Constitutive Equations for Linear Elasticity and Newtonian Fluids

I. The Generalized Hooke’s Law

The generalized Hooke’s law is a linear constitutive equation relating the Cauchy stress

to the infinitesimal strain.

C (1)

Since this is a double inner product it is sometime written as

C : (2)

and the fourth order tensor C is expressed in dyadic notation as

ijkl i j k lˆ ˆ ˆ ˆC C e e e e (3)

or

ijkl i j k lˆ ˆ ˆ ˆC C e e e e (4)

The double inner product may be written using dyadic notation as

ijkl i j k l mn m n

ijkl mn i j k m l n

ijkl mn i j km ln

ijkl kl i j

ˆ ˆ ˆ ˆ ˆ ˆC e e e e : e e

ˆ ˆ ˆ ˆ ˆ ˆC e e e e e e

ˆ ˆC e e

ˆ ˆC e e

(5)

Hence

ij ijkl kl

C (6)

Equation (6) is the usual starting point for discussing the constitutive equations of linear

elasticity and is referred to as the generalized Hooke’s law. ijkl

C is often termed the

elastic tensor or sometimes it is termed Hooke’s elastic tensor. ijkl

C is a fourth order

tensor with 81 components. Since ij ji

27 of the components are redundant as this

forcesijkl jikl

C C . Further redundancies are found if we recall that kl lk , meaning that

we must have ijkl ijlk

C C and 18 more coefficients may be eliminated. This leaves us with

36 independent components in the tensorijkl

C . Further reductions depend on material

symmetries or properties.

II. Reduction of the Number of Coefficients in the Generalized

Hooke’s Law for Hyperelastic Materials

The material models used in linear elasticity are based upon the hyperelasticity model,

i.e. there exists a strain energy function

ijkl ij kl

1W C

2 (7)

and the stress is the derivative with respect to the strain of the strain energy function

mn

mn

W

(8)

This results in a further reduction of the number of independent coefficients as follows.

First we perform the differentiation

ijkl ij kl

mn

mn

ijkl ij kl

kl ij

mn mn

kl mnkl ij ijmn

mnkl klmn kl

C1

2

C

2

1C C

2

1C C

2

(9)

We also have the generalized Hooke’s law (equation (6))

mn mnkl kl

C (10)

Comparison of equations (9) and (10) indicates that

mnkl klmn

C C (11)

And this results in a further reduction from 36 independent coefficients to 21 independent

coefficients. This reduction is seen more easily by using contracted notation. Contracted

notation allows us to write the generalized Hooke’s law using matrices as follows. First

we make the notational transformation:

1 111 11

2 222 22

3 333 33

4 423 23

5 513 13

6 612 12

(12)

Next, we write the generalized Hooke’s law using this notational transformation

1 11 12 13 14 15 16 1

2 21 22 23 24 25 26 2

3 31 32 33 34 35 36 3

4 41 42 43 44 45 46 4

5 51 52 53 54 55 56 5

6 61 62 63 64 65 66 6

c c c c c c

c c c c c c

c c c c c c

c c c c c c

c c c c c c

c c c c c c

(13)

or

k ki i

C (14)

The strain energy now is written

ki k i

1W C

2 (15)

and the stress is the derivative with respect to the strain of the strain energy function

m

m

W

(16)

Next, we perform the differentiation

ki k i

m

m

ki i k

k i

m m

k mk i im

mk km k

C1

2

C

2

1C C

2

1C C

2

(17)

We also have the generalized Hooke’s law (equation(14))

k ki i

C (18)

Comparison of equations (17) and (18) indicates that

mk km

C C (19)

And the reduction from 36 independent coefficients to 21 independent coefficients is

easily seen in the matrix form of the generalized Hooke’s law:

1 11 12 13 14 15 16 1

2 12 22 23 24 25 26 2

3 13 23 33 34 35 36 3

4 14 24 34 44 45 46 4

5 15 25 35 45 55 56 5

6 16 26 36 46 56 66 6

c c c c c c

c c c c c c

c c c c c c

c c c c c c

c c c c c c

c c c c c c

(20)

III. Reduction of the Number of Coefficients in the Generalized

Hooke’s Law for Hyperelastic Materials with Symmetry

Recall that Hooke’s Elastic tensor is a fourth order tensor

ijkl i j k lˆ ˆ ˆ ˆC C e e e e (21)

It may be shown that when subjected to an orthogonal transformation, Q, this fourth order

tensor transforms as

i j k l pqrs pi qj rk sl

C C Q Q Q Q (22)

This transformation may be used to exploit material symmetries with a view towards

reducing the number of independent constants to describe a material constitutive equation

using Hooke’s law.

