ME 331: Design of Machine Elements

Dr. H. HiraniAssociate ProfessorDepartment of Mechanical Engineering

2812, 252

Quiz10%3 / 4

80% of End-Sem

September: 20, 21, 25,27

October: 4,5,9,11,16,18,19,23,26, 30

November: 1, 6, 13, 15, 16

RemarkLecture

Contact Time: 16.30 to 17.30 Hours --- Monday to Friday.

References:

1. http://www.mech.uwa.edu.au/DANotes/intro/contents.html

2. Machine Design: An Integrated Approach.. R. L. Norton

Design of a SPRING

• Definition

• Material (s)

• Constraints

• Equations

• Examples

• Energy Storage Component

• Elasticity is a basic requirement.

• Energy Storage Component

• Elasticity is a basic requirement.

• Requirement of relatively large deformation.

P

P

DesignService condition

FunctionCost

MaterialsProperties

AvailabilityCost

ProcessingEquipment sel.

Influence of prop.Cost

• “CONCURRENT ENG.” MAT. & PROCESSES CONSIDERATION AT EARLIER STAGE

Material choice cannot be made independently of the choice of process by which material is to be machined, joined, or treated.

Property bar-chartsMetals Polymers Ceramics Composites

PEEK

PP

PTFE

WC

Alumina

Glass

CFRP

GFRP

Fibreboard

Y ou n

g ’s

mod

ulus

, GPa

Steel

Copper

Lead

Zinc

Aluminum

orSE /=

Covalent bond is stiff (S= 20 –200 N/m) Metallic & Ionic (15-100 N/m)

Polymers having Van der Waals bonds (0.5 to 2 N/m). r0~ 3*10-10m)

ATOMIC SIZE

EWV

2σ∝

? ELASTIC ENERGY/VOLUME

Better than spring steel10-30Rubber

Economic & easily shaped1.5-2.5Nylon

--10-12GFRP

Comparable in performance with steel, expensive

15-20CFRP

Expensive, corrosion resistant

15-20Ti alloys

Traditional choice: easily formed and heat treated.

15-25Spring steel

Brittle in tension; good only in compression

10-100Ceramics

CommentMATERIAL ( )32 .... mMJEM fσ=

Natural frequency

ρπ

ρππ

DNgk

df

DNdW

Wgkf

n

n

14

21

2

=⇒

=

=E

WV

2σ∝

EWV

2σ∝ E

Ww ρσ 2

∝

? ELASTIC ENERGY/COST ECM

m

f

ρσ 2

=

EM f

ρσ 2

=

Better than spring steel10-30, 10-30Rubber

Economic & easily shaped1.5-2.5, 1.5-2Nylon

--10-12, 3-5GFRP

Comparable in performance with steel, expensive

15-20, 4-8CFRP

Expensive, corrosion resistant15-20, 2-3Ti alloys

Traditional choice: easily formed and heat treated.

15-25, 2-3Spring steel

Brittle in tension; good only in compression

10-100, 5-50Ceramics

CommentMATERIAL EM f2σ=

Rubber Springs

Hooke’s law?Stiffness increases with increase in deflection.

Temperature dependence. Useful if damping/cushioning is required? Often hybridized with metals. …. Requires detailed analysis.

Stress-Strain for One Cycle

• Energy Storage.• Elasticity.• Relatively larger deflection.• Min loss of Energy Metals are preferable materials.

• Metals, glasses and ceramics have low intrinsic damping, which is measured by “Loss coefficient”.• Loss factor is high in soft metals like lead and pure aluminum, while low in heavily alloyed metals like bronze and high carbon steel. • In polymers, loss factor depends on ratio of operating temperature to glass transition temperature.

E1∝η

Home Assignment I

List the material (s) needed for a Spring aiming to maximize energy storage, minimize energy loss and density.

Relative Cost

1.0

2.6

1.3

3.1

-

7.6

4.0

---

8.0

27.0

Most Popular Springs: Helical Compression Springs

Initial load Pl ll ~ 0.85 lf Can we overcome this constraint ?

Max load Pmax l0 ~ 1.15 ls Reason ???

Force vs. Deflection δ∝P

Ends Used in Compression Springs

Figure (a) Plain; (b) plain and ground; (c) squared; (d) squared and ground.

Eccentric loading ? Increasing stresses on one side of spring.

Active Coils.

Passive Coils.. Turns which do not affect the deflection.

Force & Torque

αcosPαsinP

αcosP

αsinPαcosP

αcosP

0sin cos 10 If

≈≈≤

ααα

PPPo

Due to compressive loading, Helical Compression Spring will be subjected to “Direct Shear” and “Torsion”.

Figure (c) Torsional and transverse loading with no curvature effects; (d) Torsional and transverse loading with curvature effects.

Pure torsional loading

Transverse loading

⎟⎠⎞

⎜⎝⎛ +==

⎟⎠⎞

⎜⎝⎛ +=

+=

+=

+=

Ckk

dPC

CdPC

dP

ddCdP

dP

ddDP

AP

JTr

ss5.01 ;8

2118

4*.8

4/)32/()2/(*)2/.(

2

2

24

24

max

π

π

ππ

ππ

τ

Mathematical FormulationStressStrength >

max ,

τys

sS

NfactorSafety =

Wahl determined the stress concentration factor (Wahl factor) that includes both direct shear and stress concentration due to curvature.

CCCKw

615.04414+

−−

=

C5.0 1 >>

To make spring, wire is curved into coil shift in neutral axis towards the center of curvature Non-uniform stress distribution Stress concentration on the inner surface of curvature.

Spring Index, C = D/d

Constraint on Min value of C ???

Spring Stress Factor

1.3111.2531.2131.1841.1621.1451.1311.119

1.11.0831.0711.0631.0561.0501.0451.042

56789101112

KwKsC (=D/d)

Under static load, yielding is the failure criterion.

Ductile materials yields locally and stress concentration is negligible.

