[ Music ]
Hope you cannot solve the problems. Some of you [inaudible]. Depends how bad it
was. I may subtract one point or not. But be careful there. You have to put this
because if you don't put it then the slope[inaudible] two unknowns. Remember
that. When you put the slope like this, one unknown. This is very important.
So, therefore call this one force b. So, the force b is going in that
direction, which is one unknown. The same thing here. This one has [inaudible]
going in the direction of the right. Therefore,there must be perpendicular to
this direction. Another force up or downward,that's your choice. But this force
has the reverse direction. So, this is 4 and 3. So, that's the second unknown.
So, these are the two unknowns for this system and all you have to
do -- you had to do was to draw the free body diagram. [Inaudible] if this was
p, so you didn't have to put any moment there. But
it says in the text that this is not p actually. It is connected. Is that
connected or not? It's welded together. Therefore, at
this point is point a [inaudible] so you put here the force a in the vector form
[inaudible]. And then you put here ma. How many unknowns do you have? One, 2,
and 3 and so should be solvable. This was question
number 1 but I just want to test you to see whether you are understanding the
idea behind it, free body diagram, first,second, the type of connections that we
have. This connection, you haven't seen it before so you had to think about
it [inaudible]the principle. The principle was that if you have
freedom, you don't put the action there. If there is no freedom, then you put
that there. This too, the rod, actually take the rod
here. If it was being connected here, this rod could rotate. Right? [Inaudible]
cannot rotate at that point. Is that correct or not? Since it cannot rotate,
therefore it has to be -- moment is for rotation.
Forces are for transitional movement. Is that correct or not? Yes? So, that was
part one which has four part. Part two, many people did it correctly. However, I
did not put in this one this support here. It was a ball and socket inaudible?
Yes or no?Therefore, you have to put ax, ay, usually regardless of
what we see over there. We put everything in the positive direction. Then, also
we had there -- here a roller. When it is a contact
or a roller, if it is an even contact, I know over there it was a contact or a
roller, what did I get a contact? But it is no friction. If it is no friction,
there it says a smooth contact. Yes or no? That means there is no correction --
there is no friction. Therefore, all I have to put underneath here I have to put
only sc1.Is that correct or not? For c? Is that correct or
not? Here we had the [inaudible]. Is that correct or not? The bearing in general
have unknown. Three forces and two moment. But,again, in the text of problem we
said there is no moment. Therefore, all you have to put three forces and then I
said there is no axial force in either direction of [inaudible] so you have to
eliminate that.So, then all you have to do correctly. This is by and
that would be bz. Is that correct or not? So, this would be the correct free
body diagram. This is the correct free body diagram.
Now, all the connection was cut out replaced with a purport forces and of course
no moment in this one. Is that correct or not?Now, the question is, "How do we
solve this problem?" So, how do we solve it?It's very simple. This is x, y, and
z axis that you have there. Of course, the first three equation are your force
equation. Is that correct or not? Which you can write it. It's very
simple. It's such a short problem. Many people did OK with this one, but some of
you still have problem taking moment about x-axis,
y-axis, z-axis. If you cannot get this problem, which is very simple problem,
you will have a problem in future with all your courses that you are going to
take related to the design and strength of material, stress analysis, et
cetera, et cetera.[Inaudible] sigma if x equal to zero as you see there. Sigma
of x you have to write this. Many people bypassed that. Sigma if x
equal to zero. You have only ax. Nothing else. Therefore, ax become equal to
zero. It's obviously some of you put [inaudible].
Don't eliminate that because ax is supposed to be there. Is that correct? The
value of that is zero. Then you write sigma if y equal to zero. Sigma if y -- as
you see there. We have ay then we have plus by and then plus cy minus 900 equal
to zero. Nothing can be achieved from here. And sigma if z equal to zero, which
means az plus bz.That's only two force. Must be equal to zero. Very
simple one. That should not take more than 10 minute to finish both problem.
Many of you, of course, finished on time or before
time, which was good. Now, coming today the other three equations. Now, the
other three equation is taking moment about point a.
Notice, if you write it -- still some of you have problem even writing this
scenario. Summation of the moment about point a equals
zero become a vector operation. Is that correct or not? Because you have to use
r cross product. Usually when you take the moment
about the point in space, it will be I, j, and k. Everybody understand that. You
end up with three determined. You end up with r,but I asked you not to use that
fortunately. Many of you except one or two people -- all of you used the correct
method and that correct method is go beyond r cross product, take the moment
about x-axis,y-axis, and which is the end result. We should become
three standard equations. You don't have to worry about rjk anymore. Everybody
understand that. Because eventually -- actually if
you -- what you don't see is this. When you wrote this, this is as if writing
summation of the forces equal to zero. Everybody see
that? That is a? Vector formula. Yes or no? Which you have to write a -- vector
a plus vector b, c, d, equal to zero. Is that correct or not? Here, you expand
that into this sigma if x equal [inaudible]sigma if z. Sigma if a is the same
way. This is a general equation are these two in vector form. Is that correct or
not? When you expand that, what do we get? Sigma and x equal to
zero all the i's, correct? So, why not do it that way? Sigma m about x-axis
equal to zero. Let's see what we get. Does any of these
three have any moment about it [inaudible]? No. Does this one have a moment
about x-axis? Yes. Because away from the x-axis. The
distance is .4 so it is 900 times .4 negative. So, you write 900 times .4 which
is Newton meter, negative. Then nothing here,nothing there. You only have this
one. Cy times? Point 6. So cy times .6. Many of you did this correctly. And that
one is plus moment because it's going this way. A few of you did not. So,
therefore, equal to zero. Immediately you calculate cy equal to 600
because 360 times .6, 600 Newton. So, this one was that. Is that correct or not?
