77-1 Spur Gear

The following data is given for a

The following data is given for a pair of spur gears withpair of spur gears with  

20 20

 full depth involute teeth.  full depth involute teeth. Number

Number of of teeth teeth on on pinion= pinion= 24,24, Number

Number of of teeth teeth on on gear= gear= 56,56, Speed

Speed of of pinion= pinion= 200 200 rpm,rpm, !odule=

!odule= "mm,"mm, #a$e

#a$e width= width= "0mm."0mm. %oth the gears are made o

%oth the gears are made of steel with an ultimate tensile strength of, 600 N&mmf steel with an ultimate tensile strength of, 600 N&mm22_{. 'sing the. 'sing the} velo$it( fa$tor to a$$ount for the d(nami$ load, $al$ulate

velo$it( fa$tor to a$$ount for the d(nami$ load, $al$ulate ) %eam strength*

) %eam strength* 2) +elo$it( fa$tor* and 2) +elo$it( fa$tor* and

") ated power that the gears

") ated power that the gears $an transmit without bending failure, if the fa$tor of safet( is .5.$an transmit without bending failure, if the fa$tor of safet( is .5.

Solution-ata ata given-φ  φ  = =   20 20

full depth involutes, full depth involutes,  Number of teeth on pinion /

 Number of teeth on pinion / p p) = 24,) = 24,  Number of teeth on gear /

 Number of teeth on gear /gg) = 56,) = 56,  Speed of pinion /n  Speed of pinion /n p p) = 200 rpm,) = 200 rpm,  !odule /m) = "mm,  !odule /m) = "mm,  #a$e width /b) = "0  #a$e width /b) = "0 mm,mm, Servi$e fa$tor /1 Servi$e fa$tor /1ss) = .5,) = .5,

'ltimate tensile strength /S

'ltimate tensile strength /Sutut) = 600) = 600

2 2 mm mm  N   N  .. s both gear and

s both gear and pinion are made of same mpinion are made of same material, then pinion is a vehiaterial, then pinion is a vehi$le member. $le member. So theSo the design is based on pinion.

design is based on pinion.

3et,

3et, modulemodule

 g   g   g   g   p  p  p  p  z   z  d  d   z   z  d  d  m m))

==

==

//   So,   So,  p  p  p  p mmz z  d  d 

==

.. = " = "

××

24 = 2mm24 = 2mm  g   g   g   g  mmz z  d  d 

==

.. = " = "

××

56 = 6mm56 = 6mm /)

/) %eam s%eam stretrengtngth /Sh /S b b)- )-Y  Y  b b m m S  S _bb

==

σ  σ  _bb

××

××

××

.. ..

( ( ))

 A A

2 200 " 600 " mm  N  S _ut  b

=

=

=

σ   s  20

full depth involute s(stem,

∴

 

 

 

 





 

 

−

=

 p  z  Y  π   0.54 0.72

 

 

 



 

 

₋

=

24 72 . 0 54 . 0 π  

(

0.54

−

0.6"

)

=

π   6 . 0

×

=

π  

∴

Y 

=

0."64

So, beam strength

Y  b m S _b

=

σ  _b

×

×

×

"64 . 0 "0 " 200

×

×

×

=

=

b S  640 N /2) +elo$it( fa$tor / v C 

)-8it$h line velo$it(

( )

000 60

×

=

d  pnp v π   000 60 200 2

×

×

×

=

π   v

=

4.524ms

s the pit$h line velo$it( /v) 9 0  s m

,

So, +elo$it( fa$tor, v C _v

+

=

" "

524 . 4 " "

+

=

v C 

₌

 0."7 /") 8ower /8)-%eam strength,  s eff   b  p f   S 

=

.

So, :ffe$tive load,

5 .  640 ) /

=

=

 s b eff    f   S   p = 4"20 N 3et, 8ower 60 000 2 ) /

×

=

n T   P  p π   

( )

 B :ffe$tive load t  v  s eff   p C  C   p )

=

/   So, eff    s v t  p C  C   p

=

5 .  4"20 "7 . 0

×

=

Tangential load, t   p

₌

_{4.26 N}  ;ere,  p t  d  T   p

=

2 Therefore, 2  p t d   p T 

 =

2 2 26 . 4

×

=

mm  N  T 

=

4"".26 .

000 60 2 ) /

×

=

n T   P  π   p 000 60 26 . 4"" 200 2

×

×

×

×

=

π   W   P 

=

547.5  KW   P 

=

5.7 :!83: 25>4

 pair of spur gear is to be designed for a ma$hine operating at 240rpm, driven b( an ?@, 200rpm ele$tri$ motor. 1entre distan$e between the aAis of the gear shaft ma( be approAimatel( "00mm. Starting tor<ue of the motor ma( be taBen as 50C of the rated tor<ue. %oth pinion and gear are made of steel with teeth of 20

0

 pressure angle, full>depth in volute. Spe$if( surfa$e hardness and hardness of gears. To a$$ount for d(nami$ load, taBe velo$it( fa$tor- (= 0.54 >

0.912

 z , where  is the number of teeth on gear.

