Okay, good afternoon everybody. I hope you're doing well before we really get started. I guess I would like confirmation that you can hear me can people hear me.I'm seeing some yeses coming thank you. I appreciate that today's lecture is gonna be on stress we're going to move into sort of the next chapter of the class really this is the first chapter everything up to this point has been preliminary we're now moving into the phase of the class we're going to where we're going to talk about elasticity theory that's the beginning of our engineering structural mechanics work and then the next phase after that.

Will be engineering theories or engineering modeling techniques which takes elasticity theory and makes it a little bit easier actually elasticity theory tends to be very difficult mathematically for reasons which you'll see we then make some approximations and assumptions which can make it easier that'll be the second part portion of the course and then the last portion of the course at the end we're going to use approximation techniques that will eventually lead us to understand things like the finite element method.

Which is the dominant structural technique in in engineering today, so that's what's up for the day.We'll go ahead and get started here.

Now I'm quite certain that all of you have.Experienced stress and strain at least a little bit see if I can adjust the size of this screen here. I don't know exactly why it's out of phase or out of size. I should say.

Hmm.

Okay. I'm sorry about that. I'm just going to pause for a moment and switch to presenting off my other screen. I think that will do a better job of showing us what we need to see here on these slides.

All right, we'll recycle here.

That I think looks a little bit better.Okay so so anyway you guys have all been exposed to stress and strain a little bit in your past work on structural mechanics, however, you're gonna learn quickly in this class that it's a little bit more complex than what we have perhaps presented it to you now the most simplistic description that you've seen before is the one that's shown here on the screen that is we take a resultant force and we divide it by a reference area.

Over which that force applies and this is what you've likely seen as our description of stress and past classes now, of course what we're looking at here is a normal resultant force, so it's pulled normal to the surface that's exposed you've probably also shoot seen shear stress which will introduce more fully in a moment, but this again very common description of stress.

Now what's interesting about this is that we're assuming the way we've defined this that this axially load carrying member actually has a uniform stress over the cross-section that turns out to not be a great assumption in fact in a lot of our structures.

Let me introduce you to a new way of envisioning what stress represents.The image that I've presented here at the left. I like to refer to this as my stress potato you take a an arbitrary volume in this case. I've drawn it to be sort of in an unusual shape the the body we're going to call it a body in this class so that body then is has external surfaces and internal surfaces.

I don't know how well you're able to see my mouse but here I'm drawing a circle oh that is the external surface included including this dotted line over this three-dimensional body and then what I have represented here in the middle is an exposed internal surface.So what we're doing is we're imagining that this body has been cut at least cut in an imaginary sense.

It now has two internal surfaces which amount to one plane we've got the surface that you can exceed that's that you can see in this exposed with the normal that you see drawn here in red, of course, if we were really taking a cut on this system and we had separated it into two bodies, there would be another surface with a normal pointing in the exact opposite direction of this on the other exposed half body, but that is what we're doing. 

So we're taking an imaginary cut in some.Sort of arbitrary body in this case represented by my stressed potato. We've exposed an internal surface and that surface now can be described as having a surface normal.Now the resultant force over this entire area that is the sum of all the stresses times they're differential areas. 

So really those are differential forces those differential forces are all vectors, you sum them up by multiplying them by their area and what you get is a resultant force here shown by the vector F.Which I will circle here. It's obviously a blue vector. So that now is an internal resultant force.

Over an area that was exposed by this imaginary cut.

However, this imaginary cut that we took was entirely arbitrary. There was no reason to choose the one that we selected. We could have made cuts in other planes exposed other internal surfaces and had we done so we would have had different resultant force vectors.

So,That's an important observation just that we could have made other cuts in our exposing of internal surfaces in the body. Now, we also have to recognize that the loads pass through the body without concern for the plane of observation meaning, they don't really care which way we're pretending that this body has been cut in half. 

They don't care how we've exposed internal surfaces. They just recognize these loads that are passing through they recognize they pass through the body in an equilibrium sense.Consequently if we expose different planes the different planes feel different intensities of stress.

I'll show you a little bit more about that on the next slide.

We also need to recognize that the area of this cutting plane.Really needs to be associated with the vector that defines the plane.Okay, we've defined the normal vector.For this imaginary cut and that normal vector now determines what area is present if we had chosen to use another area we would have or another cut we would have had a different area.

So we're gonna now define our our area as a vector.You can see vector represented by the underbar here under a a is equal to the area a this is a magnitude now dotted with a normal vector.

Okay, so we recognize the area changes with the normal vector selection, this is the internal plane selection we recognize that the forces that are passed across that plane themselves are also something that varies associated, you know, according to the plane that we choose so both the magnitudes and the areas change as a function of the exposed internal surface in our imaginary cut.

Okay now the last thing we need to recognize is this cut that we've made is arbitrary and we can make further cuts, in fact we can cut ultimately down to a very small differential area within the body.That differential area then has a differential force vector associated with it that has normal and shear components. 

I didn't mention the normal and shear components previously, so there's your normal there's your shear.

So in the end what we've done is we taken a body a structure.We've loaded it loads are passing through the body we imagine taking cuts within the body and by doing so we expose internal surfaces that have forces force resultants on them we can do that in the limit of going to a differential area very small area being exposed that has a differential force and through exposing these differential forces and differential areas, we ultimately compute this complete description of stress.

