8
Review of
Continuum Mechanics:
Field Equations
8–1
Chapter 8: REVIEW OF CONTINUUM MECHANICS: FIELD EQUATIONS
TABLE OF CONTENTS
Page
§8.1
§8.2
§8.3
§8.4
§8.5
§8.6
§8.
§8.
Kinematic Equations
. . . . . . . . . . . . . . . . . .
§8.1.1
Polar Decomposition
. . . . . . . . . . . . . .
§8.1.2
Stretch Tensors . . . . . . . . . . . . . . . . .
Strain Measures
. . . . . . . . . . . . . . . . . . .
§8.2.1
Green-Lagrange Strain Measure
. . . . . . . . . . .
§8.2.2
Almansi Strain Measure
. . . . . . . . . . . . .
§8.2.3
Generalized Strain Measures: The Seth-Hill Family . . . . .
§8.2.4
Eulerian Strain Measures . . . . . . . . . . . . .
Stress Measures . . . . . . . . . . . . . . . . . . . .
§8.3.1
Cauchy Stress . . . . . . . . . . . . . . . . .
§8.3.2
PK2 Stress Measure . . . . . . . . . . . . . . .
§8.3.3
Other Conjugate Stress Measures
. . . . . . . . . .
§8.3.4
Recovering Cauchy Stresses
. . . . . . . . . . . .
§8.3.5
Stress and Strain Transformations . . . . . . . . . .
Constitutive Equations
. . . . . . . . . . . . . . . . .
Strain Energy
. . . . . . . . . . . . . . . . . . . .
§8.5.1
Strain Energy in Terms of GL Strains
. . . . . . . . .
§8.5.2
Recovering Stresses and Strains from Energy Density
. . .
§8.5.3
*Strain Energy in Terms of Seth-Hill Strains . . . . . . .
*Hyperelastic Solid Materials
. . . . . . . . . . . . . .
§8.6.1
*The Mooney-Rivlin Internal Energy
. . . . . . . . .
§8.6.2
*HSM Stress-Stretch Laws
. . . . . . . . . . . .
§8.6.3
*HSM Homogeneous Bar Extension . . . . . . . . . .
§8.6.4
*GL Strains as Quadratic Forms in Gradients
. . . . . .
Notes and Bibliography
. . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . .
8–2
8–3
8–3
8–3
8–5
8–5
8–7
8–8
8–10
8–11
8–11
8–11
8–12
8–12
8–14
8–15
8–15
8–15
8–16
8–16
8–17
8–17
8–18
8–19
8–21
8–21
8–22
§8.1
KINEMATIC EQUATIONS
This Chapter continues the review of nonlinear continuum mechanics. The focus is on the field
equations, which require the introduction of strain and stress measures. Some of the material is
transcribed from the 1966 thesis [213], which made use of the Biot measure.
§8.1. Kinematic Equations
This section continues §7.4 by connecting deformation measures to the displacement field.
§8.1.1. Polar Decomposition
Tensors F and G = F − I are the building blocks of various deformation measures used in nonlinear
continuum mechanics. The whole subject is dominated by the polar decomposition theorem:1 a
body particle motion can be expressed as a pure deformation followed by a rigid rotation, or as a
rigid rotation followed by a pure deformation. Mathematically this can be written as multiplicative
decompositions:
F = R U = V R.
(8.1)
Here R is a proper orthogonal rotation matrix (R−1 = RT , det(R) = +1) whereas U and V
are symmetric positive-definite matrices called the right and left stretch tensors, respectively.2
Accordingly, R U and V R will be called the right and left polar decompositions, respectively.
If the motion is√a pure rotation R, U = V = I. Premultiplying (8.1) by FT = U RT gives U2 √
= FT F
and thus U = FT F. Postmultiplying (8.1) by FT = RT V gives V2 = F FT and thus V = F FT .
Upon taking square roots3 to get either U or V, the rotation can be computed as either
R = F U−1 = F (FT F)−1/2 ,
R = V−1 F = (F FT )−1/2 F,
or
(8.2)
respectively. It is possible to compute R directly without getting U or V first, by using Grioli’s
theorem as explained in Appendix G.
Obviously U = RT V R and V = R U RT , which shows that U and V share the same eigenvalues.
These will be denoted by λ1 , λ2 and λ3 , which are called the principal stretches. On account of their
physical meaning, all 3 must be positive. For interpretation see Example 7.5 of previous Chapter.
The orthonormalized eigenvectors of U and V are denoted by φi and ψi , respectively, which
represent principal directions. Denoting Λ = diag[λ1 , λ2 , λ3 ], the spectral decompositions may
be written (in both indicial and matrix form) as
U = λi φi φiT = ΦT Λ Φ,
V = λi ψi ψiT = ΨT Λ Ψ.
(8.3)
Here Φ and Ψ are 3 × 3 proper orthogonal, direction-cosine matrices formed by stacking φi and
ψi as columns. Equating Λ = Φ U ΦT = Φ RT V R ΦT = Ψ V ΨT shows that Φ RT = Ψ and
Φ = Ψ R, so the principal directions are linked by the rigid rotation R. The columns of Φ and Ψ
define principal directions in the reference and current configurations, respectively. Accordingly,
they are often called the Lagrangian and Eulerian principal direction matrices.
A rigid translation may be interweaved (before, amidst or after) with the stretch-rotation sequence,
but is not reflected in the deformation gradient. See Example 7.7 of previous Chapter.
1
2
3
This is the extension of the polar representation of complex numbers to arbitrary dimensions. Discovered for general
matrices by Autonne in 1902 [42]; for the detailed story see [387, §3.0]. For elasticity its roots go back to Cauchy.
Some authors call U and V the right and left Biot-stretch tensors, e.g., [534].
The principal square root of a symmetric, positive-definite matrix such as F FT or FT F is unique.
8–3
Chapter 8: REVIEW OF CONTINUUM MECHANICS: FIELD EQUATIONS
§8.1.2. Stretch Tensors
Recall from §7.4.3 the definitions of C R and C L , which are further expanded with their square
roots:
C R = FT F = UT U = U2 ,
C L = F F T = V V T = V2 .
(8.4)
C R and C L are symmetric positive definite matrices called the right and left stretch tensors, respectively.4 To get U and V as their square roots one must solve for (directly or indirectly) their
eigensystem. The common eigenvalues of C R and C L are λ21 , λ22 and λ23 .
If R = I, C R = C L and U = V. This no-rotation case is illustrated in Example 8.1.
The invariants of either stretch tensor are denoted by I1 = λ21 + λ22 + λ23 , I2 = λ1 λ2 + λ2 λ3 + λ3 λ1 ,
and I3 = λ1 λ2 λ3 . These are often used in isotropic constitutive equations that account for finite
strains; for example those considered in §8.6.
Example 8.1. Simple extension. This is a continuation of Example 7.5 of previous Chapter. Since the
deformation gradient F shown in (7.19) is diagonal, the polar decomposition is trivial:
C R = FT F =
λ21
0
0
0
λ22
0
0
0
λ23
= CL ,
U=
C R = F = FT =
λ1
0
0
0
λ2
0
0
0
λ3
= V,
R = I.
(8.5)
Example 8.2. Simple shear. This is a continuation of Example 7.6. The deformation gradient F shown in
(7.21) is not diagonal, and neither are C R and C L :
C R = FT F =
1 + γ2
γ
0
γ
1
0
0
0
1
,
C L = F FT =
γ
1 + γ2
0
1
γ
0
The Green-Lagrange strains introduced in §8.2.1 are e X X = 12 γ 2 , e X Y = eY X =
eigenvalues of C R , which are the squares of the principal stretches, are
λ21 = 1 + 12 γ 2 + γ
1 + 14 γ 2 ,
1 2
λ22 = λ−2
1 = 1 + 2γ − γ
1 + 14 γ 2 ,
0
0
1
1
γ,
2
.
(8.6)
others zero. The
λ23 = 1.
(8.7)
Note that λ1 λ2 λ3 = 1 since the motion is isochoric. The associated orthonormalized eigenvectors are the
columns of the matrix


Φ=

1
2
1
2
−
1
γ/
2
+
1
γ/
2
4+
γ2
4+
γ2
−
0
1
2
1
2
+ 12 γ / 4 + γ 2
−
1
γ/
2
0
4+
γ2

