Crystal-Plasticity FundamentalsHenry R.Piehler,Carnegie Mellon UniversityTHE PURPOSE of this article is to enable the reader to understand through examples the fundamentals of crystal plasticity.The rich historical development of crystal plasticity is traced in some detail to enable the reader to appreciate and critically analyze more sophisti-cated recent approaches to crystal-plasticity modeling.While most of the examples involve cubic metals deforming by rate-insensitive plastic flow,the concepts outlined here can be generalized to other crystal structures and loading conditions as well.Schmid’s LawCrystal plasticity has as its origin Schmid’s law,which states that crystallographic slip is initiated when a critical resolved shear stress on a slip plane in a slip direction is reached.As shown in Fig.1,crystallographic slip initiates in uniaxial tension when the resolved shear stresson the slip plane in the slip direction reaches a critical value k (Ref 1).This critical resolved shear stress criterion can be expressed as:t ns ¼s cos l cos f ¼k(Eq 1)where t ns is the shear stress on the slip plane,with normal n in the slip direction s;s is the applied uniaxial stress;cos l is the cosine of the angle between the tensile axis and the slip direction;cos j is the cosine of the angle between the tensile axis and the slip plane normal;and kis the critical resolved shear stress.If ^l ,^s ,and ^nare unit vectors along the tensile axis l,the slip direction s,and the slip plane normal n,cos l and cos f can be found from the dot products ^l ^s and ^l ^n.For cubic metals,it is convenient to expressthe unit vectors ^l ,^s ,and ^nin terms of the unit vectors ^i 1,^i 2,and ^i 3along the [100],[010],and [001]crystallographic axes,respectively.For example,if a uniaxial stress is applied in the [001]direction of a face-centered cubic (fcc)single crystal,slip will occur on the (111)plane in the ½0 11 direction when the normal stress s reaches k/cos l cos f (Eq 1).For slip to occur on this (111)½0 11 slip system in response to a uniaxial normal stress applied along the [001]direction:^l ¼^i 3;^n ¼1ffiffiffi3p ½^11þ^i 2þ^13 ;and ^s ¼1ffiffiffi2p ½À^i 2þ^i 3 Performing the dot products ^l ^s and ^l ^n yields:s ½001 ¼ffiffiffi6p k ¼2:45kIt should be noted that the yield stress must be at least 2times the critical resolved shear stress,the minimum value achievable when both the slip plane normal and the slip direction are ori-ented at 45 to the uniaxial stress axis.As seen from Eq 1and indicated in Table 1,Schmid’s law is identical for fcc metals deforming on {111}<110>systems and body-centered cubic metals deforming on {110}<111>systems;only the identity of f and l are interchanged.One of the earliest crystal-plasticity calcula-tions was the determination by Sachs (Ref 2)of the uniaxial yield stress of an isotropic fccmetal in terms of the critical resolved shearstress k.Sachs’calculation was an average over the stereographic triangle of the uniaxial stresses necessary to initiate slip on the most highly stressed {111}<110>slip system(s).This isos-tress model predicted that the uniaxial yield stress of an isotropic fcc metal should equal 2.22k.It is also noteworthy that the single-crystal yield stress in the [111]direction is equal to:1:5ffiffiffi6p k50%higher than the isostress yield stress for the [001]orientation calculated earlier and 65%above the Sachs average.It is this depen-dence of the single-crystal yield stress on crys-tallographic orientation that provides the basis for texture hardening (Ref 3).Generalized Schmid’s LawSchmid’s law for the response of crystals to a uniaxial normal stress can be generalized to three dimensions to include both normal and shear stresses.A critical resolved shear stress criterion can again be used,this time in