1.4 Principal Strains

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An infinitesimal vector dX with size dS is transformed by the motion of the body into dx

with size ds satisfying

ds2 − dS2 = EKL dXK dXL (1.96)

or

ds2 − dS2

dS2

= EKLNKNL (1.97)

where

NK := dXK

dS

(1.98)

is a unit vector satisfying

NKNLGKL = dXKdXLGKL

dS2

= 1. (1.99)

The expression in Equation (1.97) is a measure for the relative change of length of a fiber

originally parallel to N. The question we want to look into now is the following: in which

directions is this change of length maximal? This boils down to maximizing Equation (1.97)

subject to the constraint Equation (1.99). The variables are the components of N. Following

the usual procedure of calculus, finding an extremum reduces to setting the derivative of

the target function with respect to the variables to zero. The target function is

F(N) = EKLNKNL −             E(NKNLGKL − 1) (1.100)

where   E is a Lagrange multiplier. Hence,

∂

∂NM

[EKLNKNL −            E(NKNLGKL − 1)] = 0

_

EMLNL + EKMNK −            ENLGML −    ENKGKM = 0

_

(EKM −          EGKM)NK = 0. (1.101)

14 DISPLACEMENTS, STRAIN, STRESS AND ENERGY

This is a classical eigenvalue problem (generalized for curvilinear coordinates). Since E

is a symmetric tensor, the eigenvalues   E are real and the corresponding eigenvectors are

mutually orthogonal (or can be made orthogonal). Indeed, suppose that           E is a complex

eigenvalue, then

EKMNK =      EGKMNK. (1.102)

Premultiplying with the complex conjugate of N yields

N

M

EKMNK =      EN

M

GKLNK. (1.103)

Since E is symmetric and real, one obtains

N

M

EKMNK = NMEKMN

K = NKEMKN

M

= NKEKMN

M (1.104)

and similar forG. Consequently,N

M

EKMNK andN

M

GKMNK are real and E must be real

because of Equation (1.103) and the positive-definiteness of G. Because of Equation (1.102),

the eigenvectors N are real too.

To prove that the eigenvectors are mutually orthogonal, consider two distinct eigenvalues

            1 and   2 with two corresponding eigenvectors N1 and N2. Then

EKMNK

1

=          1GKMNK

1 (1.105)

and

EKMNK

2

=          2GKMNK

2 . (1.106)

Multiplying Equation (1.105) with NM

2 and Equation (1.106) with NM

1 and subtracting both

yields

NM

2 EKMNK

1

− NM

1 EKMNK

2

=          1NM

2 GKMNK

1

−          2NM

2 GMKNK

1 (1.107)

or

NM

2 (EKM − EMK)NK

1

=          1NM

2 GKMNK

1

−          2NM

1 GKLNK

2 . (1.108)

Since both E and G are symmetric, this yields

0 = (    1 −       2)NM

2 GKMNK

1 . (1.109)

            1 and   2 are assumed to be distinct, which means

NK

1 GKMNM

2

= 0 (1.110)

or

N1 · N2 = 0. (1.111)

This completes the proof.

DISPLACEMENTS, STRAIN, STRESS AND ENERGY 15

The eigenvalues are the solution of a third-order nonlinear equation expressing that the

determinant of the matrix in Equation (1.101) has to satisfy

det(EKM −      EGKM) = 0 ⇔ det(EK

M

−          EδK

M) = 0 (1.112)

for the equation to have nontrivial solutions. Since the extremal strains have a physical

relevance and are independent of the coordinate system, the coefficients of Equation (1.112)

are invariants. Indeed, Equation (1.112) can be written as

−          3

E

+ I1E   2

E

− I2E   E + I3E = 0 (1.113)

where

I1E = δK

LEL

K

= trE (1.114)

I2E = 1

2

            I 2

1E

− tr(E2) (1.115)

I3E = detE = 1

3!

eLMP eKNQEL

KEM

NEP

Q (1.116)

are the first, second and third invariant of E. The expression trE stands for the trace

of E, eLMP and eKNQ are the alternating symbols: eKLM = 1 for KLM = 123 or any

cyclic rotation thereof, eKLM = −1 for KLM = 321 or any cyclic rotation, else eKLM = 0.

