7/21/2019 Peeling Processkkm
1/41
ANALYSIS OF PEELING
7/21/2019 Peeling Processkkm
2/41
CUTTING ANALYSIS
Most of the time cutting force acting on a tool is measured experimentally. But it is alsoimportant to predict quantity of cutting force and how different cutting parameters are affectingcutting force even before setting up the machining operation due to following reassert.
In order to design of mechanical structure of cutting machine which will withstandcutting force and thrust force effectively.
To determine power consumption during machining process. This will belt in selectingsuitable motor drive.
To increase productivity.
t 1 = undeformed chipthickness
useallyt 2 = deformedchipthickness t
3 >t
= tool ralceangle
Next we can begin to consider cutting force, chip thic ness.
!irst, consider the physical geometry of cutting.
= 40
t 1 = 0.5 mm/0.0005 m
7/21/2019 Peeling Processkkm
3/41
t 2 = 0.7 mm/0.0007 m
That need to measure the two perpendicular cutting forces are hori"ontal and perpendicular#
$ !or wood on metal % &.'(&.) % &.*
F N
= tan =
tan = 0.4
= 21.8
7/21/2019 Peeling Processkkm
4/41
F N
= tan =
tan = 0.4
= tan 1 0.4 = 21.8
r e= thecuttingratio
r e=t 1t 2
r e=0.0005
0.0007
r e= 0.7143
t 1 = h sin
t 2 = h sin ( )
r e= t 1t 2 = hsin
h sin ( )= sin cos cos +sin sin
r e cos cos +sin sin = sin
r e cos cos sin
+r e sin sin
sin = 1
7/21/2019 Peeling Processkkm
5/41
r e cos sin
= 1 r e sin
tan = r e cos 1 r e sin
tan = 0.7143cos401 0.7143sin 40
= 45.33
F t cos ( )
= F s
cos (+ )
F s= As
For shear force
s= F s A
,
h= t 1
sin
h= 0.0005sin 45.33
7.03 10 4
A= 7.03 10 4 (1 )
= F s A
= 3.64 Mpareference
3.61 10 3
3.64 Mpa = F s
7.03 10 4
7/21/2019 Peeling Processkkm
6/41
F s= 2559.17 [ N ]
F c=2559.17 cos (40 21.8 )
cos (45.33 40 +21.8 )
F c= 2731.7 [ N ]
= F ! r
= 2731.7 0.05
= 136.585 [ Nm]
+eciding that the cutting blade should have && rpm
2 " N
60
2 " (300 )
60
31.42
-ower % torque x angular velocity
# = $
#= 31.42 (136.585 )
7/21/2019 Peeling Processkkm
7/41
k% #= 4.289 [k% ] let say 4.5
7/21/2019 Peeling Processkkm
8/41
&adiusof the drum'lade , r d= 0.05 [m] .
hedistance themiddle theouter diameter of the drum = 0.01 [m] .
Assuming thediameter of the sugarcane 'efore peeling = 0.024 [m] .
(o the radius of the sugarcane , r s= 0.012 [m] .
)ength A* = 0.05 +0.012 = 0.062 [m] .
)engthA! = 0.06 [m]+
7/21/2019 Peeling Processkkm
9/41
cos = 0.660.062
= 15.6
F !- = F ! cos75 = 2731.7cos75 = 707.02 [ N ]
F !. = F ! sin75 = 2731.7sin 75 = 2638.62 [ N ]
For Vertical
7/21/2019 Peeling Processkkm
10/41
$ = 2638.621.16
= 2274.67 Neglet the mass of the drum
The cutting force developed into a uniform cutting force along the shaft in order to peel thewhole length of sugarcane.
GEAR SELECTION METHOD
FBD
Contact point of gear 20
&! = radiusof pitch
#= $
7/21/2019 Peeling Processkkm
11/41
4.289 [k% ]=(31.4 )
= 136.59 [ Nm]
F t = &c =
136.590.05
F t = 2.7318 [kN ]/ertical
$ cos = F t
$ = F t cos
= 2.7318 [kN ]cos20
$ = 2907.12 [ N ]
F &= 0 cos
F &= 2907.12sin 20
F &= 994.29 [ N ]hori1ontal
Simple ear !electio" proce#$re
7/21/2019 Peeling Processkkm
12/41
Select a number of teeth for the pinion and the gear to give the require gearratio. Use either the standard teeth numbers as listed in table 6.5 or as listedin stock gear catalogues.
