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Section 5 design of plane hydrostatic bearings
Section
s
Design of Plane Hydrostatic
USE OF THE DESIGN CHARTS The designer proceeds from one instruction to the next taking decisions or doing arithmetic as instructed. If at any stage he is dissatisfied with a result, for example, if the flowrate is inconveniently high, he is free to adjust any of his earlier choices and if he repeats the procedure from the point of adjustment he will obtain an optimized bearing for his new conditions. Thus within two to three hours the designer can compare a number of designs, each of which is optimized for the conditions laid down by his basic decisions. The use of each design chart is illustrated by an example. After the designer has specified his bearing-pad dimensions, he proceeds to the design of his restrictor or compensating device. The following notes discuss the points to be considered in the selection of the various parameters and give some insight into the theory and assumptions on which the procedures are based. The procedures also include the theory discussed in Section 8 on minimisation of power and temperature rise. The designer’s initial concern is to ensure the following conditions: The flow rate q should not be too great. The bearing should have sufficient stiffness. The bearing should be capable of withstanding the applied loads. That is to say for a given minimum film thickness the bearing thrust W should be greater than the applied load. This section contains simple methods for determining flow rate q, stiffness X and bearing thrust W. At higher bearing speeds it is often necessary to optimize the pad design for minimum power at the speed of operation. The designer uses the calculation procedures laid out in this section, together with the optimization procedures discussed in Section 8. The design of plane pad bearings will be illustrated for three common pad shapes. These are Circular thrust bearing Rectangular pads Thin-land bearing There are a number of other bearing shapes for which the effective area and the flow shape factor are not quite so easily determined. These all come under the category of thick-land bearings and it is necessary to use a computer to obtain accurate data for effective area and flow shape factor. This data has already been presented in Section 2. It is important to note that the effective areas and flow shape factors quoted in Section 2 should be used with the design procedures for flow rate q, stiffness A, and bearing thrust W.
Bearings pr = Recess pressure h = Bearing clearance n = Dynamic viscosity is = A dimensionless factor called the flow rate factor which is related to the shape of the particular bearing It is often convenient to use the equation condition h = ho. Equation 4b may then be written e&o3 Q
qo=y
related
to the design
(15)
0
where Go = 625 where 13 =pspro Flow rate from a circular thrust bearing The flow shape factor for a circular plane bearing by the expression
is given
B=
l7 (16) log, R ( RO ) not however necessary to calculate B from Equation 16 B is plotted on Fig 4b. In certain designs it is necesto resort to annular recess bearings, as for example a shaft extends through the thrust bearing. 6
It is since sary when
The circular plane bearing with annular oil feed (Fig 71is easily analysed by consideration of the previous case: B can be read off from Figs 7b and 7c and inserted into Equation 16 without the need for extra arithmetic. Flow rate from a rectangular plane bearing with radiused recess corners The expression for flow through a pad having the geometry shown in Fig 27 is plotted on Fig 28. It was derived by
P"1
CHOICE OF LAND WIDTH For very low speeds there is an optimum land width for bearing pads. This optimum is based on the minimum power needed to support a given load on a given total area and corresponds to a minimum of I% At higher speeds another term for friction must be taken into account. However the land width need not be reduced (See Section 8). For a circular thrust bearing at zero speed the optimum radius ratio is 1.89. For various other shapes the optimum is obtained when the recess perimeter is between 0.4 and 0. 8 of the bearing perimeter, as is indicated by the recommended design ranges on Figs 4-12.
W = PS A e.E
FLOW RATE In general the flow rate q from a bearing Equation 4b ghs q=PrT where q = Flow rate
is given by
Fig 27
Plane rectangular corners
pad with radiussed
TRIBOLOGY
February
recess
1969
41
b/a
‘I
c .l
0.9 d’3 0.1
.2 ‘.3
Fig 28b Leakage coefficients for planerectangular pad with radiussed recess corners - d/a = 0.1 --- d/a = 0.4
I.5
Ia Any shape constant
Ae=1/2
[Total area
+ recess
area]
0=
[Total
- recess
area]
area
with land
thin width
12 12
o.4--J---J 1
1.5
2 bla
3
Fig 28a Load coefficients for plane rectangular radiussed recess corners
Fig 29 5
10
00
pad with
Loxham, who assumed that the corners have a resistance equal to one quarter the resistance of a circular pad having the same radii, and that each of the sides could be treated as uniform flow across a slot. The flow coefficient g is given by B = (a + b + 4dl loge?+)6~
CII (1-U
loge 9
Tbinlimdbearingpads
Thin-land bearing pads (Fig 29) are used for two purposes. 1 To obtain the maximum effective area within a given total area. This is often the case where there are limi42
TRIBOLOGY
February
1969
Thin-land
bearing
tations to the space available and flow rate is of secondary importance. At higher speeds to reduce the flow resistance and hence 2 allow oil of higher viscosity than could otherwise be used. The flow shape factor and effective area for thin-land bearings are easily calculated, whatever the shape, The flow shape factor is given by B = Total area - recess area (181 12ta = land area 12ta where t = land width
BEARING THRUST AND STIFFNESS FOR SINGLE PLANE BEARINGS If the applied load W is known, the recess pressure pr 1.5 determined. Conversely if the recess pressure is known the
bearing
thrust
is determined
1
since
W = pr . A, where A, = A.Circular Central
Load at the design condition
pr = pp, = 0.5 x 600 = 3001bf/ina
W = PrAe A, = ?#
Plane Bearings admission
From
= 34sina
Fig 4 and choosing
z = 2.0, K is 0.542
(19) A = 2 Annular oil feed Effective area = A, = A,, outer - A,, inner In Equation 19 A,, outer is the effective area from Fig 4 using R, and R, for the outer land and A,, inner is the effective area from Fig 4 using R, and R, for the inner land. These values have been calculated and are given in Fig ‘7b in Section 2. Rectangular plane bearing Effective area =
with radiused
( (a + b - 4d)c + ab - 4d2 + w\ 2
\
recess
corners.
