Introduction
This is part 3 of my series on Linear Viscoelasticity. The focus of this article is on how to make the theory suitable for multiaxial deformations (not just uniaxial!), and how to make it suitable for large strains.
Multiaxial Linear Viscoelasticity
The starting point is the results from Part 1 in this series.
\( \sigma(t) =\displaystyle \int_0^{t} E_R(t – \tau) \dot{\varepsilon}(\tau) d\tau. \)
(1)
The problem with this equation is that FEA is typically 3D, so we need a 3D equation! To fix this problem, let’s start by writing down the stress-strain equation for a linear elastic material. Eq. (2) is for uniaxial loading, and Eq. (3) is for a general multiaxial strain state. The deviatoric and volumetric strains in Eq. (3) are given by \(\boldsymbol{\varepsilon}_{\text{dev}} = \boldsymbol{\varepsilon} – (1/3)\text{tr}[\boldsymbol{\varepsilon}] \mathbf{I}\), and \( \boldsymbol{\varepsilon}_{\text{vol}} = (1/3) \text{tr}[\boldsymbol{\varepsilon}] \mathbf{I}\).
\( \sigma =\varepsilon E\)
(2)
\( \boldsymbol{\sigma} = 2 \mu \boldsymbol{\varepsilon}_{\text{dev}} + \kappa \boldsymbol{\varepsilon}_{\text{vol}}\)
(3)
This gives us the hint that the linear viscoelastic stress-strain response for general multiaxial loading should be written as the sum of two terms: one for shear and one for volumetric deformations. The following equations gives the answer:
\( \boldsymbol{\sigma}(t) = \displaystyle\int_0^{t} 2\mu_R(t – \tau) \dot{\boldsymbol{\varepsilon}}_{\text{dev}} (\tau) d\tau + \int_0^{t} \kappa_R(t – \tau) \dot{\boldsymbol{\varepsilon}}_{\text{vol}} (\tau) d\tau \)
(4)
This equation shows that we need to experimentally determine two (2) functions: a shear relaxation modulus \(\mu_R(t)\), and a volumetric relaxation modulus \(\kappa_R(t)\).
Large Strain Linear Viscoelasticity
There is nothing that prevents you from using Equations (1) – (4) also for large deformation problems, but it is more common (and recommended!) to use hyperelasticity instead of linear viscoelasticity when analyzing large strain problems. The stress-strain response for linear elasticity is \( \sigma = \varepsilon E\), and for hyperelasticity it is \( \sigma = \sigma_{hyp}(\varepsilon)\). As I discussed in Part 2 of this series, it is often a good idea to integrate Eq. (1) by parts to get:
\(\sigma(t) = \displaystyle E_0 \varepsilon(t) – \int_0^t \dot{g}_R(t-\tau) E_0\varepsilon(\tau) d\tau\)
(5)
The cool thing now is that we can simply replace \(E_0 \varepsilon(t)\) by \(\sigma_{hyp}(t)\) giving:
\( \sigma(t) = \displaystyle\sigma_{\text{hyp}}(\varepsilon(t)) – \int_0^t \dot{g}_R(t-\tau) \sigma_{\text{hyp}}(\varepsilon(\tau)) d\tau \)
(6)
This equation show that the stress is given by the instantaneous hyperelastic response minus a linear viscoelastic relaxation stress. Also note that a common normalized relaxation modulus is a Prony series term (but more on that later…):
\( g_R(t) = (1-g) + g e^{-t/\tau_0}\)
(2)
The two topics presented in this article can be combined, but I will leave that as an exercise for you.
