Section 2.5
2.5Plasticity
In the viscous fluid model, the dissipation function was quadratic (i.e. homogeneous of degree two) in the strain rates. As a result the stresses were proportional to the strain rates; the proportionality constant being a viscosity (with dimensions ). As a result, if the strain rate is doubled, so is the stress. A characterizing feature of plastic deformations is that they are rate – independent. If the strain rates are doubled the stress is unaltered. To achieve such models one must choose rate of dissipation functions which are homogeneous of degree one in the strain rates.
2.5.1One Dimensional Models
Perfectly Plastic Material
Consider the dissipation function
(2.5.1)
where is an internal variable. Y is a positive constant, so that the non-negativity of the dissipation function is assured. Using the standard quadratic free energy function , one then has
(2.5.2)
Equating coefficients,
(2.5.3)
where is the signum function; for , for ; it can take any value in at . Thus
(2.5.4)
and either
(2.5.5)
The internal variable is the plastic strain and one has either
(a) - elastic loading/unloading (2.5.6a)
or
(b) - plastic yielding (2.5.6b)
This model is hence linearly elastic/ perfectly plastic. In tension or compression, the material first deforms elastically () until the yield stress is reached, then plastic deformation occurs at constant stress. A tension/compression cycle of loading is illustrated in Fig. 2.5.1; the magnitude of the internal variable plastic strain at point p in the cycle is shown.
Figure 2.5.1: Stress-Strain Response for a Perfectly Plastic Model
Isotropic Hardening Material
One can introduce linear isotropic hardening, by replacing Y by (Y+H), so that the yield stress increases linearly with plastic strain. H is the hardening modulus:
(2.5.7)
Then 2.5.6b is replaced with
- plastic yielding (2.5.8)
The slope K of the stress-strain line during this plastic phase, the tangent modulus, is obtained by differentiating these equations with respect to time and eliminating ; with , one arrives at the incremental response
, , (2.5.9)
The response of the material is illustrated in Fig. 2.5.2. In tension or compression, the material deforms elastically until and then plasticity occurs with tangent modulus .
One peculiar feature of this model occurs when the material is loaded first in tension, as to point p in Fig. 2.5.2, unloaded, and then loaded into compression, as to point q in Fig. 2.5.2. Because , , the K in 2.5.9 is negative, a decrease in strain leads to an increase in stress; this softening behaviour is sketched in the figure. This can be overcome by introducing a new internal variable through with dissipation . Then and the compressional behaviour is similar to the response in tension.
Figure 2.5.2: Stress-Strain Response for an Isotropic Hardening Material
Kinematic Hardening
Akinematic hardening material can be modelled by introducing a second term in the free energy function:
(2.5.10)
Following the same procedure as before, one now has
(a) - elastic loading/unloading (2.5.11a)
or
(b) - plastic yielding (2.5.11b)
Note that,in the isotropic hardening case, the hardening term is in the dissipation function and hence represents energy that is dissipated. In contrast the extra term needed to model kinematic hardening is in the free energy function and hence results ina stored energy. However, since the term involving His a function of the plastic, not elastic, strains, this energy can only be recovered by reversing the plastic strains.
The tangent modulus and incremental response are now
, , (2.5.12)
The response is illustrated in Fig. 2.5.3. Note that the size of the yield zone remains at , as indicated in the figure; the centre of this yield zone is shifted from the origin by an amount , which is called the back stress.
Figure 2.5.3: Stress-Strain Response for a Kinematically Hardening Model
2.5.2General Formulation
Consider now a general plasticity model in which the free energy is a function of the strains and internal variable, , and the dissipation is of the form
(2.5.13)
Note that, unlike the viscoelastic materials, the dissipation here is not a function of the strain rates . The dissipation is a homogeneous function of degree one in the so, using Euler’s theorem, Eqns. 2.2.5-6,
(2.5.14)
Substituting 2.5.14 into the isothermal thermodynamic statement 2.2.1 leads to
(2.5.15)
As before, define the quasi-conservative stress and dissipative stress through
(2.5.16)
so that 2.5.15 can be expressed as
(2.5.15)
Since the and are independent,
(2.5.16)
The stresses are independent of the strain rate, so the term inside the first bracket is zero. The dissipative stresses do depend on the but, applying Ziegler’s orthogonality principle, the term inside the second brackets can also be taken to be zero. Hence
(2.5.17)
Yield Function and Flow Rule
The dissipation is a function of and the dissipation acts as a potential for the dissipative stresses, through . As discussed in §1.A, one can construct a new function w, the Legendre transform of , such that w is a function of the dissipative stresses and acts as a potential for the , through . Thus
(2.5.18)
However, since is homogeneous of degree one in the rates, it turns out that w must be identically zero (see the Appendix to this Chapter, §2.A). Thus
(2.5.19)
Since , it can only be determined to within an arbitrary multiplicative constant, say, and it is therefore helpful to introduce a new function f through
(2.5.20)
with . This new function f turns out to be the yield function (see later).