R, = $$! R
= 2
= 6. 14in2 = nR,a = 1.4in
7 0. 7in
The design gap ho = 0.002in siderations. From Fig 4, B = 0.75 PB qo +, = 600 X (0.002j3
(20)
log, +$-
= #
will be ensured
X 0.5 X 0.75
=o
277.4ina is 1 gallon so the flowrate q. = 0.36 x i77 a - O.O78gal/min
area)
(21)
The relationship between bearing thrust W or recess pressure pr and bearing clearance h is dependent on bearing stiffness and therefore on the control device employed. This was explained in Section 3. To simplify and facilitate the designer’s work, bearing thrust, stiffness, and bearing gap are presented non-dimensionally for single plane bearings in Figs 30-38. These figures cover capillary, orifice and constant-flow compensation using the equations in Table 2 The equation2 are considerably simplified at the design condition when X = 1 and pr = p
36in3,sec
5x10-6
These values are shown on Fig 28b Thin-land bearings of any shape (Fig 29) Effective area A, = l/2 (Total area + recess
from flow con-
2
is therefore
At this point it may be realised that the flow rate is unacceptable. For example the flow rate may be too high for the pump which it is desired to use. There are a number of ways round this, such as designing for thicker lands, or increasing the bearing size and reducing the supply pressure in proportion. Alternatively if the bearing area is increased it is possible to reduce p to about 0.2 minimum which has the advantage that a higher maximum load may be applied. However the most sensitive parameter controlling flow is the gap, because it appears to the third power. From Fig 31, the dimensionless stiffness h is 0.75 for p=o.5
The actual Table 2 Thrust, stiffness and gap equations for single plane bearings (W = 6,) Capillary
Orifice
Constant
~J3%.X=§O
pc7,-- l-
Pr P?
82 1-P
3
Diaphragm
Pr=
P 3 N.B.& ? 1
o xo3+O$$
O* 75 = 750 OOOlbf/in
When x = 0.4, h = 0.0020 x 0.4 = 0.0008in From ‘Fig 30, the dimensionless thrust W is 0.94 The maximum applied load is therefore W’ = psAe . W’ W = 600 x 3.33 x 0.94 = 18801bf Comparing W at z = 0.4 with the design load it is seen that the maximum applied load is less than twice the
&3x2.1-fl pi3 + Jq2 1-3 Xf = 3&(1 - p,,
is therefore
hO
flOW
1 Pr = 1 + LA.ftJ I!
stiffness
six = 6&
1 - f& 2 - Pr
xx
= 3p,
Example 5.1 A circular capillary-compensated plane bearing with central oil admission must support a load of 10001bf. The bearing gap is to be 0.002in and the bearing is to operate at zero speed. Determine 1 Suitable dimensions for the bearing pad 2 The operating stiffness 3
The load the bearing
will support
when z is reduced
to
0.4
The load when It has been decided sure for the pump absolute viscosity 4
h = 0.0004in. that p is to be 0.5 and a convenient presis 6001bf/ina. The oil to be used has an of q = 5 x 10-a reyns at 100°F.
0
0.5
1.0
1.5
Bearing
gap,
2.0
2.5
3.
X L h0
Fig 30
Capillary-compensated
single plane bearing TRIBOLOGY
February
1969
43
design load and the dimensionless stiffness is reduced 0.42 compared with the design value of 0.75 Taking h = 0.0004in as the minimum allowable film thickness
4
,-=!?L!!Eooo4 0.2 0.0020 W’ = 600 x 3.33 x 0.99 = 19601bf However the dimensionless stiffness is now reduced to 0.1. Apparently the minimum allowable film thickness is not a good criterion for deciding the maximum allowable load. A better approach is to consider the minimum z. For 6 = 0.5 a suitable range for X is fro-m 0.4 to 1. 6. If however 6 = 0.2, a suitable range for X is from 0.2 to 1. I
to
Ps
,p=0.7
W =
G ; 0.6 P ‘C & 0.4
m
3 Stiffness
Thrust PsAe*m
\\\\\wp=O+
I$ 0.8 13
1.5 gap
1.0 Bearing
x=
.2,0 k.
2.5
X-
Fig 33
Orifice-controlled
single
plane bearing
2.0
1.5
Bearing
gap,
IA1
10
E c.Gi
0.5
F= h/ hO
Fig 31
Capillary-compensated
0
single plane bearing
0
0.5
1.0 Bearing
1.5 gap X= $
2.0 0
Fig 34
-1
Orifice-controlled
single
plane bearing
p=o.2
p -0.3
p = 0.4
/\
0.41
1.5
A3
IX t iii E iz
I//////
1
0.5 Stiffness
0.2 1
h = P
0 0.4 Bearing
Fig 32 44
Capillary-compensated TFUBOLOGY
February
0.6 thrust
0.8
0
02
k=cr
single plane bearing 1969
0.4 Bearing
Fig 35
Orifice-controlled
thrust
0.6 $;l;E
single plane bearing
0.8
1.0
0
1.0
0.5
Bearing
2.0
1.5 gap
2.5
3.0
ii = + 0
Pm= maximum tlow can
Fig 36
pressure at be maintained
Constant-flow-controlled
which
constant
single plane bearing 0.8
-Bearing
5 3C.I k h
4
Fig 38
Constant-flow-controlled
&
Pm*, -.
A= X
h0
DOUBLE PLANE BEARINGS HAVING EQUAL OPPOSED PADS The design of bearing systems having equal opposed-pad bearings is simpler in practice than for single plane bearings. The data are given in Figs 39-47. The effective area of a double plane bearing is twice the effective area of a single plane bearing, that is Ae = A,,,
1.0
1.5 gapTi
single plane bearing
-ih
Stifness
Bearing
+ A,,2
where 1 and 2 are subscripts for the individual pads. The supported load W and the stiffness X are given by the following expressions
20 =+ 0
Pm= maximum flow can
Fig 37
pressure at be maintained
Constant-flow-controlled
1.0
W=p,
9
0.3
-0.4
thrust
which
single
(22) constant
plane bearing
CHOICE OF DESIGN PARAMETER B = 2 FOR SINGLE PLANE BEARINGS At the design condition the maximum stiffness for plane bearings is obtained when fl = 0.5. If @ is increased, both the stiffness and the load range is reduced, and so values greater than 0.5 are to be avoided. If B is reduced the non-dimensional design stiffness is again reduced. However if the same applied load is resisted by increasing the bearing area the actual stiffness is increased and so is the load range. The stiffness is much less uniform and will drop more rapidly if the load is reduced. The preferred range for 0 is therefore from 0.2 to 0.5. In certain cases p may be increased up to 0.75. To obtain a greater load range there are two possibilities. 1 Increase the bearing area by 2V2 times and decrease the design parameter to 0.2 2 Use opposed-pad bearings and greatly increase the bearing area The latter method has the advantage that stiffness will be much higher and more nearly uniform.
