? Why, probably they see me doing this they have a recipe
in mind, everybody understand what I’m saying that they follow– OK, you did that, there
was a fixed support, you put c1 equal to zero. See, I’ll out c1 equal to zero, c2 equal to
zero, if that’s the case and you understood it, there is no doubt that you should use
it for b as well, yes or no? Because at b also, because that’s a fixed support, yes
or no? And b also theta equal to zero and y equal to zero. Some of you y equal to zero,
some of you– you can use it, I told you can solve it with that one. So far we got rid
of c1 and c2. But don’t forget that we have four more or no less than two equation of
equilibrium, two equation of equilibrium is ra plus rb equal to the load or sigma ma or
sigma mb equal to zero, yes or no? I need two more equation. Where are that coming from?
From two other boundary condition and two other boundary condition, this is the key
actually equation, those two was in some of your homework as well. Don’t you have an rb
here which prevent this from going up and down, yes or no? Don’t you have here an mb
there which prevent this from the rotation. That is that the key to this problem, then
these are the one I’m looking for. This is the one you get. This, you get the point,
this you don’t. And this here, which many of you missed, an x equal to 5, k2 equal to?
Zero, which gives me the following equation. You go there and put x equals to 5, so you
get this equation. Anyhow, ra5 to the power of 2 over 2 minus ma times 5 minus 2, 5 minus
1 it become 4 to the power of 3 minus 2, 5 minus 3, it become 2 to the power of 3. And
you simplify that equal to zero or you get the way I have it here. You had a different
rate, 12 one half ra or 25 ra if you wish, minus– let’s do it this way, 25ra minus 10ma,
equal to 224. So that’s equation one or any version of that is correct. Then next equation
I need actually, I don’t need to use any equilibrium equation. This was simpler than what you’ve
thought. So, an x equal to 5 meter, also y equal to? Zero. Zero. Then I use this equation, this become
two equations for? Two unknown. Two unknown. So therefore, I go there, so
I put this second equation. Let’s make it short. Anyway the second equation, I give
you the– as when you put this one into the last equation, again, this become 5542 and
just simplify that one. That become 125ra minus 75ma, become equal to 720 or some of
you divide that by 6 or whatever, that’s OK too. Usually I try to make it round number
because it’s easier. Is that correct or not? Never divide it by 6 or 3 because you get
all decimal points. Is that correct or not? As you see, all I have to do, multiply this
by 5. Is that correct or not? And let’s subtract this two together, this becomes 50 and then
subtract it, and 75 minus– it become really simple. So, if you calculate that, ma become
equal to 60 kilometer and ra become equal to 15.4 kilometer. And then in order to calculate
rb and mb, what should I do? I should use the equation of? Equilibrium. Equilibrium. Everybody understand that. Yes
or no? The– And the funny part is about 10 people in the class and I ask you to calculate
reaction, you write ra and rb but you didn’t write ma and mb. And I said, please, have
some respect for m because m is as important as ra’s. Still, this is the funny part, it’s
still all of that– all of the static, still people do not recognize the moment reaction
as part of a deal. And that’s why some of you have a little bit of problem of static
there. Is that correct or not? The moments are there, the moment of the reaction. Anyhow,
you write sigma here– but ra is actually equal to 15.4 and the total load is 24 so
the balance of it goes to rb. Is that correct or not? Yes? So, anyhow, rb become equal to
8.6 and then you take moment about here, you have to take ma into account, mb into account.
So mb also become equal 11, I suppose, 11 kilometer and meter. So that problem that
result that they– anyhow, still some– anybody have not got there Joshua, Cesar Moreno [assumed
spelling], Cesar? OK. Nick, Nick Kate [assumed spelling], yeah, OK. Josh Fa [assumed spelling],
Josh? No? Nick? OK. Joshua, Gary? OK. And Brian Rose [assumed spelling]. They’re missing.
Anyhow, let’s continue with the discussion. We have lots of work to do because they have
to finish this chapter. We are about one class behind, right, remember? Go back to the–
your notes please. Everybody put notes aside. Go back into your note. Look at the state
of strain that I gave you. Please put this aside everybody. Go to the previous page in
your note because I ran out of time that’s why I have to repeat and I don’t want to repeat.
Look at epsilon x. Where was epsilon x? When I started the practice drawing the more circle
where a strain. What was epsilon x? Epsilon x was? Y. What was epsilon y? 110. 110? What was gamma? 180. 180. Now, remember, in the stress we draw
a square or cubic and we show the sigma. But– in strain are not like that. This is representation
of a strain. This is original. You see, this is what I said that result of– look at it.
I said the result of what? Strain transformation of class exercise, that’s what we did. Yes
or no? Last time, what was the strain I started with? This one, look at this one, how much
is that? Class 34. See, this length which was one, after a strain become one, 340 micro
which is 1.00034. Everyday understand what I’m saying that. Yes? What is this? What’s
epsilon y? Hundred ten, if it was one unit because epsilon is the amount of the formation
pair one unit of length. Yes or no? So one become one plus epsilon. Yes or no? So that
is 110. What is this angle in change of this angle because this was originally 90 degree.
Now, is 90 minus gamma which is a positive scenario I gave it to you last time which
is how much? Hundred eighty. Yes or no? This is original state of those command, show me
like that. I have seen people look like that. They put here something like this. And they
put here epsilon x as if this is sigma. This is not sigma. Everybody understand that this
is change in the length. Is that correct or not? You can guess it looks like that but
you cannot put it like that. However, this I would not ask you to do many times but I
want you to realize this is what’s going on. Is that correct or not? Yes? So then, what
did we get for? What did we get for the principal strain? Go back again. What did we work for
principal strain? Come on guys. Take a look. Well how much epsilon a and epsilon b was?
Look at this one. Look at how many degree? Remember how many degree did we have? What’s
theta p equal to? Nineteen degree, yes or no? You forgot. Theta p was? Look at this,
theta p was? Nineteen degrees. So, we shift it– this is what’s positive, counterclockwise.
So, we went from x, 19. Look at 2 theta p was 38 degree. It is in your note. So, we
went up 19 degree and this is one principal strain and this is one– there is no change
in angular. So, what was epsilon a? Epsilon, was a positive, 370. It means it is expansion.
Yes or no? Yes. And epsilon y, what was epsilon b, epsilon maximum, now epsilon minimum. Epsilon
minimum was 80. Yes or no? If it is minus, then you make it smaller. Yes or no? Because
minus means compression, plus means tension. This is the state of stress, principal stress.
And principal stress, I’m sorry. Yes or no? What do you think this is? This is maximum
gamma, in plane gamma because I have to add to this either 40 by 5 degree up or 45 degree
from here down. Is that correct or not? Which we think it’s here, 64 degree. Look at it.
