TUGAS

DESIGN ELEMEN MESIN 1

DISUSUN OLEH

Nama : Donny Marfin Surbakti

Nim : 120401116

UNIVERSITAS SUMATERA UTARA

T.A (2013/2014)

SHAFT(POROS)

1.Dik : N = 400 rpm

P = 10 kN

τ = 40 Mpa

Dit : d = ? (diameter poros)

Solusi : P =2πTN60

10×1000 = 2π ×T ×400

60

T = 238,8 N.m

T = τd /2 × π32 × d4

T = τ × π16 × d3

238,8 = 40 × 106 × 3,1416 × d3

3,12 ×10-2 m = d

d = 31,2 mm dikatakan 35 mm

2. Dik : Poros berongga

P = 600 kW = 600 × 103 W

N = 500 rpm

do = 2di

T = 20%

Dit : di = ? (diameter dalam)

do = ? (diameter luar)

Solusi : P =2πTN60 , Tɤ =

τr

T = τd o 2

π32 (do

4 – di

4)

T = τ π

16d o (do4

– di4)

6 x 105 W = 2π xT x 50060

T = 360x 100000

2π x 500

T = 0,115 x 105 N.m

Taug = 0,115 x 105

= 11,5 x 103

T = 20100 x 0,115 x105

= 2,3 x 103 N.m

13,8 x 103 = 62,4 x10 6 x 3,1416d o (16do

4 – 16di

4)

2,253 x 10-3 di = 15 di4

di = 0,053m =53,16mm dikatakan 50mm

d0 = 2 x 50mm = 100mm

3. Dik : T = 4750 N.m

τ = 50 Mpa = 50 x 106 Pa

k = did 0 = 0,4

Dit : di ? (Diameter poros dalam)

do ?(Diameter poros luar)

Solusi : T = τ π16 do3 ( 1- k4)

T 16τπ x

1(1−k 4) = do

3

4750 .1650.10 00000.3,14 x

1(1−o ,4) = do

3

0,000484

0,9744 = do3

7,9 x 10-2 = do

do = 79,2mm

4. Dik : P = 10 kW =10 x 103 W

N = 2000rpm

di = 30mm

do = 38mm

Dit : τmax ?

Solusi : P =2πTN60

104 = =2π x2000T60 30mm

T = 47,7 Nm

T = π32 x τmax x

(d o 4 – d i 4)d o 4mm

47,7 x 32x 0,0383,14 (208,5−81 )0,0000001 = τmax

τmax = 14,5 Mpa

5. Dik : σ y = 700 Mpa

M = 10 x 10-3Nm

T = 30 x 103Nm

F s = 2

Dit : d?

Solusi : σ=¿ σ yF s = 7002 = 350 MPa

Te = √M .M+T .T

√1000000+3000000

= 3,2 x 104

Me = ½ (104 + 3,2.104) = 2,1 x 104

Me = π32 σx d3

2,1 x 104 = 3,14/32 xσ x d3

d = 84,8mm

6.Dik : N = 200rpm

P = 20 kW = 20 x 103W

τ = 42 Mpa

W = 900N

Dit : d ?

Solusi : 900N

A B

3m

P =2πTN60

2 x 104 = 2π 200T60

T = 955 Nm

M = WxL

4 = 900 x3

4 = 675 Nm

Te = √M .M+T .T

= √955.955+675.675

Te = 1169,4

1169,4 = π16 x τ x d3

d = 3√18710,4÷3,14 x 42x 1000000

d = 0,052m

d = 52,1mm

Me = ½ (675+1169,4) = π32 x σx d3

922,2 = π32 x 56 x 106 x d3

d = 0,0551m

d = 55mm

Dari perhitungan diatas,maka disimpulkan diameter poros adalah 55mm.

7. dik ; dpulley = 400 mm

Lporos =500 mm

Rasio torsi = 2,5

τ = 80 x 106 Pa

P = 45 x 103 W

N = 900 rpm

Dit ; dporos =. . . .?

