# Strength of Materials II: Deflection of Beams, Superposition Method (12 of 19)

First of all, it is determinant. There’s nothing
about– indeterminant about it. Remember, this is a pin on one side, a roller on the
other side. This is obviously from the static class, you know that. That’s the determinant
problem. So you can calculate for RA and RC in this case, is that correct? Although, a
couple of people make a mistake there, but nevertheless, one is 180, the other one is
145. So you do not have to worry about the static of the problem. The static of the problem
is already being solved in this problem. I’m just asking you to practice something which
was essential, which some of you didn’t do that. Notice, the best, most important part
of this problem is this. From A to B, there is no load, shear is constant, moment is linear,
yes or no? From B to C, the moment is a parabola, correct or not? How can this be expressed
as one function? So it is two different function. Everybody, understand that. And if we want
to answer the entire question, we have to set it up for two part. Notice, some of your
homework were two part. I did one example or two in class, two-part scenarios. Some
of your homework need not require two-part, and I’m going to explain that one more time.
I don’t want you to make a mistake there. So one of the problem was like that, that
you had here, a load there and a load here, something like that. You remember that I’m
talking about last homework. Is that correct or not? There, you only have to do A and B
because no question was asked about C, because if you want to find the deflection under the
C, then you have to do A, B and B, C. Since the question was at the beginning of the homework
set, therefore, they did not ask you to do anything about C. So I ask you, you can bring
this one here and treat it as one beam which I suppose you didn’t. There was another problem
that it was like this, and then load was like that which was symmetrical, yes or no? If
it is symmetrical, then you have to do it between– to expand from here to here. Because
it’s symmetrical, so you only set it up with half. Everybody, understand it. However, I
gave you another problem which was the load here or somewhere like that. Remember that
problem? And that problem you have to do it left-hand side and the right-hand side. You
must have done it. So this should not be a problem you, is that correct? But unfortunately,
about 10, 15, 20% of the class did not do that. Is that correct? They only looked at
the second part as if doesn’t– the first part doesn’t exist, which ruined the whole
system. Is that correct or not? Nevertheless, this is in two-part. One part is between A
to B. I mean, if you missed that, you missed the whole boat. Everybody– But still I gave
you eight points or 10 points. It depends what you did, which is not really relevant,
but I gave you some credit. Everybody, understand. I thought maybe you missed something. But
nevertheless, this has to be set out between A to B, totally different between B to C.
Another reason what– why– What was the reason for us to go through the singularity function?
What was singularity function that you love it so much? Remember last, last week when
we were talking about singularity function, which I asked you to do it as a homework?
I hope all of you have done your singularity problem, have done your superposition problem,
maybe a question on here or two. I saw many people during the office hours, which you
questions were about the superposition method. Is that understood? Nevertheless, what I’m
saying that in order to bypass that, if we– I said it many, many times. I’ve asked you
many, many times. If I said I have a note here, I have a note here, I have a note here,
I have a note here, how many times do we have to this into system? All of you said one,
two, three, four, five. Remember all that discussion? Here, you have to break the system
into two. Is that correct or– That is cardinal. I mean you cannot do this problem without
looking at it that way. And you’re doing only one part or the other part is not going to
get you anywhere. Is that correct or not? Yes. First of all, it’s determinant. Second,
if I’m interested of deflection here or theta here, I have to do it in two-part. In other
words, even if I put A, B, there are unknown that cannot be solved. Anyhow, EID2Y over
DX squared become equal to moment at this section. What’s the moment at that– It could
not be simpler than that. It’s 180 times X, yes or no? Correct? Plus or minus? Minus. Plus. Plus. How many people do think put minus there?
Because they have not got out of the static yet. This static, we are not talking about
180 times that. We are talking about the moment in the body of the beam, who are going to
be reacting to that, is that correct or not? So this moment is negative. That moment–
This is the M we are talking about, yes or no? How many of– You’re drawing it there
but then you are not using your static property. Is that correct or not? Anyhow, another 15%
of the class or 20% put here minus. It is 180 times X, is that correct or not? Action
is negative. The reaction to that, the sum of these two must be equal to zero. This is
RA, this is the MX that I’m looking for. Is that correct or not? That MX– If this is
10 and this is 2, 10 times 2 is 20 going that way. This M should be 20 going that way, therefore
it should be positive. This is the one, it goes there. Is that correct? This is in internal
moment which causes the stress. Is that correct or not? In the [inaudible] beam, correct? Yeah. No problem there but still there are people
who make a– How can you go through ME 218 and not recognizing this fact? That all the
stresses, you put it’s equal to MC over I or tau equal TC over R. J, those are all internal,
not the external. Understood? Yes or no? You put it’s equal P over M, first lecture of
ME 218. That is not P, that is N, remember? That N means normal force in the body of the–
Right, correct? Everybody understand what I’m saying that? I don’t know. Still, you
are quiet guys. I have a rod here. I have P1. I have P2. I have P3. I don’t care about
this one. I want to find this N. Everybody– That N gives me a stress at that point, yes
or no? I may have 50P there, I don’t care. I have only one N. And that N is the normal
force. I may have this force, this force, this force, this force. Then I have here one
shear. Is that correct or not? And also, as the result of all of that, I have one M. Everybody,
this is Chapter 6. This is Chapter– I’m sorry, this is Chapter 5, sigma equal to MC over
I. This is tau equal to VQ over IT, is that correct or not? And this is Chapter 1. Is
that correct or not? In that respect. So why you are making mistake, you are putting minus
here. You are taking the moment of this RA above this point. Is that correct– That’s
a no-no. Nevertheless, everybody understood there, yes? Now, let’s go to the other one,
EID2Y over DX squared which is totally different format. These are the format of this beam.
It’s going to bend like this. Then I said it in class. Actually, I showed you in two-color.
I said, remember, if you go like that, you may end up like that. Let’s say it ends up
like this. This is my exaggeration. Is that correct or not? Yes? This girl is this girl.
Is that understood? Yes? This black girl is this girl. Is that correct or not? Yes? That
Y represent from a to B. The other Y represent from B to C which will be this elastic curve,
this is the Y. Expressed as a function of X like any other parabola or whatever. Is
that correct or not? Yes? You getting the idea that’s why you should not make this kind
of mistake. Anyhow, I don’t want to put here. Here, I kind of put the color on top. So this
is region, purple region, let’s put it this way. This is the purple region and that’s
the black region. Is that correct or not? Nevertheless, I hope that you understand what
I’m talking about after you go to the quiz. Of course, that’s a little bit too late. So,
you got– Again, I ask you to use the same system of coordinate. Now, some of you recognize
that, that if I make a cut here, this was cut 1. For cut 1, this is the M. Yes or no?
