Baixe Soluções do Capítulo 11 do Livro Shigley's Mechanical Engineering Design, 10ª Edição e outras Exercícios em PDF para Engenharia Mecânica, somente na Docsity!
th
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
6
10
D D D
D
R
L n
x Ans
L L
L
The design radial load is
1.2 2.5 3.0 kN D
F = =
Eq. (11-9):
1/
10 1/1.
0.02 4.459 0.02 ln 1/ 0.
C
C
10 = 24.3 kN Ans.
Table 11-2: Choose an 02-35 mm bearing with C 10 = 25.5 kN. Ans.
Eq. (11-21):
3
exp 0..
R Ans
______________________________________________________________________________
11-2 For the angular-contact 02-series ball bearing as described, the rating life multiple is
6
10
D D D
D
R
L n
x
L L
L
The design radial load is
1.4 725 ( ) 1015 lbf 4.52 kN
D
F = = =
Eq. (11-9):
1/
10 1/1.
0.02 4.459 0.02 ln 1/ 0.
10 930 lbf 48.6 kN
C
Table 11-2: Select an 02-60 mm bearing with C 10
= 55.9 kN. Ans.
Eq. (11-21):
3
exp 0..
R Ans
______________________________________________________________________________
th
11-3 For the straight-roller 03-series bearing selection, x D
= 1248 rating lives from Prob. 11-
solution.
1.4 2235 3129 lbf 13.92 kN D
F = = =
3/
10
13.92 118 kN
C
Table 11-3: Select an 03-60 mm bearing with C 10
= 123 kN. Ans.
Eq. (11-21):
10/
exp 0..
R Ans
______________________________________________________________________________
11-4 The combined reliability of the two bearings selected in Probs. 11-2 and 11-3 is
R = 0.945 0.917 =0.867 Ans.
We can choose a reliability goal of 0.90 = 0.95for 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 R 1
. Then set the reliability
goal of the second as
2
1
R
R
or vice versa. This gives three pairs of selections to compare in terms of cost, geometry
implications, etc.
______________________________________________________________________________
11-5 Establish a reliability goal of 0.90 = 0.95for each bearing. For an 02-series angular
contact ball bearing,
1/
10 1/1.
0.02 4.439 ln 1/ 0.
12822 lbf 57.1 kN
C
Select an 02-65 mm angular-contact bearing with C 10 = 63.7 kN.
3
exp 0.
A
R
th
11-9 F
D
= 800 lbf, Λ D
= 12 000 hours, n D
= 350 rev/min, R = 0.
Eq. (11-3):
1/ 1/
10 6
800 5050 lbf
a
D D
D
R
n
C F Ans
L
L
______________________________________________________________________________
11-10 F
D
= 4 kN, Λ D
= 8 000 hours, n D
= 500 rev/min, R = 0.
Eq. (11-3):
1/ 1/
10 6
4 24.9 kN
a
D D
D
R
n
C F Ans
L
L
______________________________________________________________________________
11-11 F
D
= 650 lbf, n D
= 400 rev/min, R = 0.
D = (5 years)(40 h/week)(52 week/year) = 10 400 hours
Assume an application factor of one. The multiple of rating life is
6
D
D
R
L
x
L
Eq. (11-9):
1/
10 1/1.
0.02 4.439 ln 1/ 0.
C
= 4800 lbf Ans.
______________________________________________________________________________
11-12 F
D
= 9 kN, L D
8
rev, R = 0.
Assume an application factor of one. The multiple of rating life is
8
6
D
D
R
L
x
L
Eq. (11-9):
1/
10 1/1.
0.02 4.439 ln 1/ 0.
C
= 69.2 kN Ans.
______________________________________________________________________________
11-13 F
D
= 11 kips, Λ D
= 20 000 hours, n D
= 200 rev/min, R = 0.
Assume an application factor of one. Use the Weibull parameters for Manufacturer 2 in
Table 11-6.
The multiple of rating life is
th
6
D
D
R
L
x
L
Eq. (11-9):
1/
10 1/1.
0.02 4.439 ln 1/ 0.
C
= 113 kips Ans.
______________________________________________________________________________
11-14 From the solution to Prob. 3-68, the ground reaction force carried by the bearing at C is
R
C
= F
D = 178 lbf. Use the Weibull parameters for Manufacturer 2 in Table 11-6.
6
D
D
R
L
x
L
Eq. (11-10):
1/
10 1/
0 0
a
D
f D b
D
x
C a F
x θ x R
1/
10 1/1.
2590 lbf.
C
Ans
______________________________________________________________________________
11-15 From the solution to Prob. 3-69, the ground reaction force carried by the bearing at C is
R
C
= F
D = 1.794 kN. Use the Weibull parameters for Manufacturer 2 in Table 11-6.