Select a materials. ( his !ill limited to those listed in the catalog" Select a module# M
Calculate the pitch diameter# d= Mn
Calculate the pitch line velocit$# / =d2
N 2 " 60
Calculate d$namic factor# 2 / =( 66 +/ ) Calculate the transmitted load# % t =
po0er/
Calculate an acceptable face !idth using the %e!is formula in the form#
F = % t 2 / m3 4 p
7/21/2019 Peeling Processkkm
13/41
HORI%ONTAL
F ! cos20 = 12256.74 cos20
N
F !- = 11517.57 ]
VERTICAL
F ! cos20 = 12256.74 sin 20
F !. = 4192.05 [ N ]
elect a module % '
elect a materials, based on catalogues
+ecide the pitch diameter % &./m0/&cm0/&&mm
1alculate the number of teeth d= Mn
100= (2)n
n= 50 teeth
1alculate the pitch line velocity,
/ = d2
N 2 " 60
/ = " (0.1 ) (300 )60
7/21/2019 Peeling Processkkm
14/41
/ = 1.57 m /s
+ynamic factor ,
2 / =( 66 +1.57 ) 2 / = 0.7926
!alculatethe transmitted load , F t = 1385.6
% t = #.
% t =42891.57
% t = 2731.89 N
1alculate an acceptable face width using the 2ewis formula in the form of,
F =
0 t 2 / m 3 4 p
F = 2731.84(0.7926 )(0.002 )(0.39860 )(345 Mpa )
F = 12.53 mm
o its o to use the catalogues based gear based on table, the suitable gear is SG&'() .
Si"ce *ot+ p$lle, ! o" t+e !ame po!itio" a"# !ame !i-e. a" le o/ 0rap 1 2 1 3
7/21/2019 Peeling Processkkm
15/41
2.3log [ 1 2 ]= 52.3log
[ 1
2 ]= 0.42 (" )
1 = 3.75 2 +(1 )
= ( 1 2 )r
136.59 =( 1 2)0.05
7/21/2019 Peeling Processkkm
16/41
1 2 = 2731.8 (2 )
Sub equation (2) into equation (1)
3.75 2= 2731.8 [ N ]
2.75 2= 2731.8 [ N ]
2 = 993.38 N
1 = 3.75 (993.38 )= 3.72 kN
4 = 1 A
16 Mpa= 3.72 kN A
A= 0.0002325 m2 (minimum )
Selecte# *elt !i-e
%ide = 29 mm let say 30 mm
7/21/2019 Peeling Processkkm
17/41
hickness = 8 mm
MAIN PEELING SHAFT
VERTICAL
6fMat* = 0 7
4739.66 (1.205 ) & A (1.16 )+2638.6 (0.58 ) 2731.8 (0.03 )= 0
& A. = 6172.2 N
6fy= 0 7
4739.6 +6172.2 2638.6 + &* 2731.8 = 0
&*. = 3937.86
7/21/2019 Peeling Processkkm
18/41
HORI%ONTAL
6fMat* = 0 7
& A- (1.16 ) 707.02 (0.58 )+994.29 (0.03 )= 0
& A- = 327.8 N
6fy= 0 7
327.8 707.02 + &*- 994.29 = 0
7/21/2019 Peeling Processkkm
19/41
&*- = 1373.51 N
T+e re!$lta"t *e"#i" mome"t at 4B5
M M
( *. )2+(*- )2
M *=
M *= ( 81.9 )2 +(29.84 )2
M *= 87.17 [ N +m]
T+e re!$lta"t mome"t at 4C5
7/21/2019 Peeling Processkkm
20/41
M M
(!. )2+(!- )2
M ! =
M ! = (237.8 )2 +(88.14 )2
M ! = 253.6 [ N + m]
T+i! t+e ma6im$m *e"#i" mome"t i! at re!$lta"t o/ 4C5
M ma8= 253.6 [ N + m]
Tor7$e acti" o" t+e !+a/t 1 89:;(
7/21/2019 Peeling Processkkm
21/41
T+e e7$i=ale"t *e"#i" mome"t
M e=12 ( M + M
2
+ 2
)=12 ( M + e)
12 (288.05 +253.6 )
270.83 [ N + m]
4 = M c :
F + (=4 ma84 all
4 all =320 10 6
1.5
4 all = 213.33 Mpa
213.33 10 6= 270.83
" 32
d 3
d= 23.47 [mm ]let say 25 [mm]
Based on both analysis between the equivalent twisting moment and the equivalent bendingmoment, we choose the maximum diameter, that is '34mm5.
