{finish spatial power of tractions}
Example 1
A Compressible Neo-Hookean Material undergoing
Homogeneous Stretch and Rotation
Motion
Consider the deformation of a cubic block of material, with sides of length 1, initially
occupying the region 0  X 1 , X 2 , X 3  1 . The subsequent motion of the material is
given by
f (X, t )  x   X 1  tX 2 e1   tX 1  X 2 e 2  X 3e 3
This is a two-dimensional motion, with the material stretching and rotating about the
X 3 axis, as illustrated.
t 0
t2
t 1
The spatial description of the motion is obtained by inverting these equations:
f 1 (x, t )  X 
x1  tx2
tx  x
e1  1 2 2 e 2  x3e 3
2
1 t
1 t
Deformation
The deformation gradient is
 1 t 0
x 
F
  t 1 0 ,
X 
 0 0 1
F 1
 1
1  t 2
 t

2
1  t
 0

t
1 t2
1
1 t2
0


0

0

1

This deformation is homogeneous, meaning that the deformation is the same for all
points (F is independent of position). The Jacobian determinant is
J
dv
 det F  1  t 2
dV
The right Cauchy-Green tensor is
1  t 2
0

T
C  F F   0
1 t2
 0
0

0

0
1
For example, the stretch  for a line element (in the reference configuration) in the
direction dX  e1 is given by 2  e1  C  e1  C11  1  t 2 . In this simple case of a
homogeneous deformation, this stretch for line elements in the direction e1 is the
same for all points in the material.
The left Cauchy-Green tensor is b  F  F T and equals C is this example. It follows
that the stretch of a line element (in the current configuration) in the direction dx is
given by
1  t 2
0
1
dx 

 0
1 t2
2 dx 
0
 0
0
 dx
0 
dx
1
For example, in the direction dx  e1  2e 2 at t  2 , the stretch is
5.
The deformation can be decomposed into stretch tensors through F  R  U or
F  v  R , where U and v are, respectively, the right (material) and left (spatial)
stretch tensors. Since C is diagonal,
 1 t2

U C 0
 0

and
0
1 t2
0
0

0
1

R  FU
1
1


2
 1 t
t
 
 1 t2

0



0

0

1


t
1 t
1
2
1 t2
0
which is proper orthogonal, R  R T  I, det R  1. The deformation can thus be
decomposed into a pure stretch along the principal (material) axes, which here are just
the axes e1 , e 2 , e 3 , and then followed by a rotation about the X 3 axis. For example, at
t  1 , the rotation is 45 o clockwise.
The left stretch tensor is now obtained from v  F  R T and it turns out to be the same
as U (and b  v 2 ). Thus the deformation can also be decomposed into, first, a rigidbody rotation, followed by a pure stretch to the subsequent configuration. For
example, considering the configuration at t  1 (see figure above), first rotate
clockwise by 45 o ; then the rotated line elements dx (1)  (1 / 2 )e1  1 / 2 e 2 ,

dx
( 2)
 2e1 map to the final positions v  dx
(1)
 e1  e 2 and v  dx

( 2)
 2e1 .
Nanson’s formula, nds  JF T  NdS , relates unit normals in the reference and current
configurations. Each of the six sides of the block has surface area dS  1 and
ds  1  t 2 . For example, the unit normal N  e 2 is mapped to the unit normal
n  te1  e 2  / 1  t 2 through
 n1 
n 
 2
 n3 
N  e2
n
t
1  t2
 1
 1 t2

t
1  t 2  1  t 2 
2
 1 t
 0


e1 
1
1  t2

t
1 t2
1
1 t2
0

0
 0 
0 1

1 0

e2
Rates of Deformation
The velocity is (in the material description – the material coordinates are constant)
V ( X, t ) 
dx d
  X 1  tX 2 e1   tX 1  X 2 e 2  X 3 e 3   X 2 e1  X 1e 2
dt dt
The velocity in the spatial description can be obtained by substituting into the above
expression the motion X  f 1 (x, t ) , to get
v(x, t ) 
tx1  x2
 x  tx
e1  1 2 2 e 2
2
1 t
1 t
These two velocities are the same, they are the velocities of material particles in the
material, expressed in two different ways.
The acceleration is zero, A( X, t )  dV / dt  0 . The acceleration can also be obtained
directly from the spatial velocity using the material time derivative,
a(x, t ) 
dv(x, t ) v(x, t )

 grad v  v
dt
t
which turns out to be zero.
The spatial velocity gradient is now
 t
 1 t2

1
l  grad v  
2
 1 t
 0

1
1 t2
t
1 t2
0

0

0

0

This can be decomposed into its symmetric and skew-symmetric parts: the rate of
deformation
d
 t
1  t 2

 0

 0





 0

1
 
2
 1 t
 0

1
l  lT
2
0
t
1 t2
0

0

0

0

and the spin
w
1
l  lT
2

1
1 t2
0
0

0

0

0

The skew-symmetric spin can be written in the form of a vector, the angular velocity,
ω   w23e1  w13e 2  w12e 3

