ES 240
Solid Mechanics
Z. Suo
Finite Deformation
References
匡震邦, 非线性连续介质力学, 上海交通大学出版社, 2002. My favorite textbook on nonlinear
continuum mechanics, written by the man who taught me the subject in 1985.
G.A. Holzapfel, Nonlinear Solid Mechanics, Wiley, 2000.
黄筑平，连续介质力学基础，高等教育出版社， 2003.
T. Belytschko, W.K. Liu, B. Moran, Nonlinear Finite Elements for Continua and Structures,
Wiley, 2000.
L.E. Malvern, Introduction to the Mechanics of Continuous Medium, Prentice-Hall, 1969.
C. Trusdell and W. Noll, The Non-linear Field Theories of Mechanics, 3rd edition, Springer, 2004.
C. Trusdell and R.A. Toupin, The Classical Field Theories, in Encyclopedia of Physics, Volume
III/1, Pringer-Verlag, Berlin, 226-793 (1960).
L.R.G. Treloar, The Physics of Rubber Elasticity, 3rd ed. 1975. Reissued in 2005.
Be wise, linearize. Following the advice of George Carrier, we have been mostly
looking at Hookean materials and infinitesimal deformation. We have mixed the 3 ingredients in
solid mechanics (deformation geometry, momentum balance, material law) without fussing over
subtleties. The results are fascinating and useful. Now we wish to go nonlinear, hopefully also
with wisdom. We will refine the ingredients by considering non-Hookean materials and finite
deformation. The two refinements need not be mixed. For example, a viscoelastic material is
non-Hookean, but deformation of such a material can be infinitesimal. In this brief introduction
to finite deformation, we will outline some of the fundamental considerations, and describe a few
illustrative phenomena.
Finite deformation.
When a structure deforms,
Newton’s law holds true in every deformed state. We have often
violated this law. For example, in analyzing a truss, we have
balanced forces as if the truss did not deform.
You might think that a structure suffering a small strain,
say less than 1%, entitles you to neglect the change in geometry
when you balance forces. A counter example is familiar to you.
Upon buckling, the strains in a column are indeed small, but you
must enforce equilibrium in the deflected state of the column.
Mechanics of deformation is a tricky business. We proceed with
caution. The essential point is this: we must enforce Newton’s
law in every deformed state, and justify any simplification on this
basis.
reference state
W
current state
Non-Hookean materials. Moving nonlinear and inelastic will take us in many directions.
For example, when deformation is large, the force may vary nonlinearly with the elongation. As
another example, we’ve already looked at time-dependent behavior of materials, such as
viscoelasticity. we also have daily experience of metals. After elastic deformation, upon
unloading, a metal recovers its shape. After plastic deformation, upon unloading, the metal does
not fully recover its shape. If we apply an axial force to a metal bar, and measure its length, the
force-length relation is linear for elastic deformation, and is nonlinear for plastic deformation.
During unloading, the metal bar deforms elastically. After plastic loading and elastic unloading,
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the force-length relation is not a one-to-one relation,
but is history-dependent. Of course, a viscoelastic
material is also history-dependent.
To analyze finite, history-dependent
deformation of a structure, a general approach is
to evolve the state incrementally, and enforce
Newton’s law in every state.
loading
unloading
A rod under axial load. We also proceed
with our subject incrementally, beginning with a
rod in incremental states of uniaxial stress.
L0
Initially, the rod is unstressed, and has crosssectional area A and length L . The rod is then subject to an axial force P, and deforms to crosssectional area a and length l. We next examine the 3 ingredients in solid mechanics.
Strain measures. Any state can be used as a reference state. For example, we can take
the initial, unstressed state as the reference state. Define the engineering strain by the
elongation divided by the reference length:
elongation
lL
.
e

length in the reference state
L
Another strain measure is defined as follows. Deform the material from a current length l
by a small amount to l  dl . Define the increment in the strain, d , as the increment in the
length of the rod divided by the current length of the rod, namely,
increment in length
dl
d 
 .
length in the current state
l
This equation defines the increment of natural strain. Integrating from L to l, we obtain that
l
  log   .
L
Yet one more strain measure, the Lagrange strain, is defined as
l 2  L2

.
2L2
This definition is hard to motivate in 1D. But if you take the view that any increasing function of
l / L is a suitable measure of strain, then no motivation is really needed.
Indeed, even the ratio l / L itself has a name: the stretch is defined as the length of the
rod in the current state divided by the length of the rod in the reference state:
length in current state
l

 .
length in reference state L
There seems to be no lack of human ingenuity to form a dimensionless quantity out of
two lengths L and l. Needless to say, all these strain measures contain the same information. For
example, every one of the measures defined above is an increasing function of stretch:
2  1
e    1,   log  ,  
.
2
Because they are all one-to-one functions, any one measure can be taken to be “basic” and then
used to express all the rest. For example, we can express all measures in terms of the
engineering strain:
L
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  e  1,   log 1  e ,   e 2  e
Z. Suo
1
2
Thus, when we call e the engineering strain, we do not mean that e is unscientific or crude or
unnatural strain. We just need a name. When the strain is small, namely, e  1 , the three
measures are approximately equal,     e .
Later on, we will provide motivations for some of these definitions, but these motivations
are probably elaborate ways to express preferences of individual people.
Stress measures. Work done by a force. When dealing with finite deformation, we
must be specific about the area used in defining the stress. Define the nominal stress, s, as the
force in the current state divided by the area in the reference state:
force in the current state
P
s
 .
area in the reference state A
When the rod elongates from length l to length l  dl , the force P does work Pdl .
Recall that P  sA and dl  Lde , so that the work done by the force is
Pdl  ALsde .
Since AL is the volume of the rod in the reference state, we note that
increment of work in the current state
sde 
.
volume in the reference state
We say that the nominal stress and the engineering strain are work-conjugate. Also note that
de  d . Consequently, the stretch is also work-conjugate to the nominal stress.
Define the true stress,  , as the force in the current state divided by the area in the
current state, namely,
force in the current state P

 .
area in the current state
a
Recall that dl  ld . The work done by the force is
Pdl  ald .
Since al is the current volume of the bar, we note that
increment of work in the current state
d 
.
volume in the current state
That is, the true stress is work-conjugate to the natural strain.
Given a measure of strain, we can define its work-conjugate stress. For example,
consider the Lagrange strain,  . Subject to an increment in the strain, d , the force acting on
the element does the work. Denote
increment of work in the curent state
S 
.
volume in the reference state
This expression defines a new measure of stress, S.
This new stress measure does not have a “simpler” interpretation than its status as the
work conjugate to the Lagrange strain. Indeed, if we are liberal about the definition of strain
measures, without being obsessive about “motivating” each measure, we may as well take a
liberal view to call the work conjugate of each strain a stress measure, and name the stress
measure after a mechanician who can no longer protest. You can easily invent and name other
stress measures, but the above stress measures have already got names:
  : true stress or the Cauchy stress.
 s : nominal stress or the first Piola-Kirchhoff stress.
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S : the second Piola-Kirchhoff stress.
Recall the relations among the measures of strain:
e    1,   log  ,  
We obtain the relations among their increments:
de  d , d 
d
2  1
2
, d  d .