Then I have to use for cut 2, which is now, this or that? Is that correct or not? Some
of you use this part but you made a mistake here– or not a mistake, you used something
that I did not recommend them to you, which is correct. Some of you changed your X like
that. I don’t know. I have to look to see what– I’m saying this because some did it
like that, because they want to go this way as if it is simpler. They saw it in actually
in the Hibbeler book, they would recommend that which I always change that one. I said
never use that method because if you get into lots of math, this is the only difference
there. You follow the system that I’ll show you in class, the math become much, much simpler.
Everybody, understand that. You change that, you can do it. There is nothing wrong with
that. But remember, if you are at this point, let’s say X here is equal to 3, at this point,
X equal to 2. Everybody, understand what I’m saying there. So you can use this method if
you are– you’ve tried to use it. I ask you not to do, plus the fact that all the quiz,
I said, use a center of coordinate, yes or no? If that’s the center of coordinate, then
this stands from here to here. Let me draw the free-body diagram. This was the free-body
diagram of part AB. Now, the free-body diagram of part BC is like that, guys, 125, this is
the load, this is 180. You want to put that as for the last, last time everybody is correct
there. You should not view people make a mistake. This is not B. Some people call this point
B or C or E or D, I don’t know. That’s not the point. That is the distance of? X. X. So this becomes shear at point X and this
is M at point– and it has to be this way. You cannot put in negative. If you put it
in negative, everybody– everything will be in negative format, yes or no? This is the
positive direction we are assuming for our shear and moment diagram. Yes or no? Correct?
So the question is, what is MX? Now, many people had a problem there. I am going to
write it and you see if it’s correct. Equal to 180 times X, that does not change. It’s
still 180 times X. This load is 125 times this distance which is not X, it is X minus?
What’s– How much is that? X minus? One. One. So the other one is PX minus 1, yes or
no? But whether it’s plus or minus, again, we have to go over the sign conversion. Notice,
the action is positive, the reaction to that should be negative. So that– this goes plus,
this goes minus, yes or no? But at the same time, at the same point, aside from there,
you have a uniform load. You see, uniform load should be a parabola. WX squared over
2, yes? But X start from here, not there. Everybody, understand that. Therefore, it
will be minus the same thing, W times X minus 1 to the power 2 over 2. So that is your system
that you use it in your singularity method. If any of you use a little bit of singularity,
you will realize that’s the case, is that correct or not? Now, the second or the third
cardinal rule, which is very, very important, especially for this problem, I ask you not
to expand this. Remember that? If you expand that, you get the answer, but you’ll get different
solve the problem but with twice as much math. If you keep this one in this system, which
is similar to singularity, that’s what I’m saying there, you will– your math reduces.
So let’s keep it in this format and go forward. So this is the moment for section 1, correct
or not? That’s the moment for section 2. As you see, this is totally different from that
one. So this curve and this black curve have nothing in common to each other except when
they join each other, is that correct or no, which we get to it in a minute. So there’s
less process there, EI theta– let’s call it theta 1 or black area, equal 280X squared
over 2 plus C1, then EIY1 equal 280X3 over 6 plus C1X plus C3, that’s the integration
of that part, correct or not, which many people did correctly. This one become EI– sorry,
theta 280X to the power minus PX minus Y to the power of 2 minus WX minus 1 to the power
of 3 over 6. What is the W? What’s given to you? P. What’s given to you? You can put the
number there. Everybody, understand that. Many of you did, which is perfectly OK. Is
that correct or not? But please do not expand it. Some of you did, you could never finish
the math in 15 minutes, 20 minutes because you did not follow the– that rule that I
asked you to. Then next one is EI– you see in a minute what is going to happen– Y2 equal
280X3 over 6 minus PX minus Y to the power of 3 over 6 minus WX minus 1 to the power
of 4 over– or this one has plus 3, that’s C3. And this one has C3X. You cannot see from
that corner. I’ll put it here, plus C3X plus C4. I have four boundary condition I have
to resolve. Is that– a far constant of integration. C1, C3, C3 and C4 system I’ll show you. Now,
at X, look at this. This is the boundary condition, but it does not apply to the purple one, apply
to the– some of you put it all over the place. Look at your notes, what you have written.
What X equal to zero Y equal to zero, you apply it here, and again you apply it here.
Did that makes sense? Yes? Correct or not? Y equal to 0 belongs to beginning of black
line. Is that correct? X equals– this– even if it doesn’t have X equal to 0, X start from
1 to 5. You have a curve that going from here to here. I mean why should I use this boundary
condition for that, I have no clue. Ten people used it. Is that correct or not? One-fifth
of the class. OK, good. So you understand what I am saying there, yes or no? So at X
equal to zero, Y1 equal to zero which is the black one, yes or no, which many of you did
correctly. So that means zero, zero, zero, C3 equal to zero. So C– the funny part is
this, the funny part is this. They put here X equal to zero then they said C4 equal to
zero. Look, if I put X equal to zero, this is zero. Is this zero? This is minus 1. And
this is plus 3, is that is zero is minus 1, also minus 1. So you have to– you know, how
can the C4 become zero? Because there’s something remembering that everything– because this
one, everything is X. This one, everything is X minus 1. Even if you put X equal to zero,
C4 is not become zero. Everybody, understand what I’m saying there. But some people just
close their eye, they just go ahead. But don’t do that. You are– It is there in front of
you. Look at it, decide, and then you know you are making a mistake. Is that correct
or not? Anyhow, at X equal to zero applies to this line, this is the black line, that
is the– this is the one. This is the purple line, yes or no? So this is equal to zero.
Then what else do we have here? At X equal to 5, Y2 equal 5 meter, Y2 equal to zero which
apply to this one, yes or no? Let’s use that at last. But what is important here, this
is the lecture. It goes on about, how about that point? At that point, theta 1 equal to
theta 2, Y1 equal– and if you use this method, now it just can become a little bit simpler.
So in between the two, this is the borderline. At X equal to 1 meter, you should write theta
1 equal to theta. Now, if you expand it, you get into lots of math. If you don’t expand
it, look what happened here. Here, you get 180 times 1 to the power of 2 over 2 plus
C1 equal to 181 to the power of 2 over 2 minus what? Look what happened? If not expanded,
minus? Zero. Minus? Zero. Plus? C3. C3. Look, this is the method I show you, but
you did not follow it. This and this drops out, then C1 become equal to C3. If you expand
it, C1 is not equal to C3. Everybody, understand again. Remember, I’m emphasizing this for
you to know. C1 and C3, I give you the number. You see, I have to do this much math extra
to get to this. Everybody, understand what I’m saying. It’s impossible to do it any other
way. And also, write it down. So that is the– that one. So I’m going to erase that because
I do not have C3 already gone to zero because we already established that one. With this
one, is this equal to that? Is that the case? Now, again, the 1X equal to 1 meter, Y1 must
be equal to? One. One too because these two, they are joining
together only at that point. Not at X. X means all of this. X means all of that. Everybody,
understand that. At X equal to 1, again, I have to write it. I don’t have time. So as
you see, this and this going to– drops out. This one becomes C1 times 1, is that correct
or not? And then there’s the one because as you see here, this is– it goes– drops out.
