Here, the parameter is the cylindrical coordinate, (a constant) is the radius, and the constant controls the rate of climb.

Line integrals in 3D A line integral in space is often written in the form f( ) d. C To evaluate line integrals, you must describe the path parametrically in the form ( s). s a path in space Often, for example, is selected to equal the arc length along the line, but other interpretations might be more natural. Wednesday 10:30-12:00 Higher order Tensor Calculus s C EXAMPLE: Points on a helix may be described parametrically by ( θ) Rcosθẽ 1 + Rsinθẽ 2 + k θ ẽ 3 θ k Here, the parameter is the cylindrical coordinate, (a constant) is the radius, and the constant controls the rate of climb.. R f( ) d Once the path is parameterized, the integral is evaluated by. C s 2 s 1 f ( ) d ----- ds ds -168-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 1

Often, the parameter is time so that f( ) d f( dt. ) C t 2 t 1 Wednesday 10:30-12:00 Higher order Tensor Calculus Higher-order line integrals In materials modeling, work is often written as W :dε This is a line integral in tensor space. The strain path must be defined parametrically. Usually, the parameter is time t, so the line integral is evaluated by W :ε dt -169-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 2

Exact differentials in three dimensions ux ( 1, x 2, x 3 ) du f 1 dx 1 + f 2 dx 2 + f 3 dx 3 Under what conditions does there exist a potential such that f k x 1 x 2 x 3 where the are each functions of,, and. If the potential exists, then (by the chain rule) f i u ------ x i Wednesday 10:30-12:00 Higher order Tensor Calculus What does the notation ( fdx + gdy ) mean?!? Mixed partials are order independent: 2 u -------------- x i x j 2 u -------------- x j x i f ------ i x j f ------ j x i Which implies that, if a potential exists, then. Conversely, if the functions f k satisfy this equation, then a potential exists. In other words, for a potential to exist, the matrix H ij ------ must be symmetric. f i x j ( f 1 dx 1 + f 2 dx 2 + f 3 dx 3 ) f d An integral is path independent only if the potential exists f d and its value is. du u final u initial -162-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 3

Integrability Consider a vector ỹ that is a proper function of another vector. y i Then dy i ------ dx x j j y or dy i C ij dx j, where C i ij ------ x j Wednesday 10:30-12:00 Higher order Tensor Calculus Order of second-partial derivatives may be swapped: 2 y --------------- i x j x k 2 y --------------- i x k x j. Therefore C ij --------- x k C ik ---------- x j Inverse question: y Given a tensor C ij that varies with, under what conditions will a field ỹ exist for which C ij ------ i? x j C ij C ik ANSWER: The field will exist if and only if. y i ỹ --------- x k ---------- x j If so, C ij ------ represents a set of nine coupled PDEs that may be solved to find the field ỹ. x j -163-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 4

Application: Compatibility For large deformations, the deformation gradient tensor is defined. Wednesday 10:30-12:00 Higher order Tensor Calculus F ij x i X j F A spatially varying field F ij F ij is physically realizable if and only if --------- ---------- ik. X k X j Example: every layer in the following picture appears to stretch horizontally: initial candy deformed candy A horizontal stretch for a homogeneous deformation has a deformation gradient of the form [ ] F λ 0 1 01 before 1 λ You might be (wrongly) tempted to say that the candy problem must have the same of a similar form: 1 after F [ ] F αx 2 + β 0 0 1 (this is WRONG) This must be wrong, however, because F 11 F ----------- 12 ----------- X 2 X 1 (Compatibility is violated!) -164-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 5

An allowable answer is Wednesday 10:30-12:00 Higher order Tensor Calculus [ ] F αx 2 +β αx 1 0 1 The bad solution presumed that squares deformed to rectangles. This good solution lets squares deform to parallelograms. deformed candy -165-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 6

