Mechanical Engineering Design Chapter 11 Solutions
Short Description
Shigley's 9th Edition...
Description
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 L 60D nD 60 25000 350 xD D 525 Ans. LR L10 106 The design radial load is FD 1.2 2.5 3.0 kN 1/3
Eq. (11-6):
525 C10 3.0 1/1.483 0.02 4.459 0.02 ln 1/ 0.9
C 10 = 24.3 kN Table 11-2:
Ans.
Choose an 02-35 mm bearing with C 10 = 25.5 kN.
Ans.
1.483 3 525 3 / 25.5 0.02 Eq. (11-18): R exp Ans. 0.920 4.459 0.02 ______________________________________________________________________________
11-2
For the angular-contact 02-series ball bearing as described, the rating life multiple is L 60D nD 60 40 000 520 xD D 1248 LR L10 106 The design radial load is
FD 1.4 725 1015 lbf 4.52 kN Eq. (11-6): 1/3
1248 C10 1015 1/1.483 0.02 4.459 0.02 ln 1/ 0.9 10 930 lbf 48.6 kN
Select an 02-60 mm bearing with C 10 = 55.9 kN. Ans. 1.483 3 1248 4.52 / 55.9 0.02 Eq. (11-18): R exp Ans. 0.945 4.439 ______________________________________________________________________________ Table 11-2:
Chapter 11, Page 1/28
11-3
For the straight-roller 03-series bearing selection, x D = 1248 rating lives from Prob. 11-2 solution. FD 1.4 2235 3129 lbf 13.92 kN 1248 C10 13.92 1
Table 11-3:
3/10
118 kN
Select an 03-60 mm bearing with C 10 = 123 kN.
Ans.
1.483 10/3 1248 13.92 /123 0.02 Eq. (11-18): R exp 0.917 Ans. 4.459 0.02 ______________________________________________________________________________
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 0.95 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 R 1 . Then set the reliability goal of the second as R2
0.90 R1
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 contact ball bearing,
0.90 0.95 for each bearing. For an 02-series angular 1/3
1248 C10 1015 1/1.483 0.02 4.439 ln 1/ 0.95 12822 lbf 57.1 kN Select an 02-65 mm angular-contact bearing with C 10 = 63.7 kN. 1.483 3 1248 4.52 / 63.7 0.02 RA exp 0.962 4.439
Chapter 11, Page 2/28
For an 03-series straight roller bearing, 3/10
1248 C10 13.92 1/1.483 0.02 4.439 ln 1/ 0.95
136.5 kN
Select an 03-65 mm straight-roller bearing with C 10 = 138 kN. 1248 13.92 /138 10/3 0.02 1.483 RB exp 0.953 4.439
The overall reliability is R = (0.962)(0.953) = 0.917, which exceeds the goal. ______________________________________________________________________________ 11-6
For the straight cylindrical roller bearing specified with a service factor of 1, R = 0.95 and F R = 20 kN. 60D nD 60 8000 950 L xD D 456 106 LR L10 3/10
456 C10 20 145 kN Ans. 1/1.483 0.02 4.439 ln 1/ 0.95 ______________________________________________________________________________
11-7
Both bearings need to be rated in terms of the same catalog rating system in order to compare them. Using a rating life of one million revolutions, both bearings can be rated in terms of a Basic Load Rating. 1/ a
Eq. (11-3):
L C A FA A LR 8.96 kN
1/ a
n 60 FA A A LR
3000 500 60 2.0 106
1/3
Bearing B already is rated at one million revolutions, so C B = 7.0 kN. Since C A > C B , bearing A can carry the larger load. Ans. ______________________________________________________________________________ 11-8
F D = 2 kN, L D = 109 rev, R = 0.90 1/ a
1/3
LD 109 Eq. (11-3): C10 FD 2 6 20 kN Ans. 10 LR ______________________________________________________________________________
Chapter 11, Page 3/28
11-9
F D = 800 lbf, D = 12 000 hours, n D = 350 rev/min, R = 0.90
12 000 350 60 n 60 Eq. (11-3): C10 FD D D 800 5050 lbf Ans 106 LR ______________________________________________________________________________ 1/ a
1/3
11-10 F D = 4 kN, D = 8 000 hours, n D = 500 rev/min, R = 0.90 8 000 500 60 n 60 Eq. (11-3): C10 FD D D 4 24.9 kN Ans 106 LR ______________________________________________________________________________ 1/ a
1/3
11-11 F D = 650 lbf, n D = 400 rev/min, R = 0.95
D = (5 years)(40 h/week)(52 week/year) = 10 400 hours
Assume an application factor of one. The multiple of rating life is xD
LD 10 400 400 60 249.6 LR 106 1/3
249.6 C10 1 650 Eq. (11-6): 1/1.483 0.02 4.439 ln 1/ 0.95 4800 lbf Ans. ______________________________________________________________________________
11-12 F D = 9 kN, L D = 108 rev, R = 0.99 Assume an application factor of one. The multiple of rating life is xD
LD 108 100 LR 106 1/3
100 C10 1 9 Eq. (11-6): 1/1.483 0.02 4.439 ln 1/ 0.99 69.2 kN Ans. ______________________________________________________________________________
11-13 F D = 11 kips, D = 20 000 hours, n D = 200 rev/min, R = 0.99 Assume an application factor of one. Use the Weibull parameters for Manufacturer 2 on p. 608.