This is zero, this is zero and there– this is what you get. Here, you get– let’s put
it there, 181 to the power 3 over 6 plus C1 times 1 equal to 181 to the power 3 divided
by 6. Zero, zero plus C3 times 1 plus C4. Notice, now, this and this drops out. This
is the way I did it in class 2. C1, because it’s equal to C3, C1 and C3 going to drop,
then C4 become equal to? Zero. Zero. In this respect only, C4 become equal
to zero and that is the idea there. So, this one is going. Now, the only question is remaining,
what is the value of C1 and C3? So somewhat you’ve got it right. And that one, we’ll have
to use this one. Is that correct or not? Yes? So at X equal to 5Y2, now this is Y2. So Y2
becomes zero equal to 185 to the power of 3 over 6 minus P which was 16 times 4 to the
power of 3 over 6 minus W which was 4, I believe– 450, I’m sorry. Minus 50 times 4 to the power
of 4 divided 24 plus C3 times 5 equal to zero. We solved these three, as you see, become
only one equal for one unknown and that’s why it become very simple. C3 which was equal
to C1 become equal to minus 376.7, something like that kilonewton meter. That’s it. So
that is the math. Any other method that you use, you expand that. You use from this side.
You use– change your X’s, all of that could work. But it’s not a 15-minute job. Everybody,
understand what I’m saying. So you have to use [inaudible] to the other one. Anyhow,
some of you may have gotten the other one. If you expand it, C3 become equal to minus
430, C4 become 18.7. Of course, C1 will be the same C1. Everybody, understand– you’re
solving three equation for three unknown, for C1, C3, C3, it’s impossible to solve.
Correct? I keep asking you in class, do not expand. Yes or no? Correct? And if you expand
it, [inaudible] that was the method. Anyhow, going back to the last part of the question
which probably I did not give you too many points for that and if you even didn’t do
it, I gave you credit for that one. I see some of you are surprised to get better grade
than you expected, yes or no? Correct? Therefore, it was calculating Y at B. Now, what is Y
at B? It’s this one or that one? Is that correct? Obviously, you have Y at B is common between
the two, so why not use this one? Is that correct or not? Yes? So therefore, Y at B
become 1 over EI, EI. And how will you– you have to put, put this equation, X equal to
what, [inaudible] 181 to the power divided by 6, C1 is already given, times 1. You simplify
that one and ends up to be equal to minus 346.7 over EI. Now, here is that unit problem
that I have to mention it one more time. Again, the unit problem was like that. This, you
put it the way it’s given. You don’t change anything. You don’t change this into millimeter
or kilometer turn into meter. Let’s do it the way it was given to us. Minus 346.7 divided
by EI. What’s the E equal to? E equal to 200g Pascal, yes or no, which is 200 times 10 to
the power of 9 newton per meter squared, correct or not? Yes? Multiply it by I, how about if
the I is given? I equal to 900 times 10 to the power of minus 6, multiplied by 900 times
10 to the power of minus 6 millimeters– or meters or the power of 4. Yes or no? Correct?
So what’s the total unit? I gave you that in class. It’s equal to newton meter squared,
yes or no? Correct? Did we use here meter? Yes. Correct? So I don’t have to– the only
thing different is this is newton and this one was kilometer. So in order to mash that,
I have to multiply that to 10 to the power of 3. So kilonewton become newton. Is that
correct or not? Yes. However, the answer is still is in, what? Meter. Meter. And I don’t want meter, I want the
millimeter. So I should multiply– notice what happened there. So what I’m saying that
this has newton meter to the power of the– I’m sorry, kilonewton meter to the power of
3. This, as you see, has newton meter to the power of 2. Yes or no? That’s the unit. So
this one, I have to multiply it by 10 to the power of 3. So kilonewton become newton. Then
meter 3 over meter 2, this, this, this drops out. I have one meter, but I don’t want more
in meter. Who wants to go meter, right? It’s millimeter. So you multiply it by 10 to the
power of 3 to convert meter in here. So answer will become in millimeter. As I said, we didn’t
go that far, I didn’t go that far, you didn’t go that far, but that’s OK. Minus 1.93 millimeter,
that’s the answer. The same thing for theta. Theta, you calculate that based on [inaudible].
Is that correct or not? Yes? OK. I went a little bit more detail because I wanted you
to see what is in involved for the future. So please, there are two facts that you cannot
make a mistake. First of all, you have to recognize the difference between determinant
and indeterminant. And then recognize between one– to avoid this, as I said before, we
went through the– you can now write this as a singularity system, do it, solve it again,
everybody, but that requires only one moment equation from entirety. Is that– Isn’t that
what your homework was all about? Correct or not? I’m hoping that all of you have done
that and you know exactly what was involved in it. So that was lots of questions to be
answered here. And unfortunately, as usual, the average was not that good. It was really
about 11, which is OK, but it should be much better. Now, going back today, I’ll have to
do a couple of more, couple of more superposition method. One thing that I forgot to mention,
so I’m going to do one problem for you, so bear with me, it’s this problem. So let’s
do it this way so we can put numbers there or you can– Let’s do this problem now. Here
is a problem like that which has a load here. Like this, 30 inch, and a load here 125 pounds
and this is 15 pounds per inch. And this is 10 inches. So 0.8, B, C. It could be one of
your homework or it could be similar to your homework. I don’t know that but I’m doing
that. Is this one of your homework, is similar to your homework, is that– Exact homework,
this one? It changes, yes or no? Let’s do it anyway,
I’ll put it on the board. Is that correct or not? Now, notice this system is not indeterminant.
I just show you previously, what’s the difference between determinant and indeterminant. Last
week, Thursday, I did some determinant problem in superposition, one indeterminant. I’m going
to do one more today. However, this is determinant. One fixed support, that’s it. Is that correct
or not [inaudible]. So it’s nothing indeterminant. However, it depends what the question is asked.
There are going to be these questions with us. I don’t know what. Let’s say calculate
YC. These are the question we want to ask or worry about for theta C or YB. So I’m going
to do one or two of them. Not all because it’s time-consuming. All of this could be
a question raised in this problem because reaction is already there. This is determinant,
I can’t find a reaction. Is that correct or not? Yes. That’s not a question. If it was
indeterminant, that would have been a different story. Then you have to work out this system
to get the reaction first. That’s what you are doing which I’m going to discuss today.
