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Fourth order tensor

Submitted by ramdas chennamsetti on

Hi all,

I have a fundamental question on Tensors. The length of a vector (firts order tensor) is independent of the reference co-ordinate system. In case of second order tensor (stress/strain), the invariants (I1, I2, I3) are independent of the co-ordinate system.

If I consider 4th order tensor (of course 3rd order also), say Cijkl, what parameters are constant? (Like length in vector and invariants in second order tensors).

Thanks in advance,

- Ramdas

Temesgen Markos

Hi Ramadas,

One straight forward way of producing invariants is taking contractions until all the indices become dummy. For a second order tensor T_{ij}, for example, you can do one contraction T_{ii} and you get the first invariant. Similary if you take the square T^2_{ik} = T_{ij}T_{jk} and take a contraction T_{ij}T_{ji} you get the trace of the square, which is another invariant. Do the same thing to fourth order tensors. There are more ways to contract a fourth order tensor. C_{ijij}, C_{iijj}, C_{ijji}, C_{iiii} are the ones I can think of. Ofcourse you can take powers of the tensor and do contractions to get more. The contraction is basically multiplication by the kronecker delta. You can also think of multiplication by the Levi-civita symbol. For example, det(T) = 1/6e_{ijk}e_{pqr}T_{ip}T_{jq}T_{kr}.

This is true if you have orthonormal basis. Otherwise use the appropriate definitions, for example, of the Kronecker delta and follow the same procedure.

Mon, 09/14/2009 - 16:26 Permalink
Temesgen Markos

Hi Ramdas,

I don't know a book which specifically deals with tensor invariants.  As far as tensor analysis is concerned, my favourite is the first chapter from Ogden's book "Nonlinear Elastic Deformations". The later chapters touch on invariance, symmetry etc from the perspective of elasticity in general and Green elasticity in particular. 

Thu, 09/17/2009 - 04:40 Permalink
arash_yavari

Dear Ramdas:



What you're looking for is called an integrity basis. Invariance under a change of coordinates is equivalent to invariance under SO(3) (the orientation preserving subgroup of O(3)). Then you want to see what polynomial functions of your fourth-order tensor are invariant under the action of this group. In the case of isotropic tensor functions, one can reduce the dependency of the function of interest to an irreducible basis invariants. As you mentioned, for second-order tensors there are three  principal invariants. Equivalently you can rewrite these in terms of basic invariants. Denoting your second-order tensor by A, these are traces of A, A^2=A.A, and A^3=A.A.A. In the case of fourth-order tensors there are six basic (and principal) invariants. Denoting the fourth order tensor by C, these are traces of C^i, i=1,...,6.



Look at the following short paper: J. Betten, Integrity basis for a second-order and a fourth-order tensor, International journal of mathematics and mathematical sciences 5(1), 87-96, 1982.



Regards,

Arash 

Wed, 09/16/2009 - 18:23 Permalink
ramdas chennamsetti

Dear Arash,

It looks I need to understand more on Tensors. I went through which is required for stress and strain analysis. I downloaded that paper and go through. It looks it is not sufficient.

Could you please suggest some good book exclusively on 'Tensors'?

With regards,

- Ramdas 

Thu, 09/17/2009 - 00:49 Permalink
ramdas chennamsetti

Hi Biswajit Sir,

Just now I have seen that book in Google books. Nice one.Do you have soft copy of this? If so, could you please share?

With regards,

- Ramdas

Sat, 09/19/2009 - 10:13 Permalink
Biswajit Banerjee

Hi Ramdas,

I don't have a soft copy of the book.  I don't think a legal soft copy exists *cough*rapidshare*cough*.  A good alternative is Prof. Brannon's detailed notes.   

I haven't been knighted yet; so the "Sir" upadhi isn't really necessary :)

-- Biswajit 

 

Tue, 09/22/2009 - 02:53 Permalink
Arun Krishnan

Hi all,

Nice discussion going on here. Here's my 5 cents. These are a few links on tensor analysis (all of them introductory) which I found useful when I was learning about tensors. 

1) (NASA's brief introductory guide) http://www.grc.nasa.gov/WWW/K-12/Numbers/Math/documents/Tensors_TM20022…

2) (Ruslan Sharipov) http://arxiv.org/abs/math.HO/0403252

3) (Joakim Strandberg) http://medlem.spray.se/gorgelo/tensors.pdf

-Arun

Sat, 09/19/2009 - 18:15 Permalink
rrahman

Dear Ramdas

You may be interested in the following paper on 4th order tensors and their manipulations in continuum mechanics:

 "Fourth-order tensors - tensor differentiation with applications to continuum mechanics. Part I: Classical tensor analysis"  by O. Kintzel, Y. Baar.

Besides that I would suggest you to have some concepts of symmetric gruops and tensor products also. You may like to read:

"Algebra" by Artin

 Thanks

Thu, 10/01/2009 - 17:58 Permalink
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Wed, 03/31/2010 - 08:26 Permalink
szeptynski

I've accidentaly found the topic - it's quite old, however the invariants of symmetric(!) 4th order tensors are the subject of my research - maybe someone will find those information useful.