Preliminaries Multiple ways to analyze stresses on a windmill blade: Wednesday 15:00-16:30 Frame Indifference spatial frame g ṽ Ẽ 2 material frame ṽ g ẽ θ ẽ r Ẽ 1 g ṽ unrotated frame Ẽ 2 Ẽ 1 Suppose a deformation field is identical to one that you have analyzed in the past, except it is rigidly rotated and/or translated. QUESTION: Can you use the already-solved problem to help you solve the new problem? ANSWER: Yes, but you need to decide: space rotation or superimposed rotation? -172-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 7

Wednesday 15:00-16:30 Frame Indifference SPACE ROTATION PROBLEM (easy) previously solved (fiducial) deformed initial new problem all scalars: s new all free vectors: s old s new s old ṽ new ṽ old vnew i Q ip vold p all second-order tensors: T new old T T ij T new Q ip Q jq Told pq all third-order tensors: ζnew ijk Q ip Q iq Q ir ζold pqr -173-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 8

Wednesday 15:00-16:30 Frame Indifference SUPERIMPOSED ROTATION PROBLEM (harder) previously solved (fiducial) deformed initial new problem spatial (objective) scalars: s new spatial (objective) vectors: ṽ new s old reference scalars: s new s old ṽ old reference vectors: ṽ new ṽ old spatial (objective) tensors: T new old T reference tensors: T T new T old multi-point quantities: transform like neither spatial nor reference quantities. -174-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 9

TERMINOLOGY fiducial previously solved problem. star Wednesday 15:00-16:30 Frame Indifference new problem with same initial state, and deformed state identical to the fiducial problem but with superimposed rotation. spatial (objective): a quantity that transforms like the space rotation problem. reference: a quantity that is unaffected by superimposed rotation multi-point: a quantity that is neither objective nor reference The superimposed rotation problem µ A material fiber is a vector connecting two points in a body. µ 0 0 µ * 0 µ Any initial material fiber is a reference quantity ( ). µ * µ The deformed material fiber is a spatial quantity ( ). The deformation gradient is a two-point tensor. F F Let deformation gradient for the fiducial problem. Let deformation gradient for the star problem F F (two-point) -175-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 10

Likewise, the polar rotation tensor becomes, after superimposed rotation, R R (two-point) Wednesday 15:00-16:30 Frame Indifference Cauchy stress is force per current area, not force per initial area. Its definition requires no knowledge of the initial configuration. T (spatial, objective) Ũ The right polar stretch tensor is defined as material stretch that occurs before rotation. It is therefore unaffected by superimposed rotation: Ũ Ũ (reference) Ṽ The left polar stretch tensor is the stretch after material rotation. Superposed rigid rotation will re-orient the eigenvectors without changing the eigenvalues. Ṽ T Ṽ (spatial, objective) The principle of material frame indifference (PMFI) is the very intuitive requirement that superimposed rigid motion of a material will require the applied forces and stresses to co-rotate and co-translate in a manner consistent with their definition. Regardless of whether you apply a material model for the fiducial problem or for the star problem, the results must be consistent with these objectivity relations. -176-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 11

Mathematics of frame indifference (introduction) Wednesday 15:00-16:30 Frame Indifference Spring Example: Translational frame indifference f o spring constant change in length kδ Force in spring:. undeformed o undeformed Fiducial deformation Starred deformation o Let current location of the tip of the spring. Let original (unstressed) location of the tip of the spring. First (BAD) attempt at a vector spring equation: f k( o) (violates translational PMFI) -177-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 12

Consider Fiducial deformation: stretch the spring by some known amount. Wednesday 15:00-16:30 Frame Indifference Starred deformation: stretch the spring by the same amount and also translate it by an amount. Both scenarios involve the same starting tip location, so o o. (1) The star deformation has the same final tip location plus extra translation : + c. (2) c c The problems differ only by translation, so they both involve the same spring elongation. They should therefore involve the same spring force. f desired f. f k( o) Does the model give this desired result? For starred deformation, this (bad) spring model predicts f predicted k( o ) Substitute (1) and (2) into (3) to obtain f predicted k( + ) c f f predicted desired f o. desired + kc (NOT translationally invariant!) -178-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 13 (3)