Let me take you to that mathematically now.So assume we have this differential area now. I'm going to call it delta. A.Of the internally exposed surface on this imaginary cut and of course it has a corresponding differential force we can define something called the normal stress magnitude we're going to reference that magnitude with the symbol sigma.

And we're going to define sigma this is the stress normal stress as the limit as the differential area approaches zero of the change in differential force divided by the change in differential area.Really this is a derivative it's the f n d a it's the derivative of the normal force with respect to the area.

And we're ultimately going to call this a component of a vector that vector is going to be represented by the letter t where eventually going to call that a traction vector.So t n represents the normal component of stress on this exposed surface and ultimately it comes down to being a derivative you can think of it as a limit as as the area changes of how much the force changes.

Similarly we can define a shear stress magnitude in this case, we're going to use the symbol towel.And in the limit as delta. A goes to zero of the shear force component it's change over the differential area change which amounts to being a derivative so this is approximately equal to df s d a.

We're going to again call that a component of a vector in this case the traction vector t sub s this the shear component of that vector.

Alright I'm going to pause for just a moment and take any questions. I know this is starting to perhaps be a little bit more complicated than you've seen in the past.

All right. I'm not seeing any questions in the chat window.So here. I do see some now coming through some questions yes, who is that go ahead.

Okay, so the,

The.

I guess it's not entirely clear what you're asking, so here the area is a vector area.Which is a magnitude times a vector direction this this symbol here, okay, so representing dot since this is not a vector itself this is just a multiplication sign.Is that is that ultimately the question David.

Yeah.Yeah, thanks for asking that clarifying question that maybe is a notational change that I should make on that particular slide. I agree that it's confusing the way I represented it.Okay, so I mentioned that each of these things so this we had a shear component let me go back to our stressed potato again, so the resultant force had a sheer component it was a vector it had a sheer component then it had a normal component. 

Hence our traction vector also has a normal component and a shear component.So, I will write the traction vector here t as being the sum of the normal component and the shear component.

And then ultimately the the resultant force is the area in magnitude multiplied by the traction vector at that point.

Now what I think can be a little confusing is this whole description that I gave of the the magnitude of that vector being now a function of the plane that we chose. Okay, we're choosing a normal vector and the magnitude of that force is now a component, you know, the components I should say of that force vector are now a function of the plane that we chose. 

So I I I'd like to show you this slide where I think it becomes a little bit more clear exactly what we're doing.Okay, I'm showing you now a an axially loaded member where I have on the left hand side applied a uniform stress.Now the rest of the member is not drawn here. 

I'll just draw it momentarily, okay. So, of course, we could have done exactly the same thing on the right-hand side.

Now, I've taken my imaginary cut and what you see here in black and then dotted lines indicating this imaginary cut that I've made on the internal surface of my axially loaded member. I think it's relatively easy to recognize that equilibrium requires there to be a reaction force.The force on the left hand side is this uniform stress times whatever area was present here. 

So we're gonna assume it's a unit area. So it's the integral of this pressure gives us a force that is going off to the left and on the right hand side we have a force vector resultant that is in magnitude large enough to offset the that same force on the left hand side.

Now, I mentioned I can choose to make my cut on any plane.I didn't have to choose the plane that was normal so in this axially loaded member I can still choose a plane like I have in this second portion of the image that is at a different orientation with respect to the sort of axial direction of this axially loaded member. 

So here I've now made a cut that has a 45 degree plane.To that axial direction.

Now, of course the left hand side of this has not changed.So what do we know? We know that the reaction force on our now newly exposed surface is identical. So the length of this vector here this blue vector and this vector here are identical.However, relative to the surface the surface up here on top had a normal that was in the same direction as that vector whereas here on this 45 degree plane the components of this reaction force vector this this integrated internal load vector, it has to have components one of them being normal. 

I'm highlighting now and one of them being in a shear direction relative to that surface which I've highlighted.

So the magnitude of the normal component has changed from here to here the reaction vector overall has not changed but the magnitude the component of that vector that's associated with the normal direction to our internal surface. That magnitude has changed as has the shear vector.And then last but not least I there's no reason why I couldn't have made a cut in this case horizontally so I've exposed the internal surface along the entire length. 

Now, of course, we have a normal vector that's in our y direction we'll call it for lack of a better descriptor. So in the vertical direction,Now if you were to take and and calculate internal reaction forces on this surface, you can see that the left hand side is now balanced by stresses that are present at the right hand side and so actually on this exposed surface there is a well in fact still a reaction or I should say a resultant vector, it just happens to be of magnitude 0.

So if we took this cut we'd have the largest normal component in this cut we have somewhat less of a normal component, but we also have a shear component and in this cut we have no normal component and no sheer component.So this is a pretty simplistic description that sort of illustrates in ways that I believe everybody can understand how the normal that we choose now influences the traction vector that comes out of that normal selection.

Okay, now what you have heard of as the stress state and we've you know up until this point in your career, maybe have just refer to this as sigma or maybe if we got a little bit more advanced we said we have normal sigma and sheer towel to get the complete description of the stress date.

