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In order to capture the material response more accurately, a combination of Maxwell and Kelvin-Voigt models is used and is referred to as the Standard Linear Solid (SLS). A representative SLS (Kelvin form) is shown in figure below.
Figure: The Kelvin-Voigt Model
This model consists of spring ${{E}_{1}}$ which is placed in series with a Kelvin-Voigt model represented by spring ${{E}_{2}}$ and dashpot $\eta $. This allows the spring ${{E}_{1}}$ to be activated instantaneously during loading and unloading accompanied by a time dependent response of spring ${{E}_{2}}$ due to the load transfer through spring-dashpot Kelvin-Voigt element.
The applied stress $\sigma $ produces strain in spring ${{E}_{1}}$ and the stress-strain relation for this spring is governed by,
\begin{equation} \label{eq:1}
\[\sigma ={{E}_{1}}{{\varepsilon }_{1}}\
\end{equation}
The derivative of (\ref{eq:1}) gives the strain rate,
\begin{equation} \label{eq:2}
\[{{\dot{\varepsilon }}_{1}}=\frac{{\dot{\sigma }}}{{{E}_{1}}}\
\end{equation}
This applied stress $\sigma $ is also distributed in the Kelvin Voigt element such that,
\begin{equation} \label{eq:3}
\[\sigma ={{\sigma }_{1}}+{{\sigma }_{2}}\
\end{equation}
The stresses and strains in Kelvin-Voigt element has the following relationship,
\begin{equation} \label{eq:4}
\[{{\sigma }_{1}}={{E}_{2}}{{\varepsilon }_{2}}\
\end{equation}
and,
\begin{equation} \label{eq:5}
\[{{\sigma }_{2}}=\eta {{\dot{\varepsilon }}_{2}}\
\end{equation}
The time derivative of the total strain $\left( \varepsilon ={{\varepsilon }_{1}}+{{\varepsilon }_{2}} \right)$ in the Standard Linear Element is,
\begin{equation} \label{eq:6}
\[\dot{\varepsilon }={{\dot{\varepsilon }}_{1}}+{{\dot{\varepsilon }}_{2}}\
\end{equation}
By substituting the strain derivatives from (\ref{eq:2}) and (\ref{eq:5}) in (\ref{eq:6}) we can obtain
\begin{equation} \label{eq:7}
\[\dot{\varepsilon }=\frac{{\dot{\sigma }}}{{{E}_{1}}}+\frac{\sigma }{\eta }-\frac{{{E}_{2}}}{\eta }{{\varepsilon }_{2}}\
\end{equation}
Now by substituting $\left( {{\varepsilon }_{2}}=\varepsilon -{{\varepsilon }_{1}} \right)$ in (\ref{eq:7}) in conjunction with (\ref{eq:1}) and rearranging we obtain the following differential equation associated with the Standard Linear Solid,
\begin{equation} \label{eq:8}
\[\frac{\eta }{\left( {{E}_{1}}+{{E}_{2}} \right)}\dot{\sigma }+\sigma =\frac{{{E}_{1}}}{\left( {{E}_{1}}+{{E}_{2}} \right)}\eta \dot{\varepsilon }+\frac{{{E}_{1}}{{E}_{2}}}{\left( {{E}_{1}}+{{E}_{2}} \right)}\varepsilon \
\end{equation}
The differential equation (\ref{eq:8}) now can be solved for given loading and unloading cases of known/prescribed stress or strain by using Laplace transform or numerically (e.g. MATLAB ode45 function etc.).
Creep Behavior in Standard Linear Solid Model
If a constant stress ${{\sigma }_{0}}$ is applied instantaneously at time t=0, the stress rate $\dot{\sigma }=0$ for (t > 0). Hence by substituting $\dot{\sigma }=0$ in (\ref{eq:8}) we obtain,
\begin{equation} \label{eq:9}
\[\dot{\varepsilon }=\sigma \frac{\left( {{E}_{1}}+{{E}_{2}} \right)}{{{E}_{1}}\eta }-\frac{{{E}_{2}}}{\eta }\varepsilon \
\end{equation}
With the initial condition, $\varepsilon (0)={{\sigma }_{o}}/{{E}_{1}}$ the solution of (\ref{eq:9}) is,
\begin{equation} \label{eq:10}
\[\varepsilon (t)={{\sigma }_{o}}{{J}_{creep}}(t)\
\end{equation}
where,
\begin{equation} \label{eq:11}
\[{{J}_{creep}}=\frac{{{e}^{-\left( {{E}_{2}}/\eta \right)t}}}{{{E}_{1}}}+\frac{{{E}_{1}}+{{E}_{2}}}{{{E}_{1}}{{E}_{2}}}\left( 1-{{e}^{-\left( {{E}_{2}}/\eta \right)t}} \right)\
\end{equation}
is the creep compliance for the SLS.
Relaxation Behavior in Standard Linear Solid Model
When a constant strain ${{\varepsilon }_{0}}$is applied instantaneously at time t=0, the strain rate $\dot{\varepsilon }=0$ for (t > 0). Hence by substituting $\dot{\varepsilon }=0$ in (\ref{eq:8}) we obtain,
\begin{equation} \label{eq:12}
\[\dot{\sigma }=-\sigma \frac{\left( {{E}_{1}}+{{E}_{2}} \right)}{\eta }+\frac{{{E}_{1}}{{E}_{2}}}{\eta }\varepsilon \
\end{equation}
With the initial condition, $\sigma (0)={{E}_{1}}{{\varepsilon }_{o}}$ the solution of (\ref{eq:12}) is,
\begin{equation} \label{eq:13}
\[\sigma (t)={{\varepsilon }_{o}}{{E}_{relax}}(t)\
\end{equation}
where,
\begin{equation} \label{eq:14}
\[{{E}_{relax}}=\frac{E_{1}^{2}}{{{E}_{1}}+{{E}_{2}}}{{e}^{-\left( {{E}_{1}}+{{E}_{2}}/\eta \right)t}}+\frac{{{E}_{1}}{{E}_{2}}}{{{E}_{1}}+{{E}_{2}}}\
\end{equation}
is the relaxation modulus for the SLS.
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