Lectures-Part II

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

3/17/2015 1

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.

3/17/2015 20

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


( )2

21

21 ⎟

⎟⎠

⎞⎜⎜⎝

⎛ −−

= Q

QQ

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

+

+

+


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

3/17/2015 38

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



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 ≤≤

3/17/2015 72


3/17/2015 73


3/17/2015 74


Pressed brass cage

3/17/2015 75


E indicates reinforced ball set. TN9 indicates injection molded snap type cage of glass fibre reinforced polyamide

3/17/2015 76


3/17/2015 77


3/17/2015 78


3/17/2015 79

3/17/2015 80

– 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

,...



......

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.

3/17/2015 88

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

3/17/2015 89

3/17/2015 90

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

3/17/2015

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

3/17/2015

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.


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.


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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Lectures-Part II

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Transcript of Lectures-Part II