OK. Now, how do we solve this problem? Obviously, I have to break this into two system, is that
correct or not? So one system is simply a load here, 125 pounds but the length is how
many inches? The length is 40 inches. What I have not– or I actually overlooked last
time but I should have because that’s part– you could have done more homework that way
is this being which is only loaded up to 30 inch and the other 10 is free like that. Is
that correct or not? Yes? Now, I need to have the, again, the table on the board. So that’s
why I have it here to show you what happened there. OK, you have the table in your hand.
We’ll get it in a minute. We’ll come on. So this is– This changed into two system, yes
or no? Is that correct or not? Is it on or not? So it takes a couple of– All right.
OK, it’s coming up. Is that correct or not? So yeah, [inaudible]. Is that correct or not?
Let’s say we are calculating YC. How do you calculate YC? YC of this system equal to YC
of system 1 plus YC of system 2. OK, let’s write it down. YC equal to YC of system 1
which is the top beam, I hope every can see from there, plus YC of system 2. Now, YC equal
to– I’ll put it out here. Is that– What is our system 1? System 1 is this one. Yes
or no? What’s the YC? It’s given. That’s the one. Is that correct or not? Yes? The length
is how much? The length is 14 inches. Is that correct or not? Yes. And the load is 125 and
like, exactly like that, and that is the deflection at that point, is that what I want? I want–
don’t want the deflection at B, I need the deflection at C. And that’s in the table.
Yes or no? I do not have to use the curve, is that correct or not? Therefore, that’s
equal to, look at that, that’s equal to minus 125 times what? Times 40 to the power of 3.
I’m just putting number there. Is that correct or not? Yes? Are you with me? Mm-hmm. Minus pm cubic divided 3EI, yes or no? Correct?
Divided by 3EI. For the time being, I put the EI there and then at the end, we can add
it if you want to. Is that correct? Sometimes, we solve everything in terms of EI, I put
there EI. Now, we come to– this is the reason I’m doing this problem. Now, we come to what?
Part– Two. — 2. Now, this is the beam I have. This is
30 inches and this is 10 inches. Yes or no? What do I have– What– This is point B, this
is point C. I suggest you always draw out your beam, don’t do it in your head. That’s
one of the cause of mistake. Then define your action here. What do you need here? Here,
I needed this and that was on the table, yes or no for this question. For this question,
actually, I need YB, which is not in the table. Is that correct? Or it’s not here, then I
have to use the curve at X equal to 30 inch, is that correct or not, L equal to 4. So I’m
just discussing the whole. Now, what do I do for this part which is only bending from
here to here, yes or no? What do I need? Do I need YB? Some people do that. This is a
cardinal mistake again. They take YB and add it to this one. Is that correct? No. No. Look what happened here. I want you to
understand that. This is a beam loaded like that. Everybody see what happen. The entire
beam goes into curvature, yes or no? Now, I have here load, uniform load, or single
load does not make any difference. Everybody should understand that. When I bend this,
look what happen. From here to here, it goes through the? Curvature. Curvature. Is the rest of it has a curvature? No. Or a straight line? Everybody see that? Yes,
a straight line but it is tangential this curve, is that correct or not? So what happened
here, you want to put this in your note. So this one become– this is in exaggeration
of course, become a line tangent to that, and you are not looking for this BB prime.
You are looking for distance, CC prime. Is that correct or not? Yes? How do we calculate
that? That’s the question. What is given in the table? N. N equal to how much, first of all. N equal
to not 40 inch. N equal to? Thirty. Thirty inch. So this is given. Let’s write
it down. This is– This deflection which is BB prime. BB prime is given, is that correct
or not? B– Everybody following what I’m saying that BB prime is under the load, the load
finishes there. The rest of the beam doesn’t matter. Is that correct or not? Yes? That
is for the length of 30, not 40. So I’m putting here 30 inch again. So I get minus 30 inch
times– Oh, sorry, minus W. W is how much? W was? Fifteen. Fifteen. Minus 15, length 30 to the power
4 divided by 48EI, yes or no? Is that correct or 8EI? What’s that? 8EI. 8EI, OK. 8EI. Is that what I want? No, that’s
part of it. Yes or no? That portion is equal to that portion. Yes or no? So that is, let’s
say, C1. See, C1 already given. I need that but this is geometry. Everybody, understand
this is the routine that you are– you have learn. This angle is what? One more time.
That angle is theta at B, yes or no? Yes? So if I give you this is theta B, and I give
you this is 10 inch, can you calculate that one geometrically, yes or no? It’s a triangle
but the angle is given, the length is given, is that correct or no? Yes? So I put it here
in [inaudible]. This is what’s going to happen. If this happened to be a straight line, please
write it down. And this is theta, and you are looking for distance D depends on this
distance A. Everybody, the more down you go, the further it goes down. So you write it
like that. Tangent of theta equal to theta, why? Why can’t I write tangent theta equal
to theta, because theta is? Small. Very small. It’s less than one degree. Everybody,
understand. Other ones– If it’s 30 degree, I cannot do that. Everybody, understand that.
Since theta is very small, tangent theta equal to theta equal to D over A. Is that correct
or not? I’m not looking for A. A is given. Theta is looking– we are looking for D. Therefore
D become equal to A times theta. This formula, you’re going to have to remember it. It’s
all going to happen many time in your homework. When you want to extend the beam further off
the load, you have to use a triangle rule, is that correct or not? Yes? So therefore,
what do I need at that point? What– This is C1, this C is C1, is that D, is that correct
or not? What’s the A value is [inaudible] but I need the theta at that point. Where
is theta at that point? Look at it. Theta is how much? You see that’s the slope. Is
that correct or not? Slope at that point, that point is minus WL3 divided by 6EI. So
you write a slope there minus W30 to the power of 3 divided 6EI which is the theta. You need
to multiply it by this. You need to multiply it by 10. Is that correct or not, to get the
distance. So multiply it by 10, the unit force, et cetera, and this is 8. So that’s what I
put there. Theta and A. You have to use this technique anytime you go beyond the load.
Everybody, understand that. Correct? Look, another homework problem. You see, you are
all quiet. You see that if you are looking for YC, when you break this one, you have
this load at C. But when you do this one, this one the load is here, you want– this
goes like that and it goes a straight line. Is that– Because it’s not simple as that.
Is that understood by everybody? So I don’t want you to do any mistake that you go beyond
the load after the load. If this beam is going to expand, it is going to– it is going to
be a triangle rule. Is that correct or not? Which consist of– One more time, this and
this theta and this length, that’s all you need. This, which is already– I’ll put it
there. See that this is case 1. For case 2, I have this combination. First, I calculate
BB1 then I want CC prime, yes or no? But CC prime equal to theta times 10 or tangent theta
times 10, is that correct or not? Yes? Everybody is so quiet. Yes or no? Yes. What’s the difficulty? Did you get it? The big brother is watching. The big brother watching. Did you get it? It’s so easy [inaudible]. So easy. You know what, that’s what I’m saying.
