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. Written by the man who taught me
the subject in 1985 in Xian Jiaotong University.
黄筑平，连续介质力学基础，高等教育出版社， 2003.
G.A. Holzapfel, Nonlinear Solid Mechanics, Wiley, 2000.
C. Trusdell and W. Noll, The Non-linear Field Theories of Mechanics, 3rd edition, Springer, 2004.
Be wise, linearize. Following the advice of George Carrier, we have stayed linear in
this course so far. We have been mostly looking at infinitesimal deformation and Hookean
materials. We have mixed the three ingredients: deformation geometry, force balance, and
material model. The resulting theory is linear. We have learned fascinating and useful
phenomena: stress concentration around a hole, vibration of a beam, refraction of a wave, etc.
We have also learned to use commercial finite element code to analyze phenomena with great
complexity. We can go on with this linear theory and do a lot more.
In the remaining two weeks of this course, however, we wish to go nonlinear, hopefully
also with wisdom. We will refine the three ingredients by considering finite deformation and
non-Hookean materials. We can mix the refined ingredients, or mix a refined ingredient with
unrefined ones. For example, as we have already done, a viscoelastic material is non-Hookean,
but deformation of such a material can be infinitesimal. We will outline basic ideas of finite
deformation, and describe phenomena that show how finite deformation makes difference.
Finite deformation. When a structure deforms, each
state of deformation must obey Newton’s law. This principle has
often been violated in our past work. For example, in analyzing
a truss, we have balanced forces in the deformed truss using the
shape of the undeformed truss, neglecting the deformation.
This negligence is often justified on the ground that reference state
deformation in most engineering structures is small. You might
think that a structure suffering a small strain, say less than 1%,
entitles you to neglect the change in shape when you balance
forces. A counter example is familiar to you. When a column
buckles, the strain in the column is indeed small, but you must
enforce equilibrium in the deflected state of the column.
The essential point is this: we must enforce Newton’s
W
law in every deformed state of a structure, and use this correct
principle as a basis to justify any simplification.
This
current state
consideration alone requires us to examine finite deformation,
even when the strain is small everywhere in the structure.
Non-Hookean behavior of materials. Hooke’s law says that the displacement of a
material is a linear function of the applied force. This law is an idealization, and obviously
contradicts with many daily experiences. Pursuing non-Hookean materials will take us in many
directions. Here are some examples.
Nonlinear elasticity. When a force causes a material to deform by a large amount, the
displacement may be a nonlinear function of the force. We will talk more about nonlinear
elasticity in this set of notes.
Viscoelasticity. We have already looked at time-dependent material behavior, such as
viscoelasticity.
Plasticity. After elastic deformation, upon unloading, a metal recovers its shape. After
plastic deformation, upon unloading, the metal does not fully recover its shape. Say we apply an
axial force to a metal bar, and measure its length. The experimental record of the force-length
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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, the
force-length relation is not one-to-one, but is history-dependent. The theory of plasticity will
be taught in a separate course.
Deformation in response to diverse stimuli.
P
loading
A material may deform in response to stimuli other
than a mechanical force. For example, we have
already analyzed deformation caused by change in
temperature. Indeed, a material may deform in
response to electric field, moisture, light, ionic
unloading
concentration, etc. We will pick up this fascinating
subject in a subsequent course, ES 241, Advanced
Elasticity, http://imechanica.org/node/725.
Thus, a structure may undergo finite,
l
history-dependent deformation in response to
diverse stimuli. A general approach is to evolve the
L
state of the structure incrementally, and enforce
the laws of physics in every state.
Why do we need stress and strain? We also proceed with our subject incrementally,
beginning with the simplest structure: a rod. When the rod is unstressed, the cross-sectional
area is A and the length is L . The rod is then subject to an axial force P, and deforms to crosssectional area a and length l. The experimentalist records, among other things, a curve relating
the force and the length.
The force-length curve gives us some idea of the behavior of the material. However, we
cannot use the curve directly to predict the behavior of another structure made of the same
material. To do so, we need to construct a material model that is independent of the specific
shape of the rod. As a step forward, we divide the elongation by the length of the rod, and divide
the force by the cross-sectional area of the rod. This practice is sensible provided the
deformation of the rod is homogenous, and the resulting stress-strain relation is independent
of the size of the rod.
This practice is economic. In combination with other ingredients, the stress-strain
relation can predict the behavior of any structure made of the same material, even when the
structure is of a shape other than a rod, and when deformation is inhomogeneous. We can
divide the body into many small volumes, and assume that each small volume undergoes
homogeneous deformation.
In the following paragraphs, we will first define strain and stress, and describe the stressstrain relations for particular materials. We then analyze several phenomena.
Strain. We take the initial, unstressed state as the reference state. Define the
engineering strain by the elongation divided by the length of the rod in the reference state:
elongation
l−L
e=
=
.
length in the reference state
L
Another type of strain 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, dε , as the increment in the
length of the rod divided by the current length of the rod, namely,
dl
increment in length
dε =
=
.
length in the current state l
This equation defines the increment of natural strain. Integrating from L to l, we obtain that
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⎛l ⎞
⎝L⎠
Yet one more definition of strain, the Lagrange strain, is defined as
2
⎤
1 ⎡⎛ l ⎞
η = ⎢⎜ ⎟ − 1⎥ .
2 ⎢⎝ L ⎠
⎥⎦
⎣
This definition is hard to motivate in one dimension. But if you take the view that any one-toone function of l / L is a definition of strain, then no motivation is needed.
Indeed, even the ratio l / L itself has a name: stretch. Define stretch 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
= .
λ=
length in reference state L
There seems to be no lack of ingenuity to invent yet another definition. All these
definitions contain the same information. For example, every one of the definition above is a
function of stretch:
λ2 − 1
e = λ − 1, ε = log λ , η =
.
2
Because they are all one-to-one functions, any one definition can be taken to be “basic” and then
used to express all the other definitions. For example, we can express all definitions in terms of
the engineering strain:
1
λ = e + 1, ε = log (1 + e ), η = e 2 + e
2
When deformation is small, namely, e << 1 , the three definitions of strain are approximately
equal, ε ≈ η ≈ e .
When we call e the engineering strain, we do not mean that e is quick-and-dirty, or
unscientific, or crude, or unnatural. We just need a name for the quantity (l − L )/ L . Later on,
we will describe motivations for some of the definitions, but these motivations are just elaborate
ways to express preferences of individual people. The motivations, however elaborate, should
not obscure a simple fact: you can use any one-to-one function of l / L to define strain.
ε = log⎜ ⎟ .
Stress. 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 applied to the rod in the
current state divided by the cross-sectional area of the rod in the reference state:
P
force in the current state
s=
= .
area in the reference state A
Define the true stress, σ , as the force in the current state divided by the area in the
current state, namely,
force in the current state P
= .
σ=
area in the current state a
You should not be misled by the name, or influenced by the prejudice of your teachers. The true
stress is no truer than the nominal stress. They are just different definitions of stress, and we
need to have different names to tell them apart.
Work conjugates. It helps if our definitions of stress and strain ease the calculation of
work. When the rod elongates from length l to length l + δl , the force P does work Pδl . Recall
one pair of definitions of stress and strain:
P = sA , l = λL .
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Consequently the work done by the force is
Pdl = ALsδλ .
