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IIS = »2 + »2 + »2 (2.60b)
1 2 3
IIIS = »1 »2 »3 (2.60c)


2.2.4 HIGHER ORDER TENSORS
It will be seen later in Chapter 5 that the second-order stress and strain ten-
sors are related via a fourth-order constitutive or material behavior tensor.
In order to appreciate fourth-order tensors it is necessary to examine the
intermediate third-order tensors. Inevitably things are likely to get more
di¬cult and because an appreciation of this section is not immediately nec-
essary, it is suggested that what follows be read just prior to Chapter 5.
19
2.2 VECTOR AND TENSOR ALGEBRA



Although there are several ways in which higher-order tensors can be de-
¬ned, the procedure used here is a simple extension of De¬nition (2.15), em-
ployed for second-order tensors. In this way, a third-order tensor is de¬ned
as a linear map from an arbitrary vector u to a second-order tensor S as,
Au = S (2.61)
In particular, the tensor product of three arbitrary vectors yields a third-
order tensor u — v — w de¬ned in a manner similar to Equation (2.28) so
that any vector x is mapped to,
(u — v — w)x = (w · x)(u — v) (2.62)
This de¬nition combined with the fact that second-order tensors can be
expressed as linear combinations of dyadic products leads to the following
expressions for the tensor product of a vector and a second-order tensor as,
(S — v)u = (u · v)S (2.63a)
(v — S)u = v — (Su) (2.63b)
The 27 third-order tensors ei — ej — ek for i, j, k = 1, 2, 3, which are
obtained by taking the tensor products of the three Cartesian base vectors,
constitute a basis in which any third-order tensor A can be expressed as,
3
A= Aijk ei — ej — ek (2.64)
i,j,k=1

where a manipulation similar to that employed in Equation (2.33) for second-
order tensors will verify that the components of A are,
Aijk = (ei — ej ) : A ek (2.65)
An example of a third-order tensor of interest in continuum mechanics
is the alternating tensor E, which is de¬ned in such a way that any vector
w is mapped to,
Ew = ’W w (2.66)
where W w is the second-order tensor de¬ned in Equation (2.22). The com-
ponents of E follow from Equations (2.65“66) and (2.22) as,
Eijk = (ei — ej ) : (Eek )
= ’(ei — ej ) : W ek
= ’ei · (W ek ej )
= ei · (ej —ek ) (2.67)
20 MATHEMATICAL PRELIMINARIES



The result of the above triple product is zero if any indices are repeated,
1 if the permutation {i, j, k} is even, and ’1 otherwise. Note that exactly
the same components would be obtained if a di¬erent Cartesian base were
used, that is, E is an isotropic third-order tensor. Using these {0, 1, ’1}
components, the tensor E can be expressed as,

E = e1 — e2 — e3 + e3 — e1 — e2 + e2 — e3 — e1
’ e3 — e2 — e1 ’ e1 — e3 — e2 ’ e2 — e1 — e3 (2.68)

The double contraction of a third-order tensor and a second-order tensor
is de¬ned in such a way that for the dyadic product of any two vectors u
and v a new vector is obtained as,

A : (u — v) = (Av)u (2.69)

Note that the result of the double contraction is a vector and not a scalar.
In fact, third-order tensors can be alternatively de¬ned as linear maps from
second-order tensors to vectors as shown in the above double contraction.
For example, applying the above de¬nition to the alternating tensor gives,

E : (u — v) = (Ev)u
= ’W v u
= u—v (2.70)

This result, instead of Equation (2.66), could have been used to de¬ne the
alternating tensor E.
In general, the double contraction of a third-order tensor by a second-
order tensor S can be evaluated in terms of their respective components
using De¬nition (2.69) together with Equations (2.32) and (2.65) to give,

3
A:S= Aijk Sjk ei (2.71)
i,j,k=1


Additional properties of the double contraction are given below without
proof,

(u — v — w) : (x — y) = (x · v)(y · w)u (2.72a)
(u — S) : T = (S : T )u (2.72b)
(S — u) : T = ST u (2.72c)
21
2.2 VECTOR AND TENSOR ALGEBRA




EXAMPLE 2.6: Proof of Equation (2.71)
In order to prove Equation (2.71) we ¬rst express both tensors A and
S in terms of their components as,
3 3
A= Aijk ei — ej — ek ; S= Slm el — em
i,j,k=1 l,m=1

Taking the double contraction now gives,
3
A:S= Aijk Slm (ei — ej — ek ) : (el — em )
i,j,k,l,m=1
3
= Aijk Slm [(ei — ej — ek )em ]el
i,j,k,l,m=1
3
= Aijk Slm (ek · em )(ej · el )ei
i,j,k,l,m=1
3
= Aijk Slm δkm δjl ei
i,j,k,l,m=1
3
= Aijk Sjk ei
i,j,k=1



