lectures-part ii
TRANSCRIPT
STRESS
(a) Normal, tensile (b) normal, compressive; (c) shear; (d) bending; (e) torsion; (f) combined
JyTI
yMAP
b
sct
=
=
=
τ
σ
σ ,,
Elementary equations. No discontinuity in cross-section. Holes, shoulders, keyways, etc.
Critical section
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Finite element model to calculate stresses
High concentration of elements are required to estimate stress level.
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AB
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Axial Load on Plate with Hole
avg
maxtK
factorion concentrat Stress
σσ
=
Plate with cross-sectional plane
Half of plate with stress distribution.
Stress Concentration
hdbP
)(avg −=σ
Geometric discontinuity increases the stress. Stress concentration is a highly localized effect.
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Stress concentration factor for rectangular plate with central hole.
EX: A 50mm wide and 5mm high rectangular plate has a 5mm diameter central hole. Allowable stress is 300 MPa. Find the max. tensile force that can be applied.
Ans: d/b = 0.1; Kt=2.7
A = (50-5)×5
P = 25 kN
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Stress concentration factor under axial load for rectangular plate with fillet
EX: Assume H=45mm, h=25mm, and fillet radius r=5mm. Find stress concentration factor.
Ans: ~1.8
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Stress concentration factor under axial load for rectangular plate with groove
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Stress concentration factor under axial load for round bar with fillet
Gap between lines decrease with increase in r/d ratio.
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Stress concentration factor for round bar with groove3/17/2015 9
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Ex: Assuming 80 MPa as allowable strength of plate material, determine the plate thickness
Maximum stress near fillet
Maximum stress near hole
Allowable
Kt=1.8 Kt=2.1
bbfillet300
3050008.1 =⎟⎟
⎠
⎞⎜⎜⎝
⎛=σ
( ) bbhole700
153050001.2 =⎟⎟
⎠
⎞⎜⎜⎝
⎛−
=σ
80=allowableσ b=8.75 mm
Stress concentration factor under bending for rectangular plate with fillet
EX: Assume H=45mm, h=25mm, and fillet radius r=5mm. Find stress concentration factor.
Ans: ~1.5
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Stress concentration factor under bending for rectangular plate with central hole
Concentration factor for thick plate with central hole is higher compared to thin plate with same size hole.
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Stress concentration factor under bending for rectangular plate with groove
Decrease in Kt for r/h > 0.25 is negligible.
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Stress concentration factor under bending for round bar with fillet3/17/2015 14
Stress concentration factor under bending for round bar with groove3/17/2015 15
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Ex: Assuming 100MPa as allowable stress, determine the shaft dia, d.
Due to symmetry, reaction force at each bearing = 1250 N.Stress concentration will occur at the fillet.Kt=1.6
( )( )33
35012503232dd
Mavg ππ
σ ×==
( )( )
10035012502.516.1 3max =×
==davg π
σσ Diameter d=41.5 mm
Stress concentration factor under torsion for round bar with fillet
Stress concentration under torsion loading is relatively low.
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Stress concentration factor under torsion for round bar with groove3/17/2015 18
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Notch Sensitivity“Metals can accommodate stress concentration by deforming & redistributing load more evenly”.Some materials are not fully sensitive to the presence of geometrical irregularities (notch) and hence for those materials a reduced value of Kt can be used. Notch sensitivity
parameter q = 0 means stress concentration (Kf ) factor = 1; and q=1 means Kf = Kt.
11
−
−=
t
f
KK
q
Notch SensitivityIn case of loading/unloading …. Materials show better sensitivity (lesser q)…..
If notch sensitivity data is not available, it is conservative to use Ktin fatigue calculations.
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11
−
−=
t
f
KK
qtf KK ≤≤1
ra
q+
==1
1senstivitynotch
Neuber’s constant for steels
aSut (ksi) (in0.5)
50 .13
55 .118
60 .108
70 .093
80 .08
90 .07
100 .062
110 .055
120 .049
130 .044
140 .039
160 .031
180 .024
200 .018
220 .013
240 .009
Neuber’s constant for Annealed Aluminum
Sut (ksi) (in0.5)
15 .341
20 .264
25 .217
30 .18
35 .152
40 .126
45 .111
a
89.025./062.01
1
1
1ksi) 100strength ultimate andinch 0.25 radius
fillet (havingbar step steelfor senstivityNotch
=+
=+
=
==
ra
q
Kf is commonly used for fatigue loading cases.
Cast irons have a very low q (≅ 0.2) value. For such materials 'Fracture Mechanics' techniques
1 ksi = 6.895 MPa
Material selection for a plate having central hole and is subjected to Tensile force
EX: A 50mm wide (b) and h mm high rectangular plate has a 5mm diameter central hole. Length of plate is equivalent to 100mm. Select a lightest but strong material which bear tensile force P = 25 kN.Ans: Mass = ρ × (50-5)× h × 100*10-9 ; A = (50-5)× h
( ) ( ) MPahhhdb
PKt1500
550250007.2 =−
=−
=σ
σρ
σρ
6
6-
10*6750M or,
1500 10*4.500M or,
−=
=
⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛
σρ
1010 log00675.0
log M
d/b = 0.1; Kt=2.7;
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⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛
σρ
1010 log00675.0
log M( ) ( )σρ 101010 loglog
00675.0log −=⎟
⎠⎞
⎜⎝⎛ M
( ) ( ) ⎟⎠⎞
⎜⎝⎛+=
M00675.0logloglog 101010 ρσ M < 0.025 kg=> -0.57
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mmheh
8.0389.11500=⇒==σ
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Commonly available. Economic.
Stress concentration ???
Mass reduction ????
Avg. density = 1645 kg/m3
Mass = 0.0059 kg
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“Factor of Safety”FOS is a ratio of two quantities that have same units:
Strength/stress Critical load/applied loadLoad to fail part/expected service loadMaximum cycles/applied cyclesMaximum safe speed/operating speed.
NOTE: FOS is deterministic. Rational assessment of the risks associated with a particular design. Data are statistical and there is a need to use Probabilistic approach.
Is design of element/assembly safe?
FoS < 1 ?
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Variation in Material Strength (MPa)Material (AISI, rolled)
Range Mean St. Deviation
1080 865 - 975 920 18.33
1095 865 - 1070 967.5 34.17
1030 495 - 610 522.5 19.17
1040 565 - 690 627.5 20.83
1050 650 - 800 725.0 25.00
1060 725 - 900 812.5 29.17
Random variables: Friction, Load, Strength, Dimensions, environment, ….
