Tài liệu shi20396 chương 14 ppt

Chapter 14

14-1

d = N

P = 22

6 = 3.667 in

Table 14-2: Y = 0.331

V = πdn

12 = π(3.667)(1200)

12 = 1152 ft/min

Eq. (14-4b): Kv = 1200 + 1152

1200 = 1.96

Wt = T

d/2 = 63 025H

nd/2 = 63 025(15)

1200(3.667/2) = 429.7 lbf

Eq. (14-7):

σ = KvWt P

FY = 1.96(429.7)(6)

2(0.331) = 7633 psi = 7.63 kpsi Ans.

14-2

d = 16

12 = 1.333 in, Y = 0.296

V = π(1.333)(700)

12 = 244.3 ft/min

Eq. (14-4b): Kv = 1200 + 244.3

1200 = 1.204

Wt = 63 025H

nd/2 = 63 025(1.5)

700(1.333/2) = 202.6 lbf

Eq. (14-7):

σ = KvWt P

FY = 1.204(202.6)(12)

0.75(0.296) = 13 185 psi = 13.2 kpsi Ans.

14-3

d = mN = 1.25(18) = 22.5 mm, Y = 0.309

V = π(22.5)(10−3)(1800)

60 = 2.121 m/s

Eq. (14-6b): Kv = 6.1 + 2.121

6.1 = 1.348

Wt = 60H

πdn = 60(0.5)(103)

π(22.5)(10−3)(1800) = 235.8 N

Eq. (14-8): σ = KvWt

FmY = 1.348(235.8)

12(1.25)(0.309) = 68.6 MPa Ans.

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Chapter 14 361

14-4

d = 5(15) = 75 mm, Y = 0.290

V = π(75)(10−3)(200)

60 = 0.7854 m/s

Assume steel and apply Eq. (14-6b):

Kv = 6.1 + 0.7854

6.1 = 1.129

Wt = 60H

πdn = 60(5)(103)

π(75)(10−3)(200) = 6366 N

Eq. (14-8): σ = KvWt

FmY = 1.129(6366)

60(5)(0.290) = 82.6 MPa Ans.

14-5

d = 1(16) = 16 mm, Y = 0.296

V = π(16)(10−3)(400)

60 = 0.335 m/s

Assume steel and apply Eq. (14-6b):

Kv = 6.1 + 0.335

6.1 = 1.055

Wt = 60H

πdn = 60(0.15)(103)

π(16)(10−3)(400) = 447.6 N

Eq. (14-8): F = KvWt

σmY = 1.055(447.6)

150(1)(0.296) = 10.6 mm

From Table A-17, use F = 11 mm Ans.

14-6

d = 1.5(17) = 25.5 mm, Y = 0.303

V = π(25.5)(10−3)(400)

60 = 0.534 m/s

Eq. (14-6b): Kv = 6.1 + 0.534

6.1 = 1.088

Wt = 60H

πdn = 60(0.25)(103)

π(25.5)(10−3)(400) = 468 N

Eq. (14-8): F = KvWt

σmY = 1.088(468)

75(1.5)(0.303) = 14.9 mm

Use F = 15 mm Ans.

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362 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design

14-7

d = 24

5 = 4.8 in, Y = 0.337

V = π(4.8)(50)

12 = 62.83 ft/min

Eq. (14-4b): Kv = 1200 + 62.83

1200 = 1.052

Wt = 63 025H

nd/2 = 63 025(6)

50(4.8/2) = 3151 lbf

Eq. (14-7): F = KvWt P

σY = 1.052(3151)(5)

20(103)(0.337) = 2.46 in

Use F = 2.5 in Ans.

14-8

d = 16

5 = 3.2 in, Y = 0.296

V = π(3.2)(600)

12 = 502.7 ft/min

Eq. (14-4b): Kv = 1200 + 502.7

1200 = 1.419

Wt = 63 025(15)

600(3.2/2) = 984.8 lbf

Eq. (14-7): F = KvWt P

σY = 1.419(984.8)(5)

10(103)(0.296) = 2.38 in

Use F = 2.5 in Ans.

14-9 Try P = 8 which gives d = 18/8 = 2.25 in and Y = 0.309.

V = π(2.25)(600)

12 = 353.4 ft/min

Eq. (14-4b): Kv = 1200 + 353.4

1200 = 1.295

Wt = 63 025(2.5)

600(2.25/2) = 233.4 lbf

Eq. (14-7): F = KvWt P

σY = 1.295(233.4)(8)

10(103)(0.309) = 0.783 in

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