19.1 Relaxation Function

The Maxwell Chain model is based on a relaxation function E(t,$\tau$) . The relation between stresses and strains follows from

$$\boldsymbol\sigma \int_{{-\infty}}^{{t}} \tau \bar{{\mathbf{D}}} \dot{{\boldsymbol{\varepsilon}}} \tau$$ (19.1)

Here $\bar{{\mathbf{D}}}$ is a dimensionless matrix that relates the three-dimensional deformation states to the one-dimensional relaxation function by using Poisson's ratio $\nu$ .

$$\bar{{\mathbf{D}}} {\dfrac{{1}}{{(1+\nu)(1-2\nu)}}} \left[\vphantom{ \begin{array}{cccccc} 1-\nu & \nu & \nu & 0 & 0 ... ...\\ [\medskipamount] 0 & 0 & 0 & 0 & 0 & \dfrac{1-2\nu}{2} \end{array} }\right. \begin{array}{cccccc} 1-\nu & \nu & \nu & 0 & 0 & 0 \\ [\medskip... ...}{2} & 0 \\ [\medskipamount] 0 & 0 & 0 & 0 & 0 & \dfrac{1-2\nu}{2} \end{array} \left.\vphantom{ \begin{array}{cccccc} 1-\nu & \nu & \nu & 0 & 0 ... ...\\ [\medskipamount] 0 & 0 & 0 & 0 & 0 & \dfrac{1-2\nu}{2} \end{array} }\right]$$ (19.2)

To get the Maxwell model, the relaxation function is expanded in a truncated Dirichlet series, resulting in the following exponential series.

$$\tau \sum_{{\alpha=0}}^{{n}} \alpha \tau {\tfrac{{t-\tau}}{{\lambda_{\alpha}}}}$$ (19.3)

For a one-dimensional situation this relaxation function can be physically interpreted as a parallel chain of springs and dampers as in Figure 19.1.

Figure 19.1: Maxwell Chain

In (19.3) E_$\alpha$($\tau$) indicates that the stiffness of the model can be time dependent, for instance due to temperature or maturity influences. The stiffness of the spring E_$\alpha$ and the viscosity of the damper $\eta_{{\alpha}}^{}$ in the physical model determine the relaxation time $\lambda_{{\alpha}}^{}$ .

$$\lambda_{{\alpha}}^{} {\frac{{\eta_{\alpha}}}{{E_{\alpha}}}}$$ (19.4)

Often the first element in the Maxwell Chain is represented by one single spring or equivalently with $\eta_{{0}}^{}$ = $\infty$ .

If we substitute (19.3) into (19.1) for time t and time t + $\Delta$t , interchange the order of summation and integration, and assume that nothing has happened from time t = - $\infty$ until t = 0 , we get

$$\boldsymbol\sigma \bar{{\mathbf{D}}} \sum_{{\alpha=0}}^{{n}} \int_{{0}}^{{t}} \alpha \tau {\tfrac{{t-\tau}}{{\lambda_{\alpha}}}} \dot{{\boldsymbol{\varepsilon}}} \tau$$ (19.5)

$$\boldsymbol\sigma \Delta \bar{{\mathbf{D}}} \sum_{{\alpha=0}}^{{n}} \int_{{0}}^{{t+\Delta t}} \alpha \tau {\tfrac{{t+\Delta t-\tau}}{{\lambda_{\alpha}}}} \dot{{\boldsymbol{\varepsilon}}} \tau$$ (19.6)

We can calculate the stress increment by subtracting (19.5) from (19.6) where the integral from 0 to t + $\Delta$t is split into a part from 0 to t and a part from t to t + $\Delta$t . The partial stress in every element of the Maxwell Chain is called $\boldsymbol\sigma$_$\alpha$ .

$$\boldsymbol\sigma \alpha \bar{{\mathbf{D}}} \int_{{0}}^{{t}} \alpha \tau {\tfrac{{t-\tau}}{{\lambda_{\alpha}}}} \dot{{\boldsymbol{\varepsilon}}} \tau$$ (19.7)

If we assume a constant strain rate from t to t + $\Delta$t the stress increment follows from

$$\Delta \boldsymbol\sigma \sum_{{\alpha=0}}^{{n}} \left(\vphantom{ 1 - e^{-\tfrac{\Delta t}{\lambda_{\alpha}}} }\right. {\tfrac{{\Delta t}}{{\lambda_{\alpha}}}} \left.\vphantom{ 1 - e^{-\tfrac{\Delta t}{\lambda_{\alpha}}} }\right) \left(\vphantom{ \frac{E_{\alpha}(t^{*})\lambda_{\alpha}}{\Delta ... ...bf{D}} \Delta\boldsymbol{\varepsilon}- \boldsymbol{\sigma}_{\alpha}(t) }\right. {\frac{{E_{\alpha}(t^{*})\lambda_{\alpha}}}{{\Delta t}}} \bar{{\mathbf{D}}} \Delta \boldsymbol\varepsilon \boldsymbol\sigma \alpha \left.\vphantom{ \frac{E_{\alpha}(t^{*})\lambda_{\alpha}}{\Delta ... ...bf{D}} \Delta\boldsymbol{\varepsilon}- \boldsymbol{\sigma}_{\alpha}(t) }\right)$$ (19.8)

Here t^* is a sampling point, usually halfway the time increment. This is only relevant if Young's modulus E changes during the analysis.