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4.4 Derivatives of Invariants and Principal Stretches

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4.4.1 Derivatives of the invariants

In the previous section, the derivatives of the reduced invariants and (reduced) principal

stretches with respect to C were used. Recall the definitions of the reduced invariants:

I 1 = I

−1/3

3 I1 (4.145)

I 2 = I

−2/3

3 I2 (4.146)

J el = I

1/2

3 /J th. (4.147)

Differentiation yields

∂I 1

∂CKL

= −13

−4/3

3 I1

∂I3

∂CKL

+ I

−1/3

∂I1

∂CKL

(4.148)

∂I 2

∂CKL

= −23

−5/3

3 I2

∂I3

∂CKL

+ I

−2/3

∂I2

∂CKL

(4.149)

∂Jel

∂CKL

= 12

−1/2

J th

∂I3

∂CKL

(4.150)

and for the second derivatives,

∂2I 1

∂CKLCMN

= 49

−7/3

3 I1

∂I3

∂CKL

∂I3

∂CMN

− 13

−4/3

_ ∂I1

∂CMN

∂I3

∂CKL

+ ∂I1

∂CKL

∂I3

∂CMN

− 13

−4/3

3 I1

∂2I3

∂CKLCMN

+ I

−1/3

∂2I1

∂CKLCMN

(4.151)

∂2I 2

∂CKLCMN

= 10

9 I

−8/3

3 I2

∂I3

∂CKL

∂I3

∂CMN

− 23

−5/3

_ ∂I2

∂CMN

∂I3

∂CKL

+ ∂I2

∂CKL

∂I3

∂CMN

− 23

−5/3

3 I2

∂2I3

∂CKLCMN

+ I

−2/3

∂2I2

∂CKLCMN

(4.152)

∂2J el

∂CKLCMN

= −14

−3/2

J th

∂I3

∂CKL

∂I3

∂CMN

+ 12

−1/2

J th

∂2I3

∂CKLCMN

. (4.153)

Equations (4.148) to (4.153) yield the derivatives of the reduced invariants as a function

of the derivatives of the invariants. The latter yields (Equations (1.507) to (1.509))

∂I1

∂CKL

= GKL (4.154)

200 HYPERELASTIC MATERIALS

∂I2

∂CKL

= I1GKL − CPQGPKGQL (4.155)

∂I3

∂CKL

= I3C

−1KL (4.156)

∂2I1

∂CKLCMN

= 0 (4.157)

∂2I2

∂CKLCMN

= ∂I1

∂CMN

GKL − ∂CPQ

∂CMN

GPKGQL

= GMNGKL − 12

(GMKGNL + GMLGNK) (4.158)

∂2I3

∂CKLCMN

= I3C

−1KL

−1MN + I3

∂C

−1KL

∂CMN

. (4.159)

Since

−1KL

CLA = δK

A (4.160)

differentiation yields

∂C

−1KL

∂CMN

CLA + C

−1KL ∂CLA

CMN

= 0 (4.161)

which leads to

∂C

−1KL

∂CMN

CLA = −12

−1KL _δ M

L δ N

+ δ N

L δ M

= −12

_CKMδ N

+ CKNδ M

_ . (4.162)

Multiplication of both sides with C

−1AB yields

∂C

−1KB

∂CMN

= −12

−1KM

−1NB + C

−1KN

−1MB_ . (4.163)

Accordingly, Equation (4.159) can be rewritten as

∂2I3

∂CKLCMN

= I3 _C

−1KL

−1MN − 12

−1

KMC

−1

NL + C

−1

KNC

−1

ML_ . (4.164)

4.4.2 Derivatives of the principal stretches

The derivatives of the reduced principal stretches can be obtained in a similar way (for an

alternative formulation see (Simo and Taylor 1991)). Starting from

λi = I

−1/6

3 λi = J

−1/3λi (4.165)

HYPERELASTIC MATERIALS 201

one obtains

∂λi

∂CKL

= −16

−7/6

3 λi

∂I3

∂CKL

+ I

−1/6

∂λi

∂CKL

(4.166)

∂2λi

∂CKL∂CMN

= 7

36 I

−13/6

3 λi

∂I3

∂CKL

∂I3

∂CMN

− 16

−7/6

∂λi

∂CMN

∂I3

∂CKL

− 16

−7/6

3 λi

∂2I3

∂CKL∂CMN

− 16

−7/6

∂λi

∂CKL

∂I3

∂CMN

+ I

−1/6

∂2λi

∂CKL∂CMN

. (4.167)

