Let's go through the problem step-by-step, explaining each part clearly. --- # Question Breakdown A cylindrical steel tank withpherical ends is with the following: - Diameter of the tank \( D = 1.25\, \text{m} \) - Wall thickness \( t_w = 22\, \text{mm} = .022\, \text{m} \) - Internal pressure \( p = 175\, \text{kPa} = 1.75 \times 10^6\, \text{Pa} \) ### Goal: Calculate the following: 1. Maximum hoop (circumferential) stress \( \sigma_{h} \) in the hemispherical heads. 2. Maximum hoop (circumferential) stress \( \sigma_{c} \) in the cylindrical section. 3. Tensile stress \( \sigma_{t} \) acting perpendicular to the weld joints. 4. Maximum shear stress \( \tau_{max} \) in the hemispherical heads. 5. Maximum shear stress \( \tau_{max} \) in the cylindrical portion. --- # Step 1: Basic Data and Assumptions - **Inner radius of the tank:** \[ r_{inner} = \frac{D}{2} = \frac{1.25}{2} = .625\, \text{m} \] - **Outer radius:** \[ r_{outer} = r_{inner} + t_w = .625 + .022 = .647\, \text{m} \] --- # Step 2: Find the **hoop stress** in the **cylinder** and **hemispherical head** ## (a) **Maximum hoop (circumferential) stress \( \sigma_{h} \) in the hemispherical head** ### Formula for hoop stress in spherical pressure vessel: \[ \sigma_{h} = \frac{p \times r_{inner}}{2 \times t_w} \] **Note:** For hemispherical heads, the stress is similar to that in a spherical vessel. ### Calculation: \[ \sigma_{h} = \frac{(1.75 \times 10^6) \times .625}{2 \times .022} \] \[ \sigma_{h} = \frac{1.09375 \times 10^6}{.044} \] \[ \sigma_{h} \approx 24,868,182\, \text{Pa} \approx 24.87\, \text{MPa} \] --- ## (b) **Maximum hoop stress \( \sigma_{c} \) in the cylindrical section** ### Formula for hoop stress in a thin-walled cylinder: \[ \sigma_{c} = \frac{p \times r_{inner}}{t_w} \] **Note:** The hoop stress in a cylindrical shell is twice that in a spherical shell under the same internal pressure. ### Calculation: \[ \sigma_{c} = \frac{(1.75 \times 10^6) \times .625}{.022} \] \[ \sigma_{c} = 2 \times 24.87\, \text{MPa} \approx 49.73\, \text{MPa} \] --- # Step 3: Find **tensile stress \( \sigma_t \)** acting perpendicular to weld joints - The tensile stress in the tank's wall is essentially the hoop stress in the cylindrical shell, which is: \[ \sigma_t = \sigma_c \approx 49.73\, \text{MPa} \] --- # Step 4: Find **Maximum shear stresses** ### Shear stress in a thin-walled pressure vessel: \[ \tau_{max} = \frac{\sigma_{c} - \sigma_{h}}{2} \] This is because the maximum shear occurs where the principal stresses are \( \sigma_{c} \) and \( \sigma_{h} \). ### Calculation: \[ \tau_{max} = \frac{49.73\, \text{MPa} - 24.87\, \text{MPa}}{2} = \frac{24.86\, \text{MPa}}{2} \approx 12.43\, \text{MPa} \] --- # **Summary of Results:** | Part | Stress / Shear | Value | |---------|----------------|---------| | (a) Hoop stress in hemispherical head \( \sigma_{h} \) | MPa | **24.87 MPa** | | (b) Hoop stress in cylindrical shell \( \sigma_{c} \) | MPa | **49.73 MPa** | | (c) Tensile stress \( \sigma_t \) | MPa | **49.73 MPa** | | (d) Maximum shear stress in hemispherical head \( \tau_{max} \) | MPa | **12.43 MPa** | | (e) Maximum shear stress in cylindrical portion \( \tau_{max} \) | MPa | **12.43 MPa** | --- # Final notes: - The stresses are based on thin-walled assumptions. - Actual maximum stresses may vary slightly due to factors like material properties and welding effects. Let me know if you'd like a more detailed explanation on any part!