✨ VIPSolutions

Certainly! Let’s solve the problem step by step according to your guidelines. --- ### 1. **Introduction** This problem revolves around the design of a pressure cylinder subjected to internal pressure. We are required to determine the minimum wall thickness necessary to ensure that the cylinder can withstand this pressure without exceeding its material’s shear strength. --- ### 2. **Given Data** (arranged vertically) - Outer diameter of the cylinder, \( d_o = 200 \) mm - Shear strength of material, \( \tau_{max} = 25 \) MPa - Internal pressure, \( p = 4 \) MPa --- ### 3. **Explanation of Solution Approach** To ensure safety, the maximum induced hoop (circumferential) stress in the cylinder wall due to the internal pressure should not exceed the material's shear strength. For thin-walled cylinders, the hoop stress (\( \sigma_h \)) can be calculated using the formula: \[ \sigma_h = \frac{p \cdot r_i}{t} \] where: - \( p \) = internal pressure, - \( r_i \) = internal radius, - \( t \) = wall thickness. But since the outer diameter is given, we express the internal radius as \( r_i = \frac{d_o}{2} - t \). We set the maximum hoop stress equal to the shear strength to solve for the minimum required thickness. --- ### 4. **Calculation with Explanation** #### Step 1: Express variables in consistent units - \( d_o = 200 \) mm \( = 0.2 \) m - \( \tau_{max} = 25 \) MPa - \( p = 4 \) MPa #### Step 2: Relation between internal and external diameter Let the wall thickness be \( t \). - Outer radius, \( r_o = \frac{d_o}{2} = 100 \) mm - Inner radius, \( r_i = r_o - t = 100 - t \) mm #### Step 3: For thin-walled assumption (since not specified, but given dimensions suggest it) - The hoop stress formula: \[ \sigma_h = \frac{p \cdot r_i}{t} \] Set \( \sigma_h = \tau_{max} \): \[ \frac{p \cdot r_i}{t} = \tau_{max} \] \[ t = \frac{p \cdot r_i}{\tau_{max}} \] #### Step 4: Substitute values - \( p = 4 \) MPa - \( r_i = 100 - t \) mm So, \[ t = \frac{4 \times (100 - t)}{25} \] (Since all values are in mm and MPa, units are consistent.) #### Step 5: Solve for \( t \) Multiply both sides by 25: \[ 25t = 4 \times (100 - t) \] \[ 25t = 400 - 4t \] \[ 25t + 4t = 400 \] \[ 29t = 400 \] \[ t = \frac{400}{29} \] \[ t \approx 13.79 \text{ mm} \] --- ### 5. **Conclusion** The required minimum wall thickness of the cylinder to safely withstand an internal pressure of 4 MPa, given the material's shear strength of 25 MPa and an outer diameter of 200 mm, is **approximately 13.8 mm** (rounded to one decimal place). --- **If you need further explanation about thick-walled assumptions or want to consider safety factors, let me know!**