notes on lesson
DESCRIPTION
Notes on Lesson. Failure Theories. Stress in machine components should be accurately computed. Designer must understand material limits to ensure a safe design. Design Factor. Factor of Safety (N) - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: Notes on Lesson](https://reader035.vdocuments.mx/reader035/viewer/2022062408/56813b89550346895da4b61f/html5/thumbnails/1.jpg)
Notes on Lesson
Faculty Name : V. KIRAN KUMAR Code : AU28
Subject Name : DESIGN OF MACHINE
ELEMENTS
Code : ME2303
Year : III Semester : V
Degree & Branch : B.E. – AUTOMOBILE Section : -
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Failure Theories
Stress in machine components should be accurately computed.
Designer must understand material limits to ensure a safe design.
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Design Factor Factor of Safety (N)
Suitable values depend on inherent danger, certainty of calculations, certainty of material properties, etc.
FailureComponent at Stress
Stress ExpectedN
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Static Stresses - Brittle Materials Percent elongation < 5%
• for parts in tension
• for parts in compression
• for parts with general stress
tus
N
cus
N
ct uu ssN211
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Example
The Gray Cast Iron (Grade 40) cylinder carries an axial compressive load of 75,000 lbs and a torque of 20,000 in lbs. Compute the resulting design factor.
Ø4.00”
Ø5.00”
R0.25”
R0.25”
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Static Stresses - Ductile Materials
Percent elongation > 5% Distortion Energy Theory
Define von Mises Stress
• For nominal stress
• For localized stress
'ys
N
'us
N
2122
21'
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Static Stresses - Ductile Materials
Percent elongation > 5% Maximum Shear Stress Theory
• For nominal stress
• For localized stress
maxmax 2yys ss
N
maxuss
N
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Example
Specify a diameter for the middle portion of the rod, if it is to be made from AISI 1040-hot rolled steel.
5000 lbs
450
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Example
For the seat support shown, specify a standard structural tube to resist static loads shown. The tube has properties similar to AISI 1020 hot-rolled steel. Use a design factor of 3.
200 lb
400 lb
20”14”
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Repeated Loads
Time
Stress alt
mean
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Example
The notched bar is machined from AISI 1020 steel. This bar is subjected to a load that varies from 2000 lb to 3000 lb. Determine the mean and alternating nominal stresses.
0.1” R
1”
1.25”
.75”
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Fatigue Strength
R.R. Moore Test
103 104 105 106 107 108
Endurance Strength, sn
Cycles to Failure, N (log)
Alternating Stress, a
Motor
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Endurance Strength
sn = Endurance strength Listed in tables
If no information is available, use
• sn 0.5 su (Steel)
• sn 0.4 su (Aluminum)
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Adjusted Endurance Strength
The data from the standard R.R. Moore test is adjusted for a particular application.
sn’ = Adjusted endurance strength
= (Cs) (Cm) (Cst) (CR) (sn)
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Size and Stress Type Factors
– Cs = Size Factor
• D< 0.4 in Cs = 1.0
• 0.4 < D 2.0 in Cs = (D/0.3)-0.068
• 2.0 < D 10.0 in Cs = D-0.19
For rectangular sections, D=.808(h b)1/2
– Cst = Stress Type Factor
• = 1.0 for bending• = 0.80 for axial tension• = 0.50 for torsion
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Material and Reliability Factor– Cm = Material Factor
• = 1.0 for wrought steel• = 0.80 for cast steel• = 0.70 for cast iron
– CR = Reliability Factor
• 50% CR = 1.0
• 90% CR = 0.90
• 99% CR = 0.81
• 99.9% CR = 0.75
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Example
The notched bar is machined from AISI 1020 steel. This bar is subjected to a load that varies from 2000 lb to 3000 lb. Determine the endurance limit of the material.
0.1” R
1”
1.25”
.75”
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Repeated Stresses - Ductile Materials
Distortion Energy Theory
Define repeated von Mises Stress
• Solderberg criterion
n
at
y
m
s
K
sN '
''1
mmmmm 2122
21'
aaaaa 2122
21'
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Repeated Stresses - Ductile Materials
Maximum Shear Stress Theory
sn
at
sy
m
s
K
sN '
)()(1 maxmax
• ssy = 0.5 sy
• s’sn = 0.5 sn
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Example
The notched bar is machined from AISI 1020 steel. This bar is subjected to a load that varies from 2000 lb to 3000 lb. Comment on the robustness of the design.