5 = l/2
(& + X2)
where Pr.1 and Pr, 2, ,il and x, are obtained plane data and using !?, = 2 -x1
from the single
CHOICE OF DESIGN PARAMETER P FOR DOUBLE PLANE BEARINGS The optimum @ is 0.5 and in most designs it is not necessary to chose any other value for double plane bearings. However in certain cases where the flow rate is rather greater than desirable, it is pssible to reduce the flow rate by reducing fl towards 0.2 With pressure-sensing control devices the maximum allowable load will not be lower than for capillary compensation and therefore the same value may be used. Example 5.2 A capillary-compensated double plane bearing is required which will not deflect more than 0. OOOlin under an applied load of 1001bf. The available supply pressure is 5001bf/inz and in the design condition there is negligible load. The design bearing gap is to be ho, = ho, = O.OOlOin. Determine I Suitable bearing dimensions for circular bearing pads 2 The maximum allowable applied The pressure ratio fl is to be 0.5 TRIBOLOGY
load
February
1969
45
Bearing dimensions. 1 Stiffness must be at least && From Fig 40, x0 = 0.75 * Actual stiffness is given by
A0 = p
The maximum allowable W’ = psA,. i@
= lOslbf/in
load is therefore
= 500 x 3.4 x 0.264 = 4501bf Note that at this load the stiffness only 20%
. x0
106 = zg.
applied
has been reduced
by
0.75
So that A, must be greater than 2. 67in2 and A,, be greater than 1.335ina. From Fig 4, read A = 0.54 at E= 2.0
h= Ps Ae.1
A = A $ = w Ro = ,/$=
must
= 2. 5in2 approx
ho
0.9in
Suitable dimensions R, = l.Oin R = 0.5in and actual effective
are therefore
area is therefore
A, = A7i = 3.4in2
A deflection
of 0. OOOlin corresponds
to
E, = 0.0010 - 0.0001 0.001 = 0.9 From Fig 39, W= 0.0’75 w=
‘psAe.
x 0.075 = 1271bf
w = 500 X 3.4
Load Examining Figs 39 to 47 shows that for 6 between 0.2 and 0.5, stiffness is increasingly reduced for loads in excess of 0.6 of the theoretical maximum load. From Fig 39, Wulttmate = 0.44 A safe figure is therefore w’ = 0.6 x 0.44
2
= 0.264
I
I
0.5
1.0 gap ‘;i=
Bearing
Fig 40
Capillary-compensated
0.2 0.2
04 Bearing
0.6 4 gap F,=T;
0.6
1.0
2.
I ^
\\YiY\
0
1.5
double plane bearing
1.01
0
I h Fo
0.L Supported
Bearing
load
Load
wz
p, A,.%
0.6 - ^ 0.6 W =I&( p,-p2 I
0
Fig 39 46
Capillary-compensated TRIBOLOGY
February
double plane bearing 1969
Fig 41
Capillary-compensated
double plane bearing
1,
!-
,0=0.6
1.0
p.o.5 8=0.2
0.6
p-o.3 p-04
PS
’-’ t&w ‘/2”e’T&3q V2Ae,&
Ps Stiffness
p=o.1 14
D I%
P=0.75
0.6
A=
PsAe -. ho Suaaortcd
--rb--
W=ps
W
----.---load
A,.W
z.lli 04
0 0
0.2
04
0.6
Bearing
Fig 42
Orifice-controlled
gap
2= !!l ho
0.6
0.2
1
double plane bearing
04
0.6
Supported
Fig 44
load
Orifice-controlled
.
1.0
m =$$,-&I
double plane bearing
-1hl af Stiffness
h=
upported
0.5
-0
1.0 iz, = !I! ho pressure at which can be maintained
Bearing pm Bearing
Fig 43
Orifice-controlled
gap
R=$
double plane bearing
Fig 45
=
maximum flow
Constant-flow
load
W=
gap
controlled
constant
double plane bearing
TRIBOLOGY
February
1969
4’7
Stiffness
A=p
A m-7 ho
Pm = maximum pressure at which constant flow can be maintained
0.5
Q.5
1.0 Bearing
2.0
% -
0* 0
-9 h gap XT
0.1 Supported
0
pm=
Fig 46
Constant-flow
controlled
COMPLEX ARRANGEMENTS COMPENSATION)
double plane bearing
OF PLANE
Procedure (Refer to Fig 46): 1 Calculate pad thrusts under the design condition to ensure a suitable /3 and A, for each pad. 2 Calculate approximate dimensionless deflections for extreme operating loads using the design stiffness values. 3 If the dimensionless deflections are excessive the alternatives are to choose a new A, and a new ,9 or to rearrange the bearing layout and repeat steps 1 and 2. Assuming applied loads F , F, and F, the first stage of the analysis is to equate appl iyed loads and resolved bearing thrusts. The applied loads must be related to one set of coordinates centred at some point 0. For example the gravity load is represented by a direct load Ty = G and a moment is represented by F, = G X Rg. In this way we can show that: TRIBOLOGY
February
1969
load
w=
Maximum pressure can be maintained
Constant-flow
controlled
al
0.4
1/2h-p2] which
constant
flow
double plane bearing
PADS (CAPILLARY
Complex bearing arrangements can be analysed without too much difficulty for a two-dimensional problem with four pads. Analysis can be carried out for even more pads if there is summetry or if some pads are arranged perpendicularly to others. Given a set of bearing deflections it is a simple matter to find the bearing thrusts. In practical situations it is usually the other way round. One has to find the deflections from a lmown set of applied loads. Unfortunately there is no simple method. The designer overcomes this problem by using the following approximate analysis for deflections and ensuring a margin of safety. The procedure is for capillary compensation and will ensure a safe design even for other types of compensation, since load capacity will not be greatly altered and operating stiffness will be improved.
