Thrust bearings
Thrust bearings
Support the axial thrust
Support the
axial thrust
of both horizontal as well as vertical
of both horizontal as well as vertical
shafts
shafts
Functions are to prevent
Functions are to
prevent
the shaft from
the shaft from drifting
drifting
in the axial
in the axial
direction and to
direction and to transfer thrust loads
transfer thrust loads
applied on the shaft
applied on the shaft
Vertical
Vertical
thrust bearings also need to
thrust bearings also need to support the weight
support the weight
of
of
the shaft and
the shaft and any components at
any components attached to it
tached to it
The moving surface exerted against a thrust bearing may be
The moving surface exerted against a thrust bearing may be
the
the are
are
a of t
a of t
he end
he end
of th
of th
e shaf
e shaf
t
t
or
or
the area of a
the area of a collar
collar
attached at any point to the shaft
attached at any point to the shaft
Types of thrust bearings
Plain thrust: Consists of a stationary flat bearing
surface against which the flat end of a rotating
shaft is permitted to bear
ROTOR
Bearing surface Flat end of rotor
Thrust bearing- flat land type
They handle light loads for simple positioning of rotors
They are usually used in conjunction with other types of
thrust bearings
They carry 10 to 20% of the overall axial load
Bearing surface sometimes incorporated with oil grooves that
help store and distribute oil over the surface
ROTOR
Oil grooves for storing and distributing oil over the surface
Thrust bearing- step type
Step bearing: Consists of a raised or stepped bearing surface
upon which the lower end of a vertical shaft or spindle rotates
The entire assembly is submerged in lubricant
Stepped bearings are either designed to undergo
hydrodynamic lubrication or are lubricated hydrostatically
(external pump)
ROTOR Bearing Wedge formation or pressurized oil supplyThrust bearing- hydrostatic type
These depend on an external pump to provide oil under
pressure to form a load-bearing film between surfaces
Used in equipment with extremely low speeds as a
hydrodynamic film cannot form
ROTOR
Oil under pressure, supplied by pump
Thrust bearing- collar type
Collar type
Shaft Bearing surface CollarShaft moves in axial direction too
Shaft rotates
Loads are borne by the
bearing surface that comes in contact with the collar which is attached to the shaft
Thrust bearing- tilting pad type (Michell type)
The surfaces are at an angle to each other
One surface is usually stationary while the
other moves
Undergoes hydrodynamic lubrication,
therefore formation of a wedge of lubricant
under pressure
The amount of pressure build up depends on
the speed of motion and viscosity
Thrust bearing- Tilting pad type
Shaft
Collar
Tilting pad rotates around the pivot (angle
of tilt varies) Pivot
Axial loads from machinery being driven In this case thrust from propeller
Oil wedge Direction of
rotation
•Back thrust from water to propeller causes axial loading on the shaft
• Axial loads are opposed by pressure buildup in the wedge
•Gives a damping effect
Bearing plate Propeller
Tilting thrust bearings- basic geometry
U h1 h h2 X Zh
1= distance of separation at leading edge
h
2= distance of separation at trailing edge
U = velocity of lower pad in the x – direction
B = bearing breadth
The film thickness “h” at any point is given by:
B
x
h
h
h
h
(
)
2 1 1 Leading edge Trailing edge B xHeight ratios
)
1
(
2 1 K h h
B x K K h h 1 2 3 oh
h
h
U
6
dx
dp
U h1 h h2 X Z Let or , thereforeThe expression for pressure gradient was derived earlier as
Where p is the pressure
is the coefficient of dynamic viscosity
ho is the separation distance at max. pressure
U is the velocity of the bottom surface
Top surface is stationary
2 2 1 h h h K
Making the equation non-dimensional
3 2 3 2 2)
/
1
(
)
/
1
(
6
h
B
Kx
K
h
B
Kx
K
h
U
dx
dp
3 2 2 2)
/
1
(
)
/
1
(
6
K
Kx
B
Adx
B
Kx
K
dx
dp
U
h
Let A = h
o/h
2such that h
o= Ah
2Substituting this and the value of h in terms of x we get
On rearranging we get:
Let x* = x/B, a dimensionless length, so that
* 3 * 2 * * 2 2
)
1
(
)
1
(
6
K
Kx
dx
A
Kx
K
dx
dp
U
h
Pressure distribution equation
)
12
...(
dx
)
Kx
K
1
(
A
)
Kx
K
1
(
dx
dp
* 2 * 3 * * * Now h
22/U
B has the dimensions of (pressure)
-1so it is possible to
write (h
22/6U
B)p as p*, the non-dimensional pressure. The equation
therefore becomes
This is Reynold’s equation in non-dimensional form applied to
inclined pads. Integration gives the pressure distribution. On
integration we get:
)
13
...(
C
)
Kx
K
1
(
A
)
Kx
K
1
(
dx
p
* * 2 * * Applying boundary conditions
C
)
Kx
K
1
(
A
)
Kx
K
1
(
dx
p
* * 2 * * A and C are constants of integration. In order to evaluate them the value of pressure is required at two specific positions. This, in the case of a pad, is taken as the ambient pressure at the leading and trailing edges, where the pressure curve starts and stops. These pressures are usually considered as zero. Therefore the conditions are:
p = 0 at x = 0, and x = B
Non-dimensionalizing we get, p* = 0 at x* = 0 and x* = 1 (since x* = x/B) First putting p* = 0 at x* = 0, we get:
C
)
K
1
(
K
2
A
)
K
1
(
K
1
0
2 Obtaining the constants of integration
C
K
2
A
K
1
0
K
2
)
K
1
(
2
A
)
K
2
(
K
1
C
Then putting p* = 0 at x* = 1, we get:
The above two equations can be solved to give: and Thus:
2 * * *)
Kx
K
1
(
K
1
)
Kx
K
1
(
K
2
1
)
K
2
(
K
1
p
Which can be simplified to give:
)
14
...(
)
K
2
(
)
Kx
K
1
(
)
x
1
(
Kx
p
* 2 * * * Maximum pressure
)
2
(
)
1
(
2
2K
K
A
h
h
o * o ox
K
2
K
1
B
x
The max. dimensionless pressure p
o* occurs when dp/dx =
0, h = h
o, and x = x
o.Now,
Therefore
and
4
(
1
K
)(
2
K
)
K
p
o* Integration of the pressure across the bearing gives the load carried per unit length, W/L
So which can be defined as the
non-dimensional load W*. Thus Which reduces to
Load carried
* B 0 1 0 * 2 ' o 2dx
p
h
B
U
6
pdx
L
W
1 0 * * 2 2 2 6 .dx
p
L
B
U
h
W
K
2
2
K
)
K
1
(
log
K
1
W
* e 1 0 * * 2 * e 2 *x
)
K
2
(
K
)
K
1
(
)
Kx
K
1
)(
K
2
(
K
)
Kx
K
1
(
log
K
K
1
W
(as x* = x/B)Tilting pad bearing- expression for load
2 2 2 * . 6 / B h U L W W L
W
U
W
B
h
/
)
6
(
* 2 2
Now
Therefore
)
14
...(
/
2
2
)
1
(
log
6
1/2 2 L W U K K K K B h e
This equation was first derived by Reynold’s for a fixed
inclined surface
Height variation with pivot point
The ratio h
1/h
2= (1+K) is determined by the position of the
pivot point
Velocity U h1 h h2 X Z Pivot pointUpper pad rotates around the pivot point
•The position of the pivot point is found by taking moments about the
leading edge.
•For stability it should be at the center of pressure x
