Me6503 Dme Unit 5 Study Notes 2015
Me6503 Dme Unit 5 Study Notes 2015
Me6503 Dme Unit 5 Study Notes 2015
Prepared by:
R. SENDIL KUMAR, AP/MECH
Unit 5 - DESIGN OF BEARING AND MISCELLANEOUS ELEMENTS
PART - A
1. Classify the types of bearings. (MAY/JUNE 2010)
i. Depending upon the type of load coming upon the shaft:
a. Radial bearing
b. Thrust bearings.
ii. Depending upon the nature of contact:
a. Sliding contact
b. Rolling contact bearings or Antifriction bearings.
2. What are the required properties of bearing materials? (MAY/JUNE 2010)
Bearing material should have the following properties.
i. High compressive strength
ii. Low coefficient of friction
iii. High thermal conductivity
iv. High resistance to corrosion
v. Sufficient fatigue strength
vi. It should be soft with a low modulus of elasticity
vii. Bearing materials should not get weld easily to the journal material
5. State any points to be considered for selection of bearings.(or) List any six types of bearing
materials. (NOV/DEC 2010)
Lead based babbit, tin based babbit, leaded bronze, copper lead alloy, gun metal, phosphor
bronze.
6. Discuss the forces acting on the connecting rod. Or Under what force the big end bolts and
caps are designed. (NOV/DEC 2011) (NOV/DEC 2012)
The combined effect of (i)load on the piston due to the gas pressure and due to inertia of the
reciprocating parts, and(ii)the friction of the piston rings, piston, piston rod and cross
head.1.inertia of the connecting rod.2.the friction force in the gudgeon and crank pin bearing.
7. List the basic assumption used in the theory of hydrodynamic lubrication?
(NOV/DEC 2011)
a) The lubricant obeys newtons law of viscous flow.
b) The pressure is assumed to be constant throughout the film thickness.
c) The lubricant is assumed to be incompressible.
d) The viscosity is assumed to be constant throughout the film thickness.
e) The flow is one dimensional.
8. Classify the sliding contact bearings according to the thickness of layer of the lubricant
between the bearing and journal. (MAY/ JUNE 2012)
1. Thick film bearing
2. Thin film bearing
Solution:
t 5t
4t
a = 11t2
419 4
Ixx = t
12
k xx2 = 3.18 t2.
ii. Load due to burning of gas (FG): From PSG DDB Pg. No. 7.122
d2 1002
FG = p 3 23561.94 N
4 4
= 6 x 23561.94 = 141371.67 N
c a
Fcr 2
l
1 c
k xx
330 11t 2
141371.67 2
1 300
1
7500 3.18t 2
330 11t 2 3630t 2
141371.67 2
3.77 t 3.77
1 2
t t2
3630t 4
141371.67
t 2 3.77
141371.67(t 2 3.77) 3630t 4
3630t 4 141371.67t 2 532971.196 0
3630 t 4 38.95t 2 146.824 0
Height of I section = 5t = 5 x 7 = 35 mm
Width of I section = 4t = 4 x 7 = 28 mm
L1
1.75 (Assume Pb = 13 N/mm2)
d1
FG = L1 x d1 x Pb
23561.94 = 1.75d1 x d1 x 13
d1= 32.18 mm = 33 mm
L1 = 1.75(33) = 57.75 = 58 mm
L2
1.375 (Assume Pb = 8 N/mm2)
d2
FG = L2 x d2 x Pb
L2 = 1.375(47) = 64.63= 65 mm
L1
1.75
d1
p 2 cos 2
Fi r cos
g l
r
2 N 2 2000
209.44rad / sec
60 60
l = 300 mm = 0.3 m
140
r = radius of crank = stroke length / 2 = = 70 mm = 0.07 m
2
20 1
2
Fi
(209.44) 0.07 1
9.81 0.3
0.07
Fi 7720.736 N
W.K.T
d c2
Fi n
4
d c2
7720.736 4 100
4
dc 4.95mm 5mm
d 5
Diameter of bolt d c 5.95 6mm
0.84 0.84
Fi x
Bending moment mc = (x=1.5d2)
6
7220.736 1.5 47
6
90718.65Nmm
btc2
Modulus, Z = (b=L2)
6
tc = 8.35 mm
tc = 8.5 mm
2. Design a journal bearing for a centrifugal pump with the following data:
Diameter of the journal = 150 mm
Load on bearing = 40 kN
Speed of journal = 900 rpm (NOV/DEC 2007 & MAY/JUNE 2012)
Given:
D = 150 mm, W = 40 kN, n = 900 rpm, Application = Centrifugal pump
Solution:
Zn
p min = 2844.5
L
Take D = 1.5, L = 1.5 D = 1.5 x 150 = 225 mm
iv. From PSG DDB Pg. No. 7.32, Diameter clearance C = 150 microns
= 150 x 10-3 mm
C 150103
Clearance ratio, D = 150 = 1 x 10-3
Zn 40900
P = 11.85 = 3037.97
It is higher than the minimum value given in PSG DDB Pg. No. 7.31.
vii. Calculation of .
