Kinematics 1 A body is a set of

Kinematics 1 A body is a set of material particles : The concept of material particles is primitive. So it can not be defined but its properties can be described. In particular two particles can not be in the position at the same time. Therefore the mapping function is invertible

Kinematics 2 Operative description of the motion of a body

Kinematics 3 Material time derivative: This is the derivative following the material particle, i. e. along its path. The definition is not dependent upon a coordinate system Where A is a generic local quantity: scalar, vector or tensor

Kinematics 4 In Eulerian description the velocity in a given point of space at a given time, belonging to the region of space occupied by a body at time t, is given by the one of the material particle that at this time is passing in that position In particular: A can be any scalar field, spatial vector field or second-order tensor field In case of material time derivative: Scalar field

Kinematics 5

Kinematics 6 In case of material time derivative we can write (see also slides of notes for a more rigorous notation and relationships with Eulerian fields): Spatial vector field : Second order tensor field See notes on curvilinear coordinate section slides for details on gradient operator (slide 78) and notes on material derivative for more details on chain rule and its physical implications here

Kinematics 7 Some definitions i-sime component of field velocity at time t at unknown z(α)

Kinematics 8 Frame indifference Two frames of reference are in general rototraslating (plus possible reflection) between each other. In the case they are stationary each other the change of frame is trivial, since it becomes (if two different bases are adopted in them) the classic change of base. In other words each frame of reference creates a correspondence between physical reality and Euclidean point space and real time axis. So a change of frame is a mapping of space time onto itself which preserves distances and time intervals (including also time order) in the sense that in each frame of reference I can see the same physical event associated to a different Euclidean point and time t, but among two different frames the previous relationships of preservation must hold

Kinematics 9 Frame indifference A frame of reference for position is a set of objects whose mutual distances remain unchanged during the period of observation and which do not all lie in the same plane On the bases of properties required above: This is trivial see literature of Noll for a rigorous approach (omitted here for the sake of brevity). In the following we give a clear explanation of the 17. 1 based upon classic physical considerations. In particular beside the time shift factor issue, two frames of reference are in general rototraslating (plus possible reflections) between each other. In this respect we will see that c(t) could be seen as position in the starred frame of origin of axis adopted for the unstarred frame. Note that in the particular case the two frames are stationary each other the change of frame is trivial, since it reduces to (if two different bases are adopted in them) the classic change of base for spatial vectors, (plus, for the points, the translation if two different origins are also adopted) 2

Kinematics 10 Frame indifference Change of frame preserves time interval (and temporal order) and distance among points (and so angles) The vectors b(i) are seen as rotated by Q in b*(i)

Kinematics 11 Frame indifference The free spatial vector z-z(o) solidal with unstarred frame is viewed as rotate by Q in the starred one. Q is the orthogonal operator by which the old unstarred frame is seen in starred one at a given t. Each change of frame at a given t can be seen as due to an orthogonal operator (it can be proved see Noll), and obviously each orthogonal operator can be seen as associated to a change of frame at a given t. So Q is the rotation (plus possible reflections see on the next slide *) operator I must apply to the vector that in the unstarred frame is seen as z-z(o) to see it in the starred. In particular to see in starred frame the same point corresponding to point z in unstarred frame we have to rotate of Q z-z(0) and than sum the point z*(o), which correspond to z(o) seen in starred to take into account also translation. ( Note that I can directly think to z(o) as the origin O of arbitrary axis in unstarred and z*(0) 0 -0* ( O* is the origin of arbirtrary axis in starred frame) think as z-0 ( named z in figure) and z(o)-0 ( named z(o) in figure)are seen and apply linearity of Q