The eigenvalues are called principal strains and the corresponding direction Ni are called

principal directions. They are obtained by solving Equation (1.101) in which the solutions

of Equation (1.113) are substituted. For the solution of Equation (1.113), which is a cubic

equation, see (Simo and Hughes 1997) or (Abramowitz and Stegun 1972).

Since E and C differ by the metric tensor, Equation (1.101) can also be written as

(CKM −          CGKM)NK = 0 (1.117)

where

            C = 2   E + 1. (1.118)

Consequently, the eigenvectors of C and E are the same and the eigenvalues are directly

related by Equation (1.118). In what follows,   i denotes the eigenvalues of C, that is,

            i =        iC. The calculation of the eigenvectors Ni is somewhat tedious. Sometimes, it is

more advantageous to calculate the tensors Ni ⊗ Ni , which play the role of a tensorial

basis. Here, Ni (index up) are the one-forms obtained by raising the index of N:

Ni = G_ · Ni (1.119)

and satisfy (Equation (1.110)

Ni · Nj = δi

j . (1.120)

The one-forms {Ni } are the dual basis of the vectors {Ni }.

16 DISPLACEMENTS, STRAIN, STRESS AND ENERGY

Theorem 1.4.1 Let C be a symmetric covariant second-order tensor in R3,     i its eigenvalues

and Ni the corresponding eigenvectors, then

C =

3

_

i=1

            iMi (1.121)

where

Mi = Ni ⊗ Ni (1.122)

and Ni are the one-forms dual to Ni .

Proof.

_

i

            iMi · Nl =_

i

            i (Ni ⊗ Ni ) · Nl

=_

i

            iNi (Ni · Nl)

=_

i

            iNiδi

l

=          lNl =    lG_ · Nl , ∀l (1.123)

where G_ is the covariant metric tensor and an underscore or a summation sign remove

implicit summation. Consequently, C and i        iMi have the same eigenvalues and eigenvectors

and are identical. Since C is a covariant tensor, it is logical that it is made up of

one-forms and not of vectors.

An interesting property is

C · C =

_

_

i

            iMi_

·

_

j

            jMj

=_

i

_

j

            i           j (Ni ⊗ Ni ) · (Nj ⊗ Nj )

=_

i

_

j

            i           j (Ni ⊗ Nj )(Ni · Nj )

=_

i

            2

i (Ni ⊗ Ni) =_

i

            2

iMi . (1.124)

This property allows for the following simple calculation of Mi. Since





M1 +M2 +M3 = G

            1M1 +             2M2 +             3M3 = C

            21

M1 +   22

M2 +   23

M3 = C2

(1.125)

one obtains

Mi = 1

D

            C2 − (I1C −    i )C + I3C       

−1

i G (1.126)

DISPLACEMENTS, STRAIN, STRESS AND ENERGY 17

where

Di = (   i −        j )(       i −        k) (1.127)

for j, k _= i.

If two eigenvalues are identical, for example,     =         1 =       2 _=     3 one obtains instead

of Equation (1.125),





(M1 +M2) +M3 = G

            (M1 +M2) +    3M3 = C

            2(M1 +M2) + 23

M3 = C2

(1.128)

Discarding the third equation, one finds

M1 +M2 =      3G − C

            3 −      

(1.129)

M3 = C −        G

            3 −      

. (1.130)

This means that M1 and M2 are not known individually, only their sum can be derived. For

three equal eigenvalues, the set in Equation (1.125) reduces to the first equation (Itskov

2001). The tensors M1, M2 and M3 are sometimes called structural tensors. They are

genuine tensors of rank two subject to Equation (1.122) and the normality condition of Ni .

