Nonsymmetric Examples: Benites-Kriz ______________________________________________________ determination of eigenvalues by jacobi's method, where n = 3 itmax = 50 eps1 = 0.100E-09 eps2 = 0.100E-09 eps3 = 0.100E-04 the starting matrix a(1,1)...a(n,n) is 0.500E+01 -0.300E+01 0.200E+01 -0.300E+01 0.800E+01 0.400E+01 0.200E+01 0.400E+01 -0.900E+01 iter = 1 sigma1 = 0.170E+03 sigma2 = 0.227E+03 iter = 2 sigma1 = 0.227E+03 sigma2 = 0.228E+03 convergence has occured, where iter = 3 s = 0.228E+03 sigma2 = 0.228E+03 ____________________ eigenvalues eigen(1)...eigen(n) are 0.417E+01 0.102E+02 -0.103E+02 eignevectors are arranged respectively in columns. matrix is transposed when calculating Euler angles 0.873E+00 0.457E+00 0.171E+00 0.414E+00 -0.879E+00 0.236E+00 0.258E+00 -0.136E+00 -0.956E+00 Final stress tensor reconstructed by superposition of the principal stress state plus the transformation of effective moment/volume tensor as off-diagonal shear stress terms 0.417E+01 0.000E+00 0.000E+00 0.000E+00 0.102E+02 0.000E+00 0.000E+00 0.000E+00 -0.103E+02 ------------------------------------------------- Results of symmetric stress tensor transformation -- for eigenvalue = 1, row sum = 0.0 -- -- confirm: for row =1 sum = -0.666E-15 -- confirm: for row =2 sum = 0.444E-15 -- confirm: for row =3 sum = -0.444E-15 -- for eigenvalue = 2, row sum = 0.0 -- -- confirm: for row =1 sum = 0.151E-13 -- confirm: for row =2 sum = 0.233E-13 -- confirm: for row =3 sum = -0.902E-13 -- for eigenvalue = 3, row sum = 0.0 -- -- confirm: for row =1 sum = 0.433E-13 -- confirm: for row =2 sum = -0.826E-13 -- confirm: for row =3 sum = -0.158E-13 eigenvalues eigen(1)...eigen(n) are 0.417E+01 0.102E+02 -0.103E+02 eignevectors are arranged respectively in columns. matrix is transposed when calculating Euler angles 0.873E+00 0.457E+00 0.171E+00 0.414E+00 -0.879E+00 0.236E+00 0.258E+00 -0.136E+00 -0.956E+00 tensor transformation matrix, a(i,j) is 0.873E+00 0.414E+00 0.258E+00 0.457E+00 -0.879E+00 -0.136E+00 0.171E+00 0.236E+00 -0.956E+00 angles (degrees) between rotated-unrotated axes ( X->x' Y->y' Z->z' ) 29.201726 151.529722 163.036394 euler angles (degrees) ( x y z ) -13.679150 -10.129442 -25.213471 ___________________________________________________________ CHECK: reconstruct principal stress state by a second order tensor transformation. Transpose eigenvectors = transformation matrix: e.g. off-diagonal shear stress components of the transformed stresses go to zero and the diagonal components recover the eigenvalues 0.417E+01 -0.500E-15 0.000E+00 -0.500E-15 0.102E+02 0.946E-13 0.444E-15 0.946E-13 -0.103E+02 ______________________________________________________ determination of eigenvalues by jacobi's method, where n = 3 itmax = 50 eps1 = 0.100E-09 eps2 = 0.100E-09 eps3 = 0.100E-04 the starting matrix a(1,1)...a(n,n) is 0.500E+01 -0.300E+01 0.200E+01 -0.300E+01 0.800E+01 0.400E+01 0.200E+01 0.400E+01 0.900E+01 iter = 1 sigma1 = 0.170E+03 sigma2 = 0.226E+03 iter = 2 sigma1 = 0.226E+03 sigma2 = 0.228E+03 convergence