Let's break down the problem **step by step**: --- ## **Given Data** - **Rod AB:** Brass, Length = 3 ft = 36 in - **Rod BC:** Aluminum, Length = 2 ft = 24 in - **Cross-sectional area:** \(A = 1.6 \text{ in}^\) -Modulus ofity:** \(E_{} = 10,000 \text{ ksi} = 10,000,000 \text{ psi}\) \(E_{br} = 16,000 \text{ ksi} = 16,000,000 \text{ psi}\) - **Thermal Expansion Coefficient:** \(\alpha_{al} = 12.2 \times 10^{-6}/^\circ F\) \(\alpha_{br} = 10.4 \times 10^{-6}/^\circ F\) - **Temperature change:** \(T_ = 18^\circ F\) \(T_1 = 145^\circ F\) \(\Delta T = T_1 - T_ = 127^\circ F\) - **Fixed at A and C; joined by collar at B. No stress at \(T_\).** --- ## **Step 1: Free Thermal Expansion** If the rods were **free to expand**, their length would increase by: \[ \Delta L = \alpha \cdot L \cdot \Delta T \] ### **Brass Rod (AB):** \[ \Delta L_{br,free} = \alpha_{br} \cdot L_{br} \cdot \Delta T \\ = (10.4 \times 10^{-6}) \times 36 \times 127 \\ = .0475 \text{ in} \] ### **Aluminum Rod (BC):** \[ \Delta L_{al,free} = \alpha_{al} \cdot L_{al} \cdot \Delta T \\ = (12.2 \times 10^{-6}) \times 24 \times 127 \\ = .0372 \text{ in} \] --- ## **Step 2: Compatibility (Displacement)** Since **A and C are fixed**, the total expansion must be **zero** (the system can't move as a whole). But the collar at B can move, so let's **let the collar displacement be \(\delta\)** (positive to the right). - Brass rod AB will be compressed (since it wants to expand more) - Aluminum rod BC will stretched (since it wants to expand less) **Compatibility equation:** \[ (\text{Free expansion of AB}) - \delta = (\text{Actual expansion of AB}) \] \[ (\text{Free expansion of BC}) + \delta = (\text{Actual expansion of BC}) \] But the sum of actual expansions must be zero (because ends are fixed): \[ (\text{Actual expansion of AB}) + (\text{Actual expansion of BC}) = \] So, \[ (\Delta L_{br,free} - \delta) + (\Delta L_{al,free} + \delta) = \] \[ \Delta L_{br,free} + \Delta L_{al,free} = \] Wait, this would be true only if the collar could not move, but since the collar is free to move, we have to account for the **internal forces**. --- ## **Step 3: Force Equilibrium** Let the **force in AB = F_{br}** (compression), **force in BC = F_{al}** (tension), and by Newton's third law, they are equal in magnitude: \[ F_{br} = F_{al} = F \] The **actual deformation** due to force (ignoring temperature): \[ \delta_{br,force} = -\frac{F L_{br}}{A E_{br}} \] \[ \delta_{al,force} = +\frac{F L_{al}}{A E_{al}} \] **Net collar displacement:** \[ \delta = (\Delta L_{br,free} + \delta_{br,force}) = -(\Delta L_{al,free} + \delta_{al,force}) \] But since all the expansion must be taken up by the internal forces (since ends are fixed): \[ (\Delta L_{br,free} + \delta_{br,force}) + (\Delta L_{al,free} + \delta_{al,force}) = \] Plug in the values: \[ (\Delta L_{br,free} + \delta_{br,force}) + (\Delta L_{al,free} + \delta_{al,force}) = \] \[ (\alpha_{br} L_{br} \Delta T - \frac{F L_{br}}{A E_{br}}) + (\alpha_{al} L_{al} \Delta T + \frac{F L_{al}}{A E_{al}}) = \] --- ## **Step 4: Solve for Internal Force \(F\)** \[ \alpha_{br} L_{br} \Delta T + \alpha_{al} L_{al} \Delta T - \frac{F L_{br}}{A E_{br}} + \frac{F L_{al}}{A E_{al}} = \] \[ (\alpha_{br} L_{br} \Delta T + \alpha_{al} L_{al} \Delta T) = \frac{F L_{br}}{A E_{br}} - \frac{F L_{al}}{A E_{al}} \] \[ (\alpha_{br} L_{br} + \alpha_{al} L_{al}) \Delta T = F \left( \frac{L_{br}}{A E_{br}} + \frac{L_{al}}{A E_{al}} \right) \] \[ F = \frac{(\alpha_{br} L_{br} + \alpha_{al} L_{al}) \Delta T} {\frac{L_{br}}{A E_{br}} + \frac{L_{al}}{A E_{al}}} \] --- ## **Step 5: Plug in Values** - \(\alpha_{br} = 10.4 \times 10^{-6}/^\circ F\) - \(L_{br} = 36 \text{ in}\) - \(\alpha_{al} = 12.2 \times 10^{-6}/^\circ F\) - \(L_{al} = 24 \text{ in}\) - \(\Delta T = 127^\circ F\) - \(A = 1.6 \text{ in}^2\) - \(E_{br} = 16,000,000 \text{ psi}\) - \(E_{al} = 10,000,000 \text{ psi}\) ### **Numerator:** \[ (\alpha_{br} L_{br} + \alpha_{al} L_{al}) \Delta T = [(10.4 \times 10^{-6}) \times 36 + (12.2 \times 10^{-6}) \times 24] \times 127 \] Calculate each term: - \(10.4 \times 10^{-6} \times 36 = .0003744\) - \(12.2 \times 10^{-6} \times 24 = .0002928\) - Sum: \(.0003744 + .0002928 = .0006672\) Now multiply by \(\Delta T = 127\): - \(.0006672 \times 127 = .0847344\) ### **Denominator:** \[ \frac{L_{br}}{A E_{br}} + \frac{L_{al}}{A E_{al}} \] - \(\frac{36}{1.6 \times 16,000,000} = \frac{36}{25,600,000} = 1.40625 \times 10^{-6}\) - \(\frac{24}{1.6 \times 10,000,000} = \frac{24}{16,000,000} = 1.5 \times 10^{-6}\) - Sum: \(1.40625 \times 10^{-6} + 1.5 \times 10^{-6} = 2.90625 \times 10^{-6}\) ### **Now, Calculate Force \(F\):** \[ F = \frac{.0847344}{2.90625 \times 10^{-6}} = 29,156 \text{ lbs} \] --- ## **Step 6: Find Average Normal Stress** \[ \sigma = \frac{F}{A} \] - For both rods, since force is same by Newton's 3rd law: \[ \sigma = \frac{29,156}{1.6} = 18,222 \text{ psi} \] - **AB (Brass) is in compression.** - **BC (Aluminum) is in tension.** --- ## **Step 7: Collar Displacement** \[ \delta = \text{Expansion of AB (with stress)} = \Delta L_{br,free} - \frac{F L_{br}}{A E_{br}} \] \[ \Delta L_{br,free} = .0475 \text{ in} \] \[ \frac{F L_{br}}{A E_{br}} = \frac{29,156 \times 36}{1.6 \times 16,000,000} \\ = \frac{1,049,616}{25,600,000} = .041 \text{ in} \] \[ \delta_{AB} = .0475 - .041 = .0065 \text{ in (to the right)} \] --- ## **Final Answers** ### **1. Average Normal Stress in Each Member:** - **AB (Brass):** \( \boxed{18,200 \text{ psi (compression)}} \) - **BC (Aluminum):** \( \boxed{18,200 \text{ psi (tension)}} \) ### **2. Collar Displacement and Direction:** - **Displacement:** \( \boxed{.0065 \text{ in}} \) - **Direction:** **To the right** (since brass wants to expand more) --- ### **Summary Table** | Rod | Stress (psi) | Nature | |-----------|--------------|-------------| | AB (Brass)| 18,200 | Compression | | BC (Alum) | 18,200 | Tension | - **Collar moves right by .0065 in.** --- **Let me know if you need any clarifications or further breakdown!**