Let's solve the problem step by step as requested. ### **Step 1: Properties of ASTM Class 30 Cast Iron** From Table A-24 of Shigley’s textbook (commonly used values for ASTM Class 30 Cast Iron): - Ultimate tensile strength, \( S_{ut} = 31 \) ksi - Ultimate compressive strength, \( S_{uc} = 110 \) ksi ### **Step 2: Modified Mohr Criterion** The **Modified Mohr Criterion** for brittle materials (like cast iron) is: \[ \text{Failure will occur if the largest Mohr circle intersects the failure envelope, i.e.} \] For a state of plane stress (\(\sigma_x, \sigma_y, \tau_{xy}\)), the principal stresses are: \[ \sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} \] Let \(\sigma_1 \geq \sigma_2\). The **factor of safety** \( n \) is computed as: \[ \text{If } \sigma_1 \geq 0 \text{ and } \sigma_2 \leq 0: \quad \frac{|\sigma_1|}{S_{ut}/n} + \frac{|\sigma_2|}{S_{uc}/n} = 1 \] \[ n = \left( \frac{\sigma_1}{S_{ut}} - \frac{\sigma_2}{S_{uc}} \right)^{-1} \] ### **Step 3: Solve Each Case** --- #### **Case I** \[ \sigma_x = 15\, \text{ksi},\quad \sigma_y = -50\, \text{ksi},\quad \tau_{xy} = 0 \text{ ksi} \] Principal stresses: \[ \sigma_{1,2} = \frac{15 + (-50)}{2} \pm \sqrt{\left(\frac{15 - (-50)}{2}\right)^2} \] \[ = -17.5 \pm 32.5 \] \[ \sigma_1 = 15\, \text{ksi},\quad \sigma_2 = -50\, \text{ksi} \] Now, apply the criterion: \[ n = \left( \frac{15}{31} - \frac{-50}{110} \right)^{-1} \] \[ = \left(0.4839 + 0.4545\right)^{-1} \] \[ = (0.9384)^{-1} = 1.07 \] --- #### **Case II** \[ \sigma_x = 15\, \text{ksi},\quad \sigma_y = 0\, \text{ksi},\quad \tau_{xy} = -10\, \text{ksi} \] Principal stresses: \[ \sigma_{1,2} = \frac{15 + 0}{2} \pm \sqrt{\left(\frac{15 - 0}{2}\right)^2 + (-10)^2} \] \[ = 7.5 \pm \sqrt{56.25 + 100} = 7.5 \pm \sqrt{156.25} \] \[ = 7.5 \pm 12.5 \] \[ \sigma_1 = 20\, \text{ksi},\quad \sigma_2 = -5\, \text{ksi} \] Apply the criterion: \[ n = \left( \frac{20}{31} - \frac{-5}{110} \right)^{-1} \] \[ = \left(0.6452 + 0.0455\right)^{-1} \] \[ = (0.6907)^{-1} = 1.45 \] --- #### **Case III** \[ \sigma_x = -10\, \text{ksi},\quad \sigma_y = -25\, \text{ksi},\quad \tau_{xy} = -10\, \text{ksi} \] Principal stresses: \[ \sigma_{1,2} = \frac{-10 + (-25)}{2} \pm \sqrt{\left(\frac{-10 - (-25)}{2}\right)^2 + (-10)^2} \] \[ = -17.5 \pm \sqrt{56.25 + 100} = -17.5 \pm 12.5 \] \[ \sigma_1 = -5\, \text{ksi},\quad \sigma_2 = -30\, \text{ksi} \] Both are compressive. For two compressive principal stresses, use: \[ n = \frac{S_{uc}}{|\sigma_2|} \] \[ n = \frac{110}{30} = 3.67 \] --- ## **Final Answers (Rounded to two decimal places)** 1. **Case I:**    \( n = 1.07 \) 2. **Case II:**    \( n = 1.45 \) 3. **Case III:**    \( n = 3.67 \) --- **Let me know if you need the calculations explained in more detail!**