Epsilon average, look at your– Epsilon average was 225for both x and y. Everybody following
me? And then gamma was how much? Gamma max was in plane, was 225. So, this is original
state of a strain. This is principal strain. This is maximum, and what is this? You can
read it, exactly. What is that that I didn’t do? But I asked you to do. See, it was a question
but I run out of time. What was that? Thirty degrees. Rotation of? Thirty. Thirty degree here which on the more circle,
you have to go 60 degree. You take your x and y wherever it is, you go 60 degree. We
have done that in the past so I don’t have to do it. Is that correct or not? Yes? So,
this is what we do. This is the presentation. I do not usually ask you to do it but it’s
good for you to understand it because this is essential for you to see how this works,
because some of your homework at least gives you some idea. If you cannot plot it, that’s
OK. I do not expect you to do it. However, this is the format of it. Is that correct
or not? Yes? Now, how do we– Now we got to the point now. I want to leave it at that
and I would– will go to what we are really doing. Here is the following. So, let’s go
back a few pages. Go to the second page of the hand out. The first page is the outline.
We already talk about it. Let’s go back here. You see this is what we are doing in general.
We talk about– this is MA 218. We talk about this frame and gammas and plane stress which
is more circle. Stress transformation which was chapter 5. Now we are. We are doing exactly
the same but this time we are doing with the strain. Is that correct or not? Now, what
is happening is the following. Go to the next page. Here we go. These are– Really this
is a chart. This is my chart. So you can use it anyway you want to. It’s not– I haven’t
seen it in any book. However, you can– these are these, call it design process. So put
it here in design, put here design process. Of course, you cannot write it if it is in
your computer. I have an extra copy there, you can take it. Oh you can write it? Yeah. OK, good. OK, this is what we did in MA 218.
We– When we went through this, for all the chapter, we were able to calculate sigma x,
sigma y and gamma xy, yes or no? At the end of MA 218 we were calculating sigma x, sigma
y, gamma xy due to the bending, due to the torsion, et cetera, et cetera. This is the
design process. Then we don’t stop there. What did we do in last chapter in MA 218?
We put it through the Mohr’s circle to end up with what? Sigma maximum, sigma minimum,
principal stresses and principal maximum, shear stress, absolute. I didn’t put absolute
here because that means we have to draw three circle, yes or no? We went through that as
well. Then of course if I need that– epsilon was this, I have to [inaudible]. If I need
the epsilon, I can use the Hooke’s law. Remember Hooke’s law, what’s the Hooke’s law about?
Yeah. Everybody remember Hooke’s law, I put it on the board last time. Epsilon x was equal
to what, sigma x over e minus nu sigma y over e. We don’t have sigma z. Is that correct
or no? Yes? You want to write it again, go ahead write it. Sigma y equal to minus nu
sigma x over e plus sigma y over e, generally when we don’t have sigma z, this is what will
be epsilon x, epsilon y. Of course we have epsilon c or epsilon z and that is minus nu
sigma x over e minus nu sigma y over e. Everybody knows that. So if you are taking a plate and
pulling it and pulling it, the thickness becomes much smaller because of the two negative,
negative. Or if you’re pushing it and pushing it, the thickness become larger. That’s the
handle, that’s the– if you want to practice that, this is MA 218 handout, so you can take
a look there. But you don’t– I’m sure you know that or you must know that. So that’s–
There, we have three equation. Gamma xy equal to what, equal to tau xy divided by a modulus
of rigidity and also the other two that because these are the major parts, so I’m putting
here. So the Hooke’s relation as these– it refers here to this one. This is the Hooke’s
law. You can add sigma z to it if you want to but if it is three-dimensional there are
sigma z over e sigma z that comes from the third term and the other two gamma but usually
isn’t there. Now, this is but– Remember, there is always that relationship between
sigma and epsilon, yes or no? So if I have all the sigma I can calculate all the epsilon.
But if I have all the epsilon, I can calculate all the sigma. There’s no problem there. I’m
sitting in my office. I can calculate that. But the point here in the design process,
we go from here to here. Is that correct or no? Then we make a model of that. Is that
correct or no? And we measure the strength this time. But this time we end up with that?
Epsilon x, y and gamma xy through the measurement. And then we repeat the same process of course
everything should match. Is that correct or no? If the two match together, then our design
is valid design or it is good design. Other words, I have to go modify my design to see
what I miss or use a different method or something is wrong. Is that correct or no? So that design
is no longer good because I’m not– my assumption is not good, maybe. Or I’m missing something
there. Everybody understand that? So the process are parallel as you see that. The design process
is exactly the same. It needed to draw more circle. It is always through this Hooke’s
law which is in– you see I can go back and forth. And that so I have this, I can find
this, I have this then that’s no problem there but this two parallel process to each other,
yes or no? Now, the next question is how do we measure epsilon? So this is what I want
to do, this again now new battery. So how do we come up if I want to start from here,
from here I need to have, we call it gauges. Now, the gauges are shown in this page. So
let’s go back again, this is a gauge. The picture of a gauge, this is it. This is the
picture of the gauge. Everybody see that? The gauge is lots of wire here, you see how
long this wire is, and everybody and goes at the end here some of you have taken the
lab you know about that, our tag. Everybody knows from psychic classes that when there
are wires there and the length of the wire changes that resistant changes, yes or no?
So this gauges have lots of wirings, it’s very tiny wire there. So they are– When you
put it– stick it on the top of material and load it, the length become either longer or
shorter. So the wire– So that measures in a way measured the strain for you. You see
you don’t have to know if they have calibrated ahead of time for you. It connected to the
machine, the machine gives you the epsilon. Is that correct or no? So those are the gauges,
our electrical wire you connect that to the system and you wrote the epsilon. The only
problem is this. That if you want to write it down please, this is the lecture today.
Because you have to use this constantly in all your next setup, homework you have to
use that. I don’t think I need that anymore for the time being. So let’s put this one
off for the time being, see whether we need it later on. So because we need the board.
Now let’s say this is the– your instrument and then you managed to put one strain gauge
here which is in the x direction and manage one strain gauge here with the y direction.