48
Fig 47
0.3
0.2
z 0 x c rY Fig 46
A complex
arrangement
of plane pads
Fy = W, cos By1 + W, cos Byz + W, cos By3 + w, cos eyr Fx = W, cos ox1 + W, cos ox2 + W, cos ox3 + w, cos ex4 F, = W,R,
+ W,R,
+ W,R,
+ w,R,
(24)
Then substitute
W, = PsAe,B2 etc. This yields three equations with four unknown A, and four unknown p. It is therefore necessary to choose five of these values arbitrarily. Choose Ael, bz, Ae3, A,, and one p and solve the three equations for the three remaining 0. If the 0’s do not lie in the range from 0.2 to 0.75 some rearrangement of the areas will be required.
g&.J+ -.-,IY 1 /
To Analyse Deflections Chance in load Deflection = Average stiffness ho&c
- W for small
deflections
2in
X0
This mav be stated in dimensionless
form as
L3in
so that w’ - W = psAso(l
(25)
-x)
Equations 4-14 can be rewritten for extreme loading to yield approximate dimensionless deflections in the following form. The dash with a load symbolizes extreme loading. I?; - Fy = (w; - wl)
+ (wj - w3) cos ex3 + (w; - w4) cos ex4 F; - F, = (W; - W,)R,
+ (Wi - W,)R,
+ (w; - W,)R,
+ (Wi - WJR,
(26)
Substituting in Equations 26 from Equation 25 Wi -W, = psAe,~,(l - xi), etc yields three equations
l-zii,=h
G,
in
02
cos
+-+TpL ho1 cos By1
01
0x2
ho2
cos
By2
02
etc (27) It is preferable that each bearing pad should preload another so that larger bearing area can be employed and deflections will be small. In Example (5. 3) bearing pad (1) is not preloaded and the load range is limited. In many cases, for economic or other reasons, it is not possible to employ bearings in opposition. Under such circumstances the following rules tend to give confidence. CHOICE OF BEARING PARAMETER 6 FOR COMPLEX PLANE ARRANGEMENTS The optimum 6 is 0.5 but where greater load bearing capacity is required without serious loss of stiffness: 1 For bearings which will operate at clearances less than the design value: Choose 6 in the range 0.2 to 0.4 2
I Fig 49
Design
example
5.3-complex
pad arrangement
For bearings which will operate at clearances than the design value: Choose 6 in the range 0.7 to 0.75
Supply pressure ps = 5001bf/in2 e5 = ainsl Choose A = 4ina; A,, = 16ina; Ae3 = 16ina; Ae4 = 6in2; A Analysing for light loads as a design condition: y-direction: Fy = W, cos yr + W, cos y2 + W, cos ys +w,-ww,
and W, = psAen 1000 = 2000@,
+ 46, - 46,)
8, + 46, - 483 = 0.5 x-direction F,=W,-Ws 0 = 4000(6, -8.J o4 = p, and choose a value of 0.5
+--L+p
x
-6in
=w,
four unknown values of (1 - X). To reduce the number of unknowns is often facilitated by symmetry in the bearing layout, as in the following design example. However for the general case the number of unknowns is reduced to three by solving in terms of the three co-ordinate movements. x
k2in
cos ey3 + (w; - w,) cos ey4
FI( - Fx = (W; - W,) COSexi + (W; - Wa) COSexa
01 cm
wi
CO6eyl + (w; - w2) COSey2
+ (w; -w,)
l-zl=h
4
greater
s-direction F, = aw, - 2w, + 2wa - 2w, + 2w, Since W, = W, 3000 = 2000 (ap, - ap, + 86s) p1 - p2 + b3 = 31~~ = 0.1875 Since there are two equations in three unknowns may be chased arbitrarily, fi2 = 0.4
one value
Solving for x and y directions simultaneously p3 = 0.33 Pl = 0.26 Analysing deflections approximately from deflection
=
Chanee in load average stiffness
ho - h = y l-X=-
approx.
i# -w
X0 Eaamole 5.3 For complex bearing arrangement (Fig 49) 1 Light loading condition A gravity load only of 1OOOlbf ie Fy = 1OOOlbf F, = 30001bf in 2 Extreme loading condition Fy = 40001bf F, = 20001bf F, = 60001bf in
w’ -w
= psAe,io(l
- Z)
y-direction F; - Fy = (W; - Wi) cos ByI + (W; - W,) cos By2 +(wj
- w,)
cos ey3
3000 = 2000 [O. 57(1 - Zl)
+ 4 x 0.72(1
-X2)
- 4 x 0.65(1 - x3)] Since x3 = 2 - z2 the equation simplifies 0.57(1 -xl) + 5.48(1 -a,) = 1.5 TRIBOLOGY
February
to
1969
49
x-direction 2000 = 4000(0.75(1 l-84
= +$
Summary: -X4)
- 0.75(1 -X5)
= 0.333
z-direction 5000 = 8(W; -WI)
- 2(W; - Ws) + 2(W; - W,)
- 2(W:, - W,) + 2(w; - W5) 5000 = 2000[8 x 0.5771 --xl) - 4 X 2 x 0.72(1 --x2) + 4 x 2 x 0.65(1 -8s) - 2 x 2 X 0.75(1 -x4) + 2 x 2 x 0.75(1 -X,)1 2.5 = 4.56(1
--xl)
4.56(1 -xl)
- 10.96(1
Solving simultaneously 1 -7fa = 0.137 1 -xl
Section
- 10.96(1 -a,)
-&)
Pz p3 pq P5
w, x, x, x,
= = = =
0.4 0.33 0.5 0.5
= = = =
0.86 1.14 0.6’7 1.33
Inspection of Fig. 30, for single plane bearings and Fig. 39, for double plane bearings will reveal any doubt as to the safety of the design. For example in the case of the first pad p1 = 0.26 and-under extreme loading the approximated gap is reduced to X, = 0.24. However average stiffness over this range is better than the stiffness assumed and the deflection will therefore be slightly less in practice.
= 4.5
for the y and z directions
Partial
Approximated clearance extreme loading %, = 0.24
- 6(1 -i,)
= 0.76
6
P fil = 0.26
Hydrostatic
Journal
Bearings
Hydrostatic bearings can be used to support a loaded shaft for both rotational and reciprocating motion. When the applied load is unidirectional there is no need for a full hydrostatic journal bearing of the type described in the preceding sections, and a partial pad may be used as shown in Fig 50. These bearings are designed in a similar fashion to the plane bearings described in Section 5. The necessary design information regarding the pad coefficients is given in Fig 51-53. These bearings can be used in combination to take side thrust as well, as shown in Fig 54. The methods of combination are as given for plane bearings in Section 5.