Solusi

100mm 100mm500mm

A BI II

I dan II

T1 T3

d =400mm

τ = P x602.π .N =

450.000 x602. π .900 = 478 Nm

Tinjau pulley I

T 1T 2

= 2,5 ; T1 = 2,5 T2

T1 = 2,5 (4780 ) = 11.950 N

τ =( T1 – T2 )r

τ =( 2,5 T2 – T2 )r

478 = 0,5 T2 x 0,2

T2 =4780 N

Tinjau pulley II

T 3T 4

= 2,5 ; T3 = 2,5 T4

T3 = T2 = 4780 N T4 = T1 = 11.950 N

Beban pada pulley I

WI = T1 + T2

= 11.950 + 4.780

= 16.730 N

Beban pada pulley II

WII = T3 + T4

= 11.950 + 4.780

= 16.730 N

∑ fx = 0 ∑MA = 0

WI + WII = RA + RB WI . 0,1 + WII . 0,4 = RB x 0,5

(16.730)2 = RA + RB 1.673 + 6692 = 0,5 RB

RA + RB = 33.460 RB = 16.730

Momen pada pulley I

16.730 x 0,1 = 1.673 Nm

Momen pada pulley II

16.730 x 0,1 = 1.673 Nm

τe = √M 2+T 2

= √(1.673)2+(478)2

= 1.739 Nm

τe = π

16 x τ x d3

1.739 = 3,1416 x 80 x 106 x d3

RA RB

WIIWI

d = 0,048 m = 48 mm dikatakan 50 mm

8 . dik ; shaft material 40℃ dan steel

T = 15 Nm

σyt = 380 MPa

Ts = 1,5

Dit ; d =…..?

Solusi ;

W = T1 + T2 = 2000 N

A B

pulley shaft

200mm 200mm

300mm

T1 = 1200 N

T2 = 800 N

T1,T2

∑ f = 0

RAU + RBU = 2000 N

∑MA = 0

RBU = 2000x 0,20,4 = 1000 N

RAU = 2000 – 1000 = 1000 N

T = (T1 – T2) x 0,3 = 400 x 0,3 = 120 Nm

T = 15 Nm = Ft x 0,3

Ft = 50 N gaya yang bekerja pada sumbu x

RAU RBU

W

∑ f = 0

RAH + RBH = 50

∑MA = 0

RBH x 0,4 + Ft x 0,2 = 0

0,4 RBH = 10

RBH = 25 RAH = 50 – 25 = 25 N

MU = 2000x 0,44 = 200

MH = 50x 0,44 = 5

M = √(200)2+(5)2 = 200

τe = √M 2+T 2

= √(120)2+(200)2 = 233 Nm

τe = π

16 x 380x 1.000 .000x d xd xd1,5

d = 0,0167 m

RAT RBT

Ft

d = 16,7 mm dikatakan 20 mm

9 . dik ; P = 30 kW = 30 x 103 W dpulley = 1 m

N = 160 rpm wpulley = 1600 N

T1 = 2,5 T2 τ = 56 MPa

Dit ; d = . . .?

Solusi ;

P = 2πTN60

T ¿ 3x 100.000 x602. π .160 = 1791 Nm

T = (T1 – T2)r

1791 = (2,5 T2 – T2 ) X 0,5

T2 = 2388 N T1 = 4776 N

A r

T1 T2 w

w1

150mm

W = T1 + T2 = 7164 N

Wtotal = 7164 + 1600 = 8764 N

RAU = 8764 N

M = 8764 x 0.15 = 1314,6 Nm

τe = √M 2+T 2

= √(1314,6)2+(1791)2

= 2221,6 Nm

τe = π

16 x τ x d3

2221,6 = π

16 x 56 x 106 x d3

d = 0,0586 m

d = 58,6 mm dikatakan 60 mm

Wtotal

RAU

10 . dik ; P = 75 Kw 75 x 103 W P1 = 2P2

N = 500 rpm Q1 = 2Q2

Pk = 220 mm τ = 45 MPa

R0 = 160 mm

Dit ; d = . . .?

