![]() |
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Chapter 3
Constitutive
Models
3.2 Linear elastic material behavior
You are probably familiar with the behavior of a linear elastic material from introductory materials courses.
3.2.1 Isotropic, linear elastic material behavior
If you conduct a uniaxial tensile test on almost any material, and keep the stress levels sufficiently low, you will observe the following behavior:
3.2.2 Stress
Before writing down stress
You probably already know the stress
Here, E
and
The inverse relationship can be expressed as
HEALTH WARNING: Note the factor of 2 in the strain vector. Most texts, and most FEM codes use this factor of two, but not all. In addition, shear strains and stresses are often listed in a different order in the strain and stress vectors. For isotropic materials this makes no difference, but you need to be careful when listing material constants for anisotropic materials (see below).
We can write this expression in a much more convenient form using index notation. Verify for yourself that the matrix expression above is equivalent to
The inverse relation is
The stress-strain relations are often expressed using the elastic modulus tensor
In terms
of elastic constants,
3.2.3 Reduced stress-strain equations for plane deformation of isotropic solids
For plane strain or plane stress deformations, some strain or stress components are always zero (by definition) so the stress-strain laws can be simplified.Â
In index notation
where
Greek subscripts
3.2.4 Representative values for density, and elastic constants of isotropic solids
Most of the data in the table below were taken from the excellent introductory text `Engineering Materials,’ by M.F. Ashby and D.R.H. Jones, Pergamon Press. The remainder are from random web pages…
Note the units
3.2.5
Other Elastic Constants
|
|
Lame Modulus
|
Shear Modulus
|
Young’s Modulus
|
Poisson’s Ratio
|
Bulk Modulus
|
|
|
|
|
|
|
|
|
Irrational |
|
Irrational |
Irrational |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
3.2.6 Physical Interpretation of elastic constants for isotropic solids
It is important to have a feel for the physical
significance of the two elastic constants E
and .
  Young’s
modulus E is the slope of the stress
strain
curve in uniaxial tension. It has
dimensions of stress (
 ) and is usually large
 for steel,
.
You can think of E as a measure of
the stiffness of the solid. The larger the value of E, the stiffer the solid. For
a stable material, E>0.
 Poisson’s ratio
 is the ratio of lateral to longitudinal
strain in uniaxial tensile stress. It is dimensionless and typically ranges
from 0.2
0.49,
and is around 0.3 for most metals. For
a stable material,
.
It is a measure of the compressibility of the solid. Â If
,
the solid is incompressible
 its volume remains constant, no matter how
it is deformed. If
,
then stretching a specimen causes no lateral contraction. Some bizarre materials have
 -- if
you stretch a round bar of such a material, the bar increases in diameter!!
 The bulk
modulus quantifies the resistance of the solid to volume changes. It has a large value (usually bigger than E).
 The shear
modulus quantifies its resistance to volume preserving shear deformations. Its value is usually somewhat smaller than E.Â
3.2.7 Strain Energy Density for Isotropic Solids
Note the following observations
 If you
deform a block of material, you do work on it (or, in some cases, it may do
work on you…)Â
 In an
elastic material, the work done during loading is stored as recoverable
strain energy in the solid. If you
unload the material, the specimen does work on you, and when it reaches its
initial configuration you come out even.
 The
work done to deform a specimen depends only on the state of strain at the end
of the test. It is independent of the
history of loading.Â
Based on these observations, we define the strain energy density of a solid as the work done per unit volume to deform a material from a stress free reference state to a loaded state.
To write down an expression for the strain energy density, it is convenient to separate the strain into two parts
where, for an isotropic solid,
represents the strain due to thermal expansion (known as thermal strain), and
is the strain due to mechanical loading (known as elastic strain).
Work is done on the specimen only during mechanical loading. It is straightforward to show that the strain energy density is
You can also re-write this as
Observe that
3.2.8
Stress-strain relation for a general anisotropic linear elastic material  the elastic stiffness and compliance tensors
The
simple isotropic model described in the preceding section is unable to
describe the response of some materials accurately, even though the material
may deform elastically. This is
because some materials do have a characteristic orientation. For example, in a block of wood, the grain
is oriented in a particular direction in the specimen. The block will be stiffer if it is loaded
parallel to the grain than if it is loaded perpendicular to the grain. The same observation applies to fiber
reinforced composite materials. Generally, single crystal specimens of a
material will also be anisotropic  this is important when modeling stress
effects in small structures such as microelectronic circuits. Even
polycrystalline metals may be anisotropic, because a preferred texture may
form in the specimen during manufacture.
