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    SWINBURNE UNIVERSITY OF TECHNOLOGY (SARAWAK CAMPUS)

    FACULTY OF ENGINEERING AND INDUSTRIAL SCIENCE

    HES3350 Machine DesignSemester 1, 2012

    Assignment 2: Gear Specification with Consideration of Bending Fatigue and Surface

    Durability

    By

    Group 9

    Stephen, P. Y. Bong (4209168)

    Kueh Min Hui (4209974)

    Mohd Azmudin Bin Isa @ Danial (4209028)

    Jimmy Huong How Lee (4209761)

    Lecturer: Dr. Soon Kok Heng

    Due Date: 4 pm, 27th

    April 2012 (Thursday)

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    1. OBJECTIVEThe principal objective of this assignment is to select appropriate spur gears for a two-stage

    compound reverted gear train from the gear catalog provided. Apart from that, the computations of

    the maximum module, the face width, and the material used for the gears with the consideration of

    bending fatigue and surface durability are also the purposes of this assignment as well.

    2. INTRODUCTIONWith the operating parameters provided such as the amount of power to be delivered, the number of

    teeth for Gear 1, 2, 3 and 4, the input speed, the maximum gearbox height, the clearance and wall

    thickness, and the pressure angle; we are required to determine the maximum module and select

    appropriate spur gears for the two-stage compound reverted gear train from the gear catalog given

    based on the prerequisite stated which is the safety factor should not exceed 2.2. In addition, the

    determination of the face width and the material used, and the identification of the most critical gear

    are incorporated in this report as well. The spur gears are selected from the gear catalog with the

    consideration of bending fatigue and surface durability.

    Since the assignment by all means, cannot be tantamount to a real life design project as most of the

    selection of operating parameters such as the dynamic factor and the surface factor for the

    computations of bending stresses and endurance limit respectively are based on experimental data

    or charts provided and appropriate assumptions for instance the surfaces in contact (bore and shaft)

    are frictionless and thus there has no power losses through the transmission. By neglecting the

    power losses due to friction and surface roughness as a result of various manufacturing

    methodology and heat treatment processes, we only centralize our design based on the bending

    fatigue and surface durability in the selection of spur gears.

    3. MAIN CONTENT3.1 Parameters and Schematic Diagram for the Two-stage Compound Reverted Gear

    Train in the Gearbox

    The schematic diagram of the gearbox is shown in Fig. 1 below:

    Fig. 1: Schematic of the gearbox

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    The operating parameters given are tabulated in Table 1 below:

    Power to be delivered, 9 kWN1 &N3 42 Teeth

    N2 &N4 86 Teeth

    Input speed, 1 2300 rpm

    Maximum Gearbox Height,H 600 mmClearance and Wall Thickness 20 mm

    Pressure angle for spur gear, 20Expected life for 40 hr./week, 50 week/year operation 3.25 years

    Table 1: Operating parameters

    where the Maximum Gearbox Height,His given by the following equation:

    ThicknessWallClearance2

    22

    412 +++++= m

    dddH

    where m is the module and d1 = mN1, d2 = mN2 and d4 = mN4.

    3.2 Calculations and Selection of Maximum Module and Face WidthBased on the operating parameters given, the module of the gear train can be determined by using

    the relation of gears diameters and maximum height of gearbox:

    ThicknessWallClearance222

    412 +++++= m

    dddH

    SinceN2 =N4, therefore, d2 = d4 and the equation above can be expressed as:

    ThicknessWallClearance22

    3

    2

    ThicknessWallClearance222

    21

    21

    2

    ++++=

    +++++=

    mdd

    H

    mdd

    dH

    Substituting d1 = mN1 and d2 = mN2 into equation above yields:

    22

    3

    2

    Thickness)Wall(Clearance

    21++

    +=NN

    Hm

    Since the KHK Gear Catalog does not have spur gears with 86 teeth, therefore, spur gears with 85

    teeth has been selected in the analysis. Given that N1 = 42 Teeth,N2 = 85 Teeth,H= 600 mm and

    Clearance + Wall Thickness = 20 mm, hence the module can be computed as follows:

    Toothmm3.8538=

    ++

    =

    ++

    +=

    22

    Teeth)85(3

    2

    Teeth42

    mm20mm600

    22

    3

    2

    )ThicknessWallClearance(

    21 NN

    Hm

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    Based on the computations of module above, the module employed for the two-stage compound

    reverted gear train must not exceed m = 3.85388 mm/Tooth. In addition, according to KHK Gear

    Catalog, spur gears with module of 2 have the closest number of teeth (N= 85 Teeth) to the number

    of teeth for gears 2 and 4 given in this assignment. As a result, steel spur gears with 85 Teeth and

    module 2 has been chosen in the analysis of gear tooth strength and surface durability.

