Chapter 2 Kinematics: Deformation and Flow

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Chapter 2

Kinematics: Deformation and Flow

2.1 Introduction Lecture 5 We need a suitable mathematical framework in order to describe the behaviour of continua. Our everyday experience tells us that lumps of matter can both move (change in position) and deform (change in shape), so we need to be able to quantify both these eﬀects. Hopefully, you are already be familiar with the kinematics of individual particles from basic mechanics. In classical particle mechanics, the position of the particle is described by a position vectorr(t) measured from a chosen 3 origin in three-dimensional Euclidean space, � . The position is a function of one-dimensional continuous time,t [0, ), and contains all the information that we need to describe the motion ∈ ∞ of the particle.

Ωt Ω0 χt

r R

Figure 2.1: A regionΩ 0 is mapped to the regionΩ t by the mappingχ t which carries the material pointr Ω to the pointR Ω . ∈ 0 ∈ t In continuum mechanics, we need to account for the motion of all the “particles” within the 1 material. Consider a continuous body that initially occupies a region Ω0 of three-dimensional 3 Euclidean space � , with volume and surface∂ . The body can be regarded as a V 0 V 0 ≡S 0 1If you want to get fancy you can call this an open subset.

32 collection of particles, called material points, each of which2 is described by a position vector 3 r=x K eK from ourfixed origin in a Cartesian coordinate system . We shall term the configuration Ω0 the undeformed configuration. At a later timet, the same body occupies a different region of space,Ω , with volume and surface . The material points within the regionΩ are now t V t S t t described by a position vectorR=X K eK from the same origin in the same Cartesian coordinate 4 system . We shall term the configurationΩ t the deformed configuration, see Figure 2.1. The change in configuration fromΩ toΩ can be described by a functionχ :Ω Ω , 0 t t 0 → t sometimes called a deformation map. The position at timet is given byR=χ t(r) χ(r,t), whereχ:Ω [0, ) Ω is a continuous map. In other words, each material point≡ 5 in the 0 × ∞ → t undeformed configurationΩ 0 is carried to a (material) point in the deformed configurationΩ t. On physical grounds, we expect that (a) matter cannot be destroyed and (b) matter does not interpenetrate. A deformation map will be consistent with these conditions if it is one-to-one and the Jacobian of the mapping remains non zero. The Jacobian of the mapping is the determinant of ∇ the matrix,F= r χt, (note that the gradient is taken with respect to the undeformed coordinates r) whose components in the Cartesian basis are given by

∂XI FIJ = , (2.1) ∂xJ and so our physical constraints demand that

detF= 0. (2.2) 6 In fact, we impose the stronger condition that the Jacobian of the mapping remains positive, which ensures that material lines preserve their relative orientations: a body cannot be deformed onto its mirror image. If condition (2.2) is satisﬁed then a (local) inverse mapping can be constructed that gives the initial position as a function of the current position,r=χ −1(R) χ −1(R,t). t ≡ Example 2.1. An example deformation Is the mapping given by ( χ = 2x + 3x , χ(x): 1 1 2 χ2 = 1x1 + 2x2. physically admissible? If so, sketch the deformed unit squarex I [0, 1] that follows after application of the mapping. ∈ Solution 2.1. The deformed position is given by X 2 3 x 1 = 1 , X2 1 2 x2 and detF=2 2 3 1=1> 0, so the mapping is physically admissible. The deformed unit square can be× obtained− × by thinking about the deformation of the corners and the knowledge that

2We could label the position vectorr by another index to distinguish different material points, but the notation is less cumbersome if we treat each specific value ofr as the definition of the different material points. 3Even when we consider non-Cartesian coordinates later, it’s easiest to define those coordinates relative to the fixed Cartesian system. Note that because this is a Cartesian coordinate system we have not bothered to indicate the contravariant nature of the components. 4For further generality, we should assume a different orthonormal coordinate system for the deformed configuration R=X K eK , but the notation becomes even more cluttered without really aiding understanding. 5From the classical particle mechanics viewpoint, there is a single material point (the particle) with position vectorx(t) =R=χ t(r)=χ t(x(0)). the deformation is homogeneous (the matrix entries are not themselves functions ofx). The corners map as follows

(0, 0) (0, 0); (1, 0) (2, 1); (0, 1) (3, 2); (1, 1) (5, 3), → → → → and the undeformed and deformed regions are shown in Figure 2.2.

e1 Figure 2.2: A unit square (solid boundary) is deformed into a quadrilateral region (dashed boundary) by the mappingχ.

2.2 Lagrangian (Material) and Eulerian (Spatial) Descrip- tions

Given the existence of the continuous mappingχ, we can write (each Cartesian coordinate of) a material point in the deformed position as a function of (the Cartesian coordinates of) the same material point in the undeformed position and the timet,

XK (t) =χ(x J , t)·eK , which we can write in vector form asR(r,t). (2.3)

In equation (2.3) the current position is treated as a function of the original position, which is the independent variable. This is called a Lagrangian or material description. Any otherfields are also treated as functions of the original position. In the Lagrangian description, we follow the evolution of particular material particles with time6. It is more common to use a Lagrangian description in solid mechanics where the initial undeformed geometry is often simple, but the deformations become complex. Alternatively, we could use the inverse mapping to write (each Cartesian coordinates of) the undeformed position of a material point as a function of (the Cartesian coordinates of) its current position −1 xK (t) =χ (XJ , t)·eK , orr(R,t). (2.4) Here, the current position is the independent variable and this is called an Eulerian or spatial description. Any otherfields are treated as functions of thefixed positionR. In the Eulerian

6The Lagrangian description is the viewpoint used in classical particle mechanics because we are only interested in the behaviour of each individual particle. description, we observe the changes over time at afixed point in space 7. It is more common to use an Eulerian description influid mechanics, if the original location of particularfluid particles is not of interest. The transformation of an object from an Eulerian (current) description to the Lagrangian (ref- erence) description is sometimes called a pullback (terminology taken from differential geometry). The converse operation of transforming from the Lagrangian to the Eulerian description is called a pushforward. For simplicity, we have defined the deformed and undeformed positions in the same global Cartesian coordinate system, but this is not necessary. It is perfectly possible to use different coordinate systems to represent the deformed and undeformed positions, which may be useful in certain special problems: e.g. a cube being deformed into a cylinder. However, the use of general coordinates would bring in the complication of covariant and contravariant transformations. We shall consider general curvilinear coordinates in what follows after the initial development in Cartesians. The curvilinear coordinates associated with the Lagrangian viewpoint will be denoted byξ i and those in the Eulerianχ i.

2.3 Displacement, Velocity and Acceleration

Newton’s laws of mechanics were originally formulated for individual particles and the displacement, velocity and acceleration of each particle are derived from its position as a function of time. In continuum mechanics we must also deﬁne these quantities based on the position of each material point as a function of time (the Lagrangian viewpoint), but we can then reinterpret them as functions of the absolute spatial position (the Eulerian viewpoint) if more convenient.

2.3.1 Displacement

The change in position of a material point within the body between configurationsΩ 0 andΩ t is given by the displacement vectorfield u=R r, (2.5) − which can be treated as a function of eitherR orr; i.e. from the Eulerian or Lagrangian viewpoints, respectively. For consistency of notation we shall writeU(R,t) =R r(R,t) for the displacement in the Eulerian representation andu(r,t) =R(r,t) r in the Lagrangian.− − Example 2.2. Displacement of a moving block A block of material that initially occupies the regionr [0, 1] [0, 1] [0, 1] moves in such a way that its position is given by ∈ × × R(r,t) =r(1 +t), (2.6) at timet 0. Find the displacement of the block in both the Eulerian and Lagrangian representa- tions. ≥ 7The Eulerian description makes little sense in classical particle mechanics because in general motion the particle is unlikely to be located at our chosenfixed position for very long. Imagine trying to throw a ball in front of a fixed (Eulerian) video camera: the ball will only appear on a short section of the video, or not at all if we do not throw accurately. The reason why an Eulerian view makes more sense in continuum mechanics is that some part of the continuum will generally be located at our chosenfixed point, but exactly which part (which material points) changes with time. Solution 2.2. We are given the relationshipR(r,t), so the displacement in the Lagrangian viewpoint is straightforward to determine

u(r,t) =R(r,t) r=tr. − For the Eulerian viewpoint, we need to invert the relationship (2.6) which givesr=R/(1 +t) and therefore the displacement is t U(R,t) = R. 1 +t A potential problem with the Eulerian viewpoint is that it does not actually make sense unless the pointR is within the body at timet.

