Let's break down the problem using the information provided. --- ## **Given Data** - **Material:** AISI 104 cold-drawn steel - **Bar dimensions:** - Width = 25 mm - Thickness = 10 mm - Central hole diameter = 6 mm - **Axial load fluctuates:** 3 kN to 28 kN - **Goal:** Find - Fatigue factor of safety (using Modified Goodman for infinite life) - Yielding factor of safety - Number of cycles to failure --- ## **Step 1: Calculate the Stresses** ### **Net Area at Hole** \[ A = (25\ \text{mm} - 6\ \text{mm}) \times 10\ \text{mm} = 19 \times 10 = 190\ \text{mm}^2 = 190 \times 10^{-6}\ \text{m}^2 \] ### **Mean and Alternating Loads** \[ P_{\text{max}} = 28\ \text{kN} = 28000\ \text{N} \] \[ P_{\text{min}} = 3\ \text{kN} = 300\ \text{N} \] \[ P_m = \frac{P_{\text{max}} + P_{\text{min}}}{2} = \frac{28000 + 300}{2} = 15500\ \text{N} \] \[ P_a = \frac{P_{\text{max}} - P_{\text{min}}}{2} = \frac{28000 - 300}{2} = 12500\ \text{N} \] ### **Mean and Alternating Stresses** \[ \sigma_m = \frac{P_m}{A} = \frac{15500}{190 \times 10^{-6}} = 81.58\ \text{MPa} \] \[ \sigma_a = \frac{P_a}{A} = \frac{12500}{190 \times 10^{-6}} = 65.79\ \text{MPa} \] --- ## **Step 2: Material Properties for AISI 104 Cold-Drawn Steel** Typical values: - Ultimate strength: \( S_u \approx 670\ \text{MPa} \) - Yield strength: \( S_y \approx 460\ \text{MPa} \) - Endurance limit: \( S_e' = .5 S_u \approx 335\ \text{MPa} \) ### **Endurance Limit Modifiers** Assuming typical values: - Surface finish (cold drawn): \( k_a \approx .85 \) - Size factor (\(d < 8\) mm, but net section is 19 mm): \( k_b \approx .85 \) - Load (axial): \( k_c = .7 \) - Temperature: \( k_d = 1 \) - Reliability: \( k_e = .814 \) (for 99% reliability) \[ S_e = S_e' \cdot k_a \cdot k_b \cdot k_c \cdot k_d \cdot k_e \] \[ S_e = 335 \times .85 \times .85 \times .7 \times 1 \times .814 \approx 115\ \text{MPa} \] --- ## **Step 3: Modified Goodman Criterion** The equation: \[ \frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_u} = \frac{1}{n} \] Plug in the numbers: \[ \frac{65.79}{115} + \frac{81.58}{670} = \frac{1}{n} \] \[ .572 + .122 = \frac{1}{n} \] \[ .694 = \frac{1}{n} \Longrightarrow n = 1.44 \] **Fatigue factor of safety:** \[ \boxed{1.44} \] --- ## **Step 4: Yielding Factor of Safety** \[ n_y = \frac{S_y}{\sigma_{\text{max}}} \] \[ \sigma_{\text{max}} = \frac{P_{\text{max}}}{A} = 147.37\ \text{MPa} \] \[ n_y = \frac{460}{147.37} = 3.12 \] **Yielding factor of safety:** \[ \boxed{3.12} \] --- ## **Step 5: Number of Cycles to Failure** Since the fatigue factor of safety is above 1 **(n = 1.44)**, infinite life is expected using the Modified Goodman criterion. **Number of cycles to failure:** \[ \boxed{\infty} \] --- ## **Summary Table** | Item | Value | |------------------------|----------| | Fatigue factor of safety (\(n\)) | 1.44 | | Yielding factor of safety (\(n_y\)) | 3.12 | | Number of cycles to failure | \(\infty\) | --- ### **If you want cycles for a lower \( n \):** If the factor of safety were less than 1, you would use the S-N curve for AISI 104 steel. But in this case, as calculated, the part is **safe for infinite life** under the given loading. --- ### **Let me know if you need any further breakdown!**