    According to KHK Gear Catalog, Steel Spur Gears with 42 and 85 Teeth with module of 2 haveface widths of 20 mm (See KHK Gear Catalog in Appendix):

    mm20Toothmm20

    Teeth42146)pp.Catalog,Gear(KHK

    31=

    =

    == b

    m

    NN

    mm20Toothmm20

    Teeth85156)pp.Catalog,Gear(KHK

    42=

    =

    == b

    m

    NN

    Therefore, b1 = b2 = b3 = b4 = 20 mm.

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    3.3 Analysis of Gear Tooth Strength3.3.1 Bending Fatigue Analysis by Using Steel Spur Gear with Module 2For m = 2 mm/Tooth,

    ( )( )( )( )

    ( )( )

    ( )( ) mm170Teeth58Toothmm2

    mm84Teeth42Toothmm2

    mm170Teeth58Toothmm2mm84Teeth42Toothmm2

    44

    33

    22

    11

    ===

    ===

    ===

    ===

    mNd

    mNd

    mNdmNd

    The gear ratio for a pair of spur gears (pinion and gear) is given by:

    p

    g

    g

    p

    d

    d=

    where

    g = Angular velocity of gearp = Angular velocity of piniondg = Diameter of gear

    dp = Diameter of pinion

    By using the relation of gear ratio above, the angular velocity of gear 2 can be obtained:

    rpm1136.47=

    =

    == mm170

    mm48rpm2300

    2

    112

    1

    2

    2

    1

    d

    d

    d

    d

    Since gear 2 and gear 3 are located on the same shaft, therefore,

    rpm1136.47== 23

    The speed ratio for double-reduction gear train is given by:

    rpm561.55=

    =

    ==

    8585

    4242rpm2300

    42

    3114

    31

    42

    4

    1

    NN

    NN

    NN

    NN

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    Stage 1 (Gear 1 & Gear 2)

    Pitch Line Velocity,60000

    dV=

    sm10.12=

    ===

    ===1Stage

    222

    11

    1

    sm12.0160000

    rpm)7mm)(1136.4170(

    60000:2Gear

    sm12.0160000

    rpm)mm)(230084(

    60000:1GearV

    dV

    d

    V

    Tangential Forces, Ft= /V

    N889.33=

    ===

    ===

    1Stage,

    1Stage2

    1Stage1

    N33.889sm10.12W9000:2Gear

    N33.889sm10.12W9000:1Gear

    t

    t

    tF

    VWF

    VWF

    Gear Tooth Bending Stress, movt KKK

    bJm

    F=

    From Figure 14-6 (Budynas, R. G. & Nisbett, J. K., pp. 753), the Geometry Factor for spur

    gears of 42 Teeth and 85 Teeth areJ1, 30.2725 andJ2,40.295.

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    minft1992.13===m.30430

    ft1

    min1

    s60

    s

    m12.10sm12.101StageV

    Velocity (Dynamic) Factor,78

    78 VKv

    += (Highest precision, shaved and ground)

    1.2539=+

    =+

    ==78

    13.199278

    78

    78 1StageVKK vv

    From TABLE 15.1 (Juvinall, R. C. & Marshek, K. M., pp. 615), the Overload Correction

    Factor, Ko = 1 (Uniform source of machinery and power)

    From TABLE 15.2 (Juvinall, R. C. & Marshek, K. M., pp. 616), the Mounting Factor,

    Km = 1.3 (Accurate mounting, small bearing clearance, minimum deflection)