So the a gauge and b gauge give you epsilon x and epsilon y. So I need it– To start with,
I need epsilon x, I need epsilon y and I need gamma xy. Is that correct? In order to start
my process up, strain gauge process, I need epsilon x. OK, I have epsilon x. I need epsilon
y. I read epsilon y. But how about gamma? Gamma was as I showed you earlier that was
the reason I show those picture to you. Gamma is the change in the length. Gamma is change
in the format of the shape, shape of this. So I cannot measure that. Is that correct
or no? How do we do that? You see, how do we do that through the process that I gave
you? Instead of measuring gamma which is impossible to measure, we do that. We put another gauge
here. Let’s say at 45 degree, which we call it– First of all, we need three gauges. In
order to come up with epsilon x, epsilon y and gamma xy, obviously we need three gauges
not one. Is that correct or no? One gives me epsilon x, one gives me epsilon– hopefully
this one gives me gamma. Let’s see how it works. Is that correct or not? Yes? So, they
come– usually they come in three and we call it rosette. So when there are three gauges
together, we call it rosette. So please– we have 45 degree rosette, we have 60 degree
rosette which we call it delta. We have some other kind of rosette or we can now put it
on arbitrary location, I can explain that to you in a minute. Is that correct? But if
we buy this and install it all by system, so I have x, I have y and I have epsilon at
45 degree. But what was the equation you wrote last time? Remember the general equation that
you should– all of you remember? Epsilon x prime was equal to what? This is the key
equation guys which started– we started the lecture a week ago with this one. What’s epsilon
x cosine square theta, you remember that class? Epsilon y times sine plus gamma is? Two gamma and 2. Not 2. But 2 gamma over 2 but we don’t need
the 2 and 2. That was for the purpose of matching it with the stress, is that correct, to use
the same idea. But gamma xy sine theta and? You know, I added 2 and you subtract the 2
to make it– it looks like a stress plane strip. That’s all I did. Yes, is that correct
or no? But this was the equation that came up from the process. So let’s use it for gamma.
So here you say epsilon at 45 degree because that’s where you are measuring, yes or no?
x prime, your x prime is at 45 degree equal to epsilon x cosine of how much? Cosine of
45 which is the square root of 2 over 2 squared, plus epsilon y square root of 2 over 2 square
plus gamma xy square root of 2 over 2, square root of 2 over 2. Those are sine and cosine
of 45. As you see, this ends up to be one half, this ends up to be one half, of course
this 2 ends up to be one half. Therefore, from here I can calculate gamma xy, gamma
xy which is here. This 2 and 2 two goes there become 2 epsilon 45 minus because there are
going to the other side, minus epsilon x plus epsilon y. So through this equation, or through
your choice you have two choices, either use directly this or if you see that, that’s the
result of this everyone but this is for, only 45 degree. If I have a delta rosette which
is 60 degree, I cannot use that for that. Is that correct or no? Yes? Because the–
it’s 45 degree by, right, by anyhow by the way we discuss it. Here is the most common
rosette that you can buy, this is the– almost anybody uses that. Is that correct or no?
You come up with the epsilon x, epsilon y and gamma xy. However, you don’t have to do
that. If by any chance, you cannot do that. As I said, the next one is the 60 degree rosette
which is– looks something like that. So x, so it looks like something like that. But
then I will explain it to you what happen there in a minute. So you don’t have to worry
about it. Those are the detail of it. But let’s say that you don’t– cannot put your
gauges in any particular direction. If I put my first gauge as theta 1 angle, theta 1 angle
could be 25 degree, 35, whatever. I put a gauge at theta 1. I put another gauge at angle
of theta 2. This is gauge a. This is gauge b. And I put another one– I don’t have even
to put it that way. I put that one even gauge c at that direction but remember, this angle
state is like a static. So, this is not theta 3. That angle, always– you always measure
it from the positive x. Remember the static problem that we always– because that– remember
the sign would change if you use the other one. So, theta 1, remember, in general. That’s
why on the graph I also said theta 1, theta 2 and theta 3 or the– at those chart that
I gave you. So, if this is the case, you see I have still three gauges. But I have– Did
the direction of theta 1, theta 2 and I gave it to you. Theta 1, let’s say 32– 30 degree,
theta 2 may be 45 degree, theta 3 may be 120 degree. Everybody understands very well, I
put, it doesn’t matter. However, I can write this equation three times. Everybody understands
that. Yes? For theta equal to theta 1, for theta equal to theta 2 for theta equal to
theta 3 and I measure all that which is on this side. It become three equation for three
because here it was one equation, one unknown because I already had epsilon x and epsilon
y. So routinely, this become like that. So, you write it like that. Epsilon theta 1 which
you measure equal to epsilon x cosine square of theta 1 plus epsilon y sine square of theta
2 plus gamma xy sine theta cosine theta– there’s 1 to cosine theta. So, you put theta
what is given. So, cosine become a number you squared it. So as you see this is one
equation for epsilon x, epsilon y and gamma. Then, you write it again for theta 2. Theta
2 may be 52degree. I don’t care, 60 degrees. So, you write epsilon, theta 2 because you
have measured that remember, a, b, c is measured from your experience. So, you write epsilon
x, cosine square of theta 2 plus epsilon y sine square of theta 2, plus gamma xy, sine
theta 2, cosine theta 2. And you write it one more time for epsilon. Theta 3 as you
see it this become three question for three unknown. Your three unknown is epsilon x,
epsilon y, and gamma. But, usually that’s not the case because most of the time you
try to put your gauges along the x and y as you will see in some of your example. I mean,
this is silly to put it on. Unless you don’t have a way to do it or you want it to put
it at one– zero degree end up to be at 5 degree so you want to adjust your note so
that’s theta 1. Is that correct or no? Yes? You had a question. Somebody had a question,
right. Sine theta– oh, this one is all 1, this one
all 2. So, thank you. This is all theta 1, theta 1, theta 1, theta1, this is theta 2,
theta 2 and then the last one is theta 3, theta 3. So, three equations. I didn’t have
even to write it but I wrote it. Then by that, then we can calculate everything. Now, let’s
go to a couple of example and see whatever we can do. You need this. You need three gauges.
At any direction, we’ll do it. This is the bottom line. And the last thing you want to
connect, you have to connect your knowledge of this MA 218 with the knowledge of strain.
There is nothing new there. Everybody should be able to do it. What, the only problem that
I’ve seen in the past is people are afraid of the strain. But they are not afraid as
much of the stress but they are both parallel. Remember that when you have the stress, you
can’t find the strain. Now, let’s to go through some example one by one. Strain number one
is to go to the next page. You will see that problem. So, you see that problem there. Maybe
I should put that one down again, so OK. So down again. So here, we go the next problem.
Let me see. What happened? It’s not showing anymore. But, that’s OK. I pulled it out so
I cannot show you anymore. Anyhow, you have it on your handout. Look at the three gauges
there. I don’t need the board with this one anymore. So, look at the three gauges there.
I says look at problem number 1 or quiz number 6. Look at that, do everybody see that? Yes,
that’s the one. Everybody should look at this problem with three gauges like that. This
three gauges like that. I do the essential part of it to show you how this work. Already,
you got the message. I’m sure that you know exactly what to do here but let’s say the
three gauges are like that. It is given to you. Is that correct or not? Everybody see
that? It says on the surface of a structure or component. In a space vehicle, the strains
are monitoring and we monitoring the strain informing this in this information to the
computer to check it all the time. And we found out that epsilon a equal to what, 1100
micro. Yes or no? What is epsilon a? Epsilon a is? Epsilon a actually is epsilon x equal to 1100
micro. Yes or no? Correct or not? Yes? Are you looking at your homework? Are you looking
at the quiz number 6? Everybody? All right. So, what’s the epsilon y? So, where is yours? Go get one. There’s one on the table. There
are a couple or more left on the table. We taught that. You are not going to understand
what I’m talking about by writing this number down, guys. Zero. If I were you, I would not
even understand a part of what we are ever talking about. You see, this is not the way
to handle this course. You have to have– I gave you many chances to have a copy of
this one. Anyhow, epsilon a is epsilon x is 1100. Is it epsilon y is given? Yes. Yes. Epsilon y equal to how much? 200. Epsilon y is equal to 200 micro. We are looking
for gamma xy. Yes or no? This is not given, instead, what is given? Epsilon in the direction
of this c, is what you see there? This gauge, this gauge and this gauge. Is that correct?