Since AL is the volume of the rod in the reference state, we note that
increment of work in the current state
sδλ =
.
volume in the reference state
We say that the nominal stress and the stretch are work-conjugate. Also note that δe = δλ , so
that the nominal stress is also work-conjugate to the engineering strain.
Recall another pair of definitions of stress and strain:
P = σa , dl = ldε .
The work done by the force is
Pδl = alσδε .
Since al is the current volume of the bar, we note that
increment of work in the current state
.
σδε =
volume in the current state
That is, the true stress is work-conjugate to the natural strain.
Given any definition of strain, we can define its work-conjugate stress. For example,
consider the Lagrange strain, η . Subject to an increment in the strain, δη , the force acting on
the element does the work. Denote
increment of work in the curent state
Sδη =
.
volume in the reference state
This expression gives a new definition of stress, S. This new definition does not have a “simpler”
interpretation than its status as the work conjugate to the Lagrange strain.
Are these alternative definitions necessary? If we are liberal about the definition
of strain, without being obsessive about “motivating” each definition, we may as well take a
liberal view to call the work conjugate of each definition of strain a definition of stress, and name
our definition after a mechanician who can no longer protest.
We can easily invent and name definitions, but all the above definitions have already had
names:
• σ : true stress or the Cauchy stress.
• s : nominal stress or the first Piola-Kirchhoff stress.
• S : the second Piola-Kirchhoff stress.
Recall the relations among the definitions of strain:
λ2 − 1
e = λ − 1, ε = log λ , η =
2
We obtain the relations among their increments:
δλ
, δη = λδλ .
λ
Consequently, the three definitions of stress are related as
σa = sA, s = λS .
Are these alternative definitions necessary? I have my thoughts, but you should form
your own opinions. The question perhaps boils 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 agony in studying the subject by knowing
that textbooks of nonlinear continuum mechanics are full of alternatives at every turn. These
alternatives hide behind elaborate symbols and motivations. They may offer tantalizing sights.
You will just have to look beyond them for matters of substance.
δe = δλ , δε =
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Material models. Using a rod of given material, our experiment records a force-length
curve. We then convert the curve to a stress-strain relation by using some definitions of stress
and strain. Any definitions serve the purpose. All that matters is that we tell other people which
definitions we have used.
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:
σ = Kε N
where K and N 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 ε = 1 . Representative values for K are
100 MPa – 1GPa.
For a rubber, the stress-strain data may be fit to a relation known as the neo-Hookean
law:
s = µ (λ − λ−2 ) .
Representative values for µ are 0.1 MPa – 10 MPa.
When a material undergoes large deformation, volumetric strain is often negligible
compared to tensile strain. Consequently, such a material may be taken to be incompressible.
To see the difference in material behavior, we should use the same definition of stress
and strain for both materials. Recall that σ = sλ (for incompressible materials) and ε = log λ .
In terms of the nominal stress and the stretch, the two material models are
N
(
log λ )
s=K
, for metals
λ
s = µ (λ − λ−2 ) , for rubbers
In terms of the true stress and the natural strain, the two material models are
σ = Kε N , for metals
σ = µ [exp(2ε ) − exp(− ε )] , for rubbers
This change of variables makes evident a key difference: in tension, the stress in rubbers rises
more steeply than in metals. We will return to this difference shortly.
Exercise. Use the 3 ingredients outlined about to obtain the force-deflection relation
for the truss sketched in the beginning of the notes. Assuming that all three members of the
truss are made of rubber bands, and that deformation is large.
Necking in a bar. Let us mix the newly refined 3 ingredients to analyze a specific
phenomenon: necking. Subject to a tensile force, a metal bar first elongates to a homogenous
state of strain. At some level of strain, a small part of the rod thins down preferentially, forming
a neck. By contrast, a rubber band under tension usually deforms homogeneously, and does not
form a neck. We would like to understand these observations.
Here is a summary of the 3 ingredients:
• Deformation geometry: ε = log (l / L ) .
• Force balance: P = σa .
• Material model: For a metal bar under uniaxial tension, the true stress relates to the
natural strain as σ = σ (ε ) . We will assume that the volume of the rod is constant during
deformation, AL = al , or a = A exp(− ε ) .
Mixing the three ingredients, we obtain the force as a function of strain:
P = Aσ (ε )exp(− ε ) .
For metals, we use material model σ = Kε N and obtain that
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P = AKε N exp(− ε ) .
Plot P as a function of ε . In plotting the figure, I’ve set 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 σ = σ (ε ) . At the same time,
the elongation reduces the cross-sectional area, an effect known as geometric softening. For
small deformation, P → 0 as ε → 0 ; material hardening prevails, and the force increases as the
bar elongates. For large deformation, so long as the stress-strain relation σ (ε ) increases slower
than exp(ε ) , P → 0 as ε → ∞ , 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
ε
0.5
1
1.5
2
To determine the peak force, note that
dP
⎛ dσ (ε )
⎞
= A exp(− ε )⎜
−σ ⎟ .
d
dε
ε
⎝
⎠
Consequently, the force P peaks when the true stress equals the tangent modulus:
dσ (ε )
= σ (ε ) .
dε
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 strain
εc = 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:
σ = µ λ2 − λ−1 = µ [exp(2ε ) − exp(− ε )] .
Thus, at a large tensile strain, the true stress increases exponentially with the natural strain, so
that the force P = Aσ (ε )exp(− ε ) always increases with the strain. The rubber band will not form
a neck under uniaxial tension.
(
)
Exercise. Argue that necking sets in at a stretch where nominal stress is maximum.
Thus, in terms of the nominal stress and the stretch, the Considère condition is
ds (λ )
= 0.
dλ
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Apply this condition to metals and rubbers. This exercise shows that the same physical
condition may take different forms when we use different definitions of stress and strain. In this
case, the use of nominal stress simplifies the discussion.
Inhomogeneous, time-dependent deformation. Consider inhomogeneous and
time-dependent deformation of a bar. The 3 ingredients of solid mechanics take the following
forms.
Deformation geometry. Name each material particle by the coordinate X of its place
when the bar is in the reference state. In the current state t, the material particle X moves to a
place with coordinate x ( X , t ) . 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
λ=
.
length in reference state
Thus,
x ( X + dX , t ) − x ( X , t )
λ=
.
dX
The stretch in the bar is given by
∂x ( X , t )
λ=
.
∂X
Force balance. Let ρ ( X ) be the mass in the current state divided by the volume in the
reference state. The conservation of mass dictates that the mass density is time-independent.
We allow the variation of the density along the rod.
Let the nominal stress be s( X , t ) . Consider a differential element of the rod from X to
X + dX . Draw the free-body diagram of the segment when the rod is in the current state. On
the left end of the segment, the axial force is As ( X , t ) . On the right end of the segment, the axial
force is As ( X + dX , t ) . The mass of the segment is ρ ( X ) AdX , and the acceleration ∂ 2 x ( X , t )/ dt 2 .
Applying Newton’s second law to the segment of the rod in the current state, we obtain that
∂s ( X , t )
∂ 2 x (X , t )
.
= ρ (X )
∂X
∂t 2
Material model. The material is taken to be nonlinearly elastic with the stress-stretch
relation
s = g(λ ) .
This function is measured by applying to the bar a uniaxial force, which causes a state of
homogeneous deformation.
We assume that the same function is valid locally for
inhomogeneous deformation of the bar.
Wave in a pre-stressed bar. We now mix the three ingredients for a specific a
phenomenon. A bar is in a homogenous state of uniaxial stress. We then launch into the bar a
longitudinal wave of small amplitude.