Chapter 11, Page 4/28
The multiple of rating life is xD
LD 20 000 200 60 240 LR 106 1/3
240 C10 111 Eq. (11-6): 1/1.483 0.02 4.439 ln 1/ 0.99 113 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 on p. 608. xD
LD 15000 1200 60 1080 LR 106 1/ a
Eq. (11-7):
xD C10 a f FD 1/ b x0 x0 1 RD
1/3
1080 C10 1.2 178 1/1.483 0.02 4.459 0.02 1 0.95 2590 lbf 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 on p. 608. xD
LD 15000 1200 60 1080 LR 106 1/ a
Eq. (11-7):
xD C10 a f FD 1/ b x0 x0 1 RD
1/3
1080 C10 1.2 1.794 1/1.483 0.02 4.459 0.02 1 0.95 26.1 kN Ans. ______________________________________________________________________________
11-16 From the solution to Prob. 3-70, R Cz = –327.99 lbf, R Cy = –127.27 lbf 1/ 2
2 2 RC FD 327.99 127.27 351.8 lbf Use the Weibull parameters for Manufacturer 2 on p. 608.
Chapter 11, Page 5/28
xD
LD 15000 1200 60 1080 LR 106 1/ a
Eq. (11-7):
xD C10 a f FD 1/ b xo xo 1 RD
1/3
1080 C10 1.2 351.8 1/1.483 0.02 4.459 0.02 1 0.95 5110 lbf Ans. ______________________________________________________________________________
11-17 From the solution to Prob. 3-71, R Cz = –150.7 N, R Cy = –86.10 N 1/2
2 2 RC FD 150.7 86.10 173.6 N Use the Weibull parameters for Manufacturer 2 on p. 608.
xD
LD 15000 1200 60 1080 106 LR 1/ a
Eq. (11-7):
xD C10 a f FD 1/ b x0 x0 1 RD
1/3
1080 C10 1.2 173.6 1/1.483 0.02 4.459 0.02 1 0.95 2520 N Ans. ______________________________________________________________________________
11-18 From the solution to Prob. 3-77, R Az = 444 N, R Ay = 2384 N
RA FD 4442 23842
1/2
2425 N 2.425 kN
Use the Weibull parameters for Manufacturer 2 on p. 608. The design speed is equal to the speed of shaft AD, d 125 nD F ni 191 95.5 rev/min dC 250 xD
LD 12 000 95.5 60 68.76 LR 106 1/ a
Eq. (11-7):
xD C10 a f FD 1/ b x0 x0 1 RD
Chapter 11, Page 6/28
1/3
68.76 C10 1 2.425 1/1.483 0.02 4.459 0.02 1 0.95 11.7 kN Ans. ______________________________________________________________________________
11-19 From the solution to Prob. 3-79, R Az = 54.0 lbf, R Ay = 140 lbf
RA FD 54.02 1402
1/2
150.1 lbf
Use the Weibull parameters for Manufacturer 2 on p. 608. The design speed is equal to the speed of shaft AD, d 10 nD F ni 280 560 rev/min dC 5 xD Eq. (11-7):
LD 14 000 560 60 470.4 LR 106
xD C10 a f FD 1/ b x0 x0 1 RD
1/ a
3/10
470.4 C10 1150.1 1/1.483 0.02 4.459 0.02 1 0.98 1320 lbf Ans. ______________________________________________________________________________
11-20 (a) Fa 3 kN, Fr 7 kN, nD 500 rev/min, V 1.2 From Table 11-2, with a 65 mm bore, C 0 = 34.0 kN. F a / C 0 = 3 / 34 = 0.088 From Table 11-1, 0.28 e 3.0. Fa 3 0.357 VFr 1.2 7 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 = 1.53.
Eq. (11-9):
Fe X iVFr Yi Fa 0.56 1.2 7 1.53 3 9.29 kN F e > F r so use F e .
