8Review of

Continuum Mechanics:Field Equations

8–1

Chapter 8: REVIEW OF CONTINUUM MECHANICS: FIELD EQUATIONS

TABLE OF CONTENTS

Page§8.1 Kinematic Equations . . . . . . . . . . . . . . . . . . 8–3

§8.1.1 Polar Decomposition . . . . . . . . . . . . . . 8–3§8.1.2 Stretch Tensors . . . . . . . . . . . . . . . . . 8–3

§8.2 Strain Measures . . . . . . . . . . . . . . . . . . . 8–5§8.2.1 Green-Lagrange Strain Measure . . . . . . . . . . . 8–6§8.2.2 Almansi Strain Measure . . . . . . . . . . . . . 8–7§8.2.3 Generalized Strain Measures: The Seth-Hill Family . . . . . 8–8§8.2.4 Eulerian Strain Measures . . . . . . . . . . . . . 8–10

§8.3 Stress Measures . . . . . . . . . . . . . . . . . . . . 8–11§8.3.1 Cauchy Stress . . . . . . . . . . . . . . . . . 8–11§8.3.2 PK2 Stress Measure . . . . . . . . . . . . . . . 8–11§8.3.3 Other Conjugate Stress Measures . . . . . . . . . . 8–12§8.3.4 Recovering Cauchy Stresses . . . . . . . . . . . . 8–12§8.3.5 Stress and Strain Transformations . . . . . . . . . . 8–14

§8.4 Constitutive Equations . . . . . . . . . . . . . . . . . 8–15

§8.5 Strain Energy . . . . . . . . . . . . . . . . . . . . 8–15§8.5.1 Strain Energy in Terms of GL Strains . . . . . . . . . 8–15§8.5.2 Recovering Stresses and Strains from Energy Density . . . 8–16§8.5.3 *Strain Energy in Terms of Seth-Hill Strains . . . . . . . 8–16

§8.6 *Hyperelastic Solid Materials . . . . . . . . . . . . . . 8–17§8.6.1 *The Mooney-Rivlin Internal Energy . . . . . . . . . 8–17§8.6.2 *HSM Stress-Stretch Laws . . . . . . . . . . . . 8–18§8.6.3 *HSM Homogeneous Bar Extension . . . . . . . . . . 8–19§8.6.4 *GL Strains as Quadratic Forms in Gradients . . . . . . 8–21

§8. Notes and Bibliography . . . . . . . . . . . . . . . . . 8–21

§8. Exercises . . . . . . . . . . . . . . . . . . . . . . 8–22

8–2

§8.1 KINEMATIC EQUATIONS

This Chapter continues the review of nonlinear continuum mechanics. The focus is on the fieldequations, which require the introduction of strain and stress measures. Some of the material istranscribed from the 1966 thesis [221], 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 nonlinearcontinuum mechanics. The whole subject is dominated by the polar decomposition theorem:1 abody particle motion can be expressed as a pure deformation followed by a rigid rotation, or as arigid rotation followed by a pure deformation. Mathematically this can be written as multiplicativedecompositions:

F = R U = V R. (8.1)

Here R is a proper orthogonal rotation matrix (R−1 = RT , det(R) = +1) whereas U and Vare 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 Fand 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, or R = V−1 F = (F FT )−1/2 F, (8.2)

respectively. It is possible to compute R directly without getting U or V first, by using Grioli’stheorem 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 theirphysical 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, whichrepresent principal directions. Denoting Λ = diag[λ1, λ2, λ3], the spectral decompositions maybe written (in both indicial and matrix form) as

U = λi φiφTi = ΦT ΛΦ, V = λi ψiψ

Ti = ΨT ΛΨ. (8.3)

Here Φ and Ψ are 3 × 3 proper orthogonal, direction-cosine matrices formed by stacking φi andψi as columns. Equating Λ = Φ U ΦT = Φ RT V R ΦT = Ψ V ΨT shows that Φ RT = Ψ andΦ = Ψ R, so the principal directions are linked by the rigid rotation R. The columns of Φ and Ψdefine 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 This is the extension of the polar representation of complex numbers to arbitrary dimensions. Discovered for generalmatrices by Autonne in 1902 [44]; for the detailed story see [407, §3.0]. For elasticity its roots go back to Cauchy.

2 Some authors call U and V the right and left Biot-stretch tensors, e.g., [561].3 The principal square root of a symmetric, positive-definite matrix such as F FT or FT F is unique.

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Chapter 8: REVIEW OF CONTINUUM MECHANICS: FIELD EQUATIONS

§8.1.2. Stretch Tensors

Recall from §7.4.3 the definitions of CR and CL , which are further expanded with their squareroots:

CR = FT F = UT U = U2, CL = F FT = V VT = V2. (8.4)

CR and CL are symmetric positive definite matrices called the right and left Cauchy-Green stretchtensors, respectively.4 To get U and V as their square roots one must solve (directly or indirectly)for their eigensystem. The common eigenvalues of CR and CL are denoted λ2

1, λ22 and λ2

3.

If R = I, CR = CL and U = V. This no-rotation case is illustrated in Example 8.1.

The invariants of either stretch tensor are denoted by I1 = λ21 +λ2

2 +λ23, I2 = λ1 λ2 +λ2 λ3 +λ3 λ1,

and I3 = λ1 λ2 λ3. These are often used in isotropic constitutive equations that account for finitestrains; for example the hyperelasticity model considered in §8.6.

Example 8.1. Simple extension. This is a continuation of Example 7.5 of previous Chapter. Since thedeformation gradient F shown in (7.19) is diagonal, the polar decomposition is trivial:

CR = FT F =[

λ21 0 0

0 λ22 0

0 0 λ23

]= CL , U =

√CR = F = FT =

[λ1 0 00 λ2 00 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 CR and CL :

CR = FT F =[

1 γ 0γ 1 + γ 2 00 0 1

], CL = F FT =

[1 + γ 2 γ 0

γ 1 00 0 1

]. (8.6)

The Green-Lagrange strains introduced in §8.2.1 are eY Y = 12 γ 2, eXY = eY X = 1

2 γ , others zero. Theeigenvalues of CR , which are the squares of the principal stretches, are

λ21 = 1 + 1

2 γ 2 + γ

√1 + 1

4 γ 2, λ22 = λ−2

1 = 1 + 12 γ 2 − γ

√1 + 1

4 γ 2, λ23 = 1. (8.7)

Note that λ1λ2λ3 = 1 since the motion is isochoric. The associated orthonormalized eigenvectors are thecolumns of the matrix

Φ =

√

12 − 1

2 γ /√

4 + γ 2 −√

12 + 1

2 γ /√

4 + γ 2 0√12 + 1

2 γ /√

4 + γ 2

√12 − 1

2 γ /√

4 + γ 2 0

0 0 1

(8.8)

Completing the polar decomposition and using it to evaluate other strain measures is the subject of Exercise 8.3.As regards the choice of U among 8 alternatives, see Exercise 8.5.

