[ Music ]
Number one, I haven't -- I only have seen
one correct answer, so -- but I gave it only
four points, you know. I said it is fixed.
It's fixed?
Yes, it's not pinned. If the rod and the slider
-- notice in your handout they are all pinned
together, so you put one force. Here I said
it is --
Fixed.
-- fixed, but it can't --
Slide open.
-- slide. Is that correct, or not? So now
you can figure it out. There should be a [inaudible]
there, in other words.
[ Inaudible Conversations ]
That's exactly what I was asking. Yes, if
-- guys, all right. Let's -- guys, if there
is a slider -- notice on all your sliders,
the one is in the book, if there is a rod
and a slider there is a pin here. This rod
is pinned, yes or no? If you have a slider
here and this rod is being fixed here. You
see the difference between the two? Yes or
no? Can this rotate above that point? This
one can rotate. Everybody see that? So that's
the difference between this item, which is
-- this is the slider going on top of that,
and then this is a pin, and the rod is like
that, you see? That -- this rod rotates. This
rod doesn't. That's the only thing what I'm
asking.
I thought --
Never mind that some people make the mistake
on the other part, too. But this was the main
part. Okay, all right then, we'll discuss
it when I get those back, but that's not the
point, because this many people missed it.
Do we get partial credit if we put --
Oh, of course. No, you don't miss all four
points. You get one, two, or three, or four.
I purposely only put four points to bring
to your attention that the reaction is -- you
should be careful what you are designing with
that in future. Whatever you design, the reaction
is based on the -- if there is a freedom,
nothing there. Is that correct, or not? If
there is -- you are restricting a motion for
whatever reason, you put a reaction there.
So that was just -- so now, going to the trusses,
any questions about the trusses? Are you okay
[inaudible] section? Yes?
Could you write out the answers to number
five?
Answer to number five? If I have it here I
will put it down there. So I hope that I have
solution. I don't have my solution with me,
so sorry for that. So may put it on the Blackboard.
For number five?
Yes.
That's the only one?
Well, so for.
So far [chuckle]. So I take a check. I usually
have my solution with me, but today I didn't
bring it because -- okay, anybody has done
it correctly?
Five?
Five. Did you get the answer in the back of
-- I mean, in the solution manual [laughter]?
Well, that's the fact if you have the solution
manual, so you don't have it, you don't have
it. That's okay. Anyhow I will give it to
you. That doesn't matter. So are you okay
with the [inaudible] section? Or one more
example?
We have to come over --
What?
You started --
You didn't finish it. You just started it.
Okay, but I told you how to do it [laughter].
So we're all set, then.
That's all we want, you know. Well, what else
can you do? I said take the [inaudible] about
here and there, and that was it. No, I'm kidding.
So let's put another example down on the board.
It takes more time drawing the example on
the board, especially with a camera, but then
to discuss the solution for that. So I'm going
to do it anyway. So this is a typical trusses.
So let's say put it that this is the long
one. So here is, of course, as usual pin,
pin. So how many [inaudible] is that? One,
two, three, four, and 1/2, and 1/2. So here
we go like this. There we go like that. Then
we divide into four parts. So we go like this,
and this one goes like that. That one goes
like that. This one goes like that. And here
is a roller at point L. Okay, here is a pin
at point A. All right, the distances are as
follows, or you can -- these are all, again,
assumed to be all pin collection. You see
there are some zero-force members. You can
see it here. This one is a zero-force -- if
there is no force there. Let's see whether
we have any force there or not. So here we
put here a force here, a force there. Here
at the joint, as usual, not on the member.
So 1.2 kilometers each. And the distances
are one, two, to four meters each here, which
makes it 16 meters. And 2.25 meters. As you
see, it's very long, so if you have a very
large auditorium or a sport arena you want
to put something like that on top of that.
It is a -- of course, if you want to do method
of joint there are lots of members. You can
notice now it's -- for example, this member
is a zero-force member. This member is zero
-- but that's beside the point. We don't want
to discuss that, but that's one way of starting
this solution. Of course, we have to -- first,
step one always is to find a reaction. Is
that correct? No. We have -- how many reactions
do we have again, as usual? At A we have AX
and --
AY.
AY, and you know, there's so many points we
should call that. So let's [inaudible] called
A, B, C, D, E, F, G -- I don't know which
one -- this goes to the -- okay, let's finish
it. G, H, I, J, K, L. So therefore, there
is a reaction at -- two reactions at A. Of
course, AX equals to zero. We don't have any
horizontal forces in this -- at this truss.
So method of joint would be very lengthy,
and then if we make a mistake it would carry
over to other solutions. So we have step one,
finding the reaction here. Obviously, AX must
be equal to?
Zero.
Zero. Do I need to take a moment? Or can you
see what the reaction is at each point? I
purposely mentioned this before for you not
to do extra work, because this is symmetrical,
so I didn't put here. This is symmetrical
with respect to the sizes. Is that correct
or not? If this was 225, this length was 125,
would not have been a symmetrical scenario.
The loading is symmetrical, so whatever you
have at A should be exactly equal to what
you have at?
L.
L, as far as the reaction is concerned. So
that becomes 1/2 a load. Is that correct?
So 1.2, 1.2, one, two, three, four, five.
That's 6 kilometers divided by 2 each side.
It has 3 kilometers because of being symmetrical
scenario. And of course in this scenario,
as I said it before, you only solve 1/2 of
the problem. You don't have to solve the other
side because the other side will be symmetrical.
In this method we are going to use method
of section. The key element is not to cut
the truss more than three members, so depends
which three members you want to consider.
Sometimes in the homework or quizzes I will
tell you what to do. In general, if it is
your own decision you can cut it here if you
want. That's three member cut, is that cut,
or here, or here. Everybody see what [inaudible].
Let's say that the question is to find the
forces in this member, DF, DG, and EG. Is
that correct or not? As I said it before,
all you have to do is cut it into 1/2, then
you can consider either the left-hand side
or the?
Right-hand.
Doesn't make any difference because the forces
in these three members are equal and?
Opposite.
Opposite. Everybody understand what I'm saying
that, correct? Therefore, you have to cut
it and redraw it. You have to -- but for me
it is easy because I can erase the rest of
it. Is that correct or not? So if I want to
look at the free body diagram of left-hand
side I -- after I follow the reaction, step
one, this is step two. I'm not using method
of joint. I'm using method of section. I divide
that into two. I cut only three members. So
I have to look at the free body diagram. I
had both choices, either left or?
Right.
Right, so I looked to see which one is simpler.
This one has less members, so I'll go this
one. Is that correct or not? And then I have
to put forces in those three members. So the
forces on those three members, as usual we
have to put it in?
Tension.
Tension mode. That's another thing. Tension
mode is what? Going away from the joint. It
has nothing to do with static [inaudible].
Notice this one is going in that direction.
This one is going in that direction. This
one is going in that direction. Is that correct
or -- and those forces are F, E, G. Forces
F -- I don't need to draw that anymore, so
I could put it there or here. Doesn't make
any difference. This one is F -- let's [inaudible]
-- this is F, D, G. And this one is F, D -- F,
D, F. All put in tension mode. Okay, now the
next step is to finding this [inaudible].
This I leave it up to you. There is many way
of doing. Of course, last time one of you
said immediately you have to take the moment
about point D. Of course, we have to [inaudible],
but -- and I did not give you the height here.