0.1” R
1”
1.25”
.75”
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Example
Comment on the robustness of a 1-1/4” round bar made from AISI 1213 C-D steel. It carries a constant tensile load of 1500 lbs, a bending load that varies from 0 to 800 lbs at the senter of the 48” length and a constant torque of 1200 in lbs.
48”
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Connect power transmission components.
Inherently subjected to transverse loads and torsion.
Shafts
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Shaft Forces Gears
As before
Wr
Wt
T
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Shaft Forces
Chains
Fslack = 0
Ftight
D
TFtight
2
D
T
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Shaft Forces
V-belts
Fslack
Ftight
D
TFtight
5.2
D
T
D
TFslack 2
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Shaft Forces
Flat belts
Fslack
Ftight
D
TFtight
3
D
T
D
TFslack
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Material Properties For steady load (torsion)
sys=.5sy
For fatique load ( bending)
sn’=cs cR sn
cT = 1 (bending)
cm = 1 (wrought steel)
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Stress Concentrations
Keyseats– Sled Runner Kt = 1.6
– Profile Kt = 2.0
– Woodruff Kt = 1.5
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Stress Concentrations
Shoulders– Sharp, Bearing (r/d .03) Kt = 2.5
– Round, Gear Bore (r/d .17) Kt = 1.5
Grooves– Retaining Rings Kt = 1.5
Try not to let Kt’s overlap. Leave .10 - .15” between
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Strength Analysis
Bending stress
Torsion stress
S
MK
I
cMK tt
S
T
J
rT
2
32
3DS
c
I
r
J2For round sections
For round sections
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Strength Analysis
Mohr’s circle and Solderberg
S
sTsMK
N
ynt22 /
43
'/1
Suggested Design Factors:
• N=2 smooth operation
• N=3 typical industrial operation
• N=4 shock or impact loading
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Minimum Acceptable Diameter
The designer must size the shaft.– Solve for appropriate diameters
22 /
4
3'/
32ynt sTsMK
ND
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Example
Determine a suitable diameter for a shaft made from AISI 1144 OQT 1000. It is subjected to a reversing bending moment of 3000 ft lbs and a steady torque of 1800 ft lbs. The shaft has a profile keyway.
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Example
The shaft shown is part of a grain drying system At A, a 34 lb. propeller-type fan requires 12 hp
when rotating at 475 rpm. A flat belt pulley at D delivers 3.5 hp to a screw
conveyor handling the grain. All power comes to the shaft through the v-belt at
C.
Using AISI 1144 cold drawn steel, determine the minimum acceptable diameter at C.
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Example
12” 10” 10” 4”
A B C D E
Sheave C150
Sheave D
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Components used to securely mount power transmitting elements on a shaft.
Shafts Accessories
Axial
Rotational
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Keys Allow torque to be transferred from a shaft
to a power transmitting element (gear, sprocket, sheave, etc.)
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Key Design Use a soft, low strength material
(ie, low carbon steel)
Standard size H=W=1/4 D Design length
based on strength
H
L W
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Standard Key Sizes
Shaft Dia. (in) W (in)
2
22 WDHDS
T S
HW
.005.2
22
inWDHD
T
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Key Design
Key Shear
Failure Theory
Length
yDWs
TNL
4
DLW
T
A
F 2
TD
LWssN yy
42
D
T
D
TF
2
2/
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Example
Specify a key for a gear (grade 40, gray cast iron) to be mounted on a shaft (AISI 1144, hot rolled) with a 2.00 in. diameter. The gear transmits 21000 lb-in of torque and has a hub length of 4 in.
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Retaining Rings Also known as snap rings Provides a removable shoulder to lock
components on shafts or in bores. Made of spring steel, with a high shear
strength. Stamped, bent-wire, and spiral-wound.
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Retaining Ring Selection
Based on shaft diameter & thrust force
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Set Screws Setscrews are fasteners that hold collars, pulleys, or gears on shafts. They are categorized by drive type and point style.
