Let's break down the problem **step by step**: --- ## **Given Data** - **Diameters:** - Inner tube: 50 mm (inner), 55 mm (outer) - Outer tube: 55 mm (inner), 60 mm (outer) - **Diametral Interference:** \( \delta = .062 \) mm - **Torque applied:** \( T = 900 \) N·m - **Bending moment applied:** \( M = 675 \) N·m (not directly needed for radial/tangential stress at interface) - **Yield strength:** \( S_y = 415 \) MPa We want **radial** and **tangential (hoop)** stress at the **outer radius of the inner member** (i.e., at radius = 27.5 mm). --- ## **Step 1: Stresses Due to Shrink Fit (Interference Fit)** ### **A. Find the Pressure at the Interface** Let: - \( r_1 = 25 \) mm (inner radius of inner tube) - \( r_2 = 27.5 \) mm (outer of inner, inner of outer) - \( r_3 = 30 \) mm (outer radius of outer tube) The **interference** causes a **contact pressure** \( p \) at the interface \( r_2 \). The **radial displacement** at the interface for both tubes is: - For **inner tube** (expands out): \( \delta_1 = \frac{p\, r_2}{E_{steel}} \left( \frac{r_2^2 + r_1^2}{r_2^2 - r_1^2} \right) \) - For **outer tube** (contracts in): \( \delta_2 = \frac{p\, r_2}{E_{steel}} \left( \frac{r_3^2 + r_2^2}{r_3^2 - r_2^2} \right) \) The **total interference** is: \[ \delta = \delta_1 + \delta_2 \] \[ \Rightarrow \delta = \frac{p\, r_2}{E} \left( \frac{r_2^2 + r_1^2}{r_2^2 - r_1^2} + \frac{r_3^2 + r_2^2}{r_3^2 - r_2^2} \right) \] - \( \delta = .062 \) mm = \( .000062 \) m - \( r_1 = .025 \) m, \( r_2 = .0275 \) m, \( r_3 = .03 \) m - \( E_{steel} \approx 210 \times 10^9 \) Pa Plug in the values: First, calculate the geometric terms: \[ \frac{r_2^2 + r_1^2}{r_2^2 - r_1^2} = \frac{(.0275^2 + .025^2)}{(.0275^2 - .025^2)} \] \[ = \frac{(.00075625 + .000625)}{(.00075625 - .000625)} = \frac{.00138125}{.00013125} \approx 10.525 \] \[ \frac{r_3^2 + r_2^2}{r_3^2 - r_2^2} = \frac{(.03^2 + .0275^2)}{(.03^2 - .0275^2)} \] \[ = \frac{(.0009 + .00075625)}{(.0009 - .00075625)} = \frac{.00165625}{.00014375} \approx 11.525 \] So the sum: \[ 10.525 + 11.525 = 22.05 \] Now, substitute into the interference equation: \[ .000062 = \frac{p \times .0275}{210 \times 10^9} \times 22.05 \] \[ .000062 = \frac{p \times .606375}{210 \times 10^9} \] \[ p = \frac{.000062 \times 210 \times 10^9}{.0275 \times 22.05} \] First, calculate denominator: \[ .0275 \times 22.05 = .606375 \] \[ p = \frac{.000062 \times 210 \times 10^9}{.606375} \] \[ .000062 \times 210 \times 10^9 = .000062 \times 210000000000 = 13,020,000 \] \[ p = \frac{13,020,000}{.606375} \approx 21,483,000 \text{ Pa } = 21.48 \text{ MPa} \] --- ## **Step 2: Radial and Tangential (Hoop) Stresses Due to Pressure** For a **thick-walled cylinder**: \[ \sigma_r(r) = A - \frac{B}{r^2} \] \[ \sigma_\theta(r) = A + \frac{B}{r^2} \] For the **inner cylinder** (\( r_1 < r < r_2 \)), boundary