has occured, where iter = 3 s = 0.228E+03 sigma2 = 0.228E+03 ____________________ eigenvalues eigen(1)...eigen(n) are 0.114E+01 0.829E+01 0.126E+02 eignevectors are arranged respectively in columns. matrix is transposed when calculating Euler angles 0.681E+00 0.728E+00 -0.808E-01 0.568E+00 -0.455E+00 0.686E+00 -0.462E+00 0.513E+00 0.723E+00 Final stress tensor reconstructed by superposition of the principal stress state plus the transformation of effective moment/volume tensor as off-diagonal shear stress terms 0.114E+01 0.000E+00 0.000E+00 0.000E+00 0.829E+01 0.000E+00 0.000E+00 0.000E+00 0.126E+02 ------------------------------------------------- Results of symmetric stress tensor transformation -- for eigenvalue = 1, row sum = 0.0 -- -- confirm: for row =1 sum = 0.191E-12 -- confirm: for row =2 sum = -0.118E-12 -- confirm: for row =3 sum = 0.134E-12 -- for eigenvalue = 2, row sum = 0.0 -- -- confirm: for row =1 sum = 0.183E-12 -- confirm: for row =2 sum = 0.147E-12 -- confirm: for row =3 sum = -0.118E-12 -- for eigenvalue = 3, row sum = 0.0 -- -- confirm: for row =1 sum = 0.888E-15 -- confirm: for row =2 sum = 0.577E-14 -- confirm: for row =3 sum = 0.488E-14 eigenvalues eigen(1)...eigen(n) are 0.114E+01 0.829E+01 0.126E+02 eignevectors are arranged respectively in columns. matrix is transposed when calculating Euler angles 0.681E+00 0.728E+00 -0.808E-01 0.568E+00 -0.455E+00 0.686E+00 -0.462E+00 0.513E+00 0.723E+00 tensor transformation matrix, a(i,j) is 0.681E+00 0.568E+00 -0.462E+00 0.728E+00 -0.455E+00 0.513E+00 -0.808E-01 0.686E+00 0.723E+00 angles (degrees) between rotated-unrotated axes ( X->x' Y->y' Z->z' ) 47.065383 117.082149 43.682020 euler angles (degrees) ( x y z ) -43.307821 -6.371867 -51.270577 ___________________________________________________________ CHECK: reconstruct principal stress state by a second order tensor transformation. Transpose eigenvectors = transformation matrix: e.g. off-diagonal shear stress components of the transformed stresses go to zero and the diagonal components recover the eigenvalues 0.114E+01 0.261E-12 0.888E-15 0.261E-12 0.829E+01 -0.444E-15 0.444E-15 -0.444E-15 0.126E+02 ______________________________________________________ determination of eigenvalues by jacobi's method, where n = 3 itmax = 50 eps1 = 0.100E-09 eps2 = 0.100E-09 eps3 = 0.100E-04 the starting matrix a(1,1)...a(n,n) is 0.500E+01 -0.300E+01 0.400E+01 -0.300E+01 0.800E+01 0.200E+01 0.200E+01 0.400E+01 -0.900E+01 s(1,3) > s(3,1) and emo(2)= 0.200E+01 s(3,2) > s(2,3) and emo(1)= 0.200E+01 Nonsymmetric stress tensor detected. Stress tensor decomposed into a symmetric stress tensor and an equivalent nonsymmetric off-diagonal shear stress moment/volume first order tensor (vector) The symmetric stress tensor ss(i,j): 0.500E+01 -0.300E+01 0.200E+01 -0.300E+01 0.800E+01 0.200E+01 0.200E+01 0.200E+01 -0.900E+01 The equivalent nonsymmetric off-diagonal shear stress moment/volume first order tensor (vector) 0.200E+01 0.200E+01 0.000E+00 iter = 1 sigma1 = 0.170E+03 sigma2 = 