Probability density function
( )
( ) 1
21
2
21
=
=
∫∞+
∞−
⎟⎟⎠
⎞⎜⎜⎝
⎛ −−
dSSf
eSf s
SS
s
σ
πσ
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1
2
−
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−
=∑ ∑
N
Nss i
i
sσ
Mean value worst value Probabilistic value
Cumulative probability!!
Sigma, two sigma, three sigma….
coefficient of variation (COV).. 0.25 unacceptable
Probability density function
Ex: Measured ultimate tensile strength data of nine specimen are: 433 MPa, 444, 454, 457, 470, 476, 481, 493, and 510 MPa. Find the values of mean and std. dev. Assuming normal distribution find the probability density function.
( )
( ) 1
234.241
34.2467.468
2
34.2467.468
21
=
=
==
∫∞+
∞−
⎟⎠⎞
⎜⎝⎛ −
−
dSSf
eSf
MPaMPa
S
s
s
π
σμ
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1
2
−
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−
=∑ ∑
N
Nss i
i
sσ
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EX. NOMINAL SHAFT DIA. 4.5mmNUMBER OF SPECIMEN 34
4.58mm0.0097
d
d
σμ
4.59,4.34,4.5796,4.50, 4.582,4.5847……………4.5948
6
4.5294
0.0987
1
2
−
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−
=∑ ∑
N
Ndd i
i
dσ
⎟⎟⎠
⎞⎜⎜⎝
⎛ −−
= d
did
d
edf σμ
πσ21
21)( NOTE: Variation in
stress level occurs due to variation in geometric dimensions.
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Prob: A steel bar is subjected to compressive load. Statistics of load are (6500, 420) N. Statistics of area are (0.64, 0.06) m2. Estimate the statistics of stress.
Ans: (10156, 1156.4) Pa.
AP
=∴σ
2/1
22
22
stressofdeviation Standard
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛∂∂
+⎟⎠⎞
⎜⎝⎛∂∂
= AP APσσσσσ σ
Ref: Probabilistic Mechanical Design, Edward B. Haugen, 1980.
FoS vs Statistical DesignDeterministic. Lack of information. Conceptual design Based on availability of data. Find the mean and standard-deviation values of dependent variables. Actual design.
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22σ
σ
σσσ
μμμσ
σ
+=
−=−=
−=
y
y
sQ
SQy
y
orSQ
SQ
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( ) ( )MPaMPaS y 15,184 & 32,270:arebar tensilea of Stress andStrength :Ex
== σ
Ref: Probabilistic Mechanical Design, Edward B. Haugen, 1980.
( )2
21
21 ⎟
⎟⎠
⎞⎜⎜⎝
⎛ −−
= Q
Q
eQf σμ
πσ
Q
QQZ
σμ−
=
variablenormalLet
dzePz
f2
43.2 2
21 −−
∞−∫= π
R = 1-0.0075 ????
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xQyxQ
xyQyxQxCQ
CxQCQ
1===
±=+=
==
x
yx
yx
yx
x
x
CCC
μ
μμ
μμ
μμμ
μ
1
±+
ALGEBRAIC MEAN STD. DEVIATIONFUNCTIONS
2
22222
2222
22
0
xx
yyxxy
yxxy
yx
x
xC
μσ
μσμσμ
σμσμ
σσ
σσ
+
+
+
Ref: Probabilistic Mechanical Design, Edward B. Haugen, 1980.
Example: Tie-rod forged steel component.DIA.=(30, 1.2) mmSTRENGTH 300 MPa to 540 MPaLOAD 30000 ± 9600 N
?? FACTOR OF SAFETYa. BASED ON MEAN VALUE
±
( )
MPaS
mmANP
420
707304
;.....30000 22
=
===π
9.9
4328.42
=∴
=∴
SF
MPaAP
b. BASED ON Worst Case
( )
MPaSNP
mmA
MIN
MAX
MIN
30039600
4.5474.264
22
==
==π
APPLIED STRESS
15.4342.72
300
342.724.547
39600
≈=
==
MINSF
MPa
c. NORMAL DISTRIBUTION
( ) ( )( ) ( )( ) ( )mmd
MPaSNP
d
S
P
2.1;30,
40;420,3200;30000,
=
=
=
σ
σ
σ
3457.94.405672.3770
−=−
=Z
( ) 22 707304
mmA ==π
MPaQ
MPaAP
5672.3774328.42420
4328.42
=−=
==∴σ
2/1
22
22
stressofdeviation Standard
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛∂∂
+⎟⎠⎞
⎜⎝⎛∂∂
= dP dPσσσσσ σ
2/1
22
32
2 8
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −+= d
P
dP
Aσ
πσσ σ
( )[ ] 66.52.1*83.25262.42/122 =−+=σσ
4.4066.540 22 =+=Qσ
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Ex: Consider a structural member with S( ) subjected to a static load that develops a stress σ( ). Find the SAFETY of member.
Mean value approach, FOS = 40/30. 100% SAFE
Worst case approach, FOS = 22/54. 40% SAFE
6,40=sμ
8,30=σμ
830
==
σ
σ
σμ
640
==
s
s
σμ10,10 == QQ σμ
1086
10304022 =+=
=−=
Q
Q
σ
μ
0
0
≥
≥−
Q
Sy σ
110
100 0at
variableNormal
−=−
==
−=
−=
ZQ
sQQxZ
Qσμ
1086
10304022 =+=
=−=
Q
Q
σ
μ
In the present case Probability of failure is 0.1587 & reliability is .8413.
Selecting stronger material (mean value of strength = 50 units)!!!!.. CALCULATE Reliability
Reducing standard
deviation !!
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Z-Table provides probabilityof failure
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Useful
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Ex: A round 1018 steel rod having yield strength (540, 40) MPa is subjected to tensile load (220, 18) kN. Determine the diameter of rod reliability of 0.999 (z = -3.09).
