Is it possible to decompose a 4th-order tensor Dijkl with minor symmetry (i.e., Dijkl=Djikl=Dijlk) into Dijkl=AijBkl? And does Dijkl have its polar decomposition form like a 2-order tensor ? Welcome to discuss this question.
The answer is no. A counter example is the simplest tensor of elastic constants C_{ijkl}, i.e. that of an isotropic material. C_{ijkl} in this case is written in the form M_{ij}N_{kl}+P_{ik}Q_{jl}. You may first want to look at isotropic fourth-order tensors for which there is a representation theorem. You can look this up in Gurtin's book on continuum mechanics or the book by Chadwick.
Thanks for Arash's reply. However setting a typical example here, the 4th order tensor of elasto-plastic stiffness C_{ijkl}, multiplied by 2nd-order tensor of strain increment, dEps{ij}, results in a new 2nd-order tensor of stress increment, dSig_{kl}. Physically, C_{ijkl} produces two effects on dEps_{kl}: One is stretching the principal values of dEps_{kl} into those of dSig_{kl}, and the other is ratating principal directions of dEps_{kl} to be aligned with those of dSig_{kl}. Therefore theoretically, C_{ijkl} might be decomposed into the stretching part and the rotatation part. I attempted to separate C_{ijkl} into pure stretching tensor and pure rotation tensor, but I got no answer. However, I feel it is possible to decompose the 4th-fourth C_{ijkl}, although I am not sure what is the final form of decomposition. The problem really puzzled me for a long time. Thanks a lot.
You can always decompose the deformation gradient into a stretch and a rotation part; this is the polar decomposition theorem. In my opinion, it is meaningless to decompose C into a "stretch" and a "rotation" part. Why do you need to do this?
Thanks for your question. Only for study the plastic non-coxaility.
Non-coaxaility is one of important behaviors of geomaterials, epecially when subjected to non-proportional loading. Non-coaxaility is traditionally refered to the non-coincidence between the principal directions of stress tensor,Sig_{ij}, and the principal directions of plastic strain increment tensor,depsp_{kl}. Then, a problem is arising: How should we measure the degree of non-coaxiality?
I should emphesize here that tradiitonal plastic flow theory always produces coaxial depsp_{kl} since the plastic potential function or the yield function is generally defined as a isotropic function of sig_{ij}, expressed by three stress invariants.(Refer to my paper 'Three-Dimensional Noncoaxial Plasticity Modeling of Shear Band Formation in Geomaterials', Journal of Engineering Mechanics, 134(4):322-329)
Here my investigation of non-coaxaility has a slight difference from traditional non-coaxiality defined above. I would like to study the non-coaxaility between Dsig_{ij}(instead of sig_{ij}) and Deps_{kl}, which should be fisrt-step work on non-coaxiality.
What do you mean by "non-proportional loading"? Is your model defined in terms of stress invariants (i.e. isotropic functions of stress tensor) only? If yes, how can it be applied to cyclic loading? Can your model capture the hysteresis under loading/unloading conditions and phenomena such as Massing rule in one-dimensional loading?
Thanks for more concerns. Proportional loading has a stress
history in which the deviatoric stress components are kept in constant ratio to
each other. If stress paths do not fit into the definition, the loading are
called non-proportional loading. As you said, non-proportional loading is
commonly associated with cyclic loading. However, most cases of static loading
are also of non-proportional loading only if the number of independent stress
components is more than three. In fact, most of loading cases have
non-proportional instead of proportional loading history for practical
engineering problems(e.g., most cases of geotechnical engineering).
My present model is an extended static version from
classical models(defined in terms of stress invariants) with its
capacities of accounting for
non-proportional static loading problems. In my opinion, however, dynamic
models may be easily available if we incorporate some appropriate kinematic
hardening rule into static models. In this sense, a static model and a dynamic
model make little difference.
Finally back to my question. In fact, my early posted question
may be simplified as how to decompose a non-symmetrical 2nd-order
tensor. As we know, a given 2nd order tensor C_{ij}, multiplied by a given
vector, m_{j}, produces a new vector n_{i}. Just likewise what I meant in last post
comment. C_{ij} will stretch the length of m_{j}, then rotates the
direction of m_{j} to be aligned with n_{i}. How can we separate C_{ij} into
the pure stretching part and the pure rotating part? To puzzle me, I found that
U_{kj} and R_{ik} available from polar decomposition of C_{ij}, do not
match this question.
