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Chapter 14
14-1
d
=
N
P =
22
6
=
3
.
667 in
Table 14-2:
Y
=
0
.
331
V
= π
12 = π
dn
(3
.
667)(1200)
12
=
1152 ft/min
Eq. (14-4 b ):
K v =
1200
1152
1200 =
+
1
.
96
W t
=
T
d
/
2 =
63 025 H
nd
/
2 =
63 025(15)
1200(3
.
667
/
2) =
429
.
7 lbf
Eq. (14-7):
σ =
K v W t P
FY =
1
.
96(429
331) =
.
7)(6)
7633 psi
=
7
.
63 kpsi Ans.
2(0
.
14-2
16
12 =
d
=
1
.
333 in, Y
=
0
.
296
V
= π
(1
.
333)(700)
12
=
244
.
3 ft/min
Eq. (14-4 b ):
K v =
1200
+
244
.
3
=
1
.
204
1200
W t
=
63 025 H
nd /
=
63 025(1
.
5)
=
202
.
6 lbf
2
700(1
.
333
/
2)
Eq. (14-7):
σ =
K v W t P
FY =
1
.
204(202
.
6)(12)
=
13 185 psi
=
13
.
2 kpsi Ans.
0
.
75(0
.
296)
14-3
d
=
mN
=
1
.
25(18)
=
22
.
5mm, Y
=
0
.
309
V
= π
(22
.
5)(10 − 3 )(1800)
60
=
2
.
121 m/s
Eq. (14-6 b ):
K v =
6
.
1
+
2
.
121
=
1
.
348
6
.
1
W t
=
60 H
π dn =
60(0
.
5)(10 3 )
=
235
.
8N
π
(22
.
5)(10 − 3 )(1800)
Eq. (14-8):
σ =
K v W t
FmY =
1
.
348(235
.
8)
=
68
.
6MPa Ans.
12(1
.
25)(0
.
309)
672655620.002.png
350
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
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. (14-6 b ):
K v =
6
.
1
+
0
.
7854
=
1
.
129
6
.
1
W t
=
60 H
π dn =
60(5)(10 3 )
=
6366 N
π
(75)(10 − 3 )(200)
Eq. (14-8):
σ =
K v W t
FmY =
129(6366)
60(5)(0
.
=
82
.
6MPa Ans.
.
290)
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. (14-6 b ):
K v =
6
.
1
+
0
.
335
=
1
.
055
6
.
1
W t
=
60 H
π dn =
60(0
.
15)(10 3 )
=
447
.
6N
π
(16)(10 − 3 )(400)
Eq. (14-8):
F
=
K v W t
σ mY =
1
.
055(447
.
6)
=
10
.
6mm
150(1)(0
.
296)
From Table A-17, use F =
11 mm Ans.
14-6
d =
1
.
5(17)
=
25
.
5mm, Y
=
0
.
303
V
= π
(25
.
5)(10 − 3 )(400)
60
=
0
.
534 m/s
Eq. (14-6 b ):
K v =
6
.
1
+
0
.
534
=
1
.
088
6
.
1
W t
=
60 H
π dn =
60(0
.
25)(10 3 )
=
468 N
π
(25
.
5)(10 − 3 )(400)
Eq. (14-8):
F
=
K v W t
σ mY =
088(468)
75(1
1
.
=
14
.
9mm
.
5)(0
.
303)
Use F =
15 mm Ans.
1
672655620.003.png
Chapter 14
351
14-7
d
=
24
5
=
4
.
8in, Y
=
0
.
337
V
= π
(4
.
8)(50)
12
=
62
.
83 ft/min
Eq. (14-4 b ):
K v =
1200
+
62
.
83
=
1
.
052
1200
W t
=
63 025 H
nd /
2
=
63 025(6)
50(4
.
8
/
2)
=
3151 lbf
Eq. (14-7):
F
=
K v W t P
σ Y =
1
052(3151)(5)
20(10 3 )(0
.
=
2
.
46 in
.
337)
Use F =
2
.
5in Ans.
14-8
16
5 =
d =
3
.
2in, Y
=
0
.
296
V
= π
(3
.
2)(600)
12
=
502
.
7 ft/min
Eq. (14-4 b ):
K v =
1200
+
502
.
7
=
1
.
419
1200
W t
=
63 025(15)
600(3
.
2
/
2) =
984
.
8 lbf
Eq. (14-7):
F =
K v W t P
σ
Y =
1
.
419(984
.
296) =
2
.
38 in
10(10 3 )(0
.
Use F
=
2
.
5in 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. (14-4 b ):
K v =
1200
+
353
.
4
=
1
.
295
1200
W t
=
63 025(2
.
5)
=
233
.
4 lbf
600(2
.
