Certainly! Let's **break down the problem** and solve it step by step. --- ## **Problem Restatement**A stepped round shaft** is made from AISI 414 QT steel and is subjected to a shock load (suddenly applied force) as shown in the diagram. You need to: . **Determine the maximum allowable force \( P \)** 2. **Apply stress concentration factors where needed** --- ## **Step 1: Understanding the Shaft Geometry** - **Larger diameter (middle):** \( d_1 = 20 \) mm - **Smaller diameter (ends):** \( d_2 = 10 \) mm - **Fillet radius (step):** \( R = 1 \) mm - **Hole diameter:** \( d_h = 8 \) mm --- ## **Step 2: Material Properties for AISI 414 QT Steel** Typical values (please check your textbook or data sheet for your specific application): - **Ultimate tensile strength, \( S_u \):** ~ 655 MPa - **Yield strength, \( S_y \):** ~ 415 MPa For shock loading, **use a high factor of safety** (often 2 or more). --- ## **Step 3: Critical Section for Failure** The most likely **failure point** is at the **small diameter section** with the hole, because: - It has the **smallest cross-section** - There is a **stress concentration** due to the hole --- ## **Step 4: Net Area at the Hole** The net cross-sectional area at the hole is: \[ A_{net} = \frac{\pi}{4} d_2^2 - d_2 \cdot d_h \] Where: - \( d_2 = 10 \) mm - \( d_h = 8 \) mm Calculate: \[ A_{net} = \frac{\pi}{4} (10^2) - 10 \times 8 = \frac{\pi}{4} \times 100 - 80 \approx 78.54 - 80 = -1.46\, \text{mm}^2 \] This is **not correct** (the area cannot be negative)! Let's correct the formula: - The net area is the area of the shaft minus the area of the hole (as a circle): \[ A_{net} = \frac{\pi}{4} (d_2^2 - d_h^2) \] \[ A_{net} = \frac{\pi}{4} (10^2 - 8^2) = \frac{\pi}{4} (100 - 64) = \frac{\pi}{4} \times 36 \approx 28.27\, \text{mm}^2 \] --- ## **Step 5: Stress Concentration Factor (Kt)** There are two stress raisers: 1. **Hole:** For a circular hole in a shaft, \( K_t \approx 2. \) 2. **Shoulder fillet:** For a diameter change with fillet, use stress concentration charts (Peterson's, etc.) Let's focus on the **hole**, which is the most critical. --- ## **Step 6: Allowable Stress for Shock Loading** For shock (suddenly applied loads), use a **factor of safety** (\( N \)), say \( N = 2 \): Allowable stress: \[ \sigma_{allow} = \frac{S_y}{N} \] \[ \sigma_{allow} = \frac{415}{2} = 207.5\, \text{MPa} \] But **this is the nominal stress**; the maximum local stress is: \[ \sigma_{max} = K_t \cdot \sigma_{nom} \] So: \[ \sigma_{nom} = \frac{\sigma_{allow}}{K_t} \] \[ \sigma_{nom} = \frac{207.5}{2} = 103.75\, \text{MPa} \] --- ## **Step 7: Maximum Allowable Force \( P \)** \[ \sigma_{nom} = \frac{P}{A_{net}} \] \[ P = \sigma_{nom} \cdot A_{net} \] \[ P = 103.75\, \text{MPa} \times 28.27\, \text{mm}^2 \] Convert units (1 MPa = 1 N/mm²): \[ P = 103.75 \times 28.27 = 2,934\, \text{N} \] --- ## **Final Answer** ### **The maximum allowable force \( P \) is approximately \(\boxed{2,930\, \text{N}}\)** **Assumptions:** - Factor of safety = 2 (can adjust if needed) - Kt for a hole = 2 (check chart for more accuracy) - Only the critical section at the hole is considered --- ## **Summary Table** | Step | Value | |---------------------|---------------| | Net area \(A_{net}\)| 28.27 mm² | | Stress concentration| \(K_t = 2\) | | Allowable stress | 207.5 MPa | | Nominal stress | 103.75 MPa | | Maximum force \(P\) | 2,930 N | --- ## **Tips** - Use stress concentration charts for more accurate Kt values. - Double-check material properties for your specific steel grade. - Adjust the factor of safety based on application. --- **If you have more details (e.g., exact loading conditions, or if you want to consider the fillet), I can refine the answer for you!**