Kinematics 11 bis Frame indifference Let think to a vector spatial solidal with a frame associated to a rotating platform. It will be seen as constant in this frame, while it will be rotatining in a frame solidal to a “fixed” frame. At a given time t (corresponding to a given t*) it can be seen as the same vector, by adopting the same base in the two frames, but after a dt =dt*in the fixed frame it will be seen rotated. This argumentation is valid for any not mutually stationary couple of frames. In alternative if at the time t we assume that a new frame solidal to b* is created (starting from frame solidal to b) by a reflection of this last trough a plane , a vector v solidal to b is seen at t as the reflection Qref(v) in b* Qref being such reflection. In particular a vector v solidal to b* is actually the vector Qref(v) (inverse of Qref is itself) as must be seen in b. More in general we can think that b is also rotranslating respect to b* (Actually this corresponds to an observer which observe the reflected image from a mirror on the plane, and this mirror rototranslate respect to b*) As a matter of fact we can introduce a third frame a rototranslating solidal to b so a vector v solidal to b is always seen in a as Qref(v). The system a is only rototranslating respect b* and therefore the vector Qref(v) is in turn seen in b* as Q(Qref(v)) with Q defined as above , i. e. the rotation which takes b* into a (which at time t were overimposed). See also next slide* A=(QQref )t=Qreft. Qt=Qref-1 Q-1=(QQref )-1 so it is orthogonal with det =-1 from properties of det. So a spatial vector v solidal to b is seen in b* as A(v) A being A not proper orthogonal transformation. Finally the translation of origin OO* must be added to see the same point P-0=v as described above. Note that since det is a continuum function of components and we impose Q=Q(t) continuos (it means its components continuous with time) if a frame is just rototranslating (det=1) is as a function of time it can not suddenly undergoes a reflection at a given t, since composing with reflection det becomes suddenly =-1 and this is not possible by chain rule it must be continuous. But if we initially impose as above a reflection then we can compose it with rototranslation (since det=-1 constant)

Kinematics 12 Frame indifference Remember that Qt =Q-1 **In operative terms, on the bases of the argumentation expressed in previous slides and on the right, a vector v solidal to the unstarred frame is seen a a time t (corresponding to a t*=t-a) as a Q(t)v in the starred frame where Q(t) is an orthogonal transformation. In particular we can always fix at at a t o in the past two overimpoised orthonormal bases b and b* respectively solidal to frame named b and frame b* and after a generic time deltat=deltat* at time t (corresponding to t*=t-a)the base b=b* (at to) is seen transformed (rotraslated plus possible reflections) by frame b*. We can define the linear transformation Q(t) mapping bi=b*i at to into the vectorbi (t) at t i=1, 2, 3 (which is the base b seen in starred frame at t). By linearityit is easy to see that this transformation mantains the scalar product (i. evit is orthogonal) and that it is unique since map a base b* into another base b(t) ( b as seen in frame b* at t. Note that in frame b* the base b* is seen fixed , i. e. not function of t* and so of t). An arbitrary vector v solidal to frame b is given by components fixed c 1, c 2, c 3 in b, (in frame b b is seen not t dependent)while in frame b* it is seen at t as: c 1 b 1(t)+c 2 b 2(t)+c 3 b 3(t)=u(t) =Q(t) (c 1 b*1(t)+c 2 b*2(t)+c 3 b*3(t))=Q(t) (c 1 b 1(to)+c 2 b 2(to)+c 3 b*3(to)) =Q(t)v where c 1 b 1(to)+c 2 b 2(to)+c 3 b*3(to)) represents the vector v as seen fixed in frame b. (first equality by linearity and definition of Q(t)) Note that a reflection transformation is orthogonal and sym. So its inverse is the tensor itself. Note that defining a reflection as Q(e)=-Q(e) and identity in the plane ort. to e we can always associate to such Q a tern cartesian formed by e and 2 vectors in plane ort. to e

Kinematics 12 bis Frame indifference Remember that Qt =Q-1 * iterating the procedure of previous slide we can extend the considerations of the previous slide to the composition of a general number of proper and not proper (i. e. reflections) tensors. Note that any vector in an arbitrary frame f 1 can be seen as Qv with Q arbitrary orthogonal tensor by another frame. For this reason the frame indifference is required for any Q as will be shown in the following slides. Think to the two frames (f 1 and f 2) at time t f 2 is defined as rotated plus possible reflections of Qt referred to f 1. So f 1 is seen by f 2 as transformed by operator Q at t and a vector v in it is seen as Qv by f 2. In particular think to two ortonormal bases respectively associated to f 1 and f 2(b , b*) coincident at a given arbitrary time tp in the past and after a deltat, at time t, b* can be imposed as the transfomed of b by the Qt. . So b is the transfomed by Q of b* at time t. As matter of fact vector v at time tp has same components c 1, c 2, c 3 in b and b* and they are fixed in b. In b* this solidal vector v at t has seen as Qv