    MPa132.99=== )3.1)(1)(2539.1(Tooth)mm)(2mm)(0.272520(

    N33.889

    11

    1Stage,

    1 mov

    tKKK

    mJb

    F

    MPa122.85=== )3.1)(1)(2539.1(Tooth)mm(2mm)(0.295)20(

    N33.889

    22

    1Stage,

    2 mov

    tKKK

    mJb

    F

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    Stage 2 (Gear 3 & Gear 4)

    Pitch Line Velocity,60000

    dV=

    sm4.8895=

    ===

    ===2Stage

    444

    33

    3

    sm9985.460000

    rpm)mm)(561.55170(

    60000:4Gear

    sm9985.460000

    rpm)7mm)(1136.484(

    60000:3GearV

    dV

    d

    V

    Tangential Forces, Ft= /V

    N1800.54=

    ===

    ===

    2Stage,

    2Stage4

    2Stage3

    N54.1800sm4.9985W9000:4Gear

    N54.1800sm4.9985W9000:3Gear

    t

    t

    tF

    VWF

    VWF

    Gear Tooth Bending Stress, movt KKK

    bJm

    F=

    From Figure 14-6 (Budynas, R. G. & Nisbett, J. K., pp. 753), the Geometry Factor for spur

    gears of 42 Teeth and 85 Teeth areJ1, 30.2725 andJ2,40.295.

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    minft983.96===m.30430

    ft1

    min1

    s60

    s

    m9985.4sm9985.42StageV

    Velocity (Dynamic) Factor,78

    78 VKv

    += (Highest precision, shaved and ground)

    1.1844=+

    =+

    ==78

    96.98378

    78

    78 1StageVKK vv

    From TABLE 15.1 (Juvinall, R. C. & Marshek, K. M., pp. 615), the Overload Correction

    Factor, Ko = 1 (Uniform source of machinery and power)

    From TABLE 15.2 (Juvinall, R. C. & Marshek, K. M., pp. 616), the Mounting Factor,

    Km = 1.3 (Accurate mounting, small bearing clearance, minimum deflection)

    MPa269.19=== )3.1)(1)(2539.1(Tooth)mm)(2mm)(0.272520(

    N54.1800

    33

    2Stage,

    3 mov

    tKKK

    mJb

    F

    MPa248.73=== )3.1)(1)(2539.1(Tooth)mm(2mm)(0.295)20(

    N54.1800

    44

    2Stage,

    4 mov

    tKKK

    mJb

    F

    By comparing the four bending stresses above, the maximum bending stress is max = 3 = 269.19

    MPa.

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    The gear in the gear train which is subjected to the highest tooth bending stresses is the one which

    has the largest possibility to fail in advance. Based on the stresses computed above, max = 269.19MPa and therefore the critical gear is Gear 3.

    Refined Analysis of Gear-Tooth Bending Strength Endurance Limit

    msKKKCCCSS trSGLnn =

    whereun SS 5.0=

    Since the Steel Spur Gears are made of S45C Steel with a maximum Brineel Hardness of 194 HB,

    therefore, the tensile strength, Su, can be determined by using the relation:

    Su = Kh(HB) = (3.45)(194) MPa = 669.3 MPa (Kh = 3.45 in SI Units)

    ( ) MPa65.334MPa3.6695.05.0 === un SS

    Load Factor, CL = 1 (Bending loads)

    )5(1

    in.Teeth7.12

    mm25.4

    in.1mm170

    Teeth58

    in.Teeth7.12

    mm25.4in.1mm84

    Teeth42

    in.Teeth7.12

    mm25.4

    in.1mm170

    Teeth58

    in.Teeth7.12

    mm25.4

    in.1mm84

    Teeth42

    4

    44

    3

    33

    2

    22

    1

    11

    >=

    =

    ==

    =

    ==

    =

    ==

    =

    ==

    PC

    D

    NP

    D

    NP

    D

    NP

    D

    NP

    G

    From TABLE 15.3 (Juvinall, R. C. & Marshek, K. M., pp. 615), the Reliability Correction

    Factor, Kr= 1 (50% Reliability)

    Temperature Factor, Kt= 1 (Assumed the temperature lower than 160 F)

    Mean Stress Factor, Kms = 1 (Gear 3 is idler)

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    From FIGURE 8.13 (Juvinall, R. C. & Marshek, K. M., pp. 301), the Surface Factor,

    CS= 0.76 (Machined or cold-drawn)

    Therefore, the endurance limit is computed as follows:

    MPa254.334=

    =

    =

    1)76)(1)(1)(MPa)(1)(0.(334.65

    msKKKCCCSS trSGLnn

    Safety Factor, 0.9448===MPa19.269

    MPa334.254SF

    max

    nS

    Since the safety factor obtained above is less than 2.2 in which the pre-requisites is not

    satisfied. Therefore, Steel Spur Gear with module of 2 cannot be employed in the double-reduction gear train. As a result, Steel Spur Gear with module of 3 has been selected for the

    application.

