(c) Calculate the tensile yield stress based on a critical resolved shear stress that we will (arbitrarily) set at 100 MPa. (c) MPa.

Transcription

1 27-750, A.D. Rollett Due: 20 th Ot., Homework 5, Volume Frations, Single and Multiple Slip Crystal Plastiity Note the 2 extra redit questions (at the end). Q1. [40 points] Single Slip: Calulating Shmid Fators (Partially taken from Lin Hu s answers.) In a single rystal tensile test on Ni, the orientation of the rystal is given as ( , 38.33, ) in Bunge Euler angles. Assume that an axisymmetri tensile stress is applied along the sample 3 (Z) diretion. Note that we apply stress boundary onditions. (a) Determine whih rystal diretion is parallel to the tensile axis. Give your answer in the form of (hkl)[uvw] and simplify the numbers to be single digit integers. You should be able to modify the spreadsheet or program that you wrote in earlier homeworks to produe an answer. The Miller indies for this rystal orientation are: ( ) [ ], based on a standard onversion of Euler angles to matrix, to Miller indies. Then, obviously, the tensile axis is // [ ]. (b) Identify whih ombination slip plane and diretion is ative, and give the Shmid fator. (b) ( ) [ ]; Note: Ni is f and an slip on any {111} plane in any <110> diretion. You will have to find out whih is the most highly stressed slip system, i.e. find the largest Shmid fator. This value of the Shmid fator is what you should use to determine the tensile yield stress beause it determines whih slip system will be ativated first. () Calulate the tensile yield stress based on a ritial resolved shear stress that we will (arbitrarily) set at 100 MPa. () MPa. SteveSintay_Pole_Figures-26May11.xls; ShmidFators-2011.xlsx.

2 You must submit a opy of the table of results showing how you alulated your answer whih must display the maximum Shmid fator and the indies of the ativated slip system. A suggestion for how to proeed is to use a spreadsheet (e.g. Exel), or a Math program suh as Mathematia or Maple, and make a list of all possible ombinations of slip plane (111, -111, 1-11, -1-11) and slip diretion (e.g. 111 is orthogonal to 110, -101 and 0-11), taking only positive versions of eah (unit) vetor. This will give you a table with 12 rows, one for eah slip system (3 X 4 = 12 ombinations of plane and diretion). Inlude a olumn with the dot produts of the plane and diretion and explain why this is useful. Then alulate the dot produts of the tensile axis with eah ombination of plane+diretion in turn in order to obtain osφ and osλ, respetively (2 more olumns). Then alulate the Shmid fator as (osφ*osλ) (1 more olumn). Finally, identify the row with the largest absolute value of the Shmid fator in it (i.e. positive or negative). {You an expand the table to inlude the negatives of eah slip diretion in addition: this will give you 24 rows (e.g. 111 is orthogonal to 110, -101, 0-11, 1-10, 10-1 and 01-1). If you use the 24 row version, you will find that you obtain a pair of positive and negative Shmid fators for eah pair of positive and negative slip diretions. This positive/negative pairing orresponds to positive and negative diretions of slip.} Q2. [40 points] Bishop-Hill Model: Calulation of Taylor Fators for Multiple Slip Assume that slip ours on {111} planes and in <110> diretions. Assume that an axisymmetri tensile strain is applied along the sample 3 (Z) diretion. Calulate the following quantities: (a) the index of the ative stress state (from the Bishop-Hill list);