SD3'TEDN-!a$hine rpm= 240

:le$tri$ motor rpm= 200 Speed ratio= 5

Sa(, d p _{= pit$h $ir$le dia. Df pinion}

Then, pit$h $ir$le dia. Df gear = 5 d p ₌ dg

1enter distan$e= "00mm= d _p+_dg 2 Dr, 600= d p _F dg _{= d} p _{F5 d} p Dr, d p _=00mm d_{g  =500mm}

Speed, += 2_{π ×}1200 60 ×  0.1 2  = 6.2" m&s +elo$it( fa$tor, 1v = 3 3+_{v  =} 3 3 + 6.283  = 0."2" Servi$e fa$tor, 1s = startingtorque rated torque  = .5

8t, tangential tooth load =

 KW ×1000 v =

8000

6.283  = 2".2 N

:ffe$tive load,  peff   _{= 8t}

Cs

Cv = 2".2G

1.5

0.323  = 57" N

%eam strength, Sb _{= mb σ b H}

#a$e width, b = 0m @here, m is module 3ewis form fa$tor H= I(

= I

(

0.154− 0.912  z

)

#a$tor of safet( = 2 Sb _{=  p}eff  _{G#DS = 57"G2 = 26 N} σ b= σ _ut  3  = 600 3  = 200!pa 26 = bmH σ b _{= 0mG200GmGH} m2 _(=5.7"

!odule of 4mm $an be safel( $hosen for pinion and gear. d _{p = 4G25=00mm}

;N:SS TaBing Sw ₌ Sb _{= 26N} b=0m= 40mm d _{p =00mm} atio fa$tor, J = 2 _z g  z_g+ _zb = 2_×125 125_×25− 250 150 = .66 S_{b = b d p J?}  26= 40G00G.66G? 

3oad stress fa$tor, ?=

11826 4000_×1.667 = ." = 0.6

(

BHN 100

 )

2 %;N = """ Surfa$e hardness is """ %;N

 pair of straight teeth spur gears, having 20K involutes full depth teeth is to transmit 2 Bw at "00r.p.m of the pinion. The speed ratio is "-. The allowable stati$ stresses for gear of $ast iron and pinion of steel are ln❑  60!8a and 05!8a respe$tivel(. ssume the following,

 Number of teeth on pinion =6 #a$e width =4times module +elo$it( fa$tor/ cv¿ ₌

4.5

4.5+_v ,  v being the pit$h the line velo$it( in m&s-and tooth form

fa$tor H=0.54

−0.912

etermine the module fa$e width and pit$h diameter of gears, $he$B the gears for wear for  wear, given σ =600 "#a $ϵ  p=200%N /mm

2

∧ϵ 

g=100%N /mm

2

,sBet$h the gears.

Solution-Liven data- ∅=20& σ '(

=

60 N 

/

mm 2   8=2B@=2 × 10 3 W  _σ '#=105 _N /_mm2  N  _p=300_{r  p  m )}   #=16   +.= ) ₍ )  p =3 _b=14_m  * _p=200_×103 _N /_mm2 _σ CS=600 _N /_mm2  *g=100×10 3  N /_mm2 !odule

@e Bnow that pit$h line velo$it(, + =π  p N  #

60 =

πm×16_×300

60 =251mm/s=0.251m/s

 @e taBe from the table C S=1

W ₎ = # + ×C S = 12_×103 0.251_m ×1 = 47.8_×103

m N  and velo$it( fa$tor 

+=

4.5 4.5+_v=

4.5 4.5+0.251_m

 - _p=0.154−0.912 ) _g =0.154− 0.912 16 =0.097  -_g=0.154−0.912 ) ₍ =0.154− 0.912 3_×16=0.135 σ _'#× - _p=105_×0.097=10.085 nd σ C(× . (=60×0.135=8.1

Sin$e

(

σ C(×. (

)

 _{is less than / σ C(}× . ₍¿ _{.there for the gear is weaBer, now using 3ewis} e<uation to the gear, we have

w₎ =_σ  W( b  π  m  -( 47.8_×103 m =60

(

4.5 4.5+0.251_m

)

1.4m×πm×0.135 = 1603_m2 4.5+0.251_m 4.5F0.25m=0.0""5 m3

Solving this e<uation b( hit and trial method @e find that

m=5.6 sa( 6mm. #a$e width

@e Bnow that fa$e width. b =4m=4 ×6=84mm /ns  8it$h diameter of gear.