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    3.3.2 Bending Fatigue Analysis by Using Steel Spur Gear with Module 3 (N2 =N4 = 90)For m = 3 mm/Tooth,

    ( )( )

    ( )( )

    ( )( )

    ( )( ) mm270Teeth09Toothmm3

    mm126Teeth42Toothmm3

    mm270Teeth09Toothmm3

    mm126Teeth42Toothmm3

    44

    33

    22

    11

    ===

    ===

    ===

    ===

    mNd

    mNd

    mNd

    mNd

    The gear ratio for a pair of spur gears (pinion and gear) is given by:

    p

    g

    g

    p

    d

    d=

    where

    g = Angular velocity of gearp = Angular velocity of piniondg = Diameter of gear

    dp = Diameter of pinion

    By using the relation of gear ratio above, the angular velocity of gear 2 can be obtained:

    rpm1073.33=

    =

    ==

    mm270

    mm261rpm2300

    2

    112

    1

    2

    2

    1

    d

    d

    d

    d

    Since gear 2 and gear 3 are located on the same shaft, therefore,

    rpm1073.33== 23

    The speed ratio for double-reduction gear train is given by:

    rpm500.89=

    =

    ==

    0909

    4242rpm2300

    42

    3114

    31

    42

    4

    1

    NN

    NN

    NN

    NN

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    Stage 1 (Gear 1 & Gear 2)

    Pitch Line Velocity,60000

    dV=

    sm15.17=

    ===

    ===1Stage

    222

    11

    1

    sm17.1560000

    rpm)3mm)(1073.3270(

    60000:2Gear

    sm17.1560000

    rpm)mm)(2300126(

    60000:1GearV

    dV

    d

    V

    Tangential Forces, Ft= /V

    N593.28=

    ===

    ===

    1Stage,

    1Stage2

    1Stage1

    N28.593sm15.17W9000:2Gear

    N28.593sm15.17W9000:1Geart

    t

    tF

    VWF

    VWF

    Gear Tooth Bending Stress, movt KKK

    bJm

    F=

    From Figure 14-6 (Budynas, R. G. & Nisbett, J. K., pp. 753), the Geometry Factor for spur

    gears of 42 Teeth and 85 Teeth areJ1, 30.2725 andJ2,40.2975.

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    minft2991.13===m.30430

    ft1

    min1

    s60

    s

    m17.15sm17.151StageV

    Velocity (Dynamic) Factor,78

    78 VKv

    += (Highest precision, shaved and ground)

    1.3043=+

    =+

    ==78

    13.199278

    78

    78 1StageVKK vv

    From TABLE 15.1 (Juvinall, R. C. & Marshek, K. M., pp. 615), the Overload Correction

    Factor, Ko = 1 (Uniform source of machinery and power)

    From TABLE 15.2 (Juvinall, R. C. & Marshek, K. M., pp. 616), the Mounting Factor,

    Km = 1.3 (Accurate mounting, small bearing clearance, minimum deflection)

    MPa41.02=== )3.1)(1)(3043.1(Tooth)mm)(3mm)(0.272530(

    N93.285

    11

    1Stage,

    1 mov

    tKKK

    mJb

    F

    MPa37.57=== )3.1)(1)(3043.1(Tooth)mm)(3mm)(0.297530(

    N93.285

    22

    1Stage,

    2 mov

    tKKK

    mJb

    F

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    Stage 2 (Gear 3 & Gear 4)

