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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