Ans.
(b) Use Eq. (11-7) 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 on p. 608.
Chapter 11, Page 7/28
xD
LD 10 000 500 60 300 LR 106 1/ a
xD Eq. (11-7): C10 a f FD 1/ b x0 x0 1 RD
300 C10 1 9.29 1/1.483 0.02 4.459 0.02 1 0.95 73.4 kN
1/3
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) Fa 2 kN, Fr 5 kN, nD 400 rev/min, V 1 From Table 11-2, 30 mm bore, C 10 = 19.5 kN, C 0 = 10.0 kN F a / C 0 = 2 / 10 = 0.2 From Table 11-1, 0.34 e 0.38. Fa 2 0.4 VFr 1 5
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 = 1.27. Ans. Eq. (11-9): Fe X iVFr Yi Fa 0.56 1 5 1.27 2 5.34 kN F e > F r so use F e . (b) Solve Eq. (11-7) for x D . C xD 10 a f FD
a
1/ b x0 x0 1 RD 3
19.5 0.02 4.459 0.02 1 0.99 1/1.483 xD 1 5.34 xD 10.66
xD
LD D nD 60 LR 106
Chapter 11, Page 8/28
D
10.66 10 444 h
xD 106
6
Ans. nD 60 400 60 ______________________________________________________________________________
11-22 Fr 8 kN, R 0.9, LD 109 rev 1/ a
Eq. (11-3):
L C10 FD D LR
1/3
109 8 6 10
80 kN
From Table 11-2, select the 85 mm bore. Ans. ______________________________________________________________________________ 11-23 Fr 8 kN, Fa 2 kN, V 1, R 0.99 Use the Weibull parameters for Manufacturer 2 on p. 608. xD
LD 10 000 400 60 240 LR 106
First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63 Eq. (11-9):
Fe 0.56 1 8 1.63 2 7.74 kN F e < F r , so just use F r as the design load.
Eq. (11-7):
xD C10 a f FD 1/ b xo xo 1 RD
1/ a
1/3
240 C10 1 8 82.5 kN 1/1.483 0.02 4.459 0.02 1 0.99 From Table 11-2, try 85 mm bore with C 10 = 83.2 kN, C 0 = 53.0 kN Iterate the previous process:
F a / C 0 = 2 / 53 = 0.038 0.22 e 0.24 Fa 2 0.25 e VFr 1 8 Interpolate Table 11-1 with F a / C 0 = 0.038 to obtain X 2 = 0.56 and Y 2 = 1.89. Table 11-1:
Eq. (11-9):
Fe 0.56(1)8 1.89(2) 8.26 > Fr
Eq. (11-7):
240 C10 1 8.26 1/1.483 0.02 4.459 0.02 1 0.99
1/3
85.2 kN
Chapter 11, Page 9/28
Table 11-2: Move up to the 90 mm bore with C 10 = 95.6 kN, C 0 = 62.0 kN. Iterate again: F a / C 0 = 2 / 62 = 0.032 Again, 0.22 e 0.24 Fa 2 0.25 e VFr 1 8 Interpolate Table 11-1 with F a / C 0 = 0.032 to obtain X 2 = 0.56 and Y 2 = 1.95. Table 11-1:
Eq. (11-9):
Fe 0.56(1)8 1.95(2) 8.38 > Fr 1/3
240 Eq. (11-7): C10 1 8.38 86.4 kN 1/1.483 0.02 4.459 0.02 1 0.99 The 90 mm bore is acceptable. Ans. ______________________________________________________________________________
11-24 Fr 8 kN, Fa 3 kN, V 1.2, R 0.9, LD 108 rev First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63 Eq. (11-9):
Fe 0.56 1.2 8 1.63 3 10.3 kN Fe Fr 1/ a
Eq. (11-3):
L C10 Fe D LR
1/3
108 10.3 6 10
47.8 kN
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 = 3 / 28 = 0.107 0.28 e 0.30 Fa 3 0.313 e VFr 1.2 8 Interpolate Table 11-1 with F a / C 0 = 0.107 to obtain X 2 = 0.56 and Y 2 = 1.46 Table 11-1:
Eq. (11-9):
Fe 0.56 1.2 8 1.46 3 9.76 kN > Fr 1/3
108 Eq. (11-3): C10 9.76 6 45.3 kN 10 From Table 11-2, we have converged on the 60 mm bearing. Ans. ______________________________________________________________________________
Chapter 11, Page 10/28
11-25 Fr 10 kN, Fa 5 kN, V 1, R 0.95 Use the Weibull parameters for Manufacturer 2 on p. 608. xD
LD 12 000 300 60 216 LR 106
First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63 Eq. (11-9):
Fe 0.56 110 1.63 5 13.75 kN F e > F r , so use F e as the design load.