33.25
10 10 3037.97
1
1103
0.0025
0.0126
Hg and Hd
Hg = .w .v Watts
w in Newton,
Dn
v = 60 in m/min,
D in meters,
n in rpm
0.15900
Hg = 0.0126 x 4000 x 60 = 3562.56 W
t 182 L D
Hd = k
L in meters
D in meters
K constant, assume = 0.484 heat dissipation
16182 0.2250.15
Hd 0.484
Hd = 80.61 W
Here Hg > Hd so artificial cooling is required to carry away the excess heat.
Summary of Design
Given: D= 100 mm, Radial clearance = 0.12 mm, W = 50kN, L=100 mm, n = 1440 rpm, Z=16
centipoise = 16 x 10-3 Ns/m2.
Solution:
16103 1440
60
2
s= 5106
0.24
100
s = 0.013
From PSG DDB Pg. No. 7.40, for = 360o, s = 0.013 and corresponding to L/D =1,
2 ho
the minimum film thickness variable = C = 0.071
0.071C
ho = 2 = 8.52 x 10-3 mm = 0.00852 mm
CD = 1, 1 CD = 0.24
100 = 2.4 x 10
-3
Hg = w v
1001440
= 2.4 x 10-3 x 50,000 x 60
Hg = 904.8 W
4. Select a bearing for a 40 mm diameter shaft rotates at 400 rpm. Due to a bevel gear
mounted in the shaft. The bearing will have to withstand a 5000 N radial load of the bearing
thrust load. The life of the bearing expected to be at least 1000 hrs.
Solution:
Select Series 62 and for d = 40 mm, From PSG DDB Pg. No. 4.13, bearing basic design no.
SKF 6208. The values of Co, C are
For X and Y values, from PSG DDB Pg. No. 4.4, Fa and e are given
Fa/Co e
0.13 0.31
0.12 0.06
0.25 0.37
0.12 0.06
0.012, and 0.006
10 10
Fa 3000 F
For 0.1875 0.19 , For value of a 0.19 0.13 5(0.012)
Co 16000 Co
Fa 3000
0.6 > e,
Fr 5000
So, X value = 0.56, s value from PSG DDB Pg. No. 4.2
Y value = 1.2
P = 7680 N
From PSG DDB Pg. No. 4.6 (Ball bearing), For 400 rpm and 1000 hrs life
C/P = 2.88
C
2.88
7680
This dynamic load is less than the tabulated (allowable) value i.e. 22800 N. So the suitable
bearing designation is SKF 6208.
5. Select a suitable ball bearing to support the overhung countershaft. The shaft is 60 mm
diameter and rotating at 1250 rpm. The bearing is to have 99% reliability corresponding to a
life of 4000 hrs. The bearing is subjected to an equivalent radial load of 6000N.
Solution:
1/ b
L ln(1 / p)
From PSG DDB Pg. No. 4.2
L '10 ln(1/ p10 )
Here, ln(1/ p10 ) = 0.1053, L = 4000 hrs, b = 1.34, p = 0.99
4000
(0.09544) 0.7463
L '10
L10 = 23.093 hrs
From PSG DDB Pg. No. 4.6, For life 23.093 hrs and 1250 rpm,
C
The value is 12.40
P 12.40 =C/P
C 2500 hrs
= 12.4
P
C = 12.4 x Fr
C = 12.4 x 6000
C = 74400N
1250 rpm
Select the bearing for C = 74400 N or C = 7440 kgf, and the diameter of the shaft is 60
mm. (From PSG DDB Pg. No. 4.15, series 64)
Result:
SKF 6412 is suitable bearing,
Co = 7100 kgf,
C = 8450 kgf.
6. A 70mm machine shaft is to supported at ends. If operates continuously for 8 hrs per day
,320 days per year for 8 years the load of speed cycle for one of the hearing are given below,
Solution:
P2=(0.563000+1.21000)1.5
=4320w
P3=(0.564000+1.42000)1.5
=7560w
1/3
rom pg 4.2:
= (4740)3x600x0.25+(4320)3x800x0.25+(7560)3x900x0.5
600+800+900
Fm = 4618.16 N
N = n1t1+n2t2+n3t3
= 600x0.25+800x0.25+900x0.5
N = 800 rpm
= 8x320x8
C= 9.83x4618.16 = 45396 N
7. A single row deep groove ball bearing no: 6002 is subjected to an axial thrust load of
1000N and a radial load of 2200N. find the expected life that 50% of the bearing will
complete under this condition.