The height here is [inaudible] equal to 3
meters as well. So as you see, in this direction
if I want this direction, this is three height
and four run. Is that correct or not? So immediately
this one has a four run and three rise, four
meter and three meter. This, too, of course,
is for [inaudible]. So you have many options
here. One option is to take the moment about
point?
D.
D, so you write sigma MDS, sigma M at or about
D or at point D equal to zero. And as you
see, there is nothing there to worry about.
This force, and this force, and this force
have no moment. F, E, G times 3. F, E, G times
3. Of course, it will be in kilometer and
meter. We'll put the unit in front. And that
is plus. It's going that way. This one and
this one -- this one has no moment. This one
has a moment of plus 1.2 times 4 meters. Is
going also plus. So this is plus, this is
plus. This one is negative, minus 3 kilometers
times 6.25 equal to zero. You, indeed, get
-- we get F, EC, EG equal to F, EG becomes
-- by calculating that one becomes plus 4.65
kilometers. And therefore, since it came plus
it means the tension that I have chosen is
correct. Yes?
So we eliminated A of X from the symmetry,
correct?
AX was zero. But no, not because of symmetry.
Actually, because there was no horizontal
load. If there was a horizontal load it would
not be symmetrical. Actually because there
was not horizontal load. If there was a horizontal
load it would not be symmetrical. Did you
get the answer? Actually that's a very good
question. Sometimes people make a mistake
here. Let me elaborate in that one. When there
is a horizontal load there, the system will
not be symmetrical. All the -- look, this
-- I have seen this mistake, and I'm glad
you asked that question. Let's go to any structure,
not that one. Let's say that you have a structure
like that. Everything is symmetrical. Is that
correct or not? Similar to that. Let's say
you have a load here and a load there. Is
that correct or not? Right now if all the
dimensions like that are symmetrical, this
reaction and that reaction, vertically it's
going to be equal, yes or no? But as soon
as I put -- either everything is symmetrical.
If for some reason I put a force here, horizontal,
is this symmetrical?
No.
No, because the reaction here -- now because
this pushes this way the reaction here is
larger -- even the vertical reaction. The
reaction here is smaller because this force
is creating a moment that way. Is that correct
or -- and the system wants to react to that.
Therefore -- so this is no longer -- the reason
that this was reaction actually exactly, because
I could have AX and LX equal and?
Opposite.
That again becomes symmetrical. Everybody
understand? But if I only put one load on
one side, that's not symmetrical. All right,
anyway, is that understood, the question and
the answer? Okay, all right. Anyhow, so this
is very simple scenario, and therefore, we
go from there. So the rest of it is using
sigma FX and sigma FY. Everybody see's it's
very simple to calculate that? So you can
use, for example, sigma FY first, because
this one doesn't have any Y component. Is
that correct or not? Sigma FY equal to -- so
let's finish the problem. Equal to zero. Here
you have 3 kilometers going up, minus 1.2,
minus 1.2. This is minus, minus, plus. And
this one is minus 3/5 of FDG equal to zero.
And FDG becomes equal to -- [Inaudible] become
1 kilometer. Means that is also in tension.
Okay? Yes, the next is writing sigma, FX equal
to zero. And finally the last force -- the
last force will be FDF. So when you do that,
sigma FX equal to sigma FX, equal to zero.
Notice we have nothing there except starting
from top is FDF plus 4/5 of -- plus 4/5 of
FDG, but FDG is 1. And then we have FEG. FEG
is plus 4.65 equal to zero. FDF ends up to
be equal to minus 5.45 kilometers. And obviously
this one is in?
Compression.
Compression. Just ask you a question, what
do you think happened to the other -- this
-- all of these members on top and all the
members on the bottom? Can you guess it? I
want to see whether -- when you count the
strength of material you have some idea there.
Can you guess? The top one ends up to being?
Compression? The bottom one ends up to being?
Tension.
Tension.
[Inaudible] split?
Is it performing like that? Is it the formation
of truss because of the load there? Look at
this side. It's as if it's getting crushed.
Everybody on this side getting?
Exposure.
So ahead of time you have some idea what's
going to happen. Is that correct or not? Of
course, when you do more of this then you
learn it. Everybody understand? It's not necessary
[inaudible] for a static student to know all
about this, but later on in strength of materials
classes and advanced classes you are going
to design every piece of this truss, for example.
Members are two-force member. That's week
one to three in ME218, we are designing all
the members and all the joints. We are designing
the pin. The pins -- or maybe [inaudible]
stresses there, normal stresses there. We
design the size of that. We have to tell you
all about the strength of material, what material
to use, why, etcetera, etcetera. Everybody
understand? To come up with reasonable size
and reasonable factor of safety. These are
all future courses. But notice that I have
to find the forces first to design something.
Every -- and that's why static becomes so
important. Static is the cornerstone of every
problem. You -- any problem in ME218, practically,
starts with static first. Every object has
to be first -- you have to find the forces
to design that object for those forces and
moment. Everybody understand what I am saying
there? If you don't know what moment is in
that inch, if there is any, how do you design
that object to resist that moment or to resist
that force? That comes later on, but the issue
remains the same. Many of you have come to
the office hours. You saw that all the students
in ME218, most of their questions, not all
of them -- actually, 50% of their question
is all about the static, and they have some
sort of problem. For example, this morning
we were talking about eccentric loading, how
the eccentric loading behaved with different
structures, how do we take that into account?
It's all about moment, about the X axis, and
Y axis, and Z axis, etcetera, etcetera. Of
course, some can do it very easily, some have
a little bit of problem, etcetera, etcetera.
Anyhow, that's the end of the trusses. We
want to go to the frame. Anymore questions
there? You got any questions? Did you see
-- no. So in the book you can -- other alternative
is available. Rather than using sigma FX you
may want to extend these two lines to get
to this point, because these two lines are
going to intersect there. Yes or no? They'll
take the moment about this point. Everybody
understand that? Because then this force and
this force will be having no moment at that
point. That's the one I guess they are using
in the solution manual, or in the book example.
Is that correct or not? So you have that -- but
it is so simple you don't have to worry about
it. Yes? Okay. Now frames. Now let's let me
explain what the frames are. These are frames,
but they are not actual frames. So -- I always
have a little -- I have to explain that a
little bit [inaudible] to tell you what are
these -- I call it special frame. In the book
usually they call it frame. They are frames.
So what are they? Write it down, definition
again, the goal and the solution. If we can
write that in three paragraphs so we can go
forward with the example. Now the frame of
our structure with several members -- please
write it down. Frames are structure with several
members. Some of the members are pin-connected.
That's why I call them a special frame. Some
of the members are pin-connected, not all.
You don't have to write the rest of my discussion.
If they are not pin-connected, all the members
are fix-bolted together or welded together.
That becomes a regular frame, frame of every
building that you see, large building if they've
got a steel frame or concrete frame. That
is very complex problem. As I said, it has
many, many unknowns, and you cannot touch
it until two, three years down the road. Near
senior year or even graduate work, to be able
to find out how do we design this frame? Of
course, civil engineer they give you -- in
the third year or fourth year they give you
a program, and if you put everything in the
program the problem solves it for you. That
doesn't mean that you are solving it, but
however, you get the answer that you want.
