Let's break this down step by step as requested: --- ## Problem Recap Given: - **Inner cylinder:** - Inside diameter = 100 mm - Common (interface) diameter = 180 mm - **Outer cylinder:** - Common (interface) diameter = 180 mm - Outside diameter = 280 mm - **After shrink fitting:** - Intermediate (interface) pressure at 180 mm = ? - **After internal pressure of 120 MPa is applied:** - Resultant radial stress at 180 mm = 44.908 MPa ### Required: 1. **Resultant hoop stress in the inner cylinder.** 2. **Intermediate pressure at common diameter due to shrinkage.** --- ## Step 1: Find the Intermediate Pressure due to Shrinkage (Part 4.2) Let: - \( r_1 = 50 \) mm (inner radius) - \( r_2 = 90 \) mm (common/interface radius) - \( r_3 = 140 \) mm (outer radius) The Lame’s equation for radial pressure at the interface due to shrinkage: \[ p = \frac{\sigma_r}{\ln(r_2/r_1) + \ln(r_3/r_2)} \] But, usually, we use superposition with press fit and internal pressure. Given the resultant radial stress at the interface after pressure applied, we have: \[ \text{Radial stress due to shrink fit} + \text{Radial stress due to internal pressure} = \text{Given resultant radial stress} \] Given: - Internal pressure = 120 MPa - Resultant radial stress at interface = 44.908 MPa ### Example Calculation Radial stress at interface due to internal pressure (using Lame’s formula for thick cylinders): \[ \sigma_{r, \text{int}} = \frac{p_i r_1^2 (r_2^2 - r^2)}{r_2^2 (r_2^2 - r_1^2)} \] At \( r = r_2 \), the radial stress due to internal pressure is **not zero** for compound cylinders: But typically, for compound cylinders, the pressure at the common surface due to shrink fit is called the "intermediate pressure" (\( p \)), and it is found by: \[ p = \text{(Resultant radial stress at interface)} - \text{(Radial stress at interface due to internal pressure alone)} \] Let's say the radial stress at interface due to 120 MPa is \( \sigma_{r, \text{internal}} \). **Plug in numbers:** \[ p_{internal} = \frac{120 \times 50^2 \times (90^2 - 90^2)}{90^2 (90^2 - 50^2)} = \] But this is zero, so the resultant radial stress at the interface is just the shrink fit pressure. So, \[ \boxed{p = 44.908\ \text{MPa}} \] --- ## Step 2: Find Resultant Hoop Stress in Inner Cylinder (Part 4.1) The hoop stress at inner radius due to internal pressure: \[ \sigma_{\theta, \text{max}} = \frac{p_i r_1^2 (r_2^2 + r_1^2)}{r_2^2 - r_1^2} \] Where: - \( p_i = 120\ \text{MPa} \) - \( r_1 = 50\ \text{mm} \) - \( r_2 = 90\ \text{mm} \) \[ \sigma_{\theta, \text{max}} = \frac{120 \times 50^2 \times (90^2 + 50^2)}{90^2 - 50^2} \] \[ = \frac{120 \times 250 \times (810 + 250)}{810 - 250} \] \[ = \frac{120 \times 250 \times 10600}{560} \] \[ = \frac{120 \times 250 \times 10600}{560} \] Calculate numerator: \( 120 \times 250 = 300,000 \) \( 300,000 \times 10,600 = 3,180,000,000 \) Divide by 560: \( \frac{3,180,000,000}{560} = 567,857 \) MPa But this is too high; let's check the standard formula: \[ \sigma_{\theta, \text{max}} = \frac{p_i r_1^2 (r_2^2 + r_1^2)}{(r_2^2 - r_1^2) r_1^2} \] \[ = \frac{p_i (r_2^2 + r_1^2)}{r_2^2 - r_1^2} \] Plug numbers: - \( r_1 = 50 \) - \( r_2 = 90 \) - \( p_i = 120 \) \[ r_2^2 + r_1^2 = 810 + 250 = 10600 \] \[ r_2^2 - r_1^2 = 810 - 250 = 560 \] So, \[ \sigma_{\theta, \text{max}} = \frac{120 \times 10600}{560} = \frac{1,272,000}{560} \approx 227\ \text{MPa} \] --- ## **Summary Table** | Quantity | Value | |--------------------------------------------------------|------------| | **4.1 Resultant hoop stress in inner cylinder** | **227 MPa**| | **4.2 Intermediate pressure at common diameter** | **44.908 MPa** | --- ### **Final Answers** 1. **Resultant hoop stress in the inner cylinder:** \(\boxed{227\ \text{MPa}}\) 2. **Intermediate pressure at the common diameter due to shrinkage:** \(\boxed{44.908\ \text{MPa}}\) --- **Let me know if you want the detailed equations for each step!**