Solusi ;

P = 2πTN60

T = 60x 75 x10002.3,14.500 = 1432,12 Nm

T = (Q1 – Q2)RC

1432,12 = (2Q2 – Q2).0,22

Q2 = 6514 N Q1 = 13028 N

T = (P2 – P1)R0

A BD

C

600mm

500mm

Q1 Q2

300mm

P1

P2

C

1432,12 =( P2 -2P2).0,16

P2 = -8957 N P1 = -17914 N

W pada pulley C = T1 + T2 =19542 N

W pada pulley D = -26871 N

Gaya – gaya vertical yang terjadi pada poros ;

∑ f Y = 0

RAU + RBU =19542

∑MA = 0

19542 . 0,6 = RBU . 1,2

RBU = 9771 RAU = 9771

MCU =19542x 1,2

4 = 5862 Nm

MDU = 9771 x 0,3 = 2931,3 Nm

Gaya – gaya horizontal yang terjadi pada poros

RAu RBu

wc

∑ f X = 0 ∑MA = 0

-RAH + RBH = ND RBH . 1,2 = 26871 . 1,5

-RAH + RBH = 26871 RBH = 33588,75 N

RAH = 6717,75 N

MCH = 6717,75 x 0,3 = 4030,65 N

MDH = 33588,75 x 0,3 = 10076,6 N

MC = √MCH 2+MCU 2

= √4030,652+58622

= 7114 Nm

MD = √MDH 2+MDU 2

= √10076,62+2931,32

= 10494 Nm

MD ¿MC

τ e = √104942+74332

RAH RBH

wD

C D

= 10591 Nm

τ e = π16 x 45 x 106 x d3

10591 = 3,1416 x 45 x 106 x d3

d = 0,106 m

= 106 mm dikatakan 110 mm

KEY AND COUPLING

1 . Dik ; dporos = 80 mm

Power of max shear stress = 63 MPa

Wide key , w = 20 mm

Syarat σ key tidak melebihi 42 MPa t

Dit ; length key = ….? d

Solusi ; w

1 . Tinjau poros 2 . tinjau pasak

T = π16 xτ s x d3 torsi ditransmisikan ke pasak

= π16 x 63 x 803 T = Ft x d2

= 6,330240 Nmm = w x l xτ d2

= 6,33 x 106 Nmm 6,33 x 106 = l x20 x63 x 802

l = 125,5 mm

2 . dik ; dporos = 30 mm

τmax = 80 MPa

l = 4 w

σ key tidak melebihi 50 MPa

Dit = dimensi poros …?

Solusi ;

1 . tinjau poros 2 . tinjau poros

T = π16 x τmax x d3 torsi ditransmisikan ke pasak

T = π16 x 80 x 303 T = Ft x d2

T = 423.900 Nmm T = w x l xτx d2

T = 4,24 X 105 Nmm 4,24 x 105 = l4 x 50 x 302

l = 47,5 mm

3 . dik ; dporos = 25 mm

N = 600 rpm

P = 30 Kw

σ yt danτ yt = 650 MPa dan 353 MPa

F.s = 3

Dit ; carilah pasak yang sesuai dengan poros …!

Solusi ;

T = P X602πN

T = 30.000 X 602.π .600 = 477,7 Nm = 478 Nm = 478 x 103 Nmm

τ= 3532x 3 = 58,83 MPa

Torsi dipindahkan ke pasak melalui poros

T = X l x w d2 dimana w = d4 = 254

478 x 103 = 58,83 x l x 254x

252

l = 102 mm

kesimpulan ; setelaj torsi dan tegangan telah di perhitungkan,maka poros membutuhkan pasak dengan lebar = 6,25 mm dan panjang = 102mm

4 . dik ; P = 40 kW = 40.000 W , muff coupling

N = 120 rpm

τ s = τ k = 30 MPa

σ s = σ k = 80 MPa

τ = 15 MPa

Asumsikan max torsi = 25% T + T

Dit ; design dan periksa ….!