To obtain the derivative of the principal stretches with respect to C, we start from the

characteristic equation

λ6 − I1λ4 + I2λ2 − I3 = 0. (4.168)

Taking the first derivative with respect to C, one obtains

6λ

∂λ

∂CKL

− ∂I1

∂CKL

λ4 − 4I1λ3 ∂λ

∂CKL

+ ∂I2

∂CKL

λ2 + 2λI2

∂λ

∂CKL

− ∂I3

∂CKL

= 0 (4.169)

yielding

∂λ

∂CKL

= _

λ4 ∂I1

∂CKL

− λ2 ∂I2

∂CKL

+ ∂I3

∂CKL

/ _6λ5 − 4I1λ3 + 2λI2_ . (4.170)

Taking the second derivative of Equation (4.168) yields

∂2λ

∂CKL∂CMN

= _−30λ4 + 12I1λ2 − 2I2_ ∂λ

∂CKL

∂λ

∂CMN

+ 4λ3 _ ∂I1

∂CKL

∂λ

∂CMN

+ ∂I1

∂CMN

∂λ

∂CKL

_ + λ4 ∂2I1

∂CKL∂CMN

− λ2 ∂2I2

∂CKL∂CMN

− 2λ

_ ∂I2

∂CKL

∂λ

∂CMN

+ ∂I2

∂CMN

∂λ

∂CKL

+ ∂2I3

∂CKL∂CMN

/ _6λ5 − 4I1λ3 + 2λI2_ . (4.171)

Equations (4.170) and (4.171) only apply on condition that the denominator is not zero.

The denominator is the derivative of the characteristic equation, which can also be written

L = 0 ⇔ (λ2 − λ21

)(λ2 − λ22

)(λ2 − λ23

) = 0 (4.172)

and a zero denominator for λi signifies

∂L

∂λ

____

λ=λi

= 0 (4.173)

202 HYPERELASTIC MATERIALS

which means that λ2

i is at least a double root if we exclude λi = 0. To obtain the derivatives

for double and triple roots, a different approach has to be taken (Itskov 2001). The

eigenvalues of C and its invariants are related by

_1 + _2 + _3 = I1 (4.174)

_1_2 + _1_3 + _2_3 = I2 (4.175)

_1_2_3 = I3. (4.176)

Taking the derivative with respect to C, one obtains





1 1 1

_2 + _3 _1 + _3 _1 + _2

_2_3 _1_3 _1_2









_1,C

_2,C

_3,C









I1,C

I2,C

I3,C





. (4.177)

Three cases can be distinguished

1. If _1 _= _2 _= _3 _= _1, then the solution of Equation (4.177) yields

_i,C = _2

i I1,C − _iI2,C + I3,C

3_2

− 2I1_i + I2

(4.178)

which agrees with Equation (4.170) since λi,C = 2λiλi,C. Expanding the derivatives

of the invariants (Equations (1.507)–(1.509)),

∂I1

∂C

= G_ (4.179)

∂I2

∂C

= I1G_ − G_ · C · G_ (4.180)

∂I3

∂C

= I3C

−1 (4.181)

and taking Equations (4.174) and (4.175) into account, Equation (4.178) can also be

written as

∂_i

∂C

= _i(_i − I1)G_ + _iG_ · C · G_ + I3C

−1

(_i − _j )(_i − _k)

. (4.182)

Since

G_ = G_ · C · C

−1 (4.183)

G_ · C · G_ = G_ · C2 · C

−1 (4.184)

and

−1 = G_ · G · C

−1 (4.185)

HYPERELASTIC MATERIALS 203

one finds by comparison with Equation (1.126),

∂_i

∂C

= _iG_ ·Mi · C

−1. (4.186)

Writing C

−1 in terms of the structural tensors

−1 =_

−1

j Nj ⊗ Nj (4.187)

Equation (4.186) can be further simplified to

_i,C = _iG_ · (Ni ⊗ Ni ) ·_

−1

j (Nj ⊗ Nj )

= _iG__

−1

j (Ni ⊗ Nj )Ni · Nj

= _iG__

−1

i (Ni ⊗ Ni )

= Ni ⊗ Ni

= Mi . (4.188)

It is a remarkably simple expression: for three distinct eigenvalues, the derivatives

of the eigenvalues are the corresponding contravariant structural tensors. Using

Equation (4.188), one also obtains a very elegant expression for the principal stresses

in an Ogden material. Indeed,

S = 2

∂_

∂_i

_i,C = 2

∂_

∂_i

Mi . (4.189)