response to all six components of stress,chosen initially for conve-nience along the cubic axes of an fcc crystal.These stresses along the cubic axes include the three normal stresses,s 11,s 22,s 33,and the three independent shear stresses,s 12=s 21,s 23=s 32,and s 31=s 31.These stresses can be referred to as s ij ,where i and j =1,2,3.Expressions for the resolved shear stresses on the twelve crystallographic slip systems (and their negatives)are written in terms of the six independent stresses referred to the cubic axes of the fcc crystal.The notation used by Bishop and Hill (Ref 4,5)and Bishop (Ref 6)for the twelve slip systems is shown in Fig.2and Table 1,where the individual slip planes and their corresponding slip directions are identified,along with the shear strains associated with these systems.The generalized Schmid’s law is obtained using a tensor transformation to sum the contributions to each of the twelve shear stresses from the stresses s ij along the cubic axes of the fcc crystal.This generalized Schmid’s law for the critical resolved shear stress t ns on each of the twelve fcc slip systemsis:Fig.1Representation of Schmid’s critical resolved shear stress criterion:t ns =s cos l cos j =kt ns ¼l n1l s1s 11þl n2l s2s 22þl n3l s3s 33þl n1l s2s 12þl n2l s3s 23þl n3l s1s 31þl n2l s1s 21þl n3l s 2s 32þl n1l s3s 13(Eq 2a)ort ns ¼X 3j ¼1X 3i ¼1l n i l s j s ij(Eq 2b)where l n i and l s j are the direction cosines of the angle between the particular slip plane normal n and the i th cubic axis and the direction cosine of the angle between the particular slip direc-tion s and the j th cubic axis.Equation 2(a)can also be written using the summation convention that the repeated indices i and j are summed from 1to 3,or:t ns ¼l n i l s j s ij(Eq 2c)These direction cosines can again be foundfrom the dot products l n i =^n^i i and l s j =^s ^i j .Calculating the generalized Schmid’s law for the ð111Þ½0 11 system considered previously gives:l n1¼l n2¼l n3¼1ffiffiffi3p andl s1¼0;l s2¼À1ffiffiffi2p ;l s3¼1ffiffiffi2p and the generalized Schmid’s law for this systembecomes:t ns ¼1ffiffiffi6p ðÀs 22þs 33Às 12þs 23Às 32þs 13ÞSince s 23=s 32,this reduces to:t ns ¼1ffiffiffi6p ðs 22þs 33Às 12þs 13ÞThis ð111Þ½0 11 system is designated as Àa 1by Bishop and Hill (Ref 4,Table 1),whose proce-dure will subsequently make use of all 12of these generalized Schmid’s law expressions.Notice that the critical resolved shear stress reduces to the same result calculated previously for uniaxial tension along the [001]direction if only s 33is applied.Taylor ModelG.I.Taylor (Ref 7)noted that the isostressmodel used by Sachs to calculate the uniaxial yield stress of an isotropic fcc aggregate in terms of the critical resolved shear stress k failed to satisfy the strain compatibility require-ments among grains for plastic deformation of an isotropic material deformed in uniaxial ten-sion.These isostrain uniaxial tension require-ments are that,for a given plastic strain d e along the length l of a single crystal,the normal strains along two axes perpendicular to l are equal to Àd e /2(from constancy of volume requirements),and all shear strains are equal to zero.Taylor then proceeded to calculate the uniaxial yield stress of an isotropic fcc aggre-gate in terms of the critical resolved shear stress using his isostrain model,which is described next.Taylor’s approach sought to determine the crystallographic shear strains resulting from the operation of a particular set of five indepen-dent slip systems that will accommodate the uniaxial tension isostrain requirements imposed along the primed axes associated with a partic-ular fcc crystal orientation.The first step is to find the plastic strain increments referred to the cubic axes of an fcc single crystal resulting