Given:
Axial load = Fx = 2.4 kN
Radial load = Fy = 3.35 kN
Fz = 1.5 kN
Radial load = 3.35 + 1.5 = 4.85 kN
Bearing life = 10000 hours
Shaft speed = 1400 rpm
Shaft diameter = 38 mm
Reliability = 92 %
Ka = 1
Solution:
Let us assume the shaft to be of 40 mm in diameter to select the bearing from the catalogue.
To begin with, lets start with the lightest bearing.
From catalogue, d = 40 mm, D = 52 mm, B = 7 mm
Dynamic load capacity = C = 4160 N
Static load capacity = C0 = 3350 N
Therefore, (Fa/C0) = (2400 / 3350) = 0.716 and (Fa/Fr) = 0.49
Lets assume, X = 0.56 and Y = 1.5
Now, P = X.Fr + Y.Fa = (0.56 x 4850) + (1.5 x 2400) = 6316 N
We know,
L10 = (60 x N x L10h) / 106
= (60 x 1400 x 10000) / 106
L10 = 840 million revolutions
Also,
C = P (L10)1/3
C = 6316 x (840)1/3
C = 59593.91 N
From calculated figures, bearing 6408 can be selected for the given application.
Hence, according to the able, dimensions of the bearing are
Inner diameter = d = 40 mm
Outer diameter = D = 110 mm
Width of bearing = B = 27 mm
Dynamic load capacity = C = 63700 N
Static load capacity = 36500 N
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