 So, let's talk about failure criteria right now. Let's consider a uniaxial tensile specimen, and I know you already looked at that in developing a more circle when we looked at, if I section it in different directions, we will get different stresses. But let's look at a uniaxial tensile specimen in terms of how it fails, okay? So if we mount this uniaxial tensile specimen in a testing machine, so it's cylindrical specimen gripped at the two ends, and we load it under displacement control and plot force versus elongation, what we'll see is initially the force displacement behavior will be linear, we expect that, a linear elastic material. However, at some point it's going to plastically deform, and this isn't force versus strain or stress versus strain, it's force versus elongation, so be aware of that, but in a lot of materials we tend to see a strain hardening effect, so once it starts to plastically deform, we can still increase in the amount of force that that structure can carry. We call this strain hardening. At some point you're going to reach a maximum, necking will start to occur, so we will localize the deformation cross-section of that cylinder will reduce and the force begins to drop off, and that will eventually lead to complete failure of the specimen. Now if we take a close look at the actual failure mode there, and I have a few specimens zoomed up on that, we see that it's not failing in that plane of maximum principal stress. We already showed that at Mohr's circle for uniaxial tensile test means that the uniaxial applied stress is your maximum principal stress. We see that there is an angle, and you can actually measure it, and it is approximately plus or minus 45 degrees, so here's a nice cup, a sheer cup that is formed that is on a 45 degree, sometimes you get 45 through most of the specimen, so it's some combination of planes in the positive and negative 45 degree orientation. What is going on here is if we recall the Mohr's circle for uniaxial loaded, so we already looked at this before, your stress is P divided by A, so that's our maximum principal stress, stress in the y direction is zero, and so we get the following Mohr's circle with a radius of P divided by 2A, which happens to also be our maximum sheer stress, so it's just the Mohr's circle for that stress element. If we look at what's happening on that 45 degree plane, this is the point, this is the surface corresponding to the cross section through the cylinder, because that's where our uniaxial tensile stress is, and if we rotate an angle of 45 in physical space, we'll give us 90 degrees, sorry, 2 theta on the Mohr's circle, so this theta refers to the theta in physical space, obviously that's 90 degrees on the Mohr's circle, we see that at negative 45 or positive 45, we're at the maximum sheer stresses, so we get the maximum sheer stress, which in a uniaxial test, this isn't a general formula, this is just the formula for the uniaxial test, is sigma 1 divided by 2. We know that in a uniaxial test, we derive a stress allowable called the yield stress, so we look at what stress does yielding occur, that is a normal stress, so that occurs when sigma 1, because that's our stress in the x direction, is equal to our yield stress, so if we combine these two, we can say, well, sheer stress seems to be dominating that failure mode, our failure is occurring on a plus minus 45 degree plane, so failure is likely a result of sheer stress reaching a limit, so this is our engineering yield stress, we just do a test, calculate a stress, but it's not really related to the failure mode, the failure mode is related to sheer, so we can combine these and come up, this is the classical Tresca yield criterion that says, a material will fail when the maximum sheer stress in any orientation reaches the yield stress divided by 2. This is a failure criteria for ductile materials, what we see in here is we were taking, the yield stress is an engineering value, we do a test, we get that normal stress value, but then we look at how does the structure fail and try and better align those parameters, so in a ductile material, if you go back to your materials course, what is yielding? Yielding is dislocation movement, and dislocations move along consistent slip planes through sheer, so that deformation and failure mechanism is dominated by sheer stress, so it makes sense physically that sheer is driving that failure. If you have a brittle material, however, if I take a composite, well composites may be a bad example because it's got multiple components, take a ceramic, if I made a cylindrical specimen out of ceramic, pulled it in tension, it would break right in that perpendicular plane, there would be no plus minus 45, because sheer has absolutely nothing to do with the failure mode. Brittle materials fail just due to exceeding the pulling force between atoms, so it's when the normal stress reaches a certain maximum value, so that suggests another failure criteria, it's in the book but you don't have to memorize it or remember it, but you have a different failure criterion for that different class of material. Now, looking at this failure criterion, we can kind of come up with an interesting confusion that is often come across by students, and that is what happens if you have a biaxial loaded specimen. So when we had a uniaxial loaded specimen, we saw our Mohr's circle looks like this, our maximum sheer stress is just sigma x divided by 2, and so our yielding will occur when the sheer stress reaches the yield stress divided by 2. But how is the Mohr's circle going to look like in a biaxial loaded specimen? Well, if I look at this stress state, what can I say about it already? What property does it have? There's no sheer, so it's still the principal stresses. So it will look similar to this except sigma x still remains the same, I haven't put numbers but let's assume it's the same value, sigma y is half of sigma x over 2, so this principal stress is going to translate to the location of the center. So the Mohr's circle is going to shrink. If you think about the Tresca yield criterion, it's saying, well, when the radius of that Mohr's circle is half the yield stress divided by 2, we have failure. So this seems to indicate that if you biaxial load a structure, it's become stronger, which isn't very intuitive. This is where, although we do Mohr's circles in two dimensions, this is where you have to sometimes consider three dimensions. So if I think about this instead of as a square but as a cube with the z direction coming in and out of the screen, what is my stress in that direction? It's zero. We have no stress acting in and out of the page and there's also no sheer stress acting there. So I actually have a third principal stress that is equal to zero. And in any plane stress problem, that will always occur. We're looking at rotation within the plane, find the orientation of the axes where I have no shear. Always in that third dimension, I will have a principal stress of zero because of that plane stress state. Now in fully 3D problems, it gets a little bit more complex, that might have a value and again I said the Mohr's circle translates into a Mohr's sphere