Everybody understands it. Writing a program
or knowing how to set up that structure as
solve all the unknown is an issue that you
learn. If you are interested to go to the
graduate work that's road you can take to
find out how that's being done. You need to
take a few math courses. You need to take
a few engineering courses to be able to set
it up as a metrics, and then metrics you have
to solve that N equation. And N unknown in
general, but again, you do it by the help
of computer. So you have to be able to use
the computer to write the program, to read
the program, change the program, and [inaudible]
that. There are lots of methods available
for civil and mechanical engineer non-stop
program that previously engineers like you,
when you do that field and start doing that
from [inaudible] the first time the computer
came out, and then got more advanced, more
advanced. There are programs here. The department
has it. They put it in your use, you just
put the data in, you get the answer as usual.
But somehow somebody has written that program.
Everybody understand that? So that is the
idea of the frame. Now this frame -- what
did you write down? This frame is a special
frame. Has several members -- or pin connectors,
yes or no? Now what's the goal of this objective?
What's the [inaudible]? We want to find out
the forces acting -- write it down. The goal
is to find the forces acting at the end of
each member. The goal is to find the forces
at the end of each member. So that's the goal,
which is a little bit different than trusses,
because trusses are two-force?
Members.
Member. Frames are not two-force members.
They may or may not have two-force members
in it. Everybody understand that. In general,
they are not. Therefore, the members can bend,
twist, etcetera. And you don't care about
that. You are not looking for the [inaudible]
static part of it. So the solution, to reach
to that goal you need a solution. Solution,
step one is the same as before. Step one is
consider the free body diagram of the entire
frame. Consider the free body diagram of entire
frame. Are you writing it down? Solution,
consider the free body diagram of entire frame.
Use equilibrium of rigid body -- like the
previous chapter of the book -- use equilibrium
of rigid body to find the reactions, in parentheses,
if possible, because sometimes it's possible,
sometimes it's not, especially about the frame.
So far in all the trusses I gave you, finding
this reaction was possible. In frame some
of them are possible, some of them are not.
But don't worry about it, we still will be
able to solve it. Step two, which is a most
crucial point, step two, separate all the
members 
at the joint with the pin -- separate all
the members at the joint -- or at the pin
joint. That's better, right? At the pin joint,
and consider free body diagram of each and
every member. Consider free body diagram of
each and every member. Okay, this was only
three paragraphs. One was definition of the
frame, the special frame. Framework has several
members, like trusses, some of them are pin-connected,
not all. If all were pin connected and the
loads were at the joints, then that would
have bent this truss. If I draw something
like that -- this is for the purpose of discussion.
You want to write it down. So if I draw here
a member like that, here it's fixed. Here
is a rod going like this. And here is a roller.
And I believe I want to show you that. Okay,
so we'll go there. And let's say that I have
here put a pin here, and here I put 500-pound
load there, and put -- I put here 500 pound-load
here. And let's say that we have -- distance
is given to you, five feet and three feet.
And the vertical distance is given, four feet.
And let's say that this 500 pounds is right
at the middle of this [inaudible]. Height
is two and two. Now is this a frame or is
it a truss? You have to recognize from is
it a truss or is it a frame? Obviously the
discussion is the frame. It must be frame,
yes or no? That's [inaudible] frame. Why?
Because it's --
Why does not qualify for a truss? What was
the -- I want to remind yourself -- I want
you to remind yourself of the definition of
a truss. What is there that does not qualify
this structure as a truss?
Well, then there's a force acting --
There are the force at the middle of the member.
That was one of the conditions. Second, all
the members should be -- see, this is just
the purpose of discussion, what you have written
down. This is really -- this is very simple
frame. Everybody [inaudible] frame is not
going to be that simple. Is that correct or
not? But I purposely violated the two conditions.
One, the force is on the member. It's not
on the joint. This one is on the joint. This
one is on the member. And all the members
should be pin-connected. That's what is fix
[inaudible] to them all. So that is not qualified
for the -- now if I want to calculate -- let's
put A, B, C. Now look what happens. Actually,
this is exactly what I said before. Let's
put here A, B, and let's see what I put here
[inaudible] C -- okay, the C. What are step
one? Step one is finding the reaction if possible.
Let's see whether it is possible or not. So
again, we have to erase or disconnect the
structure from all its?
Connections.
Connections, remember that. A, C, the connection
[inaudible]. Now at here I had a [inaudible],
so what should I put there? Now I know everybody
knows at least about that part because I notice
you did it -- a [inaudible] would be what
force?
Optical.
Optical. You are preventing this from going
up and down. Okay, so you have here C1. Okay,
what should I put at A. This was a fixed support.
It's not going this way. So I need to have
AX. It's not going that way. I have to put
A -- A1, and then what? Because it is like
this -- remember, its fixed. Remember the
fixed versus pin. It's a big difference between
the two. Everybody see that? Yes, fixed versus
pin. [Inaudible] report there I should have
also AMA. How many [inaudible] do I have?
Three.
Not three.
Four.
Four. How many equations do I have?
Three.
Three. I can write the way CAX equals to zero
again. Everybody understands. But that's one
equation, sigma FX equal to zero, AX equal
to zero. Actually, that was your quiz. Some
of you didn't write it down. You put the A
there, but since sigma FX must be equal to
zero, AX must be equal to?
Zero.
Zero. Your quiz was like that, too. Is that
correct or not? Well, you have to write it
down. You have to -- but I don't know. I may
put here a horizontal force then. Everybody
understand? If there was a horizontal force
here at B, I would have an AX there. Is that
correct or not? However, this is the frame,
of course. Step one, sigma FX equal to zero,
AX equal to zero. Well, how many equations
left? Two equations left. How many unknowns
do I have? One, two, and?
Three.
Three, so it's still not solved. I can write
AY plus CY equal to 1000. Everybody understand
that? And then I can take the moment about
here, or there, or there, doesn't matter.
But this [inaudible] A comes to the picture.
CY comes to the picture. AY comes to the picture.
No matter what I do it's just math. I have
only two equations. Remember, one thing you
cannot do, take the moment about A, and take
the moment about C, and then take the moment
about B and make a mistake somewhere. Then
you come out with two, three different independent
equations, then you solve it. Is that correct
or not? Yes? And it happens in ME218, too.
Believe it. I give them in indeterminate problem
-- we call this indeterminate frame. [Inaudible]
is static you are able to solve it. That's
a different hub. This has one special scenario.
If this was fixed like that you could never
do it with the static at all. Everybody understand?
But this, we call it a special frame why?
Because there is a?
Pin.
Pin there, so okay. So because of that pin
there, actually by doing that we've changed
it without you realizing from an indeterminate
problem to a?
Determinate.
Determinate problem while doing step two.
Now what's the step two? What did you write
down? See, step one we're already following
that. It's -- we can continue with doing that.
So you can -- but doesn't get us anywhere.
The next equation you want to write. Please
write it down just to finish. So you write
sigma FY equal zero. That means AY plus CY
-- I'm sorry -- plus CY must be equal to 1000
pounds. Is that correct or not? Yes? What's
the next equation? The next -- where do we
want to take the moment about?
A.
A, okay. So you write sigma M -- this is not
good anymore. Let's get a better one. Sigma
M at point A equal to zero. Is that what -- just
remember what I'm write -- how I'm writing
it. Now don't forget MA since it's another
second time or third time. Some people write
it and stop writing like that. They write
it like that. First of all, some of you even
don't put summation there, which is the worst
behavior. Some of you write sig MA equal to?