And this angle being 30 degree, so this is a, this is b apparently, this is epsilon b,
this is the b and this is the c gauges. Is that correct or not? So, of course I use that
for x, that’s for I– so, I said it earlier, I used that one for gamma. Yes or no? So,
therefore, I write here. But I cannot use the equation of 45 degree rosette because
this is not a 45 degree. I have two general equation. So, epsilon at this, epsilon c which
is equal to how much? Epsilon c is given equal to this capital C. What’s given equal to also
200 micro, yes or no? Correct? So, 200 micro equal to epsilon x. Epsilon x, look at the
equation. I erase the equation but I should have left it there. Epsilon x, epsilon is
1100 times cosine of 150 square. Yes or no? Yeah. Now, notice here. Here if you put 30 here,
cosine of 30 if it is negative or positive does not make any different because it’s a
square so no changes there. So, if you– is it up 150 you put 30 no problem there. Everybody
understand that? Yes? Plus and minus become squared it become– the same thing for epsilon
y. So, epsilon y which is 200 times sine of 150 square, so let’s write it down. Again,
if this is a negative it becomes square. However, if you make a mistake, there is a big difference
in gamma because 130 and 150, sine of 30 and cosine 30, both are positive. But sine of
150 is positive. Cosine is negative. Everybody understand what I’m saying that? Therefore
if you make a mistake, there you are totally doing a wrong problem. Is that correct or
not? Therefore, definitely will be cosine of 150 degree and sine of 150 degree, and
one of them will be negative. Is that correct or not? Yes. Correct? Yes. So, that’s what I’m referring to. I hope everybody
understand that. This is 30 degrees, sine is positive, cosine is positive 150 degree
which is this number, sine is positive but cosine is negative. Everybody understands
that. So automatic would come into the– your sine. Anyhow, so you must use this angle.
So, this is you theta, this is what I’m saying here. So, please be careful. Anyhow, because
I have seen many mistakes. From here you calculate gamma xy and gamma xy become equal to 1560
micro which is plus. Then it depends what is being asked. This is the same problem that
I did last time. So, I should not do it again. But briefly, I’m going to have to explain
it a few minutes because I have lots of other problems to discuss with you. So, I’m going
to briefly mention that. So, let’s say we are looking for principal strain and maximum
shear strain. Is that correct or not? Principal which are normal strain and maximum shear
strain which is the gamma. Is that correct? I have to plot that Mohr’s circle with the
same sign convention we did before, so this become something like that. So, I’m not going
to do it but I’m going to give you an answer. So, epsilon on average, become equal to 650
micro. First of all, you divide this by 2, so you go like that because you need gamma
over 2. Gamma over 2 is 1560 divided by 2, so that is 780 micro radian. So, that is your
gamma over 2 for your problem. Then you plot it, you plot the circle, et cetera, et cetera.
I can show that what happened there at the end, so this become your circle. So, this
is what is going to happen. So, this become epsilon minimum, and this become epsilon maximum,
and this will be epsilon, this will be gamma over 2 with that arrow, remember that. And
all become equal to– anyway, you use the graphical method or equation doesn’t make
anything but this become 900 micro, and epsilon a or epsilon max become 1550 micro and epsilon
b or epsilon minimum, these are principal boundaries but I call it a and b small because
I don’t want you to make it with this a and b because this was gauges 1, gauges b, this
is principal. Everybody understand. The 2 point that I’m showing here and there which
usually I show it at a and b but that one become minus 200 micro. Now, what happened
there is like that, you go here, you plot this point. Plotting that point is 1100, you
go here, 1100, then you go 780, you have to go 780 down here, so that will be your x.
Then you go 200 here you go 780, so that will be your y. Then you connect these together,
that would be your c, this would be your r. This is 900 divided by 2, so this is 450,
this is 450. And we calculate 450 and 1100, that gives you the r, et cetera, et cetera.
Wish we have done it many time. Is that correct or not? So, I should not go into that detail.
Now, we go that one. So we calculate that one. Now the last part I want to mention it
one more time because– already. What is the in plane gamma or in plane maximum shear strain? R. It’s not the radius. Gamma over 2 is the radius.
Is that correct or not? So, please write it down. Gamma over 2, x, y plane which is this
plane which is related to x and y, is equal to the radius, correct? The radius was? Nine hundred. Nine hundred, 900 micro. And then, gamma max
in plane x, y, in plane become 1800 micro which is larger than what we started and here
which was 1560. Now, how do we calculate absolute? Look at it, I’m asking for absolute maximum
shear strain. I mentioned that last time. Answer? Is it plane strain or plane stress by the
way? Don’t forget about what I did here. Remember, I said plane stress is common. Plane strain
object must be between? Two digit. Is this surface that I put a gauge
on it, is it between two digit plane or is it open to the surface? Is that– So, sigma
z must be equal to? Zero. Zero. Sigma z must be equal to zero. Epsilon
c must not be zero. Is that correct or not? Because at the surface, look at this, this
is the surface of your car. You are putting a gauge here, your gauge here. The car is
being pressed this way, this way is that car. Is this side getting larger or smaller? Obviously
not. This is a plane of stress, is that correct? We use the idea to plot the Mohr’s circle.
But our real problem always are plane stress unless you put your object between two rigid
body which I told you last time to all your cookies. Is that correct or not? The soft
material between the two harder material or relatively rigid body. Is that correct or
not? We don’t have that often unless you put a plastic material inside a concrete hole
and everybody– they try to push it, which doesn’t have any room for expansion. Everybody
understands in the z [inaudible] and that’s sort of all you get the plate, soft material,
rubber, you might put it between two piece of metal, is that relative to that metal,
it can be considered rigid. So, therefore, there is no epsilon. You can’t push it this
way, push it this way. It wants to go up. You don’t let it go up, epsilon is zero but
sigma z is not zero, but that’s very not common. Is that correct? If I’ve seen the surface
of the shuttle there it must be plane of stress. Is that correct or not? Plane of stress mean
what? Means that sigma z equals to zero but epsilon z is not equal to zero, is that correct
or not? Yes. What was epsilon z? I gave it to you last time. Epsilon z was equal to?