MPad
MPad
MPaMPa s
22
s
4/18000;
4/220000
40;540Given
πσ
πμ
σμ
σσ ==
==
Q
Q
Z
Z
Q
Q
where
dZeRQ
Z
σμ
πσμ
−=
=−
= ∫+∞
−
0
21
Z
21;
0
2
2
22
2
7200040
880000540
⎟⎟⎠
⎞⎜⎜⎝
⎛+=
−=
d
d
Q
Q
πσ
πμ
2
2
22 880000540720004009.3
dd ππ−=⎟⎟
⎠
⎞⎜⎜⎝
⎛+ d = 26 mm
Assuming st. dev. Of d is zero
Example: Stress developed in a machine element is given by:
Given P = (1500, 50) N, Strength = (129, 3) MPa, L1=(150, 3) mm, L2=(100, 2) mm. Assume std. dev. of d is 1.5% mean value of d. k = 0.003811.
Determine distribution of d if the maximum probability of machine-element-failure is 0.001
∑ ⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
= =ni xi
ix12
2
:by expressed isfunction complex a ofdeviation Standard σφσμ
φ
( )( )22
21
3 344/ LLkdP +=σ
( ) ( ) ( ) ( )
[ ]
3
2/13
2/1
22
32
2
3
22
42
2
3
2/1
22
2
22
1
22
22
1136200
290472614204183012291.11
002.085216003.0170430015.04136355022724
21
d
d
dd
dd
LLdP
e
de
LLdP
μσ
μσ
μμ
μμσ
σσσσσσσσσ
σ
σ
σ
σ
=
+++=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+⎟⎠⎞
⎜⎝⎛∂∂
+⎟⎠⎞
⎜⎝⎛∂∂
= Statistically independent
43
( )( )3
22
21
3
34087000344/
d
LLdP k
μμ
μμμμμ
σ
σ
=
+=
( )
( )
m 001.0 m 06686.0
11031417482.11363000
113620063
340870006129009.3
2
3
2
32
21
2
32
3
==
⎟⎟⎠
⎞⎜⎜⎝
⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛+
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+
−−=−=
d
d
dd
d
d
e
eZ
σμ
μμ
μ
μ Calculating FOS = Strength/stress
FOS =129/114=1.13
Probabilistic Approach to DesignConventional design follows “Deterministic Approach” by disregarding statistical nature of
Material propertiesComponent dimensionsExternally applied load
Deterministic design considers uncertainties byApplying “Factor of safety” Considering absolute worst case
There is a growing trend toward using a probabilistic approach to better quantify uncertainty and thereby increase reliability
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Failure of Machine ElementThere are only two ways in which an element fails:
ObsolescenceLoss of function
Element losses its utility due to:Change in important dimension due to wear.
Change in dimension due to yielding (distortion)
Breakage (fracture).
Jamming (friction)
Yielding (distortion)
Wear
FractureJamming3/17/2015 47
Design of Components before Minor II
ShaftsKeysCouplingsBearings
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These components are employed to separate rotating/sliding elements from stationary/ relatively-stationary elements.
Bearing Classification
Anti-Friction !!
Misnomer
Rolling Element Bearings
“Rotation is always easier than linear motion”. Low friction & moderate lubricant requirements
are two important advantages of rolling bearings.
If you can buy it, don’t make it!Bearing selection….
Four main components ?
Bearing Terminology
RollerNeedle roller
Cylindrical Roller BearingHigher coefficient of friction because of small diameter rollers and rubbing action against each other
max
max
).06.4/(
).37.4/(
WZF
WZF
rollerr
ballr
=
=
Outer ring or Cup
Inner ring or Cone
Cage
Radial (Load) Bearing (s)
Thrust (Load) Bearing (s)
Combined (Load) Bearing (s)
Ball Cylindrical roller Angular contact ball Tapered roller Sphericalroller
Spherical roller Ball Cylindrical roller
Rolling Element BearingsLoad Direction Misalignment
CapacityRadial Axial Both High Med Low
Deep groove ball y y y
Cylindrical Roller y Some types
y
Needle y y
Taper Roller y y y y
Self Aligning Ball y y y
Self Aligning Spherical Roller
y y y
Angular contact ball
y y y
Thrust ball/roller y y
Equivalent load: P = V X Fr + Y Fa
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Equivalent Dynamic Load
Equivalent load: P = V X Fr + Y Fa
eFFYFXPeF
rar
r
>+=≤=
/Fwhen/ Fwhen F Pringinner of rotation Assuming
a
ar
e is a dimensionlessratio, indicating axial
load lower than a certain limit does not
affect total load
Value of e depends on arrangement & static load capacity (CO ) of bearing
V Rotation factor
X Radial factor
Fr Applied radial load
Y Thrust factor
Fa Applied thrust load
Basic Dynamic Load Rating: CRadial load (thrust load for thrust bearings) which a group of identical bearings with stationary outer rings can theoretically endure one million revolutions of inner ring.
Static Load Rating: C0
Radial load causing permanent deflection greater than 0.01% of ball dia.
http://www.skf.com/group/products/bearings-units-housings/ball-bearings/deep-groove-ball-bearings/single-row-deep-groove-ball-bearings/single-row/index.html
Bearing type Inner ring Single row Double row eRotating Stationary Fa/VFr > e Fa/VFr ≤ e Fa/VFr > e
Deep groove ball bearing
Fa/C0 V V X Y X Y X Y.014.028.056.084.11.17.28.42.56
1 1.2 0.56 2.301.991.711.551.451.311.151.041.00
1 0 0.56 2.301.991.711.551.451.311.151.041.00
.19
.22
.26
.28
.3
.34
.38
.42
.44Angular contact ball bearing
2025303540
1 1.2 .43.41.39.37.35
1.0.87.76.66.57
1 1.09.92.78.66.55
.70
.67
.63
.60
.57
1.631.441.241.07.93
.57
.68
.80
.951.14
Self aligning ball bearing
1 1 .4 .4 cotα
1 .42 cotα
.65 .65 cotα
1.5 tanα
Important pointsBearing lubrication: Oil/Grease –Permissible speed
Cage- to avoid sliding & collisionFull complement bearing …. Slow speed rigidity
Bearing selection…. For given shaft diaRotating ring (interference)… non-rotating (transition). Rotation factor !!!Bearing life …….. Dynamic load capacity… Applied load (3 to 10%).Equivalent load: X & Y factors depend upon bearing type.
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Bearing Life
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Lundberg Palmgren Approach
In ideal case, bearings fail by surface-fatigue.
Dynamic load rating (catalogue C0 reading) is the load which 90% (reliability=0.9) of a group of identical bearings will sustain to the minimum of 106 cycles.