1
e3
1 t 2
which clearly corresponds to a clockwise rotation of material particles about the e 3
axis.
Note also that
div v 
v1 v2 v3
2t



x1 x2 x3 1  t 2
so that the rate of change of the volume ratio dv / dV is
J  Jdiv v  2t
Stress
Suppose now that the Cauchy stress is
1 2 / 3 2


t
0
0
 ln J  3 J



1

σ  J 1 
0
 ln J  J  2 / 3t 2
0
3


2 2 / 3 2 

0
0
 ln J  J t 

3

The Equations of Motion
The equations of motion in the spatial description are div σ  b  v and it can be
seen that the Cauchy stress is uniform throughout the material at any time instant,
div σ   ij / x j  o . There are also no accelerations and so there are no body
forces.
The PK1 stress is
P  Jσ  F T
1 2 / 3 2

t
  ln J  3 J

1 t2

  ln J  1 J  2 / 3 t 2

3
  t
1 t2


0



1
3
t
1 t2
1
 ln J  J  2 / 3t 2
3
1 t2
 ln J  J  2 / 3t 2
0


0




0


2
 ln J  J  2 / 3t 2 
3



The equations of motion in the material description are Div P  B   0 A and these
three terms are also zero.
Balance of Mechanical Energy
Material Description
The PK2 stress is
S  JF 1  σ  F T
1 2 / 3 2

t
 ln J  3 J

1 t2




0


0



0
1
3
1 t2
 ln J  J  2 / 3t 2
0


0




0


2
 ln J  J  2 / 3t 2 
3



Also,
 t 0 0
E  F T  d  F  0 t 0


0 0 0
and then the stress power is
  2t
S:E
1 t2
1 2 / 3 2 

t 
 ln J  3 J

The same result can be obtained by using the alternative stress power expression,
P : F .
The stress power is independent of position within the material, and the volume in the
undeformed configuration V is unity so the total stress power is
2t 
1

 ln J  J 2 / 3t 2 
2 
3


 S : E dV  S : E  1  t
V
Look now at the power of the power of the surface tractions:  T  VdS . The velocity
dS
is zero over the front and back faces (the sides with unit normals  e 3 ), and is nonzero over the other four sides (with normals in the reference configuration  e1 ,  e 2 ).
The PK1 traction is given by T  P  N and is shown in the figure for these four faces,
and given below (the PK1 stress is uniform throughout the material).
T( 2)  P  e 2  P12 e1  P22 e 2
T(3)  P  (e1 )  T(1)
T(1)  P  e1  P11e1  P21e 2
T( 4)  P  (e 2 )  T( 2)
1 2 / 3 2 

t e1  te 2 ,
 ln J  3 J


1 
1


 ln J  J  2 / 3t 2 te1  e 2 ,
2 
3
1 t 

T(1) 
T( 2 )
1
1 t2
Substituting in the velocity V  X 2 e1  X 1e 2 now leads to
T(3)  T(1)
T( 4 )  T( 2 )
 T  VdS   T   X
(1)
dS
e  e 2 dS (1) 
 T  e
2 1
dS(1)
( 2)
1
 X 1e 2 dS ( 2 )
dS( 2 )

 T   X e dS
( 3)
2 1
( 3)

dS( 3 )
 T   X e dS
( 4)
1 2
( 4)
dS( 4 )
1 1
1 1
   T(1)   X 2 e1  e 2 dX 2 dX 3    T( 2 )  e1  X 1e 2 dX 1 dX 3
0 0
0 0
1 1
1 1
0 0
0 0
   T(1)   X 2 e1 dX 2 dX 3    T( 2 )   X 1e 2 dX 1 dX 3
1 
1


1 
 1 
 T(1)   e1  e 2   T( 2 )   e1  e 2   T(1)   e1   T( 2 )    e 2 
2 
2


2 
 2 
2t 
1


 ln J  J  2 / 3t 2 
2 
3
1 t 

which equals the stress power, and so the mechanical energy is balanced:
dV
 T  V dS   B  V dV   S : E dV   V  dt
S
V
V
dV
V
Spatial Description
In the spatial description, the stress power is
σ : d  J 1
2t
1 t2
1 2 / 3 2 


ln
J

J t 

3

The total stress power in the material, with dv  JdV , is then
2t 
1

 ln J  J 2 / 3t 2 
2 
3


 σ : ddv  σ : d dv  Jσ : d  1  t
v
v
In the spatial description, the power of the surface forces is  t  v ds . Again, we only
s
need to look at four sides, since the velocity is zero over the front and back. The unit
normals to these four faces are shown in the figure.
n (3)  n (1)
1
n ( 2) 
1 t 2
te1  e 2 
n ( 4 )  n ( 2 )
1
n (1) 
1 t 2
e1  te 2 
The tractions acting over these surfaces are then obtained from Cauchy’s law,
t  σ  n . Examining the first surface, the velocities of particles there are those for
particles with material coordinates X 1  1 , so, eliminating x1 ,
X1 
 tx  x 
x1  tx2  1  t 2 , v (1)   1 2 2  e1  e 2  t  x2 e1  e 2
 1  t  X1 1
x1  tx2
1 
1 t 2
The integration over
1
1
 dX   dx  1 is straightforward. The integration over
 dX in the reference configuration corresponds to the
integration  dl in the current configuration. With
3
0
0
3
1
2
0
dl
dx2
dx1
dl  (dx1 ) 2  (dx 2 ) 2  1  t 2 dx 2 over this surface, we have
 dl 
1t
1  t 2  dx 2
t
The double integral in the reference confuguration
to the integral
  f (x) J dx dx
2
3
J 23 
  f (X)dX
2
dX 3 can be converted
where J is the Jacobian for the change of variable:
X 2 / x2
X 2 / x3
X 3 / x2
X 3 / x3