Fig 50 50
Partial
journal
TRIBOLOGY
pad bearing
February
1969
Fig 54
Partial
journal
bearing
with side thrusts
s
Design of Plane Hydrostatic
USE OF THE DESIGN CHARTS The designer proceeds from one instruction to the next taking decisions or doing arithmetic as instructed. If at any stage he is dissatisfied with a result, for example, if the flowrate is inconveniently high, he is free to adjust any of his earlier choices and if he repeats the procedure from the point of adjustment he will obtain an optimized bearing for his new conditions. Thus within two to three hours the designer can compare a number of designs, each of which is optimized for the conditions laid down by his basic decisions. The use of each design chart is illustrated by an example. After the designer has specified his bearing-pad dimensions, he proceeds to the design of his restrictor or compensating device. The following notes discuss the points to be considered in the selection of the various parameters and give some insight into the theory and assumptions on which the procedures are based. The procedures also include the theory discussed in Section 8 on minimisation of power and temperature rise. The designer’s initial concern is to ensure the following conditions: The flow rate q should not be too great. The bearing should have sufficient stiffness. The bearing should be capable of withstanding the applied loads. That is to say for a given minimum film thickness the bearing thrust W should be greater than the applied load. This section contains simple methods for determining flow rate q, stiffness X and bearing thrust W. At higher bearing speeds it is often necessary to optimize the pad design for minimum power at the speed of operation. The designer uses the calculation procedures laid out in this section, together with the optimization procedures discussed in Section 8. The design of plane pad bearings will be illustrated for three common pad shapes. These are Circular thrust bearing Rectangular pads Thin-land bearing There are a number of other bearing shapes for which the effective area and the flow shape factor are not quite so easily determined. These all come under the category of thick-land bearings and it is necessary to use a computer to obtain accurate data for effective area and flow shape factor. This data has already been presented in Section 2. It is important to note that the effective areas and flow shape factors quoted in Section 2 should be used with the design procedures for flow rate q, stiffness A, and bearing thrust W.
Bearings pr = Recess pressure h = Bearing clearance n = Dynamic viscosity is = A dimensionless factor called the flow rate factor which is related to the shape of the particular bearing It is often convenient to use the equation condition h = ho. Equation 4b may then be written e&o3 Q
qo=y
related
to the design
(15)
0
where Go = 625 where 13 =pspro Flow rate from a circular thrust bearing The flow shape factor for a circular plane bearing by the expression
is given
B=
l7 (16) log, R ( RO ) not however necessary to calculate B from Equation 16 B is plotted on Fig 4b. In certain designs it is necesto resort to annular recess bearings, as for example a shaft extends through the thrust bearing. 6
It is since sary when
The circular plane bearing with annular oil feed (Fig 71is easily analysed by consideration of the previous case: B can be read off from Figs 7b and 7c and inserted into Equation 16 without the need for extra arithmetic. Flow rate from a rectangular plane bearing with radiused recess corners The expression for flow through a pad having the geometry shown in Fig 27 is plotted on Fig 28. It was derived by
P"1
CHOICE OF LAND WIDTH For very low speeds there is an optimum land width for bearing pads. This optimum is based on the minimum power needed to support a given load on a given total area and corresponds to a minimum of I% At higher speeds another term for friction must be taken into account. However the land width need not be reduced (See Section 8). For a circular thrust bearing at zero speed the optimum radius ratio is 1.89. For various other shapes the optimum is obtained when the recess perimeter is between 0.4 and 0. 8 of the bearing perimeter, as is indicated by the recommended design ranges on Figs 4-12.
W = PS A e.E
FLOW RATE In general the flow rate q from a bearing Equation 4b ghs q=PrT where q = Flow rate
is given by
Fig 27
Plane rectangular corners
pad with radiussed
TRIBOLOGY
February
recess
1969
41
b/a
‘I
c .l
0.9 d’3 0.1
.2 ‘.3
Fig 28b Leakage coefficients for planerectangular pad with radiussed recess corners - d/a = 0.1 --- d/a = 0.4
I.5
Ia Any shape constant
Ae=1/2
[Total area
+ recess
area]
0=
[Total
- recess
area]
area
with land
thin width
12 12
o.4--J---J 1
1.5
2 bla
3
Fig 28a Load coefficients for plane rectangular radiussed recess corners
Fig 29 5
10
00
pad with
Loxham, who assumed that the corners have a resistance equal to one quarter the resistance of a circular pad having the same radii, and that each of the sides could be treated as uniform flow across a slot. The flow coefficient g is given by B = (a + b + 4dl loge?+)6~
CII (1-U
loge 9
Tbinlimdbearingpads
Thin-land bearing pads (Fig 29) are used for two purposes. 1 To obtain the maximum effective area within a given total area. This is often the case where there are limi42
TRIBOLOGY
February
1969
Thin-land
bearing
tations to the space available and flow rate is of secondary importance. At higher speeds to reduce the flow resistance and hence 2 allow oil of higher viscosity than could otherwise be used. The flow shape factor and effective area for thin-land bearings are easily calculated, whatever the shape, The flow shape factor is given by B = Total area - recess area (181 12ta = land area 12ta where t = land width
BEARING THRUST AND STIFFNESS FOR SINGLE PLANE BEARINGS If the applied load W is known, the recess pressure pr 1.5 determined. Conversely if the recess pressure is known the
bearing
thrust
is determined
1
since
W = pr . A, where A, = A.Circular Central
Load at the design condition
pr = pp, = 0.5 x 600 = 3001bf/ina
W = PrAe A, = ?#
Plane Bearings admission
From
= 34sina
Fig 4 and choosing
z = 2.0, K is 0.542
(19) A = 2 Annular oil feed Effective area = A, = A,, outer - A,, inner In Equation 19 A,, outer is the effective area from Fig 4 using R, and R, for the outer land and A,, inner is the effective area from Fig 4 using R, and R, for the inner land. These values have been calculated and are given in Fig ‘7b in Section 2. Rectangular plane bearing Effective area =
with radiused
( (a + b - 4d)c + ab - 4d2 + w\ 2
\
recess
corners.
