Using Equation (3.19a) a similar expression involving the Almansi strain

tensor can be derived as,

1 ds2 ’ dS 2

= n · en (3.24)

ds2

2

where n is a unit vector in the direction of dx.

12 KINEMATICS

EXAMPLE 3.4: Physical interpretation of strain tensors

Refering to Example 3.2 the magnitude of the elemental vector dx2 is

ds2 = 4.51/2 . Using (3.23) the scalar value of Green™s strain associated

with the elemental material vector dX 2 is,

1 ds2 ’ dS 2 7

µG = =

dS 2

2 4

Again using (3.23) and Example 3.3 the same strain can be determined

from Green™s strain tensor E as,

103 7

0

µG = N T EN = [0, 1] =

437 1 4

Using (3.24) the scalar value of the Almansi strain associated with the

elemental spatial vector dx2 is,

1 ds2 ’ dS 2 7

µA = =

ds2

2 18

Alternatively, again using (3.24) and Example 3.3 the same strain is

determined from the Almansi strain tensor e as,

1

√

1110 7

9 2

µA = nT en = √ , √ =

1

2 2 18 9 ’4 18

√

2

Remark 3.5.2. In terms of the language of pull back and push forward,

the material and spatial strain measures can be related through the operator

φ— . Precisely how this operator works in this case can be discovered by

recognizing, because of their de¬nitions, the equality,

dx1 · e dx2 = dX 1 · E dX 2 (3.25)

for any corresponding pairs of elemental vectors. Recalling Equations (3.12“

13) enables the push forward and pull back operations to be written as,

Push forward

e = φ— [E] = F ’T EF ’1 (3.26a)

Pull back

E = φ’1 [e] = F T eF (3.26b)

—

13

3.6 POLAR DECOMPOSITION

3.6 POLAR DECOMPOSITION

The deformation gradient tensor F discussed in the previous sections trans-

forms a material vector dX into the corresponding spatial vector dx. The

crucial role of F is further disclosed in terms of its decomposition into stretch

and rotation components. The use of the physical terminology stretch and

rotation will become clearer later. For the moment, from a purely mathe-

matical point of view, the tensor F is expressed as the product of a rotation

tensor R times a stretch tensor U to de¬ne the polar decomposition as,

F = RU (3.27)

For the purpose of evaluating these tensors, recall the de¬nition of the right

Cauchy“Green tensor C as,

C = F T F = U T RT R U (3.28)

Given that R is an orthogonal rotation tensor as de¬ned in Equation (2.27),

that is, RT R = I, and choosing U to be a symmetric tensor gives a unique

de¬nition of the material stretch tensor U in terms of C as,

U2 = UU = C (3.29)

In order to actually obtain U from this equation, it is ¬rst necessary to

evaluate the principal directions of C, denoted here by the eigenvector triad

{N 1 , N 2 , N 3 } and their corresponding eigenvalues »2 , »2 , and »2 , which

1 2 3

enable C to be expressed as,

3

»2 N ± — N ±

C= (3.30)

±

±=1

where, because of the symmetry of C, the triad {N 1 , N 2 , N 3 } are orthog-

onal unit vectors. Combining Equations (3.29) and (3.30), the material

stretch tensor U can be easily obtained as,

3

U= »± N ± — N ± (3.31)

±=1

Once the stretch tensor U is known, the rotation tensor R can be easily

evaluated from Equation (3.27) as R = F U ’1 .

In terms of this polar decomposition, typical material and spatial ele-

mental vectors are related as,

dx = F dX = R(U dX) (3.32)

In the above equation, the material vector dX is ¬rst stretched to give

U dX and then rotated to the spatial con¬guration by R. Note that U is a

14 KINEMATICS

material tensor whereas R transforms material vectors into spatial vectors

and is therefore, like F , a two-point tensor.