conditions: - At \( r_1 \): \( \sigma_r(r_1) = \) (free) - At \( r_2 \): \( \sigma_r(r_2) = -p \) (outward is positive, so negative for pressure inwards) Solve for \( A \) and \( B \): \[ \begin{cases} A - \frac{B}{r_1^2} = \\ A - \frac{B}{r_2^2} = -p \end{cases} \] Subtract equations: \[ \left[ A - \frac{B}{r_1^2} \right] - \left[ A - \frac{B}{r_2^2} \right] = p \] \[ \frac{B}{r_2^2} - \frac{B}{r_1^2} = p \] \[ B \left( \frac{1}{r_2^2} - \frac{1}{r_1^2} \right) = p \] \[ B = \frac{p}{\left(\frac{1}{r_2^2} - \frac{1}{r_1^2}\right)} \] Now, \( A = \frac{B}{r_1^2} \). Plug in the values: - \( r_1 = .025 \), \( r_2 = .0275 \) - \( p = 21.48 \) MPa = \( 21,480,000 \) Pa First, compute \( \frac{1}{r_2^2} - \frac{1}{r_1^2} \): \[ r_1^2 = .000625,\quad r_2^2 = .00075625 \] \[ \frac{1}{r_2^2} = \frac{1}{.00075625} \approx 1,322.31 \] \[ \frac{1}{r_1^2} = \frac{1}{.000625} = 1,600 \] \[ \frac{1}{r_2^2} - \frac{1}{r_1^2} = 1,322.31 - 1,600 = -277.69 \] So, \[ B = \frac{21,480,000}{-277.69} \approx -77,370 \text{ N/m} \] Now, \[ A = \frac{B}{r_1^2} = \frac{-77,370}{.000625} \approx -123,792,000 \text{ Pa} \] --- ### **At \( r = r_2 \) (outer radius of inner member):** #### **Radial Stress:** \[ \sigma_r(r_2) = A - \frac{B}{r_2^2} \] \[ = -123,792,000 - \frac{-77,370}{.00075625} = -123,792,000 + 102,376,000 = -21,416,000 \text{ Pa} = -21.4 \text{ MPa} \] This matches our interface pressure (as expected). #### **Tangential (Hoop) Stress:** \[ \sigma_\theta(r_2) = A + \frac{B}{r_2^2} \] \[ = -123,792,000 + \frac{-77,370}{.00075625} = -123,792,000 - 102,376,000 = -226,168,000 \text{ Pa} = -226.17 \text{ MPa} \] --- ## **Step 3: Stresses Due to Torque (Torsion)** Shear stress at outer radius due to torque: \[ \tau = \frac{T r}{J} \] Where \( J \) for a hollow cylinder: \[ J = \frac{\pi}{2} (r_2^4 - r_1^4) \] \[ r = r_2 = .0275 \text{ m} \] \[ T = 900 \text{ N}\cdot\text{m} \] Calculate \( J \): \[ r_2^4 = (.0275)^4 = 5.716 \times 10^{-7} \] \[ r_1^4 = (.025)^4 = 3.90625 \times 10^{-7} \] \[ J = \frac{\pi}{2} (5.716 - 3.906)\times 10^{-7} = \frac{\pi}{2} (1.81 \times 10^{-7}) \approx 2.846 \times 10^{-7} \] \[ \tau = \frac{900 \times .0275}{2.846 \times 10^{-7}} = \frac{24.75}{2.846 \times 10^{-7}} \approx 86,926,000 \text{ Pa} = 86.93 \text{ MPa} \] --- ## **Final Answers** ### **Radial Stress at the Inner Member's Outer Radius:** \[ \boxed{-21.4 \ \text{MPa}} \] ### **Tangential (Hoop) Stress at the Inner Member's Outer Radius:** \[ \boxed{-226.2 \ \text{MPa}} \] ### **Shear Stress (from Torque) at the Inner Member's Outer Radius:** \[ \boxed{86.9 \ \text{MPa}} \] --- ### **Summary Table** | Stress Type | Value (MPa) | |-------------------|------------| | Radial | -21.4 | | Tangential (Hoop) | -226.2 | | Shear (Torsion) | 86.9 | --- If you need **combined/maximum stress** or **principal stresses**, let me know!