Let the bar be in the homogenous deformation of stretch λ0 . Superimposed on the
homogenous deformation is an inhomogeneous, time-dependent field of displacement u( X , t ) .
In the current state t, the material particle X moves to a place with the coordinate
x ( X , t ) = λ0 X + u( X , t ) .
The stretch is the sum of the homogenous stretch due to the pre-stress and the an
inhomogeneous stretch due to the disturbance:
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∂u( X , t )
.
∂X
We assume that the amplitude of the disturbance is small, namely,
∂u( X , t )
<< 1 .
∂X
The material is taken to be nonlinearly elastic, characterized by a function between the
stress and the stretch s = g(λ ) . We expand the function s = f (λ ) into the Taylor series around
λ0 :
s(λ ) = g(λ0 ) + g′(λ0 )(λ − λ0 ) .
Because the deviation from the homogeneous stretch, λ − λ0 , is small, the expansion is carried to
the term linear in the deviation. The quantity f ′(λ ) is the tangent modulus of the material.
A mix of the three ingredients leads to the equation of motion:
∂ 2 u( X , t )
∂ 2 u( X , t )
ρ
.
g′(λ0 )
=
∂X 2
∂t 2
This is the familiar wave equation that evolves the disturbance u( X , t ) . Consequently, the
disturbance propagates in the bar at the seed:
g′(λ0 )
.
c=
λ = λ0 +
ρ
The wave slows down when the tangent modulus reduces.
Considère reconsidered. The above analysis is sensible only when the tangent
modulus is positive, g′(λ0 ) > 0 . When the tangent modulus vanishes, g′(λ0 ) = 0 , a timeindependent inhomogeneous deformation becomes possible. Observe that the condition of
vanishing tangent modulus is the same as the Considère condition.
A column subject to a compressive axial force.
The connection between
instability and dynamics also arises in a pair of more familiar phenomena: vibration and
buckling. Demonstrate in class the effect of axial force on frequency in class using a column with
a variable length.
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
+ Pw = 0 .
EI
∂X 2
The function w = sin (πX / L ) satisfies the equilibrium equation and the boundary conditions,
provided that the axial force is given by
EI
Pc = π 2 2 .
L
This is the Euler condition for bucking.
If the column vibrates in the transverse direction, the equation of motion for deflection
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w( X , t ) is
EI
∂2w
∂2w
∂4w
ρ
.
A
P
=
−
+
∂t 2
∂X 2
∂X 4
Try the solution of the form
⎛ πX ⎞
w( X , t ) = sin⎜
⎟ sin ωt .
⎝ L ⎠
This form satisfies the boundary conditions of simply supported ends. This form also satisfies
the equation of motion, provided the frequency is
EI ⎛
P
⎜1 −
4 ⎜
ρAL ⎝ Pc
⎞
⎟⎟ .
⎠
The frequency decreases when the compressive force increases, and vanishes when the axial
force approaches the critical load for buckling.
ω =π2
Inflation of a balloon. Aneurysm. When a spherical balloon is in the unstressed
state, the radius of the balloon is R, and the thickness of its skin is H. When the balloon is
subject to a pressure p, the radius of the balloon expands to r, and the thickness of the skin
reduces to h. We would like to relate the radius r and the pressure p.
Deformation geometry. In the current state, the stretches in the circumferential
directions are
λ1 = λ2 = r / R ,
while the stretch in the direction of thickness is
λ3 = h / H .
The material is taken to be incompressible, so that
hr 2 = HR 2 ,
or
λ3 = λ1−2 .
Force balance. The stress normal to the skin is between p and zero, and is small
compared to the stress in the circumferential directions. Consequently, we assume that the skin
is in a state of equal-biaxial stress:
σ 1 = σ 2 > 0, σ 3 = 0
Draw the free-body diagram of a half balloon in the current state, and balance the forces:
2πrhσ 1 = πr 2 p . Thus
2h
σ1 .
p=
r
Because r >> h , we confirm that σ 1 >> p .
Material model. Suppose that we have tested the material, and obtained the stressstretch relation in the state of uniaxial stress:
σ = g(λ ) .
The skin, however, is in a state of equal-biaxial stress: (σ 1 ,σ 1 , 0) . 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. Thus, we can superimpose (− σ 1 ,−σ 1 ,−σ 1 ) on the
material particle, so that the state of stress state in the particle becomes (0,0,−σ 1 ) . This is a
state of uniaxial stress. Thus, the material model becomes
− σ 1 = g(λ3 ) .
Mixing the three ingredients, we obtain that
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H −3
λ1 g(λ3 ) .
R
Assume the skin is made of a neo-Hookean material, with the stress-stretch relation in the state
of uniaxial stress:
g(λ ) = µ (λ2 − λ−1 ) .
p(λ1 ) = −2
Recall that λ1 = r / R and λ3 = λ1−2 . We obtain that
⎡⎛ r ⎞ −1 ⎛ r ⎞ −7 ⎤
⎢⎜ ⎟ − ⎜ ⎟ ⎥ .
⎝ R ⎠ ⎥⎦
⎢⎣⎝ R ⎠
This is the desired relation between the pressure and the radius. Sketch the function p(r / R ) .
When the balloon is unstretched, r / R = 1 , the pressure is zero, p(1 ) = 0 . As the balloon expands,
the material stiffens, but the skin thins. For the neo-Hookean material the above calculation
shows that p (∞ ) = 0. When the stretch is large enough, the geometric thinning prevails over
material stiffening, and the pressure reaches a peak. This peak is reached at the critical radius
rc = 7 1 / 6 R ≈ 1.38 R .
The peak pressure is
pc = 1.2 µH / R .
When the pressure approaches this peak value, this model predicts that the balloon will rupture.
The critical radius above seems to be too small compared to our experience with balloons.
It is likely that the materials for balloons do not obey the Neo-Hookean law, and stiffen more
steeply. In class a student pointed out that a balloon is usually not spherical, and has a
cylindrical tube. The tube may affect the behavior of inflation.
For the neo-Hookean material, the homogenous deformation under uniaxial stress is
stable for all stretches, but the homogenous deformation of a pressurized spherical balloon is
unstable when the stretch reaches 1.38.
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).
p=
2 Hµ
R
Exercise. When the balloon is subject to a constant pressure, allow the balloon to
oscillate by changing its radius with time. Determine the natural frequency. When does the
natural frequency vanish? Link this condition to the condition of instability.
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 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. We next analyze the phenomenon allowing finite deformation.
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. In the current state, the
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radius of the cavity become a, and the material particle R moves to a place at a distance r from
the center of the cavity. The function r (R ) characterizes the field of deformation. We assume
that the material is incompressible, so that
r 3 − a3 = R3 − A3 .
Consequently, once the current radius of the cavity a is determined, the entire field of
deformation r (R ) is determined. Because the independent variable in the equilibrium equation
is the current position r, we express R in terms of r:
1/3
R(r ) = (r 3 + A 3 − a 3 )
The hoop stretch is
r
λθ = .
R
Incompressibility dictates that the radial stretch relates to the hoop stretch as
λr = λθ−2 .
Insert the expression of R(r ) , we express the radial stretch as a function of r:
2/3
⎛ r 3 + A3 − a3 ⎞
⎟⎟ .