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., [578],while B−1 is denoted by c in pre-1970 publications. However B is reserved here for the strain-displacement matrixin finite elements. The names of Cauchy and Green are attached for historical reasons: CL (actually its small-straininverse) was introduced by Cauchy in 1827 and CR by Green in 1841. To compound the genealogical confusion, thename Piola is often associated with C−1

L and that of Finger with both C−1R and C−1

L .

8–4

§8.2 STRAIN MEASURES

Example 8.3. Two-Dimensional Decomposition. For the 2D plane strain case, in which λ3 = 1, the polardecomposition 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 g12 0g21 1 + g22 00 0 1

]=

[c −s 0s c 00 0 1

] [1 + eB

11 eB12 0

eB12 1 + eB

22 00 0 1

]= R U. (8.9)

Here gi j are displacement gradients, whereas the eBi j are the Biot strain components introduced formally in

§8.2.3 A simple hand computation [98, pp. 8–9] yields

t = g21 − g12

2 + g11 + g22, c = 1√

1 + t2, s = t√

1 + t2, eB

11 = g11 c + g21 s + c − 1,

eB22 = g22 c − g12 s + c − 1, eB

12 = eB21 = 1

2 (g21 + g12) c + 12 (g22 − g11) s.

(8.10)

One may verify that this R minimizes ||F − R||F , in which ||.||F denotes the Frobenius matrix norm, if Fhas full rank. This theorem was proven by Grioli in 1940; see [828, p. 290]. In the 3D case R depends on 3independent rotational parameters, which complicates the computations.5

Expanding (8.10) in Taylor series about g11 = g12 = g21 = g22 = 0 and retaining terms up to the second ordergives

eB11 = g11 + 1

8 (3g221 − 2g12g21 − g2

12),

eB22 = g22 + 1

8 (3g212 − 2g12g21 − g2

21),

eB12 = eB

21 = 12 (g21 + g12) + 1

4 (g11g12 − g12g22 + g21g22).

(8.11)

Remark 8.1. Grioli’s theorem is important in establishing “a best-fit rotation matrix” in the corotationalkinematic description of nonlinear FEM, since it allows determination of R without need to find U first.

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. Strain Measures

Principal stretches are fine as deformation measures, but to establish constitutive equations thataccount for geometric nonlinearities it is convenient to introduce finite strain measures.6 Desirableattributes for such measures are listed in Figure 8.1. These were enunciated by Hill in 1978 [400]

5 The extension to arbitrary matrices is discussed in Higham’s book [395, p. 197], which totally ignores Grioli’s work.6 Truesdell [827, §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 physicalcomponents are dimensionless, a nearly rigid displacement may be specified by a series expansion in the strains, in whichall terms beyond some specified order are neglected.”

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Chapter 8: REVIEW OF CONTINUUM MECHANICS: FIELD EQUATIONS

as ingredients in the construction of strain families. The last one (#4) is not strictly necessary, but isconvenient 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 havea measure that vanishes for rigid motions. Thus either U − I or log(U) are appropriate strainmeasures. As discussed later in §8.2.3, some of these are difficult to express analytically in termsof the displacement gradients because of the need of solving an eigenproblem that delivers theprincipal stretches to get U. Two finite strain measures that bypass that hurdle are described next.

§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

eG = 12 (CR − I) = 1

2

(FT F − I

) = 12

(U2 − I

) = 12 (G + GT ) + 1

2 GT G =[ eX X eXY eX Z

eY X eY Y eY Z

eZ X eZY eZ Z

],

(8.12)

This measure does not explicitly require U. Its indicial form in Cartesian coordinates is

eGi j = 1

2

(∂ui

∂ X j+ ∂u j

∂ Xi+ ∂uk

∂ Xi

∂uk

∂ X j

). (8.13)

On passing to the engineering notation X1 → X , X2 → Y , X3 → Z , the full form of (8.13) is

eGX X = ∂u X

∂ X+ 1

2

[(∂u X

∂ X

)2

+(

∂uY

∂ X

)2

+(

∂uZ

∂ X

)2]

,

eGY Y = ∂uY

∂Y+ 1

2

[(∂u X

∂Y

)2

+(

∂uY

∂Y

)2

+(

∂uZ

∂Y

)2]

,

eGZ Z = ∂uZ

∂ Z+ 1

2

[(∂u X

∂ Z

)2

+(

∂uY

∂ Z

)2

+(

∂uZ

∂ Z

)2]

,

eGY Z = 1

2

(∂uY

∂ Z+ ∂uZ

∂Y

)+ 1

2

[∂u X

∂Y

∂u X

∂ Z+ ∂uY

∂Y

∂uY

∂ Z+ ∂uZ

∂Y

∂uZ

∂ Z

]= eG

ZY ,

eGZ X = 1

2

(∂uZ

∂ X+ ∂u X

∂ Z

)+ 1

2

[∂u X

∂ Z

∂u X

∂ X+ ∂uY

∂ Z

∂uY

∂ X+ ∂uZ

∂ Z

∂uZ

∂ X

]= eG

X Z ,

eGXY = 1

2

(∂u X

∂Y+ ∂uY

∂ X

)+ 1

2

[∂u X

∂ X

∂u X

∂Y+ ∂uY

∂ X

∂uY

∂Y+ ∂uZ

∂ X

∂uZ

∂Y

]= eG

Y X .