Yes? The next one is either taking moment about the y-axis or moment about the
z-axis. Let's take the moment about y-axis. So,this one sigma m. About the y. Of
course, I'm not changing the y. Usually, if you are change this xy to another
xy, I will say that. But that I'm using the same original xy, z-axis,
so it passes through the a. I don't have to measure. Unless I change it, I do
not have to specify. Is that correct or not? Never --
just because it shows in the picture. So, therefore, as you see, none of them
have any moment about y-axis. This is parallel, yes
or no? This is parallel. This is parallel. The only one is bz times .8. So, it
becomes bz times .8 and it's going negative --actually it's beside the point
because the sum, that's equal to zero. That means
what? That means bz must be equal to?Zero.Zero. Some of you were afraid. You
stopped there because you didn't know what to do. Because you were afraid to
finish the problem. This equal to zero. It means bz must be equal to? Zero. How
else can it be? So, therefore bz must be equal to zero. As
soon as bz equal to zero, there is only az left from this equation. Everybody
understand? Therefore, az must be equal to?Zero.Zero. So, that is zero. That was
a simple problem that I was surprised even I gave it to you [laughing]. Yes or
no? Is that correct or not? That last one is taking moment about the?
Z-axis.Z-axis. Of course, I did it purposely because I want to check to see at
least this minimum is there or not. Sigma m about the z-
axis. Let's write it down. Z-axis, which means you are not setting equal to
zero. So, therefore, this is the z-axis. Then it has
this one is 900 times what? Nine hundred times this .4. And it is negative. So,
therefore, 900 times .4, negative Newton meter. All
of them are Newton meter. Then this one also has a moment, by times? By times
point? Eight. Positive. And this one is parallel. And
this one also have a moment, cy times 1 point? Two. So, plus cy. But cy was 600.
Times 1.2 equal to zero. You immediately calculate
ey. Ey was [inaudible]. ey was equal to? Minus 450 Newton. As soon as you find
that ey equal to from this equation, you would find
ay. So, [inaudible] equation one. So, this is [inaudible]. So, this is all your
answer in the box hopefully. This is how you do your quizzes or homework for
somebody to check. And then from equation sigma ay.
So, we have ay plus by, which is minus 450, plus cy, which was plus 600. And
minus 900 equal to zero. Or ay become equal to 750
Newton. So, that was your quiz. The average was better than usual, of course. It
was about 17. Should be 16, 17. OK. All right.Any questions? Any questions? Not
about the quiz.This quiz was [inaudible]. We are done with the quiz. Any
questions about trusses and frames so far? Couple of points. You have to
be very careful. We discussed that in [inaudible] during the office hours. There
is a big difference. Again, I mentioned that in the last class lecture. But I
don't know if you remember it or not. There are a
couple of the point you want to consider before you
go into the frame analysis. There is a big difference. I mentioned it last time
because you could have a problem there. Let's say
there is a member like that and I show it like this. This is a member like that.
This member being on top of that other one. Is that correct or not? A couple of
you -- I'm not saying this is true for everybody. Couple of member came and they
have problem with the trusses that I gave you before. This is not a three
member. This is a two member. What is this member is one piece. This member
-- this is true. This is a pin. That means every time you see a pin, you have to
see how it goes -- it looks like three members there. Let's put it this way.
This is member ab, dc, cb. If this was the case,this was the three member
scenario as I showed you last time. If within pin, like that. Everybody see what
I'm talking about? All three member has to be pin at the same time, which
become like a truss member. That truss member -- any time we have a joint, all
the member were pin connected. Yes or no? Here, the
scenario is different. So, you have to be very careful. This is two member
because this member has a point a, it has a point b, and
a point c. Therefore, member ac actually is a three force member, but this is
bc. This is totally different. You see a lot of them
in frames. Look at the picture I did last time. So, I asked you that question in
class. Many of you of course noticed that so you gave me the correct answer. I
asked you how many member are there, you said 3 member. Yes or no? So that's the
pulley and we did it. Is that correct or not? Yes? So, pay attention to that
system. And second question is of course obvious. We are discussing it
many times before. You don't have to put forces here bx and by if you are not
disconnecting it. Is that correct or not? Any time
the object being disconnected, then you put the forces there. So, I saw a couple
of people also having the same problem. They disconnect the support. Look at
this member. See, I want you to understand this.Don't get me wrong. We are going
to discuss a little bit of that today. This member inside this member, there are
lots of forces there. Actually there are forces and moments
which we call this internal force and internal moment, which causes this member
to bend or twist or get elongated a little bit or
get shorter, which all of them will come -- you have to handle that in strength
of material classes. So, all the member going through the formation. Yes or no?
We are not yet there. Everybody understand. We never did in the static classes.
We always work on the outside. You see? This is on the outside. Everybody see
that? These are all the connection. The only time you went inside the
member, that was only one occasion and what [inaudible] was that? When we did
the trusses. Everybody understand in trusses, we went
inside the member and find out that -- whether there are a two force member or
not. And fortunately all of them end up to be a 2
force member. Look at this. Oh, I erased it. Look at this member. I purposely
explained that in class to see if this member connected to another member, pin,
pin, this member is a 2 force member. Is that correct or not? But take this
member and put it out. Is this member sitting on a support like that? As soon as
you put this on this member is going to? Bend down. As you see. This
is not a 2 force member. This is the 3 force member or 4 force member or it
depends where it is. Is that correct or not? There is a
big difference between a 2 force member and the rest of the members. The rest of
the member you cannot touch them now yet, until you go to more advanced
classes, which is the strength and later on other
structure classes for similar [inaudible]. Everybody
understand that. Therefore, we're only dealing with so far with the 2 force
member. And if we are dealing with a 3 force member or
et cetera as in the case of frame, you are only end up with the forces at the?
