problema 7-2
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Transcript of problema 7-2
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7/26/2019 problema 7-2
1/15
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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7/26/2019 problema 7-2
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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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7/26/2019 problema 7-2
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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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7/26/2019 problema 7-2
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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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7/26/2019 problema 7-2
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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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7/26/2019 problema 7-2
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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*