7/21/2019 Peeling Processkkm
22/41
VERTICAL DEFLECTION OF THE SHAFT
: =" 4 (r
4
)
: = " 4
(0.0125 4 )
: = 1.9175 10 8 m4
; stell = 200
7/21/2019 Peeling Processkkm
23/41
;: (d 2 yd82 )= 4739.66 < 8>+6172.2 < 8 0.045 > 2274.672 < 8 0.045 2 + 2274.672 < 8 1.205 2+2731.8
;: (dyd8)= 4739.662 < 82+6172.22 < 8 0.045 2 2274.676 < 8 0.045 3 +2274.676 < 8 1.205 3 +2731.82
455.3 < 8 1.205 >3+ A8+*94.78 < 8 1.205 >4+
1028.7 < 8 0.045 >3 94.78 < 8 0.045 4 + 8>3+
;: ( y)= 789.94
&hen 8= 0.045, y= 0
0.045 >3 +0.045 A+*0 = 789.94
0.045 A+* = 0.07198 (1)
&hen 8= 1.205, y= 0
94.78 4+1.205 A+*1028.7 3
1.205 >3+0= 789.94
1.205 A+* = 51.94 (2 )
0.045 A+* =
0.07198(
3
)
0.05423 A+1.205 * = 0.08674
7/21/2019 Peeling Processkkm
24/41
1.16 * = 2.4237
* = 2.09 ..into e=uation (1)
0.045 A+2.09 * = 0.07198
A= 44.83
94.78 < 8 1.205 >4 44.83 < 8>+2.091028.7 < 8 0.045 >3 94.78 < 8 0.045 4 +
8>3 + ;: ( y)= 789.94
' '. '.) '.6 '.* + +. +.)
,'.'+
,'.'+
,'.'+
'
'
'
'
'
'.'+
-e ection vs /osition
-istance (m"
-e ection (m"
7/21/2019 Peeling Processkkm
25/41
0fter 1ptimi2ing
' '. '.) '.6 '.* + +. +.)
,'.'+
,'.'+
,'.'+
'
'
'
'
'
'.'+
-e ection vs /osition
0fter 1ptimi2ation
-istance (m"
-e ection (m"
HORI%ONTAL DEFLECTION OF SHAFT
7/21/2019 Peeling Processkkm
26/41
: = " 4 (r 4)
: = " 4
(0.0125 4 )
: = 1.9175 10 8 m4
; stell = 200 1373.5 < 8 1.16 > ;: (dyd8)= 609.56 < 83 327.82 < 82 1373.52 < 8 1.16 2+ A
;: ( y)= 25.4 < 84 54.63 < 83 228.9 < 8 1.16 3 + A8+*
&hen 8= 0, y= 0
* = 0
7/21/2019 Peeling Processkkm
27/41
&hen 8= 1.16, y= 0
0= 25.4
7/21/2019 Peeling Processkkm
28/41
' '. '.) '.6 '.* + +. +.)
