Okay, so outside of those announcements the rest of the lecture is going to be dedicated to lecture content. Of course last time we were here we were working on an example where we were computing different tractions in the plane. Oh, hold on, just a moment here.

You know, we were using this example here where I had a very simplistic uniform stress in the x direction, so sigma xx was equal to one all other stresses in this.Coordinate system were zero and I was asking you to calculate n and s that is the normal component of the traction vector in the shear component of the traction vector and we went through and did this very simplistic example, this is the notes that we had from west last time and we computed the normal component we computed the shear component we did it both for the very simplistic case where the the new coordinate system was the original coordinate system, of course, we found the same result as we had previously here in we had.

The example where we're using now this new coordinate system oriented 45 degrees to the other coordinate system and we were able to calculate with that system also the normal and shear components. I think we got through the entirety of the example, although after lecture and now having had a couple of days past. 

I guess. I'm not a hundred percent confident that we completed at all. So if,If we didn't complete it all, please let me know and we can pick up and and do anything that we may have left off. Now based on the feedback I got from you guys during class time. 

I've updated the the representations of multiplication so as to not confuse the dot product with multiplication and also I added some vector symbols over the IJK unit vectors just to make sure that those were clear and so I think that or at least I hope that that won't be confusing where I could understand.

Where it might have been confusing previously.

Okay, so did we get this far can I get a thumbs up from you guys? Are we did we complete this exercise or did we not?

I'm not sure. I heard that very well. Tyler, could you repeat it?

Okay, so starting at this slide or was it one slider two back?

Okay, so what I'm hearing from you is that and I added this sentence here to the slides that there were previously there. So now this is where we ended this result of having the total force being equal to one is consistent with not only our expectation from the setup but also our intuition for the problem. 

So that's good. So, okay, so picking up then where we left off if we're going to compute the normal component of the traction vector.And of course we also then maybe interested in computing the component of stress the normal stress associated with the normal direction, which I'm going to call sigma n n. 

That calculation is going to be relatively straightforward based on what we have already computed. So we had already computed the traction vector.The traction vector, of course in this context is TN.

And then the so that was the square root of one over the square root of two zero zero and now if we dot that with the normal vector, we're going to end up with of this very simplistic calculation. So one over root two times one over root, two plus zero times, whatever plus zero times whatever it's going to end up being one half. 

So one half Pascal is the normal component.Of stress sigma n n oriented on this new normal face.

Now, of course the normal component of the traction vector is that magnitude keeping in mind that sigma n n is a magnitude.So the normal component of the traction vector is sigma and n times the normal direction.So 0.5 times n and it gives us now a vector because all traction vectors are vectors so the normal component of the traction vector is also a vector. 

So in this case n is going to be equal to 1 over 2 root 2 times 1 1 0 the vector 1 1 0. So 1 over 2 root 2 is a magnitude and that that normalizes essentially and then multiplies by 1/2 to get that the magnitude of the vector to be correct.

Now, of course, we also need to or was we were requested to compute also the shear component of the traction vector.And we could similarly of course calculate the shear stress sigma and s on this normal face.And this calculation is relatively easy also.Keeping in mind that the traction vector is the sum of the components.

If we know one of the components and the traction vector we can compute the other component so here in this in this case we can easily compute s.The shear component as just being a t minus n, so the original traction vector minus the normal component.

So I'm getting a question from Tyler thanks for the question. Tyler I see it it says how do you go from TN being a column in the last lecture to a row in this lecture, so let me just jump back there to answer his question.

So when you're doing a dot product.You what you need to think about a dot product is you can expand the dot product into the the idea of the the traction vector being one half or I should say one over root two times the i direction plus zero times the j direction plus zero times the k direction.

And then you can look at the normal component to the vector as being another vector so what I'm suggesting is you could have written these vectors out in their sort of summation sense and then the dot product and that's functionally equal to in this case what amounts to t transpose times n so if we're representing vectors three by one vectors as three by one in order to get the dot product we have to do the transposition to make that matrix mathematics work out.

So I hope that clarifies what I had done there, it's I it it's just a.Simple very simple way of doing that problem.

So anyway, here we go again attraction minus the normal.So there's the t which is one over to i plus zero j plus zero k minus the n vector which I've also written one over two root two i plus one over two root two j plus zero k and of course that makes s quite simple also one over two root two i minus one over two root two j plus zero k.

Now if you think about that and you looked back at the.Components that we drew on the image.You would have seen that you had a component in the positive x direction for this year tensor and a component in the negative y direction for the shear tensor and so that is consistent again with expectation as we see here in this computed result.

Okay, so we're gonna do a slightly more complicated example, so previously our goal was to do an intuitive example that would be very easily understood and you know just work quite well with your intuition now. I'm gonna present you with one that's perhaps a little bit more realistic to what you might need in the real world that is we have a stress tensor that has multiple components within it that are non-zero.

And of course we're looking at an arbitrary plane rather than a plane associated with.A rotation about one of the principal axes so but you're gonna find that this is just as easy as the last one it's not gonna be any more difficult so we'll go through it fairly quickly the stress tensor in this case. 

I'm going to give as 50 20 minus 10 20 minus 30 zero minus 10 zero 40.Of course along the diagonal or the normal components the off diagonal components are shear components and you remember of course the subscripts x x x y x z y x. y y z and so on.

