Articles

# Forces Acting on a Frame

In this problem we’re being asked to find
the forces acting at the five points A-E on this frame. There’s a uniform distributed load acting
downward on member CDE, and that load has a value of 2 kN per meter of length. There’s a load acting horizontally at member
ABC, and that varies from 0 at point A, and it increases vertically, with a linear increase
up to a maximum value of 2kN per meter at the top. To solve this problem let’s begin by drawing
a free body diagram, and we’ll have to do three free body diagrams, but let’s start
with this two force member BD. So here I’ve draw it, point B is in the lower
left, point D is in the upper right, and let’s make the assumption that this two force member
is under tension, and that a force, D, acts to the upper right, and a force B acts to
the lower left. And these forces, because it’s a two force
member, the forces have to be equal and opposite. If they weren’t equal and opposite it would
suggest that the member was either rotating or accelerating. So I’ve drawn the two arrows pointing opposite
of each other, and I know that D has to equal B. Already we started out we’ve got two unknowns,
but we have one equation already, D=B, and as we work our way through this problem I’ll
identify unknowns, and I’ll identify all of the equations that we can write, and you’ll
see at the end we’ll end up with seven unknowns, but seven equations, and we’ll be able to
solve all the values that we need. So start out with two unknowns already and
only one equation. So I got that out of the way, let’s continue
by drawing a fee body diagram of this member at the top, member CDE, and here’s that free
body diagram, we’ve got CD and E, and what we’ve got is a pin connection at point c,
a pin connection at D, and this roller at point E. So with a pin connection at point
C there’s some vertical component of force, and we’ll call that Cy, and some horizontal
component of force, and let’s call Cx, and at location D there’s a member BD is pulling
to the lower left at point D, and we can show this is a 45 degree angle, this angle in here
has to be 45 degrees. The reason it’s 45 degrees is because point
D is 1 meter away from point C, and point B is one meter below point C, so this angle
in here is 45 degrees. So we’ll call this force D, and at point E
this roller, because a roller can only exert a normal force, at point E that force acts
only to the left, or in the negative x direction. There’s one other force, a downward force
resulting from this distributed load of 2 kN per meter, and that force ,because this
is a uniform force, that resultant force will act right in the middle of member CDE, and
that value, let’s call it R1, it acts straight downward. R1 is equal to 2 kN per meter times the length
of CDE, which is 1 meter plus 1.5 meters, and that says that the resultant force is
equal to 5 kN, and it acts a distance right in the middle of CDE, and that distance is
1.25 meters. So we’ve got three more unknowns, the first
one would be Cy is unknown number 3, Cx is unknown number 4, D we’ve already identified
as an unknown, and the fifth unknown is E. So we’ve got five unknowns and we’ve only
written one equation so far. So let’s write three more equations, which
are going to be the force balances in the x and in the y direction, and a moment balance. The way I do this to keep my notation straight
is I’ll label my three equations, the sum of forces in the x-direction, sum of forces
in the y-direction, and the sum of moments about some point. So if I do the sum of forces in the x-direction
I’ve got Cx acting to the right, I’ve got -D, because it’s acting to the left, times
the cosine of 45 degrees; -E because it’s acting to the left, and those three forces,
when I sum them together, have to equal 0 because nothing’s accelerating. And similarly the sum of forces in the y-direction
is equal to Cy acting upwards minus D sin 45 minus R1, again has to equal 0. And now let’s do the sum of moments, and I
can choose any point on my free body diagram about which to calculate the moments, I am
going to choose point C because it knocks out Cx and Cy, and I am going to use a convention
that a counter-clockwise moment is positive, and a clockwise moment is negative. So let’s start out, from point C we’ve got
point D is acting 1 meter to the right of point C, and it’s a clockwise moment, so it’ll
be negative 1 meter, and we want the downward component of that, so it will be negative
1 meter times D times the sin of 45 degrees, and then I’ve got the moment due to R1, it’s
also clockwise, so that’s negative 1.25 meters, the distance of where the resultant force
acts from point C, times R1, and those two moments will sum to zero. And I now have a total of 5 unknowns, and