End of the member as you saw we did last time. Is
that correct or not? I asked you. You wrote it down. Find out forces acting at
the end of each member. Yes or no? And that was the
idea. Now, let's do a couple of more examples. So, for that reason, we need to
go through problem -- one of the problems in your
hand-out which we call it machine. It is in your hand-out. So, probably if I get
a copy of that one. So. Showing it on the board,it is a little bit -- here is
the problem you want to show [inaudible]. So, I hope you can be able to show
that. All right. This is[inaudible]. Because when I explain it on the board, it
doesn't look as good as the picture there. Is that correct or not? I should
have done that earlier. All right. OK. Good. So, this is the machine that we
want to find out. This is in your hand-out. This is --
again, this is a frame. I don't have to do that one because it's very similar to
the one we did in the last week. We don't have time to do that. Is that correct
or not? Very similar. All you have to[inaudible] 2 force member there. There is
a pulley there.Is that correct or not? Let's do this one. Is that correct or
not? What's the idea here? Everybody can see what the idea is there
first? What are we trying to do? Is there wire there somewhere? It's like a
cutter. Yes or no? That's why we call it a machine.
This machine is designed to break it -- there is a wire between two planes.
Everybody see that there? Yes? You have your hand-out
with you guys? Come on, guys. So, what are you getting out of this discussion
then if you don't -- oh, you have it there. On your
phone. OK. Have it in front of you. Everybody. You should have this picture in
front of you because I gave -- I asked you to bring
all your hand-out with you. If not -- makes sense the discussion we have there.
There are how many members? There are lots of members there. Yes or no?
Correct? In the machine, the same process. The machine
-- when we go through the process as a frame. Is that correct or not? The only
thing is different we don't usually find the reaction because there is no
reaction needed. All we have to do break it into several member. Look at member
one. Look at member g,b, a. Everybody see what I'm talking about? Yes or
no? Correct? Going to step 2 of the process. There is a member sitting there.
And there is a b -- point b, point a. Point a,according to the picture, is the
pin connected here. First, I draw it here,again, for the benefit
of [inaudible]. Then, there is a member sitting here like that going that way.
And at this point you are applying the force of 20 pound to this handle, which
has 1.4 -- let's put it this way -- 1.4 feet here and this leg is .2 feet. And
that angle is 60 degree. And I draw the free body diagram of this member and
come up with some reasonable answer? Of course, yes.
Because at a is what? At a is? Pin. Everybody see
that? So, I cut the pin out. Replace it with? Two forces, ax, and ay. Let me get
my solution. This is such a simple problem. I can
[inaudible]. So, here it is, ax, and ay. You can put it on top there. Then, look
at this b. When I cut this member, this seems to be connected to. Yes or no?
This is a pin connection. Yes? Should I put bx, by?
No.Or -- OK. Look at bc member. The bc member, this member bc, at the other end
also there is a pin. So, it is pin pin connected, no
load in between. Therefore, you always look [inaudible]. In all the frame, all
the members usually are not a 2 force member except
1 or 2. There are maybe 1 or 2. Sometimes there is not. Everybody understands
that? You need to pick up those first in order to
finish the problem quickly. Is that correct or not? So, bc happen to be a 2
force member. It is horizontal. So, all I have to put
only 1 force and really that's the key. Force, [inaudible] force bc. Is that
correct or not? How many unknowns do I have? Three.
Now, the goal of this problem is to find the forces applied on the? Wire. Is
that correct or not? You see, we are having here a
machine. There are 3, 4, part. But, eventually, the forces will be on the wire
and the wire will be cut. Is that correct or -- this
is a cutter? Yes? So, this is the force fbc. Can I apply force at bc first? By
taking moment about point? A. So, what we do then,
then all we have to do here, summation of the moment at or about a equal to
zero. So, we get 20 times what? Now, do I need to break
20 into 2 components? Yes or no? Twenty -- look at the picture. Twenty is
perpendicular to that rod. So, there is no need to break
it into component. We have seen that before. Twenty times what? What's the
distance? Shortest distance? Shortest distance is 1.4
plus .2. So, it gets 1.4 plus .2 so it is 1.6 feet. The moment is positive or
negative. It is? Positive. Then we have fbc times the
distance. Now, here, you can do 2 things. You can break this into component and
that's easier to use .02. Is that correct or not?
So, we have here 2 component. One is perpendicular to the rod. The other one is?
This is the picture is not too -- OK. So, it shows
that -- the angle is not correct, but nevertheless you have 2 component. One is
fbc cosine 60 degrees. The other one is fbc times?
Sine. So, that leaves fbc times sine of 60 degree, which is the vertical
component here perpendicular to the rod. Multiplied by .2
feet. So, that is -- the [inaudible] is going that way. Therefore it is
negative. And as the result, this should be equal to zero.
We would calculate fbc, fbc ends up to be equal to -- in magnitude end up to be
equal to 184.8 pound. Increase quite a bit. Or we
can call it three digit if you want to go three digit, that would be 185. Is
that correct or no? Good enough as far as [inaudible].
What's the next job? The next job is go to the? That little kidney shape, we
call it load, [inaudible] cd. Everybody understand
that? So, we have this picture here. This one was a pin connection. The end of
this member connect to that one which was like this.
So, here I have to put force fbc. But now force fbc going in this direction.