1
1 t 2
The limits of integration are obtained
Thus the complete expression of the power of the tractions is
 t  v ds 
s
1 1t
1 
1
t  v (1) dx 2 dx3 
2    (1)
1  t  0 t
1 t2

1
1 t2
1 1 t
 t
( 2)
 v ( 2 ) dx1 dx3
0 t
1 1
  t (3)  v (3) dx2 dx3 
0 0
1
1 t2
1 1
t
( 4)
0 0

 v ( 4 ) dx1 dx3 

The integral over the first surface is then
 t (1)  v (1) ds 
s
1
1 t2
1 1 t
 t
J 1

(1  t 2 ) 3 / 2


(1)
 v (1) dx 2 dx3
0 t
J 1
(1  t 2 ) 3 / 2
1 1t
  e
 te 2   t  x 2 e1  e 2 dx 2 dx3
1
0 t
1 1t
  2t  x dx dx
2
2
3
0 t
J 1
t  1 / 2
(1  t 2 ) 3 / 2
where    ln J  J 2 / 3t 2 / 3 .
1
1 t2
1 1 t
 t
0 t
(1)


1
1 t2
1 1 t
J 1
 v (1) dx1 dx3 
(1  t 2 ) 3 / 2
1 1
  t (3)  v (3) dx2 dx3 
0 0

  te
1
 e 2   e1   x1  t e 2 dx1 dx3
0 t
1 1 t
J 1
(1  t 2 ) 3 / 2
  2t  x dx dx
1
1
3
0 t
J 1
t  1 / 2
(1  t 2 ) 3 / 2
J 1
(1  t 2 ) 3 / 2
   e
J 1
(1  t 2 ) 3 / 2
1 1
1 1
1
 te 2   x 2 e1 dx 2 dx3
0 0
   x dx dx
2
2
3
0 0
J 1
 1 / 2

(1  t 2 ) 3 / 2
1
1 t2
1 1
t
0 0
( 4)
J 1
 v ( 4 ) dx1 dx3 
(1  t 2 ) 3 / 2


J 1
(1  t 2 ) 3 / 2
1 1
   te
1
 e 2    x1e 2 dx1 dx3
0 0
1 1
  x dx dx
1
0 0
J 1
 1 / 2
(1  t 2 ) 3 / 2
1
3
and the total is
 t  v ds  J
s
1


2t
 ln J  J 2 / 3t 2 / 3
2 3/ 2
(1  t )
and so the mechanical energy is balanced in the spatial description:
dv
 t  v ds   b  v dv   σ : ddv   v  dt dv
s
v
v
v
Strain Energy and Conservative Force System
The above Cauchy and PK2 stresses were actually derived from the strain energy
function


1
1
2
W   ln J    J  2 / 3 I b  3
2
2
This is the strain energy of a compressible Neo-Hookean material. It is a hyperelastic
model (since the stress is obtained from a strain energy function) of an isotropic
material (since the Cauchy stress is a function of the left Cauchy-Green strain b only).
Here, J  det F . We write this in terms of the third invariant of b, III b  det b :
III b  J 2 . Differentiating W now gives J / III b   III b / III b  1 / 2 J  , so
 W

 W W 
 W  2
σ  2 J 1 
III b  I  2 J 1 

I b b  2 J 1 
b
 III b

 I b II b 
 II b 
 W J 2 
 W 
 2 J 1 
J  I  2 J 1 
b

J

III

I
b


 b

1


 J 1  ln J I  J 2 / 3  b  trb I 
3



The PK2 stress can also be obtained directly through
 W W 
W
W
S  2

I b I  2
C2
III b C 1
II b
III b
 I b II b 
1


  ln JC 1  J  2 / 3  I  trC C 1 
3


leading to the PK2 stress given earlier.
1
1
2
Explicitly, from W   ln J    J  2 / 3 I b  3 , the time derivative of the strain
2
2
energy is
1
d

 1
W  J 1 J  ln J  J  2 / 3 trb   J  2 / 3 trb 
3
dt

 2
Now J  Jdiv v so
J 1 J  div v 
2t
1 t2
and
trb  3  2t 2
d
trb   4t
dt
so
2t
W 
1 t2
1 2 / 3 2 

t 
 ln J  3 J

It can be seen that this time derivative of the strain energy is precisely the stress
power, as expected for a hyperelastic material:
  W  1 U
Jσ : d  S : E
0
Here, U is a strain energy per unit mass, whereas W is the strain energy per unit
(reference) volume.