( )
Speed60000,1000
PC
hoursin life Bearing
bearingsroller for 3
10bearings ballfor 3
10 3322116
a
aaaa
a
a
LPLPLPC
⎟⎟⎠
⎞⎜⎜⎝
⎛=⇒
=
=
===
Failure probability (%)
Factora1
1054321
10.620.530.440.330.21
Example: Radial load = 2 224 N, Speed = 1500 rpmDesired life= 8 hours/day, 5 day/weeks for 5 years, Shock factor = 1.5. For shaft dia of 25 mm.
C > 2224*1.5*(10400*1500*60/106)1/a
C > 32, 633 N for BALL BEARINGSC > 25, 978 N for ROLLER BEARINGS
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http://www.skf.com/portal/skf/home/products?newlink=first&lang=en
In order of increasing outside bearing diameter
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Deep Groove Ball Bearing
RSH Sheet steel reinforced contact seal of acrylonitrile-butadiene rubber (NBR) on one side of the bearing. L stand for low friction.
3/17/2015 65
Suffix
61804 61804-2Z 61804-2RS1
Deep Groove Ball Bearing (DGBB): Both rings possess deep grooves. Bearing can support high radial forces as well as some axial forces. There are single-row & double row DGBB. Widely used in industry.
Cage/Separator: Ensures uniform spacing and prevents mutual contact of rolling elements.
Shield: Profiles sheet steel discs pressed into the grooves of outer ring and forming gap-type seals with the inner-ring shoulders.
Seals: Often made of elastic rubber. Bearings sealed on both sides are grease filled and in –normal working conditions the grease filling lasts the entire service life of the bearings.
Angular Contact Ball Bearing (ACBB): Raceways are so arranged that forces are transmitted from one raceway to other under certain contact-angle between line of action of the force & radial planes. Due to CA, ACBB are better suited to sustain high axial loads than DGBB.
66
3/17/2015 67
Designation – International Organization for Standardization
Multiply by 5 to get bore in mmd<10mm… 618/8 (d=8mm)d>500 mm… 511/530 (d=530mm)
00 = 10mm01= 12mm02 = 15mm03 = 17mm
Each rolling bearing is designed by a code that clearly indicates construction, dimensions, tolerances and bearing clearance.
3/17/2015 68
InstrumentBall
bearing
0 Double row angular contact ball bearings1 Self-aligning ball bearings2 Spherical roller bearings3 Taper roller bearings4 Double row deep groove ball bearings5 Thrust ball bearings6 Single row deep groove ball bearings7 Single row angular contact ball bearings8 Cylindrical roller thrust bearingsHK needle roller bearings with open endsK Needle roller and cage thrust assembliesN Cylindrical roller bearingsA second and sometimes a third letter are used to identify the configuration of the flanges, e.g. NJ, NU, NUP; double or multi-row cylindrical roller bearing designations always start with NN. QJ Four-point contact ball bearings
3/17/2015 69
Example: Assume radial and axial loads on a bearing are 7500N and 4500N respectively. Rotating shaft dia = 70 mm. Select suitable single row deep groove ball bearing.
Bearing type Inner ring
Single row eRotating Fa/VFr > e
Deep groove ball bearing
Fa/C0 V X Y.014.028.056.084.11.17.28.42.56
1 0.56 2.301.991.711.551.451.311.151.041.00
.19
.22
.26
.28
.3
.34
.38
.42
.44Fa/Fr = 0.6; Fa/C0=4500/31000 X = 0.56, Y= 1.37; P=10365
Fa/C0=4500/68000 X = 0.56, Y= 1.65; P=116253/17/2015 70
0.1452
Fa/Fr = 0.6; Fa/C0=4500/31000Fa/C0=4500/68000
0.0662
3/17/2015 71
Life consideration( )
Speed60000,1000
PC
hoursin life Bearing
bearingsroller for 3
10bearings ballfor 3
10 3322116
a
aaaa
a
a
LPLPLPC
⎟⎟⎠
⎞⎜⎜⎝
⎛=⇒
=
=
===
Example: Assume radial and axial loads on a bearing are 7500N and 4500N respectively. Shaft dia = 70 mm. Select a deep groove ball bearing. Consider shaft rotates at 1000 rpm and expected bearing life = 3000 hours
6014 631410365 1162539700 111000
937 14509
CPC 1.003.0 ≤≤
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Deep Groove Ball Bearing
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Deep Groove Ball Bearing
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Deep Groove Ball Bearing
Pressed brass cage
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Deep Groove Ball Bearing
E indicates reinforced ball set. TN9 indicates injection molded snap type cage of glass fibre reinforced polyamide
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Deep Groove Ball Bearing
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Deep Groove Ball Bearing
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Deep Groove Ball Bearing
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– single row angular contact ball bearings
– double row (O, X, Tandem) angular contact ball bearings
– four-point contact ball bearings
In SKF catalogue contact angle α =40° is designated by suffix B. Similarly contact angles of 25° and 30° are designated with suffixes AC and A respectively.
Tandem arrangement Back to back Face to face
Bearing type
Single row Double row eFa/VFr > e Fa/VFr ≤ e Fa/VFr > e
α(°)
X Y X Y X Y
Angular contact ball bearing
2025303540
.43
.41
.39
.37
.35
1.0.87.76.66.57
1 1.09.92.78.66.55
.70
.67
.63
.60
.57
1.631.441.241.07.93
.57
.68
.80
.951.14
Self aligning ball bearing
.4 .4 cotα
1 .42 cotα
.65 .65 cotα
1.5 tanα
Example: A radial load of 3000N combined with thrust load of 2500N is to be carried on a 6214 ball bearing for shaft rotating at 1000 rpm. Determine equivalent radial load to be used for calculating fatigue life. Compare life of 6214 bearing with that for a 7214 (nominal contact angle 30°).
Step 1: C0 for 6214 is 45kN and 7214 is 60 kN. C for 6214 is 63.7 kN and 7214 is 71.5 kNStep 2: Fa/VFr > e
Bearing type Single row, Fa/VFr > e eDeep groove ball bearing
Fa/C0 X Y.056 0.56 1.71 .26
Angular contact ball bearing
30 .39 .76 0.8
Step 4: Life for 6214 will be 7192 hours and for 7214, life=124,420 Hours
Speed60000,1000
PC hoursin life Bearing
3
⎟⎠⎞
⎜⎝⎛=
Appropriate selection of bearing….