Eq. (11-7):
xD C10 a f FD 1/ b x0 x0 1 RD
1/ a
1/3
216 C10 113.75 1/1.483 0.02 4.459 0.02 1 0.95
97.4 kN
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 = 5 / 69.5 = 0.072 Table 11-1:
0.27 e 0.28 Fa 5 0.5 e VFr 110
Interpolate Table 11-1 with F a / C 0 = 0.072 to obtain X 2 = 0.56 and Y 2 = 1.62 1.63 Since this is where we started, we will converge back to the same bearing. The 95 mm bore meets the requirements. Ans. ______________________________________________________________________________ 11-26 Note to the Instructor. In the first printing of the 9th edition, the design life was incorrectly given to be 109 rev and will be corrected to 108 rev in subsequent printings. We apologize for the inconvenience. Fr 9 kN, Fa 3 kN, V 1.2, R 0.99 Use the Weibull parameters for Manufacturer 2 on p. 608. xD
LD 108 100 LR 106
First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63
Chapter 11, Page 11/28
Eq. (11-9):
Fe 0.56 1.2 9 1.63 3 10.9 kN F e > F r , so use F e as the design load.
Eq. (11-7):
xD C10 a f FD 1/ b x0 x0 1 RD
1/ a
1/3
100 C10 110.9 1/1.483 0.02 4.459 0.02 1 0.99
83.9 kN
From Table 11-2, try 90 mm bore with C 10 = 95.6 kN, C 0 = 62.0 kN. Try this bearing. Iterate the previous process: F a / C 0 = 3 / 62 = 0.048 0.24 e 0.26 Fa 3 0.278 e VFr 1.2 9 Interpolate Table 11-1 with F a / C 0 = 0.048 to obtain X 2 = 0.56 and Y 2 = 1.79 Table 11-1:
Eq. (11-9):
Fe 0.56 1.2 9 1.79 3 11.4 kN Fr
11.4 83.9 87.7 kN 10.9 From Table 11-2, this converges back to the same bearing. The 90 mm bore meets the requirements. Ans. ______________________________________________________________________________ C10
11-27 (a) nD 1200 rev/min, LD 15 kh, R 0.95, a f 1.2 From Prob. 3-72, R Cy = 183.1 lbf, R Cz = –861.5 lbf. 1/2
2 RC FD 183.12 861.5 881 lbf 15000 1200 60 L xD D 1080 LR 106
1/3
1080 Eq. (11-7): C10 1.2 881 1/1.483 0.02 4.439 1 0.95 12800 lbf 12.8 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. ______________________________________________________________________________
Chapter 11, Page 12/28
11-28 (a) nD 1200 rev/min, LD 15 kh, R 0.95, a f 1.2 From Prob. 3-72, R Oy = –208.5 lbf, R Oz = 259.3 lbf. 1/ 2
2 RC FD 259.32 208.5 333 lbf 15000 1200 60 L xD D 1080 106 LR
1/3
1080 Eq. (11-7): C10 1.2 333 1/1.483 0.02 4.439 1 0.95 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) nD 900 rev/min, LD 12 kh, R 0.98, a f 1.2 From Prob. 3-73, R Cy = 8.319 kN, R Cz = –10.830 kN. 2 1/ 2
RC FD 8.319 2 10.830 13.7 kN 12 000 900 60 L xD D 648 LR 106 1/3
648 Eq. (11-7): C10 1.2 13.7 204 kN Ans. 1/1.483 0.02 4.439 1 0.98 (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) nD 900 rev/min, LD 12 kh, R 0.98, a f 1.2 From Prob. 3-73, R Oy = 5083 N, R Oz = 494 N.
RC FD 50832 4942
xD
1/2
5106 N 5.1 kN
LD 12 000 900 60 648 LR 106 1/3
648 Eq. (11-7): C10 1.2 5.1 76.1 kN Ans. 1/1.483 0.02 4.439 1 0.98 (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. ______________________________________________________________________________ Chapter 11, Page 13/28
11-31 Assume concentrated forces as shown.
