So here's the generic beam situation that I've, I, I laid out last time.

We looked at the free body diagram of a

differential element, and we came up with the following relationships.

We found that the negative value of the load at any point equals

the, the slope or the rate of the change of the shear diagram.

And we found that the change in shear between two

points equals the negative the area under the load curve.

And so what I'd like to do now is to apply those that theory, if you will, to

this beam, as, as loaded, as, as shown. And come up with the shear force diagram.

So we know on this, on this beam, all of these applied external forces

and moments. The only thing we don't know at this point

is the reactions at the roller constraint and the pin constraint.

So I'd like you to start off

by determining the external reactions at point

C here at the roller, and point F at the pin on your own.

And by now you ought to be

real experts of this, given your background from the first course

introduction engineering mechanics, and what we've done so far in this course.

So you should be able to knock this out quite,

quite handily when you have that done come on back.

Okay here is the result you should have come up with

and, and the equations of equilibrium are used to get them.

I found Fs of Y a 11,000 pounds out here on the right.

And Cs of Y is equal to 21,000 and since there is no

forces in the x direction, there was no X reaction force at point F.

Okay, now knowing those there's, there's my beam with all the loads and,

and this moment being applied.

We're going to work on the bending moment diagram in the next modules.

But for this case now, we'll use these relationships for the shear diagram.

And so for the shear diagram, as we come along and we're looking at

the material to the right, we have zero up until we get to the end of the beam.

So we start off at zero, and at the end of the beam, we see that we have this 10,000

pound shear force.

And it's acting counterclockwise on the right side

of my material, just like this situation over here.

So that's going to be negative.

So the shear force immediately drops down, to minus 10,000.

And now we don't have any change in the shear between points

A and C.

There's no other shear forces being applied.

There is a moment, but that's not going to affect the shear force.

And so, we go out to point C here and the shear force

diagram stays at minus 10,000 pounds. At that

point, we have our external reaction from the roller that's up

21,000 pounds.

And again we're, we're, we keep looking to the right on the material.

So that's up, it's going to be a positive shear force, and so we're going to change

from minus 10,000 up to A positive 11,000.

And then we're going to go from point C now to point D.

During that

interval, we know that the change in shear is going

to be equal to the area under the load curve.

And I found the area under the load curve, it's just a triangular area.

And the total load under the area under the load curve is going to be 6000 pounds.

So, we know that at point D the shear force is going to be 6,000 pounds less

than 11,000, or that will be 5,000 pounds.

Now the shape of that curve in between those

two points, we know at any point along that distance

minus the value of the, the load is equal

to the chan, or the slope of the shear diagram.

And so the slope of the shear diagram that,

the, the, the load here is, is zero and

it goes up to minus 6,000 so it gets

greater and greater and greater in a negative sense.

And so here we

start off with a slope of zero, and it becomes

more and more negative until we get down to 4,000.

And So we've integrated a ramp function here.

So this becomes a parabola function. I'm going to label that a P for parabola.

By the way, down here in going from point A to point C.

That was a straight line, so I'm just going to label that with S.

Now again between D and E we have no shear forces and

so,

And then going from E to F, the area under the load curve is 16,000.

So that means the change in shear will be from 5,000 down to minus 11,000.

We're going to be down here at minus 11,000.

And

between those two points, the negative of the value of load stays constant

at 4,000 so the, the slope of the shear diagram is going to stay constant.

And it's going to go down with a straight line here.

And, so that's a straight, curve.

And then finally we're going to go up at the

end 11,000 that's positive, and it closes off at zero.

And you should close

off at zero on both ends of your shear diagram.

The only other thing you might want to do is you might want to find this distance

where, the, shear force ended up switching through and was zero.

And so let's use these similar triangles to find that distance.

We'll use this triangle, and this triangle.

And so we've got

this side, this length is four.

And we're going to compare it to this length here, which we'll call x.

So x is to four as this drop was 16,000

actually, we'll just call it 16.

We'll keep the thousands off, that'll be easy, just to 5,000.

I mean you could do 16 to 16,000 to 5,000, or five to

16, either way is, is fine. If you solve for x, then you find this

distance x is 1.25 feet. The total distance is four.

So that means this distance is 2.75 feet. So now we know the value of the shear.

The internal value

of the shear force anywhere along the beam,

and we would use that in designing the beam.

You can see on your own, where would be the cases

where you'd have the maximum shear that you'd have to design for?

Take a minute and think about that.

Come on back.

' Kay, those, those locations, the, the worst case is here

[COUGH]

at point C, where I have a shear force of 11,000.

Or here at point F, where I have a shear force of minus 11,000.

Here's another large shear force.

you can see we actually have no shear, as I said, at this point right here.

But we've got a very good sense of how

this shear varies over the entire length of the beam.

And so this is just a clean copy of what I just did so that you can see it.

And that's it for shear force diagrams.