Let's break down the problem step by step. ### **Given:** - Horizontal force of **50 kN** at node 2. - Nodes **1 and 4** are fixed. - Lengths of elements: **1 m** each. - **Modulus of Elasticity (E):** - Elements 1 and 2: **200 GPa** - Element 3: **100 GPa** - **Area (A):** - Elements 1 and 2: **0.001 m²** - Element 3: **0.002 m²** ### **Required:** Determine the reactions at nodes 1 and 4. --- ### **Step 1: Stiffness Calculation** The stiffness \( k \) of each bar (assuming axial deformation) is: \[ k = \frac{EA}{L} \] **For Element 1 and 2:** \[ k_1 = k_2 = \frac{200 \times 10^9 \times 0.001}{1} = 200 \times 10^6 \, \text{N/m} \] **For Element 3:** \[ k_3 = \frac{100 \times 10^9 \times 0.002}{1} = 200 \times 10^6 \, \text{N/m} \] So, all elements have the **same stiffness**: \( 200 \times 10^6 \, \text{N/m} \). --- ### **Step 2: System Setup** Let's denote the displacements at nodes 2 and 3 as \( u_2 \) and \( u_3 \). Nodes 1 and 4 are fixed (\( u_1 = u_4 = 0 \)). **Forces at nodes:** - Node 2: \( +50 \) kN (to the right) - Node 3: \( 0 \) - Node 1 and 4: Reactions (to be found) --- ### **Step 3: Write Equilibrium Equations** #### **Node 2:** \[ k_1(u_2 - u_1) + k_2(u_2 - u_3) = 50{,}000 \] \[ 200 \times 10^6 (u_2 - 0) + 200 \times 10^6 (u_2 - u_3) = 50{,}000 \] \[ 200u_2 + 200u_2 - 200u_3 = 0.05 \quad \text{(divide by }10^6\text{)} \] \[ 400u_2 - 200u_3 = 0.05 \] #### **Node 3:** \[ k_2(u_3 - u_2) + k_3(u_3 - u_4) = 0 \] \[ 200 \times 10^6 (u_3 - u_2) + 200 \times 10^6 (u_3 - 0) = 0 \] \[ 200u_3 - 200u_2 + 200u_3 = 0 \] \[ 400u_3 - 200u_2 = 0 \] --- ### **Step 4: Solve the Equations** From node 3 equation: \[ 400u_3 - 200u_2 = 0 \implies u_3 = 0.5u_2 \] Substitute \( u_3 \) in the node 2 equation: \[ 400u_2 - 200(0.5u_2) = 0.05 \] \[ 400u_2 - 100u_2 = 0.05 \] \[ 300u_2 = 0.05 \] \[ u_2 = \frac{0.05}{300} = 0.0001667 \text{ m} \] \[ u_3 = 0.5 \times 0.0001667 = 0.0000833 \text{ m} \] --- ### **Step 5: Find Reactions** **Reaction at Node 1 (left):** \[ R_1 = k_1(u_1 - u_2) = 200 \times 10^6 (0 - 0.0001667) = -33,333 \text{ N} = -33.3 \text{ kN} \] **Reaction at Node 4 (right):** \[ R_4 = k_3(u_4 - u_3) = 200 \times 10^6 (0 - 0.0000833) = -16,666 \text{ N} = -16.7 \text{ kN} \] Negative sign indicates reactions are to the left (opposing the applied force). --- ### **Step 6: Match with Options** - **Reaction at node 1: ≈ 33.3 kN** - **Reaction at node 4: ≈ 16.7 kN** So, the **correct answer is:** > Reaction at node 1 = 33.3 kN, Reaction at node 4 = 16.7 kN --- **Let me know if you need a step-by-step explanation or have further questions!**

Based on the calculations: - Reaction at node 1 ≈ 33.3 kN (acting to the left) - Reaction at node 4 ≈ 16.7 kN (acting to the left) **Therefore, the correct option corresponds to these reaction values.**