Let's break down and solve the problem step by step. --- ## **Problem Restatement** **Given:** - 03-series cylindrical roller bearing, inner ring rotating. - Radial load \( F = 2235 \) lbf. - Need to find catalog rating \( C_{10} \) to enter Table 11-3. - Reliability goal: \( R = 0.90 \). - Choose a bearing and estimate its reliability in use. --- ## **Step 1: Understand Key Terms** ### **Catalog Rating \( C_{10} \)** - \( C_{10} \): Basic dynamic load rating (lbf). - It's the load a bearing can endure for 1 million revolutions with 90% reliability. ### **Reliability Goal** - Reliability \( R \): Probability the bearing will last a certain number of revolutions/hours at a stated load. --- ## **Step 2: Life Equation for Roller Bearings** The life equation for roller bearings is: \[ L_{10} = \left( \frac{C_{10}}{F} \right)^{a} \] Where: - \( L_{10} \) = Life (in millions of revolutions) at 90% reliability - \( C_{10} \) = Basic dynamic load rating (lbf) - \( F \) = Applied radial load (lbf) - \( a \) = 10/3 (for roller bearings) --- ## **Step 3: Adjust for Reliability** To **adjust for different reliabilities**, use: \[ L_{R} = L_{10} \left[ \frac{ \ln(1/R)}{\ln(0.9)} \right] \] Where: - \( L_{R} \) = Life for reliability \( R \) - \( L_{10} \) = Life at 90% reliability But since our reliability goal is 0.90, \( L_R = L_{10} \). --- ## **Step 4: Find \( C_{10} \) for the Given Load** We need to **find \( C_{10} \)** for the given load and desired life (use Table 11-3 for life in hours or revolutions). But the problem statement does **not specify** the required life (in hours or revolutions). If this is a follow-up from Problem 11-3, you need the life value from that problem. If not, let's assume a standard life of 1 million revolutions (as is typical for catalog ratings). So, \[ L_{10} = 1 \text{ (million rev)} \] Plug values into the life equation and solve for \( C_{10} \): \[ L_{10} = \left( \frac{C_{10}}{F} \right)^{a} \] \[ 1 = \left( \frac{C_{10}}{2235} \right)^{10/3} \] Take both sides to the power of \( 3/10 \): \[ 1^{3/10} = \frac{C_{10}}{2235} \] \[ C_{10} = 2235 \text{ lbf} \] --- ## **Step 5: Choose a Bearing and Estimate Reliability** Now, **check Table 11-3** for a cylindrical roller bearing, 03-series, with \( C_{10} \geq 2235 \) lbf. - Select the bearing with a \( C_{10} \) value just above 2235 lbf. - The reliability at \( L_{10} \) is **90%** (by definition). If you use a bearing with higher \( C_{10} \), reliability at the same load and life will be higher. --- ## **Step 6: Estimate Reliability for a Specific Bearing** If you select a bearing with a catalog rating \( C_{10, \text{catalog}} \): \[ L_{10} = \left( \frac{C_{10, \text{catalog}}}{2235} \right)^{10/3} \] If your actual application life \( L_{\text{app}} \) is given (from 11-3 or another source), use: \[ L_{10} = L_{\text{app}} \] \[ R = 1 - e^{\left( \frac{L_{10}}{L_{\text{app}}} \ln(0.1) \right)} \] But with only this data, reliability at \( C_{10} \) and 1 million rev is 90%. --- ## **Final Answers (Summary Table)** | Step | Value/Equation | Result (for your data) | |---------------------------------|--------------------------------------|--------------------------| | Load (\( F \)) | — | 2235 lbf | | Required Life (\( L_{10} \)) | — | 1 million rev | | Catalog Rating (\( C_{10} \)) | \( C_{10} = F \) | **2235 lbf** | | Reliability at \( C_{10} \) | (by definition) | **0.90 (90%)** | | Bearing Selection | Table 11-3, \( C_{10} \geq 2235 \) | — | --- ## **In Summary** 1. **Required catalog rating \( C_{10} \) is 2235 lbf** for 1 million revolutions at 90% reliability. 2. **Choose a bearing** from Table 11-3 with \( C_{10} \geq 2235 \) lbf. 3. **Estimated reliability in use:** 90% at 1 million revolutions and 2235 lbf load. If you have more data from Problem 11-3 (like actual hours or rpm), plug that in for more details! Otherwise, this is the general solution. --- Let me know if you have more info (like rpm or life in hours), and I can help further!