6
D
D
R
L
x
L
Eq. (11-10):
1/
10 1/
0 0
a
D
f D b
D
x
C a F
x θ x R
1/
10 1/1.
26.1 kN.
C
Ans
______________________________________________________________________________
11-16 From the solution to Prob. 3-70, R Cz = –327.99 lbf, R Cy = –127.27 lbf
1/ 2 2
327.99 127.27 351.8 lbf
C D
R F
Use the Weibull parameters for Manufacturer 2 in Table 11-6.
6
D
D
R
L
x
L
th
11-19 From the solution to Prob. 3-79, R Az = 54.0 lbf, R Ay = 140 lbf
1/ 2 2
54.0 140 150.1 lbf A D
R = F = + =
Use the Weibull parameters for Manufacturer 2 in Table 11-6. The design speed is equal
to the speed of shaft AD ,
280 560 rev/min
F
D i
C
d
n n
d
6
D
D
R
L
x
L
Eq. (11-10):
1/
10 1/
0 0
a
D
f D b
D
x
C a F
x θ x R
3/
10 1/1.
1320 lbf.
C
Ans
______________________________________________________________________________
11-20 (a) 3 kN, 7 kN, 500 rev/min, 1. a r D
F = F = n = V =
From Table 11-2, with a 65 mm bore, C 0
= 34.0 kN.
F
a
/ C
0
From Table 11-1, 0.28 ≤ e ≤ 3.0.
a
r
F
VF
Since this is greater than e , interpolating Table 11-1 with F a
/ C
0 = 0.088, we obtain
X
2
= 0.56 and Y 2
Eq. (11-12): ( 0.56 )( 1.2) ( 7 ) ( 1.53) ( 3 ) 9.29 kN
e i r i a
F = X VF + Y F = + = Ans.
F
e
> F
r so use F e
(b) Use Eq. (11-10) to determine the necessary rated load the bearing should have to
carry the equivalent radial load for the desired life and reliability. Use the Weibull
parameters for Manufacturer 2 in Table 11-6.
6
D
D
R
L
x
L
th
Eq. (11-10):
1/
10 1/
0 0
a
D
f D b
D
x
C a F
x θ x R
1/
10 1/1.
73.4 kN
C
From Table 11-2, the 65 mm bearing is rated for 55.9 kN, which is less than the
necessary rating to meet the specifications. This bearing should not be expected to meet
the load, life, and reliability goals. Ans.
______________________________________________________________________________
11-21 (a) 2 kN, 5 kN, 400 rev/min, 1 a r D
F = F = n = V =
From Table 11-2, 30 mm bore, C 10
= 19.5 kN, C 0
= 10.0 kN
F
a
/ C
0
From Table 11-1, 0.34 ≤ e ≤ 0.38.
a
r
F
VF
Since this is greater than e , interpolating Table 11-1, with F a
/ C
0 = 0.2, we obtain X 2
0.56 and Y 2
Eq. (11-12):
0.56 1 5 1.27 2 5.34 kN e i r i a
F = X VF + Y F = + = Ans.
F
e
> F
r
so use F e
(b) Solve Eq. (11-10) for x D
1/ 10
0 0
a
b
D D
f D
C
x x x R
a F
θ
3
D
x
D
x =
6
D D D
D
R
n L
x
L
L
6 6
444 h.
D
D
D
x
Ans
n
L = = =
th
F
a
/ C
0
Table 11-1: Again, 0.22 ≤ e ≤ 0.
a
r
F
e
VF
Interpolate Table 11-1 with F a
/ C
0 = 0.032 to obtain X 2 = 0.56 and Y 2
Eq. (11-12): 0.56(1)8 1.95(2) 8.38 > e r
F = + = F
Eq. (11-10):
1/
10 1/1.
1 8.38 86.4 kN
C
The 90 mm bore is acceptable. Ans.
______________________________________________________________________________
8
8 kN, 3 kN, 1.2, 0.9, 10 rev r a D
F = F = V = R = L =
First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.
Eq. (11-12):
0.56 1.2 8 1.63 3 10.3 kN e
F = + =
e r
F > F
Eq. (11-3):
1/ 1/ 8
10 6
10.3 47.8 kN
a
D
e
R
L
C F
L
From Table 11-2, try 60 mm with C 10
= 47.5 kN, C 0
= 28.0 kN
Iterate the previous process:
F
a
/ C
0
Table 11-1: 0.28 ≤ e ≤ 0.
a
r
F
e
VF
Interpolate Table 11-1 with F a
/ C
0 = 0.107 to obtain X 2 = 0.56 and Y 2
Eq. (11-12):
0.56 1.2 8 1.46 3 9.76 kN > e r
F = + = F
Eq. (11-3):
1/
8
10 6
9.76 45.3 kN
C
From Table 11-2, we have converged on the 60 mm bearing. Ans.