• Step 3: Radial load for 6214 bearing is 5955N & for 7214 bearing radial load is 3070.
Fr X Y Fa
Deep groove ball bearing 3000 0.56 1.71 2500Angular contact ball bearing 3000 0.39 0.76 2500
Equivalent load under Variable Loading
Bearing operates at 1000 rpm, applied load of 500 N for 100 hours, then bearing operates at 1200 rpm, 250 N for 250 hours….
In such situation it is advisable to find an equivalent load using
rotationsofNumberL,L,L
a
aLLL
LPLPLPPaaaa
,...
bearingsroller for 3
10bearings ballfor 3
......
321
1
321
332211
=
=
⎟⎟⎠
⎞⎜⎜⎝
⎛+++
+++=
( )( )
( ) aaaa
aaaa
fPfPfPP
LLLLLLPLPLPP
1
332211
1
321
332211
...
......
thenlife,expectedL IF
+++=
⎟⎟⎠
⎞⎜⎜⎝
⎛+++
+++=
=
Question 1: A single row cylindrical roller bearing N 205 ECP is subjected to pure radial load of 2800 N and rotational speed = 1500 rpm. Assuming bearing is subjected to light shocks. Estimate the bearing life for reliability = 0.99.
Question 2a: Select a suitable deep groove ball bearing for a shaft of 30 mm dia rotating at 2000 rpm. Bearing needs to support a radial load of 2000 N and axial load of 400 N. Minimum desirable bearing life is 5000 Hours.
2b: Will the selected bearing be able to survive for more than 10,000 Hours? Justify your answer.
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Shafts
Example: A hollow shaft must carry torque of 3400 N.m at shearing stress of 55 MPa. Assume di=0.65 do. Calculate value of outside diameter.
ANS: 72.6 mm
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Torsion of circular shaft
( )44max
3max
1r
16 or,
16L
T, Torqueby ends at the loadedsection crossuniformofshaft circular ain Stress
1
io
o
dddT
dT
JrT
rG
−=
==⇒
=
πτ
πτ
φτ
3/17/2015 91
Torsion of circular shaft
Example: Design a shaft so that angular deformation should not exceed 1° in a length of 1.8 m. Permissible shear stress = 83 MPa and modulus of rigidity = 77 GPa.
ANS: d=222.34 mm
NOTE: Design of shaft consists of determining correct shaft diameter from strength and rigidity considerations.
3/17/2015 92
A complete shaft design has much interdependence on the design of the components:
Steps size; Threads; Loading; etc.
Stress & Strength
Deflection & Stiffness
Vibration
In shaft design, usual practice is locate the critical areas, size these to meet The strength requirements, and then Size the rest of the shaft to meet Requirements of shaft-supported element, manufacturing steps, or mass criterion.localized
3/17/2015 93
Maximum static stressOften shafts carry combined loads of bending and torque.
( ) ( )223max
2
3
2
3max
22
max
16or
16232or
2
TMd
dT
dM
+=
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=
+⎟⎠⎞
⎜⎝⎛=
πτ
ππτ
τστ
( ) ( ) ⎥⎦⎤
⎢⎣⎡ ++=
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+=
+⎟⎠⎞
⎜⎝⎛+=
223max
2
3
2
33max
22
max
16or
16232
232or
22
TMMd
dT
dM
dM
πσ
πππσ
τσσσ
Maximum shear stress theory
Maximum normal stress theory
Stress Concentration ????
Design of shafts for fluctuating loads
factors.ion concentrat stress fatigue torsionalareK and K where fsmfs
JrTK
JrTK m
fsmma
fsa == ττ
factors.ion concentrat stress fatigue bending areK and K where
moment gfluctuatin tosubjected isshaft If
fmf
IrMK
IrMK m
fmma
fa == σσ
Theory of failure ???? Distortion energy (von-Mises) theory
AFK
AFK m
taxialma
taxiala == __
load axial gfluctuatin tosubjected isshaft If
σσ
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3/17/2015 95
Design for fluctuating bending, fluctuating axial & fluctuating torsion
( )( ) 22
_
22_
3
3
stresses Mises-von
maxialmmm
aaxialaaa
τσσσ
τσσσ
++=′
++=′
22
22
3
3
stresses Mises-von
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+=′
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+=′
JrTK
AFK
IrMK
JrTK
AFK
IrMK
mfsmmtmfmm
afsatafa
σ
σ
3/17/2015 96
Goodman Line Criterion
21?1''
≥≥=+ NorNNSS ut
m
e
a σσ
22
22
3
3
where
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+=′
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+=′
JrTK
AFK
IrMK
JrTK
AFK
IrMK
mfsmmtmfmm
afsatafa
σ
σ
ysma S≤+ '' σσ
Static failure ?Fatigue failure ?
3/17/2015 97
Example: Assume D=42mm, d= 28mm, fillet radius =2.8mm, M = ± 142.4 N.m, T= 124.3 N.m, Sut = 735 MPa, Sy=574 MPa, required reliability =99%.
• Find stress concentration factors
• Find endurance limit and factors reducing value of endurance limit - Find endurance strength
• Use Goodman criterion, Find factor of safety.
• Check safety against yielding
3/17/2015 98
5.12842
1.028
8.2
==
==
dDdr
68.1== fmf KK
3/17/2015 99
5.12842
1.028
8.2
==
==
dDdr
38.1== fsmfs KK
( ) 787.0MPain == butfinish SaK
Finishing method
Constant a Exponent b
Machined or cold drawn 4.51 -0.265
( ) 870.02824.1
5179.224.1107.0
107.0
==
≤≤=−
−
size
size
K
mmddK
MPaSSTorsionSSAxialSSbendingSS
ute
ute
ute
ute
5.3675.029.045.05.0SteelFor
==′⇒=′=′=′
Probability of survival, %
Reliability factor, kr
99 0.814
MPaSS
e
e
205814.0*870.0*787.0*5.367
==
3/17/2015 100
Goodman Line Criterion
NSS ut
m
e
a 1''
=+σσ
( )
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛++=′
=+⎟⎟⎠
⎞⎜⎜⎝
⎛+=′
JrTK
JrTK
IrMK
IrMK
mfsmmfsmm
afafa
3300
030
where
22
22
σ
σ
ysma S≤+ '' σσ
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3/17/2015 102
Design (ASME method) for fully reversed bending and steady torsion
61
22
2
3
2
3
a
22
22
f
16332
116332
shaft solidfor &for sexpression substitute
13
3 know we
1
N safety, offactor a Introduce
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=
=⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛
=⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛⇒
=
=⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛
y
mfsmf
e
aff
y
mfsmf
e
aff
m
y
mf
e
af
yys
ys
mf
e
af
STKN
SMKNd
SdTKN
SdMKN
SN
SN
SS
SN
SN
ππ
ππ
τσ
τσ
τσ
13
1
N safety ofFactor
22
22
f
=⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛
=⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛
=
y
mf
e
af
ys
mf
e
af
SN
SN
SN
SN
τσ
τσ
73.1=fN
3/17/2015 104
Example: Design a shaft that must transmit 2 hp at 1725 rpm. Shaft is loaded with a spur gear and a sheave. Assume stress concentration for 2.25 for step radii in bending, 1.57 for step radii in torsion, and 2.5 at keyways. Assume corrected endurance strength = 50 MPa and yield strength is 150 MPa.