Consider first a monoclinic material. This material has one plane of symmetry. Let us

choose that plane to be the 1 2

x x plane and let Q define the transformation

1 1 2 2 3 3

x x , x x , x x (23)

or

1 0 0

Q 0 1 0

0 0 1

(24)

where

klmn mnkl

C C (25)

Since this transformation is a reflection about a plane of symmetry, Hooke’s elastic

tensor must have the property

i j k l ijkl

C C (26)

where

i j k l pqrs pi qj rk sl

C C Q Q Q Q (27)

Consider the element 1111

C , we must have

1 1 1 1 pqrs p1 q1 r1 s1

C C Q Q Q Q (28)

The only non-zero contributions are when p = q = r = s = 1 ,thus

1 1 1 1 1111 11 11 11 11 1111

C C Q Q Q Q C (29)

and this is obviously true. Next consider 1123

C , we must have

1 1 2 3 1123

C C (30)

where

1 1 2 3 pqrs p1 q1 r 2 s3

C C Q Q Q Q (31)

The only non-zero contributions are p =1, q = 1, r =2, s = 3 ,thus we have

1 1 2 3 1123 11 11 22 33 1123

C C Q Q Q Q C (32)

Equations (30) and (32) can only be satisfied if 1123

C 0 . Continuing in this manner we

would find eight zero elements and the generalized Hooke’s law for the monoclinic case

would reduce to

1111 1122 1133 111211 11

2211 2222 2233 221222 22

3311 3322 3333 331233 33

2323 231323 23

2313 131313 13

1112 2212 3312 121212 12

c c c 0 0 c

c c c 0 0 c

c c c 0 0 c

0 0 0 c c 0

0 0 0 c c 0

c c c 0 0 c

(33)

Or, using the other notation to:

1 111 12 13 16

2 212 22 23 26

3 313 23 33 36

4 444 45

5 545 55

6 616 26 36 66

C C C 0 0 C

C C C 0 0 C

C C C 0 0 C

0 0 0 C C 0

0 0 0 C C 0

C C C 0 0 C

(34)

Note that shear strains may cause normal stresses for this material. Consider next an

orthotropic material. This material has three planes of symmetry. An example of a

material with three planes of symmetry is wood. Shown in the following figure is a cross-

section cut along the axis of the fibers thus this cross-section is a plane of symmetry. The

other planes of symmetry should be clear from the figure.

Starting with the 1 2

x x symmetry plane we would obtain again the result given by

equation (33). Assuming next that the 2 3

x x plane is a plane of symmetry we would use

the transformation

x1

x2

x3

1 0 0

Q 0 1 0

0 0 1

(35)

and we would find that

16 26 36 45

C C C C 0 (36)

Finally, if we use the 1 3

x x plane we would obtain no new information. This result might

lead us to be so bold as to suggest that materials possessing two planes of symmetry

necessarily possess a third plane of symmetry. The resulting contracted form of Hooke’s

law for an orthotropic material contains 13-4=9 independent constants and is:

1 111 12 13

2 212 22 23

3 313 23 33

4 444

5 555

6 666

C C C 0 0 0

C C C 0 0 0

C C C 0 0 0

0 0 0 C 0 0

0 0 0 0 C 0

0 0 0 0 0 C

(37)

Note that unlike the monoclinic material the orthotropic material will not generate normal

stresses in the presence of shear strains only. This expression may be inverted to get the

strains in terms of the stresses and the coefficients may now be written using the usual

engineering constants.

1312

1 2 1

2312

1 12 2 2

2 213 23

3 1 2 3 3

4 4

235 5

6 6

13

12

10 0 0

E E E

10 0 0

E E E

10 0 0

E E E

10 0 0 0 0

G

10 0 0 0 0

G

10 0 0 0 0

G

(38)

The reader should be able to see the usual interpretation of the Young’s moduli and the

Poisson ratios.