Zero.
Now you're laughing, because you know what
I'm -- if you know MA equal to zero, that
means this MA must be equal to?
Zero.
Zero. Many times I have told you not do that
summation of the moment at point A. Exactly
that's what you have to do. Is that correct
or not? Yes? That is nothing to do with [inaudible]
you're sitting [inaudible] this is exactly
as saying summation of the moment about point
C equal to zero MA [inaudible] your equation.
Is that correct or -- therefore, starting
from left, AX doesn't have any moment about
A. AY doesn't have it, but you already have
a?
Moment.
Moment sitting there. Is that correct or not?
That's your -- you're drawing positive, so
that is MA. Then you come here, you write
500 times what? Times -- 500 times?
Five.
Five. Minus 500 times 5. Then you write minus
500 times what? Minus --
Six and 1/2.
Five-hundred times -- this is at the middle,
so it's become 1 1/2 times 5, 6 and?
Half.
Half. And then if you leave it at that of
course you can solve for MA, but that's not
true. There you have CY times -- that's right.
Plus CY times?
Eight.
Eight. Equal -- as you see, still you have
here equation -- this is one. This is equation
two. This is equation three. Any other version
of it is the same thing, but still you have
one, two, and?
Three.
Three unknown, yeah. Everybody -- now if by
mistake you don't incorporate that into your
system, forget about MA, of course this becomes
determinate. Everybody -- and as it -- this
is not a fixed support. That's the pin support.
Is that correct? Now, what's the step two?
What did I ask you to do? So let's do exactly
like that. Let's see, this is just for their
definition of what you wrote there in order
to understand what you have written down.
Step two is separate this from? Separate all
the members from each other at the joint.
It's at the joint which has the pin. We only
have one pin, so it's very simple. So we have
not two B, one like that with the big support
here. Is that correct or not? Yes? Let's leave
it like that. All right, and we have another
member like this with the force right at the
middle, 500 pounds. And here I have -- I cannot
find that so I have to put my reaction there.
This is my unknown reaction. Oh, sorry. [Inaudible]
should be [inaudible] that was C port, this
is B. This is C. This is A. This is B. Okay,
I separate it, correct? So what else should
I do? First of all, here at AI -- no, it's
A at AY and M, but I'm not worried about that.
Let's go to B. Now what should I put a B there?
What's [inaudible], what did I separated?
What did I cut? Forget about the 500 pounds
first, what did I cut?
The pin.
It was a pin. What is action on the pin? It
is AX and AY or B2 forces horizontally and?
Vertically.
Vertically. So if I decided to put here -- in
fact, this was a pin. In the pin I should
put here BX and?
BY.
BY. I'm just putting it in this purposely
because I'll show you it doesn't make any
difference which way you put. Is that correct
or not? This -- on the other side I have to
put?
The opposite.
Equal and?
Opposite.
But we don't ask for [inaudible] when you
cut it your force then will be equal and opposite.
So you commit to one. You can commit to this
one, BX, and this one. The other one becomes
equal and?
Opposite.
Now I have committed to this B, so the other
B I have no choice than to put it BX at? BY.
Be careful about that. This is going to happen
in all the frames. Is that -- this is the
static. This is something we learned in the
past. Is that correct? When you cut a member
many times I [inaudible] the force [inaudible]
that member are upward or?
Downward.
Downward. Equal or opposite if you have only
force, and moment is the same thing. What
is left, then, that I didn't put there?
The --
This one, yes or no?
Yeah.
Question -- this is a question that comes
on the [inaudible] frame? Where should I put
it?
Both.
On both?
In both? Okay, don't write it down. Okay,
these are -- this is exactly what you are
supposed to do with your homework. I put here
a 500 here downward and a 500 here. Yes or
no? Now let's reassemble this. If I reassemble
this I should get that. Is that correct or
not? Reassembling this BX and BY going to?
Cancel.
Cancel each other, because they're going to
connect these two together. Upward or?
Downward.
Downward. Equal or opposite if you have only
force, and moment is the same thing. What
is left, then, that I didn't put there?
The --
This one, yes or no?
Yeah.
Question -- this is a question that comes
on the [inaudible] frame? Where should I put
it?
Both.
On both?
In both? Okay, don't write it down. Okay,
these are -- this is exactly what you are
supposed to do with your homework. I put here
a 500 here downward and a 500 here. Yes or
no? Now let's reassemble this. If I reassemble
this I should get that. Is that correct or
not? Reassembling this BX and BY going to?
Cancel.
Cancel each other, because they're going to
connect these two together. BY at BY. Then
I have 500 here, 500 there. It becomes?
Thousand.
How much do we have there? So that is wrong
decision. Yes or no? So what should I do?
Half and half?
Yeah.
That's too much work. Why not putting it in
one, not the other one? No, what I'm telling
you is this, you really don't know that. And
believe me, it does not make any difference.
And if you have two items like this -- actually
I can do it here. You see here, this is two
items like that, pinned together. Actually
-- oh, you cannot even see me where I put
my hand. This is my low. Is it in this member?
In this member? Or at the middle? Doesn't
make any difference. Should it make any difference?
I'm pushing this point down. Would it make
any difference if my finger is here, or here,
or at the middle? You are saying put at the
middle, half and half. I am saying put it
in the rear or there, both is the same thing.
Don't worry about it. The only thing changes
is the value of BY. Everybody at CY and the
other one will not have any effect. Try that
at home if you don't believe me. So put it
in the -- on the one or the other one. Is
that correct or not? So I don't know which
one I put -- where did I put it? I put it
on the -- on this [inaudible]. So I put here
500 pounds here. So did you write it down?
When there are forces at joint and you are
dividing that joint into two you can put that
force on either member. I have given this
type of question in the final, remember that.
Either member, so does not make any difference.
All it changes is the value of EY, which you
don't care. Everybody understand that, yes?
Okay, now how many unknown do you have? This
is very interesting to look at this scenario
now. How many unknown do I have? Now I have
increase here. I had four, yes or no?
No.
No, two more added, correct? How many unknown
do we have in [inaudible] -- I mean total?
Six.
Six. How many equations can I write for this?
Three.
How many equations can I write for that?
Three.
Three and 3 is?
Six.
Six. I should be able to solve it. Everybody
see how that -- because there was a pin there.
Now if there was not a pin there I could not
do that because I would have a moment here.
Everybody -- and that moment kills it again.
Everybody -- this is why I call it a special
frame. It's a special frame. I can do it at
the pin. Is that correct or not? Now you see
it's very simple. What do you think the value
of CY -- if I take the moment here what the
value of CY to see how much static have you
left? What do you think the value of CY should
be here if this is -- this distance is half
of this distance?
Two-fifty.
Two-fifty? Is that -- do you accept that?
Yeah.
You see it or not? Five-hundred times A, CY
times 2A. So what's the value of CY?
Two-fifty.
Two-fifty, correct? Come on guys, you know
that much. So take the -- in this one, write
it down. In member BC take the moment at point
B equal to zero. No remember what we had here.
This distance is three -- you said it before,
1 1/2 and 1 1/2. Even that was -- when I [inaudible]
here you said 6 1/2 there, yes or no? Correct?
