7 Viscoelasticity and viscoplasticity in three-dimensional space

In the previous chapters on viscoelasticity and viscoplasticity, Linear Viscoelasticity and Viscoplasticity, we have made use of the additivity of strains and, particularly, of the additivity of strain rates; see for example relations (5.10). The important question we want to answer in this Chapter is: How can such approaches be generalized to the case of three dimensions?

7.1Multiplicative decomposition of the deformation gradient tensor¶

In Kinematics on kinematics, the deformation gradient tensor has been introduced. Particularly, according to (1.2), it was shown that the deformation gradient tensor maps an infinitesimal line element $d\V{x}_0$ from the reference configuration onto the corresponding line element $d\V{x}$ in the current configuration,

Given the relation $\V{x} = \V{x}(\V{x}_0)$ between the position of a mass point in the reference state, $\V{x}_0$, and the current state, $\V{x}$, it was shown that the deformation gradient is given by

Imagine now that you perform two deformations in a sequence (see Figure 7.1).

First, the line element $d\V{x}_0$ in the reference state is deformed into the corresponding line element in the so-called intermediate configuration, $d\V{x}_\rI$, by a deformation gradient $\T{F}_1$

After that first step, the material is deformed further, with a deformation described by the deformation gradient tensor $\T{F}_2$. The line element $d\V{x}_\rI$ is then mapped onto the corresponding line element in the current configuration, $d\V{x}$, according to

where we have used (7.3) in the second equality. This implies that the entire deformation gradient tensor $\T{F}$ describing the effect of both deformation steps combined is given by

Particular attention should be paid to the fact that the deformation applied first stands on the right end of the product; in other words, the right hand side of (7.5) must be read from right to left.

The composition of different deformation processes is useful for the 3d modeling of viscoelasticity and viscoplasticity, as shown in the following (see also Figure 7.2).

Consider a material that shows both elastic and viscous material behavior. The deformation gradient $\T{F}$ can be considered as a representation of the deformation state you apply in the experiment. It describes the deformation from an initial state to the current state, following a certain deformation path. That current state of deformation has encoded both elastic as well as viscous (= deformation-rate dependent) aspects, as described also in the previous chapters with the Maxwell model. In order to split the total deformation gradient into these two contributions, one can decide to unload the material infinitely rapidly, since in this case the rate-dependent processes are “switched off” (meaning: they are inactive), and the infinitely rapid unloading only affects the elastic contribution. If one uses this procedure to unload until the stress-free state is reached, on calls the reached configuration intermediate. The deformation gradient tensor describing that process of infinitely rapid unloading is called $\T{F}^{-1}_\re$. In terms of line elements, we thus have the following scenario:

where in the second equality of (7.7) we have used (7.6). The object $\T{F}_\re$ is called the elastic part of the deformation gradient tensor, since it represents the elastic effects, as described above, because the rate-dependent effects are inactive. Conversely, the plastic part of the deformation gradient tensor is defined as

According to (7.7), $\T{F}_\rp$ maps the line element from the reference configuration, $d\V{x}_0$, to the corresponding line element in the (stress-free) intermediate configuration, $d\V{x}_\rI$. Eq. (7.8) also implies that the total deformation gradient tensor is decomposed according to the multiplicative split

In this formulation, one observes immediately that, if the $\T{F}$-deformed state is followed by infinitely rapid unloading -- which corresponds to multiplying (7.9) with $\T{F}^{-1}_\re$ from the left, the remaining deformation is the plastic part, $\T{F}_\rp$.

7.2Special cases of the multiplicative decomposition¶

How can the multiplicative split (7.9) be simplified in the case of small deformations? To that end, we assume that both the elastic and the plastic parts of the deformation gradient tensor as well as the total deformation gradient tensor can be written in the form

with the assumption that $(\T{\Delta})_{ij} \ll 1$, $(\T{\Delta}_\re)_{ij} \ll 1$ and $(\T{\Delta}_\rp)_{ij} \ll 1$. Inserting these expressions into the general large-deformation relation (7.9) leads to

Therefore, the additive combination of deformations is recovered for arbitrary but small deformations.