Consider next a transversely isotropic material. This material has three planes of

symmetry and there is an axis about which arbitrary rotations do not change the material

properties. Such a case is shown below (typically a fiber embedded isotropic material – a

composite material). The axis about which the properties are isotropic is3

x and the

orthogonal rotation for this case would be

cos sin 0

Q sin cos 0

0 0 1

(39)

Applying this transformation to Hooke’s elastic tensor for the orthotropic case will result

in the four conditions:

11 22 13 23 44 55 66 11 12

1C C , C C , C C , C C C

2 (40)

The resulting contracted form of Hooke’s law for a transversely isotropic material

contains 9-4 = 5 independent constants and is:

x1

x2

x3

11 12 131 1

12 11 132 2

13 13 333 3

444 4

445 5

6 611 12

C C C 0 0 0

C C C 0 0 0

C C C 0 0 0

0 0 0 C 0 0

0 0 0 0 C 0

10 0 0 0 0 C C

2

(41)

Note that this form is only for the case where the axis of symmetry is the

3x axis.

Finally consider next an isotropic material. This material will be isotropic with respect to

rotations about the three axes1 2 3

x , x , x . Note that the material cannot contain fibers with

different properties for this type of symmetry to exist. The orthogonal transformation for

the rotation about the 1

x axis is

1 0 0

Q 0 cos sin

0 sin cos

(42)

Applying this transformation to Hooke’s elasticity tensor obtained from the transversely

isotropic case we get the conditions

22 33 12 13 55 66 66 22 23

1C C , C C , C C , C C C

2 (43)

If we then proceed to apply the orthogonal transformation corresponding to the rotation

about the 2

x axis we would find no new information. This result might lead us to be so

bold as to suggest that materials possessing two axes of symmetry necessarily possess a

third axes of symmetry. Combining this result with the result given by equation (40) and

eliminating the redundancy we have the seven constraints

11 22 33 12 13 23 44 55 66 11 12

1C C C , C C C , C C C C C

2 (44)

Thus Hooke’s elasticity tensor will have 9-7 = 2 independent constants. The constants are

represented using the Lame constants , as follows

11 22 33

12 13 23

44 55 66 11 12

C C C 2

C C C

1C C C C C

2

(45)

The resulting contracted form of Hooke’s law for a isotropic material containing 2

independent constants is:

1 1

2 2

3 3

4 4

5 5

6 6

2 0 0 0

2 0 0 0

2 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

(46)

Hooke’s law for an isotropic material may also be written using indicial notation as

ij kk ij ij

2 (47)

And Hooke’s elasticity tensor for an isotropic material takes the form

ijkl ij kl ik jl il jkC (48)

Recall that the most general fourth order isotropic tensor may be written in terms of three

linearly independent terms as

ij kl ik jl il jk (49)

The symmetry of the stress tensor, ij ji

, reduces this expression to the preceding

expression.

Hooke’s law may be inverted to obtain the strains in terms of the stresses with the result

ij kk ij ij

1

2 3 2 2

(50)

Where 3 2 0 . This expression may be used to define Young’s modulus and

Poisson’s ratio in terms of the Lame constants. To see this consider 11

11 11 22 33 11

22 33 11

22 33 11

22 33 11

1( )

2 3 2 2

1( )

2 3 2 2 2 3 2

2 ( )3 2 3 2

1( )

E E

(51)

where

3 2

E

(52)

and

2

(53)

The usual from of Hooke’s law for isotropic linear elastic bodies may be found by adding

and subtracting the same term to the left and right hand sides of equation (51). The result

is

11 22 33 11 11

1( )

E E

(54)

A similar set of operations in the other directions results in the usual form of Hooke’s law

for isotropic linear elastic bodies.

ij kk ij ij

1

E E

(55)

Finally, equation (47) may be written using Young’s modulus and Poisson’s ratio as

ij kk ij ij

E E

1 1 2 1

(56)

Example 1

Plane stress is define as a state of stress where zz zy zx

0 . Determine the form

that Hooke’s law must take for this case – note that zz

0 . Show that xz yz

, 0 and

find formulas for xx yy xy zz

, , , given xx yy xy

, , . Show

xx xx

yy yy2

xy xy

1 0E

1 01

10 0

2

(57)

Note that the engineering strain, xy

=xy

2

Solution

The pertinent form of the constitutive equation is

ij kk ij ij

1

E E

Expanded for this case we get the six strains:

xx xx yy xx

yy xx yy yy

zz xx yy

xy xy

xz xz

yz yz

1( 1 ) ( )

