---
title: "Required rating of a tapered roller bearing"
subject: "Machine Design I"
chapter: 6
key: "semester_6/machine_design_i:tapered-roller-bearing-capacity"
source: https://www.bctengineer.com/q/semester_6/machine_design_i/tapered-roller-bearing-capacity
site: BCT Engineering
---

# Required rating of a tapered roller bearing

Machine Design I · Chapter 6: Rolling contact bearing

- Asked in: year not given, 6 marks
- Video: [BEARINGS BASICS and Bearing Life for Mechanical Design in 10 Minutes!](https://www.youtube.com/watch?v=aU4CVZo3wgk) — Less Boring Lectures
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/machine_design_i/tapered-roller-bearing-capacity
- Whole chapter: https://www.bctengineer.com/md/semester_6/machine_design_i/chapter-6

## Question

A tapered roller bearing on a shaft running at 600 rpm (inner ring rotating) carries a radial load of 5 kN and a thrust load of 2 kN. The required L10 life is 10000 hours. Take e = 0.35, and for Fa/(V Fr) > e use X = 0.4 and Y = 1.7; use a service factor of 1.2. Find the dynamic load rating required and state how the bearing is then chosen from a catalogue.

## Short answer

Fa/(V Fr)=0.4 > e, using X=0.4, Y=1.7. P=0.4*5+1.7*2=5.4 kN; with service factor 1.2, Pe=6.48 kN. L10=(60*600*10000)/10^6=360 million rev. Required dynamic capacity C=Pe L10^(3/10)=6.48*360^0.3=37.9 kN.

## Answer

#### Equivalent load

$$
\frac{F_a}{V F_r} = \frac{2}{1 \times 5} = 0.40 > e = 0.35
$$

So $X = 0.4$ and $Y = 1.7$.

$$
\begin{aligned} P &= X V F_r + Y F_a = 0.4 \times 5 + 1.7 \times 2 = 5.4\ \text{kN} \\ P_e &= 1.2 \times 5.4 = 6.48\ \text{kN} \end{aligned}
$$

#### Required life in million revolutions

$$
L_{10} = \frac{60\, n\, L_{10h}}{10^6} = \frac{60 \times 600 \times 10\,000}{10^6} = 360\ \text{million revolutions}
$$

#### Dynamic rating

For a roller bearing the exponent is $k = 10/3$:

$$
C = P_e\, L_{10}^{1/k} = 6.48 \times 360^{0.3} = 6.48 \times 5.846 = 37.9\ \text{kN}
$$

#### Choosing the bearing

From the manufacturer's catalogue, select a tapered roller bearing whose bore equals the shaft diameter and whose basic dynamic rating is not less than 37.9 kN. Then check that:

- the static rating $C_0$ exceeds the maximum static load;
- the limiting speed is above 600 rpm;
- the $X$ and $Y$ values and $e$ of the selected bearing match those assumed (if they differ, recompute $P$ and $C$).

Tapered bearings are normally mounted in opposed pairs, and each radial load induces an axial thrust that must be added to the external thrust when the loads are split between the two bearings.

**Answer: Required C = 37.9 kN (P = 5.4 kN, 6.48 kN with service factor).**

## Check yourself

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

1. Required dynamic rating uses exponent
   A. k = 3
   B. k = 10/3 for roller bearings
   C. k = 2
   D. k = 1/3
   Answer: B. C = Pe L10^(1/k). (Key point: Exponent)

2. Required L10 for 10000 h at 600 rpm is
   A. 36 million
   B. 3600 million
   C. 360 million revolutions
   D. 6 million
   Answer: C. 60x600x10000/10^6 = 360. (Key point: Life in revolutions)

3. Pe with X=0.4, Y=1.7, service factor 1.2 is
   A. 5.4 kN
   B. 8.8 kN
   C. 7.0 kN
   D. 6.48 kN
   Answer: D. P = 5.4 kN, times 1.2 = 6.48 kN. (Key point: Equivalent load)

4. Choose a bearing whose
   A. Bore equals shaft diameter and rating C is not less than 37.9 kN
   B. Rating is less than 37.9 kN
   C. Bore is any size
   D. Static rating alone is checked
   Answer: A. Also check C0, limiting speed and assumed X, Y. (Key point: Catalogue choice)

## Related in Machine Design I

- [Life of a deep groove ball bearing](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Aball-bearing-life-check) — Machine Design I, chapter 6
- [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

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