R, = $$! R
= 2
= 6. 14in2 = nR,a = 1.4in
7 0. 7in
The design gap ho = 0.002in siderations. From Fig 4, B = 0.75 PB qo +, = 600 X (0.002j3
(20)
log, +$-
= #
will be ensured
X 0.5 X 0.75
=o
277.4ina is 1 gallon so the flowrate q. = 0.36 x i77 a - O.O78gal/min
area)
(21)
The relationship between bearing thrust W or recess pressure pr and bearing clearance h is dependent on bearing stiffness and therefore on the control device employed. This was explained in Section 3. To simplify and facilitate the designer’s work, bearing thrust, stiffness, and bearing gap are presented non-dimensionally for single plane bearings in Figs 30-38. These figures cover capillary, orifice and constant-flow compensation using the equations in Table 2 The equation2 are considerably simplified at the design condition when X = 1 and pr = p
36in3,sec
5x10-6
These values are shown on Fig 28b Thin-land bearings of any shape (Fig 29) Effective area A, = l/2 (Total area + recess
from flow con-
2
is therefore
At this point it may be realised that the flow rate is unacceptable. For example the flow rate may be too high for the pump which it is desired to use. There are a number of ways round this, such as designing for thicker lands, or increasing the bearing size and reducing the supply pressure in proportion. Alternatively if the bearing area is increased it is possible to reduce p to about 0.2 minimum which has the advantage that a higher maximum load may be applied. However the most sensitive parameter controlling flow is the gap, because it appears to the third power. From Fig 31, the dimensionless stiffness h is 0.75 for p=o.5
The actual Table 2 Thrust, stiffness and gap equations for single plane bearings (W = 6,) Capillary
Orifice
Constant
~J3%.X=§O
pc7,-- l-
Pr P?
82 1-P
3
Diaphragm
Pr=
P 3 N.B.& ? 1
o xo3+O$$
O* 75 = 750 OOOlbf/in
When x = 0.4, h = 0.0020 x 0.4 = 0.0008in From ‘Fig 30, the dimensionless thrust W is 0.94 The maximum applied load is therefore W’ = psAe . W’ W = 600 x 3.33 x 0.94 = 18801bf Comparing W at z = 0.4 with the design load it is seen that the maximum applied load is less than twice the
&3x2.1-fl pi3 + Jq2 1-3 Xf = 3&(1 - p,,
is therefore
hO
flOW
1 Pr = 1 + LA.ftJ I!
stiffness
six = 6&
1 - f& 2 - Pr
xx
= 3p,
Example 5.1 A circular capillary-compensated plane bearing with central oil admission must support a load of 10001bf. The bearing gap is to be 0.002in and the bearing is to operate at zero speed. Determine 1 Suitable dimensions for the bearing pad 2 The operating stiffness 3
The load the bearing
will support
when z is reduced
to
0.4
The load when It has been decided sure for the pump absolute viscosity 4
h = 0.0004in. that p is to be 0.5 and a convenient presis 6001bf/ina. The oil to be used has an of q = 5 x 10-a reyns at 100°F.
0
0.5
1.0
1.5
Bearing
gap,
2.0
2.5
3.
X L h0
Fig 30
Capillary-compensated
single plane bearing TRIBOLOGY
February
1969
43
design load and the dimensionless stiffness is reduced 0.42 compared with the design value of 0.75 Taking h = 0.0004in as the minimum allowable film thickness
4
,-=!?L!!Eooo4 0.2 0.0020 W’ = 600 x 3.33 x 0.99 = 19601bf However the dimensionless stiffness is now reduced to 0.1. Apparently the minimum allowable film thickness is not a good criterion for deciding the maximum allowable load. A better approach is to consider the minimum z. For 6 = 0.5 a suitable range for X is fro-m 0.4 to 1. 6. If however 6 = 0.2, a suitable range for X is from 0.2 to 1. I
to
Ps
,p=0.7
W =
G ; 0.6 P ‘C & 0.4
m
3 Stiffness
Thrust PsAe*m
\\\\\wp=O+
I$ 0.8 13
1.5 gap
1.0 Bearing
x=
.2,0 k.
2.5
X-
Fig 33
Orifice-controlled
single
plane bearing
2.0
1.5
Bearing
gap,
IA1
10
E c.Gi
0.5
F= h/ hO
Fig 31
Capillary-compensated
0
single plane bearing
0
0.5
1.0 Bearing
1.5 gap X= $
2.0 0
Fig 34
-1
Orifice-controlled
single
plane bearing
p=o.2
p -0.3
p = 0.4
/\
0.41
1.5
A3
IX t iii E iz
I//////
1
0.5 Stiffness
0.2 1
h = P
0 0.4 Bearing
Fig 32 44
Capillary-compensated TFUBOLOGY
February
0.6 thrust
0.8
0
02
k=cr
single plane bearing 1969
0.4 Bearing
Fig 35
Orifice-controlled
thrust
0.6 $;l;E
single plane bearing
0.8
1.0
0
1.0
0.5
Bearing
2.0
1.5 gap
2.5
3.0
ii = + 0
Pm= maximum tlow can
Fig 36
pressure at be maintained
Constant-flow-controlled
which
constant
single plane bearing 0.8
-Bearing
5 3C.I k h
4
Fig 38
Constant-flow-controlled
&
Pm*, -.
A= X
h0
DOUBLE PLANE BEARINGS HAVING EQUAL OPPOSED PADS The design of bearing systems having equal opposed-pad bearings is simpler in practice than for single plane bearings. The data are given in Figs 39-47. The effective area of a double plane bearing is twice the effective area of a single plane bearing, that is Ae = A,,,
1.0
1.5 gapTi
single plane bearing
-ih
Stifness
Bearing
+ A,,2
where 1 and 2 are subscripts for the individual pads. The supported load W and the stiffness X are given by the following expressions
20 =+ 0
Pm= maximum flow can
Fig 37
pressure at be maintained
Constant-flow-controlled
1.0
W=p,
9
0.3
-0.4
thrust
which
single
(22) constant
plane bearing
CHOICE OF DESIGN PARAMETER B = 2 FOR SINGLE PLANE BEARINGS At the design condition the maximum stiffness for plane bearings is obtained when fl = 0.5. If @ is increased, both the stiffness and the load range is reduced, and so values greater than 0.5 are to be avoided. If B is reduced the non-dimensional design stiffness is again reduced. However if the same applied load is resisted by increasing the bearing area the actual stiffness is increased and so is the load range. The stiffness is much less uniform and will drop more rapidly if the load is reduced. The preferred range for 0 is therefore from 0.2 to 0.5. In certain cases p may be increased up to 0.75. To obtain a greater load range there are two possibilities. 1 Increase the bearing area by 2V2 times and decrease the design parameter to 0.2 2 Use opposed-pad bearings and greatly increase the bearing area The latter method has the advantage that stiffness will be much higher and more nearly uniform.