Consequently, the three measures of stress are related as
V 2
 S , s  S .
v
All stress measures are linear in force applied to the cross-sectional area, but differ by geometric
factors.
Material laws. For a metal undergoing large, plastic deformation, the stress-strain curve
(without unloading) is often fit to a power law in terms of the true stress and the natural strain:
    K N
where K and  are parameters to fit experimental data. Some representative values: N = 0.150.25 for aluminum, N = 0.3-0.35 for copper, N = 0.45-0.55 for stainless steel. K has the
dimension of stress; it represents the true stress at strain   1 . Representative values for K are
100 MPa – 1GPa. At large deformation, volumetric strain is negligible compared to tensile strain.
Consequently, the material is often taken to be incompressible.
Rubbers are often assumed to obey the neo-Hookean law (more details later). For a
rubber rod in uniaxial states of stress, the stress-strain data are fit to
     2  1 .
Recall that   1  e . For small strains, e  1 , the above reduces to
  3e .
Thus, we interpret 3 as Young’s modulus and  the shear modulus. Rubbers are nearly
incompressible, so that Poisson’s ratio is taken to be ½. Representative values for  are 1 MPa
– 100 MPa.



Are these alternatives necessary? Now we have described the 3 ingredients for a rod
under uniaxial tension. Even in this simplest setting, for each ingredient we have given several
alternative descriptions. Some alternatives are necessary; for example, metals and rubbers
behave differently. But the difference in their force-displacement relations does not justify us to
use different stress and strain measures to describe different materials. In fact, to see the
difference in material behavior, we would like to use the same stress and strain measures for both
materials. For example, we can use the natural strain to describe the stress-strain relation for
rubbers:
    exp 2   exp   .
This change of variable immediately brings out a key insight: in tension, the stress in rubbers
rises more steeply than in metals. We will return to this insight shortly.
Are these alternative stress and strain measures necessary? I have my own thoughts, but
you should form your own opinions. The question perhaps boil down to something no more
profound than asking, “Is it necessary to know many alternative roads to Boston Common?”
Whatever your opinions are, however, it may alleviate some of your pains in studying the subject
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by knowing that textbooks of nonlinear continuum mechanics are full of equivalent alternatives
at every turn. These alternatives often hide behind forests of notation and verbiage, and may
offer some tantalizing sights. You will just have to look beyond them for matters of consequence.
Exercise. Use the 3 ingredients outlined about to obtain the force-deflection relation for
the truss sketched in the beginning of the notes. Assuming all three members of the truss are
made of rubber bands, and that deformation is large.
Necking in a bar. Considère condition. Let us try to apply the newly refined 3
ingredients to a specific phenomenon: necking. Subject to a tensile force, a metal bar first
elongates uniformly and then, at some strain, a small part of the rod starts to thin down
preferentially, forming a neck. By contrast, a rubber band under tension usually does not form a
neck. We would like to interpret these observations. To do so we must be explicitly specify the
measures of stress and strain.
Here is a summary of the 3 ingredients, using a specific set of alternatives:
 Force balance: P  a
 Material law: For a metal bar under uniaxial tension, the true stress relates to the natural
strain as      .
 Deformation geometry:   log l / L . We will assume that the volume of the rod is
constant during deformation, AL  al , or a  A exp    .
Put the three ingredients together, and we obtain the force as a function of strain:
P  a  A  exp   .
Plot P as a function of . In plotting the figure, I’ve set   K N , with N = 0.5. Observe the
two competing factors: material hardening and geometric softening. As the bar elongates, the
material hardens, as reflected by the hardening exponent in the stress-strain relation      .
At the same time, the elongation reduces the cross-sectional area, an effect known as geometric
softening. For small deformation, P  0 as   0 ; material hardening prevails, and the force
increases as the bar elongates. For large deformation, so long as the stress-strain relation   
increases slower than exp   , P  0 as    , geometric softening prevails, and the force
drops as the bar elongates.
geometric
softening
0.4
P
AK
0.3
0.2
material
hardening
0.1
0.5
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To determine the peak force, note that
dP
 

 A exp   
 .
d
 

Consequently, the force P peaks when the true stress equals the tangent modulus:
d  
    .
d
This equation, known as the Considère condition, determines the strain at which the force peaks.
For the power-law material, the force peaks at the critical strain
c  N .
When the metal bar is loaded beyond this critical strain, deformation becomes
nonuniform, with a segment of the bar elongates at a higher strain than the rest of the bar. That is,
a neck forms in the bar. For an analysis of the nonuniform deformation, see Needleman (1972, A
numerical study of necking in circular cylindrical bar, Journal of the Mechanics and Physics of
Solids, 20, 111). ABAQUS can be used to study the necking process.
For a rubber band, assume the material is Neo-Hookean:
   2  1   exp 2   exp    .
Thus, at a large tensile strain, the true stress increases exponentially with the natural strain, so
that the force P  A  exp    always increases with the strain. The rubber band will not
form a neck under uniaxial tension.