2.3.2 Velocity The velocity of a material point is the rate of change of its displacement with time. The Lagrangian (material) coordinate of each material point remainsﬁxed, so the velocity,v, in the Lagrangian representation is simply

∂u(r,t) ∂(R(r,t) r) ∂R(r,t) v(r,t) = = − = , (2.7) ∂t rfixed ∂t r ∂t r becauser is heldfixed during the differentiation 8. Note that the velocity moves with the material points, which is exactly the same as in classical particle mechanics. In the Eulerian framework, the velocity must be defined as a function of a specificfixed point in space. The problem is that different material points will pass through the chosen spatial point at different times. Thus, the Eulerian velocity must be calculated byfinding the material coordinate that corresponds to thefixed spatial location at chosen instant in time,r(R,t), so that

∂U(R,t) ∂R V(R,t) = = =v(r(R,t), t). (2.8) ∂t r ∂t r Example 2.3. Calculation of the velocity for a rotating block A cube of material that initially occupies the regionr [ 1, 1] [ 1, 1] [ 1, 1] rotates about the ∈ − × − × − x3-axis with constant angular velocity so that its position at timet is given by       X1 cost sint0 x1  X  =  sint cost0   x  . (2.9) 2 − 2 X3 0 0 1 x3 Calculate both the Lagrangian and Eulerian velocities of the block. Solution 2.3. The Lagrangian velocity is simple to calculate       v sint cost0 x ∂R 1 − 1 v(r,t) = ,  v  =  cost sint0   x  , (2.10) ∂t ⇒ 2 − − 2 v3 0 0 0 x3 which is, as expected, the same as the velocity of a particle initially at (x1, x2) rotating about the origin in thex x plane. 1 − 2 8This is indicated by the vertical bar next to the partial derivative, which should not be confused with the covariant derivative. In order to convert to the Eulerian viewpoint we need tofindr(R), which follows after inversion of the relationship (2.9):       x1 cost sint0 X1    −    x2 = sint cost0 X2 . (2.11) x3 0 0 1 X3 Thus the Eulerian velocityV(R,t) =v(r(R,t), t) is obtained by substituting the relationship (2.11) into the expression for the Lagrangian velocity (2.10)         V sint cost0 cost sint0 X 1 − − 1  V  =  cost sint0   sint cost0   X  2 − − 2 V3 0 0 0 0 0 1 X3       0 1 0 X1 X2 =  1 0 0   X  =  X  , − 2 − 1 0 0 0 X3 0 which remains constant. In other words the velocity of the rotating block at afixed point in space is constant. Different material points pass through our chosen point but the velocity of each is always the same when it does so.

2.3.3 Acceleration The acceleration of a material point is the rate of change of its velocity with time. Once again, computing the acceleration in the Lagrangian representation is very simple

2 2 ∂v(r,t) ∂ u(r,t) ∂ R(r,t) a(r,t) = = = . (2.12) ∂t r ∂t2 r ∂t2 r In the Eulerian framework, the velocity at afixed point in space can change through two different mechanisms: (i) the material velocity changes with time; or (ii) the material point (with a specific velocity) is carried past thefixed point in space. The second mechanism is a consequence of the motion of the continuum and is known convection or advection. The two terms arise quite naturally in the computation of the material acceleration at afixed spatial location because for afixed material coordinate, the positionR is also a function of time

∂V(R,t) ∂V(R(r,t), t) A(R,t) = = . (2.13) ∂t r ∂t r After application of the chain rule, equation (2.13) becomes

V V R V V V ∂ ∂ · ∂ ∂ · ∂ ∂ ·∇ A(R,t) = + = +V = +V R V, (2.14) ∂t R ∂R t ∂t r ∂t R ∂R t ∂t or in component form ∂VI ∂VI AI (R,t) = +V J . (2.15) ∂t ∂XJ Note that the gradient is taken with respect to the Eulerian (deformed) coordinatesR. Thefirst term∂V/∂t corresponds to the mechanism (i) above, whereas the second termV·∇ R V corresponds to the mechanism (ii) and is the advective term. The advective term is nonlinear inV , which leads to many of the complex phenomena observed in continua9. 9Note that this nonlinearity only arises when observations are made in the Eulerian viewpoint, so it is definitely observer dependent. 2.3.4 Material Derivative Lecture 6 The argument used to determine the acceleration is rather general and can be used to define the rates of change of any property that is carried with the continuum. For a scalarfieldφ(r,t) =Φ(R(r,t), t) the material derivative is the rate of change of the quantity keeping the Lagrangian coordinater fixed. The usual notation for the material derivative isDφ/Dt and in the Lagrangian framework, the material derivative and the partial derivative coincide Dφ ∂φ = (2.16) Dt ∂t In the Eulerian framework, however,

DΦ ∂Φ ∂Φ ∂XK ∂Φ ∂Φ ∂Φ ·∇ = + = +V K = +V R Φ. (2.17) Dt ∂t ∂XK ∂t ∂t ∂XK ∂t

Curvilinear coordinates If we use general (time-independent) curvilinear coordinatesχ i as as Eulerian coordinates, then R(χi) and equation (2.17) becomes

Dφ ∂φ ∂φ ∂χi ∂X ∂φ ∂χi ∂φ ∂φ ∂φ = + K = + V = +V i , i K i i Dt ∂t ∂χ ∂XK ∂t ∂t ∂XK ∂χ ∂t ∂χ where we have use equation (1.16b) to determine the componentsV i in the covariant basis corre- i i i sponding to the coordinatesχ . The velocity vector isV=V Gi, whereG i =∂R/∂χ are the covariant base vectors in the deformed position with respect to Eulerian curvilinear coordinatesχ i. It follows that the acceleration in the Eulerian framework in general coordinates is given by

D(V jG ) ∂(V jG ) ∂(V jG ) j = j +V i j . Dt ∂t ∂χi

Assuming that the base vectors areﬁxed in time, which is to be expected for a (sensible) Eulerian coordinate system, we obtain ! DV ∂V i = +V jV i G , Dt ∂t ||j i where here indicates covariant diﬀerentiation in the Eulerian viewpoint with respect to the ||j i · Eulerian coordinatesχ , i.e. using the the metric tensorG i j =G i Gj. As you would expect, this is simply the tensor component representation of the expression DV ∂V = +V·∇ V; Dt ∂t R a coordinate-system-independent vector expression.

2.4 Deformation

Thus far, we have considered the extension of concepts in particle mechanics, displacement, velocity and acceleration, to the continuum setting, but the motion of isolated individual points does not tell us whether the continuum has changed its shape, or deformed. The shape will change if material points move relative to one another, which means that we need to be able to measure distances. In the body moves rigidly then the distance between every pair of material points will remain the same. A body undergoes a deformation if the distance between any pair of material points changes. Hence, quantiﬁcation of deformation requires the study of the evolution of material line elements, see Figure 2.3.

R(r+dr) r+dr χ t R dr d R(r) r

Ω0 Ωt

Figure 2.3: A regionΩ 0 is mapped to the regionΩ t by the mappingχ t which carries the material pointr Ω 0 to the pointR(r) Ω t and the material pointr+dr toR(r+dr). The undeformed line element∈ dr is carried to the∈ deformed line elementdR.