A
more general stressstrain
relation is needed to describe anisotropic solids.Â
The most general linear stressstrain
relation has the form
Here,
 is a fourth order tensor (horrors!), known
as the elastic stiffness tensor, and
 is the thermal expansion coefficient tensor.
The stress strain relation is invertible:
where
 is known as the elastic compliance tensor
At
first sight it appears that the stiffness tensor has 81 components. Imagine having to measure and keep track of
81 material properties! Fortunately,  must have the following symmetries
This
reduces the number of material constants to 21. The compliance tensor has the
same symmetries as .
To see the origin of the symmetries of ,
note that
 The
stress tensor is symmetric, which is only possible if
 If a
strain energy density exists for the material, the elastic stiffness tensor
must satisfy
 The
previous two symmetries imply
,
since
 and .
To see that ,
note that by definition
and recall further that the stress is the derivative of the strain energy density with respect to strain
Combining these,
Now, note that
so that
These symmetries allow us to write the stress-strain relations in a more compact matrix form as
where ,
etc are the elastic stiffnesses of
the material. The inverse has the form
where ,
etc are the elastic compliances of
the material.
To satisfy Drucker stability, the eigenvalues of the elastic stiffness and compliance matrices must all be greater than zero.
HEALTH WARNING: The shear strain and shear stress components are not always listed in the order given when defining the elastic and compliance matrices. The conventions used here are common and are particularly convenient in analytical calculations involving anisotropic solids. But many sources use other conventions. Be careful to enter material data in the correct order when specifying properties for anisotropic solids.
3.2.9 Physical Interpretation of the Anisotropic Elastic Constants.
|
It
is easiest to interpret ,
rather than
. Imagine applying a uniaxial stress, say
,
to an anisotropic specimen. In
general, this would induce both extensional and shear deformation in the
solid, as shown in the figure.
The strain induced by  the uniaxial stress would be
All the constants have dimensions . The constant
 looks like a uniaxial compliance, (like
 ), while the ratios
 are generalized versions of Poisson’s ratio:
they quantify the lateral contraction of a uniaxial tensile specimen.  The shear terms are new
 in an isotropic material, no shear strain is
induced by uniaxial tension.
3.2.10 Strain energy density for anisotropic, linear elastic solids
The strain energy density of an anisotropic material is
3.2.11 Basis change formulas for anisotropic elastic constants
|
The
material constants  or
 for a particular material are usually
specified in a basis with coordinate axes aligned with particular symmetry
planes (if any) in the material. When
solving problems involving anisotropic materials it is frequently necessary
to transform these values to a coordinate system that is oriented in some
convenient way relative to the boundaries of the solid. Since
 is a fourth rank tensor, the basis change
formulas are highly tedious, unfortunately.Â
Suppose
that the components of the stiffness tensor are given in a basis ,
and we wish to determine its components in a second basis,
. We define the usual transformation tensor
 with components
,
or in matrix form
This
is an orthogonal matrix satisfying .
In practice, the matrix can be computed in terms of the angles between the
basis vectors. It is straightforward to show that stress, strain, thermal
expansion and elasticity tensors transform as
The basis change formula for the elasticity tensor in matrix form can be expressed as
where the basis change matrix K is computed as
and the modulo function satisfies
Although these expressions look cumbersome they are quite convenient for computer implementation.
The basis change for the compliance tensor follows as
where
The proof of these expressions is merely tiresome algebra and will not be given here. Ting’s book `Anisotropic Elasticity: Theory and Applications’ OUP (1996) has a nice clear discussion.
For
the particular case of rotation through an angle  in a counterclockwise sense about the
 axes, respectively, the rotation matrix
reduces to
 Â
    Â
where
.
The inverse matrix
 can be obtained simply by changing the sign
of the angle
 in each rotation matrix. Clearly, applying the three rotations
successively can produce an arbitrary orientation change.
For an isotropic material, the elastic stress-strain relations, the elasticity matrices and thermal expansion coefficient are unaffected by basis changes.
3.2.12 The effect of material symmetry on stress-strain relations for anisotropic materials
|
A general anisotropic solid has 21 independent elastic constants. Note that in general, tensile stress may induce shear strain, and shear stress may cause extension.
If a material has a symmetry plane, then applying stress normal or parallel to this plane induces only extension in direction normal and parallel to the plane
For
example, suppose the material contains a single symmetry plane, and let    be normal to this plane.
Then
the components of the elastic stiffnes matrix  (
 ). (symmetrical terms also vanish, of
course). This leaves 13 independent constants.Â
Similar restrictions on the thermal expansion coefficient can be determined using symmetry conditions. Details are left as an exercise.
In the following sections, we list the stress-strain relations for anisotropic materials with various numbers of symmetry planes.
