Sure, let's break down the problem step by step. --- ## **Problem Restatement** A stepped bar shown in the figure is loaded axially. - **Material:** AISI 102 cold-drawn steel (yield strength = 370 MPa) - **Small radius:** \( r_1 = 8 \) mm - **Large radius:** \( r_2 = 16 \) mm - **Diameter of larger section:** \( d = 28 \) mm - **Fillet radius:** \( r_f = 2 \) mm You are to calculate the **maximum axial force P** that can be applied without causing yielding, considering the stress concentration at the fillet. --- ## **Step 1: Find Stress Concentration Factor \( K_t \)** For a stepped shaft in tension, the stress concentration factor (\( K_t \)) depends on: - The **fillet radius to small diameter ratio** \((r_f/r_1)\) - The **small-to-large diameter ratio** \((r_1/r_2)\) From the image: - \( r_1 = 8 \) mm (small radius) - Fillet radius \( r_f = 2 \) mm - Large radius \( r_2 = 16 \) mm So: - \( r_f/r_1 = 2/8 = .25 \) - \( r_1/r_2 = 8/16 = .5 \) From standard **stress concentration charts** (Peterson, Shigley's), for a stepped shaft in tension: - For \( r_1/r_2 = .5 \) and \( r_f/r_1 = .25 \), \( K_t \approx 1.7 \) (you can confirm via table, but this is a typical value). --- ## **Step 2: Calculate Nominal Stress** The smallest area is at section with radius \( r_1 \): \[ A = \pi r_1^2 = \pi (8\, \text{mm})^2 = \pi (64)\, \text{mm}^2 = 201.06\, \text{mm}^2 \] Nominal tensile stress at the step: \[ \sigma_{\text{nom}} = \frac{P}{A} \] --- ## **Step 3: Apply Stress Concentration Factor** The **maximum stress** at the fillet is: \[ \sigma_{\text{max}} = K_t \cdot \sigma_{\text{nom}} \] To avoid yielding: \[ \sigma_{\text{max}} \leq \sigma_{\text{yield}} \] \[ K_t \frac{P}{A} \leq \sigma_{\text{yield}} \] \[ P \leq \frac{\sigma_{\text{yield}} \cdot A}{K_t} \] --- ## **Step 4: Substitute the Values** - \( \sigma_{\text{yield}} = 370 \) MPa = 370 N/mm² - \( A = 201.06 \) mm² - \( K_t = 1.7 \) \[ P \leq \frac{370 \times 201.06}{1.7} \] \[ P \leq \frac{74,392.2}{1.7} \] \[ P \leq 43,760 \text{ N} \] --- ## **Step 5: Final Answer** \[ \boxed{P_{\text{max}} = 43,760 \text{ N} } \] --- ### **Summary of Steps** 1. **Found the stress concentration factor** \( K_t = 1.7 \). 2. **Calculated the smallest area** \( A = 201.06 \) mm². 3. **Used yield condition:** \( \sigma_{\text{max}} = K_t \cdot \sigma_{\text{nom}} \leq \sigma_{\text{yield}} \). 4. **Solved for P:** \( P_{\text{max}} = \frac{\sigma_{\text{yield}} \cdot A}{K_t} \). 5. **Final answer:** \( \boxed{43,760 \text{ N}} \). --- If you have the chart/table, you can look up the exact \( K_t \) for your ratios, but 1.7 is a typical value for this geometry. Let me know if you'd like to see the chart or discuss safety factors!