Equal at? Opposite. This direction is correct, what
does it mean? It means this member is under tension or compression? Just to
refresh your memory. When you put it this way and that
way, the member in between is in? Tension. That's right. So, therefore, you put
here at that force is 800 -- 184 -- I'm sorry --point 8 pound going this way.
So, that's the amount of force you have there.Then, you come out to see what you
have here. Here was the connection with the ground. Therefore, I have to put
only normal force. No friction. So, I put here normal force of -- what did
I do? You can call it -- I call it ne because that force was ne. There was a pin
here attached to the wall. So, I cut that and put
here dx and? Dy. All right. So, here is dx and dy. Again, I have to put all the
dimension there so be careful what you are writing
here from the picture. This dimension is half a foot or 6 inches. This vertical
direction also is given. Half a foot and half a foot all in the picture
give [inaudible]. D is for the pin connection. This is a
contact and this is a 2 force member. So, again,we -- how many unknowns do we
have? Three unknown. Now, sometimes when you read
the question they ask you to find all the forces in
every member. Notice, in this case I did not calculate ax and ay because my goal
is only to find the force in the wire -- on the wire. Yes or no? So, I don't
care about ax and ay. But what I'm asking you, read
your question carefully. Sometimes in the question
they ask you to find all the forces in every member. Is that correct or not? In
that case, you have to calculate ax and ay. I'm going forward without doing
extra work, but be careful about the type of question you may get. OK. Here all
I need -- not dx and dy. All I need is value of n equal to the other member. Is
that correct? Look at the picture. That's why I show the picture there
for everybody to see. So, therefore, here I take the moment, again, about point
d. Look how simple this is. Sigma m at or about d
equal to zero. Obviously this distance is .5. This distance is .5. You don't
have to do anything. This one is plus. This one is?
Minus. So, what's the value of ne? The value of ne must be 184.8. Everybody see
what I'm talking about? Or what I just said -- I repeat myself. Therefore
without that one is 180.48 pound multiplied by this
distance of [inaudible] moment about point d.Therefore, it's .5 and it is plus.
This one is ne times .5 feet, which is this distance and it is negative. That
should be equal --that's what I said a minute ago. However, since this is equal
to that, ne must be also equal to 184.8. Correct? Yes? So, ne end up
equal to 184.8 pound. Finally, we get to the top part of this cutter. Is that
correct? Really, if you look at it, it's a larger
version of the [inaudible] that you are using in the paper. You are doing the
same way. You're applying pressure there. Is that
correct or not? So, top part of this shape, which is something like that, here
is a pin connection. That is f. Point f is pin connected to the lower part. Here
is the blade. Doesn't matter where the blade is here. Here -- somewhere here you
have the wire here. Now, be careful here. Wire is here. The wire is going to be
on the cutting force, you have to press it. But what force I put
here equal and opposite. We went through this many many times. So, we are not
showing the free body diagram of the wire. We are
showing the free body diagram of upper blade. So, that blade we get a force
going up like that. This is the force we are looking
for. Is that correct or not? So, let's call it whatever you want to.
R.OK. We all it r there. OK? And then what else do we have? At f we have fx and
fy because this was a pin connection. And here somewhere was the contact force
on top of that with that one. That's where the contact is. That's point e. At
point e here we have 184.8 pound. See how simple this process is, but drawing it
takes much more time than solving it. Everybody understand what I'm
saying? Yes? Again, there's 3 unknowns and look at the dimension. This dimension
is 2.5 feet and this dimension is 6 inches or half
a foot. And, again, I'm taking sigma m about point f equal to zero. Obviously,
184.8 times 3 feet. And it is negative plus r times
2.5 foot equal to zero or r for a result on force acting on nr or r or whatever
you want to call it. Acting on the wire will be 222
pounds. What did we achieve here through this machine? You take n -- let's say
you're a civil engineer and you are pouring some concrete in some of the
building, concrete everybody has seen outside. There are
lots of [inaudible] there. Steel bar, which are this diameter, small
diameter [inaudible]. You want to cut them, you cut it by
hand -- of course, this is the cutter. Yes or no?
So, what was the result? What did we achieve here? We have seen this before.
Remember, the pulley system that we applied 50 pounds
on one end and the other side 500 pounds or 3,000 pounds we could go up and
down? That's the same thing. Our 20 pound force on the
handle turned into 222 pound force on the wire. You are the same [inaudible].
Twenty pound there become 200 -- 11 times more. Why?
Because of the size of this machine. You make this machine much larger, the
force become much larger. That's why we go to the Home
Depot, you buy a small plier for 5, $6, it doesn't do any good for you. You want
the larger one, you have to pay 50, $60. That does
a better job because of the distance. It all depends on the force distance. This
is the exercise of force distance. Everybody see that? Nothing new there. This
is a machine -- all the machine falls in the same
category. What did I do? Break it into different member. Consider free body
diagram of every end each member, come up with the
forces on those members. Isn't that what we did last
week on the frame too? Yes or no? Correct? Now, let's look at problem number 122
in your book, which is also a cutter, but this time it is a [inaudible]. Maybe
you want to put that one here. Are you OK? Just to show that picture there.