(8.14)

If the nonlinear portion (that enclosed in square brackets) of these expressions is neglected, oneobtains the infinitesimal strains εxx , εyy , εzz , εxy = 1

2γxy , εyz = 12γyz , and εzx = 1

2γzx , already

7 Introduced for finite strains by Green (1841) and St.Venant (1844). The name of Lagrange is attached because of theconnection to the Lagrangian kinematic description since it involves CR ; see §7.4.3. The minor variation “Green-Lagrangian” is also in use. Many authors call it just the Green strain tensor. A few label it (more properly) asGreen-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 [271]. For future use in finite element work we shall arrange thecomponents (8.14) as a strain 6-vector configured as

e =

e1

e2

e3

e4

e5

e6

=

eX X

eY Y

eZ Z

eY Z + eZY

eZ X + eX Z

eXY + eY X

=

eX X

eY Y

eZ Z

2eY Z

2eZ X

2eXY

, (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 CR by CL there to get

eG = 12 (CL − I) = 1

2

(F FT − I

) = 12 (V2 − I) = 1

2 (G + GT ) + 12 G GT . (8.16)

This “Green-Euler” measure is rarely used8, but it must be chosen if the polar decomposition isF = V R; that is, the body is first rotated and then stretched. See Example 8.4. If R = I, bothGreen 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 Almansi-Lagrange strain, which is defined by

eA = 12 (I − C−1

R ) = 12 (I − U−2). (8.17)

Like (8.12) this measure avoids the solution of an eigenproblem but requires inversion of F to getthe xi partials from F−1, since C−1

R = F−1F−T .

An Eulerian version of the Almansi strain,9 denoted by eA, is obtained by replacing CR by CL in(8.17) to get

eA = 12

(I − C−1

L

) = 12

(I − (F FT )−1

) = 12

(I − V−2) . (8.18)

Its indicial form in Cartesian coordinates is

eAi j = 1

2

(∂ui

∂x j+ ∂u j

∂xi− ∂uk

∂xi

∂uk

∂x j

). (8.19)

This variant is far more popular; in fact, the Lagrangian form is rarely used aside of its appearancein the SH family introduced in §8.2.3. The Eulerian version is mandatory if the polar decompositionis F = V R; that is, the body is first rotated and then stretched. See Example 8.4.

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.

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Chapter 8: REVIEW OF CONTINUUM MECHANICS: FIELD EQUATIONS

Example 8.4. Combined stretch, rotation and rigid translation. This is a continuation of Example 7.7. Theright polar decomposition F = R U can be directly lifted from (7.24): the first and second square matricesin the right hand side of that equation are R and U, respectively, and V can be computed as R U RT .Thus thestretch tensors are

CR = FT F = U2 =[

λ21 0 0

0 λ22 0

0 0 1

], CL = F FT = V2 =

[λ2

1 c2 + λ22 s2 (λ2

1 − λ22) s c 0

(λ21 − λ2

2) s c λ22 c2 + λ2

1 s2 00 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 stretchtensors we can compute the Green-Lagrange strain tensor and its Eulerian counterpart:

eG = 12

[λ2

1 − 1 0 00 λ2

2 − 1 00 0 0

], eG = 1

2

[λ2

1 c2 + λ22 s2 − 1 (λ2

1 − λ22) s c 0

(λ21 − λ2

2) s c λ22 c2 + λ2

1 s2 − 1 00 0 0

]. (8.21)

For the Almansi-Lagrange strain and its Eulerian counterpart we get

eA = 12

[1 − λ−2

1 0 00 1 − λ−2

2 00 0 0

],

eA = 1

4 λ21λ

22

[2 λ2

1λ22 − λ2

1 − λ22 + (λ2

1 − λ22) c2 (λ2

1 − λ22) s2 0

(λ21 − λ2

2) s2 2 λ21λ

22 − λ2

1 − λ22 − (λ2

1 − λ22) c2 0

0 0 0

],

(8.22)

in which c2 = cos 2ψ and s2 = sin 2ψ . Both eG and eA 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 appliedby rotation followed by stretch, the polar decomposition becomes F = V R. Now eG and eA are diagonaland 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 nonlinearanalysis. Several others have been proposed for use in particular scenarios. To introduce them intoto it is illuminating to look at the one-dimensional (1D) case first. Consider a line element oflengths L0 and L in the reference and current configurations, respectively. The length change is�L = L − L0. Define

λ = L/L0, λ = L0/L , g = λ − 1 = �L/L0, g = 1 − λ = 1 − (1/λ) = g/(1 + g). (8.23)

We call λ, λ, g and g the stretch, inverse stretch, displacement gradient , and complementary dis-placement 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 theycan be seen to be members of a one-parameter form, which defines the Seth-Hill family listed inthe 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 Green-Lagrange and Almansi measures correspond to m = 2 and m = −2, respectively. The case m = 0

8–8

§8.2 STRAIN MEASURES

Name (m=index 1D finite strain measure Taylor series expansion Commentsin SH family) about g=0

Green (m = 2) eG = (λ2−1) = g + g2 g + g2/2 Widely used since the start of geometrically nonlinear FEM = (1−λ2)/ λ2 = (1−g2/2)/(1−g)

Biot (m = 1) eB = λ − 1 = g g Generalization of "engineering strain" to finite deformations. = 1/λ −1 = g/(1 − g) Becoming increasingly popular

Hencky (m = 0) eH = log λ = log (1+g) g − g2/2 + g3/3 − ... Used in finite elastoplasticity. Sometimes replaced by log-less = − log λ = log (1/(1−g)) approximations (see below) Swainger eS = 1−1/λ = g/(1 + g) g − g2 + g3 − ... Counterpart of Biot measure(m = −1) = 1 − λ = g

Almansi (m=−2) eA = (1−1/λ ) = (g + g2 )/(1+g) g − 3g2/2 + 2g3 − ... Counterpart of Green measure = (1− λ ) = (g− g2)/(1−g) Seth-Hill family e(m) = (λm−1) = ((1+g)m−1) g − (m−1) g2/2 Includes above five measures(arbitrary m) for m = 2, 1, 0, −1, −2, resp. + (m−1)(m−2) g3/6 − ...

Midpoint eM = 2(λ−1)/(λ+1) = g/(1+g/2) g − g2/2 + g3/4 − ... (1,1) Pade approximant to Hencky strain that avoid logs. = 2(1−λ)/(λ+1) = g/(1−g/2) Not part of the Seth-Hill family

Bazant eZ = (eB + eS) = (g+g2/2)/(1+g) g − g2/2 + g3/2 − ... Another log-free approximant to Hencky strain; accuracy similar to eM. Not part of the Seth-Hill family

12

12

12

12

12

12

12

12

12

1m

1m

^ ^

^ ^

^

^ ^

^ ^ ^

^^

^

^ ^^ ^

^ ^

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. Bluebackground: 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 isused often in finite strain elastoplasticity as well as for foam materials.

The multi dimensional (tensor) versions are obtained through obvious mappings: λ → U = C1/2R ,

λ → U−1 = C−1/2R , 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 CR = F FT or itsinverse do not require an eigensolution. This happens for m = 2 (Green-Lagrange) and m = −2(Almansi-Lagrange). For other m values, the eigensolution of CR is usually required.