Differentiating 2.5.20 with respect to the dissipative stresses gives
(2.5.21)
which is the flow rule, Fig. 2.5.4.
Figure 2.5.4: Yield Surface in Dissipative Stress Space
The requirement that the dissipation be non-negative implies that f must satisfy
(2.5.22)
Formulation in terms of the Gibbs Energy
Thus far, the formulation has been presented in terms of the free energy. Here, the problem is presented in terms of the Gibb’s energy, which is a function of the stresses and internal variables:
(2.5.23)
Using the relations and , one has
(2.5.24)
and so
(2.5.25)
The incremental response can be obtained by differentiating this further:
(2.5.26)
The first term here corresponds to the strain rate at constant ; it can be regarded as the elastic strain. The second term is the strain at constant stress.
The dissipation is now taken to be a function of the stresses:
(2.5.27)
As before, one can define now the quasi-conservative stress and dissipative stress through
(2.5.28)
with .
De-coupled Materials
In most applications, one can express the Gibb’s energy in the form
(2.5.29)
so that the only term involving both the stress and internal variable is linear in the latter. Further, if is linear in the stresses, then one can choose (for a suitable choice of ) then
(2.5.30)
This ensures that the coefficient of in 2.5.26 is a function of (and not of ), implying that the instantaneous elastic response () is independent of - such a material is usually termed de-coupled.
In this case
(2.5.31)
with clearly playing the role of a plastic strain and the elastic strain is
(2.5.32)
Also,
(2.5.33)
The corresponding form of the free energy for a de-coupled material is
(2.5.34)
with
(2.5.35)
In this case,
(2.5.36)
The Shift Stress
Note that, from 2.5.28, 2.5.30,
(2.5.37)
One can define the shift stress :
(2.5.38)
Note that the elastic strain is a function of the stress, and the shift stress is a function of the internal variables, .
Plastic Work
The plastic work is defined to be . The difference between the plastic work and the dissipation is then and so they are equal only in the case of .
Yield Surface in Stress Space
Using the Gibb’s energy formulation, construct now a new function w through a Legendre transformation:
(2.5.39)
with and again
(2.5.40)
Since , the dissipative stress can be expressed as a function of the stress and internal variables, and so the yield function f can be expressed as , the superscript implying that f here is a function of the true stress rather than the dissipative stress. Differentiating leads to
(2.5.41)
which must equal
(2.5.42)
Assuming the de-coupled form 2.5.30 for the Gibb’s function, so that
(2.5.43)
equating coefficients of leads to
(2.5.44)
The plastic strain rates are normal to the yield surface in dissipative stress space, Fig. 2.5.4. Associated plastic flow occurs when the plastic strain rates are normal to the yield surface in true stress space, i.e.
(2.5.45)
and, from 2.5.44, this occurs when, that is, when the yield function is independent of the stresses (or, a special case, if and are always parallel). Further, only if , that is, the dissipation is independent of the true stress.
2.5.3One Dimensional Models Re-visited
Using the de-coupled formulation, the free energy and Gibb’s energy functions are
(2.5.46)
Introduce a general dissipation of the form
(2.5.47)
The dissipative stress is
(2.5.48)
and the yield surface can be taken to be
(2.5.49)
with
(2.5.50)
The stress and dissipative stress are
(2.5.51)
and
(2.5.52)
There are two cases:
(i) . Here, the dissipative stresses do not have to equal .