That's problem number 122. All right, then. There we go. So I have to draw that
picture. I don't want to draw that picture in the short time that we have
here. Draw the picture there, but the discussion we need to discuss every
problem like that. So, everybody have that problem in
front of you? It is a cutter. Let me briefly show what's happening here. Now,
the reason I'm showing there because this is a little
bit different than -- there is a handle like this. And this point is called
point e. I'm trying to show the same here. Here, this
point c. This point c is connected through a -- pin c. And then this one is
connected to a member that goes all the way down to
here, for example. This is a 2 force member again. This connected to a blade
like that. Here is the blade. Goes like this. And the
blade is inside the cylinder. I showed that at the little distance for you to
see. This is the wall and this is the cylinder that
goes through the wall. And the end of this blade comes -- let's say you have
here a table. There is a plate here [inaudible] like
that. And that blade is sitting on a table like this. So, as soon as I put a
force here, this force is also perpendicular[inaudible]. I have to
draw [inaudible]. So, this force is perpendicular to the
rod and I put here a 400 Newton force there. And this one will disappear because
this is on top of that. And these are the information that is given here. That
angle is given 30 degree. That means the slope of that line. This length from
point a to point b is given 300 millimeter. This length from point b to
point c is given 45 millimeter. This is point d. D to b given equal to 60
millimeter. And d to b is given -- this is not to scale.
This d to b also is given 25 millimeter. Sorry. This is not correct. I have to
change that a little bit. I know something was wrong
here. So, let's make this angle going like this. OK. Point c is here. So, I put
that -- this dimension, I put it back again. It's
difficult to show this picture. This is 45 millimeter. So, from d to b is 25
millimeter. From b to c is 30 millimeter. Again, this
is not in scale because it is hard to show. But how many member do we have? As
you see, there is a member abc, there is a member
bd. Is that correct or not? And there is a blade. And when you push this bar,
look what happen. This is a machine. You push this
one, member bd become under compression. So, push this blade and blade cut this
plate. So, we are -- the intention is to cutting this middle plate, this plate,
which is on top of the table here. There is a table here. Down here and that
plate, you can slide it underneath the blade and you cut it. Is that correct or
not? Yes? How much force are we applying to the -- this is in your book so
you don't have to really try. I did it for the purpose of if somebody else wants
to listen outside this class to this. Is that correct or not? Yes? Again, this
is a machine. The idea is transferring that load to the blade. Obviously,
again, the load is going to be increased more than 400 depending on the -- what
should I draw here? Now,there is a point here and that's what the reason I
am showing that. Because of this blade sliding inside this cylinder. Is that
correct or not? It is a cylinder or a wall or something that this blade can go
up and down. Is that correct or not? Therefore,first part of it exactly what I
just did here.First part of it, drawing the free body diagram of a, b, and c.
OK. That part is very simple. So, you draw that part like this --
or the way it was. OK. And you put here at a, you put your forces there, 400
meter. And at b, this is the whole point -- at b all
you have to do is draw a force in compression format. In compression format, let
me put it in red because that's unknown.Therefore, in compression format
increase must be going in that direction. Yes or no? Correct? And then at c you
have to -- you are going to put cx and cy. All the dimension is given. So it is
what we need is this dimension. This dimension is given 30 millimeter.
And this leg from a to b is given equal to 300 millimeter. And the height here
is given also equal to 45 millimeter. And what is
important here, the slope of this f is given also. Look at the picture you have
there. The slope is 25 and 60. Is that correct or
not? Which is 5, 12, and 13. Is that correct or not? Therefore, the slope of
this line, the way I've drawn it is not correct, but
this is 5 run and 12 rise, 25 millimeter versus 60 millimeter. This is 60
millimeter. This is 25 millimeter. As I said, this is not
in scale. Is that correct or not? Yes? So, that's the slope. How many unknown do
we have? Three unknown. Can we solve for b? I'm not going to do that because we
have done that so many times. What is important here, writing your -- drawing
your free body diagram correctly, specifying all the unknown and known forces,
which we did here, taking moment about point c? Of course, you have
to break that into 2 component, 5 over 13. Let's call this one f. Maybe we can
do that. So, let's do at least that point. So what
happened here you have 400? Of course 400, you have to find the vertical
distance and the horizontal distance. But giving that
angle [inaudible] to 30 degree. Remember, that angle is also equal to that angle
is also equal to 30 degrees which makes this perpendicular to here.
Nevertheless, we don't want to go through detail of it
because this detail is a little bit lengthy. It is
just everybody can do that. Is that correct or not? You find the vertical
distance and the horizontal distance from point a to
point c. Is that correct? This distance -- is a little geometry. You add this
height, which is this 300, versus this 30. Everybody
see that sine or cosine -- sine of 30 degrees. You can calculate that. Add to
this and break this into component. Find the moment
here. Break this into companion. Find the moment here. But the net result is
that's what really not my point here. The net result
is after you do that. This f force, magnitude force become equal to 3,098
millimeter. So, increase by that amount. Then what happen
here next is my -- why I'm doing this problem for you. Because this is a little
bit different than previous problem. Then what happens next, guys? Now, OK, I
find this force on this rod. Is that correct or not? I have to come here to this
plate. Is that correct or not? What should I do about the plate? Is there a
question there?Please, ask. Is it OK?Yeah. I just couldn't read your
handwriting.No. Handwriting. Yeah. OK. So f equal to -- I set f equal to 3,098
Newton. OK?All right? Now, what do I -- my question is -- guys
-- what should I do next? Free body diagram of the? Of? Blade. Yes or no? OK.
Very. So, how do we draw the free body diagram? You
start doing it and while I'm operating the board, you start doing it yourself.
There is a question there I want you to answer. That
question is part of this problem -- this new stuff there. Tell me what you do
after -- OK. I can see what you said is correct. OK.
I am drawing the free body diagram of the blade. Fine. This is free body diagram
of the blade. Is that correct or not? So, this is
point -- where was that? That point was point d. So, what should I put at point
d? The force equal and opposite, but it is in compressive format so it should be
going toward the d. Is that correct or not?So, that force is how much? 3,098
Newtons. OK. 3,098 Newtons. What else?Pressure on the [inaudible].