So [inaudible] simple. These two have no moments,
so 500 times 1/2, going negative, plus CY
times 3, going positive equal to zero. As
you see, this is the ratio of [inaudible].
That's what I asked you to do it in your head.
So obviously CY must be equal to 500 divided
by?
Two.
Two, so therefore, it will be 250 pounds going
up. What's the value of BY now?
Two-fifty.
BY will be 250, not going down. It should
be going?
Up.
Up, very good. Sigma FY [inaudible] so we
are doing it by inspection, okay? So simple.
Sigma FY equal to zero gives me BY equal to
minus 250 pounds or going up. What's the value
of BX? The value of BX is equal to?
Zero.
Zero. There is no other horizontal force.
Is that correct or not? Yes? Sigma FX, we
are going backward. Moments equal Y, and that's
equal X, because that's the way it looks like.
Is that correct or not? Yes? Sigma FX equal
to zero. BX equal to zero. So that one disappeared,
so that one I don't need to put it there.
In this problem it becomes like that. This
one, this is the answer. So this one is gone,
and this one becomes 250. This is 250 up.
This must be 250?
Down.
Down. Everybody -- or the value of that one,
so please write it down. The value of that
one is minus. This is the best, because you
already committed to that. So you don't want
to go and change it. This happened in your
trusses as well. The value of that one is
minus 250. Is that correct or not? BX is equal
to zero, so the two Bs are equal to zero,
because if this B equal to zero, that B -- X
must be also equal to?
Zero.
Zero. So this was just for demonstration of
what I told you. So end up only with 750 pounds
there. Everybody understand? Or if you put
it 500 here, the value of BY would be changing,
the same thing. The value of BY here becomes
750, that becomes 750. That's the same. So
doesn't matter where you put it. Now is this
difficult to solve? Of course it's not. Yes
or no? AX must be equal to?
Zero.
Zero. AY must be equal to?
Seven-fifty.
Seven-fifty. The moment up here would be which
way? Now let's be careful --
The distance.
If this is five feet, is that correct or not?
The moment here at this support should be
going that way or should be going that way?
Well, be careful what you are answering.
Down.
You see, that -- two things here. Exactly
the same mistake that I saw this morning in
ME219, that you are consider the action. I'm
not looking for the action. I'm looking for
the?
Reaction.
Reaction. Action plus reaction must be equal
to?
Zero.
Zero. So you can do it by inspection if you
want. However, if this action going that way
-- notice I put another moment going this
one as reaction. These two are going in the?
Same.
Same direction. It cannot be equal to zero.
Is that correct or not? Be careful about that.
When you write sum of the moment equal to
zero it would eventually end up in the?
Other.
Other direction. Everyone -- opposite to action.
So in other words, action plus reaction must
be equal to?
Zero.
Zero, because this is such a simple case I'm
doing it this way. So this is the answer -- final
answer, 750. And this becomes 750 times -- becomes
3750 pounds per moment, and this is pounds.
So this is the [inaudible] B. That's -- so
as you see a problem what's solved? Notice
two, three things. Step one was not complete.
I could not finish step one because I could
not find a reaction. I had too many reaction.
Everybody -- however, this was a special frame.
Why a special frame? I had pin. I do it like
that, then I got it. Now that we did that,
let's go to your example in your handout,
which is much more sophisticated. So go to
problem number one. There are several examples
in your handout. Everybody has a copy. I had
a few copies. They are here. So let's do the
one that is more sophisticated than this.
And it has a pulley, and I promised some of
you to take care of the pulley as well. So
as you see, everything that you wrote in those
three paragraphs applied here. Is that correct
or not? Everything that you wrote now should
have a meaning for you, so. Now the only thing
now we have to be careful about in each frame
or each machine is to be not to making any
mistake or not to make any unreasonable assumption,
and then you should be okay. So let's look
at the picture itself. So I have to draw one
more time the frame here on the board for
you guys. Of course, you have it in your handout.
You don't have to redraw this. I have to do
it that for everybody to see that. So where
is my handout? Can I get a copy of your handout?
Yeah, I believe I have one here. Don't worry.
I'll get it.
I have one left. Okay, we are done with the
first page, and now we are on the second page.
We are on the third page. So there are a -- five,
six, seven frames here. Everybody see that?
So I'm going to do this one. We cannot do
this one. This for the final question. So
you can do -- because we don't know about
this uniform load or linear load here. I talk
about this later on. Let's do this problem
on the right-hand side which has been a previous
quest. Is that correct or not? Everybody with
me? Yes? Okay, so now let's draw that frame
and see what we have there. So this frame
actually you will see the frame become -- if
you do it in the proper format, so the frame
becomes really very simple to solve it. So
here we go. Dispense of losing this so quickly.
So here it was. There is the vertical member
there, and A, B, C, and D here. And there
is another horizontal member going here. Is
pin-connected here, and pin-connected here
like this. Pin-connected, these two together,
pin, pin. And here is also pin-connected to
the floor. So this is a pin-connection there.
And there is a pulley here, and there is a
weight hanging from here. Let's say put the
weight of the -- I have not given in the picture
there, but let's put the weight equal to 40
pounds. Forty-pound bucket here with 40 pound
weight. And that rope also goes to [inaudible],
okay? Goes something like that. Goes -- this
end should be a little bit bigger, but let's
put it there somewhat [inaudible]. Pulley
goes like that. The rope goes like this over
there. Just this -- this is not in the scale,
no. What I have given you is this, this three-dimension
is given, each one equal to six inches. Six
inches, six inches, six inches. This is the
pin. Every member is pinned. That there is
a -- of course, you would see that in the
back, or you don't see this one or [inaudible].
This one in front, but you don't see -- this
is all dash [inaudible]. Either in front of
it or in the back of it, doesn't matter if
it's in front or the back. They put it in
front. So you see this, okay. You don't see
that, okay. That's fine. So that goes there.
And the -- this radius is given equal to three
inches. And then we call this one A, B, C,
D, E, and G. This is your frame. How many
members are there? Let's decide on that one.
First things first. How many members is there?
So I don't want anybody to make any mistakes
between this frame and the truss member, because
there is a big difference between the two.
How many members do you see here?
Four.
Four? Which are the four?
Well, does the pulley count?
This is the member B, yes or no? Correct,
there is a member C, D, and EG. Is that what
you said or not?
No.
No, very good. This is actually not two members.
This is?
One.
One member. I want you to understand what's
the difference between a truss member and
a frame. In trusses if I have a point like
E, this member, and that member, and this
member, all three of them are going to be?
Pins.
Pinned together, because all the members should
be pin-connected. Here -- hold on. Here -- I
will answer your question. Here, this is what
we are saying. I want you to see what's the
difference. Here there is a solid member.
There is a pin here, and this one is being
pinned on top of the other one. So in other
words, this is a member, and this is a member.
So with that in mind, there is a member CEG,
yes or no? What about this one? Okay, this
is one piece. Is that correct? These two are
attached on top of them, so we have member
ABCD. I have member CEG. I have member --
BE.
BE. Is that correct? And also you can consider
pulley as the fourth member. Pulley is also
a member, but it's not a straight line. It
is a member. Is that -- so it is fourth member.
What was your question?
Are we disregarding the roller at D?
What?
Are we disregarding the roller at D?
