Let's solve the spring assemblage problem step by step as requested: --- ### **Step 1: Degrees of Freedom (DOF)** #### **Explanation:** - Each node in a 1D spring system can potentially move along the axis (horizontal in this case). - In the figure, there are 4 nodes (let’s name them: 1, 2, 3, and 4). - Node 1 and Node 4 are attached to walls (fixed), so their displacements are zero: \( u_1 = 0 \) and \( u_4 = 0 \). - Nodes 2 and 3 are free to move. #### **Calculation:** - **Total nodes:** 4 - **Fixed nodes:** 2 (nodes 1 and 4) - **Free nodes:** 2 (nodes 2 and 3) #### **Answer:** **Degrees of Freedom = 2 (at nodes 2 and 3)** --- ### **Step 2: Displacements of Nodes 2, 3, and 4** #### **Explanation:** - We need to set up and solve the equilibrium equations using the **stiffness matrix method**. - Let \( k_1 = 10{,}000 \) N/m, \( k_2 = 20{,}000 \) N/m, \( k_3 = 10{,}000 \) N/m. - Let \( u_1, u_2, u_3, u_4 \) be the displacements at nodes 1, 2, 3, and 4. ##### **Global Stiffness Matrix:** - For 4 nodes, the global system is: \[ \begin{bmatrix} k_1 & -k_1 & 0 & 0 \\ -k_1 & k_1 + k_2 & -k_2 & 0 \\ 0 & -k_2 & k_2 + k_3 & -k_3 \\ 0 & 0 & -k_3 & k_3 \\ \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\ u_3 \\ u_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 450 \\ 0 \\ 0 \end{bmatrix} \] - \( u_1 = 0 \) (fixed), \( u_4 = 0 \) (fixed): So the reduced system for \( u_2 \) and \( u_3 \): \[ \begin{bmatrix} k_1 + k_2 & -k_2 \\ -k_2 & k_2 + k_3 \\ \end{bmatrix} \begin{bmatrix} u_2 \\ u_3 \end{bmatrix} = \begin{bmatrix} 450 \\ 0 \end{bmatrix} \] Substitute values: \[ \begin{bmatrix} 10,000 + 20,000 & -20,000 \\ -20,000 & 20,000 + 10,000 \\ \end{bmatrix} \begin{bmatrix} u_2 \\ u_3 \end{bmatrix} = \begin{bmatrix} 450 \\ 0 \end{bmatrix} \] \[ \begin{bmatrix} 30,000 & -20,000 \\ -20,000 & 30,000 \\ \end{bmatrix} \begin{bmatrix} u_2 \\ u_3 \end{bmatrix} = \begin{bmatrix} 450 \\ 0 \end{bmatrix} \] ##### **Solving the Equations:** From the second row: \[ -20,000u_2 + 30,000u_3 = 0 \implies u_2 = 1.5u_3 \] Now substitute into the first equation: \[ 30,000u_2 - 20,000u_3 = 450 \\ 30,000(1.5u_3) - 20,000u_3 = 450 \\ 45,000u_3 - 20,000u_3 = 450 \\ 25,000u_3 = 450 \\ u_3 = \frac{450}{25,000} = 0.018\ \text{m} \] \[ u_2 = 1.5 \times 0.018 = 0.027\ \text{m} \] \[ u_4 = 0 \] #### **Answer:** - \( u_2 = 0.027 \) m - \( u_3 = 0.018 \) m - \( u_4 = 0 \) m --- ### **Step 3: Reaction Forces at Nodes 1, 2, 3, 4** #### **Explanation:** - Reaction at node 1: \( R_1 = k_1(u_2 - u_1) \) - Reaction at node 2: internal, but let's compute the net force for completeness. - Reaction at node 3: internal. - Reaction at node 4: \( R_4 = k_3(u_3 - u_4) \) #### **Calculation:** - \( u_1 = 0 \), \( u_4 = 0 \) - \( R_1 = k_1 (u_2 - u_1) = 10,000 \times (0.027 - 0) = 270 \) N (to the left) - \( R_2 = \) net force at node 2 = \( k_1(u_1-u_2) + k_2(u_3-u_2) + \text{external force} \) Internal equilibrium (should sum to zero): \( k_1(u_1-u_2) + k_2(u_3-u_2) + 450 \) \( = 10,000(0-0.027) + 20,000(0.018-0.027) + 450 \) \( = -270 + 20,000 \times (-0.009) + 450 \) \( = -270 - 180 + 450 = 0 \) N (as expected, equilibrium at free node) - \( R_3 = k_2(u_2-u_3) + k_3(u_4-u_3) \) \( = 20,000(0.027-0.018) + 10,000(0-0.018) \) \( = 20,000(0.009) - 180 \) \( = 180 - 180 = 0 \) N - \( R_4 = k_3(u_3-u_4) = 10,000(0.018 - 0) = 180 \) N (to the right, at fixed support) #### **Answer:** - **Node 1:** 270 N (reaction, leftward) - **Node 2:** 0 N (free node, equilibrium) - **Node 3:** 0 N (free node, equilibrium) - **Node 4:** 180 N (reaction, rightward) --- ### **Step 4: Force in Each Element** #### **Explanation:** - Force in spring 1 (\( F_1 \)): \( k_1(u_2-u_1) = 10,000 \times 0.027 = 270 \) N - Force in spring 2 (\( F_2 \)): \( k_2(u_3-u_2) = 20,000 \times (0.018-0.027) = 20,000 \times (-0.009) = -180 \) N (compression) - Force in spring 3 (\( F_3 \)): \( k_3(u_4-u_3) = 10,000 \times (0-0.018) = -180 \) N (compression) #### **Answer:** - **Spring 1:** 270 N (tension) - **Spring 2:** 180 N (compression) - **Spring 3:** 180 N (compression) --- ## **Final Summary Table** | Step | Quantity | Value | |------|------------------------|------------------------------| | 1 | Degrees of Freedom | 2 | | 2 | Displacement \( u_2 \) | \( 0.027 \) m | | | Displacement \( u_3 \) | \( 0.018 \) m | | | Displacement \( u_4 \) | \( 0 \) m | | 3 | Reaction at Node 1 | 270 N | | | Reaction at Node 2 | 0 N | | | Reaction at Node 3 | 0 N | | | Reaction at Node 4 | 180 N | | 4 | Force in Spring 1 | 270 N (tension) | | | Force in Spring 2 | 180 N (compression) | | | Force in Spring 3 | 180 N (compression) | --- **If you need the full matrix solution or any step detailed further, let me know!**