The plane that I would like you to compute the normal and sheer components for is shown here in this image what I'm doing is I'm taking a plane that has in the x direction a magnitude of 3 offset a y direction magnitude 4 and z direction magnitude 5 from the origin so this plane is you know pointing off in the in the positive x y and z directions.

Each of the angles is a little bit different relative to the principal axes or I should say the Cartesian axes.

All right, so now you've seen how to do it, how do we do it?You of course remember from last time that the traction vector is simply using a calculated using this couchy stress formula that is the stress tensor dotted with the normal vector so the only thing difficult in in this problem is the you know, essentially finding the normal vector that's really the challenge here in this problem.

So how do you go about doing that what's is anybody have a suggestion?Feel free to type in the chat box. I'm gonna give you just a minute to do so.

So Ethan gives us a good.Good response. Ethan says take the cross product of A and B, you kind of I think saw the hint a Tyler says the same good job guys. I just drawing them drawing those two vectors on there, maybe gave you a hint.So if we take A and B.

Riga, did you have a question your your your mic is on but I'm not hearing you.

Okay, so A and B, of course were identified if we look back at that image that I showed on the prior slide the vector A.Would be equal to zero in the eye direction minus four in the j direction plus five in the k direction.And the vector vector b would be three i minus four j plus zero, k.

And of course to compute the normal to that you may recall is the cross product but it's not just the cross product we actually then would like to normalize that unit I should say that normal vector to be of unit length because we're only trying to find the plane so it ends up being a cross b over the magnitude of a cross b.

And so this is a relatively straightforward and simple calculation a cross b is equal to minus four times three times j cross i.Plus minus four times five times k cross j.Plus five times three times k cross i.And you guys can probably do that in your sleep you're probably using your right hand rule to do so and trying to remember the ultimately whether it's a plus sign or a negative sign when you do the k cross i or i cross k but you know, I'm sure you can do it no problem and in the end what we get here is a vector a cross b is equal to 20 i plus 12 k plus 15 j.

Of course the magnitude of that is just a sum of the squares square rooted which works out to be the square root of 769 or about 27.73.And hence we can take the.

The overall magnitude is going to be that there it is now if you're doing this on your homework, you can either leave it in this form or you could write out the the decimal version of it and I would accept either one as long as they're both correct.

Okay so from that of course, then we're looking back at that couchy formula so what you're seeing here is the couchy formula the traction vector is equal to the stress tensor.Dotted with the normal vector now here. I've you know the stress tensors written in the square brackets the normal vector. 

I left in its original form where it's now written as a vector 2015 12 times 1 over the square root of 769, so the the 1 over square root on the right hand side just normalizes the length of that normal vector.

And so we can compute very directly the traction components TX TY and TZ is 42.55 minus 1.8 and 10.1 all in mega. Pascal units.

And again you're you know in your homework you're allowed to write this any any way that's consistent and clear, you know, you could write it as a vector you can write it out in this summative form.You know directly showing the components it needs to be clear, of course that it is a vector and you do need to include units on it when it's appropriate but I'll accept really any form as long as it's appropriate.

Now like we did with our simplistic example, we are going to do the same thing with this example because of course I asked you to compute not the traction vector itself, but rather the components of the traction vector.So the magnitude of the component to the traction vector, that is sigma and n and of course the n direction both of those need to be known we know n already.

What we can do to calculate sigma n n is take that traction vector t n and dot it with n.So this is the component of the traction vector associated with the normal direction again. I did the transposition put it as a 1 by 3 matrix that's t n then multiplied it by n which I left in this form that has the 2015 12 and then the,Unitizing constant on the right hand side.

And the end result is a computation that says we have 34.08 megapascals.As our magnitude sigma and n.

So Nicole is asking a question. Nicole asks is TN just the traction vector.Or does the n mean it has something to do with the normal. So the the scheme that I'm using here the notational scheme that I'm using here is that this is the traction vector associated with the normal n.

So t sub n is the traction vector associated with n. That distinguishes it from other traction vectors that may be present on different normal planes, which of course can can and do exist as we saw last time. And in fact, we're going to spend the next half of this lecture once we finish these examples sort of opening our mind to finding traction vectors on arbitrary planes in a way that is extremizing. 

We're going to try and minimize or maximize stress and that becomes really an important calculation extremely important calculation for.Us. But we'll get there in a few minutes.So sigma and and now is known that is the magnitude of that component of stress.And so from that of course we can calculate the normal component of the traction vector at just multiplies that times the direction so n is equal to 34.08 and.

So Iran's asking is it 42.55 minus? 1.8 plus 10.10, yes, the answer is that's plus 10.10 the plus is you know, this is now a matrix. It's a 3 by I'm sorry 1 by 3 row, I dropped the pluses in front of both the 42 and the 10 but yeah, that is supposed to be 10.

Okay, now the shear component again, how do we do that? Well, it's not complicated. It's just essentially the same thing that we did with our more intuitive example.

So all we need to do is subtract the traction vector or I should say the normal vector from the traction vector, so t n minus n is going to be equal to s we can express TN as in this form 42.55. I minus 1.80 j plus 10 point 1 0 k minus the normal vector, which is now written in somewhat long form here.

