Trapezoidal or truncated conical pivot bearing

TRAPEZOIDAL OR TRUNCATED CONICAL PIVOT BEARING

Consider a truncated conical pivot bearing as shown in Fig.6.22. By using the similar procedure used for conical pivot bearing, the torque required to overcome the friction for truncated conical pivot bearing may be obtained as below.

(i) For uniform pressure: 

(ii) For uniform wear:

1. Summary

Table 6.1 summarises the torque required to overcome the friction by the bearing surface for various types of bearings.

Table 6.1. Formulae summary

Example 6.17 

In a thrust bearing, the external and internal diameters of the contacting surfages are 320 mm and 200 mm respectively. The total axial load is 80 kN and the intensity of pressure is 350 kN/m2. The shaft rotates at 400 rpm. Taking the coefficient of friction as 0.06, calculate the power lost in overcoming the friction and the number of collars required.

Given data: 

d1 = 320 mm or r1 = 160 mm = 0.16 m; d2 = 200 mm or r2 = 100 mm = 0.1 m; W = 80 kN; p = 350 kN/m2 = 350 × 103 N/m2; N = 400 rpm; μ = 0.06.

Solution: ω = 2πN/60 = 2π (400)/60 = 41.89 rad/s

Power lost in overcoming friction (P):

Since the power lost in overcoming the friction is to be determined and also assumption is not mentioned, therefore for safer design uniform pressure theory may be assumed.

We know that the total frictional torque transmitted for uniform pressure,

Number of collars required (n):

We know that the intensity of pressure,

Example 6.18

The thrust of a propeller shaft in marine engine is taken up by a number of collars integral with the shaft which is 300 mm in diameter. The thrust on the shaft is 200 kN and the speed is 75 rpm. Taking μ = 0.05 and assuming intensity of pressure as uniform and equal to 0.3 N/mm2, find the external diameter of a the collars and the number of collars required, if the power lost in friction is not to exceed 16 kW. 

[A.U., Nov/Dec 2011] 

Given Data: 

d2 = 300 mm or r2 = 150 mm = 0.15 m; W = 200 kN = 200 × 103 N;

N = 75 rpm; μ = 0.05; p = 0.3 N/mm2 = 0.3 × 106 N/m2; P = 16 kW = 16 × 103 W.

Solution:

External diameter of the collar (d1):

We know that the torque transmitted, for uniform pressure,

On simplification, we get

(r1)2 - 0.155 r1 - 0.0233 = 0

Solving this quadratic equation, we get

r1 = 0.2485 m or -0.0935 m

... (neglecting the negative value and considering the positive value) 

d1 = 2r1 = 2 × 0.2485 = 0.497 m or 497 mm Ans.

Number of collars (n):

We know that intensity of pressure,