3 (b) the Taylor fator (to at least 2 deimal plaes) for rystals with the following orientations: 2.1 ( 0 0 1) [1 0 0] 2.2 ( ) [1 0 0] 2.3 ( ) [1 0 0] 2.4 (-2 3 3) [3 2 0] 2.5 ( ) [3 2 0] 2.6 ( ) [ ] 2.7 ( ) [ ]. Question Orientation 2.1 ( 0 0 1) [1 0 0] 2.3 ( ) [1 0 0] 2.4 (-2 3 3) [3 2 0] Taylor Stress State Fator , -1.0, 0.0, 0.0, 0.0, * 0.0, 0.0, 0.0, 1.0, 0.0, (to be determined) * orreted 28 Apr 06 Answers are provided (above) for a few of the ases so that you an hek your method. The meaning of the Miller indies is the onventional one disussed earlier in the lass: (hkl) // sample-z (=sample-3), [uvw] // sample-x (= sample-1). Steps to follow in order to obtain a solution: 1. Write down the strain inrement in vetor form. 2. Convert the strain vetor to a strain tensor (3x3 matrix) form: be areful with fators of 2 if you have non-zero off-diagonal terms. 3. Calulate the von Mises equivalent strain (Slide set Aniso-4 slide #9). 4. Calulate the orientation matrix from the Miller indies speified. 5. Rotate the strain tensor into rystal oordinates (no need to symmetrize the answer). 6. For eah of the 28 multi-axial stress states (listed below, to save you the trouble of transribing them), alulate the work inrement; be areful to use the formula appropriate to the strain tensor alulated in step 5 above. 7. Identify whih stress state yields the largest (+ or -) absolute work inrement ( dw ). 8. Calulate the Taylor fator as the work inrement divided by the von Mises equivalent strain. Note that it is interesting to ompute the Shmid fators for the same orientations and ompare them to the Taylor fators. (This is not required, but if you developed a suitable spreadsheet for Q1, you should find it easy to perform the alulations). Bishop & Hill stress states (Fortran style); fator of 6 not inluded: stress(1,1)=1. stress(1,2)=-1.

4 stress(1,3)=0. stress(1,4)=0. stress(1,5)=0. stress(1,6)=0. # 1 stress(2,1)=0. stress(2,2)=1. stress(2,3)=-1. stress(2,4)=0. stress(2,5)=0. stress(2,6)=0. # 2 stress(3,1)=-1. stress(3,2)=0. stress(3,3)=1. stress(3,4)=0. stress(3,5)=0. stress(3,6)=0. # 3 stress(4,1)=0. stress(4,2)=0. stress(4,3)=0. stress(4,4)=1. stress(4,5)=0. stress(4,6)=0. # 4 stress(5,1)=0. stress(5,2)=0. stress(5,3)=0. stress(5,4)=0. stress(5,5)=1. stress(5,6)=0. # 5 stress(6,1)=0. stress(6,2)=0. stress(6,3)=0. stress(6,4)=0. stress(6,5)=0. stress(6,6)=1. # 6 stress(7,1)=0.5 stress(7,2)=-1. stress(7,3)=0.5 stress(7,4)=0. stress(7,5)=0.5 stress(7,6)=0. # 7 stress(8,1)=0.5 stress(8,2)=-1. stress(8,3)=0.5 stress(8,4)=0. stress(8,5)=-0.5 stress(8,6)=0. # 8 stress(9,1)=-1. stress(9,2)=0.5 stress(9,3)=0.5 stress(9,4)=0.5 stress(9,5)=0. stress(9,6)=0. # 9 stress(10,1)=-1. stress(10,2)=0.5 stress(10,3)=0.5 stress(10,4)=-0.5

5 stress(10,5)=0. stress(10,6)=0. # 10 stress(11,1)=0.5 stress(11,2)=0.5 stress(11,3)=-1. stress(11,4)=0. stress(11,5)=0. stress(11,6)=0.5 # 11 stress(12,1)=0.5 stress(12,2)=0.5 stress(12,3)=-1. stress(12,4)=0. stress(12,5)=0. stress(12,6)=-0.5 stress(13,1)=0.5 stress(13,2)=0. stress(13,3)=-0.5 stress(13,4)=0.5 stress(13,5)=0. stress(13,6)=0.5 stress(14,1)=0.5 stress(14,2)=0. stress(14,3)=-.5 stress(14,4)=-0.5 stress(14,5)=0. stress(14,6)=0.5 stress(15,1)=0.5 stress(15,2)=0. stress(15,3)=-0.5 stress(15,4)=0.5 stress(15,5)=0. stress(15,6)=-0.5 stress(16,1)=0.5 stress(16,2)=0. stress(16,3)=-0.5 stress(16,4)=-0.5 stress(16,5)=0. stress(16,6)=-0.5 stress(17,1)=0. stress(17,2)=-0.5 stress(17,3)=0.5 stress(17,4)=0. stress(17,5)=0.5 stress(17,6)=0.5 stress(18,1)=0. stress(18,2)=-0.5 stress(18,3)=0.5 stress(18,4)=0. stress(18,5)=-0.5 stress(18,6)=0.5 stress(19,1)=0. stress(19,2)=-0.5 stress(19,3)=0.5 stress(19,4)=0. stress(19,5)=0.5 stress(19,6)=-0.5