E E

1( 2 ) ( )

E E

( 3 ) ( )E

1( 4 )

E

1( 5 ) 0

E

1( 6 ) 0

E

Solving (1) and (2) for xx

and yy

we obtain

xx xx yy2

yy yy xx2

E

( 1 )

E

( 1 )

And from (6) we have

xy xy xy

E E

1 2 1

Putting these into matrix form we get the desired result

xx xx

yy yy2

xy xy

1 0E

1 01

10 0

2

End of Example 1

Similarly, plane strain is defined as a state of strain where zz xz yz

, , 0 , note that

zz0 but

xz yz, 0 .

xx yy xy, , may be found given

xx yy xy, , with the matrix

formula:

xx xx

yy yy

xy xy

1 0E

1 01 1 2

1 20 0

2

(58)

Example 2

Develop the appropriate form of Hooke’s law in terms of the Lame’ constants for plane

strain and for plane stress.

Solution

(i) Plane Strain - we have, using equation (47)

11 11 11 22 33 11 11 22

0

22 22 11 22 33 22 11 22

0

33 33 11 22 33 11 22

0 0

12 12

13

23

2 2

2 2

2

2

0

0

(59)

(i) Plane Stress - we have, using equation (47)

11 11 11 22 33

22 22 11 22 33

33 22 11 22 33

12 12

13

23

2

2

0 2

2

0

0

Eliminating 33 using equation expressing

330

33 11 222

(60)

This may be substituted into the previous expression to arrive at Hooke’s law for plane

stress in terms of Lame’ constants:

11 11 11 22 11 11 22

22 22 11 22 22 11 22

33

12 12

13

23

22 2

2

22 2

2

0

2

0

0

(61)

End of Example 2

Note the definition 2

2

. The reader should be careful that the appropriate Lame

constants are implemented in solutions.

Example 3

Show that Hooke’s law for an isotropic media (ij kk ij ij

2 ) may be inverted to

give

ij kk ij ij

1

2 ( 3 2 ) 2

Solution

Starting with ij kk ij ij

2 we expand to obtain

xx xx yy zz

yy yy xx zz

zz zz xx yy

xy xy

xz xz

yz yz

( 1 ) 2

( 2 ) 2

( 3 ) 2

( 4 ) 2

( 5 ) 2

( 6 ) 2

Solution of equations (4) through (6) for the shear strains is trivial. Solution of equations

(1) through (3) for the normal strains will requires some algebra. The first step is to add

the three equations to get the result

kk kk3 2

The next step is to use this to eliminate kk in the expression

ij kk ij ij2 , i.e.

kk

ij kk ij ij ij ij2 2

3 2

The final step is to solve for the strains

ij kk

ij ij2 2 3 2

This is the desired result

End of Example 3

IV. Hooke’s Law for Isotropic Linear Materials with Thermal Effects

Included.

We consider here only the simplest case of a linear isotropic material subject to a

temperature field that is different then the reference temperature. The only non-zero

strains are given by

xx o

yy o

zz o

T T

T T

T T

(62)

whereo

T is the reference temperature and T is the temperature field and is the thermal

expansion coefficient. The temperature field may be found from solution of the

conduction equation

2Tc k T

t

(63)

Using equation (55) the temperature effects are included in Hooke’s law as

ij ij kk ij ij

1T

E E

(64)

This simplest formulation results in the temperature field solution being uncoupled from

the displacement and stress solution. This is why in a typical linear FEA package thermal

stress problems require the solution of the conduction equation as an input to the stress

and deformation solution. It is possible for large deformation and for different material

models for the energy equation to include mechanical variables in which case the above

mentioned decoupled solution is not possible.

IV. Newtonian Fluids

The constitutive equation for a Newtonian fluid may be expressed as

C : D (65)

Recall again that the most general fourth order isotropic tensor may be written in terms of

three linearly independent terms as

ijkl ij kl ik jl il jk

C (66)

The symmetry of the stress tensor, ij ji

, reduces this expression to

ijkl ij kl ik jl il jkC (67)

An additional assumption that well describes Newtonian fluids is Stokes hypothesis:

2

3 (68)

This restriction reduces the number of coefficients to one resulting in the constitutive

equation for a Newtonian viscous fluid:

ij kk ij ij

D 2 D (69)

Or, after implementing Stoke’s hypothesis:

ij kk ij ij

2D 2 D

3 (70)