and it's difficult to visualize so we won't look at that. But that three dimensional aspect comes into play. So if we consider that as a cube rather than as just a simple square, we know that sigma z is equal to zero, that's our plane stress assumption and the shear stress is acting on this face are also zero. So that is the definition of a principal stress and we will obtain our third principal stress is equal to zero in plane stress problems. So what happens then is if we're considering the problem as three dimensional, we have three different planes that you can create a Mohr's circle for. This sort of should be yellow but the predictor is not so good. This face here is the yx plane which we've already calculated the Mohr's circle for. That was the sigma x and sigma y which happened to be sigma x over 2 and that gave us this small Mohr's circle. So that we already did. But you could then change your perspective to a different plane. So if we consider the z y plane, so now we're looking at this face, sigma y is there but sigma z is equal to zero. So it's an element that looks like this and it's also because we have no shear stresses in this case, our principal stress plane. So we have a maximum principal stress at sigma y which is this point here and in the other direction is zero. So we're going to get a second Mohr's circle that represents the y z plane. Now the more interesting one in this case is if you look at the x z plane, now we have our uniaxial load sigma x in the z direction we have nothing. This is indeed that original, before we look at the biaxial load, the original Mohr's circle. So principal stress of sigma 1, principal stress of 0 and we get a third Mohr's circle that has that maximum shear stress. But now our 45 degree rotation is not within the y x plane, it's within the x z plane. So it still fails at that 45 degree angle, it's just rotating in three dimensional space. So what we see from here, this is the exact, if we looked at just the uniaxial loaded case, this large Mohr's circle is the same one. So biaxial loading it didn't change the point at which it fails, our maximum shear stress will still be the same. But it changes the three dimensional orientation of that failure plane. So you have to be careful in looking at Mohr's circle if you're looking at this plane which is our original plane, we had this smaller Mohr's circle, but there was this larger one hidden because of our principal stress of 0. So how you identify that will depend on how you can visualize it. Some people like formulas as a crutch, so I always say effectively you have three possible maximum shear stress, you have to check which one is largest. It's always the difference of all the principal stresses and sigma 3 is always 0 for a plane stress problem which is what will limit ourselves to in this course. So if we go back to this, what if sigma y is equal to 0? As I said, this Mohr's circle that we produce is smaller, but we have these other two hidden Mohr's circles that intersect with our principal stress of 0 and the other two principal stresses. So when we look at that hidden principle, that hidden Mohr's circle, we still see that our maximum shear stress is sigma x divided by 2 so we would get failure at the same applied stress. So in fact the biaxial load doesn't change the stress in the x direction at which failure will occur, it just changes the orientation of the failure plane. This was the three shear stresses. What we can actually do is plot how this would look in the principal stress coordinate system and what I mean by that is if I make a plot showing values of sigma 1 and sigma 2, sigma 3 is always equal to 0 because this is for plane stress, we can create what is known as a failure surface. Failure surface is just a nice plot that you make. If you're inside the surface, then you're below your limit for failure. If you're outside of it, your structure has failed. So if we look at this criterion, tau max is sigma yield over 2. If I'm in the positive, sigma 1, sigma 2 quadrant, so both sigma 1 and sigma 2 are positive. If both of them are positive, I absolutely have this hidden Mohr's circle. So if I have that hidden Mohr's circle, it doesn't matter what the value of sigma 1 and sigma 2 are. I'm always going to have this maximum stress that is sigma 1, maximum shear stress that is sigma 1 divided by 2. Therefore I will have limits with either sigma 1 exceeding the yield or sigma 2 exceeding the yield. If either one of those exceeds yield, then I would fail. Similarly, if I look at the negative quadrant, I get the mirror image of this. So both my principal stresses are negative and if they're both negative, I will always have this hidden circle defined by the maximum principal stress divided by 2. So my limits will be when the maximum principal stress, whether it be sigma 2 or sigma 1, reaches yield. The other two quadrants are a little bit different. If we look at when sigma 1 is positive and sigma 2 is negative, the Mohr's circle we plot is the one where we have the maximum shear stress. The other two Mohr's circles are smaller. They're contained within that original Mohr's circle. Therefore my maximum shear stress will be sigma 1 minus sigma 2 divided by 2. Don't divide by 0, that breaks mathematics. So when that is equal to yield stress over 2, so we basically get that the absolute value of sigma 1 minus sigma 2 is equal to the yield stress. So that will give us a straight line defining failure. And similarly, we get the mirror image of it. They're both 1's positive, 1's negative. So the absolute value of 1 minus the other when that's equal to the yield stress, we have failure. So this produces this failure surface and really what you need to do is just consider whatever your principal stress 1 and 2 is, 3 is always 0 for plane stress problems. If I'm within the blue area of this, I'm fine. It's just a graphical representation of this formula. But I like to show it on because it shows the regions where these hidden Mohr's circles occur. So the hidden Mohr's circles will occur, or there's always hidden Mohr's circles, but the hidden one is always maximum if your two calculated principal stresses are non-zero and have the same sign. So in this quadrant and this quadrant. So that changes how you calculate the limit. So there is that Von Mises yield criterion for ductile materials. It's in the textbook if you're interested to look at it. There are brittle failure criterion for brittle materials that relate more to the normal stress. For composites, one of the most common ones is a side woo, or the side woo hill, or the side hill. There's many different variations on it. But there are many, many, many more, especially when you get into composite materials. There are hundreds of failure criteria trying to describe the complex failure in composites. So as I said, in this course, we are going to limit ourselves to the Tresca yield criterion. That's the only failure criterion I'll ask you in the exam. Just a word of a warning. When the old exams get posted up, in previous years we did also cover Von Mises stress. So you might see a question that references the Von Mises stress criterion. Just ignore that, okay?