Notice that

C ·Mi = C · (Ni ⊗ Ni ) = (C · Ni)Ni =            i (G · Ni )Ni = iG ·Mi (1.131)

and

C : Mi = C : (Ni ⊗ Ni ) = Ni · C · Ni = Ni      i · (G · Ni ) =    i (1.132)

since Equation (1.117) is equivalent to

C · Ni =           iG · Ni (1.133)

and the double contraction or inner product of two second-order tensors a ⊗ b and c ⊗ d

is defined by

(a ⊗ b) : (c ⊗ d) = (a · c)(b · d) = tr _(a ⊗ b)T · (c ⊗ d)_ . (1.134)

One finds that the eigenvalues λi of F satisfy

λi = _   i (1.135)

because of Equation (1.42). Defining

ni := F · Ni/λi (1.136)

one can write

F =_

i

λi (ni ⊗ Ni ). (1.137)

18 DISPLACEMENTS, STRAIN, STRESS AND ENERGY

The normals Ni along the principal directions in the material frame are mapped into the

normals ni in the spatial frame, strained by an amount λi . Notice that Equation (1.136)

actually defines the right-hand side of the eigenvalue problem for F. Furthermore, since

F is a two-point tensor, it cannot map a vector into a multiple of itself.

Not only are {Ni } mutually orthogonal but {ni } are also a mutually orthogonal set of

vectors. Indeed,

ni · nj = 1

λiλj

Ni · FT · F · Nj

= 1

λiλj

Ni · C · Nj

=

λj

λi

Ni · Nj =

λj

λi

δij . (1.138)

Hence, in each material point, there exist three mutually orthogonal vectors, the deformation

of which is extremal and yields again three mutually orthogonal vectors. The vectors {ni } are

the eigenvectors of the inverse of the left Cauchy–Green tensor b

−1. Indeed, substituting

Equation (1.136) into Equation (1.117) yields

C · F

−1 · ni =           CG · F

−1 · ni . (1.139)

Substituting C (Equation (1.42)) leads to

FT · g · ni =      CG · F

−1 · ni (1.140)

or

1

            C

g · ni = F

−T · G · F

−1 · ni (1.141)

which is equivalent to

1

            C

g · ni = b

−1 · ni . (1.142)

At this point, the polar decomposition theorem should be mentioned because of its

physical relevance. It states that the deformation gradient F can be written as the product

of an orthogonal matrix R and a symmetric tensor U, called the right-stretch tensor.

Accordingly,

F = R · U (1.143)

where

RT = R

−1 (1.144)

and

U = UT. (1.145)

DISPLACEMENTS, STRAIN, STRESS AND ENERGY 19

Since

C = FT · F = UT · RT · R · U

= UT · U = U · U = U2 (1.146)

U and F have the same eigenvalues equal to the square root of the eigenvalues of C.

Since C is positive-definite (Equation (1.41)), U is also positive-definite. Furthermore, the

eigenvectors of C and U are identical. We have

U =_

i

λiNi ⊗ Ni =_

i

_          iNi ⊗ Ni (1.147)

and

R =_

i

ni ⊗ Ni . (1.148)

Indeed,

R · U =_

i

(ni ⊗ Ni )_

j

λj (Nj ⊗ Nj )

=_

i

_

j

λjni ⊗ Nj (Ni · Nj )

=_

i

λini ⊗ Ni = F. (1.149)

In a similar way, one can decompose F into

F = V · R. (1.150)

V is the left-stretch tensor.

Equation (1.143) shows that the motion can be locally decomposed into a pure stretch

along the principal directions followed by a rotation. It should be emphasized that a pure

stretch is guaranteed for the principal directions only. For all other directions N, the product

U · N will involve some rotation, unless some of the principal values coincide. Furthermore,

R is not constant in space. Consequently, R denotes a microscopic rotation in the material

point of interest and not a macroscopic rotation.

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