0


0
(8.8)
1
Completing the polar decomposition and using it to evaluate other strain measures is the subject of Exercise 8.3.
4
In the literature the notation C = FT F and B = F FT is more common. Occasionally A is used for FT F, e.g., [551],
while B−1 is denoted by c in pre-1970 publications. However B is reserved here for the strain-displacement matrix in
finite elements. Some authors call C R and C L the Green and Cauchy stretch tensors, respectively, for historical reasons:
C L (actually its small-strain inverse) was introduced by Cauchy in 1827 and C R by Green in 1841. The names Piola
−1
and Finger are often associated with C−1
R and C L , respectively.
8–4
§8.2
STRAIN MEASURES
Example 8.3. Two-Dimensional Decomposition. For the 2D plane strain case, in which λ3 = 1, the polar
decomposition can be easily worked out in closed form. Let the rotation angle about Z be ψ (positive CCW).
Denote c = cos ψ, s = sin ψ, and t = s/c = tan ψ. The full form of the product F = R U can be written
F=
1 + g11
g21
0
g12
1 + g22
0
0
0
1
=
c
s
0
−s
c
0
0
0
1
B
1 + e11
B
e12
0
B
e12
B
1 + e22
0
0
0
1
= R U.
(8.9)
Here gi j are displacement gradients, whereas the eiBj are the Biot strain components introduced formally in
§8.2.3 A simple hand computation [95, pp. 8–9] yields
t=
B
e22
g21 − g12
,
2 + g11 + g22
c= √
= g22 c − g12 s + c − 1,
1
,
s= √
t
,
B
e11
= g11 c + g21 s + c − 1,
1+
1+
B
B
1
e12 = e21 = 2 (g21 + g12 ) c + 12 (g22 − g11 ) s.
t2
t2
(8.10)
One may verify that this R minimizes ||F − R|| F , in which ||.|| F denotes the Frobenius matrix norm, if F
has full rank. This theorem was proven by Grioli in 1940; see [792, p. 290]. In the 3D case R depends on 3
independent rotational parameters, which complicates the computations.
Expanding (8.10) in Taylor series about g11 = g12 = g21 = g22 = 0 and retaining terms up to the second
order gives
B
2
2
= g11 + 18 (3g21
− 2g12 g21 − g12
),
e11
B
2
2
e22
= g22 + 18 (3g12
− 2g12 g21 − g21
),
(8.11)
B
B
= e21
= 12 (g21 + g12 ) + 14 (g11 g12 − g12 g22 + g21 g22 ).
e12
Example 8.4. Grioli’s Theorem. It may be verified that the rotation R minimizes ||F − R|| F , in which ||.|| F
denotes the Frobenius matrix norm, if F has full rank. This theorem was proven by Grioli in 1940; see [792,
p. 290].5 This property is important in establishng “a best-fit rotation matrix” in the corotational kinematic
description of nonlinear FEM, since it allows determination of R without need to find U first.
§8.2. Strain Measures
Principal stretches are fine as deformation measures, but to establish constitutive equations that
account for geometric nonlinearities it is convenient to introduce finite strain measures.6 Desirable
attributes for such measures are listed in Figure 8.1. These were enunciated by Hill in 1978 [381]
as ingredients in the construction of strain families. The last one (#4) is not strictly necessary, but is
convenient for smooth transition to linear elasticity as deformations become infinitesimally small.
To convert a stretch tensor to a strain tensor one substracts I from it or takes its log, so as to have
a measure that vanishes for rigid motions. Thus either U − I or log(U) are appropriate strain
measures. As discussed later in §8.2.3, some of these are difficult to express analytically in terms
of the displacement gradients because of the need of solving an eigenproblem that delivers the
principal stretches to get U. Two finite strain measures that bypass that hurdle are described next.
5
6
The extension to arbitrary matrices is discussed in Higham’s book [378, p. 197], which totally ignores Grioli’s work.
Truesdell [791, §15] argues the case for strains as follows: “Strain tensors are convenient for approximations, because
their vanishing is a necessary and sufficient condition for a rigid displacement, and consequently, since their physical
components are dimensionless, a nearly rigid displacement may be specified by a series expansion in the strains, in which
all terms beyond some specified order are neglected.”
8–5
Chapter 8: REVIEW OF CONTINUUM MECHANICS: FIELD EQUATIONS
1. Should be a symmetric second-order tensor
2. Must identically vanish for rigid motions
3. Should be a continuous, unique and monotonic function
of displacement gradients
4. Should reduce to the infinitesimal strain measure if the
deformations become "small"
Figure 8.1. Desirable attributes of finite strain measures (after R. Hill)
§8.2.1. Green-Lagrange Strain Measure
Traditionally the most popular strain measure used in FEM work has been the Green-Lagrange
(GL) strain tensor.7 Its three-dimensional expression in Cartesian coordinates is
eX X eX Y eX Z
eG = 12 (C R − I) = 12 FT F − I = 12 U2 − I = 12 (G + GT ) + 12 GT G = eY X eY Y eY Z ,
eZ X eZ Y eZ Z
(8.12)
This measure does not explicitly require U. Its indicial form in Cartesian coordinates is
∂u j
∂u k ∂u k
∂u i
G
1
ei j = 2
+
+
.
(8.13)
∂Xj
∂ Xi
∂ Xi ∂ X j
On passing to the engineering notation X 1 → X , X 2 → Y , X 3 → Z , the full form of (8.13) is
2 2 2 ∂u
∂u
∂u
∂u
X
X
Y
Z
e GX X =
+
+
+ 12
,
∂X
∂X
∂X
∂X
∂u Y
∂u Y 2
∂u Z 2
∂u X 2
G
1
+
+
+2
,
eY Y =
∂Y
∂Y
∂Y
∂Y
2 2 2 ∂u
∂u
∂u
∂u
Z
X
Y
Z
+ 12
+
+
,
e GZZ =
∂Z
∂Z
∂Z
∂Z
(8.14)
∂u Z
∂u Y ∂u Y
∂u Z ∂u Z
∂u Y
∂u X ∂u X
+
+ 12
+
+
= e GZY ,
eYGZ = 12
∂Z
∂Y
∂Y ∂ Z
∂Y ∂ Z
∂Y ∂ Z
∂u Z
∂u X
∂u Y ∂u Y
∂u Z ∂u Z
G
1
1 ∂u X ∂u X
+
+2
+
+
= e GX Z ,
eZ X = 2
∂X
∂Z
∂Z ∂X
∂Z ∂X
∂Z ∂X
∂u Y
∂u Y ∂u Y
∂u Z ∂u Z
∂u X
G
1
1 ∂u X ∂u X
+
+2
+
+
= eYGX .
eX Y = 2
∂Y
∂X
∂ X ∂Y
∂ X ∂Y
∂ X ∂Y
If the nonlinear portion (that enclosed in square brackets) of these expressions is neglected, one
obtains the infinitesimal strains x x , yy , zz , x y = 12 γx y , yz = 12 γ yz , and zx = 12 γzx , already
7
Introduced for finite strains by Green (1841) and St.Venant (1844). The name of Lagrange is attached because of the
connection to the Lagrangian kinematic description since it involves C R ; see §7.4.3. The minor variation “GreenLagrangian” is also in use. Many authors call it just the Green strain tensor. A few label it (more properly) as
Green-St.Venant, which has the virtue of rebalancing both sides of the Channel. For its derivation see Exercise 7.1.
8–6
§8.2 STRAIN MEASURES
encountered in linear elasticity [263]. For future use in finite element work we shall arrange the
components (8.14) as a strain 6-vector configured as

 
 

e1
eX X
eX X
eY Y
 e2  
  eY Y 
  
 

eZ Z
 e3  
  eZ Z 
e= =
=
,
 e4   eY Z + e Z Y   2eY Z 
  
 

