Certainly! Let's break down and solve the problem step by step. --- # **Step-by-Step Solution: Load-Carrying Capacity of the Bearing** ## **Given Data** - **Shaft diameter, d:** 12 mm - **Shaft speed, n:** 1725 rpm - **Bearing type:** Flanged ball bearing (02-series deep groove) - **Rating life, L₁₀:** 30,000 hours - **Load:** Light-shock loading (negligible thrust load) - **Both bearings A and B are identical** - **Inner ring rotates** --- ## **Step 1: Find the Basic Dynamic Load Rating, C** The **basic dynamic load rating (C)** is the constant load that a group of identical bearings can endure for a rating life of 1,000,000 revolutions with 90% reliability. For ball bearings, the life equation is: \[ L_{10} = \left(\frac{C}{P}\right)^p \] Where: - \( L_{10} \) = basic rating life (in millions of revolutions) - \( C \) = basic dynamic load rating (N) - \( P \) = equivalent dynamic bearing load (N) - \( p \) = 3 (for ball bearings) But, our rating life is given **in hours**, so we need to convert it to **millions of revolutions**. --- ## **Step 2: Convert Life from Hours to Millions of Revolutions** \[ L_{10} \text{ (rev)} = \text{Life (hours)} \times \text{Speed (rpm)} \times 60 \] \[ L_{10} \text{ (rev)} = 30,000 \text{ hr} \times 1725 \text{ rpm} \times 60 \text{ min/hr} \] \[ = 30,000 \times 1725 \times 60 \] \[ = 3,105,000,000 \text{ revolutions} \] Convert to **millions**: \[ L_{10} (\text{million rev}) = \frac{3,105,000,000}{1,000,000} = 3,105 \text{ million rev} \] --- ## **Step 3: Find the Load-Carrying Capacity (P)** Rearrange the life equation to solve for \( P \): \[ L_{10} = \left(\frac{C}{P}\right)^p \implies P = \frac{C}{L_{10}^{1/p}} \] But we need to **find C** for a 12 mm, 02-series deep groove ball bearing. --- ## **Step 4: Find C for 6201 Ball Bearing** For a 12 mm 02-series deep groove ball bearing (also called 6201), standard catalog data gives: - **C** ≈ 7,100 N --- ## **Step 5: Calculate Allowable Equivalent Load (P)** \[ P = \frac{C}{L_{10}^{1/p}} \] \[ p = 3 \] \[ L_{10} = 3,105 \] \[ C = 7,100\, \text{N} \] \[ P = \frac{7,100}{(3,105)^{1/3}} \] Calculate \( (3,105)^{1/3} \): \[ (3,105)^{1/3} \approx 14.6 \] \[ P = \frac{7,100}{14.6} \approx 486 \, \text{N} \] --- ## **Step 6: Final Answer** **The maximum equivalent dynamic load \(P\) that the bearing can support for a life of 30,000 hours at 1725 rpm is:** \[ \boxed{486 \ \text{N}} \] --- ## **Summary Table** | Parameter | Value | |--------------------|---------------------| | Shaft Diameter | 12 mm | | Bearing Catalog C | 7,100 N | | Life | 30,000 hours | | Speed | 1725 rpm | | Calculated \(P\) | **486 N** | --- ### **Conclusion** - For the given life and speed, a 6201 (12 mm, 02-series) deep groove ball bearing can carry up to **486 N** of dynamic load. - This assumes light-shock, negligible thrust, and correct catalog C value. --- **Let me know if you have a different bearing number or need clarification on any step!**