3.2.13 Stress-strain relations for linear elastic orthotropic materials
|
An orthotropic material has three mutually perpendicular symmetry planes. This type of material has 9 independent material constants. With basis vectors perpendicular to the symmetry plane, the elastic stiffness matrix has the form
This relationship is sometimes expressed in inverse form, in terms of generalized Young’s moduli and Poisson’s ratios (which have the same significance as Young’s modulus and Poisson’s ratio for uniaxial loading along the three basis vectors) as follows                                                                  Â
Here the generalized
Poisson’s ratios are not symmetric but instead satisfy  (no sums). This ensures that the stiffness
matrix is symmetric.
The engineering constants are related to the components of the compliance tensor by
or in inverse form
For an orthotropic material thermal expansion cannot induce shear (in this basis) but the expansion in the three directions need not be equal. Consequently the thermal expansion coefficient tensor has the form
3.2.14 Stress-strain relations for linear elastic Transversely Isotropic Material
A special case of an orthotropic solid is one
that contains a plane of isotropy (this implies that the solid can be rotated
with respect to the loading direction about one axis without measurable
effect on the solid’s response).Â
Choose  perpendicular to this symmetry plane. Then, transverse isotropy requires that
,
,
,
,
so that the stiffness matrix has the form                         Â
The engineering constants must satisfy
and the compliance matrix has the form
where
. As before the Poisson’s ratios are not
symmetric, but satisfy
The engineering constants and stiffnesses are related by
For this material the two thermal expansion coefficients in the symmetry plane must be equal, so the thermal expansion coefficient tensor has the form
3.2.15 Representative values for elastic constants of transversely isotropic hexagonal close packed crystals
|
Hexagonal close-packed crystals are an example of
transversely isotropic materials. The  axis must be taken to be perpendicular to
the basal (0001) plane of the crystal, as shown in the picture. Since the plane perpendicular to
 is isotropic the orientation of
 and
 is arbitrary.
A table of values of stiffnesses (taken from Freund and Suresh, Thin Film Materials, CUP 2003) is listed below. F&S list the original sources for their data on page 163.
|
|
|
|
|
|
Be |
292.3 |
336.4 |
162.5 |
26.7 |
14 |
C |
1160 |
46.6 |
2.3 |
290 |
109 |
Cd |
115.8 |
51.4 |
20.4 |
39.8 |
40.6 |
Co |
307 |
358.1 |
78.3 |
165 |
103 |
Hf |
181.1 |
196.9 |
55.7 |
77.2 |
66.1 |
Mg |
59.7 |
61.7 |
16.4 |
26.2 |
21.7 |
Ti |
162.4 |
180.7 |
46.7 |
92 |
69 |
Zn |
161 |
61 |
38.3 |
34.2 |
50.1 |
Zr |
143.4 |
164.8 |
32 |
72.8 |
65.3 |
ZnO |
209.7 |
210.9 |
42.5 |
121.1 |
105.1 |
The engineering constants can be calculated to be
|
|
|
|
|
|
|
(GPa) |
Be |
289.38 |
335.17 |
0.09 |
0.04 |
0.04 |
162.50 |
132.80 |
C |
903.69 |
30.21 |
0.04 |
0.08 |
2.25 |
2.30 |
435.00 |
Cd |
83.02 |
30.21 |
0.09 |
0.26 |
0.72 |
20.40 |
38.00 |
Co |
211.30 |
313.15 |
0.49 |
0.22 |
0.15 |
78.30 |
71.00 |
Hf |
139.87 |
163.07 |
0.35 |
0.26 |
0.22 |
55.70 |
51.95 |
Mg |
45.45 |
50.74 |
0.36 |
0.25 |
0.23 |
16.40 |
16.75 |
Ti |
104.37 |
143.27 |
0.48 |
0.27 |
0.20 |
46.70 |
35.20 |
Zn |
119.45 |
35.28 |
-0.06 |
0.26 |
0.87 |
38.30 |
63.40 |
Zr |
98.79 |
125.35 |
0.40 |
0.30 |
0.24 |
32.00 |
35.30 |
ZnO |
127.30 |
144.12 |
0.44 |
0.32 |
0.28 |
42.50 |
44.30 |
3.2.16 Linear elastic stress-strain relations for cubic materials
|
A huge number of materials have cubic symmetry  all the FCC and BCC metals, for
example. The constitutive law for such
a material is particularly simple, and can be parameterized by only 3
material constants. Pick basis vectors
perpendicular to the symmetry planes, as shown.
