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This is module four of Mechanics of Materials part three on beam bending.

Today's learning outcomes are to continue to determine the internal shear forces and

bending moments in multiforce members.

And to sketch a shear force diagram from multiforce member.

And once again, we're going to go back and look at

this topic as I presented it in my course, Applications in Engineering Mechanics.

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So, here's the generic beam situation that 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 slope of the rate of change of the shear diagram.

And we found that the change in shear between two points

equals negative the area under the load curve.

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

to this beam as loaded as shown and come up with a shear force diagram.

So we know 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 at this giving 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 handily.

When you have that done, come on back.

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Okay, here's the result you should have come up with and

the equations of equilibrium I used to get them.

I found f sub y is 11,000 pounds out here on the right.

And c sub y is equal to 21,000.

And since there's no forces in the x direction, there was no x reaction for

set point F.

2:07

Okay. Now, knowing those,

there's my beam with all the loads 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 sheer 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 counter clockwise on

the right side of might material just like this situation over here.

So, that's going to be negative.

So, a shear force immediately drops down to -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 here.

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

21,000 pounds and again, we keep looking to the right on the material.

So that's up and it's going to be a positive sheer force and so

we're going to change from -10,000 up to a positive 11,000.

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

During that interval, we

know that that change in sheer is going to be equal to the area under the load curve.

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

And the total load under the area under the load curve is going to be

6,000 pounds.

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

than 11,000 or that'll be 5,000 pounds.

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Now, the shape of the curve in between those two points,

we know at any point along that distance, minus the value of

the load is equal to the change or the slope with the shear diagram.

And so, the slope of the shear diagram the load here is 0 and

goes up to minus 6000, so it gets greater, greater, and

greater in a negative sense, and so here we started off with a slope of 0 and

it becomes more 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 as 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 shared forces and

so no shear force changes.

And so we're going to go from D to E as a straight line.

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 -11,000.

We're going to be down here at -11,000, and

between those two points, the negative value loads stay constant at 4000 so that

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

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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 at 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 sheer force ended up switching through and was zero.

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

We'll use this triangle, and this triangle.

And so we've got.

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This side, this length is 4 and

we're going to compare it to this length here,

which we'll call x.

So x is to 4 as this drop was 16,000.

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Actually, we'll just call it 16.

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

I mean, you could do 16,000 to 5,000 or 5 to 16, wither way, is fine.

If you solve for x, then you find this distance x is 1.25 feet.

The total distance is 4, so that means this distance is 2.75 feet.

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So now we know the value of the sheer, the internal value of the sheer 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 would have the maximum

shear that you would design for it, take a minute and think about that.

Come on back.

7:46

Okay those locations, the worst case is here at point C where I have a shear force

of 11,000 or here at point f where I have a sheer force of -11,000.

Here's another large sheer force, you can see we actually have no sheer as I

said at this point right here, but we've got a very good sense of how this sheer

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 sheer force diagrams.

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