e5
eZ X + eX Z
2e Z X
e6
e X Y + eY X
2e X Y
(8.15)
where the G superscript has been omitted. This operation is called tensor casting or just casting.
An Eulerian counterpart of (8.12) denoted by eG , is obtained by replacing C R by C L there to get
eG =
1
2
(C L − I) =
1
2
F FT − I = 12 (V2 − I) = 12 (G + GT ) + 12 G GT .
(8.16)
This “Green-Euler” measure is rarely used8 , but it must be chosen if the polar decomposition is
F = V R; that is, the body is first rotated and then stretched. See Example 8.5. If R = I, both
Green strain variants coalesce; this is always the case in 1D.
§8.2.2. Almansi Strain Measure
Another finite strain measure that does not require the explicit computation of U is the AlmansiLagrange strain, which is defined by
−2
1
e A = 12 (I − C−1
R ) = 2 (I − U ).
(8.17)
Like (8.12) this measure avoids the solution of an eigenproblem but requires inversion of F to get
−1 −T
the xi partials from F−1 , since C−1
.
R =F F
e A , is obtained by replacing C R by C L in
An Eulerian version of the Almansi strain,9 denoted by (8.17) to get
1
e A = 12 I − C−1
= 2 I − (F FT )−1 = 12 I − V−2 .
(8.18)
L
Its indicial form in Cartesian coordinates is
∂u j
∂u k ∂u k
∂u i
A
1
ei j = 2
+
−
.
∂x j
∂ xi
∂ xi ∂ x j
(8.19)
This variant is far more popular; in fact, the Lagrangian form is rarely used aside of its appearance
in the SH family introduced in §8.2.3. The Eulerian version is mandatory if the polar decomposition
is F = V R; that is, the body is first rotated and then stretched. See Example 8.5.
8
As of January 17, the only Google hit on “Green-Euler strain” is this Chapter.
9
Often called the Almansi-Hamel strain measure in the literature. Introduced for infinitesimal strains by Cauchy (1827),
and for finite strains by Almansi (1911) and Hamel (1912). For its derivation see Exercise 7.2.
8–7
Chapter 8: REVIEW OF CONTINUUM MECHANICS: FIELD EQUATIONS
Example 8.5. Combined stretch, rotation and rigid translation. This is a continuation of Example 7.7. The
right polar decomposition F = R U can be directly lifted from (7.24): the first and second square matrices
in the right hand side of that equation are R and U, respectively, and V can be computed as R U RT .Thus the
stretch tensors are
C R = FT F = U2 =
λ21
0
0
0
λ22
0
0
0
1
,
C L = F FT = V2 =
λ21 c2 + λ22 s 2
(λ21 − λ22 ) s c
0
(λ21 − λ22 ) s c
λ22 c2 + λ21 s 2
0
0
0
1
(8.20)
in which c = cos ψ and s = sin ψ. Note that V for arbitrary ψ is not diagonal if λ1 = λ2 . Having both stretch
tensors we can compute the Green-Lagrange strain tensor and its Eulerian counterpart:
eG =
1
2
λ21 − 1
0
0
0
λ22 − 1 0
0
0
0
, eG =
1
2
λ21 c2 + λ22 s 2 − 1
(λ21 − λ22 ) s c
0
(λ21 − λ22 ) s c
2 2
λ2 c + λ21 s 2 − 1
0
0
0
0
.
(8.21)
For the Almansi-Lagrange strain and its Eulerian counterpart we get
1 − λ−2
0
0
1
A
1
,
e = 2
0
0
1 − λ−2
2
0
0
0
2 2
2 λ1 λ2 − λ21 − λ22 + (λ21 − λ22 ) c2
(λ21 − λ22 ) s2
0
1
eA =
2 λ21 λ22 − λ21 − λ22 − (λ21 − λ22 ) c2 0 ,
(λ21 − λ22 ) s2
4 λ21 λ22
0
0
0
(8.22)
in which c2 = cos 2ψ and s2 = sin 2ψ. Both eG and e A are diagonal and thus reflect the initial stretch correctly.
On the other hand their Eulerian counterparts are garbage since they are contaminated by the rotation.
Do we conclude from this example that the Eulerian measures are useless? Not at all. If the motion is applied
eG and e A are diagonal
by rotation followed by stretch, the polar decomposition becomes F = V R. Now and reflect the stretch correctly, whereas their Lagrangian cousins are garbage. Conclusion: the choice of
“physically correct” finite strain measure is strongly linked as to how the multiplicative decomposition is built.
§8.2.3. Generalized Strain Measures: The Seth-Hill Family
The Green and Almansi strain measures are not the only ones used in geometrically nonlinear
analysis. Several others have been proposed for use in particular scenarios. To introduce them in
toto it is illuminating to look at the one-dimensional (1D) case first. Consider a line element of
lengths L 0 and L in the reference and current configurations, respectively. The length change is
L = L − L 0 . Define
λ = L/L 0 , λ̂ = L 0 /L , g = λ − 1 = L/L 0 , ĝ = 1 − λ̂ = 1 − (1/λ) = g/(1 + g). (8.23)
We call λ, λ̂, g and ĝ the stretch, inverse stretch, displacement gradient , and complementary displacement gradient, respectively. Using (8.23) several 1D finite strain measures can be constructed.
Five popular ones are listed in the second column of the table in Figure 8.2. By inspection they
can be seen to be members of a one-parameter form, which defines the Seth-Hill family listed in
the sixth row. The parameter, called the index of the measure, is denoted by m.
The five instances are generated by setting m = 2, 1, 0, −1 and −2, respectively. The GreenLagrange and Almansi measures correspond to m = 2 and m = −2, respectively. The case m = 0
8–8
§8.2 STRAIN MEASURES
Name (m=index
in SH family)
1D finite strain measure
Green (m = 2)
eG =
1
2
(λ2−1) = g +
=
1
2
^2 ^2
Biot (m = 1)
eB
1 2
2g
Taylor series expansion
about g=0
Comments
g + g2/2
Widely used since the start of
geometrically nonlinear FEM
g
Generalization of "engineering
strain" to finite deformations.
Becoming increasingly popular
g − g2/2 + g3/3 − ...
Used in finite elastoplasticity.
Sometimes replaced by log-less
approximations (see below)
^
(1−λ )/ λ = (1−g^ 2/2)/(1−g)
=λ−1 =g
^
^
= 1/λ −1 = g/(1
− ^g)
Hencky (m = 0)
eH = log λ = log (1+g)
^
^
= − log λ = log (1/(1−g))
Swainger
(m = −1)
eS = 1−1/λ = g/(1 + g)
Almansi (m=−2)
eA = 12 (1−1/λ ) = (g + 12 g2 )/(1+g)
g − g2 + g3 − ...
Counterpart of Biot measure
g − 3g2/2 + 2g3 − ...
Counterpart of Green measure
g − (m−1) g2/2
Includes above five measures
for m = 2, 1, 0, −1, −2, resp.
^
= 1 − λ = g^
^
= 12 (1− λ ) =
1
2
^ 1 g
^ 2)/(1−g)
^
(g−
2
Seth-Hill family
(arbitrary m)
e(m) = m1 (λm−1) = m1 ((1+g)m−1)
Midpoint
eM = 2(λ−1)/(λ+1) = g/(1+g/2)
+ (m−1)(m−2)
^
g3/6
g − g2/2 + g3/4 − ...
(1,1) Pade approximant to
Hencky strain that avoid logs.
Not part of the Seth-Hill family
g − g2/2 + g3/2 − ...
Another log-free approximant to
Hencky strain; accuracy similar to eM.
Not part of the Seth-Hill family
^
^
^
= 2(1−λ)/(λ+1) = g/(1−g/2)
Bazant
eZ =
1
2
(eB + eS) = (g+g2/2)/(1+g)
− ...
Figure 8.2. A catalog of finite strain measures in the one-dimensional case. Yellow background:
specific instances of the Seth-Hill (SH) family, which is listed with green background. Blue
background: two measures that approximate the Hencky strain but do not belong to the SH family.
No distinction between Lagrangian and Eulerian versions are needed in 1D, since there is no rotation.
Thus those qualifiers are not attached to the Green and Almansi names.
requires taking a 0/0 limit, and log functions emerge. This Hencky strain or logarithmic strain is
used often in finite strain elastoplasticity as well as for foam materials.
1/2
The multi dimensional (tensor) versions are obtained through obvious mappings: λ → U = C R ,
−1/2
λ̂ → U−1 = C R , 1 → I, etc. This produces the forms given in the second column of Figure 8.3.
Here a “computability distinction” emerges. Forms obtained directly from C R = F FT or its
inverse do not require an eigensolution. This happens for m = 2 (Green-Lagrange) and m = −2
(Almansi-Lagrange). For other m values, the eigensolution of C R is usually required.
The midpoint and Bazant strains, listed in the last two rows, can be used as log-free approximants
to the Hencky strain. See Remark 8.3 below. Those two measures do not belong to the SH family,
but do fulfill the conditions set out by Hill in [380], which are summarized in Figure 8.1.
Remark 8.1. The Seth-Hill (SH) family was proposed by Seth [697] for one dimensional strain measures.
The idea was tensorially generalized by Hill [380], who defined explicit admissability conditions. Those are
summarized in Figure 8.1. Doyle and Ericksen had presented that family for arbitrary m almost a decade
earlier [196] but the idea, buried in an tensor-laced long chapter, did not attract attention. By contrast, Seth used
high-school mathematics in a short article, and got his point across. For references to the strain measure names,
see [95,551,658,791,792,877]. The history is confusing and naming attributions often do scant justice.
8–9
Chapter 8: REVIEW OF CONTINUUM MECHANICS: FIELD EQUATIONS
Name (m is index 2D and 3D tensor forms (only those dependent
in SH family)
on CR and U are listed). Note: Gsym= (G + GT)/2
Green-Lagrange e_G =
(m=2)
1
2
(U2 − I) = 12 (CR − I) = Gsym+ 12 GT G
Biot (m=1)
e_B = U − I = CR1/2 − I = ( I + 2Gsym+ G G)
Hencky (m=0)
T
_eH = log(U) = 12 log(CR )= 12 log( I+ 2Gsym+ G G)
T
−1/2
Swainger (m=−1) e_S = I − U−1= I − CR
1/2
−I
−1/2
= I − ( I+ 2Gsym + GT G)
−1
Almansi-Lagrange e_A = 12 (I − U−2) = 12 (I − CR )
(m=−2)
= 12 (I − (I + 2Gsym+ GT G)−2)
m
m/2
eM = 2 (U − I) (I + U)−1
Bazant
eZ = (eB + eS)/2 = (U − U−1)/2
Rational function of displacement gradients.
Easy to compute, widely used in FEM
Generalizes "engineering strains" to finite strains.
Needs eigenanalysis of CR to get U
Prone to singularities, so often replaced by log-free
approximations. Needs eigenanalysis of CR to get U
Counterpart of Biot. Rarely used in Lagrangian
form. Needs eigenanalysis of CR to get U
Counterpart of Green. Rarely used in Lagrangian
form. Needs inversion of CR but no eigennalysis.
Seth-Hill family _e(m) = m1 (U − I) = m1 (CR − I)
Midpoint
Comments
Family includes above 5 measures. All assume that
F = R U. If F = V R, replace CR by CL, and U by V
A (1,1) Pade approximant of eH , correct to 2nd order.
Avoids logs but needs U and an inversion.
Another log-free approximant to eH, slightly inferior
to eM (see Taylor expansions in previous figure)
Figure 8.3. A catalog of finite strain measures in multiple space dimensions. Yellow background:
specific instances of the Seth-Hill (SH) family, which is listed with green background. Blue background:
two measures that approximate the Hencky strain but do not belong to the SH family. Note: index m receives
various names in the literature, and is sometimes replaced by k = m/2.
Remark 8.2. No distinction between Lagrangian and Eulerian forms need to be made in 1D, so those qualifiers
are not attached to the Green and Almansi measures, as done instead in Figure 8.3. The same would be true
in multiple dimensions if R = I (no rotation) since if so C R and C L coalesce.
Remark 8.3. The Hencky strain measure has long been a popular choice in finite elastoplasticity as well
as some crystalline models; see the recent expository article [534]. It causes problems, however, in FEM
computations because of 0/0 singularities for zero strain. Accordingly log-free approximants such as those
listed in the last two rows of Figures 8.2 and 8.3 may be preferred. Those two agree with the Hencky measure
up to second order in the displacement gradients. Third order approximations are possible: a particular simple
one is the linear combination e B + e S /3 − eG /3.