λr = ⎜⎜
r3
⎝
⎠
Force balance. Consider the material in the current state. 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 is subject to a state of triaxial stress:
σ r (r ),σ θ (r ) = σ φ (r ) .
We balance force for a material particle in the current state, so that
σ − σθ
dσ r (r )
+2 r
=0.
dr
r
Material Model. Suppose that we have tested the material. The stress-stretch curve in
the state of uniaxial stress is
σ = g(λ ) .
Around the cavity, a material particle is under a triaxial stress state: one radial component and
two hoop components: ( σ r , σ θ , σ θ ). 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 (− σ θ ,−σ θ ,−σ θ ) on the
material particle, so that the stress state of the particle becomes (σ r − σ θ ,0,0 ) . This is a uniaxial
stress state. Thus,
σ r − σ θ = g(λr ) .
Because of the spherical symmetry of this phenomenon, we will only need the stress-strain curve
determined in the state of uniaxial stress to analyze the process of cavitation.
Mixing the three ingredients, and integrating the equation of force balance from the
surface of the cavity to the remote point, we obtain that
∞
S = −2
∫
1
g(λr )dα
α
,
where α = r / a , and
⎛ α 3 + ( A / a )3 − 1 ⎞
⎟ .
λr = ⎜⎜
⎟
α3
⎠
⎝
This pair of equations relate the remotely applied stress S to the stretch of the cavity, a/A.
2/3
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5
Large Deformation
Small Deformation
4.5
4
3.5
S/µ
3
2.5
2
1.5
1
0.5
0
0
1
2
3
4
5
a/A
6
7
8
9
10
.
For the Neo-Hookean material, the stress-stretch relation in the state of uniaxial stress is
g(λ ) = µ λ2 − λ−1 .
The integral can be evaluated analytically, giving
(
)
−1
−4
5
1⎛a⎞
⎛a⎞
= − 2⎜ ⎟ − ⎜ ⎟
µ 2 ⎝ A⎠
2⎝ A⎠
Sketch this result and compare it with the Lame solution, which assumes Hookean material and
infinitesimal deformation. The cavity can expand indefinitely when the remote stress is still
finite. The remote stress needed to cause the cavity to expand indefinitely ( a / A → ∞ ) is called
the cavitation limit, and is
5
Sc = µ
2
for a Neo-Hookean material.
Cavitation limit also exist for other material models. To see this, set a / A → ∞ , so that
S
2/3
1 ⎞
⎛
.
3 ⎟
⎝ α ⎠
We need to see if the above integral is finite. As α → ∞ , the material is under hydrostatic state,
and λ r → 1 , so that g (λ r ) → E (λ r − 1) , were E is Young’s modulus. Thus
λr = ⎜ 1 −
2/3
⎛⎛
⎞
1 ⎞
2E
⎜
g(λr ) → E ⎜ 1 − 3 ⎟ − 1 ⎟ → − 3 .
⎜⎝ α ⎠
⎟
3α
⎝
⎠
The integral converges.
Cavitation was analyzed for metals by R. Hill (1949, Journal of Applied Mechanics16,
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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
cavitation in metals, see Y. Huang, J.W. Hutchinson and V. Tvergaard (1991, Journal of the
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 the 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.
Free energy. A 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 + δl , the force does work Pδl . We’ll only consider the isothermal conditions, so that
that we’ll drop the dependence on temperature. Assume that the material is elastic, so that the
work done by the force is converted into the Helmholtz free energy F of the rod, namely,
δF = Pδl .
Recall the definitions of the nominal stress s and the stretch λ :
P = sA , l = λL .
Define the free-energy density by
energy in the current state
W=
.
volume of the reference state
Divide the equation δF = Pδl by the volume of rod in the reference state, and we obtain that
dW = sdλ .
We can measure experimentally the nominal stress as a function of stretch, s (λ ) , and
then integrate the curve s (λ ) to obtain W (λ ) , known as the strain-energy function.
Alternatively, we can obtain an expression of W (λ ) from some theoretical considerations, and
then obtain s (λ ) by
dW (λ )
s (λ ) =
.
dλ
For the neo-Hookean model, s = µ (λ − λ−2 ) , so that
1
W (λ ) = µ (λ2 + 2λ−1 − 3) .
2
The number 3 is included to follow the convention W (1 ) = 0 .
For a nonlinearly elastic material obeying the power law, σ = Kε N , we have
N
s = K (log λ ) λ−1 , so that
K
(log λ )N +1 .
W (λ ) =
N +1
Stability of a state of equilibrium. A rod is pulled by a force. When the force is held
fixed, the rod may reach a state of equilibrium, where the strain is homogeneous in the rod. Will
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this state of equilibrium be stable? The material is taken to be elastic, characterized by the freeenergy function W (λ ) . Think of the fixed force as a weight hanging to the rod. The rod and the
weight in combination form a thermodynamic system, with the free energy
G = ALW (λ ) − PL (λ − 1 ) .
Here A and L are the cross-sectional area and the length of the rod in the reference state, and P
is the fixed force.
A stable stats of equilibrium minimizes the free-energy function G (λ ) . The state of
equilibrium is determined by dG (λ )/ dλ = 0 , or
dW (λ ) P
= .
dλ
A
The state of equilibrium minimizes G (λ ) if d 2G (λ )/ dλ2 > 0 , or
The critical condition is
d 2W (λ )
> 0.
dλ2
d 2W (λ )
= 0.
dλ2
This critical condition is reached at the peak of the curve s (λ ) .
Exercise. For the nonlinearly elastic material obeying the power law, σ = Kε N ,
determine the critical force Pc . Plot the function G (λ ) for a value of P larger than Pc . Plot the
function G (λ ) for a value of P smaller than Pc . Comment on these plots.
Exercise. Show that the free energy of the spherical balloon is
4π 3
G = 4πR 2 HW −
r p.
3
Derive the equation of equilibrium and the critical condition.
Other modes of instability. Of all possible states, the stable state of equilibrium
minimizes the free energy. This statement is a basic principle of thermodynamics. But the
principle can be hard to implement: it is hard to seek a minimizer among all possible states.
Say we try to find the tallest human being ever lived. It will not do if we just look among all
living Americans. We have to do a world-wide search. Should we extend our search beyond the
Earth? Should we go back to history to look among dead people? Then we do not have good
records. If we do decide to go beyond the Earth and back to history, it will be controversial at
what point we wish to call something human.
Thus, in searching for the state that minimizes the free energy, we will have to be clear
what we mean by all possible states. For the rod, we have restricted our search among all
states of homogenous deformation, characterized by the stretch λ . In particular, we have
excluded inhomogeneous field. We have formulated a problem solvable in the classroom, but
have lost our contact with the experimental observation of necking.
If we broaden our search, will we find other states of equilibrium? Yes. Here are some of
other models of instability.
Buckle. When we restrict the states to be homogenous, the homogenous state is stable
when the force is compressive. When we broaden the search to include bending states, the rod
buckles when the compressive force reaches a critical value.
Wrinkle. For rod not very long compared to its width, the homogenous state can be
unstable against perturbation of the shape of the surface, even when the compressive force is
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below the critical load of Euler buckling. M. A. Biot, Surface instability of rubber in compression,
Appl. Sci. Res. A 12, 168 (1963).
Crease. In all above analysis, the search has
been restricted among fields of displacement whose
amplitude only slightly deviates from the
homogeneous state. Experiments have shown that a
block of rubber form creases at even a lower
compressive forces. A. N. Gent, I. S. Cho, Surface
instabilities in compressed or bent rubber blocks,
Rubber Chemistry and Technology 72, 253 (1999).
