MAE Continuum Mechanics Course Notes

Transcription

1 Course Notes Brandon Runnels Contents LECTURE 1 0 Introduction Motivation Notation Sets Proof notation Tensor Analysis Index notation and the Einstein summation convention Vector equality Inner product Kronecker delta Components of a vector Norm of a vector Permutation tensor Cross product Notation LECTURE Mappings and tensors Second order tensors Index notation Dyadic product Tensor components Higher order tensors Transpose Trace (first invariant) Determinant (third invariant) Inverse The special orthogonal group Tensor calculus Gradient LECTURE Divergence Laplacian Curl Gateaux derivatives Notation Evaluating derivatives The divergence theorem LECTURE Curvilinear coordinates All content , Brandon Runnels 1

2 Course Notes The metric tensor Orthonormalized basis Change of basis Calculus in curvilinear coordinates Gradient Divergence Curl Tensor transformation rules LECTURE 5 2 Kinematics of Deformation Eulerian and Lagrangian frames Time-dependent deformation LECTURE The material derivative Kinematics of local deformation Metric changes Change of length Change of angle LECTURE Determinant identities Change of volume Change of area Covariance and contravariance of vectors Tensor decomposition Eigenvalues LECTURE Symmetric and positive definite tensors Spectral theorem (symmetric tensors) Spectral theorem (general case) Functions of tensors Polar decomposition LECTURE Principal deformations Compatibility Continuous case LECTURE Discontinuous case (Hadamard) Other deformation measures Linearized kinematics Linearized metric changes LECTURE Small strain compatibility The spatial/eulerian picture LECTURE 12 3 Conservation Laws Conservation of Mass Control volume Conservation of linear momentum Forces, tractions, and stress tensors LECTURE Balance laws LECTURE Conservation of angular momentum LECTURE Conservation of energy Energetic quantities All content , Brandon Runnels 2

3 Course Notes LECTURE Balance laws Power-conjugate pairs Second law of thermodynamics Introduction to statistical thermodynamics and entropy LECTURE Internal entropy generation Continuum formulation Review and summary LECTURE 18 4 Constitutive Theory Introduction to the calculus of variations Stationarity condition LECTURE Variational formulation of linear momentum balance Material frame indifference Elastic modulus tensor Elastic material models Useful identities Pseudo-Linear Compressible neo-hookean LECTURE Internal constraints Review of Lagrange multipliers Examples of internal constraints Lagrange multipliers in the variational formulation of balance laws LECTURE Linearized constitutive theory Major & minor symmetry and Voigt notation Material symmetry The Cauchy-Navier equation and linear elastodynamics Thermodynamics of solids and the Coleman-Noll framework LECTURE 22 5 Computational mechanics The finite element method Shape functions Weak formulation Numerical quadrature LECTURE Linearized kinematics D linearized elasticity Newton s method Finite kinematics Computational fluid dynamics About These notes are for the personal use of students who are enrolled in or have taken MAE5100 at the University of Colorado Colorado Springs in the Spring 2015 semester. Please do not share or redistribute these notes without permission. All content , Brandon Runnels 3

4 Course Notes Nomenclature A a a i α B B b β β C C d δ ij E E() e i ε ɛ ijk g GL(n) F G G i g i Lagrangian acceleration Eulerian acceleration Covariant basis vectors The rotation vector For all The left Cauchy-Green deformation tensor Material body force Spatial body force The displacement gradient tensor Reciprocal temperature The right Cauchy-Green deformation tensor The set of complex numbers subset of The rate of strain tensor The Kronecker delta The Green-Lagrange strain tensor Internal Energy Standard Cartesian unit vectors The small strain tensor The Levi-Civita alternator there exists in Metric tensor The general linear group in n dimensions. The deformation gradient Angular momentum vector Unit vectors for undeformed configuration Unit vectors for deformed configuration H, h Outward heat flux vectors J K L(R m, R n ) l λ The Jacobian Kinetic energy The set of m n tensors The spatial velocity gradient tensor Stretch ratio {λ i } Principal stretches M Moment {N I } Material principal directions {n i } Spatial principal directions O(n) P P D () P E () φ Q R R n r ρ S S S n, s n SO(n) σ T Θ Orthogonal group A set, typically R 3, denoting a body The microcanonical partition function The boundary of a body The first Piola-Kirchhoff stress tensor Deformation power External power Deformation mapping Heat Lagrangian mass density The set of n-dimensional vectors The infinitesimal rotation tensor Eulerian mass density The second Piola-Kirchoff stress tensor Entropy Internal heat generation Special orthogonal group (rotation tensors) The Cauchy stress tensor Temperature (Lagrangian frame) Temperature (Eulerian frame) U, u Material and spatial intensive internal energy u V v w ω X x Z Displacement Lagrangian velocity Eulerian velocity The spin tensor The vorticity vector Material position Spatial position The set of integers All content , Brandon Runnels 4

5 Lecture 1 Introduction and index notation 0 Introduction Welcome to continuum mechanics. In this course we will develop the mathematical framework for describing precisely the deformation of solids, fluids, and gasses, and describing the physical laws that govern their motion. We will develop the mathematical formulation of the equations of motion, elasticity, viscoelasticity, plasticity, etc. The course will be organized in the following way: (1) Tensor analysis: index notation, tensor algebra and calculus, curvilinear coordinates and transformation rules. (2) Kinematics of deformation: deformation mappings, local deformation, metric changes, decompositions, compatibility, linearized kinematics. (3) Balance laws: conservation of mass, linear momentum, angular momentum, energy. (4) Constitutive modeling and the thermodynamics of solids: constitutive models, second law of thermodynamics and dissipative systems. (5) Computational mechanics: finite elements, Galerkin method, Ritz method. 0.1 Motivation Consider a bar subjected to a tensile load as shown in the following figure. How do we describe this process? Undeformed l 0 Governing equations Kinematics: ɛ = l l0 l 0 Deformed l A f Balance law: σ = f A Constitutive model: σ = E ε The above equations are fairly straightforward for this simple system, and we are familiar with them from statics and mechanics of materials. But we are in the business of mechanics of bodies with arbitrary shape, loading, constraints, etc: Undeformed Deformed What is ε, σ, E for this complex case? How do we formulate our equations of kinematics, balance laws, and consitituve models here? Can we use what we know about the system to determine what the deformed configuration is under applied loads and displacements? All content , Brandon Runnels 1.1

