I'm Jim Kerns welcome back to our
examples of statics problems at
Lawrence Tech.
Today we're going to look at finding
the
forces in certain members of
the truss shown at the right.
it’s kind of an odd looking truss but we will
look at it and see if we can figure out.
We’ll use the method of sections in this
case since
all three of those forces that we want
to find
lie across one section. So our problem
statement is to find the forces in the
members
EF EH
and GH. And you know obviously we can
draw a section through there
and find the forces across that
cut line and that gives us everything we
need.  Okay.
But before we continue just a comment
Here;
I have a 700
Newton load here at point C
which acts down through here. This
member here from F to C
is actually in theory, a zero force
member.
And I did it this way to give you an
example
- even though this is a, you know,
according to our rules
of analyzing trusses, a 0 force
member,
you wouldn't want to build the truss
without that member because one of our
other assumptions is this is just a pin
down here at that joint.
So if we had a pin their and this member
was gone
We'd essentially be balancing our 700
Newton force on a pinned member and would just
fall over, okay?
So we need this element here from F to C
To supply structural stability even
though
it's not carrying significant loads
through the truss.
We wanted to find those forces, the
obvious thing
to do is the method sections. Again this
is, you know,
a way I would give you a hint by that I
want to use a method to sections. If I
give you
forces and members all across a
common section.
We could use a method of joints, you know,
We’d have to start here
at A
and then we could find that forces a D
and then we find the forces a G
Then we find the forces that E and,
you know, then we could
find all these,
that's kinda long way around. So we'll
 use the method of sections.
First step is, I wanna find the reaction
forces because,
assuming that the stress is stable, any
Sub-part of that truss that I cut out - this
piece of the truss is going to be stable.
And we can apply our rules a statics
but to do that we need to know the force
right here at A.
So if this is 700 Newton's downward,
we're gonna start by finding the sum of
the moments about point B, and that will
give us this reaction force over here
At A. The sum of the moments about B 
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00:03:03,689 --> 00:03:08,909
Equals Zero. I’ve got my 700 Newton forces
right here, it's pressing downward,
that's going to give me a counter
clockwise rotation around B so that’s positive
700 for the magnitude
And the distance,
The perpendicular distance between that
where that force in applied
and point B there, is three
meters
seven hundred times 3 and then
I have my force
at A which is 21 meters away, and that's
going to be a
clockwise rotation. That’s negative-
um the reaction force
at A times the distance that’s
twenty-one meters.
And if I solve that, I get
my reaction force at A = 100
Newtons. And that's enough for me to
continue the problem.
But as long as we're here up when we do
This,
I'm going to do this both sections just
to show you can go both ways and get
the same answer.
So let's find the reaction force at B as
Well,
And that is the sum of the forces Y = 0
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00:04:20,480 --> 00:04:24,190
and that's going to be
equal to my
100 Newtons at A,
plus my reaction B,
minus our 700 Newtons downward.
And, so Rb equals
600 Newtons. And we can also see
from inspection, there are no forces
at all in the
x-direction so there is no
external reaction forces in the
X-direction.
The X direction forces that A and B
are
are going to be equal to 0. So here, I’ve cut
the truss into two sections,
I have penciled in our
reaction forces
and I've drawn in some assumptions about
the forces at our section.
I'm going to assume that that EH across
the top
and GH on the bottom are both
in compression. So they're both going to
apply force to the left here on the
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00:05:21,870 --> 00:05:25,110
left section. And I'm assuming that
did
EH, this diagonal, is in tension
so that's going to give me a force
downward. And I did that because
obviously if I just look at all the
y-direction forces, I see that
I've got an upward force, I'm gonna need
a downward force somewhere, so I just
assumed that.
And on the right, you get
the same, essentially the same thing,
where my assumption of compression gives me a
force to the right here,
A force upwards there, and a force
to the right there. So if we just start with
our left hand Section
we can see that there's only 2
elements here where there is any
vertical forces - one
is the 100 Newton
reaction force at A, and the other is the
force here on EH which is on the
diagonal. So I can do the
Sum of hte forces in the y-direction
equals 0
and that's going to be equal to my 100
minus the force in EH. And this
is a 45
degree angle they're all 45s from
horizontal -
just the way I drew it - I didn't label it
but
I guess add that in as I go.
and in these sections here are
3 meter intervals, I failed to mention at
the beginning -
if you're wondering where some of these numbers
come from
So the sum of the 
forces in the y-direction = 0,
that's equal to 100 minus the
force
Force in EH times the sin of that
45 degree angle to give us the
vertical component.