(ii) . Here, the dissipative stresses must equal and, from 2.5.51b, one has
(2.5.53)
so that
(2.5.54)
Then
(2.5.55)
From 2.5.52,
(2.5.56)
If the partial derivatives in 2.5.54 are zero, then (and ). Otherwise, using 2.5.55-6,
(2.5.57)
Also, using both 2.5.55, 2.5.57,
(2.5.58)
Perfectly Plastic Material
As discussed in §2.5.1, the free energy is andthe true stress is
(2.5.59)
The Gibb’s function is
(2.5.60)
Note that this is of the form 2.5.46 with . The dissipation is and, from 2.5.35, the yield function is
(2.5.61)
with. From 2.5.51-2,
(2.5.62)
When a (tensile) stress is applied, and 2.5.61 is satisfied with . When the stress reaches Y, then , and, from 2.5.53,, as in Eqns. 2.5.6.
Fig. 2.5.1 is re-drawn in Fig. 2.5.5, together with the response in dissipative stress / plastic strain space.
Figure 2.5.5: Stress-Strain Response for a Perfectly Plastic Model; (a) true stress space, (b) dissipative stress space
Isotropic Hardening Material
As discussed in §2.5.1, isotropic hardening can be modelled using the dissipation 2.5.7, . The Gibb’s energy is again given by 2.5.60. The dissipative stress is given by 2.5.48,
(2.5.63)
During plastic flow, the stress is given by 2.5.53, which gives 2.5.8, and the plastic modulus is given by 2.5.57, which is 2.5.9.
Fig. 2.5.2 is re-drawn in Fig. 2.5.6, together with the response in dissipative stress / plastic strain space.
Figure 2.5.6: Stress-Strain Response for an Isotropic Hardening Model; (a) true stress space, (b) dissipative stress space
Kinematic Hardening
With the free energy now given by 2.5.10b,
(2.5.64)
the Gibb’s energy is
(2.5.65)
From 2.5.51,
(2.5.66)
From 2.5.53, during plastic flow,
(2.5.67)
The shift stress is given by 2.5.38:
(2.5.68)
2.5.4Von-Mises Elastoplasticity
2.5.5The Dissipation Potential
Once the free energy is specified, many of the quantities mentioned above can be evaluated. However, one needs to specify in Eqn. 2.3.3. Such expressions describe how the internal variables (and dissipation mechanisms) evolve over time and are known as evolution equations. The evolution equations must be chosen in such a way that the dissipation inequality 2.3.5 is always satisfied. At thermodynamic equilibrium, one has and hence may be viewed as the rate at which tends towards its equilibrium state.
One way to ensure 2.3.5 is always satisfied is simply to choose some evolution equation and check a posteriori that it is indeed satisfied. A more efficient and general way of proceeding is to assume that the fluxes can be expressed in the form
(2.3.10)
Here, is called the dissipation potential, and depends on the and possibly some other variables. Substitution into 2.3. then leads to an alternative inequality which must be satisfied:
(2.3.11)
Within this potential approach, there are a number of ways to ensure that 2.3.11 is automatically satisfied, as detailed next.
Onsager’s Linear Theory
The inequality 2.3.11 is automatically satisfied if one chooses for the dissipation potential the linear form
(2.3.12)
where A is a positive definite second order tensorand, for the present purposes, is a vector of thermodynamic forces corresponding to a set of scalar internal variables: , . To see this, split A into its symmetric and skew parts, leading to (using the relation Part III, Eqn. 1.9.25e)
(2.3.13)
Substitution into 2.3.11 leads to
(2.3.14)
which is satisfied sinceA is positive definite. Examining 2.3.10, the appropriate evolution equation is the linear form
(2.3.15)
where is a positive definite symmetric matrix. The symmetry property of L, the relations , are known as Onsager’s reciprocal relations. This can be generalised to the case of a tensor in which case where now the fourth-order tensor A contains the major symmetries.
Homogeneous Potential Function
In many important practical cases, for example plasticity theory, one can assume that the dissipation potential is a homogeneous function of degree n, that is, analogous to 2.2.5, for a scalar, vector or tensor,
(2.3.16)
for scalar k. Using Euler’s theorem, 2.2.5-6, then leads to
(2.3.17)
The dissipation inequality 2.3.11 now simply reads
(2.3.18)
and is satisfied provided .
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Solid Mechanics Part IV Kelly