0.204E+03 iter = 2 sigma1 = 0.204E+03 sigma2 = 0.204E+03 convergence has occured, where iter = 3 s = 0.204E+03 sigma2 = 0.204E+03 ____________________ eigenvalues eigen(1)...eigen(n) are 0.374E+01 0.988E+01 -0.962E+01 eignevectors are arranged respectively in columns. matrix is transposed when calculating Euler angles 0.843E+00 0.513E+00 -0.162E+00 0.495E+00 -0.858E+00 -0.139E+00 0.210E+00 -0.366E-01 0.977E+00 ------------ Nonzero off-diagonal shear stresses are the transformed equivalent moment/volume vector using the same transposed eigenvector matrix used for the stress transformation 0.268E+01 -0.691E+00 -0.601E+00 Final stress tensor reconstructed by superposition of the principal stress state plus the transformation of effective moment/volume tensor as off-diagonal shear stress terms 0.374E+01 -0.601E+00 0.000E+00 0.000E+00 0.988E+01 0.000E+00 -0.691E+00 0.268E+01 -0.962E+01 ------------------------------------------------- Results of symmetric stress tensor transformation -- for eigenvalue = 1, row sum = 0.0 -- -- confirm: for row =1 sum = -0.278E-13 -- confirm: for row =2 sum = -0.227E-13 -- confirm: for row =3 sum = 0.168E-12 -- for eigenvalue = 2, row sum = 0.0 -- -- confirm: for row =1 sum = -0.236E-15 -- confirm: for row =2 sum = -0.112E-14 -- confirm: for row =3 sum = 0.111E-15 -- for eigenvalue = 3, row sum = 0.0 -- -- confirm: for row =1 sum = 0.145E-12 -- confirm: for row =2 sum = 0.839E-13 -- confirm: for row =3 sum = 0.369E-13 eigenvalues eigen(1)...eigen(n) are 0.374E+01 0.988E+01 -0.962E+01 eignevectors are arranged respectively in columns. matrix is transposed when calculating Euler angles 0.843E+00 0.513E+00 -0.162E+00 0.495E+00 -0.858E+00 -0.139E+00 0.210E+00 -0.366E-01 0.977E+00 tensor transformation matrix, a(i,j) is 0.843E+00 0.495E+00 0.210E+00 0.513E+00 -0.858E+00 -0.366E-01 -0.162E+00 -0.139E+00 0.977E+00 angles (degrees) between rotated-unrotated axes ( X->x' Y->y' Z->z' ) 32.522200 149.075983 12.313241 euler angles (degrees) ( x y z ) 7.963144 -9.422187 -29.979750 ___________________________________________________________ CHECK: reconstruct principal stress state by a second order tensor transformation. Transpose eigenvectors = transformation matrix: e.g. off-diagonal shear stress components of the transformed stresses go to zero and the diagonal components recover the eigenvalues 0.374E+01 -0.125E-15 0.171E-12 -0.847E-15 0.988E+01 0.722E-15 0.171E-12 0.722E-15 -0.962E+01 ______________________________________________________ determination of eigenvalues by jacobi's method, where n = 3 itmax = 50 eps1 = 0.100E-09 eps2 = 0.100E-09 eps3 = 0.100E-04 the starting matrix a(1,1)...a(n,n) is 0.500E+01 -0.300E+01 0.400E+01 -0.300E+01 0.800E+01 0.200E+01 0.200E+01 0.400E+01 0.900E+01 s(1,3) > s(3,1) and emo(2)= 0.200E+01 s(3,2) > s(2,3) and emo(1)= 0.200E+01 Nonsymmetric stress tensor detected. Stress tensor decomposed into a symmetric stress tensor and an equivalent nonsymmetric off-diagonal shear stress moment/volume first order tensor (vector) The