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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?

Typology: Study Guides, Projects, Research

2019/2020

Uploaded on 01/11/2020

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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 022-201388
2019
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Download Transmission System Design: Sabuk V and Rantai Rol and more Study Guides, Projects, Research Marine Engineering in PDF only on Docsity!

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 =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 =

(

)

  1. Daya yang ditransmisikan

Po

2

Po = Po 1

  • Po 2

Po 2

Po

1

Po

1

  1. Jumlah sabuk N
N =

pd

P kɵ

  1. Lebar pulli

B = 2f + ( n-1 ) e

  1. Syarat
V ¿ 30

11,38 ¿ 30 baik

dk + Dk

¿ c

207,35 ¿ 450 baik

  1. Panjang sabuk

L = 2c +

π

(dp + Dp) +

(

Dpdp

4 e

)

2

π

(

)

2

= 1755,68 ≈ ( L = 1778 ) B
  1. 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
  1. Kesimpulan

Sabuk :B

Puli

dp =

Dp = 269,

Jumlah sabuk = 8

Jarak sumbu poros

C = 1743 -

Gaya sabuk

  1. Kecepatan

v = 11,38 m/s

  1. 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

  1. Berat Puli

Wp = πμ D

2

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,

  1. Daya Rencana

Pd = 24 Kw

  1. Pemilihan Rantai

Pd = 24 Kw

n

1

= 1500 rpm

Rangkaian 1

  1. Jumlah Gigi
Z1 ≥ 13
Z2 =1,5.13 = 19,5 ≈ 21

P =19,

FB=

Fu =

No 60-

  1. 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

  1. kecepatan
V =
P Z

1

n 1

= 6,191 m/s

Gaya keliling

F =

102_. pd_

v

= 395,37 kg

Faktor koreksi

Sf =

  1. 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)

  1. kecepatan
V = 6,
  1. gaya keliling F
F = 395,
  1. sudut bantalan

ϑ = 180- 57,

Dpdp

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

Rr

C

β= arc tg

  1. 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,

  1. Daya rencana

Pd = P.Fc = 24 Kw

  1. Momen Rencana

T = 9,74 x

5

Pd

n 1

= 9,74 x

5

= 15584 Kg.mm

  1. Tegangan geser ijin

τaa =

σu

Sf 1. Sf 2.9,

Kg

mm

2

  1. Diameter Poros
  1. Tekanan Kecepatan

PV = P.v =0,016.8,635=0,

PV ( PV ) a

  1. Daya yang diserap
PH=
M

2

. W .V

=3,1 Kw

  1. 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)
  1. Tegangan Geser Ijin

τaa =

σu

Sf 1. Sf 2.9,

  1. Momen Rencana

τa =

5

n 1

5

  1. Diameter Poros

d ¿ ¿

d ¿ ¿

d 33,

d=

  1. Beban bantalan

Wo = √¿ ¿

  1. Beban Rencana

W = Wo. fw

  1. Bahan Bantalan = Kuningan

Pa=(0,7-2) Kg/ mm

2

(Pv)a = (0,1-0,2)

L/d = (2-3)

  1. Panjang Bantalan
L

π. w. n

( Pv ) a. 60000

L
L ≥ 623,
L=

L/d = 624/240 =2,6 baik

  1. Tekanan yang terjadi
P=

w

L. d

− 3

P ≤ ( 0,7− 2 )

− 3

baik

  1. Kecepatan
V =

π. d. n

m

s