in a spur gear drive for a stone crusher
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8/12/2019 In a Spur Gear Drive for a Stone Crusher
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In a spur gear drive for a stone crusher, the gears are made of C40 steel. The pinion
is transmitting 30kW at 1200 r.p.m. The gear ratio is 3. ear is to !ork " hours per
da#, si$ da#s a !eek and for 3 #ears. %esign the drive.
P = 30 kW, N1= 1200 rpm, i = 3
Since the pinion and gear are made of same material, therefore we have to do the design
of pinion alone.
&aterial selection Pinion and gear are made of ! "0 Steel,
#ss$me s$rface hardness % 3&0
'ear life = ( ) &2 ) * ) 3 = +"(( ho$rs
ife in n$m-ers of ccles, N = *0 ) +"(( )1200 = &3./ ) 10+ ccles
!alc$lation of initial design tor$e t
4esign tor$e t = t. 5. 5 d = 1.3 ) 23(.+3 = 310.3" Nm
Calculation of 'e(, )*+, )*c-
6e = 2.1& ) 10& N7mm2
n = 2 5 8 = 1.& , N= &3./) 10+ccles9 5 -l=0.+, 8$= *30 N7mm2
2
1 7&.**21203&.0 mm N
u =+=
− σ σ
1".1
−= σ σ
σ nK K bl
b = 111.23 N7mm2
!: = 2*.& ;:! = && 5 cl = 0.&(&
[ ] Kcl HRC R
C c =σ = (&2.*" N7mm2
Center distance #ss$me < =0.3
[ ]
[ ]3
2
+".0=1>
+≥
ψ σ i
t M
eq
E
c
ia = 1&& mm
?1 = 1+, ?2 = i.?1 = 3 ) 1+ = &1
21
2
z z
a
m
+= = ".&* mm standard m = & mm
evised Center distance a = 1+0 mm
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Calculation of +, d1, v and /p-
- = <.a = 0.3 ) 1+0 = &1 mm
d1= m.?1= & ) 1+ = (& mm
Picth line velocit >v =21
11
z z
N d
+
π
= &.3" m7s
< p = -7d1 = 0.* 5 =1.03 5 d = 1." evised %esign tor(ue )&t 344.24 m
Check for ending-
abmy
t M i
b
=1> +=σ = (*.+( N7mm2 @ 8 -
Check for !ear strength-
ib
i
t M eq E
a
i
c
1
=1>+".0
++=σ = +*&./ N7mm2 @ 8c
helical gear !ith 30o heli$ angle has to transmit 3 kW at 100 rpm !ith a speed
reduction ratio of 2.. If the pinion has 24 teeth, %esign the gear for 20 o full depth
teeth and service is continuous. ssume 40i2Cr1&o2" material for +oth pinion
and !heel.Selection of material properties
8$= 1&&0 N7mm2 S$rface hardness && :! and core hardness%3&0A;N
!alc$lation of design stresses
n = 2.& 5 8 = 1.& N= 10000 ) *0 ) 1&00 =/0 ) 10+ccles9 5 -l=0.+
1
".1
3−
= σ
σ
σ
nK
bl K
b
= 1+3.1N7mm2
2
7&.**21203&.01 mm N u =+=− σ σ
!: = 2*.& ;:! = && 5 cl = 0.&(&
[ ]cl
K HRC R
C c =σ =(&2.*N7mm2
t = 2(/.** N.m 6 = 2.1)10& i = 2.& < =0.&
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• !alc$lation of a)ial mod$le
m)= 2a7>E? = (.01mm Standard m)=10mm
:evise a and o-tain d1
a = 0.& m)>E? = 32& mm
Pitch circle diameter of worm = d1= m) = 110 mm
Ds=D17cosF tanF= C7 = % lead angle F = 1&.2&o
D1 = Pitch line velocit of the worm
=
×1000*0
11nd π
= (.3 m7s
:evised 8c = 12+ N7mm2
%etermination of induced stresses
10
31=7>
7
&"0
+=
t M
a
q z
q z cσ
= 120 N7mm2@ 8c
v x
t
bqzym
M 3
/.1=σ = 2.33 N7mm2 @ 8 -
=tan>
tan/&.0
ρ ν
ν η
+= = 0.("/
6eat emoval nal#sis
>1GH ) Inp$t Power = 22*0 Nm
%esign a 12 speed gear +o$ for an all geared headstock of a lathe. &a$imum and
minimum speeds are 700 rpm and 2 rpm respectivel#. The drive is from an electric
motor giving 2.2 kW at 1440 rpm.
Progression ratio1
1
min
ma)
−
=
N
n
nϕ = 1.33& Where N B No of Speeds in gear -o)
Jrom :"0 series Speed >Skip " speeds B 2&,33.&, "&, *0, (0, 10*, 1"0, 1/0, 2&0, 33&, "&0 and*00 rpm
Str$ct$ral form$la for 12 speeds
= 3>1 2>3 2>*
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machine tool gear +o$ po!ered +# 11 kW, 890 rpm motor is to give 14 spindle
speeds ranging from 20 to 400 rpm. %ra! the kinematic arrangement and the speed
chart.
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