Kinematics 13 Frame indifference As seen if starred frame rotates (and reflects) as a function of t Q=Q(t) and if translates as a function of t also we have z*o(t), i. e. we have transform the spatial vector plus the translation as z(o) is seen in b*. Think z(o) as an arbitrary origin for instance See slide 49 for an operative example of determination of relatioship for components

Kinematics 14 Frame indifference So they are seen as the same base In the corresponding frames of reference Associated means solidal to So ei and e*i allow to identify the change of frame as a change of base, in general function of time t. In particular the operator Q is the operator that maps ei* into ei (see also slide 49). As a matter of fact if at a given t (corresponding to a t*) ei* and ei are the same base initially located at the same point O=O* after a dt=dt* they are rototraslated (plus reflections) to each other. Now let think to an objective point of space P that is described just a t (corresponding to t*) by the vector solidal to unstarred ei z-0. Now let we see it at t applied to o* in ei* (in the sense that we are using the component of z-0 in ei and we will use them now in e*i. So this vector in e*i named r will not represent in general the same objective spatial vector z-0). Now if we apply the Q mapping e*i into ei to this vector r in e*i, it by linearity becomes the spatial vector obiective. Finally to obtain also the objective point P we need to sum also the translation at t (corresponding to t*) O-O* of unstarred frame to the starred frame

Kinematics 15 Frame indifference

Kinematics 16 Frame indifference 1. 2. 1 -15 Q is orthogonal and using associative properties of tensors A(Bv)=(AB)v A, B tensor of second order and b vector

Kinematics 17 Frame indifference For instance by its definition it is possible to show that a gradient of a scalar field is frame invariant (proof omitted for the sake of brevity) Since this holds at any t corresponding to a t*, it is valid also at t 0 corresponding to t 0* in which z=zk so z*k is the reference configuration in unstarred seen in starred frame. So 12. 2 -2 defines the map in starred frame with z*k at t 0* as reference configuration. So zk and z*k label the same particle In this respect we can say that a law T=T(z, t) with z=X(zk, t) is frame invariant if T*(z*, t*)=Q(t)T(z, t)Q(t)t with z*=X*(zk*, t*) =z 0(t)*+Q(t)[X(z, t)-z 0] and with t*=t-a In general if in a frame I see a material particle at a point P=z-0 as fixed, but this frame is moving with respect to starred frame (see example of rotating platform frame respect to a “fixed” frame) this material particle must be seen consistently move in the other frame due to relative rototraslation (plus reflection) between the two frames. More in general if the material particle is also moving with respect to the first frame by X (zk, t) , its motion seen by the other frame will be given by 1. 2. 2 -2. (Composition of the two motions). This motion will be in general described in starred frame also by X*(z*k, t*).

Kinematics 18 Note that v is velocity of the motion respect to z o (which is an arbitrary vector solidal to unstarred frame, i. e. the arbitrary origin) If we follow a material particle zk in unstarred frame its motion is given by X(zk, t) and 1. 2. 2. -2 provides the relationship it holds when it is seen in as X*(zk*, t*) in starred frame. Both member of equation by t*(t) can be seen only function of t and they are equal at any t, so their derivative in t must be equal. But derive in t the left side is equivalent to derive at the corresponding t*by chain rule (since t*=t-a), so we obtain at first side of equation the velocity v* in starred frame of material particle (labelled z*k at t*0) at t* and at second side appears the velocity v of the material particle (labelled zk at t 0 corresponding to t 0*, i. e. the same particle) at the corrresponding t. Since this result holds for any arbitrary material particle it holds also for the one that in unstarred frame pass for a z 1 assigned at t. So the velocity v in z 1 is related to the velocity seen in starred frame v* in the z 1* corresponding to z 1 at the corresponding t* by this relationship.