6 stress(20,1)=0. stress(20,2)=-0.5 stress(20,3)=0.5 stress(20,4)=0. stress(20,5)=-0.5 stress(20,6)=-0.5 stress(21,1)=-0.5 stress(21,2)=0.5 stress(21,3)=0. stress(21,4)=0.5 stress(21,5)=0.5 stress(21,6)=0. stress(22,1)=-0.5 stress(22,2)=0.5 stress(22,3)=0. stress(22,4)=-0.5 stress(22,5)=0.5 stress(22,6)=0. stress(23,1)=-0.5 stress(23,2)=0.5 stress(23,3)=0. stress(23,4)=0.5 stress(23,5)=-0.5 stress(23,6)=0. stress(24,1)=-0.5 stress(24,2)=0.5 stress(24,3)=0. stress(24,4)=-0.5 stress(24,5)=-0.5 stress(24,6)=0. stress(25,1)=0. stress(25,2)=0. stress(25,3)=0. stress(25,4)=0.5 stress(25,5)=0.5 stress(25,6)=-0.5 stress(26,1)=0. stress(26,2)=0. stress(26,3)=0. stress(26,4)=0.5 stress(26,5)=-0.5 stress(26,6)=0.5 stress(27,1)=0. stress(27,2)=0. stress(27,3)=0. stress(27,4)=-0.5 stress(27,5)=0.5 stress(27,6)=0.5 stress(28,1)=0. stress(28,2)=0. stress(28,3)=0. stress(28,4)=0.5 stress(28,5)=0.5 stress(28,6)=0.5

7 Answer: See the desription in Hosford, Ch. 3. his definition of the Taylor fator is rather obsure but the von Mises equivalent strain approah seems to work well enough. The Taylor fators appear to be idential with those in the books. The following results were obtained with Bishop_Hill.f90. Question Orientation Taylor Fator Stress State 3.1 ( 0 0 1) [1 0 0] , -1.0, 0.0, 0.0, 0.0, ( ) [1 0 0] , -1.0,.0,.0,.0, ( ) [1 0 0] 3.585* 0.0, 0.0, 0.0, 1.0, 0.0, (-2 3 3) [3 2 0] ,.5,.0,.5, -.5, ( ) [3 2 0] ,.5,.0,.5, -.5, ( ) [ ] ( ) [ ] * orreted 28 Apr 06 Q3. [10 points] The objetive of this question is to show you how to perform alulations of volume fration given an Orientation Distribution (and vie versa in Q3). A disrete OD is defined on a grid that has ells at every 10 in (Bunge) Euler angles in ubi-orthorhombi symmetry with a range of 0-90 in eah angle. The ells are entered on their oordinates, as disussed in the slides and as used in popla, so be areful of how to treat the ells at 90 and at 0 beause the orientation spae volume (in the sense of Euler angle spae, or dg integrated over some subspae, dω) assoiated with ells at the edges is different. The intensities are in multiples of a random intensity. (φ 1,Φ,φ 2 ) Intensity ( 0, 10, 0) 540. (30, 60, 20) 9.05 (40, 70, 30) 8.3 a) Calulate the number of ells 10 x 10 x 10 = Be areful here, the ells are split at edges at 5 degree inrements. Therefore, they are just half ells, whih is why we have 10 ells in eah dimension. b) Calulate the total orientation spae volume, Ω, of the spae (units of degrees 2 ) Ω = 8100 ( ^2). ) Calulate the orientation spae volumes, dω, assoiated with eah of the 3 ells (units of degrees 2 ). In general, the orientation spae volume is given by dω =10 x os(φ)x10, exept for the first one whih is divided twie by two beause both phi-1 and phi-2 are equal to

