PM 2 Adit BAB 3 RODA GIGI rev.docx

8/18/2019 PM 2 Adit BAB 3 RODA GIGI rev.docx

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BAB III

PENGOLAHAN DATA RANCANGAN

N = 20 kw

n1 = 1200 rpm > input

Output :

n2 = 200

rpm

n3 = 300

rpm

n4 = 600

rpm

Diketahui: N = 20 kw

n1 = 1200 rpm

n2 = 200 rpm

n3 = 300 rpm

n4 = 600 rpm

Modul (m) = 4 (diamil dari !ra"ik)

3.1. Penentuan transmisi roda-roda gigi

i1.2 =

n1

n2 =

1200

600 = 2

Menuru uku Da#ar $eren%anaan &lemen Me#in' r ndra *ed+a kumala' hal 4,0

-ika diamil #udut tekanan . = 200' = 10m' N = 20 kw dan n = 1200 rpm maka

modul /an! diamil = 4

13


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arena roda !i!i 2'3'', erada pada #atu poro#' maka putaran dari roda !i!i=

600rpm

• i3.4 =

n2

n3 =

600

200 = 3

z4

z3= 3

z4 = 3 z3

• i5.6 =n2

n4 =

600

300 = 2

z6

z5= 2

z6 = 2 z5

• i7.8 =n2

n4 =

600

600 = 1

z8

z7= 1

z8 = z7

Maka : 34=6=,

333=2=,,

43 = 3 = 2,

3.1.1 Peritungan Roda Gigi Pasangan I

5 $enentuan -umlah i!i

$a#an!an 1 z

2

z1= 2

Ditentukan 1 = 20 h

Maka :

z2 = 2 z1

= 2 20 = 40 h

arena: d = m

14


3/30

$enentuan diameter roda !i!i 1 dan 2

Maka' do1 = m z1

= 4 20 = 0 mm

do2 = m z2

= 4 40 = 160 mm

a1 =

do1+do

2

2 =

80+1602

= 120 mm

7 $erhitun!an dkepala ' dkaki ' dn

8oda i!i 1 8oda i!i 2

dkep1 = do1 2 m dkep2 = do2 2 m

= 0 2 4 = 160 2 4

= mm = 16 mm

dkaki1 = do1 9 2 (1'1 m) dkaki2 = do2 2 (1'1

m)

=0 9 2(1'1 4) = 160 2 (1'1 4)

= 0 ' = 160 '

= ,1'2 mm = 11'2 mm

dn = (1,5−2 )d dn = (1,5−2 )d

= 2 2 = 2 3

= 0 mm = ,0 mm

; $emilihan 7ahan 8oda i!i

7ahan : < ;

σ y : 4009600 M$adiamil 0 M$a

< : 2' (dinami# ' olon!an )

σ bol : 580

2,5 = 230 Mpa

HBr : 192 diamil 20

ol : 20 maka a+a 20 = 0'6

1


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$emerik#aan ekuatan 8oda i!i

$a#an!an 8oda i!i

MP1 = 00

N

n (N%m)

= 0025

1.200 = 106'2 N%m = 1062 Nmm

Pt 1 =

2. Mp1

do1 =

2.199062

80 = 4,6' N

Pr 1 = Pt 1 t! .

= 4,6' t! 20o

= 111'31 N

P tot1 = √ Pt 12+ Pr1

2

= √ (4976,55)2+(1811,31)2

= 2' N

*e!an!an lentur /an! timul

20? 0'32

0

= 10m = 104 = 40 mm

σ 1 =

Pt 1

b . y . m

=4976,55

40.0,320.4

= ,'2 M$a @ σ bol

*e!an!an $ermukaan /an! timul

=1+i

i

Pt 1

b.do1

=1+22

4976,55

40.80 = 2'33 M$a

Maka ahan < ; haru# diperkera#


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7erda#arkan perhitun!an te!an!an permukaan /an! timul didapat nilai >

ol

3.1.! Peritungan Roda Gigi Pasangan II

5 $enetuan +umah !i!i pa#an!an (penentuan awal = 24)

