1. **Problem 1(a): Calculate the temperature increase $\Delta T$ at which the two bars come into contact.** Given: - Brass length $L_b = 20$ mm, area $A_b = 65$ mm$^2$ - Magnesium length $L_m = 30$ mm, area $A_m = 130$ mm$^2$ - Initial gap $d = 0.1$ mm - Young's modulus: $E_b = 103$ GN/m$^2 = 103 \times 10^3$ N/mm$^2$, $E_m = 45$ GN/m$^2 = 45 \times 10^3$ N/mm$^2$ - Thermal expansion coefficients: $\alpha_b = 19.1 \times 10^{-6}$/°C, $\alpha_m = 26.1 \times 10^{-6}$/°C **Step 1:** Calculate free thermal expansions of each bar for temperature increase $\Delta T$: $$\Delta L_b = L_b \alpha_b \Delta T$$ $$\Delta L_m = L_m \alpha_m \Delta T$$ **Step 2:** The bars come into contact when the sum of their expansions equals the initial gap plus any elastic deformation. At contact, the gap closes, so: $$\Delta L_b + \Delta L_m = d + \delta_b + \delta_m$$ Since bars are fixed at ends, they develop internal forces causing elastic deformations $\delta_b$ and $\delta_m$: $$\delta_b = \frac{P L_b}{E_b A_b}, \quad \delta_m = \frac{P L_m}{E_m A_m}$$ **Step 3:** Force equilibrium: internal forces in bars are equal and opposite, so $P$ is the internal force. **Step 4:** Total elongation including elastic deformation: $$L_b \alpha_b \Delta T - \frac{P L_b}{E_b A_b} + L_m \alpha_m \Delta T + \frac{P L_m}{E_m A_m} = d$$ Rearranged: $$\Delta T (L_b \alpha_b + L_m \alpha_m) = d + P \left( \frac{L_b}{E_b A_b} - \frac{L_m}{E_m A_m} \right)$$ **Step 5:** At contact, bars just touch, so internal force $P=0$ (no stress yet), thus: $$\Delta T = \frac{d}{L_b \alpha_b + L_m \alpha_m}$$ Calculate denominator: $$L_b \alpha_b + L_m \alpha_m = 20 \times 19.1 \times 10^{-6} + 30 \times 26.1 \times 10^{-6} = 0.000382 + 0.000783 = 0.001165$$ Calculate $\Delta T$: $$\Delta T = \frac{0.1}{0.001165} \approx 85.84\, ^\circ C$$ --- 2. **Problem 1(b): Calculate the stress in the magnesium bar when $\Delta T = 150$ °C.** **Step 1:** Calculate free thermal expansions: $$\Delta L_b = 20 \times 19.1 \times 10^{-6} \times 150 = 0.0573\, \text{mm}$$ $$\Delta L_m = 30 \times 26.1 \times 10^{-6} \times 150 = 0.11745\, \text{mm}$$ **Step 2:** Total free expansion difference: $$\Delta L_m - \Delta L_b = 0.11745 - 0.0573 = 0.06015\, \text{mm}$$ **Step 3:** The bars are fixed, so they develop internal force $P$ causing elastic deformation: $$\delta_b = \frac{P L_b}{E_b A_b}, \quad \delta_m = \frac{P L_m}{E_m A_m}$$ **Step 4:** Compatibility condition: $$\Delta L_m - \delta_m = \Delta L_b + \delta_b + d$$ Rearranged: $$\Delta L_m - \Delta L_b - d = \delta_b + \delta_m = P \left( \frac{L_b}{E_b A_b} + \frac{L_m}{E_m A_m} \right)$$ Calculate left side: $$0.06015 - 0.1 = -0.03985\, \text{mm}$$ Calculate flexibility term: $$\frac{L_b}{E_b A_b} = \frac{20}{103 \times 10^3 \times 65} = \frac{20}{6,695,000} \approx 2.987 \times 10^{-6}$$ $$\frac{L_m}{E_m A_m} = \frac{30}{45 \times 10^3 \times 130} = \frac{30}{5,850,000} \approx 5.128 \times 10^{-6}$$ Sum: $$2.987 \times 10^{-6} + 5.128 \times 10^{-6} = 8.115 \times 10^{-6}$$ **Step 5:** Solve for $P$: $$P = \frac{-0.03985}{8.115 \times 10^{-6}} \approx -4913.5\, \text{N}$$ Negative sign indicates compression in magnesium. **Step 6:** Calculate stress in magnesium: $$\sigma_m = \frac{P}{A_m} = \frac{-4913.5}{130} \approx -37.8\, \text{N/mm}^2 = -37.8\, \text{MPa}$$ --- 3. **Problem 2(i): Find the force resisted by each bar using the flexibility method.