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Solutions Mechanical Engineering Design Shigley 7th Edition (1)

Apr 04, 2018

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  • 7/31/2019 Solutions Mechanical Engineering Design Shigley 7th Edition (1)

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    12-1 Given d m a x = 1.000 in and b m i n = 1.0015 i n , the mini mum radial clearance is

    Also

    Eq. (12-7):

    Fig. 12-16:

    Fig. 12-19:

    Fig. 12-18:

    Fig. 12-20:

    cmin

    b mi n d n 1 .0015 1 .000

    2 2

    l/d = 1

    r = 1.000/2 = 0.500

    r/c = 0.500/0.00075 = 667

    N = 1100/60 = 18.33 rev/s

    P = W/(ld) = 250 /[( 1)( 1)] = 250 psi

    8(10- 6

    )(18.33) '

    0.000 75 in

    S (6672

    )250

    h00/c

    Q/(rcNl)

    fr/c

    Qs / Q

    0. 261

    0.595

    3.98

    5.8

    0.5

    f 0.0087

    h0 = 0.595(0.00075) = 0.000466 in Ans.

    5.8 5.8- ^

    r/c 667

    The power loss in Btu/s is

    2:TfWrN 2n (0.0087)(250)(0.5)(18.33)H778(12)778(12)

    = 0.0134 Btu/s Ans.

    Q = 3.98rcNl = 3.98(0.5)(0.000 75)(18.33)(1) = 0.0274 in 3/s

    Qs = 0.5(0.0274) = 0.0137 in3

    /s Ans.

    _ _ _

    _ _

    _

    _

    _ _

    bmin - dmax 1.252 - 1.250

    c m i n = 2 = 2 = 0.001 in

    r =.

    1 .25/2 = 0.625 in

    r/c = 0.625/0.001 = 625

    N = 1150/60 = 19.167 rev/s

    400

    P = = 128 psi1 .25(2.5)

    l/d = 2.5/1.25 = 2

    (6252

    )(10)(10- 6

    )(19.167)S = = 0.585

    128

  • 7/31/2019 Solutions Mechanical Engineering Design Shigley 7th Edition (1)

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    313

    The interpolation formula of Eq. (12-16) w i l l have to be used. From Figs. 12-16, 12-21,

    and 12-19

    For //d = oo, ho/e = 0.96, PI p m a x = 0.84,

    //d = 1, ho/e = 0.77, P /pmax = 0.52,

    1

    l/d = , ho/e = 0.54, P /p m a x = 0.42,2

    1l/d = , ho/e = 0.31, P / p m a x = 0.28,

    Q

    reNl

    Q

    reNlQ

    reNl

    Q

    3.09

    3.6

    4.4

    5.254 rcNl

    Equation (12-16) is easily programmed by code or by using a spreadsheet. The results are:

    l/d yiy

    1 / 2 y 1 / 4

    ho/e 2 0.96 0.77 0.54 0.31 0.88

    pP/pPmax 2 0.84 0.52 0.42 0.28 0.64

    Q / reNl 2 3.09 3.60 4.40 5.25 3.28

    ho = 0.88(0.001) = 0.000 88 in Ans.

    128p

    max0.64

    200 psi Ans.

    Q = 3.28(0.625)(0.001)(19.167)(2.5) = 0.098 in3/s Ans.

    12-3

    e

    min

    bmin - dmax 3.005 - 3.000

    2 2

    r =. 3.000/2 = 1 .500 in

    l/d = 1 .5/3 = 0.5

    r/e = 1 .5/0.0025 = 600

    N = 600/60 = 10 rev/s

    800

    0.0025 in

    P1 . 5(3)

    Fig. 12-12: SAE 10, ' = 1.75 /zreyn

    S = (6002

    )

    177.78 psi

    1.75(10 - 6 )(10)

    177.780.0354

    Figs. 12-16 and 12-21: ho/e = 0.11, P /p m a x = 0.21

    ho = 0.11(0.0025) = 0.000 275 in Ans.

    pmax = 177.78/0.21 = 847 psi Ans.

    _ _

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    314 Instructor's Solution Manual to Accompany Mechanical Engineering Design

    Fig. 12-12: SA E 40, = 4.5 reyn

    ( 4.5 \) = 0.09101 .75ho/e = 0.19, P /p m a x = 0.275

    ho = 0.19(0.0025) = 0.000475 in Ans.

    pmax = 177.78/0.275 = 646 psi Ans.