Solusi ;

1 . tinjau poros

T =P X602πN = 40 X 1000 x60

2 . π .120 =3185 x 105 Nmm

Tmax = 0,25 T + T = 0,25 X 3185 X 105 + 3185 X 105

= 3981,25 X 103 Nmm

Tmax = 3981,25 x 103 = π16x 30x d3

d = 87,7 mm berdasarkan table diameter poros yang tersedia adalah d = 90 mm

2 . design muff

D = 2d + 13 =

= 2 x 90 + 13 = 180 +13

= 193 mm

Berdasarkan table diameter muff ( D ) yang tersedia adalah D = 195 mm

L = 3,5 X d = panjang muff kopling

L = 3,5 x 90 = 315 mm

Pengecekan tegangan T = π16 X τ c (D4 – d4)/D

3981,25 X 103 = π16 X τ c (1954 – 904) / 195

τ c = 2,866 MPa

Kesimpulan ; hasil perhitungan tegangan yaitu 2,866 MPa adalah lebih kecil dari tegangan geser yang diketahui yaitu 15 MPa ,

2,866 MPa < 15 MPa muff coupling dalam keadaan aman

3 . design key

Berdassarkan table standard untuk poros ukuran diameter 90 mm harus menggunakan pasak berukuran w = 28 mm t = 16 mm l = 315 mm

Pengecekan tegangan yang terjadi pada pasak

a . T = t2 x l x σ c x d2 3981,25 x 103 = 162 x 315 x σ c x 90

2

σ c = 35,1 MPa

b . T = w x l x τ x d2 3981,25 x 103 = 28 x 315 xτ x 902

τ k = 10 MPa

Kesimpulan ;tegangan penggerus σ c =35,1 MPa lebih kecil dari tegangan penggerus yang diizinkan 80 MPa tegangan geser τ k = 10 MPa

Lebih kecil dari tegangan geser yang diizinkan 30 MPa status kondisi pasak adalah aman

5 . dik ; torque T = 1300 Nm

τ s = allowable shear stress for shaft and key = 40 MPa

n = number of table = 4

σb = 70 MPa

μ = 0,3

Dit ; design dan perhitungan dimensi

Solusi ;

1 . tinjau shaft

T = π16 x τ s x d3 1300 x 103 Nmm = π16 x 40 x d3 d = 54,9 mm

Berdasarkan table diameter poros yang tersedia adalah d = 55 mm

2 . tinjau muff

Diameter muff

D = 2d + 13 = 2 x 55 + 13 = 123 mm

Berdasarkan table diameter muff coupling yang tersedi adalah

D = 125 mm

Panjang muff coupling

L = 3,5d

L = 3,5 x 55 = 192,5 mm

3 . tinjau baut

T = μ x π16π x (db)2 x σ t x n x d

1300 = 0,3 π16 π x ( db )2 x 70 x 10-3 x 4 x 55

1300 = 2,84 (db)2

Db = 21,4 mm , berdasarkan table yang tersedia ukuran baut adalah

= 24mm

BEARING

1 . for the deep groove 02 series ball bearing with R = 0,90,the design

life XD , in multiples of rating life is

XD = = 30.000x 300 x60

106

¿¿ = 540

The design radial load FD is

FD = 1,2 X 1.898 = 2.278 Kn

From Eq (11.6)

C10 = 2.278 { -[

540

0,02+

4439[ ln( 10,9 )]

1

1489¿

¿]}1/3

= 18,59 Kn

Table 11.2 choose a 02-30 mm with C10 =19,5 Kn

Eq 11-18

R = exp { -[540( 227819,5 )

3

−0,02

4439]1483 }

= 0,919

2 . for the angular contact 02 series ball bearing as described, the rating

life multiple is

XD = 50.000x 480 x 60

106

¿¿ = 1440

The design load is radial and equal to

FD = 1.4 x 610 =3.80 kN

Eq. 11-6

C10 =854 {50.000 x 480 x60

0.02+4.439 [ ln( 10.9 )]

11.483

¿

}

= 43 Kn

Table 11-2 select a 02-55 mm with C10 = 46.2 Kn

Using Eq 11-18

R = exp {-[1440( 3.8

46.2 )3

−0.02

¿4.439

¿ ]1.483 }

3 . for the straight roller 03 series bearing selection ,xD =1440 rating

lives from prob 11-2

solution ;

