## 4 Verification and Discussion of the Example
### Viscoelastic Displacement of the Rod Subject to Uniaxial Tension
#### 4.2.2. Analytical Solution of Viscoelastic Displacement
Consequently, we have the following:  
\[E(s)=E_{\infty}+\frac{E_{1}s}{s+\frac{1}{\tau_{1}}} \tag{32}\]  
\[\mu(s)=\mu_{0}+\frac{\mu_{1}}{\tau_{1}\Big{(}s+\frac{1}{\tau_{1}}\Big{)}} \tag{33}\]  
Based on this framework, we derive the following:  
\[\frac{1}{s}\cdot\frac{\mu(s)}{E(s)}=\frac{\mu_{0}}{(E_{\infty}+E_{1})}\Bigg{[} \frac{1}{s+\frac{E_{\infty}}{(E_{\infty}+E_{1})\tau_{1}}}+\frac{1}{\tau_{1}} \cdot\frac{1}{s\Big{(}s+\frac{E_{\infty}}{(E_{\infty}+E_{1})\tau_{1}}\Big{)} }\Bigg{]}+\frac{\mu_{1}}{\tau_{1}(E_{\infty}+E_{1})}\cdot\Bigg{[}\frac{1}{s \Big{(}s+\frac{E_{\infty}}{(E_{\infty}+E_{1})\tau_{1}}\Big{)}}\Bigg{]} \tag{34}\]  
Substituting the above equation into Equation (30) and introducing \(E_{0}=E_{\infty}+E_{1}\) yields the following:  
\[w_{1}(s)=\frac{P}{2}\Big{(}x^{2}+y^{2}\Big{)}\frac{\mu_{0}}{E_{0}}\Bigg{[} \frac{1}{s+\frac{E_{\infty}}{E_{0}\tau_{1}}}+\frac{1}{\tau_{1}}\cdot\frac{1}{ s\Big{(}s+\frac{E_{\infty}}{E_{0}\tau_{1}}\Big{)}}\Bigg{]}+\frac{\mu_{1}}{ \tau_{1}E_{0}}\cdot\Bigg{[}\frac{1}{s\Big{(}s+\frac{E_{\infty}}{E_{0}\tau_{1}} \Big{)}}\Bigg{]} \tag{35}\]  
Applying the inverse Laplace transform to Equation (35) yields the following:  
\[w_{1}(t)=\frac{P}{2}\Big{(}x^{2}+y^{2}\Big{)}\cdot\frac{\mu_{0}}{E_{0}}\bigg{[} e^{-\frac{E_{\infty}}{E_{0}\tau_{1}}t}+\frac{E_{0}}{E_{\infty}}\bigg{(}1-e^{- \frac{E_{\infty}}{E_{0}\tau_{1}}t}\bigg{)}\bigg{]}+\frac{P}{2}\Big{(}x^{2}+y^ {2}\Big{)}\cdot\frac{\mu_{1}}{E_{\infty}}\bigg{(}1-e^{-\frac{E_{\infty}}{E_{0} \tau_{1}}t}\bigg{)} \tag{36}\]  
Similarly, applying the inverse Laplace transform to Equation (31) yields the following:  
\[w(s)=\frac{P}{2}\Big{[}\frac{\mu(s)}{sE(s)}\Big{(}x^{2}+y^{2}\Big{)}+\frac{z^{ 2}-L^{2}}{sE(s)}\Big{]}=w_{1}(s)+w_{2}(s) \tag{29}\]0  
From this, the inverse Laplace transform of Equation (29) can be derived as follows:  
\[w(s)=\frac{P}{2}\Big{[}\frac{\mu(s)}{sE(s)}\Big{(}x^{2}+y^{2}\Big{)}+\frac{z^{ 2}-L^{2}}{sE(s)}\Big{]}=w_{1}(s)+w_{2}(s) \tag{29}\]1  
where  
\[w(s)=\frac{P}{2}\Big{[}\frac{\mu(s)}{sE(s)}\Big{(}x^{2}+y^{2}\Big{)}+\frac{z^{ 2}-L^{2}}{sE(s)}\Big{]}=w_{1}(s)+w_{2}(s) \tag{29}\]2  
\[w(s)=\frac{P}{2}\Big{[}\frac{\mu(s)}{sE(s)}\Big{(}x^{2}+y^{2}\Big{)}+\frac{z^{ 2}-L^{2}}{sE(s)}\Big{]}=w_{1}(s)+w_{2}(s) \tag{29}\]3  
In this example, the analytical solution for viscoelastic displacement, accounting for both Poisson's ratio and elastic modulus variations over time, can be obtained by substituting the respective parameters as follows:  
\[w(s)=\frac{P}{2}\Big{[}\frac{\mu(s)}{sE(s)}\Big{(}x^{2}+y^{2}\Big{)}+\frac{z^{ 2}-L^{2}}{sE(s)}\Big{]}=w_{1}(s)+w_{2}(s) \tag{29}\]4  
where  
\[w(s)=\frac{P}{2}\Big{[}\frac{\mu(s)}{sE(s)}\Big{(}x^{2}+y^{2}\Big{)}+\frac{z^{ 2}-L^{2}}{sE(s)}\Big{]}=w_{1}(s)+w_{2}(s) \tag{29}\]5\[w(s)=\frac{P}{2}\Big{[}\frac{\mu(s)}{sE(s)}\Big{(}x^{2}+y^{2}\Big{)}+\frac{z^{ 2}-L^{2}}{sE(s)}\Big{]}=w_{1}(s)+w_{2}(s) \tag{29}\]6  
Based on the algorithm proposed in this paper, the viscoelastic displacement of the sectioned straight bar under self-weight is calculated. The time interval is set to 1 s, and the finite element mesh is partitioned, as shown in \[w(s)=\frac{P}{2}\Big{[}\frac{\mu(s)}{sE(s)}\Big{(}x^{2}+y^{2}\Big{)}+\frac{z^{ 2}-L^{2}}{sE(s)}\Big{]}=w_{1}(s)+w_{2}(s) \tag{29}\]7 The calculation results of the viscoelastic displacement at the free end corner within 50 s are shown in Figure 6, where the analytical solution obtained by Equation (41) is also provided. From the calculation results, it can be observed that for the sectioned straight bar under self-weight, at the position of the free end corner, the instantaneous elastic displacement due to loading is approximately 0.5 mm. Subsequently, due to the viscous flow of the material, the viscoelastic displacement gradually increases with time, reaching 2.84 mm at 50 s, which is approximately five times the initial displacement. A comparison with the analytical solution shows that the maximum error of the algorithm proposed in this paper is about 0.3%, indicating good computational accuracy.