I calculated it for you. Do you remember that? Yes? Negative epsilon? That’s right. I use the– I set and use the
equations all of everything I already discussed that. But I gave you also equation that epsilon
c equal to epsilon z equal to minus nu 1 minus nu 2 times epsilon, it– I can’t put a, b
or x, y or what because they are all constant. The sum is constant. Is that correct or not?
Yes? So what will be epsilon c, can you guess it? No. The sign of it, can you guess it? What
is epsilon max? Is it plus or minus? Minus. So it means lots of tension in this direction.
What’s the epsilon minimum, it is? Minus. Extension or compression? Compression. Compression, means you have an object that
being stretched this way by sigma x, is that correct, a lot? But in this y, it is compressed
but this is much little with– compared because you have this one. This side is going to get
the smaller or larger? Larger. See, if this was not there, and you are stretching
it, this side becomes smaller. Is that correct or not? The effect of this is less than effect
of that. Is that correct or not? So, in other words, epsilon z or epsilon c equal to minus
nu sigma x over e minus nu sigma y over e. Is that correct or not? Sigma y, x is tension
so you get huge negative. Sigma y is in? Compression. What’s the smaller quantity? Of course, minus
time. Minus become? Plus, but does not overcome that. Is that correct? So, I expect epsilon–
when I see this I expect my epsilon c to be? Negative. Negative number, is that understood? Everybody
understand about that? However, if this was small tension but this was not compression,
this side would go into become larger. Is that correct or not? The z take this become
larger because I’m putting more pressure on it than more tension. Is that correct or not?
Therefore, in that case, epsilon c would that been become? So, you have some idea ahead
of time what you expect but let’s see whether that happened here. So, epsilon c equal to
epsilon z equal to minis 0.29, did I give you points enough there? Or point three, did I give you 0.03? Thirty-five
both sides. Point 35 for this problem, 1 minus 0.35, 1550 minus– what did I put here? I
think this is 250. What did I put? You know, this become maximum, did I gave you maximum
or minimum? No, not yet? Yeah. This is 1550. This is 250, I have it here.
I don’t know. So that’s 250, yes. So minus 250, so you put minus 250 here. So you do
it together, it becomes minus 700 micro. So what I’m saying that, if you are looking for
absolute maximum, gamma will not be in the x, y plane which is this circle, will be in
the x and z frame because your– this become minus 700, and then your second circle become
like that and your third circle become like that. We have done– so you need the diameter
of the larger circle, yes or no? Correct? It’s the same thing we did for stress. So–
But the point is epsilon c is not equal. Any time, you’re bumping off your car, is epsilon
c equal to zero? You stretch it this way, you stretch it, and it’s going to change a
little bit. Of course, you don’t see it by eye. Everybody see– because these are all
micro. Two hundred micro, what is that? Point 00002, everybody, four zero, 2 that’s not
something — Yes? What? We can’t, no, you can’t. Obviously, the whole
idea is the Poisson ratio. The relationship between epsilon z and epsilon x is also Poisson
ratio. If you — but if you stretch something in one direction, that was one of those early
question. That was not a good question, that’s what I’m saying there. Look what you are asking
me. What is Poisson ratio? Well, because one of the home works that did
give you mu. Oh, I gave you g and e, yes? Yeah. Then you have– you use the formula to calculate
that. Is that– what we can’t– mu– the problem is not solved. Because nu is the whole key.
If I stretch something in one direction, nu part– nu Poisson of it goes to the whole
Hooke’s law that I repeat it. I give it to you as the handout. This all depends on the
value of Poisson ratio. However, if in a problem they were not given which I’m going to do
within a minute. And so there is a relationship between g and e and– Nu. — nu. Is that what you are referring to?
Yes. So, your question is different. If the problem is not– you said can I solve this
problem with nu? No you cannot solve it without nu. All of it. If g is given and e is given,
there is a relationship like that. I was going to do it through another example. That’s the
relationship between g and e and u. So you can get it from that equation. Because nu
again is not measurable, g and e, you can measure it in the lab, g by the torsion, e
by the tension. So you measure– that’s not what you have taken. You know what I’m talking
about, yes or no? Correct? Before the lab– great. For this problem, why did he said epsilon
c equal to epsilon z? Because that’s what– no changes that. Epsilon
c and z are not– this is only rotation of xy. Epsilon c given here is 200 micro. OK. This is reference to xyz axis. This is
the gauge abc. That’s a capital C. Those are abc, that’s why I’ve changed the capital to–
those have nothing to do with this. OK. Yes? Those are principle. The three principle.
You see, this is xyz become abc, small abc. It has nothing to do with this one. This is
the measurement. That’s why I’ve changed– actually exactly what I did, I call it here
epsilon a or something. Yeah. Epsilon max, epsilon minimum, more than. This is x reference
to x, y, z axis there. Alright. Yeah. You know that, you know it already. Is that the
correct one? Let’s do some more example now. Let’s go to problem number six in your handout.
There are many on the handout. This is now you have to be very careful about this type
of problem. So, exactly what you have to do depends on your knowledge of, again, back
MA218. So go to handout. There are six problem there on the handout. And problem number 6.
So, I wish I could get that one, look I took it out. So it isn’t sure. So, let’s see whether
this come back again. So, if you have current back again, yes you can do it. OK. So just
while I’m taking it out. So– OK. It’s week seven– alright let’s go back again. Look
at this problem. This problem we did that one. This problem we did already. That’s the
one with a nu, just talked about it. And these are your homework. We do that. Let’s go through
this problem. Problem number six. We kind of have to do that. This is what, on and–
ok. It will come in a bit. You have it in your handout. You know I’ll put this for the
benefit of camera. What you all have in guys in your handout. So, go to problem number
six in you handout. And take a look at problem– handout problem number six. Take a look, read
the question and can you answer that quiz. That was your quiz, today can you answer that
question. If yes then you can go home all. Alright. Come on guys. The only thing I can
recommend here to you is this guide. Although we are talking all about the strain and the
strain relationship, but you should not look at it this way. You should go back to MA218,
look at it as state of a stress. So everything depends. So, everything and we talked about
it in this lecture, it depends on the state of stress first. Yes or no. Look– I started
the lecture by saying that we may or may not have a sigma x, yes or no? Correct. Then we
may or may not have a sigma y. And then we may or may not have a tau xy– I put it all
positive, yes or no? In all your homework problem, MA218, 219, you end up with this
is which is a plane stress. Is that correct or not? Sequence z of course equals to zero.
Yes or no? Yes? Yeah. All right. What is this? Of course it’s a
plane of stress. Is that correct or no? Now here is the question. We have put a gauge
there, gauge a and b. The gauge reading of gauge a is equal to it’s given, 100 micro.
Yes or no? So, let’s put it there. The gauge, this is the system, this is a force p, a torque
t and here at that point, I put here a gauge like that and a gauge like that. And that
direction is how much? That this is x, this is y and that direction is, how many degree?