sgsg
sgsgA
FFRFRFRF
FFRqFpFbRM
35.06.00
35.14.00
121
22
+−=⇒=+++∑ =
−−=⇒=++∑ =
Fr
Fg
Ref: Machine Design: An Integrated Approach.. R. L. Norton
3/17/2015 105
( ) ( )( ) ( )( ) ( )( ) ( ) NRR
NRR
NRRNRR
yy
xx
yy
xx
66035.01106.07.3216235.0406.0
44035.11104.07.23416235.1404.0
11
11
22
22
=⇒+−−==⇒+−=
=⇒−−−=−=⇒−−=
Fr
Fg
( ) ( ) ( ) ( ) 1111
112
111
1721621277.234 504007.32
172127 500
sections at various plane in xzfunction Moment
⟩−⟨+⟩−⟨−+⟩−⟨+⟩−⟨=
⟩−⟨+⟩−⟨+⟩−⟨+⟩−⟨=
zzzzM
zFzRzFzRM
xz
sxxgxxxz
( ) ( ) ( ) 111
112
111
12744 50110066
172127 500
sections at various plane yzin function Moment
⟩−⟨+⟩−⟨−+⟩−⟨=
⟩−⟨+⟩−⟨+⟩−⟨+⟩−⟨=
zzzM
zFzRzFzRM
yz
syygyyyz
Calculate moment
mNM
mNM
mNMmNMmNMmNM
yzD
yzC
yzB
xzD
xzC
xzB
.088.0
.088.0
.508.2.0769.1.2329.7.2426.1
−=
−=
====
0805.1;2334.7;7989.2 === DCB MMM
61
22
3
61
22
2
61
22
1
16332
16332
16332
methodASME Using
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=
y
mfsmf
e
Dff
y
mfsmf
e
Cff
y
mfsmf
e
Bff
STKN
SMKNd
STKN
SMKNd
STKN
SMKNd
ππ
ππ
ππ
( )
mNT
T
m
m
.3.8
1725602
746*2
=
⎟⎠⎞⎜
⎝⎛
=π
ANS: d1=11.7 mm
d2=15.0 mm
d3=09.8 mm
As per available drawing d1>d2. Therefore select d3=10mm, d2=17mm, and d1=20 mm. 107
Corrected endurance strength !!
3/17/2015 108
Design for fluctuating bending & fluctuating torsion
( ) ( ) ( ) ( ) 31
2222
f
f
75.075.032
N1
line Goodman in Nsafety offactor Using
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧ +
++
=
′+
′=
ut
mfsmmfm
e
afsaff
ut
m
e
a
STKMK
STKMKN
d
SS
π
σσ
Example: Design a shaft to support attachments shown in Figure. Torque & moment on shaft are both varying with time in repeated fashion, i.e., their alternating & mean components are of equal magnitude. Mean & alternating components of torque are both 17 N.m. There is no axial loads. Assume stress concentration for 2.25 for step radii in bending, 1.57 for step radii in torsion, and 2.5 at keyways. Assume corrected endurance strength = 50 MPa and ultimate strength is 250 MPa.
3/17/2015 109
225050
.17.1.1.3.7.8.2
=======
f
ut
e
D
C
B
NMPaS
MPaSmNT
mNMmNMmNM
( ) ( )
( ) ( )[ ] 6/122
31
22
75.00063.0
1175.032
TKMKd
SSTKMKd
fsf
utefsf
+=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎭⎬⎫
⎩⎨⎧
++=π
ANS: d1=28.6 mm
d2=30.2 mm
d3=28.3 mm
As per available drawing d1>d2. Therefore select d3=29mm, d2=31 mm, and d1=33 mm.
3/17/2015 110
Deflection/Stiffness consideration
Critical speed Unstable shaft.
3/17/2015 111
t = t1
t = 0
t4 > t1
t = t4
t = t1
t = 0
t = t4
t = t1
t = 0
t4 > t1
t = t4
t = t1
t = 0
3/17/2015 112
EquilibriumThis image cannot currently be displayed.
Speed at which shaft is unstable Critical.Continuous increase in deflection without upper bound.
Shaft rotates slightly off of its true centerline. Such rotor deflection is referred to as mode shape. First/Second mode….
3/17/2015 113
Critical Speed of Shaft
Speed at which shaft is unstable.Continuous increase in deflection without upper bound.
0 1 2 3 4 5 6 7 8 9 100
.5
1
1.5
2
2.5
3
3.5
4Tranmissibility as function of ζ
Frequency ratio
Tram
issi
bilit
y
3/17/2015 114
TransmissibilityRatio (T) of output to input. T is function of operating frequency.T > 1 Amplification. Max amplification when forcing frequency (ω) and natural frequency of system (ωn) coincides.T < 1 Isolation (i.e. passenger compartment from automobile chassis).
Minimize natural frequency of system. Often transmissibility is referred as “Q factor”.
3/17/2015 115
Critical Speed of Shaft
Estimation of critical speed helps to decide maximum operating speed. Rate of increase in amplitude estimates the allowable time to shoot the speed above critical speed.
)sin(0 tFkxxcxm ω=++ &&&
0 .2 .4 .6 .8 1 1.2 1.41
1.52
2.53
3.54
4.55
5.5Tranmissibility as function of ζ
Frequency ratio
Tram
issi
bilit
y
Frequency of external force to Natural frequency of system.
