Let's break down the solution step by step for each part: --- ## 3a. **Free-Body Diagram of the Crank** - **Forces:** The only external force on the crank is the 75 lb vertical load at point A (the pedal). - **Reactions:** At B (where the crank attaches to the axle), there will be a vertical reaction force and a moment reaction (because it is cantilevered). - **Diagram:** - Draw the crank as a horizontal beam, fixed at B. - Arrow down at A, labeled “75 lb.” - At B, show a vertical reaction force upward and a moment reaction (curved arrow). **[Free-body diagram not shown here, but you should draw as described above.]** --- ## 3b. **Shear, Moment, and Deflection Diagrams** ### Shear Diagram, \( V(x) \) - For a cantilever beam of length \( L = 4 \) in with a downward force \( P = 75 \) lb at the end: - Shear is constant along the beam, \( V(x) = -75 \) lb (downward) from \( x = \) to \( x = 4 \) in. ### Moment Diagram, \( M(x) \) - \( M(x) = -P \cdot (L - x) \) - At B (\( x = \)): \( M_B = -75 \times 4 = -300 \) in-lb - At A (\( x = 4 \)): \( M_A = \) ### Elastic Curve, \( v(x) \) - The deflection equation for a cantilever beam with end load: \[ v(x) = \frac{P}{6EI} (3Lx^2 - x^3) \] where \( E \) is Young’s modulus, \( I \) is the moment of inertia. --- ## 3c. **End Deflection Due to Applied Load** Given: - \( P = 75 \) lb - \( L = 4 \) in - Material: 6061 Aluminum Alloy (\( E \approx 10,000,000 \) psi) - **Moment of inertia \( I \):** - Section is rectangular: width = .4 in, height = .2 in \[ I = \frac{1}{12} b h^3 = \frac{1}{12} \times .4 \times (.2)^3 = \frac{1}{12} \times .4 \times .008 = .000267 \text{ in}^4 \] - **Deflection at end:** \[ \delta_{max} = v(L) = \frac{P L^3}{3EI} \] \[ \delta_{max} = \frac{75 \times 4^3}{3 \times 10,000,000 \times .000267} \] \[ = \frac{75 \times 64}{3 \times 10,000,000 \times .000267} \] \[ = \frac{480}{801} \] \[ = .599 \text{ in} \] --- ## 3d. **Stresses at Point C** - **Bending stress:** \[ \sigma_x = \frac{M y}{I} \] - \( M = 75 \times 4 = 300 \) in-lb = 360 in-lb (convert to lb-in if necessary; here units are consistent) - \( y = .1 \) in (distance from neutral axis to outer fiber; half the height) - \( I = .000267 \) in\(^4\) \[ \sigma_x = \frac{300 \times .1}{.000267} = \frac{30}{.000267} \approx 112,360 \text{ psi} \] - **Shear stress:** \[ \tau_{xy} = \frac{V Q}{I b} \] For a rectangular section at neutral axis: - \( V = 75 \) lb - \( Q = \) area above neutral axis × distance to centroid \( = (.2 \times .2/2) \times (.2/4) = .02 \times .05 = .001 \) - \( b = .4 \) in \[ \tau_{xy} = \frac{75 \times .001}{.000267 \times .4} \approx \frac{.075}{.0001068} \approx 702 \text{ psi} \] - **Principal Stresses:** \[ \sigma_{1,2} = \frac{\sigma_x}{2} \pm \sqrt{\left(\frac{\sigma_x}{2}\right)^2 + \tau_{xy}^2} \] Since \( \sigma_y = \): \[ \sigma_{1,2} = \frac{\sigma_x}{2} \pm \sqrt{\left(\frac{\sigma_x}{2}\right)^2 + \tau_{xy}^2} \] Plug in values: \[ \sigma_{1,2} = \frac{112360}{2} \pm \sqrt{\left(\frac{112360}{2}\right)^2 + 702^2} \] \[ = 56180 \pm \sqrt{56180^2 + 702^2} \] \[ \approx 56180 \pm 56185 \] So, - \( \sigma_1 \approx 112,365 \) psi - \( \sigma_2 \approx -5 \) psi --- ## 3e. **Von Mises Stress and Factor of Safety** - **Von Mises Stress (principal form):** \[ \sigma' = \sqrt{\sigma_1^2 + \sigma_2^2 - \sigma_1 \sigma_2} \] \[ = \sqrt{(112,365)^2 + (-5)^2 - (112,365)(-5)} \] \[ = \sqrt{1.2638 \times 10^{10} + 25 + 561825} \] \[ = \sqrt{1.2638 \times 10^{10} + 561850} \] The 561,850 is negligible compared to \( 1.2638 \times 10^{10} \). \[ \approx \sqrt{1.2638 \times 10^{10}} \approx 112,465 \text{ psi} \] - **Yield Strength of 6061 Aluminum Alloy:** \( \sigma_{y} \approx 40,000 \) psi - **Factor of Safety:** \[ n = \frac{\sigma_{y}}{\sigma'} \] \[ n = \frac{40,000}{112,465} \approx .36 \] ### **Interpretation** - The factor of safety is less than 1, which means the part is not safe for the given load and geometry. - **Other Points:** The highest stress will be at point C (fixed end, outer fiber), so this is the most likely location for failure. --- ## **Summary Table** | Step | Value/Result | |-----------|-------------| | Deflection at A (\(\delta_{max}\)) | .599 in | | Max Bending Stress (\(\sigma_x\)) | 112,360 psi | | Max Shear Stress (\(\tau_{xy}\)) | 702 psi | | Principal Stresses (\(\sigma_1, \sigma_2\)) | 112,365 psi, -5 psi | | Von Mises Stress (\(\sigma'\)) | 112,465 psi | | Yield Strength (6061 Al) | 40,000 psi | | Factor of Safety | **.36 (unsafe)** | --- ### **Conclusion** - The crank will **fail** under this load and geometry. - To improve safety, increase the section size or use a material with higher yield strength. If you need more details on any step or want the diagrams, let me know!