Forty degree. Forty five. Oh 45, whatever is there. I don’t see if I’ll
be– Is that correct or not? Yes? OK. The question is, if epsilon, this is again the
reference what you just asked India and like y a couple of minutes ago. This is epsilon
a and epsilon b. Epsilon a is measured after the loading. Before the loading, of course,
there is no stress, no strain. After the loading, epsilon is a is 100 micro. Epsilon b is minus
55 micro new, Poisson ratio is a given equal to 0.29 for this problem, 0.29. It is all
there. Modulus of elasticity is 30 times 10 to the power of 6 psi or 30,000 ksi, that’s
the same thing. It’s there. And the question is, what is p and what is t? Come on guys.
Now, did I give you the location of the gauge? Notice this is r gauge, two gauges actually,
c and I put it there. Is this important where I put my gauge? No. Don’t say no. You have
to have an answer for this. You see, although you said no, they knew the answer or you just
said no because everybody says no? Which is true? Which is the true case? No because– No why? –you didn’t put any of this. Why? You didn’t put any lengths? Why? OK. So, can I put it anywhere? So in this case– That’s what I’m asking you. Yes. So, why? OK. You say yes. I guess I don’t know. So, that’s the question. You see, by saying
that no or yes, what’s your reasoning? And that’s exactly what I’m giving you a quiz
for. The quiz I ask for all the reason. If you don’t know reason, why you are putting
theta equals to zero here? Then we need to cut or change a to b, don’t put that b, then
I realized that you did not 100% are there that I want you to be. Is that correct or
not? That’s exactly my point. Yes. That’s not a good question. I mean, the answer
is correct in a way but that’s not the way I’m looking at this one, it goes back to basic.
When you have lecture 1 MA 218, when you have a p here in this cross section sigma is everywhere.
Yes or no? So, does it make any difference whether it’s on top or on the bottom or on
this side? No. Is that correct or not? That is p. However for the torque, when we have
the torque here, what the shear stress was outside? Wasn’t that outside equal to constant
of tc over j? So, it makes a difference? No. No. However, if this was a beam, that’s what
some of your homework is. And I put here a load. Does it make a difference where I put
my gauge? OK. Of course it makes a big difference now. See,
this stance that I’m looking for, so you have to go to your background and check that here.
Normal stress is uniformly distributed in the cross section, including on the surface.
Yes or no? It’s going tension, all tension. Yes or no? Here or here or here or here or
here is all the same, except I don’t want to put my gauges near the support because
I’m afraid of stress concentration. Remember the stress concentration ideas, so I’m trying
to stay away from that. Is that correct? And what is the route on the tension every sigma
in every cross section? This is all making the story short, both at the tau and the sigma
is all similar in all the cross-section. That’s the answer that I was looking for. However,
if it become a beam, not even the distance x become important. This way, y also is important.
Yes or no? Mx changes the value of the m. Yes or no? Oh this is MA 218, the further
I go the more moment do I have. So, I have two know where I’m putting it. Yes or no?
This space sigma equal to what? Sigma equal to m, y over? You know, all the neutral access
the effect is zero as you go up is tension or comprehension of one side is the tension,
one side is compression. Is that correct or not? Vice versa, so, the definition is where
you put it here is totally dependent on all the state of stress at that point. Is that
correct or not? Which will take us to some of your homework? I don’t know whether you
look at those homework or not, I’m going explain it before I did here. It does not make any
difference, everybody agreed. All the answer I heard is correct because it’s uniformly
distributed. Yes or no? Therefore, the location does not matter. Let’s finish the problem,
then I get– go back to that one. Now, first thing first, how does my state of stress look
like that? Go back. This is the whole idea. This is interesting. If I give you the p and
I give you the t and I give you the dimension, by the way the diameter of this rod is 1 and
a half inch, which is written there, everybody can calculate the state of stress because
this is very elementary for you. Are there a normal stress, the p over a and tc over
j. Yes or no? So therefore, in every point here or there or their including point c,
you have a normal stress. Sigma equal to p over a because it is in tension. Yes or no?
Correct? So sigma x equal to p over a and then tau, which the way it looks like because
it’s going this way. So, the tau will be like that. We have gone through that. This is very
basic. The tau, which is tau xy equal to tc over j. I have to give it minus because this
is going downward. Is that correct or not? Yes? Correct? Yes. This would be the state of a stress. Correct?
OK. Now, is this going to create epsilon x? Now, the key as you saw in that graph is the
Hooke’s law. What’s the Hooke’s law? And see, I have to use more board here. The Hooke’s
law is the following. This is in general. Here is particular to that problem. So I’m
going to erase that, and then epsilon x equal to what guys? Sigma x over e minus nu what? Sigma z There is no sigma z. Is that correct or no?
Do we have a sigma y here? No. No. So, therefore this become equals. This
cannot be anymore elementary than that. Everybody understand what I’m saying that the relationship
between a stress and a strain is as I’ve showed you in that graph earlier, I can go back again
and look at it everywhere, which was everywhere. I go back before looking at that one more
time. Look at it. Hooke’s law, Hooke’s law, Hooke’s law. Anytime I have Hooke’s law, I
can go from the stress to strain. You have seen it many times in there or vice versa.
Is that correct or not? That is the idea I’m asking you to use in this set of homework
which is very simple. Now, if you look at it carefully, it becomes very simple. However,
you have to remind yourself. Anytime I have sigma, I can put it in the Hooke’s law to
come to the epsilon or I have epsilon, I put it to the Hooke’s law to get to– that’s as
simple as that. They are parallel to each other. Is that correct or no? Nevertheless,
we go there so therefore– now sigma x also equal to in this problem only p over a. I’m
looking for p. So, if I give– I have sigma, I can solve it. But sigma equal to e times
epsilon because there is no sigma y. There is no sigma z. Is that correct or not? Therefore,
sigma x, equal to epsilon x times modulus of elasticity, epsilon x was given equal 200
times 10 to the power of minus 6, multiplied by 30 times 10 to the power of 6 psi. So that
this and this drops out. So it becomes 3000 psi. So, the sigma x is 3000 psi. There–
It doesn’t take a genius. This is nature of MA218, to calculate the value of? P. P. Is that correct or not? So, therefore,
you calculate the p equals to sigma times the area and sigma is 3000 psi and the area
is pi times radius squared which is 0.75 squared. You can calculate that one because the diameter
was given. This is a circular shaft and you calculate the p to be equal to 55,299 or 5300
pound. So, for 5300 pound loading here, epsilon a will be reading that 100 micro. Is that
correct or not? Yes? You could have done by– I could put the load here. You could calculate
epsilon in reverse. Yes or no? Now, the next question is what is the value of torque? OK.
What’s the torque? Look at these guys. Torque is related to tau is equal to tc over j. Yes
or no? But what is tau? What do I need there? What is the relationship between stress and
the strain? Shear strain? This is Shear stress and shear strain, gamma. Gamma was equal to,
I wrote it a few minutes ago. Gamma xy was equal to what? Tau over modulus of rigidity.
