Analysis of the rolling contact bearings

Let us analyze the rolling contact bearings for strength. Initial dates are:

– type designation of the bearing and its sizes (# and d × D × B);

– rotational speed n of the bearing inner ring;

– components of reacting forces in supports R_xA, R_yA, R_xC, R_yC and R_zA;

– basic load rating C_r and static load rating C ₀ for radial ball bearings and angular-contact bearings with pressure angle α≤18º (tables 9.3 and 9.4);

– basic load rating C_r and axial load parameter e for angular- contact bearings with pressure angle α>18º (tables 9.4 and 11.1);

– basic load rating C_r, axial load parameter e and axial load factor Y for tapered roller bearings (table 9.5).

In our case for the output shaft we will use angular-contact boll bearings 36214 (70×125×24); n = 93.6 rpm; R_xA = 7491.5 N,
R_yA = 36.26 N, R_xC = 9835.5 N, R_yC = 861.74 N and R_zA = F_a = 346.7 N;
C_r =80.2 kN, C ₀ = 54.8 kN.

11.1. Determine the total radial reacting forces which act to the bearings

; .

11.2. Determine the total axial forces acting to the bearings

11.2.1. Calculate additional axial forces S ₁ and S ₂ that develop as a result of action of radial forces F_{r 1} and F_{r 2}

S ₁ = F_{r 1}× e’; S ₂ = F_{r 2}× e’,

where – e’=e for radial ball bearings and angular contact ball bearings;

e’ =0.83× e for tapered roller bearings.

It is necessary to note that for radial ball bearings and angular contact ball bearings with pressure angle α≤18º axial load parameter e is determined by table 11.1 depending upon ratio F_a/C _o.

In our case F_a/C _o = 346.7/57800 = 0.006; e’=e = 0.3 (table 11.1); S ₁ = F_{r 1}× e’= 7491.6∙0.3 = 2247.5 N; S ₂ = F_{r 2}× e’ = 9870.2∙0.3= 2961.1 N.

Fa

11.2.2. Plot the analytical model of the shaft and show all forces acting on the shaft in the axial direction (Fig. 11.1).

Fig. 11.1. Forces acting to the shaft in the axial direction.