---
title: "Life of a deep groove ball bearing"
subject: "Machine Design I"
chapter: 6
key: "semester_6/machine_design_i:ball-bearing-life-check"
source: https://www.bctengineer.com/q/semester_6/machine_design_i/ball-bearing-life-check
site: BCT Engineering
---

# Life of a deep groove ball bearing

Machine Design I · Chapter 6: Rolling contact bearing

- Asked in: year not given, 6 marks
- Video: [Problem on ball bearing (rolling contact bearing) using data book](https://www.youtube.com/watch?v=GevX7x_PxWE) — Reference Book
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/machine_design_i/ball-bearing-life-check
- Whole chapter: https://www.bctengineer.com/md/semester_6/machine_design_i/chapter-6

## Question

A deep groove ball bearing has basic dynamic load rating C = 29.5 kN and static rating C0 = 15 kN. It carries a radial load of 3.2 kN and an axial load of 1.1 kN at 1450 rpm, with the inner ring rotating (V = 1). Take e = 0.26; for Fa/(V Fr) > e use X = 0.56 and Y = 1.64. The service factor for moderate shock is 1.3. Find the equivalent load, the L10 life in million revolutions and in hours.

## Short answer

Fa/(V Fr)=0.344 > e, use X=0.56, Y=1.64. P=0.56*3.2+1.64*1.1=3.596 kN; with service factor 1.3, Pe=4.675 kN. L10=(29.5/4.675)^3=251 million rev; L10h=251e6/(60*1450)=2888 h.

## Answer

#### Check the load ratio

$$
\frac{F_a}{V F_r} = \frac{1.1}{1 \times 3.2} = 0.344 > e = 0.26
$$

So $X = 0.56$ and $Y = 1.64$.

#### Equivalent load

$$
\begin{aligned} P &= X V F_r + Y F_a = 0.56 \times 1 \times 3.2 + 1.64 \times 1.1 \\ &= 1.792 + 1.804 = 3.596\ \text{kN} \end{aligned}
$$

With the service factor 1.3:

$$
P_e = 1.3 \times 3.596 = 4.675\ \text{kN}
$$

#### Life

For a ball bearing $k = 3$:

$$
L_{10} = \left(\frac{C}{P_e}\right)^3 = \left(\frac{29.5}{4.675}\right)^3 = (6.310)^3 = 251.3\ \text{million revolutions}
$$

$$
L_{10h} = \frac{251.3 \times 10^6}{60 \times 1450} = 2888\ \text{hours}
$$

The life is about 2900 hours (about 120 days of continuous running), which is short for a continuous-duty machine. A bearing of higher rating would be selected for 10000 hours or more.

**Answer: P = 3.6 kN (4.68 kN with service factor); L10 = 251 million revolutions = about 2890 hours.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Fa/(V Fr) = 0.344 vs e = 0.26, so
   A. X = 1, Y = 0
   B. X = 0.56, Y = 1.64 are used
   C. X = 0.56, Y = 0
   D. X = 1.64, Y = 0.56
   Answer: B. The ratio exceeds e so the axial load counts. (Key point: X and Y)

2. P without service factor is
   A. 4.675 kN
   B. 3.2 kN
   C. 3.596 kN
   D. 1.8 kN
   Answer: C. 0.56x3.2 + 1.64x1.1 = 3.596; with factor 1.3 it is 4.675 kN. (Key point: Equivalent load)

3. L10 in million revolutions is
   A. 63
   B. 29.5
   C. 1500
   D. 251 (since k = 3, (29.5/4.675)^3)
   Answer: D. 6.31^3 = 251. (Key point: Life)

4. L10h = 10^6 L10/(60 n) gives
   A. About 2890 hours
   B. About 289 hours
   C. About 28900 hours
   D. About 120 hours
   Answer: A. 251.3e6/(60x1450) = 2888 h. (Key point: Hours)

## Related in Machine Design I

- [Bearing life, load rating and equivalent load](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Abearing-life-load-rating) — Machine Design I, chapter 6
- [Required rating of a tapered roller bearing](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Atapered-roller-bearing-capacity) — Machine Design I, chapter 6
- [Dynamic and wear load check, spur pair](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Abuckingham-dynamic-wear-check) — Machine Design I, chapter 9
- [Journal bearing design with film thickness check](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Ajournal-bearing-minimum-film-thickness) — Machine Design I, chapter 7

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