Pressure on the plate. That's right. Because this blade -- you're trying to cut
-- see, if this is separated, you are trying to cut this plate here. Is that
correct or not? This is where the cut is going to
happen. Is that correct or not? So, you're pressing
it down. There is an r force or n force at this coordinate. So, that is n. I'll
call it n or r. I don't know what they call it here. So --
Yeah. The Newton force [inaudible]?That's right, but you can see that this is
not -- that's exactly what I want you to think about it. This is a. There is a
normal force there. But this is very -- inside the cylinder -- there
is [inaudible].Where should I put my n? To the left? To the right?
And why? If you know the answer, so I didn't have to do this. Yes or no? Where
is that normal? Because this blade, notice, is touching in general this one and
that one. Yes or no? It's not [inaudible]inside that wall. It is sliding up and
down. Yes or no?The blade cannot be [inaudible]. It has to go down [inaudible]
something. Yes or no? Where is the normal force? See? That's
exactly. Is that like that or not? It cannot be like that because if like that,
they cancel each other. If that's correct -- that's
exactly my point. Only reason I'm showing you this. Right. Everybody said. So,
the normal force would be only this direction. You
see here there is a horizontal force there. That is the answer to your question
[laughing]. However, what happens here is this.This is the [inaudible]. This
blade, you can pull it or you can push. When you put it this way, we said it is
a [inaudible]. But it is going to touch the other wall. It is going to be a
small [inaudible] crack there. Everybody -- that [inaudible] separate this
wall, the normal force would be here when you pull it. Now you are pushing it.
So it separates here but contact here. So, you have
to put only normal force on the contact side which is the -- which I'm sure you
can. Is that correct or not? So, that is another n
on the wall and this is m on the plate. Is that correct or not? And that is the
free body diagram of the plate and you should be able to calculate all the
forces because all the direction is here. Is that
correct or not? So, therefore, this is free body diagram of a -- we call
this -- because the dimension -- this dimension is not
given so that is a -- I think they consider it as --
as a vertical because the horizontal force must be equal to horizontal force.
Vertical force must be equal to -- vertical. We don't
have to worry about anything else. Is that correct or not? Therefore, this angle
is given. And remember that was the slope of the-- the slope of that was? 5,
12. So, 12, 13 of that one equals to n wall, 5 vertical. That point equal to --
sorry -- 12, 13 --that one equal to mp. So, mp minus 12, 13 of 3,098, which is
the vertical component going down negative. This is going up. That's
our answer equal to zero. And mp become equal to 2,860 Newton. That is how much
pressure you have -- the only, as I said, reason I
did this problem is showing you how this normal force works. Is that correct?
And the other problem on your hand-out is the same
thing. Again, you can take a look at all the picture I gave you. I already gave
you the hand-out there so you can look at the hand-out. And that should take
care of this type of problem, all the machine[inaudible]. Now, I want to tell
you something else[inaudible] new hand-out that I gave you. Look at chapter 7 in
your static book.We are not going to cover chapter 7 in this class
because chapter 7 is all about internal forces in a member. And also is about
the shear and moment diagram. Actually this morning
in static, the same class here, this morning I was talking about shear and
moment diagram. Which is all about static but it was in
a strength of material class. Everybody understands. So, you are skipping
chapter 7. However, I always mention this to student
prepare them for future classes to see the difference between internal forces
and external. Notice this member, you said it is under? Two force members, under
pure tension. So what -- pure compression in this case. What happened here?
Please write down in your notes, this is just a brief explanation of chapter 7.
I have given 4 pages.Therefore, if you want to do self-study, you do
it. It's not required for this class. But I'm preparing you for future class,
which is ME280. Is that correct or not? For you to
understand what we did in this chapter. What we did in trusses, we went inside
the [inaudible]. What we did for the frame, we stayed outside. Everybody
understand that. What that means is this. Don't get me
wrong. This member, if I cut it here or cut it here or cut this member, all of
them -- there are some internal action. Due to those internal action, which
could be a force or could be a moment, object is going to go through the
formation. Those the formation are the subject of the strength of material.
Class, because if no object is actually rigid, you put here beam here, in the
static, we consider a load here, we find the reaction
here and reaction. But put a load somewhere in this beam. Somewhere, doesn't
have to be -- see how it happened to this. Right?
Actually this morning we were talking about why this is bending, how much it's
going down, where is the stresses, et cetera, et cetera, and how we draw the
shear and moment diagram for this because in the body, there are forces and we
call them differently.So, that's the explanation -- beginning of explanation of
chapter 7. It is very important for static student to understand that
before they go to future classes. So, for that reason, I'm going to spend about
5, 10 minutes. It really helps you a lot so you understand what you are going to
face. Although chapter 7 is not part of this course. Everybody understand that.
Let's look at the beam like this. Of course the beam, when it is
here, [inaudible] and a pin. As soon as I put here, let's say, a load of 900
pound here, this no longer is a 2 force member. This is a 3 force member. Yes or
no?Correct? And let's give it some dimension. OK. Let's
say this is equal to 4 feet and this is equal to 2 feet. So, the next job for
you in general is to find the reaction, which is not
a big deal. Everybody by now can do it. Is that correct or not? You have a pin
at the left. You have a roller at the right. And there is no horizontal force.
Actually put even the -- let's put -- let's change
that. Sorry, guys. Let's change that to something
that I can get the horizontal force as well. So, let's change that to let's say
900 pound force, but let's give it a slope of 3 run
and 4 rise at point b. OK? So, it makes it a little bit more [inaudible]. Frank?
