14_ch 06 Mechanical Design budynas_SM_ch06

14_ch 06 Mechanical Design budynas_SM_ch06 - Ans 6-22 The...

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

View Full Document Right Arrow Icon
160 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Fatigue: n f = 605 244 . 4 = 2 . 48 Thus, the spring is likely to fail in fatigue at the inner radius. Ans. At Outer Radius ( σ o ) a = 456 . 9 152 . 3 2 = 152 . 3MPa ( σ o ) m = 456 . 9 + 152 . 3 2 = 304 . 6MPa Yield load line: σ m = 152 . 3 + σ a Langer line: σ m = 1504 σ a = 152 . 3 + σ a Intersection: σ a = 675 . 9MPa, σ m = 828 . 2MPa n y = 675 . 9 152 . 3 = 4 . 44 Fatigue line: σ a = [1 ( σ m / S ut ) 2 ] S e = σ m 152 . 3 605 ± 1 ² σ m 1671 ³ 2 ´ = σ m 152 . 3 σ 2 m + 4615 . 3 σ m 3 . 4951(10 6 ) = 0 σ m = 4615 . 3 + µ 4615 . 3 2 + 4(3 . 4951)(10 6 ) 2 = 662 . 2MPa σ a = 662 . 2 152 . 3 = 509 . 9MPa n f = 509 . 9 152 . 3 = 3 . 35 Thus, the spring is not likely to fail in fatigue at the outer radius.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Ans. 6-22 The solution at the inner radius is the same as in Prob. 6-21. At the outer radius, the yield solution is the same. Fatigue line: σ a = ² 1 − σ m S ut ³ S e = σ m − 152 . 3 605 ¶ 1 − σ m 1671 · = σ m − 152 . 3 1 . 362 σ m = 757 . 3 ⇒ σ m = 556 . 0 MPa σ a = 556 . − 152 . 3 = 403 . 7 MPa n f = 403 . 7 152 . 3 = 2 . 65 Ans ....
View Full Document

This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

Ask a homework question - tutors are online