Pz 8 28 224 lbf Py 8 35 280 lbf T 224 2 448 lbf in T x 448 1.5 F cos 20 0 448 F 318 lbf 1.5 0.940
M Oz 5.75Py 11.5RAy 14.25F sin 20 0 5.75 280 11.5RAy 14.25 318 0.342 0 RAy 5.24 lbf M Oy 5.75 Pz 11.5 RAz 14.25 F cos 20 0
5.75 224 11.5RAz 14.25 318 0.940 0 1/2
2 2 RA 482 5.24 z z z F RO Pz RA F cos 20 0
RAz 482 lbf;
482 lbf
ROz 224 482 318 0.940 0 ROz 40.9 lbf
F y ROy Py RAy F sin 20 0 ROy 280 5.24 318 0.342 0 ROy 166 lbf 1/2
2 2 RO 40.9 166
171 lbf
So the reaction at A governs. Reliability Goal: 0.92 0.96
FD 1.2 482 578 lbf xD 35 000 350 60 / 106 735 1/3
735 C10 578 1/1.483 0.02 4.459 0.02 ln 1/ 0.96 6431 lbf 28.6 kN
From Table 11-2, a 40 mm bore angular contact bearing is sufficient with a rating of Chapter 11, Page 14/28
31.9 kN. Ans. ______________________________________________________________________________ 11-32 For a combined reliability goal of 0.95, use 0.95 0.975 for the individual bearings.
xD
40 000 420 60 106
1008
The resultant of the given forces are R O = [(–387)2 + 4672]1/2 = 607 lbf R B = [3162 + (–1615)2]1/2 = 1646 lbf At O: 1/3
1008 C10 1.2 607 Eq. (11-6): 1/1.483 0.02 4.459 0.02 ln 1/ 0.975 9978 lbf 44.4 kN
From Table 11-2, select an 02-55 mm angular-contact ball bearing with a basic load rating of 46.2 kN. Ans. At B: 3/10
Eq. (11-6):
1008 C10 1.2 1646 1/1.483 0.02 4.459 0.02 ln 1/ 0.975 20827 lbf 92.7 kN
From Table 11-3, select an 02-75 mm or 03-55 mm cylindrical roller. Ans. ______________________________________________________________________________ 11-33 The reliability of the individual bearings is R 0.98 0.9899
Chapter 11, Page 15/28
From statics, T = (270 50) = (P 1 P 2 )125 = (P 1 0.15 P 1 )125 P 1 = 310.6 N, P 2 = 0.15 (310.6) = 46.6 N P 1 + P 2 = 357.2 N FAy 357.2 sin 45 252.6 N FAz
M F M F
850REy 300(252.6) 0 REy 89.2 N
z O
y
252.6 89.2 ROy 0 ROy 163.4 N 850REz 700(320) 300(252.6) 0 REz 174.4 N
y O
z
174.4 320 252.6 ROz 0 ROz 107 N
RO
163.4
2
RE
89.2
174.4 196 N
2
107 2 195 N 2
The radial loads are nearly the same at O and E. We can use the same bearing at both locations. xD
60 000 1500 60 106
5400 1/3
Eq. (11-6):
5400 C10 1 0.196 1/1.483 0.02 4.439 ln 1/ 0.9899
5.7 kN
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 0.980 T 12(240 cos 20 ) 2706 lbf in F
2706 498 lbf 6 cos 25
In xy-plane:
M Oz 16(82.1) 30(210) 42 RCy 0
Chapter 11, Page 16/28
RCy 181 lbf ROy 82.1 210 181 111.1 lbf
In xz-plane:
M Oy 16(226) 30(451) 42 RCz 0 RCz 236 lbf ROz 226 451 236 11 lbf
181
RO 111.12 112 RC
xD
2
236 2
1/ 2
112 lbf
Ans.
1/ 2
297 lbf
Ans.
50000 300 60 106
900 1/3
C10 O
900 1.2 112 1/1.483 0.02 4.439 ln 1/ 0.980 1860 lbf 8.28 kN
1/3
C10 C
900 1.2 297 1/1.483 0.02 4.439 ln 1/ 0.980 4932 lbf 21.9 kN
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) 30000 500 60 xD 900 106
Fr 36 2 67 2
1/ 2
76.1 lbf 0.339 kN
Try a reliability of 1 to see if it is readily obtainable with the available bearings.
Chapter 11, Page 17/28
3/10
Eq. (11-6):
900 C10 1.2 0.339 1/1.483 0.02 4.439 ln 1/1.0
10.1 kN
The smallest capacity bearing from Table 11-3 has a rated capacity of 16.8 kN. Therefore, we select the 02-25 mm straight cylindrical roller bearing. Ans. Bearing at A (angular-contact ball) With a reliability of 1 for bearing B, we can achieve the combined reliability goal of 0.99 if bearing A has a reliability of 0.99.