8 zero, i.e. they are at the edge of the spae = 0.757, 15.1, & 16.4, respetively. d) Calulate the volume frations, dv, assoiated with eah of the 3 ells This is given by dv = f(g) x dω / Ω = 0.050, 0.017, 0.017, respetively. e) Compare the ratios of the intensities, f, to the ratios of the volume frations what do you notie? The ratio of intensities is about 60:1:1 whereas the ratio of the volume frations is 3:1:1. Note: other ells in the OD must have non-zero intensity beause the three volume frations alulated do not add up to one. (You might want to amuse yourself by alulating how large the intensities would have to be in order for these 3 ells to aount for all the volume.) The answer is f(g) = , 107.6, Q4. [10 points] The objetive of this question is to show you how to perform alulations the Orientation Distribution given values of volume frations. Some of the values are given so that you an hek your work. It is important, therefore to show your working. You have performed an analysis of eletron diffration patterns in the transmission eletron mirosope from a polyrystalline speimen. The orientations and areas that you indexed are as follows. Later on in the ourse we will see from a study of stereology that area frations are equal to volume frations. ( φ 1, Φ, φ 2 ) Area of grain (31, 3, 1) 3.2 (31, 4, 3) 1.6 (28, 0, 2) 1.9 (33, 3, 0) 6.6 (10, 31, 38) 4.0 ( 8, 29, 42) 1.7 (14, 27, 41) 12.4 (12, 31, 43) 11.3 ( 1, 81, 50) 5.3 ( 3, 78, 48) 9.1 (71, 46, 11) 3.3 (68, 53, 13) 5.3 (70, 51, 8) 1.4 Using the same 10 x10 x10 grid as above, bin the areas (remember that we use ellentered oordinates) what are the intensities in the orresponding OD ells? Areas are treated identially to volumes. Also list the angles of the ells that ontain the points, with the volume of orientation spae (dω) assoiated with eah ell.

9 For example, some of the points fall into the ell for {70, 50, 10 }, for whih the intensity is 90 (MRD). Name of file shown in sreenshot above. Q-A. [20 points, EXTRA CREDIT] Shmid fator maximization Assume that slip ours on {111} planes and in <110> diretions. (a) For a tensile axis loated inside the standard stereographi triangle, what are the indies of the ative slip diretion and ative slip plane normal? Diretion: [uvw] = [0 1 1]; Plane: (hkl) = (1-1 1). Note an older version of the homework had as the triangle, in whih ase the answer is [101]/diretion, (-111)/plane. (b) Compute the indies of the diretion in the standard triangle that has a Shmid fator of exatly 0.5. Leave your answer in the form of radials (i.e. involving only simple frations, suh as ½, and square roots, suh as 2). The vetor should be a unit vetor and, however, you obtain an answer, you must demonstrate that it is orret by showing that the dot produt with both the slip diretion and the slip plane normal is 1/ 2. Hint: the value of 0.5 suggests ertain values for the angles (and therefore diretion osines) between the tensile axis and the plane normal and diretion. It is aeptable to use a mathematis pakage suh as Mathematia to find a solution but, if you do, be sure to demonstrate that your answer is orret. [uvw] = [ 1/ 6, (0.5 1/ 6), ( / 6)], or, [ 0.408, 0.092, 0.908] Chek that it s a unit vetor:

10 [ 1/ 6, (0.5 1/ 6), ( / 6)] 2 = 1/ /6-1/ /6 + 1/ 6 = /6 = 1. Proof: take dot produts with 1/ 2[0, 1, 1] and with 1/ 3( 1, -1, 1) to hek that the diretion osines are both 1/ 2. os(phi) = 1/ 2 {0.5 1/ / 6} = 1/ 2 os(lambda) = 1/ 3 [ 1/ 6 - (0.5 1/ 6) + ( / 6)] = 1/ 3 {3/ 6} = 3/( 3*6)} = 3/ 6 = 3/( 3* 2) = 1/ 2. Q.E.D. Also valid: [ 28 2, 2, 14 ] (normalized to be a unit vetor). Q-B [50 points extra redit]: determining the list of 28 stress states used in Bishop-Hill theory. Note I am suggesting this question for extra redit but I do not know for sure that it will work, so it is rather open-ended. Step 1. Generate a list of the 96 ombinations of 5 slips systems that are linearly independent. It is suggested that you write a program that generates all possible ombinations and use the non-zero determinant test to weed out the non-independent ombinations. Step 2. Chek that, out of the list of 96, when you ompute the orresponding deviatori stress states (assuming onstant CRSS on all 5 slip systems), only 28 disrete states remain (taking positive and negative versions of the same state to be only one result). Note that proeeding in this fashion should also yield the sets of possible ative slip systems (and there should be multiple sets for eah of the 28 Bishop-Hill stress states) program BishopHill gfortran -gdwarf-2 -O0 -o BishopHill Bishop_Hill.f90 solve the Bishop Hill model, adr 28 iv 01 updated for rystal orientation, 21 iv 05 formatted to f90, x 11 impliit none real salar real rtmp real stress,strain,dw,dwmax integer index,i,j,k,l dimension stress(28,6),strain(6) real a(3,3),hkl(3),uvw(3),t1(3) real eps(3,3),eps_xtal(3,3) real strain_xtal(6) real rtmp1,rtmp2,tayf real evm,enorm real sqr3 real tayf2 sqr3 = sqrt(3.) write(*,*) 'Welome to the Reid version of the BH algorithm'