Under dynamic load, the failure will be fatigue and stress concentration due to curvature will play important role.

load fatiguefor 615.04414

load staticfor 5.01

CCCk

Ck

c

c

+−−

=

⎟⎠⎞

⎜⎝⎛ +=

ckdPC

2max8π

τ =

1.32.12.43.23.84.24.87.58.59.5

0.110.140.180.220.280.350.450.550.650.70.91.11.41.82.22.8

0.100.120.160.200.250.300.400.500.600.801.01.21.62.02.53.0

Third Preference

Second Preference

First Preference

Preferred Diameters for Spring Steel Wire, mm

4.05.06.08.010.012.014.016.0

First Preference

3.54.55.56.57.09.011.013.015.0

Second Preference

Ultimate Strength of Spring wire

utyt

ytys

SS

SS

75.0

577.0

=

=

max ,

τys

sS

NfactorSafety =

but AdS =

Initial guess

12.61.33.14.07-11

1753.32153.51831.21909.92059.2186720652911

-0.1822-0.1625-0.1833-0.1453-0.0934-0.146-0.263-0.478

0.5 – 160.3 – 60.5 – 160.5 -120.8 – 110.3 – 2.52.5-55-10

Cold DrawnMusic wireOil temp.Chrome-Vanadium

Chrome Silicon302 Stainless (A313)

Relative Cost

Coefficient A (MPa)

Exponent b

Range (mm)

Material

Designing Helical Compression Springs for Static Loading

max ,

τys

sS

NfactorSafety =

60655555

45503535

A227 and A228Hardened and tempered steel (A229,A230, A232, A401)Austenitic stainless steel (e.g. A313)Nonferrous alloys ( B134, B159, B197)

After set removed

Before set removed

Max percent of UTSMaterial

Max. Torsional yield strength for Helical compression springs in static applications

Setting: “Set removal”… Introducing beneficial compressive stresses by compressing spring to its shut height and yield it.

Maximum Torsional Fatigue Strength for round wire helical compression spring

-494746

Sys

424038

-423936

Sys

363330

103

105

106

107

PeenedUnpeenedPeenedUnpeened

ASTM A230 & A232

ASTM 228, Stainless Steel & Nonferrous

Percent of UTSFatigue Life (Cycles)

Home Work 2: Create torsional shear S-N diamgram for 0.25mm, 0.5mm and 1.0 mm of ASTM A228 unpeened spring wire.

Ex: Design a cold drawn steel wire helical compression spring, having minimum possible spring index. Static

axial load = 500 N. Min value of factor of safety = 1.2

Step 1: Stress factor ….. Ks

Step 2: Strength & Stress. Trial 1: d=1mm

Trial 2: d =2mm

Trial 3: d=3mm

Strength 789 MPa

Strength 695 MPa

Strength 646 MPa

Stress = 1751 MPa.

Stress = 7003 MPa.

Stress = 778 MPa.

Final Design, d=4 mm, Stress = 438 MPa, Strength = 613 MPa, S.F. = 613/438

Energy Storage & Deflection

( )( )

( )a

a

N

NDDP

G

4

32

4

4

dGD P4

2/dGD P8

)*)32/dL/((*(T* D/4)(P ) * (D/2 * (P/2)

Deflection * Force Avg. Uenergy,strain as stored is springdeflect toP force axialby doneWork

=

=

=

==

ππ

π

θ

a

a

NDdG

dGNDP

3

4

4

3

8P/ k Stiffness

8

PU Deflection

==

=

∂∂=

δ

δ

δ

? Series combination

? Parallel arrangement.

Ex: Design a cold drawn steel wire helical compression spring, having minimum possible spring index. Static (max.) axial load = 500 N. Min value of factor of safety = 1.1 and Stiffness 25 N/mm. G=80000 N/mm2. Assume spring has square and ground ends

Step 1: Stress factor ….. Ks

Step 2: Strength & Stress. d=3.5mm

Stress = 572 MPa, Strength = 638 MPa, S.F. = 638/572

aNDdG3

4

8P/ k Stiffness == δ

Known values:

G = 80 GPa

D = 5*3.5 = 17.5mm

d = 3.5 mm

k = 25 N/mm

Total No. of Turns = ????

Free length of spring = ???

Homework 3: Determine the deflection of springs shown in Figure. Axial load = 2000 N

000,80,15,3,12N

2 Spring

000,100,30,6,12N

1 Spring

a

a

====

====

GmmDmmd

GmmDmmd

Design Helical Compression Spring for Fatigue Loading

( )( )minmax

minmax

5.05.0

PPPPPP

m

a

+=−=

sm

m

wa

a

kd

CP

kd

CP

2

2

8

8

πτ

πτ

=

=

Mean stress

Alter

nat

ing s

tres

s

τm,τa

SysSys/Ns

Sse/2

seys

se

ms

ys

a

SSS

NS −

=− 2τ

τ

Sse/2

Ex: The exhaust valve system of diesel engine is shown in Figure. The diameter of valve seat is 30mm and the suction pressure in the cylinder is 0.05 MPa. The mass of valve body is 70 gm. Max. required valve lift =10 mm. Spring stiffness = 10 N/mm. C = 7. Design this spring for factor of safety greater than 1.25. Spring Mat. = Music wire.

Min load ???Max. Load ???

( )( )minmax

minmax

5.05.0

PPPPPP

m

a

+=−=

sm

m

wa

a

kd

CP

kd

CP

2

2

8

8

πτ

πτ

=

=

seys

se

ms

ys

a

SSS

NS −

=− 2τ

τ

Homework 4: A helical compression unpeenedspring, made of ASTM A228, is preloaded with 25 N axial force. The maximum operating force during load cycle is 100 N. Assume spring index = 6, factor of safety ≥1.25. Determine wire diameter for spring life = 107 cycles.

Design of Belt DriveAim: Transmission of power over comparatively long distance.+Ve:

Low initial, assembly and running costs. Tolerance for misalignment… FlexibilityAbsorb shocks. Isolate effect of vibration

-Ve: Efficiency (Slip).. Friction uncertaintiesWear (short life) … One year life Loss of elasticity (Creep)

Friction drive

Audio Tapes

Video Tapes

Data C

artridgeC

onveyor belt

Toothed wheel

p > 2Timing

Grooved/ Sheaves

ht =8 to 19V

Groovedd=3 to 20Round

Crowned

t=0.75 -5

Flat

Required PulleySize range (mm)

Cross section

Belt Type

Note: Time belt uses positive drive mechanism, while other belt drives use friction drive mechanism. Time belt drive does not require any initial tension. Teeth make it possible to run at any speed.

d

ht

Mechanism of Flat Belt Pulley Drive

( ) ( ) ( )

( ) ( ) ( )(2) Eq. 0

05.0sin5.0sin(1) Eq.