Kinematics 19 If T=identity tensor I tensor in b Therefore 1. 2. 2 -7 velocity is not an objective vector. In addition to tranformation law there are two terms : one related to velocity of translation of b* trought the velocity of translation of and one associated to derivative of Q (taking into account therefore velocity of rotation plus possible reflection of b* referred to b)

Kinematics 20 It is possible to show that (proof omitted here for the sake of brevity): = See A. 7. 3 -4 and A. 7. 3 -6 (slide 88) of curvilinear coordinate system for definition of this tr operator: Third order tensor defined trought its coordinates in a given cartesian system. In this case it is defined in a right handed e* It can be proved that to be frame invariant once it has been defined for instance in e, in e* it will be given by: Where Q is tranformation from e to e*. See slides 79 -81 85 98 100 and 102 of section curvilinear coordinate systems for more details

Kinematics 21 Mass conservation Conservation of mass: the mass of a body is time-independent Mass is a primitive concept, in the sense it is not defined, but its properties can be defined. In particular it is postulated that: Mass and Volume are a frame-independent scalars (and consequently also the density of mass)

Kinematics 22 Transport theorem We want to determine now a local conditions to express the conservation of mass: to this aim we can use the divergenge theorem, but we need to take into account that the configuration of the body is time dependent. This issue involves the use of Transport theorem Note that we will use in the following extension to tensor fields of relationships that in basic courses are geneally provided only for vector fields (such as as the Green’transformations and the use of chain rule in derivation of material time derivatives). These extensions are well-established and can be easily understood working in ad hoc bases such as orthonormal ones associate to rectangular Cartesian coordinate systems.

Kinematics 23 Transport theorem Theorem of integration of composite function in 3 D generalized to vector and tensor value fields. (generalization can be easily understod by working on the components for instance rectangular Cartesian ones) At fixed time (transforming integral) spatial region occupied by the of body in reference configuration R(m)k is constant : we can interchange differentation and integration. Note that in reference domain the material derivative is the partial time derivative F is the deformation gradient here espressed in orthonormal base The determinant of tensor F can be expressed by determinant of associated representation matrix

Kinematics 24 Transport theorem Note in R(m)k the material derivative are partial time derivative at zk. Coming back to R(m) by inverse mapping at any t the time derivativen become formally calculated at corresponding z at that t by composite function. But these derivatives are by definition the material derivative Eulerian at that z at that t. In this respect they can be also expressed in the usual Eulerian form expressed in previous slides Here d. V is in reference configuration. Here I can interchange differentiation and integral Here d. V is in current configuration We are omitting time dependence since these considerations are done at any fixed t Last equality follows by applying theorem of integration of composite function to coming back to current configuration. In doing this we have to multiply the integrand by: : In any z point by theorem of inverse function in 3 D where:

Kinematics 25 Transport theorem In other words in the previous consideration J can be thought as the ratio between the d. V in the new configuration (the one in which we want calculate the new integral) and the d. Vo (the differential volume in the old configuration before to change the variable of integration). So it is evident that for the inverse transformation Jinv is = 1/J, where J is the one associated to the direct transformation The following relationship holds: Note that in case of second-order tensor field Ψ in expliciting the material time derivative it is involved the gradient of Ψ (third-order tensor fields) applied to v (obtaining a secon-order tensor field) plus (Ψ divv ) (a second order tensor). Finally div(Ψ v) is the divergence of a third-order tensor field (which gives second-order tensor). So Ψ v in 1. 3. 2 -10 is a third order tensor ( different from Ψ(v) which is a vector) Note general: if an operator is well-defined it provides an object (tensor, vector or scalar) in a coordinate system (ad so in the associated system of bases) and this object will be objective (coordinate indifferent), in the sense that the same operator in an another coordinate system will provide the same object (as it must appear in this new system). Furthermore if two tensors or two vectors are equal it is equivalent to say that they have the same components in a same coordinate systems. Therefore we can determine identities directly in a given coordinate system (for instance rectangular Cartesian). Since the coordinates will change accordingly by changing the base. Note in this consideration we are focused on obyects (vector, tensor) which are already assumed base indifferent (not simple matrixes), but we are not dealing with the issue of frame-indifference that is more general, since the issue of base indifference is the trivial case of two reference frames stationary each other.

Kinematics 26 Transport theorem More about this relationship: an intuitive approach:

Kinematics 27 Transport theorem Green ‘s transformation This result can be deduced by applying definition of continuity to the integrand applying B. O. 3 -6 in a volume containing a point in which integrand is hipothetycally different from zero

