TUGAS ELEMEN MESIN
Disusun oleh :
Nama : Dewi Srirejeki Lestari (181211041)
Riki Gunawan (181211060)
Kelas : 2 MB
POLITEKNIK NEGERI BANDUNG
Jalan Gegerkalong Hilir Desa Ciwaruga 6468 Bandung
Telepon 022-2013789 Faks 022-201388
2019
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Transmission System Design: Sabuk V and Rantai Rol, Study Guides, Projects, Research of Marine Engineering
The design details of a transmission system with two components: Sabuk V and Rantai Rol. The system uses a motor with a power of 20 Kw and a rotational speed of 2800 rpm. The rotational speed is reduced at each stage using a belt and a chain drive. calculations for the diameter of pulleys, gears, and the length of the belts and chains. It also discusses the power transmitted and the forces acting on the system.
What you will learn
- What is the power transmitted at each stage of this system?
- How long are the belts and chains in this transmission system?
- What is the power of the motor used in this transmission system?
- What are the diameters of the pulleys and gears used in this system?
- What are the rotational speeds of the motor and the reduced speeds at each stage?
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2019/2020
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TUGAS ELEMEN MESIN
Disusun oleh :
Nama : Dewi Srirejeki Lestari (181211041)
Riki Gunawan (181211060)
Kelas : 2 MB
POLITEKNIK NEGERI BANDUNG
Jalan Gegerkalong Hilir Desa Ciwaruga 6468 Bandung
Telepon 022-2013789 Faks 022-
Soal
Sebuah mesin digerakkan oleh motor dengan daya 20 Kw dengan putaran 2800 rpm .Putaran ini
direduksi oleh sabuk v menjadi 1500 rpm pada tingkat pertama. Pada tingkat berikutnya putaran
direduksi oleh rantai rol menjadi 1000rpm.Bahan Poros St60(
σ
u
= 600 MPa ¿ dan bantalan
kuningan dengan factor koreksi daya fc=1,3 ;
sf
1
= 6 ; sf
2
= 2 ; cb =2,2 ; Kt =2,5 ;
fw =1,3 ;μ =0,06. Jarak sumbu poros pada sabuk V 450mm , jarak sumbu poros transmisi pada
sabuk rantai rol 800mm. Rencanakan ukuran poros, bantalan ,puli, dan sabuk pada system diatas.
K ɵ =
(
)
- Daya yang ditransmisikan
Po
2
Po = Po 1
- Po 2
Po 2
Po
1
Po
1
- Jumlah sabuk N
N =
pd
P kɵ
- Lebar pulli
B = 2f + ( n-1 ) e
- Syarat
V ¿ 30
11,38 ¿ 30 baik
dk + Dk
¿ c
207,35 ¿ 450 baik
- Panjang sabuk
L = 2c +
π
(dp + Dp) +
(
Dp − dp
4 e
)
2
π
(
)
2
= 1755,68 ≈ ( L = 1778 ) B
- Jarak sumbu poros sebenarnya
b = 2L –
π (269,7 – 145)
= 2(1778) - π (269.7 -145)
C =
b + √
b
2
daerah penyetelan
∆ C (L= 1778) ∆ C
i
= 35 ∆ C
t
C
- 50
- Kesimpulan