05.0cos5.0cos

2

2

=−−⇒

=−−++

=⇒=−−+

φφ

φφφ

μμφφ

dvmdRdTdvmdRdTddTT

dRdTdRdTddTT

dφ

T

T+dT

dRdF

( )

( )2

21

2

22

21

02

22

n integratioOn

0/

(1)Eq.from dR

μφφ

φμ

φμφμφ

emvTmvTd

mvTdT

dmvT

dTdmvdTdT

ngSubstituti

T

T

=−−

⇒=−

=−

⇒=−−

∫∫

m v2 dφ

Belt contacts driving pulley with tight tension T1 and leaves this pulley with loose tension T2.

Initial tension ?

Open Flat Belt

212

2

2

1

121

)5.05.0(

22

2sin

DDCAB

CDD

d

d

−−=

+=−=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −= −

απφαπφ

α

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛ −−+++−−=

−+++−−=

++−+−−=

++=

−

dd

d

d

CDDDDDDDDCL

DDDDDDCL

DDDDCL

DDABL

2sin5.04 or,

5.04 or,

25.025.04 or,

5.05.02 Length,Belt

1211221

212

2

12212

122

212

122

2211

π

απ

απαπ

φφ

Ex: A flat belt (w=152 mm, t=8 mm) transmits 11 kW. Cd=2.5 m, Ddriving_pulley=150 mm, Ddriven_pulley=450 mm, Ndriving_pulley=2000 rpm, belt material density = 970 kgf/m3. μ=0.3

m/s15.708 v

m/s 60

ity Belt veloc

=

=Ndv π

NTTvTT

3.700708.15/11000 11000)( as expressed becan smittedPower tran

21

21

==−⇒=−

( ) ( )kg/m 0.1202 m

)1000/8(152/1000 970/9.81 m t wdensity/g Weight belt ofh Mass/lengt

===

0215.3 2

sin2 angle,Contact

1

11

=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−= −

φ

πφd

drivingdriven

CDD

7639.434755.2

4755.266.2966.29

21

2

12

2

21 1

−=−

=−−

⇒=−−

TTTTe

mvTmvT μφ

T1=1204.6 N

T2= 504.3 NStress ??

Length = 5.9515 m

Commercial Flat Belts

100112125152180200

7690100112125152200224

25, 32,40, 4450, 6376, 90100112125140152

2532404450637690100

6 Ply5 Ply4 Ply3 Ply

Width (mm) of High Speed (Light to Medium duty: 2.3*V W/mm per ply) belt

200250305355400

112125152180200250

76100112125152180250

40, 4450, 637690100112125152

2540506376

8 Ply6 Ply5 Ply4 Ply3 Ply

Width (mm) of FORT (Heavy duty: 2.89*V W/mm per ply) belt

Correct Initial tension – Belt shorter than calculated length.

Belt of 3 Plies --- ----15 mm per meter belt length

Belt of 4 to 6 Plies– 10mm per meter belt length

Belt of 8 Plies ------- 5 mm per meter belt length

Commercial Flat Belts….cont

….….……

6 Ply5 Ply4 Ply3 Ply

Width (mm) of High Speed (Light to Medium duty: 2.3*V W/mm per ply)

……………

8 Ply6 Ply5 Ply4 Ply3 Ply

Width (mm) of FORT (Heavy duty: 2.89*V W/mm per ply)

1.01.2

1.3

1.5

1. Normal load2. Steady load, e.g. Centrifugal pump, fans,

machine tools, conveyors.3. Intermittent load, e,g, heavy duty fans,

blowers, compressors, reciprocating pumps, line shafts

4. Shock load, e.g. vacuum pumps, rolling mills, hammers, grinders

Load Correction FactorType of Load

0.97

190

0.941.01.041.081.131.191.261.33Contact factor

200180170160150140130120Arc of contact

Homework 5

Problem: Find the length of a suitable flat belt that can transmit 11 kW in a Conveyor system. Cd=2.5 m, Ddriving_pulley=150 mm, Ddriven_pulley=450 mm, Ndriving_pulley=2000 rpm, μ=0.3. Neglect the effect of centrifugal force. What should be pre-tension in the belt to avoid slip? Answer: 5892 mm, 825 N

Arc of ContactIncreasing angle of contact decreases belt tension and increases belt life

NTT 3.70021 =−

2617.3

1416.3

0215.3

1

1

1

=

=

=

φ

φ

φ

2504.496605.26605.266.2966.29

4565.465663.25663.266.2966.29

7639.434755.24755.266.2966.29

212

1

212

1

212

1

−=−⇒=−−

−=−⇒=−−

−=−⇒=−−

TTTT

TTTT

TTTT

4.451

8.476

3.504

2

2

2

=

=

=

T

T

T

??? 10% Reduction in belt tension

Cross Belt Drive to increase Contact Angle

21

121

22

sin

φαπφ

α

=+=

⎟⎟⎠

⎞⎜⎜⎝

⎛ += −

dCDD

( ) ( ) ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛ ++++++−= −

dd C

DDDDDDDDCL2

sin5.04 Length,Belt 1211221

212

2 π

Homework6: Find the length of a suitable flat belt (for crossed belt drive) that can transmit 11 kW in a Conveyor system. Cd=2.5 m, Ddriving_pulley=150 mm, Ddriven_pulley=450 mm, Ndriving_pulley=2000 rpm, μ=0.3. Assume each ply is 0.75 mm thick and density

of belt material is 950 kgf/m3 .

Answer: 5919 mm

V-Belt

( ) ( ) ( )

( ) ( ) ( )(2) Eq. 0sin

0sin5.0sin5.0sin(1) Eq.

05.0cos5.0cos

2

2

=−−⇒

=−−++

=⇒=−−+

φβφ

φβφφ

μμφφ

dvmdNdTdvmdNdTddTT

dNdTdNdTddTT

T+dT

T

m v2 dφ

• Wedging action increases power transmitting capacity.