6 Course Notes - Lecture Notation Continuum mechanics is built on the mathematical framework of differential geometry. As a result, the convention is to use standard math notation when formulating continuum mechanics. Additionally, as we will see, it is generally necessary to maintain a level of mathematical precision beyond that typically found in engineering disciplines Sets A set is a collection of objects. Examples: The integers Z = {..., 1, 0, 1, 2, 3,...} The real numbers R The complex numbers C n-dimensional vectors R n To indicate that an item is in a set, we use the symbol. For instance, x R 3 (0.1) indicates that x is a 3D vector. To indicate that a set is a subset, we use the symbol. For instance Z R (0.2) indicates that the integers are a subset of the real numbers. Another common use is to denote a 3D body: is an arbitrary region in 3D space Proof notation R 3 (0.3) We will not be doing any serious proofs in this course, but we frequently use some of the proof notation to simplify definitions and theorems. is read there exists. For instance, x R 3 states that there is at least one item in the R 3 set, or that the R 3 set is not empty. is read for all. For instance, Example: tells us that the statement x x = 0 is true for every possible vector. x x = 0 x R n (0.4) x R 3 a R 3 s.t. x a = 0 (0.5) can be read for all 3D vectors there exists another 3D vector such that their difference is equal to zero. 1 Tensor Analysis Let us consider the space of three dimensional vectors, R 3. A vector r R 3 can be represented in two different ways: in terms of its components, or in terms of basis vectors {g 1, g 2, g 3 } All content , Brandon Runnels 1.2

7 Course Notes - Lecture 1 z c [ ] ab r = c ĝ 3 r = r 1 ĝ 1 + r 2 ĝ 2 + r 3 ĝ 3 a b y ĝ 2 x ĝ 1 For example, we may have [ ] 10 g 1 = 0 [ ] 01 g 2 = 0 [ ] 00 g 3 = 1 (1.1) where we see that the basis vectors correspond to the familiar î, ĵ, ˆk notation. For maximum generality, however, we do not define the basis vectors explicitly. In subsequent sections will talk about changes of basis. Note also that we have dropped the familiar x, y, z notation in favor of 1, 2, 3. In general, we will stick with this convention exclusively; the reason for this will become apparent in the next section. 1.1 Index notation and the Einstein summation convention Let us consider r defined in the previous section as We can write this more simply using summation notation: r = r 1 g 1 + r 2 g 2 + r 3 g 3 (1.2) r = 3 r i g i (1.3) i=1 It turns out that we write sums like this a lot, and it becomes cumbersome to write the summation symbol every time. Thus, we introduce the Einstein summation convention, and we drop the explicit sum. This allows us to simply write r = r i g i (1.4) This leads us to define the rules of the summation convention. For a vector equation in R n, expressed using index notation: Rule 1: Rule 2: An index appearing once in a term must appear in every term in the equation, and is not summed. It is referred to as a free index. An index appearing twice must be summed from 1 to n. It is referred to as a dummy index. (Dummy indices can be changed arbitrarily, that is, e.g. r i g i = r j g j ) Rule 3: No index may appear more than twice in any term. (If an index does appear more than twice, we go back to using a summation symbol. Alternatively, if an index appears twice but is not a dummy index, we use parentheses to denote this, e.g. u i = λ (i) v (i). Fortunately, these cases are pretty rare. Usually, when a rule gets broken, it means that some algebra got messed up.) These rules may seem a bit strange, and they usually take a little bit of time to get used to. To help solidify them, let us look at a couple of algebraic examples. (Fun fact: the Einstein summation convention was introduced by Albert Einstein to simplify the equations of general relativity. In fact, the formulation of continuum mechanics has a number of similarities to general relativity.) All content , Brandon Runnels 1.3

8 Course Notes - Lecture Vector equality Consider two vectors u, v R n with components u 1, u 2,..., v 1, v 2,... In invariant/symbolic notation, we say the two vectors are equal if u = v or u i g i = v i g i (1.5) This tells us that each component of the vector is equal; in other words, u i = v i (1.6) Does this obey the summation convention? Yes it does: i is a free index that appears exactly once in every term of the equation Inner product Let us again consider u, v R n. The inner product (or dot product ) is defined as u v = u T v = u v cos θ (1.7) where is the magnitude of and θ is the angle between the two vectors. In matrix notation, we evaluate this as v 1 u T v = [u 1 u 2... u n] v n 2... = u 1 v 1 + u 2 v u n v n = u i v i = u i v i (1.8) v n i=1 Note that there are no free indices, only dummy indices Kronecker delta The Kronecker Delta is defined in the following way: δ ij = { 1 i = j 0 i j (1.9) Consider the basis vectors that we described above. We know that they are orthonormal, so g i g j is 1 if i = j and 0 otherwise; that is, g i g j = δ ij (1.10) Let us use this technology in the context of the dot product. Let u = u i g i, v = v i g i. Then we might write the dot product as u v = (u i g i ) (v i g i ) (1.11) But wait: this breaks one of our rules, that an index cannot repeat more than twice. To fix this, we will replace the is in the second term with js: Now, let us distribute these terms: u v = (u i g i ) (v j g j ) (1.12) u v = u i v j (g i g j ) = u i v j δ ij (1.13) This term has two summed indices, so if we expand it out, we would have n 2 terms. However, we know that only the terms where i = j survive. Thus, the effect of the Kronecker Delta is to turn one of the dummy indices into the other: in this case, if we sum over j u i v j δ ij = u i v i (1.14) All content , Brandon Runnels 1.4