Fr 36 2 2122
1/ 2
215 lbf 0.957 kN
Fa 555 lbf 2.47 kN Trial #1: Tentatively select an 02-85 mm angular-contact with C 10 = 90.4 kN and C 0 = 63.0 kN. Fa 2.47 0.0392 C0 63.0 30000 500 60 xD 900 106
Table 11-1:
Interpolating, X 2 = 0.56, Y 2 = 1.88
Eq. (11-9):
Fe 0.56 0.957 1.88 2.47 5.18 kN 1/3
Eq. (11-6):
900 C10 1.2 5.18 1/1.483 0.02 4.439 ln 1/ 0.99 99.54 kN 90.4 kN
Trial #2: Tentatively select a 02-90 mm angular-contact ball with C 10 = 106 kN and C 0 = 73.5 kN. Fa 2.47 0.0336 C0 73.5
Table 11-1:
Interpolating, X 2 = 0.56, Y 2 = 1.93
Fe 0.56 0.957 1.93 2.47 5.30 kN 1/3
900 C10 1.2 5.30 1/1.483 0.02 4.439 ln 1/ 0.99
102 kN < 106 kN O.K.
Chapter 11, Page 18/28
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.
x 1
For F = 18 kN,
115 2000 60 106
13.8
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. (20-25)]: x A ln 1/ 0.90
1/ b
(1)
xB ln 1/ 0.20
1/ b
and x B /x A is in the same ratio as 600/115. Eliminating θ, b
ln ln 1/ 0.20 / ln 1/ 0.90 ln 600 /115
1.65
Ans.
Solving for θ in Eq. (1),
xA ln 1/ RA
1/1.65
1 ln 1/ 0.90
1/1.65
3.91
Ans.
Chapter 11, Page 19/28
Therefore, for the data at hand, x 1.65 R exp 3.91 Check R at point B: x B = (600/115) = 5.217 5.217 1.65 R exp 0.20 3.91 Note also, for point 2 on the R = 0.20 line,
log 5.217 log 1 log xm 2 log 13.8
xm 2 72
______________________________________________________________________________ 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 (0.99)1/6 = 0.9983. Shaft a
502
FAr 2392 1112 FBr
2
1/ 2
264 lbf 1.175 kN
1/ 2
10752
1186 lbf 5.28 kN
Thus the bearing at B controls. xD
10 000 1200 60 106
720
0.02 4.439 ln 1/ 0.9983
1/1.483
720 C10 1.2 5.28 0.080 26
0.080 26
0.3
97.2 kN
Select either an 02-80 mm with C 10 = 106 kN or an 03-55 mm with C 10 = 102 kN. Shaft b
393
FCr 874 2 2274 2 FDr
2
657 2
1/ 2
1/ 2
2436 lbf
766 lbf
or or
Ans.
10.84 kN 3.41 kN
The bearing at C controls.
Chapter 11, Page 20/28
xD
10 000 240 60 106
144
144 C10 1.2 10.84 0.080 26
0.3
123 kN
Select either an 02-90 mm with C 10 = 142 kN or an 03-60 mm with C 10 = 123 kN. Shaft c
417
FEr 11132 23852 FFr
8952
2
1/ 2
1/ 2
2632 lbf
987 lbf
or
or
Ans.
11.71 kN
4.39 kN
The bearing at E controls. xD
10 000 80 60 106
48
48 C10 1.2 11.71 0.080 26
0.3
95.7 kN
Select an 02-80 mm with C 10 = 106 kN or an 03-60 mm with C 10 = 123 kN. Ans. ______________________________________________________________________________ 11-38 Express Eq. (11-1) as F1a L1 C10a L10 K
For a ball bearing, a = 3 and for an 02-30 mm angular contact bearing, C 10 = 20.3 kN.
K 20.3 106 8.365 109 3
At a load of 18 kN, life L 1 is given by:
9 K 8.365 10 L1 a 1.434 106 rev 3 18 F1 For a load of 30 kN, life L 2 is:
L2
0.310 10 rev 30
8.365 109
6
3
In this case, Eq. (6-57) – the Palmgren-Miner cycle-ratio summation rule – can be expressed as
Chapter 11, Page 21/28
l1 l2 1 L1 L2
Substituting, l2 200 000 1 6 1.434 10 0.310 106 l2
0.267 10 rev Ans. 6
______________________________________________________________________________ 11-39 Total life in revolutions Let: l = total turns f 1 = fraction of turns at F 1 f 2 = fraction of turns at F 2 From the solution of Prob. 11-38, L 1 = 1.434(106) rev and L 2 = 0.310(106) rev. Palmgren-Miner rule: l1 l2 fl f l 1 2 1 L1 L2 L1 L2
from which l
1 f1 / L1 f 2 / L2
l
0.40 / 1.434 10 0.60 / 0.310 10
451 585 rev
1
6
6
Ans.
