Sabuk-v_muhammad Nurrizal Zabawi_160511609258.docx

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Nama

: Muhammad Nurrizal Zabawi

NIM

: 160511609258

Prodi / Off

: PTM A2 2016

Perencanaan Sabuk V

1.

P = 3 kW; n1 = 1450 rpm ; n2 = 850; i = 1450/850 = 1,7 ; C = 300 mm

2.

Fc = 1,0

3.

Pd = 1,0 x 3 = 3 kW

4.

T1 = 9,74 x 105 x (3/1450) = 2015,17 kg.mm T2 = 9,74 x 105 x (3/850) = 3437,64 kg.mm

5.

Bahan poros S30C-D, 𝜎b = 58 kg/mm2, Sf1 = 6, Sf2 = 2 (dengan alur pasak),

πœπ‘Ž =58/(6x2) = 4,83 kg/mm2, Kt = 2 untuk beban tumbukan, dan Cb = 2 untuk lenturan. 6.

ds1 = {(5,1/4,83) x 2 x 2 x 2015,17}1/3 = 20,42 οƒ  21 mm (baik) ds2 = {(5,1/4,83) x 2 x 2 x 3237,64}1/3 = 23,9 οƒ  24 mm (baik)

7.

Penampang sabuk-V: tipe A

8.

dmin = 95 mm

9.

dp =95 x 1,7 = 161,5 mm dkepala = 95 + 2 x 4,5 = 104 mm Dp = 161,5 + 2 x 4,5 = 170,5 mm 5

5

dB = 3 ds1 + 10 = 3 x 21 + 10 = 45 mm 5

5

DB = 3 ds1 + 10 = 3 x 24 + 10 = 50 mm 10. v =

πœ‹ . 𝑑𝑝 . 𝑛1 60 .1000

=

πœ‹ . 161,5 . 1450 60 .1000

= 12,26 m/s

11. 12,26 m/s < 30 m/s, Baik. 12. 300 -

104+170,5 2

= 162,75 mm, Baik.

13. Dipakai tipe standar. 𝐢

1

P0 = (dp.n) {(C1 (dp.n)-0,09 – (𝑑𝑝2 ) – C3 (dp.n)2} -C2.n x{1- 𝐢 } 5

𝐢2

1

= (120,7 x 1,450) {(C1 (dp.n)-0,09 – (𝑑𝑝) – C3 (dp.n)2} -C2.n x{1- 𝐢 } 5

πœ‹

1

14. L = 2C + 2 (𝑑𝑝 + 𝐷𝑝) + 4𝐢 (Dp – dp)2 πœ‹

= 2 x 300 + 2 (120,7 + 129,7) + 4 .

1 300

(129,7 – 120,7)2 = 453,33 mm

15. Nomor nominal sabuk-V: No. 18 L = 457 16. b = 2L – 3,14 (Dp + dp) = 2 x 457 – 3,14 (129,7 + 120,7) = 127,74 mm C=

𝑏+ βˆšπ‘2 βˆ’8(π·π‘βˆ’π‘‘π‘)2

=

8

17. πœƒ = 1800 18. 𝑁 = 𝑃

𝑃𝑑

0

. 𝐾0

57 (π·π‘βˆ’π‘‘π‘) 𝐢

=𝑃

0

3 . 0,97

127,74 + √(127,74)2 βˆ’8(129,7βˆ’120,7)2 8

= 1800 -

57 (129,7βˆ’120,7) 31,6

= 31,6 mm

= 163,770 οƒ πΎπœƒ = 0,97

= οƒ  n buah

19. βˆ†πΆπ‘– = 20 mm, βˆ†πΆπ‘‘ = 25 mm 20. Tipe A, No. 18, n buah, dk = 80 mm, Dp = 129,7 mm Lubang poros x mm, x mm π‘šπ‘š Jarak sumbu poros 31,6+25 βˆ’20 π‘šπ‘š

Jika dipakai sabuk sempit: 7) Penampang sabuk-V: 9) dp = x mm, Dp = x 10) v =

πœ‹ . 𝑑𝑝 . 𝑛1 60 .1000

=x

11) x < 35, baik. 12) C -

(𝑑𝑝+𝐷𝑝+ 60 .1000

= x m/s 𝐢

1

13) Po = (dp.n) {(C1 (dp.n)-0,09 – (𝑑𝑝2 ) – C3 (dp.n)2} -C2.n x{1- 𝐢 } = 5

πœ‹

1

14) L = 2C + 2 (𝑑𝑝 + 𝐷𝑝) + 4𝐢 (Dp – dp)2 = 15)

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