9 Course Notes - Lecture Components of a vector To extract a specific component of a vector, we can dot with the corresponding unit vector. That is: Norm of a vector As you may recall, the norm of a vector is given by In index notation, this becomes Permutation tensor u g i = (u j g j ) g i = u j (g i g j ) = u j δ ij = u i (1.15) u = u u (1.16) u = u i u i (1.17) The next order of business is to introduce the cross product in tensor notation. To work with cross products, we need to introduce a new bit of machinery, called the permutation tensor, also referred to as the Levi-Civita tensor. (Note: it s not actually a tensor. However, it is frequently referred to as one, so we will stick with convention here.) Here it is: 1 ijk = 123, 231, 312 = "even permutation" ɛ ijk = 1 ijk = 321, 132, 213 = "odd permutation" (1.18) 0 otherwise Let s make a couple of notes here: ɛ ijk is zero if any of the two indices take the same value. Flipping two indices changes the sign of ε; that is, e.g. These identities will come in handy in the future Cross product Let us consider the cross product of unit vectors. We know that ɛ ijk = ɛ jik = ɛ ikj = ɛ kji (1.19) g 1 g 2 = g 3 g 2 g 1 = g 3 (1.20) g 2 g 3 = g 1 g 3 g 2 = g 1 (1.21) g 3 g 1 = g 2 g 1 g 3 = g 2 (1.22) Let us attempt to express this using the permutation tensor. Try: Does this work? Let s plug in i = 1, j = 2. Then we have g i g j = ɛ ijk g k (1.23) g 1 g 2 = ɛ 12k g k = 0 ɛ 121 g ɛ 122 g 2 + ɛ 123 g 3 = g 3 (1.24) as expected. Plugging in other values for i, j shows that we can recover all of the identities expressed above. Now, let us see what happens when we take the cross product between u, v: Alternatively, u v = (u i g i ) (v j g j ) = u i v j (g i g j ) = u i v i ɛ ijk g k (1.25) (u v) k = ɛ ijk u i v j (1.26) All content , Brandon Runnels 1.5

10 Course Notes - Lecture Notation Let us clarify some of the notation that we have been using: In general: Invariant/Symbolic Notation Full Component Notation Termwise Index Notation Equality u = v u i g i = v i g i u i = v i Dot product u v u i v i u i v i Cross Product u v ɛ ijk u i v j g k ɛ ijk u j v k Example (u v) w u k v k w i g i u k v k w i Invariant/symbolic notation is independent of coordinate system, which means that invariant expressions are more general. However, there are some operations that are too complex to be represented in invariant notation, and it can more easily get confusing. Full component notation is slightly less general than invariant notation, but is the best for working in almost any coordinate system, especially ones with non-constant unit vectors. Termwise index notation is very nimble and convenient when working in a constant, orthonormal coordinate system. This is frequently what we use, so we will use it a lot. However, it is dangerous to use when unit vectors are non-constant. All content , Brandon Runnels 1.6

11 Lecture 2 Mappings and tensors 1.2 Mappings and tensors A mapping is a machine that takes a thing of one type and turns it into a thing of another type. For instance, f (x) = x 2 takes a real number and turns it into a positive real number. We use the notation to denote a mapping; in this case, if x U, then f (x) V. A linear mapping f : R n R m is a mapping that satisfies f (α x) = α f (x) x R n, α R f : U V (1.27) f (x + y) = f (x) + f (y) x, y R n Second order tensors A second order tensor (S.O.T.) is a linear mapping from vector spaces to vector spaces. The set of second order tensors mapping n-dimensional vectors to m-dimensional vectors is referred to as L(R n, R n ). We are familiar with thinking of them as n m matrices: For example, if A L(R n, R m ) and u R n, v R m then we could write v 1 A 11 A A 1n u 1 v 2 A 21 A A 2n u 2 v = A u. v m = A m1 A m2... A mn. u n (1.28) We can write all possible linear mappings from R n to R m in matrix form. Therefore, in general, we can think of second order tensors as being similar to matrices. Let us make a few notes: If m = n the matrix is said to be square. For the most part, we will work with square matrices. If u = A u then A = I is said to be the identity mapping. The difference between tensors and matrices is subtle. A matrix is just a collection of numbers, but a tensor is something that must be transformed with a change of coordinates. Thus we say that tensors have transformation properties but matrices do not Index notation Index notation makes it very convenient to write second order tensors and tensor-vector multiplication. In the above example, we have v 1 A 11 u 1 + A 12 u A 1n u n v 2 A 21 u 1 + A 22 u A 2n u n and so we can write =. v m. A m1 u 1 + A m2 u A mn u n (1.29) v i = A ij u j (1.30) All content , Brandon Runnels 2.1

12 Course Notes - Lecture Dyadic product We have expressed tensor-vector multiplication using invariant notation and termwise index notation. How can we express a tensor using full component notation? To do this, we introduct the dyadic product: u 1 u 1 v 1 u 1 v 2... u 1 v n u v = u v T u 2 = [v 1 v 2... v u 2 v 1 u 2 v 2... u 2 v n n] = (1.31). u n u n v 1 u n v 2... u n v n In tensor notation, we simply write (u v) ij = u i v j (1.32) We can use the dyadic product with unit vectors to extract a specific component of a tensor. That is, is the zero matrix except for a 1 in the ij column. For example: [ ] 01 g 2 g 3 = [0 0 1] = 0 Therefore, we can express a tensor A as: How do we write a tensor operating on a vector? Suppose A acts on v: g i g j (1.33) [ 0 0 ] (1.34) A = A ij g i g j (1.35) Av = (A ij g i g j )(v k g k ) = A ij v k (g i g j )g k (1.36) What do we do with this? Recall that we can write u v as uv T. Then we have Substituting, we get (g i g j )g k = g i g T j g k = g i (g j g k ) = g i δ jk (1.37) Av = A ij v k g i δ jk = A ij v j g i (1.38) as expected. We will take this opportunity to reiterate the identity we described earlier: namely, for u, v, w R n, This is easily seen using index notation: Tensor components We can extract components of a tensor A in the following way: (u v) w = u (v w) (1.39) (u i v j )w j = u i (v j w j ) (1.40) g i A g j = g i (A pq g p g q ) g j = A pq g i (g p g q ) g j = A pq g i g p (g q g j ) = A pq δ ip δ jq = A ij (1.41) Higher order tensors We can express higher order tensors in the following way: A = A ij...k g i g j... g k (1.42) When we start working with contitutive theory, we will frequently see fourth-order tensors (the elasticity tensor). I don t know of any cases where we work with anything higher than fourth order. All content , Brandon Runnels 2.2