Total life in loading cycles 4 min at 2000 rev/min = 8000 rev/cycle 6 min at 2000 rev/min = 12 000 rev/cycle Total rev/cycle = 8000 + 12 000 = 20 000 451 585 rev 22.58 cycles 20 000 rev/cycle
Ans.
Chapter 11, Page 22/28
Total life in hours min 22.58 cycles 10 3.76 h Ans. cycle 60 min/h ______________________________________________________________________________ FrA 560 lbf FrB 1095 lbf Fae 200 lbf
11-40
xD
LD 40 000 400 60 10.67 LR 90 106
R 0.90 0.949 Eq. (11-15): Eq. (11-15):
0.47 FrA 0.47 560 175.5 lbf KA 1.5 0.47 FrB 0.47 1095 FiB 343.1 lbf KB 1.5 FiA
FiA ? FiB Fae
175.5 lbf 343.1 200 543.1 lbf, so Eq. (11-16) applies. We will size bearing B first since its induced load will affect bearing A, but is not itself affected by the induced load from bearing A [see Eq. (11-16)]. From Eq. (11-16b), F eB = F rB = 1095 lbf.
Eq. (11-7):
10.67 FRB 1.4 1095 4.48 1 0.949 1/1.5
3/10
3607 lbf
Ans.
Select cone 32305, cup 32305, with 0.9843 in bore, and rated at 3910 lbf with K = 1.95. Ans. With bearing B selected, we re-evaluate the induced load from bearing B using the actual value for K. 0.47 FrB 0.47 1095 Eq. (11-15): FiB 263.9 lbf KB 1.95 Find the equivalent radial load for bearing A from Eq. (11-16), which still applies. Eq. (11-16a): FeA 0.4 FrA K A FiB Fae
FeA 0.4 560 1.5 263.9 200 920 lbf
Chapter 11, Page 23/28
FeA FrA Eq. (11-7):
10.67 FRA 1.4 920 4.48 1 0.949 1/1.5
3/10
3030 lbf
Tentatively select cone M86643, cup M86610, with 1 in bore, and rated at 3250 lbf with K = 1.07. Iterating with the new value for K, we get F eA = 702 lbf and F rA = 2312 lbf. Ans. By using a bearing with a lower K, the rated load decreased significantly, providing a higher than requested reliability. Further examination with different combinations of bearing choices could yield additional acceptable solutions. ______________________________________________________________________________ 11-41 The thrust load on shaft CD is from the axial component of the force transmitted through the bevel gear, and is directed toward bearing C. By observation of Fig. 11-14, direct mounted bearings would allow bearing C to carry the thrust load. Ans. From the solution to Prob. 3-74, the axial thrust load is F ae = 362.8 lbf, and the bearing radial forces are F Cx = 287.2 lbf, F Cz = 500.9 lbf, F Dx = 194.4 lbf, and F Dz = 307.1 lbf. Thus, the radial forces are FrC 287.22 500.9 2 577 lbf FrD 194.4 2 307.12 363 lbf
The induced loads are 0.47 FrC 0.47 577 181 lbf Eq. (11-15): FiC 1.5 KC 0.47 FrD 0.47 363 114 lbf Eq. (11-15): FiD KD 1.5 Check the condition on whether to apply Eq. (11-16) or Eq. (11-17), where bearings C and D are substituted, respectively, for labels A and B in the equations. FiC ? FiD Fae 181 lbf 114 362.8 476.8 lbf, so Eq.(11-16) applies Eq. (11-16a): FeC 0.4 FrC KC FiD Fae
0.4 577 1.5 114 362.8 946 lbf FrC , so use FeC Assume for tapered roller bearings that the specifications for Manufacturer 1 on p. 608 are applicable.
Chapter 11, Page 24/28
xD
LD 108 1.11 LR 90 106
R 0.90 0.949 1.11 Eq. (11-7): FRC 1 946 1/1.5 4.48 1 0.949 Eq. (11-16b): FeD FrD 363 lbf
3/10
1130 lbf
Ans.