This mode of instability has been analyzed recently
by E. Hohlfeld (PhD Thesis) and L. Mahadevan. In
creasing, the amplitude of the field deviates greatly
from the homogenous state, even though the spatial
extent of an initial crease can be small. This mode of
instability can also be analyzed by minimizing the free energy (Hong, Zhao, Suo).
Rupture. A rubber band in tension does not form a neck, but does rupture when the
stretch is very large.
Multiaxial stress. To broaden our search, we next explore states of multiaxial stress.
Of course, the only way to really know stress-strain relations is to run tests, 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 models. As an example, here we attempt to describe this
art for rubbers.
Hyperelastic materials. 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 λ1 , λ2 and λ3 , and the corresponding nominal stresses be
s1, s2 and s3. The complete stress-strain relations involve three functions s1 (λ1 , λ2 , λ3 ) ,
s2 (λ1 , λ2 , λ3 ) and s3 (λ1 , λ2 , λ3 ) . 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.
Consider a rectangular block of a material, lengths L1 , L2 and L3 in a reference state
state. In a current state, the block is subject to forces P1 , P2 and P3 on its faces, and is deformed
into a block of lengths l1 , l2 and l3 . When lengths in the current state change, each of the forces
does work, respectively, P1 dl1 , P2 dl2 , P3 dl3 . We assume that the sum of the work done by the
external forces equals the change in the free energy. Let W be the free energy of the block in the
current state divided by the volume of the block in the reference state. Thus,
L1 L2 L3 dW = P1 dl1 + P2 dl2 + P3 dl3
Dividing this equation by the volume of the block in the reference state, L1 L2 L3 , and recalling
the definitions of nominal stress and stretch, we obtain that
dW = s1 dλ1 + s2 dλ2 + s3 dλ3 .
The free-energy density is take to be a function of the three stretches, W (λ1 , λ2 , λ3 ) .
Consequently, the stresses are the partial derivatives:
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s1 =
Z. Suo
∂W (λ1 , λ2 , λ3 )
∂W (λ1 , λ2 , λ3 )
∂W (λ1 , λ2 , λ3 )
, s2 =
, s3 =
.
∂λ3
∂λ2
∂λ1
An elastic material whose stress-strain relation is derivable from a free-energy function is
known as a hyperelastic material. By contrast, an elastic material whose stress-strain relation is
not derivable from a free-energy function is called a hypoelastic material.
The nominal stress on one face is
P
s1 = 1 .
L2 L3
The true stress on the same face is
P1
.
l2l3
Consequently, the true stress relates to the nominal stress by
s
σ1 = 1 .
σ1 =
λ2 λ3
The true stress is derivable from the strain-energy function:
∂W (λ1 , λ2 , λ3 )
σ1 =
.
λ2λ3∂λ1
The other two components of true stress can be similarly obtained. Note that the true stress is
not work-conjugate to the stretch.
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.
A block of a material, of lengths L1 , L2 and L3 in the undeformed state, is deformed into
a rectangle of lengths l1 , l2 and l3 . If the material is incompressible, the volume of the block
must remain unchanged, namely, L1 L2 L3 = l1l2l3 , or
λ1 λ2 λ3 = 1 .
The incompressibility places a constraint among the three stretches: they cannot vary
independently. We may regard λ1 and λ2 as independent variables, so that
λ3 = (λ1 λ2 ) .
−1
Consequently, the strain-energy density is a function of two independent variables, W (λ1 , λ2 ) .
Consequently, only biaxial tests need be done to fully characterize the material.
−1
Inserting the constraint λ3 = (λ1 λ2 ) into the expression dW = s1 dλ1 + s2dλ2 + s3dλ3 , we
obtain that
dW = (s1 − λ1−2λ2−1 s3 )dλ1 + (s2 − λ2−2λ1−1 s3 )dλ2 .
Once the function W (λ1 , λ2 ) is determined, the stress-stretch relation is given by differentiation:
∂W (λ1 , λ2 )
∂W (λ1 , λ2 )
s1 − λ1−2λ2−1 s3 =
, s2 − λ2−2 λ1−1 s3 =
.
∂λ1
∂λ2
These relations, together with the incompressibility condition λ1 λ2 λ3 = 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
σ 1 = λ1 s1 , σ 2 = λ2 s2 , σ 3 = λ3 s3 .
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Consequently, the stress-strain relations become
∂W (λ1 , λ2 )
∂W (λ1 , λ2 )
σ 1 − σ 3 = λ1
, σ 2 − σ 3 = λ2
.
∂λ1
∂λ2
Free-energy functions for rubbers. The function W (λ1 , λ2 ) can be determined by
subjecting a sheet of rubber under various states of biaxial stress. 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=
µ
(λ
2
+ λ22 + λ23 − 3 ) .
2
1
Inserting the constraint λ3 = (λ1 λ2 ) , we obtain the stress-stretch relations:
−1
σ 1 − σ 3 = µ (λ21 − λ23 ) ,
σ 2 − σ 3 = µ (λ22 − λ23 ) .
A neo-Hookean material is characterized by a single elastic constant, µ . The constant
may be determined experimentally. For example, consider a rod in a state of uniaxial stress:
σ 1 = σ ,σ 2 = σ 3 = 0 .
Let the stretch along the axis of loading be λ1 = λ . Incompressibility dictates that the stretches
in the directions transverse to the loading axis be λ2 = λ3 = λ−1 / 2 . Inserting into the above stressstretch relation, we obtain the relation under the uniaxial stress:
σ = µ (λ2 − λ−1 ) .
Recall that stretch relates to the engineering strain as
λ =l/L =1+e .
When the strain is small, namely, e << 1 , the above stress-stretch relation reduces to
σ = 3µe .
Thus, we interpret 3µ as Young’s modulus and µ 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 µ = 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 (λ21 + λ22 + λ23 − 3) + c2 (λ1−2 + λ2−2 + λ3−2 − 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=
µ
∑ α (λ
n
αn
1
n
)
+ λα2 + λα3 − 3 ,
n
n
n
where α n may have any values, positive or negative, and are not necessarily integers, and µ n
are constants. The stress-stretch relations are
σ 1 − σ 3 = µn λα1 − λα3 ,
∑ (
= ∑ µ (λ
n
n
n
σ2 −σ3
αn
n
2
− λα3
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.
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See Erwan Verron (http://imechanica.org/node/4167) for a bibliography of the literature
on the mechanics of rubbers up to 2003.
Exercise. Study the inflation of a spherical balloon using various material models.
Exercise. Study the inflation of a cylindrical balloon. Can your solution interpret the
following observation? This experiment can be done readily with a party balloon. Chater, E.,
Hutchinson, J.W., "On the Propagation of Bulges and Buckles." Journal of Applied Mechanics,
51, 1-9 (1984).
http://www.doitpoms.ac.uk/tlplib/bioelasticity/demo.php
A body is a field of material particles. The above formulation is convenient if the
principal directions of stretch are obvious from the symmetry of the problem. We next
formulate the theory without invoking the principal directions.
We view a body as a field of material particles. Each material particle can be named any
way we want. For example, we can use an English letter, a Chinese character, or a color. To talk
about how close two particles are, we choose to name each material particle by its coordinate X
when the body is in a particular state. We will call this particular state of the body the
reference state. Often we choose the undeformed state the reference state. However, even
without external loading, a body may be under a field of residual stress. Thus, we may not be
able to always set the reference state as the undeformed state. Rather, any state of the body may
be used as a reference state.
