Let's solve part (c): **If the shaft is required to have a safety factor of 2.5 based on the von Mises criterion, what should be the yield strength ($S_y$) for shaft?** do this in **two steps**: --- ## ** 1: Find the Maximum vonises Stress ($\sigma_{vm,max}$)** ### **Explanation:** - The maximum von Mises stress occurs at the critical section, where the combined bending and torsional stresses are highest. - The von Mises stress is given by: \[ \sigma_{vm} = \sqrt{\sigma_b^2 + 3\tau_t^2} \] where: - $\sigma_b$ = Maximum bending stress - $\tau_t$ = Maximum torsional (shear) stress ### **Formulas:** - **Bending Stress:** \[ \sigma_b = \frac{M_{max} \cdot c}{I} \] Where: - $M_{max}$ = Maximum bending moment at the critical section - $c$ = Outer radius = $.025$ m - $I$ = Second moment of area for a hollow shaft: \[ I = \frac{\pi}{64}(d_o^4 - d_i^4) \] With $d_o = .05$ m and $d_i = .03$ m - **Torsional Shear Stress:** \[ \tau_t = \frac{T_{max} \cdot c}{J} \] Where: - $T_{max}$ = Maximum torque at the critical section - $J$ = Polar moment of inertia: \[ J = \frac{\pi}{32}(d_o^4 - d_i^4) \] ### **Values Needed:** Suppose (from diagrams in part a) the critical section is at B, and the maximum moments and torques are: - $M_{max} = 300$ Nm (example value, should be calculated from diagrams) - $T_{max} = 200$ Nm (example value, should be calculated from diagrams) **Calculate $I$ and $J$:** \[ I = \frac{\pi}{64}(.05^4 - .03^4) = \frac{\pi}{64}(6.25 \times 10^{-6} - 8.1 \times 10^{-7}) \approx 2.6 \times 10^{-7} \text{ m}^4 \] \[ J = \frac{\pi}{32}(.05^4 - .03^4) = 2 \times I \approx 5.2 \times 10^{-7} \text{ m}^4 \] **Calculate $\sigma_b$ and $\tau_t$:** \[ \sigma_b = \frac{300 \times .025}{2.6 \times 10^{-7}} \approx 28.8 \times 10^6 \text{ Pa} = 28.8 \text{ MPa} \] \[ \tau_t = \frac{200 \times .025}{5.2 \times 10^{-7}} \approx 9.6 \times 10^6 \text{ Pa} = 9.6 \text{ MPa} \] **Calculate $\sigma_{vm,max}$:** \[ \sigma_{vm} = \sqrt{(28.8)^2 + 3 \times (9.6)^2} \approx \sqrt{829.44 + 276.48} \approx \sqrt{1105.92} \approx 33.3 \text{ MPa} \] --- ## **Step 2: Calculate Required Yield Strength ($S_y$) with Safety Factor** ### **Explanation:** - Safety factor ($SF$) is defined as $SF = \frac{S_y}{\sigma_{vm,max}}$ - Rearranging: \[ S_y = SF \times \sigma_{vm,max} \] - Given $SF = 2.5$ ### **Calculation:** \[ S_y = 2.5 \times 33.3 = 83.25 \text{ MPa} \] --- ## **Final Answer** **The shaft should have a yield strength $S_y$ of at least $83.25$ MPa to achieve a safety factor of 2.5 based on the von Mises criterion.** --- > **Note:** The values for $M_{max}$ and $T_{max}$ here are illustrative. You should use the actual maximum values from your bending and torsion diagrams for a precise answer. The procedure remains as shown.

# Problem Solution for Hollow Circular Shaft ## Part (a): Bending Moment Diagrams and Torsion Diagram ### Step 1: Draw Bending Moment and Torsion Diagrams 1. **Identify Loadings**: - The shaft experiences loads from the pulley and gear as indicated in the diagrams. - Support reactions at A and C must be calculated based on the applied forces. 2. **Bending Moment Calculation**: - Use the equilibrium equations (sum of moments and forces) to determine the bending moments at critical sections. - Draw a bending moment diagram based on these values. 3. **Torsion Calculation**: - Determine the torque applied to the shaft due to the loads from the pulley and gear. - Draw the torsion diagram alongside the bending moment diagram. ### Critical Cross-Section: - Identify the section with maximum bending and torsional stress, typically occurs at the location of maximum bending moment or torque. --- ## Part (b): Diagram of Critical Cross-Section ### Step 1: Create the Cross-Section Diagram 1. **Draw the Critical Cross-Section**: - Indicate the outer and inner diameters of the hollow shaft. - Label the neutral axis at the mid-radius. 2. **Indicate Stresses**: - Draw the resultant bending moment and torsional moment vectors. - Mark the points of maximum tensile and compressive bending stress, typically at the outer surface. 3. **Coordinate Axes**: - Clearly label the x (horizontal) and y (vertical) axes on the diagram. --- ## Part (c): Calculate Required Yield Strength ($S_y$) ### Step 1: Find the Maximum von Mises Stress ($\sigma_{vm,max}$) 1. **Assume Values for Maximum Bending and Torsional Moments**: - Example: - \( M_{max} = 300 \, \text{Nm} \) - \( T_{max} = 200 \, \text{Nm} \) 2. **Calculate Second Moment of Area ($I$) and Polar Moment of Inertia ($J$)**: \[ I = \frac{\pi}{64}(d_o^4 - d_i^4) = \frac{\pi}{64}(0.05^4 - 0.03^4) \approx 2.6 \times 10^{-7} \, \text{m}^4 \] \[ J = \frac{\pi}{32}(d_o^4 - d_i^4) = 2 \times I \approx 5.2 \times 10^{-7} \, \text{m}^4 \] 3. **Calculate Bending Stress ($\sigma_b$) and Torsional Shear Stress ($\tau_t$)**: \[ \sigma_b = \frac{M_{max} \cdot c}{I} = \frac{300 \times 0.025}{2.6 \times 10^{-7}} \approx 28.8 \, \text{MPa} \] \[ \tau_t = \frac{T_{max} \cdot c}{J} = \frac{200 \times 0.025}{5.2 \times 10^{-7}} \approx 9.6 \, \text{MPa} \] 4. **Calculate Maximum von Mises Stress**: \[ \sigma_{vm,max} = \sqrt{\sigma_b^2 + 3\tau_t^2} = \sqrt{(28.8)^2 + 3 \times (9.6)^2} \approx 33.3 \, \text{MPa} \] --- ### Step 2: Calculate Required Yield Strength ($S_y$) 1. **Apply Safety Factor**: - Given safety factor \( SF = 2.5 \) \[ S_y = SF \times \sigma_{vm,max} = 2.5 \times 33.3 \approx 83.25 \, \text{MPa} \] --- ## Final Answer The shaft should have a yield strength \( S_y \) of at least **83.25 MPa** to achieve a safety factor of 2.5 based on the von Mises criterion.