At E?
D.
D? Oh, no, I don't. I forgot. Of course, I
cannot do -- disregard. You have to look at
the picture, and there was a roller here as
well. That's -- we'll get to that one, because
I was so anxious to explain the [inaudible]
to you, so that's better. So how many unknown
reactions do I have? We want to go with step
one. How many unknown reactions do we have
there? Is it solvable or not? Remember what
you wrote there in previous example, it was
not solvable. [Inaudible] in this one, is
this -- can you solve the reaction or not?
Don't forget that. Do that always first, because
it makes it much simpler later on. The step
two becomes more simple. How many reactions
do you have at A?
Two.
Two. How many reactions you have at B?
One.
One. How many reactions is total?
Three.
Three, actually without D this would be moving.
Is that correct or not? So therefore -- so
can we solve it? So let's put that. Again,
I am going to erase that as usual and put
here for you AX. If you don't have time to
do it you can put it -- you have to draw this
without the D -- or this was point D. And
I'm going to put here D going that way, although
it may be wrong, because AX and D cannot be
on the same direction. Is that correct or
not? Yes? One of them has to be -- so call
it DX, because there is no DY, because that's
what the roller. How many unknowns do we have?
Three.
Three. So I should be able to solve it. Is
that correct or -- and this is not like the
other one. Many times you can do it. Many
-- sometimes you can't. The other one was
we couldn't, but we went to step two. That's
what the point is. Here, always do step one.
So step on, as if it was previous homework.
So you take the moment C for M at or above
point?
A.
A equal to zero. Very good. So you have DX
times what? DX times 18. Plus or minus? DX
times --
Minus.
-- eighteen is going to be?
Minus.
Negative. Then we have 40 times what? Oh,
I didn't put your [inaudible] this time. I
shift here. So I know always something missing.
This time mention also I should be giving
it to you. Those dimension is?
Eighty and two.
Okay, well, I have it here. No, I just want
to be sure. Those are given at?
Eight and eight.
Six of -- eight and eight. Eight, eight. Now
then I have -- I don't have to break this
apart. I have only 40 [inaudible]. Notice
this rope is not being cut. So you don't need
to put 1/4 here, 1/4 there. Everybody -- if
you put it you do extra work for nothing.
The forces are equal and?
Opposite.
Opposite. You have that. Then we have 40 times
what? Forty times?
Seventeen point five.
Be careful here. When this was the quiz some
people made mistakes. It's not 18. It is?
Seventeen point five.
Sixteen.
It's not sixteen. It is 16 plus three. It
is 19, so that's right. Nineteen, and it is
negative as well, equal to zero. So DX or
D adds up to be equal to -- if we are going
to finish it shortly. See, this is very good.
Y is 42.2 pounds. So it means the direction
that I have chosen is incorrect. Then we go
to sigma FX. Sigma FX equal to zero. I don't
have to write that. Everybody knows. Sigma
F equal to zero. Gives me AX equal to 42.2
pounds. And AY, I don't have even to solve
it. AY must be equal to 40. This is so simple.
So AY must be equal to 40 pounds. This is
correct direction, 42.2 pounds. And this one
has the value of equal to minus?
Forty-two.
Forty-two, two pounds. Everybody -- so the
[inaudible] is going that direction. So step
one is done, finished. Is that correct or
-- now we go to step?
Two.
Two. The idea is to find the forces acting
at -- here, B, because these two are done
out of here. Everybody understand that? A
and D are done. Where force acting at point
CBEG on every member. Is that correct? Which
forces our equal and?
Opposite.
Opposite. Now what all we have to do? Draw
each member separately. So please draw that
in any scale you want. Now at D I know what
I have. At D I have a force of 42.2 pounds
here. I suggest for you to make -- not to
make any mistake, always put your point there,
and put each point there. Every point that
you see, put it there in order not to make
any mistake. This is one turn, one turn, so
this is point A. This is point B. This is
point C. That's point D. And there is a cable
[inaudible]. Everybody understands that, yes
or no? Because when I separate this member
-- now look what we have done in the past.
When I separate this member from all its connections,
look what has been connected. [Inaudible]
support was disconnected, so I put there forces
there. Yes or no? I should put. Did I put
it there yet or not? So let's put AX and AY.
AX was equal to 42.2 pounds, and AY is equal
to?
Forty.
Forty pounds. So we have A was connected and
put their appropriate forces there. Here,
D is? Disconnected. I put the appropriate
forces there. Now I have here a tension [inaudible].
Some people forget about that. If that tension
was potentially [inaudible] in that table?
Forty.
If there is no friction, that one is 40 pounds.
That's correct. So that's 40 pounds. And then
we come to the C, and to come to the B. At
C I have no choice to put here, CX and?
CY.
CY. Okay, I put here CX and CY. Now in order
not to make a mistake I would suggest that
you try to do your frame as much as you can
similar to the picture that it is there. So
if this is the C, this must be the C on the
other member. Is that correct? So the other
member is CEG, and the G there is a pulley.
I purposely didn't put the pulley there. You
can put it there if you wish, or you can separate
it. I'm going to separate that because easier.
I tell you in a minute why. And I'm drawing
the pulley separately here. This is the pulley.
Forty, 40. Is that correct or not? Because
when I cut the pulley I have here one cable
here, one cable there. There is a G there.
Is that current or not? So the center of the
-- this is G. And this is also G, so now we
can -- as I said, it's four members. And there
there is a member here between B to E. Yes
or no? Okay, what should I put -- so at C
we decided to have CX, CY, on the other side
I should put? CX, CY equal and?
Opposite.
Opposite. Please do not forget that. This
is CX, and that is CY. What should I put at
on the member at B?
[Inaudible] members.
BX and BY?
No?
No. This is the two-force?
No.
No. If you put B and X [inaudible] it's still
unsolvable, but you have to go through twice
as much [inaudible]. You are advisable always
to look for a two-force?
Member.
Member first. BE is a two-force?
Member.
Member. It could be pure tension or pure compression.
What do you think? Is it pure tension or pure
compression? What do you think?
Compression.
Compression, very good. So if you are pulling
this down, this is going to be compressed.
Is that correct or not? So that -- however,
if you are not too sure, assume it is in -- under?
Tension.
Tension, because that's the rule. Is that
-- I don't know what I have chosen here. I
assume this is absolutely under compression,
so I can assume it under compression. Everybody
under it, because I already know that's so.
Your assumption was correct. So here is under
compression, yes or no? You are crushing it.
You are pushing it. Is that correct or not?
So what should I put at B? Notice if this
is force B, the force B must be equal to force
B equal and?
Opposite.
Opposite. So what should I put here? A force
equal and?
Opposite.
Opposite, which is going like that. And I
put here equal and upward. It looks like that.
What's the slope of that? No, slope of that
given there. The slope of it is four wrong
and three right. Is that correct? Eight, six,
which is the same, just for simplicity. So
that's it. Notice what happened here. This
one I don't need, because this one you really
don't need it. I can't [inaudible] because
B is there. This is B. And that equal to E.
And this member is become irrelevant. So you
show it there for you to see what's happened
there, especially if it is quiz or the homework.
But notice that I don't care about that as
long as I find the force B or force C, which
had the [inaudible]. Yes or no? Now as you
look at it, how many unknown do you have here?
Actually this problem becomes very simple.
