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Â Welcome to Module 21 of Mechanics and Materials Part IV.

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

Â We've done deflections, we've now

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

Â This is a culmination of all my mechanics and

Â materials courses, the parts one through four.

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

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

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

Â And so, our learning outcome is to combine,

Â to solve a combined static loading problem.

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

Â I've got an engineering structure.

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

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

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

Â elastic region for all the loading conditions.

Â There the loading condition that's given.

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

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

Â and then add them together by superposition.

Â And so, how might we start?

Â 1:24

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

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

Â moments are at that cut that goes through point A.

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

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

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

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

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

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

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

Â M sub RY, and M sub RZ.

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

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

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

Â sum forces factorially, and set them equal to zero.

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

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

Â Plus R sub y in the j direction,

Â plus R sub Z in the K direction

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

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

Â and then we have plus 4,000 Newtons.

Â In the, well I put Newtons in there.

Â Let's take that out.

Â [LAUGH] Newtons.

Â 4:09

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

Â the center of our engineering member.

Â But we do have M sub Rx in the i direction

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

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

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

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

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

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

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

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

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

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

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

Â And all that has to equal 0.

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

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

Â plus moment reaction about the y axis plus

Â 1,600,000 all in the j direction.

Â Plus moment reaction about z,

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

Â I match components, I get M sub

Â Rx equals -800,000

Â Newton millimeters.

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

Â And finally M sub Rz equals

Â 400,000 Newton millimeters.

Â And so, here are the total results.

Â Those are my force reactions.

Â And those are my moment reactions.

Â Okay now with those force reactions and moment reactions,

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

Â And so let's start with torsion.

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

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

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

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

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

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

Â I'm going to have my moment of reaction about

Â the z axis is shown there in that direction.

Â Here is my point A, right at the top.

Â And we have our axis.

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

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

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

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

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

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

Â 8:20

the torque times row which is the area,

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

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

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

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

Â about Z axis is 400,000

Â newton millimeters times row,

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

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

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

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

Â And so we find that the sheer stress, then,

Â is equal to 2.037 newtons per millimeter squared,

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

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

Â 9:33

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

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

Â With the stress block and

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

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

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

Â this direction here, this direction here and

Â stress due to torsion is pure sheer in this case and

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

Â 11:11

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

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

Â equal to 800,000 in this direction.

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

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

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

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

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

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

Â a flexial stress due to that torque or moment.

Â So we've got sigma due to m

Â sub ry is equal to zero.

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

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

Â 12:25

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

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

Â Whoops.

Â Sigma about the z axis.

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

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

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

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

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

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

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

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

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

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

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

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

Â times 50 millimeters to the fourth.

Â 13:41

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

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

Â And so that value is 8.149 NPA or

Â newtons megapascals or newtons per millimeter squared.

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

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

Â So this is the Z direction.

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

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

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

Â this is going to be in tension.

Â And so, this is tension.

Â And I can draw a stress block again.

Â 15:22

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

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

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

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

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

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

Â the cross section.

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

Â stress contribution to this overall combined loading is quite small.

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

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

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

Â 16:20

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

Â We have our coordinates Y up and

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

Â this 2,000 Newton force.

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

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

Â it's in the negative X direction.

Â I can now inappropriately but

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

Â 17:25

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

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

Â If you look at a reference

Â out to the centroid,

Â this distance will be 4/3 r/pi,

Â and the area = pi r squared over two.

Â And so, if I put those values in,

Â I'm going to have 4r over 3pi,

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

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

Â And then we're going to divide by I.

Â 18:44

And if you put in those values,

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

Â It's 0.00679 Newtons per

Â millimeter squared again, all this was,

Â let me write down where r equals 50

Â millimeters which was the radius.

Â Finally let's draw our stress block at A.

Â My X direction will be In this direction,

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

Â And the sheer stress caused by this transversed

Â 21:35

Equal to 0.510 Megapascals.

Â All right.

Â So, we took care of torsion shear,

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

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

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

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

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

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

Â Z is in this direction.

Â 23:38

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

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

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

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

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

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

Â the shear stress.

Â So we're going to go over here,

Â 8.659, down, so

Â -8.659 and -2.044.

Â We have two points.

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

Â 24:50

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

Â Let's start by finding the center.

Â That's going to be 8.659 divided by 2,

Â or 4.33 and zero,

Â which means my radius now is equal

Â to the square root of 4.33

Â squared + 2.044 squared.

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

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

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

Â rounding to three significant digits,

Â Newtons per millimeter squared, or 44.79 megapascals.

Â And so that's one of our answers.

Â 26:01

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

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

Â which is 4.33 plus the radius, which is 4.79 or

Â a tensile stress 9.12 megapascals

Â tension for our first principal stress.

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

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

Â or -0.46, which means it's

Â 0.46 megapascals in compression.

Â 27:23

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

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

Â the maximum shear stress at point A.

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

Â as far as mechanics as materials are concerned.

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

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

Â But if you can understand this,

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

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