11 write(*,*) 'Enter the strain as de11,de22,2*de23,2*de31,2*de12' read(*,*) strain(1),strain(2),strain(4),strain(5),strain(6) strain(3)=0.-strain(1)-strain(2) write(*,*) 'Full strain vetor: ' print"(2x,6(2x,f7.3))", (strain(i),i=1,6) eps(1,1) = strain(1) eps(2,2) = strain(2) eps(3,3) = strain(3) eps(2,3) = strain(4)/2. eps(3,1) = strain(5)/2. eps(1,2) = strain(6)/2. eps(3,2) = strain(4)/2. eps(1,3) = strain(5)/2. eps(2,1) = strain(6)/2. evm = sqrt((2./9.*((strain(1)-strain(2))**2 + 1 (strain(2)-strain(3))**2 + (strain(3)-strain(1))**2)) + 2 ((eps(1,2)**2 + eps(1,3)**2 + eps(2,3)**2)/3.)) evm = (2./sqr3)*sqrt(0.5*(eps(1,1)**2 + eps(2,2)**2 + eps(3,3)**2 & + eps(2,1)**2 + eps(3,1)**2 + eps(3,2)**2 + & eps(1,2)**2 + eps(1,3)**2 + eps(2,3)**2)) print"(' eps vm = ',f7.3)",evm write(*,*) write(*,*) 'Enter (hkl) // sample.3 as 3 numbers' read(*,*) hkl(1),hkl(2),hkl(3) write(*,"( 'Plane HKL = ',3(f8.1))") hkl all venorm(hkl) 10 write(*,*) 'Enter [uvw] // sample.1 as 3 numbers' read(*,*) uvw(1),uvw(2),uvw(3) write(*,"( 'Plane UVW = ',3(f8.1))") uvw all venorm(uvw) rtmp=salar(hkl,uvw) if(abs(rtmp).gt.1e-4) then write(*,*) 'Sorry, uvw is not perp. to hkl; try again' goto 10 endif all vepro2(hkl,uvw,t1) all venorm(t1) do 20, i=1,3 a(i,1)=uvw(i) a(i,2)=t1(i) a(i,3)=hkl(i) a(i,3)=t1(i) a(i,2)=hkl(i) 20 enddo standard arrangement, not Canova matrix print*,' Orientation Matrix: ' do i=1,3 print"('[ ',3(1x,f7.3),' ]')", (a(i,j),j=1,3) enddo

