Certainly! Let's go through the problem step-by-step. --- # Problem Breakdown **Given Data:** - Type bearing: O2 series single deep-gro ball bearing Bore diameter,d_\): 30 mm Internal ring rotates: 500utions per minuterpm) -ings: - Axial load \(F_a): 2N (200 N) Radial load, \(F_r): 5 kN (500 N) - Task: 1. Determine the equivalent radial load, \(F_e\), that will be experienced by this bearing. 2. Determine the predicted life (L10 life) based on this load. --- # Step 1: Understanding the Bearing's Equivalent Load The equivalent load for a ball bearing combining axial and radial loads is given by: \[ F_e = X F_r + Y F_a \] Where: - \(F_r\): Radial load - \(F_a\): Axial load - \(X, Y\): Factors depending on the bearing type and load conditions. --- # Step 2: Find the Factors \(X\) and \(Y\) For deep-groove ball bearings: - When the axial load is small relative to the radial load, \(X \approx 1\), \(Y \approx \). - When axial load is significant, the factors are different. In this case: \[ \frac{F_a}{F_r} = \frac{200}{500} = .4 \] Referring to standard bearing tables (like those in the bearing catalog or textbooks), for deep-groove ball bearings: | \(F_a/F_r\) | \(X\) | \(Y\) | |-------------|--------|--------| | to .2 | 1. | | | .2 to .6 | .56 | .44 | | > .6 | .33 | .67 | Since \(.4\) falls in the .2 to .6 range, we use: \[ X = .56,\quad Y = .44 \] --- # Step 3: Calculate the Equivalent Radial Load \(F_e\) \[ F_e = X F_r + Y F_a \] Plugging in the values: \[ F_e = (.56)(500\, \mathrm{N}) + (.44)(200\, \mathrm{N}) \] Calculate each term: \[ .56 \times 500 = 280\, \mathrm{N} \] \[ .44 \times 200 = 880\, \mathrm{N} \] Sum: \[ F_e = 280 + 880 = 368\, \mathrm{N} \] --- # **Final Result for \(F_e\):** \[ \boxed{ F_e = 368\, \mathrm{N} } \] --- # Step 4: Determine Bearing Life (L10 life) The basic life equation for ball bearings is: \[ L_{10} = \left( \frac{C}{F_e} \right)^3 \] Where: - \(L_{10}\): Life in millions of revolutions, - \(C\): Dynamic load rating (depends on bearing specifics; typically obtained from bearing catalogs). --- # Step 5: Find the Bearing's Dynamic Load Rating \(C\) Since \(C\) is not provided in the problem, typical values for deep-groove ball bearings are used or given in catalogs. Let's assume: \[ C = 10\, \mathrm{kN} = 10,000\, \mathrm{N} \] (You should replace this with actual data if available.) --- # Step 6: Calculate Life in Revolutions \[ L_{10} = \left( \frac{C}{F_e} \right)^3 = \left( \frac{10,000}{368} \right)^3 \] Calculate the ratio: \[ \frac{10,000}{368} \approx 2.717 \] Now cube this: \[ L_{10} \approx (2.717)^3 \approx 20.1 \text{ million revolutions} \] --- # Step 7: Convert to Hours of Operation Given rotational speed: \[ N = 500\, \mathrm{rpm} \] Revolutions per minute to revolutions per hour: \[ 500\, \mathrm{rpm} \times 60\, \mathrm{min} = 30,000\, \mathrm{rev/hr} \] Number of hours to reach \(L_{10}\): \[ \text{Hours} = \frac{L_{10} \times 10^6}{30,000} = \frac{20.1 \times 10^6}{30,000} \approx 670\, \mathrm{hours} \] --- # **Summary of Results:** - **Equivalent Radial Load, \(F_e\):** **368 N** - **Estimated L10 Life in revolutions:** approximately **20.1 million revolutions** - **Estimated operational hours at 500 rpm:** approximately **670 hours** --- **Note:** The exact bearing life depends heavily on the actual dynamic load rating \(C\), which should be obtained from the bearing manufacturer for precise calculations. --- Let me know if you'd like further clarification or assistance!