3. Red Sea University Faculty of Engineering Department of Mechanical Engineering
Hydrodynamic Pad Bearing
Bearing Parameters Predicted from Reynolds Equation
Moataz Abdelgadir Ali Abdelgadir

4. PAD BEARINGS
Pad bearings consist of a pad sliding over a smooth surface
They are widely used in machinery to sustain thrust loads from shafts.
Examples: Thrust from the propeller shaft in a ship.
Thrust on turbines in hydropower plants.

5. Pad Bearings - Geometry
P. B infinitely long linear wedge with L/B > 3
‘L’ & ‘B’ : pad length and width parallel to the sliding direction
The bottom surface is moving in the direction shown with velocity U,
Due to motion, the lubricant is dragged into the wedge which results in pressurization of the lubricant within the wedge.
h1’ and ‘h0’ the maximum and minimum film thicknesses

6. The film thickness The film thickness ‘h’ is expressed as a function
Which can be simplified to
Where K = (h1 - h0)/h0 is often known in the literature as ”the convergence ratio”.
(1)
(2)

7. Pressure Distribution
The pressure distribution can be calculated by integrating the following Reynolds equation over the specific film geometry (already seen).
The variables ‘h’ & ‘x’ are related by the following ( by differentiation of film lubricant (2)).
(3)
(4)

8. Pressure Distribution
By substituting equation (4) into (3) and after simplifying and separating variables, we obtain:
Equation (5) can be integrated to give.
(5)
(6)

9. Pressure Distribution
We apply the B.C. taken from the bearing's inlet and outlet:
The constants C and h are :
h is the film thickness where the pressure is maximum
(7) _
(8) _

10. Pressure Distribution
Finally the pressure distribution is given by:
(9)

11. Load Capacity
The total load that a bearing will support at a specific film geometry is obtained by integrating the pressure distribution over the specific bearing area.
substituting for ‘p’, equation (9), and the variable ‘x’ has to be replaced by the other variable ‘h’ gives
(10)

12. Load Capacity
integrating of equation (1) yields.
(11)

13. Friction Force
The friction force generated in the bearing due to the shearing of the lubricant is obtained by integrating the shear stress ‘’ over the bearing area
where du/dz is obtained by differentiating the velocity equation where it is, in the bearing, considered that the bottom surface is moving while the top surface remains stationary
And the friction force is equal to
(12)
(13)
(14)

14. Friction Force
After differentiation of (13) and substitution in (12) & (14) we obtain the friction force per unit length:
Now, the friction force on the lower moving surface (ie where z = 0)
Equ (16) is integrated and gives, the friction force per unit length
(15)
(16)
(17)

15. Coefficient of Friction
By definition the coefficient of friction is expressed as a ratio of the friction and normal forces acting on the surface
substituting for F/L and W/L and simplifying already found
(18)
(19)

16. Lubricant Flow Rate
It is an important design parameter since enough lubricant must be supplied to the bearing to fully separate the surfaces by a hydrodynamic film.
If an excess of lubricant is supplied, however, then secondary frictional losses such as churning of the lubricant become significant.
This effect can ever overweigh the direct bearing frictional power loss.
Precise calculation of lubricant flow is necessary to prevent overheating of the bearing from either lack of lubricant or excessive churning.

17. Lubricant Flow Rate No side leakage (y’ direction). qy = 0
lubricant flow is obtained by integrating the flow per unit length ‘qx’ over the length of the bearing.
Where qx is
The B.C. conditions are
substituting for qx’ & ‘h’ with applying B.C., the integration gives the lubricant flow per unit length
(20)

18. Lubricant Flow Rate
From equation (20), It is clear that the lubricant flow is determined by : -
sliding speed and film geometry
but not by : -
viscosity or length in the direction of sliding.

19. Non-dimensional parameters
It is useful to find the Pad bearing parameters (pressure, friction, coefficient and others..) in terms of bearing geometry and operating parameters such as the velocity ‘U’ and lubricant viscosity ‘η’.
A convenient method of finding the controlling influence of these parameters is to introduce non-dimensional parameters.

20. Non-dimensional parameters
Equation of pressure
where the non-dimensional pressure ‘p*’ is
quick estimation of hydrodynamic pressure, for example, whether the pad material will suffer plastic deformation
(21)

21. Non-dimensional parameters
Equation of load
where the non-dimensional load ‘W*’ is
Bearing geometry can now be optimized to give maximum load capacity. (optimum value for ‘K’ is obtained : K= 1.2 i.e. h1/h0 = 2.2)
(22)

22. Non-dimensional parameters
Equation of Friction Force
where the non-dimensional friction force ‘F*’ is
The bearing geometry can be optimized to give a minimum friction force, but it is more useful to optimize the bearing to find the minimum coefficient of friction since this provides the most efficient bearing geometry for any imposed load
(23)

23. Non-dimensional parameters
Equation of Friction coefficient
where the non-dimensional friction of friction ‘*’ is
The optimum bearing geometry which gives a minimum value of coefficient of friction is K = 1.55
it is the optimum convergence ratio for a minimum coefficient of friction
(24)

24. Summary
the maximum load capacity occurs at K = 1.2 but the minimum coefficient of friction is obtained when K = 1.55.
Hence, K must be compromised between the two values, i.e. 1.2 < K < 1.55 to give the optimum performance.

25. Example
A plane bearing plate is traversed by a very wide, 150 mm long, plane inclined slipper moving at 1.5m· s−1. The clearance between slipper and bearing plate is mm at the toe and mm at the heel. If the load to be sustained by the bearing divided by the width is 500 kN ·m−1, determine the viscosity of the lubricant required, the power consumed divided by the width of the bearing, the maximum pressure in the lubricant and the position of the centre of pressure.

26. Power consumed divided by the width of the bearing, Pmax
Given
h1 = mm
h0 = mm
B = 150 mm
U = 1.5 m·s−1
W/L = 500 kN·m−1
Find ,
Power consumed divided by the width of the bearing,
Pmax
Position of Pmax _

27. Solution K = (h1 - h0)/h0 K = (0.075 - 0.025)/ 0.025=2
Using the following equ.
Substitute U with negative value
500= *6*-1.5*(0.15)2/(2*25*10-3)2*
[-ln(2+1)+2*2/(2+2)]
 = Pa.s
Friction force per unit length
F/L= -1.5* *0.15[6/(2+2)-4*ln(2+1)/2]
=392.8 N
Power=1.5* 392.8=589 W

28. Solution
The position of the centre of pressure (i.e. the point of application of this resultant force) may be determined by calculating the total moment of the distributed force over the slipper about some convenient axis and then dividing the result by the total force (Ans = 91.1 mm).
Use next figure