And that gives me EH
equals 100 over
Sin(4), and that’s equal to about 141
Newton's. And that is in tension.
I can do the sum of the forces
In the x-direction = 0; 
assuming compression, I've got
EF, and that’s
what I assumed when I drew those arrows
there. Plus 
GH -  actually that those should be
negative
because if they're in compression
they're pointing to left. Negative EF
minus GH plus
EH
Cosine ( the x component)of 45,
Okay? And I've got
to unknowns in there because I know,
%eh
I know EH - that magnitude, but I don't know the
other two. So we're not quite ready for that
Equation. But let's jump down and do
the sum of the moments at some point.
I could do the sum of the moments about
point E here,
then I only have one unknown.
I'm going to do the sum of the moments
about point H
which is out here somewhere; okay?
it’s of my diagram, but that's okay
because
The sum of the moments about any point in
space has to be equal 0
for this to be in equilibrium.
So the sum of the moments about point H
equal zero. I picked that point
because the force
in GH doesn't cause any moment,
The force in EH doesn't cause any moment,
so I can ignore those. But I have the
force from the reaction it is causing a
clockwise moment that’s equal minus
100 times 12
Meters. And I have the force
EF here which is going to result in a
um counterclockwise moment. So that's plus
EF times the distance which is three
Meters.
If I solve for EF I get
400 Newton's in compression,
and I can substitute this back up into
that equation. I now know
EF and EH, all I need to find
is GH. And I get
GH equals
my EH times
Cos(45)
minus GH -
not GH – EF,
And when I plug those numbers in there I
I get -300;  so that's negative;  so my
assumption
about compression is wrong.
So what I really have is GH
equal
300 Newton's in tension.
And that's really if, you're thinking
about a truss
in general, if the loading is downward
the top of the truss is gonna be
In compression, the bottom the truss is
going to be
in tension, and the other members can be
in tension or compression depending on
what's going on.
So that's my answer for those three
things -
there  - there - 
and there. And we’ll now work on
doing it for the right hand side.
So here’s the right hand section,
I got my 700 Newton applied force there,
I got the 600 Newton reaction force, and
I've got the forces in the elements
that cross the border
here. I'm assuming that that EF, the top
one, is in compression
and GH
is in tension, the bottom one, because
that’s
a typical force for a section of a
truss like this.
And just looking at this, I'm also gonna
assume that
EH, on the diagonal, is in tension as
well because
my net force is downwards, I need some
net
upward force. And I can do the sum of
the forces in the Y direction 
equal 0; to keep the whole thing
in static equilibrium. So I've got
my 600
Newton force from the reaction at B.
I've got a -700 Newton force from my
applied force, plus EH
times the sin of
45 degrees. That gives me
EH equals 100
Newtons in tension, which is the same
answer we had on the previous example.
Now for the sum the moments, I can't
do some the forces X yet,
we don’t have everything, I could pick a
points off my section like I did last
time.
But I'm going to use H again because
it's a point
on my section. When I pick H, this force GH:
I don't care what it is because it goes
through H, 
so it's not in my equation. Same with
this force here EH
that doesn't contribute any moment, even
though I know what it is.
So my only unknown left then is the EF
there
that I need to solve for. So if I do 
the sum of the moments H
 got
equal 0 and it’s
equal to -700 times
this distance here, which
is six meters from H
perpendicular to the line a force there.
And that's
clockwise so it's negative. And
plus 600 Newtons and this perpendicular
distance here is nine meters
and that's counterclockwise so that's
Positive. Minus EF,
 because it's giving me a clockwise
moment.
Times this distance here which is three
meters
And
working that out 
I get 400 Newtons 
in compression; same as the other side.
And now I can do the sum of the forces
in the Y direction – oops!
Sum of the forces in the X direction.
= 0
And that's going to be equal to my
EF
which is 400 Newtons
minus GH which is still unknown,
minus
141, the force in EH, times the
Cosine of 45 degrees.
And solving at out I come up with
GH equals 300
Newton's in tension,
which is the same answer we got before.
So we can pick either side, we can go,
 you know, works in both directions.
The thing to be careful, you know, if
you're working with this right hand
section is don't forget
to include this in all your calculations
because you have these two forces,
to external forces to that truss that
you have to account for.
Other than that,  that’s all intended to cover.
Thanks for watching, and I'll catch on
the flip side