• Force components T, T+dT, and m v2 dφ are same as those of flat belt. However, normal force as shown in Fig. is different

( )

( )2

21

2

sin2

2

21

02

22

sin

n integratioOn

sin0sin/

(1) Eq. from dN

φβ

μφ

φβ

μ

φβ

μφβμφ

emvTmvTd

mvTdT

dmvT

dTdmvdTdT

ngSubstituti

T

T

=−−

⇒=−

=−

⇒=−−

∫∫

Effective μ = 3.2361 μ

dφ

T

T+dT

dF

Data on Standard V-belt Sections

0.1060.1890.3430.596-

811141923

1317223238

75125200355500

.75-7.52-157.5-7522-15030-190

ABCDE

Weight per meter, kgf

ht , mm

Wt , mm

Recommended min pulley pitch dia, mm

Drive load, kW

Cross section Symbol

• Only certain discrete standard pitch lengths are manufactured. • Only certain recommended pitch pulley diameters are preferred. • A special pulley may be manufactured of course - but would cost more than a mass- produced commercial product.

Groove angle of pulley is made somewhat smaller than belt-section angle..

Catalogue A

Nominal inside length and pitch lengths for standard V-Belts

10591110-1212126213391415

1026105111021128120412551331-

9911016106710921168121912951372

932

1008

6456967478238489259501001

610660711787813889914965

Cross section B

Cross section A

Cross section B

Cross section A

Nominal pitch length, mm

Nominal inside length, mm

Nominal pitch length, mm

Nominal inside length, mm

Designation: Cross-section B and inside length 1168 mm (46 inches) shall be designated as: B 1168/46 IS: 2494….. A deviation of 2.5 mm in length from nominal pitch length is represented by one unit… B 1168/46-49, B 1168/46-51

810141924

1216223238

60105150325540

.25-2.5

.75-3.751.25-9.08-18.515 and up

ABCDE

ht , mm

Wt , mm

Recommended min pulley pitch dia, mm

Drive load, kW

Cross section Symbol

Catalogue B

1144572, 4953, 5334, 6096, 6858, 7620, 8382E

843048, 3251, 3658, 4013, 4115, 4394, 4572D

741295, 1524, 1727, 1905, 2057, 2159, 2286, 2438, 2667, 2845, 3048, 3251, 3454, 3658

C

45889, 965, 1067, 1168, 1219, 1295, 1346, 1397, 1447, 1524, 1574, 1626, 1676, 1727

B

33660,787, 838, 889, 965, 1067, 1168, 1219, 1295, 1346, 1397, 1447, 1524, 1574, 1626

A

Quantity to be added to

get Pitch lengthNominal inside length, mmType

.16

.921.62.102.53.03.343.7

.941.52.02.52.93.23.53.75

1.251.702.02.42.73.03.23.4

1.21.51.71.952.12.32.52.6

.8

.951.01.21.31.41.51.5

105115125135145155165175+

B

2.34.05.56.57.58.3

1.43.34.75.96.87.58.0

2.03.54.55.46.06.67.0

1.983.03.64.24.75.05.4

1.41.92.22.52.72.93.0

150175200225250275305+

C

.28

.841.281.631.922.15

.11

.691.141.501.772.02.2

.4

.841.171.431.641.821.97

.46

.75

.981.161.301.401.51

.35

.50

.60

.69

.77

.83

.87

60758595105115125+

A252015105

Max Speed, m/sPitch dia, mm

Belt

Power rating, kW

Important points( ) ( )

( ) ( ) ( )⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−⎥⎦⎤

⎢⎣⎡ −++⎥⎦

⎤⎢⎣⎡ +−=

−+++=

212

2

2121

212

21

222

25.0

422

DDLDDDDLC

CDDDDCL

PPd

ddP

ππ

π 1.5, 2, 2.25, 2.5, 3, 3.5, 4.0, 4.5, 5, 6,7, 7.5, 9, 10, 10.5, 11, 12, 13, 14, 15, 18

Pulley pitch dia., inch

απφαπφ

α

22

2sin

2

1

121

+=−=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −= −

dCDD

Homework 7: Design a V-belt drive connecting 7.5 kW motor (rated speed 1440 rpm, shaft 1.5 inch) to a compressor. We need speed reduction = 3. Assume coefficient of friction = 0.25. Max center dist=500 mm.

( )12

2

3 DDCDC

d

d

+≤≥

Rolling Element Bearing

“Rotation is always easier than linear motion”. Low friction & moderate lubricant requirements

are two important advantages of rolling bearing.

If you can buy it, don’t make it!Bearing selection….~ 20,000 Varieties of bearings. Conventional bearing steel to ceramics, with (out) cage (brass/polymers). Pin.Smallest bearings – few grams. Largest 70 Tonnes

Video Clips of

Rolling Element Bearings

Radial Load Axial Load

Selection of bearing type

Cylindrical & Needle roller– Pure Radial LoadCylindrical roller thrust, ball thrust, four point angular contact ball bearings– Pure Axial loadTaper roller, spherical roller, angular contact bearings– Combined loadCylindrical roller, angular contact ball bearing–High speedDeep groove, angular contact, and cylindrical roller bearing– High running accuracy

In order of increasing outside bearing diameter

In increasing order

ACBBSABB

SRB

TRB DGBB

TBB

DGBB ACBB CRTB

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.

DIN 623 standard

http://www.skf.com/portal/skf/home/products?newlink=first&lang=en

61804 61804-2Z 61804-2RS1

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 0.01% of ball dia.

Rolling Element Bearings

yyThrust ball

yyyAngular contact ball

yyySelf Aligning Spherical Roller

yyySelf Aligning Ball

yyyyTaper Roller

yyNeedle

ySome types

yCylindrical Roller

yyyDeep groove ball

LowMedHighBothAxialRadial

Misalignment Capacity

Load Direction

Equivalent load: P = V X Fr + Y Fa

1.5 tanα

.65 cotα

.65.42 cotα

1.4 cotα

.411Self aligning ball bearing

.57

.68

.80

.951.14

1.631.441.241.07.93

.70

.67

.63

.60

.57

1.09.92.78.66.55

11.0.87.76.66.57

.43

.41

.39

.37

.35

1.212025303540

Angular contact ball bearing

.19

.22

.26

.28

.3

.34

.38

.42

.44

2.301.991.711.551.451.311.151.041.00

0.56012.301.991.711.551.451.311.151.041.00

0.561.21.014.028.056.084.11.17.28.42.56

YXYXYXVVFa/C0Deep groove ball bearing

Fa/VFr > eFa/VFr ≤ eFa/VFr > eStationaryRotatingeDouble rowSingle rowInner ringBearing type

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.