FD = 1.4(1650) = 10.279 kN

C10 = 10.279(14401 )3/10 = 91.1 Kn

Table 11-3 select a 03-55 mm with C10 = 102 Kn

Using Eq.11-18

R = exp {-[1440( 10.28

102 )10

3−0.02

¿4.439

¿

]1.483}

= 0.942

4 . We can choose a reliability goal of √0.90 = 0.95 for each bearing. We make the selections, find the existing reliabilities, multiply them together, and observe that the reliability goal is exceeded due to the roundup of capacity upon table entry. Another possibility is to use the reliability of one bearing, say R1. Then set the reliability goal of the second as

R2 = 0.90R1

or vice versa. This gives three pairs of selections to compare in terms of cost, geometry implications, etc.

5 . Establish a reliability goal of √0.90 = 0.95 for each bearing. For a 02-series angular contact ball bearing,

C10 = 854 {1440

0.02+4.439 [ ln( 10.9 )]

11.483

¿

}1/3

Select a 02-60 mm angular contact bearing with C10 = 55.9 Kn

RA = exp{-[1440( 3.8

55.9 )3

¿−0.024.439

¿]1.483} = 0.969

For a 03-series straight-roller bearing,

C10 = 10.279{1440

0.02+4.439 [ ln( 10.95 )]

11.483

¿

}3/10 = 105.2 Kn

Select a 03-60 mm straight-roller bearing with C10 = 123 kN.

RB = exp { - [1440( 10.28

123 )3

10¿−0.024.439

¿

]1.483} = 0.977

The overall reliability is R = 0.969(0.977) = 0.947, which exceeds the goal. Note, usingRA from this problem, and RB from Prob. 11-3, R = 0.969(0.942) = 0.913, which stillexceeds the goal. Likewise, using RB from this problem, and RA from Prob. 11-2,R = 0.927(0.977) = 0.906.The point is that the designer has choices. Discover them before making the selection decision.Did the answer to Prob. 11-4 uncover the possibilities?

6 . The reliability of the individual bearings is R =√0.999= 0.9995

From staticsRy

O = - 163.4 N, RzO = 107 N, RO = 195 N

RyE = - 89.2 N, RZ

E = -174.4 N, RE =196 N

XD =60000 (1200 )(60)

106

¿¿ = 4320

C10 ={ 4340

0.02+4.439 [ ln( 10.9995 )]

11.483

¿

}1/3

=8.9 kNA 02-25 mm deep-groove ball bearing has a basic load rating of 14.0 kN which is ample.An extra-light bearing could also be investigated.

7. Choose a 02-series ball bearing from manufacturer #2, having a service factor of 1. For

y

Ry

O

Rz

O

O

z

A

C

FC

x

E

300

400

150

RYE

FY

A

Fz

A

Fr = 8 kN and Fa = 4 kN

XD = 5000 (900 )(60)

106

¿¿ = 270

Eq 11-5

C10 = 8 { 270

0.02+4.439 [ ln( 10.09 )]

11.483

¿

}1/3

Trial #1: From Table (11-2) make a tentative selection of a deep- groove 02-70 mm with C0 = 37.5 kN.

FaCD = 437.5 = 0.107

TABLE 11-1

Fa/(V Fr ) = 0.5 > e X2 = 0.56, Y2 = 1.46

Eq. (11-9): Fe = 0.56(1)(8) + 1.46(4) = 10.32 Kn Eq. (11-6): For R = 0.90,C10 = 10.32( 270

1 )1/3 = 66.7 kN > 61.8 kNTrial #2: From Table 11-2 choose a 02-80 mm having C10 = 70.2 and C0 = 45.0.Check:

FaCD = 445 = 0.089

Table 11-1: X2 = 0.56, Y2 = 1.53 Fe = 0.56(8) + 1.53(4) = 10.60 kN Eq. (11-6): C10 = 10.32( 270

1 )1/3 = 66.7 kN > 61.8 kN ∴ Selection stands. Decision: Specify a 02-80 mm deep-groove ball bearing.\

8 . For the straight cylindrical roller bearing specified with a service factor of 1, R = 0.90 and Fr = 12 kN

XD = 4000 (750 )(60)

106

¿¿ = 180

C10 = 12 ( 1801 )1/3 = 57 kN

9 . Given:

Fr A = 560 lbf or 2.492 kN

Fr B = 1095 lbf or 4.873 kN

Trial #1: Use KA = KB = 1.5 and from Table 11-6 choose an indirect mounting.