from the imposition of the plastic strains imposed along primed specimen axes.If the imposed strain increments along the primed specimen axes are designated as d e k 0l 0using the summation convention,the strain incre-ments along cubic axes can be written in terms of these imposed strains as:d e ij ¼l i k 00l j l 0d e k 0l 0(Eq 3)The results of this isostrain requirement for uni-axial tension along various orientations can then be used to relate the plastic strains along the cubic axes and the contributions of each of the twelve possible slip systems (four <111>planes each containing three {110}slip direc-tions),that is:d e ij ¼X 4n ¼1X 3s ¼1l i n l j s d e ns(Eq 4a)where d e ns is 1of 12true plastic strains asso-ciated with the operation of particular slip systemwith slip plane normal ^n and a slip direction ^s .The true shear strain d e ns is equal to ½the simple shear strain g ns ,which contains a rotation equal to (½)d g ns .This relationship between crystallo-graphic or simple shear,pure or tensorial shear,and rotation is shown in ing the nomen-clature of Bishop and Hill contained in Table 1,the relationships between the five independent components of strain referred to the cubic axes and the12crystallographic simple shear strains a i (Eq 4a)become:d e 11¼1ffiffiffi6p ðÀ2a 2þ2a 3À2a 5þ2a 6À2a 8þ2a 9À2a 11þ2a 12Þd e 22¼1ffiffiffi6p ðÀ2a 1À2a 3þ2a 4À2a 6þ2a 7À2a 9þ2a 10À2a 12Þd e 33¼1ffiffiffi6p ðþ2a 1þ2a 2À2a 4À2a 5À2a 7À2a 8À2a 10þ2a 11Þd e 12¼1ffiffiffi6p ðþa 1Àa 2þa 4Àa 5Àa 7þa 8Àa 10þa 11Þd e 23¼1ffiffiffi6p ðþa 2Àa 3Àa 5þa 6þa 8Àa 9Àa 11þa 12Þd e 31¼1ffiffiffi6p ðþa 1þa 3þa 4Àa 6þa 7Àa 9Àa 10þa 12Þ(Eq 4b)Since there are five independent components of strain e ij along the cubic axes (reduced from six by the constancy of volume requirement d e 11+d e 22+d e 33=0),at least five slip systems (a’s)must operate for Eq 4(b)to be inverted to find the amounts of shear on particular slip systems that must operate to accommodate the strains imposed along the cubic axes.This requirement for the simultaneous operation of five independent slip systems to accommodate an arbitrary strain state in an fcc crystal was first pointed out by von Mises (Ref 8).Since one can find more than one set of five independent shear strains to accommodate a given imposed strain state,the solution for e ij in terms of a i (Eq 4b)is not unique.Taylor selected the particular set of five inde-pendent crystallographic simple shear strains a i that accommodated the imposed macroscopic strain state e k 00l 0with a minimum of the total shear strain (or shear strain energy):X 5i À1a iTherefore,to calculate the value of the uni-axial yield stress for an isotropic fcc crystalline aggregate using Taylor’s isostrainminimumFig.2Schematic representation of Bishop-Hill slip system notationTable 1Slip system notation used by Bishop and Hill for cubic crystalsSlip plane (or direction)(111)ð 1 11Þð 111Þð1 11ÞSlip direction (or plane)½01 1 ½ 101 ½1 10 ½0 1 1 ½101 ½ 110 ½01 1 ½101 ½ 1 10 ½0 1 1 ½ 101 ½110 Shear straina 1,a 2,a 3b 1,b 2,b 3c 1,c 2,c 3d 1,d 2,d 3Slip system designation(a 1),(a 2),(a 3)(b 1),(b 2),(b 3)(c 1),(c 2),(c 3)(d 1),(d 2),(d 3)2/Fundamentals of the Modeling of Microstructure and Texture Evolutionshear criterion,one starts by imposing the plastic strain increment d e ll =d e along the sample ten-sile axis and Àd e /2along two axes perpendicular to this axis.This strain state referred to the sam-ple axes is then transformed to the cubic axes of the fcc crystal using Eq 3.The particular set of five independent crystallographic shear strains a i that accommodate the imposed strains is found next by looking for a set of five a i ’s that satisfy Eq 4(b)and,at the same time,minimize the total crystallographic shear:X 5i À1a iKnowing the minimum total