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Standard Set Screw Sizes
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Set Screw Holding
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Pins
A pin is placed in double shear Holds torsion and axial loads
ysD
NTd
8
D
d
Hole is made slightly smaller than the pin (FN1 fit)
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Example
Specify a pin for a gear (grade 40, gray cast iron) to be mounted on a shaft (AISI 1144, hot rolled) with a 2.00 in. diameter. The gear transmits 21000 lb-in of torque and has a hub length of 4 in.
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Roll Pins
Easier disassembly
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Collars
Creates a shoulder on shaft without increasing stock size.
Held with either set screw or friction (clamped)
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Mechanical Couplings Couplings are used to join two shafts Rigid couplings are simple and low cost.
But they demand almost perfect alignment of the mating shafts.
Misalignment causes undue forces and accelerated wear on the shafts, coupling, shaft bearings, or machine housing.
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Mechanical Couplings
In connecting two shafts, misalignment is the rule rather than the exception. It comes from such sources as bearing wear, structural deflection, thermal expansion, or settling machine foundations.
When misalignment is expected, a flexible coupling must be used.
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Mechanical Couplings Selection factors include:
Amount of torque (or power & speed)
Shaft Size
Misalignment tolerance
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Fasteners, Powers Screws, Connections
Helical thread screw was an important invention.
Power Screw, transmit angular motion to liner motion
Transmit large or produce large axial force
It is always desired to reduce number of screws
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Definition of important Terminologies
Major diameter d, Minor diameter dr Mean dia or pitch diameter dp
Lead l, distance the nut moves for one turn rotation
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Single and Double threaded screws
Double threaded screws are stronger and moves faster
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Screw Designations
United National Standard UNS
International Standard Organization
Roots and crest can be either flat or round
Pitch diameter produce same width in the thread and space,
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Coarse thread Designated by UNC
Fine Thread UNF, is more resistance to loosening, because of its small helix angle.
They are used when Vibration is present Class of screw, defines its fit, Class 1 fits have
widest tolerances, Class 2 is the most commonly used
Class three for very precision application Example:1in-12 UNRF-2A-LH, A for Ext. Thread
and B for Internal, R root radius Metric M10x1.5 10 diameter mm major
diameter,1.5 pitch
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Some important Data for UNC, UNF and M threads Lets Look at the Table 8-1 on Page 398
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Square and Acme Threads are used for the power screw
Preferred pitch for Acme Thread
d, in 1/4 5/16 3/8 1/2 5/8 3/4 7/8 1 1 1/4
p,in 1/16 1/14 1/12 1/10 1/8 1/6 1/6 1/5 1/5
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Mechanics of Power Screws
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Used in design to change the angular motion to linear motion, Could you recall recent failure of power screw leading to significant causalities
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What is the relationship between the applied torque on power screw and lifting force F
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Torque for single flat thread
)sec
sec(
2
fld
fdlFdT
m
mmR
)(2
)(2
fld
lfdFdT
fld
fdlFdT
m
mmL
m
mmR
If the thread as an angle α, the torque will be
Wedging action, it increases friction
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Stresses in the power Screw
pnd
F
A
V
pnd
F
pnd
Fd
T
tr
trb
tmB
3
2
3
6
2/
163
Shear stress in the base of the screw
Bearing stress
Bending stress at the root of the screw
Shear stress in the thread
nt number of engaged thread
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Loading to the fasteners and their Failure considerations
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Bolts are used to clamp two or more parts
It causes pre tension in the bolt Grip length is the total thickness of parts and washersl
ld
lt
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t2
ldh
L’ effective grip= h+t2 if t2<d
=h=d/2 for t2 d
lt=L’- ld
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Failure of bolted or riveted joints
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Type of Joints Lap Joint (single Joint) But Joint
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Example 1
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Example 2
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Example 2
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Example 3
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Weld
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Weld under Bending
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Springs
Used to: Exert force Store energy
Flexible machine elements
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Spring Rate
Effective springs have a linear deflection curve. Slope of the spring deflection curve is the rate
Force
Deflection
k1
L
Fk
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Example
A compression spring with a rate of 20 lb/in is loaded with 6 lbs and has a length of 1.5 in. Determine the unloaded spring length (free length)
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Geometry
Wire diameter, Dw (Standard gages)
Mean Diameter, Dm
Dm = Do - Dw
Do
Di
Dw
L
Dw
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Spring Parameters
Spring index
C > 5 (manufacturing limits) Active coils, Na
= N for plain ends
= N-1 for ground ends
= N-2 for closed ends
w
m
D
DC
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Deflection
Deflection for helical springs
w
a
w
am
GD
NFC
GD
NFD 3
4
3 88
Spring rate for helical springs
a
w
NC
GDk
38
G = Shear modulus
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Example
A helical compression spring is formed from 35 gage music wire with 10-1/4 turns and an O.D. if 0.850 in. It’s ends are squared. The free length is 2 inches. Determine the force to press the spring solid.