42.15.0/
>>dnω
ω
Estimating Natural Frequency
3/17/2015 116
mkfn π2
1=
2
221
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
mc
mkfd π
( )( )
mk
xkxmtXxassume
xxkWFxmWxk
n
n
i
i
=
−=−
=
+−==
=
∑
ω
ω
ω2
sin
&&
3/17/2015 117
Damped Natural Frequency
0=++ kxxcxm &&&
2
2,1
2
2,1
22
22
⎟⎟⎠
⎞⎜⎜⎝
⎛−±−=
−⎟⎟⎠
⎞⎜⎜⎝
⎛±−=
mc
mki
mcs
mk
mc
mcs
2
221
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
mc
mkfd π
kx C dx/dt
ste= xassume
3/17/2015 118
How to estimate Transmissibility ?
( )anglephaseamplitudeX
tXxtFkxxcxm
==−==++
φφω
ωsin
)sin(0&&&
( )( ) ( )( ) ( )( ) )sin(sincossin 02 tFtXktXctXm ωφωφωωφωω =−+−+−−
( ) ( ) ( ) )sin(cossin 02 tXFtctmk ωφωωφωω =−+−−
( ) ( )2220 ωω cmkXF +−= ( ) ( )
0
22
FXckX
Tω+
=
( ) ( )( ) ( )222
22
ωω
ω
cmkX
ckXT+−
+=
0 .02 .04 .06 .08 .1 .12 .14 .16 .18 .2-1
-.75-.5
-.250
.25.5
.751
Force and displacement
time
harm
onic
func
tion
3/17/2015 119
( ) ( )2220 ωω cmkXF +−=
( ) ( )0
22
FXckX
Tω+
=
( ) ( )( ) ( )222
22
ωω
ω
cmkX
ckXT+−
+=
How to estimate Transmissibility ?
Ratio (T) of output to input.
kx C dx/dt
222
2
1
1
⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛ −
⎟⎠⎞
⎜⎝⎛+
=
kc
km
kc
Tωω
ω
3/17/2015 120
Mathematical Model of Transmissibility
22
2
2
2
21
21
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛−
⎟⎟⎠
⎞⎜⎜⎝
⎛+
=
kckmc
kckmc
T
cn
c
ωωω
ω
22
2
2
2
21
21
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛−
⎟⎟⎠
⎞⎜⎜⎝
⎛+
=
km
cc
km
cc
T
cn
c
ωωω
ω
( )( ) ( )
cn cc
T
==
+−
+=
ςωωλ
ςλλ
ςλ222
2
21
21
3/17/2015 121
0 .5 1 1.5 2 2.5 3 3.50
.51
1.52
2.53
3.54
4.55
5.5ζ = 0.1
Frequency ratio
Tram
issi
bilit
y
0 .5 1 1.5 2 2.5 3 3.5.3.4.5.6.7.8.91
1.11.21.31.41.5
ζ = 0.5
Frequency ratio
Tram
issi
bilit
y0 .5 1 1.5 2 2.5 3 3.5
.5
.6
.7
.8
.9
1
1.1
1.2ζ = 1
Frequency ratio
Tram
issi
bilit
y
0 .5 1 1.5 2 2.5 3 3.5.7
.75
.8
.85
.9
.95
1
1.05
1.1ζ = 1.5
Frequency ratio
Tram
issi
bilit
y
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0 1 2 3 4 5 6 7 8 9 100
.5
1
1.5
2
2.5
3
3.5
4Tranmissibility as function of ζ
Frequency ratio
Tram
issi
bilit
y
Robust design ???
Vibration free design ????
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Question: Determine the natural frequency of mass M on the end of a cantilever beam (length L) of negligible mass
M
kP
==IE3
LP Pload edconcentrata
under beam cantilever of Deflection
3
δ
δ
MLIEf
Mkf
n
n
33
21
21
π
π
=
=
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Question: Determine the natural frequency of mass M on the end of a simply supported beam (length L) of negligible mass
M
kP
==IE84
LP Pload edconcentrata
under beam supportedsimply of Deflection
3
δ
δ
MLIEf
MLIEf
Mkf
nn
n
333
2448
21
21
ππ
π
=⇒=
=Four times !!!!
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Dunkerley equation
222
21
2
1...111
ncr ωωωω+++=
existsn massonly if speed naturalexists 2 massonly if speed natural
exists 1 massonly if speed natural
2
1
===
nωωω
• Estimates fundamental critical speed of shaft carrying a number of components (gears, pulley, coupling, etc.)
• Estimate critical speed of each individual subsystem by direct formula.
• To obtain critical speed of shaft-system, combine individual frequencies using following equation.
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Ex: Assume shaft (negligible mass, dia =40mm) shown in Fig. is made of steel with E=207 GPa. Mass of each disk is 13.5 kg. Estimate fundamental frequency of system.
Dis
k 1
Dis
k 2
Dis
k 3
200
• Disk 1 and Disk 3 have same mass and are symmetrically placed, therefore natural frequency of disk 1 and disk 3 will be same.
• Find natural frequencies of Disk 1 & Disk 2
200 200 200
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kP
==IE84
LPusing estimatedbecan2Disk offrequency Natural
3
δ
MLIEf
MLIEf
Mkf
nn
n
333
2448
21
21
ππ
π
=⇒=
=
kP
=⎟⎠⎞
⎜⎝⎛ −−+= )8.0/()4^2.0(8.0*)2^2.0(*2)3^2.0(*3)3^2.0(
800600
IE6P
using estimated be can 1 Disk offrequency Natural
δ
1.255=fn l
a b
⎥⎦
⎤⎢⎣
⎡−−+−−=
lxaalxaxaxx
lb
IEP
x
3233 23
6δ
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23
22
21
21111ωωωω
++=cr
3only Disk if speed natural 2only Disk if speed natural
1only Disk if speed natural
3
2
1
===
ωωω
Using Dunkerley equation
22
21
2121fffcr
+=
QuestionA simply supported 25-mm diameter (E=207 GPa, Specific weight = 75 kN/m^3) uniform steel shaft is 600 mm long.
Find the lowest critical speed of the shaft.If aim is to double the critical speed, find the new diameter.What will be the critical speed if the shaft length is reduced from 600mm to 400 mm.
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Shafts are meant to transmit power through sheave (pulley: Flat, V), gear (spur, helical, bevel, etc.), sprocket, etc. Shaft is positioned on supports (bearings, etc.).
