Section 2.4
2.4IsothermalLinear Viscoelasticity
In this section, small-strain isothermal linear viscoelasticity is considered. In the first sub-section, the elementary Maxwell and Kelvin modelsare considered. Following that, the thermomechanics of linear viscoelastic materials is discussed more generally.
2.4.1The Maxwell Model
The one-dimensional Maxwell model (see Part I, §7.3.2) consists of a Spring in series with a dashpot. Introduce an internal variable, representing the inelastic strain in the model. If is the total strain, is the elastic strain. Assume that the free energy is a quadratic function of the elastic strain (not the total strain) and that the dissipation is proportional to the square of the inelastic strain-rate, so
(2.4.1)
The basic thermodynamic statement for isothermal processes is (as in 2.2.1)
(2.4.2)
Substituting in 2.4.1 leads to
(2.4.3)
The first term on the left does not involve and the second term does not involve . Since the strain-rate and internal variable rate are independent, one has,evaluating the partial derivatives,[1]
(2.4.4)
Then (assuming Ziegler’s hypothesis holds for the second of these), one has
(2.4.5)
These are the thermomechanical equations for the Maxwell model. A number of ways of proceeding from these equations will be discussed next.
Explicit Constitutive Equation
The constitutive equation in terms of stress and strain (and their rates) only can be determined by eliminating . This can be achieved by first differentiating 2.4.5a, , and substituting this into 2.4.5b (and also an expression for from 2.4.5a):
(2.4.6)
This can now be solved for given stress and stress rate (a stress-driven formulation), or given strain rate (a strain-driven formulation).
Evolution Equations
Eqn. 2.4.5b is an evolution equation for the internal variable :
(2.4.7)
where is the relaxation time:
(2.4.8)
Given the strain, the evolution equation can be solved subject to the (physical) condition that as .
Eqn. 2.4.7 is a strain-driven evolution equation. The stress-driven version can be obtained from 2.4.5:
(2.4.9)
Functional Form
The Maxwell model can be expressed in terms of functionals (integrals). The evolution equation 2.4.7 is linear and, using the condition , the solution can be represented in the closed-form convolution representation {▲Problem 1}
(2.4.10)
which can be integrated by parts to give
(2.4.11)
The internal variable can be eliminated using 2.4.5a, leading to the hereditary integral
(2.4.12)
where the relaxation modulus functionis given by
(2.4.13)
Directly from 2.4.10 and 2.4.5a, one has the alternative to 2.4.12:
(2.4.14)
Following the same procedure, the stress-driven functional can be derived from 2.4.9 {▲Problem 2}:
(2.4.15)
where thecreepcompliance functionis given by
(2.4.16)
Equilibrium Conditions
Equilibrium is reached when the rate of change of the internal variable is zero, . Note that, from 2.4.9, equilibrium of the Maxwell model only occurs when it is stress-free.
Also, from ,
(2.4.17)
The first term on the right here is the stress rate at equilibrium:
(2.4.18)
which for the Maxwell model reads . The second term on the right hand side of 2.4.16 gives the stress rate at constant strain, i.e. the response to a stress relaxation test.
The Gibbs Free Energy
Sometimes it is useful to work with stress as the independent variable. To that end, the Gibbs energy is {▲Problem 3}
(2.4.19)
The Legendre transform relations between and G are
(2.4.20)
which can be seen to be consistent with the earlier result 2.4.5, .
The thermomechanical statement 2.4.3 can be expressed as
(2.4.21)
Thus
(2.4.22)
Each of these terms is zero; the first gives 2.4.19a; with the orthogonality hypothesis, the second gives
(2.4.23)
Analogous to 2.4.16,
(2.4.24)
The first term gives the strain rate at equilibrium. The second term gives the strain-rate at constant stress, i.e. the response to a creep test.
{{ Maximise Entropy Production }}
2.4.2The Kelvin Model
Etc etc
2.4.3A General Formulation for Viscoelastic Materials
Consider now generally a linear viscoelastic material undergoing isothermal small-strains . The thermodynamic state can be generally specified by a constitutive equation of the form (and similarly for stress and entropy)
(2.4.24)
where is an internal variablewhich represents an as yet unspecified, internal, microscale process through which energy is dissipated. The internal variable is taken to be a second order tensor, but it could just as well be a scalar or vector, and there can be more than one internal variable describing the dissipative processes.
