Constitutive Models Relations between Stress and Strain
3.2 Linear elastic material behavior
You are probably familiar with the behavior of a linear elastic material from introductory materials courses.
If you conduct a uniaxial tensile test on almost any material, and keep the stress levels sufficiently low, you will observe the following behavior:
The specimen deforms reversibly: If you remove the loads, the solid returns to its original shape.
The strain in the specimen depends only on the stress applied to it it doesn’t depend on the rate of loading, or the history of loading.
For most materials, the stress is a linear function of strain, as shown in the picture above. Because the strains are small, this is true whatever stress measure is adopted (Cauchy stress or nominal stress), and is true whatever strain measure is adopted (Lagrange strain or infinitesimal strain).
For most, but not all, materials, the material has no characteristic orientation. Thus, if you cut a tensile specimen out of a block of material, as shown in the figure, the the stressstrain curve will be independent of the orientation of the specimen relative to the block of material. Such materials are said to be isotropic.
If you heat a specimen of the material, increasing its temperature uniformly, it will generally change its shape slightly. If the material is isotropic (no preferred material orientation) and homogeneous, then the specimen will simply increase in size, without shape change.
Before writing down stressstrain relations, we need to decide what strain and stress measures we want to use. Because the model only works for small shape changes
Deformation is characterized using the infinitesimal strain tensor defined in Section 2.1.7. This is convenient for calculations, but has the disadvantage that linear elastic constitutive equations can only be used if the solid experiences small rotations, as well as small shape changes.
All stress measures are taken to be equal. We can use the Cauchy stress as the stress measure.
You probably already know the stressstrain relations for an isotropic, linear elastic solid. They are repeated below for convenience.
Here, E and are Young’s modulus and Poisson’s ratio, is the coefficient of thermal expansion, and is the increase in temperature of the solid. The remaining relations can be deduced from the fact that both and are symmetric.
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 or the elastic compliance tensor as
In terms of elastic constants, and are
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.
For a plane strain deformation . The stress strain laws are therefore
In index notation
where Greek subscripts can have values 1 or 2.
For a plane stress deformation
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 values of E are given in ; the G stands for Giga, and is short for . The units for density are in - that’s Mega grams. One mega gram is 1000 kg.
Young’s modulus and Poisson’s ratio are the most common properties used to characterize elastic solids, but other measures are also used. For example, we define the shear modulus, bulk modulus and Lame modulus of an elastic solid as follows:
A nice table relating all the possible combinations of moduli to all other possible combinations is given below. Enjoy!
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 stressstrain 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 0.20.49, and is around 0.3 for most metals. For a stable material, . It is a measure of the compressibility of the solid. 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!!
Thermal expansion coefficient quantifies the change in volume of a material if it is heated in the absence of stress. It has dimensions of (degrees Kelvin)-1 and is usually very small. For steel,
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.
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…)
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.
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
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.
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
Now, note 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.
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.
The strain energy density of an anisotropic material is
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.
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
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
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.
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.
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.
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
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
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). Choose perpendicular to this symmetry plane. Then, transverse isotropy requires that , , , , so that the stiffness matrix has the form
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
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.
The engineering constants can be calculated to be
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.
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. For the material is isotropic.
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
(c) A.F. Bower, 2008