25
/
2)
Eq. (14-7):
F
=
K v W t P
σ Y =
1
.
295(233
.
4)(8)
=
0
.
783 in
10(10 3 )(0
.
309)
8)(5)
672655620.004.png
352
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Using coarse integer pitches from Table 13-2, the following table is formed.
P
d
V
K v
W t
F
2
9.000
1413.717
2.178
58.356
0.082
3
6.000
942.478
1.785
87.535
0.152
4
4.500
706.858
1.589
116.713
0.240
6
3.000
471.239
1.393
175.069
0.473
8
2.250
353.429
1.295
233.426
0.782
10
1.800
282.743
1.236
291.782
1.167
12
1.500
235.619
1.196
350.139
1.627
16
1.125
176.715
1.147
466.852
2.773
Other considerations may dictate the selection. Good candidates are P
=
8( F
=
7
/
8in)
and P
=
10 ( F
=
1
.
25 in)
.
Ans.
14-10 Try m
=
2mmwhich gives d
=
2(18)
=
36 mm and Y
=
0
.
309
.
V
= π
(36)(10 − 3 )(900)
60
=
1
.
696 m/s
Eq. (14-6 b ):
K v =
6
.
1
+
1
.
696
=
1
.
278
6
.
1
W t
=
60(1
.
5)(10 3 )
=
884 N
π
(36)(10 − 3 )(900)
Eq. (14-8):
F
=
278(884)
75(2)(0
1
.
=
24
.
4mm
.
309)
Using the preferred module sizes from Table 13-2:
m
d
V
K v
W t
F
1.00
18.0
0.848
1.139
1768.388
86.917
1.25
22.5
1.060
1.174
1414.711
57.324
1.50
27.0
1.272
1.209
1178.926
40.987
2.00
36.0
1.696
1.278
884.194
24.382
3.00
54.0
2.545
1.417
589.463
12.015
4.00
72.0
3.393
1.556
442.097
7.422
5.00
90.0
4.241
1.695
353.678
5.174
6.00
108.0
5.089
1.834
294.731
3.888
8.00
144.0
6.786
2.112
221.049
2.519
10.00
180.0
8.482
2.391
176.839
1.824
12.00
216.0
10.179
2.669
147.366
1.414
16.00
288.0
13.572
3.225
110.524
0.961
20.00
360.0
16.965
3.781
88.419
0.721
25.00
450.0
21.206
4.476
70.736
0.547
32.00
576.0
27.143
5.450
55.262
0.406
40.00
720.0
33.929
6.562
44.210
0.313
50.00
900.0
42.412
7.953
35.368
0.243
Other design considerations may dictate the size selection. For the present design,
m
2mm( F
=
25 mm) is a good selection. Ans.
=
672655620.005.png
Chapter 14
353
14-11
d P =
22
6 =
3
.
667 in, d G =
60
6 =
10 in
V
= π
(3
.
667)(1200)
12
=
1152 ft/min
Eq. (14-4 b ):
K v =
1200
1152
1200 =
+
1
.
96
W t
=
63 025(15)
1200(3
.
667
/
2) =
429
.
7 lbf
2100 psi [Note: using Eq. (14-13) can result in wide variation in
C p due to wide variation in cast iron properties]
Table 14-8: C p =
Eq. (14-12): r 1 =
3
.
667 sin 20°
2
=
0
.
627 in, r 2 =
10 sin 20°
2
=
1
.
710 in
C p K v W t
F cos
1
r 1 +
1
r 2
1 / 2
Eq. (14-14):
σ C =−
φ
2100 1
1
0
1 / 2
.
7)
2 cos 20°
96(429
.
1
=−
627 +
.
1
.
710
=−
65
.
6(10 3 ) psi
=−
65
.
6 kpsi Ans.
14-12
16
12 =
48
12 =
d P =
1
.
333 in, d G =
4in
V
= π
(1
.
333)(700)
12
=
244
.
3 ft/min
Eq. (14-4 b ):
K v =
1200
+
244
.
3
=
1
.
204
1200
W t
=
63 025(1
.
5)
2) =
202
.
6 lbf
700(1
.
333
/
Table 14-8: C p = 2100 √ psi (see note in Prob. 14-11 solution)
Eq. (14-12): r 1 =
1
.
333 sin 20°
2
=
0
.
228 in, r 2 =
4 sin 20°
2 =
0
.
684 in
Eq. (14-14):
2100 1
1
0
1 / 2
.
6)
F cos 20°
.
1
σ C =−
228 +
=−
100(10 3 )
.
0
.
684
F
=
2100
100(10 3 )
2 1
.
202(202
.
6)
1
0
228 +
1
=
0
.
668 in
cos 20°
.
0
.
684
Use F = 0 . 75 in Ans.
202(202
672655620.001.png

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