And from that we can compute s itself.17.97 i minus 20.23 j minus 4.65 k and we may also be interested in knowing its magnitude so that we can compare the magnitude to the magnitude of the normal component and here we see that the magnitude is 27.46 megapascals.

So of course now we know the magnitude and the direction of both the shear and normal components of the traction vector.These are good outcomes.Going to be useful to us going forward.

Okay, so we've done it you did it on an exam you've got your answer at least you think you do what are you going to do next?

So as an engineer yeah as an engineer you want to do a sanity check right you want to see if your answer makes sense it's often difficult to know if it's correct but it's often easy to know it's incorrect and so you want to check at least to know if it's incorrect so does anybody have any ideas how you might do that for this problem?

It's good idea all right, so we could do that in fact. I think that's not the first thing I did what would you get? I?

Correct yeah so so that's not the first one here at least in the order of my slides. I would suggest instead to check for orthogonality, so the dot product being equal to zero essentially does confirm that we have orthogonality.So there you go, that's that.The other thing of course that you could do is what?

Actually I thought I did the other one first apparently. I did the dot product first.

Well I won't won't leave you hanging the other thing you could have done as taken the cross product of the two vectors and then you should have gotten the third vector that would make sense also another thing that you can do is you can verify the magnitude being correct.

And we can go ahead and do that.So all of those things, you know, I just want to encourage you guys as you're going through this course and of course then subsequently in your professional lives when you are providing people answers to important questions to think about how you can confirm that your answer is at least not blatantly wrong and then of course.

It's really good idea often to address the same problem from a deaf couple of different perspectives to make sure that the answer that you're giving is right or at least feasible.

Okay, so believe it or not that was the end of lecture for so we're about 30 minutes behind what I had expected us to be at this point but that's okay, we're gonna just keep on plugging away. I'm gonna move on down now to lecture five.

Well, so you can.You want to make sure that if you have these unit vectors in multiple directions that the cross product of the unit vector is cross product of any two vectors should give you the third that's really the suggestion that I meant to be making there.

I guess really about perhaps to your point in this context that's less obvious and less critical but since you know, the two directions of n and s what you could do with the cross product here is you could compute that third component and make sure it was also zero using a cross product.

Okay so now moving on to lecture five this is part two of stress and the internal equilibrium. I actually don't think we're gonna get through the entirety of this lecture so we'll probably have one more lecture that includes stress and internal equilibrium at least but we'll go ahead and get started with this one as we are now.

So what we've now done is we've learned how to compute traction vectors traction components, now it's worth noting these are just some comments really about what attraction is.Note that you've heard that word many times. I'm sure so I've been watching the Tour de France I'm enjoy watching that and one of the things that as a cyclist, they're constantly concerned about if you've ever watched that race is traction.

You know, when it's wet and raining they tend to lose their wheels and and go down in big crashes there's another great video if you're a cyclist it's there's a race called the mountain of hack although they use a different word and there's hundreds of bike racers going down a mountain of ice and the the 2018 version of that one of the cyclists lost his traction and and ended up starting chain of incidents that led to oh at least a hundred people crashing at the bottom of this major hill so.

Anyway the point there is that you've heard this word before you probably never really thought much about it, but that word traction is derived from this concept of traction the traction vector so this is usually referencing some sort of external surface boundary condition where a stress is applied okay, so for a bike you're gonna have a normal component of the traction that holds the bike up out of the ground keeping it from penetrating the ground and then you're gonna have a sheer component that keeps the wheel moving straight as opposed to sliding out under.

Neath.I will mention that this is an extremely important aspect of the concepts of traction but it's not the only application and we're gonna get to the next really important application here in the next few slides.Anyway, so this idea of attraction what is it what physically does it represent well it ultimately represents the intensity of stress at a given point.

Over a an area of given orientation.Okay now again usually referencing external surfaces but not exclusively.

I'll mention that if I give you a homework problem or an exam problem as boundary conditions. I'm typically going to give you displacement boundary conditions or tractions or maybe point forces and those are what are going to be specified that's external boundary conditions as a sort of practical note, it would be rare that I would give you both tractions and displacements.

And that's because if I'm doing so I may actually be over constraining the problem.Or alternatively. I mean, certainly in all cases when you have a solution you have the ability to compute both displacements and tractions on those surfaces, but it would be rare that both would be given as part of the problem statement.

Now I've hinted a couple of times that this idea that external surfaces are not the only thing that tractions are useful for and that will find the ability to calculate shear and normal components of tractions to be extremely important well, here's why.We have this concept called principal stress.

Suppose we're interested in finding the plane. I either normal to a given plane.That extremeizes the normal stress component, so it makes sigma n and either as large as it could be or as small as it could be.Maximizes or minimizes.Now before we get into how we do that, it's worth asking why we would do that.

Why do we care? Why do we search the different planes that are possible? Of course, there's an infinite number of planes. Why do we search amongst the infinite number of planes for the plane that maximizes or minimizes the normal component of stress?

Go ahead Joel.Experience that.