Now, we need to recognize that we must be able to describe a resultant traction vector on any surface that's exposed within the body.Okay, so I'm going to pause again for questions. I hope I might get a few.

Yes.

Okay, so that's a good question. Can you tell me who was asking that?Okay, thanks Caden. So as we've now exposed so we're looking now at the third one we've exposed this internal surface.Okay, we need to do the sum of all forces on this internal surface. So, I'm going to take the bottom half. 

We see that we have a uniform stress here.On the left hand side times an associated area. If we look at the right hand side now because there's no cut on that, you know that interrupts the flow that goes through here. Now, we have a uniform stress on the right hand side and it has a an associated area. 

So this area times this distributed force plus this area times this distributed force these two things identically balance each other out. So even if I drew a vector here.We would know that the components of that vector actually have to be zero both in the normal direction and in the shear direction.

So does that make sense?

Well it's more it's more applying equilibrium and you can think about it as interrupting the flow of the stress, we we will talk about flow as load is flowing through the structure. The reality is is that you can actually think of it as a flow if you so choose to but really this is just application of equilibrium. 

You know, you need to even though so what we're doing here is we're saying all of the forces on this lower body and I'm now going to draw a box around this entire body have to be an equilibrium.Okay, we know the the boundary condition here we know the boundary condition here. 

We know the boundary condition here on the bottom's got to be zero and if we sum up all of those forces we the the the boundary condition on the top it also contributes to the forces and it just forced balanced drives it to zero in this case.

Okay, so I hope now. I have convinced you.That these internal planes these internal imaginary cuts that we can make each of them has an associated traction vector as loads pass through the body.Now the vectors themselves, I'm going to claim must be related to a more fundamental set of quantities which from now forward we will refer to as the stress tensor.

Okay. I want you to stop thinking of stress as a magnitude and start thinking of it as a tensor. Now, what do I mean by that?

Okay. Consider that we took that stress potato.And we began to take imaginary cuts.

I need to start bringing props with me to class I guess if I were if this were normal lecture, I actually would have brought in brat Bratton I would have brought props with me and and I would have held them up and sort of hand waved with them a little bit and said, all right, let's imagine cutting it in this surface and now let's imagine cutting it in this surface imagine cutting it in this surface and you continue to cut until you get a differential parallel pipe it. 

So you're getting a a small cube.With parallel sides.This is a very small portion a differential volume interior to our stressed potato.So the sides it has of length dx in the x direction it's dy in the y direction and dz in the z direction. We refer to this small differential element as a parallel pipe.

We're going to choose to conveniently align these parallel faces with the coordinate system defined by x y and z. So it's a cartesian coordinate system in this example. Most of the work we'll do the semester will will be in Cartesian coordinate systems, but eventually we're going to move away to different coordinate systems.

Now we've exposed by by taking these imaginary cuts actually this parallelo pipe, it has six internally exposed faces.And load is passing through these this cube through the faces of the cube.You can think of the external tractions of the cube as being the boundary conditions on the cube with load passing through the cube.

So looking just at one face. I'm going to call it the x normal face.So plus x direction.We have now on this face three components of a vector. We have a normal component we have a shear component that's oriented with the y direction we have a shear component that's oriented with the z direction.

Now, this is really no different than what I had done here with our in this case, we would have only had two dimensions because the way it was drawn, so we would have had an x component and a y component.Or a in you know, with the exposed surface of the one in the middle we could have defined our coordinate system to be x in one direction y in the other direction relative to the normal and again we would have x and x and y components but when we move to three dimensions, of course, now we have three components rather than two.

So on this face there is a traction with three components again, it's aligned with the coordinate system.We have to look at how we name our components and by convention we name them associated with the normal direction of the face.And we use subscripts.So let me go back to this image for a moment the first subscript is the normal direction for the face so on this particular face, they are all x normal faces.

All of these components are associated with that face the x normal face positive x normal face, the second subscript is now the direction of the component force or a component traction.So we have an x normal face with a x direction component we have an x normal face with a y direction component and we have an x normal face with a z direction component now. 

I've drawn the arrows the associated with the assumed component directions and will eventually let minus signs switch those directions around if that is appropriate.So going back to this slide with the text on it again, we have a normal direction the first subscript describes the normal direction of the face the second subscript describes the direction in which that traction acts for that particular component.

In all of these cases the components of stress have units of force per area.

And we'll call these the components of the stress tensor.

Now by convention a positive stress.Is defined when that stress acts in a positive direction on the positive face.Or alternatively we'll also call it a positive stress when the stress acts in the negative direction on a negative face.Let me see if I can draw that for you. 

I'm doing this with my mouse for so forgive my.Less than stellar artistic ability.

Here we have a normal face that's positive in the positive direction.So here's our positive normal. If we happen to have a normal associated if if we had chosen to make a cut where we had a normal that was facing in the negative direction if the component to force was in the negative direction, we still cancel the negatives and call it positive. 

So either of these would be called by convention a positive stress.

Now similarly so I'm doing this in two dimensions.

Here's my positive normal if I have a force component that is attraction component in this direction.It is on a positive face.In a positive direction. In this case the positive y direction, so if we call this sigma x y.It's positive if it's oriented this way if it were on a negative normal face and it were in the negative direction.

This would also be a positive sigma x y.