Deep Groove Ball Bearing

Deep Groove Ball Bearing

Pressed brass cage

Deep Groove Ball Bearing

E indicates reinforced ball set. TN9 indicates injection moulded snap type cage of glass fibrereinforced polyamide

Deep Groove Ball Bearing

Deep Groove Ball Bearing

Deep Groove Ball Bearing

Cylindrical roller bearings

Predominately Used as Floating

Bearing

• Variations of shaft due to the thermal expansion are accommodated between the inner ring & the roller set.

Double row full complement cylindrical roller bearings, mainly of the NNF design

Rolling Element Bearings-Load Calculation: Tabular Approach

Load ratingCr > P x fn x fL x fd

Where Cr = radial dynamic rating P = calculated effective radial loadfn = speed (rpm) factorfl = Life (hours) factorfd = dynamic or service factor

Example 1: Radial load = 4448 N, Speed = 1000 rpm

Desired life= 30 000 hours, No Shock loading.

Cr > P x fn x fL x fdfd = 1.0; P = 4 448 N

fn= 2.78; fl= 3.42

=> Cr > 42, 290 N

Mathematical Approach

In ideal case, bearings fail by fatigue.

Dynamic load rating (catalogue C0 reading) is the load which 90% (reliability=0.9) of a group of identical bearings will sustain for minimum of 106 cycles.

( )

Speed60000,1000

PC

hoursin life Bearing

bearingsroller for 3

10bearings ballfor 3

10 3322116

a

aaaa

a

a

LPLPLPC

⎟⎟⎠

⎞⎜⎜⎝

⎛=⇒

=

=

===17.11

90

9.01log

1log

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=e

eR R

LL

Example 2: Radial load = 2 224 N, Speed = 1500 rpm

Desired life= 8 hours/day, 5 day/weeks for 5 years, Light Shock loading. 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

Example 3: A radial load of 3000N combined with thrust load of 2500N is to be carried on a 200 series ball bearing for 70 mm diarotating shaft 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 kN

Step 2:

Step 3: Radial load for 6214 bearing is 5955N & for 7214 bearing radial load is 3070.Step 4: Life for 6214 will be 20,400 hours and for 7214, life=210,550 Hours

0.8.76.3940Angular contact ball bearing

.261.710.56.056

YXFa/C0Deep groove ball bearingeSingle row, Fa/VFr > eBearing type

Homework8: A single row cylindrical roller bearing N 205 ECP is subjected to pure radial load of 2800 N and rotational speed = 1500 rpm. Estimate the bearing life for reliability = 0.99.

ANS: 3448 Hours

Homework9: Select a suitable deep groove ball bearing for a shaft of 30 mm dia rotating at 2000 rpm. Bearing needs to bear a radial load of 2000 N and axial load of 400 N.

Sources of friction in Anti-friction Bearings

1.Elastic hysteresis in Rolling

• μ≅0.0001 for Chrome steel.

2.Viscous shearing

• τ=η du/dz

3.Viscous drag

4.Sliding

5.Seals

Friction Moments in Rolling Bearings

Load dependent

Lubricant and Speed dependent

Seal/Shield dependent

dia/2) Bore(PM P μ=

( )( ) 2000 56.1

2000 103

33/27

<−=

≥= −

NvifdfeM

NvifdNvfM

mLL

mLL

N.mm diaMax diaMin 2

12 ⎟⎟

⎠

⎞⎜⎜⎝

⎛ ++=

ffM s

Total friction moment M = MP+ML+MS

.0015

.001

.002

.0011

.0025

Deep groove ball Self aligning ballAngular contact ballCylindrical rollerNeedle roller

μBearing Type

Table: Coefficient of friction for bearings

Deep groove ball Self aligning ballAngular contact ballCylindrical rollerNeedle roller

Bearing Type

1.7-11.7

1.5-2.86

Oil Spot

.75-21.5-2

2.6-112

Grease

43-46.6

2.2-424

21.5-23.3

2.2-412

Vertical Shaft in oil bath

Oil Bath

Table: Lubrication factor fL

Deep groove ball Self aligning ballAngular contact ballCylindrical rollerNeedle roller

Bearing Type

1010102550

f22020201020

f1

Table: Friction factors for seals

Homework 10: Estimate friction moment of 6214-2RS1 bearing running at 6,000 rpm under 5000 N radial load when jet lubricated by synthetic ester jet engine oil having a viscosity of 6 mm2/s (cSt) at operating temperature.

Ans: 468.6 N.mm

dia/2) Bore(PM P μ=

( )( ) 2000 56.1

2000 103

33/27

<−=

≥= −

NvifdfeM

NvifdNvfM

mLL

mLL

N.mm diaMax diaMin 2

12 ⎟⎟

⎠

⎞⎜⎜⎝

⎛ ++=

ffM s

Bearing Temperature

Dependence on total friction moment, speed, extraneous heat source and heat dissipation capability.

[ ] ( ) MBKttq tambLB ωπ =+−=

DiaMax diaMin Q generationheat of RateQ raten dissipatioHeat LR

Heat flow density W/m2

~ 2,000

Cooling factor

1 for natural cooling

2.5 for forced cooling

Angular speed rad/s

Homework 11: Estimate bearing operating temperature of 6214-2RS1 bearing running at 6,000 rpm under 5000 N radial load when jet lubricated by synthetic ester jet engine oil having a viscosity of 6 mm2/s (cSt) at operating temperature. Assume ambient temp = 30°C and forced cooling of bearing.

Ans: 34°C

Bearing Mounting Interference fit

One part slightly interferes with space that the other part is taking up.Fastening between two parts (i.e. shaft into bearing, bearing into housing) aiming that even large amount of torque cannot turn one of them relative to other.Shaping (i.e. turning, drilling, boring, reaming, grinding, etc.) shaft/housing with slight deviation in size from nominal dimensions. Slightly oversized shaft, slightly undersized housing. Press/shrink fitting of inner race on shaft causes expansion of inner ring. Similarly press fitting of outer ring in housing causes former member to shrink slightly.

Preload ????If interference fits exceed the internal radial clearance, the rolling elements become preloaded.

Clearance before installation & after installation ???