Great thanks Scott very good answer. The only thing that I would minorly tweak about that answer is that you use force and I would say instead stress but yeah, absolutely you're a hundred percent right on the nose. So,An example of that would be that you know, if you have cracks that form they tend to propagate tangentially to the maximum normal stress component, so when you're pulling on something it'll either fracture and and have a fast fracture that'll open and and catastrophically fail or in in the case that it fractures slowly it'll at least open that crack perpendicular to the maximum normal stress component typically unless there's some sort of material grain or something that it would then instead follow.

So very good answer, thank you for that one. So, of course, then the next question is, how do we do it? How do we find it?And the answer to that question is you need to think back to vector. I'm gonna say step back from tensors to vectors.And think about how you can find a component of vector that maximizes the length of the vector in that coordinate system.

I don't know if you're able to see me. I'm I do have of course my video going I'm holding up the pen that I'm using for the class. I'm gonna use that as my vector. I have my right hand rule coordinate system. I'm only gonna use two fingers to represent because this is essentially a two-dimensional problem for now but at this concept generalizes to three dimensions. 

So in the XY space, I've got this vector that's aligned with the direction.It has components in both the X and the Y direction.My goal would then be to find an orientation by rotating my coordinate system.That extremizes the length of that vector in the coordinate system. And I think you guys would agree again thinking of it as a sum of vector components that I have really two choices in this context. 

I either have this orientation.Where what I had originally drawn as the y axis is now aligned with that vector or alternatively. I have this orientation where what I had originally aligned as the x-axis is oriented with those coordinate systems. Now either of those two extremize.Though the component of the vector in that orientation.

And what's notable is the other component in this orientation is zero.Is zero so to find the maximum normal component to in fact all we really need to do is find the the component where the shear stress is zero now conceptually they met that may not be easier for you to take a guess at how to do it but in practice it's actually much easier mathematically, so we're gonna go ahead and do it.

So here's our traction vector. I'm writing it as a we're trying to find the orientation, so I'm finding a new coordinate system where the shear component s is equal to zero.Of course we remember that the traction vector itself. TN is equal to the sum.Of n plus s.

And because s now is zero, we know that the traction vector itself is simply going to be equal to n now. I see a question coming in.Jeremy asks to be clear the length of the vector is not changing just the way do we describe it to have a maximum axis value so yes that's correct, of course a vector what what I was doing with my fingers here was a is a vector transformation and conceptually we're going to generalize that to tensors higher ranked vectors, so it's a in in the context of stress, it's a second rank tensor.

But either way the vector stays the same just the components of that vector the the coefficients associated with the direction vectors change as you change the the orientation so we're finding the orientation where those component coefficients are the largest with the same vector, so thanks for the question, it's a good one.

Okay so getting back to where we were.Here we have the traction vector is then equal to n the normal vector.Which of course we also know is going to be equal to the magnitude of n so the two vertical bars they're represent magnitude of n now multiplied by the normal direction.

So t n is now equal to a magnitude times a normal direction.

And last but not least on this slide. TN is there for going to be equal to something we're going to call sigma p times n.Now sigma p is just the magnitude.Of that normal component of stress.And of course it is now oriented in the end direction.

So this this magnitude sigma P, we're going to call it.The principal stress.

Now the orientation of that principal stress, which will now call n.We're going to say is a principal orientation or also called principal axis.

So it's important to recognize we have not only a principal stress magnitude, but also a direction both are equally important.To be able to identify.

Okay, so that's conceptually how we're going to do it.Now, I want to actually accomplish that goal.So, here's how we accomplish that goal.Of course the traction vector itself just using that calcy equation, again, you'll see it comes up quite frequently. It is the stress tensor dotted with normal vector. 

Now, the only thing that's funky here is we don't know the normal vector yet, okay, but nevertheless we know that the stress tensor dotted with this unknown vector is going to be equal to the traction component TN which we also know is going to be equal to the magnitude of the traction times its direction. 

So right hand side sigma p times n.Okay now we can do a little bit of mathematical manipulation, we move the sigma pn to the left hand side, so sigma dot n minus sigma p times n.Is equal to zero zero on the right hand side of vector.And looking one step further we can see that that sigma p needs to be multiplied by an identity matrix in order to collect the terms, so sigma the three by three matrix the tensor minus sigma p times i the identity matrix three by three dot n is equal to zero.

Okay now so I've set the problem up.I wonder if anybody recognizes this problem is this a problem that you've ever seen before.

Ethan says it looks like we're finding an eigenvalue great thanks for not leaving me hanging for too long there, you're absolutely right.

So what we're finding is an eigenvalue. Now, this is actually by more than coincidence, this is the very definition of the eigenvalue problem with sigma P. There's a missing symbol there with sigma P representing the eigenvalues of sigma.

This is how we set them up when you were studying how to solve eigenvalue problems. Sigma X X now is the component minus sigma P and then of course the other terms the normal terms are all subtracting sigma p the shear terms are left alone. We now have this unknown normal vector LM and N, which is going to multiply that and then the end result is zero zero zero on the right hand side.

So if you went through and you did this manipulation.You'll recognize that nontrivial solutions and this goes back to your days as a freshman or sophomore non-trivial solutions to this equation can be determined by setting the determinant of that matrix, sigma minus sigma PI equal to zero.Now when I say nontrivial solutions, I want to make sure that that's clear. 

Can anybody propose a trivial solution?

