```markdown # Problem 4-82 The three bars are made of A-36 steel and form a pin-connected truss. If the truss is constructed when \( T_1 = 50^\circ F \), determine the vertical displacement of joint A when \( T_2 = 150^\circ F \). Each bar has a cross-sectional area of 2 in\(^2\). ### Given Data: - Material: A-36 steel - Cross-sectional area of each bar: 2 in\(^2\) - Initial temperature: \( T_1 = 50^\circ F \) - Final temperature: \( T_2 = 150^\circ F \) - Geometry: - Vertical height: 4 ft - Horizontal spans: 3 ft each ### Diagram: A pin-connected truss with joints labeled A, B, C, D, and the bars AB, BC, and AC. --- ## Step 1: Determine the change in length due to temperature \[ (\delta T)_{AB} - (\delta T)_{AB} = (\delta T)_{AD} + (\delta T)_{AD} \] However, \[ \delta_{AB} = \delta'_{AB} \cos \theta \] and \[ \delta'_{AB} = \frac{\delta_{AB}}{\cos \theta} = \frac{5}{4} \delta_{AB} \] Substituting into Eq. (1): \[ \frac{5}{4} (\delta T)_{AB} - \frac{5}{4} (\delta T)_{AB} = (\delta T)_{AD} + (\delta T)_{AD} \] which simplifies to: \[ \frac{5}{4} \left[ 6.60 \times 10^{-6} (150^\circ - 50^\circ) (5)(12) - \frac{F_{AB} (5)(12)}{2(29)(10^3)} \right] \] plus \[ \frac{F_{AD} (4)(12)}{2(29)(10^3)} \] leading to: \[ 239.25 - 6.25 F_{AB} = 153.12 + 4 F_{AD} \] or, \[ 4 F_{AD} + 6.25 F_{AB} = 86.13 \] --- ## Step 2: Equilibrium equations \[ \pm \sum F_x = : \quad \frac{3}{5} F_{AC} - \frac{3}{5} F_{AB} = \Rightarrow F_{AC} = F_{AB} \] \[ \pm \sum F_y = : \quad F_{AD} - 2 \left( \frac{4}{5} F_{AB} \right) = \] which yields: \[ F_{AD} = 1.6 F_{AB} \] --- ## Step 3: Solving equations From equations (2) and (3): \[ F_{AB} = 6.8086 \text{ kip} \] \[ F_{AD} = 10.8939 \text{ kip} \] Calculating the vertical displacement: \[ (\delta \omega)_r = (\delta T)_{AD} + (\delta T)_{AB} \] \[ = 6.60 \times 10^{-6} (150^\circ - 50^\circ) (4)(12) + \frac{10.8939 (4)(12)}{2(29)(10^3)} \] \[ = .0407 \text{ in} \] ## **Final Answer:** \[ \boxed{ \text{Vertical displacement of joint A} \approx \mathbf{.0407 \text{ inches}} } \] ```

```markdown # Problem 4-81 The three bars are made of A-36 steel and form a pin-connected truss. If the truss is constructed when \( T_1 = 50^\circ F \), determine the force in each bar when \( T_2 = 10^\circ F \). Each bar has a cross-sectional area of 2 in\(^2\). ### Given Data: - Material: A-36 steel - Cross-sectional area of each bar: 2 in\(^2\) - Initial temperature: \( T_1 = 50^\circ F \) - Final temperature: \( T_2 = 10^\circ F \) - Geometry: - Vertical height: 3 ft - Horizontal spans: 3 ft each ### Diagram: A pin-connected truss with joints labeled A, B, C, D, and the bars AB, BC, and AC. --- ## Step 1: Determine the change in length due to temperature The change in length can be represented as: \[ \Delta L = L \cdot \alpha \cdot \Delta T \] where: - \( \alpha \) = coefficient of thermal expansion for A-36 steel - \( \Delta T = T_2 - T_1 \) Assuming \( \alpha \approx 6.6 \times 10^{-6} \, \text{in/in}^\circ F \): \[ \Delta T = 10^\circ F - 50^\circ F = -40^\circ F \] --- ## Step 2: Calculating forces in the bars Using the equilibrium equations, the forces in the bars can be expressed as: 1. For horizontal equilibrium: \[ \sum F_x = 0 \quad \Rightarrow \quad F_{AB} \cos \theta + F_{BC} \cos \theta = 0 \] 2. For vertical equilibrium: \[ \sum F_y = 0 \quad \Rightarrow \quad F_{AB} \sin \theta + F_{BC} \sin \theta - F_{AD} = 0 \] Given that: - The length of each bar is 5 ft. - The angle \( \theta \) can be determined from the geometry of the truss. --- ## Step 3: Solve for forces Using the derived equations and substituting values: From \( F_{AB} \) and \( F_{AD} \): After rearranging and solving the system of equations: \[ F_{AB} = 654 \text{ kip} \] \[ F_{AD} = 409 \text{ kip} \] ### **Final Answer:** \[ \boxed{ \text{Force in bar AB} = 654 \text{ kip}, \quad \text{Force in bar AD} = 409 \text{ kip} } ```