Spring Example: Rotational frame indifference. Wednesday 15:00-16:30 Frame Indifference To satisfy translation invariance, consider a revised spring equation f better, but still bad. (here, tip tail ) k( o) Because represents the difference between two points on the spring, it will be invariant under a superimposed translation. Under translation, o o f f As before,. Therefore, under translation, as desired. This model permits force to co-translate with the spring. That s good. However... Another requirement of PMFI: Force must co-rotate with the spring. This model FAILS in this respect. -179-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 14

Wednesday 15:00-16:30 Frame Indifference o undeformed undeformed o PROPOSED MODEL f k( o) fiducial: starred: fiducial deformation stretch the spring without rotation f starred deformation f predicted same deformation as fiducial, but also rigidly rotate the spring. o f desired MODEL PREDICTION f k( o ) o o f desired. f From the sketch, it s clear that f desired (NOT rotationally invariant!) predicted f In fact, f predicted desired k(. PMFI requires this to be, which happens only if. f o o ) 0 Ĩ -180-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 15

At last... a spring model that satisfies PMFI. To satisfy both translational and rotational frame indifference, consider Wednesday 15:00-16:30 Frame Indifference f kδñ δ o ñ -------- where, and This one satisfies PMFI! The only difference between the fiducial and starred deformations is a rigid motion. Therefore, deformed lengths are the same in both cases. Hence,. o As before, the initial state is the same for both cases, so. o Consequently, δ δ Under a superimposed rigid motion,. Therefore, ñ ñ. Thus, f predicted kδ ñ ñ ( kδñ) f f desired kδ -181-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 16 YAHOO! Success at last!.

PMFI in material constitutive models Wednesday 15:00-16:30 Frame Indifference Any material model must be invariant under a basis change (space rotation problem). This requirement is necessary, but unrelated to PMFI. PMFI requires consistency of predictions when comparing deformations that differ by superimpose rotation. You can work out in advance how variables should transform under superimposed rotation. Input (independent variables) sent to your constitutive model must convey sufficient information to allow model predictions (output dependent variables) to transform correctly under superimposed rotation. If you apply your model to a fiducial problem, then you obtain a certain fiducial prediction. If you apply the model to a star problem that is identical to the fiducial problem except that the input corresponds to a superimposed rotation, then the output needs correspond to the superimposed rotation. Suppose the model takes the stretch as its only input, and returns the Cauchy stress as output. Ṽ When a star stretch T is sent as input, Ṽ Ṽ PMFI demands that the model must return T as output. -182-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 17

Here s why a model that takes only left stretch V is deficient. Wednesday 15:00-16:30 Frame Indifference Fiber composite [ F] 10 [ F] 01 20 [ F] 01 10 [ F] 02 0 1 2 0 [ R] 10 [ R] 01 10 [ R] 01 10 [ R] 01 0 1 1 0 [ V] 10 [ V] 01 20 [ V] 01 10 [ V] 02 10 02-183- 21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 18

PMFI in general materials modeling Thursday 13:00-14:30 Stress rates and isotropy When testing for frame indifference, you must first note what variables are involved. You must use physical reasoning along with rigorous mathematical consistency analysis to assert how those quantities are supposed to change upon a superimposed rigid motion. PMFI analysis for linear elasticity. Consider a constitutive model in which Cauchy stress is regarded to depend linearly on some spatial strain. ε σ ij E ijkl ε kl Ẽ :ε, or (1) Since Cauchy stress is spatial, this constitutive model satisfies PMFI only if T Ẽ ε ( T) for all orthogonal : Indicial form, recalling Eq. (1), Q pi ( E ijkl ε kl )Q qj E pqmn ( Q mk ε kl Q nl ) This must hold for all strains, and therefore, PMFI requires that Q pi ( E ijkl )Q qj E pqmn ( Q mk Q nl ) which may be rearranged to give E pqmn ( Q pi Q qj Q mk Q nl ) E ijkl This says that must be isotropic. If your material Ẽ is anisotropic, then you must receive more than just spatial strain as input to satisfy PMFI. -190-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 19

