Let's break down the problem step by step: --- # Given Data - **Bending Moment, \( M \):** \( 110~\text{lbf} \cdot \text{in} \) - **Torsion, \( T \):** \( 800~\text{lbf} \cdot \text{in} \) - **Ultimate Strength, \( S_{ut} \):** \( 118~\text{kpsi} \) - **Yield Strength, \( S_y \):** \( 102~\text{kpsi} \) - **Fracture Strength, \( \tilde{\sigma}_f \):** \( 185~\text{kpsi} \) - **Room temperature** - **Infinite life required, safety factor \( n = 1.5 \)** - **Reliability = 99%** - **Shoulder fillet: \( r/d = .1 \)** --- # (a) First Estimate of the Small Diameter \( d \) (DE-Goodman Criterion) ## **Step 1: Calculate Stresses** For a solid shaft: - **Bending stress:** \( \sigma = \frac{32M}{\pi d^3} \) - **Torsional stress:** \( \tau = \frac{16T}{\pi d^3} \) ### Convert units: - 1 kpsi = 100 psi - 1 lbf·in = 1 lbf·in ### **Stress Concentration Factors:** Assume: - \( K_t \) (bending, from charts for shoulder fillet, \( r/d = .1 \)): approx. **1.7** - \( K_{ts} \) (torsion): approx. **1.5** ### **Alternating and Mean Stress:** Assume fully reversed bending and steady torsion: - \( \sigma_a = K_t \frac{32M}{\pi d^3} \) - \( \tau_m = K_{ts} \frac{16T}{\pi d^3} \) ## **Step 2: Endurance Limit** - \( S_e' = .5 S_{ut} = .5 \times 118 = 59~\text{kpsi} \) - Modify for: - Surface finish (assume machined): \( k_a \approx .85 \) - Size (for \( d < 2 \) in): \( k_b \approx .85 \) - Reliability (99%): \( k_e = .814 \) - Temperature, load: \( k_c = k_d = 1 \) So: \[ S_e = S_e' \cdot k_a \cdot k_b \cdot k_c \cdot k_d \cdot k_e = 59 \times .85 \times .85 \times 1 \times 1 \times .814 = 34.5~\text{kpsi} \] ## **Step 3: Goodman Criterion** \[ \frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_{ut}} = \frac{1}{n} \] For bending + steady torsion: - \( \sigma_a = \) fully reversed bending (alternating) - \( \sigma_m = \) mean stress from torsion But for combined loading: \[ \sigma_a' = \sqrt{\sigma_a^2 + 3\tau_a^2} = \sigma_a \] \[ \sigma_m' = \sqrt{\sigma_m^2 + 3\tau_m^2} = \sqrt{^2 + 3 \left( K_{ts} \frac{16T}{\pi d^3} \right)^2 } \] But since bending is fully reversed, \( \sigma_m = \), so only mean torsion stress: \[ \sigma_m' = \sqrt{3} K_{ts} \frac{16T}{\pi d^3} \] \[ \sigma_a = K_t \frac{32M}{\pi d^3} \] Plug into Goodman: \[ \frac{\sigma_a}{S_e} + \frac{\sigma_m'}{S_{ut}} = \frac{1}{n} \] ## **Step 4: Plug in Numbers** Let \( d \) be in inches. \[ \sigma_a = 1.7 \frac{32 \times 110}{\pi d^3} = 1.7 \frac{35200}{\pi d^3} \] \[ \sigma_m' = \sqrt{3} \times 1.5 \frac{16 \times 800}{\pi d^3} = 2.598 \times 1.5 \frac{12800}{\pi d^3} = 3.897 \frac{12800}{\pi d^3} \] So: \[ \frac{1.7 \times 35200}{\pi d^3 \times 34,500} + \frac{3.897 \times 12800}{\pi d^3 \times 118,000} = \frac{1}{1.5} = .6667 \] Calculate each term: - \( 1.7 \times 35200 = 59,840 \) - \( 3.897 \times 12800 = 49,881.6 \) So: \[ \frac{59,840}{\pi d^3 \times 34,500} + \frac{49,881.6}{\pi d^3 \times 118,000} = .6667 \] \[ \frac{59,840}{108,300\pi d^3} + \frac{49,881.6}{369,400\pi d^3} = .6667 \] Combine over common denominator: \[ \frac{59,840 \times 369,400 + 49,881.6 \times 108,300}{108,300\pi d^3 \times 369,400} = .6667 \] Calculate: - \( 59,840 \times 369,400 = 22,114,096,000 \) - \( 49,881.6 \times 108,300 = 5,398,758,080 \) - Sum: \( 27,512,854,080 \) Denominator: \( 108,300 \times 369,400 = 40,010,820,000 \) So: \[ \frac{27,512,854,080}{40,010,820,000 \pi d^3} = .6667 \] \[ \frac{27,512,854,080}{.6667 \times 40,010,820,000 \pi} = d^3 \] \[ .6667 \times 40,010,820,000 \approx 26,673,880,000 \] \[ \frac{27,512,854,080}{26,673,880,000 \pi} = d^3 \] \[ \frac{27,512,854,080}{83,819,522,812} = d^3 \] \[ d^3 \approx .328 \] \[ d \approx (.328)^{1/3} \approx .69~\text{in} \] **Round up to standard shaft size:** Nearest standard size is **.75 in**. --- ## **Final Answer (a):** > **First estimate for diameter \( d = .75~\text{in} \) (rounded to nearest standard size).** --- ### Do you want to proceed to part (b) with this result?