Hyperelastic materials. We next explore nonlinear stress-strain relations under
multiaxial states of stress. Of course, the only way to really know such relations is to run tests
for a given history of state of stress, but this would be too time-consuming and quickly become
impractical. We’ll have to reduce the number of tests by some approximations. The art of
making such compromise between accuracy and labor is known as formulating constitutive laws.
As an example, here we attempt to describe this art for rubbers.
A rubber rod, length L and cross-sectional area A in the unstressed state, is stretched by
force P to length l and cross-sectional area a. When the rod extends from length l to length
l  dl , the force does work Pdl . Recall that P  sA and dl  Ld , so that the work is
Pdl  ALsd .
Since AL is the volume of the rod in the undeformed state, we note that
increment of work in the current state
sd 
.
volume of the reference state
We say that the nominal stress is work-conjugate to the stretch. Define the nominal energy
density by
energy in the current state
W 
.
volume of the reference state
Assuming that the work done by the force is fully stored as energy in the rod, we obtain that
dW  sd .
Note that W should be considered the Helmholtz free energy. We’ll only consider the isothermal
conditions, so that that we’ll drop the dependence on temperature.
We can measure experimentally the nominal stress as a function of stretch, s  , and
then integrate the curve s  to obtain W   , , known as the strain-energy function.
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Alternatively, we can obtain an expression of W   from some theoretical considerations, and
then obtain s  by
dW  
s   
.
d
For an isotropic material, we can represent a material particle by a rectangular block cut
in the orientation of the three principal stresses. Let the block be stretched in the three directions
by 1,  2 and  3, and the corresponding nominal stresses be s1, s2 and s3. The complete stressstrain relations involve three functions s1 1 , 2 , 3  , s 2 1 , 2 , 3  and s3 1 , 2 , 3  . In general,
we can run tests to determine these functions. However, as we indicated above, running test
alone would be too time-consuming. Instead, we will formulate a constitutive law on the basis of
the assumption that work done by the forces is all stored in the block as energy.
When the stretches increase by d1 , d2 and d3 , following the same line of reasoning
as given above, we conclude that the free energy changes by
dW  s1d1  s 2 d2  s3 d3 .
Now the strain energy is a function of the three stretches, W 1 , 2 , 3  . Once the function
W 1 , 2 , 3  is determined by experiment, say, the stresses are calculated from the partial
derivatives:
W 1 , 2 , 3 
W 1 , 2 , 3 
W 1 , 2 , 3 
.
s1 
, s2 
, s3 
1
2
3
An elastic material whose stress-strain relation is derivable from a strain-energy function
is known as a hyperelastic material. An elastic material whose stress-strain relation is not
derivable from a strain-energy function is called a hypoelastic material.
A rectangular block of a material, lengths L1 , L2 and L3 in the undeformed state, is
subject to forces P1 , P2 and P3 on its faces, and is deformed into a block of lengths l1 , l 2 and l 3 .
The nominal stress on one face is
P
s1  1 .
L2 L3
The true stress on the same face is
P
1  1 .
l 2 l3
Consequently, the true stress relates to the nominal stress by
s
1  1 .
 2 3
The true stress is derivable from the strain-energy function:
W 1 , 2 , 3 
.
1 
2 3 1
The other two true stress components can be similarly obtained.
Incompressible, isotropic, hyperelastic material. When a material undergoes large
deformation, the amount of volumetric deformation is often small compared to the overall
deformation. Consequently, we may neglect the volumetric deformation, and assume that the
material is incompressible.
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A block of a material, of lengths L1 , L2 and L3 in the undeformed state, is deformed into
a rectangle of lengths l1 , l 2 and l 3 . If the material is incompressible, the volume of the block
must remain unchanged, namely, L1 L2 L3  l1l 2 l3 , or
123  1 .
The incompressibility places a constraint among the three stretches: they cannot vary
independently. We may regard 1 and  2 as independent variables, so that
3  12 1 .
Consequently, the strain-energy density is a function of two independent variables, W 1 , 2  .
Consequently, only biaxial tests need be done to fully characterize the material.
1
Inserting the constraint 3  12  into the expression dW  s1d1  s 2 d2  s3 d3 , we
obtain that
dW  s1  12 21 s3 d1  s 2  22 11 s3 d2 .
Once the function W 1 , 2  is determined, the stress-stretch relation is given by differentiation:
W 1 , 2 
W 1 , 2 
, s 2  2 2 11 s3 
.
s1  12 21 s3 
1
2
These relations, together with the incompressibility condition 123  1 , replace Hooke’s law
and serve as the stress-strain relations for incompressible, isotropic and hyperelastic materials.
For incompressible materials, the true stresses relate to the nominal stresses as
 1  1 s1 ,  2  2 s 2 ,  3  3 s3 .
Consequently, the stress-strain relations become
W 1 , 2 
W 1 , 2 
,  2   3  2
.
 1   3  1
1
2
Stress-strain relations for rubbers. The function W 1 , 2  can be determined by
subjecting a sheet of rubber under biaxial stress states. The form of the function is sometimes
inspired by theoretical considerations. Here are three often used forms.
The Neo-Hookean materials are materials whose energy density is expressed by

W  12  22  32  3 .
2
1
Inserting the constraint 3  12  , we obtain the stress-stretch relations:
 1   3   12  32  ,
 2   3   22  32  .
A neo-Hookean material is characterized by a single elastic constant,  . The constant
may be determined experimentally. For example, consider a rod in a state of uniaxial stress:
 1   , 2   3  0 .
Let the stretch along the axis of loading be 1   . Incompressibility dictates that the stretches
in the directions transverse to the loading axis be  2  3  1 / 2 . Inserting into the above stressstretch relation, we obtain the relation under the uniaxial stress:
   2  1 .
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Recall that stretch relates to the engineering strain as
  l / L  1 e .
When the strain is small, namely, e  1, the above stress-stretch relation reduces to
  3e .
Thus, we interpret 3 as Young’s modulus and  the shear modulus.
The form of strain-energy function of Neo-Hookean materials has also emerged from a
model in statistical mechanics. The model gives   NkT , where N is the number of polymer
chains per unit volume, and kT the temperature in units of energy. See Treloar.
The Mooney materials are those that can be fitted by
W  c1 12  22  32  3  c 2 12  22  32  3.
M. Mooney (1940) A theory of large elastic deformation, Journal of Applied Physics 11, 582-592.
The Ogden materials are those that can be fitted by a series of more terms
W 
n
n 

1  2  3  3 ,
n
n
n
n
where  n may have any values, positive or negative, and are not necessarily integers, and  n are
constants. The stress-stretch relations are
 1   3    n 1 n  3 n ,

   
n
 2 3
n
n
2
 3 n

.
n
R.W. Odgen (1972) Large deformation isotropic elasticity – on the correlation of theory and
experiment for incompressible rubberlike solids. Proceedings of the Royal Society of London
A326, 567-583.
For the literature on other functional forms, theoretical motivations, and experimental
data, see Treloar for a classic treatment, and Section 6.5 of Holzapfel for an update.
Inflation of a balloon (or bursting of an aneurysm?). A spherical balloon, radius R
and thickness H in the undeformed state, is subject to a pressure p, and expands to radius r while
reduces its thickness to h. We would like to determine the relationship between the radius and
the pressure.
As always in solid mechanics, we invoke the 3 ingredients.
Deformation geometry. The stretch in the thickness direction is 3  h / H . The
stretches in the circumferential directions are 1  2  r / R , which we will denote as  . The
material of the skin is taken to be incompressible, so that hr 2  HR 2 , or 3  2 .
Force balance. The stress normal to the skin is between p and zero, and is small
compared to the stress in the circumferential direction,  . Draw the free-body diagram of a half
balloon, and balance the forces, 2rh  r 2 p . Thus
2h
p
.
r
Because r  h , we confirm that   p .
Materials law. The material law will take the form
     .
Put the three ingredients together, and we obtain that
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2 H 3
    .
R
This is the desired relation between the pressure and the current radius, if we take   r / R as
the normalized current radius. Assume the skin is made of a neo-Hookean material. under the
stretches 1  2   , 3  2 , so that    2  4 . Sketch the function p  . Note
p


