Let's solve the problem step by step: --- ## **Given Data** - **Lengths:** \( L_{al} = 3 \) ft \( = 36 \) in (Aluminum, AB) \( L_{br} = 2 \) ft \( = 24 \) in (Brass, BC) - **Cross-sectional area:** \( A = 1.6 \) in² - **Young's Modulus:** \( E_{al} = 10,000 \) ksi \( E_{br} = 16,000 \) ksi - **Thermal Expansion Coefficients:** \( \alpha_{al} = 12.2 \times 10^{-6} \)/°F \( \alpha_{br} = 10.4 \times 10^{-6} \)/°F - **Temperature change:** \( \Delta T = T_1 - T_ = 145^\circ F - 18^\circ F = 127^\circ F \) - **Both ends fixed at A and C, joined at B** --- ## **Step 1: Free Thermal Expansion** If the rods were free to expand, their expansions would be: \[ \delta_{al, free} = \alpha_{al} \cdot L_{al} \cdot \Delta T \] \[ \delta_{br, free} = \alpha_{br} \cdot L_{br} \cdot \Delta T \] Plug in the values: \[ \delta_{al, free} = (12.2 \times 10^{-6})(36)(127) = .0558 \text{ in} \] \[ \delta_{br, free} = (10.4 \times 10^{-6})(24)(127) = .0317 \text{ in} \] --- ## **Step 2: Compatibility Condition** Since the ends are fixed at A and C, the net expansion from A to C is zero. The rods will push against each other at B, causing compressive and tensile stresses such that: \[ \delta_{al} + \delta_{br} = \] Where: \[ \delta_{al} = \delta_{al, free} + \delta_{al, mech} \] \[ \delta_{br} = \delta_{br, free} + \delta_{br, mech} \] But since the system is constrained: \[ (\delta_{al, free} + \delta_{al, mech}) + (\delta_{br, free} + \delta_{br, mech}) = \] \[ (\delta_{al, free} + \delta_{br, free}) + (\delta_{al, mech} + \delta_{br, mech}) = \] \[ \delta_{al, mech} + \delta_{br, mech} = -(\delta_{al, free} + \delta_{br, free}) = -.0875 \text{ in} \] --- ## **Step 3: Mechanical Deformations** Mechanical deformations due to internal force \( P \): \[ \delta_{al, mech} = \frac{P L_{al}}{A E_{al}} \] \[ \delta_{br, mech} = -\frac{P L_{br}}{A E_{br}} \] (\( +P \) for tension in aluminum, \( -P \) for compression in brass, as they act in opposite directions) So, \[ \frac{P L_{al}}{A E_{al}} - \frac{P L_{br}}{A E_{br}} = -.0875 \] \[ P \left( \frac{L_{al}}{A E_{al}} + \frac{L_{br}}{A E_{br}} \right) = -.0875 \] Plug in values (\( L_{al} = 36 \) in, \( L_{br} = 24 \) in, \( A = 1.6 \) in², \( E_{al} = 10,000 \times 100 = 10^7 \) psi, \( E_{br} = 16,000 \times 100 = 1.6 \times 10^7 \) psi): \[ \frac{L_{al}}{A E_{al}} = \frac{36}{1.6 \times 10,000 \times 100} = \frac{36}{16,000,000} = 2.25 \times 10^{-6} \] \[ \frac{L_{br}}{A E_{br}} = \frac{24}{1.6 \times 16,000 \times 100} = \frac{24}{25,600,000} = .9375 \times 10^{-6} \] Sum: \[ 2.25 \times 10^{-6} + .9375 \times 10^{-6} = 3.1875 \times 10^{-6} \] So, \[ P = \frac{-.0875}{3.1875 \times 10^{-6}} = -27,450 \text{ lbs} \] --- ## **Step 4: Average Normal Stresses** \[ \sigma_{al} = \frac{P}{A} = \frac{-27,450}{1.6} = -17,156 \text{ psi (compression)} \] \[ \sigma_{br} = \frac{-P}{A} = \frac{27,450}{1.6} = 17,156 \text{ psi (tension)} \] *Negative sign: Aluminum is in compression, brass in tension.* --- ## **Step 5: Collar Displacement** Collar B moves by the **mechanical deformation of the aluminum rod** (since both ends are fixed, but B can move relative to A): \[ \text{Displacement of collar B} = \delta_{al, free} + \delta_{al, mech} \] But since the whole system is fixed, the net movement is: \[ \delta_B = \delta_{al, free} + \frac{P L_{al}}{A E_{al}} \] \[ = .0558 + \frac{(-27,450) \times 36}{1.6 \times 10,000 \times 100} \] \[ = .0558 + \frac{-988,200}{16,000,000} \] \[ = .0558 - .0618 = -.006 \text{ in} \] **Negative sign means B moves to the left (toward A).** --- ## **Final Answers** ### **1. Average Normal Stress** - **Aluminum (AB):** \( \boxed{-17,200 \text{ psi (compression)}} \) - **Brass (BC):** \( \boxed{17,200 \text{ psi (tension)}} \) ### **2. Collar Displacement** - **Magnitude:** \( \boxed{.006 \text{ in}} \) - **Direction:** **To the left (toward A)** --- Let me know if you would like a more detailed explanation for any step!