Consider a line elementdr that connects two material points in the undeformed domain. If the position vector to one end of the line isr, then the position vector to the other end isr+dr. If we take the Lagrangian (material) viewpoint, the corresponding endpoints in the deformed domain are given byR(r) andR(r+dr), respectively 10. Thus, the line element in the deformed domain is given by dR=R(r+dr) R(r). − If we now assume that dr 1, then we can use Taylor’s theorem to write | |≪ ∂R ∂R dR R(r)+ (r) dr R(r)= (r) dr F(r) dr, (2.18) ≈ ∂r · − ∂r · ≡ · or in component form in the global Cartesian coordinates

∂XI dRI = drJ F IJ drJ . (2.19) ∂xJ ≡

The matrixF of componentsF IJ is a representation of a quantity called the (material) deformation gradient tensor. It describes the mapping from undeformed line elements to deformed line elements. In fact in this representationF IJ is something called a two-point tensor because theI-th index refers

10The dependence on timet will be suppressed in this section because it does not affect instantaneous measures of deformation. to the deformed coordinate system, whereas theJ-th index refers to the undeformed. We have chosen the same global coordinate system, so the distinction may seem irrelevant11, but we must remember that transformation of the undeformed and deformed coordinates will affect different components of the tensor. 2 The length of the undeformed line element is given by ds where ds =dr·dr=dr K drK and 2 the length of the deformed line element is dS where dS =dR·dR=dR K dRK . Thus, a measure of whether the line element has changed in length is given by dS2 ds 2 =F dr F dr dr dr = (F F δ δ ) dr dr (c δ ) dr dr ; (2.20) − KI I KJ J − K K KI KJ − KI KJ I J ≡ IJ − IJ I J and the matrixc=F T F is (a representation of) the right12 Cauchy–Green deformation tensor. Its components represent the square of the lengths of the deformed material line elements relative to the undeformed lengths, i.e. from the Lagrangian viewpoint. Note thatc is symmetric and positive 2 definite because drI cIJ drJ = dS > 0, for all non-zerodr. From equation (2.20) if the line element does not change in length thenc IJ δ IJ = 0 for all I,J, which motivates the definition of the Green–Lagrange strain tensor − 1 1 ∂XK ∂XK eIJ = (cIJ δ IJ ) = δ IJ . (2.21) 2 − 2 ∂xI ∂xJ −

The components ofe IJ represent the changes in lengths of material line elements from the La- grangian perspective. An alternative approach is to start from the Eulerian viewpoint, in which case dr=r(R+dR) r(R), − and using Taylor’s theorem as above we obtain ∂r dr (R) dR H(R) dR, (2.22) ≈ ∂R · ≡ · or in component form ∂xI drI = dRJ H IJ dRJ . (2.23) ∂XJ ≡

The matrixH with componentsH IJ is (a representation of) the (spatial) deformation gradient tensor. Comparing equations (2.19) and (2.23) shows thatH=F −1. Thus, the equivalent to equation (2.20) that quantiﬁes the change in length is 2 2 dS ds = dRK dRK H KI dRI HKJ dRJ = (δKI δKJ H KI HKJ ) dRI dRJ (δ IJ C IJ ) dRI dRJ . − − − ≡ − (2.24) The matrixC=H T H=F −T F−1 = (FFT )−1 =B −1 is known as (a representation of) the Cauchy deformation tensor and is the inverse of the Finger deformation tensor13. The components of C represent the square of lengths of the undeformed material elements relative to the deformed lengths, i.e. from the Eulerian viewpoint. We can deﬁne the corresponding Eulerian (Almansi) strain tensor 1 1 ∂xK ∂xK EIJ = (δIJ C IJ ) = δIJ , (2.25) 2 − 2 − ∂XI ∂XJ whose components represent the change in lengths of material line elements from the Eulerian perspective.

∂X 11The distinction is clear if we writeR=X e , in which caseF = I , but there are already too many K K IJ ∂xJ overbars in this section! 12The left Cauchy–Green deformation tensor is given byB=FF T and is also called the Finger tensor. 13Thefinal possible deformation tensor isb=c −1 =HH T and was introduced by Piola. Curvilinear Coordinates If the undeformed position is parametrised by a set of general Lagrangian coordinatesr(ξ i), the undeformed length of a material line will depend on the metric tensor. In general, lengths relative to general coordinates vary with position, whereas in Cartesian coordinates relative lengths are independent of absolute position. The undeformed material line vector is given by ∂r dr=r(ξ i +dξ i) r(ξ i) dξi =g dξi, − ≈ ∂ξi i whereg i is the covariant base vector in the undeformed configuration with respect to the Lagrangian coordinates. Similarly, the deformed material line vector is given by ∂R dR=R(ξ i +dξ i) R(ξ i) dξi =G dξi, − ≈ ∂ξi i whereG i is the covariant base vector with respect to the Lagrangian coordinates in the deformed configuration14. Hence,

dS2 ds2 =dR·dR dr·dr=G ·G dξidξj g ·g dξidξj = (G g )dξ idξj, − − i j − i j ij − ij · · whereg ij =g i gj is the (undeformed) Lagrangian metric tensor andG ij =G i Gj is the (deformed) Lagrangian metric tensor. The Green–Lagrange strain tensor relative to these curvilinear coordinates is therefore 1 γ = (G g ); (2.26) ij 2 ij − ij and 1 ∂XK ∂XK ∂xK ∂xK 1 ∂XK ∂XK ∂xI ∂xJ ∂xI ∂xJ γij = i j i j = δ IJ i j =e IJ i j , (2.27) 2 ∂ξ ∂ξ − ∂ξ ∂ξ 2 ∂xI ∂xJ − ∂ξ ∂ξ ∂ξ ∂ξ which demonstrates thatγ ij are indeed the components of the Green–Lagrange tensor, after the appropriate (covariant) tensor transformation of the undeformed coordinates; and that the deformed metric tensor is obtained from the right Cauchy–Green deformation tensor under (covariant) transformation of the undeformed coordinates. Alternatively, 1 ∂xK ∂xK ∂XI ∂XJ ∂XI ∂XJ γij = δIJ i j =E IJ i j , (2.28) 2 − ∂XI ∂XJ ∂ξ ∂ξ ∂ξ ∂ξ and the (curvilinear) Green–Lagrange strain tensor can be obtained from the Almansi strain tensor after a covariant transformation of the deformed coordinates. Of course this means that the Carte- sian Green–Lagrange strain tensor is the Almansi strain tensor after a change in coordinates from those in the deformed body to the undeformed, i. e. the Almansi tensor is the pushforward of the Green–Lagrange tensor: ∂xK ∂xL −T −1 EIJ =e KL , orE=F eF . (2.29) ∂XI ∂XJ 14Note that the components of the covariant base vector in the global Cartesian basis ∂X I ∂X I ∂xJ ∂xJ [G ]I = = =F =F , j ∂ξj ∂xJ ∂ξj IJ ∂ξj Ij and are therefore equivalent to the deformation gradient tensor after covariant transformation of the undeformed coordinate (the second index). The definition ofF means that equation (1.30) defines the pushforward for covariant and contravariant tensors. Comparison with that equation reveals thatE should be denotedE ♭ to avoid mistakes when working in general coordinates. The fact that all these tensors are equivalent should be no surprise — they all represent the same physical measure: half the difference between the change in square lengths of material line elements.

