Let's solve the problem step by step: Given: - Two different stress cycles: 1. \( -200 < \sigma_1 < 500 \) MPa for 2000 cycles 2. \( -100 < \sigma_2 < 100 \) MPa for 1000 cycles - Bar Properties: - Ultimate strength, \( S_u = 650 \) MPa - Yield strength, \( S_y = 600 \) MPa - Fatigue data table is given for various stress amplitudes and corresponding failure cycles. #### Step 1: Calculate the mean and amplitude of both cycles **Cycle 1**: \[ \sigma_{m1} = \frac{500 + (-200)}{2} = 150 \text{ MPa} \] \[ \sigma_{a1} = \frac{500 - (-200)}{2} = 350 \text{ MPa} \] **Cycle 2**: \[ \sigma_{m2} = \frac{100 + (-100)}{2} = 0 \text{ MPa} \] \[ \sigma_{a2} = \frac{100 - (-100)}{2} = 100 \text{ MPa} \] #### Step 2: Find the fatigue life for each cycle From the fatigue data table: - At 350 MPa (not directly given, but between 250 MPa and 400 MPa) - At 100 MPa (5000 cycles) Let's interpolate for 350 MPa: - 250 MPa → 3000 cycles - 400 MPa → 2500 cycles \[ \text{Slope} = \frac{2500-3000}{400-250} = \frac{-500}{150} = -3.33 \text{ cycles per MPa} \] \[ \Delta \sigma = 350 - 250 = 100 \text{ MPa} \] \[ \text{Cycles at 350 MPa} = 3000 + (-3.33) \times 100 = 3000 - 333 = 2667 \text{ cycles} \] So: - \( N_1 = 2667 \) cycles for cycle 1 - \( N_2 = 5000 \) cycles for cycle 2 #### Step 3: Use Miner's Rule Miner's Rule: \[ \frac{n_1}{N_1} + \frac{n_2}{N_2} = 1 \] where \( n_1, n_2 \) are the actual cycles in one duty cycle \[ \frac{2000}{2667} + \frac{1000}{5000} = 0.75 + 0.2 = 0.95 \] Let the number of duty cycles before failure be \( n \): \[ n \left( \frac{2000}{2667} + \frac{1000}{5000} \right) = 1 \] \[ n \times 0.95 = 1 \] \[ n = \frac{1}{0.95} = 1.05 \] #### Step 4: Choose the Correct Option So, the bar will fail just after the completion of 1 duty cycle (since \( n \) is just over 1). **Correct Option:** > n = 1, bar will fail just after completion of 1 duty cycle.