5 = l/2
(& + X2)
where Pr.1 and Pr, 2, ,il and x, are obtained plane data and using !?, = 2 -x1
from the single
CHOICE OF DESIGN PARAMETER P FOR DOUBLE PLANE BEARINGS The optimum @ is 0.5 and in most designs it is not necessary to chose any other value for double plane bearings. However in certain cases where the flow rate is rather greater than desirable, it is pssible to reduce the flow rate by reducing fl towards 0.2 With pressure-sensing control devices the maximum allowable load will not be lower than for capillary compensation and therefore the same value may be used. Example 5.2 A capillary-compensated double plane bearing is required which will not deflect more than 0. OOOlin under an applied load of 1001bf. The available supply pressure is 5001bf/inz and in the design condition there is negligible load. The design bearing gap is to be ho, = ho, = O.OOlOin. Determine I Suitable bearing dimensions for circular bearing pads 2 The maximum allowable applied The pressure ratio fl is to be 0.5 TRIBOLOGY
load
February
1969
45
Bearing dimensions. 1 Stiffness must be at least && From Fig 40, x0 = 0.75 * Actual stiffness is given by
A0 = p
The maximum allowable W’ = psA,. i@
= lOslbf/in
load is therefore
= 500 x 3.4 x 0.264 = 4501bf Note that at this load the stiffness only 20%
. x0
106 = zg.
applied
has been reduced
by
0.75
So that A, must be greater than 2. 67in2 and A,, be greater than 1.335ina. From Fig 4, read A = 0.54 at E= 2.0
h= Ps Ae.1
A = A $ = w Ro = ,/$=
must
= 2. 5in2 approx
ho
0.9in
Suitable dimensions R, = l.Oin R = 0.5in and actual effective
are therefore
area is therefore
A, = A7i = 3.4in2
A deflection
of 0. OOOlin corresponds
to
E, = 0.0010 - 0.0001 0.001 = 0.9 From Fig 39, W= 0.0’75 w=
‘psAe.
x 0.075 = 1271bf
w = 500 X 3.4
Load Examining Figs 39 to 47 shows that for 6 between 0.2 and 0.5, stiffness is increasingly reduced for loads in excess of 0.6 of the theoretical maximum load. From Fig 39, Wulttmate = 0.44 A safe figure is therefore w’ = 0.6 x 0.44
2
= 0.264
I
I
0.5
1.0 gap ‘;i=
Bearing
Fig 40
Capillary-compensated
0.2 0.2
04 Bearing
0.6 4 gap F,=T;
0.6
1.0
2.
I ^
\\YiY\
0
1.5
double plane bearing
1.01
0
I h Fo
0.L Supported
Bearing
load
Load
wz
p, A,.%
0.6 - ^ 0.6 W =I&( p,-p2 I
0
Fig 39 46
Capillary-compensated TRIBOLOGY
February
double plane bearing 1969
Fig 41
Capillary-compensated
double plane bearing
1,
!-
,0=0.6
1.0
p.o.5 8=0.2
0.6
p-o.3 p-04
PS
’-’ t&w ‘/2”e’T&3q V2Ae,&
Ps Stiffness
p=o.1 14
D I%
P=0.75
0.6
A=
PsAe -. ho Suaaortcd
--rb--
W=ps
W
----.---load
A,.W
z.lli 04
0 0
0.2
04
0.6
Bearing
Fig 42
Orifice-controlled
gap
2= !!l ho
0.6
0.2
1
double plane bearing
04
0.6
Supported
Fig 44
load
Orifice-controlled
.
1.0
m =$$,-&I
double plane bearing
-1hl af Stiffness
h=
upported
0.5
-0
1.0 iz, = !I! ho pressure at which can be maintained
Bearing pm Bearing
Fig 43
Orifice-controlled
gap
R=$
double plane bearing
Fig 45
=
maximum flow
Constant-flow
load
W=
gap
controlled
constant
double plane bearing
TRIBOLOGY
February
1969
4’7
Stiffness
A=p
A m-7 ho
Pm = maximum pressure at which constant flow can be maintained
0.5
Q.5
1.0 Bearing
2.0
% -
0* 0
-9 h gap XT
0.1 Supported
0
pm=
Fig 46
Constant-flow
controlled
COMPLEX ARRANGEMENTS COMPENSATION)
double plane bearing
OF PLANE
Procedure (Refer to Fig 46): 1 Calculate pad thrusts under the design condition to ensure a suitable /3 and A, for each pad. 2 Calculate approximate dimensionless deflections for extreme operating loads using the design stiffness values. 3 If the dimensionless deflections are excessive the alternatives are to choose a new A, and a new ,9 or to rearrange the bearing layout and repeat steps 1 and 2. Assuming applied loads F , F, and F, the first stage of the analysis is to equate appl iyed loads and resolved bearing thrusts. The applied loads must be related to one set of coordinates centred at some point 0. For example the gravity load is represented by a direct load Ty = G and a moment is represented by F, = G X Rg. In this way we can show that: TRIBOLOGY
February
1969
load
w=
Maximum pressure can be maintained
Constant-flow
controlled
al
0.4
1/2h-p2] which
constant
flow
double plane bearing
PADS (CAPILLARY
Complex bearing arrangements can be analysed without too much difficulty for a two-dimensional problem with four pads. Analysis can be carried out for even more pads if there is summetry or if some pads are arranged perpendicularly to others. Given a set of bearing deflections it is a simple matter to find the bearing thrusts. In practical situations it is usually the other way round. One has to find the deflections from a lmown set of applied loads. Unfortunately there is no simple method. The designer overcomes this problem by using the following approximate analysis for deflections and ensuring a margin of safety. The procedure is for capillary compensation and will ensure a safe design even for other types of compensation, since load capacity will not be greatly altered and operating stiffness will be improved.
