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F ’T N ± = n±
»±
1
F ’1 n± = N±
»±
F T n± = »± N ±


EXAMPLE 3.6: Polar decomposition (ii)
Building on Example 3.5 the physical meaning of the stretches »± and
rotation R can easily be illustrated. Using the deformation gradient
F the principal material vectors N 1 and N 2 deform (push forward) to
give the orthogonal spatial vectors φ— [N 1 ] and φ— [N 2 ] as,
’0.4715 ’1.1843
φ— [N 1 ] = ; φ— [N 2 ] = ;
2.2219 ’0.2513
φ— [N 1 ] · φ— [N 2 ] = 0
However these two vectors may alternatively emerge by, ¬rstly, stretch-
ing the material vectors N 1 and N 2 to give,
1.9046 ’0.6597
»1 N 1 = ; »2 N 2 =
1.2377 1.0152
and, secondly, rotating these stretched vectors using the rotation tensor
R [see (3.44a)],
0.3590 ’0.9333 1.9046 ’0.4715
φ— [N 1 ] = R»1 N 1 = =
0.9333 0.3590 1.2377 2.2219
similarly for φ— [N 2 ]. Hence the deformation of the eigenvectors N 1
and N 2 associated with F , at a particular material position, can be
interpreted as a stretch followed by a rotation of (about) 69—¦ . Finally,
it is easy to con¬rm (3.39) that the spatial unit vectors n± = RN ± .



Equations (3.44a“b) can be interpreted in terms of the push forward of
vectors in the initial con¬guration to vectors in the current con¬guration.
Likewise, (3.44c“d) can be interpreted as alternative pull back operations.


Remark 3.6.1. The Lagrangian and Eulerian strain tensors, de¬ned in
18 KINEMATICS



X 3, x 3


d x3
`


p

d x1
d X3
d x2

P
d X2 time = t
d X1
X1 , x1
X 2 ,x2


time = 0

FIGURE 3.5 Volume change.



Section 3.5, can now be expressed in terms of U and V as,

3
1 12
E = (U 2 ’ I) = » ’ 1 N± — N± (3.45)

2
±=1
3
1 1
e = (I ’ V ’2 ) = 1 ’ »’2 n± — n± (3.46)
±
2 2
±=1

These expressions motivate the de¬nition of generalized material and spatial
strain measures of order n as,

3
1 1n
(n)
= (U n ’ I) = » ’ 1 N± — N± (3.47)
E

n
±=1
3
1 1
(n)
= (I ’ V ’n ) = 1 ’ »’n n± — n± (3.48)
e ±
n n
±=1

e(’n) = RE (n) RT (3.49)

In particular, the case n ’ 0 gives the material and spatial logarithmic
19
3.7 VOLUME CHANGE



strain tensors,
3
(0)
= ln »± N ± — N ± = ln U (3.50)
E
±=1
3
e(0) = ln »± n± — n± = ln V (3.51)
±=1



3.7 VOLUME CHANGE

Consider an in¬nitesimal volume element in the material con¬guration with
edges parallel to the Cartesian axes given by dX 1 = dX1 E 1 , dX 2 = dX2 E 2 ,
and dX 3 = dX3 E 3 , where E 1 , E 2 , and E 3 are the orthogonal unit vectors
(see Figure 3.5). The elemental material volume dV de¬ned by these three
vectors is clearly given as,
dV = dX1 dX2 dX3 (3.52)
In order to obtain the corresponding deformed volume, dv, in the spatial
con¬guration, note ¬rst that the spatial vectors obtained by pushing forward
the previous material vectors are,
‚φ
dx1 = F dX 1 = dX1 (3.53)
‚X1
‚φ
dx2 = F dX 2 = dX2 (3.54)
‚X2
‚φ
dx3 = F dX 3 = dX3 (3.55)
‚X3
The triple product of these elemental vectors gives the deformed volume as,
‚φ ‚φ ‚φ
dv = dx1 · (dx2 —dx3 ) = dX1 dX2 dX3 (3.56)
· —
‚X1 ‚X2 ‚X3
Noting that the above triple product is the determinant of F gives the
volume change in terms of the Jacobian J as,
dv = J dV ; J = det F (3.57)
Finally, the element of mass can be related to the volume element in
terms of the initial and current densities as,
dm = ρ0 dV = ρ dv (3.58)
Hence, the conservation of mass or continuity equation can be expressed as,
ρ0 = ρJ (3.59)
20 KINEMATICS



3.8 DISTORTIONAL COMPONENT OF THE
DEFORMATION GRADIENT

When dealing with incompressible and nearly incompressible materials it
is necessary to separate the volumetric from the distortional (or isochoric)
components of the deformation. Such a separation must ensure that the
ˆ
distortional component, namely F , does not imply any change in volume.
Noting that the determinant of the deformation gradient gives the vol-
ˆ
ume ratio, the determinant of must
F
therefore satisfy,

ˆ
det F = 1 (3.60)

ˆ
This condition can be achieved by choosing F as,

ˆ
F = J ’1/3 F (3.61)

The fact that Condition (3.60) is satis¬ed is demonstrated as,

ˆ
det F = det(J ’1/3 F )
= (J ’1/3 )3 det F
= J ’1 J
=1 (3.62)

The deformation gradient F can now be expressed in terms of the volumetric
ˆ
and distortional components, J and F , respectively, as,

ˆ
F = J 1/3 F (3.63)

This decomposition is illustrated for a two-dimensional case in Figure 3.6.
Similar decompositions can be obtained for other strain-based tensors
such as the right Cauchy“Green tensor C by de¬ning its distortional com-
ˆ
ponent C as,

ˆT ˆ
ˆ
C=F F (3.64)

ˆ
Substituting for F from Equation (3.61) gives an alternative expression for
ˆ
C as,

ˆ det C = J 2
C = (det C)’1/3 C; (3.65)
21
3.8 DEFORMATION GRADIENT



X2 x2




F



ˆ p
F


P X1 x1

FIGURE 3.6 Distortional component of F .



EXAMPLE 3.7: Distortional component of F
ˆ
Again using Example 3.5 the function of the isochoric component F of
F can be demonstrated. However to proceed correctly it is necessary
to introduce the third X3 dimension into the formulation, giving,
® 
1 ’5 0
1
F = °8 4 0»; J = det F = 2.75
4
0 04
ˆ
from which F , is found as,
® 
0.1784 ’0.8922 0
1
ˆ ’3
F = J F = ° 1.4276 0.7138 0»
0 0 0.7138
Without loss of generality consider the isochoric deformation at X = O
of the orthogonal unit material vectors N 1 = (0.8385, 0.5449, 0)T , N 2 =
(’0.5449, 0.8385, 0)T and N 3 = (0, 0, 1)T , for which the associated
elemental material volume is dV = 1. After deformation the material
unit vectors push forward to give,
®  ® 
’0.3366 ’0.8453
ˆ ˆ
ˆ ˆ
n1 = F N 1 = ° 1.5856 » ; n2 = F N 2 = ° ’0.1794 » ;
0 0

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