Let us consider another special case. Assuming simple shear for $\T{F}_\re$, and $\T{F}_\rp$ (in $(1,2)$-plane, for example), we can write

with elastic $\gamma_\re$ and plastic $\gamma_\rp$ shear strains, respectively. In this case, we find for the total deformation gradient tensor according to (7.9),

In other words, if both the elastic and the plastic parts of the deformation gradient tensor are representative of simple shear, then also the total deformation gradient tensor represents simple shear in the $(1,2)$-plane, with a shear strain given by $\gamma = \gamma_\re + \gamma_\rp$. This also holds true for large shear deformation.

7.3Free energy: revisited¶

In Hyperelasticity on hyperelasticity, it has been discussed in detail, that the free energy depends on the state of deformation. In the general 3d setting discussed in this Chapter, there are three measures for the state of deformation ($\T{F}$, $\T{F}_\re$, and $\T{F}_\rp$). However, as (7.9) shows, they are not mutually independent, but just two of them are. Therefore, one can ask the question which two measures of deformation should be chosen as arguments of the free energy function. As a matter of fact, this choice historically depends on the specific field of application:

• In metals, it is common to choose the total deformation gradient tensor $\T{F}$ and its plastic part $\T{F}_\rp$ as arguments of the free energy. The free energy then contains typically three contributions, related to elasticity, kinematic hardening, and dislocations,

  $$A(\T{F}, \T{F}_\rp) = A_{\rm elasticity}(\T{F} \inp \T{F}^{-1}_\rp) + A_{\rm kin. hardening}(\T{F}_\rp) + A_{\rm dislocations}(\T{F}_\rp, {\rm curl}\T{F}_\rp) \ ,$$

  (7.14)

  where these three contributions depend on $\T{F}$ and $\T{F}_\rp$ as indicated by the respective arguments. Note that the elastic contribution depends only on the elastic part of the deformation gradient tensor, $\T{F}_\re = \T{F} \inp \T{F}^{-1}_\rp$, according to (7.9). Furthermore, in the last contribution, ${\rm curl}\T{F}_\rp$ (“curl” is English for “rotatie” in Dutch) is a measure of the dislocation density.^[1]

• In solid polymers, it is common to choose the total deformation gradient tensor $\T{F}$ and its elastic part $\T{F}_\re$ as arguments of the free energy. The free energy then contains typically two contributions, related to the local segmental contributions (glassy or crystalline) and a chain-network contribution (rubber elasticity due to the underlying network of entangled chains),

  $$A(\T{F}, \T{F}_\rp) = A_{\rm local \ segmental}(\T{F}_\re) + A_{\rm chain \ network}(\T{F}) \ ,$$

  (7.15)

  where these two contributions depend on $\T{F}$ and $\T{F}_\re$ as indicated by the respective arguments (see also Figure 7.3). The local-segmental contribution is the analogue of the elastic contribution in metals, depending on $\T{F}_\re$ only. Significant differences with metals come in the other contribution, due to the different origin of the large-deformation behavior.

7.4Evolution equations¶

In Linear Viscoelasticity on linear viscoelasticity, we have used the additivity of strain rates, see (5.10). If we change nomenclature and call the Hookean element “elastic” and the Newtonian element “plastic”, we arrive at

How can this relation be generalized to the 3d case?

Let us start by the multiplicative decomposition (7.9). We then find for the time-derivative of the total deformation gradient tensor,

which already has an additive form resembling (7.16). In order to proceed, we recall that in Kinematics the evolution of the total deformation gradient tensor has been written in the form

with $\T{L}$ the velocity gradient tensor defined by

and $\V{v}$ denotes the velocity.

Using (7.18) with the velocity gradient (7.19), still two time-derivatives in (7.17) are undetermined, and thus (7.17) can not yet be used for practical purposes. To proceed, one can introduce, in analogy to (7.18), the plastic velocity gradient tensor $\T{L}_{\rp, \rI}$ according to

One can show that $\T{L}_{\rp, \rI}$ acts in the intermediate configuration: Multiplication of (7.20) with a (time-constant) line segment in the reference state $d\V{x}_0$, one obtains with $d\V{x}_\rI = \T{F}_\rp \inp d\V{x}_0$ the segment-evolution $\dot{d\V{x}}_\rI = \T{L}_{\rp, \rI} \inp d\V{x}_\rI$. Since $d\V{x}_\rI$ is a line-segment in the intermediate configuration, it is clear that $\T{L}_{\rp, \rI}$ operates on the intermediate configuration.