Tau xy divide it by– the only question is, that’s why probably you were asking. Here,
I have given you new and modulus of elasticity but I did not give you the modulus of rigidity.
I was going to bring it here. You asked earlier. Is that correct or not? Therefore g equal
to what? First, let’s calculate the g based on these two, g become equal to e. Again,
divide that to one plus nu so I give you that 11.6 times 10 to the power of 6 psi. So if
I have the g, if I have the gamma, I have the g like the other one. I can calculate
the tau. If I have the tau, I can calculate the t, by noting reverse order. Is that–
Everything is in reverse order. It’s nothing new. You all know the answer. Is that correct
or not? Yes? But in– instead of going forward, we are going sort of backward. Everybody understand
that? We are having the testing and we want to see what kind of loading caused this kind
of gauges, is that kind. Even if it’d break, we want some time to do that, what kind of
forces this kind of damage to the structure, which is by itself is another art. Is that
correct or not? Yes? All right, anyhow. So what should we do now? OK. Now, what should
I do? You see here, I had epsilon x. So what do I need? Look at the problem. I need the
gamma. But gamma is not measurable. Instead of that, what do I have? I have that epsilon
at? At 45 degree, didn’t I ask you to what to do to get that? Yes or no? It’s again,
it’s reverse. So you can use the equation that I gave you. Epsilon of 45 degree equal
2, you can use that equation. Or, I’m sorry, gamma xy equal to 2 epsilon or 45 degree minus
epsilon x plus epsilon y. I just gave you a while ago. Yes or no? Yes? Correct? Yes. All right. So let’s use it that one. Become
two times, what’s epsilon 45 degree? Epsilon 45 degree is? It– We read it because it’s needed there
instead. It is? Negative 55. Negative 55. So it is minus 55 minus epsilon
x. Epsilon x is? Epsilon a is? One hundred. One hundred plus what? Epsilon y, what’s epsilon
y? A. A? A. You see, that’s the mistake, common
mistake. Look, how many of you immediately, you said epsilon y is zero because I didn’t
measure it? You all assumed epsilon y is zero? Yes or no? Right and you know again as that– that’s
what I was referring earlier. That is the Poisson ratio, yes or no? If that is an epsilon
x, nu Poisson goes to epsilon? I don’t need to do measurement because I only have two
unknown, if I had another unknown there, I needed three measurement. Everybody understand
what I’m saying there? However, yes, write it down. I called it the 4 with the epsilon
y equal 2 minus nu sigma x over e, plus sigma y over e, minus the other one, the other one
is zero, sigma y you all just said those. So it actually become very standard minus
nu tau epsilon x, which is really the definition of the Poisson ratio. Whatever you are producing,
what direction, nu Poisson? What’s that other one? Is that correct or not? Yes? So therefore,
if epsilon x was 100, then epsilon y become equal 2, minus 0.29 times 100. So it is minus
295. So do not jump to the conclusion. This is always true. It is not measured but it
is there. I can calculate that. Is that correct or not? I did not measure because I could
calculate. One measurement was enough. Actually, if you measure it, it should be showing negative
29. Is that correct or not? If all the instrument is correct. Therefore, you put here minus
29, not zero and you calculate gamma xy and gamma xy become equal to gamma xy become equal
to somewhere here. Minus 180 times 10 to the power of minus 6 or you can write it in micro,
micro radian. We don’t have to put it, so minus 8, 181. As soon as you have the gamma,
it is needless to say, you immediately can calculate tau xy. So you calculate tau xy.
Tau xy, equal to g times gamma, g was 11.6, times 10 to the power 6, multiplied by minus
181, times 10 to the power of minus 6. But notice again, if you apply this 6 business
that I always ask you to use, tau become minus 2100 psi. And it become minus, notice t is
positive but tau is negative. And that’s the answer that came by itself, so because the
measurement was like that. Yes? What is the unit for the gamma right there? Gamma, it hasn’t– doesn’t have a unit because
it’s a micro radian. But in radian– but this is unitless, it’s just inch over an inch millimeter
about this radian. Yes? Oh, yes. All right. The same thing with epsilon, epsilon
doesn’t have any– its just inch over inch, millimeter over millimeter. It’s the ratio
that we need. Anyhow, so therefore no don’t worry. I have the most important problem yet
there. Yet there, still we have that. Did you see how this problem is solved, OK? Then
we take– that takes us to the next problem which is this problem. So you have homework
like that in the book or homework like that again. I’ve mention that for your beam. If
there is a beam here, there is a load here where you put your gauges is very important.
Is that correct or not? So you have to go to a state of a stress. Find what we did earlier.
This is you have homework like that. You missed it, some of you. You had homework like that.
I thought in this class, I gave a problem like that. It was a p here. Is that correct
or not? Yes? Yes? Now, you’re getting similar problem. That part of it is not a question.
That part of it, you already know. You have a homework like this. I’m going to mention
to you because I don’t have yet. There is here, a force p and there is here a force
q. Is that correct or not? And then you’ve measure something here, somewhere here, you
measure your strain and then you want to calculate the value of p and q which is exactly this
problem, a little bit different. Because that one now, it is bending involved as you see.
OK? Now, look what happened here. If you cannot do that problem, certainly you cannot this
problem. That’s why I gave you earlier those homework in order that you for at least through
the quiz to learn it. Is that correct or not? Now, what you have to do here, you have to
draw the free-body diagram either the upper part or the lower part. So I’m aggressive
because we are out of time. I’m just saying that this one is one is p. Yes or no? This
one is q. This is your n. This is your shear. Is that correct? You have a shear force in
this cross section. You have normal forces. There is cross section. Also, you have depending
on this lpl. You also have a– moment, so you put your gauge here or here, or here,
or here. That’s much a big difference, depends where you are. Is that correct? But you have
to be able to come up with your state of stress. That defines your strain. But if you have
done it in the past, it should not be a problem. But if you haven’t done it in the past or
you couldn’t do it in the past, you have a problem with the static which many of you
did. Remember in this class especially– yes? You could not even bring those forces down
here which is very simple as you know by now, everybody knows that. However, I don’t know
which homework they gave you, they– whether they put it here or here. I just– if you
are here, that’s sigma equal to q over area, minus q over area, minus this is pl, everybody
under– minus m, y over i, et cetera. So what I’m saying that, you may have a sigma y in
the form of compression. If you are on this site, compression to compression, if you are
in this site, compression and tension. So it– let’s go to this site, so you have definitely
compression minus q over a, minus my over I because we are not at the end, we are not
here. We are somewhere near the centroid of the section, depend on the center of that.
You also have a shear stress like that. That shear stress this time is not tc over j. It
is vq over it, et cetera, et cetera. But if you quite frankly look at it, this is exactly
what you see there. Everybody understand that, this is one stress and one shear. One normal
stress and one shear stress. That’s also is one normal stress and shear. What is the x?
