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Welcome to Module 21 of Mechanics and Materials Part IV.

We're moving along and getting close to the end of the course.

We've done deflections, we've now

completed buckling, and today we're going to look at combined static loading.

This is a culmination of all my mechanics and

materials courses, the parts one through four.

And so let me warn you in advance that this is going to be a rather long module

but you can go through it, stop it, take it a step at a time, and

be able to digest and understand how to do this type of analysis.

And so, our learning outcome is to combine,

to solve a combined static loading problem.

This is the problem I'm going to look at.

I've got an engineering structure.

I want to find the principal stresses and the max sheer stress at point A,

which is a point at the top of the beam right in the center.

And I am going to assume that the material in the structure remains in the linear

elastic region for all the loading conditions.

There the loading condition that's given.

And that's important because when we do combine static loading,

this linear elasticity will allow us to analyze each type of loading individually,

and then add them together by superposition.

And so, how might we start?

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And what you would say is, okay, to start,

we're going to have to make a cut and find out what the internal forces and

moments are at that cut that goes through point A.

And so, do that cut and then draw the free body diagram and

if you draw the free body diagram, this is what you get.

You've got your external forces, and then we have,

since this is a, when we cut it we've got,

it's a completely fixed condition between the two parts.

And so we have three force reactions, RX, RY and RZ.

And we have three moment reactions that prohibit rotation, M RX,

M sub RY, and M sub RZ.

So, let's go ahead now and find the forced reactions, and do that on your own.

Come on back, and let's do it together.

And so, to find the forced reactions, we're going to, at the cut,

sum forces factorially, and set them equal to zero.

And so I'm going to have R sub x in the i direction, we're going to use

a vector form, because that's the easiest way to do it for the 3D problem.

Plus R sub y in the j direction,

plus R sub Z in the K direction

plus now we've got 2,000 newtons in the I direction,

now I'm going to leave the units off, we can put the units in at the end,

and then we have plus 4,000 Newtons.

In the, well I put Newtons in there.

Let's take that out.

[LAUGH] Newtons.

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cause any moment because their line of action goes through the point in

the center of our engineering member.

But we do have M sub Rx in the i direction

plus the moment reaction in the y direction and the j direction,

and plus the moment reaction about the z axis in the k direction.

Plus then we have the two moment, if I'm looking at the outer side,

if I'm looking at the right hand side here of my cut, I have

the moments due to these two forces, which we're going to use as R cross F.

R is the distance from the point about which we're rotating out to

the line of action of the forces, and so R will be, in this case,

we're going out 800 millimeters in the z direction or the k direction.

And then we're gpmma go up 200 millimeters in the j direction.

And we're going to cross that with the forces acting at that point,

which are 2000 in the i direction plus 4,000 in the K direction.

And all that has to equal 0.

If you do that math and match components, you get M,

moment reaction about the x axis plus 800,000 In the i direction,

plus moment reaction about the y axis plus

1,600,000 all in the j direction.

Plus moment reaction about z,

-400,000, all in the k direction equals zero.

I match components, I get M sub

Rx equals -800,000

Newton millimeters.

M sub Ry equals -1,600,000 Newton millimeters.

And finally M sub Rz equals

400,000 Newton millimeters.

And so, here are the total results.

Those are my force reactions.

And those are my moment reactions.

Okay now with those force reactions and moment reactions,

we're going to get various loading conditions at Point A.

And so let's start with torsion.

And so for torsion if I take this cut, and this is difficult, so

you may have to try to take a piece of pool noodle or whatever and

look at this, but we're looking in this direction at this cut.

We see that the moment reaction in the Z direction is 400,000, do it in

millimeters and so by the righthand rule that's going to be this direction.

And so if I draw a cross section looking back out in this direction,

I'm going to have my moment of reaction about

the z axis is shown there in that direction.

Here is my point A, right at the top.

And we have our axis.

Up is y, and if I'm looking back in this direction,

then I'm going to get this axis is x.

And so let's first calculate the stresses due to torsion out at point A.

And try to do that on your own, and then come on back.

We know that the shear stress due to torsion is going to be,

if you review back to my Mechanics and Material part two course,

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the torque times row which is the area,

or excuse me the radius from the center out to the point we're looking at.

And that's point A which is at the outside, which is going

to be 50 millimeters divided by J where J is the polar moment of inertia.

So, I've got the torque, which in this case for

about Z axis is 400,000

newton millimeters times row,

again that's at the outer surface at A's that's 50 millimeters divided by J,

you go back and look at my part two course in mechanics materials,

we would calculate J for this circular cross-section as pi over two

times the radius, which is 50 millimeters to the fourth.

And so we find that the sheer stress, then,

is equal to 2.037 newtons per millimeter squared,

which is the same as megapascals, and so I can draw a stress block now.

So let's draw a stress block here, at point A.

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And we see at the, let's put our axis on here first.

We have our axis to the right, which is z, and our axis up, which is x.

With the stress block and

our coordinates if we look at this inner edge here of point A,

the stress due to torsion will be in the negative X direction.

And so, it's going to be down like this, which means it's up over here,

this direction here, this direction here and

stress due to torsion is pure sheer in this case and

the tal is equal to 2.037 megapascals or newtons per millimeter squared.

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And so, I've got my bending moment about the x axis

is in the negative x direction, so it's going to be

equal to 800,000 in this direction.

And the bending about the y axis is going to be down, and it's 1,

it's negative, so it's in the negative y direction,

so It's going to be 1,600,000 newton millimeters.