p0  0 and p = 0. As the balloon expands, the material stiffens, but the skin thins. At
certain stretch, the geometric thinning prevails over material stiffening, and the pressure reaches
a peak. This peak is reached at c  7 1 / 4 .
What will happen after this peak? See an analysis by Needleman (1977, Inflation of
spherical rubber balloons, International Journal of Solids and Structures 13, 409-421). Also see
experiments by D.K. Bogen and T.A. McMahon (1979, Do cardiac aneurysms blow out?
Biophysics Journal, 27, 301-316)
Exercise. Study the problem for a cylindrical balloon.
Inhomogeneous field. When the state of deformation in a body is inhomogeneous, we
must establish differential equations. Balancing forces on a material particle in the current state,
we find the equilibrium equation in the familiar form:
 ij
 0.
x j
The deformation geometry is complicated in general. However, for a problem with such a
symmetry that the principal directions of stretch are obvious, when an element of length in the
reference state, dL, becomes dl in the current state, the stretch is
dl

dL
We have now established the 3 ingredients in solid mechanics for finite deformation.
Cavitation. Consider a cavity in an infinite medium, subject to a remote hydrostatic
tension S. For a brittle material, the cavity concentrates stress, so that a crack may emanate from
the cavity. In such a case, deformation is typically small when the stress reaches a critical value,
so that we can analyze the problem assuming that the material is linearly elastic and deformation
is infinitesimal. Formulated this way, the problem is known as the Lame problem, as we have
seen before. The key result of the analysis is that the hoop stress at the surface of cavity is 3/2
times the remote stress.
For a material capable of large deformation, such as a ductile metal or a rubber, however,
the cavity may cause another mode of failure. Under the hydrostatic stress remote from the
cavity, the cavity may expand indefinitely when the applied stress reaches some finite value, a
phenomenon known as cavitation. To study this phenomenon, we need to determine the radius
of the cavity as a function of the remote tension. The Lame solution clearly gives an erroneous
prediction in this case because it says that the radius of cavity increases linearly with the applied
stress.
Deformation geometry. In the unstressed state, let the radius of the cavity be A, and the
distance between a material particle and the center of the cavity be R. The deformation make the
radius of the cavity become a, and move the material particle to a new position r. Deformation
of the medium is characterized by the function r R  . The hoop stretch is
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  r / R .
The radial stretch is
r  dr R / dR .
We assume that the material is incompressible, so that
r 3  a 3  R 3  A3 .
Consequently, once the current radius of the cavity a is determined, the entire field of
deformation is determined. Because the independent variable in the equilibrium equation is
current position r, we express the R in terms of r:

Rr   r 3  A3  a 3
Incompressibility dictates that

1/ 3
r  2  R / r 2 ,
Insert the expression of Rr  , we express the stretch as a function of r:
2/3
 r 3  A3  a 3 
 .
 r  
r3


Force balance. The stress in the medium is nonuniform: remote from the cavity, the
state of stress is hydrostatic; near the cavity, the state of stress is equal-biaxial. A material
particle at location r in the current state is subject to a triaxial state of stress,  r r ,   r     r  .
We balance force for a material particle in the deformed state, so that
 
d r
2 r
 0.
dr
r
This equation takes the same form as that used in the Lame problem.
Material law. Suppose that we have conducted a test of the material under unaxial stress
states. The uniaxial stress-stretch curve is
  g   .
In the medium around the cavity, a material particle is under a triaxial stress state: one radial
component and two hoop components: (  r ,   ,   ). Because we have assumed that the
material is incompressible, superposing a hydrostatic stress on the material particle will not
change the state of deformation of the particle. For example, we can superimpose
   ,  ,   on the material particle, so that the stress state of the particle becomes
 r    ,0,0 .
This is a uniaxial stress state. Thus,
 r     g r  .
Putting the three ingredients together, and integrating from the surface of the cavity to the
remote point, we obtain that

g  r d
S  2 
,

1
where   r / a ,
  3   A / a 3  1 

r  
3




and for Neo-Hookean material,
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
g     2  1 .
This set of equations define S as a function of a/A. The integral can be evaluated analytically,
giving
1
4
5
1 a 
a
  2     .
 2  A
2 A
Sketch this result and compare it with the linear elastic solution. The cavity can expand
indefinitely when the applied stress is still finite. The stress needed to cause the cavity to expand
indefinitely is called the cavitation limit. The cavitation limit for a Neo-Hookean material is
S c  5 / 2 .
Cavitation limit should exist for other material laws. To see this, set A / a  0 , so that
S
2/3
  3 1
r   3 
  
We need to see if the above integral is finite. As    , the material is under hydrostatic state,
and r  1 , so that g r   Er  1 , were E is Young’s modulus. Thus
2/3


1 
2E

g r   E 1  3   1   3 .
  

