Let's analyze and solve the problem step by step! --- ## **Step 1: Understanding the Kelvin Model (Voigt Model)** The Kelvin (Voigt) model is a common viscoelastic model consisting of a **spring** (stiffness \( k \)) and a **dashpot** (viscosity \( \eta \)) connected in **parallel**. ### **Constitutive Equation** The **total force** \( F \) applied is shared by both the spring and dashpot: \[ F = F_{\text{spring}} + F_{\text{dashpot}} \] \[ F = kU + \eta \frac{dU}{dt} \] where: - \( U \) = displacement, - \( k \) = spring constant, - \( \eta \) = viscosity coefficient. --- ## **Step 2: Applying the Sudden Load** A **weight \( F \)** is *suddenly placed* on the model. This means a step load is applied at \( t = 0 \): \[ F(t) = \begin{cases} 0 & t < 0 \\ F & t \ge 0 \end{cases} \] --- ## **Step 3: Differential Equation for Displacement** The governing equation is: \[ F = kU + \eta \frac{dU}{dt} \] For \( t \ge 0 \), \( F \) is constant. --- ## **Step 4: Steady-State (Final) Displacement** We are asked for the **final displacement** \( U_{\infty} \), i.e., as \( t \to \infty \). ### **At Steady-State:** - The dashpot (damper) only resists changes in displacement. - As \( t \to \infty \), \( \frac{dU}{dt} \to 0 \) (displacement becomes constant). Plug into the equation: \[ F = kU_{\infty} + \eta \cdot 0 \] \[ F = kU_{\infty} \] \[ U_{\infty} = \frac{F}{k} \] --- ## **Step 5: Final Answer** ### **Final Displacement:** \[ \boxed{U_{\infty} = \frac{F}{k}} \] --- ## **Explanation** - **Initially** (at \( t = 0 \)), the dashpot resists deformation, so \( U = 0 \). - **Over time**, the displacement increases as the spring and dashpot deform. - **At steady-state** (long after the load is applied), the dashpot no longer resists (rate of change is zero), and only the spring supports the load. --- ### **Summary Table** | Time | Displacement \( U \) | |------------------|---------------------------| | Immediately (\( t=0 \)) | 0 | | After long time (\( t \to \infty \)) | \( \frac{F}{k} \) | --- ### **Physical Meaning** - The final displacement is **identical to that of a spring alone**. The dashpot only affects the **rate of deformation**, not the final position. --- **If you need the full time-dependent solution, let me know!**