Let's break down and solve the given problem step by step: --- ## **Problem Restatement** A deep groove ball bearing 601ZZ supports a rotating shaft, subjected to high-to-moderate shock loads. - **Shaft speed:** 600 rpm - **Radial load:** 210 N - **Reliability:** 90% - **Find:** Estimate the **potential life of the bearing in hours**. --- ## **Step 1: Bearing Life Formula** From the image: \[ L_{10} = \left( \frac{C}{P} \right)^3 \] Where: - \( L_{10} \) = Bearing life at 90% reliability (in millions of revolutions) - \( C \) = Dynamic load rating (N) *(to be found from bearing catalog)* - \( P \) = Equivalent dynamic bearing load (N) *(here, just the radial load = 210 N)* Life in hours: \[ L_{10h} = \frac{10^6}{60N} \cdot L_{10} \] Where: - \( N \) = Rotational speed (rpm) --- ## **Step 2: Find the Dynamic Load Rating \( C \) for 601ZZ** From standard bearing catalogs: - For **601ZZ**, \( C \approx 28,000\,\text{N} \) (If you have a different value, use that; but 28,000 N is a typical value.) --- ## **Step 3: Substitute Values** ### **Calculate \( L_{10} \) (in millions of revolutions):** \[ L_{10} = \left( \frac{C}{P} \right)^3 = \left( \frac{28,000}{2,100} \right)^3 \] Calculate \( \frac{28,000}{2,100} = 13.333 \) \[ L_{10} = (13.333)^3 = 2,370.37 \] So, the bearing can last **2,370.37 million revolutions** at 90% reliability. --- ## **Step 4: Convert to Hours** \[ L_{10h} = \frac{10^6}{60N} \cdot L_{10} \] Plug in values: - \( N = 600\, \text{rpm} \) - \( L_{10} = 2,370.37 \) \[ L_{10h} = \frac{10^6}{60 \times 600} \times 2,370.37 \] \[ L_{10h} = \frac{10^6}{36,000} \times 2,370.37 \] \[ L_{10h} = 27.78 \times 2,370.37 \] \[ L_{10h} \approx 65,830 \text{ hours} \] --- ## **Step 5: State Assumptions** - The load is purely radial. - Load is constant (no variation). - The bearing is properly lubricated and installed. - The catalog value for \( C \) is accurate. - Standard reliability factor for 90% is used. - Shock loading does **not** require an additional factor (if it did, apply \( P = X \cdot \text{Radial Load} \), where \( X \) is a factor \( >1 \)). --- ## **Step 6: Final Answer** ### **Bearing Life at 90% Reliability:** \[ \boxed{65,830 \ \text{hours}} \] --- ### **Summary of Steps** 1. **Find \( C \) from catalog**: \( 28,000 \) N. 2. **Calculate \( L_{10} \)**: \( (28,000/2,100)^3 = 2,370.37 \) million revolutions. 3. **Convert to hours**: \( 65,830 \) hours. If you have any more details (such as a higher shock load factor), let me know!