If we're on the negative face, but instead went in the positive direction, then that would be negative.Okay. I hope all of that's relatively easy if it's not feel free to interrupt me as we move forward.

Now to make this even more general as I stated we have six faces. So we have a positive x normal face. We have a positive y normal face. We have a positive z normal face.By arguments that we made a few slides ago, each one of those faces has its own traction vector.

So the one that I've already shown you is x x y x z we have we have one.Let me view this image so that.It's a little bit easier to see perhaps. I don't know if that helped at all. We have the in fact. I don't think it helped at all.

We have the y y face so it's on a y positive y face direction in y so the normal stress is called sigma y y here we have sigma y x, so it's in the normal y normal direction, but the traction vectors creating a force in the x direction and similarly z and then last but not least, of course, we have the z normal and then z y and z x. 

So these are actually three different traction vectors which I've written down here, so the traction on the x face is sigma x x times the unit vector in that x.Direction which we'll call i Sigma x y times the unit vector in the j to x the y direction called j and then sigma x z times the unit vector in the z direction called k.

Three stresses or three components of stress for each traction vector three faces, so we actually now have nine stress components x x y x z y x. y y y z z x z y and z z.

All of these are still vectors.Again the components have magnitude of force per unit area, it's a component times the direction component times direction component times direction.

Okay so at this parallel pipe it this is a point in space it's a differential volume that we've imagined taking that full body and taking successively smaller and smaller cuts until we've exposed a parallel pipet or a cube interior to the body of the of the load carrying member there are three different planes and x normal a y normal and a z normal plane and it has three different traction vectors totaling nine components the nine components now can be written more succinctly in a matrix format that I'm showing here.

And we're going to now begin to refer to this as a tensor at the point.

That tensor is called the stress tensor so the stress tensor previously you've thought of stress likely just as a nor is as the normal component one magnitude or maybe two magnitudes you had some normal components and you had some sheer now we recognize that in a more general sense we have to have three tensor or three traction vectors nine components, so stress is actually in the most general sense a nine component tensor now what am I saying I'm saying tensor note. 

I'm not saying vector will get into the details of that in a in just a moment.Very briefly though it instead of having just one direction associated with it we actually have two subscripts or two directions to two unit vectors ultimately define different components of stress.

Now this is a bit redundant to what I've already said each stress component is named using two subscripts the first describes the normal direction to the face the second describes the direction in which the component acts on that face so more specifically we can we can refer to this as sigma ij we're using two latin indices as subscripts, so it's stress on the plane i in the direction j sigma xx is a normal stress, so when you see a repeated direction,We now recognize that is what we used to think of as normal stress.

We still think of it as normal stress but now we recognize there are actually three normal stresses XX YY or ZZ now periodically you might just see this written as sigma X or even even more greatly simplified and written as sigma if it was just a uniaxial stress present however, I don't particularly like those notational schemes. 

I think they drop some of the rich value that comes into recognizing that there's actually nine components.Of this tensor.As I'm also had previously alluded to so now if we have two subscripts that have different letters associated with them, so XZ.By the way, we're gonna not not strongly differentiate between towel XZ and sigma XZ those things are functionally the same.

So we have XZ we have XY we have YZ those are now called shear stresses the reason why you have seen towel written at times in your past experiences because usually when we're thinking about a two-dimensional stress field, then we might just call them sigma representing the normal and towel representing the shear but you know in this class we're gonna largely drop that representation of tau with one exception, which I'll allow.

To later in the semester and we're just gonna call them all sigma so they're all uniformly sigma with a two subscripts and then if that's x z y z or x or or y x z y z or x y those are the shears.And the positive stresses are what I have shown.

Okay questions about that.

Okay, so now I I called this a tensor.And I'm sure you've maybe heard that word once or twice before maybe not put a lot of thought into it so a natural question is what is a tensor a tensor simply describes a linear relationship and a geometric space where you have components and they're associated with directions so a vector is actually a tensor so you guys have heard that word many many times a vector is a magnitude and then the direction that it's associated with.

When we talk about tensors more generally we need to now think in higher dimensions so in this example, we actually needed two directions in order to define what component we're talking about we had an x x two directions we had an x y so this is what we would call a second rank tensor stress is a second ranked tensor.

So if you hear me say the word rank the rank of a tensor describes the number of directions that must be assigned when describing that quantity.

So the simplest tensor is actually a scalar. It's a rank 0 tensor. It's a magnitude with 0 directions.Now one way to think of it is how many components exist on a scalar well, it's 1 in this case, it's 3 to the power 0.So a scalar is just represented by 1 magnitude.

Vectors which were all quite familiar with in this more general terminology is called a rank 1 tensor again, it has a magnitude and one direction. So it has three components in a three-dimensional space two components in a 2D space. Most of the work will do is in a 3D space.

The word that we would use to describe a rank two tensor would be to call it a dyad. I often just forget that word and instead call it rank two it has a magnitude and two directions.So there are three to the power two components that's of course nine components. 

We can go a step higher to a triad which is a rank three tensor magnitude and three directions three to the third is twenty seven components. This can be generalized to saying a rank end tensor, which is a magnitude and end directions and so that's three to the end components. 