0.47 FrAKA < ? > 0.47 FrB

KB – (-1)(0)

0.47(2.492)1.5 < ? > 0.47(4.873)

1.5

0.781 < 1.527 Therefore use the upper line of Table 11-6.

FaA = FaB = 0.47 FrBKB = 1.527 Kn

PA = 0.4Fr A + KAFaA = 0.4(2.492) + 1.5(1.527) = 3.29 kN

PB = Fr B = 4.873 kN

Fig. 11-16: fT = 0.8Fig. 11-17: fV = 1.07

Thus, a3l = fT fV = 0.8(1.07) = 0.856Individual reliability: Ri =√0.9 = 0.95Eq. (11-17):

(C10)A = 1.4(3.29)[ 40000 (400 )(60)

4.48 (0.856 ) (1−0.95 )2/3(90)¿¿ ]0.3 = 11.40 kN

(C10)B = 1.4(4.873) [ 40000 (400 )(60)

4.48 (0.856 ) (1−0.95 )2/3(90)¿¿ ]0.3 = 16.88 kN

From Fig. 11-15, choose cone 32 305 and cup 32 305 which provide Fr = 17.4 kN andK = 1.95. With K = 1.95 for both bearings, a second trial validates the choice of cone32 305 and cup 32 305. Ans.

10 . Shafts subjected to thrust can be constrained by bearings, one of which supports the thrust.

The shaft floats within the endplay of the second (Roller) bearing. Since the thrust forcehere is larger than any radial load, the bearing absorbing the thrust is heavily loaded comparedto the other bearing. The second bearing is thus oversized and does not contributemeasurably to the chance of failure. This is predictable. The

reliability goal is not √0.99 but 0.99 for the ball bearing. The reliability of the roller is 1. Beginning here saves effort.Bearing at A (Ball)

Fr = (362 + 2122)1/2 = 215 lbf = 0.957 kNFa = 555 lbf = 2.47 kNTrial #1:

Tentatively select a 02-85 mm angular-contact with C10 = 90.4 kN and C0 = 63.0 kN. FaCD = 2.47

63 = 0.0392

XD = 25000 (600 )(60)

106

¿¿ = 900

Table 11-1: X2 = 0.56, Y2 = 1.88

Fe = 0.56(0.957) + 1.88(2.47) = 5.18 kNFD = fAFe = 1.3(5.18) = 6.73 kN

C10 = 6.73 { 900

0.02+4.439 [ ln( 10.99 )]

11.483

¿

}1/3

= 107.7 kN > 90.4 kN

Trial #2:Tentatively select a 02-95 mm angular-contact ball with C10 = 121 kN and C0 = 85 kN.

FaCD = 2.4785 = 0.029

Table 11-1: Y2 = 1.98Fe = 0.56(0.957) + 1.98(2.47) = 5.43 kNFD = 1.3(5.43) = 7.05 kN

C10 = 7.05 { 900

0.02+4.439 [ ln( 10.99 )]

11.483

¿

}1/3

= 113 kN < 121 kNSelect a 02-95 mm angular-contact ball bearing. Ans.Bearing at B (Roller): Any bearing will do since R = 1. Let’s prove it. From Eq. (11-18)When

( af FDC10 )3 XD < XO R = 1

The smallest 02-series roller has a C10 = 16.8 kN for a basic load rating.

( 0.42716.8 )3 (900) < ? > (0.02)

0.0148 < 0.02 ∴ R = 1

Spotting this early avoided rework from√0.99 = 0.995.Any 02-series roller bearing will do. Same bore or outside diameter is a common choice.(Why?)