13 Course Notes - Lecture Transpose The transpose of a tensor is identitcal to the transpose of a matrix: the ij term is swapped with the ji term. How can we express this in index notation? A tensor A is called symmetric iff A tensor is called antisymmetric iff (A T ) ij = A ji (1.43) A ij = A ji (1.44) A ij = A ji (1.45) (Note that the diagonal terms in an antisymmetric tensor must be zero.) Any tensor can be decomposed into its symmetric and antisymmetric parts in the following way: A = 1 2 (A + A) (AT A T ) = 1 2 (A + AT ) + 1 }{{} 2 (A AT ) }{{} symmetric antisymmetric (1.46) Trace (first invariant) The trace of a tensor is defined in the folowing way. For u, v R n, Let A L(R n, R n ). Then the trace is given by tr(u v) = u v (1.47) tr(a) = tr(a ij g i g j ) = A ij tr(g i g j ) = A ij (g i g j ) = A ij δ ij = A ii (1.48) One can think of this as the sum of the diagonal terms in the tensor. Note: the trace of a tensor is called the first invariant of the tensor. This means that the trace of the tensor does not change under rotation. The significance of this will become apparent later on Determinant (third invariant) A quantity that we use frequently is the determinant of a tensor. For A L(R 3, R 3 ), A 11 A 12 A 13 det(a) = A 21 A 22 A 23 A 31 A 32 A 33 = A 11A 22 A 33 + A 12 A 23 A 31 + A 13 A 21 A 32 A 11 A 23 A 32 A 12 A 21 A 33 A 13 A 22 A (1.49) 31 How can we represent this using index notation? In 3D, we can write it as Alternatively, we can write it in a slightly more satisfying way as Some things to note: for A, B L(R n, R n ) det(a) = ɛ ijk A 1i A 2j A 3k (1.50) det(a) = 1 6 ɛ ijkɛ pqr A ip A jq A kr (1.51) det(a) = det(a T ) (1.52) det(ab) = det(a) det(b) (1.53) In three dimensions, the determinant is the third invariant of the tensor, which means it is the third of three important quantities that do not change under rotation. All content , Brandon Runnels 2.3

14 Course Notes - Lecture Inverse Let A L(R n, R n ). The inverse of A, A 1 satisfies, u R n, (A 1 )A u = I u = u (1.54) If the inverse of a tensor exists it is said to be invertible. One can prove that a tensor A is invertible iff det(a) 0. In index notation, For composite mappings, A 1 ij A jk = I ik = δ ik (1.55) (AB) 1 = B 1 A 1 (1.56) What is the determinant of the inverse of a tensor? For A defined above, we know that AA 1 = I. So we can say that det(i) = 1 = det(aa 1 ) = det(a) det(a 1 ) (1.57) so det(a 1 ) = 1 det(a) (1.58) What if det(a) = 0? Then A 1 cannot exist, so its determinant is naturally ill-defined The special orthogonal group Consider the set of tensors A L(R 3, R 3 ) such that A T A = I. We call this group of tensors the orthogonal group, and we denote it as S(n) where n is the dimensions. So for A S(n) we see that A 1 = A T 1/ det(a) = det(a 1 ) = det(a T ) = det(a). This implies that det(a) = ±1 Now, let us consider only those tensors in the orthogonal group that satisfy det(a) = 1. We will call this the special orthogonal group, and denote it by SO(d). It turns out that SO(d) is exactly the same as the group of all rotation matrices. In 3D, we will refer to tensors that are in SO(3) very frequently. 1.3 Tensor calculus Tensor calculus is the language of continuum mechanics. So far we have talked a lot about vectors and tensors. Now, we are going to talk about vector and tensor fields. There are three kinds of fields that we will use a lot: Scalar fields f : R n R; e.g. temperature, pressure Vector fields v : R n R n ; e.g. displacement, velocity Tensor fields T : R n L(R n, R n ); e.g. stress, strain We will be looking at a wide variety of differentiation operations on these types of fields. Note: for this section, we are assuming that we are working in a constant Cartesian basis. That is, we assume that the basis vectors g i are constants. This is not always true! When we work with curvilinear coordinates, we will have to be very careful about taking derivatives of basis vectors. All content , Brandon Runnels 2.4

15 Course Notes - Lecture Gradient Suppose we have a scalar field φ : R n R, φ(x). We define the gradient of φ to be grad(φ) = φ x 1 g 1 + φ x 2 g φ x n g n = φ x i g i (1.59) Note that gradient operator turns a scalar field into a vector field. Now consider a vector field u : R n R n. The gradient of u is defined to be Note that the gradient operator here turns a vector field into a tensor field. We can generalize this in the following way: grad(u) = u x i g i = u i x j g i g j (1.60) grad( ) = ( ) x i g i (1.61) where we drop the dyadic product if is a scalar field. In general, we only generally care about gradients on scalar and vector fields. However, there are some models that depend on tensor field gradients, such as strain gradient plasticity and ductile fracture. All content , Brandon Runnels 2.5

