Here's a summarized and organized version of the handwritten solution for clarity: --- # Solution Summary ## Given Data - Cross-sectional area, \(A = 500\, \text{mm}^2\) - Modulus of Elasticity, \(E = 200\, \text{GPa}\) --- ## Step 1: Free Body Diagram - Coordinates and dimensions are drawn, with points \(A, B, C, D\). - Dimensions: - \(AB = 1.6\, \text{m}\) - \(AD = 1.2\, \text{m}\) - \(DC = .9\, \text{m}\) - \(BC = 4.8\, \text{m}\) - \(DE = 2.5\, \text{m}\) - Vertical and horizontal components are labeled. --- ## Step 2: Equations of Equilibrium ### Horizontal (x-direction): \[ \sum F_x = \Rightarrow A_x - D_x = \quad \Rightarrow A_x = D_x = 5\, \text{kN} \] ### Vertical (y-direction): \[ \sum F_y = \Rightarrow -4.8 + A_y = \Rightarrow A_y = 4.8\, \text{kN} \] --- ## Step 3: Joint Method for Force Components ### At Point A: - \(\tan \theta_1 = \frac{1.2}{1.6} \Rightarrow \theta_1 = 36.07^\circ\) - Horizontal component: \[ F_{AB} \cos 36.07^\circ = 5\, \text{kN} \] \[ F_{AB} = \frac{5}{\cos 36.07^\circ} \approx 6.25\, \text{kN} \] ### At Point D: - \(\tan \theta_2 = \frac{1.2}{1.6} \Rightarrow \theta_2 = 36.07^\circ\) - Vertical component: \[ F_{DB} \sin 36.07^\circ = 1.75\, \text{kN} \] \[ F_{DB} = \frac{1.75}{\sin 36.07^\circ} \approx 3.00\, \text{kN} \] --- ## Step 4: Force Calculation - \(F_{DC}\) from equilibrium: \[ F_{DC} = -3.60\, \text{kN} \] - Components at C: - \(\theta_3 = 53.13^\circ\) - \(F_{CB} = 6.00\, \text{kN}\) --- ## Step 5: Axial Strain and Stress - Axial deformation: \[ \delta = \frac{F \times L}{A \times E} \] - Using the force \(F_{AB} = 6.25\, \text{kN}\): \[ \text{Stress } \sigma = \frac{F}{A} \approx \frac{6.25 \times 10^3}{500 \times 10^{-6}} = 12.5\, \text{MPa} \] --- ## Step 6: Deflection Calculations - Using the equilibrium equations, deflections are found via the method of joints and applying trigonometric relations. --- ## Final Note: - The diagram and forces are used to determine the internal forces in each member. - The displacement and stress are calculated based on these forces and material properties. --- If you need a detailed step-by-step calculation or a specific part clarified, please specify!

# Solution Summary ## Given Data - Cross-sectional area, \( A = 500\, \text{mm}^2 \) - Modulus of Elasticity, \( E = 200\, \text{GPa} \) --- ## Step 1: Free Body Diagram - **Coordinates and dimensions**: - Points: \( A, B, C, D \) - Dimensions: - \( AB = 1.6\, \text{m} \) - \( AD = 1.0\, \text{m} \) - \( DC = 0.9\, \text{m} \) - \( BC = 4.8\, \text{m} \) --- ## Step 2: Equations of Equilibrium ### Horizontal (x-direction): \[ \sum F_x = A_x + D_x = 0 \] ### Vertical (y-direction): \[ \sum F_y = -4.8 + A_y + D_y = 0 \] From which we get: - \( A_y = 4.8\, \text{kN} \) - \( D_y = 0\, \text{kN} \) --- ## Step 3: Joint Method for Force Components ### At Point A: - **Angle Calculation**: \[ \tan \theta_1 = \frac{1.0}{1.6} \Rightarrow \theta_1 = 36.87^\circ \] - **Horizontal Component**: \[ F_{AB} \cos 36.87^\circ = 4.8 \quad \Rightarrow \quad F_{AB} = \frac{4.8}{\cos 36.87^\circ} \approx 6.00\, \text{kN} \] ### At Point D: - **Vertical Component**: \[ F_{DB} \sin 36.87^\circ = 1.0\, \text{kN} \quad \Rightarrow \quad F_{DB} = \frac{1.0}{\sin 36.87^\circ} \approx 1.6\, \text{kN} \] --- ## Step 4: Force Calculation - **Equilibrium at C**: \[ F_{DC} = -3.0\, \text{kN} \] --- ## Step 5: Axial Strain and Stress - **Axial Deformation**: \[ \delta = \frac{F \times L}{A \times E} \] - **Stress Calculation**: \[ \sigma = \frac{F}{A} \approx \frac{6.00 \times 10^3}{500 \times 10^{-6}} = 12.0\, \text{MPa} \] --- ## Step 6: Deflection Calculations - **Deflection**: - Using equilibrium equations and trigonometric relations based on forces. --- ## Final Note - The internal forces in each member are determined using the diagram and equilibrium conditions. - Stress and deformation are calculated based on the forces and material properties. If further breakdown or specifics are needed, please specify!