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[SOUND] This is module seven of

Mechanics of Materials part IV.

Today's learning outcome is to employ singularity functions for

beam deflection problems.

And so, as a review we have a differential equation for the elastic curve of a beam.

If we can have an equation for the moment,

we can find the deflections by integrating when we've done that in the past.

However, if the moment equation is difficult to write,

this method becomes very time consuming, and

because you have to do matching conditions of slope and deflection at each junction.

And so what we said is we're going to us a mathematical

function called singularity functions and I defined those last module.

And so let's look at an actual system,

a real world beam with a variety of loads being applied.

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And so if I go from left to right and

I look at the moment anywhere along the beam, If I cut the beam between A and

B, and I look at the left hand side you'll recall from my previous courses that

the moment would be clockwise, so it would be a positive moment.

And it would be equal to Ay times the moment arm,

which would be the distance we've gone out in the x direction.

At point B now, so that was true for

x between x and x1 but then we get to point B.

At point B now and beyond point B, between B and C,

we have the moment if we look to the left A sub y times it's moment arm,

which will be x, plus now M sub B, this applied moment at point B.

We continue on.

If now we go, and that's true, that last section was between B and C.

Now we're going to look between x2 and

x3 where we now have a point load also applied.

And so we've got Ay times the moment arm x plus MB

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plus now P and its moment arm is whatever x is minus x2.

And then if we go into the distributive load portion which is between x3 and

x4, actually x equals L, we now have all of this added together.

We have A sub x times A sub y times x plus

MB plus P times its moment arm which is x minus x2.

And then we have, when we get into the distributed load, minus,

because it's going to be a negative moment being applied.

And in fact there was also a negative moment applied because of this point force

here and there was a minus sign here.

But we got minus W which is the distributed load times x- x3 which is

the portion of the distributed load that we're actually

looking at at what ever cut we're at.

Times its moment arm, which will be x- x3 divided by two,

which is in the middle of that distributed load.

That goes all the way over to where x equals L.

And we don't have to include D sub y,

because at x equals L the singularity function is equal to zero.

We don't get to that point,

we get right up to D sub y and we don't have to include that point though.

And so now we can write this full equation in terms of singularity functions.

And so we've got M equals between zero and x1 we've got A sub yx,

but that's actually is included all the way up through x equals L.

And then we have M sub B, which is only included beyond x equals 1.

So when x is less than x1, then this doesn't come in.

When it does, when this singularity term is included it's raised to the zero

power which means that the singularity function is equal to one or just MB.

So we've got MB, MB, and MB.

Now between x2 and

x3 when we add P we've got negative because it causes a negative moment.

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P times x minus x2 singularity functions raised to the first power,

and that means that it's just x minus x2 like we have here.

And then finally, we've got between x3 and

L are our distribute load, which is going to be W over two negative again.

Times x minus 3 minus x minus 3,

or in singularly function form x minus x3 squared.

And so anything less than any x value less than x3,

that distributed load does not contribute.

And so you can see now that this bottom equation

takes all of these sections of the beam and puts them together by using

the singularity functions into one expression for the moment.

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And this is the expression again.

So let's use this result

to generalize what the singularity function terms look like.

And so F will represent the magnitude of the load, x is going to be the point,

x sub LOAD is going to be the point where the load is applied.

And then n is the integer which will describe the load.

And so for an applied moment, like we had here at M sub B,

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the magnitude of the load is, we'll just say generically M sub a.

In this case it was M sub B, but generically M sub a, and

it was applied at point A, which in this case was x1.

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the zero factorial, zero factorial being one.

Then for a point load, which we have here, we have the magnitude of the point load,

which would be generically P sub b times x- b,

where b is the location that it would be applied, which is here x2.

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For the distributed load we have w,

the magnitude of the load times x minus c in the singularity function.

Which c is where the distributed load starts being applied which is x3

in this example and then we square that and

we divide by 2 factorial, and 2 factorial is equal to 2.

If you continue this process you'd find for a ramp load that

the magnitude of the load would be the change in the w over the change x.

The change in the load over the change in x,

times x minus d where d is the location that the ramp load would start.

And then it would be cubed over 3 factorial.

Three factorial is six, and this continues on.

And so this is for distributive loads that start before the beam.

We use super position of these different types of loads.

And so let's look at an example of when the distributed load or

ramp loader whatever loads stops before the end of the beam.

And so here we have this situation shown here, as we go along

when we reached x1 we have a ramp load so we're going to use this form.

And so this form, it's going to be a negative moment.

And it's going to be delta w, the change in the load,

over change in x times x- x1 cubed again, over 3 factorial.

But what that's going,

that term is going to give us a ramp that goes all the way to the end of the beam.

But we only wanted a ramp that goes out to x2.

So we're going to have to use super position and

we're going to have to subtract out this portion that's dashed.

So let's start by subtracting out,

and in this case it would be now a positive moment,

a ramp that starts at x2 to the end of the beam and that would be this portion.

And then we'll also subtract out a straight distributed load from x2

to the end of the beam, which would be this term, which

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is delta w from x equals x2 out to the end of the beam,

and that's this portion shown in brown.

And so this and this balances out and

we are left with what we had originally which is just this portion of the load.