In the last segment we've seen 3D linearized elasticity using coordinate notation, in strong form. What I'd like to do to start off this segment is for completeness put down the direct notation version of the strong form of 3D linearized elasticity. Okay? So, let's start then with linearized elasticity In three dimensions. Okay. Direct notation Okay, the setting is as before, we have our bases, e1 e2, e3. We have our body, right? Omega. We have our position vector of a point there. Right? And as we've done before, we observe that for boundary conditions, we have Dirichlet boundary conditions for each spatial dimension, right? So this is where we have u sub i equals u. G sub i for a specific i, right? And in the complimentary part, we have attraction boundary condition, right, or the Neumann boundary condition for a specific i. Okay, and remember that the whole business with the fact that we are now looking at a vector problem, right. Our, our solution the solution that we're looking for is a vector solution with three components. This is what specifies that when we look at the boundary of the body, we need to specify boundary conditions separately for each component. So, it being linearized elasticity, being an elliptic problem, it means that we do need to have boundary conditions at every point of the boundary but what can happen is that perhaps over the maze part of this boundary we specified maybe the u1 component of the, of displacement over the blue part of the boundary, which is complement to the maize part. We may be specifying therefore, the T1 component of the traction. All right? However over a different part, maybe this part of the boundary, we specify the u2 component of the, of the displacement, and on its compliment, which is facing me, we specify the t2 component of contraction. Observe that the way we've partitioned the boundary for the two different cases for spatial dimension 1 and 2 are overlapping, all right? Okay? And, and likewise for, for spatial dimension three, it may be a completely different sort of partitioning. Okay. Whoever, on the partitioning for each spatial dimension must satisfy the requirement that the Neumann and Dirichlet boundaries are disjoint. Okay, as I've indicated in the figure here. Okay. So we have, of course, that that the boundary is equal to the Dirichlet boundary union, the traction bond, the Neumann boundary and furthermore ,for each such decomposition, the intersection is disjoint, right? Sorry, the intersection is empty, right? Those boundary subsets are disjoined. Okay. Given all this, here is what we want to do. What we want to say, is, given, ug sub i, t bar sub i, the vector f. Now, I'm using direct notation for the vector f, okay? And the cosiderative relation sigma equals c, contracted with epsilon. Note that I'm using direct notation here before. The contraction of, the fourth order tensor, e is c, and there is, and there, the stream tensor epsilon. Okay, given all this, and the kinematic condition the kinematic relation. Epsilon equals the symmetric part of the gradient of u, okay? Given all of this, find. U, okay. Such that. Okay we have divergence of sigma plus f equals zero vector in omega, okay? And then we also have, e sub i equals ug sub i on that Dirichlet boundary, okay? Note that we cannot write out the Dirichlet boundary condition in direct notation simply because the boundary subsets are different and each component is specified. Possibly in a different boundary subset. Right, and the same thing happens for the traction condition. For the traction condition, we can however write sigma n, which is a vector. We're seeing that its ith component equals t bar i on partial omega t bar i, okay? All right one thing to note here is that This notation, delta sigma, is sometimes written as divergent of sigma, right? And, it is, sometimes also interpreted as taking this operator, the gradient operator, and dotting it with sigma. Okay, which is also the same thing that's expressed here. Okay. Alright, this is the end as far as direct notation is concerned. Perhaps the one other thing to remember, is that, when we write in direct notation, C contracted with epsilon, this is, a second order tensor, okay? This is a second order tensor with free indices i j, okay? And this is, by definition, this is C. I j k l epsilon k l. Okay. All right. So this is direct notation. This is an explanation of direct notation using coordinate notation. All right. Now, however, in order to continue, in order to develop our, formulation, I'm going to prefer to use coordinate notation because it makes all the manipulation of indices much more transparent. Right. Long before we get into that however, we need to see a little more about the constitutive relations. Okay, so let's do that. We have sigma, I'll, I'll use coordinate notation. Sigma i j equals C i j k l epsilon k l. Okay. Let's say a few things about this. C i j k l is a fourth-order tensor, it's called a fourth-order elasticity tensor. Okay, for linearized elasticity, it is constant with respect to. Wrt is short for with respect to. Right. C i j k l is constant with respect to epsilon. All right? With respect of the epsilon tensor. All right? So this is what we mean by linearized elasticity. And the context of linearized elasticity, this is what we get. Essentially what that means is that the relation between sigma and epsilon is a linear one. Okay? The relation on this, in this equation is a linear one. When we have linearized elasticity, and that's translated by the fact that C is a constant with retrospect to epsilon. Okay. I want to say a few more things about the properties of C. C has what we call as major symmetry. 