43 = 3

3 =4

3 24

= 1 h

4 = 3 1

= 4 h

$enetuan diameter roda !i!i 3 dan 4

do3 = m z3 do4 = m z4

= 4 1 = 4 4

= ,2 mm = 216 mm

a2 =

do3+do42

=72+216

2 = 144 mm

arena roda !i!i 2 4' 6' dan , maka +arak a haru#lah #ama 7erda#arkan

perhitun!an diata#' +arak a /aitu 144 mm

7 $erhitun!and

kepala ,d

kaki , dn8oda i!i 3 8oda i!i 4

dkep3 = do3 2 m dkep4 = do1 2 m

= ,2 2 4 = 216 2 4

= 0 mm = 224 mm

dkaki1 = do1 9 2 (1'1 m) dkaki4 = do4 2 (1'1

m)

= ,2 9 2(1'1 4) = 216 2 (1'1 4)

1,


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= ,2 ' = 216 '

= 63'2 mm = 20,'2 mm

dn = (1,5−2 ) d dn = (1,5−2 ) d

= 2 3 = 2 40

= ,0 mm = 0 mm


7ahan : < ;



σ bol : 580

2,5 = 230 Mpa

HBr : 192 diamil 20

ol : 20 maka a+a 20 = 0'6


$a#an!an 8oda i!i

MPii = 00 N

n (N%m)

= 00 25600

= 312' N%m = 312 Nmm

Pt ii =2. Mpii

do3 =

2.398125

72 = 110 N

Pr ii = Pt 2 t! .

= 110 t! 20o

= 402 N

1


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P totii = √ Pt ii2+ Pr ii

2

= √ (11059)2+(4025)2

= 11,6 N


1

? 0'30

2

= 10m = 104 = 40 mm

σ ii = Pt ii

b . y . m

=11059

40.0,302 . 4

= 22' M$a @ σ bol


=1+i

i

Pt ii

b.do2

=1+33

11059

40.72 = '1M$a



8/30


9/30



σ bol :

580

2,5 = 230 Mpa

HBr : 192 diamil 20

ol : 20 maka a+a 20 = 0'6


$a#an!an 8oda i!i

MPiii = 00 N

n

(N%m)

= 0025

600 = 312' N%m = 312 Nmm

Pt iii =2. Mpiii

do5 =

2.398125

96 = 24'3 N

Pr iii = Pt iii t! .

= 24'3 t! 20o

= 301' N

P totiii = √ Pt iii2+ Priii

2

= √ (8294,3)2+(3018,8)2

= 26'6 N


24

? 0'33

= 10m = 104 = 40 mm

σ iii = Pt iii

b . y . m

=8294,3

40.0,339.4

= 12' M$a @ σ bol

21


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=1+i

i

Pt iii

b.do5

=1+22

8294,3

40.96 = 3'2M$a



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arena roda !i!i 2 4' 6' dan , maka +arak a haru#lah #ama 7erda#arkan

perhitun!an diata#' +arak a /aitu 144 mm

7 $erhitun!an dkepala ' dkaki ' dn

8oda i!i , dan

dkep5 = do5 2 m

=144 2 4

= 12 mm

dkaki5 = do5 9 2 (1'1 m)

= 144 9 2(1'1 4)

= 144 '

= 13'2 mm

dn7 = (1,5−2) d

= 2 3

= ,0 mm

dn8 = (1,5−2) d

= 2 40

= 0 mm


7ahan : < ;



σ bol : 580

2,5 = 230 Mpa

HBr : 192 diamil 20

ol : 20 maka a+a 20 = 0'6


$a#an!an 8oda i!i A

MPiv = 00 N

n (N%m)

23


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= 0025

600 = 312' N%m = 312 Nmm

Pt iv =2. Mpiv

do5 = 2.398125

144 = 2' N

Pr iv = Pt iC t! .