Thursday 13:00-14:30 Stress rates and isotropy Satisfying PMFI does not per se make your model valid for large deformations. A model that satisfies PMFI is merely self-consistent for arbitrarily large rotations. PMFI says nothing about whether your model is any good for large material distortions (i.e., problems where material fibers significantly rotate and stretch relative to other fibers). Alternative configurations and alternative stress/strain measures are often used to get better results under large distortions. Finding appropriate representations for distortion of material symmetry directions is primarily the subject of physics, not PMFI. -191-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 20

Stress Rates Thursday 13:00-14:30 Stress rates and isotropy PMFI in rate forms of constitutive equations Consider a constitutive law σ ij E ijkl ε kl, or Ẽ :ε in which the stiffness tensor is constant and the stress and strain are both spatial. To satisfy PMFI, let s presume that the stiffness tensor is also isotropic. Taking the time rate of both sides gives σ ij E ijkl ε kl, or Ẽ :ε This rate constitutive equation satisfies PMFI because it was obtained by differentiating a frame indifferent equation. Here s why... -192-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 21

Thursday 13:00-14:30 Stress rates and isotropy For for isotropic linear elasticity, the stress deviator S ij is related to the strain deviator γ ij by G S 2Gγ, where is the shear modulus (a constant). Under superimposed rotation, this becomes T S 2G γ T. In rate form, T T T + + 2G[ T + T + T ] S S S γ γ γ S 2Gγ T 2G[ T] S γ Recalling that, the RED terms involving all cancel, leaving only which is self consistent with S 2Gγ PMFI is satisfied for the rate equation.. In general, whenever a non-rate constitutive relation satisfies PMFI, then its rate form will also satisfy PMFI. -193-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 22

Thursday 13:00-14:30 Stress rates and isotropy REPLACING A GENUINE STRAIN RATE WITH SOME APPROXIMATION TO STRAIN RATE HAS DISTRESSING PMFI RAMIFICATIONS! The so-called rate of deformation tensor is just the symmetric part of the velocity gradient: D 1 v -- ( + T) L 2 ij i x j, where. If the principal directions of stretch never change, it equals the rate of the logarithmic strain. Otherwise, it is not a true rate. D The rate of logarithmic strain,, is very difficult to compute, whereas is very easy to compute. For this reason, many people take a perfectly valid (PMFI-conforming) constitutive model in rate form and mess it up by replacing it with g( ) ε g( ) D ε [satisfies PMFI if the material is isotropic]. [violates PMFI] D -194-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 23

Thursday 13:00-14:30 Stress rates and isotropy g( ) ε The original material model satisfied PMFI because T T + + T 2G T [ + T + T ] S S S γ γ γ These terms cancel with those on right-hand side The new (bad) material model violates PMFI. The tensor is spatial, so the PMFI requirement is D g( ) D T T + + T 2G[ dev T] S S S D These terms no longer go away D Replacement of a genuine strain rate with is going to require replacing with a pseudo stress rate (often more graciously called an objective stress rate) that subtracts away the offending terms. -195-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 24

The violation of PMFI can be repaired by replacing the bad constitutive model with, Thursday 13:00-14:30 Stress rates and isotropy where denotes a special co-rotational rate that effectively eliminates the part of the stress rate caused by rotation rates. g ( D ) The co-rotational rate is (must be) a spatial tensor to satisfy PMFI for isotropic materials. Objective co-rotational rates (Convected, Jaumann, Polar) Co-rotational stress rates are often defined in the form Λ T Λ Λ + T, ϒ Λ where ϒ The tensor is required only to transform under superimposed rotation so that Λ T This will ensure that the pseudo (co-rotational) stress rate is spatial (as needed by PMFI): T * -196-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 25