'Kay? And it's important that we understand these properties betw, because we are going to use them we, develop our formulation. Okay. Major symmetry. What this means is that C i j k l equals C k l i j. Okay, and here's why. It follows. From. The fact. That. When one is dealing with linearized elasticity, linearized elasticity itself is built upon the idea that, there exists a function. That backward looking E is, the symbol for there exists, okay? There exists a function, psi, okay, which is a mapping from S3. S3 being the space of symmetric, second order tensor. Okay. There's a mapping from that space S3 to the space of positive release. Okay? All right? So S3 is symmetric second order tensors. It's, it's that space. Okay. Now, there exists a function psi which is mapping from S3 to R, it's in fact call, written as psi function of epsilon. Okay? And this is the strain energy density function. All right. We saw an explicit form of strain energy density function when we developed our variational methods for linear elasticity in 1D. All right, you consider it a quadratic form, there. We'll get into, into the particular form for psi in a, in a little bit, but let's just, for now, all we need to know is that this exists, okay? It can be shown then, that, right, one can show that C i j k l is the second derivative of psi with respect to epsilon i j, and epsilon k i. Okay? In fact, linearized elasticity, since it says that C is a constant, also says that, since C is constant, with respect to Epsilon, okay? It implies that psi is, at most, what? In fact, psi has to have a certain form. Can you think of it? Psi of epsilon is quadratic. Okay? So this second derivative of, psi with respect with epsilon, can be taken and, in general, it's not zero. Okay. Because C i j k l has that relation to the second derivative of psi with respect to epsilon, it follows that. C, I, J, K, L, which is equal to second derivative of psi with respect to epsilon I, J, epsilon K, L Is also secondary root of psi with respect of epsilon kl epsilon ij. What condition do we need to have that to be true? It is that in general psi is, let's just say here smooth enough. Okay, with respect to Epsilon. It can be more precise but for our purposes that's all we need to do. All right, if psi is smooth enough we can interchange the order of derivatives of epsilon, I, G and epsilon, K, L, right? But then if you look at the second, or rather the third member of this equation, this by definition is just Cklig. Right, and what we've called our sum major symmetry has just been proven. Okay? So, so the fact that C has major symmetry, it really follows from the existence of the strain energy function, strain energy density function, all right? It emerges however, that C has further symmetries, okay? C also has minor symmetries. Okay, which is that C, ij kl equals C, j i k l. Okay? And this follows because the stress Is symmetric. Okay, it turns out that the stress that we are dealing with here is properly called the Cauchy stress, but actually that's a detail that is somewhat lost in the context on linearized elasticity. Okay, it gets more important if we we're doing non-linear elasticity which we are not. Anyway, the Cauchy stress is symmetric. Okay? Which is that sigma i j equals sigma j i. Can you think of why that last relations holds? It follows from the balance of angular momentum, for this version of theory of elasticity. Okay? Well, if that is the case what we have is sigma ig which from our constitutive relation to cig kl epsilon kl and note that the kl and the Cs have been contracted out with the kl and the Cs of epsilon leaving free in the Cs ig. Okay? But this is also sigma ji which from definition is C ji kl, epsilon kl. Okay? These two conditions then imply that C ijkl equals C jikl, right, which is this condition of minor symmetry that we'd put forth, okay? C has the other kind of minor symmetry as well. Okay? So the other kind of minor symmetry is that C ijkl equals C ijlk. Okay? And why does this come about? This comes about because epsilon kl is one half, by definition, it's one half uk comma l plus ul comma k, right, and remember what that is, that is just one half derivative of uk with respect to xl, which is the first element of the parenthesis, plus derivative of ul with respect to xk, okay? Well, if this is the case, it follows that epsilon kl equals epsilon lk, okay? But then what that says is that sigma ij, which is C ig kl epsilon k l, is also C ij kl epsilon lk. Right? Okay? Now, if you look at, the order of the indices here. Because of this contraction, right? Okay? And recognizing that l and k are just dummy indices. Okay, it follows that this last term can equally be written as C ij, okay, lk, epsilon kl, all I've done there is to switch the order of the, switch the position of the key and l indices. Okay, it doesn't matter because they are dummy indices and they are being contracted out anyway. But then if you compare this and that, okay, it implies that C ij kl equals C, ij lk. Okay. Which is what we'd set out to prove for the second of our minor symmetries, right, this one. Okay. That sort of sums up what we needed to see about the symmetries of our elasticity tensor c, we will use, these symmetries in a very important way when we develop our formulations for finite elements. All right, we'll end the segment here.