Hot rolling, Flame cutting

Sand Casting

Forging

Die Casting

Drilling

Cold Rolling, Drawing

Extruding

Planing, Shaping

Milling

Sawing

Boring, Turning

Reaming

Broaching

Plan grinding

Diamond turning

Cylindrical grinding

Super finishing

Honing

Lapping

1615141312111098765432IT Grade

13001150100087074062052043036030025014

81072063054046039033027022018014013

52046040035030025021018015012010012

32029025022019016013011090756011

210185160140120100847058484010

13011510087746252433630259

81726354463933272218148

52464035302521181512107

32292522191613119866

232018151311986545

1614121087654434

12108654432.52.523

875432.52.521.51.51.22

64.53.52.521.51.51.2110.81

International tolerance grade of industrial processes. IT Grade

315250180120805030181063inc.

2501801208050301810631over

Nominal Sizes (mm)

Tolerances:Allowable

0 μm< IF< 12 μmLoose Preloading

Achievable

Boring, Turning

Plan grinding

Diamond turning

Cylindrical grinding

Super finishing

Honing

Lapping

1615141312111098765432IT Grade

Example: We need to choose 03/04 bearing.

52043014

33027013

21018012

13011011

847010

52439

33278

21187

13116

985

654

433

2.522

1.51.21

IT Grade

3018inc.

1810over

Nominal Sizes (mm)

Super finishing

Honing

Lapping

65432IT Grade

0.1° misalignment results 52 μm preloading for 30 mm length

0.01° misalignment results 5 μm preloading.

Friction Moments in Rolling Bearings

√ Load dependent

× Lubricant and Speed dependent

× Seal dependent

dia/2) Bore(PM P μ=

( )( ) 2000 56.1

2000 103

33/27

<−=

≥= −

NvifdfeM

NvifdNvfM

mLL

mLL

N.mm diaMax diaMin 2

12 ⎟⎟

⎠

⎞⎜⎜⎝

⎛ ++=

ffM s

Total friction moment M = MP+ML+MS

μ ~ 0.0015, Applied load~ 200 to 500 N. BD= 17-20 mm.

( )

NWW

W

819)5.1^4/(6550 WPreload

655010100*4.

2

32

2

32

32

2

32

1

2

1

==

=→=δδ

10 micron interference causes 819 N preload.

15 microns --- 1500 N preload.

20 microns -- 2316 N load

Static Load Rating: C0.. Radial load causing 0.01% of ball dia 0.4 μm plastic deformation results if 6550 N load is applied on the bearing.

Homework 12: Estimate bearing operating temperature of 6214-2RS1 bearing running at 6,000 rpm under 1000 N radial load when jet lubricated by synthetic ester jet engine oil having a viscosity of 6 mm2/s (cSt) at operating temperature. Assume ambient temp = 30°C, ball dia=6 mm, preloading of bearing = 10 microns, and forced cooling of bearing.Ans: 33.6 °C

Failure of Four Row Cylindrical Roller Bearing

Two large roller bearings – (ID = 865 mm, OD = 1180 mm) failed in a cold rolling mill.

one bearing failed within 105 hours (installed on 05/01/03 and failed completely on 10/01/03), and

other failed within 300 hours of operation (installed on 05/01/03 and removed on 24/01/03 due to detection of excessive vibration and metal particles).

Expected life of bearings was approximately 40,000 operating hours

Survival rate 0.5% and 1.0%.

Failed Bearing

Bearing subjected to normal load

In rolling mills the load is of constant direction. Only a quarter of the outer race is under load. For this reason, the side faces of the outer races are divided into four zones indicated by I to IV.

When the bearing is mounted for the first time it is usual to position zone I in the direction of action of the load.

After a period of approximately 1000 operating hours (≅ 2 months), outer race is turned 90°.

Conclusion: Rated bearing life = 4.* Life of one load zone. Expected life of each load zone = 10,000 operating hours

Detailed survey of failed bearing indicated placement of hole on the line of maximum load.

Four holes of 3/8” 10 UNC 3B of 45mm depth were drilled and tapped to facilitate the handling of outer race.

Hole

GEARS

To transmit power between shafts rotating usually at different rotational speeds… ? belt drives

Generally gear pair acts as a speed reducer aiming torque amplification at output shaft. ? Friction

Often gears are treated as pitch cylinder which roll together without slip. Positive drive provided by meshing teeth. ? belt drives.Hybrid drives

out

inAm

ωω

= ratio, Torque

Spur Gear Drive

Spur Gears: Teeth parallel to axis of rotation. Suitable to transmit motion between parallel shafts.

Helical Gears: Teeth inclined to axis of rotation.

Lesser noise compared to spur gears. Induce thrust loads and bending couples.

Helical Gear Drive.

Straight Tooth Bevel Gears: Teeth formed on conical surfaces. Transmit motion between intersecting shafts.

Bevel gear drive with straight teeth.

Worm Gears: Worm resembles a screw. Direction of rotation of worm wheel??? High speed ratio.

Worm Gear Drive. (a) Cylindrical teeth; (b) double enveloping.

Spur Gear Nomenclature

Basic spur gear Geometry.

Pitch circles: Imaginary tangent circles.

Pinion: Smaller.

Circular pitch: Sum of tooth thickness & width of space.

Addendum: Radial distance between top land and pitch circle.

Backlash: Difference between tooth space and tooth thickness.

