14_ch 06 Mechanical Design budynas_SM_ch06

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

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

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.
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

{[ snackBarMessage ]}

Ask a homework question - tutors are online