HYDROSTATIC LUBRICATION

Circular Footstep Bearing

The lubricant(oil) is pumped at a high pressure to the pocket (circular recess). Oil flows radially outward through the narrow gap. The pressure distribution (which varies from pumping pressure at the recess to atmospheric pressure) on the shaft lift the load W.

Preliminary Theory (Slot Leakage formula)

Consider steady viscous flow between very close parallel plates, under the pressure gradient dp/dx. It can be shown that the velocity and shear stress distribution are given by u = (1/2)(-dp/dx)[(t/2)2 - y2] and = (dp/dx)y

The discharge per unit width q = (t3/12)(-dp/dx). This is known as slot leakage formula.

This formula can be used to analyse the footstep bearing. Consider a shaft of outer radius r2 located over a plane pad with a circular recess of radius r1. The chamber is supplied with oil at pressure p0. The radial flow rate through the section at radius r is

Q = q(2pir) = (pit3/6)(-dp/dx) r.

For the continuity Q should be a constant (ie. independent of r). Then by integrating this equation, the pressure distribution can be obtained, with the boundary condition at r = r2 p = 0, as

p = (6Q/pit3)loge[r2/r] ; r1 r r2 .

At r = r1 , p = p0. Therefore p0 = (6Q/pit3)loge[r2/r1]

Then p = p0loge[r2/r]/loge[r2/r1]

The load carrying capacity W =pdA =p(2pirdr) = pip0(r22 - r12)/{2loge[r2/r1]} = (3Q/t3)(r22 - r12).

HYDRODYNAMIC LUBRICATION

It is known that a journal bearing operates efficiently when it is brought up to a certain speed. This is due to hydrodynamic action. A film of oil which completely separates the surfaces is maintained by relative motion of the two surfaces. The thrust bearing also operates on the same principle.

(a) Thrust Bearings

Tilted Pad Bearing

The simplest trust bearing consists of a slider and a bearing pad, which is slightly inclined to it.

The flow rate per unit width across the section x-x is Q = Ut/2 + (t3/12)(-dp/dx) .

Since p = 0 at x = b/2 , there must be a point - b/2 < x < b/2 where the pressure is maximum (ie. dp/dx = 0). Let at t = t0 , x = x0 and dp/dx = 0. Then Q = Ut0/2 and for the continuity Ut0/2=Ut/2+(t3/12)(-dp/dx).

Load W

Flow In at p = p0

Flow Out at p = 0

Flow Out

Circular Recess (p = p0)

p0

Q r2

r1

p0

r1 r2

Slider

Bearing Pad

Therefore dp/dx = 6U(t - t0)/t3. This is known as Reynolds Equation in one dimension.

Now using the relation (h - t)/x = 2e/b and the boundary conditions at t = h e, p = 0; the above equation can be integrated to show that t0 = (h2 - e2)/h , x0 = eb/2h ,

The load carrying capacity per unit width of the pad is

The frictional force acting on the slider per unit width (drag force) is given by

If / is the virtual coefficient of friction for the slider then

The frictional force acting on the bearing pad F0 = Fcos - Wsin ; where = tan-1[2e/b] is the slope of the pad.

If / = tan; F0 = Wcos[tan - tan].

Therefore it is clear that when > F0 becomes negative. That is the direction of flow of fluid near the pad is reversed. In order to prevent this situation . Therefore the critical slope *=. By using the expressions for tan=/ and tan, it can be shown that at the critical slope e/h = 0.8586. Thus for a given h and b 1.7172(h/b).

Rayleigh Step Bearing

The flow rate per unit width acrossany cross section is given by

Q = Ut/2 + (t3/12)(-dp/dx).

Then for the continuity Q = Ut1/2 + (t13/12)(-dp/dx)1 = Ut2/2 + (t23/12)(-dp/dx)2 = constant.

Therefore (-dp/dx)1 = constant and (-dp/dx)2 = constant. Let the maximum pressure is po. Then (dp/dx)1 = po/b1 and (dp/dx)2 = -po/b1. Substitution of these expressions in the expressions for the flow rate leads to

po = 6Ub1b2(t1 - t2)/(b2t13 + b1t23).

The load carrying capacity per unit width of the pad is

The frictional force acting on the slider per unit width (drag force) is given by

Condition for the maximum load:

Let b1+b2 = b = constant and x = t1/t2 and y = b1/b2. Then W = [3Ub2/t12]y(x - 1)/[(1+y)(y+x3)]. For a given t2 , W is maximum when W/y = 0 and W/x = 0. These two conditions simplify to x3 = y2 and y = (3-2x)x2. And the solutions are x=1.866; y=2.549. Then Wmax = 0.206U(b/t2)2.

( )[ ] and xehb

xb2hUbe3

= the2he

Ub3 = p

2222

2222

44 [ ] ( ) . e - h2h

Ube3 = p = p 22t = tmax 0

.

h2e

-

e-he+hlog

e2Ub3

= dx p = We2

2b/2b/2-

.

2h3e

-

e-he+h

e

Ub2 = dx

dydu

= dx = Fe

b/2b/2-

b/2b/2-

log

.

2loglog34

'

he

-

e-he+h

2h3e

-

e-he+h

be

=

WF

= ee

Slider U

Bearing Pad

( )tb + tb

)t - t(bb)b + bU(3 = t+tp2

1 = dx p = W 3

21312

212121210

)b+b(0

21

( ).

tb + tbt - tbb3

+ t

b +

t

bU = dx dxdp

2t

+ t

U = dx = F 3

21312

212

21

2

2

1

1)b+b(0

)b+b(0

2121

Journal Bearings

Journal bearing is the most common type of hydrodynamic bearing in use. It consists of a circular shaft (or Journal) rotating inside a circular bush. If the journal and the bush are not co-axial, the bearing can carry a transverse load.

Eccentricity Ratio = e/h In general 0.4 < < 0.8 Note: Small may result in instabilities of the bearing operation. Large may result in some degree of metal to metal contact (surface irregularity problems). Clearance ratio = h/r = (R - r)/r In general 110-3