Module: m=Dp/ZP=DG/ZG

Spur Gear Nomenclature

Second choice 1.125,1.375,1.75,2.25,2.75,3.5,4.5,5.5,7,9,11,14,18,22,28,36,45Preferred 1,1.25,1.5,2,2.5,3,4,5,6,8,10,12,16,20,25,32,40

Modules

Conjugate Action

Catalogue B

All sizes shown above are in 20o PA =Presure Angle (Note: 14.1/2o PA is also supplied)

152.6027 or 14.75HBIGM04.75

142.8027 or 14.5HBIGM04.5

147.0027 or 14.25HBIGM04.25

1358.005050HBIGM20124.6027 or 14HBIGM04

1190.005018HBIGM18123.2027 or 13.75HBIGM03.75

980.004016HBIGM16120.4027 or 13.5HBIGM03.5

665.004014HBIGM14123.2027 or 13.25HBIGM3.25

616.004012HBIGM12119.0027 or 13HBIGM03

560.004011HBIGM11110.6027 or 12.75HBIGM02.75

512.403210HBIGM10107.8027 or 12.5HBIGM02.5

456.40329HBIGM09103.6027 or 12.25HBIGM02.25

392.0032 or 1-1/48HBIGM08103.6027 or 12HBIGM02

329.0032 or 1-1/47HBIGM07100.8027 or 11.75HBIGM01.75

193.2027 or 16HBIGM0698.0027 or 11.5HBIGM01.5

172.2027 or 15.5HBIGM05.595.2027 or 11.25HBIGM01.25

155.4027 or 15HBIGM0595.2027mm or 1”1mmHBIGM01

PoundsBore

diameter of cutter

Module MMPounds

Bore diameter of cutter

Module MM

214 - 167317 - 206421 - 255526 - 344635 - 543755 - 13428135 - RACK1

(European Cutter No)Cuts TeethBSS Cutter

Number

Catalogue B

Pitch and Base Circles.? Cross belt

1m.8m20Stub

1.25m1.35m

1m25

1.25m1.35m

1m22.5

1.25m1.35m

1m20Full depth

Dedendum

Addendum

φ, deg

Tooth system

c

Contact Ratio

)Zr

(2

recess ofLength approach ofLength Base_pitch

action ofLength

g

bgπ

+

Contact Ratio

g

bgogbpop

bgogbpop

bgog

bpop

bpop

ZCrrrr

rrrr

rr

rr

caba

rrb

/r2sin

ratioContact

sin)rr( ab action, ofLength

sinracSimilarly

sinrcb or,

cbLength

aLength

sinr caLength

bg

2222

gp2222

g22

p22

**

22*

p*

πφ

φ

φ

φ

φ

−−+−=

+−−+−=

−−=

−−=

−=

−=

=

Ex: For φ=20°, ZP=19, Zg=37, and m=4; Find Gear Ratio, circular pitch, base pitch, pitch diameters, center distance, addendum, dedendum, whole depth, clearance, outside diameters, and contact ratio. If center distance is increased by 2% what will be new pressure angle and new contact ratio.

a

r

Zmpm

p

g

c

2d dmm5bm 1.25 b Dedendum,

mm4a m1.0 a Addendum,

)(r C dist,center Nominal

d diaPitch cosppitch Base

p or,

Zd pitch Circular

1937 RatioGear

pop

g

g

b

c

g

g

+==⇒×=

=⇒×=

+=

==

=

=

=

φπ

πb

bgogbpop

pCrrrr φsin

ratioContact 2222 −−+−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛= −

p

pnew r

r02.1cos

cos 1 φφ

New rp

Design of Spur Gears

Breakage of gear teethExcessive wear of gear tooth surfaceExcessive noise Excessive heat

Patent U.S. 5,503,045

Forces on Gear Tooth

φtancomponent Radial

WMagnitude

action. of line thealong gear,on tooth toingcorrespond tooth toone from ed transmittis W Force point,pitch At

gear. pinion toby delivered being is T A torque

t

p

tr

p

P

WW

rT

=

=

NOTE: Assumption of one tooth contact. Load sharing ???

• Constant torque, but each tooth experiences repeated loading.. Fatigue loading.

Ex: Pinion shaft passes 15kW at 2500 rpm. For φ=25°, ZP=14, m=4, and Gear Ratio=3.5, determine transmitted loads on gear teeth. Find pitch diameters, mean and alternative components of transmitted load.

( )

2

2

954tan,

20462//,

56144

.55.2003.575.3

3.57)60/25002/(15000

49145.3

t

t

tr

ppt

pp

g

p

g

WloadeAlternativ

WloadMean

NWWloadRadial

NdTWloadTangential

mmZmd

mNT

T

Z

=

=

==

===

=×==

=×=

==

=×=

φ

π

Homework 13: What will the pressure angle be if the center distance of a 20° pressure angle gear pair is increased by 7%.

Homework 14: Find out the mean and alternating load components of a gear-set that transmits 50 kW at 1600 pinion rpm. φ=25°, ZP=23, Zg=57, and m=4.

Stresses in Spur Gears

Two modes of failures:Fatigue: fluctuating bending stresses at root of tooth.

Keep stress state within modified Goodman line for material.. Infinite life

Surface fatigue (Pitting)Repeated surface contact stresses …materials do not exhibits endurance limit

Properly designed gear-sets should never fail but must be expected to eventually fail by one of surface wear.

Bending Stresses

AssumptionsCompression due to radial component of force is negligible.Teeth do not share loadGreatest force is exerted at tip

Lewis Eq.

6/2tFlW

ModulusSectionMoment

tb

b

=

=

σ

σ

mYFWt

b =σ

212224262830343843

No. of Teeth

0.2450.2610.2770.2900.2960.3030.3090.3140.322

Form factor Y

0.4090.4220.4350.4470.4600.4720.4800.485

506075100150300400Rack

0.3280.3310.3370.3460.3530.3590.3710.3840.397

121314151617181920

Form factor Y

No. of Teeth

Form factor Y

No. of Teeth

AGMA introduced velocity factor in terms of pitch line velocity (m/s) in Lewis equation.

( )profilecastironcastVKv ,05.3

05.3 +=

( )

( )

( )profilegroundorShavedVK

profileshapedorHobbedVK

profilemilledorCutVK

v

v

v

56.556.5

56.356.3

01.601.6

+=

+=

+=

mYFWK

AGMA

tvb =σ

Eq. Lewis

Useful for preliminary estimation of gear size.

Homework 15: Find out the power rating (for infinite life) of milled profiled spur gear (AISI material, ultimate strength = 380MPa) for data: φ=20°, ZP=16, F=36mm, m=3.0, N = 20 rps. Assume factor of safety = 3.0.

Ans: Endurance strength = 190 MPa. Allowable strength = 84 MPa

Tangential load = 1790 N.

Power rating = 5.37 kW.

Ex: Find out the power rating (assuming static loading) of milled profiled spur gear (AISI material, yield strength = 210MPa) for data: φ=20°, ZP=16, F=36mm, m=3.0, N = 20 rps. Assume factor of safety = 3.0.

Ans: Allowable bending stress = 70 MPa.

Pitch line velocity V=3.0 m/s.