Key joint (flat key, taper pin), coupling, or interference fit is required to mount required elements on the shaft.
Keys: ASME defines a key “demountable element which when assembled into keyways, provides a positive means for transmitting torque between shaft and hub.
Keyways
Square Rectangular
Gib-head taper
Key Joint
131
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Key:Primary function: Transmit torque from shaft to hub of mating element and vice-versa.Prevent relative rotation between shaft & joining element.
Keyway: A recess or slot on shaft and/or hub to accommodate key.Keyway results in stress concentration (initial value ~ 2.5) in shaft and hub.
Flat Key Assembly
NOTE: Key is inexpensive and relatively easy to replace if keyway is undamaged. Therefore relatively softer compared to keyway material (having lesser strength) is used for the key.
Parallel or taper key
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Two modes of failures:1. Shear Failure (AS = w.l)
Fatigue failure2. Crushing/Bearing Failure
(Abearing=(h/2.l) static failure
Assumptions: Axis, shear force at top/bottom/sides, turning couple, uniform contact
Ductile failure
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Failure of KEYRounding of corners
Standard keysShaft diameter (mm)
Key width* Height (mm*mm)
8<d≤10 3*310<d ≤12 4*412<d ≤17 5*517<d ≤22 6*622<d ≤30 8*730<d ≤38 10*838<d ≤44 12*844<d ≤50 14*950<d ≤58 16*1058<d ≤65 18*1165<d ≤75 20*1275<d ≤85 22*1485<d ≤95 25*14
NOTES: Parallel key is placed with half of its height in the shaft and half in hub.
Parallel keys are machined with negative tolerances.
Key-fit (backlashimpact & high stresses) can be of concern when torque loading is alternating from positive to negative each cycle.
Length of key should not be too long otherwise twisting of key must be accounted.
135Sizing of KEY based on Shaft size (30 to 40°) …… Remaining variable
LENGTH
RadiusTorque
=
lwA
lhA
s
c
=
=
forceshear subject to Area2
force ecompressiv subject to Area
Example: Design a key for fixing a gear on a shaft of 25mm diameter. Shaft transmits 15 kW power at 720 rpm to the gear. Assume yield strength of key material is 150 MPa.
Referring to Table of standard keys, width of key = 8mm and height is 7 mm.
136
RadiusTorque
=
Example: Design a key for fixing a gear on a shaft of 25mm diameter. Shaft transmits 15 kW power at 720 rpm to the gear. Assume yield strength of key material is 150 MPa.
Referring to Table of standard keys, width of key = 7mm and height is 8 mm.
137
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Example: Design a key to transmit 475 N.m torque from the shaft to hub. Shaft diameter is 50 mm. Assume yield strength of key material is 100 MPa.
Referring to Table of standard keys, width of key = 14mm and height is 9 mm.
Tangential force = 475/(0.5*0.05)=19000 N
Length to avoid crushing failure: 19000/(100*0.5*9)=42.2 mm
Length to avoid shear failure: 23.5 mm
lwA
lhA
s
c
=
=
forceshear subject to Area2
force ecompressiv subject to Area
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Keeping width of key = 09mm and height is 14 mm.
Length to avoid crushing failure: 19000/(100*0.5*14)=27.15 mm
Length to avoid shear failure: 36.6 mm
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Two keys at 90° phase– Kennedy Key
Key-fit (backlash impact & high stresses) can be of concern when torque loading is alternating from positive to negative each cycle.
41.4% area in Increase50% torquein Decrease
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SplinesKeys integral part of shaft.
Required to allow axial movement between shaft & hub (i.e. gear shifting mechanism).
Spline area AA = 0.5*(D-d)*L*N
Torque arm RmRm=0.25*(D+d)
NLdDdDTAreaArmTorque
Torque
c
c
**)(*5.0*)(25.0
*_
−+=
=
σ
σ
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Stress concentration in Keyways
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Splines: (number of splines* minor diameter *
outside diameter)
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Ex: A Standard medium duty splined connection (10*72*82) is used for the gear and shaft assembly to transmit 30 kW at 300 rpm. Permissible compressive stress is 100 MPa. Find minimum length of hub and force required for shifting the gear. Assume μ=0.03.
)10*100(*10*10*)7282(*125.0)60/300**2/(1000*30
**)(*5.0*)(25.0
6622 −−=
−+=
π
σ
L
NLdDdDT
c
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CouplingsCoupling is an element (a device) that joins two rotating shafts to each other.
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CouplingsMost common application is joining 2-shafts of 2-separately (modular) built or purchased units so that a new assembly can be formed
Motor-pumpMotor-gearbox
“Oldham’s coupling(parallel offset shafts), Hooke’s coupling(shafts having intersecting axes) and rigid/flexible parallel coupling
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Objectives of coupling are:Should be capable of transmitting torque from driving shaft to driven shaft.Should keep two shafts in proper alignment.Should be easy to assemble and disassemble.Maintain zero relative motion between parallel shafts… Sleeve coupling
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Design of Sleeve couplingAlso known as “Muff coupling” and “box coupling”.Uses “sunk key”, “grab-screw” or “interference fit”.
⎩⎨⎧
=+=
≤dL
mmdDmmd
5.3)132(
70 For
( )4416
dDDT−
=π
τ
In practice
Torsional shear stress
Iterations!!
+ve drive
Low power
Fluctuating torque
Stress concentration?
Example: Design a muff coupling to connect two steel shafts transmitting 25 kW power at 360 rpm. Maximum allowable compression/tensile stresses in shaft and key are 100 MPa. Coupling is made of grey cast iron, which should not be stressed beyond 35 MPa (tensile strength).
( ) mmdmmde
Td
mNTT
40or 8.386100*577.
16diameter Shaft
.15.663
603602
25000 Torque,
31
==⇒⎥⎦
⎤⎢⎣
⎡=
=⇒⎟⎠⎞
⎜⎝⎛
=
π
π
Choosing key of 12*8 mm.
Standard keysShaft diameter (mm) Key width* Height (mm)38<d ≤44 12*844<d ≤50 14*9
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Minimum value of outer diameter of coupling is = Shaft dia + height of keyDmin = 40+8 48 mm.
( ) MPaDDDT 52.416 stressshear 4
min4 =⇒−
= τπ
τ
mmLmmD
dLmmdD
mmd14093
5.3)132(
70 as=⇒=⇒
⎩⎨⎧
=+=
<
Coupling is safe
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