Look how many unknown you have here.
Three.
Can you solve it?
Yes.
Yes. Now go to here. How many unknown you
have here? You have more than three unknown.
Let's see what happened there. Now what happened
here, you -- there is two options. Write it
down. Option one, you can keep this -- I'm
giving you [inaudible] because I want to erase
it. You can keep the pulley here with the
force here of 40 and put the force 40 and
solve it like that. Is that correct or not?
Yes, this is solvable. How many unknown do
you have?
Three.
Three. All right, that did not separate the
G from each other. Everybody sees that. But
it is still solvable. Don't forget, all you
have to do, take the one with the [inaudible]
point?
C.
C, and this is the slope of that. It's given.
I'm going to do it in a form [inaudible] four,
three. Is that correct or not? And then the
only unknown is that one, because that's 40
and that's 40. But for the future, because
there is a technique here that I want you
to learn it. I'd rather you separate these
two from each other. Is that correct or not?
Yes, the way I had it originally, because
sometimes these forces are not going in this
vertical and horizontal [inaudible]. Is that
correct or not? Now we go to here. Is this
in equilibrium the way I have drawn it?
No.
Of course not, because I have not put anything
at G. Is that correct or -- at G I have to
put what?
GY.
GX? No. Both GX and?
GY.
GX and?
GY.
GY. What's the value of GY? Look at it.
Forty pounds.
What's the value of GX?
Forty pounds.
What should I put on this G? Equal -- no,
equal -- GX and GY equal and?
Opposite.
Opposite. So what happened if this become
equal to -- this 40 you said, yes or no? Going
to the right I should put here 40 going to
the?
Left.
Left, and this one is going up. I should put
40 on the down. So notice these two actions,
without looking at G has transferred to this
one. Is that correct or not? So that's what
you should do. From now on [inaudible] you
have a tension there on the pulley, transfer
it to the? To the pin on the other rod. You
are done. You don't have to worry about it.
Now for this one it was very simple. Both
case was very simple. Notice again one more
time, this is 40 going to the right because
of that, yes or no? This is 40 going up because
of that here. I put 40 going to the left,
40 going down, which is exactly the action
of these two tensions. Everybody -- what is
the big outside it is on there? No. This problem
does not make any difference. However, usually
you will see this. Please write it down in
your notes. You see here a pulley, and let's
say that -- somebody asked me, actually, in
this class before in one of your homework.
There is a tension here. Let's say the tension
is going like that. Tension is going at the
angle of 30 degrees. Some student -- I don't
know if it was in this class or in the other
class. I asked -- asked me, "How do I calculate
that point," or, "How do -- where is that
point?" I said, "You have to go with this
angle," etcetera. So I think it was in the
other class. Is that correct or not? Yes?
Yeah.
Notice as long as this pulley is frictionless,
this tension must be equal to that tension,
yes or no? But I don't care about all of this.
All I do it is this. So I put it here. I put
here a tension going down, and a tension going
at what degree? At?
Thirty degrees.
See, I transferred this action right to the
center. I don't worry about GX and GY. This
technique helps you to set up your -- all
the pulleys when are in there, as long as
-- of course, as long as there are?
Friction.
Friction, because when we go to chapter eight,
many pulleys have the friction, so there are
some changes there. Everybody -- the team
will not be equal because of the friction.
Is that correct or not? So is that -- so this
is become a technique that you should put
it aside. Put it in notes. Put this in a box.
So in this case, this is down, this is left.
So I put here down and left. If this was going
to -- table was going at 50, 60 degrees I'll
put it right there. Is that correct? Which
is much simpler. That's what I said. [Inaudible]
harder, this -- doing this, this is my two
free body diagrams. The other two disappeared.
Is that correct or not? That is -- are they
solvable? Of course they are, look at this
one. For this one, take the moment above sigma
M at or about point C equal to zero. Of course,
you write -- in your notes you write this.
This is the free body diagram of [inaudible]
CEG. You've got other ones. [Inaudible] I
would get lost in your writing because you
are writing this equation for this member,
for that member. So you should put it under
the member, or separate the member, put it
in a box. Everybody -- somebody can follow
your writing. Is that correct or not? Otherwise,
when you want to go back there you would not
see what happened. So sigma MC equal to zero.
This one doesn't have any moment. You have
40 times 16. This was eight inches and eight
inches. So 40 times 16 inches. The moment
of 40 about point C. This one doesn't have
any moment. The horizontal component doesn't
have any moment. Only then -- well, as you
see, it's very simple. I don't have to do
it, but I'll do it anyhow, 3/5 of the E -- that's
the vertical component, multiplied by -- this
stands for [inaudible] A, and that one is
-- also is positive. The other one was?
Negative.
Negative. Equal to zero. So E ends up to be
equal to? Simply E ends up to be equal to
133.333.3, or you can leave it at 133. So
this one is solved as soon as E becomes equal
to this force E become equal to this one point
E, 133.3. Notice immediately you can calculate
CX and?
CY.
CY. Put it here, calculate. Of course, actually
what you do is -- really this is so simple.
You will be so [inaudible]. Your frame, it
takes twice as much or three times as much
to draw the free body diagram of each member,
but immediately by doing couple of the small
moment here and there, sometimes you have
to take the moment above this member, sometimes
the other member. Depends how many forces
you have. It's very simple operation. It's
[inaudible] is that correct or not? Yes?
Yes.
Next one, okay, there are few here. Just what
is important here, guys? Notice what is important
here is not sigma MX, no MA equal to zero
sigma, FX equal to zero sigma [inaudible].
I'm assuming everybody now knows that part.
What is here involved?
Free body.
The free body?
Diagram.
Diagram. Actually, this is the key for every
problem [inaudible] from now on. For that
reason, there are three problems down here
which I would like you to look at it at home.
And if you have any question ask me how to
draw those free body diagrams. Notice I did
not -- purposely I did not give you the dimension,
because I don't want you to solve it. Solving
it, everybody right now should be do, because
we have gone through chapter one, two, three,
four, especially four. We had this bunch of
homework. You have used it so many times,
yes or no? However, let's do the problem number
one together, the one on the left-hand corner.
This one. The one on this left-hand corner
that you want to draw the free body diagram.
First of all, tell me what is this for? I
think we've discussed that once before. I
don't know whether we did or not. What do
you think this problem is? Engineering-wise
what are we trying to do, that I'm asking
you [inaudible]? Assume that cross-section
-- do you see a circle there? That's a pipe.
Could be a pipeline. Is that correct? Could
be water pipe, or could be oil pipe, or under
pressure or not, that's beside the point.
What are we trying to do? We are holding it
against the slope, so let's say that you are
Alaskan oil line or pipeline, and here is
the hill. And the pipe wants to go here. This
is the pipe [inaudible] you are putting outside,
not in the ground. And you want to support
that. To support that you put here a pin connection
here. This is pin connection to the ground.
And here is pin connection. And the slope
goes, of course, like that. And then you have
another rod connecting here to here. Is that
correct or not? So holding this pipe, you
know, so there is a weight involved here.
Is that correct or not? Let's say that every
-- this is the ground. Of course, this touches
the ground as well, so you [inaudible] like
that. So it's [inaudible]. Is that correct
or not? Yes? And therefore, let's say every
50 foot you -- of the pipeline you put one
of these supports. Is that correct or not?