Accordingly (Equation (1.132)),

_jS = S : Mj = 2

∂_

∂_j

(4.190)

or (Equation (1.428))

λjσ = 2

∂_

∂_j

= 2

∂_

∂ ln_j

= 1

∂_

∂ ln λj

. (4.191)

2. If two eigenvalues are equal, for example, _ = _1 = _2 _= _3 Equation (4.177)

reduces to





1 1 1

_ + _3 _ + _3 2_

__3 __3 _2









_1,C

_2,C

_3,C









I1,C

I2,C

I3,C





(4.192)

204 HYPERELASTIC MATERIALS

and column 1 and 2 are identical: the system is singular. It can be reduced to

1 1

_ + _3 2_

__1,C + _2,C

_3,C

_ = _I1,C

I2,C

. (4.193)

The solution satisfies

_1,C + _2,C = 2_I1,C − I2,C

_ − _3

(4.194)

_3,C = I2,C − (_ + _3)I1,C

_ − _3

. (4.195)

It is not difficult to prove that

_1,C + _2,C = G_ −M3 = M1 +M2 (4.196)

_3,C = M3 (4.197)

where M1 +M2 and M3 satisfy Equation (1.129) and Equation (1.130).

3. For three equal eigenvalues _ = _1 = _2 = _3, Equation (4.193) reduces to one

single equation:

_1,C + _2,C + _3,C = I1,C = G_. (4.198)

Now let us take a look at the second derivatives of λi . Instead of using Equation (4.171),

one can also express it through the second derivative of _i = λ2

i :

λi,CC = 1

√

_i,CC − 1

4_i

√

_i,C ⊗ _i,C (4.199)

obtained by differentiating

λi,C = 1

√

_i,C (4.200)

with respect to C. Again, three cases can be distinguished

1. For _1 _= _2 _= _3 _= _1 one obtains (Equation (4.188))

_i,CC = Mi,C. (4.201)

An expression forMi,C is found by differentiating Equation (1.125) leading to (notice

that Mi,C = G_ ·Mi

· G_)





1 1 1

_1 _2 _3

_21

_22

_23





















(4.202)

HYPERELASTIC MATERIALS 205

where

A = C,C −_

Mi ⊗Mi (4.203)

B = C2

− 2_

_iMi ⊗Mi (4.204)

and

C,C := ∂C

∂C

(4.205)

and similar expressions for the other terms. Straightforward calculation yields for

C,C and C2

,C,

∂CKL

∂CPQ

= 12

(δP

KδQ

+ δP

LδQ

K) =: II (4.206)

and

∂CKLCNMGLN

∂CPQ

= 12

(δP

KCQ

+ δQ

KCP

M) + 12

(δQ

MC P

+ δP

MC Q

K ). (4.207)

Notice the following shorthand notation:

(II )IJ

KL := (Iδ)IJ

KL := 12

(δI

KδJ

+ δI

LδJ

K) (4.208)

(IG)IJKL := 12

(GIKGJL + GILGJK) (4.209)

G_ )IJKL := 12

(GIKGJL + GILGJK), (4.210)

and similarly for other tensor fields. The solution of Equation (4.202) amounts to

= 1

[B − (I1 − _i )A] (4.211)

where

Di = (_i − _j )(_i − _k), i = 1, 2, 3; j, k _= i. (4.212)

2. For _ = _1 = _2 _= _3, Equation (4.202) reduces to

1 1

_ _3

_M1

+M2

= _0

_ (4.213)

leading to

+M2

= −A/(_3 − _) (4.214)

= A/(_3 − _). (4.215)

206 HYPERELASTIC MATERIALS

_ = _1 = _2

Figure 4.7 Tangent ambiguity for identical eigenvalues

3. For _ = _1 = _2 = _3, one obtains in a similar way

+M2

+M3

= 0. (4.216)

Notice that for double or triple roots, the derivatives of _i (Equation (4.196) and

Equation (4.198)) and Mi (Equation (4.214) and Equation (4.216)) are not known separately:

only the sum is known. This is not surprising, since double roots cannot be

distinguished, and consequently it is not clear whether tangent 1 or tangent 2 applies

(Figure 4.7).