______________________________________________________________________________
11-25 10 kN, 5 kN, 1, 0. r a
F = F = V = R =
Use the Weibull parameters for Manufacturer 2 in Table 11-6.
th
6
D
D
R
L
x
L
First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.
Eq. (11-12):
0.56 1 10 1.63 5 13.75 kN e
F = + =
F
e
> F
r
, so use F e
as the design load.
Eq. (11-10):
1/
10 1/
0 0
a
D
f D b
D
x
C a F
x θ x R
1/
10 1/1.
1 13.75 97.4 kN
C
From Table 11-2, try 95 mm bore with C 10
= 108 kN, C 0
= 69.5 kN
Iterate the previous process:
F
a
/ C
0
Table 11-1: 0.27 ≤ e ≤ 0.
a
r
F
e
VF
Interpolate Table 11-1 with F a
/ C
0
= 0.072 to obtain X 2
= 0.56 and Y 2
= 1.62 B 1.
Since this is where we started, we will converge back to the same bearing. The 95 mm
bore meets the requirements. Ans.
______________________________________________________________________________
11-26 9 kN, 3 kN, 1.2, 0. r a
F = F = V = R =
Use the Weibull parameters for Manufacturer 2 in Table 11-6.
8
6
D
D
R
L
x
L
First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.
Eq. (11-12):
0.56 1.2 9 1.63 3 10.9 kN e
F = + =
F
e
> F
r
, so use F e
as the design load.
Eq. (11-10):
1/
10 1/
0 0
a
D
f D b
D
x
C a F
x θ x R
th
6
D
D
R
L
x
L
Eq. (11-10):
1/
10 1/1.
C
= 4837 lbf =4.84 kips Ans.
(b) Results will vary depending on the specific bearing manufacturer selected. A general
engineering components search site such as www.globalspec.com might be useful as
a starting point.
______________________________________________________________________________
11-29 (a) 900 rev/min, 12 kh, 0.98, 1. D D f
n = L = R = a =
From Prob. 3-73, R Cy = 8.319 kN, R Cz = –10.830 kN.
1/
2 2
8.319 10.830 13.7 kN C D
R F
= = + − =
6
D
D
R
L
x
L
Eq. (11-10):
1/
10 1/1.
1.2 13.7 204 kN.
C Ans
(b) Results will vary depending on the specific bearing manufacturer selected. A general
engineering components search site such as www.globalspec.com might be useful as
a starting point.
______________________________________________________________________________
11-30 (a) 900 rev/min, 12 kh, 0.98, 1. D D f
n = L = R = a =
From Prob. 3-73, R Oy
= 5083 N, R
Oz
= 494 N.
1/ 2 2
5083 494 5106 N 5.1 kN C D
R = F = + = =
6
D
D
R
L
x
L
Eq. (11-10):
1/
10 1/1.
1.2 5.1 76.1 kN.
C Ans
(b) Results will vary depending on the specific bearing manufacturer selected. A general
engineering components search site such as www.globalspec.com might be useful as
a starting point.
______________________________________
________________________________
________
11-31 Assume concentrated forces as shown.
th
8 28 224 lbf z
P = =
8 35 280 lbf
y
P = =
T = 224 2 = 448 lbf in⋅
448 1.5 cos 20 0
x
Σ T = − + F =
o
318 lbf
F = =
5.75 11.5 14.25 sin 20 0
z y
O y A
Σ M = P + R − F =
o
y
A
+ R − =
5.24 lbf
y
A
R = −
5.75 11.5 14.25 cos 20 0
y z
O z A
Σ M = − P − R − F =
o
z
A
− − R − =
1/ 2 2
482 lbf; 482 5.24 482 lbf
z
A A
R R
cos 20 0
z z z
O z A
Σ F = R + P + R + F =
o
z
O
R + − + =
40.9 lbf
z
O
R = −
sin 20 0
y y y
O y A
Σ F = R + P + R − F =
o
y
O
R + − − =
166 lbf
y
O
R = −
1/ 2 2
40.9 166 171 lbf O
R
So the reaction at A governs.
Reliability Goal: 0.92 =0.
1.2 482 578 lbf D
F = =
6
35 000 350 60 / 10 735 D
x = =
1/
10 1/1.
0.02 4.459 0.02 ln 1/ 0.