All right. So, now at a -- I call this point a and
this point b and point c. Notice I'm erasing all the -- or cutting all the
support and putting reaction there. So, if I put a
reaction here, I should put here ax and I should put here ay and I should put
here only cy because that's a roller. Is that correct
or not? Can I calculate those lambda? OK. So, get your calculator because I just
did that. So, first of all, of course break this into 3/5 and 4/5. So 1/5 of 900
is how much? Hundred -- this is when put myself[inaudible]. How much?
180.180. And 3 times 180 is? 540.540, yeah.OK. So, we have here 580 horizontal.
So, I break that. I show the result in now.540 pound and how much vertical? 4 of
them --another 180. Yes? 360, 720? OK. You do it with calculator. I have to do
it in my head. OK. I was faster than you [laughing]. 720.
Is that correct or not? Yes? All right. Now, 720 -- so this force ax -- this is
so simple problem -- ax become equal to 540. I don't have to -- so, this part of
it is so simple, 540 pound equal and opposite.Then, in order to find here, I
have to find ay and cy by taking moment about that. Now, give me that one. So,
please, ay equal to 700 -- 1/3 of 720. How much is 1/3 of 720?
240 How much?240 240? OK. 240? Ay become 240 pound. And cy become 485. Correct?
For a total of 720 -- a ratio of 2 to 1. This is the reaction that
you have done -- not in this chapter -- you have done it not in chapter 7, not
in chapter 6. You did it in chapter 4, rigid body.
Is that correct? Everybody should be able to do. That's not the point. The point
is this. These are all now become external. All the reaction -- action and --
all the reaction is external. Now, if you want to
go in the body and calculate forces in the body,this is not a 2 force member. If
it was 2 force member, it would have been pure tension and pure compression like
this member. Is that correct or not? This is not a 2 force member. There is more
to it than tension and compression. Let's see what happen here.
Let's cut this -- this beam, which is like this actually has a dimension -- cut
it at 3 feet distance. For example, cut it and find
out what are the internal forces at that point. Is that correct? Like what you
see in chapter 1. Look, there is a beam there. They
cut it. Is that correct or not? It's in your hand-out. But this is an example.
There is a [inaudible]. Is that correct or not? If I
cut the beam at that point, now, this is the body of the beam. I have here ax
which was 540 pound. I have here ay which is 240 pound. Here, let's call it
point -- I call it a, b, c. Let's call this point
point b for example. Is that correct or not? Yes? So,
what you do you do at b? What do you put [inaudible] at d? What should I put at
d? Dx? Dy? And a moment to find out what are given.
Some of them may be zero. Some of them may not. Is that correct or not? Yes? So,
this dx -- write [inaudible] -- this was 540 going
this way. Notice if I put here a force going that way, in general, I put it all
in positive direction. This is cx. This is cy. And
that is mc. Is that correct or not? Yes? Cx, cy -- I know. So, we call it
different. Yes or no? They call it differently. Is that
correct or not? Oh, b. Oh, I'm sorry, b. OK. That's what you said. OK. So, dx,
by, and mb. Is that correct or not? Yes? What's the
value of dx?Negative 540.Negative 540 means this [inaudible] is under
compression of 540. In future, we don't call it dx. We call it normal forces.
Let's just added something new. So, there is a normal force in this member with
a magnitude of 540 going in the compression mode. Is that
correct? That's what you were doing in the trusses. That's what we were
determining. But no there is something else vertically. Is
there -- this is going to go up. This force must be going down. So, dy must be
equal to minus 540 pound which is going down at the
value of that from now on strength of material become the shear forces because
that causes the member to shear off. Everybody understand this. This is normal
force. This is shear force in general. And this is a? Bending moment. That
cause -- that's what you saw here. You saw only part of it. We don't see the
other part. That's what you just saw there. One more time. With the force there
is the same exactly almost here. Put it here to back that part. Is that correct
or not? Yes? Look. It's bent. Right now it's not bent. As soon as you put the
force there, it is start bending. At this point,why it is bending because of
this action. Everybody understand that. So, these are the internal. We talked
all about it. They are not in this form. They are form of express. We talk
about it all 10 weeks of strength of material one and 10 weeks of strength
material 2. Then a few -- we all understand it. Now, can
I calculate mp? Everybody understand that. To calculate mp you can take the
moment about point a or point b and calculate that. But
the point is there are all these actions there for 2D. But look at the 3D. In 3D
it's a little bit different. I am going to give you an example of a 3D briefly.
I'm not going to do it because it's not necessary to do that. However, in 3D we
don't have -- here is 2D, 2 forces and 1 moment equivalent to your 3 equation of
equilibrium. Like sigma fx, sigma fy, and sigma m about the z-axis.
Yes or no? But in 3D problem what do we get?Six.Six. You get fx, fy, and fz. And
then you get mx, my, and mz. Therefore, if I have -- let's look what I picture
here just brief.I'm not going to do that. If you want more explanation, you can
an come to my office and I'm going to explain it to you in detail
if you want any further. But you don't need that in here [inaudible] next
quarter if you are going to go into ME280. Is that
correct or not? Let's say there is a rod here bent like that. Let's say this is
3 meters long and let's say this is 1 meter long.
And then you put here a 2 kilometer force vertically. And you put here a 5
kilometer force horizontally at this end, which I'm
going to call ma, b -- this is b and [inaudible] at point c we want to find out
the internal forces there. Is that correct or not?
