1. **Problem Statement:** Determine the internal forces and average normal stresses in the aluminium and steel rods subjected to a 50 kN load at point B. 2. **Given Data:** - Aluminium rod length $L_{al} = 300$ mm - Steel rod length $L_{st} = 200$ mm - Load at B, $P = 50,000$ N - Aluminium cross-sectional area $A_{al} = 645$ mm$^2$ - Steel cross-sectional area $A_{st} = 1290$ mm$^2$ - Modulus of elasticity for aluminium $E_{al} = 69$ GPa = $69 \times 10^3$ MPa - Modulus of elasticity for steel $E_{st} = 200$ GPa = $200 \times 10^3$ MPa 3. **Assumptions and Formulae:** - Both rods are in axial tension/compression. - The rods are fixed at walls A and C, and the load is applied at B. - The rods deform compatibly, so the total elongation of aluminium plus steel rods equals zero (since walls are rigid). - Axial deformation formula: $$\delta = \frac{PL}{AE}$$ - Compatibility condition: $$\delta_{al} + \delta_{st} = 0$$ - Internal forces: $F_{al}$ in aluminium, $F_{st}$ in steel. - Equilibrium at point B: $$F_{al} + F_{st} = P$$ 4. **Step 1: Write compatibility equation:** $$\frac{F_{al} L_{al}}{A_{al} E_{al}} + \frac{F_{st} L_{st}}{A_{st} E_{st}} = 0$$ 5. **Step 2: Write equilibrium equation:** $$F_{al} + F_{st} = 50,000$$ 6. **Step 3: Express $F_{st}$ from equilibrium:** $$F_{st} = 50,000 - F_{al}$$ 7. **Step 4: Substitute $F_{st}$ into compatibility:** $$\frac{F_{al} \times 300}{645 \times 69,000} + \frac{(50,000 - F_{al}) \times 200}{1290 \times 200,000} = 0$$ Calculate denominators: - Aluminium denominator: $645 \times 69,000 = 44,505,000$ - Steel denominator: $1290 \times 200,000 = 258,000,000$ Rewrite: $$\frac{300}{44,505,000} F_{al} + \frac{200}{258,000,000} (50,000 - F_{al}) = 0$$ 8. **Step 5: Multiply out terms:** $$\frac{300}{44,505,000} F_{al} + \frac{200 \times 50,000}{258,000,000} - \frac{200}{258,000,000} F_{al} = 0$$ Calculate constants: - $\frac{300}{44,505,000} \approx 6.74 \times 10^{-6}$ - $\frac{200}{258,000,000} \approx 7.75 \times 10^{-7}$ - $\frac{200 \times 50,000}{258,000,000} = \frac{10,000,000}{258,000,000} \approx 0.03876$ Rewrite: $$6.74 \times 10^{-6} F_{al} + 0.03876 - 7.75 \times 10^{-7} F_{al} = 0$$ 9. **Step 6: Combine $F_{al}$ terms:** $$ (6.74 \times 10^{-6} - 7.75 \times 10^{-7}) F_{al} = -0.03876$$ Calculate coefficient: $$6.74 \times 10^{-6} - 7.75 \times 10^{-7} = 5.965 \times 10^{-6}$$ 10. **Step 7: Solve for $F_{al}$:** $$F_{al} = \frac{-0.03876}{5.965 \times 10^{-6}} = -6497.9 \text{ N}$$ 11. **Step 8: Find $F_{st}$:** $$F_{st} = 50,000 - (-6497.9) = 56,497.9 \text{ N}$$ 12. **Step 9: Calculate average normal stresses:** - Aluminium: $$\sigma_{al} = \frac{F_{al}}{A_{al}} = \frac{-6497.9}{645} = -10.08 \text{ MPa}$$ - Steel: $$\sigma_{st} = \frac{F_{st}}{A_{st}} = \frac{56,497.9}{1290} = 43.81 \text{ MPa}$$ 13. **Interpretation:** - Negative force and stress in aluminium indicate compression. - Positive force and stress in steel indicate tension. **Final answers:** - Internal force in aluminium rod: $F_{al} = -6498$ N (compression) - Internal force in steel rod: $F_{st} = 56,498$ N (tension) - Average normal stress in aluminium: $\sigma_{al} = -10.08$ MPa - Average normal stress in steel: $\sigma_{st} = 43.81$ MPa