Turns out instructional mechanics and fluid mechanics, we're going to actually encounter rank four tensors quite frequently, so three to the power.Four would be eighty-one components and we'll see that in this class.

So this word tensor.What is it? Well, just in your mind think of it as a generalization of a vector to a higher number of dimensions.

Now.Without going into all the details.I will mention mathematically speaking that a tensor follows certain transformational rules.These are essentially mathematical rules that govern how tensor can be calculated in different coordinate systems.Now, that's something that's not unique, you know, you've experienced that with vectors. If I were to hold up my hand and actually I have to hold up both hands. 

I'm going to hold up. Now, I have to be thoughtful about this because I recognize that you guys are seeing me from a different direction. So,

I'm holding up my right hand so that my coordinate system X Y and Z in a Cartesian coordinate system defines the right hand rule. Now, I encourage you guys to think about this because I think when it projects the screen it may project it a little bit differently.So if I take a vector.

And I put it in this three-dimensional space if I reorient my coordinate system.So here. I have my my thumb is aligned with that direction. I know this is quite difficult for you guys to see but I'm gonna do my best.Here my index finger would be aligned with this direction here, my middle finger would be aligned with that direction as I reorient the magnitude of the components in the x and y and z directions change the vector never changed. 

So it's magnitude was preserved irrespective of that transformation that I did as I observed it from different corn directions.Same thing happens to stress.The magnitude of the stress in some sense needs to be conserved as you observe it from different coordinate systems.

So it's not in fact higher rank tensors second-order tensors must conserve magnitude and more will get into what that more is as we go forward into the next lecture.

Okay, are there any questions?

All right now.We've talked about this traction vector, we've talked about the stress tensor.We related the two of them by examining that parallel pipe it that little cube infinitesimal cube that was oriented with respect to our x y and z coordinate frame.However, really to be able to consider traction vectors on arbitrary planes not directly aligned with x y or z we need to do a little mathematics.

So, that's what I am showing you here.With this image.I've got an x y and z direction in our Cartesian space that parallel pipette that I had previously that cube. I'm now imagining one more cut. Take my cube cut it one more time.And I've defined a new internal plane that new internal plane has the normal N.

Now.In our prior discussion where we were looking at the cube we recognize that the outer surfaces of that cube had traction vectors associated with it. So now these happen to be drawn on the negative faces TY on the negative face and the negative direction TZ again, negative negative T X negative negative.

Our new internal face with the normal n has yet another traction vector, which we're going to call TN.

Now, we have four tractions T n T x TY and TZ.We can determine what T n is the traction on this new arbitrary plane from the other tractions.So I'm going to highlight that so the traction plane can be determined from the tensor by passing a cutting plane through the equilibrium element and then ultimately summing the forces.

Okay to really do this the easiest way to to do it is to define a normal so the normal n.Is going to have three components to it. We're going to call it L I M J and N. K.Now it turns out that these LM and N are direction cosines, so they're cosine of the angle swept between the N direction and the X Cartesian direction.

Cosine of the angle swapped between the N-direction and the Cartesian Y direction and cosine of the angle swept between the n-direction and the z-direction.So this normal.Is defined by the angles of the plane that has been cut.

Now as a vector.If if the magnitude of the normal vector is of length 1, then you realize of course then that there's going to be some constraint on the magnitudes of lm and n the sum of the squares of those magnitudes will have to be equal to 1 no none of those individual magnitudes can exceed 1.

Okay this cut plane now it has a differential area. DA.So, that's the surface of this internally exposed.Imaginary cut.So if the cut plane has an area of DA then we can determine what the areas of the other faces are so really after taking it down from six external surfaces making a cut we now have four total external surfaces one associated with T X over here one associated with. 

Ty one associated with TZ the area of the largest surface, the one that is the normal N surface that has magnitude DA and then just using.The direction cosines we can compute the areas of the other triangular surface. They are DA times L D. A times N and D. 

A times N.

So in this case, we have the magnitude DA times the normal vector dotted with the unit vector i j and k are unit vectors in the original Cartesian frame.

Okay, so let me go back to this for just a second where the image is slightly larger.Okay this now newly exposed sub-volume this differential volume has to be an equilibrium.So the sum of all the forces on that internal body have to be equal to zero.So, I'll put a zero on the right-hand side.

And then I will take the vectors so we have TN which is this vector here times its area remember its whole area was DA okay? I'm gonna drop the d because ultimately the D the differential area would be existing in all of these.So if the traction vector times a minus notice that this one's pointing in the positive direction, this one's pointing in the negative direction, so minus the traction vector tx times its area.

Minus the traction vector y times its area minus the traction vector z times its area of the sum of all of those has to be equal to zero. If this element is to be an equilibrium.So from that, of course we can divide by our area and we can ultimately get that the traction vector n.

Is the sum of tx times L.T y times m.And tz times n again those are direction cosines.

We could equivalently write this as components.Of a vector so TN is going to be equal to T and X times the I direction.Plus. Tn Y times the J direction plus TNZ times the K direction. Again, these are subscripts you recognize this is the n normal face the component of it in the x direction and normal face times the component of the y direction and normal face times the component in the z direction. 

So this is the magnitude of that component which now we will see is related to TX times L and so in the end these components are TNX is equal to sigma xx times L.Plus sigma, xy times m, plus sigma xz times n.So this component is all of this stuff here.

