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Chapter 11
11-1 For the deep-groove 02-series ball bearing with R
=
0.90, the design life x D , in multiples
of rating life, is
x D
=
30 000(300)(60)
10 6
=
540 Ans.
The design radial load F D is
F D
=
1
.
2(1
.
898)
=
2
.
278 kN
From Eq. (11-6),
C 10 =
2
.
278
540
9)] 1 / 1 . 483 1 / 3
0
.
02
+
4
.
439[ln(1
/
0
.
=
18
.
59 kN Ans.
Table 11-2: Choose a 02-30 mm with C 10 =
19
.
5 kN. Ans.
Eq. (11-18):
exp
1 . 483
540(2
.
278
/
19
.
5) 3
−
0
.
02
R
=
−
4
.
439
=
0
.
919 Ans.
11-2 For the Angular-contact 02-series ball bearing as described, the rating life multiple is
x D
=
50 000(480)(60)
10 6
=
1440
The design load is radial and equal to
F D
=
1
.
4(610)
=
854 lbf
=
3
.
80 kN
Eq. (11-6):
C 10 =
854
1440
9)] 1 / 1 . 483 1 / 3
0
.
02
+
4
.
439[ln(1
/
0
.
=
9665 lbf
=
43
.
0kN
Table 11-2: Select a 02-55 mm with C 10 =
46
.
2 kN. Ans.
Using Eq. (11-18),
exp
1 . 483
1440(3
.
8
/
46
.
2) 3
−
0
.
02
R
=
−
4
.
439
=
0
.
927 Ans.
672655979.005.png
290
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
11-3 For the straight-Roller 03-series bearing selection, x D =
1440 rating lives from Prob. 11-2
solution.
F D =
1
.
4(1650)
=
2310 lbf
=
10
.
279 kN
279 1440
1
3 / 10
C 10 =
10
.
=
91
.
1kN
Table 11-3: Select a 03-55 mm with C 10 =
102 kN. Ans.
Using Eq. (11-18),
exp
1 . 483
1440(10
.
28
/
102) 10 / 3
−
0
.
02
R
=
−
=
0
.
942 Ans.
4
.
439
11-4 We can choose a reliability goal of √ 0
95 for each bearing. We make the selec-
tions, 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 R 1 .
.
90
=
0
.
Then set the relia-
bility goal of the second as
90
R 1
or vice versa. This gives three pairs of selections to compare in terms of cost, geometry im-
plications, etc.
R 2 =
0
.
11-5 Establish a reliability goal of √ 0
.
90
=
0
.
95 for each bearing. For a 02-series angular con-
tact ball bearing,
C 10 =
854
1440
95)] 1 / 1 . 483 1 / 3
0
.
02
+
4
.
439[ln(1
/
0
.
=
11 315 lbf
=
50
.
4kN
Select a 02-60 mm angular-contact bearing with C 10 =
55
.
9 kN.
exp
1 . 483
1440(3
.
8
/
55
.
9) 3
−
0
.
02
R A =
−
=
0
.
969
4
.
439
For a 03-series straight-roller bearing,
C 10 =
10
.
279
1440
95)] 1 / 1 . 483 3 / 10
=
105
.
2kN
0
.
02
+
4
.
439[ln(1
/
0
.
Select a 03-60 mm straight-roller bearing with C 10 =
123 kN.
exp
1440(10
.
28
/
123) 10 / 3
−
0
.
02
1 . 483
R B =
−
=
0
.
977
4
.
439
672655979.006.png
Chapter 11
291
947, which exceeds the goal. Note, using
R A from this problem, and R B from Prob. 11-3, R
=
0
.
969(0
.
977)
=
0
.
913, which still
exceeds the goal. Likewise, using R B from this problem, and R A from Prob. 11-2,
R
=
0
.
969(0
.
942)
=
0
.
.
The point is that the designer has choices. Discover them before making the selection de-
cision. Did the answer to Prob. 11-4 uncover the possibilities?
=
0
.
927(0
.
977)
=
0
.
906
11-6 Choose a 02-series ball bearing from manufacturer #2, having a service factor of 1. For
F r
=
8kNand F a
=
4kN
x D
=
5000(900)(60)
10 6
=
270
Eq. (11-5):
8
90)] 1 / 1 . 483 1 / 3
C 10 =
270
=
51
.
8kN
0
.
02
+
4
.
439[ln(1
/
0
.
Trial #1: From Table (11-2) make a tentative selection of a deep-groove 02-70 mm with
C 0 =
37
.
5kN.
F a
C 0 =
4
37
.