12 do i = 1,3 do j = 1,3 eps_xtal(i,j) = 0. do k =1,3 do l = 1,3 eps_xtal(i,j) = eps_xtal(i,j) + a(i,k)*a(j,l)*eps(k,l) enddo L enddo K enddo J enddo I print*,' Strain in Xtal oords: ' do i=1,3 print"('[ ',3(1x,f7.2),' ]')", (eps_xtal(i,j),j=1,3) enddo evm = (2./sqr3)*sqrt(0.5*(eps_xtal(1,1)**2 + eps_xtal(2,2)**2 & + eps_xtal(3,3)**2 + & eps_xtal(2,1)**2 + eps_xtal(3,1)**2 + eps_xtal(3,2)**2 + & eps_xtal(1,2)**2 + eps_xtal(1,3)**2 + eps_xtal(2,3)**2)) print"(' eps[xtal] vm = ',f7.3)",evm enorm = sqrt((eps_xtal(1,1)**2 + eps_xtal(2,2)**2 + eps_xtal(3,3)**2 + & eps_xtal(2,1)**2 + eps_xtal(3,1)**2 + eps_xtal(3,2)**2 + & eps_xtal(1,2)**2 + eps_xtal(1,3)**2 + eps_xtal(2,3)**2)) print"(' eps[xtal] vm = ',f7.3)",evm strain_xtal(1) = eps_xtal(1,1) strain_xtal(2) = eps_xtal(2,2) strain_xtal(3) = eps_xtal(3,3) strain_xtal(4) = eps_xtal(2,3) + eps_xtal(3,2) strain_xtal(5) = eps_xtal(3,1) + eps_xtal(1,3) strain_xtal(6) = eps_xtal(1,2) + eps_xtal(2,1) print* print*,'strain in rystal axes: ' print"(2x,6(2x,f7.3))", (strain_xtal(i),i=1,6) stress(1,1)=1. stress(1,2)=-1. stress(1,3)=0. stress(1,4)=0. stress(1,5)=0. stress(1,6)=0. # 1 stress(2,1)=0. stress(2,2)=1. stress(2,3)=-1. stress(2,4)=0. stress(2,5)=0. stress(2,6)=0. # 2 stress(3,1)=-1.

13 stress(3,2)=0. stress(3,3)=1. stress(3,4)=0. stress(3,5)=0. stress(3,6)=0. # 3 stress(4,1)=0. stress(4,2)=0. stress(4,3)=0. stress(4,4)=1. stress(4,5)=0. stress(4,6)=0. # 4 stress(5,1)=0. stress(5,2)=0. stress(5,3)=0. stress(5,4)=0. stress(5,5)=1. stress(5,6)=0. # 5 stress(6,1)=0. stress(6,2)=0. stress(6,3)=0. stress(6,4)=0. stress(6,5)=0. stress(6,6)=1. # 6 stress(7,1)=0.5 stress(7,2)=-1. stress(7,3)=0.5 stress(7,4)=0. stress(7,5)=0.5 stress(7,6)=0. # 7 stress(8,1)=0.5 stress(8,2)=-1. stress(8,3)=0.5 stress(8,4)=0. stress(8,5)=-0.5 stress(8,6)=0. # 8 stress(9,1)=-1. stress(9,2)=0.5 stress(9,3)=0.5 stress(9,4)=0.5 stress(9,5)=0. stress(9,6)=0. # 9 stress(10,1)=-1. stress(10,2)=0.5 stress(10,3)=0.5 stress(10,4)=-0.5 stress(10,5)=0.

14 stress(10,6)=0. # 10 stress(11,1)=0.5 stress(11,2)=0.5 stress(11,3)=-1. stress(11,4)=0. stress(11,5)=0. stress(11,6)=0.5 # 11 stress(12,1)=0.5 stress(12,2)=0.5 stress(12,3)=-1. stress(12,4)=0. stress(12,5)=0. stress(12,6)=-0.5 stress(13,1)=0.5 stress(13,2)=0. stress(13,3)=-0.5 stress(13,4)=0.5 stress(13,5)=0. stress(13,6)=0.5 stress(14,1)=0.5 stress(14,2)=0. stress(14,3)=-.5 stress(14,4)=-0.5 stress(14,5)=0. stress(14,6)=0.5 stress(15,1)=0.5 stress(15,2)=0. stress(15,3)=-0.5 stress(15,4)=0.5 stress(15,5)=0. stress(15,6)=-0.5 stress(16,1)=0.5 stress(16,2)=0. stress(16,3)=-0.5 stress(16,4)=-0.5 stress(16,5)=0. stress(16,6)=-0.5 stress(17,1)=0. stress(17,2)=-0.5 stress(17,3)=0.5 stress(17,4)=0. stress(17,5)=0.5 stress(17,6)=0.5 stress(18,1)=0. stress(18,2)=-0.5

