JEFF HANSON: This
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below, and get the help that you need today. Now, on with the video. [MUSIC PLAYING] Hey, gang. Welcome back. We're talking today
about trusses. We're talking about the
method of joints, probably one of my most very popular
videos on all of YouTube here. And I'm going to show you how
to solve truss problems using the method of joints. So we have a little
simple truss here. And it says, find the force
in each member of the truss and state whether that force is
in tension or in compression. So this is the first time that
we're going to be asking you, is the member in
tension or compression? Remember, a truss
is made completely of two force members. And two force
members either exist in tension or compression. So step one, we need to
identify what are those members. So how many members
are in that truss? What do you think? Here's what I think. There's 1, 2, 3, 4, 5. This is not one
member down here. Every member terminates
at every joint. Let's just write them all down. So we got we got member AB. We got member BC. We got member AD. And then we got BD
and, finally, CD. 1, 2, 3, 4, 5. 1, 2, 3, 4, 5-- good. So that's what
we're looking for. So how do we find
out what's going on? So what we're talking about
now are internal forces. These-- this 8 and this 5-- are external forces. And those guys are
causing some force. If I cut that member
in half-- whoop-- and I looked inside of it, there
is some forces inside of there that's generated by these guys. And that's what we're finding. That's what AB is is the force
that's inside of member AB. And I'm going to show
you how we get it. It's quite easy, using
the method of joints. Let's write a recipe, OK? So number one, step one-- find global equilibrium. And that's just a
fancy way to say, find the reactions
at the supports. So step one, I'm going
to look at this thing and say, OK, what's going on? I've got a pin connection
at A. So at A, I'm going to have two forces,
one like this and like this. So an Ay and an A x. And let's see. A x, which way does it go? Well, the 5 goes that way. So the x better go that way. Ay, which way does it go? Now, remember, if you don't
understand which way it goes, it's OK to guess. I don't know. I'm going to guess
that it goes down. I might be wrong. But when I solve for Ay,
if I get a negative, hey, I was backwards. And then over here,
at C, there's a Cy. So I've got a first
step, step one. I've got to find
A x, Ay, and Cy. And we do that by looking at-- you guessed it--
the moment at A. So sum of the moments at
point A, this being positive, equals 0. What do I have? I put my finger at A. I've got
5 that rotates me clockwise. So that guy's negative, so
minus 5 times how far away? 4. And then I got 8-- oh, he also
rotates me negative, minus 8-- times how far away? 3. And then I got Cy,
which rotates me-- oh, that's counterclockwise,
that's positive-- so plus Cy times how far away? 4. So let's see what we get here. We got 20 plus 24 equals 4 Cy. So what is that, 44? So 44 divided by 4 equals
Cy, which equals 11. So Cy is 11
kilonewtons going up. What's the total downward
force on the system? 8. So what does Ay have to do? It does have to go
down, doesn't it? And as a matter of fact,
it has to go down, Ay. And then here it is. I'm going to write it down. Here's what I was
doing in my head. I was doing the sum of the
force in the y in my head, which is 11 positive--
going uphill-- and then minus 8
and then minus Ay. So move the Ay to
the other side. Ay is equal to 3 kilonewtons. And of course, the sum
of the forces in the x tells me that A x has got
to be equal to 5, right? So this guy, 5 kilonewtons,
this guy 3 kilonewtons, and this guy-- I'm doing him in red-- 11 kilonewtons. We have completed step one. We found global equilibrium. We're good to go. So step two-- step two. Select a joint. We're going to pick a
joint to start working on. However, there's a
special note here. Do not pick a joint with
more than two unknowns. Why two unknowns? Because we're going
to analyze a particle. These are little
chapter 3 problems. Remember the chapter 3 problems. So let's say we
start off at joint A. The next step is step three-- three-- draw FBD of the joint. And then, I guess,
step four is solve-- whoops. And then, if there's a step
five, it would be repeat. So let's start at joint
A. If I start at joint A, if I take my free-body
diagram cookie cutter and I select joint A, I'm
only going to cut through two members that I don't know. That's only two unknowns. So joint A would look like this. There he is. It's got something like
this, something like that, one like this,
and one like that. This guy is 5 kilonewtons. This guy, going
downhill, 3 kilonewtons. Now, this guy up
here is member AD. And this guy over
here is member AB. Well, this one's
pretty easy, isn't it? Because for that joint to be in
equilibrium-- he's not moving, so the sum of the
forces has to be 0-- then the up stuff has
to equal the down stuff, the left stuff has to
equal the right stuff. So if that guy is
going that way, this one has to go this way. And if that guy's going down,
then this one has to go up. And this is pretty easy. It tells me that AB has to
also be equal to 5 kilonewtons, and AD has to be 3 kilonewtons. So right away, I know something. I know AB is 5 kilonewtons. And I know that AD
is 3 kilonewtons. Now, here's the one thing that's
different in this chapter. You have to tell
me, is this member AD, is that member,
and member AB, are they in tension
or compression? So let's talk about
that for a second. If you are the member and
I come up and I squeeze-- [GRUNTS]---- and you're
in compression, what are you going to do to me? [GRUNTS] I'm going to
push back on you, right? This is a free-body
diagram of a joint. So a compression member
always pushes on the joint. A tension member,
he's being stretched. And so he's going to
pull on the joints. So is this force pulling on
the joint or pushing on it? It's pulling. So forces going away from
a joint are always tension. Away-- always tension. That's not a free-body diagram--
it looks like, oh, the member's getting squeezed-- no, wait. That is not a free-body
diagram of a member. That's a free-body