The history-dependence of the material is described by the changes in over time, according to an evolution equation of the form
(2.4.2)
Viscoelastic materials are characterised by the fact that, for any fixed strain , there exists an equilibrium state, at which the internal variable takes on its (unchanging) equilibrium value , which depends on that strain :
(2.4.3)
and that, for any given strain, the solution of the differential equation tends to the equilibrium value, . For any given strain, then, the difference can be regarded as the “distance” of the material from its equilibrium state.
Similarly, for any given stress , the material will tend toward an equilibrium state where .
The Dissipation
The dissipation function for an isothermal viscoelastic material is generally be of the form
(2.4.4)
2.4.4A Broad Class of Viscoelastic Materials
A very broad class of viscoelastic materialscan be defined by a dissipation function of the form
(2.4.5)
Here, the functions and are called dissipative stresses. They must be chosen such that the dissipation inequality is satisfied. This definition encompasses a wide range of behaviour. One practical case which it does not allow for is where one might want coupled terms of the form ; this case will be discussed later on.
Thermodynamic Relations
The basic thermodynamic statement for isothermal processes is (as in 2.2.1)
(2.4.6)
Substituting in 2.4.1 and 2.4.5 leads to
(2.4.7)
It is helpful now to introduce the quasi-conservative stresses and , which are defined through
(2.4.8)
so that 2.4.7 can be expressed as
(2.4.9)
The first term does not contain and the second term does not contain . Since the strain-rate and internal variable rate are independent,[2]
(2.4.10)
Again, as with equations 2.3.8 for the viscous fluid, one cannot assume that the terms in brackets are zero, since the dissipative stresses depend on the strain rate and the dissipative stresses depend on the internal variable rates . Assuming Ziegler’s hypothesis again holds, one has
(2.4.11)
The total stress is thus the sum of two terms: a quasi-conservative stress associated with the free energy function (as for the stress in an elastic material) and a dissipative stress . The second of 2.4.11 states that the quasi-conservative and dissipative stresses associated with the internal variable are equal:
(2.4.12)
showing that the dissipative stress in 2.4.5 associated with the internal variable can actually be determined from the free energy.
Quadratic Free Energy
In order to generate linear models, one can expand the free energy in a Taylor series, retaining only those terms up to quadratic, as in §2.2.3 for the linear thermoelasticity. Arbitrarily taking , at the reference state, one has
(2.4.13)
The quasi-conservative stresses are
(2.4.14)
Assuming the material is stress free in the reference state, .
The reference state is an equilibrium state, which is defined as a state where
(i)
(ii) the dissipative stresses are zero
The dissipative stresses are
(2.4.15)
With this zero at the reference state, one also has . Thus the free energy can be expressed as (arbitrarily taking at the reference state)
(2.4.16)
where the fourth-order “stiffness” tensors are
(2.4.17)
and the quasi-conservative stresses are
(2.4.18)
As for the stiffness tensor discussed in relation to elastic materials, §2.2.1, since the strain energy is positive, the stiffness tensors are positive definite and so are invertible and, further, the tensors and possess the minor and major symmetries.
System Equations
Substituting 2.4.18 into 2.4.11 then leads to the equations governing the response of the material:
(2.4.19)
A Sub-Class of Viscoelastic Material
Eqns. 2.5.19 are a system of differential equations in and , treating as a known load. In general, they are non-linear because of the possibility of non-linear terms in the dissipative stress expressions.
A sub-class of viscoelastic material can be defined by dissipative stresses which are linear in the rates:
(2.4.20)
The functions A and B can be regarded as viscosities, since they relate stresses to strain rates. This still produces a non-linear system. A linear system is guaranteed when the further classification is made whereby A and B are constant viscosities. In that case, 2.4.19 becomes
(2.4.21)
As will be seen, this system includes all the classical spring-dashpot linear viscoelastic rheological models.
From 2.4.5, the dissipation is
(2.4.22)
which shows that the tensors A and B are positive definite (they are zero in the reversible case), so that they are invertible.
It is usually further assumed that they are symmetric,
(2.4.23)
This assumption is known as Onsager’s reciprocal relations.