Kinematics 28 Transport theorem: Alternative approach more rigorous on 1. 3. 2 -8 : In other words the gradient of function z(zk, t) (i. e. F(zk, t)), evaluated in (zk, t) is the inverse tensor of the gradient of the function zk (z, t), evaluated in the point z(zk, t) Chain rule. Note that zi is a different function in the two domains (here the same symbol is used to simplify the notation). Note that the function on the left reduces actually to an identity It is also possible to show that (proof omitted for the sake of brevity): at a given (zk, t) Furthermore: inverting order derivation for any i, j (i. e. Schwartz theorem) and by definition of velocity with diadic terms constant and linearity of derivative Chain rule. Note that vi is a different function in the two domains (here the same symbol is used to simplify the notation)

Kinematics 29 Transport theorem: Alternative approach more rigorous on 1. 3. 2 -8 tr(AB)=Ajr. Brj Chain rule Finally we obtain: recoupling last two terms for given i, j (i. e coming back to the original expression) Now we can come back to transport theorem

Kinematics 30 Transport theorem In conclusion (coming back to B. 0. 3. 1): By Green’s transformation we obtain : Note that in case of tensor fields the first term is a tensor field or equivalenty can be seen as the tensor multiplied for the scalar flux of v. Similar considerations hold for vector fields As anticipated all these relationships for tensor and vector field can be derived in rectangular Cartesian coordinate (details omitted for brevity)

Kinematics 31 Transport theorem Generalization: In other words we can follow any fictious system by introducing its mapping function x(X, t). Obviously the velocity field to be used in the relationships shown must be the one derived by this mapping function.

Kinematics 32 Transport theorem In other words in the development of the proof of transport theorem we can decide to follow a generic map of transformation as a function of time of an initial configuration and consequently we will obtain a velocity field associated to it. In particular the velocity flux at the boundary of this system will provide the change of volume with the time associated to this system. Typical example of this approach is given by following as a function of time the region of space occupied by a polymer which is exposed to a penetrant phase whose sorption in the polymer induces swelling of the polymer-penetrant mixture

Kinematics 33 Differential mass balance Since this condition is valid for any body and a part of a body is a body itself the integrand needs to be equal to zero at any time, otherwise by hypothesys that the integrand is a continuous function there will exist a volume d. V within the volume R(m) in which the 1. 3. 3 -2 is not satisfied, but this volume d. V is at that istant associated to a body (part of body associated to the total volume R(m) ): Equivalently:

Kinematics 34 Differential mass balance It is possible to prove that the differential mass balance equation is frameindifferent (proof omitted for the sake of brevity) Not necessary since material derivative is sum of two terms Alternative form of the generalized transport theorem : represents the local value of a quantity for unit of the mass that we are following

Kinematics 35 Proof of alternative form of transport theorem

Kinematics 36 Transport theorem for systems containing a dividing surface

Kinematics 37 surface Transport theorem for system containing a dividing Note that in the following we will refer to material motion to identify the one of the body (defined inside R+ and R-) to which is overimposed the motion of the discontinuity surface. This “material” motion can be more in general the one of system also ideal characterized by its smooth mapping functions (inside here R+and R-) as seen previously. (for instance the one following a polymer mass in an open polymer-penetrant mixture). The term material is here adopted to distinguish the motion of body from the one of the two “fictive” systems R+ and R- moving with the discontinuity surface velocity u along it. As 1. 3. 5 -4 will show if the discontinuity is steady, since the integrand in TTR of a volumetric integral will be discontinuous on a surface, (with finite values) the TTR collapses to the classic case (and the two mapping functions inside R+ and R- define the material motion of the whole body). However in order to apply Green’transformation we need sum and subtract ad hoc values in R+ and R- side of discontinuity to obtain 1. 3. 5 -6

Kinematics 38 surface Transport theorem for system containing a dividing It is not closed to mass fluxes relative to u since it does not move with mass 1. 3. 5 -2 is based upon the formulation of TTR in terms of flux at boundary + intrinsic derivative (formulation of TTR 1. 3. 2 -11) and considering the real term of flux at dividing discontinuous surface. The proof rigorous is not trivial and based upon the fact that 1. 3. 2 -11 can be deduced directly in a more general form under less restrictive assumptions of smoothness involving Jacobian of mapping of trasformation used in previous argumentations (see more on next slide)

Kinematics 39 Transport theorem for system containing a dividing surface I can choose a volume small enought to obtain the partition in figure In general with the velocity field material In R+ and Rexcept on dividing surface Conditions of more general validity of 1. 3. 2 -11 (assuring also integrability due to continuous therefore finite limits) Note that the use 192. 3 (1. 3. 2 -11) in this case involves also not trivial extension of 1. 3. 2 -11 in presence of the discontinuity of v field on the curves intersection between the dividing surface and external surface (this extension is based upon the fact that the curves are of null measurability in the contest of surface integral)

