# Structural System Analysis Let's solve the problem step by step as requested. --- ## Step 1: **Global Stiffness Matrix** We have 3 elements and 4 nodes. **Element Stiffness Values:** - \( k_1 = 100 \times 10^3 \) kN/m (between Node 1 and 2) - \( k_2 = 200 \times 10^3 \) kN/m (between Node 2 and 3) - \( k_3 = 140 \times 10^3 \) kN/m (between Node 3 and 4) ### **Global Stiffness Matrix \( K \):** Each element adds to the global matrix at its connected nodes. The 4x4 matrix is: \[ K = \begin{bmatrix} k_1 & -k_1 & & \\ -k_1 & k_1 + k_2 & -k_2 & \\ & -k_2 & k_2 + k_3 & -k_3 \\ & & -k_3 & k_3 \\ \end{bmatrix} \] Substitute values: \[ K = \begin{bmatrix} 100 & -100 & & \\ -100 & 300 & -200 & \\ & -200 & 340 & -140 \\ & & -140 & 140 \\ \end{bmatrix} \times 10^3 \text{ kN/m} \] --- ## Step 2: **Solving for Displacements and Forces** ### **Boundary Conditions & Loads** - Node 1 and Node 4 are fixed (\(u_1 = u_4 = \)) - Node 2: Solen (applies force, let's call it \(F_2\)) - Node 3: 100 kN force to the right (\(F_3 = 100\) kN) #### **Reduced System (Nodes 2 & 3 only):** Stiffness matrix for free nodes: \[ \begin{bmatrix} 300 & -200 \\ -200 & 340 \\ \end{bmatrix} \begin{bmatrix} u_2 \\ u_3 \\ \end{bmatrix} = \begin{bmatrix} F_2 \\ 100 \\ \end{bmatrix} \] #### **Example Calculation (Assume \(F_2 = -50\) kN for illustration):** \[ \begin{bmatrix} 300 & -200 \\ -200 & 340 \\ \end{bmatrix} \begin{bmatrix} u_2 \\ u_3 \\ \end{bmatrix} = \begin{bmatrix} -50 \\ 100 \\ \end{bmatrix} \] Solve using matrix algebra: \[ u_2 = \frac{(-50 \times 340) - (100 \times -200)}{(300 \times 340) - (200 \times 200)} \] \[ u_3 = \frac{(300 \times 100) - (-50 \times -200)}{(300 \times 340) - (200 \times 200)} \] Calculate denominator: \[ (300 \times 340) - (200 \times 200) = 102000 - 40000 = 62000 \] Calculate numerators: \[ u_2: (-50 \times 340) - (100 \times -200) = -17000 + 20000 = 300 \] \[ u_3: (300 \times 100) - (-50 \times -200) = 30000 - 10000 = 20000 \] So, \[ u_2 = \frac{300}{62000} \approx .048 \text{ mm} \] \[ u_3 = \frac{20000}{62000} \approx .323 \text{ mm} \] --- ### **Reaction Forces at Nodes 1 and 4** Use global stiffness matrix: \[ F_1 = k_1(u_1 - u_2) = 100 \times 10^3 ( - .048) = -480 \text{ kN} \] \[ F_4 = k_3(u_4 - u_3) = 140 \times 10^3 ( - .323) = -45220 \text{ kN} \] --- ### **Force in Each Member** \[ \text{Member 1: } F_1 = k_1 (u_2 - u_1) = 100 \times 10^3 (.048 - ) \] \[ = 480 \text{ kN} \] \[ \text{Member 2: } F_2 = k_2 (u_3 - u_2) = 200 \times 10^3 (.323 - .048) \] \[ = 200 \times 10^3 \times .275 = 55000 \text{ kN} \] \[ \text{Member 3: } F_3 = k_3 (u_4 - u_3) = 140 \times 10^3 ( - .323) \] \[ = -45220 \text{ kN} \] --- ## **Graphical Representation** Below is a schematic: ```plaintext [Fixed]--(k1)--[2]--(k2)--[3]--(k3)--[Fixed] 1 2 3 4 ``` - Node 2: Solen applies force - Node 3: 100 kN load to right --- ## **Summary Table** | Member | Stiffness (kN/m) | Force (kN) | |:--------:|:----------------:|:----------:| | 1 | 100,000 | 4,800 | | 2 | 200,000 | 55,000 | | 3 | 140,000 | 45,220 | --- ## **Final Answers** 1. **Global stiffness matrix**: As shown above. 2. **Displacements of nodes 2 & 3**: \( u_2 \approx .048 \) mm, \( u_3 \approx .323 \) mm (example, depends on actual \( F_2 \)). 3. **Reaction forces at nodes 1 & 4**: \( F_1 = -4,800 \) kN, \( F_4 = -45,220 \) kN. 4. **Force in members**: See table above. --- **Note:** - The direction/sign of forces and displacements depends on the assumed force directions and sign conventions. - If specific load values differ, recalculate accordingly. - If you have the actual value for the force at node 2, substitute and recalculate. If you want the MATLAB or Python code for solving the matrix, let me know!