Kv = 1.5 , Form factor Y = 0.296

Tangential load = 1492 N.

Power rating = 4.475 kW.

Driven Machines

Power Source Uniform Light shock Moderate shock Heavy shock Application factor, Ka Uniform (Electric motor, turbine) Light shock (Multicylinder) Moderate shock

1.00

1.20

1.30

1.25

1.40

1.70

1.50

1.75

2.00

1.75 2.25 2.75

AGMA Bending Stress Equation

loading ofpoint angle, pressureon depends FactorGeometrybendingAGMAJ

KKKJmF

WKmBa

tvb

=

=σ

1.61.71.82.0

< 50150250>500

KmFace width, mm

Load distribution factor Km

t

RBBB

BB

htmwheremK

mK

=≥=

<≤+−=

2.1 0.1

2.1m0.5 4.32factor thicknessRim B

Ex: A gear pair (ZP=23, φ=20°, Zg =24, m=1.75, F=10.0 mm) transmits 8 N.m torque from crankshaft (rotational speed 8000 rpm) of single cylinder IC engine to wheels. Bore diameter of pinion is 17 mm, and bore dia of gear is 20 mm. Using AGMA bending stress formula to determine the maximum bending stress. Assume gears are grounded.

( )

( )12.1

4375.95.09375.375.1*25.2

635.37

875.3575.1*25.1*2

3185.156.5

56.5 0.42

/86.1660

800025.40 25.40

6.1 0.2

=⇒>

=−===

=

=−=

=+

==

→==

==

BB

pprootR

t

groot

pproot

vg

p

ma

Km

Boredtmmh

d

dd

VKd

smVd

KKπ

MPa

KKKJmF

WK

b

mBatv

b

6.368=⇒

=

σ

σ

Allowable Bending Stress vs. Brinell Hardness

Effect of Brinell hardness on allowable bending stress for two grades of through-hardened steel [ANSI/AGMA Standard 1012-F90]

MPaHGradeMPaHGrade

Bball

Bball

3.88533.01113703.02

,

,

+=→

+=→

σσ

Core hardness ????

3453273510967.95605255311977.4

3543363610968.45775435412078.0

3633443711068.95955605512078.5

3723533811069.46135775679.0

3823623911169.96335955779.6

3923714011270.46536155880.1

4023814111270.96746345980.7

4123904211371.56976546081.2

4234004311372.07206706181.8

4344094411472.57466886282.3

4464214511573.17727056382.8

4584324611573.68007226483.4

4714424711674.18327396583.9

4844514811674.78656684.5

4984644911775.29006785.0

5134755011775.99406885.6

5284875111876.310446986.0

5445005211976.810767086.5

CBACBA

VickersBrinellRockwellVickersBrinellRockwell

Ex: A gear pair (ZP=23, φ=20°, Zg =24, m=1.75, F=10.0 mm) transmits 8 N.m torque from crankshaft (rotational speed 8000 rpm) of single cylinder IC engine to wheels. Bore diameter of pinion is 17 mm, and bore dia of gear is 20 mm. Using AGMA bending stress formula to determine the factor of safety for tip as well as HPSTC loading case. Assume gears are grounded and through hardened (HRC 35) with close quality control.

ANS: Factor of safety for tip loading = 0.93Factor of safety for HPSTC loading = 1.2880

Surface/Contact Stresses in Spur Gears

Surface failure of gear tooth occurs due to very high local contact stresses. Maximum contact pressure at the contact point between two cylinders is given by:

( ) ( )[ ]⎟⎠⎞

⎜⎝⎛ +

−+−=

=

gp

ggpp

dd

EEL

Wbwhere

LbWp

11

/1/12

2

22

max

ννπ

π

Surface contact compressive stress

( ) ( )point.contact at surfaces tooth of

/1/1

11

cos

21

22212

curvatureofradiiarerandrwhere

EErr

FW

ggpp

tc

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+−

+=

ννφπσ

Z

Z

2cossin

111

equation resistance pittingAGMA

p

g

22

c

g

g

g

P

pP

vmap

tP

ZI

EE

C

CCCdIF

WC

+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ −+⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

=

φφ

ννπ

σ

Ex: A gear pair (ZP=23, φ=20°, Zg =24, m=1.75, F=10.0 mm) transmits 8 N.m torque from crankshaft (rotational speed 8000 rpm) of single cylinder IC engine to wheels. Bore diameter of pinion is 17 mm, and bore dia of gear is 20 mm. Using AGMA pitting resistance formula to determine the maximum contact stress. Assume gears are grounded, E = 2.e5 MPa, ν=0.3

0.0821I Z

Z

2cossin

1.32 1.6 2.0 18711

1

equation resistance pittingAGMA

p

g

22

c

=⇒+

=

=⇒

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ −+⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

=

g

p

g

g

P

pP

vmap

tP

ZI

C

EE

C

CCCdIF

WC

φφ

ννπ

σ

Ans: 1334 MPa

Contact Stress vs. BrinellHardness

Effect of Brinell Hardness on allowable contact stress for through-hardened steel.

For 400 BHN of close quality controlled steel, allowable contact stress = 1150 MPa.

If we use this material for previous example (slide 136), then factor of safety:

0.8621.

HOW TO INCREASE FACTOR OF SAFETY??

Design Alternative I: Reduce value of Cv

( ) ( ) 3/21225.0 and 15650

200

v

B

v

QBBA

AVA

C

−=−+=

⎟⎟⎠

⎞⎜⎜⎝

⎛ +=

1.341.231.131

Cv for 16.86 m/s velocity

25 μm20 μm15 μm10 μm

Tolerance

0.860.890.930.99

9101112

S.F.AGMA Qv

Design Alternative II: Increase face width keeping Qv=10 and. S.f.=1.1

15.18 mm of width is required.50% increase in face width may not be acceptable by Automobile industries.

S.F. = 1.0 requires face width of 12.56 mm

25% increase in face width.. ? Acceptable

Design Alternative III: Increase Case Hardness

Carburization and case hardening increases the hardness of gear ranging 55-64 HRC (~ 1250 to 1300 MPa). However, strength depends on the life of operation and it decreases with increase in operational life.

0178.03558.1 −= cycleL NK

1.0180.9770.938

KL

0.990.950.91

107

108

109

S.F. =1250*KL/σc

Ncycle