Let's break down and solve the problem step by step using the information from your image. --- ## **Problem Statement:** A simply supported 50 mm bar is subjected to variable stresses during a duty cycle: 1. \(-200 < \sigma < 500 \text{ MPa}\) for **2000 cycles** 2. \(-100 < \sigma < 100 \text{ MPa}\) for **1000 cycles** **Bar Properties:** - Ultimate strength, \(S_u = 650 \text{ MPa}\) - Yield strength, \(S_y = 600 \text{ MPa}\) **Fatigue Test Data:** | Fatigue Test Data (\(\sigma_a\)) | Fatigue Failure Cycle (Nf) | | -------------------------------- | -------------------------- | | 50 MPa | 10,000 | | 100 MPa | 5,000 | | 150 MPa | 4,000 | | 250 MPa | 3,000 | | 400 MPa | 2,500 | | 500 MPa | 2,000 | --- ## **Steps to Solve:** ### **1. Identify Stress Ranges & Cycles:** #### **Case 1:** \(-200 < \sigma < 500 \text{ MPa}\) for 2000 cycles - Mean stress (\(\sigma_m\)): \(\frac{500 + (-200)}{2} = 150 \text{ MPa}\) - Stress amplitude (\(\sigma_a\)): \(\frac{500 - (-200)}{2} = 350 \text{ MPa}\) #### **Case 2:** \(-100 < \sigma < 100 \text{ MPa}\) for 1000 cycles - Mean stress (\(\sigma_m\)): \(\frac{100 + (-100)}{2} = 0 \text{ MPa}\) - Stress amplitude (\(\sigma_a\)): \(\frac{100 - (-100)}{2} = 100 \text{ MPa}\) --- ### **2. Find Fatigue Life for Each Case:** From the table, interpolate or use values for cycles to failure at the respective amplitude. - For **350 MPa**, interpolate between 250 MPa (3000 cycles) and 400 MPa (2500 cycles): \[ \text{Nf at 350 MPa} = 3000 + \frac{2500 - 3000}{400 - 250} \times (350 - 250) \] \[ = 3000 + \frac{-500}{150} \times 100 \] \[ = 3000 - 333.3 = 2666.7 \text{ cycles (approx.)} \] - For **100 MPa**, Nf = 5000 cycles. --- ### **3. Apply Miner's Rule (Linear Damage Rule):** \[ \text{Total Damage} = \sum \frac{n_i}{N_{fi}} \] Where \(n_i\) = number of cycles at stress amplitude \(i\), \(N_{fi}\) = cycles to failure at that amplitude. \[ D = \frac{2000}{2667} + \frac{1000}{5000} \] \[ = 0.75 + 0.2 = 0.95 \] --- ### **4. Find Number of Duty Cycles to Failure:** Let \(n\) be the number of duty cycles. Each duty cycle = 2000 cycles at 350 MPa + 1000 cycles at 100 MPa. \[ D_{\text{total}} = n \left(\frac{2000}{2667} + \frac{1000}{5000}\right) \] Set \(D_{\text{total}} = 1\): \[ n \times 0.95 = 1 \implies n = \frac{1}{0.95} \approx 1.05 \] --- ## **Final Answer:** **The bar will just fail after approximately 1 duty cycle.** If you need to consider the Goodman criterion (for mean stress), let me know and I can provide the adjustment. --- **Let me know if you'd like to see the Goodman correction or have any other questions!**