16 Lecture 3 Tensor calculus Divergence For a vector field u : R n R n, the divergence is defined to be div(u) = u x i g i = u j x i g j g i = u j x i δ ij = u i x i (1.62) Note that the divergence turns a vector field into a scalar field. Now, consider a tensor field T : R n L(R n, R n ). The divergence of the tensor field is div(t ) = dt g i = dt jk dx i dx i (g j g k )g i = dt jk dx i g j (g k g i ) = dt jk g j δ ik = dt ji g j (1.63) dx i dx i Note that the divergence turns a tensor field into a vector field. We can generalize this in the same way we generalized the gradient: div( ) = ( ) x i g i (1.64) where we drop the dot for tensor fields. Note also that we cannot take the divergence of a scalar field Laplacian The Laplacian is the composition of the gradient and the divergence operators on a scalar field: for φ : R R: In Cartesian coordinates, this comes out to be Note that we do not write x 2 i Curl φ = div(grad(φ)) (1.65) φ = 2 φ x i x i (1.66) in order to be in keeping with the summation convention. The curl operator is defined on a vector field u : R 3 R 3 in the following way: curl(u) = u x i g i = u j x i g j g i = u j x i ɛ jik g k = u j x i ɛ ijk g k (1.67) Note that the curl acts on a vector field and produces a vector field, so we cannot take the curl of a scalar field. In generalized form we have Gateaux derivatives curl( ) = ( ) x i g i (1.68) A more general type of derivative is the Gateaux derivative, which will prove very useful later on. Consider some field (scalar, vector, or tensor, etc.) φ : R n V (where V is the set of scalars, vectors, tensors, etc.) and a vector v R n. The Gateaux derivative is defined as that is, the derivative is taken with respect to ε which is then set to 0. Dφ(x) v = d dε φ(x + εv) ε 0 (1.69) All content , Brandon Runnels 3.1

17 Course Notes - Lecture 3 Example 1.1 Let φ(x) = x i x i = x 2, and compute Dφ(x) v. We evaluate this simply by substituting φ into (1.69): d Dφ(x) v = lim ε 0 dε ((x d i + εv i )(x i + εv i )) = lim ε 0 dε (x ix i + 2εx i v i + ε 2 v i v i ) = lim (2x i v i + 2εv i v i ) = 2x i v i (1.70) ε Notation Here it is important to make a couple of remarks about notation. (1) We avoid the use of the operator (e.g. for divergence, for curl) because it is difficult to impossible to express certain vector operations. (2) We work with a lot of derivatives in continuum mechanics and it frequently becomes cumbersome to write x i. Therefore, we adopt comma notation: x i ( ) = ( ),i (1.71) For example, we can write grad/div/curl compactly for scalar/vector/tensor fields φ, v, T grad(φ) i = φ,i div(v) = v i,i curl(v) i = ɛ ijk v k,j φ = φ,ii (1.72) grad(v) ij = v i,j div(t ) i = T ij,j (1.73) (3) We will occasionally use the symbol to denote differentiation. Examples of usage include: x x Evaluating derivatives θ θ i x i (1.74) How do we evaluate a derivative with respect to x in terms of x? For instance, how would we compute the gradient of φ(x) = x x? In index notation, we have x i x j = δ ij Let us look at a couple of examples: Example 1.2 or, in symbolic notation, x x = I (1.75) Compute the gradient of φ(x) = x x. The first step is to write φ in index notation: φ(x) = x k x k. Then, we use the formula: grad(φ) = (x k x k ) g i (1.76) x i We can use the product rule exactly like we would normally: = (δ ik x k + x k δ ik ) g i = 2δ ik x k g i (1.77) Summing over i we get = 2x i g i = 2x (1.78) All content , Brandon Runnels 3.2

18 Course Notes - Lecture 3 Example 1.3 Let φ : R 3 R. Show that Dφ(x) v = grad(φ) v. To do this we again evaluate the Gateaux derivative: We use the chain rule exactly as we would normally: φ(x + ε v) = lim ε 0 (x i + ε x i ) d Dφ(x) v = lim φ(x + ε v) (1.79) ε 0 dε d(x i + ε v i ) dε φ(x + ε v) = lim ε 0 (x i + ε x i ) v i = φ v i = grad(φ) v (1.80) x i In addition to taking derivatives with respect to vectors (such as x) we may take derivatives with respect to tensors. For example, given φ : L(R n, R n ) R, we may wish to calculate Similarly to the case of vectors, we have Example 1.4 dt ij dt pq = δ ip δ jq dφ(t ) dt ij (1.81) or, in symbolic notation, Let A : L(R n, R n ) L(R n, R n ) with A(T ) = T T T. Find the derivative of A with respect to T : dt dt = I I (1.82) A ij = (T ki T kj ) = T ki T kj T kj + T ki = δ kp δ iq T kj + T ki δ kp δ jq = T pj δ iq + T pi δ jq (1.83) T pq T pq T pq T pq How do we represent this in symbolic notation? It s actually rather tricky, and is much easier to leave things in index notation. Note: when taking vector or tensor derivatives, it is always important to use a fresh new free index. Do not reuse existing ones. 1.4 The divergence theorem We have developed enough machinery to introduce the single most important theorem in all of continuum mechanics: the divergence theorem. V(x) n Theorem 1.1 (The divergence theorem). Let R n, and V : R 3 R 3 be a differentiable vector field: Then: div(v) dv = V n da V i,i dv = V i n i da (1.84) All content , Brandon Runnels 3.3

