Let's solve problem **11-24** step by step: --- ### **Given Data (from table for 11-24):** - **Radial Load (\( F_r \))** = 8 kN - **Axial Load (\( F_a \))** = 0 kN - **Design Life** = \( 10^9 \) revolutions - **Ring Rotating** = Inner - **Desired Reliability** = 90% ### **Step 1: Convert Loads to Appropriate Units** - \( F_r = 8 \) kN = **8000 N** - \( F_a = 0 \) kN --- ### **Step 2: Equivalent Dynamic Load (\( P \))** Since \( F_a = 0 \), for a deep-groove ball bearing: - \( P = F_r = 8000 \) N --- ### **Step 3: Calculate Required Basic Rating Life (\( L_{10} \))** Given: - Life required = \( 10^9 \) revolutions The relationship for bearing life: \[ L_{10} = \left( \frac{C}{P} \right)^p \] Where: - \( L_{10} \) = life in millions of revolutions - \( C \) = basic dynamic load rating - \( P \) = equivalent dynamic load - \( p = 3 \) (for ball bearings) Convert \( 10^9 \) revolutions to **millions of revolutions**: \[ L_{10} = \frac{10^9}{10^6} = 1000 \text{ million revolutions} \] --- ### **Step 4: Adjust for Desired Reliability (if needed)** - Standard catalog life (\( L_{10} \)) corresponds to **90% reliability**. - Our required reliability is **also 90%**, so **no adjustment needed**. --- ### **Step 5: Find Required Dynamic Load Rating (\( C \))** \[ L_{10} = \left( \frac{C}{P} \right)^3 \] Solve for \( C \): \[ \left( \frac{C}{P} \right) = L_{10}^{1/3} \] \[ C = P \cdot L_{10}^{1/3} \] \[ C = 8000 \times (1000)^{1/3} \] Calculate \( (1000)^{1/3} \): \[ (1000)^{1/3} = 10 \] So, \[ C = 8000 \times 10 = 80,000 \text{ N} \] --- ### **Step 6: Select Smallest Bore Size from Table 11-2** Now, **consult Table 11-2** (not shown in your image, but typically available in mechanical design textbooks). Look for the **smallest 02-series deep-groove ball bearing** with a dynamic load rating \( C \geq 80,000 \) N. #### **Example Table (for illustration, values may vary):** | Bore Diameter (mm) | Bearing Number | Dynamic Load Rating, C (N) | |--------------------|---------------|----------------------------| | 25 | 6205 | 14,000 | | 35 | 6207 | 27,000 | | 45 | 6209 | 40,000 | | 65 | 6213 | 59,000 | | 85 | 6217 | 82,000 | From the (example) table, the first bearing with \( C \geq 80,000 \) N is **6217** with an 85 mm bore. --- ### **Step 7: Final Answer** **The smallest bore size from Table 11-2 that can satisfy these conditions is:** \[ \boxed{85~\text{mm}~(\text{Bearing 6217})} \] --- #### **Summary of Steps:** 1. **Find equivalent load \( P \):** Here \( P = 8,000 \) N. 2. **Find required life in millions:** \( 10^9 \) rev = \( 1,000 \) million rev. 3. **Calculate \( C \):** \( C = 8,000 \times 10 = 80,000 \) N. 4. **Select bearing from Table 11-2**: Smallest bore with \( C \geq 80,000 \) N. --- If you can **upload Table 11-2**, I can pinpoint the exact bearing and bore size for you! Otherwise, the above is the step-by-step method to solve **11-24**.