6431 lbf 28.6 kN
C
From Table 11-2, a 40 mm bore angular contact bearing is sufficient with a rating of
31.9 kN. Ans.
th
T = (270 − 50) = ( P
1
− P
2
= ( P
1
− 0.15 P
1
P
1
= 310.6 N,
P
2
= 0.15 (310.6) = 46.6 N
P
1
+ P
2
= 357.2 N 357.2 sin 45 252.6 N
y z
A A
F = = = F
o
850 300(252.6) 0 89.2 N
z y y
O E E
M = R + = ⇒ R = −
∑
252.6 89.2 0 163.4 N
y y y
O O
F = − + R = ⇒ R = −
∑
850 700(320) 300(252.6) 0 174.4 N
y z z
O E E
M = − R − + = ⇒ R = −
∑
174.4 320 252.6 0 107 N
z z z
O O
F = − + − + R = ⇒ R =
∑
2 2
2 2
163.4 107 195 N
89.2 174.4 196 N
O
E
R
R
The radial loads are nearly the same at O and E. We can use the same bearing at both
locations.
6
D
x = =
Eq. (11-9):
1/
10 1/1.
1 0.196 5.7 kN
0.02 4.439 ln 1/ 0.
C
From Table 11-2, select an 02-12 mm deep-groove ball bearing with a basic load rating
of 6.89 kN. Ans.
______________________________________________________________________________
11-34 R = 0.96 =0.
T = 12(240 cos 20 ) = 2706 lbf ⋅in
o
498 lbf
6 cos 25
F = =
o
In xy -plane:
z y
O C
Σ M = − − + R =
181 lbf
y
C
R =
th
82.1 210 181 111.1 lbf
y
O
R = + − =
In xz -plane:
y z
O C
Σ M = − − R =
236 lbf
z
C
R = −
226 451 236 11 lbf
z
O
R = − + =
1/ 2 2 2
111.1 11 112 lbf. O
R = + = Ans
1/ 2 2 2
181 236 297 lbf. C
R = + = Ans
6
D
x = =
1/
10 1/1.
0.02 4.439 ln 1/ 0.
1860 lbf 8.28 kN
O
C
1/
10 1/1.
0.02 4.439 ln 1/ 0.
4932 lbf 21.9 kN
C
C
Bearing at O : Choose a deep-groove 02-17 mm. Ans.
Bearing at C : Choose a deep-groove 02-35 mm. Ans.
______________________________________________________________________________
11-35 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
force here is larger than any radial load, the bearing absorbing the thrust (bearing A ) is
heavily loaded compared to bearing B. Bearing B is thus likely to be oversized and may
not contribute measurably to the chance of failure. If this is the case, we may be able to
obtain the desired combined reliability with bearing A having a reliability near 0.99 and
bearing B having a reliability near 1. This would allow for bearing A to have a lower
capacity than if it needed to achieve a reliability of 0.99. To determine if this is the
case, we will start with bearing B.
Bearing B (straight roller bearing)
6
D
x = =
1/ 2 2 2
36 67 76.1 lbf 0.339 kN r
F = + = =
Try a reliability of 1 to see if it is readily obtainable with the available bearings.
th
Select an 02-90 mm angular-contact ball bearing. Ans.
______________________________________________________________________________
11-36 We have some data. Let’s estimate parameters b and θ from it. In Fig. 11-5, we will use
line AB. In this case, B is to the right of A.
For F = 18 kN,
6 1
x = =
This establishes point 1 on the R = 0.90 line.
The R = 0.20 locus is above and parallel to the R = 0.90 locus. For the two-parameter
Weibull distribution, x 0 = 0 and points A and B are related by [see Eq. (11-8)]:
1/
ln 1/ 0.
b
A
x = θ
1/
ln 1/ 0.
b
B
x = θ
and x B / x A is in the same ratio as 600/115. Eliminating θ ,
ln ln 1/ 0.20 / ln 1/ 0.
ln 600 /
b Ans
Solving for θ in Eq. (1),
1/1.65 1/1.
ln 1/ ln 1/ 0.
A
A
x
Ans
R
θ = = =
th
Therefore, for the data at hand,
exp
x
R
Check R at point B : x B
exp 0.
R
Note also, for point 2 on the R = 0.20 line,
2
log 5.217 log 1 log log 13.
m
− = x −
2
m
x =
______________________________________________________________________________
11-37 This problem is rich in useful variations. Here is one.
Decision : Make straight roller bearings identical on a given shaft. Use a reliability goal of
1/
Shaft a
1/
2 2
239 111 264 lbf 1.175 kN
r
A
F = + = =
1/
2 2
502 1075 1186 lbf 5.28 kN
r
B
F = + = =
Thus the bearing at B controls.
6
D
x = =
1/1.
0.02 + 4.439 ln 1/ 0.9983 =0.080 26
10
1.2 5.28 97.2 kN
C
Select an 02-80 mm with C 10
= 106 kN. Ans.
Shaft b
1/
2 2
874 2274 2436 lbf or 10.84 kN
r
C
F = + =
1/ 2 2
393 657 766 lbf or 3.41 kN
r
D
F = + =
The bearing at C controls.
