problema 7-2

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  • 7/26/2019 problema 7-2

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    Ecuacin de ED-

    Goodman

    d= {16n/ [1/Se{4(KfMa)2+3(KfsTa)2}1/2

    +1/Sut{4(KfMm)2+ 3(KfsTm)2}1/2]}1/3

    ec. !" (#a$. 3%6)

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    Donde:

    Se= lmite de resistencia a la fatigan= factor de diseo de la fatiga

    Kf= factor de concentracin del esfuerzo por fatiga

    de la exin

    Ma= Momento exionante alternanteTa= Par de torsin alternante

    Kfs= factor de concentracin del esfuerzo porfatiga

    de la exinSut= esistencia a la tensin

    Mm= Momento exionante medio

    Tm= Par de torsin medio

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    &n e' cas de un ee $*at* cn f'e,*-n ts*-n cnstantes Ta=

    Mm= 0

    Ec. 7-8 (p. 356):

    d= {16n/ [1/Se{4(KfMa)2+3(Kfs0)

    2}1/2+1/Sut{4(Kf0)2+

    3(KfsTm)2}1/2]}1/3

    d= { 16n/ [ 1/Se{ 4(KfMa)2}1/2+ 1/Sut{3(Kfs Tm)

    2}1/2]}1/3

    d= { 16n/ [ 1/Se{ 41/2(KfMa)2/2}+ 1/Sut{ 31/2(Kfs Tm)2/2}]}1/3

    d={ 16n/ [ 2(KfMa)/Se+31/2(KfsTm)/Sut]}

    1/3

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    Datos:

    Se= 27 kpsi = 27.000 psi Kt= Kf = 1,7

    T= 32!0 "#. p$"% Kts= Kfs= 1,5

    &a= 3651 "#. p$"%

    '= 1,5

    S$t= 68 kpsi = 68.000 psi

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    eem#'a 's dats

    d={ 16(1.%)/ [ 2(1.,36%1'. #u'$)/2000 './#u'$2+31/2(1.%,3240 '. #u'$)/6"000 '. /#u'$2]}1/3

    d= {.6354 [0.4%5% #u'$3+ 0.1235 #u'$3]}1/3

    d= {4.4%"5 #u'$3}1/3

    d= 1.64% =1.6% #u'$

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    m n=1.% se ca'cu'a un nue7 d de*d a 8ue se

    taaa cn est*mac*nes

    16% 1009

    , 1.%

    ,= 1.6%,1.% = 0024%= 002%

    100

    d= 1.6%!0.02%= 162% (nue7 d)

    :aa ca'cu'a e' d*;met m;s $ande

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    :aa ca'cu'a Kt (/d=0.1

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    Kt= 16

    :aa ca'cu'a 8 (= 0.16 Sut= 6" >#s* 'a cu7a m;s cecana Sut=100)

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    8= 0."2

    ;'cu' de Kf

    Kf= 1 + 8(Kt? 1)

    Kf= 1 + 0."2(1.6!1) = 1.452

    ;'cu' de Kfs (detem*n Kts en e' $;f*c)

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    Kts= 1.3%

    ;'cu' de 8s (#aa = 0.16)

    8s= 0.5%

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    Kfs= 1 + 8s(Kts! 1)

    Kfs= 1+ 0.5%(1.3%!1) = 1.33

    Ka= 0.""3

    :aa ca'cu'a Kus e' $;f*c 6!20)

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    K= (d/0.3)!0.10

    K= (1.62%/0.3)!0.10= 0."3%

    Se= KaKKc KdKeKfSe@

    m Kc=Kd= Ke=1

    Se= KaKKfSe@= 0.""3 , 0."3% , 0% , 6"

    Se= 2%.1 >#s*

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    ;'cu' de Aa@

    Aa@ = {(32KfMa/d

    3)2+ 3(16KfsTa/ d3)2}1/2

    m Ta= 0 8ueda

    Aa@= {(32KfMa/d

    3)2}1/2

    A

    a

    @

    = 32KfMa/d3

    Aa@= 32 , 1.45 , 3641/ (1.62%)3= 12510 #s*

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    a'cu' de Am@

    Am

    @= {(32KfMm/d3)2+ 3(16KfsTm/ d

    3)2}1/2

    m Mm= 0 8ueda

    Am

    @= [3(16KfsTm/d3)2]1/2

    Am@= 31/2, 16, 133, 3240 /(1.62%)3= ""%5 #s*