Sabuk :B
Puli
dp =
Dp = 269,
Jumlah sabuk = 8
Jarak sumbu poros
C = 1743 -
Gaya sabuk
- Kecepatan
v = 11,38 m/s
- Gaya Keliling
F =
102 pd
v
=215,11 Kg
S 1
S 2
= e
μθ
S 1
S 2
(0,3.2,86 )
→ S 1 =2,35 S
2
θ =
π =2,
F=
S
1
− S
2
=922,31 Kg
Fx=
S
1 x
+ S
2 x
=27,36 Kg
- Berat Puli
Wp = πμ D
2
tγ
π
2
=63,15 Kg
B. Transmisi II (Rantai Roll)
Pd =24 KW
n
1
= 1500 rpm
n
2
= 1000 rpm
C
2
i = 1,
- Daya Rencana
Pd = 24 Kw
- Pemilihan Rantai
Pd = 24 Kw
n
1
= 1500 rpm
Rangkaian 1
- Jumlah Gigi
Z1 ≥ 13
Z2 =1,5.13 = 19,5 ≈ 21
P =19,
FB=
Fu =
No 60-
- Diameter
Diameter Lingkaran jarak bagi
dp =
P
sin
sin
Dp =
P
sin
sin
Diameter naf
dB = 19,
{
cot
}
DB = 19,
{
cot
}
Diameter luar
dk = 19,05 (0,6 + cot
Dk = 19,05 (0,6 + cot
- kecepatan
V =
P Z
1
n 1
= 6,191 m/s
Gaya keliling
F =
102_. pd_
v
= 395,37 kg
Faktor koreksi
Sf =
- Syarat
V ¿ (4-10)
6,191 ¿ (4-10) baik
C ¿
dk + Dk
Panjang = 1905
Sproket :
jumlah gigi : Z 1
= 13 Diameter: dp = 76,
Z
2
= 20 Dp = 121,
Gaya Sabuk(Transmisi rantai rol)
- kecepatan
V = 6,
- gaya keliling F
F = 395,
- sudut bantalan
ϑ = 180- 57,
Dp − dp
ϑ = 180 – 57,
= 176,54 = 3,08 rad
S
1
S
2
e
μθ
S
1
e
μθ
. S
2
(0,3_._ 3,08 )
S
2
=2,5 S
2
F = S
1
- S
2
395,47 = 2,2 S
2
– S
2
= 1,5 S
2
S
2
S
1
β = arc tg
R − r
C
β= arc tg
- Berat Puli
W
p
= V. J
π
2
¿ 6,2 Kg
C. Poros Spindel
Pemilihan ukuran poros spindle dan bantalan
Dik :
F = 20Kw Kt = 2,
n
1
= 1500 rpm Fw= 1,
σu = 600 MPa Wp=63,15 Kg
Fc = 1,2 Fx =41,
Sf1=6 Fy =532,
Sf2 = 2
Cb = 2,
- Daya rencana
Pd = P.Fc = 24 Kw
- Momen Rencana
T = 9,74 x
5
Pd
n 1
= 9,74 x
5
= 15584 Kg.mm
- Tegangan geser ijin
τaa =
σu
Sf 1. Sf 2.9,
Kg
mm
2
- Diameter Poros
- Tekanan Kecepatan
PV = P.v =0,016.8,635=0,
PV ≤ ( PV ) a
- Daya yang diserap
PH=
M
2
. W .V
=3,1 Kw
- Kesimpulan :
Poros = Bahan ST
Diameter = 110
Bantalan : Bahan Kuningan
Panjang : 330 mm
Daya yang diserap :3,1 Kw
D. Poros Transmisi
Diketahui
Wp
1
Wp 2
Fx 1
Fy 1
Fx2 = 27,
Fy 2
Fv 1
= Wp1+Fx1 = 63,15 + 532,13 = 595,
Fv 2
= Fy2-Wp2 = 922,31-6,2 = 916,
RAH = 49,
∑
H = 0
RAH – Fx 1
- Rbh - Fx 2
RBH = Fx 1
- Fx 2
– RBH
Momen Horizontal
MAH=MBH=
MCH= RAH (0,1) MDH = RBH (0,1)
- Tegangan Geser Ijin
τaa =
σu
Sf 1. Sf 2.9,
- Momen Rencana
τa =
5
n 1
5
- Diameter Poros
d ≥ ¿ ¿
d ≥ ¿ ¿
d ≥ 33,
d=
- Beban bantalan
Wo = √¿ ¿
- Beban Rencana
W = Wo. fw
- Bahan Bantalan = Kuningan
Pa=(0,7-2) Kg/ mm
2
(Pv)a = (0,1-0,2)
L/d = (2-3)
- Panjang Bantalan
L
π. w. n
( Pv ) a. 60000
L
L ≥ 623,
L=
L/d = 624/240 =2,6 baik
- Tekanan yang terjadi
P=
w
L. d
− 3
P ≤ ( 0,7− 2 )
− 3
baik
- Kecepatan
V =
π. d. n
m
s