15 stress(18,3)=0.5 stress(18,4)=0. stress(18,5)=-0.5 stress(18,6)=0.5 stress(19,1)=0. stress(19,2)=-0.5 stress(19,3)=0.5 stress(19,4)=0. stress(19,5)=0.5 stress(19,6)=-0.5 stress(20,1)=0. stress(20,2)=-0.5 stress(20,3)=0.5 stress(20,4)=0. stress(20,5)=-0.5 stress(20,6)=-0.5 stress(21,1)=-0.5 stress(21,2)=0.5 stress(21,3)=0. stress(21,4)=0.5 stress(21,5)=0.5 stress(21,6)=0. stress(22,1)=-0.5 stress(22,2)=0.5 stress(22,3)=0. stress(22,4)=-0.5 stress(22,5)=0.5 stress(22,6)=0. stress(23,1)=-0.5 stress(23,2)=0.5 stress(23,3)=0. stress(23,4)=0.5 stress(23,5)=-0.5 stress(23,6)=0. stress(24,1)=-0.5 stress(24,2)=0.5 stress(24,3)=0. stress(24,4)=-0.5 stress(24,5)=-0.5 stress(24,6)=0. stress(25,1)=0. stress(25,2)=0. stress(25,3)=0. stress(25,4)=0.5 stress(25,5)=0.5 stress(25,6)=-0.5

16 stress(26,1)=0. stress(26,2)=0. stress(26,3)=0. stress(26,4)=0.5 stress(26,5)=-0.5 stress(26,6)=0.5 stress(27,1)=0. stress(27,2)=0. stress(27,3)=0. stress(27,4)=-0.5 stress(27,5)=0.5 stress(27,6)=0.5 stress(28,1)=0. stress(28,2)=0. stress(28,3)=0. stress(28,4)=0.5 stress(28,5)=0.5 stress(28,6)=0.5 dwmax=0. index=0 do 100, i=1,28 dw = 0. dw = dw - strain_xtal(1)*stress(i,2) dw = dw + strain_xtal(2)*stress(i,1) dw = dw + strain_xtal(4)*stress(i,4) dw = dw + strain_xtal(5)*stress(i,5) dw = dw + strain_xtal(6)*stress(i,6) dw = -Bde11 + Ade22 + 2Fde23 + 2Gde31 +2Hde12 if(dw.gt.dwmax) then dwmax=dw index=i print*,'dw,i ',dw,' ',i endif if((-1.*dw).gt.dwmax) then dwmax=-1.*dw index=i+28 print*,'dw,i ',dw,' ',i endif 100 end do print* write(*,*) 'Index of the multiple slip stress state = ',index if(index.gt.28) then index = index-28 write(*,*) 'Index of the +/- stress state = ',index endif print*,'stress # A B C F G H' write(*,120) (stress(index,j),j=1,6) 120 format(6x,6(2x,f6.1))

17 rtmp1 = 0. do i = 1,6 rtmp1 = rtmp1 + stress(index,i)**2 enddo magnitude of the stress rtmp2 = 0. do i = 1,6 rtmp2 = rtmp2 + strain_xtal(i)**2 enddo magnitude of the strain in xtal oords. print*,'dwmax,rtmp1,rtmp2',dwmax,sqrt(rtmp1),sqrt(rtmp2) tayf = 2.* dwmax / sqrt(rtmp2) tayf = dwmax / evm This is the definition of Taylor fator that allows for multi-axial strain or stress states, as found in LApp print* print"('the Taylor fator = ',f7.3,' *sqrt(6)')",tayf print"(' = ',f7.3)",tayf*sqrt(6.) these next few lines were for uriosity about other normalizations of strain tayf2 = dwmax / enorm print* print"('the Taylor fator = ',f7.3,' *sqrt(6)')",tayf2 print"(' = ',f7.3)",tayf2*sqrt(6.) all exit(0) end program BishopHill subroutine venorm(ve) real ve(3),rnorm rnorm=0. do 10, i=1,3 rnorm=rnorm+ve(i)**2 10 end do if(rnorm.le.0.0) return rnorm=sqrt(rnorm) do 20, i=1,3 ve(i)=ve(i)/rnorm 20 end do return end subroutine venorm funtion salar(a,b)

18 return SCALAR PRODUCT of A and B real salar,a(3),b(3) salar=0. do 100, i=1,3 salar=salar+a(i)*b(i) 100 end do return end funtion salar subroutine vepro2(v1,v2,vout) vetor produt real v1(3),v2(3),vout(3) vout(3)=v1(1)*v2(2)-v1(2)*v2(1) vout(1)=v1(2)*v2(3)-v1(3)*v2(2) vout(2)=v1(3)*v2(1)-v1(1)*v2(3) return end subroutine vepro2