Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
120 views
in General by (240k points)
closed by
For anti-friction bearing, relation between basic dynamic capacity, equivalent load and rating life is
1. \(\frac{C}{P} = (\frac{L}{L_{10}})^{\frac{1}{K}}\)
2. \(\frac{P}{C} = (\frac{L}{L_{10}})^{\frac{1}{K}}\)
3. \(\frac{C}{P} = (\frac{L}{L_{10}})^K\)
4. \(\frac{C}{P} = (\frac{L}{L_{10}})^{2K}\)

1 Answer

0 votes
by (238k points)
selected by
 
Best answer
Correct Answer - Option 1 : \(\frac{C}{P} = (\frac{L}{L_{10}})^{\frac{1}{K}}\)

Explanation:

Dynamic Load capacity:

  • It is the radial load at which the 90% of the group of apparently identical bearings run for 1 million revolutions before the evidence of the first crack.
  • The fatigue life of bearing at which 90% of a sufficiently large group of identical bearings operating under identical conditions fails is called rated life.

Bearing rated life in million revolutions can be calculated by:

\(\frac{L}{L_{10}} = {\left( {\frac{C}{P}} \right)^k}\)

\((\frac{L}{L_{10}})^\frac{1}{k} = {\frac{C}{P}}\)

where, L = Life of bearing in Million revolution L10 = Rated life in Million revolutions, C = Dynamic load capacity/Basic load rating, P = Equivalent dynamic load

k = 3 for ball bearing, k = \(\frac{10}{3}\) for roller bearing

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...