48
Fig 47
0.3
0.2
z 0 x c rY Fig 46
A complex
arrangement
of plane pads
Fy = W, cos By1 + W, cos Byz + W, cos By3 + w, cos eyr Fx = W, cos ox1 + W, cos ox2 + W, cos ox3 + w, cos ex4 F, = W,R,
+ W,R,
+ W,R,
+ w,R,
(24)
Then substitute
W, = PsAe,B2 etc. This yields three equations with four unknown A, and four unknown p. It is therefore necessary to choose five of these values arbitrarily. Choose Ael, bz, Ae3, A,, and one p and solve the three equations for the three remaining 0. If the 0’s do not lie in the range from 0.2 to 0.75 some rearrangement of the areas will be required.
g&.J+ -.-,IY 1 /
To Analyse Deflections Chance in load Deflection = Average stiffness ho&c
- W for small
deflections
2in
X0
This mav be stated in dimensionless
form as
L3in
so that w’ - W = psAso(l
(25)
-x)
Equations 4-14 can be rewritten for extreme loading to yield approximate dimensionless deflections in the following form. The dash with a load symbolizes extreme loading. I?; - Fy = (w; - wl)
+ (wj - w3) cos ex3 + (w; - w4) cos ex4 F; - F, = (W; - W,)R,
+ (Wi - W,)R,
+ (w; - W,)R,
+ (Wi - WJR,
(26)
Substituting in Equations 26 from Equation 25 Wi -W, = psAe,~,(l - xi), etc yields three equations
l-zii,=h
G,
in
02
cos
+-+TpL ho1 cos By1
01
0x2
ho2
cos
By2
02
etc (27) It is preferable that each bearing pad should preload another so that larger bearing area can be employed and deflections will be small. In Example (5. 3) bearing pad (1) is not preloaded and the load range is limited. In many cases, for economic or other reasons, it is not possible to employ bearings in opposition. Under such circumstances the following rules tend to give confidence. CHOICE OF BEARING PARAMETER 6 FOR COMPLEX PLANE ARRANGEMENTS The optimum 6 is 0.5 but where greater load bearing capacity is required without serious loss of stiffness: 1 For bearings which will operate at clearances less than the design value: Choose 6 in the range 0.2 to 0.4 2
I Fig 49
Design
example
5.3-complex
pad arrangement
For bearings which will operate at clearances than the design value: Choose 6 in the range 0.7 to 0.75
Supply pressure ps = 5001bf/in2 e5 = ainsl Choose A = 4ina; A,, = 16ina; Ae3 = 16ina; Ae4 = 6in2; A Analysing for light loads as a design condition: y-direction: Fy = W, cos yr + W, cos y2 + W, cos ys +w,-ww,
and W, = psAen 1000 = 2000@,
+ 46, - 46,)
8, + 46, - 483 = 0.5 x-direction F,=W,-Ws 0 = 4000(6, -8.J o4 = p, and choose a value of 0.5
+--L+p
x
-6in
=w,
four unknown values of (1 - X). To reduce the number of unknowns is often facilitated by symmetry in the bearing layout, as in the following design example. However for the general case the number of unknowns is reduced to three by solving in terms of the three co-ordinate movements. x
k2in
cos ey3 + (w; - w,) cos ey4
FI( - Fx = (W; - W,) COSexi + (W; - Wa) COSexa
01 cm
wi
CO6eyl + (w; - w2) COSey2
+ (w; -w,)
l-zl=h
4
greater
s-direction F, = aw, - 2w, + 2wa - 2w, + 2w, Since W, = W, 3000 = 2000 (ap, - ap, + 86s) p1 - p2 + b3 = 31~~ = 0.1875 Since there are two equations in three unknowns may be chased arbitrarily, fi2 = 0.4
one value
Solving for x and y directions simultaneously p3 = 0.33 Pl = 0.26 Analysing deflections approximately from deflection
=
Chanee in load average stiffness
ho - h = y l-X=-
approx.
i# -w
X0 Eaamole 5.3 For complex bearing arrangement (Fig 49) 1 Light loading condition A gravity load only of 1OOOlbf ie Fy = 1OOOlbf F, = 30001bf in 2 Extreme loading condition Fy = 40001bf F, = 20001bf F, = 60001bf in
w’ -w
= psAe,io(l
- Z)
y-direction F; - Fy = (W; - Wi) cos ByI + (W; - W,) cos By2 +(wj
- w,)
cos ey3
3000 = 2000 [O. 57(1 - Zl)
+ 4 x 0.72(1
-X2)
- 4 x 0.65(1 - x3)] Since x3 = 2 - z2 the equation simplifies 0.57(1 -xl) + 5.48(1 -a,) = 1.5 TRIBOLOGY
February
to
1969
49
x-direction 2000 = 4000(0.75(1 l-84
= +$
Summary: -X4)
- 0.75(1 -X5)
= 0.333
z-direction 5000 = 8(W; -WI)
- 2(W; - Ws) + 2(W; - W,)
- 2(W:, - W,) + 2(w; - W5) 5000 = 2000[8 x 0.5771 --xl) - 4 X 2 x 0.72(1 --x2) + 4 x 2 x 0.65(1 -8s) - 2 x 2 X 0.75(1 -x4) + 2 x 2 x 0.75(1 -X,)1 2.5 = 4.56(1
--xl)
4.56(1 -xl)
- 10.96(1
Solving simultaneously 1 -7fa = 0.137 1 -xl
Section
- 10.96(1 -a,)
-&)
Pz p3 pq P5
w, x, x, x,
= = = =
0.4 0.33 0.5 0.5
= = = =
0.86 1.14 0.6’7 1.33
Inspection of Fig. 30, for single plane bearings and Fig. 39, for double plane bearings will reveal any doubt as to the safety of the design. For example in the case of the first pad p1 = 0.26 and-under extreme loading the approximated gap is reduced to X, = 0.24. However average stiffness over this range is better than the stiffness assumed and the deflection will therefore be slightly less in practice.
= 4.5
for the y and z directions
Partial
Approximated clearance extreme loading %, = 0.24
- 6(1 -i,)
= 0.76
6
P fil = 0.26
Hydrostatic
Journal
Bearings
Hydrostatic bearings can be used to support a loaded shaft for both rotational and reciprocating motion. When the applied load is unidirectional there is no need for a full hydrostatic journal bearing of the type described in the preceding sections, and a partial pad may be used as shown in Fig 50. These bearings are designed in a similar fashion to the plane bearings described in Section 5. The necessary design information regarding the pad coefficients is given in Fig 51-53. These bearings can be used in combination to take side thrust as well, as shown in Fig 54. The methods of combination are as given for plane bearings in Section 5.
Fig 50 50
Partial
journal
TRIBOLOGY
pad bearing
February
1969
Fig 54
Partial
journal
bearing
with side thrusts