19 Course Notes - Lecture 3 where is the boundary of the body. We have a similar theorem for tensor fields: Let T : R 3 L(R 3, R 3 ) be a tensor field. Then div(t ) dv = T n da (1.85) Using index notation in an Cartesian coordinate system, we can write V i,i dv = V i n i da T ij,j dv = T ij n j da (1.86) What does this mean? We are relating a volume integral of the divergence to a surface integral or, in this case, a flux integral. For the case of a vector field, the integral of the divergence over the body can be intuitively thought of as the total amount of compression/expansion in the vector field. The flux integral can be thought of as the total amount of vector field entering or leaving the body. Thus, you can think of the divergence theorem as the mathematical formulation of the statement the total compression of the vector field is equal to the rate of flux through the boundary. We will make extensive use of the divergence theorem in this course. All content , Brandon Runnels 3.4

20 Lecture 4 Curvilinear coordinates and tensor transformations 1.5 Curvilinear coordinates Up until now we have worked within a simple Cartesian basis, let us call it e i. However, it is frequently convenient to switch to a more natural coordinate system: ĝ 3 ĝ θ ĝ r r = r 1 ĝ 1 + r 2 ĝ 2 + r 3 ĝ 3 ĝ 2 ĝ 1 Consider a new set of coordinates {θ 1, θ 2, θ 3,..., θ n }. Position as a function of these coordinates is expressed as x(θ 1, θ 2,..., θ n ) (1.87) Let us define a new basis: {a 1, a 2,..., a 3 } defined as a 1 = x θ 1 a 2 = x θ 1... a i = x θ i = x j θ i e j (1.88) We will refer to {a i } as the covariant basis vectors The metric tensor The metric tensor g is defined as Notes: g ij = a i a j (1.89) The metric tensor is symmetric If the metric tensor is diagonal then the new coordinate system {a i } is said to be orthogonal If g = I then the coordinate system is said to be orthonormal Orthonormalized basis There is no guarantee that our new basis {a i } will be normalized, that is, we don t know that a i = 1. But we can make sure that they are by defining scale factors h i = a i. Then we define a new basis g i = a (i) h (i) no summation over i (1.90) Example 1.5 Cylindrical Polar Coordinates: we can specify any point using x 1, x 2, x 3, but we can also specify it using the coordinates r, θ, z. All content , Brandon Runnels 4.1

21 Course Notes - Lecture 4 z r θ Let us compute x(r, θ, z): x 1 = r cos θ x 2 = r sin θ x 3 = z (1.91) Now we can compute our basis vectors: [ ] a r = x i cos θ r e i = sin θ 0 Our scale factors are a θ = x i θ e i = [ ] r sin θ r cos θ 0 a z = x i z e i = [ ] 00 1 (1.92) h 1 = a r = 1 h 2 = a θ = r h 3 = a z = 1 (1.93) so we have g r = [ ] cos θ sin θ 0 g θ = [ ] sin θ cos θ 0 [ ] 00 g z = 1 (1.94) Important note: while {e 1, e 2, e 3 } are independent of x 1, x 2, x 3, {g i } is not necessarily independent of {θ i }. In our above example, we see that [ ] [ ] sin θ θ g cos θ r = cos θ = g θ 0 θ g θ = sin θ = g r (1.95) Change of basis Let {g i } be an orthonormal basis for R n, and let v R n. Suppose we wish to find {v i } such that v = v i g i. To do this, we use the orthogonality property of the basis: v g j = v i g i g j = v i δ ij = v j = v = (v g i ) g j (1.96) Suppose we have another basis {e i }. Then we can relate the two bases by writing e i = (e i g j ) g j g i = (g i e j ) e j (1.97) These relationships will be useful as we start discussing curvilinear coordinates. 1.6 Calculus in curvilinear coordinates Now that we have defined a framework for working in other coordinate systems, we need to know what our calculus operations look like in those systems. Before we do that, however, we need to introduce a couple of important identities. (1) Earlier we learned that g i = 1 a (i) = 1 x j e j no sum on i (1.98) h (i) h (i) θ (i) All content , Brandon Runnels 4.2

22 Course Notes - Lecture 4 How can we express e i in terms of {g j }? To do this, we ll pull a trick. Remember that {g i } forms an orthonormal basis for R n : that means that we can express any vector in terms of that basis. For instance, a vector v can be written as v = v i g i. How do we find v i? It s nothing other than v g i. So we can write v = (v g i ) g i (1.99) We can do exactly the same thing for our original basis vectors e i = (e i g j ) g j = j 1 h j (e i a j ) g j = j 1 h j (e i x k θ j e k ) g j = j 1 x k (e i e k ) g j = h j θ j }{{} δ ik j 1 h j x i θ j g j (1.100) (Note that we broke one of our summation convention rules. To compensate for this we drop the summation notation and use an explicit sum.) (2) There is an important theorem called the inverse function theorem that states: [ θ ] = x [ x ] 1 θ x = θ x θ = I or, in index notation, θ i x k = δ ij (1.101) x k θ j We will use both of these rules to derive expressions for the familiar divergence, gradient, and curl in curvilinear coordinates Gradient We want to express the gradient as computed in the above grad(f (θ)) = (f (θ)) e i = f θ j e i = f θ ( j 1 x ) i g k = x i θ j x i θ j x i h k θ k k k 1 f θ j x i h k θ j x i θ }{{ k } δ jk g k = k 1 h k f θ k g k (1.102) Notice how this is almost identical to our original expression for the gradient, except that {x i }, {e i } have been replaced with {θ i }, {g i }. The only difference is the presence of the scale factors. Let s solidify this with an example: Example 1.6 Let us continue with our example of cylindrical polar coordinates. Let f = f (r, θ, z). Then we have: grad(f ) = 1 f h r r g r + 1 f h θ θ g θ + 1 f h z z g z = f r g r + 1 f r θ g θ + f z g z (1.103) Divergence Let v = v i (θ) g i be a vector field that is defined exclusively using the {θ i }, {g i } coordinate system. What is the divergence of this vector field? We can follow the exact same procedure as when computing the gradient: div(v) = (v) e i = v θ j e i = v θ j 1 x i g k = x i θ j x i θ j x i h k θ k k k = k 1 h k v θ k g k 1 v h k θ j θ j x i }{{} J ji x i θ k }{{} J 1 ik g k = k 1 v δ jk g k (1.104) h k θ j (1.105) All content , Brandon Runnels 4.3