Yes? Let's assume this is a rod like that, but it is solid. If it is a rod like
that, I have to cut it here. Yes? And show all the
internal forces there. Is that correct or not? Yes? To do that, this is what we
are going to do. We -- to do that, you are going to
cut it here, show that end in 3D format. This is the best way to do it. So,
that's that end. Everybody see what I'm talking about.
This would be x direction. This would be y direction. And this would be the z
direction. Is that correct or not? So, here I have --
and this is point that I [inaudible] support here. I don't care about the
support. I'm here and at this end, this is point c. This
is point c. This is x, y, and z. Now, do I have a force in the direction of x? A
reaction here? If this is all the action there, do
I have a force going -- exerting [inaudible] body? If this 5 moves that way, I
should have a 5 going that way. So, I have here a 5
kilometer force go there. Is this the normal force or is it a shear force?
Normal force. You are getting hang of ME218. Any problem
-- many student ME218 have difficulty recognizing this. Everybody understand
that. Do we have a shear force going in the y direction? Look. There is a --
this is all about the stacking. The point is this
is the stacking, you have to carry it into the strength of material. Some people
have problem with that. Now with the forces.With the moment because you have
problem with the moment, you will have problem -- look what happened here. There
is a 2 kilometer going down. There should be a 2 kilometer going to
be upwards. So, there is a 2 kilometer shear force going 2 kilometer in the y
direction. This is normal in the x direction. Is that
correct or not? Do we have any shear force going in the z direction?
No.No. Because there is nothing here. If I had another force going this way or
that way, that would be showing up there. Therefore,
no shear force in the z direction. One of them is zero in other words. Do we
have a moment about x-axis?No.That is what I said we have problem with. Do we
have a moment about the x-axis?Yes.Oh, yes. That's exactly right. I don't want
to go any further into that, but you can see there is a moment about x-axis.
That moment twists this. This will turn it around. So, that moment going that
way we call it torque. Somebody ask me in this class what
is torque -- the difference between torque and the moment. One component of the
moment become the torque, the other 2 component become bending. It is there. You
don't want it or not. It is there if I have a member like that, apply this
moment to it, it is torque. Look at my top. Apply this to it, it is? Apply this
to it, it is?Bending. This is going to haunt you in ME218, so don't
tell me that I didn't tell you about all of this. Is that correct or not? Yes?
So, therefore, this moment will be the torque is that you see this twisting.
This moment is going to bend. This moment is going
to bending. And that, at the end of ME218, you will
design an object like that. You are -- you see outside any traffic light that
you easily pass by every traffic light, every one of
them has all 6 action in that. At the base of the traffic light, if there is a
traffic light like this, this is the column, this is
the light, is that correct or not? The wind -- this column has a weight. Yes or
no? Then it has the wind blowing that way. Yes or
no? At this space, look at it. You have all 6 actions there. Everybody
understand that. You have the shear force. You have the
normal force. You have the twist. And you have the bending in both -- it bends
this way or it bend that way. Is that correct or not? Everybody see what I'm
talking about. That is the subject you are going to
study in the ME218. Is that understood? Now, I have
5 minutes to start chapter 8 because we are done with chapter 7. Chapter 6 is
finished. Is that understood? You have frame. You
have machine. You have -- there are some other picture, but I'm not going to go
through that. There are part of it [inaudible] but
you only drew the small part of it, the one that is 2 force member, et cetera.
That I'm not going to ask you. But there is the plier. The plier you divided it
into several member and you solve it. That's not a big deal. Either we don't
have time to do it so we are going -- I am going to go to the next stage to
review the principles of next chapter. Again, we have -- we have skipped
chapter 5, which is the centroid. We have skipped chapter 7, not part of this
course. We are going to go chapter 8, which is the
friction. Open it today and Thursday we are going to go. Then next week we are
going to go to the centroid. And then after that we
are going to go to moment of inertia. If you come to my office this morning, I
ask a question from [inaudible] very complicated design of eccentric loading,
but 60%, 70% of the problem was centroid and moment
of inertia because in every design, you have to do
the centroid first and then do the moment of inertia. And many of those students
do not get it because they ran quickly through the
stacking without realizing the next two week is very essential for you to
understand, prepare you from today's lecture, you realize
that is preparation for future courses. Everybody understand that. Friction is
something simple. Everybody has done it. So, let's
look at friction, see what's the difference between friction and nonfriction.
So, I explained that idea to you. And the next time
we are going to start with some example. So, let's understand what is
[inaudible] friction. Let's say we have here an object with
the weight of w sitting on a ground that has a friction. And let's say from
physics, you remember that coefficient of a static
friction, mu is given .3 something like that. Is that correct or not? Yes? OK.
When I do the free body diagram of this object, so
how do I do that now? Assuming this ground is having friction. If it no
friction, w and n are equal and opposite. There are no
other forces. Remember that. Here, I put here the w. And then I put here n. Of
course n must be equal to w. Where should I put my
small f [inaudible]? Small f is equal to mu times mu static time [inaudible].
Where should I put that? This way? That way? This
way? That way? Which way? That is something I want to explain. No way. Because
there is none. Although the ground is having friction, there is no fraction
there. Everybody understand that. So, there is no
this way, that way. Both way is wrong because you
only have it when there is a horizontal force. That's right. So, that's the key
here. Since I don't have any horizontal force, this
is equilibrium. There is no f or any sort. So, therefore, that would come in the
picture if you have a horizontal force. Everybody
now know that friction is against the action of the p. It is not going that way.
Is that correct or not? Now, what should I put here? This f? Yes? Wrong again.
That's the -- that's the whole story. Now we are-- we know better about the
static [inaudible].This is no. We never put that there. We put here force f
because force f is equal to p. If p equal to zero, f --
[ Music ]