5 =
0
.
107
Table 11-1:
F a
/
( VF r )
=
0
.
5
>
e
X 2 =
0
.
56, Y 2 =
1
.
46
Eq. (11-9):
F e
=
0
.
56(1)(8)
+
1
.
46(4)
=
10
.
32 kN
Eq. (11-6): For R
=
0
.
90,
32 270
1
1 / 3
C 10 =
10
.
=
66
.
7kN
>
61
.
8kN
Trial #2: From Table 11-2 choose a 02-80 mm having C 10 =
70
.
2 and C 0 =
45
.
0
.
Check:
F a
C 0 =
4
45 =
0
.
089
Table 11-1: X 2 =
0
.
56, Y 2 =
1
.
53
F e
=
0
.
56(8)
+
1
.
53(4)
=
10
.
60 kN
Eq. (11-6):
60 270
1
1 / 3
C 10 =
10
.
=
68
.
51 kN
<
70
.
2kN
Selection stands.
Decision: Specify a 02-80 mm deep-groove ball bearing. Ans.
The overall reliability is R
∴
672655979.007.png
292
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
11-7 From Prob. 11-6, x D =
270 and the final value of F e is 10
.
60 kN.
6
96)] 1 / 1 . 483 1 / 3
C 10 =
10
.
270
=
84
.
47 kN
0
.
02
+
4
.
439[ln(1
/
0
.
Table 11-2: Choose a deep-groove ball bearing, based upon C 10 load ratings.
Trial #1:
Tentatively select a 02-90 mm.
C 10 =
95
.
6, C 0 =
62 kN
F a
C 0 =
4
62 =
0
.
0645
From Table 11-1, interpolate for Y 2 .
F a /
C 0
Y 2
0.056
1.71
0.0645
Y 2
0.070
1.63
Y 2 −
1
.
71
71 =
0
.
0645
−
0
.
056
=
0
.
607
1
.
63
−
1
.
0
.
070
−
0
.
056
Y 2 =
1
.
71
+
0
.
607(1
.
63
−
1
.
71)
=
1
.
661
F e =
0
.
56(8)
+
1
.
661(4)
=
11
.
12 kN
12
96)] 1 / 1 . 483 1 / 3
C 10 =
11
.
270
0
.
02
+
4
.
439[ln(1
/
0
.
=
88
.
61 kN
<
95
.
6kN
Bearing is OK.
Decision: Specify a deep-groove 02-90 mm ball bearing. Ans.
11-8 For the straight cylindrical roller bearing specified with a service factor of 1, R
=
0.90 and
F r =
12 kN
x D =
4000(750)(60)
10 6
=
180
12 180
1
3 / 10
C 10 =
=
57
.
0kN Ans.
672655979.008.png
Chapter 11
293
11-9
y
R y O
O
R z O
T
z
P z
R A
11 1
2
"
F
P y
20
A
R z A
B
x
2 3
4
"
T
Assume concentrated forces as shown.
P z
=
8(24)
=
192 lbf
P y
=
8(30)
=
240 lbf
T x
T
=
192(2)
=
384 lbf
·
in
=−
384
+
1
.
5 F cos 20 ◦ =
0
F
=
384
=
272 lbf
1
.
5(0
.
940)
M z O =
5
.
75 P y
+
11
.
5 R A −
14
.
25 F sin 20 ◦ =
0
;
thus
5
.
75(240)
+
11
.
5 R A −
14
.
25(272)(0
.
342)
=
0
M O =−
R A =−
4
.
73 lbf
5
.
75 P z
−
11
.
5 R z A −
14
.
25 F cos 20 ◦ =
0
;
thus
−
5
.
75(192)
−
11
.
5 R z A −
14
.
25(272)(0
.
940)
=
0
R z A =−
413 lbf;
R A
=
[(
−
413) 2
+
(
−
4
.
73) 2 ] 1 / 2
=
413 lbf
F z
=
R z O +
P z
+
R z A +
F cos 20 ◦ =
0
R z O +
192
−
413
+
272(0
.
940)
=
0
F y
R z O =−
34
.
7 lbf
=
R O +
P y
+
R A −
F sin 20 ◦ =
0
R O +
240
−
4
.
73
−
272(0
.
342)
=
0
R O =−
142 lbf
R O
=
[(
−
34
.
6) 2
+
(
−
142) 2 ] 1 / 2
=
146 lbf
So the reaction at A governs.
Reliability Goal: √ 0
.
92
=
0
.
96
F D
=
1
.
2(413)
=
496 lbf
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