It is y. And then the rest is the same. The static of the problem changed, the strain
part of it change. Let’s get– Give me five more minutes because I did this for the other
class. I need to mention it because this is very interesting problem, sorry. I have to
go somewhere too but I take five minutes of your time because this is interesting problem.
Go to problem number 3. So go to problem number– sorry that I have to– it is here. I put it
here on the board too. I hope it is still there. So go to problem number 3, give me
another five minutes. If you want to leave, you have something to do. But I suggest you
stay because this is very interesting problem. It’s simple problem but at the same time,
it’s very interesting problem. You’ve got there a bunch of problem coming like that.
Because this is a pressurized tank, yes or no, correct? Let’s go quickly there. First
of all, if it is a pressurized tank, what is sigma 1? Remember the two sigma, we had
sigma for sphere, sigma 1 was equal to sigma 2. One was longitudinal, the other one was
over their circumference of that but this is a sphere, this not a cylinder. In this
one, x and y doesn’t matter whether this is x and y or this is x and y. They put it anywhere
you want to. Is that correct or not? Because this is spherical shear, it doesn’t matter.
But this was equal to pressure inside or inside divided by 2t, yes or no? Correct? Now knowing
that, which is from the beginning of the class, that’s the pressurized tank and we put here
and didn’t put a gauge, we put a line here 20 millimeter long, originally was 20 millimeter.
After we load it, that increased by how much? That increased by 0.012 millimeter increase.
That’s plus. So that’s what we did. And the outside diameter is– the inside diameter,
outside, the r inside is equal to 1000 millimeter. The thickness of the tank is 10 millimeter.
So the ratio is more than 10, the ratio is 100, so I can use the thin-walled theory which
I have and that’s what I’m using there. Is that correct, but the ratio should have been
larger than 10, the ratio is 100. Everybody see that? Yes. So that works. You don’t have
to worry about this. So all of this is changing. Now, the question is what’s the value of the
pressure inside? So I’ll put this gauge, there is a pressure inside. I don’t know what. This
gauge tell me what’s the pressure inside. And don’t you want to know every time you
are in there, let’s say, we’re living in tolerance near the refinery, what is the pressure inside
the tank all the time, what is related, yes. So because the pressure can go up and down.
What you see is through the strain gauges. That’s what I said. This problem is interesting.
You don’t have to put the strain gauge inside. You can put it outside, yes or no? Yes. Yes? The question is if I have a reading of
what it said that we have what, there is something there. It says what? So let’s read there,
oh, that’s it. This become like that. Is that correct or no? Again, let’s go quickly. So
epsilon must be equal to delta over l. That’s the definition of epsilon, yes or no, right?
The delta is– How much this time? The delta is 0.012 millimeter. Original length was 20
millimeter. So you divide that. That becomes 600 times 10 to the power of 6, or 600 micro
plus. Why? Because it’s increasing. Notice of course everybody knew the tank is going
to expand. Yes or no? The location, not important, as I said because x and y doesn’t matter here.
Yes or no? I can put it anywhere. So this is what we did. So here is your x. So this
is the gauge epsilon x. This is a and that’s epsilon a, which is epsilon x. Is that correct
or not? Yes? Now, what’s the state of stress? Again, I said everything depends on the–
now please, when you go to this homework, start from this theta. Although we are talking
about the strain but you should think of a stress because that makes it simple a lot
for you. What’s the stress like on top of this tank? Sigma 1 and? Sigma 2 And both of them are? Equal. Equal. So that’s the state of stress. This
is the key to the problem, guys. Sigma 1, going this way or it doesn’t matter because
this is a sphere. And sigma 2 and both of them are equal to each other. Yes or no? Correct?
Therefore, what’s the epsilon x equal to? Epsilon x equal to what? Sigma x over e? Yes,
minus nu sigma y over e. Yes or no? But sigma x equal to sigma, sigma 2 equal to sigma,
the same sigma, yes or no, right? And that is– Let’s put it that way. Let’s go to that
equation like what we get. That is equal to p times ri, ri is– well how much, 1000 divided
by 2 times 10 millimeter because the thickness is 10 millimeter. So it becomes 50p. So all
I’m saying that sigma 1 equals to sigma 2, this 50p, this is 50p, so all you have to
do, e is given, nu is given, nu is given, nu is not given or is given. I don’t know
which of us the nu, nu was given. Point 3. Yeah 0.3. So this was given. Sorry to keep
you here, 0.3 is given. So you put it here everything. So epsilon x is equal to epsilon
x, we measure it, 600 times 10 to the power of minus 6, equals to 50p, times y minus nu
divided by e. Because sigma x is equal to sigma y. That’s what I did. Nu is given, e
is given, you calculate the p. Is that correct or not? Yes? The p become equal to, please
write it down. The p become 3.43 megapascal. After you calculate the p equal to [inaudible],
you can calculate sigma 1 equals to sigma 2 equals to 50 times 3.43 megapascal, become
171 megapascal. So that’s the stress of that. It’s all simple. Very simple. What is interesting
that I kept you here, let’s draw the circle for the stress and for– Strain. Strain. So you see something that you haven’t
seen in the past. OK. This is for a stress guide. For stress, this is sigma versus tau,
with the arrow. Is this sigma x or is this principal? Is this principal or not? It is not a principal. These are principal, because there is no shear
or stress. Yes or no? So this one is how much? 171. Yes or no? The other one is how much?
171. Have you seen a circle with zero radius? That is your first circle, because sigma 1
is equal to sigma 2. So your inclined shear is zero. Is that correct because your circle
is zero. Now what’s out-of-plane shear? The out-of-plane shear, what is sigma z? Is there
a sigma z anything going this way? No. No. So sigma z equals to– this is in other
words sigma max, sigma minimum both of them are 171. This sphere and that one, both principal,
yes or no? Not if you have seen in that before. But sigma z equal to zero because there is
no sigma like that. So this is your tau max x and z or y and z– it doesn’t matter, both
of them applies, equal one half of sigma equal to one half of 171. Now, let’s go to a strain.
Epsilon versus gamma– oh wait, I want you to see both of them. Now epsilon, what was
epsilon equal to? The epsilon x was equal to 600, yes or no? Yes, 600 micro. What is
epsilon y? Does that make any difference? It’s the same thing. Epsilon y also is 600
micro, so both of them here, 600 and 600. What is epsilon z? Is epsilon c zero? No. No. That equation I gave you. Do you remember
that? Epsilon c equal– because epsilon c, this is the thickness of check– tank changes,
yes or no? Actually there was one homework I asked you to calculate the changes, the
thickness of that. And that one, epsilon c equal to minus nu, 1 minus nu, epsilon a plus
epsilon b. And since both of them are positive, that one should be negative and you do that
and you draw that last circle and that is your maximum gamma. So that will be it. So–

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