    Pitch Line Velocity,60000

    dV=

    sm7.079=

    ===

    ===2Stage

    444

    33

    3

    sm079.760000

    rpm)mm)(500.89270(

    60000:4Gear

    sm079.760000

    rpm)3mm)(1073.3126(

    60000:3GearV

    dV

    d

    V

    Tangential Forces, Ft= /V

    N1271.37=

    ===

    ===

    2Stage,

    2Stage4

    2Stage3

    N37.1271sm7.079W9000:4Gear

    N37.1271sm7.079W9000:3Geart

    t

    tF

    VWF

    VWF

    Gear Tooth Bending Stress, movt KKK

    bJm

    F=

    From Figure 14-6 (Budynas, R. G. & Nisbett, J. K., pp. 753), the Geometry Factor for spur

    gears of 42 Teeth and 85 Teeth areJ1, 30.2725 andJ2,40.295.

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    minft23.23===m.30430

    ft1

    min1

    s60

    s

    m079.7sm079.72StageV

    Velocity (Dynamic) Factor,78

    78 VKv

    += (Highest precision, shaved and ground)

    1.0304=+

    =+

    ==78

    23.2378

    78

    78 2StageVKK vv

    From TABLE 15.1 (Juvinall, R. C. & Marshek, K. M., pp. 615), the Overload Correction

    Factor, Ko = 1 (Uniform source of machinery and power)

    From TABLE 15.2 (Juvinall, R. C. & Marshek, K. M., pp. 616), the Mounting Factor,

    Km = 1.3 (Accurate mounting, small bearing clearance, minimum deflection)

    MPa69.44=== )3.1)(1)(0304.1(Tooth)mm)(3mm)(0.272530(

    N37.1271

    33

    2Stage,

    3 mov

    tKKK

    mJb

    F

    MPa63.61=== )3.1)(1)(0304.1(Tooth)mm(3mm)(0.295)30(

    N34.1271

    44

    2Stage,

    4 mov

    tKKK

    mJb

    F

    By comparing the four bending stresses above, the maximum bending stress is max = 3 = 69.44

    MPa.

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    The gear in the gear train which is subjected to the highest tooth bending stresses is the one which

    has the largest possibility to fail in advance. Based on the stresses computed above, max = 269.19MPa and therefore the critical gear is Gear 3.

    Refined Analysis of Gear-Tooth Bending Strength Endurance Limit

    msKKKCCCSS trSGLnn =

    whereun SS 5.0=

    Since the Steel Spur Gears are made of S45C Steel with a maximum Brineel Hardness of 194 HB,

    therefore, the tensile strength, Su, can be determined by using the relation:

    Su = Kh(HB) = (3.45)(228) MPa = 786.6 MPa (Kh = 3.45 in SI Units)

    ( ) MPa3.393MPa86.675.05.0 === un SS

    Load Factor, CL = 1 (Bending loads)

    )5(1

    in.Teeth4667.8

    mm25.4

    in.1mm270

    Teeth58

    in.Teeth4667.8

    mm25.4in.1126mm

    Teeth42

    in.Teeth4667.8

    mm25.4

    in.1mm270

    Teeth09

    in.Teeth4667.8

    mm25.4

    in.1mm126

    Teeth42

    4

    44

    3

    33

    2

    22

    1

    11

    >=

    =

    ==

    =

    ==

    =

    ==

    =

    ==

    PC

    D

    NP

    D

    NP

    D

    NP

    D

    NP

    G

    From TABLE 15.3 (Juvinall, R. C. & Marshek, K. M., pp. 615), the Reliability Correction

    Factor, Kr= 1 (50% Reliability)

    Temperature Factor, Kt= 1 (Assumed the temperature lower than 160 F)

    Mean Stress Factor, Kms = 1 (Gear 3 is idler)

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    From FIGURE 8.13 (Juvinall, R. C. & Marshek, K. M., pp. 301), the Surface Factor,

    CS= 0.74 (Machined or cold-drawn)

    Therefore, the endurance limit is computed as follows:

    MPa291.042=

    =

    =

    1)74)(1)(1)(MPa)(1)(0.(393.3

    msKKKCCCSS trSGLnn

    Safety Factor, 4.19===MPa9.446

    MPa042.291SF

    max

    nS

    Since the safety factor is equal 4.19 (SF > 2.2), therefore, the pre-requisite is attained and Steel Spur

    Gears with module 3 and face width of 30 mm has been selected from the KHK Gear Catalog forthe double-reduction gear train.