3/10
1.11 433 lbf Ans. Eq. (11-7): FRD 1 363 1/1.5 4.48 1 0.949 ______________________________________________________________________________
11-42 The thrust load on shaft AB is from the axial component of the force transmitted through the bevel gear, and is directed to the right. By observation of Fig. 11-14, indirect mounted bearings would allow bearing A to carry the thrust load. Ans. From the solution to Prob. 3-76, the axial thrust load is F ae = 92.8 lbf, and the bearing radial forces are F Ay = 639.4 lbf, F Az = 1513.7 lbf, F By = 276.6 lbf, and F Bz = 705.7 lbf. Thus, the radial forces are FrA 639.4 2 1513.7 2 1643 lbf FrB 276.62 705.7 2 758 lbf
The induced loads are 0.47 FrA 0.47 1643 Eq. (11-15): FiA 515 lbf KA 1.5 0.47 FrB 0.47 758 Eq. (11-15): FiB 238 lbf 1.5 KB Check the condition on whether to apply Eq. (11-16) or Eq. (11-17). FiA ? FiB Fae 515 lbf 238 92.8 330.8 lbf, so Eq.(11-17) applies Notice that the induced load from bearing A is sufficiently large to cause a net axial force to the left, which must be supported by bearing B. Eq. (11-17a): FeB 0.4 FrB K B FiA Fae
0.4 758 1.5 515 92.8 937 lbf FrB , so use FeB Assume for tapered roller bearings that the specifications for Manufacturer 1 on p. 608 are applicable. Chapter 11, Page 25/28
6 LD 500 10 xD 5.56 LR 90 106
R 0.90 0.949 5.56 Eq. (11-7): FRB 1 937 4.48 1 0.949 1/1.5 Eq. (11-16b): FeA FrA 1643 lbf
3/10
1810 lbf
Ans.
3/10
5.56 3180 lbf Ans. Eq. (11-7): FRA 11643 4.48 1 0.949 1/1.5 ______________________________________________________________________________
11-43 The lower bearing is compressed by the axial load, so it is designated as bearing A. FrA 25 kN FrB 12 kN Fae 5 kN
0.47 FrA 0.47 25 7.83 kN KA 1.5 0.47 FrB 0.47 12 Eq. (11-15): FiB 3.76 kN KB 1.5 Check the condition on whether to apply Eq. (11-16) or Eq. (11-17) Eq. (11-15):
FiA
FiA ? FiB Fae 7.83 kN 3.76 5 8.76 kN, so Eq.(11-16) applies Eq. (11-16a): FeA 0.4 FrA K A FiB Fae
0.4 25 1.5 3.76 5 23.1 kN FrA, so use FrA 60 min 8 hr 5 day 52 weeks LD 250 rev/min 5 yrs yr hr day week
156 106 rev Assume for tapered roller bearings that the specifications for Manufacturer 1 on p. 608 are applicable. Eq. (11-3):
L FRA a f FD D LR
3/10
156 106 1.2 25 90 106
3/10
35.4 kN
Ans.
Eq. (11-16b): FeB FrB 12 kN
Chapter 11, Page 26/28
3/10
156 FRB 1.2 12 17.0 kN Ans. 90 ______________________________________________________________________________
Eq. (11-3):
11-44 The left bearing is compressed by the axial load, so it is properly designated as bearing A. FrA 875 lbf FrB 625 lbf Fae 250 lbf Assume K = 1.5 for each bearing for the first iteration. Obtain the induced loads. Eq. (11-15): Eq. (11-15):
0.47 FrA 0.47 875 274 lbf 1.5 KA 0.47 FrB 0.47 625 FiB 196 lbf KB 1.5 FiA
Check the condition on whether to apply Eq. (11-16) or Eq. (11-17). FiA ? FiB Fae 274 lbf 196 250 lbf, so Eq.(11-16) applies We will size bearing B first since its induced load will affect bearing A, but it is not affected by the induced load from bearing A [see Eq. (11-16)]. From Eq. (11-16b), F eB = F rB = 625 lbf.
Eq. (11-3):
L FRB a f FD D LR
3/10
90 000 150 60 1 625 90 106
3/10
FRB 1208 lbf Select cone 07100, cup 07196, with 1 in bore, and rated at 1570 lbf with K = 1.45. Ans. With bearing B selected, we re-evaluate the induced load from bearing B using the actual value for K. 0.47 FrB 0.47 625 Eq. (11-15): FiB 203 lbf KB 1.45 Find the equivalent radial load for bearing A from Eq. (11-16), which still applies.
Chapter 11, Page 27/28
Eq. (11-16a): FeA 0.4 FrA K A FiB Fae
0.4 875 1.5 203 250 1030 lbf FeA FrA Eq. (11-3):
L FRA a f FD D LR
3/10
90 000 150 60 11030 90 106
3/10
FRA 1990 lbf Any of the bearings with 1-1/8 in bore are more than adequate. Select cone 15590, cup 15520, rated at 2480 lbf with K = 1.69. Iterating with the new value for K, we get F eA = 1120 lbf and F rA = 2160 lbf. The selected bearing is still adequate. Ans. ______________________________________________________________________________
Chapter 11, Page 28/28
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