§8.2.4. Eulerian Strain Measures
According to the definition in §7.4.3, a Lagrangian strain measure can be converted to its Eulerian
counterpart by substituting C R by C L or, equivalently, by changing U to V. Doing that for the SH
family gives
1 m/2
(8.24)
CL − I .
e(m) =
m
The choices m = 2 (Green-Euler measure) and m = −2 (Almansi-Euler measure, often called
Almansi-Hamel) have been briefly described before. When should these alternative forms be
8–10
§8.3
STRESS MEASURES
chosen? The “forking rules” are simple:
(1) Go right. If the motion is built using the right polar decomposition R U, i.e.,
first deform and then rotate, use a Lagrangian measure based on C R or U.
(2) Go left. If the motion is built using the left polar decomposition V R, i.e., first
rotate and then deform, use an Eulerian measure based on C L or V.
If the strains are infinitesimal although rotations may be finite (which is often the case in the CR
kinematic descrition of finite elements) the choice is irrelevant: only R counts.
§8.3. Stress Measures
As in the case of strains, multiple stress measures are used in geometrically nonlinear structural
analysis. We quickly review here the Cauchy (true) stress as well as the so-called conjugate stresses,
which are associated with specific finite strain measures.
§8.3.1. Cauchy Stress
The Cauchy or true stress is that acting in the current configuration and measured per unit area of
material there. The notation used in linear FEM [263] is retained for it:
σx x σx y σx z
σ = σ yx σ yy σ yz ,
(8.25)
σzx σzy σzz
σ = [ σx x σ yy σzz σ yz σzx σx y ]T = [ σ1 σ2 σ3 σ4 σ5 σ6 ]T .
in which σx y = σ yx , etc. Here σ denotes the Cauchy stress tensor written as a 3 × 3 matrix of its
Cartesian components in C, whence the use of {x, y, z} in (8.25). The non-underlined symbol σ
identifies the 6-vector cast form used in FEM formulations. Two closely related measures are the
back-rotated Cauchy stress σ R and the Kirchhoff stress σ̂, defined as
σ R = RT σ R,
σ̂ = J σ in which J = det F,
(8.26)
in which R is the rotation tensor of the polar decomposition (8.1). The tensor σ̂, which is often used
in metal plasticity, is also called the weighted Cauchy stress. All of these tensors are symmetric.
By definition the Cauchy stress reflects what is actually happening in the material (hence the “true”
qualifier). Thus it is clearly of interest to an engineer calculating safety factors. Unfortunately it is
not work-conjugate to the finite strain measures commonly used in FEM analysis.
§8.3.2. PK2 Stress Measure
Given a finite strain measure, a conjugate stress measure is one that corresponds to it in the sense of
virtual work. More precisely, the dot product of stress times the material derivative strain rate is the
internal power density. Since the GL measure is common in FEM implementations, its conjugate
stress measure is of primary interest. This happens to be the second Piola-Kirchhoff stress tensor,
8–11
Chapter 8: REVIEW OF CONTINUUM MECHANICS: FIELD EQUATIONS
which is symmetric.10 In the sequel it is often abbreviated to “PK2 stress.” In terms of its Cartesian
components in the reference configuration C0 , the PK2 stress tensor is denoted as
sX X sX Y sX Z
s = sY X sY Y sY Z ,
(8.27)
sZ X sZ Y sZ Z
s = [ s X X sY Y s Z Z sY Z s Z X s X Y ]T = [ s1 s2 s3 s4 s5 s6 ]T .
in which s X Y = sY X , etc. These components are referred to the reference configuration C0 , whence
the use of {X, Y, Z } in (8.27). The PK2 and Cauchy (true) stresses are connected through the
transformations
σ = J −1 F s FT .
(8.28)
s = J F−1 σ F−T = F−1 σ̂ F−T ,
In terms of the back-rotated Cauchy stress σ R = RT σ R we have s = J U−1 σ R U−1 . The detailed
transformation from σ to s and vice-versa, when both are cast as 6-vectors, may be implemented
as matrix-vector products for computational efficiency.
If the material is isotropic, s is coaxial with U, whereas the Cauchy stress σ is coaxial with V.
In general (curvilinear) coordinates PK2 is the contravariant stress tensor conjugate to the covariant
Green-Lagrange strain tensor.
Remark 8.4. The physical meaning of the PK2 stresses is as follows: si j are stresses “pulled back” to the
reference configuration C0 and referred to area elements there.
Remark 8.5. For an invariant reference configuration, PK2 and Cauchy (true) prestresses plainly coincide
(see previous Remark). Thus σ0 ≡ s0 in such a case. However if the reference configuration is allowed to
vary often, as in the UL description or in staged analysys, the two may differ.
§8.3.3. Other Conjugate Stress Measures
Conjugate stress measures for other members of the Seth-Hill strain measure family have been
obtained — one rather recently. Several are listed in the table of Figure 8.4. For derivation details
the reader is referred to [208,336,383,551]. The expression for m = 0, found first in 1987 [383] is
extremely complicated and thus omitted from the table.
The stress conjugate to the finite midpoint strain is unknown.
A question rarely addressed in the literature: which is the strain measure conjugate to the Cauchy
stress? Answer: it does exist, but it is not admissible in the sense of observer invariance [381].
§8.3.4. Recovering Cauchy Stresses
During an incremental solution process, a nonlinear FEM program generally carries along —
usually at Gauss points — one of the conjugate stress measures listed in Figure 8.4. Sometimes it
is necessary to recover the Cauchy (true) stress for tasks such as checking strength safety factors.
Should the conjugate stress be s (PK2) or s A (Almansi-Hamel) the recovery is straightforward: just
apply the congruential transformation given in the table. For example, σ = F s FT /J . But if it is
the Biot or Swainger stress, the task is more involved, requiring three steps:
10
The first Piola -Kirchhoff stress tensor is s P K 1 = J σ F−T = F s. Its transpose s N = J FT σ is called the nominal stress
tensor. Since both of these tensors are generally unsymmetric, they are rarely used in FEM.
8–12
§8.3
Name (coeff in
Finite strain tensor (a member Conjugate stress tensor
Seth-Hill family)
of Seth-Hill family)
Green-Lagrange _e =
(m = 2)
1
2
(U 2 − I) =
1
2
(CR − I)
1/2
−1
−T
_s = J F σ
_ F
T
−1
σ
s F
_= J F _
STRESS MEASURES
Comments
Second Piola-Kirchhoff tensor.
Abbreviated to PK2 in text
Biot (m = 1)
e = U − I = CR − I
_
s B = 2 ( _s U + U _s)
_
−1
R
= 12 J (U−1 σ
_RU )
_ + σ
T
R
Biot stress. In 2nd form, σ
_ = R σ_ R
is back-rotated Cauchy stress
Hencky (m = 0)
e H = log(U) =
Omitted (see text)
For isotropic material, σ_R . Extremely
complicated for anisotropic material
Swainger
(m = −1)
eS = I − U = I − C R
B
−1
Almansi-Hamel
(m=−2)
1
2
1
log(CR )
−1/2
−1
−1
1
s A U + U s A_)
_s S = 2 ( _
R
1
_ RU )
= 2 J (U σ
_ +σ
−1
e A = 12 (I − U−2) = 12 (I − C R )
_s A = J F σ
_ F
−1
_ = J −1 F _s A F −T
σ
T
Swainger stress, rarely used. 2nd form
in terms of back-rotated Cauchy stress
Almansi stress
The Biot stress is sometimes called the Jaumann stress. The names "Hencky stress", "Swainger stress" and
"Almansi stress" are introduced by pairing with the corresponding finite strain measures. The name "von Mises"
sometimes appears along with Hencky.
If the material is isotropic, stress tensors referred to the reference configuration are coaxial with U, whereas
those referred to the current configuration (e.g., the Cauchy stress tensor) are coaxial with V.
Figure 8.4. Catalog of conjugate stress measures for selected strain measures that are members of the Seth-Hill family.
That corresponding to m = 0 is extremely complicated for anisotropic material and was first published in [383].
(1) Form F and do the polar decomposition (8.1) to get U and R.
(2) Solve an algebraic Lyapunov-type equation of the form A X + X A = B for X. Specifically
Biot: U−1 σ R + σ R U−1 = 2J s B ,
Swainger: U σ R + σ R U = 2J s S .
(8.29)
The unknown is the back-rotated Cauchy stress σ R . To solve either of (8.29) in 3D, expand
into a system of six linear equations with six unknowns [70,327]. If the stress state is 2D, the
number of unknowns reduces to three, and to one in 1D.
(3) Transform to actual Cauchy stress via σ = R σ R RT .
The following principal-directions specialization is worth working out, since it occurs in some of
the finite elements developed later and is useful for Exercises. Suppose that U is diagonal with
entries λi , i = 1, 2, 3, that both the Biot and Cauchy stress tensors are diagonal and coaxial with
U, and that the rotation matrix is the identity:
U = diag[λ1 , λ2 , λ3 ],
s B = diag[ s1B , s2B , s3B ],
σ = diag[ σ1 , σ2 , σ3 ],
R = I, (8.30)
with J = λ1 λ2 λ3 . The back-rotated Cauchy stress σ R = RT σ R = σ is also diagonal. The
Cauchy-to-Biot mapping is s B = 12 J (U−1 σ R + σ R U−1 ) = J U−1 σ since diagonal matrices
commute. We get
(8.31)
σ = J −1 diag[λ1 s1B , λ2 s2B , λ3 s3B ].
This avoids the solution of a Lyapunov equation such as the first of (8.29).
8–13
(a)
;;
;
Y, y
(b)
Elastic modulus E,
initial cross section A0
0
X, x P = κ E A 0
Z, z
L0
Ratio of predicted vs. actual
Cauchy stress
Chapter 8: REVIEW OF CONTINUUM MECHANICS: FIELD EQUATIONS
5
Almansi
4
Swainger
3
2
Hencky
1
Biot
Green
L
−0.4
−0.2
0
0.2
0.4 κ = F/E
Figure 8.5. Extension of incompressible bar under applied force: (a) problem configuration; (b) ratio of
Cauchy stress predicted by FEM to actual Cauchy stress results for several finite strain measures.
For the Green-to-Cauchy mapping under the foregoing assumptions, note that U = F = FT since
R = I and U is diagonal. The transformation is σ = J −1 F s FT = J −1 U s U, which gives
σ = J −1 diag[λ21 s1 , λ22 s2 , λ23 s3 ].
(8.32)
For the general stress measure σ (m) conjugate to the SH strain e(m) , one gets σi = J −1 λim s (m) ,
i = 1, 2, 3. See Exercise 8.10 and the next example.
Example 8.6. Axially Loaded Incompressible Bar. Here we illustrate a case where the Cauchy stress is known
under any stretch, so it can be reliably compared with FEM predictions. Consider a prismatic, homogeneous,
incompressible bar of initial length L 0 , cross section A0 and elastic modulus E. It is aligned along the X ≡ x
axis and fixed at the left end. The bar is unstrained and untressed in C0 . It stretches to L along X , moving to C
under axial force P, as pictured in Figure 8.5(a). For convenience set P = κ E A0 , in which κ is dimensionless.
The elongation predicted by a two-node bar finite element based on any SH strain measure is L =
κ, and F =
P L 0 /(E A0 ), whence the axial stretch is λ1 = 1 + L/L 0 = 1 + P/(E A0 ) = 1 + √
diag[λ1 , λ2 , λ3 ]. For isochoric (fixed volume) motion, the lateral stretches are λ2 = λ3 = 1/ λ1 , whence
J = det(F) = λ1 λ2 λ3 = 1. The area becomes A = A0 λ2 λ3 = A0 /λ1 . The exact Cauchy stress is
σexact = P/A = Pλ1 /A0 = κ (1 + κ) E.
Using the e(m) strain measure, the predicted stress in C is s (m) = E e(m) = E (λm1 − 1)/m. The associated
Cauchy stress is σ (m) = λm1 s (m) , as noted above. The ratio of predicted to actual is
r (m) =
σ (m)
σexact