If it is so easy– No, the last part, you said Y equals to 10. What? You said 10 theta equals to 10? I put it there. Oh. If– Go ahead. It’s just, this is high school.
If this is theta and I give you EI, how do you calculate the tangent delta equal D over
A, yes or no? What tangent theta equal to? Theta. Theta because it’s small. It’s one degree
or it’s less. Is that correct? Tangent theta B become equals B, become equals to A times
tangent theta, yes or no? B equals times A times theta but [inaudible] tangent theta
I’m using theta because if it is less than a degree, they are the same. Is that correct
or not? So what do I need? I need theta at that point because this is the slope of the
beam at that point. Yes or no? That’s the theta and that’s the 10 inch to get that little
job. But you are all looking as if I did some magic. I didn’t. It is just simple fact, correct
or not? Yes? And theta use these radians, does that make
a difference? Yes, of course, it’s radians, not in degree.
Of course. Come on, if you don’t remember that I did that in ME– in the static class
or ME don’t remember that take a tangent of 2 degree, change it radian, do it because
I don’t have time. Take calculator, everybody. I think I did that in ME 218. I don’t have
to do it here. But just since you asked, get a– take 2 degree, change the 2 degree to
radian, what do you get? Theta equal 2 degree change it radian. Theta become point-o-something.
Is that correct or not? What do you get? Point? 0.0349, OK? Now get sine of theta, look what
you got. I just answer it here, you just check things. Is that correct or not? OK, and get
that tangent of theta. I would like the answer 0.0349. Is that correct or not? Yes? Why?
Why sine and tangent and theta in radian, they are all the same? Why? Because the angle
is very? Small. Small, so I can use it. Yes? If you want to
use tangent theta, use it, it’s not going to– theta does it for me. Yes or no? Correct?
Because this is small theta, we did that in ME 218 to show you– actually, if the formation
is not small, I cannot repeat this for this class and then move on with the lecture. It
is such an important thing to know. Actually, everything that you learn in ME 218 works
if the formation are small. If the formation is big, none of this will work. So you cannot
apply that to a plastic, the formation. It doesn’t– It has to be an elastic material.
We are talking about steel because all the formation and all this slope, anytime you
walk into this beam, in this room [inaudible] people. Do you see that this beam is bending
like that? It’s not a cable that bends like that, is not because the cable does not follow
this rule. This is all about very minute, why, very minute theta. Everybody, understand
that. So this rule apply. Everybody see that? For angle less than 2 degree. If you go to
3 and 4, still you can use it to three-digit accuracy. But after that, you lose the accuracy.
For larger number, this is not true. Is that correct or not? For a small angle less than
2, theta and sine and tangent, they are all the same. That’s what I’m using in different
formula. Is that understood? Yes? Correct? So I’m using theta instead of tangent theta
times 10. So there’s nothing there. Anyhow, this is the end of a story. You can use that
in your system. Is that correct or not? Let’s do another problem. So I’m going to erase
that. That technique that I put in there, you should put it in the box. And then follow
it into different system because this happened a lot in your system. Yes? So for theta C, we just need [inaudible]– OK, theta at C. That’s a very good question.
Because I just want it because we’ve discussed that. Now, let’s calculate theta C. Very good
question. Now, if you understood what I did here, theta C, for this one, I have to refer
to the table. Yes or no? That’s what you are asking. Good question. So for that one, this
is the theta C, yes or no? Minus P, let’s do that. So for that one, this was YC. The
YC consists of system 1 but system 2 is this which I added one more time this through that.
Is that correct or not? Geometrically, I did it, which is– end up like that. Now, I’m
going to erase that and this time because I ask you to calculate theta C, that’s why
he’s asking. So theta C become equal to theta C, this is theta there. Is that correct? And
it’s given in the table. What– How much is it? Minus PL squared divided by? 2EI. 2EI, very good. Minus P. P is 125 times L
squared, you said, yes. L which is? Forty. Forty, that’s right, squared divided by 2EI,
correct or not? Now, coming to theta C. That was his question. What’s theta C equal to?
Theta C is into– equal to theta B, because this is a straight line, yes or no? The straight
line has the same theta, yes, correct or not? So I don’t have to adjust it, is that correct
or not? Is that what you are asking? Yes? So do we add the two? So this theta is equal to that theta, yes
or no? So theta at C is equal to theta at B. So theta at B, how much was it? Was it
minus W? How much was it? Minus W– L. L3 divided by sigma. So I don’t have to add
anything to it. That’s what I thought that was your question. Is that correct or not? Which L do we use for the second one? What L do you think I should use for the second
one? What’s the length of the beam? What is the length of the beam? The length of the
beam, according to here, is the length of the load. The load is how much? Thirty. Thirty. Yes? I’m going to do it. We already
did it before, so I don’t know why you– This is minus 50. We’re going to 50– I’m sorry,
50 30 to the power of? Three. Three. Three divided by 6EI, and that’s the asset,
that’s the thetas. This is a little bit of adjustment that you have to do according to
your table. Everybody, understand that. So this is of course with the same rule that
I said look one more time, guys. Look, if I do this, as you go further, the deflection
become more. So it depends on the theta and it depends on the 10, the length here, the
10-inch length. The 10 inch, it would become 15 inch. They also become– Y become larger.
But this is a line normal curvature, theta stays the same from beginning to the end,
yes or no? So I can use it here, I can use it there, that’s the same line and have the
same angle. Yes? So for an exam, you’ll give this too or would
you– Of course I will give it to you. You don’t
have to remember any of them. You want to remember? No. No. No, that would another course
for you need to remember all of that. Are you kidding? Of course not. I will give the–
all the table to you. But how to use the table is the question. You have to practice through
example. I think this is very simple. I cannot spend anymore time on this one. Now, I don’t
know why you are sort of quiet. This is just a geometry tangent of theta equals D over
A, calculated D. Is that correct? Yes? So that is so simple, maybe you are surprised,
I don’t know, whatever. Is that correct? Let’s do another example which I did for the other
class as well. So then, let’s put that down a little bit. So this time, this is the practice
of this. Now, be careful of what you are saying and what you are doing. In this table, I already
talked to you last time about all the significant part of this table. The rest is you. The only
problem that people have with this superposition method, they do not define their action first.
So be careful what you are doing. Every time you want to use superposition method, you
have to see what you are doing or what the question is being asked. Look at the next
question. Now, this time, I’m going to practice this part of the table. So here, maybe I went
a little bit too far and then down, so let’s go a little bit higher. OK, that’s the one
I want. OK, so this is the problem, guys. Write it down. This is not in your book. So
this is– or in handouts. If I put here a 400 newton load here and 400 newton load there
and this is 1 meter and this is 1 meter, another meter, 2 meter, this is no problem for you
to calculate the? Reaction. Reaction because this is a determinant problem.