4.4.3 Expressions for the stress and stiffness for three equal

eigenvalues

In the previous section, it was found that for three equal eigenvalues the derivatives of λi

are not known separately, only their sum can be calculated. In the present section, it will

be shown that this suffices to determine the stress and the stiffness. For λ1 = λ2 = λ3 = λ,

Equation (4.132) and Equation (4.133) reduce to

S = 2

_ n

i=1

2μi

αi

λi−1

k=1

λk,C +

i=1

(J el − 1)2i−1J el

(4.217)

and

_ = 4

_ N

i=1

2μi

αi

(αi − 1)λ

αi−2 _ 3

k=1

λk,C ⊗ λk,C

+ λ

αi−1

k=1

λk,CC

i=1

2i(2i − 1)

(J el − 1)2i−2J el

⊗ J el

+ 2i

(J el − 1)2i−1J el

,CC

. (4.218)

HYPERELASTIC MATERIALS 207

Furthermore, Equation (4.166) and Equation (4.167) now lead to

k=1

λk,C = −12

−7/6

3 λI3,C + I

−1/6

k=1

λk,C (4.219)

k=1

λk,C ⊗ λk,C = 1

12 I

−7/3

3 λ2I3,C ⊗ I3,C + I

−1/3

k=1

λk,C ⊗ λk,C

− 16

−4/3

3 λI3,C ⊗

k=1

λk,C − 16

−4/3

3 λ

k=1

λk,C ⊗ I3,C (4.220)

k=1

λk,CC = 7

12 I

−13/6

3 λI3,C ⊗ I3,C − 16

−7/6

k=1

I3,C ⊗ λk,C

− 12

−7/6

3 λI3,CC − 16

−7/6

k=1

λk,C ⊗ I3,C + I

−1/6

k=1

λk,CC.

(4.221)

In this way, the sums of the derivatives of the reduced stretches are written in terms of

the derivatives of the unreduced stretches. Now, Equations (4.198), (4.199), (4.200) and

(4.216) show that

k=1

λk,C = 1

2λ

k=1

_k,C = 1

2λG_ (4.222)

k=1

λk,CC = − 1

4λ3

k=1

_k,C ⊗ _k,C = − 1

4λ3

k=1

Mk ⊗Mk. (4.223)

For λ = λ1 = λ2 = λ3, Equation (4.202) reduces to rank one and, consequently, A = 0

k=1

Mk ⊗Mk = C,C (4.224)

which is equivalent to

k=1

Mk ⊗Mk = (G_ ⊗ G_) : C,C. (4.225)

Accordingly,

k=1

λk,CC = − 1

4λ3 (G_ ⊗ G_) : II . (4.226)

208 HYPERELASTIC MATERIALS

In a similar way, one arrives at

k=1

λk,C ⊗ λk,C = 1

4λ2

k=1

_k,C ⊗ _k,C

= 1

4λ2 (G_ ⊗ G_) : II . (4.227)

Hence, Equations (4.219) to (4.221) yield

k=1

λk,C = −12

−1/6

3 λC

−1 + I

−1/6

2λ

G_ (4.228)

k=1

λk,C ⊗ λk,C = 1

−1/3

3 λ2C

−1 ⊗ C

−1 + 1

4λ2 I

−1/3

3 I

− 1

12 I

−1/3

3 C

−1 ⊗ G_ − 1

−1/3

3 G_ ⊗ C

−1 (4.229)

k=1

λk,CC = 7

12 I

−1/6

3 λC

−1 ⊗ C

−1 − 1

12λ

−1/6

3 C

−1 ⊗ G_

− 12

λI

−1/6

3 C

−1 ⊗ C

−1 + 12

λI

−1/6

3 I

−1

− 1

12λ

−1/6

3 G_ ⊗ C

−1 − 1

4λ3 I

−1/6

3 I

G_ . (4.230)

In a similar way, the expressions for the stress and stiffness for an elastomeric foam

for ˆλ = ˆλ1 = ˆλ2 = ˆλ3 reduce to

S = 2

i=1

2μi

αi

_ˆλ

k=1

ˆλ

k,C − (J el)

−αiβi−1J el

(4.231)

_ = 8

i=1

μi

αi

(αi − 1)ˆλαi−2

k=1

ˆλ

k,C ⊗ ˆλk,C + ˆλαi−1

k=1

ˆλ

k,CC

+ (αiβi + 1)(J el)

−αiβi−2J el

⊗ J el

− (J el)

−αiβi−1J el

,CC

. (4.232)

Since J th depends on the temperature only, Equations (4.222) and (4.227) also apply

to ˆλ.

HYPERELASTIC MATERIALS 209

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