Relationship between Cartesian and curvilinear formulations It seems to be most common in the literature to work in terms of the tensorsF,c andC (orb and B) because it is more compact; but the meaning can be obscured. For example, consider a vector field that is convected with the motion of the continuum so that it obeys the same transformation rules as line elements, see also 5.2.4. If the vectorfield in the undeformed configuration is given bya then by analogy with equation§ 2.19, the vectorfield in the deformed configuration,A, is given by AI =F IJ aJ , (in global Cartesian components) orA=Fa. A natural question is how to represent this in general curvilinear coordinates using the established metric tensors. If we are completely explicit about the bases in our tensor formulation then it follows that ∂X ∂X ∂ξk ∂X ∂ξk F e e I e e I e e I e e =F IJ I J = I J = k I J = k I J ⊗ ∂xJ ⊗ ∂ξ ∂xJ ⊗ ∂ξ ⊗ ∂xJ

F=G g k and henceF −1 =g G k, ⇒ k ⊗ k ⊗ which reveals the desired connection between the two formulations. Returning to the transformation of our vectorfield we can write � AiG =F(a ig ) A iGj·G =G j· G g k ·g ai A iδj =δ j δkai A j =a j, i i ⇒ i k ⊗ i ⇒ i k i ⇒ and we might have expected. The result indicates that a vectorfield convected with continuum line elements keeps identical values of its components, provided that covariant base vectors deform with i i the motion of the continuum, i. e. ifa=a gi, thenA=a Gi. If we have a vectorfieldm, in the undeformed configuration, with different transformation properties, e. g. the deformed configuration is given byM=F −T m, then � M iG =F −T (m gi) M iGj·G =G j· Gk g ·gi m M j =G jkm , i i ⇒ i ⊗ k i ⇒ k a different type of transformation.

2.4.1 The connection between deformation and displacement From equation (2.5), the deformed position can be written as the vector sum of the undeformed Lecture 7 position and the displacement R=r+u, which means that ∂(r+u) dR dξi = (r +u ) dξi =g dξi +u dξi =dr+u dξi. (2.30) ≈ ∂ξi ,i ,i i ,i ,i Theﬁrst term in equation (2.30) is the undeformed line element and so represents a rigid-body translation; the second term contains all the information about the strain and rotation. The covariant base vectors in the deformed conﬁguration are

Gi =R ,i =g i +u ,i, so the Green–Lagrange strain tensor becomes 1 � 1 � 1 γ = G ·G g ·g = (g +u ) g +u g = g ·u +u ·g +u ·u . ij 2 i j − i j 2 i ,i · j ,j − ij 2 i ,j ,i j ,i ,j In order to simplify the scalar products, we write the displacement using the undeformed base vectors u=u gj, u =u gk, j ⇒ ,i k|i where represents the covariant derivative in the Lagrangian viewpoint with respect to Lagrangian | coordinates, i.e. using the metric tensorg ij. Then the strain tensor becomes 1 1 γ = g ·gku +u gk·g +u gk·glu = δku +u δk +u u gkl ij 2 i k|j k|i j k|i l|j 2 i k|j k|i j k|i l|j and so 1 γ = u +u +u k u . ij 2 i|j j|i |i k|j The quantity with componentsu is known as the displacement gradient tensor and can also be i|j written∇r u. In our global Cartesian coordinate system, the Green–Lagrange strain tensor has the form ⊗ 1 e = [u +u +u u ], (2.31) IJ 2 I,J J,I K,I K,J which is often seen in textbooks.

2.4.2 Interpretation of the deformed metric (right Cauchy–Green deformation) tensorG ij We can scale the inﬁnitesimal increments in the general coordinates so that dξi =n i dǫ, wheren i i represents the direction of the increment and is chosen so that the vectorn=n gi has unit length, · i · j i j n n=n gi n gj =n gijn = 1. The undeformed and deformed line elements are then functions of the directionn

i i dr(n) =g in dǫ anddR(n) =G in dǫ. We define the stretch in the directionn to be the ratio of the length of the deformed line element to the undeformed line element: p dR(n) niG njdǫ p p ij i j λ(n) = | | = = n Gijn = √nI cIJ nJ , (2.32) dr(n) nig njdǫ | | ij where the last equality is obtained on transformation from the general coordinates to the global Cartesian coordinates. Note that the stretch is invariant and is well-defined because we have already established thatn I cIJ nJ > 0 for all non-zeron. (i) i (i) (i) If we definen to be a unit vector in theξ direction, thenn i = 1/√gii (not summed)n j = 0, j=i, so p 6 (i) λ(n ) = Gii/gii (not summed), which means that the diagonal entries of the deformed metric tensor are proportional to the squares of the stretch in the coordinate directions. If the undeformed coordinates are Cartesian coordinates, for whichg ii = 1 (not summed), then the diagonal entries of the deformed metric tensor are exactly the squares of the stretches in coordinate directions. We now consider two distinct line elements in the undeformed configuration

i i dr=n gi dǫ anddq=m gi dǫ,

i i wheren gi andm gi are both unit vectors. The corresponding line elements in the deformed conﬁguration are i i dR=n Gi dǫ anddQ=m Gi dǫ. The dot product of the two deformed line elements is p p niG mj(dǫ)2 = dR dQ cosΘ= niG nj dǫ miG mj dǫ cosΘ, ij | || | ij ij whereΘ is the angle between the two deformed line elements. Thus, niG mj cosΘ= p pij ; (2.33) i j i j n Gijn m Gijm and ifθ is the angle between the two undeformed line elements

i j cosθ=n gijm .

Thus, the information about change in length and relative rotation of two line elements is entirely contained within the deformed metric tensor. Example 2.4. Rigid body motion If the body undergoes a rigid-body motion show that every component of the Green–Lagrange strain tensor is zero. Solution 2.4. In a rigid-body motion the line elements do not change lengths and the angles between any two line elements remain the same. Hence, for any unit vectorsn andm in the undeformed conﬁguration

i j i j λ(n) =n Gijn = 1,λ(m) =m Gijm = 1, (2.34a) and cosΘ = cosθ, which means that p p i j i i j i j n Gijm =n gijmj n Gijn m Gijm . (2.34b)

Using equation (2.34a) in equation (2.34b), we obtain the condition

niG mj =n ig mj n i (G g )m j = 0, ij ij ⇒ ij − ij which must be true for all possible unit vectorsn andm and so by picking the nine diﬀerent combinations on non-zero components, it follows thatG ij g ij = 0 for all i, j. Hence, every component of the Green–Lagrange strain tensor is zero. − 2.4.3 Strain Invariants and Principal Stretches At any point, the direction of maximum stretch is given by

max λ(n) subject to the constraint n =1, n | | which can be solved by the method of Lagrange multipliers. We seek the stationary points of the function L(n, µ) =λ 2(n) µ g ninj 1 =n injG µ g njni 1 , − ij − ij − ij − whereµ is an unknown Lagrange multiplier andn i are the components of the unit vector in the basisg i. The condition∂L/∂µ = 0 recovers the constraint and the three other partial derivative conditions give ∂L = 0 G nj µg nj = (G µg )nj = 0. (2.35) ∂ni ⇒ ij − ij ij − ij Thus, the maximum (or minimum) stretch is given by non-trivial solutions of the equation (2.35), which is an equation that deﬁnes the eigenvalues and eigenvectors of the deformed metric tensor. The deformed metric tensor has real components and is symmetric which means that it has real eigenvalues and mutually orthogonal eigenvectors15. The eigenvectorsv are non-trivial solutions of the equation (v)=µv orG vj = µg vj = µv , etc, (2.36) G ij ij i where represents the deformed metric tensor as a linear map and the scalarsµ are the associated eigenvalues.G Note that because the eigenvalues are scalars they do not depend on the coordinate system, but, of course, the components of the eigenvectors are coordinate-system dependent. The eigenvalues (or in fact any functions of the eigenvalues) are, therefore, scalar invariants of the deformed metric tensor. Equation (2.36) only has non-trivial solutions if

det(G µg ) = 0. ij − ij and expansion of the determinant gives a cubic equation forµ, so there are three real eigenvalues and, hence, three distinct invariants. The invariants are not unique, but those most commonly used in the literature follow from the expansion of the determinant of the mixed deformed metric tensor in which the index is raised by multiplication with undeformed covariant metric tensor16 which means that the undeformed metric tensor becomes the Kronecker delta:

gikG µδ i = µ 3 +I µ2 I µ+I . (2.37) | kj − j| − 1 − 2 3 Expanding the determinant using the alternating symbol gives

eijk(G1 µδ 1)(G2 µδ 2)(G3 µδ 3) i − i j − j k − k � � =e ijk G1G2G3 µ δ1G2G3 +δ 2G1G3 +δ 3G1G2 +µ 2 δ1δ2G3 +δ 1δ3G2 +δ 2δ3G1 µ 3δ1δ2δ3 , i j k − i j k j i k k i j i j k i k j j k i − i j k 15This is a fundamental result in linear algebra is easily proved for distinct eigenvalues by considering the appropriate dot products between different eigenvectors and also between an eigenvector and its complex conjugate. For repeated eigenvalues the proof proceeds by induction on subspaces of successively smaller dimension. 16 i ik In other words we are definingG j =g Gkj, which is a slight abuse of notation: if we worked solely in the i ik i i i deformed metric then we would defineG j =G Gjk =δ j. The fact that our definition ofG j =δ j is why we can reuse the notation without ambiguity, but it has the potential to be confusing. 6 and on comparison with equation (2.37) we obtain

I =G 1 +G 2 +G 3 =G i = trace( ), (2.38a) 1 1 2 3 i G I =G 2G3 G 2G3 +G 1G3 G 1G3 +G 1G2 G 1G2, 2 2 3 − 3 2 1 3 − 3 1 1 2 − 2 1 1 1 = GiGj G i Gj = (trace( ))2 trace( 2) . (2.38b) 2 i j − j i 2 G − G I =e ijkG1G2G3 = G i = g ikG = g ik G = G/g, (2.38c) 3 i j k | j| | kj| | || kj| whereG= G ij is the determinant of the deformed covariant metric tensor andg= g ij is the determinant| of the| undeformed covariant metric tensor. | | The use of invariants will be an essential part of constitutive modelling, because the behaviour of a material should not depend on the coordinate system. The mutual orthogonality of the eigenvectors means that they form a basis of the Euclidean space � � I and if we choose the eigenvectors to have unit length we can write the eigenbasis asv I =v . The circumﬂexes on the indices are used to indicate the eigenbasis rather than the standard Cartesian basis. The components of the deformed metric tensor in the eigenbasis are given by b G�� =v �· (v �) =v �·µ � v � =µ � v �·v � (J not summed), IJ I G J I (J) J (J) I J � � whereµ (J) is the eigenvalue associated with the eigenvectorv J . Thus the components of the deformed metric tensor in the eigenbasis form a diagonal matrix with the eigenvalues as diagonal entries ( b b � µ(I) I= J, � � � � � GI J = b b =µ (I) δI J (not summed). 0 I= J, 6 or writing the components in the eigenbasis in matrix form   µ1 0 0 =  0µ 0  , (2.39) G 2 0 0µ 3 or X3 � � � = µ(I) vI v I . G � ⊗ I=1 These eigenvalues are related to the stretch because if we decompose a unit vector into the � � eigenbasis,n=n I vI , then from equation (2.32) v u p uX3 t � � � � � � � λ(n) = nI GI J nJ = nI µ(I) nI , � I=1

and so the stretches in the direction of the eigenvectors are the square-roots17 of the associated eigenvalues p λ(v �) λ � = µ � . (2.40) I ≡ (I) (I) � The three quantitiesλ (I) are called the principal stretches and the associated eigenvectors are the principal axes of stretch.

17We could have anticipated this result because the metric tensor was formed by taking the square of the lengths of line elements. These results motivate the deﬁnition of another symmetric tensor in which the eigenvectors remain the same, but the eigenvalues coincide exactly with the stretches

X3 X3 p � � � � � � √ = λ(I) vI v I = µ(I) vI v I = . U � ⊗ � ⊗ G I=1 I=1 The deformed metric tensor can be written as the product of the deformation gradient tensor and its transpose, so in matrix form in Cartesian coordinates (although the coordinates don’t matter because the quantities are all tensors)

c=F T F=U 2.

Recall that the matrixc is the Cauchy–Green deformation tensor, which is the deformed metric tensor when the Lagrangian and Eulerian coordinates are both Cartesian. We deﬁne

R=FU −1 F=RU, ⇒ and note that becauseU is symmetricU −1 =U −T and so

RT R=(FU −1)T (FU−1) =U −T FT FU−1 =U −1U2U−1 =I,

soR is orthogonal. Hence,F can be written as the product of an orthogonal transformation (rotation) andU which is known as the (right) stretch tensor 18. The interpretation is that, in addition to rigid-body translation, any deformation consists (locally) of a rotation and three mutually orthogonal stretches — the principal stretches, which are the square-roots of the eigenvalues of the deformed metric tensor.

2.4.4 Alternative measures of strain The strain is zero whenever the principal stretches are all of unit length, so in the eigenbasis representation the matrix of components of the right stretch tensor isU=I. However, the eigenbasis is orthonormal and therefore any other Cartesian basis is obtained via orthogonal transformation so that the matrix representation of the stretch tensor becomes,

U= QUQ T = QIQT =QQ T =I.

Hence, if the strain is zero thenU=I when the stretch tensor is represented in any orthonormal coordinate system. Transforming to a general coordinate system would giveU ij =g ij in the zero strain case, whereg ij is the undeformed metric tensor. We can deﬁne general strain measures using tensor functions of the right stretch tensor ( ). In the eigenbasis is diagonal and we deﬁne ( ) by F U U F U X3 � � � ( )= f λ(I) vI v I , F U � ⊗ I=1

′ wheref(x) is any monotonically increasing function such thatf(1) = 0 andf (1) = 1. The monotonicity ensures that an increase in stretch leads to an increase in strain; the conditionf(1) = 0

18This result is often called the polar decomposition. It is unique and the orthogonal transformation does not involve any reﬂection because detR> 0, which follows from detU> 0 and detF> 0. ′ means that when there is no stretch there is no strain; and the conditionf (1) = 1 is a normalisation to ensure that all strain measures are the same when linearised and consistent with the theory of small deformations. Admissible functions include 1 f(x) = (xm 1) (m= 0) and lnx(m = 0). m − 6 In orthonormal coordinate systems, we can then write the corresponding strain measures as 1 (Um I)(m= 0) and lnU(m = 0). m − 6 The casem = 2 corresponds to the Green–Lagrange strain tensor; the casem= 2 is the − Almansi strain tensor; the casem = 1 is called the Biot strain tensor and the casem = 0 is called the Hencky or incremental strain tensor. Note that it is only the Green–Lagrange and Almansi strain tensors that can be formed without prior knowledge of the eigenvectors, so these are most commonly used in practice.

2.4.5 Deformation of surface and volume elements Lecture 8 Having characterised the deformation of material line elements, we will alsoﬁnd it useful to consider the deformation of material surfaces and volumes. This will be particularly important when developing continuum conservation, or balance, laws.