symmetric stress tensor ss(i,j): 0.500E+01 -0.300E+01 0.200E+01 -0.300E+01 0.800E+01 0.200E+01 0.200E+01 0.200E+01 0.900E+01 The equivalent nonsymmetric off-diagonal shear stress moment/volume first order tensor (vector) 0.200E+01 0.200E+01 0.000E+00 iter = 1 sigma1 = 0.170E+03 sigma2 = 0.204E+03 iter = 2 sigma1 = 0.204E+03 sigma2 = 0.204E+03 convergence has occured, where iter = 3 s = 0.204E+03 sigma2 = 0.204E+03 ____________________ eigenvalues eigen(1)...eigen(n) are 0.205E+01 0.936E+01 0.106E+02 eignevectors are arranged respectively in columns. matrix is transposed when calculating Euler angles 0.773E+00 0.625E+00 0.108E+00 0.514E+00 -0.519E+00 -0.683E+00 -0.370E+00 0.583E+00 -0.723E+00 ------------ Nonzero off-diagonal shear stresses are the transformed equivalent moment/volume vector using the same transposed eigenvector matrix used for the stress transformation 0.258E+01 0.211E+00 -0.115E+01 Final stress tensor reconstructed by superposition of the principal stress state plus the transformation of effective moment/volume tensor as off-diagonal shear stress terms 0.205E+01 -0.115E+01 0.211E+00 0.000E+00 0.936E+01 0.000E+00 0.000E+00 0.258E+01 0.106E+02 ------------------------------------------------- Results of symmetric stress tensor transformation -- for eigenvalue = 1, row sum = 0.0 -- -- confirm: for row =1 sum = 0.666E-15 -- confirm: for row =2 sum = 0.888E-15 -- confirm: for row =3 sum = -0.888E-15 -- for eigenvalue = 2, row sum = 0.0 -- -- confirm: for row =1 sum = 0.215E-12 -- confirm: for row =2 sum = -0.137E-11 -- confirm: for row =3 sum = -0.145E-11 -- for eigenvalue = 3, row sum = 0.0 -- -- confirm: for row =1 sum = 0.126E-11 -- confirm: for row =2 sum = -0.104E-11 -- confirm: for row =3 sum = 0.117E-11 eigenvalues eigen(1)...eigen(n) are 0.205E+01 0.936E+01 0.106E+02 eignevectors are arranged respectively in columns. matrix is transposed when calculating Euler angles 0.773E+00 0.625E+00 0.108E+00 0.514E+00 -0.519E+00 -0.683E+00 -0.370E+00 0.583E+00 -0.723E+00 tensor transformation matrix, a(i,j) is 0.773E+00 0.514E+00 -0.370E+00 0.625E+00 -0.519E+00 0.583E+00 0.108E+00 -0.683E+00 -0.723E+00 angles (degrees) between rotated-unrotated axes ( X->x' Y->y' Z->z' ) 39.332267 121.270807 136.279501 euler angles (degrees) ( x y z ) 43.053793 -8.482390 -44.734180 ___________________________________________________________ CHECK: reconstruct principal stress state by a second order tensor transformation. Transpose eigenvectors = transformation matrix: e.g. off-diagonal shear stress components of the transformed stresses go to zero and the diagonal components recover the eigenvalues 0.205E+01 -0.888E-15 0.000E+00 -0.888E-15 0.936E+01 0.201E-11 0.000E+00 0.201E-11 0.106E+02 ______________________________________________________ Max= 0.500E+00 Min= 0.500E+00 Scale Factor= 0.200E+02 x(1)= 0.000E+00 y(1)= 0.000E+00 z(1)= 0.000E+00 x(2)= 0.100E+02 y(2)= 0.000E+00 z(2)= 0.000E+00 x(3)= 0.000E+00 y(3)= 0.000E+00 z(3)=-0.100E+02 x(4)= 0.100E+02 y(4)= 0.000E+00 z(4)=-0.100E+02