Actually, it’s symmetrical so each reaction become 400 and 400. Is that correct? So that
will not be a question to ask, is that correct? However, if this is the system, I can ask
you to calculate theta of at A, or what else did I use for this problem. Let’s use the
same format that I used before. So here we go. If I can find it. Oh, here it is, finally
find it. So A– Oh, here I use– No. Here I used A, here I used C, and now I’m going
to change this problem by putting one more support here. This is 1 meter and that is
1 meter. So the length of the beam is 4 meter. And there, at the middle of the beam, there
is an extra support. Aside from A and C, there is another support at B, everybody see what
I’m talking about. Is this a determinant problem or is this– I don’t want you to get wrong
because when I gave you homework last time, three problem you have indeterminant problem,
the rest was determinant. To the other class, I give a quiz which was indeterminant and
few people missed it. For you, I gave you a quiz which was determinant. Some of student
in that class objected to that and said why did you give them the easy one and we had
the difficult one because some of them totally missed it, indeterminant versus the determinant.
Everybody, understand. This system right now is indeterminant, why? Because I just did
what I did. I added to its support there, yes or no? Now, you have three verticals load
and two equation of equilibrium. Sigma Y equals to zero and sigma M equals to zero which you
could never find the reaction unless you do some crazy things that happened in the other
class. Is that correct or not? Yes? That means that in the indeterminant problem is not static
equation, it’s not sufficient to solve those equation, yes or no? Correct? Everybody with
me? Now, what should I– how do I show– I want to use superposition method to find the
reaction at A, B and C? Do exactly what– Is this the support then? Yes or no? If I
take this support out, what should I put underneath it? I should put there a? Reaction. Reaction. But do I know the magnitude of that
one? No. No. So I’m going to put it there. Look exactly
what you just feeling about it and I’ll be there. Yes, if somebody come and suddenly
tell you RB equal to 100 newton, is the problem solved? If I give you– look– You have to
look at the static of the problem. This is for 400, this is 4. If this is 100, there’s
no– it doesn’t need a genius to find out what FC is. Is that correct or not? If I say
this 100, this is remaining 700, which is symmetrical, so 350 goes there, 350 goes there.
Yes? So suddenly, the system from the– this indeterminant become a determinant. The question
is what is RB? Yes or no? So that is your criteria. The question is, if I calculate
RB, then I don’t have to worry about A and C because there I can use equilibrium to solve
for the other one, that’s what I’m saying. Yes? RB equal to what? Is that the question?
Yes? How about, anytime you have a reaction, you have a boundary condition. What’s the
value of Y at that point? Zero. So why not use that as your criteria? Is that
you have to set up– If you break these into three different beam in different load when
I add those three beam together, Y and B must be equal to? Zero. Zero because this is zero. Look at it. This
beam is going to bend like this, something like that. Is that correct or not? Yes or
no? This means Y at that point, that’s exactly what it means. Y at that point must be equal
to zero, correct? Now, let’s do that. So in order to do that, we are going to break that
into three system. Let’s go a little bit quicker. So this is one. I put a 400 load here, newton.
This is 1 meter, this is 3 meter. I have to go– I call this little table. I’ll show you
the table. Is that correct or not? Yes? And then we have to go to this beam which has
a load at the middle, RB. Granted RB is not known, is that correct or not? Yes? However,
this beam is going to bend like that. Is that correct or not? This beam is most likely going
to go like that and go like that. Is that correct or not? This is heavier, this side
than the other side. Correct or not? Yes? And find out if you have another beam, you
don’t need to draw this. This dashed line just to tell you what the idea is. And this
one, there is a 400 here which will be a mirror image of the other one because it’s symmetrical.
Sometimes, some of your homework is not. So this goes down there, this goes like that.
Is that correct or not? Yes? Is this going down? No. Here is the key. This is A, B, C,
D, E. Where is D? This is point D, this is point A, look at the table. At the table,
I have this. Is that correct or not? Yes? And look at this. This Y-max or whatever given
is that at the middle or is it where I want to? No. It’s not. Is that correct or not?
This is that certain different point. This Y-max happen to be at here. You see here,
it says this is X-max, is that correct or not? Yes? Therefore, what do I want to do?
I want you to look at the curve. Is that the curve? Is that correct or not? I need– look
what we need. We need not a D, not an E, we need the formation at point B, yes or no?
Isn’t that what we want to set up? What is this B? This is YB, is that correct or not?
So YB of this beam plus Y of this beam and Y of this beam, is that correct or not? Together,
should be altogether must be equal to? Zero. Zero, correct or not? Yes, that’s the question
you want to answer. Now, you draw this or at least show me there telling the math part,
you make a couple of mistake here, that’s the other big deal. But not putting this and
not defining your system, that is– you’ll– it doesn’t get you anywhere. Is that correct
or not? Again, one more time. Since this YB pushes B up such a way at this point, you
have no deformation. When you break it into one, two, three system, the sum of the Y at
the same point B must be equal to zero. You cannot add– This is what the mistake a student
make some time. They add this maximum to that maximum, to that maximum. That’s irrelevant.
Everybody see that that’s a wrong approach. Yes? OK. So we want this. We want this call
this system 1, system 2 and system 3. We want YB of system 1 plus YB of system 2 plus YB
of system 3 together must be equal to zero. Now, let’s see whether we can read it. You
have your table in front of you. Hopefully, if you want, take a look at that. Let’s do
the one that it is easier. Look at this one. First of all, look at the table, what is the
table there. I mentioned that last time. I don’t know whether you recognize it. I mentioned
it in class. A must be larger than B. And I said if that’s not the case, you have to
reverse it. Is that correct or not? Which one is true? Which one matches this one? Not
the first one, not the second one, the third one. Yes or no? Look at this one. This is
A, yes or no? A equal to what? Three. Three meter. Yes or no? Yeah. B equal to? One. One meter. What is X? What is the distance
X? X equal to where of Y? At the middle of the beam, yes or no? X equal to what? Two
meter, correct? I need Y at the middle, yes or no? Correct? So which one– Can I use this
one? Of course, I cannot use that. Although, I cannot do, what do I need? I need for X
less than A, is that correct or not? I use this equation. That’s the value of B, that’s
the value of B, that’s the value of L. L equal to what? L equal to? Four. Four. X equal to what? X equal to 2. Everybody,
B equal to 1, A equal to 3, is that correct or not? As the picture show, that’s exact.