Tensors of any order can now be de¬ned by recursive use of Equa-
tion (2.61). For instance, a fourth-order tensor C is a linear map between
an arbitrary vector u and a third-order tensor A as,
Cu = A (2.73)
Observe that no explicit notational distinction is made between third-, fourth-
, or any higher-order tensors. Examples of fourth-order tensors are obtained
by extending the de¬nition of the tensor product of vectors in the obvious
way to give,
(u1 — u2 — u3 — u4 )v = (v · u4 )(u1 — u2 — u3 ) (2.74)
Similar tensor products of a vector and a third-order tensor and second-order
tensors are inferred from the above de¬nition as,
(A — u)v = (u · v)A (2.75a)
(u — A)v = u — (Av) (2.75b)
(T — S)v = T — (Sv) (2.75c)
22 MATHEMATICAL PRELIMINARIES



Additionally, the double contraction of a fourth- (or higher) order tensor C
with a second-order tensor is again de¬ned using Equation (2.69) to yield a
second-order tensor as,

C : (u — v) = (Cv)u (2.76)

An illustration of this would be the crucially important constitutive rela-
tionship between a second-order stress tensor σ and a second-order strain
tensor µ given as σ = C : µ, where C would be a fourth-order elasticity
tensor.
Properties similar to those listed in Equations (2.72a“c) are obtained as,

(u1 — u2 — u3 — u4 ) : (x — y) = (x · u3 )(y · u4 )(u1 — u2 ) (2.77a)
(S 1 — S 2 ) : T = (S 2 : T )S 1 (2.77b)
(A — u) : T = A(T u) (2.77c)
(u — A) : T = u — (A : T ) (2.77d)

Recalling that the double contraction of a fourth-order tensor with a second-
order tensor gives a second-order tensor, fourth-order tensors can be also
de¬ned as linear mappings between second-order tensors. For instance, the
˜
fourth-order identity tensor I and the transposition tensor I are de¬ned
in such a way that any second-order tensor S is mapped onto itself and its
transpose respectively as,

I :S=S (2.78a)
˜
I : S = ST (2.78b)

Fourth-order tensors can be expressed as linear combinations of the 81
tensor products of the Cartesian base vectors ei — ej — ek — el for i, j, k, l =
1, 2, 3 as,

3
C= Cijkl ei — ej — ek — el (2.79a)
i,j,k,l=1


where it can be proved that the components Cijkl are given as,

Cijkl = (ei — ej ) : C : (ek — el )

For instance, the components of I are obtained with the help of Equa-
23
2.2 VECTOR AND TENSOR ALGEBRA



tions (2.79) and (2.52c) as,

Iijkl = (ei — ej ) : I : (ek — el )
= (ei — ej ) : (ek — el )
= (ei · ek )(ej · el )
= δik δjl (2.80)

˜
and, similarly, the components of I are,

˜ ˜
Iijkl = (ei — ej ) : I : (ek — el )
= (ei — ej ) : (el — ek )
= (ei · el )(ej · ek )
= δil δjk (2.81)

Consequently, these two tensors can be expressed as,
3
I= (2.82a)
ei — ej — ei — ej
i,j =1

3
˜
I= (2.82b)
ei — ej — ej — ei
i,j =1

Observe that both these tensors are isotropic, as the same components would
emerge from Equations (2.80“81) regardless of the Cartesian base being
used. Isotropic fourth-order tensors are of particular interest in continuum
mechanics because they will be used to describe the elasticity tensor of
materials that exhibit equal properties in all directions. In fact, it is possible
˜
to prove that all fourth-order isotropic tensors are combinations of I, I and
the additional isotropic tensor I — I as,

˜
C = ±I — I + βI + γ I (2.83)

Examples of such combinations are the tensors S, W, and D de¬ned as,

˜
S = 1 (I + I) (2.84a)
2
˜
1
W = 2 (I ’ I) (2.84b)
D = I ’ 1I — I (2.84c)
3

which project a second-order tensor S onto its symmetric, skew, and devi-
24 MATHEMATICAL PRELIMINARIES



atoric components as,
S : S = 1 (S + S T ) (2.85a)
2
W : S = 1 (S ’ S T ) (2.85b)
2
D : S = S = S ’ 1 (trS)I (2.85c)
3

Finally, the de¬nition of symmetric second-order tensors given in the
previous section can be extended to fourth-order tensors. In this way a
fourth-order tensor C is said to be symmetric if for any arbitrary second-
order tensors S and T , the following expression is satis¬ed,
S :C :T =T :C :S (2.86)
˜
For example it is easy to show that I — I, I and I are symmetric fourth-
order tensors and consequently combinations of these three tensors such as
S, W, and D are also symmetric.

EXAMPLE 2.7: Elasticity tensor
In linear isotropic elasticity the stress tensor σ is related to the small

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