'
'
'
'
'
'
'
'
'
-e ection 3s /osition
4efore 1ptimi2ation 0fter 1ptimi2ation
/osition (m"
-e ection ( m"
MOTOR SHAFT
7/21/2019 Peeling Processkkm
29/41
6fM & A. = 0 7
4697.11 (0.121 )+ &*. (0.409 )= 0
&*. = 1389.61 N
6 fy = 0 7
4697.11 N & A. + &*. = 0
& A. = 4697.11 N +1389.61 N
6'*6. 7
2.3log
[ 1
2 ]= 5
2.3log [ 1 2 ]= 0.42 (" ) 1 = 3.75 2 +(1 )
7/21/2019 Peeling Processkkm
30/41
= ( 1 2 )r
136.59 =( 1 2 )0.05
1
2= 2731.8 (2)
Sub equation (2) into equation (1)
3.75 2= 2731.8 N
2.75 2= 2731.8 N
2 = 993.38 N
1 = 3.75 (993.38 )= 3.72 kN
Maximum bending moment based in the bending moment diagram is at point 6 that is &.378N
torque on the &./ 7 Nm
The equivalent twisting moment
ince torque % &./ 78Nm
7/21/2019 Peeling Processkkm
31/41
e= M 2+ 2
(0.57 )2+(0.137 )2
0.586 2Nm
ma8 = allo0a'le
1.5
380 106
1.5
253.33 Mpa
= r9
253.33 106= 0.586 > 103
" 16
d 3
d= 22.8 mmlet say 25 mm
he equivalent bending moment
M e=12( M + M 2 + 2)= 1
2 ( M + e)
12
(0.57 +0.586 )
0.578 2N + m
4 = M! :
F + (=4 ma84 all
4 all =320 10 6
1.5
7/21/2019 Peeling Processkkm
32/41
4 all = 213.33 Mpa
213.33 106 =
0.578 > 103
" 32
d3
d= 30.2 mm let say 35 mm
DEFLECTION AT AC MOTOR SHAFT
7/21/2019 Peeling Processkkm
33/41
;: (d 2 yd82 )= 1389.61 < 8> 6086.72 < 8 0.409 > 8 0.409 >2 + A
8>2
6086.72
2
;: (dyd8)= 1389.612 6086.72
6< 8 0.409 >3+ A8+*
8>3 ;: ( y)= 1389.61
6
&hen 8= 0, y= 0
* = 0
&hen 8= 0.409, y= 0
0.409 >3 +0.409 A ;: (0 )= 1389.61
6
A= 38.74
1014.45 < 8 0.409 >3 38.74 < 8> ;:
8>3 231.60
y=
7/21/2019 Peeling Processkkm
34/41
LENGTH OF BELT
7/21/2019 Peeling Processkkm
35/41
)ength of 'elt ,l = " (r 1+r 2 )+2 8+(r 1 r 2 )2
8 .. (terms of pulley radii )= "
2 (d1 +d 2 )+2 8+(
d 1 d 2 )24 8
>OINT OF AC MOTOR
7/21/2019 Peeling Processkkm
36/41
6 f 1= 0 7
4691.77 1962 2 F A. 2 F *. = 0 (1)
6 M at y= 0 7
4691.77 (0.504 )+19.62 +2 F A. (0.3 )= 0
F A. = 3908.4 [ N ].. into (1 )
4691.77 19.62 2 (3908.4 ) 2 F *. = 0
7/21/2019 Peeling Processkkm
37/41
F *. = /37'. 4N5 9 ? :
BOLT AND NUT
F + (= 4 ma84 allo0a'le
4 allo0a'le =240 [ M#a ]
1.5
4 allo0a'le = 160 [ M#a ]
SCREWED JOINTS 8ighl$ reliable Convenient to
assemble Cheap
ateria!
0 9 ' stainlesssteel
Tensi!e stren"th
'' 7: mm2
#$2% &ie!' stren"th
)5' 7: mm2
!;< B;2T I=> ;! M3(M/&&
4 MA> = 240 [ M#a ]
MA> = 311 [ M#a ]
7/21/2019 Peeling Processkkm
38/41
F + (= ma8 allo0a'le
allo0a'le = 311 [ M#a ]1.5
allo0a'le = 207.33 [ Mpa ]
4 = F A
d
" (2)
4
160 106= 3908.4
d= 5.58 [mm ] or bigger
7/21/2019 Peeling Processkkm
39/41
SUN? ?EY
= orque transmitted b$ the shaft.
F = angential force acting at the circumference of shaft.
d= -iameter of shaft.
l= %ength of ke$.
0 = &idth of ke$.
t = hickness of ke$.
= Shear for the material ke$.
4 c= Crushing stresses for the material ke$.
7/21/2019 Peeling Processkkm
40/41
= 136.59 [ Nm]
0 = 0.01 [m]
md= 0.04 ;
t = 0.01 [m]
l= 0.04 [m]
Consi'erin" shearin" of the e&
F = l 0
= F d2
136.59 = F 0.04
2
F = 6829.5 [ N ]
6829.5 = 0.04 0.01
N /m2
= 17073750 ;
Consi'erin" crushin" of the e&
F = l t 2
4 c
6829.5 = 0.04 0.01
2 4 c
N /m2
4 c= 34147500 ;
7/21/2019 Peeling Processkkm
41/41
The e& is equa!!& stron" in crushin" an' shearin" if
0t =
4 c2
0.010.01
= 341475002(17073750 ) proven
To *n' the !en"th of the e& to trans+it fu!! ,o-er shaft
l= 1.571 d
1.571 (0.04 )
0.06284 [m] 62 [mm]