So, of course the trivial solution here is going to be zero zero zero that is l m and n would be zero. So that would be no unit vector. Of course that's trivial it's meaningless it doesn't help us in any way, so we're trying to find the non-trivial solutions. 

So again to find those you need to have the determinant of this matrix and set it equal to zero.Now of course computing the determinant is something you may have forgotten how to do it's not terribly difficult and in your homework problems, you're gonna go through and do it so you'll quickly remember if you had forgotten but basically you take the top left corner and multiply it by the minor determinant of the bottom right to columns two rows and columns and then you subtract from that the middle top times it's minor determinants and then you add to that the right upper rights times the minor determinant of the lower left two by two make.

Sure x anyway long story short in the end what you get is a big polynomial equation. It's a third order polynomial equation in sigma p.And then you put it in this standard form where the sigma p itself has a leading coefficient of one even if it numerically previously did not you want to then divide by whatever was the leading coefficient so that sigma p.

Is of leading coefficient 1.Minus now another coefficient will call i 1 times sigma p squared plus another coefficient i2, sigma p minus a last coefficient i3.And that is going to be set equal to 0 so that's the determinant going to be equal to 0 and ultimately the roots of this characteristic equation third order characteristic equation are called principal stresses.

And by convention we order them according to their magnitude. I'm sorry not magnitude actually, they're absolute value so that sigma p is the largest and absolute value and sigma 3 is the smallest and absolute value.So one is greater than minus 10, so if you had one zero minus ten sigma p sigma one would be one zero would be sigma two minus ten would be sigma three.

Now I should mention what these i values are.

The i values we put the i in front of them because they're called in variance of the stress tensor.Okay the characteristic equation doesn't change as the stress tensor is transformed.Now again, I want to make sure everybody's clear on what I'm saying, if you're looking at my face in the video again, what is a transformation that's as we reorient our coordinate system.

The vector stays the same.Okay, it doesn't matter what the what the coordinate system is.The magnitude and direction of that vector needs to be maintained earlier in the semester. I said for higher rank tensors, we need to do not only magnitude and direction but additional things so those additional things here for the second ranked tensor are the so-called invariants these invariants.

Don't change as we observe the stress tensor from different coordinate systems.

Now these are very common things that need to be computed frequently and so you've probably heard of a couple of them, so the trace of the tensor that is the sum of the diagonal xxyzz.Is something that we call instructional mechanics the hydrostatic stress.Really all it is is the trace of the tensor so if you had a strain it would be the trace of the strain if you had you know, there are other disciplines where we would not call it hydrostatic stress but here in this context the trace of the tensor is the invariant we give it the name hydrostatic stress.

Because in in mechanics, that's something that's actually an important concept hydrostatic stress.Okay, the next one actually doesn't really have a fancy name.It's not something you're gonna see expressed too frequently.But ultimately we're gonna call it I2 and it relates the strength of the deviatoric stress relates to the strength of deviatoric stress meaning when if you take the advanced structural mechanics class AE5100 will get into this into a lot more detail but this is a driver of plasticity in a material.

Okay, especially in a ductile material.

And then last but not least is I3 this one's extremely common you've heard of it before the determinant of the tensor and that's of course the determinant of the original stress tensor not the modified stress tensor.

Okay, so now we've done it we've calculated principal stresses. It's worth noting that because stress always is a three by three matrix, so it's a second ranked tensor with nine components, it will always have.Three principal stresses. So we should always compute three sometimes that means that one of them will be zero if we have a planar stress field within no out of plane components, but nevertheless we should recognize that there are three principal stresses.

Okay, also because each of these three principal stresses have different magnitudes conceivably they could each have different principal stress directions and in fact in most cases they do.It's actually rather unusual that you would have of that not be the case. We'll cover those corner cases in a few minutes.

But for the time being we just recognize that because we have three principal stresses we have three principal directions and so I now I need more normal vectors so I'm going to use n sub. I now the subscript I is representing whether it's principal stress one principle stress two or principal stress three.

And they're now determined from this set of equation. So sigma x x minus sigma i and the remainder of the modified stress tensor ultimately dotted with l i m. i n i is going to be equal to zero zero zero.

I just want to confirm that you guys are still with me for whatever reason the the students whose faces i can see right now are all locked up, nobody's moving assume that maybe there's a network problem, can you confirm that you can hear and see me?

Grace great, thank you.

Okay, so a couple of brief notes about principal stresses and orientations so the principal vectors principal orientation vectors are normalized by convention to have unit length the unit vectors have unit length.And it turns out that they are what we have called in the past eigenvectors of the stress tensor.

So we have eigenvalues which are principal stresses we have eigenvectors which are principal orientations.I always think it's a little odd that we teach you this concept what two three years ago, and then we let you forget it before we actually use it it's a that concept that we taught you so long that was actually really critical for calculating material behavior and ultimately failure, these are failure metrics is what we're seeing a material failure metrics, and so I'm glad that we're finally using that skill that we taught you so long ago.

So what we computed were principal stress is now by definition as you saw those are the components the the normal components of the traction vector.In orientations where the shear stress was zero, so what we found was the maximum value of the normal stress the minimum value of the shear stress now let's go back to this two-dimensional vector, here's my pen.

