Let's solve the finite element analysis (FEA) problem **step-by-step** as requested: --- ## **Given Data** - \( E = 1.75 \) ksi \( = 175 \) psi - \( A = 1.25 \) in² - \( L = 1.25 \) in - \( P_2 = -350 \) lb (at node 2, negative means left) - \( P_3 = 190 \) lb (at node 3, positive means right) Segment lengths and areas: - **Element I**: \( L_1 = 1.5L = 1.875 \) in, \( A_1 = 3.5A = 4.375 \) in² - **Element II**: \( L_2 = 2.5L = 3.125 \) in, \( A_2 = 2.5A = 3.125 \) in² - **Element III**: \( L_3 = 3.5L = 4.375 \) in, \( A_3 = 1.5A = 1.875 \) in² --- ## **(a) Element Stiffness Matrices** The **axial element stiffness** for an element \( e \): \[ k_e = \frac{EA}{L} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \] **Calculate for each element:** --- ### **Element I (Nodes 1-2):** \[ E_1 = 175\, \text{psi},\ A_1 = 4.375\,\text{in}^2,\ L_1 = 1.875\,\text{in} \] \[ k_1 = \frac{175 \times 4.375}{1.875} = \frac{7656.25}{1.875} = 4083.33\ \text{lb/in} \] \[ K^{(1)} = 4083.33 \begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} \] --- ### **Element II (Nodes 2-3):** \[ A_2 = 3.125\,\text{in}^2,\ L_2 = 3.125\,\text{in} \] \[ k_2 = \frac{175 \times 3.125}{3.125} = 175\ \text{lb/in} \] \[ K^{(2)} = 175 \begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} \] --- ### **Element III (Nodes 3-4):** \[ A_3 = 1.875\,\text{in}^2,\ L_3 = 4.375\,\text{in} \] \[ k_3 = \frac{175 \times 1.875}{4.375} = \frac{3281.25}{4.375} = 750\ \text{lb/in} \] \[ K^{(3)} = 750 \begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} \] --- ## **(b) Assemble Global Stiffness Matrix** For 4 nodes, the global matrix is \( 4 \times 4 \). - **Element I:** between nodes 1 and 2 - **Element II:** between nodes 2 and 3 - **Element III:** between nodes 3 and 4 Sum the contributions: \[ K = \begin{bmatrix} 4083.33 & -4083.33 & & \\ -4083.33 & 4083.33+175 & -175 & \\ & -175 & 175+750 & -750 \\ & & -750 & 750 \end{bmatrix} \] \[ K = \begin{bmatrix} 4083.33 & -4083.33 & & \\ -4083.33 & 5833.33 & -175 & \\ & -175 & 250 & -750 \\ & & -750 & 750 \end{bmatrix} \] --- ## **(c) Boundary Conditions and Reduced Matrix** - Node 1 and Node 4 are fixed: \( u_1 = u_4 = \) - Unknowns: \( u_2, u_3 \) Reduce matrix to nodes 2 and 3: \[ \begin{bmatrix} 5833.33 & -175 \\ -175 & 250 \end{bmatrix} \begin{bmatrix} u_2 \\ u_3 \end{bmatrix} = \begin{bmatrix} -350 \\ 190 \end{bmatrix} \] --- ## **(d) Solve for Unknown Displacements** Write the system: \[ 5833.33 u_2 - 175 u_3 = -350 \tag{1} \] \[ -175 u_2 + 250 u_3 = 190 \tag{2} \] **Solve:** From (2): \[ -175 u_2 + 250 u_3 = 190 \implies 250 u_3 = 190 + 175 u_2 \implies u_3 = \frac{190 + 175 u_2}{250} \] Substitute into (1): \[ 5833.33 u_2 - 175 \left( \frac{190 + 175 u_2}{250} \right) = -350 \] \[ 5833.33 u_2 - \frac{175 \times 190}{250} - \frac{175 \times 175 u_2}{250} = -350 \] \[ 5833.33 u_2 - 133 - 1225 u_2 = -350 \] \[ (5833.33 - 1225)u_2 - 133 = -350 \] \[ 4608.33 u_2 = -350 + 133 = -217 \] \[ u_2 = \frac{-217}{4608.33} = -.471\ \text{in} \] Now, solve for \( u_3 \): \[ u_3 = \frac{190 + 175 \times (-.471)}{250} = \frac{190 - 824.25}{250} = \frac{1075.75}{250} = .43\ \text{in} \] --- ## **(e) Solve for Nodal Reactions** - At node 1: \[ R_1 = K_{11} u_1 + K_{12} u_2 + K_{13} u_3 + K_{14} u_4 \] \[ = 4083.33 \cdot + (-4083.33) \cdot (-.471) + \cdot .43 + \cdot \] \[ = 1922.78\ \text{lb} \] - At node 4: \[ R_4 = K_{41} u_1 + K_{42} u_2 + K_{43} u_3 + K_{44} u_4 \] \[ = \cdot + \cdot (-.471) + (-750) \cdot .43 + 750 \cdot \] \[ = -322.78\ \text{lb} \] --- ## **(f) Internal Forces and Stresses** ### **Element I:** \[ \delta_1 = u_2 - u_1 = -.471 - = -.471\ \text{in} \] \[ \text{Force} = k_1 \cdot \delta_1 = 4083.33 \cdot (-.471) = -1922.78\ \text{lb} \] \[ \text{Stress} = \frac{-1922.78}{4.375} = -439.49\ \text{psi} \] ### **Element II:** \[ \delta_2 = u_3 - u_2 = .43 - (-.471) = .901\ \text{in} \] \[ \text{Force} = 175 \cdot .901 = 1576.75\ \text{lb} \] \[ \text{Stress} = \frac{1576.75}{3.125} = 504.70\ \text{psi} \] ### **Element III:** \[ \delta_3 = u_4 - u_3 = - .43 = -.43\ \text{in} \] \[ \text{Force} = 750 \cdot (-.43) = -322.78\ \text{lb} \] \[ \text{Stress} = \frac{-322.78}{1.875} = -172.15\ \text{psi} \] --- ## **Final Answers Summary** - **(d) Displacements:** \( u_2 = -.471 \) in, \( u_3 = .43 \) in - **(e) Nodal Reactions:** \( R_1 = 1922.78 \) lb, \( R_4 = -322.78 \) lb - **(f) Stresses:** - \( \sigma_1 = -439.49 \) psi - \( \sigma_2 = 504.71 \) psi - \( \sigma_3 = -172.15 \) psi --- ### **If you need the matrices or results presented in a specific format, let me know!**