The midpoint and Bazant strains, listed in the last two rows, can be used as log-free approximantsto the Hencky strain. See Remark 8.4 below. Those two measures do not belong to the SH family,but do fulfill the conditions set out by Hill in [399], which are summarized in Figure 8.1.

Remark 8.2. The Seth-Hill (SH) family was proposed by Seth [732] for one dimensional strain measures.The idea was tensorially generalized by Hill [399], who defined explicit admissability conditions. Those aresummarized in Figure 8.1. Doyle and Ericksen had presented that family for arbitrary m almost a decadeearlier [202] but the idea, buried in an tensor-laced long chapter, did not attract attention. By contrast, Sethused high-school mathematics in a short article, and got his point across. For references to the strain measurenames, see [98,578,693,827,828,67]. The history is confusing and naming attributions often do scant justice.

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Chapter 8: REVIEW OF CONTINUUM MECHANICS: FIELD EQUATIONS

12

12

12

12

1212

12

12

1/2

1m

1m

−1/2

−1

R

Name (m is index 2D and 3D tensor forms (only those dependent Commentsin SH family) on CR and U are listed). Note: Gsym= (G + GT)/2

Green-Lagrange eG = (U2 − I) = (CR − I) = Gsym+ GT G Rational function of displacement gradients.

(m=2) Easy to compute, widely used in FEM

Biot (m=1) eB = U − I = CR − I = ( I + 2Gsym+ GT G)1/2 − I Generalizes "engineering strains" to finite strains.

Needs eigenanalysis of CR to get U

Hencky (m=0) eH = log(U) = log(CR )= log( I+ 2Gsym+ GT G) Prone to singularities, so often replaced by log-free

approximations. Needs eigenanalysis of CR to get U

Swainger (m=−1) eS = I − U−1= I − CR = I − ( I+ 2Gsym + GT G)−1/2 Counterpart of Biot. Rarely used in Lagrangian

form. Needs eigenanalysis of CR to get U

Almansi-Lagrange eA = (I − U−2) = (I − CR ) Counterpart of Green. Rarely used in Lagrangian

(m=−2) = (I − (I + 2Gsym+ GT G)−2) form. Needs inversion of CR but no eigennalysis.

Seth-Hill family e(m) = (U − I)m = (Cm/2− I) Family includes above 5 measures. All assume that

F = R U. If F = V R, replace CR by CL, and U by V

Midpoint eM = 2 (U − I) (I + U)−1 A (1,1) Pade approximant of eH , correct to 2nd order.

Avoids logs but needs U and an inversion.

Bazant eZ = (eB + eS)/2 = (U − U−1)/2 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 8.3. No distinction between Lagrangian and Eulerian forms need to be made in 1D, so those qualifiersare not attached to the Green and Almansi measures, as done instead in Figure 8.3. The same would be truein multiple dimensions if R = I (no rotation) since if so CR and CL (as well as U and V), coalesce.

Remark 8.4. The Hencky strain measure has long been a popular choice in finite elastoplasticity as wellas some crystalline models; see the recent expository article [561]. It causes problems, however, in FEMcomputations because of 0/0 singularities for zero strain. Accordingly log-free approximants such as thoselisted in the last two rows of Figures 8.2 and 8.3 may be preferred. Those two agree with the Hencky measureup to second order in the displacement gradients. Third order approximations are possible: a particular simpleone is the linear combination eB + eS/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 Euleriancounterpart by substituting CR by CL or, equivalently, by changing U to V. Doing that for the SHfamily gives

e(m) = 1

m

(Cm/2

L − I)

. (8.24)

The choices m = 2 (Green-Euler measure) and m = −2 (Almansi-Euler measure, often calledAlmansi-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 CR or U.

(2) Go left. If the motion is built using the left polar decomposition V R, i.e., firstrotate and then deform, use an Eulerian measure based on CL or V.

If the strains are infinitesimal although rotations may be finite (which is often the case in the CRkinematic description 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 structuralanalysis. 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 ofmaterial there. The notation used in linear FEM [271] is retained for it:

σ =[

σxx σxy σxz

σyx σyy σyz

σzx σzy σzz

],

σ = [ σxx σyy σzz σyz σzx σxy ]T = [ σ1 σ2 σ3 σ4 σ5 σ6 ]T .

(8.25)

in which σxy = σyx , etc. Here σ denotes the Cauchy stress tensor written as a 3 × 3 matrix of itsCartesian components in C, whence the use of {x, y, z} in (8.25). The non-underlined symbol σidentifies the 6-vector cast form used in FEM formulations. Two closely related measures are theback-rotated Cauchy stress σ R and the Kirchhoff stress σ, defined as

σR = RTσR, σ = J σ in which J = det F, (8.26)

in which R is the rotation tensor of the polar decomposition (8.1). The tensor σ, which is often usedin 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 isnot 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 ofvirtual work. More precisely, the dot product of stress times the material derivative strain rate is theinternal power density. Since the GL measure is common in FEM implementations, its conjugatestress 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 Cartesiancomponents in the reference configuration C0, the PK2 stress tensor is denoted as

s =[ sX X sXY sX Z

sY X sY Y sY Z

sZ X sZY sZ Z

],

s = [ sX X sY Y sZ Z sY Z sZ X sXY ]T = [ s1 s2 s3 s4 s5 s6 ]T .

(8.27)

in which sXY = sY X , etc. These components are referred to the reference configuration C0, whencethe use of {X, Y, Z} in (8.27). The PK2 and Cauchy (true) stresses are connected through thetransformations

s = J F−1 σF−T = F−1 σF−T , σ = J−1 F s FT . (8.28)

In terms of the back-rotated Cauchy stress σR = RTσR we have s = J U−1 σR U−1. The detailedtransformation from σ to s and vice-versa, when both are cast as 6-vectors, may be implementedas 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 covariantGreen-Lagrange strain tensor.

Remark 8.5. The physical meaning of the PK2 stresses is as follows: si j are stresses “pulled back” to thereference configuration C0 and referred to area elements there.

Remark 8.6. 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 tovary 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 beenobtained — one rather recently. Several are listed in the table of Figure 8.4. For derivation detailsthe reader is referred to [216,352,403,578]. The expression for m = 0, found first in 1987 [403] isextremely 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 Cauchystress? Answer: it does exist, but it is not admissible in the sense of observer invariance [400].