3


The integral must converge.
5
Large Deformation
Small Deformation
4.5
4
3.5
S/ 
3
2.5
2
1.5
1
0.5
0
0
1
2
3
4
5
a/A
6
7
8
9
10
This problem was solved for metals by R. Hill (1949, Journal of Applied Mechanics16,
259), and for rubbers by A.N. Gent and D.A. Tompkins (1969, Nucleation and growth of gas
bubbles in elastomers, Journal of Applied Physics 6, 2520-2525). For more recent studies of
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cavitation in metals, see Y. Huang, J.W. Hutchinson and V. Tvergaard (1991, Journal of
Mechanics and Physics of Solids 39, 223); A.G. Varias, Z. Suo and C.F. Shih (1992, Mode
mixity effect on the damage of a constrained ductile layer, Journal of Mechanics and Physics of
Solids 40, 485-509).
Exercise. Plot the hoop stress at the surface of the cavity as function of a/A. Compare
the result with the Lame solution.
Exercise. Plot S as a function of a/A for a power-law material. Numerical integration
might be needed.
Exercise. Study the problem for a cavity in a sphere of material of finite radius.
Exercise. Study the phenomenon of cavitation under the plane strain conditions.
Exercise. In the above formulation, we have used the equilibrium equation formulated in
terms of Cauchy stresses as functions of r. Derive the equilibrium equation in terms of the
nominal stresses as functions of R. Solve the problem using this alternative formulation.
Particle, place and deformation. The above formulation is convenient if the principal
directions are obvious from the symmetry of the problem. We next formulate the theory without
reference to the principal directions.
A body consists of a field of material particles. We name a particle by X and name its
place at time t by x. The program of continuum mechanics is to formulate equations of motion
to evolve the field of deformation xX, t  .
The displacement, velocity and the acceleration of material particle X at time t are,
respectively,
U  xX, t   X ,
V  xX, t  / t ,
A   2 xX, t  / t 2 .
We have tacitly assumed that deformation of the body preserves the identities of the
particles. Whether this assumption is valid can be determined by experimental observation. For
example, we can paint a grid on the body. After deformation, if the grid is distorted but remains
intact, then we can rightly assume that the deformation preserves the identities of the particles is
valid. If, however, after the deformation the grid disintegrates, we should not assume that the
deformation preserves the identities of the particles.
Whether a deformation of body preserves the identities of the particles depend on the size
scale we paint the grid, and the time scale over which we look at the body. A rubber, for
example, consists of cross-lined long molecules. If our grid is over a size much larger than
individual molecules, then deformation will not disintegrate the grid. By contrast, a liquid
consists of molecules that can change neighbors. A grid paint on a body of liquid, no matter how
coarse the grid is, will disintegrate over a long enough time. Similar remarks may be made for
metals. Also, in many situations, the body will grow over time. Examples include growth of
cells in a tissue, and growth of thin films when atoms diffuse into the films. The combined
growth and deformation clearly does not preserve the identities of particles.
In these notes, we will assume that deformation of the body do preserve the identities of
the particles. We can name the field of particles any way we want. For example, we can use
English letters and Chinese characters. To use calculus, we can choose to name the particles by a
continuous variable. In particular, we can name each material particle by its coordinates when
the body is in a particular state. We will call this state the reference state. Even without external
loading, a body may be under a field of residual stress. Thus, we cannot always use the
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undeformed state as the reference state. Rather, any state of the body may be used as a reference
state. Let the coordinates of a material particle in the reference state be X, which we use to name
the material particle. At time t, the same material particle moves to a position with the
coordinates xX, t  .
Deformation gradient. First consider a one-dimensional deformation, a material particle
at position X in the reference state moves to position x at time t. The deformation of the body is
described by the function x X , t  . The deformation need not be uniform; that is, the stretch may
vary from one segment of the body to another segment. In the reference state, a segment is in the
interval (X, X + dX). After deformation, the material particle at X moves to position x X , t  , and
the material particle at X + dX moves to postion x(X + dX, t). Recall that the stretch is defined as
length in current state

length in reference state
Thus,
x X  dX , t   x X , t  x X , t 
X , t  

.
dX
X
The quantity x / X is called the deformation gradient.
x(X,t)
X+dX
x(X+dX,t)
X
current state
reference state
We next generalize this idea to three dimensions. Consider two nearby material particles.
In the reference state: one particle is at position X , and the other at position X  dX . We will
call dX a material element of line. In the current state, the two particles move to, respectively,
xi X  and xi X  dX  . The material element of line dX deforms to dx  xX  dX  xX .
The two vectors, dX and dx differ in length and in orientation. Note that
xi
xi X  dX  xi X 
dX K .
X K
The tensor,
x X, t 
,
FiK X, t   i
X K
maps a material element of line in the reference state, dX , to the same material element in the
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current state, dx :
dx  FdX .
The tensor F is known as deformation gradient. The deformation will change both the length
and the orientation of the material element of line.
Lagrange strain tensor. A material element of line is dX in the reference state, and is
dx in the current state. The length of the element in the reference state, dL, is given by
dL2  dX K dX K .
The length of the same element in the current state, dl, is given by
dl 2  dxi dxi .
Define tensor E KL by
dxi dxi  dX K dX K  2E KL dX K dX L .
We can express the new tensor in terms of the deformation gradient:
1
E KL  FiK FiL   KL  ,
2
where  KL  1 when K = L, and  KL  0 when K  L . The E KL tensor is symmetric.
The Lagrange strain forms a link between finite deformation and the infinitesimal
deformation approximation. Expressed in terms of displacement field UX, t  , the Lagrange
strain is
1  U K U L U i U i 
.
E KL  


2  X L X K X K X L 
Thus, the Lagrange strain coincides with the strain obtained in the infinitesimal strain
formulation if
U i
 1 .
X K
This in turn requires that all components of linear strain and rotation be small. However, even
when all strains and rotations are small, we still need to be careful with deformation in setting up
force balance, as we remarked before.
Aside. The tensor E KL is a 3D generalization of the Lagrange strain that we have defined
in one-dimensional deformation. To see this, consider a material element of line. In the
reference state, the element is the vector
dX  MdL ,
where dL is the length of the element and M the unit vector along the direction of the element.
Divide dxi dxi  dX K dX K  2E KL dX K dX L by 2dL2 , and we obtain that
M X, t   EKL M K M L .
This gives the Lagrange strain for a material element of line in direction M when the element is
in the reference state. We will call the tensor E KL the Lagrange strain.
Aside, the stretch. The tensor of deformation gradient F is a 3D generalization of stretch.
To make this idea precise, let a material element of line in the reference state be
dX  MdL ,
where dL is the length of the element and M the unit vector along the direction of the element.
The same element in the current state is
dx  mdl ,
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where dl is the length of the element and m the unit vector along the direction of the element in
the current state.
The stretch of the material element is
dl

.
dL
From dx  FdX , we obtain that
mdl  FMdL ,
or,
m  FM
That is, for a material element of line in direction M in the reference state, FM is a vector whose
magnitude gives the stretch of the element, and whose direction gives the direction of the
element in the current state. Thus,
M X, t   FM  FiK FiL M K M L .
We have added a subscript to the stretch to remind us that the stretch is for a material element of
line in the direction M when the element is in the reference state.
Exercise. You can confirm the same relation we identified for a tensile bar:
2M  1
.
M 
2
Exercise. The tensor of Lagrange strain can also be used to calculate the engineering
shear strain. Consider two material elements of line. In the reference state, the two elements are
in two orthogonal directions M and N . In the current state, the angle between the two elements

become   . We have defined  as the engineering shear strain. Show that
2
2 E KL M K N L
.
sin  
M  N
Aside, homogenous deformation. A deformation is said to be homogenous when the
deformation gradient is the same everywhere in the body. The deformation gradient, however,
may still vary with time. Denote the deformation gradient by Ft  . For a material particle at
position X in the reference state, and at position x at time t, we have
xX, t   Ft X  ct  ,
where ct  is a rigid body translation. The velocity and acceleration of the particle are
V X, t  
dFt 
dct 
X
,
dt
dt
AX, t  
The Lagrange strain is
E KL t  
d 2 Ft 
d 2 ct 
X