Volume elements We already know from equation (1.46) that a volume element in the undeformed conﬁguration is 1 2 3 given by d 0 = √gdξ dξ dξ . The corresponding volume element in the deformed conﬁguration is V 1 2 3 given by d t = √Gdξ dξ dξ . Hence, the relationship between the two volume elements is V p d = G/gd . (2.41a) Vt V 0 From equation (2.19) i ∂XI ∂XI ∂ξ FIJ = = i , ∂xJ ∂ξ ∂xJ which means thatJ = detF= √G/√g from equation (1.44). Thus we can also write the change in volume using the determinant of the deformation gradient tensor

d = detFd =Jd . (2.41b) Vt V 0 V 0 Surface elements If we now consider a vector surface element given byda=nda in the undeformed conﬁguration, where da is the area andn is a unit normal to the surface. We can form a volume element by taking the dot product of the vector surface element and a line elementdr, so

d = dv=dr·da=n dr da. V0 I I From equation (2.41b) the corresponding deformed volume element is

d = dV=Jn dr da=N dR dA=dR·dA, Vt I I I I wheredA=NdA is the deformed vector surface element. From equation (2.19) dR I =F IJ drJ , so Jn dr da=N F dr dA, Jn da=N F dA I I I IJ J ⇒ J I IJ dA =N dA=Jn F −1 da=JF −1da NdA=JF −T nda, (2.42) ⇒ I I J JI JI J ⇒ a result known as Nanson’s relation. Converting to general tensor notation, we decompose the deformed area vectordA=NdA into the deformed contravariant basis in the Eulerian coordinatesχ i, to obtain ∂X I ∂xJ ∂X I ∂xJ ∂ξj ∂xJ ∂ξj dA = dAI =J daJ =J daJ =J daJ =J daj, (2.43) i ∂χi ∂X I ∂χi ∂χi ∂χi ∂ξj ∂χi where the undeformed area vector has been decomposed into the undeformed contravariant basis i in Lagrangian coordinates, da=da ig . Thus, the transformation is similar to the usual covariant transformation between Eulerian and Lagrangian coordinate systems, but with an additional scaling byJ to compensate for the change in area.

2.4.6 Special classes of deformation Required If the deformation gradient tensor,F IJ , is a function only of time (does not vary with position) Reading then the deformation is said to be homogeneous. In this case, the equation (2.19) (not covered in lectures) dRI (t) =F IJ (t) drJ , and integrating between two material points, at positionsa andb in the undeformed conﬁguration, gives A (t) B (t) =F (t)(a b ), (2.44) I − I IJ I − I whereA andB are the corresponding positions of the material points in the deformed conﬁguration. Equation (2.44) demonstrates that straight lines are carried to straight lines in a homogeneous deformation. Furthermore, any homogeneous deformation can be written in the form

XI =A I (t) +F IJ (t)x J . Pure strain is such that the principal axes of strain are unchanged, which means that the • rotation tensor in the polar decomposition is the identityR=I. Rigid-body deformations are such that the distance between every pair of points remains • unchanged and we saw in Example 2.4 that the strain is zero in such cases. We can also show that in this case the invariants areI 1 =I 2 = 3 andI 3 = 1. Isochoric deformation is such that the volume does not change, which means thatJ= • √I3 = 1. Uniform dilation is a homogeneous deformation in whichF = λδ , which means that all • IJ IJ the principal stretches areλ. In other words, the material is stretched equally in all directions. Uniaxial strain is a homogeneous deformation in which one of the principal stretches isλ • and the other two are 1, i.e. the material is stretched (and therefore strained) in only one direction. Simple shear is a homogeneous deformation in which all diagonal elements ofF are 1 and • one off-diagonal element is nonzero. 2.4.7 Strain compatibility conditions Optional The strain tensors are symmetric tensors which means that they have six independent components, Reading but they are determined from the displacement vector which has only three independent components. If we interpret the six equations (2.31) as a set of differential equations for the displacement, assuming that we are given the strain components, then we have an overdetermined system. It follows that there must be additional constraints satisfied by the components of a strain tensor to ensure that a meaningful displacementfield can be recovered. One approach to determine these conditions is to eliminate the displacement components from the equations (2.31) by partial differentiation and elimination, which is tedious particularly in curvilinear coordinates. A more sophisticated approach is to invoke a theorem due to Riemann that the Riemann–Christoffel tensor formed from a symmetric tensorg ij should be zero if the tensor is a metric tensor of Euclidean space. The Riemann–Christoffel tensor is motivated by considering second covariant derivatives of vectors

v = (v ) = (v ) Γ k (v ) Γ k (v ), i|jr i|j |r i|j ,r − ir k|j − jr i|k see Example Sheet 1, q. 10. Writing out the covariant derivatives gives

v =v Γ l v Γ l v Γ k v +Γ k Γl v Γ k v +Γ k Γl v . (2.45a) i|jr i,jr − ij,r l − ij l,r − ir k,j ir kj l − jr i,k jr ik l Interchangingr andj gives

v =v Γ l v Γ l v Γ k v +Γ k Γl v Γ k v +Γ k Γl v , (2.45b) i|rj i,rj − ir,j l − ir l,j − ij k,r ij kr l − rj i,k rj ik l and subtracting equation (2.45b) from equation (2.45a) gives v v = Γl Γ l +Γ k Γl Γ k Γl v R l v , i|jr − i|rj ir,j − ij,r ir kj − ij kr l ≡ .ijr l after using the symmetry properties of the Christoffel symbol and the partial derivative. The quantities on the left are the components of the type (0,3) tensors andv l are the components of l a type (0,1) tensor, which means thatR .ijr is a type (1,3) tensor called the Riemann–Christoffel l tensor. Thus ifR .ijr = 0, then covariant derivatives commute. The tensor is a measure of the curvature of space and in Cartesians coordinate in an Eulerian space the Christoffel symbols are all l zero, which means thatR .ijr = 0 in any coordinate system in our Eulerian space. A physical interpretation for compatibility conditions is that a deformed body must “fit together” so that a continuous region of space isfilled. Thus, the tensorG ij associated with the deformed material points must be a metric tensor of our Euclidean space and the required conditions are that the Riemann–Christoffel symbol associated withG ij must be zero. We can use the relationship 1 γ = (G g ) G = 2γ +g , ij 2 ij − ij ⇒ ij ij ij to determine explicit equations for the conditions on the components of the Green–Lagrange strain tensorγ ij, but these are somewhat cumbersome.

2.5 Deformation Rates

In many materials, particularlyfluids, it is the rate of deformation, rather than the deformation itself that is of most importance. For example if we take a glass of water at rest and shake it (gently) then wait, the water will come to rest and look as it did before we shook it. However, the configuration of the material points will almost certainly have changed. Thus, thefluid will have deformed significantly, but this does not affect its behaviour. The obvious “rates” are material time derivatives of objects that quantify the deformation. The general approach when computing material derivatives of complicated objects is to work in the undeformed configuration, by using (or transforming to) Lagrangian coordinates, in which case the time-derivative is equal to the partial derivative and the coordinates arefixed in time. We shall use such an approach throughout the remainder of this chapter. A measure of the rate of deformation of material lines is given by ! � � � D 2 ∂ 2 ∂ i j ∂Gi ∂Gj i j dR = dR = Gij dξ dξ = ·Gj +G i· dξ dξ , Dt | | ∂t r | | ∂t ξ ∂t ξ ∂t ξ because the Lagrangian coordinatesξ i arefixed on material line and are independent of time. The fact that time and the Lagrangian coordinates are independent means that

DG ∂G ∂2R ∂ DR � i i v Gk Gk = = i = i = ,i = Vk ,i =V k i , (2.46) Dt ∂t ξ ∂t∂ξ ∂ξ Dt || where indicates covariant diﬀerentiation in the Lagrangian deformed basis with covariant base || i vectorsG i and contravariant base vectorsG . Thus, D � � dR 2 = V Gk·G +V G ·Gk dξidξj = (V +V ) dξidξj. Dt | | k||i j k||j i i||j j||i Now because the undeformed line elementdr has constant length, we can write D � D � D � dR 2 = dR 2 dr 2 = 2γ dξidξj = 2˙γ dξidξj, Dt | | Dt | | −| | Dt ij ij where Dγ 1 ˙γ = ij = (V +V ), (2.47) ij Dt 2 i||j j||i is the (Lagrangian) rate of deformation tensor and is the material derivative of the Green–Lagrange strain tensor. Alternatively, we could have decomposed the velocity into the (curvilinear) Eulerian coordinates χj, in which case ∂v ∂χl ∂V ∂χl = = V Gk, ∂ξi ∂ξi ∂χl ∂ξi k||l and then ! D � ∂χl ∂χl dR 2 = V Gk·G + V Gk·G dξidξj, Dt | | ∂ξi k||l j ∂ξj k||l i ! ∂χl ∂χk ∂χl ∂χk = V Gm·G + V Gm·G dξidξj, ∂ξi k||l ∂ξm j ∂ξj k||l ∂ξm i

