engranajes #5
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Transcript of engranajes #5
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7/26/2019 engranajes #5
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EngranajeEjercicio # 5 luis Buelvas , Omar Vargas
5. A gear drive proposal consists of a 20 pressure angle spur pinion it!
" teet! driving a $$ toot! gear. %!e module of t!e gears is & and t!e faceidt! is $5mm.%!e pinion runs at '$5 rpm. %!e gear set is to transmit "(.5
)* and t!e load and drive can +e considered smoot!. %!e gear material
proposed is grade " steel and t!e Brinell !ardness of t!e pinion and gear is
'00 and 200, respectivel. %!e gears are to +e manufactured to num+er &
A-A /ualit standard and t!e teet! are uncroned. %!e gears are
straddle mounted it!in an enclosed unit. Assume a pinion life of "0
ccles and a relia+ilit of 0 per cent. 1alculate t!e A-A +ending and
contact stresses and t!e corresponding factors of safet for +ot! t!e pinion
and t!e gear.
Solucin.
Dimetro pin = Dp= 19*6=114mm
Dimetro engranaje = De=77*6=462mm
Relacin e !elocia = R!dientes engranaje
dientes pion =77
19=4.052
"actor e #o$recarga=k0=1
%=&*r=37520.114
602=2.238
m
s
Wt=potencia
v =14500
2.238 =6477.885N
B=(12Q)2 /3
4=(126)2 /3
4=0.8255
A=50+56 (10.8255 )=59.77
"actor inmico
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KV=( A+200VA )B
=(59.77+2002.23859.77 )0.8255
=1.284
"actor e tamao
Ks=1
Diente# #in corona
'mc=1
"=7(mm=).)7(m0.0254m
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"=2.466(1108)0.056=0.879
"actor re#i#tencia a la# picaura# en el engranaje
Ye=2.466(110
8
)!v
0.056
=2.466(110
8
)4.052
0.056
=0.9506
- "actor e con+ia$ilia al 9)/0=).(- "actor e temperatura 0o=1- "actor e #uper+icie 3r=1
'oe+iciente e ela#ticia el pin/ 191(#a)0.5
E#+uero e +le,in permi#i$le para un acero grao 1
En el pin
$fp=0.533%p+88.3=0.533300+88.3=248#a
En el engranaje
$fe=0.533%e+88.3=0.533200+88.3=195#a
- E#+uero e contacto
En el pin
$cp=2.22%p+200=2.22300+200=866#a
En el engranaje
$ce=2.22%e+200=2.22200+200=644#a
A&=8.98103
%p
%e8.29103=8.98103
300
2008.29103=5.18103
"actor e urea
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"'=1+A&(!v1 )=1+5.18103 (4.0521 )=1.015
or la +igura 7.2
0jp=).56 0je=).44
"actor geomtrico
"1=
cossin 2mn
!v
!v+1=
cos20sin 202
4.052
4.052+1=0.1288
E#+uero e +le,in 8.
$=
WtKoKvKs1Fm
KhK(
Yj
ara el pin.
$fp=
6477.88511.284110.0750.006
1.2381
0.36=63.57106a
ara el engranaje.
$fe=
6477.88511.284110.0750.006
1.2381
0.44=52.01106a
E#+uero +le,in permi#i$le.- a#umimo# S+=1
ara el pin
$f )adm=$fYn
*fYoY+=
248.20.9284110.85
=271.09#a
"actor e #eguria
*fp=271.09
63.57=4.26
ara el engranaje
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$f )adm= $fYn
*fYoY+=
1950.9712110.85
=222.8#a
"actor e #eguria
*fe=222.8
52.01=4.28
E#+uero e contacto
$c="c(WtKoKvKs1
Fd Kh"r
"t )0.5
ara el pin
$cp=1916477.88511.28411238
0.1140.0750.1288=1912964.76=584.05#a
ara el engranaje
$ce=1916477.88511.28411238
0.4620.0750.1288=290.12#a
E#+uero e contacto ami#i$le/ #uponemo# S:=1
ara el pin
$cp) adm=$h"n
*hYoY+=
8660.879110.85
=895.54#a
"actor e #eguria
*cp=895.54
584.05
=1.53
ara el engranaje
$ce)adm=$h"n
*hYoY+=
6440.9506110.85
=720.21#a
"actor e #eguria
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*ce=720.21
290.12=2.48
8eometr;a e la re#i#tencia a la +le,in.- +igura 7.2