§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 itis necessary to recover the Cauchy (true) stress for tasks such as checking strength safety factors.Should the conjugate stress be s (PK2) or sA (Almansi-Hamel) the recovery is straightforward: justapply the congruential transformation given in the table. For example, σ = F s FT /J . But if it isthe Biot or Swainger stress, the task is more involved, requiring three steps:

10 The first Piola -Kirchhoff stress tensor is sP K 1 = J σF−T = F s. Its transpose sN = J FT σ is called the nominal stresstensor. Since both of these tensors are generally unsymmetric, they are rarely used in FEM.

8–12

§8.3 STRESS MEASURES

Name (coeff in Finite strain tensor (a member Conjugate stress tensor CommentsSeth-Hill family) of Seth-Hill family)

Green-Lagrange e = (U − I) = (C − I) s = J F σ F Second Piola-Kirchhoff tensor. (m = 2) σ = J F s F Abbreviated to PK2 in text

Biot (m = 1) e = U − I = C − I s = ( s U + U s) Biot stress. In 2nd form, σ = R σ R = J (U σ + σ U ) is back-rotated Cauchy stress

Hencky (m = 0) e = log(U) = log(C ) Omitted (see text) For isotropic material, σR . Extremely complicated for anisotropic material Swainger e = I − U = I − C s = ( s U + U s ) Swainger stress, rarely used. 2nd form(m = −1) = J (U σ + σ U ) in terms of back-rotated Cauchy stress

Almansi-Hamel e = (I − U ) = (I − C ) s = J F σ F Almansi stress(m=−2) σ = J F s F

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, whereasthose referred to the current configuration (e.g., the Cauchy stress tensor) are coaxial with V.

2

12

12

12

12

1212

12

12

12

B B

H

A

A

A A

A

R

RR

R R

R

R

R

R

1/2

SS

−1−2

−1/2 −1−1_

__ _

__ _

_

_

_

_ __ _

_

_

__

_ _ _R __

−1

−1

−1

−1−1

−1

−1

−T

−T

T

T

T

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 [403].

(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 σR + σR U−1 = 2J sB, Swainger: UσR + σR U = 2J sS. (8.29)

The unknown is the back-rotated Cauchy stress σR . To solve either of (8.29) in 3D, expandinto a system of six linear equations with six unknowns [?,341]. If the stress state is 2D, thenumber of unknowns reduces to three, and to one in 1D.

(3) Transform to actual Cauchy stress via σ = RσR RT .

The following principal-directions specialization is worth working out, since it occurs in some ofthe finite elements developed later and is useful for Exercises. Suppose that U is diagonal withentries λi , i = 1, 2, 3, that both the Biot and Cauchy stress tensors are diagonal and coaxial withU, and that the rotation matrix is the identity:

U = diag[λ1, λ2, λ3], sB = diag[ s B1 , s B

2 , s B3 ], σ = diag[ σ1, σ2, σ3 ], R = I, (8.30)

with J = λ1 λ2 λ3. The back-rotated Cauchy stress σR = RT σR = σ is also diagonal. TheCauchy-to-Biot mapping is sB = 1

2 J (U−1σR + σR U−1) = J U−1 σ since diagonal matricescommute. We get

σ = J−1diag[λ1 s B1 , λ2 s B

2 , λ3 s B3 ]. (8.31)

This avoids the solution of a Lyapunov equation such as the first of (8.29).

8–13

Chapter 8: REVIEW OF CONTINUUM MECHANICS: FIELD EQUATIONS

���

X, x

Y, y 0

Z, z

0L

P = κ E A

L

Elastic modulus E,initial cross section A0

0

κ = F/E

Rat

io o

f pr

edic

ted

vs. a

ctua

l C

auch

y st

ress

(a) (b)

−0.4 −0.2 0 0.2 0.4

1

2

3

4

5

Green

Biot

Swainger

Almansi

Hencky

Figure 8.5. Extension of incompressible bar under applied force: (a) problem configuration; (b) ratio ofCauchy 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 sinceR = I and U is diagonal. The transformation is σ = J−1 F s FT = J−1 U s U, which gives

σ = J−1diag[λ21 s1, λ2

2 s2, λ23 s3]. (8.32)

For the general stress measure σ (m) conjugate to the SH strain e(m), one gets σi = J−1 λmi s(m),

i = 1, 2, 3. See Exercise 8.11 and the next example.

Example 8.5. Axially Loaded Incompressible Bar. Here we illustrate a case where the Cauchy stress is knownunder any stretch, so it can be reliably compared with FEM predictions. Consider a prismatic, homogeneous,incompressible bar of initial length L0, cross section A0 and elastic modulus E . It is aligned along the X ≡ xaxis and fixed at the left end. The bar is unstrained and untressed in C0. It stretches to L along X , moving to Cunder 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 =P L0/(E A0), whence the axial stretch is λ1 = 1 + �L/L0 = 1 + P/(E A0) = 1 + κ , and F =diag[λ1, λ2, λ3]. For isochoric (fixed volume) motion, the lateral stretches are λ2 = λ3 = 1/

√λ1, whence

J = det(F) = λ1λ2λ3 = 1. The area becomes A = A0 λ2 λ3 = A0/λ1. The exact Cauchy stress isσexact = 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=

(1 + κ)m−1

((1 + κ)m − 1

)m κ = λm−1

1 (λm1 − 1)

m (λ1 − 1)if m = 0,

log(1 + κ)κ(1 + κ)

= log λ1(λ1 − 1)λ1

if m = 0.

(8.33)

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 correspondsto 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 mustbe remembered, however, that most structural materials cannot normally be stretched beyond 1%. But formaterials such as polymers and foams, the Biot measure is certainly a winner.

The case of a compressible material is treated in Exercise 8.12.

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 ruleslearned in linear elasticity. The same is true of their conjugate stress tensors. The reason is that (bydefinition) 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 RCCframes, say {xi } and {xi }, (i = 1, 2, 3), related by xk = Tki xi , in which Tki = ∂ xk/∂xi are directioncosines. Let si j and si j denote the RCC components of a stress tensor in {xi } and {xi }, respectively.Then skm = s ji Tk j Tmi . If expressed as 3×3 matrices, this can be written as a three-matrix productof the form s = T s TT . If the stress tensors are cast as 6-vectors, the transformation becomes amatrix-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 castas 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 stressesare linearly related.11 For the Green-Lagrange and PK2 measures, the stress-strain relations willbe written, with the summation convention implied,

si = s0i + Ei j e j . (8.34)

Here ei and si denote components of the GL strain and PK2 stress vectors defined by (8.15) and thesecond of (8.27), respectively, s0

i are PK2 stresses in the reference configuration (also called initialstresses or prestresses), and Ei j are constant elastic moduli with Ei j = E ji . In full matrix notation,

s1

s2

s3

s4

s5

s6

=

s0

1s0

2s0

3s0

4s0

5s0

6

+

E11 E12 E13 E14 E15 E16

E12 E22 E23 E24 E25 E26

E13 E23 E33 E34 E35 E36

E14 E24 E34 E44 E45 E46

E15 E25 E35 E45 E55 E56

E16 E26 E36 E46 E56 E66

e1

e2

e3

e4

e5

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). (8.37).