From now on, we will use the phase “the material particle X” as shorthand for “the
material particle whose coordinate is X when the body is in the reference state”. When a body
deforms, each material particle moves from one place to another. At time t, the material particle
X occupies a place with coordinate x. The time-dependent field x (X , t ) characterizes the history
of deformation of the body. A central aim of continuum mechanics is to evolve the field of
deformation x (X, t ) by using an equation of motion.
The displacement, velocity, and acceleration of the material particle X at time t are,
respectively,
U = x (X , t ) − X ,
∂x (X , t )
V=
,
∂t
∂ 2 x (X , t )
A=
.
∂t 2
When a body deforms, does each material particle preserve its identity? We
have tacitly assumed that, when a body deforms, each material particle preserves its identity.
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 say that the deformation preserves the identity of each particle. If, however, after the
deformation the grid disintegrates, we should not assume that the deformation preserves the
identity of each particle.
Whether a deformation of body preserves the identity of the particle is subjective,
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depending 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-linked long molecules. If our grid is over a size much
larger than the individual molecular chains, then deformation will not disintegrate the grid. By
contrast, a liquid consists of molecules that can change neighbors. A grid painted 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 undergoing plastic deformation. 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 identity of each material particle.
In these notes, we will assume that the identity of each material particle is preserved as
the body deforms.
x(X,t)
X+dX
x(X+dX,t)
X
current state
reference state
Deformation gradient. Consider two nearby material particles in a body. When the
body is in the reference state, the coordinates of the two particles are X and X + dX . When the
body is in the current state at time t, the two particles occupy places x (X, t ) and x (X + dX , t ) .
The stretch is generalized to the ratio
x i (X + dX , t ) − x i (X , t )
.
dX K
This ratio is given a symbol:
∂x (X , t )
.
FiK = i
∂X K
The tensor F is known as the deformation gradient.
Stress. When a rod is pulled by a force, the nominal stress is defined as the force acting
on a cross section of the rod in the current state divided by the area of the cross section of the
rod in the reference state. We now generalize this definition to three dimensions. Consider a
material element of area. When the body is in the reference state, the material element of area is
normal to the axis X K . When the body is in the current state, the same material element of area
moves to a new place and a new orientation, and acting on the material element of area is a force.
Define the nominal stress siK as the component i of the force acting on the material element
of area in the current state divided by the area of the material element of area in the reference
state.
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s31
C=dX3
s11
B=dX2
s21
A=dX1
Force balance. 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 ρ (X )dV (X ) be the mass of
the material element of volume. Because the mass of the material element is invariant during
deformation, ρ (X ) is time-independent. By force balance we obtain that
∂siK (X , t )
∂ 2 x i (X , t )
+ Bi (X , t ) = ρ (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.
Material model.
deformation gradient:
Characterize a material by a relation between the stress and the
siK = giK (F ) .
For the time being, we do not place any restriction on the form of this relation. Such a material
is known as a hypoelastic material.
Define the tensor of tangent modulus by
∂g (F )
.
K iKjL (F ) = iK
∂F jL
For a hypoelastic material, the tangent modulus in general does not possess the major symmetry,
namely,
K iKjL ≠ K jLiK .
Summary of equations. The equations in one dimension are
∂x (X , t )
λ=
,
∂X
∂s ( X , t )
∂ 2 x (X , t )
+ B( X , t ) = ρ ( X )
,
∂X
∂t 2
s = g(λ ) .
The corresponding equations in three dimensions are
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FiK =
Z. Suo
∂x i (X , t )
,
∂X K
∂siK (X , t )
∂ 2 x i (X , t )
+ Bi (X , t ) = ρ (X )
,
∂X K
∂t 2
siK = giK (F ) .
Weak statement of force balance. Force balance is equivalent to the following weak
statement. The equation
⎛
∂∆ i
∂2 x ⎞
siK
dV = ⎜ Bi − ρ 2i ⎟∆ i dV + Ti ∆ i dA
⎜
∂X K
∂t ⎟⎠
⎝
holds for any function ∆ i (X ) . The integrals extend over the volume and surface of the body in
the reference state. This weak statement of force balance, in conjunction with the definition of
the deformation gradient,
∂x (X , t )
FiK = i
,
∂X K
and the material model,
siK = giK (F ) ,
forms the basis for the finite element method.
∫
∫
∫
Perturb an equilibrium state of finite deformation. A body is made of a material
characterized by a relation between the stress and the deformation gradient:
siK = giK (F ) .
The body is subject to a dead load B(X ) in the volume. One part of the surface of the body is
subject to dead load T (X ) , and the other part of the surface is held at prescribed place.
Let x 0 (X ) be a state of equilibrium compatible with the above prescriptions. Associated
with this state is the deformation gradient
∂x 0 (X )
,
FiK0 = i
∂X K
and the stress
0
siK
= giK (F 0 ) .
The stress field balances the forces
0
(X ) + B (X ) = 0
∂siK
i
∂X K
in the volume of the body, and
0
(X )N K (X ) = Ti (X )
siK
on the part of surface where the traction is prescribed. On the part of the surface where the
place is prescribed, x 0 (X ) is held at the prescribed place.
When the state of equilibrium, x 0 (X ) , is now perturbed by an infinitesimal deformation,
u(X , t ) , the field of deformation becomes
x (X , t ) = x 0 (X ) + u(X , t ) .
Associated with this field of deformation is the deformation gradient
∂u (X , t )
FiK (X , t ) = FiK0 (X ) + i
,
∂X K
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and the stress
siK (F ) = siK (F 0 ) + K iKjL (F 0 )
Z. Suo
∂u j (X , t )
.
∂X L
Here we have expanded the stress into the Taylor series up to terms linear in the perturbation.
Recall the equation of force balance
∂2 x
∂siK
+ Bi = ρ 2 i .
∂t
∂X K
Inserting the stress-strain relation into the equation of force balance, we obtain a homogeneous
partial differential equation:
∂u j (X , t ) ⎤
∂ ⎡
∂ 2 ui (X , t )
0
.
⎢ K iKjL (F )
⎥ = ρ (X )
∂X K ⎣
∂X L ⎦
∂t 2
The boundary conditions also homogeneous:
K iKjL (F 0 )
∂u j (X , t )
∂X L
NK = 0
on the part of the surface where the dead force T(X ) is prescribed, and
u(X , t ) = 0
on the part of the surface where the place is prescribed. These equations are solved subject to an
initial condition.
The equation of motion for the perturbation u(X , t ) looks similar to that in linear
elasticity. Thus, solutions in linear elasticity may be reinterpreted for phenomena of this type.
We give examples below.
Linear vibration around a state of equilibrium. We have already analyzed the
vibration of beam subject to an axial force. In general, we look for solution of the form
u(X , t ) = a (X )sin ωt ,
where ω is the frequency, and a(X ) is the field of amplitude. The equation of motion becomes
that
∂a j (X ) ⎤
∂ ⎡
0
2
⎢ K iKjL (F )
⎥ = − ρ (X )ω ai (X ) .
∂X K ⎣
∂X L ⎦
This equation, in conjunction with the homogeneous boundary conditions, forms an eigenvalue
problem. The solution to this eigenvalue problem determines ω and a(X ) for a set of normal
modes.