m−1
m
(1
+
κ)
(1
+
κ)
−
1
λm−1
(λm1 − 1)

1

=
mκ
m (λ1 − 1)
=

 log(1 + κ) = log λ1
if m = 0.
κ(1 + κ)
if m = 0,
(8.33)
(λ1 − 1)λ1
This ratio is plotted in Figure 8.5(b) for m = 2 (Green-Lagrange), m = 1 (Biot), m = 0 (Hencky), m = −1
(Swainger) and m = −2 (Almansi-Hamel), over the range κ = (P/E A0 ) ∈ [−1/2, 1/2], which corresponds
to λ1 ∈ [1/2, 3/2]. As can be seen only the Biot measure predicts the Cauchy stress correctly for any stretch.
For very large deformations — say, |κ| > 10% — discrepancies for other measures can be huge. It must
be remembered, however, that most structural materials cannot normally be stretched beyond 1%. But for
materials such as polymers and foams, the Biot measure is certainly a winner.
The case of a compressible material is treated in Exercise 8.11.
8–14
§8.5
STRAIN ENERGY
§8.3.5. Stress and Strain Transformations
As regards coordinate transformations, all finite strain tensors transform exactly by the same rules
learned in linear elasticity. The same is true of their conjugate stress tensors. The reason is that (by
definition) all second-order tensors expressed in RCC frames obey the same transformation rules.
We employ indicial notation here to make derivations compact. Consider two common-origin RCC
frames, say {xi } and {x̄i }, (i = 1, 2, 3), related by x̄k = Tki xi , in which Tki = ∂ x̄k /∂ xi are direction
cosines. Let si j and s̄i j denote the RCC components of a stress tensor in {xi } and {x̄i }, respectively.
Then s̄km = s ji Tk j Tmi . If expressed as 3 × 3 matrices, this can be written as a three-matrix product
of the form s̄ = T s TT . If the stress tensors are cast as 6-vectors, the transformation becomes a
matrix-vector product, in which the matrix is 6 × 6 with entries built as quadratic forms in the Ti j .
Similar rules apply to finite strain tensors. When expressing transformations between strains cast
as 6-vectors, however, one should account for the 2-factors that appear in entries 4, 5 and 6.
§8.4. Constitutive Equations
We will primarily consider constitutive behavior in which finite strains and their conjugate stresses
are linearly related.11 For the Green-Lagrange and PK2 measures, the stress-strain relations will
be written, with the summation convention implied,
si = si0 + E i j e j .
(8.34)
Here ei and si denote components of the GL strain and PK2 stress vectors defined by (8.15) and the
second of (8.27), respectively, si0 are PK2 stresses in the reference configuration (also called initial
stresses or prestresses), and E i j are constant elastic moduli with E i j = E ji . In full matrix notation,
  0 
s1
E 11
s1
0
 s2   s2   E 12
   0 
 s3   s3   E 13
 = 0+
 s4   s4   E 14
   0 
s5
s5
E 15
s6
s60
E 16