I’ve written the first equation already, here’s equation number 2, 3, and 4. Five unknowns and only four equations, so
we need to keep going with this and write another free body diagram, and let’s do a
free body diagram around that vertical member ABC. So I’ve cleaned it up a little bit and I’ve
drawn a free body diagram of ABC, and I’ve left the free body diagram of CDE in there just
for reference for right now. So let’s look at this, we’ve got at point
A there’s some unknown vertical component at A, we’ll call this Ay, and there’s a horizontal
component which we’ll call Ax. We don’t necessarily know that they act upward
and to the right, but let’s assume that they act in the positive x and y direction, and
if we get a negative sign it simply means that they act in the opposite direction. And at point B there’s a force acting to the
upper right, and this again is a 45 degree angle, and let’s call this force B, and it
acts to the upper right, because we’re going to assume that this two force member BD is
under tension, and we may get a negative number, implying that the force acts in the direction
opposite of which we proposed at the beginning. But here’s where we have to be careful for
point C, in the first free body diagram I proposed that Cy is acting upward, and C acts
to the right on member CDE, so that means if it acts upwards on member CDE it has to
act downward on member ABC, so here I am going to draw Cy acting in the downward direction,
and Cx acting to the left, and this is just Newton’s law that forces are acting equal
and opposite to one another. Now there’s one other force, the resultant
force due to this horizontal triangular distributed load, and we’ll call it R2 and we’ll say we
need to calculate where R2 is located, and it’s magnitude, but let’s just say it acts
right here. And let’s calculate the magnitude of R2. So start with that, we’ll say R2 is equal
to the integral of this distributed load, w, over the length Dy, or over the entire
length of ABC, and we need to come up with a function for w, and w, it’s a linear function,
and w will equal, it turns out, 2 kN per meter. So I am saying the maximum value 2 kN per
meter acting at C divided by the height of this thing, which is 3.6 meters, multiplied
by y. So what I get at y=0 that means we’re at point
A, and that suggests that w is equal to 0 kN, and at y=3.6 it suggests that w is equal
to 2 kN per meter at the top. So if we integrate this function of w over
the entire surface of it we’ll find that the magnitude of R2 is equal to 3.6 kN. We also need to find the distance that R2,
that this resultant force acts above point A, and I’ll call that y(bar). I can solve for y(bar) by using this relationship
that says the magnitude of the resultant force multiplied by y(bar) is equal to the moment
that this resultant force causes, so I can integrate the differential moment, which is
w times y dy, and integrate this differential moment over the entire length to solve for
y(bar). So when I plug in numbers and rearrange to
solve for y(bar) and perform the integration what I’ll find is that y(bar) is equal to
2.4 meters, and if I substitute y(bar) in my diagram I’ve got R2 acting 2.4 meters above
point A, and R2 has a magnitude of 3.6 kN. So with this free body diagram what we’ve
introduced are two more unknowns, unknown 6 is Ay, and unknown 7 is Ax. We’ve written four equations, let’s write
three more, the sum of forces and sum of moments for this third free body diagram, and then
we’ll have a total of seven equations, and seven unknowns, and at that point we can do
some linear algebra and solve the problem. And here are those final three equations,
sum of forces in the x, and in the y-direction, and again I can choose any point about which
to do a moment balance, in this case I’ve chosen point A, so in the x-direction, in
the horizontal direction, I’ve got Ax acting to the right, R2 acting to the right, plus
the horizontal component of B, which is B sin 45 minus Cx acting to the left. Similarly for the vertical components, and
in the moment about point A I’ve got a clockwise moment due to R2, which acts 2.4 meters above
point A, and I’ve got another clockwise moment acting 2.6 meters above point A. So it’s the
horizontal component is B sin 45, and I’ve got a counter-clockwise moment acting at C,
which acts a distance of 3.6 meters above point A. So here’s where the final three equations
that I need, here’s equation 5, 6, and 7, and now I’ve got seven equations and seven
unknowns. And when I solve for these I’ll find that
B is equal to negative 8.8 kN, suggesting that member BD is under compression, B and
D are equivalent, and I’ll find that Cx is equal to -2.1 kN, and Cy is equal to -1.3
kN. The force E is equal to a positive 4.1 kN,
and Ax is equal to 0.54 kN, and Ay, the last, the seventh one, is equal to 5 kN.