Constitutive Equation
The constitutive equation for the material can be obtained by eliminating from the system 2.4.21. This can be achieved by first differentiating 2.4.21a to get an expression for . Eqn. 2.4.21a can also be used to get an expression for . Substituting these expressions into 2.4.21b then gives the constitutive relation
(2.4.24)
where
(2.4.25)
Solution of the System of Equations
The constitutive relation 2.4.24 can be solved for given stress or strain (and rates). An alternative solution strategy is to attack the relations 2.4.21 directly. They can be expressed conveniently in matrix form by introducing the vector of generalized coordinates and generalized forces
(2.4.26)
so that 2.4.21 can be written as the system
(2.4.27)
It will be appreciated that this can be done for scalar, vector or tensor variables, and for any number of internal variables.
Eqn. 2.4.27 is a system of linear first-order differential equations which can be solved for the (in terms of the ) using standard methods. They can also be solved for the (in terms of the ).
When there is no temperature dependence, this reduces to
(2.5.20)
which is consistent with, for example, the isothermal 3-element models discussed in the previous section:
Standard Solid I:
Standard Solid II:
Standard Fluid I:
Standard Fluid II:
A Sub-Class of Viscoelastic Materials
The dissipation inequality (see 1.6.23b) reads
(2.4.13)
and so
(2.4.14)
For a certain sub-class of materials for which is not a function of the , i.e. , so that the dissipation is a function of the internal variable rate only, (as is the case for the Maxwell model, but not for the Kelvinor 3-element fluid models – see later), then the term inside the first bracket is zero and
, (2.4.15)
With the same definitions as before, one now has , and again .
2.4.5Kelvin (Voigt) Model
For the one-dimensional Kelvin model (see Part I, §7.3.3), one uses the free energy function of the linear elastic solid and the dissipation function for an ideal viscous fluid,
(2.4.16)
Because of the squared rate term in the dissipation function, there is dissipation of energy whether the strain rate is positive or negative, and as required.
The Kelvin model is a special case in that there is no internal variable; at each fixed strain, the material is at an equilibrium state and so there is no need for an internal variable to describe the evolution of the model toward an equilibrium state.
One has
(2.4.17)
and, from 2.4.11a,
(2.4.18)
Physically, the dissipation function is the rate of work dissipated in the dashpot. In this context, the dissipative stress is also known as the viscous stress.
Note that the dissipation function here is homogeneous of degree 2 in and so, from 2.4.7 and Euler’s theorem, 2.3.6-7, the dissipative stress can also be expressed as
(2.4.19)
Three-Dimensional Model
The simplest 3D Kelvin-type model is to make the following generalization from the 1D case:
(2.4.20)
Now
(2.4.21)
and
(2.4.22)
and the constitutive equation is
(2.4.23)
Different Kelvin models can be generated by choosing different free energy functions of the form and combining with the dissipation function 2.4.22.
Isotropic Kelvin Models
Consider the isotropic elastic strain energy function, Eqn. 2.2.22,
(2.4.24)
where K is the bulk modulus and is the deviatoric strain , and use this to generate the quasi-conservative stress (which is simply the isotropic elastic stress, Eqn. 2.2.21)
(2.4.25)
For the dissipation, one can choose
(2.4.26)
and so the dissipative stress is
(2.4.27)
This dissipation function ensures that the viscoelastic response is only associated with distortion, but not with dilatation, a property of many viscoelastic materials; a pure volume change will occur elastically.
2.4.6Maxwell Model
For the one-dimensional Maxwell model (see Part I, §7.3.2), introduce an internal variable, represents the inelastic strain in the model. If is the total strain, () is the elastic strain. Assume that the free energy is a function of the elastic strain (not the total strain), so
(2.4.28)
Then
(2.4.29)
and from 2.4.11b,
(2.4.30)
From 2.4.11a,
(2.4.31)
To eliminate , differentiate this expression: , and it follows that
(2.4.32)
It can be seen that is the viscous strain across the dashpot.
The Evolution Equation
From 2.4.30, the internal variable satisfies the evolution equation
(2.4.33)
subject to the (physical) condition that as .
Equilibrium Conditions
Equilibrium is reached when the rate of change of the internal variable is zero, . Note that, from 2.4.31, 2.4.33, equilibrium of the Maxwell model only occurs when it is stress-free.
Also, from (which is true when ),
(2.4.34)
The first term on the right here is the stress rate at equilibrium:
(2.4.35)
which for the Maxwell model reads . The second term on the right hand side of 2.4.34 gives the stress rate at constant strain, i.e. the response to a stress relaxation test.
