Let's solve **Problem 11-3** step by step: --- ## **Problem Statement** An angular-contact, inner ring rotating, 02-series ball bearing is required for an application in which the life requirement is **40 kh** (thousand hours) at **520 rev/min**. - The design radial load is **725 lbf**. - The application factor is **1.4**. - The reliability goal is **0.90**. **Tasks:** 1. Find the multiple of rating life, \( a_1 \), required and the catalog rating \( C_{10} \) with which to enter Table 11-2. 2. Choose a bearing and estimate the existing reliability in service. --- ## **Step 1: Calculate Required Life in Revolutions** Life required: \[ L_{hours} = 40,000 \text{ hours} \] Speed: \[ N = 520 \text{ rev/min} \] Convert hours to millions of revolutions (\( L_{10} \), in millions of revolutions): \[ \text{Total revs required} = 40,000 \text{ hr} \times 60 \text{ min/hr} \times 520 \text{ rev/min} \] \[ = 40,000 \times 60 \times 520 = 1,248,000,000 \text{ rev} \] \[ L_{10} = \frac{1,248,000,000}{1,000,000} = 1,248 \text{ million rev} \] --- ## **Step 2: Adjust Load for Application Factor** Design radial load \( F \): \[ F = 725 \text{ lbf} \] Application factor \( A_f \): \[ F_{eq} = A_f \times F = 1.4 \times 725 = 1,015 \text{ lbf} \] --- ## **Step 3: Find Reliability Factor \( a_1 \)** Reliability goal = 0.90 From standard bearing tables (often Table 11-1 or similar): | Reliability \( R \) | \( a_1 \) | |---------------------|-----------| | 0.90 | 0.62 | ← for our case | 0.95 | 0.44 | | 0.99 | 0.10 | --- ## **Step 4: Calculate Required Catalog Rating \( C_{10} \)** The bearing life equation for ball bearings is: \[ L_{10} = \left(\frac{C_{10}}{F_{eq}}\right)^3 \times a_1 \] We rearrange for \( C_{10} \): \[ L_{10} = (C_{10}/F_{eq})^3 \times a_1 \] \[ \frac{L_{10}}{a_1} = \left(\frac{C_{10}}{F_{eq}}\right)^3 \] \[ C_{10} = F_{eq} \left( \frac{L_{10}}{a_1} \right)^{1/3} \] Plug in values: - \( L_{10} = 1,248 \) - \( a_1 = 0.62 \) - \( F_{eq} = 1,015 \) \[ C_{10} = 1,015 \left( \frac{1,248}{0.62} \right)^{1/3} \] First, calculate \( \frac{1,248}{0.62} \): \[ \frac{1,248}{0.62} = 2,013 \] Now take the cube root: \[ (2,013)^{1/3} \approx 12.63 \] So, \[ C_{10} = 1,015 \times 12.63 \approx 12,820 \text{ lbf} \] --- ## **Step 5: Choose a Bearing and Estimate Existing Reliability** - Look up a 02-series angular-contact ball bearing with a \( C_{10} \) rating at or just above 12,820 lbf in Table 11-2 (bearing catalog or textbook). - Once found, calculate the actual reliability for the selected bearing using the same life equation, but solve for reliability: \[ a_1 = \frac{L_{10}}{(C_{10}/F_{eq})^3} \] Then, match \( a_1 \) to the reliability table to estimate the reliability in service. --- ## **Summary of Answers** ### **Step-by-Step Results** 1. **Required reliability factor \( a_1 \):** 0.62 2. **Required catalog rating \( C_{10} \):** 12,820 lbf 3. **Choose a bearing:** Select a 02-series angular-contact ball bearing with \( C_{10} \geq 12,820 \) lbf. 4. **Estimate reliability in service:** Use the chosen bearing's actual \( C_{10} \) to recalculate \( a_1 \), then use the table to estimate reliability. --- ### **Final Answers** - **(a) \( a_1 = 0.62 \), \( C_{10} = 12,820 \) lbf** - **(b) Choose a bearing with \( C_{10} \geq 12,820 \) lbf from Table 11-2. Calculate its actual reliability as described above.** If you provide the actual table values, I can help you select the exact bearing and compute the exact reliability. --- **Let me know if you need help with Problem 11-4, or want to see the reliability calculation for a specific \( C_{10} \) from your table!**