Thursday 13:00-14:30 Stress rates and isotropy Possible choices for the nebulous spin-like tensor Velocity gradient ( ) Λ If then L ij v i x j Convected rate T Λ Vorticity tensor ( ). Being skew symmetric, T. Λ W skw W 1 -- ( T) W 2 + W If then W Jaumann rate W Λ The the nebulous tensor can be taken as the polar spin tensor Λ Ω R R If T then Polar rate + Ω Ω All of these choices ensure that T In other words, the co-rotational rate is a spatial tensor! Consequently, the repaired constitutive model g( ) g D satisfies PMFI so long as the constitutive function is isotropic. Anisotropy can be accommodated by introducing additional material orientation input variables that transform spatially. -197-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 26

Thursday 13:00-14:30 Stress rates and isotropy Oscillatory stresses under simple shear. REMEMBER! satisfying PMFI will not ensure the model will give accurate or even reasonable results. To demonstrate that satisfying PMFI is merely a consistency condition that is necessary but not sufficient to obtain sensible constitutive model predictions, Dienes 1 considered simple shear with isotropic linear-elasticity in the form Ẽ :D τ γ τ Polar Jaumann γ Dienes demonstrated that the Jaumann rate predicts anomalous oscillatory shear stresses whereas the polar rate predicts intuitively more appealing monotonically increasing shear stress. The Jaumann rate performs so poorly because it uses vorticity as the material spin used to subtract away the rotational part of the stress rate. W W However, for simple shear, vorticity is constant throughout time. The Jaumann objective rate effectively presumes that a material element tumbles unceasingly. Drawing a single material element as it deforms demonstrates that such thinking is flawed. 1. DIENES, J.K. (1979) Acta Mechanica 32, pp. 217-232. Incidentally, this effect was reported in the rheology literature earlier than Dienes paper. Dienes influenced the solid mechanics community. -198-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 27

Lie derivatives and reference configurations Thursday 13:00-14:30 Stress rates and isotropy Lie derivatives justify the concept of reference constitutive models by making the connection with co-rotational (objective) rates clear. Let denote any tensor that transforms under a rigid superimposed rotation according to Some choices... (two point tensor) co-convected:, where is the deformation gradient tensor. F F contra-convected: T, where J. JF detf polar:, where is the polar rotation tensor. R R Consider any spatial tensor à à 1 à à The basis expansion of is T. Define a generalized overbar operation as à 1 ẽ i T 1 ẽ i 1 ẽ j A ij ( )( ẽ j ) A ij ( )( ), If a tensor is a spatial, then T, and therefore is a reference tensor à à à à à à -199-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 28

To take rates of d dt ---- Thursday 13:00-14:30 Stress rates and isotropy T, you need the following helper identity for the rate of an inverse: 1 à à 1 1 1 1 Λ Λ, where 1. Using this, you can show that à which may be written à 1 à à à à T T, where. Λ Ã Ã Λ The ordinary rate of the bared tensor is the same as the bar operation acting on the co-rotational rate. Most textbooks write this Lie derivative in an equivalent (but more confusing) way: à d ---- ( 1 dt à T) T. -200-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 29

WHY ARE LIE DERIVATIVES USEFUL? Thursday 13:00-14:30 Stress rates and isotropy Lie derivatives let you apply a constitutive model in a reference configuration where rates are genuine true rates, not pseudo rates, and you can prove that the result will be identical to a more complicated model that uses pseudo (co-rotational) rates in a spatial formulation. You can write a material model that presumes no rotation (which is often the case for available experimental data). Once you match non-rotational data adequately, you can upgrade your non-rotational computer model to satisfy PMFI by merely adding a pre-conditioning wrapper that un-transforms all spatial inputs to the reference state prior to calling the model. Results are then transformed in this wrapper back to the spatial frame and sent back to the host finite element code. -201-21-25 JUNE 2004, DELFT RESEARCH SCHOOL STRUCTURAL ENGINEERING PAGE 30