With the above ingredients, one can solve a 3d model in two ways:

• Either, given a velocity field $\V{v}$ and a constitutive relation for $\T{L}_{\rp, \rI}$, one solves the evolution equations for $\T{F}$, (7.18), and for $\T{F}_\rp$, (7.20). Subsequently, one can use this to calculate $\T{F}_\re$ according to the relation $\T{F}_\re = \T{F} \inp \T{F}^{-1}_\rp$.

• Or, one rewrites (7.17) into the form

  $$\begin{aligned} \dot{\T{F}}_\re &= \dot{\T{F}}\inp \T{F}^{-1}_\rp - \T{F}_\re \inp \dot{\T{F}}_\rp\inp \T{F}^{-1}_\rp \\ {}&= \T{L} \inp \T{F}\inp \T{F}^{-1}_\rp - \T{F}_\re \inp \T{L}_{\rp, \rI} \inp \T{F}_\rp\inp \T{F}^{-1}_\rp \\ {}&= \T{L} \inp \T{F}_\re - \T{F}_\re \inp \T{L}_{\rp, \rI} \ . \end{aligned}$$

  (7.21)

  Therefore, given a velocity field $\V{v}$ and a constitutive relation for $\T{L}_{\rp, \rI}$, there is also the option to solve the evolution equations for $\T{F}$, (7.18), and for $\T{F}_\re$, (7.21). Subsequently, one can use this to calculate $\T{F}_\rp$ according to the relation $\T{F}_\rp = \T{F}^{-1}_\re \inp \T{F}$.

Clearly, in both procedures, a constitutive relation for the plastic velocity gradient tensor $\T{L}_{\rp, \rI}$ is needed. This implies that one should have an idea about what physical processes take place during non-elastic deformation. Two common ansatzes for $\T{L}_{\rp, \rI}$ are discussed in the following.

7.5Plastic velocity gradient tensor for crystal plasticity¶

Imagine a crystal with dislocations that is exposed to significant simple shear. If after a certain amount of deformation the material is unloaded infinitely rapidly, one arrives at a crystal structure that differs from the initial structure (see Figure 7.4). While both initial structure and unloaded structure do not have any overall strain (meaning: no overall lattice distortion), they differ in terms of the positions of the dislocations: the dislocations have moved. This corresponds to saying that there is a certain amount of permanent (= plastic) shear of the system. It is left as an exercise to the reader (i) to determine the form of $\T{F}_\rp$ for simple (plastic) shear, (ii) to use (7.20) to determine the form of $\T{L}_{\rp, \rI}$, and (iii) to make a link with the 1d models discussed in Linear Viscoelasticity and Viscoplasticity.

Since moving dislocations result in permanent deformation, we briefly mention the corresponding “force” that drives dislocation motion; for simplicity, we only consider edge dislocations. Clearly, a shear stress drives a shear deformation, which is for example also encoded in Newton’s relation for the viscous stress, $\sigma_{xy} = \eta \dot{\gamma}$. Given a full stress tensor $\T{\sigma}$, the shear stress can be calculated for given shear direction $\V{s}$ and shear plane normal $\V{n}$ as

which in the crystal-plasticity literature is called the resolved shear stress (see Figure 7.5). The resolved shear stress is the driving force for the motion of dislocation lines on their glide planes. However, edge dislocations do not only move on glide planes, but they also climb in directions perpendicular to the glide plane. That latter process is not driven by the resolved shear stress, but rather by the diagonal components of the stress tensor: An edge dislocation can be viewed as an additionally inserted (for simplicity: inserted from the top) crystal half-plane. It does therefore not come as a surprise that compressing the crystal in the direction normal to that “inserted plane” works towards expelling the inserted half-plane. On the level of the dislocation (= the lower end of the inserted half-plane) this means that the dislocation will climb upwards.

7.6Plastic velocity gradient tensor for isotropic materials¶

Crystals are anisotropic materials, and therefore also the crystal plasticity models are anisotropic. However, there is a large class of solid materials that are isotropic. This can be materials that are isotropic on small scale and completely lack a crystal structure (such as polymer glasses). Another example is when semi-crystalline polymers (for example: polyethylene, polypropylene, isotactic polystyrene, ...) or poly-crystalline metals are studied on macroscopic scales (meaning: on scales much larger than the crystal size, and therefore an average of the crystals is observed). In all these cases, an ansatz for the plastic velocity gradient tensor is needed that reflects the material isotropy.