Kinematics 40 surface Transport theorem for system containing a dividing Finally by Green’s transformation:

Kinematics 41 Transport theorem for system containing a dividing surface So this is a jump if we calculate it in correspondence of the same normal vector To obtain 1. 3 -5 -6: To obtain 1. 3. 5 -6 we have note that in order to apply Green’s transformation we need a C(1) field of subtract to: . Therefore we sum and to: in this way on R+ and R- appear the fluxes of two C(1) fields and Green’transformation holds With the continuity values of v on surface of discontinuity in R+ and R-respectively and n+and nthe outward versor normal to the discontinuity surface respectively for R+ and R- Then we can apply the Green’s transformation in each domain to the continuous field obtained (continuous with its spatial derivative as required by Green’s tranformation). Finally we can sum the two volumetric integrals in R+ and R- obtaining the one in R (since in the total volumetric integral the discontinuity is on a surface, which is of null measurability and the discontinuous integrand approach a finite value on both the two sides) Note that if the material body mapping follows a given law the total derivative in any case is also function of the assigned motion of the dividing surface of discontinuity (think for instance the case in which the material body is steady but the total derivative must change since the discontinuity moves). In particular this makes clear that we can not write the TTR to the total volume of body disregarding the discontinuity inside. In particular to describe the real situation we have implement 192. 2

Kinematics 42 Transport theorem The formulation 1. 3 -2 -11 of TTR is intuitive since it expresses : For unity of mass V represents in this framework the Eulerian volume occupied istantaneously by the body that we are following (balance in form acc=gen-(out-in))

Kinematics 43 Mass balance in presence of surface of discontinuity By 1. 3. 5 -6: If now I consider a point within the domain R(m) (inside R+ or R-) and I assume that at this point the integrand on first integral on right side is not zero at time t, there will be exist a finite volume (applying continuity condition) within R+ or R-in which the mass balance is not satisfied (consider for instance what the Eulerian balance of mass in this volume dictates. In alternative to see this consider that, (since velocity of discontinuity and velocity fields in R+ and R- are finite, any point inside R+ or R- can be seen as belonged to a neighborhood (volume) which is inside respectively R+ or R- for a finite range deltat of time, for istance [t-deltat t] or [t t+deltat]. This volume is a a body (part of total body inside respectively R+ or R-) moving with a material body without discontinuity (note that in the contest of the problem the discontinuity moves with a motion superimposed on the two material ones). Therefore, since any part of this volume is a body, we can repeat for this the considerations done previously to obtain that the integrand is zero also in this generic point inside R+ and R- under consideration. Finally the integrand on first term on the right null inside R+ and R- assure that its integral in R(m) is zero (since it approaches finite discontinuity on a surface of null measure ( for the volumetric integral))

Kinematics 44 Mass balance in presence of surface of discontinuity Portion of a body is a body itself and I can see part of this surface as part of a subbody containg surface discontinuity The integrand is the sum of two terms: one is the restriction on the discontinuity surface of a continuous function defined in R+ and in the discontinuity surface (the last defined as prolungation for continuity of function inside R+) and one is the restriction on the discontinuity surface of a continuous function defined in R- and in the discontinuity surface (the last defined as prolungation for continuity function inside R-). Therefore, the restricted sum function is continuous (sum of two continuous functions) on the discontinuity surface (its domain). For each function of this sum this is obtained by applying the definition of continuilty by using the same neighborhood used for the function not restriced. Since the integrand is continuous on the surface (its domain) we can obtain 1. 3. 6 -4 with usual argumentation: in fact assuming that in a point of it 1. 3. 6 -4 does not hold, we can found by continuity a neighborhood in which integrand is always positive or negative. Finally calculating the integral on a the portion of surface within this neighborood the 1. 3. 6 -3 can not holds in contrast with the hypothesys. Note that in this argumentation we are assuming that, for the regular geometry assumed, the intersection of each neighborood of a point on the surface with the volumetric domain (R+ or R-) contains also a portion of surface containing the point x.