You can see now for this bending due to this moment about the y axis,

since point A lies on the neutral axis, which is right in the center of

the cross section there is not going to be a bending stress or

a flexial stress due to that torque or moment.

So we've got sigma due to m

sub ry is equal to zero.

However, we can see due to this bending moment, or torque,

about the x axis that we are going to get flexural stresses in the z direction.

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And in this case, it's going to be equal to

by my third course in mechanics and materials, part three, sigma.

Whoops.

Sigma about the z axis.

Right? Because it's bending about the x axis.

So, it's going to be either coming in or out.

And we'll talk about whether it's intention or compression in a second.

So that's going to be m sub r, x times

y over i, which is the area moment of inertia.

Y is from the center of rotation out to the point of interest, which is point A.

So, that's going to be the radius which is 50 millimeters, so I've got

800,000 newton millimeters for m sub-bar x,

y is from the neutral axis here out to the point of interest, which is up at the top.

So, that's 50 millimeters, since the diameter is 100 millimeters.

We're going to divide that by i, i for

the cross section, it's a circular cross section, is pi over 4,

times 50 millimeters to the fourth.

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I guess I left off my units up here, newtons per millimeters.

That's going to give us newtons per millimeter squared or NPA.

And so that value is 8.149 NPA or

newtons megapascals or newtons per millimeter squared.

Let's look physically here and see If that's going to be

in the Z direction if that's going to be intention or compression.

So this is the Z direction.

You see it's coming from this 4,000 newton force in the Z direction.

That's what caused this bending about the X direction, and so it's going to pull.

And so, you can see physically that at the top of the beam

this is going to be in tension.

And so, this is tension.

And I can draw a stress block again.

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Now, realize with transverse shear that we're violating the assumptions that

we made in my Mechanics and Material, part three course about calculating transverse

shear, the shear stress formula that we came up with, because the edges of

the cross section we said to use the shear formula must be parallel to the y axis and

so it wouldn't be good for circles, or triangles or semi-circles.

And we also said said that there had to be uniform shear stress across the width of

the cross section.

That's not the case here, but you're going to find the transverse shear

stress contribution to this overall combined loading is quite small.

and rather negligible and so we'll use this sheer stress formula even

though it's by rights not exactly what we should apply.

And so for transfer sheer again let's look at the cross section.

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looking back out the z axis towards the positive z axis again at the cut.

We have our coordinates Y up and

X to the left and that transverse shear stress will be caused by

this 2,000 Newton force.

That's going to be in the negative X direction.

So let's draw that on here, we have 2,000 Newtons and

it's in the negative X direction.

I can now inappropriately but

it's close enough because it's a small effect use the sheer stress formula, T

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for the outer area, we're talking about from point y here and

it'll go out and to the centroid of that outer area.

If you look at a reference

out to the centroid,

this distance will be 4/3 r/pi,

and the area = pi r squared over two.

And so, if I put those values in,

I'm going to have 4r over 3pi,

which is the Y out to the centroid of the outer area times the outer area itself.

The outer area itself which is pi r squared over two.

And then we're going to divide by I.

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And if you put in those values,

you'll find out that the cal due to the transverse shear is pretty small.

It's 0.00679 Newtons per

millimeter squared again, all this was,

let me write down where r equals 50

millimeters which was the radius.

Finally let's draw our stress block at A.

My X direction will be In this direction,

if i'm looking down on top of point A, my Z will be to the right.

And the sheer stress caused by this transversed

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Equal to 0.510 Megapascals.

All right.

So, we took care of torsion shear,

torsion stress for this loading, we took care of bending stress for

this loading, we looked at transfer shear stress to this loading and

we've looked at axial stress and we assumed that the elements stayed

in the linear elastic region so we can combine them all by superposition and

here they are torsion we found here looking at our element A.

Again, we're looking down on our element A, so X is in this direction.

Z is in this direction.

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And this is our horizontal by the Mohr's circle.

Clockwise is positive so it's going to be plus two point four four,

for the tail direction, so this is no normal stress.

And 2.004 For shear stress, that's the horizontal face.

For the vertical face, we're going to have positive 8.659 for the normal stress.

And since this is counterclockwise, it's going to be negative 2.044 for

the shear stress.

So we're going to go over here,

8.659, down, so

-8.659 and -2.044.

We have two points.

I can go ahead and now draw a line between them.

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And when we do that, we can find the radius.

Let's start by finding the center.

That's going to be 8.659 divided by 2,

or 4.33 and zero,

which means my radius now is equal

to the square root of 4.33

squared + 2.044 squared.

Or my radius of my Mohr's circle = 4.788.

And so that means tau max is going to be up here.

And it's going to be equal to the radius which is 4.79,

rounding to three significant digits,

Newtons per millimeter squared, or 44.79 megapascals.

And so that's one of our answers.

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And then finally sigma 1 will be our principal stress out here.

And it's equal to the value of sigma at the center,

which is 4.33 plus the radius, which is 4.79 or

a tensile stress 9.12 megapascals

tension for our first principal stress.

For our second principal stress it'll be the center minus the radius.

And so I've got sigma sub 2 = 4.33- 4.79,

or -0.46, which means it's

0.46 megapascals in compression.

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We combined all those together, we found the state of stress at point A.

And then we used Moh'rs circle to find the principal stresses and

the maximum shear stress at point A.

And so that's about as comprehensive as you can get

as far as mechanics as materials are concerned.

And so long module, lots of stuff, probably have to stop and

start a few times to get your head wrapped around it.

But if you can understand this,

you've got a really good handle on mechanics in materials.

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