    12-4bmin - dmax 3.006 - 3.000

    e m i n = 2 = 2 = 0.003

    r = . 3.000/2 = 1 . 5 in

    l/d = 1

    r/e = 1 .5/0.003 = 500

    N = 750/60 = 12.5rev/s

    600

    P = = 66.7 psi3(3)

    Fig. 12-14: SAE 10W, = 2.1 reyn

    '2 .1(10- 6

    ) (12.5)S = (500 2)

    66.70.0984

    From Figs. 12-16 and 12-21:

    ho/e = 0.34, P /pm a x

    = 0.395

    ho = 0.34(0.003) = 0.001 020 in Ans.

    66.7

    p m a x = = 169 psi Ans.0.395

    Fig. 12-14: SAE 20W-40, = 5.05 reyn

    ' 5.05(10- 6 ) (12.5)

    S = (500 2)66.7

    0.237

    From Figs. 12-16 and 12-21:

    ho/e = 0.57, P/pmax = 0.47

    ho = 0.57(0.003) = 0.001 71 in Ans.

    66.7

    pm a x = = 142 psi Ans.

    0.47

    _

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    315

    1 2 - 5

    bmin - dmax 2.0024 - 2c m i n = = = 0.0012 in

    2 2

    d 2r = = - = 1in , l/d = 1/2 = 0.50

    2 2

    r/c =1/0.0012 =833

    N =800/60 = 13.33 rev/s

    600P = = 300 psi

    2(1)

    Fig. 12-12: SAE 20, =3.75 reyn

    ' 3.75(10- 6

    )(13.3)S = (8332

    )300

    0.115

    From Figs. 12-16, 12-18 and 12-19:

    ho/c = 0.23, rf/c = 3.8, Q/(rcNl) =5.3

    ho = 0.23(0.0012) = 0.000 276 in Ans.

    3.8

    f = = 0.00456

    833

    The powerloss due to friction is

    2:TfWrN 2n (0.004 56)(600)( 1) (13.33)

    H = =778(12) 778(12)

    = 0.0245 Btu/s Ans.

    Q = 5.3rcNl

    = 5.3(1)(0.0012)(13.33)(1)

    = 0.0848 in3

    /s Ans.

    12-6_ bmin - dmax _ 25.04 - 25

    c min _

    2 2

    r ' d/2 = 25/2 = 12.5 mm,

    r/c = 12.5/0.02 = 625

    N = 1200/60 = 20 rev/s

    = 0.02 mm

    l/d = 1

    1250P =

    l o r = 50 MPa s, S

    252

    (6252

    )

    2 M Pa

    50(10- 3

    ) (20)2(10

    6

    )0.195

    From Figs. 12-16, 12-18 and 12-20:

    ho/c = 0.52, fr/c = 4.5, Qs/Q = 0.57

    ho = 0.52(0.02) = 0.0104 mm Ans.

    f=

    T =

    4. 5

    6250.0072

    fWr = 0.0072(1.25)(12.5) = 0.1125 N m

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    316 Instructor's Solution Manual to Accompany Mechanical Engineering Design

    The power loss due to friction is

    H = 2nTN = 2n(0.1125)(20) = 14.14 W Ans.

    Qs = 0.57 Q The side flow is 57% of Q Ans.

    12-7bmin - dmax 30.05 - 30.00

    c m i n = = = 0.025 mm

    d 30r = 15 mm

    2 2

    r 15- = = 600c 0.025

    1120N = = 18.67 rev/s

    60

    2750

    P = = 1.833 MPa30(50)

    2 ' 60(10- 3 )(18.67)

    S = (600 2)1 . 833(10

    6)0.22

    l 50= = 1.67d 30

    This //d requires use of the interpolation of Raimondi and Boyd, Eq. (12-16).

    From Fig. 12-16, the ho/c values are:

    y1/4 = 0.18, y i / 2 = 0.34, y i = 0.54, y ^ = 0.89

    hoSubstituting into Eq. (12-16), = 0.659

    c

    From Fig. 12-18, the fr/c values are:

    y 1 / 4 = 7.4, y 1 / 2 = 6.0, y 1 = 5.0, yoo = 4.0

    fr

    Substituting into Eq. (12-16), = 4.59c

    From Fig. 12-19, the Q/(rcNl) values are:

    y 1 / 4=

    5.65, y 1 / 2=

    5.05, y 1=

    4.05,=

    2.95Q

    Substituting into Eq. (12-16), = 3.605

    rcNl

    ho = 0.659(0.025) = 0.0165 mm Ans.

    f = 4.59/600 = 0.007 65 Ans.

    Q = 3.605(15)(0.025)(18.67)(50) = 1263 mm3

    /s Ans.