Okay again. I have my coordinate system we're doing it 2d not 3d.Okay, this is what we said was an orientation that maximized the normal component of the vector this was another one now it's notable that we also want to be able to find an orientation that maximizes the shear component and intuitively what are we saying so this is the orientation where that second magnitude is the largest and as I rotate around in this two-dimensional world to a 45-degree plane, this one has a you know, it has a,Equivalent normal and shear components in this two-dimensional vector and so that's the orientation that maximizes the shear component in this two-dimensional world now it's not quite as clean as that in the three-dimensional world what we find is that the maximum stress the maximum shear stress. 

I should say the maximum magnitude of the shear stress.In magnitude is going to be one-half sigma one minus sigma three and it's going to be on this these planes the the the local maximum are on the octahedral planes turns out there's eight of them that are a functionally equivalent, you know, they're 45 degrees offset.

From the principal directions, they bisect the principal planes, that's what that word means 45 degrees offset and the the maximum maximum of all those shears is going to be sigma 1 minus sigma 3 divided by 2, that's a magnitude now.Now this of course is also going to be a important mechanical concept the maximum shear stress because this very much relates to yield inductal materials, so the maximum normal stress.

Is really critical for brittle materials because that has to do with crack propagation and it turns out the maximum normal stress also is critical to materials that yield that ie ductile materials except it's a little bit backhanded meaning that you first need to calculate the shears, but we'll get into that or if we'll spend a little bit more time on that later.

Okay, so here we go. I want to do a quick example so that you guys can confidently move forward knowing you know, how to do this.So the example here given a stress state, sigma.One one zero, one, one one zero zero zero one find the principal stresses.Principal directions and maximum shear stress.

Okay, so what do we do?It's just a few slides ago.Bring it back up well that if the solution of course is we find the invariants.And then ultimately the principal stresses.

So we set the problem up we actually write.One minus sigma p one zero.One one minus sigma p zero, zero zero, one minus sigma p we take the determinant of that matrix.We take the characteristic equation that results we make sure to put it in standard form, so the leading coefficient on sigma p cubed is one.

And then we look at what the other coefficients are and here are what they are, so i one is going to be equal to three.I two will be equal to two.

And i three will be equal to the determinant of the original matrix, which works out to be equal to zero.

Now if you choose to you you can do what I've done here on this particular slide and that's just use the equations for i one i two and i three that I presented you earlier in terms of the components of the original stress tensor, sigma x x y y z z x. 

y y z x z and.You just plug in the numbers and chug that's that's what I'm doing here, but really what you conceptually need to recognize is how how to do the problem and that's ultimately to take that determine it and and reconfigure the characteristic equation.

So here we have I12 and 3, here's the equation now where we've written the characteristic equation and standard form, so sigma pq minus 3 times sigma p squared plus 2 times sigma p plus 0 is equal to 0 and we have to solve this equation. So, how do you solve that equation?

Well, you know, you've probably would have to go back to high school.Algebra perhaps to solve it or maybe freshman year of college depending on what point you managed to pass through this mathematical system. I think there are a number of good ways to do it. You can plot the roots, you can plot this equation with respect to sigma P and then find where the roots cross 0 or you can find where the value of this equation crosses zero to find the roots. 

That's one way to do it. The way I like to do it when it's simple is to just factor the equation. So here by inspect.We could factor the equation and determine that sigma p is equal to two one and zero. I've ordered them conventionally to being the largest in absolute value, so that's sigma one one and zero. 

So sigma three is equal to zero.

Now, of course we found the principal stresses, but we have not yet found the principal directions.And we know that these principal directions are important.So we'll go ahead and find the principal directions.Here what we have is and basically all you're doing is you're plugging sigma one into the original equation, so one minus two one zero one one minus two zero zero zero one minus two which of course simplifies as you can see on the screen minus one one zero, one minus one, zero zero zero minus one. 

That new matrix really a tensor.Dotted with l one m one n one the unit vector has to be equal to zero.

So, this is a rather trivial exercise.So just looking at equation three, let me go back to equation three for a moment.Equation 3, I.

That's this equation.

And if we examine that equation alone what we see immediately is that n1 is.

Zero.

Of course we can then look at the second equation.

Oops sorry about that wrong direction. That one is going to be equal to one times. L1 minus one times N1 is equal to zero.And from that we can by inspection say that L1 is equal to M1.

Okay, so we still don't know L and M, but we do know n what's the last step?

So pause for dramatic effect here, usually a student volunteers, well use the first equation.Right? So you got three equations for three unknowns.Oh David says parametrize. What does that mean? David? Be more specific.

S.Okay, and and the unknown constants are what here?So good okay, so what we really need so we need three equations and three unknowns, right? We have three unknowns. We need three equations. We've used two from the from the matrix representation.So you might intuitively think well let's use the third one, but in this case that wouldn't be correct because we know in fact that these are redundant equations that's the the whole concept of the eigenvalue problem if we had three equations in three unknowns where all three equations were unique we'd able we'd be able to solve for those three unknowns without doing the eigenvalue problem, but the definition of the eigenvalue problem is in fact that one of those equations at least one of those equation.

Is redundant. I'll go back for a moment here.So if you stare at equation one an equation, two minus one one zero, one minus one zero, keeping in mind that if we're just solving equations you.Can multiply all of the first equation by some constant.And you haven't really changed the equation in terms of solving for the 3 unknowns. 