EXAMPLE 3.5: Polar decomposition (i)

This example illustrates the decomposition of the deformation gradient

tensor

F = RU using the deformation shown below as,

x1 = 1 (4X1 + (9 ’ 3X1 ’ 5X2 ’ X1 X2 )t)

4

1

x2 = 4 (4X2 + (16 + 8X1 )t)

For X = (0, 0) and time t = 1 the deformation gradient F and right

Cauchy“Green tensor C are,

1 1 ’5 1 65 27

F= ; C=

48 4 16 27 41

from which the stretches »1 and »2 and principal material vectors N 1

and N 2 are found as,

0.8385

»1 = 2.2714; »2 = 1.2107; N1 = ;

0.5449

’0.5449

N2 =

0.8385

Hence using (3.31) and R = F U ’1 , the stretch and rotation tensors

can be found as,

1.9564 0.4846 0.3590 ’0.9333

U= ; R=

0.4846 1.5257 0.9333 0.3590

It is also possible to decompose F in terms of the same rotation tensor R

followed by a stretch in the spatial con¬guration as,

F =VR (3.33)

which can now be interpreted as ¬rst rotating the material vector dX to the

spatial con¬guration, where it is then stretched to give dx as,

dx = F dX = V (RdX) (3.34)

where the spatial stretch tensor V can be obtained in terms of U by com-

bining Equations (3.27) and (3.33) to give,

V = RU RT (3.35)

15

3.6 POLAR DECOMPOSITION

Additionally, recalling Equation (3.17) for the left Cauchy“Green or Fin-

ger tensor b gives,

b = F F T = (V R)(RT V ) = V 2 (3.36)

Consequently, if the principal directions of b are given by the orthogonal

¯¯ ¯

spatial vectors {n1 , n2 , n3 } with associated eigenvalues »2 , »2 , and »2 , then

1 2 3

the spatial stretch tensor can be expressed as,

3

¯

V= »± n± — n± (3.37)

±=1

Substituting Equation (3.31) for U into Expression (3.35) for V gives V in

terms of the vector triad in the undeformed con¬guration as,

3

V= »± (RN ± ) — (RN ± ) (3.38)

±=1

Comparing this expression with Equation (3.37) and noting that (RN ± )

remain unit vectors, it must follow that,

¯

»± = »± ; n± = RN ± ; ± = 1, 2, 3 (3.39a,b)

This equation implies that the two-point tensor R rotates the material vec-

tor triad {N 1 , N 2 , N 3 } into the spatial triad {n1 , n2 , n3 } as shown in Fig-

ure 3.4. Furthermore, the unique eigenvalues »2 , »2 , and »2 are the squares

1 2 3

of the stretches in the principal directions in the sense that taking a mate-

rial vector dX 1 of length dS1 in the direction of N 1 , its corresponding push

forward spatial vector dx1 of length ds1 is given as,

dx1 = F dX 1 = RU (dS1 N 1 ) (3.40)

Given that U N 1 = »1 N 1 and recalling Equation (3.39) gives the spatial

vector dx1 as,

dx1 = (»1 dS1 )n1 = ds1 n1 (3.41)

Hence, the stretch »1 gives the ratio between current and initial lengths as,

»1 = ds1 /dS1 (3.42)

It is instructive to express the deformation gradient tensor in terms

of the principal stretches and principal directions. To this end, substitute

Equation (3.31) for U into Equation (3.27) for F and use (3.39) to give,

3

F= »± n± — N ± (3.43)

±=1

16 KINEMATICS

n3

R

‚3 dX N3

N3

UdX dx

N2

dX

‚3 d X N3

p

d X N3 ‚2 d X N2 n2

‚2 d X N2

d X N2

‚1d X N1 ‚1d X N1

P

d X N1

N1

n1

time = t

time = 0

FIGURE 3.4 Material and spatial vector triads.

This expression clearly reveals the two-point nature of the deformation gra-

dient tensor in that it involves both the eigenvectors in the initial and ¬nal

con¬gurations.

It will be seen later that it is often convenient, and indeed more natural,

to describe the material behavior in terms of principal directions. Con-

sequently, it is pertinent to develop the relationships inherent in Equa-

tion (3.43) a little further. For this purpose, consider the mapping of the

unit vector N ± given by the tensor F , which on substituting the polar

decomposition F = RU gives,

F N ± = RU N ±

= »± RN ±

= »± n± (3.44a)

17

3.6 POLAR DECOMPOSITION

Alternative expressions relating N ± and n± can be similarly obtained as,

1