Let's break down the problem step by step: --- **Given Data:** - Outer diameter \( d_o = 14 \) in - Wall thickness \( t = 1.75 \) in - \( K_{IC} = 72\, \text{ksi} \sqrt{\text{in}} \) - Yield strength \( S_y = 170 \) ksi (for reference, but not directly needed) - Smallest detectable crack \( a = 0.5 \) in - **Find:** The internal pressure \( p \) (in psi) that would cause failure --- ### **Step 1: Find outer radius (\( r_o \)) and inner radius (\( r_i \))** \[ r_o = \frac{d_o}{2} = \frac{14}{2} = 7 \text{ in} \] \[ r_i = r_o - t = 7 - 1.75 = 5.25 \text{ in} \] --- ### **Step 2: Stress Intensity Factor Equation** From fracture mechanics: \[ K_I = \beta \sigma_t \sqrt{\pi a} \] where: - \( \sigma_t \) = hoop (tangential) stress at the outer surface - \( \beta \) = geometric factor (from chart, depends on \( r_o/t \) and \( a/t \)) - \( a \) = crack length #### **Find \( \beta \):** \[ \frac{r_o}{t} = \frac{7}{1.75} = 4 \] \[ \frac{a}{t} = \frac{0.5}{1.75} \approx 0.286 \] From the provided chart, for \( r_o/t = 4 \) and \( a/t \approx 0.29 \), interpolate between the curves. The middle curve for \( r_o/t = 4 \) at \( a/t = 0.29 \) gives approximately \( \beta \approx 1.3 \). --- ### **Step 3: Hoop Stress at Outer Surface** For a thin-walled cylinder, hoop stress at the outer surface is: \[ \sigma_t = \frac{p r_i^2}{r_o^2 - r_i^2} \] But since the crack is at the outer surface, we use the stress at \( r_o \): For a thick-walled cylinder, the tangential stress at radius \( r \) due to internal pressure \( p \) is: \[ \sigma_t (r) = \frac{p r_i^2 (r_o^2 + r^2)}{r_o^2 - r_i^2} \cdot \frac{1}{r^2} \] At \( r = r_o \): \[ \sigma_t (r_o) = \frac{p r_i^2 (r_o^2 + r_o^2)}{r_o^2 - r_i^2} \cdot \frac{1}{r_o^2} \] \[ = \frac{p r_i^2 (2 r_o^2)}{(r_o^2 - r_i^2) r_o^2} \] \[ = \frac{2 p r_i^2}{r_o^2 - r_i^2} \] --- ### **Step 4: Set \( K_I = K_{IC} \) and solve for \( p \)** \[ K_{IC} = \beta \sigma_t \sqrt{\pi a} \] \[ 72 = 1.3 \cdot \sigma_t \cdot \sqrt{\pi \cdot 0.5} \] \[ \sigma_t = \frac{72}{1.3 \cdot \sqrt{\pi \cdot 0.5}} \] Calculate \( \sqrt{\pi \times 0.5} \): \[ \pi \times 0.5 = 1.5708 \] \[ \sqrt{1.5708} = 1.2533 \] \[ 1.3 \times 1.2533 = 1.6293 \] \[ \sigma_t = \frac{72}{1.6293} = 44.21 \text{ ksi} \] --- ### **Step 5: Relate \( \sigma_t \) to \( p \):** \[ \sigma_t = \frac{2 p r_i^2}{r_o^2 - r_i^2} \] \[ p = \frac{\sigma_t (r_o^2 - r_i^2)}{2 r_i^2} \] Plug in values: \[ r_o = 7,\ r_i = 5.25 \] \[ r_o^2 = 49,\ r_i^2 = 27.5625 \] \[ r_o^2 - r_i^2 = 49 - 27.5625 = 21.4375 \] Now, convert \( \sigma_t \) to psi: \( 44.21 \times 1000 = 44210 \) psi \[ p = \frac{44210 \times 21.4375}{2 \times 27.5625} \] \[ = \frac{948,262.375}{55.125} \] \[ = 17,195.6\ \text{psi} \] --- ## **Final Answer** \[ \boxed{ p = 17,200\ \text{psi} } \] (Rounded to three significant digits, within the 2% tolerance specified.)