Plane waves superimposed on a homogeneous deformation in an infinite
body. 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 body. Assume a plane wave of small amplitude:
u(X , t ) = af (N ⋅ X − ct ) ,
where a is the unit vector in the direction of displacement, N the unit vector in the direction of
propagation, c the wave speed, and f the wave form.
The equation of motion becomes
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K iKjL (F 0 )N K N L a j = ρc 2 ai .
This is an eigenvalue problem. When the tangent modulus is positive-definite, we can find three
distinct plane waves.
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 along the axis of stress as a shear wave. Determine the
speed of this shear wave.
Bifurcation. An essential difference can lead to significant consequences. The tensor
of tangent modulus depends on the state of homogeneous deformation, K iKjL (F 0 ) , and may no
longer be positive-definite. When the tangent modulus reaches the 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.
Surface instability of rubber in compression, M. A. Biot, Appl. Sci. Res. A 12, 168
(1963). A semi-infinite block of a neo-Hookean material is in a homogeneous state of
deformation, with λ1 and λ2 being the stretches in the directions parallel to the surface of the
block, and λ3 being the stretch in the direction normal to the surface. The material is taken to
be incompressible, so that λ1 λ2 λ3 = 1 . The compression in direction 1 is taken to be more severe
than that in direction 2, so that when the surface wrinkles in one orientation, leaving λ2
unchanged. That is, the wrinkled block is in a state of generalized plane strain. Biot found the
critical condition for this instability:
λ3 / λ1 = 3.4
The analysis of this problem is analogous to that of the Rayleigh wave.
Polar decomposition. We next refine the theory. By definition, the deformation
gradient is a linear operator that maps one vector to another vector. Any linear operator can be
written as
F = RU ,
where R is an orthogonal operator, satisfying R T R = I , and U is calculated from
FT F = U 2 .
Thus, U is symmetric and positive-definite.
This theorem is given a geometric interpretation as follows.
. Deformation of a material element of line. Consider two nearby material
particles in a body. When the body is in the reference state, the coordinates of the two particles
are X and X + dX . The vector dX is a material element of line. When the body is in the
current state at time t, the two material particles occupy places x (X , t ) and x (X + dX , t ) , so that
the vector between the two particles is
dx = x (X + dX , t ) − x (X , t ) .
By definition, the deformation gradient F is a linear operator, such that
dx = FdX , dx i = FiK dX K .
The length of the element in the current state, dl, is given by
dl 2 = dx i dx i .
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Thus,
dl 2 = C KL dX K dX L ,
where
C = F T F , C KL = FiK FiL .
In the reference state, let dL be the length of the element, and M be the unit vector in the
direction of the element, so that
dX = MdL .
Recall that the stretch of an element is defined by
dl
λ=
.
dL
Thus
λ2 = C KL M K M L .
The tensor C allows us to calculate the stretch in any direction M. This tensor is known as the
Green deformation tensor.
Lagrange strain. Define the Lagrange strain E KL by
dl 2 − dL2 = E KL dX K dX L .
Recall
dL2 = dX K dX K ,
and
dl 2 = FiK FiL dX K dX L
We obtain that
1
(FiK FiL − δ KL ) ,
2
where δ KL = 1 when K = L, and δ KL = 0 when K ≠ L . The E KL tensor is symmetric.
The three tensors are related as
1
E = (C − I ), C = U 2
2
E KL =
Exercise. The Lagrange strain forms a link between finite deformation and the
infinitesimal deformation approximation. Let the displacement field be
U (X , t ) = x (X , t ) − X ,
Show that 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 balance force in the deformed state, as
remarked before.
Exercise. We can relate the general definition of the Lagrange strain to that introduced
in the beginning of the notes for a tensile bar. Consider a material element of line in a three
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dimensional body. In the reference state, the element is in the direction specified by a unit
vector M. In the current state, the stretch of the element is λ . We have defined the Lagrange
strain in one dimension by
λ2 − 1
.
η=
2
Show that the Lagrange strain of the element in direction M is given by
η = E KL M K M L .
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 stretches of the two
elements are λM and λN , and the angle between the two elements become
defined γ as the engineering shear strain. Show that
2 E KL M K N L
sin γ =
.
π
2
− γ . We have
λM λN
Deformation of a material element of volume.
Let dV = dX 1 dX 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
dx i1 = Fi 1 dX 1 .
We can write similar relations for dx 2 and dx 3 . In the current state, the volume of the element
is
dv = (dx 1 × dx 2 )⋅ dx 3 .
We can confirm that this expression is the same as
dv = det(F )dV .
Deformation of a material element of area. 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 .
Virtual work. Consider a small block in a body. In the reference state, the block is
rectangular, of lengths A, B, C . In the current state, the block deforms to some other shape.
Consider the free-body diagram of the block in the current state: a pair of forces si 1 BC act on the
pair of the material elements of area BC , a pair of forces si2CA act on the pair of the material
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elements of area CA, and a pair of forces si3 AB act on the pair of the material elements of area
AB .
When the block in the current state undergoes a virtual deformation δx i (X ) , the virtual
change in the deformation gradient is
∂δx i (X )
δFiK =
.
∂X K
The virtual change in the distance between the pair of the material elements of area BC is
δx i (X 1 + A, X 2 , X 3 ) − δx i (X 1 , X 2 , X 3 ) = δFi 1 A .
Consequently, the forces on the pair of the material elements of area do virtual work
si 1 BCδFi 1 A .
We can similarly calculate the work done by the forces acting on the pair of material elements of
area CA and those of AB . The sum of the virtual works by all the forces acting on the block is
si 1 BCδFi 1 A + si 2CAδFi 2 B + si 3 ABδFi 3C = ABCsiK δFiK .
Thus,
virtual work in the curent state
= siK δFiK .
volume in the reference state
Hyperelastic material. Let W be the free energy of the block in the current state
divided by the volume of the block in the reference state. As a material model, we assume that
the free-energy density is a function of the deformation gradient, W(F). When the block in the
current state undergoes a virtual deformation δx i (X ) , the free energy density changes by
∂W (F )
δW = W (F + δF ) − W (F ) =
δFiK .
∂FiK
For a hyperelastic material, the virtual work done by all the forces acting on the block
equals the virtual change in the free energy:
δW = siK δFiK .
A compression of the two expressions of the virtual change in the free energy gives that
∂W (F )
siK =
.
∂FiK
This equation relates the stress to the deformation gradient once the function W(F) is prescribed.
Incompressible material. Rubbers and metals can undergo large change in shape,
but very small change in volume. A commonly used idealization is to assume that such materials
are incompressible, namely,
det F = 1 .
In arriving at siK = ∂W (F )/ ∂FiK , we have tacitly assumed that each component of FiK can vary
independently. The condition of incompressibility places a constraint among the components.
To enforce this constraint, we replace the free-energy function W (F ) by a function
W (F ) − Π (det F − 1 ) ,
with Π as a Lagrange multiplier. We then allow each component of FiK to vary independently,
and obtain the stress from
∂
[W (F ) − Π(det F − 1)] .
siK =
∂FiK
Recall an identity in the calculus of matrix:
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∂ det F
= H iK det F ,
∂FiK
where H is defined by H iK FiL = δ KL and H iK F jK = δ ij .
Thus, for an incompressible material, the stress relates to the deformation gradient as
∂W (F )
siK =
− ΠH iK det F .
