Rowe 1969
Rowe 1969
Paper 1
The diaphragm valve described is of very simple construction and may be used to control opposed pad
hydrostatic bearings as an alternative to capillaries, orifices, or other types of control device. Experiments
and analysis for flat pad bearings show good agreement. Data have been computed for a range of operating
characteristics in a form suitable for use by designers. A reference bearing system is proposed for correcting
errors in long slideways in which the reference is a diaphragm controlled hydrostatic bearing. The computed
data provides information necessary to design such a system.
Diophroqm I I
Opposed
main bearing area, and a factor for the relay (if this is not
adjusted to unity).
m
valve beoring
A linearized solution was suggested by de Gast (3) which
.Diophroqm applies only for small variations from the unloaded con-
ditions. It was found that a linearized solution under-
estimates deflections in some cases and overestimates in
P, others. The following results are therefore based on exact
I L I solutions of the basic equations.
Fig. l.la. Schematic arrangement of an opposed pad
hydrostatic bearing controlled by a double diaphragm
valve Notation
A d Effective area of diaphragm valve.
Ae Bearing effective area.
/// A M o i n load
b
I 1 1 .-
0.1 0.2 0-3 0.4 C ' 0 0.1 02 0.3 0.4 (
W
w=ps (A,I+A,ZJ
a fl = 0.1. b fl = 0.2.
01 02 03 04 0;
I I I I 1
01 02 03 04 05
16
1-4
10
-=-I<"
II
%-
n
5 0.8
Fig. 1.3. Theoretical relationships between load, recess
pressure, and clearance for a diaphragm controlled <I e
bearing
'-II
0 6
0.4
0.2
0
5
g = 0.7.
The charts are given for seven values of the design pressure
ratio /3 and for various values of diaphragm stiffness of
which one value is calculated for optimum bearing stiffness.
0.1 10 x wo 0.18 1-8 X WO
0.2 5XWo 0.34 1.7 x wo
Design of diaphragm controlled opposed pad 0.3 3.33 x wo 0.42 1.4 x wo
bearings 0.4 2.5 x w0 0.54 1.35 X W O
0.5 2xwo 065 1.3 X W O
The design calculations are extremely simple. It is possible 0.6 1.67 X w0 0.79 1.32 X W O
to calculate the optimum diaphragm stiffness, but this has 0.7 1.43 x wo 0.81 1.15 x wo
1
I I
not been found to be very successful. I n practice, the
optimum thickness has been found by experiment. The
final adjustment can be made by varying the supply Opposed main bearing arrangement
pressure. This can be seen from the form of diaphragm Assuming, as before, a unity relay factor for pressure the
stiffness parameter. The required condition is supported load expression has now an additional term for
the low-pressure main bearing :
w = PsA,W+PsA,,P,-PsAm~P~
It is also known that P2 = P1-2W by definition, and
It follows that an increase in P, is equivalent to a reduction therefore
in h d .
w = P s A e W + P s A m ~ P ~ - P s A m ~ ( P ~ -(1.12)
2W)
Example: A bearing is to carry 1000 lbf with negligible
deflection. The supply pressure is to be 500 lbf/in2. At the design condition P1 = P2 = /3, w = W,,, and
Selecting /3 = 0.5, the maximum value of load para- W = 0, so that equation (1.12) reduces to a form useful
meter for negligible deflection is W = 0.15. to a designer:
wmax = PsAe W
W O = Ps(Aml-Am2) .
. (1.13)
The difference Am,-Am2 must be adjusted so that the
1000 4 bearing thrust just equals the design load condition.
= - in2
Hence
= 500~0.15 3 +
However, the total area Aml Am2may be made as large
as desired, and in this way it is possible to achieve a
and therefore each pad should have an effective area of maximum load many times greater than the design load.
5 in2. The remainder of the calculations for flow rate,
pad dimensions, etc., follow design procedure 2 from Kinematic design
reference (5).
In machine tool slideways it is usual to provide at least
two slides with support along the length of the supported
Design of the reference bearing system table. Sometimes there is a bearing at each corner of the
Single main bearing arrangement table; that is, four bearings, of which two or all are opposed
pad bearings. From considerations of kinematics these
One configuration of the system involves one main bearing
arrangements are not ideal and lead to some structural
in which the pressure is controlled from the high-
deflections. With diaphragm controlled bearings designed
pressure side of a diaphragm controlled reference bearing.
for infinite stiffness, the structure must deflect until each
The load supported, assuming unity relay factor, is
bearing is at its operating gap. If the structure is too rigid
w = PsAeW+PsAmP, . . (1.10) to allow this, the diaphragm valves will compensate to the
Since it is desirable that forces on the reference bearing limit and the resultant bearing stiffness will be low.
should be as small as possible, it may be arranged that A better arrangement is to make three of the four bearings
W = 0 at the design condition. Writing equation (1.10) diaphragm controlled and the remaining bearing con-
for the design condition when P1 = /3: trolled by capillary. Where the machine arrangement will
allow a three-point support, each formed by a diaphragm
w0 = PsAm/3 . . . (1.11) controlled bearing, this arrangement should be used.
The value P,A,/3 must be arranged to equal the design
load. EXPERIMENTS
It is not immediately apparent whether there is an Details of the test rig are shown in Fig. 1.4. The pockets
advantage in making /3 low or not. Table 1.1 shows that were machined in a 1-in square section bar so that the
the low value of /3 has an improved load range over which bearings were symmetrically arranged with respect to the
deflections are negligible. I n this case w,,, is the thrust loading point. Three displacement indicator positions are
exerted by the main bearing pad only. shown which allowed the bar to be set up for equal bearing
If insufficient load range is unobtainable from this gaps under no load.
arrangement, it is necessary to go to an opposed main By making the diaphragm valve adjustable it was
bearing arrangement. possible to vary the design pressure ratio /3. The recess
Proc lnstn Mech Engrs 1969-70 Vol184 Pt 3L
-._2 8
-
0
z
111 6
!I
z
--
3
, 4
0
c
0
w
_I
0.037in
LOAD-lbf
This equation is made dimensionless by writing The expression for Pz is derived similarly, starting from
the equation of flow rates in the low-pressure side:
xd = &
xd
B
P1-Pz = Pl
- -P, (P~-Pz)(hd-Xx,)~ ;
Bd
= p2hZ3-
77
. (1.4)
PS
leading to
hdhd
Ad =-
PsAd P 2w
Hence
P, =
1-8(1-x)
(1.7)
p1- PZ B (I-?)~
x d =- . . . (1.2a) (2-~,)3+- ’
Ad
Combining (1. la) and (1.2a), where 2-X, = X,.
2w Substituting for Pz in equation (1.7) from equation
. . . . (1.2b) (l.la>,
Considering the high-pressure side of the bearing and 2w
diaphragm valve it is possible to equate the expressions
P, = 1-P
for flow rate through each, which leads to equation (1.3) : +2W (1.14)
(2 - X d 3 +-1-PB (I+?)
At the design condition p , = PPS;x d = 0 ; and h, = ho; Equations (1.6) and (1.14) were solved as two simultaneous
hence from equation (1.3), equations in P, and X , with the restriction that x d
cannot be greater than 1. The restriction was accommo-
EL-.-P ho3 . . . (1.5) dated by an instruction of the form
B 1-P h d 3
2w 2w
Using equation (1.5) to eliminate Bd and B from if - > 1 then write =1
equation (1.3), *d /‘d