```markdown # Problem 4-82 The three bars are made of A-36 steel and form a pin-connected truss. The truss is constructed when \( T_1 = 50^\circ F \). The task is to determine the vertical displacement of joint A when \( T_2 = 150^\circ F \). Each bar has a cross-sectional area of 2 in\(^2\). ### Given Data: - **Material**: A-36 steel - **Cross-sectional area of each bar**: 2 in\(^2\) - **Initial temperature**: \( T_1 = 50^\circ F \) - **Final temperature**: \( T_2 = 150^\circ F \) - **Geometry**: - Vertical height: 4 ft - Horizontal spans: 3 ft each ### Step 1: Change in Length Due to Temperature The change in length \( \Delta L \) due to temperature can be calculated as follows: \[ \Delta L = L \cdot \alpha \cdot \Delta T \] Where: - \( \alpha \approx 6.6 \times 10^{-6} \, \text{in/in}^\circ F \) - \( \Delta T = T_2 - T_1 = 150^\circ F - 50^\circ F = 100^\circ F \) ### Step 2: Force Relationships The forces in the bars can be expressed in terms of equilibrium equations. The following relationships apply: 1. Horizontal equilibrium: \[ \sum F_x = 0 \quad \Rightarrow \quad F_{AB} \cos \theta + F_{BC} \cos \theta = 0 \] 2. Vertical equilibrium: \[ \sum F_y = 0 \quad \Rightarrow \quad F_{AD} - 2 \left( \frac{4}{5} F_{AB} \right) = 0 \] ### Step 3: Solving for Forces Solving the equations leads to: \[ F_{AB} = 6.8086 \text{ kip} \] \[ F_{AD} = 10.8939 \text{ kip} \] ### Step 4: Calculating Vertical Displacement The vertical displacement at joint A can be calculated using: \[ \delta = \frac{F_{AB} (L_{AB})}{A \cdot E} \] Where: - \( E = 29 \times 10^3 \, \text{ksi} \) - \( L_{AB} = 12 \, \text{in} \) Substituting the values into the displacement equation provides: \[ \delta = 6.60 \times 10^{-6} \times (150^\circ - 50^\circ) (4)(12) + \frac{10.8939 (4)(12)}{2(29)(10^3)} \] Calculating gives: \[ \delta \approx 0.0407 \text{ in} \] ### Final Answer: The vertical displacement of joint A is: \[ \boxed{0.0407 \text{ inches}} \] ``` ```markdown # Problem 4-81 The three bars are made of A-36 steel and form a pin-connected truss. The truss is constructed when \( T_1 = 50^\circ F \). The goal is to determine the force in each bar when \( T_2 = 10^\circ F \). Each bar has a cross-sectional area of 2 in\(^2\). ### Given Data: - **Material**: A-36 steel - **Cross-sectional area of each bar**: 2 in\(^2\) - **Initial temperature**: \( T_1 = 50^\circ F \) - **Final temperature**: \( T_2 = 10^\circ F \) - **Geometry**: - Vertical height: 3 ft - Horizontal spans: 3 ft each ### Step 1: Change in Length Due to Temperature The change in length can be expressed as: \[ \Delta L = L \cdot \alpha \cdot \Delta T \] Where: - \( \alpha \approx 6.6 \times 10^{-6} \, \text{in/in}^\circ F \) - \( \Delta T = 10^\circ F - 50^\circ F = -40^\circ F \) ### Step 2: Calculating Forces in the Bars Using equilibrium equations, the forces in the bars are defined as: 1. For horizontal equilibrium: \[ \sum F_x = 0 \quad \Rightarrow \quad F_{AB} \cos \theta + F_{BC} \cos \theta = 0 \] 2. For vertical equilibrium: \[ \sum F_y = 0 \quad \Rightarrow \quad F_{AD} - 2 \left( \frac{4}{5} F_{AB} \right) = 0 \] ### Step 3: Solve for Forces By solving the equilibrium equations: \[ F_{AB} = 654 \text{ kip} \] \[ F_{AD} = 409 \text{ kip} \] ### Final Answer: The forces in the bars are: \[ \boxed{ \text{Force in bar AB} = 654 \text{ kip}, \quad \text{Force in bar AD} = 409 \text{ kip} } ```