So that’s exactly what I’m going to do there. So the first one or the third one, I’m putting
here. YB of 3, so let’s do this one first. So that’s what I’m saying. So P. P is equal
to RB– no, it’s 400, I’m sorry. Four hundred times what? I’m not looking there because
I need another table from you. Anybody has a table? Yeah. OK. Thank you. So OK, 400,
OK, times B. B is 1 divided by what? Divided by 6EIL. So 6EIL, L equal to 4, is that correct
or not? Yes? Multiply it by X to the power of 3– 2 to the power 3 minus L to the power
of 2 which is 4 to the power of 2 minus B square 1 to the power 2, and the whole thing
multiply it by 2 and this is the bracket. Notice, people think this is positive but
it is not positive. This is a negative. I mentioned that because this quantity become
3 and it’s larger than that one. Everybody understand this quantity is larger than that
one, so the whole thing is negative, so don’t worry about it. That is this one. Is that
correct or not? To that one, I have to add this one. Yes? Is this in the table? Look
at it. Yes. Yes. That is– Four. — at the middle. This is not this one. This
is the other one. Yes or no. This is good. What do I need? I need Y-max, yes or no? Y-max
equal to, how much? PL3 divided by– Forty-eight. Forty-eight. However, P is going down, my
force is going up, so I have to change signs, obviously. Is that correct? So [inaudible]
plus RB because now I don’t know the RB. That’s the RB, become my load. Is that correct or
not? Yes? Multiply it by, here. You said that, you know, L3. L is how much? L is? Four. Four to the power of 3 divided 40EI. Now,
what do I do? Now, I did this one, I did actually backward. I should put it on there but– because
this is the [inaudible]. I did that one, I did that one, now we get to that one. Why
should I do that? What? I said– No, it is– this is two thing. First of all,
I have to reverse it, yes? So I reverse it. I go this way, is that correct or not? So
A become 3, B become 1, so I have to look at that. However, since this is at the middle,
it doesn’t make any difference. Is that correct or not? Yes? Because this is equal to that.
If it was here, then it would have been a different story. I have to add two different
number together. Since at the middle is the same, so all you have to do, multiply that
by 2 in this problem. Notice, if there was any other location, you have to change that.
Is that correct or not? Yes? So therefore, you multiply that to that become equal to
zero. As you see in this, because the static has nothing to do with the EI. EI is going
to drop out. Static never is dependent on the EI. EI would be dropped out [inaudible]
equation RB, you calculate RB. RB become equal 550 newtons. As soon as you find RB equal
to [inaudible]. And then now this become a determinant problem. Look what happen, RB
equal to 550 newton, correct or not? So all this superposition and practically every method,
all the method I give you, double integration, [inaudible], direct method, all of them, many
comes through the indeterminant problem actually solve the static for you. Is that correct
or not? After static is [inaudible], you can draw your shear and moment diagram, then you
can design the beam and that’s the end of a story. We always check the deformation but
deformation is for some machine parts, something you want to have very small deformation. And
we always be careful about that, but the formation is not the issue generally. First, the design
is the issue, is first, is that correct or not? But after you design if you don’t want
the formation go beyond certain limit as well, because– after all your equation is not going
to be valid anyway if you have large deformation. Is that correct or not? That become issue,
as you see all of these method finally gives you your static, so this now static is RA
and RB. Of course, it is symmetrical. So this one become 250 divided by two so this become
125 newton and this is 125 meter then the problem is done. Another type of problem which
is very important is the– let’s keep that one table. Here is your handout, look at the
handout. On your handout there are three problems there and one of your homework is similar
to that one, you have to understand again the concept of the superposition, so what
we are doing there. There are three handout problem. Number one and three, you are supposed
to do it. Number three actually is very simple. You have to be careful how many unknown you
have and how many equation you need. Number one, is the one you are– is your homework,
I’m going to do number two. Number two is the following problem, right, which is a beam
coming here a fixed beam this– I can show it three-dimensionally or two-dimensionally.
That’s sitting on another beam here. And this one as its roller support here and a pin support
there. This is a fixed support. This is one and a half meter. This is one and a half meter.
This is one and a half meter. This is nothing in scale and there is uniform load of– I
forgot how much was it, the uniform load of– Four Four. Yeah, OK. Four kilonewton, and uniform
load of 4 kilonewton per meter is on top of that and we want to solve this problem to
calculate the reaction at the support. Obviously, this is a special problem. This does not qualify
for any other method actually superposition method is a good method for this type of problem,
yes or no? Which happens a lot in all the engineering work, you are designing a beam
sitting in another beam. Everybody understand that? Crossing each other what’s going to
happen between the two? Is this, first of all, determinant or indeterminant? Indeterminant. Indeterminant, loss of reaction. Is that correct
or no? Now, what is the key is important as far as you guys as engineer. Notice, I have
to measure a few things here which is important. You have seen this problem. Actually, the
first problem that I did as indeterminant problem is this problem, done by that. Is
that the same problem, yes or no? Yeah. Yeah. How can you modify that? Yes or no? That’s
the answer. Yes and no, which one? No. No. Yes, yes because it looks the same, yes.
You know what the difference between this and that? That if this is point A and this
is point B and this is point A and this is point B, is this point B going down? No. Now, that’s the whole thing. This is a
rigid, this is sitting. This is a beam, elastic beam sitting of massive concrete which is
rigid. It’s not moving at all. Is that correct or not? Then here, is this happening the same
thing or no? If this beam is– If this is plastic and this is a rigid beam, yes, but
is this beam going to bend? Yes. So that point B is not going to bend. So that
respect is not the same, it looks the same but it’s not the same I want you to understand
the difference between when we put a support there and when something is moving up and
down a little bit, is that going to– point B goes down or not? It goes down. Yes, OK. How much down does it go depends
on the flexibility of the lower beam. If I put something like a plastic like this which
does not take much low to bend, it has no effect. If I put a concrete below, that is
going to crush this, is that correct or not? Everybody understand that? However, if I make
this of a steel beam very heavy, it’s going to resist the other one. How much? We don’t
know. Is there going to be a transaction of force, transaction between the two beam, yes
or no? Yes. Do you know that magnitude of that? No. No. Therefore this is the scenario, this is
something you have to figure it out that’s why I gave you a couple of this homework.
Now, you all– I call you junior engineer, did I call you junior engineer before or not?
Yeah, you are a junior engineer. You have to think like that, you have to resolve this
issue. You are going to design part of a thing that is not in your textbook, something similar
to that which should happen a lot. Is that correct or not? However, analysis of that
is this that you have here a lower beam, let’s draw the lower beam first. Is this the lower
beam? And this causes– let’s put it in color because when it is in color I put it as I
draw. Is there an R here, yes or no? Let’s put that R, that’s a load. Is the top beam
pushes the lower beam down, yes or no? Is the lower beam is going down like that, is
that correct or not? Yes, this is point C, this is point D, this is point B, correct?