∂FiK
The Lagrange multiplier Π is not a material parameter. For a given boundary-value problem
rather, Π is to be determined as part of the solution.
The free-energy density is unchanged if the body undergoes a rigid-body
rotation. The free energy is invariant if the particle undergoes a rigid body rotation. Thus, W
depends on F only through U. Recall
U 2 = C = FT F .
Equivalently, W depends on F only through
1
E = (F T F − I ) .
2
We write
W (F ) = Ω(E )
Recall that the stress is work-conjugate to the deformation gradient. Using the chain rule
in differential calculus, we obtain that
∂W (F ) ∂Ω(E ) ∂E MN
siK =
=
,
∂FiK
∂E MN ∂FiK
or
∂Ω(E )
siK = FiL
.
∂E LK
The second Piola-Kirchhoff stress. Define the tensor of the second Piola-Kirchhoff
stress, S KL , by
virtual work in the curent state
= S KLδE 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 δFiK .
Equating the two expressions for the virtual work,
S KLδE KL = siK δFiK .
Recall definition of the Lagrange strain,
1
E KL = (FiK FiL − δ KL ) .
2
We obtain that
1
δE KL = (FiLδFiK + FiK δFiL )
2
and
siK = S KL FiL .
For a hypoelastic material, the free energy is a function of the Lagrange strain, Ω(E ) .
The second Piola-Kirchhoff stress is work-conjugate to the Lagrange strain:
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S KL =
Z. Suo
∂Ω(E )
.
∂E KL
Neo-Hookean material. For a neo-Hookean material, the free-energy function takes
the form
W (F ) =
µ
FiK FiK .
2
The material is also taken to be incompressible, namely,
det F = 1 .
The stress relates to the deformation gradient as
siK = µFiK − H iK Π .
Lagrangian vs. Eulerian formulations. The above formulation uses the material
coordinate X as the independent variable. The function x (X , t ) gives the place occupied by the
material particle X at time t. This formulation is known as Lagrangian. The inverse function,
X (x , t ) , tells us which material particle is at place x and time t. The formulation using the
spatial coordinate x is known as Eulerian. In the above, we have used the Eulerian formulation
to study 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 material 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 ) ∂x i (X , t )
=
+
.
∂t
∂t
∂x i
∂t
Thus, we can calculate the substantial time rate from
∂G (X , t ) ∂g(x , t ) ∂g(x , t )
vi (x , t ) .
=
+
∂t
∂t
∂x i
Let us applies the idea to the acceleration of a material particle. In the Lagrangean
formulation, the acceleration is
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∂V (X , t ) ∂ 2 x (X , t )
.
=
∂t 2
∂t
In the Eulerian formulation, the acceleration of a material particle is
∂v (x , t ) ∂vi (x , t )
+
ai (x , t ) = i
v j (x , t ) .
∂t
∂x j
A (X , t ) =
Conservation of mass. In the Lagrange formulation, the nominal mass density is
defined by
mass in current state
ρR =
.
volume in reference state
That is, ρ R dV is the mass of a material element of volume. A subscript is added here to remind
us that the volume is in the reference state.
In the Eularian formulation, the true mass density is defined by
mass in current state
ρ=
.
volume in current state
That is, ρdv is the mass of a spatial element of volume.
The two definitions of density are related as
ρ R dV = ρdv ,
or
ρ R = ρ det F .
Conservation of mass requires that the mass of the material element of volume be timeindependent. Thus, the nominal density can only vary with material particle, ρ R (X ) , and is
time-independent. By contrast, the true density is a function of both place and time, ρ (x , t ) .
Conservation of mass requires that
∂ρ (x , t ) ∂
[ρ (x, t )vi (x, t )] = 0 .
+
∂t
∂x i
When the material is incompressible, det F = 1 , we obtain that
ρ R (X ) = ρ (x , t ) .
True strain. 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 + δl , the increment in the true strain is defined
as
δl
δε = .
l
Now consider a body undergoing a deformation x (X , t ) . Imagine a virtual displacement
δu(x ) in the current state. Consider two nearby material particles in the body. When the body
is in the reference state, the coordinates of the two particles are X and X + dX . The vector dX
is a material element of line. When the body is in the current state at time t, the two material
particles occupy places x (X , t ) and x (X + dX , t ) , so that the vector between the two particles is
dx = x (X + dX , t ) − x (X , t ) .
Consider a material element of line. In the reference state, the element is dX = MdL ,
where dL is the length of the element, and M is the unit vector in the direction of the element.
In the current state, the element becomes dx = mdl , where dl 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
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dl 2 = dx i dx i
Subject to the virtual displacement, the above equation becomes
dlδ (dl ) = d (δui )dx i .
Note that
∂ (δui )
d (δui ) =
dx j
∂x j
and
δε =
Thus,
δε =
δ (dl )
dl
.
∂ (δui )
mi m j
∂x j
This is the virtual increment of the strain of the material element of line. Only the symmetric
part
1 ⎡ ∂ (δui ) ∂ (δu j )⎤
+
⎢
⎥
∂x i ⎥⎦
2 ⎢⎣ ∂x j
will affect the virtual increment of the strain of the material element of line. This tensor
generalizes the notion of the true strain, and is given the symbol:
1 ⎡ ∂ (δui ) ∂ (δu j )⎤
+
δε ij (x , t ) = ⎢
⎥.
∂x i ⎥⎦
2 ⎢⎣ ∂x j
Rate of deformation. Let v (x , t ) be the field of true velocity in the current state. The
same line of reasoning shows that the tensor
d ij (x , t ) =
1 ⎡ ∂vi (x , t ) ∂v j (x , t ) ⎤
+
⎢
⎥
∂x i ⎥⎦
2 ⎢⎣ ∂x j
allows us to calculate the rate of change the length of material element of line, namely,
∂[dl (X , t )]
1
= dij mi m j .
dl (X , t )
∂t
The tensor dij is known as the rate of deformation.
Force balance. The true stress obeys that
⎡ ∂v (x , t ) ∂vi (x , t )
⎤
∂σ ij (x , t )
+
+ bi (x , t ) = ρ (x , t )⎢ i
v j (x , t )⎥ ,
∂x j
∂x j
⎢⎣ ∂t
⎥⎦
in the volume of the body, and
σ ij n j = t i
on the surface of the body. These are familiar equations used in fluid mechanics.
Relation between true stress and nominal stress. Recall the definition of the
nominal stress
virtual work in the curent state
= siK δFiK ,
volume in the reference state
and the definition of the true stress
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virtual work in the curent state
= σ ijδε ij .
volume in the current state
Equating the virtual work on an element of material, we obtain that
σ ijδε ij dv = siK δFiK dV .
Because the true stress is a symmetric tensor, we obtain that
∂δui (x )
σ ijδε ij = σ ij
.
∂x j
Also note that
∂δU i (X ) ∂δui (x ) ∂x j (X , t ) ∂δui (x )
=
F jK (X , t ) ,
δFiK =
=
∂x j
∂X K
∂x j
∂X K
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
s F
σ ij = iK jK .
det (F )
Newtonian fluids. The components of the stress relate to the components of the rate
of deformation as
⎡ ∂v (x , t ) ∂v j (x , t ) ⎤
+
σ ij = η ⎢ i
⎥ − pδ ij ,
∂x i ⎦⎥
⎣⎢ ∂x j
where η is the viscosity. The material is assumed to be compressible:
∂vi (x , t )
=0.
∂x i
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