And similarly we have a y component and a z component associated with that n normal direction and they take on these other magnitudes yx times l yy times m, yz times n and lastly zx times lz y times mzz times n.

So Nicole is asking, can you explain the negatives on the summing forces again? I sure can. Let me go back to this slide here. In fact, let me go back a few more slides.

So what I drew here was my parallel pipid the positive x-face and the positive direction. So the traction vector that was positively drawn on this face was positive.Positively drawn on the y-face was positive positively drawn on the z face was positive. Then I took this cube and I sliced it and what remained was the negative face is which would have had traction vectors associated with them. 

I didn't draw them here, but they certainly had those traction vectors associated with them on those negative faces and they would have had components drawn in the negative in order to be drawing positive stress components, they would have had to be negative on the negative side. Now going back over here.

Nevertheless those traction vectors are vectors. So when you sum them up, they have in this case with x it's an arrow pointing in the x negatives than the negative x direction.So it is a negative force. Even though it was components on a so it's a positive component now on a negative face.

So, it's negative. So that's where we got the negative from.So if you're summing them all up this one's pointing in the positive direction, this one's pointing in the negative direction.This one although it appears to be pointing in the positive. Y direction, it's actually pointing in the three-dimensional view as a negative z direction. 

So this is z direction over here. The force is in the negative z direction. So that's where the negative tz times a comes from.

Okay. I hope that makes it more clear in a cool.

Alright, I'm gonna pause. Is everybody with me on this?So Ron asks, so a is equal to ax equals a y equals az So no, that's not the case.AX let me go back one slide here it was.AX was equal to a times L.So it was the area of the act on the X face.

Was equal to the total differentiation differential area times the unit vector dotted with the X direction, which was equal to AL. Similarly we had an AM and an AN.So so to correct your equation, it would be.Well, it would be slightly different than what you've written.

Okay, I appreciate that you guys are asking questions. I do think that that clarifies. I'm more than confident that if one person is confused that multiple people are confused, so I don't hesitate to ask questions when there's a lack of clarity.

All right, so now I'm gonna just kind of highlight this whole set of equations here.

What I've got highlighted in the box can be more succinctly written as so you see this is three components of attraction vector and X and Y and NZ. So this is a vector on the left-hand side it's equal to something it's got to be a vector on the right-hand side. 

I can write it more easily this way.The traction vector associated with the n-face. It's a vector.Is equal to that 3 by 3 matrix the tensor dotted with a normal vector.

Okay, so this formula is called couches formula you recognize the dot product reduces the rank of something the rank 2 tensor dotted with the rank 1 vector results in a rank 1 vector.It's the dot product. You you know, if you go back to vectors that's how you get magnitudes from vectors as you dot them.

Similarly if we have if we dot a rank two tensor with the rank one vector, we reduce the the rank by one so we get a vector is equal to a vector here now the sigma represents those nine components.The n vector is now the vector that's associated with the face it's normal to that face and this is now the exposed internal traction vector associated with that face.

Okay. I hope everybody is able to follow along any questions out there. I do want to pause regularly for questions they don't all come through a chat box easily, so if you would like to ask a question verbally now would be a great time.

So that is a.That is essentially mathematically the definition of a dot product and you've never probably heard it described that way, although you have used it that way.Let me take you back in fact. I can draw right here on this screen, perhaps a good a good way to do it, so if I define an x and a y coordinate system.

And I draw a vector in that coordinate system.

Okay, so I'm gonna call that vector v.

Which in a two-dimensional space is gonna be equal to some magnitude a times i.Plus b times j.

We can calculate the x component of that.By essentially dotting v.With i.

So you're dotting it and then this is now a magnitude it's a scalar value.

So in this case, it's a really I could have written this equivalently, so v i c this was an equivalent way of writing what we had before is v x times i.Plus v y times j.This is a scalar this is a scalar the combination of the two of them when appropriately multiplied by the direction vectors.

Was a vector so v itself was a vector but when you take the dot product, you are reducing the rank to a scalar.So similarly when we had sigma.Which was a 3 by 3 tensor in a matrix sense.You can also just think of it as being a scalar that has 9 components dotted with a 3 components of vector, you're going to reduce its rank and it's going to end up being so in a matrix sense this is going to be 3 by 1.

Is equal to a 3 by 3.This is a matrix.

Multiplied by a 3 by 1.

So matrix math you probably didn't think much of it at the time but you were working with tensors and you were doing rank reductions when you multiply these vectors together.

Okay, I have to find my place again now.Okay so here we are and any further questions thanks for the question. I'm not sure who that that was but thank you oh so here's a question from Sarah so Sarah asks, so the rank is only dependent on the direction no it's actually dependent on the number of directions that you need to describe the overall quantity so stress being a rank two tensor, we need nine components and each component has two directions associated with it one is the direction of the face normal one is the direction of the acting force.

So two directions necessary to describe stress rank two tensor vectors one now when we say one direction it's components in that direction right so.Where did my?Find my work that I just did sorry. I'm a little bit turned around and must have disappeared it was on couches, oh here it is.

So a vector each component had one direction associated with it, so AI BJ a I one direction rank one.Sigma xx two directions rank, two.

