










Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
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
Typology: Study Guides, Projects, Research
1 / 18
This page cannot be seen from the preview
Don't miss anything!
Disusun oleh :
Nama : Dewi Srirejeki Lestari (181211041)
Riki Gunawan (181211060)
Kelas : 2 MB
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 =
(
)
Po
2
Po = Po 1
Po 2
Po
1
Po
1
pd
P kɵ
B = 2f + ( n-1 ) e
11,38 ¿ 30 baik
dk + Dk
¿ c
207,35 ¿ 450 baik
L = 2c +
π
(dp + Dp) +
(
Dp − dp
4 e
)
2
π
(
)
2
b = 2L –
π (269,7 – 145)
= 2(1778) - π (269.7 -145)
b + √
b
2
daerah penyetelan
i
t
Sabuk :B
Puli
dp =
Dp = 269,
Jumlah sabuk = 8
Jarak sumbu poros
Gaya sabuk
v = 11,38 m/s
102 pd
v
=215,11 Kg
= e
μθ
(0,3.2,86 )
2
θ =
π =2,
1
2
=922,31 Kg
Fx=
1 x
2 x
=27,36 Kg
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
2
i = 1,
Pd = 24 Kw
Pd = 24 Kw
n
1
= 1500 rpm
Rangkaian 1
P =19,
FB=
Fu =
No 60-
Diameter Lingkaran jarak bagi
dp =
sin
sin
Dp =
sin
sin
Diameter naf
dB = 19,
{
cot
}
{
cot
}
Diameter luar
dk = 19,05 (0,6 + cot
Dk = 19,05 (0,6 + cot
1
n 1
= 6,191 m/s
Gaya keliling
102_. pd_
v
= 395,37 kg
Faktor koreksi
Sf =
6,191 ¿ (4-10) baik
dk + Dk
Panjang = 1905
Sproket :
jumlah gigi : Z 1
= 13 Diameter: dp = 76,
2
= 20 Dp = 121,
Gaya Sabuk(Transmisi rantai rol)
ϑ = 180- 57,
Dp − dp
ϑ = 180 – 57,
= 176,54 = 3,08 rad
1
2
e
μθ
1
e
μθ
2
(0,3_._ 3,08 )
2
2
1
2
2
2
2
2
1
β = arc tg
R − r
β= arc tg
p
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,
Pd = P.Fc = 24 Kw
T = 9,74 x
5
Pd
n 1
= 9,74 x
5
= 15584 Kg.mm
τaa =
σu
Sf 1. Sf 2.9,
Kg
mm
2
PV = P.v =0,016.8,635=0,
PV ≤ ( PV ) a
2
=3,1 Kw
Poros = Bahan ST
Diameter = 110
Bantalan : Bahan Kuningan
Panjang : 330 mm
Daya yang diserap :3,1 Kw
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 – Fx 1
RBH = Fx 1
Momen Horizontal
τaa =
σu
Sf 1. Sf 2.9,
τa =
5
n 1
5
d ≥ ¿ ¿
d ≥ ¿ ¿
d ≥ 33,
d=
Wo = √¿ ¿
W = Wo. fw
Pa=(0,7-2) Kg/ mm
2
(Pv)a = (0,1-0,2)
L/d = (2-3)
π. w. n
( Pv ) a. 60000
L/d = 624/240 =2,6 baik
w
L. d
− 3
− 3
baik
π. d. n
m
s