constitutive equations for linear elasticity hooks law
TRANSCRIPT
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)