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Transcript of 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
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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
$a#an!an 1 z
2
z1= 2
Ditentukan 1 = 20 h
Maka :
z2 = 2 z1
= 2 20 = 40 h
arena: d = m
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$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$a
< : 2' (dinami# ' olon!an )
σ bol : 580
2,5 = 230 Mpa
HBr : 192 diamil 20
ol : 20 maka a+a 20 = 0'6
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$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)
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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
; $emilihan 7ahan 8oda i!i
7ahan : < ;
σ y : 4009600 M$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
$emerik#aan ekuatan 8oda i!i
$a#an!an 8oda i!i
MPii = 00 N
n (N%m)
= 00 25600
= 312' N%m = 312 Nmm
Pt ii =2. Mpii
do3 =
2.398125
72 = 110 N
Pr ii = Pt 2 t! .
= 110 t! 20o
= 402 N
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P totii = √ Pt ii2+ Pr ii
2
= √ (11059)2+(4025)2
= 11,6 N
*e!an!an lentur /an! timul
1
? 0'30
2
= 10m = 104 = 40 mm
σ ii = Pt ii
b . y . m
=11059
40.0,302 . 4
= 22' M$a @ σ bol
*e!an!an $ermukaan /an! timul
=1+i
i
Pt ii
b.do2
=1+33
11059
40.72 = '1M$a
Maka ahan < ; haru# diperkera#
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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
$emerik#aan ekuatan 8oda i!i
$a#an!an 8oda i!i
MPiii = 00 N
n
(N%m)
= 0025
600 = 312' N%m = 312 Nmm
Pt iii =2. Mpiii
do5 =
2.398125
96 = 24'3 N
Pr iii = Pt iii t! .
= 24'3 t! 20o
= 301' N
P totiii = √ Pt iii2+ Priii
2
= √ (8294,3)2+(3018,8)2
= 26'6 N
*e!an!an lentur /an! timul
24
? 0'33
= 10m = 104 = 40 mm
σ iii = Pt iii
b . y . m
=8294,3
40.0,339.4
= 12' M$a @ σ bol
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*e!an!an $ermukaan /an! timul
=1+i
i
Pt iii
b.do5
=1+22
8294,3
40.96 = 3'2M$a
Maka ahan < ; haru# diperkera#
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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
; $emilihan 7ahan 8oda i!i
7ahan : < ;
σ y : 4009600 M$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
$emerik#aan ekuatan 8oda i!i
$a#an!an 8oda i!i A
MPiv = 00 N
n (N%m)
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= 0025
600 = 312' N%m = 312 Nmm
Pt iv =2. Mpiv
do5 = 2.398125
144 = 2' N
Pr iv = Pt iC t! .
= 2' t! 20o
= 2012'6 N
P totiv = √ Pt iv2+ Pr iv
2
= √ (5529,5)2+(2012,6)2
= 4'3 N
*e!an!an lentur /an! timul
36
? 0'3
3
= 10m = 104 = 40 mm
σ iv = Pt iv
b . y .m
=5529,5
40.0,383.4
= 0'23 M$a @ σ bol
*e!an!an $ermukaan /an! timul
=1+i
i
Pt iv
b.do5
=1+11
5529,5
40.144 = 1'M$a
Maka ahan < ; haru# diperkera#
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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
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3.!. Peritungan Poros
321 $oro# 1
$emilihan 7ahan $oro# 1
7ahan : < ;
σ y : 4009600 M$adiamil 0 M$a
< : 2' (dinami# ' olon!an )
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
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=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
< : 2' (dinami# ' olon!an )
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
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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
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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 $oro# 3
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$emilihan ahan $oro# 3
7ahan : < ;
σ y : 4009600 M$adiamil 0 M$a
< : 2' (dinami# ' olon!an )
σ bol : 580
2,5 = 230 Mpa
ML5 =
f .(g++i)f +g++i
Ptot 6
=70.(90+70+85)70+90+70+85
26'6
= 40'3 Nmm
ML6 =
( f +g) .(+i)f +g++i
Ptot 6
=(70+90 ) .(70+85)70+90+70+85
26'6
= 641'6 Nmm
ML7 =
( f +g+ ) . i
f +g++ i
Ptot 7
=(20+50+30 ) .50
¿ ¿20+50+30+50
4'3
= 36201' Nmm
M tot 5 = √ Ml52+
3
4 Mp
2
=
398125¿
¿(480559,3)2+3
4¿
√ ¿
= 141' Nmm
M tot 6 = √ Ml62+
3
4 Mp
2
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=
796250¿¿
(694919 !6)2+3
4¿
√ ¿
= ,1'2 N
M tot 7 = √ Ml72+ 3
4 Mp
2
=
398125¿¿
(365201,5)2+3
4¿
√ ¿ = 02244'4 Nmm
d3 =
3
√10. M tot 1
σ bol
=3
√10.796250
230 = 32'$ 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
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% 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
332 7antalan $oro# 2
7ean &kiCalen
$ = E8 ?5$ Di!unakan 7antalan Deep roCe 7all 7earin! (D77)
= 1 46,' 0
34
$t 141' mm141'mm
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= 46,'6 N
7ean Dinami#
f l =3
√ L500
$ L % 20&000 'am ntk pmakaian 8 'am
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
*ari prhitn+an diata dipro-h data .
No /anta-an . *G// !40!
%4"!00 N
" o % 24000 N
bar /anta-an, b % 2" mm
*iamtr bor, d % "0 mm
*iamtr -ar, * % 0 mm
3
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333 7antalan $oro# 3
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
$ L % 20&000 'am ntk pmakaian 8 'am
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
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% 882!,! &3,4
0,38
% 78!,5 N
*ari prhitn+an diata dipro-h data .
No /anta-an % N2"073P
% 1"00 N
" o % 112000 N
bar /anta-an, b % "1 mm
*iamtr bor, d % "5 mm
*iamtr -ar, * % 80 mm
3,
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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
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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
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