crystallographic shear for a particular crystal,one can find the value of the uniaxial yield stress in terms of the critical resolved shear stress k by equating the strain energy from slip to the external work done:kX 5i À1a i ¼s d eors ¼kX 5i À1a i =d e ¼Mkwhere M,the ratio of the sum of the crystallo-graphic shears to the uniaxial axisymmetricextension resulting from these shears,is the Taylor factor.More generally,M is defined as the sum of the crystallographic shears per unit effective strain.Averaging the results for M over the stereo-graphic triangle using this isostrain procedure,Taylor predicted that the uniaxial yield stress of an isotropic fcc aggregate is equal to 3.06k.Taylor’s minimum shear criterion was initi-ally based on intuition but was later shown by Bishop and Hill (Ref 4,5)to be rigorously cor-rect and equivalent to their procedure described subsequently.Taylor also did not address the question as to whether or not one could find stress states that will simultaneously reach the critical resolved shear stress k on at least five independent slip systems while remaining belowk on the remaining systems,a question that againwas addressed subsequently by Bishop and Hill (Ref 4,5).Even if such stress states could be found,however,the issue of stress compatibility across grain boundaries remained.Nevertheless,Taylor’s isostrain model remains a significant advance in crystal plasticity and is also currently used as the basis for several numerical codes capable of calculating Taylor factors for arbitrary imposed strain states rather than just axisymmet-ric deformation.Bishop-Hill ProcedureLike that of Taylor (Ref 7),the procedure of Bishop and Hill (Ref 4,5)and Bishop (Ref 6)is an isostrain model.However,their stress-based approach sought to directly find stress states that could simultaneously operate at least five inde-pendent slip systems.Bishop and Hill began by examining the generalized Schmid’s law requirements for operation of the 12{111}<110>fcc slip systems.They chose to describe this yielding in terms of the new stresses:A ¼s 22Às 33;B ¼s 33Às 11;C ¼s 11Às 22;F ¼s 23;G ¼s 31;andH ¼s 12(Eq 5)which leads to the 12(24with negatives)criti-cal resolved shear stress expressions given in Table 2.These 12yield expressions (24with nega-tives)can be plotted in three separate three-dimensional stress spaces having coordinates A,G,and H;B,H,and F;and C,F,and G.The yield condition for a particular slip system is represented by a plane in one of the three three-dimensional stress spaces.These planes make equal angles to the three coordinate axes;their intersection forms three separate octahe-dral,as shown in Fig.4.The intersection of these planes along lines and at corners repre-sents polyslip stress states.One notes that there are two types of polyslip stress states:one that is situated at the vertex of an octahedron (points x 1and x 2in Fig.4)and the other that lies on a line oriented at 45 to two coordinate axes (points y 1,y 2,y 3,and y 4).The former willsimultaneously operate four slip systems,the latter two.Hence,any polyslip stress state that simultaneously operates at least five slip sys-tems will actually operate six or eight.Slip will be activated on eight slip systems if the stress state is located at two vertices (e.g.,points x 1and x 2)or at one vertex and two 45 lines (e.g.,points x 1,y 1,y 2).Six systems will be acti-vated if the stress state is located along three 45 lines (e.g.,points y 1,y 3,and y 4).