diagram of the joint. The member's in tension, and
he's pulling on the joint. So I'm going to put this. I'm going to put a T
there and a T there. And that's all you have to do
in this chapter is identify is it in tension
or in compression? Is it a T or a C? That's it. OK, now we've done joint
A. So what should we do? Let's look at joint B over here. Now, there is one
thing that we're going to have to do real quick. And we kind of need
this angle right here. Oh, that's easy, isn't it? That's a 3-4-5 triangle. So that means that this angle
right here is 36.87 degrees. Where did you get that? Well, you can use
tangent if you want. But it's so common in statics
that I just remember it. So here is joint B. I'm going
to draw joint B over here now. And joint B, what does
joint B look like? Well, here's what he looks like. He has a force
there, a force there, a force there that's
8 kilonewtons. And then it's got
this one right here. Do we know any of
the directions? Let's just label these. This is AB, and this is BC. And this one up here is BD-- BD. Look at AB. You know what? There's AB up there. So on this one, AB was pulling. So what's he going to be
doing to this joint over here? Now, if you said going
that way, that's not right. Because if one end is going
that way and the other end is going that way,
what's the beam doing? It's moving, right? If one end goes that way, the
other side has to go that way. It's a two-force member. So the two ends act in
opposite directions. So if the guy is pulling
on the joint over here, he's pulling on the
joint over there. So this goes like
this-- wah, wah. And we know how big AB is. It's 5 kilonewtons. We know this angle right here-- 36.87. Is that right? No. No, that angle is not 36.87. Wrong. I should have done that
after all, isn't it? It's 53.13. That one up there is 36.87. Woo, I'm glad you all
told me about that. I almost messed that up. 53.13 degrees-- that's better. So BD and BC, do I
know which way they go? Well, I got something
to the left. I got something going down. If I got something going
down, then BD has to go up, doesn't it? Since BC doesn't have
an upward component, this one has to go up
just to have equilibrium-- downwards, upwards. OK, now that guy's
going to the left. That guy's going to the left. So BC has to go to the right. A lot of tension members in
this truss, aren't there? We haven't had a
compression member yet. So let's see if we
can solve this truss. So this guy has two components. And you can use that
angle if you want to. Or you could do this. This side's 3. That side's 4. It's a 3-4-5 triangle. So the x side, this guy right
here, would be 3/5 of BD. And this guy over here
would be 4/5 of BD. And if you want to use
the angle, that's fine. If you put that
in your calculator and you do the cosine of
53.13, guess what you get? 0.6, which is the same as 3/5. That's a little shortcut
trick that we've learned. So here we go. We have two equations
we can write-- the sum of the forces
in the x and the sum of the forces in the y. That's all we can write. So here we go. In the x, what do we have? We have BC minus 5
and then minus 3/5 BD. And then in the y, what do
we have in the y direction? We have, going uphill, 4/5 BD. And then in the negative,
we got negative 8. So we move the 8
to the other side. And we get-- what do we get? We get 8 divided by
0.8, which is 10. So BD equals 10. And that's one of the
things that we need. And then we can substitute
that back in right there. 0.6 times 10 is 6. So negative 5 minus
6 is minus 11. Move that to the other side. So BC equals 11 kilonewtons. And which way do we draw them? They're going away
from the joint. So BC is in tension. BD is in tension. So BC is 11 kilonewtons. What did we say? Tension. And then BD is 10
kilonewtons tension. OK, we got one more to go. Let's make us some
space, and we'll do CD. We'll put him right here,
the old hand eraser trick. All right, here we go. We can do joint C. Let's
just do joint C right there. So joint C. Always, when you're working
these problems on your test, label what you're doing, so
if somebody's grading it, it's very easy to tell exactly
what you were doing there. And there's Cy, isn't he? He's 11 kilonewtons. So if that one goes up-- this is CD up here-- then this one has to
go down, doesn't it? It has to. And then this one-- that
one goes to the right, so this one has
to go to the left. And we knew that. It goes away from the
joint because he went away from the joint on
the other side. So this is BC. He was away from the joint. He was going this way over
here, so he's going this way over there, right? It has to be. And then one thing we need
more is this angle-- whoop-- right there, which is
4 this way and 4 that. It's 45 degrees, isn't it? So from all of this-- this guy has two components,
one there, one there-- I need that component, which is
the sum of the forces in the y equals 0 equals-- what do I have-- 11 going uphill. And then what do
I, going downhill? CD minus CD times
the sine of 45. And so move the 11
to the other side. Let's see. 11 divided by 0.707,
which is the sine of 45-- 15.56. So CD equals 15.56 kilonewtons. And CD, is he going away from
the joint or to the joint? He's going to the joint. So that means that's
the only compression member in the whole thing-- 15.56 kilonewtons compression. That's all there is to it, gang. You just start at one
joint, and you just start working your
way around the truss. This is not the fastest
method in the world. It's not fast at all. And what we found at
A, we carried to B. What we found at
B, we carried to C. So if you mess something
up along the way, you just carry your
error, and it just gets worse and worse and worse. So there's a lot of
places to get messed up. But what did we do here? Right here, when we started off
and we did global equilibrium right there, that's
a chapter 5 problem. Well, sum of the forces of
x, sum of the forces of y-- this thing right here is a
little old chapter 3 problem. This up here, a little
old chapter 3 problem. The one that I erased,
a chapter 3 problem. So we had to do, for
one chapter 6 problem, we had to do a chapter 5 problem
and three chapter 3 problems. So there you go. That is the method of joints. I hope this helps. Thanks.