The Gibbs Free Energy
Sometimes it is useful to work with stress as the independent variable. To that end, the Gibbs energy is {▲Problem 1}
(2.4.36)
The Legendre transform relations between and G are
(2.4.37)
which can be seen to be consistent with the earlier result 2.4.31, .
The thermomechanical statement 2.4.8 can be expressed as
(2.4.38)
Thus
(2.4.39)
Each of these terms is zero; the first gives 2.4.35a; with the orthogonality hypothesis, the second gives
(2.4.40)
or .
Analogous to 2.4.34,
(2.4.41)
The first term gives the strain rate at equilibrium. The second term gives the strain-rate at constant stress, i.e. the response to a creep test.
Three-Dimensional Model
The simplest Maxwell-type model is to make a simple generalization from the 1D case and write
(2.4.42)
Then
(2.4.43)
with , leading to
(2.4.44)
and the evolution equation for the internal viscous strains is
(2.4.45)
Different Maxwell models of this type can be generated by choosing different free energy functions of the form
2.4.7One-Dimensional Three-element Models
In this section, the four three-element linear viscoelastic models are examined. The solid models are characterised by the presence of the strain in the constitutive equations, whereas the fluid models have strain rates only.
The Standard Solid I
Considerthe following free energy and dissipation:
(2.4.46)
Then
(2.4.47)
and
(2.4.48)
These equations describe the three-element model shown in Fig. 2.4.1: the stress is that in the free-spring of stiffness , represents the strain in the Kelvin unit, and the stress is split into in the spring and in the dashpot of the Kelvin unit.
Eliminating by first differentiating 2.4.48a leads to the constitutive equation
(2.4.49)
Figure 2.4.1: the Standard Solid I
The Gibbs free energy is {▲Problem 2}
(2.4.50)
and the Legendre transform relations 2.4.37 are again seen to be satisfied.
The Standard Solid II
The free energy and dissipation
(2.4.51)
lead to {▲Problem 3}
(2.4.52)
This is the constitutive equation for the three-element model of Fig. 2.4.2.
Figure 2.4.2: the Standard Solid II
Note that the constitutive equation for this model can be transformed into the Standard Solid I using the following substitutions:
(2.4.53)
The Gibbs free energy is {▲Problem 4}
(2.4.54)
The 3-Element Fluid I
The functions
(2.4.55)
lead to the constitutive equation {▲Problem 5}
(2.4.56)
which describes the fluid model of Fig. 2.4.3, with the strain in the dashpot 2.
Figure 2.4.3: the Three-Element Fluid I
The 3-Element Fluid II
The fluid model of Fig. 2.4.4 can be generated using the formulation given in Eqns. 2.4.1-2.4.12, only the internal variable will not have a direct physical meaning as it does in the other 3-element models discussed above. The functions
(2.4.57)
with
(2.4.58)
lead to
(2.4.59)
and
(2.4.60)
Eliminating leads to the constitutive equation
(2.4.61)
Figure 2.4.4: the Three-Element Fluid II
2.4.8Generalised Maxwell Model
Consider now a single spring of stiffness in parallel with a number, , of Maxwell elements (as in the Standard Solid II), each with spring stiffness and viscosity . This model is widely used to model the viscoelastic response of materials.
Define the free energy to be the elastic stored energy in the springs and the dissipation the energy dissipated in the dashpots:
(2.4.62)
The quasi-conservative stresses are
(2.4.63)
and the dissipative stresses are
(2.4.64)
The dissipation can then be expressed as
(2.4.65)
where are the dissipative stress and internal variable vectors. Further, with the dissipation not a function of the strain rate , the dissipation can be expressed as
(2.4.66)
The total stress in the model is
(2.4.67)
where is given by
(2.4.68)
and
(2.4.69)
The evolution equations for the are
(2.4.70)
At equilibrium, one has. From the evolution equation, (so there is no strain in the Maxwell springs) and so, from 2.4.67,.
Convolution Representation
Linear elastic models can be expressed in terms of functionals (integrals).
The evolution equations
(2.4.71)
are linear and the solution can be represented in the closed-form convolution representation {▲Problem 6}
(2.4.72)
which can be integrated by parts to give
(2.4.73)
The here arerelaxation times. Substituting this back into the constitutive equation then leads to the hereditary integral{▲Problem 7}