Let's break down the solution step by: --- ## **Step 1: List Given Data** - \( F_y = 420. \) lbf - \( F_x = F_z = \) - Diameter \( D = 2.000 \) in - Height \( h = 2.500 \) in - Length \( AB = 9.00 \) in - Length \( BC = 6.00 \) in --- ## **Step 2: Identify Critical Section** The highest stresses will occur at the fixed support, **point A**. --- ## **Step 3: Calculate Resulting Moments** The force \( F_y \) acts at a distance from A: - **Total length from A to F application:** \( L = AB + BC + \frac{h}{4} \) \( L = 9.00 + 6.00 + \frac{2.5}{4} = 15.625 \) in - **Moment about A:** \( M = F_y \times L = 420. \times 15.625 = 6562.5 \) in-lbf --- ## **Step 4: Calculate Sectional Properties** - **Area, \( A \):** \( A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (2.000)^2 = 3.142 \) in² - **Moment of inertia, \( I \):** \( I = \frac{\pi}{64} D^4 = \frac{\pi}{64} (2.000)^4 = 1.571 \) in⁴ --- ## **Step 5: Calculate Stresses** ### **A. Bending Stress (\( \sigma_b \))** \[ \sigma_b = \frac{Mc}{I} \] where \( c = \frac{D}{2} = 1.000 \) in \[ \sigma_b = \frac{6562.5 \times 1.000}{1.571} = 418 \text{ psi} \] --- ### **B. Direct (Axial) Stress** Since \( F_x = \), there is **no direct axial stress**. --- ### **C. Shear Stress** Shear is negligible in bending-dominated cases for circular bars, but let's check: - **Transverse Shear Stress (\( \tau \)):** \[ \tau = \frac{4F_y}{3A} = \frac{4 \times 420}{3 \times 3.142} = \frac{168}{9.426} = 178.3 \text{ psi} \] --- ## **Step 6: Principal Stresses** At the **surface** (outer fiber at top or bottom), the stress state is: - Normal: \( \sigma_b \) (tension or compression) - Shear: \( \tau \) (from above) Principal stresses: \[ \sigma_{1,2} = \frac{\sigma_b}{2} \pm \sqrt{\left(\frac{\sigma_b}{2}\right)^2 + \tau^2} \] Plug the values: \[ \sigma_{1,2} = \frac{418}{2} \pm \sqrt{\left(\frac{418}{2}\right)^2 + (178.3)^2} \] \[ = 209 \pm \sqrt{209^2 + 178.3^2} \] \[ = 209 \pm \sqrt{4,368,100 + 31,801} \] \[ = 209 \pm \sqrt{4,399,901} \] \[ = 209 \pm 2098.8 \] So, - \( \sigma_1 = 209 + 2098.8 = 4188.8 \) psi - \( \sigma_2 = 209 - 2098.8 = -8.8 \) psi --- ## **Step 7: Maximum Shear Stress** \[ \tau_{max} = \frac{\sigma_1 - \sigma_2}{2} = \frac{4188.8 - (-8.8)}{2} = \frac{4197.6}{2} = 2098.8 \text{ psi} \] --- ## **Final Answers** ### **Principal Stresses:** - **Maximum:** \( \boxed{4188.8~\text{psi}} \) - **Minimum:** \( \boxed{-8.8~\text{psi}} \) ### **Maximum Shear Stress:** - \( \boxed{2098.8~\text{psi}} \) --- **Summary Table:** | Stress Type | Value (psi) | |---------------------|:-----------:| | Principal (max) | 4188.8 | | Principal (min) | -8.8 | | Max Shear | 2098.8 | --- **If you need further breakdown or explanation for any step, please let me know!**