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Note the similarity and the difference with the de¬nition of the Green Lagrange
strain tensor E. The tensor µF can be considered as an objective version of the
invariant tensor E. The interpretation of the Finger strain tensor reveals some
problems which are beyond the scope of the present context.
In this section a number of different strain tensors have been reviewed. It is
an interesting exercise to compare these different tensors for a few elementary
homogeneous deformations, for example for the case of uniaxial stress (see Fig.
10.2) and for pure shear. It should be noted that for (in¬nitesimally) small defor-
mations and (in¬nitesimally) small rotations (so the limiting case that F ’ I) the
difference between all treated strain tensors vanishes.
Deformation and rotation, deformation rate and spin
180

10.5 The volume change factor

Consider an arbitrary, in¬nitesimally small material element (parallellepipedum),
in the reference con¬guration spanned by three linearly independent vectors dxa ,
0
b c
dx0 and dx0 . The volume of the element is speci¬ed by dV0 and can in principle
be calculated when the vectors dxi (with i = a, b, c) are known. The vectors
0
dx in the current con¬guration, associated with the vectors dxi in the reference
i
0
con¬guration, can be determined using the deformation tensor F via
dxi = F · dxi for i = a, b, c. (10.53)
0

Based on the vectors dxi the volume dV of the material element in the cur-
rent con¬guration can be calculated. For the volumetric change ratio J it can be
shown:
dV
J= = det( F) . (10.54)
dV0
The result is independent of the originally chosen shape and orientation of the
element dV0 .
For isochoric deformation (this is a deformation such that locally the volume
of the material does not change) the volume change ratio satis¬es
J = det( F) = 1. (10.55)
For materials that are incompressible, a property which is often attributed to
biological materials (related to the high water content of the materials), the defor-
mation is locally always isochoric. For compressible materials it is possible to
prescribe an isochoric deformation by means of a special choice of the external
mechanical load.



10.6 Deformation rate and rotation velocity

In the preceding sections the current con¬guration or current state is considered
at a (¬xed) time t and compared with the reference con¬guration. Based on that,
concepts like deformation and rotation were de¬ned. In this section the attention
is focussed on (in¬nitesimally) small changes of the current state, as seen in the
time domain.
A material line segment dx (solid or ¬‚uid) in the current state at time t converts

in the line segment dx + dx dt at time t + dt, see Fig. 10.4.
It can be written (also see Chapter 7):
T

dx = L · dx with L = ∇v , (10.56)
181 10.6 Deformation rate and rotation velocity

·
dx + dx dt = dx + L · dx dt
z


…dt
dx

y
x


x

Figure 10.4
Change of a material line segment dx after a time increment dt.




where v = x speci¬es the velocity of the material and L is the velocity gradient
tensor. The tensor L is purely a current variable, not in any way related to the
reference con¬guration.
Using the deformation tensor it can be written:

dx = F · dx0 (10.57)

and therefore
™ ™ ™
dx = F · dx0 = F · F’1 · dx, (10.58)

accordingly resulting in the relation between the tensors L and F:

L = F · F’1 . (10.59)

It is common practice to decompose the velocity gradient tensor L in a symmet-
rical part D and a skew symmetrical part . The tensor D is called the rate of
deformation tensor and the tensor the rotation velocity tensor or spin tensor.
The de¬nitions are:
1 1™
1 T
™T
F · F’1 + F’T · F
L + LT = ∇v + ∇v =
D=
2 2 2
1 1 1™
T
™T
F · F’1 ’ F’T · F
L ’ LT = ∇v
= ’ ∇v =
2 2 2
(10.60)

and so

dx = ( D + ) ·dx, (10.61)

with

DT = D (10.62)
Deformation and rotation, deformation rate and spin
182

and
=’ .
T
(10.63)
For an interpretation of the symmetrical tensor D we depart from the relations that
have been derived in Section 10.3:
F · e0 = » e (10.64)
and
»2 = e0 · FT · F · e0 . (10.65)
The material time derivative of the equation for »2 can be elaborated as follows:
™T ™

2»» = e0 · F · F + FT · F · e0
™ ™
T
= e0 · I · F · F + FT · F · I · e0
™T ™
= e0 · FT · F’T · F · F + FT · F · F’1 · F · e0
™T ™
= e0 · FT · F’T · F + F · F’1 · F · e0
= »2 e · ( 2D) · e, (10.66)
eventually resulting in the simple relation:

»
= e · D · e (= ln( »)) . (10.67)
»
This equation shows that the deformation velocity tensor D completely determines
the current rate of (logarithmic) strain for an arbitrary line segment in the current
state with a direction speci¬ed by e. The analogous equation in component form
is written as

»
= ∼T D e .
e (10.68)
» ∼


The terms on the diagonal of the matrix D represent the rate of strain in the direc-
tions of the x-, y- and z-coordinates. The off-diagonal terms represent the rate of
shear.
For the interpretation of the skew symmetric spin tensor Eq. (10.61) can be

used directly. After all, it is clear that the contribution · dx to dx is always
perpendicular to dx, because for all dx:
dx · · dx = 0 because =’ ,
T
(10.69)
meaning that the contribution ·dx has to be considered as the effect of a rotation.
For the material time derivative of the volume change factor J = det (F) it can
be derived (without proof):


J = J tr( F · F’1 ) = J tr( L) = J ∇ · v = J tr( D) . (10.70)
183 Exercises

Like in previous sections we will examine the in¬‚uence of an extra rigid body
movement of the current con¬guration, also see Sections 9.7 and 10.3. Variables,
associated with the extra rotating (via the rotation tensor P = P( t) that depends
explicitly on the time) and extra translating (with » = »( t) the translation vector)
virtual current state, are again denoted by the superscript — . It can be written:
x— = P · x + »
™™ ™

v— = x — = P · x + P · x + » = P · v + P · x + » .
™ ™ (10.71)
For the velocity gradient tensor L— in the extra moving con¬guration it is found:
—T
™T ™
T
L— = ∇v — = P · ∇ v · PT + x · P + »
TT

= P · LT · PT + P · I · P

= P · L · P T + P · PT . (10.72)
An identical result is obtained via a different procedure:
™—
L— = F · ( F— )’1
™ ™
= P · F + P · F · F’1 · P’1

= P · L · P T + P · PT . (10.73)
It can be observed that the velocity gradient tensor L is neither objective, nor

invariant. The contribution P · PT in Eq. (10.73) however is skew symmetric,
after all:
™ ™T
P · PT = I ’ P · PT + P · P = 0
T
™ ™T ™
’ P · PT = ’P · P = ’ P · PT , (10.74)
meaning that the symmetrical part of the tensor L (which is the rate of deformation
tensor D) is objective:
D— = P · D · P T . (10.75)
The spin tensor is not objective.



Exercises

10.1 In a subvolume of a material continuum the deformation of the current
state, with respect to the reference state, is homogeneous. In a Cartesian
xyz-coordinate system the associated deformation tensor is given as
F = I + 3ey ey ’ 7ey ez ’ ez ey + ez ez ,
with I the unit tensor.
Deformation and rotation, deformation rate and spin
184

For a material point P within the subvolume the position vectors in the
reference state as well as the current state are given, respectively:

x 0P = ex + ey + ez , xP = 2ex + 3ey ’ 2ez .

Another point Q within the subvolume appears to be in the origin in the
current con¬guration: xQ = 0.
Calculate the position vector x0Q of the point Q in the reference state.
10.2 Within a subvolume of a material continuum the deformation tensor in the
deformed current state, with respect to the reference state, is constant. Con-
sider a vector of material points, denoted with a0 in the reference state and
with a in the current state. The angle between a0 and a is given by φ.
Prove that we can write for φ:
a 0 · F · a0
cos( φ) = .
( a0 · a0 ) ( a0 · FT · F · a0 )
10.3 Within a subvolume of a material continuum the deformation tensor in the
deformed current state, with respect to the reference state, is constant. In a
Cartesian xyz-coordinate system the following deformation tensor is given:

F = I + 4 ex ex + 2 ex ey + 2 ey ex ,

with I the unit tensor. Within the subvolume two material points P and Q
are considered. The position vectors in the reference state are given:

x 0 P = ex + e y + e z , x0Q = 2ex + 3ey + 2ez .

In addition, the position vector for the point P in the current state is given:

xP = 2ex + 2ey + 2ez .

Calculate the position vector xQ of the point Q in the current state.
10.4 The deformation of a material particle, in the current state with respect to

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