19.2.1 Kelvin Chain Model

J(t, $\displaystyle \tau$ ) = $\displaystyle \sum_{{\alpha=0}}^{{n}}$ $\displaystyle {\frac{{1}}{{E_{\alpha}(\tau)}}}$ $\displaystyle \left(\vphantom{ 1- e^{- \frac{t-\tau}{\lambda_{\alpha}}} }\right.$ 1 - e^- $\displaystyle \left.\vphantom{ 1- e^{- \frac{t-\tau}{\lambda_{\alpha}}} }\right)$

(19.15)

$\scriptstyle \alpha$

indicates that the stiffness of the model can be time dependent, for instance due to temperature or maturity influence. Physically the Dirichlet series can be interpreted as a Kelvin Chain model [Fig.19.2].

**Figure 19.2:** Kelvin Chain
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The stiffness and viscosity of the spring and damper in each part of the Kelvin Chain determine the retardation time $\lambda_{{\alpha}}^{}$ :

$\displaystyle \lambda_{{\alpha}}^{}$ = $\displaystyle {\frac{{\eta_{\alpha}}}{{E_{\alpha}}}}$

(19.16)

Substitution of (19.15) into (19.13) and (19.14) yields

$\displaystyle {\frac{{1}}{{\tilde{E}(t^{*})}}}$ = $\displaystyle {\frac{{1}}{{\Delta t}}}$ $\displaystyle \sum_{{\alpha=0}}^{{n}}$ $\displaystyle \int_{{t}}^{{t+\Delta t}}$ $\displaystyle {\frac{{1}}{{E_{\alpha}(\tau)}}}$ $\displaystyle \left(\vphantom{ 1- e^{- \frac{t+\Delta t-\tau}{\lambda_{\alpha}}} }\right.$ 1 - e^- $\displaystyle \left.\vphantom{ 1- e^{- \frac{t+\Delta t-\tau}{\lambda_{\alpha}}} }\right)$ d $\displaystyle \tau$

(19.17)

and

$\displaystyle \tilde{{\boldsymbol{\sigma}}}$ (t) = - $\displaystyle \tilde{{E}}$ (t^*) $\displaystyle \sum_{{\alpha=0}}^{{n}}$ $\displaystyle \int_{{0}}^{{t}}$ $\displaystyle {\frac{{1}}{{E_{\alpha}(\tau)}}}$ $\displaystyle \left(\vphantom{ 1 - e^{-\frac{\Delta t}{\lambda_{\alpha}}} }\right.$ 1 - e^- $\displaystyle \left.\vphantom{ 1 - e^{-\frac{\Delta t}{\lambda_{\alpha}}} }\right)$ e^- $\displaystyle \dot{{\boldsymbol{\sigma}}}$ ( $\displaystyle \tau$ ) d $\displaystyle \tau$

(19.18)

If we take for E( $\tau$ ) the value at time t^* we can integrate the remaining part of the integrals analytically.

$\displaystyle {\frac{{1}}{{\tilde{E}(t^{*})}}}$ = $\displaystyle \sum_{{\alpha=0}}^{{n}}$ $\displaystyle {\frac{{1}}{{E_{\alpha}(t^{*})}}}$ $\displaystyle \left(\vphantom{ 1 - \frac{\lambda_{\alpha}}{\Delta t} \left( 1 - e^{- \frac{\Delta t}{\lambda_{\alpha}}} \right) }\right.$ 1 - $\displaystyle {\frac{{\lambda_{\alpha}}}{{\Delta t}}}$ $\displaystyle \left(\vphantom{ 1 - e^{- \frac{\Delta t}{\lambda_{\alpha}}} }\right.$ 1 - e^- $\displaystyle \left.\vphantom{ 1 - e^{- \frac{\Delta t}{\lambda_{\alpha}}} }\right)$ $\displaystyle \left.\vphantom{ 1 - \frac{\lambda_{\alpha}}{\Delta t} \left( 1 - e^{- \frac{\Delta t}{\lambda_{\alpha}}} \right) }\right)$

(19.19)

and

$\displaystyle \tilde{{\boldsymbol{\sigma}}}$ (t) = - $\displaystyle \tilde{{E}}$ (t^*) $\displaystyle \sum_{{\alpha=0}}^{{n}}$ $\displaystyle \left(\vphantom{ 1 - e^{-\frac{\Delta t}{\lambda_{\alpha}}} }\right.$ 1 - e^- $\displaystyle \left.\vphantom{ 1 - e^{-\frac{\Delta t}{\lambda_{\alpha}}} }\right)$ $\displaystyle \tilde{{\boldsymbol{\varepsilon}}}_{{\alpha}}^{}$ (t)

(19.20)

with

$\displaystyle \tilde{{\boldsymbol{\varepsilon}}}_{{\alpha}}^{}$ (t) = $\displaystyle \int_{{0}}^{{t}}$ $\displaystyle {\frac{{1}}{{E_{\alpha}(\tau)}}}$ e^- $\displaystyle \dot{{\boldsymbol{\sigma}}}$ ( $\displaystyle \tau$ ) d $\displaystyle \tau$

(19.21)

The value of $\tilde{{\boldsymbol{\varepsilon}}}_{{\alpha}}^{}$ (t) is fully determined at the start of the increment (time t ). To find the value at the start of the next increment we write

$\displaystyle \tilde{{\boldsymbol{\varepsilon}}}_{{\alpha}}^{}$ (t + $\displaystyle \Delta$ t)	= $\displaystyle \int_{{0}}^{{t}}$ $\displaystyle {\frac{{1}}{{E_{\alpha}(\tau)}}}$ e^- $\displaystyle \dot{{\boldsymbol{\sigma}}}$ ( $\displaystyle \tau$ ) d $\displaystyle \tau$ + $\displaystyle \int_{{t}}^{{t+\Delta t}}$ $\displaystyle {\frac{{1}}{{E_{\alpha}(\tau)}}}$ e^- $\displaystyle \dot{{\boldsymbol{\sigma}}}$ ( $\displaystyle \tau$ ) d $\displaystyle \tau$
	$\displaystyle \approx$ e^- $\displaystyle \tilde{{\boldsymbol{\varepsilon}}}_{{\alpha}}^{}$ (t) + $\displaystyle {\frac{{\lambda_{\alpha} \Delta \boldsymbol{\sigma}}}{{E_{\alpha}(t^{*})\Delta t}}}$ $\displaystyle \left(\vphantom{ 1 - e^{- \frac{\Delta t}{\lambda_{\alpha}}} }\right.$ 1 - e^- $\displaystyle \left.\vphantom{ 1 - e^{- \frac{\Delta t}{\lambda_{\alpha}}} }\right)$	(19.22)