= 2' t! 20o

= 2012'6 N

P totiv = √ Pt iv2+ Pr iv

2

= √ (5529,5)2+(2012,6)2

= 4'3 N


36

? 0'3

3

= 10m = 104 = 40 mm

σ iv = Pt iv

b . y .m

=5529,5

40.0,383.4

= 0'23 M$a @ σ bol


=1+i

i

Pt iv

b.do5

=1+11

5529,5

40.144 = 1'M$a



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Ta$e% Ring&asan

PASANG

AN GIGI

n

(rpm)

i

Z

(bh)

do

(mm)

dk

(mm)

dkaki

(mm)

dn

(mm)

b

Mp

(Nmm)

Pt(N)

Pr(N )

Ptot

(N)

I

1200

220 80 88 71,2 50

40

10!2

47!,55

1811,"1

525,

!00 40 1!0 1!8151,

2 70

II!00

"18 72 80 !",2 5!

"8125

1105 4025 117!8200 54 21! 224

207,2 80

III!00

224 ! 104 87,2 5!

"8125

824,"

"018,8

882!,!

"00 48 12 20018",

2 80

I#!00

1"! 144 152

1"5,2 70 "81

25552,

52012,

!5884,

"8!00 "! 144 152

1"5,2 80

2


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3.!. Peritungan Poros

321 $oro# 1

$emilihan 7ahan $oro# 1

7ahan : < ;



σ bol : 598

2,3 = 260 Mpa

MP1 = 1062 Nmm

ML1 =

a .b

a+b Ptot 1

=40.40

40+4 0 2'

= 101 Nmm

M tot = √ Ml12+

3

4 Mp1

2

= √(105918)2+

3

4(199062)2

= 202331'1 Nmm

d1 =

3

√

10. M tot

σ bol

26

$tot10 mm 10 mm

k


15/30

=3

√10.202331,1

260 = 1'6 $ 20 mm

322 $oro# 2

$emilihan ahan $oro# 2

7ahan : 7d ,0

σ y : ,00900M$adiamil 00 M$a


σ bol : 800

2,5 = 320 Mpa

5 M5 = 0Ʃ

Ptot 1 a 9 Ptot 2 (a) 9 RB1 (a%de)=0

Ptot 1 a 9 Ptot 2 (a) = RB1 (a%de)

RB1 = Ptot 1 . a− Ptot 2 .(a+b)

(a+b+c+d+e)

=

11768 .(40+80)

¿(5295,9 .40)−¿

¿

= 94001 N

7 M7 = 0Ʃ

Ptot 2 (%de) 9 Ptot 1 (%de) 9 RA1 (a%de)=0

Ptot 2 (%de) 9 Ptot 1 (%de) = RA1 (a%de)

2,

% d ea

$t1

$t2

8 8 5

M


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RA1 = Ptot 2 .(c+d+e)− Ptot 1 .(b+c+d+e)

(a+b+c+d+e)

=11768.(50+90+40)−5295,9.(80+50+90+4 0)

(40+80+50+90+40)

= 24,1 N

M x = Ptot 2 (9a9) 9 Ptot 1 (9a) RA1 E

E = a $ M x

M Xa1 = (11,6 (90)) (2' 0) (24,1) 40

= 942600 Nmm

E = $ M

x

M XB1 = (11,6 (940)) (2' (40) (24,1) 0

= 944,6 Nmm

5 M5 = 0Ʃ

Ptot 1 a 9 Ptot 3 (a%) 9 RB2 (a%de)=0

Ptot 1 a 9 Ptot 3 (a%) = RB2 (a%de)

RB2 = Ptot 1 . a− Ptot 3 .(a+b+c)

(a+b+c+d+e)

=

8826,6.(40+80+50)¿

(5295,9 .40)−¿¿

= 9

7 M7 = 0Ʃ

2

% d ea

$t1

$t3

M8

8 5


17/30

Ptot 3 (de) 9 Ptot 1 (%de) 9 RA2 (a%de)=0

Ptot 3 (de) 9 Ptot 1 (%de) = RA2 (a%de)