E 12
E 22
E 23
E 24
E 25
E 26
E 13
E 23
E 33
E 34
E 35
E 36
E 14
E 24
E 34
E 44
E 45
E 46
E 15
E 25
E 35
E 45
E 55
E 56
 
E 16
e1
E 26   e2 
 
E 36   e3 
 ,
E 46   e4 
 
E 56
e5
E 66
e6
(8.35).
or in compact form,
s = s0 + E e.
(8.36).
If matrix E is nonsingular, inversion yields the strain-stress law
e = E−1 (s − s0 ).
11
(8.37).
This assumption, whereby the 3D Hooke’s law survives if infinitesimal strains and stresses are replaced by GL strains
and PK2 stresses, respectively, is called the St.Venant-Kirchhoff’s theory in the literature; for its history see [791, §49]. It
can be a good approximation when deformations stay small whereas displacement gradients and rotations may be large.
8–15
Chapter 8: REVIEW OF CONTINUUM MECHANICS: FIELD EQUATIONS
§8.5. Strain Energy
With all previous definitions in place, we can obtain the expression of the strain energy density U
in the current configuration reckoned per unit volume of the reference configuration.
§8.5.1. Strain Energy in Terms of GL Strains
Again we pair the GL strain measure with its conjugate (PK2) stress. Premultiplying (8.36). by
deT and path-integrating from C0 to C, as worked out in Exercise 8.2, gives
U = si0 ei + 12 (si − si0 ) ei = si0 ei + 12 ei E i j e j ,
(8.38)
in which indices i, j run from 1 to 6. In matrix form
U = s0T e + 12 eT E e.
(8.39)
If the current configuration coincides with the reference configuration, e = 0 and U = 0. It can
be observed that the strain energy density is quadratic in the GL strains. To obtain this density in
terms of displacement gradients, substitute their expressions into the above form to get
U = si0 (hiT g + gT Hi g) + 12 (gT hi + 12 gT Hi g)E i j (hTj g + 12 gT H j g) .
(8.40)
Since hi and Hi are constant, this relation shows that the strain energy density is quartic in the
displacement gradients collected in g.
The total strain energy in the current configuration is obtained by integrating the energy density
over the reference configuration:
U=
U d X dY d Z .
(8.41)
V0
This expression forms the basis for deriving linearly elastic finite elements based on the Total
Lagrangian (TL) description.
§8.5.2. Recovering Stresses and Strains from Energy Density
If U given in (8.38) is viewed as a function of GL strains e, differentiation gives
∂U
= s0 + E e = s.
∂e
(8.42)
This recovers the linear constitutive law (8.36). If E is nonsingular, formally replacing (8.37) into
(8.46) yields the complementary energy density in terms of PK2 stresses:
U ∗ = s0T E−1 (s − s0 ) + 12 (s − s0 )T E−1 (s − s0 ) = 12 (sT E−1 s − s0T E−1 s0 ),
(8.43)
in which the final simplification assumes that E is symmetric. Differentiating U ∗ with respect to s
yields e = ∂U ∗ /∂s = E−1 s, which does not agree with (8.37) unless s0 = 0. This can be reconciled
by writing e − e0 = ∂U ∗ /∂s = E−1 s, in which e0 = E−1 s0 is a fictitious “initial strain.”
8–16
§8.6 *HYPERELASTIC SOLID MATERIALS
§8.5.3. *Strain Energy in Terms of Seth-Hill Strains
The energy expressions introduced in §8.5.1 can be readily generalized to strains and (conjugate) stresses
pertaining to the Seth-Hill (SH) family. For the measure with index m, strains and stresses are collected in the
6-vectors e(m) and s(m) , respectively. Hooke’s law is assumed to remain valid for those measures:
(m)
s(m) = s(m)
.
0 + Ee
(8.44),
T (m)
U (m) = (s(m)
+ 12 (e(m) )T E e(m) .
0 ) e
(8.45)
The strain energy density becomes
The energy-differentiation stress recovery (8.42) remains unchanged with the only changes e → e(m) and
s → s(m) . But it is instructive to notice the complications that arise when e(m) is expressed in terms of stretches,
as done for hyperelastic materials in §8.6. To keep things simple we consider only the one-dimensional case
introduced in §8.2.3 with the notation (8.23), and exclude m = 0. On replacing e(m) = (λm − 1)/m in the
strain energy we get
U
(m)
=
s0(m)
λm − 1
+
m
1
2
E
λm − 1
m
2
,
m = 0.
(8.46)
Is the 1D stress-strain law s (m) = s0(m) + E e(m) recoverable through energy differentiation? This can be checked
through the chain rule:
s
∂U (m) ∂λ
=
=
∂λ ∂e(m)
(m) guess
s0(m)
λm−1
λm −1
m
m λm−1
+E
m
m2
(m)
∂λ
∂λ
m−1
(m)
=
λ
s
+E
e
. (8.47)
0
∂e(m)
∂e(m)
To get the last partial note that
λ = (1 + m e(m) )1/m ⇒
(1−m)/m
∂λ
= 1 + m e(m)
= λ1−m ,
(m)
∂e
(8.48)
whence
s (m) =
∂U (m)
.
∂e(m)
(8.49)
Ah, sono tutti contenti. The case m = 0 (Hencky measure) has to be separately verified. For the general (3D)
case, partials of scalars with respect to matrices appear, but the good old chain rule still reigns supreme.
§8.6.
*Hyperelastic Solid Materials
The linear constitutive relations (8.34) are not applicable to materials that may undergo large deformations
while remaining elastic. Those are identified as hyperelastic solid materials, or HSM. They include polymers
(e.g., rubber), foams (e.g., sponges), and some biomaterials (e.g., artery walls, ligaments). HSM may appear
in structural constituents; for instance rubber in vehicle tires, and metal foams in energy absorbers. Following
is a review of constitutive properties that are used to develop finite element models in later Chapters.
Only the isotropic case is covered. More complicated (e.g., anisotropic) constitutive equations, still formulated
within the HSM framework, can be studied in [330,551,796].
8–17
Chapter 8: REVIEW OF CONTINUUM MECHANICS: FIELD EQUATIONS
§8.6.1. *The Mooney-Rivlin Internal Energy
To computationally treat an isotropic HSM, the customary procedure is to assume a strain energy density that
is a function of the three invariants of either stretch tensor C R or C L , defined in (8.4). Those are
I1 = λ21 + λ22 + λ23 ,
I2 = λ1 λ2 + λ2 λ3 + λ3 λ1 ,
I3 = λ21 λ22 λ23 .
√
in which λi are the principal stretches. Recall that J = det(F) = λ1 λ2 λ3 = I3 .
(8.50)
Let s0i(m) , i = 1, 2, 3 denote the principal stresses in the reference state, specified in the stress measure conjugate
to the SH strain e(m) . Those will be called initial stresses.12 The total energy density U is taken as the sum of
initial-stress, deviatoric and volumetric components: U = U0 + Ud + Uv , with
U0 = s0i(m) e(m) = s0i(m)
λim −1
,
m
Ud = C1 (I1 −3) + C2 (I2 −3),
Uv = D1 (J − 1)2 .
(8.51)
The sum Udv = Ud + Uv of the preceding terms is known as the 3-parameter compressible Mooney-Rivlin
model [522,674]. This is a popular constitutive form for rubber-like materials. In this model C1 , C2 and D1
are coefficients with dimension of stress. The special case corresponding to C2 = 0 is a simplification called
the neo-Hookean material.
One simplifying assumption have been made for the initial stress term U0 : the principal directions of initial
stress and deformational stresses stay coaxial. This is sufficient for subsequent FEM developments.
Matching the stress-strain laws derived from (8.51) with the isotropic Hooke’s law in the vicinity of C0 is
messy. To simplify that operation it is beneficial to used a J -scaled form of the deviatoric term:
Ud = C 1
I1
−3 + C2
J k1
I2
−3 ,
J k2
(8.52)
The exponents k1 and k2 are adjustable real parameters. The original expression in (8.51) corresponds to
k1 = k2 = 0. A simple calculation with Mathematica reveals that the two choices
choice (I): k1 = 2/3, k2 = 4/3,
choice (II): k1 = 4/3, k2 = 1,
(8.53)
do simplify the match; however, only one will be found to be unblemished.
§8.6.2. *HSM Stress-Stretch Laws
The constitutive law become slightly simpler if the Biot strain measure and its conjugate stress are used. Reason:
the principal Biot stresses σiB and stretches λi are work-conjugate (recall that eiB = λi − 1). Accordingly,
taking the partials of the energy density gives
siB =
∂U
∂U0
∂Ud
∂Uv
0
= s0iB + sdi
+ sviB =
+
+
= s0iB + siB .
∂λi
∂λi
∂λi
∂λi
(8.54)
in which i = 1, 2, 3. The sum siB = sdiB + sviB gives the incremental Biot stresses, which represent the change
fom the reference state. Full expressions of siB can be obtained in closed form using computer algebra, but are
not necessary here. For other SH strain/stress measures, an additional scaling factor appears
si(m) = λi1−m siB .
12
(8.55)
Most of the HSM literature since 1948 has ignored initial stresses, taking C0 as an unstressed natural state. A notable
exception is Biot’s monograph [?], which treats geological and geomechanical problems where those effects, such as
gravity, are essential. Recently an uptick of interest can be noted as biomaterial modeling attracts more attention.
8–18
§8.6 *HYPERELASTIC SOLID MATERIALS
This can be remembered by taking the partial of e(m) = (λm − 1)/m with respect to λ to get λm−1 , whence the
reciprocal partial is λ1−m . This becomes 1 for Biot’s m = 1. To recover Cauchy principal stresses, use m = 0.
Of interest is how C1 , C2 and D1 relate to the linear Hooke’s law coefficients E and ν. To find that, replace
λi by 1 + ei(m) , and linearize about the reference state e(m) = 0. Taking the J -exponent choice (I) in (8.53):
k1 = 2/3 and k2 = 4/3, yields