Kinematics 45 Mass balance in presence of surface of discontinuity If there is no mass transfer along the discontinuity surface? : From this relationship and distribution property of scalar product we obtain: Selected a ξ between the two ones If the “material” motion of the body is the one associated to the conservation of the mass expressed by the density function ρ the following expression is an alternative form of TTR in presence of discontinuity: Represents the local value of quantity for unity of the mass that we are following

Kinematics 46 Mass balance in presence of surface of discontinuity This relationship follows from 1. 3. 5 -6, with and repeating the same calculations in slide 35 for first integral on right side, since mass balance for ρ holds within R+ and R- as discussed above. As also noted in developing 1. 3. 5 -6 the contribution to the volumetric integral R(m) does not depend upon the value of integrand on discontinuity surface which is of null measurability (and the discontinuous integrands on the two sides approach finite values).

Kinematics 47 Additional note on exercise 1. 3. 3 -3 Starting from 1. 3. 2 -1 (consider second expression) and with mass balance: As seen previously this derivative is calculated in reference configuration (so it is the derivative of product of partial time derivative). X is the vector in reference configuration and d. V 0 the differential volume in it.

Kinematics 48 Additional note on exercise 1. 3. 3 -3 As seen previously since this integral relationship holds for every subbody (i. e. every subvolume of R(m)k ) and by continuity of integrand it means that the integrand is equal to zero everywhere. * (which is the time partial derivative for any X ) Since at t=0 F is the identic tensor (whose determinant is 1). From this the relationship ii) holds. Note that if we divide * for the term: (which is not null since density is always >0) and to invetibility of map function at any t also relationship i) is obtained

Kinematics 49 Frame indifference. Example: for spatial vector with two orthonormal bases solidal two different frames remember that if T is a tensor (lin. transf. ) and ei a base Tij=T(ej)/i* (i* component in e* of T(ej) is rapr. matr and TX=X’ with X components of v in ei and X’ of T(v) in e* (by linearity of T is esasily proved) If (z-z 0)i is the i-sime component of (z-z 0) in base b , by linearity, the i-sime component of the same objective vector in base b* (defined here as (z*-z*0)i*) is described by i-sime matrix product: j-sime component of bi in b* is the ij term of this matrix A In case of the orthonormal Cartesian coordinate systems b, b* the matrix which represents in b the transformation orthogonal (see below why ort. ) Qb into b*is: Now let be A the matrix that gives AX=X*, X* being component of v (in b*) and X the one of v (in b) and let. be F the one doing the opposite AX=X* and X=FX* AF=FA=I matrix. (remember AX=X’ of above where. T= I tensor) So (z*-z*o)=A∙(z-z 0) (in the sense of this matrix product). Note the book defines components of an objective vector in a frame with the term “as it appears in…” So the rapresentation of Qt in b* is actually the matrix A we are using on the left in tranforming z-zo. So it is orthogonal by linear algebra see also below* Note that we are using in this argumentation the unique linear operator defined as: b*i=Q(bi) for i=1, 2, 3. Since b and b* are orthonormal it preserves the scalar product of any couple of vectors u, v so it is orthogonal (by linearity it will be <uv>=<Q(u)*Q(v)>= uivi , product of components in bi ). Coming back to the relationship above the matrix A is the matrix which represents the change of component for the identity transformation (mapping base b in itself and so the objective vector (z-zo) in itself. ). ( This is a case particular of general change of components for a tensor T between a vector u and its transformed Tu in which T=I and the same base are used to represent u and Tu). This identity application is orthogonal so its inverse is equal to its transpose. Furthermore this matrix A is also the rapresentation, of the operator Qt (as shown above)) in the base b*, which is also the rapresentation of Q-1 (in the base b*). So considering the property of rapresentation matrix of a linear operator, the matrix product A∙(z-zo)= Q-1 ∙(z-zo)( gives, using in b* the vector r* with components in b* equal to components of (z-zo) in b, also the components in b* of the vector obtained as Q-1 applied to r*. But as seen above these components are also the ones of the objective (z-zo) vector seen in b*. This is also intuitive since by applying Q-1 we are offsetting the Q transformation that I would introduce if i would use the same components of z-zo in b*. Note that if i repeat the dual procedure to get the component in b ( defined here as (z-zo)i) of a vector of components (zz*o)i in b* I will obtain (z-zo)=F ∙(z*-z*0) where: * therefore F and A are orthogonal matrixes