** Given: - Force $F = 450$ N applied downward on rigid bar - Aluminium bars (2 bars) each with area $A_a = 0.0013$ m$^2 = 1300$ mm$^2$ - Steel bar area $A_s = 0.0026$ m$^2 = 2600$ mm$^2$ - Young's moduli: $E_a = 70$ GN/m$^2 = 70,000$ N/mm$^2$, $E_s = 200$ GN/m$^2 = 200,000$ N/mm$^2$ - Length aluminium bars $L_a = 30$ mm **Step 1:** Calculate flexibility (compliance) of each bar: $$f_a = \frac{L_a}{E_a A_a} = \frac{30}{70,000 \times 1300} = \frac{30}{91,000,000} \approx 3.2967 \times 10^{-7}$$ Two aluminium bars in parallel, total flexibility: $$f_{Al} = \frac{f_a}{2} = 1.6483 \times 10^{-7}$$ Steel bar length not given, assume same length $L_s = 30$ mm (typical assumption): $$f_s = \frac{L_s}{E_s A_s} = \frac{30}{200,000 \times 2600} = \frac{30}{520,000,000} \approx 5.769 \times 10^{-8}$$ **Step 2:** Total flexibility of assembly: $$f_{total} = f_{Al} + f_s = 1.6483 \times 10^{-7} + 5.769 \times 10^{-8} = 2.2252 \times 10^{-7}$$ **Step 3:** Calculate elongation $\delta$ of assembly under force $F$: $$\delta = F \times f_{total} = 450 \times 2.2252 \times 10^{-7} = 1.0013 \times 10^{-4} \text{ mm}$$ **Step 4:** Calculate force in each bar using compatibility: $$F_{Al} = \frac{\delta}{f_a/2} = \frac{1.0013 \times 10^{-4}}{1.6483 \times 10^{-7}} \approx 607.5 \text{ N}$$ $$F_s = \frac{\delta}{f_s} = \frac{1.0013 \times 10^{-4}}{5.769 \times 10^{-8}} \approx 1735.5 \text{ N}$$ Sum forces $F_{Al} + F_s = 607.5 + 1735.5 = 2343$ N, which is inconsistent with applied $F=450$ N, so re-examine step. **Correction:** Since bars are in parallel and rigid bar applies total force $F=450$ N, forces distribute inversely proportional to flexibility: Calculate stiffness $k = 1/f$: $$k_{Al} = \frac{1}{1.6483 \times 10^{-7}} = 6.067 \times 10^{6}$$ $$k_s = \frac{1}{5.769 \times 10^{-8}} = 1.734 \times 10^{7}$$ Total stiffness: $$k_{total} = k_{Al} + k_s = 2.340 \times 10^{7}$$ Force in each bar: $$F_{Al} = F \times \frac{k_{Al}}{k_{total}} = 450 \times \frac{6.067 \times 10^{6}}{2.340 \times 10^{7}} = 116.6 \text{ N}$$ $$F_s = 450 - 116.6 = 333.4 \text{ N}$$ --- 4. **Problem 2(ii): Calculate the resultant stress in each bar.** $$\sigma_{Al} = \frac{F_{Al}}{A_a} = \frac{116.6}{1300} = 0.0897 \text{ N/mm}^2 = 0.0897 \text{ MPa}$$ $$\sigma_s = \frac{F_s}{A_s} = \frac{333.4}{2600} = 0.1282 \text{ N/mm}^2 = 0.1282 \text{ MPa}$$ --- 5. **Problem 2(iii): Calculate elongation of the assembly due to force $F$.** Use total flexibility: $$\delta = F \times f_{total} = 450 \times 2.2252 \times 10^{-7} = 1.0013 \times 10^{-4} \text{ mm}$$ --- **Final answers:** 1(a) $\Delta T \approx 85.84$ °C 1(b) Stress in magnesium bar $\sigma_m \approx -37.8$ MPa (compression) 2(i) Force resisted: Aluminium bars $116.6$ N, Steel bar $333.4$ N 2(ii) Stress: Aluminium $0.0897$ MPa, Steel $0.1282$ MPa 2(iii) Elongation $\delta \approx 1.0013 \times 10^{-4}$ mm