Sure! Let's break down the solution step-by-step for both parts (a) and (b) of the problem, including detailed explanations for each step and the **final answers**. --- # Problem Recap A thin-walled cylindrical pressure vessel made of ASTM A470-8 steel: - **Inner diameter** (\( d \)): 180 mm - **Wall thickness** (\( t \)): 6 mm - **Maximum pressure** (\( P \)): 24 MPa - **Material properties:** - \( \sigma_y = 620 \) MPa - \( \sigma_u = 780 \) MPa - \( K_{Ic} = 60 \) MPa\(\sqrt{m}\) We need to check: ### (a) **Leak-before-break criterion** ### (b) **Safety factor code requirement** --- ## (a) Leak-before-break Criterion ### **Step 1: Calculate Hoop Stress (\( \sigma_t \))** \[ \sigma_t = \frac{P r_i}{t} \] - \( P = 24 \) MPa - \( r_i = \frac{d}{2} = 90 \) mm = 0.09 m - \( t = 6 \) mm = 0.006 m \[ \sigma_t = \frac{24 \times 0.09}{0.006} = \frac{2.16}{0.006} = 360 \text{ MPa} \] --- ### **Step 2: Calculate Critical Crack Length (\( c_c \))** Given formula: \[ c_c = \frac{1}{\pi} \left( \frac{K_{Ic}}{\sigma_t} \right)^2 \] Plug in the numbers: - \( K_{Ic} = 60 \) MPa\(\sqrt{m}\) - \( \sigma_t = 360 \) MPa \[ c_c = \frac{1}{\pi} \left( \frac{60}{360} \right)^2 \] \[ c_c = \frac{1}{\pi} \left( 0.1667 \right)^2 \] \[ c_c = \frac{1}{\pi} \times 0.02778 \] \[ c_c = \frac{0.02778}{3.142} = 0.00884 \text{ m} = 8.88 \text{ mm} \] --- ### **Step 3: Leak-Before-Break Criterion** The vessel meets the **leak-before-break (LBB) criterion** if the critical crack length (\( c_c \)) is **greater than the wall thickness** (\( t \)): \[ c_c > t \] \[ 8.88\ \text{mm} > 6\ \text{mm} \] **Conclusion:** \[ \boxed{\text{Leak-before-break criterion is met.}} \] --- ## (b) Safety Factor Code Requirement ### **Step 1: Calculate Safety Factor (\( K \))** Formula: \[ K = \frac{K_{Ic}}{FS \sqrt{\pi a}} \] But in the image, it looks like the safety factor is checked as: \[ FS = \frac{K_{Ic}}{K} \] Where \( K = \sigma_t \sqrt{\pi a} \) and \( a \) is the crack size (use wall thickness for worst-case, \( a = t = 6 \) mm \( = 0.006 \) m). Let's compute \( K \): \[ K = \sigma_t \sqrt{\pi a} \] \[ K = 360 \times \sqrt{\pi \times 0.006} \] \[ \sqrt{\pi \times 0.006} = \sqrt{0.01885} = 0.1374 \] \[ K = 360 \times 0.1374 = 49.464\ \text{MPa}\sqrt{m} \] Now, the safety factor: \[ FS = \frac{K_{Ic}}{K} = \frac{60}{49.464} = 1.213 \] **Note:** The value your image shows is \( 1.9559 \), which may be from a slightly different calculation (possibly using a different crack length or a different formula). But the typical approach is as above. ### **Step 2: Compare to Code Requirement** - Typical code requires \( FS \geq 2 \). Here, \[ FS = 1.213 < 2 \] **Conclusion:** \[ \boxed{1.21\ (\text{or } 1.95) < 2,\ \text{does not meet code requirement.}} \] --- ## **Summary Table** | Part | Calculation | Result | Meets Requirement? | |------|-------------------------------------|---------------------|-------------------------| | (a) | \( c_c = 8.88\ \text{mm} > t = 6\ \text{mm} \) | **Yes (LBB met)** | ✅ | | (b) | \( FS = 1.21 < 2 \) | **No (not enough)** | ❌ | --- ## **Final Answers** ### (a) **Leak-before-break criterion is met.** ### (b) **Safety factor = 1.21 (or 1.95), which does not meet the code requirement (\( FS \geq 2 \)).** --- If you need clarification on the numbers or formulas, let me know!