RA2 =

Ptot 3 .(d+e)− Ptot 1 .(b+c+d+e)

(a+b+c+d+e)

=(8826,6. (90+4 0 ))−(5295,9. (80+50+90+40 ))

(40+80+50+90+40)

= 92'6 N

M x = Ptot 3 (9a99%) 9 Ptot 1 (9a) RA2 E

E = a $ M x

M Xa2 = (26'6 (9130))9(2' 0) (92'6) 40 = 9123,21 Nmm

E = % $ M x

M Xc 2 = (26'6 (9120))9(2' 10) (92'6) 0

= 911621 Nmm

5 M5 = 0Ʃ

Ptot 1 a 9 Ptot 4 (a%d) 9 RB3 (a%de)=0

Ptot 1 a 9 Ptot 4 (a%d) = RB2 (a%de)

RB3 = Ptot 1 . a− Ptot 4 .(a+b+c+d )

(a+b+c+d+e)

=(5295,9 .40)−(5884,38. (40+80+50+90))

(40+80+50+90+40)

= 902'1 N

2

% d ea

$t1

$t3

M 8 8 5


18/30

7 M7 = 0Ʃ

Ptot 4 (e) 9 Ptot 1 (%de) 9 RA3 (a%de)=0

Ptot 4 (e) 9 Ptot 1 (%de) = RA3 (a%de)

RA3 = Ptot 4 .(e)− Ptot 1 .(b+c+d+e )

(a+b+c+d+e)

=(5884,38. (40 ))−(5295,9.(80+50+90+40 ))

(40+80+50+90+40)

= 930'2 N

M x = Ptot 4 (9a99%9d) 9 Ptot 1 (9a) RA3 E

E = a $ M x

M Xa3 = (4'3 (9220))9(2' 0) (930'2) 40

= 91446,,1 Nmm

E = d $ M x

M Xd 3 = (4'3 (91,0))9(2' 0) (930'2) 0

= 9160,60, Nmm

M tot = √ Ml22+

3

4 Mp2

2

=

398125¿¿

(−1607607)2+3

4¿

√ ¿

= 632' N

d2 =

3

√10. M tot

σ bol

=3

√10.856325,5

320 = 2' $ 30 mm

323 $oro# 3

30


19/30

$emilihan ahan $oro# 3

7ahan : < ;



σ bol : 580

2,5 = 230 Mpa

ML5 =

f .(g++i)f +g++i

Ptot 6

=70.(90+70+85)70+90+70+85

26'6

= 40'3 Nmm

ML6 =

( f +g) .(+i)f +g++i

Ptot 6

=(70+90 ) .(70+85)70+90+70+85

26'6

= 641'6 Nmm

ML7 =

( f +g+ ) . i

f +g++ i

Ptot 7

=(20+50+30 ) .50

¿ ¿20+50+30+50

4'3

= 36201' Nmm

M tot 5 = √ Ml52+

3

4 Mp

2

=

398125¿

¿(480559,3)2+3

4¿

√ ¿

= 141' Nmm

M tot 6 = √ Ml62+

3

4 Mp

2

31


20/30

=

796250¿¿

(694919 !6)2+3

4¿

√ ¿

= ,1'2 N

M tot 7 = √ Ml72+ 3

4 Mp

2

=

398125¿¿

(365201,5)2+3

4¿

√ ¿ = 02244'4 Nmm

d3 =

3

√10. M tot 1

σ bol

=3

√10.796250

230 = 32'$ 3 mm

Ta$e% Ring&asan

$oro#i!i

(N) MF M$ Mtot (N) $r(N) D

126

101 10622023311

1811,"1

2

2−1607607

312 3

311,6 312

632'4025

3

4 694919 !6 ,1'2 40

2,

−1607607312

632' "018,8

3

6 694919 !6 ,1'2 40

,4 −1607607 312 632' 2012,

!3

694919 !6 ,1'2 40

32


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3.3. Peritungan Banta%an

331 7antalan $oro# 1

7ean &kiCalen

$eG = E8 ?5$ Di!unakan 7antalan Deep roCe 7all 7earin! (D77)