Yes. What happen to the outer beam? Can you draw the mid free-body diagram of outer beam?
This is case 1 or free-body diagram of this beam, is that correct? Or beam two? Let’s
put it– this is beam two, I put here beam two. Is that correct or not? Yes? Maybe you
want to call this then case 2 as well. Right. Now, case one is this. Is this the case 1?
Yes? Is that complete or not complete? Not complete. Not complete. What do we have at this end
then? Four score equal and– Opposite. — opposite, is that correct? This is all.
This is the representation of free-body diagram of upper beam, free-body diagram of the lower
beam. The only thing is, what’s the value of R? We don’t know. Yes? The question is
what is R? Now, if I want to calculate value of the R which is out here and that one, what
should I say? Now, the condition is set there. What is the condition now? Y at B. This cannot
go down here and that stay up there. The two of them going together, is that correct or
not? So you have to write, this is system 2, this is system 1, is that correct or not?
This is the key. You draw it like that then you say YB of system 1 which is this one,
must be equal to YB of system 2 in the same magnet. In other words, if this is one goes
down quarter of an inch, this one must be going down quarter of an inch. Yes, everybody
agree on that? Yes? However, this one is very simple to calculate. This is system 2, how
much is that? Is this in the– Look at this beam. The load is right out the middle, yes
or no? We already used it. Which one is that one, that one is? Four. That one, case four. Yes or no? How much load,
how much deflection do I have at the middle? We already used it in the other problem, minus
PL3 over 48EI but the P changes to? R. R. OK, so this one the right-hand side, this
one. I’ll put it here. Now, the right hand side is R minus R is going down, minus R.
This is the– L is how much? L equal to? Look at the L. L equal to? Three. L equal to? Three. Three? Yes or no? No, but they are not responding.
Three to the power of three or four? Three. Three. Divided by how many 6EI? Forty-eight. 48Y. Is that the deformation of this beam
at the middle, yes or no? Which is symmetrical, yes? Yes. Which is case 4 symmetrical, this is unsymmetrical.
That’s the only reason I’m doing all of this problem, yes or no, correct? So therefore
that is that the formation I want, PL3 over 48, correct? Now, we come to this one, now
is this one load or two load? Two. So I have to break it into two part is that
correct or not? Then this become equal to, now you’re getting the idea. You get this
one which is causing this to go down, yes or no? Plus, what? This one which causing
to go? Up. Up. But this total of this one must be equal
to that one, yes or no, correct? This is case 1A, let’s call it 1A. This is case 1B, is
that correct? But two of them together will be this one, yes or no, correct? Now how much
this will go down? This is– Which case is that? Do we have it there? I have to pull
it down a little bit. So let’s go– or up, sorry. Is that that one? Yes or no, correct?
That one, everybody see that? This one? You don’t see that. OK. This one yes? And how
much W do we have? This much, yes or no? Minus WL4 divided by? 8EI. 8EI. So minus W is going down. OK, minus W,
W was how much? Four kilonewton. Then we look here. So we’ll have it here. L4, what’s L4,
this one. L4, this one is 1.5. So you have to be careful here. 1.5 to the power of– Four. — 4 divided by 8EI, correct, 8EI. Yes, 8EI,
is that correct or not? Now, I have to add, this is negative, this is positive because
it’s going upward, yes or no? But this part also is in the table. I have to go a little
bit further up. Is that that one, yes or no? Is this that one, yes or no? Yes? In opposite.
This is negative, I need to put it– Positive. — plus. So plus R times length, length is
again one and a half, one and a half to the power of 3 divided by 6EI, is that– 3EI. 3EI, OK, good. 3EI. Yes. So this two must
be again equal. Now, there are– Sometimes, the problem is like that. The EI of this beam
and EI of this beam could be different. This could be aluminum, this could be steel or
vise versa, they have different sides. But that you put it here, but they eliminate or
one become twice as much the other one or one become half of that. But nevertheless,
in this case, I’m assuming EIs are the same. So assume EI same for both beam, write it
down, for both beam. It says in the problem, it’s in your handout. So if that’s the case,
EI and EI and EI will drop out. If not, you plug in the number, there will be a ratio,
is that correct? Let’s say if this is 30,000 PSI, this is 10,000 PSI. OK, so you put it
there, so a ratio of 3. Nevertheless, this is a formula for R. Is that correct or not?
You solve for R, R become equal to one and a half kilonewton. So that solve the problem
for you. As soon as you find the R equal to one and a half kilonewton, all the static
itself because this become one and a half kilonewton, then this reaction become 0.75,
that is you can find it there, this is just simple static, you know? Now, for the other
class, I had about 10 minute to give them the preview of next chapter. Now, I want you
to understand what all the chatter is about, just give me a couple of minute then I let
go. I don’t have that one. We are done with all this superposition with it, with all the
beam. We are still about one class behind or half a class behind, we can catch up. However,
next chapter, it is very important for you guys, especially for mechanical engineer.
Now, I need you to know this. In order to understand the next chapter, which I’m going
to go quickly over and there are lots of fancy problem coming at the end of it, which is
the real-life scenario problem is this. You need to know the strain, shear strain, which
is what? Gamma. You don’t– You need to know about strain which is epsilon. Not in one-dimensional,
in three-dimensional. Epsilon equal to sigma over– sigma X over E minus mu sigma Y Poisson
ratio. Shear strength, normal strength, everything that you learned in Chapter 1, everybody ME
218, plus the fact you need to know the Mohr Circle plus the fact you have to know outer
plane, inner plane, everything we talk about the Mohr Circle which you are now presumably
good at it, hopefully. And then we have to now do everything that we need for strain.
Now, we are measuring the strain because when we test a model, we measure the strain. We
never measure the stress. Everybody, understand that you are sitting, as I said for the other
class, you are sitting in the airplane. In the airplane, every part of the wing or the
body, there are strain gauges all over the place. So strain gauges measure the strain,
not the stress. Stress is in the body. You can never go inside the body but you can measure
the strain. When we get the measure of strain, we can now go back check to see what kind
of stress is called that kind of strain. So we need to draw the Mohr Circle for a strain.
We have to understand what the plane stress is, what the plane strain is, what the absolute
maximum strain is and all of that. I’ll repeat in a way, very simple repeat with one exception.
You have to understand the concept of a strain which is in Chapter 1, is that correct or
not? I hoping that you get a chance to take a look at it. I ask you to do it over the
weekend but I have no choice but to go ahead assuming you know all about that which is
in Chapter 1. Take a quick look at normal strain, shear strain, Poisson ratio and the
relationship at least to get by the–

## One Reply to “Strength of Materials II: Deflection of Beams, Superposition Method (12 of 19)”

1. gökhan yüksel says:

Cameraman
is sleeping between 54:00 and 57:30.