Re: decomposition of fourth-order tensor
The answer is no. A counter example is the simplest tensor of elastic constants C_{ijkl}, i.e. that of an isotropic material. C_{ijkl} in this case is written in the form M_{ij}N_{kl}+P_{ik}Q_{jl}. You may first want to look at isotropic fourth-order tensors for which there is a representation theorem. You can look this up in Gurtin's book on continuum mechanics or the book by Chadwick.
Regards,
Arash
more question about decomposition of 4th-order tenor
Thanks for Arash's reply. However setting a typical example here, the 4th order tensor of elasto-plastic stiffness C_{ijkl}, multiplied by 2nd-order tensor of strain increment, dEps{ij}, results in a new 2nd-order tensor of stress increment, dSig_{kl}. Physically, C_{ijkl} produces two effects on dEps_{kl}: One is stretching the principal values of dEps_{kl} into those of dSig_{kl}, and the other is ratating principal directions of dEps_{kl} to be aligned with those of dSig_{kl}. Therefore theoretically, C_{ijkl} might be decomposed into the stretching part and the rotatation part. I attempted to separate C_{ijkl} into pure stretching tensor and pure rotation tensor, but I got no answer. However, I feel it is possible to decompose the 4th-fourth C_{ijkl}, although I am not sure what is the final form of decomposition. The problem really puzzled me for a long time. Thanks a lot.
Regards,
Jiangu qian
re: C "decomposition"
Dear Jiangu:
You can always decompose the deformation gradient into a stretch and a rotation part; this is the polar decomposition theorem. In my opinion, it is meaningless to decompose C into a "stretch" and a "rotation" part. Why do you need to do this?
Regards,
Arash
In reply to re: C "decomposition" by arash_yavari
For solving non-coaxiality
Thanks for your question. Only for study the plastic non-coxaility.
Non-coaxaility is one of important behaviors of geomaterials, epecially when subjected to non-proportional loading. Non-coaxaility is traditionally refered to the non-coincidence between the principal directions of stress tensor,Sig_{ij}, and the principal directions of plastic strain increment tensor,depsp_{kl}. Then, a problem is arising: How should we measure the degree of non-coaxiality?
I should emphesize here that tradiitonal plastic flow theory always produces coaxial depsp_{kl} since the plastic potential function or the yield function is generally defined as a isotropic function of sig_{ij}, expressed by three stress invariants.(Refer to my paper 'Three-Dimensional Noncoaxial Plasticity Modeling of Shear Band Formation in Geomaterials', Journal of Engineering Mechanics, 134(4):322-329)
Here my investigation of non-coaxaility has a slight difference from traditional non-coaxiality defined above. I would like to study the non-coaxaility between Dsig_{ij}(instead of sig_{ij}) and Deps_{kl}, which should be fisrt-step work on non-coaxiality.
Regards,
Jiangu qian
In reply to For solving non-coaxiality by qianjiangu
few questions for Jiangu qian
What do you mean by "non-proportional loading"? Is your model defined in terms of stress invariants (i.e. isotropic functions of stress tensor) only? If yes, how can it be applied to cyclic loading? Can your model capture the hysteresis under loading/unloading conditions and phenomena such as Massing rule in one-dimensional loading?
In reply to few questions for Jiangu qian by Aleksandar Spa…
For Arash's question
Thanks for more concerns. Proportional loading has a stress
history in which the deviatoric stress components are kept in constant ratio to
each other. If stress paths do not fit into the definition, the loading are
called non-proportional loading. As you said, non-proportional loading is
commonly associated with cyclic loading. However, most cases of static loading
are also of non-proportional loading only if the number of independent stress
components is more than three. In fact, most of loading cases have
non-proportional instead of proportional loading history for practical
engineering problems(e.g., most cases of geotechnical engineering).
My present model is an extended static version from
classical models(defined in terms of stress invariants) with its
capacities of accounting for
non-proportional static loading problems. In my opinion, however, dynamic
models may be easily available if we incorporate some appropriate kinematic
hardening rule into static models. In this sense, a static model and a dynamic
model make little difference.
Finally back to my question. In fact, my early posted question
may be simplified as how to decompose a non-symmetrical 2nd-order
tensor. As we know, a given 2nd order tensor C_{ij}, multiplied by a given
vector, m_{j}, produces a new vector n_{i}. Just likewise what I meant in last post
comment. C_{ij} will stretch the length of m_{j}, then rotates the
direction of m_{j} to be aligned with n_{i}. How can we separate C_{ij} into
the pure stretching part and the pure rotating part? To puzzle me, I found that
U_{kj} and R_{ik} available from polar decomposition of C_{ij}, do not
match this question.
Regards,
Jiangu qian