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    317

    1 2 - 8 bmin - dmax 75.10 - 75c m i n = = = 0.05 mm

    2 2

    l/d = 36/75 = 0.5 (close enough)

    r = d/2 = 75/2 = 37.5 mm

    r/c = 37.5/0.05 = 750

    N = 720/60 = 12 rev/s

    2000

    P = = 0.741 MPa75(36)

    Fig. 12-13: SAE 20, 1 = 18.5MPa s

    18.5(10- 3

    )(12)S = (750 2)

    0.741(106)0.169

    From Figures 12-16, 12-18 and 12-21:

    ho/c = 0.29, fr/c = 5.1, P /pmax = 0.315

    ho = 0.29(0.05) = 0.0145 mm Ans.

    f = 5.1/750 = 0.0068

    T = fWr = 0.0068(2)(37.5) = 0.51 N m

    The heat loss rate equals the rate ofworkon the film

    //loss = 2:rTN = 2n (0.51)(12) = 38.5W Ans.

    p m a x = 0.741/0.315 = 2.35 MPa Ans.

    Fig. 12-13: SAE 40, 1 = 37 MPa s

    S = 0.169(37)/18.5 = 0.338

    From Figures 12-16, 12-18 and 12-21:

    ho/c = 0.42, fr/c = 8.5, P/pmax = 0.38

    ho = 0.42(0.05) = 0.021 mm Ans.

    f = 8.5/750 = 0.0113

    T = fWr = 0.0113(2)(37.5) = 0.85 N m

    H l o s s = 2n TN = 2n (0.85)(12) = 64 W Ans.

    pmax = 0.741 /0.38 = 1 .95 MPa Ans.

    _

    bmin - dmax 50.05 - 50

    c m i n = = = 0.025 mm2 2

    r = d/2 = 50/2 = 25 mm

    r/c = 25/0.025 = 1000

    l/d = 25/50 = 0.5, N = 840/60 = 14 rev/s

    2000

    P = = 1.6MPa25(50)

  • 7/31/2019 Solutions Mechanical Engineering Design Shigley 7th Edition (1)

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    318 Instructor's Solution Manual to Acc omp an y Mechani cal Engineering Design

    Fig. 12-13: SAE 30, X = 34 MPa s

    2 r34(10- 3 ) (14)

    S = (1000 )1.6(10

    6

    )

    From Figures 12-16, 12-18, 12-19 and 12-20:

    0.2975

    ho/c = 0.40, fr/c = 7.8, Qs / Q = 0.74, Q/(rcNl) = 4.9

    ho = 0.40(0.025) = 0.010 mm Ans.

    f= 7.8/1000 = 0.0078

    T = fWr = 0.0078(2)(25) = 0.39 N m

    H = 2n TN = 2n (0.39)(14) = 34.3W Ans.

    Q = 4.9rcNl = 4.9(25 )(0.025 )(14)(2 5) = 1072 mm2

    /s

    Qs = 0.74(1072) = 793 mm3

    /s Ans.

    12-10 Consider the bearings as specified by

    minimum f:

    maximum W:

    and differing only in dand d'.

    Preliminaries:

    b+tb

    b

    - 0

    Fig. 12-16:

    minimum f:

    maximum W:

    Fig. 12-12:

    Eq. (12-7):

    Forminimum f:

    IJLN / P =

    l/d = 1

    P = 700/(1.252

    ) = 448 psi

    N = 3600/60 = 60 rev/s

    S = 0.08

    S = 0.20

    1.38(10- 6

    )reyn

    1.38(10- 6

    )(60/448) = 0.185(10- 6

    )

    r

    c

    S

    XN / P

    0.08658

    0 .185(10- 6

    )

    c = 0.625/658 = 0.000 950 = 0.001 in

    _

    r

  • 7/31/2019 Solutions Mechanical Engineering Design Shigley 7th Edition (1)

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    319

    I f this iS C m i n ,

    b d = 2(0.001) = 0.002 in

    The median clearance is

    , td + tb , td + tb

    c = c m i n + 2 = 0.001 + 2

    and the clearance range forthis bearing is

    . td + tbAc =

    2

    which is a function only of the tolerances.

    Formaximum W:

    r / 0.2

    - = J ^ = 1040

    c \ 0

    I f this is c m i n

    c y 0.185(106 )

    c = 0.625/1040 = 0.000 600 = 0.0005 in

    b d' = 2c m i n = 2(0.0005) = 0.001 in

    , td + tb n^nr- .t

    d +t

    b

    c = c m i n + 2 = 0.00