## 4 Verification and Discussion of the Example
### Viscoelastic Displacement of the Rod Subject to Uniaxial Tension
#### 4.1.3 Comparison and Discussion of Simulation Results  
Following the calculation steps previously outlined, considering the case of only one element, the viscoelastic displacements for different material parameters can be obtained. As depicted in Figure 2, the graph also includes the analytical solution obtained using Equation (25). As presented in Figure 3, the relative error of two methods under different material parameter values can be calculated using the following formula:  
\[E_{error}=\frac{\left(u_{numeric}-u_{analytical}\right)\times 100}{u_{analytical}} \tag{27}\]where \(u_{numeric}\) is the axial tension displacement obtained by the numerical algorithm proposed in this study, \(u_{analytical}\) is the analytical solution to the displacement obtained according to Equation (25), and \(E_{error}\) is the relative error of the results obtained by the two methods. From Figure 2, it is evident that in the case of using only one element, the results yielded by the algorithm proposed in this paper nearly perfectly align with the analytical solution. Under constant loading, the viscoelastic displacement of the axial rod progressively increases with time, indicating the creep properties inherent to viscoelastic materials. Analyzing the error variation depicted in Figure 3, it becomes apparent that initially, the error is minimal, followed by an increase over time, reaching a peak around the relaxation time. Subsequently, the error gradually diminishes. Throughout the process, the error consistently remains negative, indicating a slight underestimation in the calculated results compared to the analytical solution. The initial ascent and subsequent descent of the error can be attributed to the rapid change in the slope of the curve near the relaxation time, as evident from the equation for the relaxation modulus of the generalized Maxwell model. This suggests a steep slope of the curve in proximity to the relaxation time, succeeded by a more gradual change beyond this range, signifying a relatively stable relaxation modulus. In this example, for a relaxation time of 3.6 s, the stress decreases by 63.2% from its initial value (\(1/e\) of the initial stress), indicative of rapid attenuation. Consequently, the material parameters undergo rapid changes near the relaxation time (or within a time range of the same order of magnitude as the relaxation time). Due to the uniform time intervals used in the calculation, the error increases within this interval. However, beyond this range, as  
Figure 3: Relative error at different relaxation times. Figure 2: Comparison of different relaxation times. the elastic modulus changes more slowly, the error decreases. To enhance the calculation accuracy, reducing the time intervals near the relaxation time and enlarging them away from the relaxation time can be contemplated. Viscoelastic Displacement of Slender Members with Equal Cross-Section under Self-Weight Effect 4.2.1. Information Detail  
As depicted in Figure 4, we consider a straight rod with an equal cross-section characterized by a unit weight. \(P=10^{-6}\)N/mm\({}^{3}\). The rod is fixed at the upper end and free at the lower end, with a length of L = 2 m (equivalent to 2000 mm) and a cross-section of 400 mm \(\times\) 400 mm. The coordinate origin is denoted as O, with the z-axis extending positively upward and the y-axis oriented horizontally and positively to the right. Considering the material as linear viscoelastic and using the generalized Maxwell model to reflect the relaxation characteristics of the material's elastic modulus as \(E_{\infty}=0.65\) MPa, \(E_{1}=3\) MPa. The initial elastic modulus is \(E_{0}=3.65\) MPa. Using the generalized Kelvin model to reflect the creep Poisson's ratio of the material, where \(\mu_{0}=0.34\), \(\mu_{1}=0.15\), and the Poisson's ratio at the final stable state is \(\mu_{\infty}=0.49\). The relaxation time is 3.6 s, i.e., \(\tau_{r}=\frac{\eta_{1}}{E_{1}}=\frac{10.8}{3}=3.6\) s, \(\tau_{i}=\eta_{i}\mu_{1}=24\times 0.15=3.6\) s = \(\tau_{r}\).