.
dt 2
dt 2
1
FiK t FiL t    KL  .
2
As expected, in homogenous deformation, the Lagrange strain is the same for all material
particles, but can vary with time.
In continuum mechanics, material laws are usually formulated theoretically and tested
experimentally for homogenous deformation. The material laws are then used, together with
PDEs, to analyze inhomogeneous deformation. This practice has been reexamined intensely in
recent years. Material laws invoking inhomogeneous deformation are known as strain-gradient
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theories.
Aside, rigid-body rotation. The homogenous deformation includes rigid-body rotation
as a special case. A rigid body rotation preserves the distance between any two material particles.
Consider two materials particles at X and Y in the reference state, and at x and y in the current
state. The distance between the material particles in the reference state is
 X K  YK  X K  YK  .
The distance between the two material particles in the current state is
xi  yi xi  yi   FiK t FiL t  X K  YK  X L  YL 
For the distance to be invariant, the deformation gradient must satisfy
FiK t FiL t    KL .
That is, the deformation gradient must be an orthogonal tensor. Note that the rigid-body rotation
will not induce any Lagrange strain.
The tensor of nominal stress. Let dV X be an element of material volume in the
reference state, and let BX, t dV X be the force in the current state acting on the material
element of volume. Let NXdAX be an element of material area in the reference state, and let
TX, t dAX be the force in the current state acting on the material element of area. Let
 XdV X be the mass of the material element of volume. Because the mass of the material
element is invariant during deformation,  X is time-independent.
Let  i X  be a triplet test function. Define a field siK X, t  such that

 i
 2 xi 

 siK X K dV    Bi   t 2  i dV   Ti  i dA
holds true for any test function  i X  . The integrals extend over the volume and surface of the
body in the reference state. The field siK X, t  is a tensor field and, as will show shortly, is
consistent with our notion of nominal stress.
Together with a material law that relates the current deformation gradient to the nominal
stress, the above definition provides the basis of using the Galerkin method to evolve the
deformation xX, t  as time progresses.
Momentum balance in the form of PDE. Applying the divergence theorem, we can
rewrite the left side of the above definition as
 siK  i i  siK

 i
siK
s
dV

iK
 X K
  X K  X K  i dV   siK N K  i dV   X K  i dV
Inserting this expression into the PVW, and insisting that the PVW hold true for arbitrary virtual
motion, we find that the tensor of nominal stress obeys that
siK X, t 
 2 xi X, t 
 Bi X, t    X 
,
X K
t 2
in the volume of the body, and
siK N K  Ti
on the surface of the body. This last expression gives an alternative definition of the tensor of
nominal stress.
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Consider a material element of area normal to the axis X K . Thus, s iK is component i of
the force acting on the element in the current state divided by the area of the element in the
reference state. We can also derive the above PDE by applying momentum balance to a freebody diagram in the current state.
s3I
dXIII
s1I
dXII
s2I
dXI
Hyperelastic material. Let us look again at this equation that defines the nominal stress,

 i
 2 xi 

s
dV

B


 iK X K
  i t 2  i dV   Ti  i dA .
When we replace the test function  i X  by a small deformation x i , the right-hand side of the
equation is the work done by the applied force and the inertial force to the body. If the material
is conservative, this work must be fully converted to the free energy of the body. The left-hand
side is siK FiK . A material particle is represented by a rectangular block of unit volume in the
reference state. In the current state, the particle is in a state of stress s iK . When the deformation
gradient changes by FiK , the stress does work to the particle siK FiK . Assume that all this work
is stored as energy in the particle, so that
dW  siK FiK
Consequently, the energy is a function of the tensor of deformation gradient, W(F). Once this
function is prescribed, the stress is given by
W F 
s iK 
.
FiK
The deformation gradient F consists of both stretch and rotation. The energy is a scalar,
and is invariant if the particle undergoes a rigid body rotation. To exclude rigid body rotation,
we can take the energy be a function of the Lagrange strain, W E . The nominal stress is
W E E MN
,
siK 
E MN FiK
or
W E 
siK  FiL
.
E LK
Exercise. For the neo-Hookean material, argue that the strain-energy function takes the
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form
W E  E JJ  pdet F   1 ,
where p is a Lagrange multiplier introduced to enforce incompressibility. Derive the relations
between the nominal stress and the the deformation gradient. Hint: Use the identity
 det F
 F T det F .
F
For example, see p. 223 of Holzapfel.
Infinitesimal, inhomogeneous deformation superimposed on a homogenous field.
Consider a homogenous deformation; that is, F is independent of X. When the homogenous
field is perturbed by an infinitesimal deformation, xX, t  , the deformation gradient is perturbed
by
xi X, t 
,
FiK 
X K
and the stress field is perturbed by
siK  K iKjLF jL .
Here K iKjL is the tensor of tangent modulus. For a hyperelastic material, this tensor can be
calculated from the strain-energy function:
s F   2W F 
K iKjL F   iK

.
F jL
F jL FiK
When F is independent of X, so is the tangent modulus. The increment of the field satisfies
momentum balance,
siK
 2xi

.
X K
t 2
Inserting the stress-strain relation into the equation of momentum balance, we obtain that
 2x j
 2xi
.
K iKjL