In this case, it is convenient to start with reformulating the evolution equation (7.21) slightly into

where the plastic velocity gradient tensor in the current configuration is defined by

It turns out that a convenient generalization of the 1d ansatz $\dot{\gstrn}_\rp(\psigma) = \psigma/\eta(\psigma)$ in Viscoplasticity is given by

where $\T{\sigma}^{\rm d}$ stands for the deviatoric part of the Cauchy stress tensor, $\T{\sigma}^{\rm d} = \T{\sigma} - \tfrac{1}{3} (\tr \T{\sigma}) \, \T{1}$. Relation (7.25) is manifestly isotropic, since there is no specific material axis (symmetry axis) that occurs in that relation. The “direction” of $\T{L}_{\rp, \rC}$ is exclusively determined by the components of the current stress state, i.e. through $\T{\sigma}$. It is now also clear that it was convenient to express the plastic velocity gradient in the current state: it is then straight forward to make a link with the Cauchy stress tensor.^[2] Furthermore, it is convenient for studying steady-state situations. Particularly, if network-contributions are neglected, the free energy (7.15) depends only on the elastic part of the deformation gradient $\T{F}_\re$, and therefore also the stress tensor $\T{\psigma}$ depends only on $\T{F}_\re$, but is independent of the total deformation $\T{F}$. Therefore, situations of constant stress (i.e. steady viscoplastic yielding) correspond to situations of constant $\T{F}_\re$, which implies the (symmetrized) steady state condition,

This is the tensorial version of the statement “the applied rate of deformation is equal to the plastic rate of deformation”.

What does that imply practically? In Kinematics we have seen that, based on $\dot{\T{F}} = \T{L}\inp \T{F}$, the evolution of the volume change $J = \det(\T{F})$ is given by $\dot{J} = J \ \tr(\T{L})$. An analogous argument can be given for the evolution of the volume change of the elastic deformation, $J_\re = \det(\T{F}_\re)$. Based on (7.23), we find that the elastic volume change evolution is given by

where in the last equality we have used $\tr(\T{L}_{\rp, \rC}) = 0$. In other words, the elastic volume change is affected only by the imposed velocity gradient, while the plastic deformation does not lead to any volume change: Relation (7.25) results in incompressible plastic deformation.

The other main issue with the constitutive relation (7.25) is that one needs to specify a constitutive relation the for viscosity $\eta$. Here we want to concentrate on the following aspect: Since the viscosity $\eta$ is a scalar, it can only depend on scalar measures of the full stress tensor $\T{\sigma}$. Which scalar measures of the stress tensor are most useful for that purpose? Clearly, it does not make sense to have the viscosity depend explicitly on, say, the $xx$-component of the stress tensor. Rather, we seek dependencies via the three invariants of the stress tensor:

It can be shown that all three of these three quantities are not affected by rotations (in the current configuration). After examination of some special cases for loading (isotropic pressure; shear; uniaxial tension), it appears to be most convenient to use the following two invariants:

• $J_1(\T{\sigma})$: determines the isotropic pressure contribution (incl. its sign);

• $J_2(\T{\sigma}^{\rm d})$ (not: $J_2(\T{\sigma})$): determines the shear stress (in shear loading), or the excess uniaxial stress (in uniaxial tension loading).

In summary, the viscosity function takes the form

Particularly, the dependence on $J_1(\T{\sigma})$ allows to make a distinction between uniaxial extension and uniaxial compression.

For the rest of this section, we assume that the viscosity depends on the stress tensor only through the invariant $J_2(\T{\sigma}^{\rm d})$. Using (7.26), this implies that yielding is observed for states with same value for $J_2(\T{\sigma}^{\rm d})$. In the case of simple shear,

one obtains $J_2(\T{\sigma}^{\rm d}) = -\psigma^2_{xy}$, with $\psigma_{xy}$ the simple-shear yield stress. In contract, in uniaxial extension,

one obtains $J_2(\T{\sigma}^{\rm d}) = -\tfrac{1}{3}\psigma^2_{\rm ex}$, with $\psigma_{\rm ex}$ the (excess) yield stress in uniaxial deformation. Since yielding must occur for the same value of $J_2(\T{\sigma}^{\rm d})$, it becomes evident that the yield stress in uniaxial extension is a factor $\sqrt{3}$ larger than that in simple shear, as shown in Figure 7.6.