11 This assumption, whereby the 3D Hooke’s law survives if infinitesimal strains and stresses are replaced by GL strainsand PK2 stresses, respectively, is called the St.Venant-Kirchhoff’s theory in the literature; for its history see [827, §49]. Itcan 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 Uin 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). bydeT and path-integrating from C0 to C, as worked out in Exercise 8.2, gives

U = s0i ei + 1

2 (si − s0i ) ei = s0

i ei + 12 ei Ei j e j , (8.38)

in which indices i, j run from 1 to 6. In matrix form

U = sT0 e + 1

2 eT E e. (8.39)

If the current configuration coincides with the reference configuration, e = 0 and U = 0. It canbe observed that the strain energy density is quadratic in the GL strains. To obtain this density interms of displacement gradients, substitute their expressions into the above form to get

U = s0i (hT

i g + gT Hi g) + 12

[(gT hi + 1

2 gT Hi g)Ei j (hTj g + 1

2 gT H j g)]. (8.40)

Since hi and Hi are constant, this relation shows that the strain energy density is quartic in thedisplacement gradients collected in g.

The total strain energy in the current configuration is obtained by integrating the energy densityover the reference configuration:

U =∫

V0

U d X dY d Z . (8.41)

This expression forms the basis for deriving linearly elastic finite elements based on the TotalLagrangian (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∂e

= s0 + E e = s. (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∗ = sT0 E−1(s − s0) + 1

2 (s − s0)T E−1 (s − s0) = 1

2 (sT E−1 s − sT0 E−1 s0), (8.43)

in which the final simplification assumes that E is symmetric. Differentiating U∗ with respect to syields e = ∂U∗/∂s = E−1 s, which does not agree with (8.37) unless s0 = 0. This can be reconciledby 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) stressespertaining to the Seth-Hill (SH) family. For the measure with index m, strains and stresses are collected in the6-vectors e(m) and s(m), respectively. Hooke’s law is assumed to remain valid for those measures:

s(m) = s(m)

0 + E e(m). (8.44),

The strain energy density becomes

U (m) = (s(m)

0 )T e(m) + 12 (e(m))T E e(m). (8.45)

The energy-differentiation stress recovery (8.42) remains unchanged with the only changes e → e(m) ands → 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 caseintroduced in §8.2.3 with the notation (8.23), and exclude m = 0. On replacing e(m) = (λm − 1)/m in thestrain energy we get

U (m) = s(m)

0

λm − 1

m+ 1

2 E

(λm − 1

m

)2

, m = 0. (8.46)

Is the 1D stress-strain law s(m) = s(m)

0 +E e(m) recoverable through energy differentiation? This can be checkedthrough the chain rule:

s(m) guess= ∂U (m)

∂λ

∂λ

∂e(m)=

(s(m)

0 mλm−1

m+ E

λm−1

m2m λm−1

)∂λ

∂e(m)= λm−1

(s(m)

0 +E e(m)) ∂λ

∂e(m). (8.47)

To get the last partial note that

λ = (1 + m e(m))1/m ⇒ ∂λ

∂e(m)= (

1 + m e(m))(1−m)/m = λ1−m, (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 deformationswhile 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 appearin structural constituents; for instance rubber in vehicle tires, and metal foams in energy absorbers. Followingis 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 formulatedwithin the HSM framework, can be studied in [345,578,832].

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 thatis a function of the three invariants of either stretch tensor CR or CL , defined in (8.4). Those are

I1 = λ21 + λ2

2 + λ23, I2 = λ1 λ2 + λ2 λ3 + λ3 λ1, I3 = λ2

1 λ22 λ2

3. (8.50)

in which λi are the principal stretches. Recall that J = det(F) = λ1 λ2 λ3 = √I3.

Let s(m)

0i , i = 1, 2, 3 denote the principal stresses in the reference state, specified in the stress measure conjugateto the SH strain e(m). Those will be called initial stresses.12 The total energy density U is taken as the sum ofinitial-stress, deviatoric and volumetric components: U = U0 + Ud + Uv , with

U0 = s(m)

0i e(m) = s(m)

0i

λmi −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-Rivlinmodel [549,709]. 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 calledthe neo-Hookean material.

One simplifying assumption have been made for the initial stress term U0: the principal directions of initialstress 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 ismessy. To simplify that operation it is beneficial to used a J -scaled form of the deviatoric term:

Ud = C1

(I1

J k1−3

)+ C2

(I2

J k2−3

), (8.52)

The exponents k1 and k2 are adjustable real parameters. The original expression in (8.51) corresponds tok1 = 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 σ B

i and stretches λi are work-conjugate (recall that eBi = λi − 1). Accordingly,

taking the partials of the energy density gives

s Bi = ∂U

∂λi= s B

0i + s0di + s B

vi = ∂U0

∂λi+ ∂Ud

∂λi+ ∂Uv

∂λi= s B

0i + �s Bi . (8.54)

in which i = 1, 2, 3. The sum �s Bi = s B

di + s Bvi gives the incremental Biot stresses, which represent the change

fom the reference state. Full expressions of s Bi 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

s(m)

i = λ1−mi s B

i . (8.55)

12 Most of the HSM literature since 1948 has ignored initial stresses, taking C0 as an unstressed natural state. A notableexception is Biot’s monograph [?], which treats geological and geomechanical problems where those effects, such asgravity, 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 thereciprocal 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, replaceλi by 1 + e(m)

i , 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

s2

s3

=

s01

s02

s03

+ 4(C1 + C2)

3

2 −1 −1

−1 2 −1

−1 −1 2

e1

e2

e3

+ D1 p

1

1

1

, (8.56)

in which p = (s1 + s2 + s3)/3 is the pressure. (Note that superscript (m) is not needed since the limit is thesame for any m.) Matching to 3D isotropic linear elasticity is found to require C1 + C2 = 1

2 E/(2+2ν) = 12 G

and D1 = 12 E/(3−6ν) = 1

2 K , where G and K denote the shear and bulk moduli, respectively. If instead wechose the J -exponent choice (II) in (8.53), namely k1 = 1/3 and k2 = 1, we get s1

s2

s3

=

s01 + C1 + C2

s02 + C1 + C2

s03 + C1 + C2

+ 3(C1 + C2)

2

2 −1 −1

−1 2 −1

−1 −1 2

e1

e2

e3

+ D1 p

1

1

1

, (8.57)

Now taking C1 + C2 = 49 G and D1 = 1

2 K matches the isotropic elasticity matrix, but the initial stresses areincorrect. Consequently choice (II) will not be considered further.