 
s1
s01
2 −1 −1
e1
1
 s2  =  s02  + 4(C1 + C2 )  −1 2 −1   e2  + D1 p  1  ,
3
1
−1 −1 2
s3
s03
e3
(8.56)
in which p = (s1 + s2 + s3 )/3 is the pressure. (Note that superscript (m) is not needed since the limit is the
same for any m.) Matching to 3D isotropic linear elasticity is found to require C1 + C2 = 12 E/(2+2ν) = 12 G
and D1 = 12 E/(3−6ν) = 12 K , where G and K denote the shear and bulk moduli, respectively. If instead we
chose the J -exponent choice (II) in (8.53), namely k1 = 1/3 and k2 = 1, we get







 
s1
s01 + C1 + C2
2 −1 −1
e1
1
 s2  =  s02 + C1 + C2  + 3(C1 + C2 )  −1 2 −1   e2  + D1 p  1  ,
2
−1 −1 2
1
s3
s03 + C1 + C2
e3
(8.57)
Now taking C1 + C2 = 49 G and D1 = 12 K matches the isotropic elasticity matrix, but the initial stresses are
incorrect. Consequently choice (II) will not be considered further.
Rubber-like materials are often assumed incompressible, whence deformational motions become isochoric
processes. If so I3 = λ21 λ22 λ23 = 1, J = 1 and one λi , for instance λ3 , can be eliminated. The volumetric
term Uv is dropped, leaving only the initial stress and deviatoric energies. The partials (8.54) now only define
deviatoric stresses, while the pressure p must be obtained from traction (or stress) boundary conditions.
§8.6.3. *HSM Homogeneous Bar Extension
Simplifications are possible for homogeneous extension of a bar with zero lateral stresses. The constitutive
law is used later in the formulation of 2-node bar elements. The axial principal stretch λ1 is taken as surviving
variable and renamed λ. The lateral stretches λ2 = λ3 are renamed λν to reminds us of the Poisson’s effect.
Only the neo-Hookean Mooney-Rivlin model will be considered here.13 This has the advantage of not requiring
additional material constants beyond those of linear elasticity. Taking C2 = 0, C1 = 12 G = 14 E/(1 + ν),
D1 = 12 K = 16 E/(1 − 2ν), k1 = 2/3, and λ2 = λ3 = λν in (8.52) and (8.52) gives
U(λ, λν ) = U0 + Udv =
s0(m)
λm −1
+
m
1
2
G
I1
−3 +
J 2/3
1
2
K (J − 1)2 .
(8.58)
Here I1 = λ2 + 2λ2ν and J = λ λ2ν still carry λν along. To reduce (8.58) to a function of λ only, λν must be
linked to λ. Two limit cases can be solved in closed form using only kinematics:
(C) The cross section remain unchanged, so λv = 1. Then I1 = λ2 + 2 and J = λ, giving
UC = U0 +
(I)
1
2
G λ−2/3 (λ2 −1) +
sC(m) = s0(m) + λ1−m
K (λ−1)2 ,
1
3
E λ+λ1/3 −λ−5/3 −1 .
(8.59)
√
The material is incompressible, so λν = 1/ λ. Then I1 = λ2 + 2/λ and J = 1, giving
U I = U0 +
13
1
2
1
2
G λ2 + 2λ−1 − 3 ,
s I(m) = s0(m) + λ1−m G λ − λ−2 .
(8.60)
Another Neo-Hookean model has been proposed in [334]: U = 12 K (log J )2 + 12 G(I1 /J 2/3 − 3). For incompressible
material it coalesces with the neo-Hookean Mooney-Rivlin model.
8–19
Chapter 8: REVIEW OF CONTINUUM MECHANICS: FIELD EQUATIONS
2
A
1.5
S
H
1
B
0.5
G
0
(b)
2
A
1.5
Axial stress ratio s /E
(a)
Axial stress ratio s /E
Axial stress ratio s /E
2
S
H
1
0.5
B
0
G
−0.5
−0.5
−1
Case (C): unchanged cross
section, five SH measures
−1.5
0.6
0.8
1
1.2
1.4
Axial stretch λ
1.6
1.8
1.5
Stress-stretch
interpolation
1
0.5
0
Invariant and
lateral stretch
interpolations
−0.5
−1
Case (I): incompressible
material, five SH measures
−1.5
2
(c)
0.6
0.8
1
1.2
1.4
Axial stretch λ
1.6
1.8
−1
Poisson's ratio =1/4
Biot stress
−1.5
2
0.6
0.8
1
1.2
1.4
Axial stretch λ
1.6
1.8
Figure 8.6. Stress-stretch response for simple extension of homogeneous prismatic HSM bar modeled by neoHooken, scaled Mooney-Rivlin model (8.58). (a) Limit case (C): no cross section change; equivalent to ν = 0;
(b) Limit case (I): incompressible material, equivalent to ν = 1/2; (c) Biot stress response for ν = 1/4 computed
by three interpolation schemes: (S), (V) and (L), described in text. Stress measures labels in (a,b): G, B, H, S,
and A stand for Green-Lagrange, Biot, Hencky, Swainger and Almansi-Hamel, respectively.
These correspond to ν = 0 and ν = 12 , respectively, in the linear elastic regime. Stress-stretch responses for
(C) and (I) are shown in Figure 8.6(a,b) for five measures and zero initial stress.
The intermediate case between (C) and (I) cannot be solved in closed form because λν is connected to λ
through roots of a sixth order polynomial that comes from setting the lateral stress to zero. Only a numerical
solution is possible. To get an analytical approximation, an expedient way out is to interpolate between the
limit cases. There are several ways to do that; three are listed below.
(S)
Interpolate the (C) and (I) stress-stretch responses: s (m) = (1 − 2ν) sC(m) + 2ν s I(m) , which expands to
s (m) = s0(m) + 13 E λ−1−m (λ − 1) (2 ν + λ1/3 (1 + λ − 2 ν (1 − λ2/3 + λ) + λ5/3 )).
(8.61)
√
(L) Interpolate the lateral stretches: λν = (1 − 2ν) + 2ν / λ. The resulting expressions are complicated
and will not be listed here.
2
(V) Interpolate the active invariants so I1 (ν) = (1 − 2ν)(λ2 +
2) + 2ν (λ + 2/λ) and J (ν) = (1 − 2ν)λ + 2ν.
Both of these result if the lateral stretch is set to λv = 1 − 2ν(1 − 1/λ). The following expressions,
derived and simplified with Mathematica, are recorded in detail since they are used later in the bar finite
element element implementation covered in Chapter 15. Introduce the abbreviations νh = 1 − 2 ν,
J = νh λ + 2 ν, U,λ = ∂U/∂λ, Udv,λλ = ∂ 2 U/∂λ2 , Udv,λ = ∂Udv /∂λ, and Udv,λλ = ∂ 2 Udv /∂λ2 . Then
U0 = s0(m) (λm − 1)/m
if m = 0,
else U0 = s0(m) log λ,
Udv = 12 G ((λ2 + 4 ν/λ + 2 νh ) J −2/3 − 3),
Udv,λ = 2 G (λ − 1) (λ + λ2 + 2 ν) (3 ν + λ (1 − 2 ν))/(3 J 5/3 λ2 ),
Udv,λλ = G (9 J 2 (λ3 + 4 ν) − 12 J λ (λ3 − 2 ν) νh + 5 λ2 (λ3 + 2 λ νh + 4 ν) νh2 )/(9 J 8/3 λ3 ),
U = U0 + Udv ,
U,λ = s0 λm−1 + Udv,λ ,
s (m) = s0(m) + λm−1 Udv,λ ,
(8.62)
U,λλ = (m − 1) s0 λm−2 s0 + Udv,λλ ,
E (m) = (m − 1) λm−2 Udv,λ + λm−1 Udv,λλ ,
in which E (m) = ∂s (m) /∂λ is a tangent elastic modulus used in Chapter 15.
The limit case plots of Figures 8.6(a,b) suggest that any interpolation method can be expected to work reasonably
well since the response curves for the same strain measure are similar in shape, indicating mild dependence
8–20
2
§8. Notes and Bibliography
on compressibility. Indeed, comparing the responses given by the three interpolation methods for 0 < ν < 12 ,
(V) and (L) can hardly be distinguished over the range λ ∈ [1/2, 2] within plot accuracy, whereas (S) shows
a tiny deviation. This can be observed in Figure 8.6(c) for ν = 14 . By contrast, the choice of stress measure
makes a big difference for moderate and large stretch values.
§8.6.4. *GL Strains as Quadratic Forms in Gradients
For the development of the TL core-congruential formulation presented in [233,237] it is useful to have a
compact matrix expression for the Green-Lagrange strain components of (8.15) in terms of the displacement
gradient cast vector (7.13). To this end, note that (8.14) may be rewritten as
e1 = g1 + 12 (g12 + g22 + g32 ),
e2 = g5 + 12 (g42 + g52 + g62 ),
e3 = g9 + 12 (g72 + g82 + g92 ),
e4 = g6 + g8 + g4 g7 + g5 g8 + g6 g9 ,
e5 = g3 + g7 + g1 g7 + g2 g8 + g3 g9 ,
e6 = g2 + g4 + g1 g4 + g2 g5 + g3 g6 .
(8.63)
These relations may be collectively embodied in the quadratic form
ei = hiT g + 12 gT Hi g,
(8.64)
where the hi are sparse 9 × 1 vectors:
h1 = [ 1 0 0 0 0 0 0 0 0 ] T ,
h2 = [ 0 0 0 0 1 0 0 0 0 ]T , . . . etc.,
and the Hi are very sparse 9 × 9 symmetric matrices:
H1 =
I3
06×3
03×6
06×6
. . . etc.
(8.65)
(8.66)
For strain measures other than Green-Lagrange’s, expressions similar to (8.64) may be constructed. But
although the hi remain the same, the Hi become generally complicated functions of the displacement gradients
— again, because of the need to solve an intermediate eigenproblem.
Notes and Bibliography
There is a huge literature in continuum mechanics, as well as several journals devoted to the topic. Many
books treat the subject as a closed one, with no connection to physical reality — those tensorial stews should be
avoided. Three readable textbooks are Fung [297], Ogden [551] and Prager [627]. The first one is unfortunately
out of print, but the others are available as Dover reprints. Both [297] and [627] cover elastoplasticity and
viscoelasticity although their treatment of finite deformations coupled with those material models is succint.
On the other hand, [551] stops with nonlinear elasticity. Sommerfeld’s textbook [721] is nicely written and
highlights physics, but is way outdated. Murnaghan’s text [529] was historically important in introducing
direct matrix notation to finite elasticity. Novozhilov’s monograph [540] has an excellent treatment of finite
strains and rotations, but is hindered by the use of full form notation; some equations span entire pages.
The long expository article [791] is worth perusing since it had significant impact on the post-1952 evolution
of the field, and contains an exhaustive list of references dating from 1676 (Hooke’s writings). The material
therein was later expanded into two comprehensive book-length surveys that appeared in the Handbuch der
Physik in 1960 [792] and 1965 [796]. These are excellent as reference sources, especially those that need to
be backtraced in a historical context.
8–21
Chapter 8: REVIEW OF CONTINUUM MECHANICS: FIELD EQUATIONS
Homework Exercises for Chapter 8
Review of Continuum Mechanics: Field Equations
EXERCISE 8.1 [A:15] Obtain the expressions of H3 and H5 in the expressions of §8.6.4.
EXERCISE 8.2 [A:15] Derive (8.38) by integration of si dei from C0 (ei = 0) to C (ei = ei ) and use of
(8.34). (The integral is path independent.)
EXERCISE 8.3 [A+N:20] Complete the development of the pure shear case started in Examples 7.6 and 8.2
by showing that the polar decomposition matrices in F = R U are