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    Assignment 2 Group 9

    HES3350 Machine Design, Semester 1, 2012 Page 19 of21

    3.4 Analysis of Surface Durability of Tooth SurfaceThe critical gear (Gear 3) has been selected for the analysis of surface durability as it is subjected to

    maximum bending stress.

    Surface Fatigue Stress,mov

    p

    t

    p

    g

    g

    p

    p

    gp

    t

    HKKK

    Ibd

    FC

    E

    v

    E

    vb

    ddF

    =

    +

    +

    =22

    11cos

    sin

    2

    sin

    2

    564.0

    Gear 3: Ft3 = 1271.37 N, b = 30 mm = 0.03 m, dp = d3 = 126 mm = 0.126,R = dg/dp = 270/126 =

    2.14. From previous calculations, Kv = 1.0304, Ko = 1 and Km = 1.3.

    From TABLE 15.4b (Juvinall, R. C. & Marshek, K. M., pp. 621), the Surface Factor, the Values of

    Elastic Coefficient Cp for Spur Gears is Cp = 191 ( )Pa10191MPa 3= .

    Geometry Factor, 1024.0114.2

    14.2

    2

    20cos20sin

    12

    cossin=

    +

    =

    +=R

    RI

    Surface Fatigue Stress, MPa123.8===m)(0.1024)m)(0.126(0.03

    (1)(1.3)N)(1.0304)37.1271(mov

    p

    t

    HKKK

    IbdF

    Surface Fatigue Strength, SH= SfeCLiCR

    The angular velocity of gear 3, 3 = 1073.33 rpm and given that the expected life for 40 hr/week,and 50 week/year operation is 3.25 years. Therefore, the Surface Fatigue Life (Cycle) is given by:

    Cycle102.4hr1

    min60years25.3

    year

    week50

    week

    hr40

    min

    rev1073.33LifeFatigueSurface

    8=

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    Assignment 2 Group 9

    HES3350 Machine Design, Semester 1, 2012 Page 20 of21

    From FIGURE 15.27 (Juvinall, R. C. & Marshek, K. M., pp. 624), the Life Factor, CLi = 0.83 for a

    surface fatigue life of 4.2(108) Cycle.

    From TABLE 15.6 (Juvinall, R. C. & Marshek, K. M., pp. 625), the Reliability Factor, CR = 1.25

    for a reliability of 50%.

    From TABLE 15.5 (Juvinall, R. C. & Marshek, K. M., pp. 624), the relation for the computation of

    the Surface Hardness, Sfe, is Sfe = 2.8(Bhn) 64 MPa.

  • 7/30/2019 HES3350 Machine Design, Assignment 2: Gear Specification with Consideration of Bending Fatigue and Surface Du

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    Assignment 2 Group 9

    Since the maximum Brineel Hardness of the Steel Spur Gear is 228 HB or 228 Bhn, therefore, the

    surface hardness can be determined by using the relation above:

    Sfe = 2.8(228 Bhn) 64 MPa = 574.4 MPa

    Thus, the Surface Fatigue Strength SHcan be computed as follows:

    SH= (574.4 MPa)(0.83)(1.25) = 595.94 MPa

    Checking of Safety Factor for Surface Durability:

    4.8===MPa123.9

    MPa94.595SFFactor,Safety

    H

    HS

    Since the Safety Factor SF = 4.8 (> 2.2), therefore, the selection and utilization of Steel Spur Gear

    with module of 3 from the KHK Gear Catalog will not results in any operating failure. As a verdict,

    based on the analysis of bending fatigue and surface durability of gear tooth strength above and asin previous section, it is ascertain that the Steel Spur Gear with Module 3 from the KHK Gear

    Manufacturer can be employed in the compound-reverted gear train. 2 models have been selected

    from the KHK Gear Catalog due to the same size of pinions and gears.

    Gear 1 & Gear 3 (Pinion):

    42TeethofNumber3ModuleGearSpurS45C:Steel

    423SS=

    Gear 2 & Gear 4 (Gear):

    90TeethofNumber3ModuleGearSpurS45C:Steel

    903SS=

    The material used for the spur gears is Steel S45C with Brineel Hardness of 228 HB or 228 Bhn.