∂χk ∂χl = (V +V ) dξidξj = 2D dχkdχl, (2.48) k||l l||k ∂ξi ∂ξj k l where 1 � D = V +V , i j 2 i||j j||i is the Eulerian rate of deformation tensor in components in the Eulerian curvilinear basis. At a given time, we can recover the Lagrangian rate of deformation tensor from the Eulerian by ♭ simply applying the appropriate covariant transformation;˙γij is the pullback ofD and is therefore equal toD ij componentwise. However, an important point is that although the Lagrangian rate of deformation tensor is the material time-derivative of the (Green-Lagrange) strain tensor. The Eulerian rate of deformation tensor, in general, does not coincide with the spatial or material time-derivative of an Eulerian strain tensor because the Eulerian coordinates themselves vary with time. In the Eulerian representation (in Cartesian coordinates for convenience) D � D DE D(dX ) D(dX ) dR 2 = (2E dX dX ) = 2 IJ dX dX + 2E I dX + 2E dX J , Dt | | Dt IJ I J Dt I J IJ Dt J IJ I Dt and the material time derivative of the Eulerian line element is not zero in general. In fact, making use of Lagrangian coordinates, D(dXI ) D ∂XI ∂ DXI = dxJ = dxJ , Dt Dt ∂xJ ∂xJ Dt because the material derivative and the Lagrangian coordinates are independent. By deﬁnition, the material derivative of the Eulerian position is the Eulerian velocity, which in Cartesian coordinates is given byV=V I eI , so

D(dXI ) ∂VI ∂VI ∂XK ∂VI = dxJ = dxJ = dXK . Dt ∂xJ ∂XK ∂xJ ∂XK Thus, D DE (2E dX dX ) = 2 IJ +E V +E V dX dX Dt IJ I J Dt IK K,J KJ K,I I J i j = 2Di j dχ dχ = 2DIJ dXI dXJ , after transformation to Cartesian coordinates. The equation is valid for arbitrary dXI and dXJ , so E˙ =D E V E V . (2.49) IJ IJ − IK K,J − KJ K,I Hence, if the Eulerian rate of deformationD IJ is zero, but the material is strainedE IJ = 0, an Eulerian observer will perceive a rate of strain of 6 E˙ = E V E V , IJ − IK K,J − KJ K,I which is a consequence of the movement of the material lines in theﬁxed Eulerian coordinate system.

2.5.1 Rates of stretch, spin & vorticity Required The Eulerian velocity gradient tensor is given by Reading (brieﬂy ∂VI covered in Li j =V i j, orL IJ = , lectures) || ∂XJ which means that the (Eulerian) deformation rate tensor is the symmetric part of the (Eulerian) velocity gradient tensor, 1 � D = L +L . (2.50) i j 2 i j j i The antisymmetric part of the (Eulerian) velocity gradient tensor is called the spin tensor 1 � W = L L , i j 2 i j − j i and so by construction

Li j =D i j +W i j. The vorticity vector is a vector associated with the spin tensor and is deﬁned by 1 � ωk =ǫ k i jW = ǫk i j V V =ǫ k i jV , j i 2 j||i − i||j j||i whereǫ i j k is the Levi-Civita symbol in the Eulerian curvilinear coordinatesχ i, which means that

ω = curlR V, in the Eulerian framework. We can now use the same arguments as in 2.4.2 to demonstrate that the the diagonal entries § DII (not summed) are the instantaneous rates of stretch along Cartesian coordinate axes and that the oﬀ-diagonal entries represent the instantaneous rate of change in angle between two coordinate lines, or one-half the rate of shearing. In addition, equation (2.46) expresses the material derivative of our Lagrangian base vectors, i.e. how lines tangent to the Lagrangian coordinates evolve with the deformation. Expressing the equation (2.46) in Cartesian coordinates gives

DG DF ∂ξk i =V Gi Ii e =V e , Dt k||i ⇒ Dt I k||i ∂X I I

k DFIi ∂ξ ∂XK DFIJ =v k i =V I i =V I,K i =V I,K FKi =V I,K FKJ , (2.51) ⇒ Dt || ∂XI || ∂ξ ⇒ Dt which demonstrates that the material derivative of the deformation gradient tensor is simply the velocity gradient tensor composed with the deformation gradient. A physical interpretation is that it is gradients (differences) in velocity that lead to changes in material deformation, which explains the above identification of the rate of deformation with the velocity gradient. We next introduce the relative deformation gradient tensor, in which we measure the deformation from the initial state to a timeτ and then the deformation from timeτ to a subsequent timet. Then ∂XI (t) ∂XI (t) ∂XK (τ) FIJ (t) = = = F˜IK (t,τ)F KJ (τ), ∂xJ ∂XK (τ) ∂xJ where F˜IK (t,τ) is called the relative deformation gradient tensor. The deformation gradient at the fixed timeτ is independent oft so that

DF DF˜ (t,τ) IJ = IK F (τ), (2.52) Dt Dt KJ and taking the limitτ t of equation (2.52) and comparing to equation (2.51) we see that →

DF˜ (t,τ) DF˜(t,τ) IK =V , or written as matrices =L, Dt I,K Dt τ=t τ=t and the instantaneous relative deformation tensor is given by the Eulerian velocity gradient tensor. We can now use the polar decomposition to write

DF˜ D(R˜U˜) = =L=D+W Dt Dt τ=t

DU˜ DR˜ R˜ + U˜ =D+W. ⇒ τ=t Dt Dt τ=t τ=t τ=t Now the relative rotation R˜ and stretch U˜ tensors both tend to the identity asτ t and there is diﬀerence between the state atτ andt. Thus, →

DU˜ DR˜ + =D+W. Dt Dt τ=t τ=t The (right) stretch tensorU is symmetric which means that

DU˜ DR˜ =D and =W, Dt Dt τ=t τ=t with the interpretation that the instantaneous material rate of change of the stretch tensor, i.e. the rate of stretch is given by the Eulerian deformation rate tensor; and the instantaneous material rate of rotation is given by the spin tensor.

2.5.2 Material derivative of volume and area elements Lecture 8 A volume element in the current (deformed) position is given by (ctd)

d = √Gdξ 1dξ2dξ3, Vt so the material derivative of the volume gives the dilation rate

Dd D√G V t = dξ1dξ2dξ2. Dt Dt We use a similar approach to that in the proof of the divergence theorem to determine

D√G 1 DG 1 ∂G DG = = ij . Dt 2√G Dt 2√G ∂Gij Dt From comparison with equation (1.56) we have

∂G =GG ij, (2.53) ∂Gij and using equation (2.46)

DG DG DG ij = i ·G + j ·G =V +V . Dt Dt j Dt i i||j j||i Hence, � D√G 1 ij 1 j i i i = GG (Vi j +V j i) = √G V j +V i = √GV i = √GV , (2.54) Dt 2√G || || 2 || || || ||i after change of coordinates from Lagrangian to Eulerian. In other words, the rate of change of a volume element is given by Dd V t =V i √Gdξ 1dξ2dξ3 =V i d , Dt ||i ||i Vt and the relative volume change, or dilation, is