The thing concernes only the 4th order tensors C satisfying symmetry conditions C_(ijkl)=C_(jikl)=C_(ijlk)=C_(klij) - so called Hooke's tensors (e.g. elasticity - stiffness S and compliance C - tensors). They have 3^4=81 components of which only 21 are independent due to symmetry of stress and strain tensors and symmetry of elastic potential differentiation. Among them we can distinguish:

-6 eigenvalues (invariants)

-12 stiffness distributors (invariants)

-3 orienting parameters (not invariant)

Those 3 last parameters are not invariants - they are used only to oriente the internal structure of the material in the considered coordinate system (e.g. Euler angles, or components of 3 orthogonal versors (+9 components -3 orthogonality conditions -3 normalization conditions)) - it is espacially clearly visible in case of orthotropy or tetragonal or cubic symmetry, however it concernes all 8 elastic symmetries.

It is important to note that elasticity tensors act in the linear space of symmetric 2nd order tensors (let's dentoe it with T2s) as a linear automorphic operator. If we consider space of "relative stresses" (dimension-less space) then stress and strain states' spaces are both equivalent to T2s. T2s is in fact 6-dimensional - that's why we can distinguish only 6 eigenvalues of C and S - Rychlewski suggested to call them the "Kelvin moduli" (it was William Thomson, baron Kelvin who introduced them first) - they are invariants.

12 stiffness distributors are defined as invariant functions of the eigenstates (or eigentensors - eigenvectors in a "vector space" of symmetric 2nd order tensors T2s) which is also sign insensitive. Eigenvectors of 2nd order tensors are simply vectors (1st order tensors) - they are uniquely given by the orientation of the principal directions of the tensor. In case of 4th rank tensors we have in general 6 eigenstates - each consists of 6 independent components. Those states has to be mutually orthogonal and normalized - it gives us 6*6-(5+4+3+2+1)-6=15 parameters. 3 of them are the above mentioned orienting parameters. 12 stiffness distributors determine the form of the eigenstates (ratios between each components of each eigenstate). E.g. some eigenstates for certain symmetries are a composition of two pure shear states of different magnitudes - the ratio of those magnitudes is a function of stiffness distributor.

As the symmetry of C and S rises the number of independent invariants decreases - some Kelvin moduli are equal one to another and some stiffness distributors "vanish".

There are also 6 other invariants I1,...,I6 of C - the coefficients of its charateristic polynomial p(L) - they can be expressed in terms of the Kelvin moduli L1,...,L6:

p(L)=L^6-I1*L^5+I2*L^4-I3*L^3+I4*L^2-I5*L+I6=(L-L1)*(L-L2)*(L-L3)*(L-L4)*(L-L5)*(L-L6) 

They can be found (as well as the Kelvin moduli and certain functions of stiffness distributors) through an eigenproblem analysis of a 6x6 matrix being a representation of C in T2s (it's better to do so in so called Mendel notation - a bit different than the classical Voigt notation). Physical interpretation of those invariants is rather unclear if any. 

Those problems were first discussed by J.Rychlewski in 1983 (in an almost unavailable small print in Russian) and then in 1984 (J.Rychlewski, "On Hooke's Law", J Appl Math Mech, 48, 3 (1984), pp. 303-314) and later independently by M.Mehrabadi and S.Cowin ("Eigentensors of linear anisotropic elastic materials", Q J Mech Appl Math, 43 (1990), pp. 15-41) (however in both papers there are some minor mistatements on the independency of eigenstates' invariants and on uniqueness of spectral decomposition) and then by S.Sutcliffe ("Spectral decomposition of the elasticity tensor", Transactions of the ASME 762, vol. 59, (1992)).

It concernes all the invariants respective for the spectral analysis of C. It may be also interesting to study so called isotropic (or harmonic) decompositon of C. I don't know this topic well, so I can only suggest some articles -it's worth seeing: 

K. Kowalczyk-Gajewska, J. Ostrowska-Maciejewska, "Review on spectral decomposition of Hooke's tensor for all symmetry groups of linear elastic material", Engineering Transactions, 57 (2009)

J.Rychlewski, "A qualitative approach to Hooke's tensors. Part 1", Archives of Mechanics, Vol. 52, 4-5 (2000)

J. Rychlewski, "A qualitative approach to Hooke's tensors. Pt. 2", Archives of Mechanics, 53, 1 (2001) 
Tue, 05/08/2012 - 08:52 Permalink
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Sat, 04/18/2015 - 09:01 Permalink
ebarbero

I found the posts here and in https://en.wikipedia.org/wiki/Invariants_of_tensors to be extremely useful and may be someone here can tell me if I am wrong about this issue (*) below. 

The Tsai-Wu failure criterion is a tensor polynomial I=I(s,F) where I is the failure index (I=1 denotes imminent failure), s is the stress tensor in an orthotropic lamina and F are the material strength parameters (one per stress component).

It seems to me that, in the notation of https://en.wikipedia.org/wiki/Invariants_of_tensors, s is the "indeterminate".

The fact that I(s,F) is invatiant to coordinate transformations means that I(s,F)=I(s',F) where s' is obtained by coordinate transformation of s.

The stress is calculated from the strain, which in turn are calculated from the loads N using the 4th order compliance tensor S=C^-1, where C is the stiffness tensor. 

Question (*) The fact that I(s,F)=I(s',F) does not mean that you can transform C into C' and S into S', without transforming the loads into N' because then you would get incorrect stress s' and incorrect I(s',F). I am wright? 

 

Sat, 10/29/2016 - 12:35 Permalink