(One ver-tex and one 45 line would violate the condition that A +B +C =0.)A summary of all permis-sible polyslip stress states is given in Table 2.The three cases considered previously corre-spond to stress states À2,À12,and À16.As can be shown with reference to the octahedral in Fig.4,stress state À2will operate the eight slip systems À(a 2),(a 3),À(b 2),(b 3),À(c 2),(c 3),À(d 2),and (d 3);stress state À12will oper-ate the eight systems À(a 2),(a 3),À(b 2),(b 3),À(c 2),(c 3),À(d 1),and (d 3);and stress state À16will operate the six systems À(b 2),(b 3),À(c 1),(c 2),À(d 1),and (d 3).A similar examina-tion of the remaining polyslip stress states in Table 3will show that the stress states 1to 12simultaneously operate eight slip systems,while stress states 13to 28operate six.The particular state of stress that acts to accommodate an imposed state of strain can be found by selecting from the 28(56with negatives)permissible fcc polyslip stress states shown in Table 3,which shows the particular stress state that maximizes the external work done.This principle of maximum work was first derived rigorously by Bishop and Hill (Ref 4,5),who also showed it to be equivalent to the minimum shear approach taken earlier by Taylor (Ref 7).The increment of work done,with reference to the cubic axes,is:d W ¼s 11de 11þs 22d e 22þs 33d e 33þ2s 12d e 12þ2s 23d e 23þ2s 31d e 31(Eq 6a)or,in Bishop-Hill notation:d W ¼ÀB de 11þA d e 22þ2F d e 23þ2G d e 31þ2H d e 12(Eq 6b)In the actual calculation procedure,the strain increments d e ij along the cubic axes arefirstFig.3Decomposition of a simple crystallographic shear into a pure or tensorial shear plus a rotationTable 2Bishop-Hill shear stress expressionsSlip systemYield expressionþÀ(a 1)A ÀG þH ¼Æffiffiffi6p k þÀ(a 2)B ÀH þF ¼Æffiffiffi6p k þÀ(a 3)C ÀF þG ¼Æffiffiffi6p k þÀ(b 1)A þG þH ¼Æffiffiffi6p k þÀ(b 2)B ÀH ÀF ¼Æffiffiffi6p k þÀ(b 3)C þF ÀG ¼Æffiffiffi6p k þÀ(c 1)A þG ÀH ¼Æffiffiffi6p k þÀ(c 2)B þH þF ¼Æffiffiffi6p k þÀ(c 3)C ÀF ÀG ¼Æffiffiffi6p k þÀ(d 1)A ÀG ÀH ¼Æffiffiffi6p k þÀ(d 2)B þH ÀF ¼Æffiffiffi6p k þÀ(d 3)C þF þG ¼Æffiffiffi6p kCrystal-Plasticity Fundamentals /3written in terms of the imposed macroscopic strain state d e k 0l 0and substituted into Eq 6(b).The values of the 28(56)permissible stress states contained in Table 2are then substituted sequentially into Eq 6(b).That particular per-missible stress state that maximizes d W for the given imposed strain state will operate the six or eight fcc slip systems that can accom-modate this imposed strain.Like Taylor before them,Bishop and Hill used an isostrain assumption to calculate the yield stress of an isotropic fcc crystalline aggre-gate in terms of the critical resolved shear stress k.Since,as shown by Bishop and Hill,their maximum work and Taylor’s minimum shear criteria are equivalent,not surprisingly their value for the uniaxial yield stress of an iso-tropic fcc crystalline aggregate was also 3.06k.Bounds for Yield Loci from Two-Dimensional Sachs and Bishop-Hill AveragesThe isostress and isostrain approaches to cal-culating the uniaxial yield stress of an isotropic fcc aggregate in terms of the critical resolved shear stress k can be generalized to two dimen-sions to calculate yield loci for single crystals or ideal textural components (Ref 9).Individual single-crystal yield loci can then be averaged to calculate yield loci for crystalline aggregates.These rather simple two-dimensional bounds are often sufficient to answer engineering ques-tions regarding texture hardening,for example.The basic approach is again illustrated for fcc metals deforming by rate-insensitive {111}<110>slip,but this general approach is appli-cable to other crystal structures with other slip behaviors as well.Isostress or lower-bound yield loci for sheets having single-crystal orientation are found by:Applying a biaxial principal stress state along the rolling and transverse directions of a single-crystal or ideal texture componentCalculating the resolved shear stresses on the 12{111}<110>slip systems resulting from this biaxial principal stress state applied along the