Given:
Fa= 1000N
Fa = 2200N
Solution:
Co = 255kgf = 255*10N
C = 440kgf = 4400N
Fa/co=1000/2550 = 0.392
.. . Equivalent load
= (0.56*2200+1.83*1000)1.3
= 3980.6N
WKT,
Ln(1/R90)
Ln(1/0.9)
L50= 4.958*1.35
8. The load on the journal bearing is 150KN due to turbine of 300mm diameter running at
1800rpm determine the following
(1) Length of the bearing if the allowable bearing pressure is 1.6N/mm2
(2) Amount of heat to be removed by the lubricant per minute if the bearing temperature
is 600c and viscosity of the oil at 600c is 0.02kg/m-s and the bearing clearance is 0.25.
(NOV/DEC 2011)
Given:
W=150KN = 150X103N
D= 300mm=0.3m
N= 1800rpm
P= 1.6N/mm2
Z=0.02kg/m-s
C= 0.25mm
Solution:
A=ld=lx300=300lmm2
1.6= = =
= 33/108 +k
. )
= 33/108 + 0.002
. .
= 0.009+0.002= 0.011
Rubbing velacity,
.
V= = = 28.3 /
Q8 = 0.11x150x103 x28.3
= 46.695 J/s or W
= 46.695kw
9). Design a journal bearing for a centrifugal pump from the following data:
Load on the journal = 20000N
Speed of the journal = 900 rpm
Type of oil is = SAE10
o
For which absolute viscosity at 55 C = 0.017 kg/ms
Ambient temperature of oil = 15.5oC
Maximum bearing pressure for the pump =1.5N/mm2
Given:
W = 20000N
N = 900rpm
To= 55oC,ta=15.5oC
Solution:
Take / =1.6
WKT,
WKT,
= x 12.24 x + 0.002
.
= 0.0051 Ans
Qg = wv
= w ( )W
.
= 0.0051x20000 ( )
= 480.7 W
Qd = CA (tb-ta)
= cld (tb-ta)W
= Qg-Qd
= 4807.7 389.3
= 91.4 W
10. A journal bearing is to be designed for a centrifugal pump for the following
Data:
Load on the journal = 12kN,
Diameter of the journal = 75mm
Speed (N) = 1440rpm
Atmospheric temperature of the oil =16
Operating temperature of the oil = 60
Absolute viscosity of oil at 60 = 0.23 kg/ms
Give the systematic design of the bearing. (MAY-JUNE-2012)
11). A single row deep groove ball bearing is subjected to a radial force of 8kN and a thrust
force of 3kN. the rotates at 1200rpm the expected life L10th of the bearing is 20000hr the
minimum acceptable diameter of having for this application. (MAY/JUNE 2012)
Given:
Fn = 8kn
Fa = 3kn
N = 1200rpm
L10h = 20,000
d = 75mm
Solution:
WKT,
C = p(L10)1/3 = (8980)(1440)1/3
= 101406.04N
The shaft of 75mm diameter, bearing no.6315 (c= 112000) is suitable for the above data for this
bearing,
Co = 72000N
, = = 0.375
And = = 0.04167
STEP 2:
= 9850N
Result:
12. Design a connecting rod for an I.C. engine running at 1800 r.p.m. and developing a
maximum pressure of 3.15 N/mm2. The diameter of the piston is 100 mm; mass of the
reciprocating parts per cylinder 2.25 kg; length of connecting rod 380 mm; stroke of piston
190 mm and compression ratio 6:1. Take a factor of safety of 6 for the design. Take length to
diameter ratio for big end bearing as 1.3 and small end bearing as 2 and the corresponding
bearing pressures as 10 N/mm2 and 15 N/mm2 . The density of material of the rod may be
taken as 8000 kg/m3 and the allowable stress in the bolts as 60 N/mm2 and in cap as 80
N/mm2. The rod is to be of I-section for which you can choose your own proportions. Draw a
Prepared by R. SENDIL KUMAR Page 19
neat dimensioned sketch showing provision for lubrication. Use Rankine formula for which
the numerator constant may be taken as 320 N/mm2 and the denominator constant 1 / 7500.
2 MARKS
1. List the basic assumption used in the theory of hydrodynamic lubrication?
(NOV/DEC 2011)
1. Design of journal bearing for 12 MW, 1000 rpm steam turbine, which is supported by two
bearings. Take atmospheric temperature as 16o C and operating temperature of oil as 60o C.
assume viscosity of oil as 23 Ns/m2. (MAY/JUNE 2012)