And so we could multiply the first equation by minus 1 and identically turn it into the second equation.And so it's redundant. It hasn't helped us. So we actually need to come up with a third unique equation and you know, there's no trick to it here with these eigenvalue problems where we're calculating unit vectors. 

The last equation for our purposes is always going to be this one. That is that the magnet magnitude of the unit vector is 1.

So we already determined that L and M.Were the same.

So, therefore we have L1 squared plus L1 squared plus 0 the third one, we already determined is equal to 1 and so from that we can determine that L1 itself.Is equal to 1 over root 2.And then we can write the whole unit vector principle direction.N1 as being equal to 1 over root 2 times 1, 1 0.

Okay so we've done it once we have one principle stress. Actually, we have all three principal stresses, but we only have one unit vector. I'll move through these other ones fairly quickly. You know, it's it's not this is not really rocket science at this point. It's pretty obvious. So the second principle direction is the one associated with sigma two I eat a second principal stress. 

So we just plug and chug one minus one one zero, one, one minus one zero zero zero one minus one and we get this matrix zero one zero one zero zero zero zero zero.Zero times l two m two n two is going to be equal to zero.We kind of do the same thing that we did last time. 

It's often easy well, you know, I set this problem up specifically to be easy, but we can look at these equations and make some bold statements right away.

So the bold statement that I'll make is that inspection of both equation one and equation two immediately gives us the magnitudes of l two and m two.

Okay, again, you can see this is this equation here.

And this equation here.We know that l two and m two are both going to be identically equal to zero then the last thing is how do we find n two?Well, of course if we looked at end to zero times l two plus zero times m, two plus zero times n two would give us zero automatically but that hasn't told us anything about what n two is again, we have a redundant equation here we're at least a useless equation because it's all zero.

And therefore we need to then use that technique of taking the magnitude of the unit vector and make sure it's equal to one so l and the m are both zero that means n two squared is equal to one and so therefore we have n two being equal to zero zero one.

Okay so at risk of getting boring and and letting you guys fall asleep. I will continue and do the third one for completeness, but we're going to move through it very quickly so the third principal stress was zero.Okay nevertheless we still need to do the calculation, so here's the calculation.

And this looks in fact quite similar to the first calculation, it's a little bit different.We had some minus signs in the first principal direction calculation that we don't have here in this in this calculation.But we can use the same process so inspection of equation three showed us that n three was equal to zero.

And then of course we can look at equation one, so let me go back and just highlight these okay, so this equation was oh shoot, I guess I've lost my ability to edit the go back and inspect equation one.And we can see in this equation we can see one times l three plus one times m, three plus zero is equal to zero and from that really we're we're seeing now l three and m three are opposite of each other in a minus sign sense, so l three is minus m, three and then of course we can place that into this same magnitude of the unit vector as equal to one calculation.

And of course the minus sign goes away when we square.So two l three squared is equal to one l three is still equal to one half.An n three now is equal to square root of one half times one minus one zero.If you recall from the first principle calculation, it was one one zero here, it's one minus one zero.

Okay, so we've essentially completed the problem.You know, again, you've gotten to the end of your exam you're fairly confident in your answers there anything you can do.To ensure to check to sanity check your result, of course, yes, there is by the way, when you get to the end of your answer, you should summarize it.

Okay, you don't want to leave unit vectors hanging out all over the place so bring them all together a good way to represent them is actually as a set of row. I'm sorry a set of column vectors so the column vectors here square of one half times one one zero zero zero square root of two and one minus one zero, so they're all succinctly in one place.

And then as I mentioned it's a good idea to sanity check, so what are we gonna do again similar to what we had done in the prior set of problems some some things stand out, so these are unit vectors they are defining a coordinate system.

So again we can think about dot products and cross products dot products have two unit vectors assuming it's not i dot i but rather is i dot j.Are gonna be zero.And of course cross products i cross j is going to be equal to k so we can do each of those and check essentially for orthogonality, so here we have n1 cross into going to be equal to n3 at least we hope after we do the calculation.

How do we do that so here's just a way of representing that cross product as a determinant of a matrix? I don't know if you've seen that representation of cross products before but it's fairly straightforward.And what we find is that n1.N2 given is a and b and up being equal to n3 this is the calculation for n3 and o by the way that checks that worked out. 

I'll go back to that slide. I know some of you are copying it down by hand. I don't want to rush you.So it's always a good idea as I said to check you know, check whether your answer is sane it's good and easy to do for.You know comparing unit vectors perhaps a little bit more difficult to do for computing the principal stress values themselves, but you could if you wanted the sanity check your your principal stress values, now that you know, the unit vectors you've gotten to the end you could go through and use the couch you stress formula you could dot the stress tensor with the unit vector you'd get attraction vector, you could dot that with the traction vector and you should be able to.

Raise it. I'm sorry you say you dot the traction vector again with the unit vector so you're going down two ranks by doing two dot products and get to the principal stress value that would be a way to sanity check that answer as well. I don't demonstrate that here in the lecture notes but it's fairly straightforward.