rolling and transverse directionsSetting the 12resolved shear stresses equal to k,the critical resolved shear stressPlotting the 12lines representing yielding in the biaxial principal stress spaceForming the lower-bound isostress yield locus from the intersection of these 12linesIn general,if the rolling,transverse,and sheet normal direction are designated as r,t,and p,the six independent macroscopic stress components are s rr ,s tt ,,s pp ,s rt ,s tp ,and s pr .The stresses t ns acting on the {111}<110>slip systems can then be written in terms of these macroscopic stresses using thetransformation:Fig.4Generalized Schmid’s law plotted as octahedra using three separate coordinate systemsTable 3Bishop-Hill stress states that simultaneously operate six or eight slip systemsStress state numberStress terms A =ffiffiffi6p kB =ffiffiffi6p kC =ffiffiffi6p k F =ffiffiffi6p kG =ffiffiffi6p kH =ffiffiffi6p k11À10000201À10003À1010004000100500001060000017½À1½0½08½À1½0À½09À1½½½0010À1½½À½0011½½À100½12½½À100À½13½0À½½0½14½0À½À½0½15½0À½½0À½16½0À½À½0À½170À½½0½½180À½½0À½½190À½½0½À½200À½½0À½À½21À½½0½½022À½½0À½½023À½½0½À½024À½½0À½À½025000½½À½26000½À½½27000À½½½28½½½4/Fundamentals of the Modeling of Microstructure and Texture Evolutiont ns ¼l nr l sr s rr þl nt l st s tt þl np l sp s pp þ2l nr l st s rtþ2l nt l sp s tp þ2l np l sr s pr(Eq 7)For example,for the ideal cube-on-corner tex-ture,t is taken along ½1 10 ,r is along ½11 2 ,and p is along [111];the corresponding unit vectors are:^t ¼1ffiffiffi2p ð^i 1À^i 2Þ;^r ¼1ffiffiffi6p ð^i 1þ^i 2À2^i 3Þ;and^p¼1ffiffiffi3p ð^i 1þ^i 2þ^i 3ÞCalculating the direction cosines using theappropriate dot products and substitution into Eq 7give the expressions in Table 4.A lower bound to the yield locus for this ideal cube-on-corner crystal can be found directly from the equations in Table 4by setting s pp =s rt =s tp =s tr =0.The operation of each slip system will be governed by a line in s rr Às tt space;the intersections of these lines form the lower bound to the yield locus for the ideal cube-on-corner crystal (Fig.5).A similar procedure is used to relate the strain incrementsalong therolling,transverse,and sheet normal directions of the ideal cube-on-corner orientation and the summed contributions from the12crystallographicshears,yielding:d e rr ¼13ffiffiffi6p ðÀ2b 1þ2b 2À3c 1þc 2þ2c 3Àd 1þ3d 2À2d 3Þd e tt ¼1ffiffiffi6p ðc 1Àc 2þd 1Àd 2Þd e pp ¼23ffiffiffi6p ðb 1Àb 2þc 2Àc 3Àd 1þd 3Þd e rt ¼13ffiffiffi2p ðÀb 1Àb 2þ2b 3Àc 1þc 3Àd 2þd 3Þd e tp ¼1=12ðÀ3a 1À3a 2þ6a 3Àb 1Àb 2þ2b 3Àc 1À3c 2þ4c 3À3d 1Àd 2þ4d 3Þd e pr ¼112ffiffiffi3p ð9a 1À9a 2þ7b 1À7b 2þ3c 1À5c 2þ2c 3þ5d 1À3d 2À2d 3Þ(Eq 8)These strain-slip equations are used subsequently.An upper bound to the yield locus for this same crystal is be found by:Imposing different ratios of the principal strains e rr and e ttTransforming these principal strains to the cubic axesUsing the maximum work principle to select the Bishop-Hill stress state that operates the slip systems necessary to accommodate this strain imposed along the cubic axesTransforming these Bishop-Hill stresses to the rolling (r),transverse (t),and sheet normal(p)coordinate system and supplementingthese stresses with the condition that s pp =0Plotting the results in s rr Às tt spaceUsing the maximum work principle,the resulting Bishop-Hill stress states for this crys-tal to accommodate biaxial principal strains in the ideal cube-on-corner crystal are No.6,24,25,and 28,the specific stress state depending on the imposed strain ratio.After performing the transformation outlined previously,there results the upper-bound yield locus shown in Fig.6.Figures 5and 6,together with the normality condition (Ref 4,5),are used to determine bounds for the biaxial stresses needed to enforce a specific strain state (in this