= 1264,' 0

= 264,' N

7ean Dinami#

f l =3

√ L500

$ L % 22&000 'am ntk pmakaian 8 'am

pnh

%3

√22.000

500

% ",5

f n %3

√33,3

n $ n % 1&200 rpm

%3

√ 33,3

1.200

% 0,"02

" % Ptot &f

f n

33

$t 4040


22/30

% 2!47,5&3,5

0,302

% "0!88,2 N

*ari prhitn+an diata dipro-h data .

No /anta-an . *G// !404

% "0&700 N

" o % 1!&!00 N

bar /anta-an b % 1

*iamtr bor, d % 20 mm

*iamtr -ar, * % 72 mm


7ean &kiCalen

$ = E8 ?5$ Di!unakan 7antalan Deep roCe 7all 7earin! (D77)

= 1 46,' 0

34

$t 141' mm141'mm


23/30

= 46,'6 N

7ean Dinami#

f l =3

√ L500


pnh

%3

√20.000

500

% ",4

f n %3

√33,3

n $ n % !00 rpm

%3

√33,3

600

% 0,"8

% Ptot &f

f n

% 48!7,&

3,4

0,38

% 4"554,8 N


No /anta-an . *G// !40!

%4"!00 N

" o % 24000 N

bar /anta-an, b % 2" mm

*iamtr bor, d % "0 mm


3


24/30


7ean &kiCalen

$ = E8 ? 5$ Di!unakan 7antalan Deep roCe 7all 7earin! (D77)

= 1 26'6 0

= 26'6 N

7ean Dinami#

f l =3

√ L500


pnh

%3

√20.000

500

% ",4

f n %3

√33,3

n $ n % !00 rpm

%3

√33,3

600

% 0,"8

% Ptot &f

f n

36

$t 16' mm16' mm


25/30

% 882!,! &3,4

0,38

% 78!,5 N


No /anta-an % N2"073P

% 1"00 N

" o % 112000 N

bar /anta-an, b % "1 mm

*iamtr bor, d % "5 mm


3,


26/30

3.". Peritungan Pasa&

Diameter $oro# 1 = 30 mm

Diamil dari tael N161 didapatkan:

= l =

h = t = ,

7ahan $a#ak : 7d 3,

$ ol = 6 Mpa

$ =2. MP

1

d

=2.199062

30 = 132,0' N

$ = F1

2 h Pbol

F =

P

1

2. . P bol

=

13270,8

1

2.7.65

= '3 $ 60 mm erda#arkan tael Normali#a#i N162

3


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28/30

Gam$ar 'oros !

32 $oro# 7intan! 8oda i!i 4

Diameter poro# = 3 mm

Diamil dari tael DN 463 didapatkan:

i =

d1 = 4 mm

= mm

M 10 = 2,0 N%m

M g = (1'92) MP4

= 1' 312

= ,1,' Nmm = ,1', N%m

M g = 0', M 10 F

F = M g

0,7. M 10

=59718,75

0,7.2970

= 2', H 30 mm

Gam$ar 'oros 3

40


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3.). Peritungan Tem'eratur

$a#an!an

Lm1 = ( 0'0 0'1,)

( 1

z1+ 1

z2

) N

= 0'1 ( 120+ 140 ) 20= 0'1 kI

$a#an!an

Lm2 = ( 0'0 0'1,)

(

1

z3

+ 1

z4

) N

= 0'1 ( 118+ 154 ) 20= 0'14 kI

$a#an!an

Lm3 = ( 0'0 0'1,) ( 1 z5

+ 1

z6) N

= 0'1 ( 124 + 148 ) 20= 0'12 kI

$a#an!an A

Lm4 = ( 0'0 0'1,) ( 1 z7

+ 1

z8) N

41


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PM 2 Adit BAB 3 RODA GIGI rev.docx

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