Let's solve the problem step by step as requested: --- ### **Step 1: Summarize Given Data** #### **Given:** - **Bearing type:** 206 ball bearing - **Loads:** - For 20% of cycle: Radial = 750 lb, Axial = 250 lb - For remaining 80%: Radial = 300 lb, Axial = 100 lb - **Speed:** 400 rpm (constant) - **Operating time:** 6 hours/day, 250 days/year - **Dynamic capacity (\(C\))**: 3360 lb --- ### **Step 2: Calculate Equivalent Radial Load \((P)\) for Each Part of Cycle** #### **Step 2.1: Find \(X\) and \(Y\) Factors for Ball Bearings** For ball bearings: - \(X\) = radial load factor - \(Y\) = thrust (axial) load factor From standard tables (typical for ball bearings, check Table 9-2 as referenced): - For \( F_a/F_r \leq e \) (where \(e\) is a given value, commonly around 0.22 for 206 bearing): - \(X = 1\), \(Y = 0\) - For \( F_a/F_r > e \): - \(X = 0.56\), \(Y = 1.63\) Let's check for each case: #### **Case 1: 20% Cycle** - \(F_r = 750\) lb, \(F_a = 250\) lb - \(F_a / F_r = 250/750 = 0.333\) #### **Case 2: 80% Cycle** - \(F_r = 300\) lb, \(F_a = 100\) lb - \(F_a / F_r = 100/300 = 0.333\) Both \( = 0.333 > e \) (assuming \(e = 0.22\)), so use: - \(X = 0.56\) - \(Y = 1.63\) --- ### **Step 3: Calculate Equivalent Radial Load (\(P\)) for Each Case** #### **General Formula:** \[ P = X F_r + Y F_a \] #### **Case 1:** \[ P_1 = 0.56 \times 750 + 1.63 \times 250 = 420 + 407.5 = 827.5 \text{ lb} \] #### **Case 2:** \[ P_2 = 0.56 \times 300 + 1.63 \times 100 = 168 + 163 = 331 \text{ lb} \] --- ### **Step 4: Calculate the Equivalent Load for the Complete Cycle (\(P_{eq}\))** Use the formula for equivalent load: \[ P_{eq} = \left[ (P_1)^p \cdot f_1 + (P_2)^p \cdot f_2 \right]^{1/p} \] where: - \(f_1 = 0.2\) (fraction of time for Case 1) - \(f_2 = 0.8\) (fraction of time for Case 2) - \(p = 3\) (for ball bearings) Calculate: \[ P_{eq} = \left[ (827.5)^3 \cdot 0.2 + (331)^3 \cdot 0.8 \right]^{1/3} \] Calculate terms: - \((827.5)^3 = 566,337,484\) - \((331)^3 = 36,241,891\) Multiply by fractions: - \(566,337,484 \times 0.2 = 113,267,497\) - \(36,241,891 \times 0.8 = 28,993,513\) Add: - \(113,267,497 + 28,993,513 = 142,261,010\) Take the cube root: \[ P_{eq} = (142,261,010)^{1/3} \approx 523 \text{ lb} \] --- ### **Step 5: Calculate Life in Revolutions and Convert to Years** #### **Life Equation:** \[ L_{10} = \left( \frac{C}{P_{eq}} \right)^p \] where: - \(C = 3360\) lb - \(P_{eq} = 523\) lb - \(p = 3\) \[ L_{10} = \left( \frac{3360}{523} \right)^3 \] \[ \frac{3360}{523} \approx 6.426 \] \[ L_{10} = 6.426^3 \approx 265.3 \text{ million revolutions} \] #### **Convert to Hours** Total revolutions per hour: \[ \text{Revolutions per hour} = 400 \text{ rpm} \times 60 = 24,000 \text{ rev/hr} \] Total hours for \(L_{10}\): \[ \text{Life (hours)} = \frac{265,300,000}{24,000} \approx 11,054 \text{ hours} \] #### **Convert to Years** Operating per year: \[ 6 \text{ hours/day} \times 250 \text{ days/year} = 1500 \text{ hours/year} \] So, \[ \text{Life (years)} = \frac{11,054}{1500} \approx 7.37 \text{ years} \] --- ## **Final Answer** ### **The rating life for the 206 bearing is approximately:** \[ \boxed{7.4 \text{ years} \text{ (rounded to 1 decimal place)}} \] --- ### **Summary Table** | Step | Value | |-----------------------|-----------------------| | Equivalent Load (\(P_{eq}\)) | 523 lb | | Life (revolutions) | 265 million rev | | Life (hours) | 11,054 hours | | Life (years) | 7.4 years |