23 Course Notes - Lecture 4 Notice again how we arrive at an almost identical formula except that we include scale factors. We must also make one very important note: how do we evaluate the following? (v) = (v i g i ) = v i g i g i + v i (1.106) θ k θ k θ k θ k In Cartesian coordinates the basis vectors are constant so their derivatives vanish. However, this is not true in most other curvilinear coordinates! To illustrate this, let s do an example: Example 1.7 Compute the divergence of a vector field in cylindrical polar coordinates: v = v r g r + v θ g θ + v z g z. div(v) = 1 h r v r g r + 1 h θ v θ g θ + 1 h z v z g z θ (v r g r + v θ g θ + v z g z ) g θ + z (v r g r + v θ g θ + v z g z ) g z = r (v r g r + v θ g θ + v z g z ) g r + 1 r ( vr = r g r + v θ r g θ + v ) z r g z g r + 1 ( vr r θ g r + v θ θ g θ + v z θ g z + v r θ g r + v θ θ g θ + v z θ g z ( vr + z g r + v θ z g θ + v ) z z g z g z = v r r + 1 ( vθ ) r θ g θ + v r g θ v θ g r g θ + v z z = v r r + 1 r ( vθ θ + v r ) + v z z ) g θ (1.107) Notice how we picked up a couple of extra terms: this is a result of our choice of coordinate system. This is a tedious process, but fortunately we only have to do it a couple of times Curl Hopefully this is starting to seem familiar. Starting with our original expression for curl and converting to curvilinear coordinates, we have curl(v) = x i (v) e i = v θ j θ j x i k 1 h k x i θ k g k = k 1 v θ j x i g k = h k θ j x i θ k k 1 v δ jk g k (1.108) h k θ j = k 1 h k v θ k g k (1.109) Once again, we see that we recover a very similar expression except for the presence of scale factors. All content , Brandon Runnels 4.4

24 Course Notes - Lecture 4 Example 1.8 Find the expression for the curl in cylindrical polar coordinates. We can reuse quite a bit of what we computed earlier; we just need to be careful about which vectors we cancel out. [ 1 v curl(v) = h r ( vr = 1 ( vr r ( vr ( vθ = ( 1 = r r g r + 1 h θ v θ g θ + 1 h z v r g r + v θ r g θ + v z r g z ) g r z g z θ g r + v θ θ g θ + v ) z θ g z + v r g θ v θ g r g θ ) g z z g r + v θ z g θ + v z z g z ) + 1 ( v r r r g z v z r g θ v z θ v θ z ) g r + ( vr z v z r ] θ g z + v ) ( z θ g vr r + v θ g z + z g θ v ) θ z g r ) ) g θ + (g z + 1 ( (r vθ ) v r r r θ ) g z (1.110) 1.7 Tensor transformation rules Suppose we have two orthonormal {e i }, {g i }. We recall that orthonormality allows us to write each basis in terms of the other: e i = (e i g p ) g p g q = (g q e i ) g i (1.111) Now, let us suppose we have a tensor A L(R n, R n ) with components A ij in the {e i } basis, that is, A = A ij e i e j. How can we express A in terms of the other basis? To do that, we simply substitute A ij e i e j = A ij [(e i g p ) g p ] [(e j g q ) g q ] = (g p e i ) }{{} Q T pi A ij (e j g q ) }{{} Q jq where Âpq are the components of A in the other basis. Or, put more simply, we have that T g p g q = Qpi A ij Q jq g p g q = Âpqg p g q (1.112) Â pq = Q T pi A ij Q jq (1.113) Note that we are not transforming the tensor itself: we are merely changing the components of the tensor to fit with the assigned basis. This is called a tensor transformation property. Let s take another look at the Q matrices. Specifically, let s look at Q T Q: [Q T Q] pq = Qpi T Q iq = (g p e i )(e i g q ) = [(g p e i ) e i ] }{{} g q = g p g q = δ pq = [I] pq (1.114) g p Because Q T Q = I, we conclude that Q T = Q 1. We can also write Q = [g 1... g n]. If the new basis is right-handed, then we have det(q) = 1. Thus, we have shown that Q SO(n), or that Q is a rotation matrix. All content , Brandon Runnels 4.5

25 Lecture 5 Kinematics of deformation, frames 2 Kinematics of Deformation We are now ready to introduce the machinery that we need to describe the deformation of solid bodies. Let us introduce a few definitions: Definition 2.1. A body is a set of material particles occupying a region in Euclidean space; generally denoted R 3. Definition 2.2. A configuration is a specific correspondance between particles of the body and points in space. Definition 2.3. A deformation mapping is an injective 1 mapping that describes a configuration of the body. We will also refer to the deformed and undeformed configurations. The following figure illustrates the general setup for describing the deformation of a solid body: φ G 3 G 1 G 2 X g 3 x = φ(x) φ() g 2 g 1 {G I } are the basis vectors in the undeformed configuration {g i } are the basis vectors in the deformed configuration. **Note: this is completely general, but we often (usually) keep the same in both configurations. φ : R 3 R 3 is the deformation mapping. X = X I G I is the location of a point in the undeformed configuration. x = x i g i = φ i (X) g i is the location of point X in the deformed configuration. Note that we adopt the convention that uppercase symbols correspond to quantities the undeformed configuration; whereas lowercase symbols correspond to quantities in the deformed configuration. We will even use uppercase and lowercase indices to indicate components in the undeformed and deformed frames. We will stick to this convention consistently, and it will be useful in helping us to keep track of which coordinate system we are in. To illustrate, let us consider the following examples: (i) Stretching of a unit cube: What is the deformation mapping for the following stretched cube? G 3 1 φ g 3 λ G 1 G 2 g 1 λ 1 λ 2 g 2 1 injective no two points can be mapped to the same location, that is, if f : U V is injective, then for x, y U, f (x) = f (y) = x = y. All content , Brandon Runnels 5.1