Rubber-like materials are often assumed incompressible, whence deformational motions become isochoricprocesses. If so I3 = λ2

1λ22λ

23 = 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 definedeviatoric 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 constitutivelaw is used later in the formulation of 2-node bar elements. The axial principal stretch λ1 is taken as survivingvariable 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 requiringadditional material constants beyond those of linear elasticity. Taking C2 = 0, C1 = 1

2 G = 14 E/(1 + ν),

D1 = 12 K = 1

6 E/(1 − 2ν), k1 = 2/3, and λ2 = λ3 = λν in (8.52) and (8.52) gives

U(λ, λν) = U0 + Udv = s(m)

0

λm−1

m+ 1

2 G

(I1

J 2/3−3

)+ 1

2 K (J − 1)2. (8.58)

Here I1 = λ2 + 2λ2ν and J = λ λ2

ν still carry λν along. To reduce (8.58) to a function of λ only, λν must belinked 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 + 12 G λ−2/3(λ2−1) + 1

2 K (λ−1)2, s(m)

C = s(m)

0 + λ1−m 13 E

(λ+λ1/3−λ−5/3−1

). (8.59)

(I) The material is incompressible, so λν = 1/√

λ. Then I1 = λ2 + 2/λ and J = 1, giving

UI = U0 + 12 G

(λ2 + 2λ−1 − 3

), s(m)

I = s(m)

0 + λ1−m G(λ − λ−2

). (8.60)

13 Another Neo-Hookean model has been proposed in [350]: U = 12 K (log J )2 + 1

2 G(I1/J 2/3 − 3). For incompressiblematerial it coalesces with the neo-Hookean Mooney-Rivlin model.

8–19

Chapter 8: REVIEW OF CONTINUUM MECHANICS: FIELD EQUATIONS

(a) (b) (c)

Axial stretch λ

Axi

al s

tres

s ra

tio s

/E

Axial stretch λ

Axi

al s

tres

s ra

tio s

/E

Axial stretch λ

Axi

al s

tres

s ra

tio s

/E

G

A S H

B

G

A SH

B

Case (C): unchanged cross section, five SH measures

Case (I): incompressiblematerial, five SH measures

Poisson's ratio =1/4Biot stress

0.6 0.8 1 1.2 1.4 1.6 1.8 2

−1.5−1

−0.50

0.51

1.52

0.6 0.8 1 1.2 1.4 1.6 1.8 2

−1.5−1

−0.50

0.51

1.52

0.6 0.8 1 1.2 1.4 1.6 1.8 2

−1.5−1

−0.50

0.51

1.52

Stress-stretchinterpolation

Invariant andlateral stretchinterpolations

Figure 8.6. Stress-stretch response for simple extension of homogeneous prismatic HSM bar modeled by neo-Hooken, 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 computedby 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 numericalsolution is possible. To get an analytical approximation, an expedient way out is to interpolate between thelimit 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) s(m)

C + 2ν s(m)

I , which expands to

s(m) = s(m)

0 + 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 complicatedand will not be listed here.

(V) Interpolate the active invariants so I1(ν) = (1−2ν)(λ2 +2)+2ν (λ2 +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 finiteelement element implementation covered in Chapter 15. Introduce the abbreviations νh = 1 − 2 ν,J = νh λ + 2 ν, U,λ = ∂U/∂λ, Udv,λλ = ∂2U/∂λ2, Udv,λ = ∂Udv/∂λ, and Udv,λλ = ∂2Udv/∂λ2. Then

U0 = s(m)

0 (λm − 1)/m if m = 0, else U0 = s(m)

0 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) ν2h )/(9 J 8/3 λ3),

U = U0 + Udv, U,λ = s0 λm−1 + Udv,λ, U,λλ = (m − 1) s0 λm−2 s0 + Udv,λλ,

s(m) = s(m)

0 + λm−1 Udv,λ, E (m) = (m − 1) λm−2 Udv,λ + λm−1 Udv,λλ,

(8.62)

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 reasonablywell since the response curves for the same strain measure are similar in shape, indicating mild dependence

8–20

§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) showsa tiny deviation. This can be observed in Figure 8.6(c) for ν = 1

4 . By contrast, the choice of stress measuremakes 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 [241,245] it is useful to have acompact matrix expression for the Green-Lagrange strain components of (8.15) in terms of the displacementgradient cast vector (7.13). To this end, note that (8.14) may be rewritten as

e1 = g1 + 12 (g2

1 + g22 + g2

3), e2 = g5 + 12 (g2

4 + g25 + g2

6),

e3 = g9 + 12 (g2

7 + g28 + g2

9), e4 = g6 + g8 + g4g7 + g5g8 + g6g9,

e5 = g3 + g7 + g1g7 + g2g8 + g3g9, e6 = g2 + g4 + g1g4 + g2g5 + g3g6.

(8.63)

These relations may be collectively embodied in the quadratic form

ei = hTi g + 1

2 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., (8.65)

and the Hi are very sparse 9 × 9 symmetric matrices:

H1 =[

I3 03×6

06×3 06×6

]. . . etc. (8.66)

For strain measures other than Green-Lagrange’s, expressions similar to (8.64) may be constructed. Butalthough 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. Manybooks treat the subject as a closed one, with no connection to physical reality — those tensorial stews should beavoided. Three readable textbooks are Fung [309], Ogden [578] and Prager [662]. The first one is unfortunatelyout of print, but the others are available as Dover reprints. Both [309] and [662] cover elastoplasticity andviscoelasticity although their treatment of finite deformations coupled with those material models is succint.On the other hand, [578] stops with nonlinear elasticity. Sommerfeld’s textbook [756] is nicely written andhighlights physics, but is way outdated. Murnaghan’s text [556] was historically important in introducingdirect matrix notation to finite elasticity. Novozhilov’s monograph [567] has an excellent treatment of finitestrains and rotations, but is hindered by the use of full form notation; some equations span entire pages.