case plane strain)as well as to illustrate the physical dif-ferences between these two bounds for the plane-strain yield stress.Applying the normality condition as shown in Fig.5,plane strain in the transverse direction (d e horizontal),one sees that the biaxial stress state needed to initiate this deformation is:s rr ¼s tt ¼3/2ffiffiffi6p kThis stress state simultaneously operates two branches of the yield locus;the first (horizontal)branch of the yield locus initiates yielding on systems À(b 1),+(b 2),+(c 3),and À(d 3);the sec-ond on systems À(c 2)and +(d 1).Before proceeding further,it should first be noted that the strains d e rr and d e tt that result from the operation of each particular system are normal to their respective branch of the yield locus.Focusing first on the horizontal branch and letting slip occur by a unit amount in any one of the systems À(b 1),+(b 2),+(c 3),and À(d 3),the strain state from Eq 8for each of these four systems is:d e rr ¼Àd e pp ¼23ffiffiffi6p d e and d e tt ¼0which is normal to the horizontal branch.Simi-larly,if slip occurs by a unit amount on either system À(c 2)or (d 1),the resulting strain is:d e rr ¼Àffiffiffi6p ;d e tt ¼ffiffiffi6p ;and d e pp¼À2ffiffiffi6p which again is perpendicular to its branch of the yield locus.This two-dimensional normality holds true even if the biaxial principal stresses activate systems that result in shear strains as well as normal strains;that is,the principal directions of stress and strain do not coincide.Plane strain in the transverse direction of the cube-on-corner crystal is achieved by combining the strain vectors normal to the two branches of the yield locus to achieve the strain state d e rr =0,d e tt =1,and d e pp =À1.This is achieved by adjusting the amounts of shear on the opera-tive systems to be:b 1¼d 3¼Àffiffiffi6p 8;b 2¼c 3¼ffiffiffi6p 8;c 2¼Àffiffiffi6p 2;andd 1¼ffiffiffi6p Substituting these values into Eq 8results in plane strain in the transverse direction,since d e rr =0,d e tt =1,and d e pp =À1.However,only two of the shear strains,d e rt and d e tp ,vanish;the remaining shear strain (d e pr )is equal to:15ffiffiffi2p In order to impose a state of transverse plane strain in the absence of shear strain on this cube-on-corner crystal,it is necessary both to alter the levels of s rr and s tt and to apply an additional shear stress s pr .After transforming Bishop-Hill stress state No.6:s 12¼ffiffiffi6p kto the r,t,and p coordinate system and imposing the condition that s pp =0,one finds that:s rr ¼ffiffiffi6p 3k ;s tt ¼5ffiffiffi6p 3k ;s pr ¼À2ffiffiffi3p 3k ;ands pp ¼s rt ¼s tp ¼0By substituting these stresses into Eq 8,one can verify that the critical resolved shear stress isTable 4Stress transformation in Bishop-Hill notationSlip systemShear stress expressionþÀ(a 1)À1/2s tp þ1/2ffiffiffi3p s pr ¼Æffiffiffi6p kþÀ(a 2)À1/2s tp À1/2ffiffiffi3p s pr ¼Æffiffiffi6p kþÀ(a 3)s tp ¼Æffiffiffi6p k þÀ(b 1)À2ffiffip s rr þ2ffiffip s pp À2ffiffip s rt À1/6s tp þ7ffiffip s pr¼Æffiffiffi6p k þÀ(b 2)23ffiffi6p s rr À23ffiffi6p s pp À23ffiffi2p s rt À1/6s tp À76ffiffi3p s pr ¼Æffiffiffi6p k þÀ(b 3)43ffiffi2p s rt þ1/3s tp ¼Æffiffiffi6p k þÀ(c 1)À1ffiffi6p s rr þ1ffiffi6p s tt À23ffiffi2p s rt À1/6s tp þ12ffiffi3p s pr ¼Æffiffiffi6p k þÀ(c 2)13ffiffi6p s rr À1ffiffi6p s tt þ23ffiffi6p s pp À1/2s tp À56ffiffi3p s pr ¼Æffiffiffi6p k þÀ(c 3)23ffiffi6p s rr À23ffiffi6p s tt þ23ffiffi2p s rt þ2/3s tp þ13ffiffi3p s pr ¼Æffiffiffi6p k þÀ(d 1)À13ffiffi6p s rr þ1ffiffi6p s tt À23ffiffi6p s pp À1/2s tp þ56ffiffi3p s pr ¼Æffiffiffi6p k þÀ(d 2)1ffiffi6p s rr À1ffiffi6p s tt À23ffiffi2p s rt À1/6s tp À12ffiffi3p s pr ¼Æffiffiffi6p k þÀ(d 3)À23ffiffi6p s rr þ23ffiffi6p s pp þ23ffiffi2p s rt þ2/3s tp À13ffiffi3p s pr ¼Æffiffiffi6p k Crystal-Plasticity Fundamentals /5。