26 Course Notes - Lecture 5 (Notice that g 1 = G 1, etc.) We identify the deformation mapping simply as x 1 = λ 1 X 1 x 2 = λ 1 X 2 x 3 = λ 1 X 3 (2.1) Or, we can describe this as x = φ(x) = [ ] λ λ 2 0 X = F X (2.2) 0 0 λ 3 (ii) Shearing of a unit cube: What is the deformation mapping for the following cube subjected to pure shear? G 3 1 φ 1 g 3 γ 1 1 G g 2 G 1 g 1 We identify the deformation mapping to be x 1 = X 1 x 2 = X 2 + γ X 3 x 3 = X 3 (2.3) Or, we can describe it as x = φ(x) = [ 1 0 ] γ 1 X = F X (2.4) (iii) An affine deformation is one in which straight lines remain straight as in the following figure: φ φ affine undeformed non-affine What form must this mapping take? Consider the line between two vectors X 1, X 2, and the resulting line between the mapped vectors x 1, x 2. αx 1 + (1 α)x 2 X 2 X 1 x 2 φ x 1 αx 1 + (1 α)x 2 All content , Brandon Runnels 5.2

27 Course Notes - Lecture 5 The set of vectors αx 1 + (1 α)x 2 are all in the line in the undeformed configuration. By the definition of affine mappings, αx 1 + (1 α)x 2 must be must form the points in the deformed line. In other words: φ(αx 1 + (1 α)x 2 ) = α φ(x 1 ) + (1 α) φ(x 2 ) X 1, X 2 (2.5) Since this must hold for all X, we conclude that the mapping must be linear. Since all linear maps can be represented in tensor form, we conclude that affine maps can be represented in the form φ(x) = F X F L(R n, R n ), φ affine (2.6) Because this is a real deformation, we know that it must be invertible. That is, we can construct a φ 1 such that X = φ 1 (x). But if x = φ(x) = F X, then the inverse mapping must be X = F 1 x = φ 1 (x) = F 1 x (2.7) This means that F must be invertible, which means that det(f ) 0. The set of all invertible matrices is called the general linear group GL(n) = {F L(R n, R n ) : det(f ) 0} L(R n, R n ) (2.8) so we say that for a mapping to be affine, F GL(n). A rigid body mapping in R n is, formally, an orientation-preserving isometry of R n. What does that mean? First, let s define the term isometry: Definition 2.4. A mapping φ : R n R n is an isometry if φ(x) = X x R n (2.9) In other words, an isometry is a mapping that does not change the length of any vector. Let us make the ansatz (i.e. starting assumption) that φ is an affine isometry, that is, φ = F X = x. What are the conditions for φ to be an isometry? φ = x = x T x = (F X) T (F X) = X T F T F X =! X T X X R n (2.10) What does this imply about F T F? It must equal the identity. So F T F = I = F T = F 1, or F O(n), the orthogonal group. What about the other part? Without going into extensive detail, orientation-preserving simply means that the body cannot be reflected or turned inside-out. This is equivalent to stating that det(f ) > 0. Thus, for φ to be orientation-preserving, F SO(3); that is, F must be a rotation. There is one more aspect of affine and rigid-body mapping that we have not yet discussed: rigid body translation. A rigid body translation mapping can simply be expressed as where x 0 is the translation vector. Thus, the general expression for a rigid body mapping is 2.1 Eulerian and Lagrangian frames Let s go back to our generalized form of the deformation mapping. φ(x) = x 0 + X (2.11) φ(x) = x 0 + F X x 0 R n, F SO(n) (2.12) All content , Brandon Runnels 5.3

28 Course Notes - Lecture 5 φ G 3 G 1 G 2 X g 3 x = φ(x) φ() g 2 g 1 As stated before, we have adopted the convention of using uppercase variables (and indices) to describe the material in the undeformed configuration, and lowercase for the deformed configuration. The reason for doing this, as we ll see, is that we will be able to formulate almost everything analagously in terms of either set of variables. Definition 2.5. The Lagrangian / material frame refers to the quantities defined in the undeformed configuration. Definition 2.6. The Eulerian / spatial frame refers to the quantities defined in the deformed configuration. As we go along and derive various equations, we will frequently formulate those equations in both the Lagrangian and Eulerian frames. (You can think of this as finding the uppercase and lowercase versions of the equations.) The following are a couple of examples of the convention that we will use. Variables and unit vectors: X are the locations of the material points in the Lagrangian frame, x = φ(x) are the locations of the points in the Eulerian frame. Calculus: X = X I G I = φ 1 I (x) G I x = x i g i = φ i (X) g i (2.13) Grad(F (X)) I = F X I Div(V(X)) I = V I X I grad(f (x)) i = f x i (2.14) div(v(x)) i = v i x i (2.15) and similarly for curl, the Laplacian, etc. Note that Div, Grad are not necessarily the same as div, grad! 2.2 Time-dependent deformation Let us now consider a body whose deformation varies with time: that is, x(t) = φ(x, t). What is the velocity of the material? Let us define the Lagrangian velocity field as Similarly, the Lagrangian acceleration as V(X, t) = t φ(x, t) V i(x, t) = t φ i(x, t) (2.16) A(X, t) = t V(X, t) A i(x, t) = t V i(x, t) (2.17) Suppose we want to get the velocity and acceleration as a function of the deformed location? To do this we define the Eulerian velocity field as and the Eulerian acceleration field as v(x, t) = V(φ 1 (x), t) v i (x, t) = V i (φ 1 (x), t) (2.18) a(x, t) = A(φ 1 (x), t) a i (x, t) = A i (φ 1 (x), t) (2.19) All content , Brandon Runnels 5.4