Sliding Contact Bearing Example 1 : A hydrostatic spherical step bearing is shown in fig. Show that the load carrying capacity of the bearing is given by 2
π Pi R ( cos ∅1−cos ∅2 ) ∅ tan( 1 ) 2 log e [ ] ∅2 tan( ) 2
W=
and the flow requirement is given by, Q=
π pi h03 ∅1 ) 2 6 μ log e [ ] ∅2 tan ( ) 2 tan (
Step 1 : Requirement of flow Consider the flow of lubricant through a slot of length R d ɵ and thickness shown in fig. 3
Δpb h Q= 12 μl (a)
∆ p =d
p
, b=2 π Rsin ɵ
, l=R d ɵ
, h= h0
h0
as
Substituting above values in equation (a),
6 μQ d ɵ π h03 sin ɵ
dp = -
p= -
6 μQ π h03
Intrgrating,
ɵ
log [tan( 2 )] + C
(b) The first boundary condition is,
ɵ=∅1
p = 0 when
C=
6 μQ π h0 3
log [tan(
∅1 2 )]
(c) From (b) and (c),
6 μQ p= π h 3 0
log e
∅1 ) 2 { ɵ } tan( ) 2 tan(
(d) The second boundary condition is, p= pi when
ɵ=∅2
Substituting above condition in equation (d),
6 μQ log e Pi = 3 { π h0
π pi h0
∅1 ) 2 } or ∅ tan( 2 ) 2 tan(
3
∅1 ) 2 Q= 6 μ log [ ] e ∅2 tan( ) 2 tan(
Step 2: Load carrying capacity
The load carrying capacity is given by, ∅1 2
W= π ( R sin ∅2) Pi
+
∫ ( 2 πR sin ɵ ) R d ɵ ( pcosɵ) ∅2
Substituting value of p from equation (d),
∅1 ) 6 μQ R 2 2 2 sin 2 ɵ π P R sin ∅ + log ∫ [ i 2 W= ɵ ] h03 ∅ tan( ) 2 tan(
2 ∅1
2
(e) Define I as,
∅1 ) 2 log ɵ d ɵ=∫ u dv I= ∫ [ ɵ ]sin2 tan( ) 2 tan(
Where,
∅1 ) 2 log ɵd ɵ u= ∫ [ ɵ ] and dv= sin2 tan ( ) 2 tan(
ɵ tan( ) 2 du= ∅ tan( 1 ) 2
∅1 ) 2 ɵ [−tan 2 ( )] 2 tan(
x
1
= - sin ɵ
v=-
cos 2 ɵ 2
Since, I=
∫ u dv
= uv -
∫ v du
x
ɵ sec 2 ( ) 2
x
1 2
dɵ
cos2 ɵ
∅1 ) 1 cos 2 ɵ 2 dɵ log [ ] 2 ∫ sin ɵ ɵ tan( ) 2
cos2 ɵ
∅1 ) 1 ɵ 2 tan ( ) log [ ] [log 2 2 ɵ tan( ) 2
tan(
1
=- 2
tan(
1
=- 2
+ 2 cos ɵ ]
∅ ∅1 ) −1 2 1 ɵ ]− log tan( )+cos ɵ ] Substituting limits, I =[ 2 cos 2 ɵ log[ ɵ 2 2 tan( ) 2 ∅
1
tan(
∅1 ) 2 [ ] ∅ tan( 2 ) 2 tan(
=- log e
sin 2 ∅2 – (cos ∅1 - cos ∅2 )
Substituting above expression in expression in equation (e),
W=
π Pi R 2( cos ∅1−cos ∅2 ) ∅ tan( 1 ) 2 log e [ ] ∅2 tan( ) 2
Example 2 : Following data is given for a full hydrodynamic bearing : Radial load = 25 kN Journal speed = 900 rpm Unit bearing pressure = 205 MPa
i
( d ¿ ratio = 1 Viscosity of lubricant = 20 cP
2
Class of fit =
H 7 e7
Calculate : (i) (ii) (iii) (iv)
Dimensions of the bearing , minimum film thickness , and requirement of oil flow. Step I : Bearing dimensions
I=d
W
p= ld
25 X 103 2.5= l2 I = d = 100 mm (i) Step II : Sommerfeld number For values of tolerances, the hole and shaft limits for
H 7 e7
running fit are as
follows : Hole limits : (100.00) and (100.035) mm Shaft limits : (100 – 0.072) and (100 – 0.107) mm If the manufacturing processes are centered, the average diameter of the bearing and journal will be 100.0175 and 99.9105 mm respectively. c=
1 ( ) (100.0175 – 99.9105) = 0.0535 mm 2
S=
r ( ) c
2
μ ns p
−9
=
50 2 ( ) 0.0535
(20 X 10 )( (2.5)
900 ) 60
= 0.1048
Step III : Minimum film thickness The values of dimensionless performance parameters are obtained by linear interpolation. For
l d ¿
= 1)
(
h0 c ) = 0.2 +
(0.1048−0.0446) (0.121−0.0446)
(0.4 – 0.2) = 0.3576
h0 = 0.3576c = 0.3576 (0.0535) = 0.0191 mm (ii) Step IV : Flow requirement
Q
(0.1048−0.0446) (0.121−0.0446)
( rc n l ) = 4.62 s Q = 4.391
(4.62 – 4.33) = 4.391
rc n s l
= 4.391 (50) (0.0535)(15)(100) 3
=17618.89
mm s
−6
= 17618.89 (60 X 10
= 1.057
)
litres min
litres min
Example 3: Following data is for the full hydra dynamic bearing.