X K X L
t 2
This equation appears to be the same as that obtained by balancing momentum in the reference
configuration. Be wise, linearize.
Waves and bifurcation. When a bar is subject to uniaxial tension, homogenous
deformation satisfies all governing equations. However, the state of homogeneous solution may
not be the only solution. Necking is an alternative solution, a bifurcation from the homogenous
deformation. In the previous analysis, we have just considered the homogenous deformation, and
have identified the onset of necking with the axial force in the force-strain curve peaks. Now we
wish to push the analysis a step further: we wish to consider the possible small perturbation from
a homogenous deformation.
Consider an infinite medium. Assume a plane wave with unit normal vector N and speed
c. The disturbance takes the form
xX, t   af N  X  ct  ,
where a is the direction of the displacement, and f is the profile of the wave.
The equation of motion becomes
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K iKjL N K N La j  c 2ai .
This is an eigenvalue problem. When the tangent modulus is positive-definite, we can find three
distinct plane waves.
An essential difference can lead to significant consequences. The tensor of tangent
modulus depends on the state of homogeneous deformation, and may no longer be positivedefinite. When the tangent modulus reaches a critical condition
det K iKjL N K N L   0
in some direction N, we can find a nonvanishing displacement a at c = 0. This equation may
be regarded as a 3D generalization of the Considère condition, which is the condition for the
onset of bifurcation.
Exercise. An infinite body of an incompressible material cannot support longitudinal
wave, but can support shear wave. Assume that the body is in a homogeneous state of uniaxial
stress. A small disturbance propagates long the axis of stress as a shear wave. Determine the
speed of this shear wave.
Considère reconsidered. A bar is in a homogenous state of uniaxial tension. Derive the
equation of motion that evolves infinitesimal disturbance along the bar. Use the equation to
examine the Considère condition.
Let the bar be in the homogenous deformation of stretch  . Superimposed on the
homogenous deformation is an inhomogeneous deformation x X , t  . The increment in the
stretch is
x X , t 
 
X
Let the increment in the nominal stress be s X , t  . Momentum balance requires that
s X , t 
 2x X , t 

.
X
t 2
We specify the material law by a function s  . Thus, the tangent modulus K is defined by
s  K  
Because the unperturbed state is homogeneous, K   is independent of X and t.
The equation of motion becomes
 2x X , t 
 2x X , t 
K  

.
X 2
t 2
This is the familiar wave equation that evolves the perturbation x X , t  . When K    0 , the
disturbance can propagate in the bar at velocity
K  
c
.

When K    0 , a time independent inhomogeneous deformation becomes possible.
We can express the tangent modulus in a familiar form. Recall the relation between the
nominal stress and strain
  
s  
.

The tangent modulus is
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K   
The condition K    0 corresponds to
Z. Suo
ds  1  d      
 

.
d
  d
 
d     

.
d

This is the same as the Considère condition.
A column subject to a compressive axial force. A column, length L, is simply
supported at the two ends. When the column has a small deflection w , we balance the moment
in the current state:
M  Pw .
(We will miss this term if we balance the moment in the reference configuration.) The momentcurvature relation is
2w
M   EI
.
X 2
The equilibrium equation becomes
2w
EI
 Pw  0 .
X 2
The function w  sin X / L satisfies the equilibrium equation and the boundary conditions,
provided that the axial force is given by
EI
Pc   2 2 .
L
This is the Euler condition for bucking.
If the column vibrates in the transverse direction, the equation of motion for deflection
w X , t  is
EI
Try the solution of the form
4w
2w
2w

P



A
.
X 4
X 2
t 2
 X 
w X , t   sin 
 sin t .
 L 
We obtain the frequency
EI 
P

.
1

AL4  Pc 
Consequently, the frequency decreases when the compressive force increases, and approaches
zero when the axial force approaches the Euler condition.
Demonstrate the effect of axial force on frequency in class using a column with a variable
length.
 2
Lagrangian vs. Eulerian formulations. The formulation using the material coordinates
X is known as the Lagrangian formulation. The inverse function, Xx, t  , tells us which material
particle is at position x and time t. The formulation using the special coordinates x is known as
the Eulerian formulation. In the above, we have used the Eulerian formulation to study
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cavitation because we know the current configuration. In general, the current configuration is
unknown, and we have used the Lagrangian formulation to state the general problem. In the
following, we list a few results related to the Eulerian formulation.
Time derivative of a function of material particle. At time t, a material particle X
moves to position xX, t  . The velocity of the particle is
xX, t 
V X, t  
.
t
If we would like to use x as the independent variable, we change the variable from X to x by
using the function Xx, t  , and then write
vx, t   VX, t  .
Let GX, t  be a function of material particle and time. For example, G can be the
temperature of material particle X at time t. The rate of change in temperature of the material
particle is
G X, t 
.
t
This rate is known as the material time-derivative. We can calculate the material time-derivative
by an alternative approach. Change the variable from X to x by using the function Xx, t  , and
write
g x, t   GX, t 
Using chain rule, we obtain that
GX, t  g x, t  g x, t  xi X, t 
.


t
t
xi
t
Thus, we can calculate the substantial time rate from
G X, t  g x, t  g x, t 


vi x, t  .
t
t
xi
In particular, the acceleration of a material particle is
v x, t  vi x, t 
ai x, t   i

v j x, t  .
t
x j
Rate of deformation. Spin. Now we would like to examine the true strain in three
dimensions. Recall the definition of the true strain in one dimension. When the length of a bar
changes by a small amount from l t  to l t  t  , the increment in the true strain is defined as
  l / l . Consider a material element of line dX  MdL in the reference state. In the
deformed state, the element becomes
dx  mdl ,
where l is the length of the element, and m is the unit vector in the direction of the element. The
length of a material element of line at time t is
dl 2  dxi dxi
Take time derivative, and we obtain that
 dl 
dl
 dvi dxi .
t
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Note that
dvi 
and
Z. Suo
vi x, t 
dx j
x j
 dl  

.
dlt
t
Thus,
 M X, t  vi x, t 

mi m j
t
x j
This is rate of natural strain of a material element. Only the symmetric part
1  v x, t  v j x, t  
d ij x, t    i


2  x j
xi 
will affect the strain rate. This tensor is known as the rate of deformation.
Another useful tensor is
1  v x, t  v j x, t  
 ij   i

.
2  x j
xi 
This tensor is known as spin.
Rigid-body motion. Consider a rigid body motion
xi X, t   liK t X K  ci t  ,
where the vector ci t  represents a rigid-body translation, and the tensor l iK is an orthogonal
tensor, namely,
liK liL   KL .
Taking time derivative, we obtain that
dl
dl
liK iL  liL iK  0
dt
dt
Multiply xi X, t   liK t X K  ci t  by liL on both sides, and we obtain that the inverse
function:
X L  liL t xi  ci t  .
The velocity field is
dl t  dc t 
Vi X, t   X K iK  i .
dt
dt
The velocity as a function of current position is
dl jK t  dci t 
vi x, t   liK t  x j  c j t 

.
dt
dt
The gradient of the velocity field is
vi x, t 
dl t 
 liK t  mK .
xm
dt
Consequently, the rate of deformation vanishes, d ij  0 , and the spin is