The long expository article [827] is worth perusing since it had significant impact on the post-1952 evolutionof the field, and contains an exhaustive list of references dating from 1676 (Hooke’s writings). The materialtherein was later expanded into two comprehensive book-length surveys that appeared in the Handbuch derPhysik in 1960 [828] and 1965 [832]. These are excellent as reference sources, especially those that need tobe 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 de′i from C0 (e′

i = 0) to C (e′i = ei ) and use of

(8.34). (The integral is path independent.)

EXERCISE 8.3 [A+N:20] Complete the development of the simple shear case started in Examples 7.6 and8.2 by showing that the polar decomposition matrices in F = R U are

R = 1

χ

1 12 γ 0

− 12 γ 1 0

0 0 χ

, U = 1

χ

1 12 γ 0

12 γ 1 + 1

2 γ 2 0

0 0 χ

, (E8.1)

in which γ = tan α and χ =√

1 + 14 γ 2. Using this result, express the SH strains components e(m)

X X , e(m)

Y Y

and γ(m)

XY = 2e(m)

XY as a function of γ , excluding m = 0. Plot these 3 components over the angular rangeα = [−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 α in thesame plot. Are the strains linear in α?

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 00 log λ2 00 0 0

]. (E8.2)

Plot the 3 components over the α range of the previous exercise.

EXERCISE 8.5 [A:15] Taking (by brute-force symbolic computation) the square roots of the CR matrix givenin (T.3) for the simple shear case yields eight candidates for the right stretch tensor U = √

CR :

U1 =[

0 −1 0−1 −γ 00 0 −1

], U2 =

[0 −1 0

−1 −γ 00 0 1

], U3 =

[0 1 01 γ 00 0 −1

], U4 =

[0 1 01 γ 00 0 1

],

U5 =[ −2/µ −γ /µ 0

−γ /µ −(2 + γ 2)/µ 00 0 −1

], U6 =

[ −2/µ −γ /µ 0−γ /µ −(2 + γ 2)/µ 0

0 0 1

],

U7 =[

2/µ γ/µ 0γ /µ (2 + γ 2)/µ 0

0 0 −1

], U8 =

[2/µ γ/µ 0γ /µ (2 + γ 2)/µ 0

0 0 1

],

(E8.3)

in which µ =√

4 + γ 2. Which one would you pick for U? (Hint: it should be p.d. for real γ .) Compare youranswer with (E8.1).

EXERCISE 8.6 [N:15] A continuation of 8.3. Assumed the following linear relation between SH stressesand strains hold: [ s(m)

X X

s(m)

Y Y

s(m)

XY

]=

[E 0 00 E 00 0 1

3 E

][ e(m)

X X

e(m)

Y Y

2e(m)

XY

]. (E8.4)

8–22

Exercises

Denote the associated forces on the faces of the cube as F (m)

X = a s(m)

X X , F (m)

Y = a s(m)

Y Y , and F (m)

XY = a s(m)

XY .Plot the dimensionless ratios ρX = FX (m)/F (m)

XY and ρY = F (m)

X /F (m)

XY for m = 2, 1, −1, −2 over the rangeθ = [−60◦, 60◦]. Present four plots, one for each separate m, showing the 2 ratios in the same plot. Thenonlinear effect of the geometry change should be evident. (These ratios are essential for designing a simpleshear experiment.)

EXERCISE 8.7 [A:20] Let L0 and L denote the length of a bar element in the reference and current con-figurations, respectively. The Green-Lagrange finite strain e = eX X , if constant over the bar, can be definedas

e = L2 − L20

2L20

. (E8.5)

Show that the definitions (E8.5) and of e = eX X in (8.14) are equivalent. Hint: the axial displacement gradientis obviously ∂u X/∂ X = (L − L0)/L0; how about ∂u X/∂Y and ∂u X/∂Y ?

EXERCISE 8.8 [A:20] Find the axial GL strain e = eX X for the non-homogeneous bar extension case treatedin Example 7.8 in terms of L0, L1, L2 and ξ . The result is useful for 3-node bar finite elements. (Hint: it is aquadratic polynomial in ξ .)

EXERCISE 8.9 [A:20] Expand in Taylor series in g, up to and including O(g2), the seven one-dimensionalfinite strain measures shown in the 2nd columns of the table in Figure 8.2, about g = 0. Verify that all measuresagree up to O(g) (the infinitesimal strain) but differ in terms O(g2) and higher.

EXERCISE 8.10 [A:25] Complete the 3rd column of the table of finite strain measures in Figure 8.2 withtensorial forms in terms of V.

EXERCISE 8.11 [A:25] A bar is in a one dimensional stress state. The axial stress computed with SH strainmeasure of index m is s(m , while all other stress components are zero. The axial stretch is λ. Show that theaxial Cauchy stress σ can be recovered as σ = λm s(m) for arbitrary m.

EXERCISE 8.12 [A:25] An extension of Example 8.5 to compressible material. Suppose that now the bar isisotropic but compressible, so that the lateral stretches are

λ2 = λ3 = 1 − 2ν + 2ν√λ1

, (E8.6)

in which ν ∈ [0, 12 ] is an extension of Poisson’s ratio to finite strains (for isochoric motions, ν = 1

2 ). Find theratio r (m)(ν) of predicted to actual Cauchy stress and discuss the cases ν = 1

4 and ν = 0. Is the Biot strainmeasure still the winner?

EXERCISE 8.13 [A:20] Let ε denote the infinitesimal strain tensor. Show that the two-term expansion ofe(m) is

e(m) = ε+ 12 ∇uT ∇u − (1 − 1

2 m) εT ε.

For which m does last term vanish?

EXERCISE 8.14 [A:30] Given CR = FT F and the invariants I1, I2 and I3, show [402] that U = c1 (c2 +c3 CR − C2

R), in which c1 = 1/(I1 I2 − I3), c2 = I1 I3, and c3 = I 21 − I2. (Hint: use the Cayley-Hamilton

theorem.) Is this expression practical?

EXERCISE 8.15 [A:40] (Advanced). Define the stress measure conjugate to the midpoint strain measure eM .defined in the last row of the table in Figure 8.2. (Publishable if successful since it is unsolved to date.)

EXERCISE 8.16 [A:40] (Advanced). Define the stress measure conjugate to the Hencky finite strain measurelog(U) in a compact form. (Publishable if successful since it is unsolved to date.)

8–23