1 Dd t i V =V = divR V, (2.55) d Dt ||i Vt the divergence of the Eulerian velocityﬁeld. The material derivative of an area element can be obtained by using the equation (2.43) to map the area element back into the Lagrangian coordinates j j j DdA i D ∂ξ DJ ∂ξ D ∂ξ = J daj = +J daj. Dt Dt ∂χi Dt ∂χi Dt ∂χi

j j k ∂ξ l ∂ξ k l =V J daj V J daj =V dA V dA . (2.56) ||k ∂χi − ||i ∂χl ||k i − ||i l p Thefirst term is obtained on using the result (2.54) divided by √g becauseJ= G/g and √g is constant under material differentiation. The second term uses the result that D ∂ξj ∂ξj = V l , (2.57) Dt ∂χi − ||i ∂χl which has a non-trivial derivation in curvilinear coordinates19 because we need to be careful about what is being heldfixed when computing all the derivatives. Firstly, we note that ! i i i D ∂χ D ∂χ ∂χ i = χi = ,j +V l ,j +V l Γ χm, (2.58) Dt ∂ξj Dt ,j ∂t ∂χl l m ,j r t 19The derivation in Cartesians is even simpler because the complexity of covariant differentiation isfinessed, D ∂XJ ∂ DXJ ∂VJ ∂VJ ∂XK ∂XK DF = = = =V J,K , or =LF; Dt ∂xI ∂xI Dt ∂xI ∂XK ∂xI ∂xI Dt and then D ∂XJ ∂xI D ∂XJ ∂xI ∂XJ D ∂xI D = + = (δJL) = 0, Dt ∂xI ∂XL Dt ∂xI ∂XL ∂xI Dt ∂XL Dt ∂XJ D ∂xI D ∂XJ ∂xI ∂XK ∂xI = = V J,K = V J,K δKL = V J,L ⇒ ∂xI Dt ∂XL − Dt ∂xI ∂XL − ∂xI ∂XL − − −1 D ∂xK ∂xK DF −1 = V J,L , or = F L. ⇒ Dt ∂XL − ∂XJ Dt − i because although the quantityχ ,j is a two-point tensor, the Lagrangian component is unaffected i by material differentiation. Thus, the material derivative ofχ ,j is the same as for a contravariant vector component. The velocity is given by

∂R(χi(r,t)) ∂R ∂χi ∂χi V= = = G , ∂t i ∂t ∂t i r ∂χ r r which means that the components of the Eulerian velocity vector in the basisG i are given by i ∂χi V = ∂t r. Thus, equation (2.58) can be rewritten as | ! i i l i i D ∂χ ∂χ ∂χ ∂χ i dχ i = ,j + ,j +V l Γ χm = ,j +V l Γ χm, (2.59) Dt ∂ξj ∂t ∂t ∂χl l m ,j dt l m ,j r r t i where d/dt is the total derivative with respect to time and, in the Eulerian framework,χ ,j is a function ofχ i(t) andt. Now, ! ! ! dχi d ∂χi(ξj, t) ∂ ∂χi ∂ ∂χi ∂V i ,j = = = = , dt dt ∂ξj ∂t ∂ξj ∂ξj ∂t ∂ξj because the Lagrangian coordinates are not functions of time and so the partial time derivative is equal to the material derivative. Equation (2.59) can therefore be written as ! i h i D ∂χ i i =V i +V l Γ χm = V i +V l Γ χm =V i χm, Dt ∂ξj ,j lm ,j ,m lm ,j ||m ,j after using the chain rule. The right-hand side has the appropriate two-point tensor transformation properties. In order to derive equation (2.57), we use the fact that ! D ∂ξk ∂χi D � = δk = 0, Dt ∂χi ∂ξj Dt j to write ! D ∂ξk ∂χi ∂ξk D ∂χi + = 0 Dt ∂χi ∂ξj ∂χi Dt ∂ξj ! k i k i k m D ∂ξ ∂χ ∂ξ D ∂χ ∂ξ i ∂χ = = V m ⇒ Dt ∂χi ∂ξj − ∂χi Dt ∂ξj − ∂χi || ∂ξj and multiplying both sides by∂ξ j/∂χl gives D ∂ξk ∂ξk = V i , ⇒ Dt ∂χl − ∂χi ||l as required. We can interpret the result for the transformation of the area element as follows

DdA i =V k dA V l dA , Dt ||k i − ||i l Material Derivative = Volume Expansion Expansion normal of Area − to Area Element. 2.6 The Reynolds Transport Theorem

The Reynolds transport theorem concerns the time derivative of integrals over material volumes (although it can be generalised to arbitrary moving volumes by considering a new set of coordinates that move with the volume in question rather than with thefluid) Z ∂ (t) d (t) d I = I = φd t. ∂t dt dt Ωt V Integratingφ(R,t) over space means that is a function only oft so the total derivative is equal to the partial derivative. Firstly we transformI the material volume to the equivalent undeformed volume Z p Z p d d 1 2 3 d I = φ G/g √gdξ dξ dξ = φ G/gd 0, dt dt Ω0 dt Ω0 V and because the undeformed volume isfixed, we can take the time derivative under the integral so ! that Z Z p d ∂ p ∂φp ∂ G/g I = φ G/g d 0 = G/g+φ d 0. dt Ω0 ∂t V Ω0 ∂t ∂t V In the undeformed configuration we take the partial time derivative withr held constant, i.e. the material derivative D/Dt=∂/∂t, and Z s ! Z s Z d Dφ G φ D√G Dφ i G Dφ i I = + d 0 = +φV i d 0 = +φV i d t, dt Ω0 Dt g √g Dt V Ω0 Dt || g V Ωt Dt || V after using equation (2.54) and transforming back to the material volume. Thus, we can write Z Z d Dφ ∇ · φd t = +φ R V d t, dt Ωt V Ωt Dt V or expanding the material derivative Z Z Z d ∂φ ·∇ ∇ · ∂φ ∇ · φd t = +V R φ+φ R V d t = + R (φV) d t; dt Ωt V Ωt ∂t V Ωt ∂t V and using the divergence theorem Z Z Z d ∂φ φd t = d t + φV·Nd t, dt Ωt V Ωt ∂t V ∂Ωt S which means that the time rate of change of a quantity within a material volume is the sum of its rate of change obtained by treating the volume asfixed and theflux of the quantity across the boundaries (due to the movement of the boundaries). In fact, the result is the multi-dimensional generalisation of the Leibnitz rule for differentiating under the integral sign when the limits are functions of the integration variable. Consider R R Z b(t+δt) b(t) ∂I dI ∂ b(t) f(x, t+δt) dx f(x, t) dx = = f(x, t) dx = lim a(t+δt) − a(t) , ∂t dt ∂t a(t) δt→0 δt from the fundamental definition of the partial derivative. Splitting up the domain of integration for thefirst integral gives R R R R a(t) b(t) b(t+δt) b(t) ∂I f(x, t+δt) dx+ f(x, t+δt) dx+ f(x, t+δt) dx f(x, t) dx = lim a(t+δt) a(t) b(t) − a(t) . ∂t δt→0 δt Now, assuming that all the limits exist we can write Z ∂I b(t) [f(x, t+δt) f(x, t)] = lim − dx ∂t a(t) δt→0 δt "Z Z # 1 a(t) b(t+δt) + lim f(x, t+δt) dx+ f(x, t+δt) dx . δ→0 δt a(t+δt) b(t) Thefirst integral is simply the integral of∂f/∂t and the others can be written using the mean value theorem as the product of the length of the integration interval and the function evaluated at some point within the interval, so Z ∂I b(t) ∂f 1 = dx + lim [a(t) a(t+δt)]f(x,τ a) + [b(t+δt) b(t)]f(x,τ b) , ∂t a(t) ∂t δt→0 δt { − − } wherea(t)<τ a < a(t+δt) andb(t)<τ b < b(t+δt). Using Taylor’s theorem to expand the terms a(t+δt) andb(t+δt) we obtain Z Z b(t) b(t) ∂I ∂f ′ ′ ∂f ′ ′ = dx + lim a (t)f(x,τ a) +b (t)f(x,τ b) = dx a (t)f(x, a) +b (t)f(x, b), ∂t a(t) ∂t δt→0 {− } a(t) ∂t − becauseτ a a andτ b b asδt 0, where the last two terms can be identified as theflux off out of the moving→ domain.→ →