Okay, so you've seen this equation in the box now already this is.Couches formula in this case, it's represented as three equations three tractions t x t y t z.These are vector components, this is going to be now equal to sigma x x times i sigma x y times j sigma x z times k. 

Three vectors each having three components one with each direction.

Now that's on on a arbitrary surface.With normal n so a question is what is the value of the stress component normal to the surface?Okay if we're to look back at this image here with the normal n direction.One of the questions we will ask ourselves repeatedly in this course explicitly or implicitly is what is the normal component of stress on that face?

What is the normal component of stress on this n normal face?

All right, so.The answer is yes, we can compute it the stress normal to that cutting plane.We're gonna call that stress and end.

So n is the normal.It's actually that traction TN dotted with the normal vector.So we have a here's that rank reduction that we were talking about.Right, so we have a three by one.Vector.TN as a vector we dot it with a normal and we'll get the component of that vector in the normal direction.

So mathematically this is going to be as simple as saying after the the dot has occurred l the magnitude of the direction cosine times t m x plus m times t and y plus n times t n z.And we had gone through and calculated each one of these magnitudes separately and a prior slide t n x t y and t and z and so when we make our substitutions the magnitude of the n n component, this is the normal stress magnitude now. 

I got to be careful it's it is a magnitude at this point.Because we took a vector dotted with a normal gives us a magnitude it's going to be sigma x x times l squared plus sigma x sigma, yy times n squared plus sigma zz times n squared plus these combinations mn, sigma, yz plus ln sigma xc plus lm sigma xy plus.

MN sigma z y plus l n sigma x yeah, you don't need me to read at all to you you see it actually has a contribution from each of the nine components x x y y z z y z x z x. y z. y z x. y x each of the stress components, there were nine of them contribute to the normal stress.

And how much do they contribute well it depends on the direction cosines associated with the plane that you have cut.

Now what one might ask why why do we care why do we care what the magnitude of the normal component of stress is?Why do you go through the trouble of computing it?

Well, it turns out that the maximum value and direction of sigma n n will often control the design of the load bearing member.

Okay, if you think about it as stress flows through a structure it's most likely to fail especially if it's a brittle failure where the magnitude of that normal stress is its largest and it will crack perpendicular to that magnitude or to the perpendicular to the direction associated with that magnitude.

So I don't know how well you can see my hands if I'm pulling normal let's see here. I'll try it this way okay my thumbs now representing stress magnitudes normal stress magnitudes.We're going to pull those apart where these magnitudes become the largest we're pulling the material apart if it's brittle it will fail in a stress a crack will begin to open like so.

Where the normal stresses mag maximized.So that's why we would like to be able to compute the stress normal on any cutting plane.

Now since we've computed the normal stress component.We also would like to be able to compute the shear stress component.

Okay, we're going to call that stress component and s.And we recognize that since.The.The magnitude of the n component is the sum of the squares the square root of sum of the squares of the normal and shear components we can do just simple mathematical manipulation to say that the shear component is.

Going to be related to the normal component and the planar traction. So the traction itself squared minus the normal component squared that under the square root sign gives us the magnitude of the shear component.

So here mathematically speaking sigma and s is equal to.T and x squared plus t and y squared plus t and z squared this is the magnitude.Of the traction minus the normal component squared.

So, of course a natural follow-up question is y why do we care about what the shear stress component is?Well, it turns out that it's also quite important in mechanical design. It actually turns out that material yields when we when you've heard of plastic deformation and yield what's going on at the atomic level is the material is sliding along slip planes and that is a sheer driven process so the sheer component.

Drives yields or failure for a ductile material. So a ductile material like most metals yield based on a shear component of stress where it's brittle materials as I mentioned on the last slide they tend to fracture and they fracture associated with normal components of stress.

So one of the things that will find as we go forward in the class is that the maximum value of the shear stress is related to the maximum value of the normal stress.And the minimum value of the normal stress. So we will want to eventually compute both the maximum and minimum values and then that will give us an indication of what the maximum shear stress is. 

Now, I haven't taught you how to do that yet. We're going to do that probably next lecture.

Okay again, I'm going to pause for questions.I hope things are relatively clear.

Okay, let's go through a couple of quick examples of computing tractions so that you can get an intuitive understanding that this actually works. Okay, not not just a description, you know, I've given you now a description of why it works but I want to convince you that it actually works out mathematically as well and that that mathematical outcome actually follows your intuition on the matter.

So, here's an example given a stress that I define for you. Compute the normal and shear stress on some specified planes.Okay, we're going to use this example that I showed you previously. I felt that that was a fairly intuitive description of the resultant force on a internal surface that we've exposed by taking an imaginary cut.

Okay, I'm going to define the stress. Again, we've got a negative face negative stress here. It's uniform.I'm going to say that it has magnitude 1. And further I'm just going to assume for simplicity that the cross-sectional area is 1.So this area is 1 this magnitude is 1. 

And so if you look at all of the forces that are present here, we can define the stress tensor 9 components as having magnitude 1 in the sigma x x location the remaining of these sigma components the stress components are all going to be 0 for this example.

And what I'm asking for is n and s So n and s in this case represent the normal component to the traction vector and the shear component of the traction vector on the internally exposed surfaces.