Okay a couple of additional thoughts on principal stress.So what are they again physically what are they they're the largest normal stresses on any plane?They're on orthogonal planes in fact so that's an important realization is that these normal stresses are each oriented in three different directions orthogonal to each other and they're extremeizations they're the the maximum values of the principal stress or the normal stress components.

This of course occurs where the shear stress is vanish.And again as I mentioned the coordinate directions are called principal axes, we've identified that previously what else can I say well? I mentioned these unique kind of corner cases where you may have principal directions that are a little bit funky now what I'm what I meant by that.

Was the following.It is certainly possible and in fact not entirely infrequent that you have two principal stress values after you've calculated the characteristic equation and you found the roots of that characteristic equation you found that two roots were the same.That means two principal stresses are equal.And what you'll find when that is the case.

Is that you're not going to find unique unit vectors associated with those two identical principal stresses.In fact what you're going to find is that two axes two principal axes are arbitrary in the plane.

Okay at some point during the semester you're going to explore that a little bit more probably in a homework set or an exam so be ready for that be looking for it be thinking about it what does it mean it means that in a particular plane the stresses are equal by axial there's you're applying stress in two different directions at equal magnitudes then as you were to rotate in that plane, if you're going to observe how that influences the components of stress, you would find that as you rotate in that plane the component magnitudes don't change.

Now if you can do it with two vectors it's conceivable that you can do it with three that does happen periodically if all principal stresses are equal then all three axes are arbitrary.And then we have this state of what's called hydrostatic stress.

Okay this actually does happen in the real world first of all if there's no stress on the structure, then the stress everywhere is zero you've got identical principal stresses of zero and so therefore you have an arbitrary axis at which that's applied that's the trivial case the less trivial case, of course is when you have let's say you drop a structure a submarine down into the Marianas trench you've got these equal stresses being applied external to a sphere if you drop the sphere down into the Marianas trench way down there you get very high stresses, but they'd be uniform.

Addict equal on all sides equal in all directions. That's the kind of unique case where that shows up.It's perhaps not common but you know it exists.

Okay, so it is 350 so we are getting close to 350. I should say we've essentially used up our 80 minutes for the day. We do have a whole another lecture essentially on stress and equilibrium. So we'll be ready to do that next Tuesday. It'll pick up at this point here.

I'll pause I'll stay on the line here for a little while take any questions that you guys may have.Otherwise have a great weekend and I'll see you next Tuesday be looking for the next homework set be looking for the returned homeworks at one. If you have any questions feel free to let me know.

Well because we have three principal stresses, we're always going to have three principal directions now the caveat to that is the caveat that I just gave that when you have repeated principal values that you then lose the uniqueness of the three principal orientations. Now what that means is that within a plane all the orientations within that plane are sort of equally valid principle orientations or by that argument equally invalid print.

Able orientations. There are arbitrary in that plane. So, you know that they're not if you were so let me say it this way if if you were given that problem on an exam.And it were it and you had repeated unit repeated roots.And then you defined for those repeated roots unique principle stress directions, you would be wrong. 

And so, I would mark you down for that. Because they're not unique they're arbitrary they could be any set in that plane. So you're it wouldn't be that you're set was totally incorrect but it's not the complete and correct answer.You're never going to have more than three if that was the other part of the question.

So Tyler asks.

Why then you wrote out the symbolic tensor for finding principal stresses, do you have the non-diagonal elements mirror each other in tensor notation?A Tyler why don't you get on unmute yourself and ask that again, it's not entirely clear what you're asking.

Something.

Okay. I'll go back to that that's a great question.

I'm trying to let me see here if I can do this more efficiently.

I think this is what you're asking. Tyler is that correct?Okay, so that's a great question. I'm glad you caught it it's not often that this is caught so what you're seeing here is the stress tensor now of course modified by subtracting the principal directions from the from the trace.

And what you're what you asked was in row one column two. I have sigma xy in row two column one. I also have sigma xy similarly three one and one three are the same and two three and three two are the same so I kind of snuck that through turns out that that is true and and correct.

However, it requires something that you don't know yet that we're going to cover next lecture.Ultimately we're going to find that you know, I've been saying we have nine components of the stress tensor which we do we have nine components, but not all of them are unique it turns out only six of them are unique and were symmetric across this this diagonal.

So that'll be a conclusion of next lecture.

So Gavin asked a question. Gavin, why don't you hang out for a few minutes after everybody else departs and you know, if it's a specific question about an email I can answer that one privately any other questions from anybody.

Okay, thank you all for attendance and and your great questions and I look forward to seeing you guys next week have a great weekend.

Yeah.

Well I haven't looked at my email since a little bit before lecture started you know the if you sent it and it was before the deadline if you sent it by email before the deadline then I'll accept it if it came after the deadline then that I'm not going to you know, I mentioned at the beginning of the semester that but I have that very strict policy of getting it in before the deadline and you know, the reason for that is you know with this many students, you know, every week every homework set that goes by somebody says well I did this or I needed that or make this exception this way or that way.

And that's inevitable that that happens every week and so this is just my response so that I don't have to then parse who's who's issue was big whose issue was small whose issue was a reasonable who's is unreasonable, you know you if if it was late then this homework set gets dropped from your grade and it's it's no big deal, this is the exception that you get for the semester.

Okay, thanks thanks for understanding Evan.