iameter of the pinion   _p=_{m )}   p=6×16=96mm /ns iameter of gear   ₍=_{m )}  (=6×16mm=288mm /ns.. atio fa$tor,

J=

2_{× +  0}

+  0+1 = 2_×3

3+1 =1.5

nd load stress fa$tor,

?= σ _m2sin_∅ 1.4

[

1  *  #

+

1  * (

]

= 6002sin 20 1.4

[

1 206_×103 + 1 100_×103

]

  =0.44F0.   =."2  N /mm 2

@e Bnow that the maAimum limiting load for wear, w_w= _

 p b  q  %  =96×84×1.5×1.32

  =576N

nd tangential load on the tooth w₎ =47.8×10 3 m = 47.8_×103 6 =76N ns...

 re$ipro$ating $ompressor is to be $onne$ted to an ele$tri$ motor with the he lp of spur gears. The distan$e between the shafts is to be 500mm. the speed of the ele$tri$ motor is 700 r.p.m. and the speed of the $ompressor shaft is desired to be 200 r.p.m. the tor<ue, to be transmitted is 5000  N>m. taBing starting tor<ue as 25C more than the normal tor<ue, determine - . !odule and fa$e

width of the gears using 20 degrees stub teeth, and 2. Number of teeth and pit$h $ir$le diameter of ea$h gear. ssume suitable values of velo$it( fa$tor and 3ewis fa$tor.

Solution = Liven data-3=500 mm

 N  _{"   = 700 r.p.m.}  N _{C   = 200 r.p.m.}

T = 5000 N>m ) _{ma1  = .25T}

. !odule and fa$e width of the gears m = module in mm, and

 b = fa$e width in mm.

sin$e the starting tor<ue is 25C more than normal tor<ue, therefore the maAimum tor<ue, ) _{ma1  = .25T = .25 G 5000 = 6250 N>m = 6250 G} 103  _N>mm

@e Bnow that velo$it( ratio,

+.. =

 N  _"   N _C   =

900

200  = 4.5

  3et,   #  _{= 8it$h $ir$le diameter of the pinion on the motor shaft, and}

 _{(  = 8it$h $ir$le diameter of the gear on the $ompressor shaft.} @e Bnow that distan$e between the shaft /3),

500 =   _# 2  F  ₍ 2  or  ,  #

+

 ,(  _{= 500 G 2 = 000} ./i) nd velo$it( ratio, +.. =  ,₍  , _#  = 4.5 or  ( = 4.5   #   .. /ii)

Substituting the value of  ( _{in e<uation /i), we have}

 , _#+4.5 _,

 #  _{= 000 or  } #  _{= 000&5.5 = 2 mm}

  nd  (  _{= 4.5  } #  _{= 4.5 G 2 = 20 mm = 0.2 m}

@e Bnow that pit$h line velo$it( of the drive,

v = π ,₍ N _C  60  = π ×0.82_×200 60  = .6 m&s velo$it( fa$tor,

C _{v  =} 3

3+_{v  =}

3

3+8.6  = 0.26

3et us taBe the allowable tensile stress for the gear material as, σ _{'(  = 40 !8a = 40 N& mm}2

@e Bnow that for 200  stub teeth, lewis fa$tor for the gear,

 -_{(  = 0.5 M} 0.841 ) ₍  = 0.5 > 0.841_{× m}  ₍ / ) ₍= ( m ) = 0.5 > 0.841_{× m} 820  = 0.5 M 0.00 m

nd maAimum tangential for$e on the gear, W _{)   =} 2) ma1

 ₍  =

2_×6250_×102

820  = 5.244 N

@e also Bnow that maAimum tangential for$e on the gear,

W _{)   =} σ _{w(  b I m  -(  = / σ '(  G} C _{v )b G I m G}  -₍ 5244 = /40 G 0.26) G 0m G I m /0.5 M 0.00m) = 200 m 2  > .44 m 3 /assuming b = 0m)

Solving this e<uation b( hit and trial method, we find that m = .75 sa( 0 mm ns.

 b = 0m = 0 G 0 = 00 mm ns.

2. Number of teeth and pit$h $ir$le diameter of ea$h gear  @e Bnow that number of teeth on the pinion,

) _{v  =}   # m  = 182 10  = .2 ) _{( =}   # m  = 820 10  = 2

En order to have the eAa$t velo$it( ratio of 4.5, we shall taBe )  _{#  =  and} ) _{(  =  ns.}

8it$h $ir$le diameter of the pinion,  _{v  = m G} ) 

 #  = 0 G  = 0 mm ns. nd pit$h $ir$le diameter of the gear,