Radial load ¿ 23 kw Journal speed load ¿ 900 rpm Unit bearing pressure ¿ 2.5cp µɸ/ratio ¿ 1
calculate: (1) Dimension of bearing (2) Minimum film thickness Solution: Given Data:
W ¿ 25 KN n ¿ 900rpm B.P ¿ 2.5pa µɸ ¿ 1 µ ¿ 20cp (1) Dimension of bearing 70(µɸ) ¿ 1 3 P ¿ 25000kpa ¿ 25000 ×10
−3 −6 Pa ¿ 25000 ×10 ×10
w w 2500 P ¿ ld = ld =2.5= 20
d ¿ 100=L
d ¿ L=100
Sometimes, Hole limits (50 −¿ 0.00) (50 +0.0030 ¿ Shaft limit (50 −¿ 0.110) (50 −0.080¿ ¿ 50.0195−41.9005
¿ 8.119
C
1 ¿ =8.119=0.0895 2
ns
¿
400 =1.5 60
(2) Minimum thickness µ=20 cp=20 ×10−3 −4
pr=20× 10
µ=20 ( 10−4 ) ho =30 micron=mh w 25000 P ¿ ld = 100 ×100
¿ 0.25
N /mm2
25 2 µ n s 20 ×10−4 ×15 + = p 0.25 c2
(
C ¿
√
)
1.92(10−3 ) 5
Assume value,
ho =0.9 c
ho 0.03 ¿ =0.9= =0.033mm C c 0.9
From table, S ¿ 1.33
√
1.92 ( 10−3 ) =0.041 C ¿ 1.133
ho c
¿ 0.8
0.03 C ¿ 0.8 =0.0375
From table, S ¿ 0.631
√
1.92 ( 10−3 ) C ¿ 0.631
¿ 0.055
Assume, hdc ¿ 0.264
√
1.92 ( 10−3 ) =0.005 C ¿ 0.264
Co-efficient of friction,
( rc ) f =5.79 f =5.79
( cr )=5.79 ( 0.05 28 ) ¿ 0.01158
Requirement of oil flow,
Q ¿ 0.75(h s)
¿ 4.7 ( 28 ) ( 0.08 ) (15) (100)(100)
2
¿ 58750.1mm /s
Q ¿ ( 58750.1 )( 106 ) ( 60 )
¿ 3.525litre /min
….Ans.
Example 4 Design journal bearing for centrifugal pump running at 1440 rpm. The diameter of journal in 100 rpm to load on each bearing 20 kn. The factor 2 N/p to many take 28 for centrifugal pump bearing is running at 75 ℃ temperature and atmosphere temperature. Is 30 C energy dissipated co-efficient is 875
Solution: Given data :
w C . Diameter clearance 0.5 Q/m. 2 m
N=1440 rpm W =120 kn
d=100 mm Tc=75 ℃
Atm .=30 ℃
Energy co-efficient ¿ 875=0.0875
C ¿ 0.1
w 20000 P ¿ ld = c
n=
1440 =24 rpm 60
Length of bearing
w 20000 P ¿ l d = 2 ( 100 ) =100
μd =
100 =1 100
l=d =100
Sequence of to section
P
¿
w 20000 = =2 N /mm2 l d 100 ( 100 )
0.01 100 C ¿ r =0.01 2 =0.5
( )
ho =22.5 m=0.022 mm
μd =
1∧ho 0.022 = =0.044 c 0.5
S=¿ 0.264
¿
n s=
( rc ) f
5.762 =3.99 μ ns
1440 =24 rpm 60
f =¿
1000 ¿ ¿ r 2 μ ns ( ) =0.264=¿ c p
−8
f =2.2× 10
∆ t=
¿
8.3 Cp(c + μ) Fy
8.3 cμ(5.79) 3.99
¿ 24.08 ℃
For,
T av=T 1 +
∆t 2
¿ 40+
24.08 2
¿ 52.04 ℃
Atmosphere temperature ¿ 30
T av=T 1 +
¿
40+
∆t 2
30 2
¿ 55 ℃
…Ans.
