1. **Stating the problem:** We need to determine the magnitude of the average normal stress developed in rods AB, CD, and EF. Rods AB and CD are made of C83400 copper, and rod EF is made of stainless steel 304 alloy. 2. **Given data:** - Force applied on each rod end: $F = 40,000$ N (40 kN) - Lengths: $L_{AB} = L_{CD} = 300$ mm, $L_{EF} = 450$ mm - Diameters: $d_{AB} = d_{CD} = 30$ mm, $d_{EF} = 40$ mm 3. **Formula for average normal stress:** The average normal stress $\sigma$ in a rod under axial load is given by: $$\sigma = \frac{F}{A}$$ where $A$ is the cross-sectional area of the rod. 4. **Calculate cross-sectional areas:** The rods are circular in cross-section, so area $A$ is: $$A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}$$ - For rods AB and CD: $$A_{AB} = A_{CD} = \frac{\pi (30)^2}{4} = \frac{\pi \times 900}{4} = 225\pi \approx 706.86\ \text{mm}^2$$ - For rod EF: $$A_{EF} = \frac{\pi (40)^2}{4} = \frac{\pi \times 1600}{4} = 400\pi \approx 1256.64\ \text{mm}^2$$ 5. **Calculate average normal stresses:** - For rods AB and CD: $$\sigma_{AB} = \sigma_{CD} = \frac{40,000}{706.86} \approx 56.58\ \text{MPa}$$ - For rod EF: $$\sigma_{EF} = \frac{40,000}{1256.64} \approx 31.83\ \text{MPa}$$ 6. **Interpretation:** The average normal stress in rods AB and CD is approximately 56.58 MPa, and in rod EF it is approximately 31.83 MPa. These values represent the tensile stress developed due to the applied 40 kN force. **Final answer:** - $\boxed{\sigma_{AB} = \sigma_{CD} \approx 56.6\ \text{MPa}}$ - $\boxed{\sigma_{EF} \approx 31.8\ \text{MPa}}$