Let's break down the problem step by step and solve it: --- ## **Given Data** - **Steel rod:** - Length = 400.05 mm - \( E_s = 200 \) GPa \( = 200 \times 10^3 \) MPa - \( A_s = 100 \) mm² - **Copper tube:** - Length = 400.00 mm - \( E_c = 120 \) GPa \( = 120 \times 10^3 \) MPa - \( A_c = 150 \) mm² - **Gap between steel and cap:** .05 mm - **Applied axial load:** \( P = 145 \) kN = \( 145,000 \) N --- ## **Step 1: Check if the gap closes with the applied load** First, **only the steel rod is compressed** until the gap closes. After the gap is closed, both steel and copper resist the load. **Let the compressive deformation needed to close the gap be \( \delta = .05 \) mm.** ### **Deformation of Steel Rod under Load \( P \):** \[ \delta_s = \frac{P L_s}{A_s E_s} \] Plug in the values: - \( L_s = 400.05 \) mm - \( A_s = 100 \) mm² - \( E_s = 200 \times 10^3 \) MPa \( = 200,000 \) N/mm² \[ \delta_s = \frac{P \times 400.05}{100 \times 200,000} \] \[ \delta_s = \frac{P \times 400.05}{200,000,000} \] Set \(\delta_s = .05\) mm and solve for \(P\): \[ .05 = \frac{P \times 400.05}{200,000,000} \] \[ P = \frac{.05 \times 200,000,000}{400.05} \] \[ P = \frac{10,000,000}{400.05} \approx 24,997~\text{N} \approx 25~\text{kN} \] **So, the minimum load to close the gap is approximately \(\boxed{25~\text{kN}}\).** --- ## **Step 2: Since \( P = 145~\text{kN} > 25~\text{kN} \), the gap **will close** and both steel and copper will carry the load.** --- ## **Step 3: Find the normal stresses in steel and copper** Let: - Steel force: \( P_s \) - Copper force: \( P_c \) \[ P_s + P_c = P = 145,000~\text{N} \] After closing the gap, **the steel and copper deform by the same amount (\(\delta\))**, since they are in parallel. ### **Deformation Relationship:** \[ \delta_s = \delta_c \] \[ \frac{P_s L_s}{A_s E_s} = \frac{P_c L_c}{A_c E_c} \] But \( L_s \approx L_c \), so we can use \( L = 400 \) mm for both. \[ \frac{P_s}{A_s E_s} = \frac{P_c}{A_c E_c} \] \[ P_s = P_c \left( \frac{A_s E_s}{A_c E_c} \right) \] Plug in values: - \( A_s E_s = 100 \times 200,000 = 200,000,000 \) - \( A_c E_c = 150 \times 120,000 = 180,000,000 \) \[ \frac{A_s E_s}{A_c E_c} = \frac{200,000,000}{180,000,000} = \frac{10}{9} \] So: \[ P_s = \frac{10}{9} P_c \] And: \[ P_s + P_c = 145,000 \] \[ \frac{10}{9}P_c + P_c = 145,000 \] \[ \frac{19}{9}P_c = 145,000 \] \[ P_c = \frac{145,000 \times 9}{19} \approx 68,684~\text{N} \] \[ P_s = 145,000 - 68,684 = 76,316~\text{N} \] --- ### **Normal Stresses:** \[ \sigma_s = \frac{P_s}{A_s} = \frac{76,316}{100} = 76.3~\text{MPa} \] \[ \sigma_c = \frac{P_c}{A_c} = \frac{68,684}{150} = 45.8~\text{MPa} \] --- ## **Final Answers** ### 1. **Minimum load to close the gap:** \[ \boxed{25~\text{kN}} \] ### 2. **Normal stresses:** - **Steel:** \(\boxed{76.3~\text{MPa}}\) - **Copper:** \(\boxed{45.8~\text{MPa}}\) --- Let me know if you want more details or clarification on any step!