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ij t   liK t 
Z. Suo
dl jK t 
.
dt
In particular, if we set the reference state at time zero, liK 0   iK , we obtain that
dl ji 0
.
ij 0 
dt
Jaumann rate. When a body undergoes a rigid-body rotation, the components of the
stress remain unchanged in the coordinate rotating with the body. However, the components of
the stress in a fixed coordinate do change. For example, a bar under uniaxial tension rotates is
initially along the x1 axis, and then rotates to be long the x 2 axis. The rotation changes the
components of the stress.
Let us calculate the rate of change in the components. Let xi be a set of coordinates fixed
in space, and x' be a set of coordinates rotating with the body. The two sets of coordinates are
related by
x'  li t xi ,
where l i is an orthogonal tensor. At time t = 0, the two sets of coordinates coincide, so that
li 0   i .
Let the stress components in the fix coordinates be  ij t  at t. The stress components in
the rotating coordinates remain invariant:
 ' t     0 .
The stress components in the fixed coordinates relate to those in the rotating coordinates by the
usual rules:
 ij t   li t l j t  ' t  .
Inserting  ' t     0 into the above expression, we obtain that
 ij t   li t l j t   0 .
Take time derivative, and we obtain that
dl j t 

dl t  
d
 ij t    li t 
 l j t  i   0
dt
dt
dt 

At time zero, li 0   i , so that
d
 ij   j 0 i 0   i 0 j 0 .
dt
There is nothing particular when we call time zero, so that the above rate is valid for any time. It
says the components of stress change even when the material element is under a rigid body
rotation.
Such a rate is unsuitable to enter material law. One modification is to remove this time
change from the stress rate, and define
d ij
 ijJ 
  j  i  i  j .
dt
This expression is known as the Jaumann rate. Such an expression and its variants have been
used to formulate rate dependent material laws.
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Principle of virtual work. The true stress tensor. Let b be the force acting on a
material element of volume in the current state divided by the volume of the element in the
current state. Let t be the force acting on a material element of area in the current state divided
by the area of the element in the current state. When the body undergoes a virtual motion, a
material particle X moves from a position x to a position x + u. The virtual displacement field
ux Associated with this virtual motion, the applied forces do virtual work
 b u dv   t u da .
i
i
i
i
The first integral extends over the volume of the body in the current state. The second integral
extends over the surface of the body in the current state.
Let  be the mass of a material element of volume divided by the volume of the element
in the current state. Associated with the virtual motion, the inertial forces do virtual work
  au i dv .
The integral extends over the volume of the body in the reference state. the acceleration is given
before.
For a bar subject to a uniaxial state of stress, where the length of the bar increases by l ,
the work done by the axial force is Pdl . Recall that P  a and dl  ld , so that the work is
Pdl  ald . Note that al is the current volume of the bar. Thus, the work done by the force in
the current state divided by the volume in the current state is d . We next generalize this idea
to the three dimensions. Associated with the virtual motion, the virtual natural strain is
1  u i x  u j x  
 ij x   


2  x j
xi 
Recall that in 1D the nominal stress is work-conjugate to stretch. Define the true stress
tensor by the work-conjugate to the tensor of natural strain, such that the internal virtual work is
  ijij dv .
The integral extends over the volume of the body in the current state.
We insist that for any virtual deformation x the principle of virtual work (PVW) hold
true:
  ijij dV   bi  ai ui dV   t iui dA .
The PVW defines the tensor of nominal stress, and is a weak statement of momentum balance.
As usual, the PVW is equivalent to a set of PDEs. We find that the tensor of nominal
stress obeys that
 v x, t  vi x, t 

 ij x, t 
 bi x, t    x, t  i

v j x, t  ,
x j
x j
 t

in the volume of the body, and
 ij n j  t i
on the surface of the body. These are familiar equations used in Fluid Mechanics.
Deforming elements of volume and area. Let dV  dX 1dX 2 dX 3 be an element of
volume in the reference state. After deformation, the line element dX 1 becomes dx1 , with
components given by
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dxi1  Fi1 dX 1 .
We can write similar relations for dx 2 and dx 2 . In the current state, the volume of the element
is
dv  dx1  dx 2  dx 3 .
We can confirm that this expression is the same as
dv  detFdV .
For an incompressible material, det F   1 .
Let an element of area be NdA in the reference state, where N is the unit vector normal
to the element and dA is the area of the element. After deformation, the element of area
becomes nda in the current state, where n is the unit vector normal to the element and da is the
area of the element. An element of length dX in the reference state becomes dx in the current
state. The relation between the volume in the reference state and the volume in the current state
gives
dx  nda  detFdX  NdA ,
or
dX K FiK ni da  det F dX K N K dA .
This relation holds for arbitrary dX , so that
FiK ni da  det F N K dA ,
or, in vector form,
F T da  det F dA .


Relations between alternative measures of stress. Recall the definition of the nominal
stress
virtual work in the curent state
 siK FiK ,
volume in the reference state
and the definition of the true stress
virtual work in the curent state
  ij ij .
volume in the current state
Equating the virtual work on an element of material, we obtain that
 ij ij dv  siK FiK dV .
Because the true stress is a symmetric tensor, we obtain that
u i x 
 ijij   ij
.
x j
Also note that
U i X  u i x  x j X, t  u i x 
FiK 


F jK X, t  ,
X K
x j
X K
x j
and that dv / dV  det F . Insisting that the two expressions for the virtual work be the same for
all virtual motion, we obtain that
siK F jK
.
 ij 
det F 
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An alternative derivation of the above relation is as follows. Let t i be the force acting on
an element of area in the current state divided by the area of the element in the current state.
Thus
Ti dA  t i da .
Recall siK N K  Ti and  ij n j  t i , as well as F jK n j da  det F N K dA . We obtain the relation
between the true and the nominal stress.
.
Define the tensor of the second Piola-Kirchhoff stress, S KL , by
virtual work in the curent state
 S KL E KL .
volume in the reference state
This expression defines a new measure of stress, S KL . Because E KL is a symmetric tensor, S KL
should also be symmetric. Recall that we have also expressed the same virtual work by siK FiK .
Equating the two expressions for the virtual work,
S KL E KL dV  siK FiK dV ,
1
1
and recalling that E KL  FiK FiL   KL  , we obtain that E KL  FiLFiK  FiK FiL  and
2
2
siK  S KL FiL .
We also have the relation
S KL F jK FiL
.
 ij 
det F 
Exercise. Formulate the principle of virtual work using the 2nd Piola-Kirchhoff stress and
the Largrange strain. Reduce the PVW to PDEs.
Chatting must be addictive and healthful, probably not just for online teenagers. With
reluctance, we will have to stop chatting about all these alternatives, and look for matters of
consequence.
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