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[SOUND] This is module seven of

Â Mechanics of Materials part IV.

Â Today's learning outcome is to employ singularity functions for

Â beam deflection problems.

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

Â If we can have an equation for the moment,

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

Â However, if the moment equation is difficult to write,

Â this method becomes very time consuming, and

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

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

Â function called singularity functions and I defined those last module.

Â And so let's look at an actual system,

Â 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

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

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

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

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

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

Â At point B now, so that was true for

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

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

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

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

Â We continue on.

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

Â Now we're going to look between x2 and

Â x3 where we now have a point load also applied.

Â 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.

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

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

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

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

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

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

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

Â here and there was a minus sign here.

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

Â the portion of the distributed load that we're actually

Â looking at at what ever cut we're at.

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

Â which is in the middle of that distributed load.

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

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

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

Â We don't get to that point,

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

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

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

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

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

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

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

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

Â So we've got MB, MB, and MB.

Â Now between x2 and

Â 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,

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

Â And then finally, we've got between x3 and

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

Â Times x minus 3 minus x minus 3,

Â or in singularly function form x minus x3 squared.

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

Â that distributed load does not contribute.

Â And so you can see now that this bottom equation

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

Â the singularity functions into one expression for the moment.

Â 5:04

And this is the expression again.

Â So let's use this result

Â to generalize what the singularity function terms look like.

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

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

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

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

Â 5:30

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

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

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

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the zero factorial, zero factorial being one.

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

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

Â 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,

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

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

Â in this example and then we square that and

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

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

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

Â The change in the load over the change in x,

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

Â And then it would be cubed over 3 factorial.

Â Three factorial is six, and this continues on.

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

Â We use super position of these different types of loads.

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

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

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

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

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

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

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

Â But what that's going,

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

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

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

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

Â So let's start by subtracting out,

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

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

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

Â 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,

Â and that's this portion shown in brown.

Â And so this and this balances out and

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

Â