Then
or in terms of engineering constants
This is virtually identical to the constitutive
law for an isotropic solid, except that the shear modulus  is not related to the Poisson’s ratio and
Young’s modulus through the usual relation given in Section 3.1.6.  In fact, the ratio
provides
a convenient measure of anisotropy.Â
For  the material is isotropic.Â
For this material the thermal expansion coefficient matrix must be isotropic.
The relationships between the elastic constants are
3.2.17 Representative values for elastic properties of cubic crystals and compounds
A table of elastic constants for various cubic crystals and compounds (modified from Simmons and Wang ‘Single Crystal Elastic Constants and Calculated Aggregate Properties’ MIT Press (1970)) is given below
Material
|
|
|
|
|
|
|
|
|
Ag |
(fcc) |
124.00 |
46.10 |
93.40 |
43.75 |
0.43 |
46.10 |
3.01 |
Al |
(fcc) |
107.30 |
28.30 |
60.90 |
63.20 |
0.36 |
28.30 |
1.22 |
Au |
(fcc) |
192.90 |
41.50 |
163.80 |
42.46 |
0.46 |
41.50 |
2.85 |
Cu |
(fcc) |
168.40 |
75.40 |
121.40 |
66.69 |
0.42 |
75.40 |
3.21 |
Ir |
(fcc) |
580.00 |
256.00 |
242.00 |
437.51 |
0.29 |
256.00 |
1.51 |
Ni |
(fcc) |
246.50 |
127.40 |
147.30 |
136.31 |
0.37 |
127.40 |
2.57 |
Pb |
(fcc) |
49.50 |
14.90 |
42.30 |
10.52 |
0.46 |
14.90 |
4.14 |
Pd |
(fcc) |
227.10 |
71.70 |
176.00 |
73.41 |
0.44 |
71.70 |
2.81 |
Pt |
(fcc) |
346.70 |
76.50 |
250.70 |
136.29 |
0.42 |
76.50 |
1.59 |
Cr |
(bcc) |
339.80 |
99.00 |
58.60 |
322.56 |
0.15 |
99.00 |
0.70 |
Fe |
(bcc) |
231.40 |
116.40 |
134.70 |
132.28 |
0.37 |
116.40 |
2.41 |
K |
(bcc) |
4.14 |
2.63 |
2.21 |
2.60 |
0.35 |
2.63 |
2.73 |
Li |
(bcc) |
13.50 |
8.78 |
11.44 |
3.00 |
0.46 |
8.78 |
8.52 |
Mo |
(bcc) |
440.80 |
121.70 |
172.40 |
343.86 |
0.28 |
121.70 |
0.91 |
Na |
(bcc) |
6.15 |
5.92 |
4.96 |
1.72 |
0.45 |
5.92 |
9.95 |
Nb |
(bcc) |
240.20 |
28.20 |
125.60 |
153.95 |
0.34 |
28.20 |
0.49 |
Ta |
(bcc) |
260.20 |
82.60 |
154.50 |
145.08 |
0.37 |
82.60 |
1.56 |
V |
(bcc) |
228.00 |
42.60 |
118.70 |
146.72 |
0.34 |
42.60 |
0.78 |
W |
(bcc) |
522.40 |
160.80 |
204.40 |
407.43 |
0.28 |
160.80 |
1.01 |
C |
(dc) |
949.00 |
521.00 |
151.00 |
907.54 |
0.14 |
521.00 |
1.31 |
Ge |
(dc) |
128.40 |
66.70 |
48.20 |
102.09 |
0.27 |
66.70 |
1.66 |
Si |
(dc) |
166.20 |
79.80 |
64.40 |
130.23 |
0.28 |
79.80 |
1.57 |
GaAs |
|
118.80 |
59.40 |
53.70 |
85.37 |
0.31 |
59.40 |
1.82 |
GaP |
|
141.20 |
70.50 |
62.50 |
102.85 |
0.31 |
70.50 |
1.79 |
InP |
|
102.20 |
46.00 |
57.60 |
60.68 |
0.36 |
46.00 |
2.06 |
KCl |
|
39.50 |
6.30 |
4.90 |
38.42 |
0.11 |
6.30 |
0.36 |
LiF |
|
114.00 |
63.60 |
47.70 |
85.86 |
0.29 |
63.60 |
1.92 |
MgO |
|
287.60 |
151.40 |
87.40 |
246.86 |
0.23 |
151.40 |
1.51 |
NaCl |
|
49.60 |
12.90 |
12.40 |
44.64 |
0.20 |
12.90 |
0.69 |
TiC |
|
500.00 |
175.00 |
113.00 |
458.34 |
0.18 |
175.00 |
0.90 |
(c) A.F. Bower, 2008
This site is made freely available for educational purposes.
You may extract parts of the text for non-commercial purposes provided that the source is
cited.
Please respect the authors copyright.