peeling processkkm

7/21/2019 Peeling Processkkm

1/41

ANALYSIS OF PEELING


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


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


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


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


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 )


7/41

k% #= 4.289 [k% ] let say 4.5


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]+


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


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

#= $


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


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


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


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


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


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


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


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


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


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:;(


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.


22/41

VERTICAL DEFLECTION OF THE SHAFT

: =" 4 (r

4

)

: = " 4

(0.0125 4 )

: = 1.9175 10 8 m4

; stell = 200


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


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"


25/41

0fter 1ptimi2ing

' '. '.) '.6 '.* + +. +.)

,'.'+

,'.'+

,'.'+

'

'

'

'

'

'.'+

-e ection vs /osition

0fter 1ptimi2ation

-istance (m"

-e ection (m"

HORI%ONTAL DEFLECTION OF SHAFT


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


27/41

&hen 8= 1.16, y= 0

0= 25.4


28/41

' '. '.) '.6 '.* + +. +.)

'

'

'

'

'

'

'

'

'

-e ection 3s /osition

4efore 1ptimi2ation 0fter 1ptimi2ation

/osition (m"

-e ection ( m"

MOTOR SHAFT


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 )


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


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


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


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=


34/41

LENGTH OF BELT


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


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


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 ]


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


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$.


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 ;


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]

peeling processkkm

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