{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

2_ch 11 Mechanical Design budynas_SM_ch11

# 2_ch 11 Mechanical Design budynas_SM_ch11 - plications etc...

This preview shows page 1. Sign up to view the full content.

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 C 10 = 10 . 279 1440 1 3 / 10 = 91 . 1 kN Table 11-3: Select a 03-55 mm with C 10 = 102 kN. Ans. Using Eq. (11-18), R = exp 1440(10 . 28 / 102) 10 / 3 0 . 02 4 . 439 1 . 483 = 0 . 942 Ans. 11-4 We can choose a reliability goal of 0 . 90 = 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 . Then set the relia- bility goal of the second as R 2 = 0 . 90 R 1 or vice versa. This gives three pairs of selections to compare in terms of cost, geometry im-
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: plications, etc. 11-5 Establish a reliability goal of √ . 90 = . 95 for each bearing. For a 02-series angular con-tact ball bearing, C 10 = 854 · 1440 . 02 + 4 . 439[ln(1 / . 95)] 1 / 1 . 483 ¸ 1 / 3 = 11 315 lbf = 50 . 4 kN Select a 02-60 mm angular-contact bearing with C 10 = 55 . 9 kN. R A = exp ³ − ´ 1440(3 . 8 / 55 . 9) 3 − . 02 4 . 439 µ 1 . 483 ¶ = . 969 For a 03-series straight-roller bearing, C 10 = 10 . 279 · 1440 . 02 + 4 . 439[ln(1 / . 95)] 1 / 1 . 483 ¸ 3 / 10 = 105 . 2 kN Select a 03-60 mm straight-roller bearing with C 10 = 123 kN. R B = exp ³ − ´ 1440(10 . 28 / 123) 10 / 3 − . 02 4 . 439 µ 1 . 483 ¶ = . 977...
View Full Document

{[ snackBarMessage ]}