So, how do we do it? Well, we do it using the the mathematics that I've just described. So, here's my solution to compute the traction vector itself.I need to know what the stress tensor is and I need to dot it with a normal vector associated with some surface.

Now the easiest one easiest one, in fact. I don't even have it really in slide form we'll just do it quickly in our heads. The easiest one is the top one. So the normal vector for the top one is i equals one.Or really l this is l m and n so the magnitude of the direction cosine the the normal surface is entirely in the x direction. 

So the normal for that would be one zero zero.And if we put a one zero zero right here.

I for some reason I'm not able to draw it so if I put one zero zero we took this dotted with that we'd get one times one plus zero times zero plus zero times zero, so you'd get a one and then zero times this and zero times this so you get one zero zero and you'd recover the most simplistic form of the traction vector that is that it has magnitude one.

And 0 and 0 in the x x y and z directions explicitly. Now, we're actually doing the computation for the second one. So for the second one, we need to ask ourselves, well what's the normal and shear component n and s for this imaginary cut in order to do that again, we need to define the normal vector associated with that plane.

So the normal vector.Because I've cut the plane that 45 degrees is going to have a component in the x direction and it's going to have a component in the y direction.Again, intuitively, we're looking at the the vector associated with this direction. Let me draw it for you here. 

So, you know if I'm defining this plane and this normal.Assuming that that's 45 degrees it has a component here and a component here.Which are in magnitude 1 over root 2 in the x direction 1 over root 2 in the y direction if the magnitude of this vector is 1.

Now, it's worth also noting that the area of this planar surface is going to be the square root of 2.So it's going to be a larger surface area than when we had the surface area, that was straight vertical. That was a magnitude 1. We can see that this one is larger and it turns out that in magnitude, it has value square root of 2.

So coal as can I return to the normal equation. I assume you're talking about this normal coal, can I move on?

I'm going to assume that's what you mean.Ping me back if that assumption is incorrect.

Okay, so here's the math, that's pretty simple. TXTYTZ.A coal wants me to move back a little bit. Know the one prior to the sheer equation.Not sure which one you're referring to.

Coal feel free to jump in and just ask verbally more with more clarity what you're asking for. Okay, says nevermind. Alright, so.Anyway, here's our calculation it's the couchy formula TXTYDZ is equal to the stress tensor represented in this case by a three by three matrix dotted with a normal vector in this case represented by a three by one matrix and so you can see the x component of that using matrix mathematics is pretty straightforward, it would be the same of course with any other form of mathematics. 

So one times one over root two plus zero times one over root, two plus zero times zero.TX is going to be equal to 1 over root 2. Pascal's. Now, this is it has units of force per unit area the X component.1 over root 2. The Y component 0 times those the Z component 0 times those.

All right, so we're going to need to sanity check that answer.

Hence our TN.

TN is equal to 1 over root 2 times. I plus 0 plus 0.Again, the traction vector is force per unit area and the total force is now the the total area times the magnet to the traction vector, so this is going to be equal to 1.Cool, I see your question. 

I'm getting back to it just a minute. So I want to make sure that this is intuitively clear to you.We have a traction vector that's itself that has a component only in one direction the other direction's are zero. So if I go back and I look at our image.

The traction vector is the big black arrow and it makes sense because it needs to offset the left hand side that it is only in the x direction the other two components are zero.

That is the y and z components.

Now, let me get back to coal it's not terribly critical that you memorize this equation. I think this is what you're now asking for.

This equation for n and it's not terribly critical that you memorize that you're gonna find that you're you're gonna move very quickly to doing this using either matrix operations or in what's called index notation in either case, you don't it's it's very infrequent that you would use it in this form this was derived this way so you could see that the normal component has contributions from each of the nine components, but no need to memorize that equation.

So Nicole says for notation, what is the difference between the square brackets and the squiggly ( really the way I'm using them the squiggly ) are representing a three by one vector and the square brackets are representing a three by three tensor, but that's not strict. That's just the way I'm representing them here.

It's not critical you could you could infer those from the equation if you chose to if you know what you're doing you can infer those the reason to do that is to make it a little bit more clear that we're talking about vector tensor vector, although I admittedly I'm a bit redundant because I also put the vector symbol over TN and over the normal so I've kind of,Made a hybrid notational scheme here where one is a vector, you know, I could actually put two arrows over the sigma representing a rank two tensor and it turns out that once you get more advanced structural mechanics, you're going to drop both of those notational schemes because they're a little bit lengthy and you'll you'll largely infer what the rank is from how the equations are being used.

Okay.It is now four o'clock.Which means our time has expired will return to doing more examples next lecture, that'll be Thursday.I'll stay on the line here for a little while so that I can take more questions if you have them otherwise. I will talk to you guys again on Thursday, hopefully I'll get some office hours visits Thursday as well, you know, I haven't seen you guys in office hours yet. 

I imagine that's because largely there hasn't been a real challenging homework set that's come out but you know don't hesitate to come in to office hours we're making appointment if you need to.Some of these things that we cover in lecture can be a little bit confusing and I want to make sure we get them clear early in the semester when they will give you the maximum benefit, so I'll see you guys. 

I'll hang out here for a few minutes but otherwise. I'll see you guys on Thursday.