1
1

R=
−γ
χ
0
γ
1
0


0
0,
1
U=
1
1
 1γ
2
χ
0
1
1
γ
2
+ 12 γ 2
0
0

0,
1
(E8.1)
(m)
in which χ = 1 + 14 γ 2 and γ = tan θ. Using this result, express the SH strains components e(m)
X X , eY Y
(m)
and γ X(m)
Y = 2e X Y as a function of γ , excluding m = 0. Plot these 3 components over the angular range
θ = [−60◦ , 60◦ ] for the cases m = 2 (Green-Lagrange), m = 1 (Biot), m = −1 (Swainger) and m = −2
(Almansi-Hamel). Present four plots, one for each separate m, showing the 3 components versus θ in the same
plot. Are the strains linear in θ ?
Note: sign error in alledged U given above corrected 2/9/16.
EXERCISE 8.4 [A+N:15] Repeat the previous exercise for the case m = 0 (Hencky or logarithmic strain).
To get the strain components it is necessary to evaluate log(U). Using the spectral decomposition given in
(8.7) and (8.8), show that
log(U) = ΦT log()Φ,
in which
log() =
log λ1
0
0
0
log λ2
0
0
0
0
.
(E8.2)
Plot the 3 components over the θ range of the previous exercise.
EXERCISE 8.5 [N:15] A continuation of 8.3. Assumed the following linear relation between SH stresses
and strains hold:
s (m) XX
sY(m)
Y
s X(m)
Y
=
E
0
0
0
E
0
0
0
1
E
3
e(m) XX
eY(m)
Y
2e(m)
XY
.
(E8.3)
(m)
(m)
(m)
= a sY(m)
Denote the associated forces on the faces of the cube as FX(m) = a s X(m)
X , FY
Y , and FX Y = a s X Y .
(m)
(m)
(m)
Plot the dimensionless ratios ρ X = FX (m)/FX Y and ρY = FX /FX Y for m = 2, 1, −1, −2 over the range
θ = [−60◦ , 60◦ ]. Present four plots, one for each separate m, showing the 2 ratios in the same plot. The
nonlinear effect of the geometry change should be evident. (These ratios are essential for designing a pure
shear experiment.)
EXERCISE 8.6 [A:20] Let L 0 and L denote the length of a bar element in the reference and current configurations, respectively. The Green-Lagrange finite strain e = e X X , if constant over the bar, can be defined
as
L 2 − L 20
.
(E8.4)
e=
2L 20
Show that the definitions (E8.4) and of e = e X X in (8.14) are equivalent. Hint: the axial displacement gradient
is obviously ∂u X /∂ X = (L − L 0 )/L 0 ; how about ∂u X /∂Y and ∂u X /∂Y ?
8–22
Exercises
EXERCISE 8.7 [A:20] Find the axial GL strain e = e X X for the non-homogeneous bar extension case treated
in Example 7.8 in terms of L 0 , L 1 , L 2 and ξ . The result is useful for 3-node bar finite elements. (Hint: it is a
quadratic polynomial in ξ .)
EXERCISE 8.8 [A:20] Expand in Taylor series in g, up to and including O(g 2 ), the seven one-dimensional
finite strain measures shown in the 2nd columns of the table in Figure 8.2, about g = 0. Verify that all measures
agree up to O(g) (the infinitesimal strain) but differ in terms O(g 2 ) and higher.
EXERCISE 8.9 [A:25] Complete the 3rd column of the table of finite strain measures in Figure 8.2 with
tensorial forms in terms of V.
EXERCISE 8.10 [A:25] A bar is in a one dimensional stress state. The axial stress computed with SH strain
measure of index m is s (m , while all other stress components are zero. The axial stretch is λ. Show that the
axial Cauchy stress σ can be recovered as σ = λm s (m) for arbitrary m.
EXERCISE 8.11 [A:25] An extension of Example 8.6 to compressible material. Suppose that now the bar is
isotropic but compressible, so that the lateral stretches are
2ν
λ2 = λ3 = 1 − 2ν + √ ,
λ1
(E8.5)
in which ν ∈ [0, 12 ] is an extension of Poisson’s ratio to finite strains (for isochoric motions, ν = 12 ). Find the
ratio r (m) (ν) of predicted to actual Cauchy stress and discuss the cases ν = 14 and ν = 0. Is the Biot strain
measure still the winner?
EXERCISE 8.12 [A:20] Let denote the infinitesimal strain tensor. Show that the two-term expansion of
e(m) is
e(m) = + 12 ∇uT ∇u − (1 − 12 m) T .
For which m does last term vanish?
EXERCISE 8.13 [A:30] Given C R = FT F and the invariants I1 , I2 and I3 , show [881] that U = c1 (c2 +
c3 C R − C2R ), in which c1 = 1/(I1 I2 − I3 ), c2 = I1 I3 , and c3 = I12 − I2 . (Hint: use the Cayley-Hamilton
theorem.) Is this expression practical?
EXERCISE 8.14 [A:40] (Advanced). Define the stress measure conjugate to the midpoint strain measure e M .
defined in the last row of the table in Figure 8.2. (Publishable if successful since it is unsolved to date.)
EXERCISE 8.15 [A:40] (Advanced). Define the stress measure conjugate to the Hencky finite strain measure
log(U) in a compact form. (Publishable if successful since it is unsolved to date.)
8–23