Let's break down the **fatigue problem** step by step. --- ## **Given Data** - **Diameter, \( d \) = 5 mm** - **Bending Moment, \( M \): 200 to 400 N·mm** - **Torsion, \( T \): -100 to 100 N·mm** - **Ultimate Strength, \( S_u = 50 \) MPa** - **Yield Strength, \( S_y = 45 \) MPa** - **Corrected Endurance Limit, \( S_e = 20 \) MPa** - **Fatigue Strength Factor, \( f = 0.8 \)** --- ## **Step 1: Calculate Stresses** ### **Section modulus:** \[ I = \frac{\pi}{64} d^4, \quad J = \frac{\pi}{32} d^4 \] \[ c = \frac{d}{2} \] ### **Bending Stress:** \[ \sigma_b = \frac{M \cdot c}{I} \] ### **Torsional Shear Stress:** \[ \tau_t = \frac{T \cdot c}{J} \] --- ## **Step 2: Stress Range Calculation** ### **Bending:** - \( M_{min} = 200 \) N·mm, \( M_{max} = 400 \) N·mm \[ I = \frac{\pi}{64} (5^4) = \frac{\pi}{64} \times 625 = 30.68 \text{ mm}^4 \] \[ c = 2.5 \text{ mm} \] \[ \sigma_{b,max} = \frac{400 \times 2.5}{30.68} = 32.6 \text{ MPa} \] \[ \sigma_{b,min} = \frac{200 \times 2.5}{30.68} = 16.3 \text{ MPa} \] ### **Torsion:** \[ J = \frac{\pi}{32} \times 625 = 61.36 \text{ mm}^4 \] \[ \tau_{t,max} = \frac{100 \times 2.5}{61.36} = 4.07 \text{ MPa} \] \[ \tau_{t,min} = \frac{-100 \times 2.5}{61.36} = -4.07 \text{ MPa} \] --- ## **Step 3: Mean and Alternating Stresses** ### **Normal (Bending):** \[ \sigma_m = \frac{\sigma_{b,max} + \sigma_{b,min}}{2} = \frac{32.6 + 16.3}{2} = 24.45 \text{ MPa} \] \[ \sigma_a = \frac{\sigma_{b,max} - \sigma_{b,min}}{2} = \frac{32.6 - 16.3}{2} = 8.15 \text{ MPa} \] ### **Shear (Torsion):** \[ \tau_m = \frac{\tau_{t,max} + \tau_{t,min}}{2} = 0 \] \[ \tau_a = \frac{\tau_{t,max} - \tau_{t,min}}{2} = \frac{4.07 - (-4.07)}{2} = 4.07 \text{ MPa} \] --- ## **Step 4: Equivalent Alternating and Mean Stresses (von Mises)** \[ \sigma_{a,eq} = \sqrt{\sigma_a^2 + 3\tau_a^2} = \sqrt{8.15^2 + 3 \times 4.07^2} = \sqrt{66.42 + 49.71} = \sqrt{116.13} = 10.78 \text{ MPa} \] \[ \sigma_{m,eq} = \sqrt{\sigma_m^2 + 3\tau_m^2} = \sqrt{24.45^2 + 0} = 24.45 \text{ MPa} \] --- ## **Step 5: Goodman Criterion** \[ \frac{\sigma_{a,eq}}{f S_e} + \frac{\sigma_{m,eq}}{S_u} = 1 \] Given \( f S_e = 0.8 \times 20 = 16 \) MPa, \( S_u = 50 \) MPa \[ \frac{10.78}{16} + \frac{24.45}{50} = 0.674 + 0.489 = 1.163 \] Since 1.163 > 1, **the cycles to failure need to be found**. --- ## **Step 6: S-N Curve (Basquin's Law)** \[ \sigma_a = S_f' (2N)^b \] But in practice, most questions give a log-log S-N curve. For the corrected endurance limit (for \( N = 10^6 \)), \( S_e = 20 \) MPa. Assume: \[ \sigma_a N^b = C, \quad \text{at } N = 10^6, \; \sigma_a = 20 \Rightarrow C = 20 \times (10^6)^b \] Set \( \sigma_{a,eq} = 10.78 \) MPa: \[ 10.78 = 20 \left(\frac{N}{10^6}\right)^b \] \[ \left(\frac{10.78}{20}\right) = \left(\frac{N}{10^6}\right)^b \] \[ 0.539 = \left(\frac{N}{10^6}\right)^b \] Take log (assuming \( b = -0.12 \) typical for steels): \[ \log_{10}(0.539) = b \log_{10}\left(\frac{N}{10^6}\right) \] \[ -0.269 = -0.12 \log_{10}\left(\frac{N}{10^6}\right) \] \[ \log_{10}\left(\frac{N}{10^6}\right) = \frac{-0.269}{-0.12} = 2.242 \] \[ N = 10^{6 + 2.242} = 10^{8.242} = 1.74 \times 10^8 \text{ cycles} \] --- ## **Step 7: Closest Option** From the options: - \( 15.409 \times 10^6 \) - \( 30.154 \times 10^6 \) - \( 120.897 \times 10^6 \) - \( 60.345 \times 10^6 \) **The closest is:** **120.897 × 10⁶ cycles** (i.e., \( 1.2 \times 10^8 \), close to our calculation). --- ## **Final Answer** ### **The correct option is:** \[ \boxed{120.897 \times 10^6 \text{ cycles}} \]