Therefore, the weight of the 50 foot of 25
feet of pipe this way and the other way goes
to each support. So you calculate W. Is that
correct or not? Yes? So W is given, and then
you want to calculate the forces on each member
later on to design this rod and that rod.
Is that correct or not? Obviously we have
done that in the past. What should be do first?
How do you come up with the free body diagram?
Disconnect everything you want --
First free body diagram of the cylinder, which
is free body diagram of particle. You can
consider, because all the forces going through
the center. So this is the free body diagram
-- correct free body diagram for this [inaudible].
I have a cylinder with the weight of -- we
have seen that in the past, yes or no? W,
correct? Yes? What else do we have here? Let's
say it is the contact force here, ground contact
forces. Correct? Which we show it with what?
Normal.
Normal force. Normal, let's call it N1. Is
that correct or not? Then there is a contact
between the pipe and this rod. So therefore,
there is another one. And we have seen this
problem before, so call it N2. So if I give
you W can you find N1 and N2? You take equilibrium
of particle. Of course you have sigma of X
equal to zero and sigma of Y equal to zero.
You get to that one. And then you come to
the rod. Rod on here somewhere of 45. Give
you all the distances, angles, etcetera. Somewhere
you have N2 now applied here perpendicular
to that rod, yes or no? Correct? And here
-- this is not a two-force member. So therefore,
at point -- what did we call it here? So here
they call this one A, this one B, this one
C. So now here is member AB. So at A I put
here AX, and let's put it here in unknown
format. AX and?
AY.
AY. Now what should I put at B? That looks
like a pin. But should I put BX, BY? Or only
one force?
One force.
Only one force.
One force, because BC is the two-force?
Member.
Member, because the way the pin design. Therefore,
I put only one force in that direction, wherever
that -- [inaudible] these will be given to
you at the angle of 10 degrees, 20 degrees.
So I put here one force in the direction of
the member. So how many unknowns do I have
there? Three. So if I give you all the numbers
you should be able to do it. This is equilibrium
of particle. This is equilibrium of [inaudible]?
Body.
Body, so all being compiled there. Now go
to the next page. Okay, let's look at that
picture there. Okay, how do I draw that for
the -- for -- there for everybody to see?
This -- we call it a machine. Is that correct
or not? Yes or no? Correct? This one. What
are we trying to achieve here? There is a
handle there. There's lots of members there.
This is -- frame is [inaudible]. This is the
frame. You may have pulley. You may not have
pulley, etcetera, etcetera. I don't have time
to do it this one. Let me do that one next
week. However, just to finish all the frame,
we are not going to have a quiz -- [inaudible]
our quiz will be Thursday next week, which
will be entire chapter, trusses and frames.
Let me give you one more frame just to ask
your idea. At least you'll be able to do all
your frame and all your trusses before Tuesday.
And there are some machine part -- machine
part is the same idea. The machine part, break
it into?
Parts.
Different members, and do it [inaudible] exactly
what I do. It's really nothing else to add,
but I'm going to do one or two examples on
Tuesday in order not to be left out. Everybody
understand? Show everybody. But however, technically
you should be able to do it, but let's look
at one more frame, because it has something
in it of interest that I have to discuss with
you which is not in your handout. So please
draw this frame. This one I'll show you how
to solve it. Everybody see what -- these are
important. Okay, if you understood what I
said then the rest of it is just writing sigma
FX and?
Sigma F.
Sigma F. And everybody should be able to do.
I should not be any spend time doing this
detail work anymore, because you already have
done that. Here has couple of other members
that I want you to be -- not similar to the
other one, but something new that you have
seen in the past, but not in this frame. So
here I have a frame -- member, I'm sorry,
pinned to the ground. Let's call it point
B. And then here is another member here like
this. And here is a fixed support. So this
member is fixed to the ground, so let's call
it A. This is point C. This is point D. The
point is attached to a cable, and this is
attached to a little pulley which has no dimension.
And here we have 50 pound force there. However,
this one is like this. This one is on top,
and little cut here. And a pin here which
is attached to the other one in the back.
Everybody understand what -- okay. What we
have is a slider there. Is that correct or
not? How many -- if I want to calculate XM
-- let's give me couple of minutes -- extra
minutes so we can finish this and explain
it at least to you. If this was a quiz, all
the dimension is given. This height, this
height, this height, this length, this length,
this length. Everybody understand that? Yes?
Now how many reactions do you have here?
One.
Two.
No, this is a -- this is not a roller. This
is the pin. And this is not a two-force member,
remember that. So you have here BX and?
BY.
BY. That's the first member. BX here and BY.
How many unknown do you have here?
Three.
Three, because it's a fixed support. AX and?
AY.
AY. AX, AY.
And then a moment.
And MA, correct. Now five unknown. Obviously
it is not?
Solvable.
Solvable. But if I go to the frame -- because
there is a pin here. Everybody understand
that? Separate these two members from each
other. I have a member like that, and I have
a vertical member. Is that correct or not?
Yes? Now again, one more time. Here I put
BX, BY. What should I put here? That's the
only reason I'm doing that. This one you already
answered me correctly, because this was a
fixed support [inaudible]. This one or that
I'm just interested here. What should I put
there? How many forces should I put there?
I explained that to you in the previous chapter.
It is there. It is in your table. You have
a freedom going this way, and no freedom going
that way. So in one of them -- doesn't matter
-- in one of them you put this -- line will
be given. If this slope is 4.3 you have to
draw a line perpendicular to that, which is
in this direction. So you put here a C going
-- if this is four, three, right, you put
here three, four because that's perpendicular
to that. So [inaudible] angle changes. And
that is unknown, yes or no? However, on the
other member you put C equal and?
Opposite.
Opposite. So if this is C for the other one,
you put exactly the same force, C equal and
opposite, with a run of three and rise of
the four. And here, of course, you put here
50. And here you put -- what do you have here?
We have AX -- sorry, have to rush it a little
bit. AX, AY, and MA. Is that correct or not?
However, this point is left. What should I
put at that point? That's point BCA -- let's
call it point E. What should I put at point
E? I already explained to you.
Down.
A force going down and a force going that
way, because [inaudible]. So there is 50 going
here, 50 going there. Is that correct or not?
Yes? Look at how many unknown do you have
here?
Three.
See that's the bottom line. After you draw
this, which problem to me suggest?
Why is that 50 going this way on --
Which 50?
Oh, never mind, never mind.
Oh, okay. That's the other end of cable, of
course. Yes. You -- when you cut it here and
you cut it here, these two are going opposite
sides, of course it's that. Everybody understand
that, yes? Now look at it guys. Can you solve
this?
Yes.
Take a moment about here. You can -- actually
you don't have to break this in -- this is
perpendicular. If you have this line, the
moment of that C times this distance. That's
their moment. Is that correct or not? And
the moment of these two, which is given, you
find C, [inaudible] BX defined. You put the
C here. You find AX. And actually that part
of it is very simple. What is the key, guys?
Free body diagram.
Free body diagram. Please, do your free body
diagram correctly. That means that if you
do free body [inaudible] correctly, if I give
you a quiz you can [inaudible]. If I give
you a quiz I know that you know how to solve
it. Everybody understand? But if your free
body diagram is wrong like your previous quiz,
then you are in trouble.
The homework is not --
No, the --
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