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Stress Analysis
Spring wire is in torsion
F
V
T2
8
wD
CFK
J
rT
Wahl factor, KAccounts for the curvature of the
wire
CC
CK
615.
44
14
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Example
A helical compression spring is formed from 35 gage music wire with 10-1/4 turns and an O.D. if 0.850 in. It’s made from A228 and the ends are squared. The free length is 2 inches.
If the spring is repeatedly compressed to 1.3 in, do you expect problems?
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Design Procedure
Select a material Compute required spring rate Estimate Dm based on size constraints
Determine required Dw (use K=1.2)
Select standard wire Verify actual stress is satisfactory. Compute number of coils required.
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Example
Design a helical compression spring to exert a force of 22 lbs when compressed to a length of 1.75 in. When its length is 3.0 in, it must exert 5 lb. The spring will be cycled rapidly. Use ASTM A401 steel wire.
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Provides support for machine elements, while allowing smooth motion.
=0.001 - 0.005
Rolling Element Bearings
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Types
Single-rowRadial Ball
Angular Contact Ball
Radial Roller
Angular Roller
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Types
Spherical Roller
Needle
Thrust
Tapered Roller
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Ball Bearings
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Stress Analysis
Contact Stress
c=300,000 is not unusual
Balls, rollers and races are made from extremely high strength steel
ex. AISI 52100
sy = 260,000 psi
su=322,000 psi
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Bearing Load/Life
Test (fatigue) data
L10 Life (cycles)
Radial Load (lbs)k
P
P
L
L
2
1
1
2
Empirical relationship:
•k=3.0 (ball)
•k=3.33 (roller)
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Example
A bearing is mounted on a shaft rotating at 1200 rpm. The bearing has been tested to have a L10 life of 300 hrs, when loaded with 500 lbs. Determine the expected L10 life, if the load is increased to 700 lbs.
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Manufacturer’s Data Vendors publish the
Basic Dynamic Load rating (C) of a bearing at an L10 life of 1 million cycles.
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Bearing Selection
Determine the design life (in cycles) Determine the design load
Pd = V R
Calculate the required basic dynamic load
Select a bearing with (C > Creq’d) and a bore that closely matches the shaft diameter.
kd
ddreq
LPC
1
6' 10
•V=1 for inner race rotation•V=1.2 for outer race rotation
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Example
Specify suitable bearings for a shaft used in an grain dryer. The shaft rotates at 1700 rpm. The required supporting loads at the bearing are
and the minimum acceptable diameter is 2.16”.
RBx=589 lbRBy=164 lb
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Mounting of Bearings
Shaft/bearing bore has a light interference fit. Housing/outer race has a slight clearance fit.
Check manufacturers catalog
Match maximum permissible fillet radius. Shaft or housing shoulders not to exceed
20% of diameter.
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Mounted Bearings
Pillow block
Bearing is inserted into a cast housing, with base or flange slots, which can be readily attached to a machine base.
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Bearings with Varying Loads Compute a weighted average load based on
duty cycle.
p
i
ip
im N
NFF
1
Fm=equivalent load
Fi= load level for condition i
Ni= cycles for condition i
p = exponent for load/life
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Example
Bearing 6211 is carrying the following load cycle, while rotating at 1700 rpm.
Stage Load (lbs) Time (min)
1 600 480
2 200 115
3 100 45
Compute the bearing L10 life in minutes.
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Radial & Thrust Loads
Calculate an equivalent load
P=VXR +YT
T=thrust loadX factors dependingY on bearing
=
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Thrust factors, Y
Deep -groove, ball bearings
X = 0.56 for all values of Y
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Example
A bearing is to carry a radial load of 650 lb and a thrust load of 270 lb. Specify a suitable single-row, deep-groove ball bearing if the shaft rotates at 1150 rpm and the design life is 20,000 hrs.