26_ch 05 Mechanical Design budynas_SM_ch05

26_ch 05 Mechanical Design budynas_SM_ch05 - r ) o = Inner...

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

View Full Document Right Arrow Icon
140 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-37 δ = 1 . 9998 2 1 . 999 2 = 0 . 0004 in Eq. (3-56) 0 . 0004 = p (1) 14 . 5(10 6 ) ± 2 2 + 1 2 2 2 1 2 + 0 . 211 ² + p (1) 30(10 6 ) ± 1 2 + 0 1 2 0 0 . 292 ² p = 2613 psi Applying Eq. (4-58) at R , ( σ t ) o = 2613 ³ 2 2 + 1 2 2 2 1 2 ´ = 4355 psi ( σ r ) o =− 2613 psi, S ut = 20 kpsi, S uc = 83 kpsi µ µ µ µ σ o σ A µ µ µ µ = 2613 4355 < 1, use Eq. (5-32a) h = S ut A = 20 / 4 . 355 = 4 . 59 Ans. 5-38 E = 30(10 6 ) psi, ν = 0 . 292, I = ( π/ 64)(2 4 1 . 5 4 ) = 0 . 5369 in 4 Eq. (3-57) can be written in terms of diameters, p = E δ d D ( d 2 o D 2 )( D 2 d 2 i ) 2 D 2 ( d 2 o d 2 i ) · = 30(10 6 ) 1 . 75 (0 . 002 46) ± (2 2 1 . 75 2 )(1 . 75 2 1 . 5 2 ) 2(1 . 75 2 )(2 2 1 . 5 2 ) ² = 2997 psi = 2 . 997 kpsi Outer member: Outer radius: ( σ t ) o = 1 . 75 2 (2 . 997) 2 2 1 . 75 2 (2) = 19 . 58 kpsi, ( σ
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: r ) o = Inner radius: ( t ) i = 1 . 75 2 (2 . 997) 2 2 1 . 75 2 1 + 2 2 1 . 75 2 = 22 . 58 kpsi, ( r ) i = 2 . 997 kpsi Bending: r o : ( x ) o = 6 . 000(2 / 2) . 5369 = 11 . 18 kpsi r i : ( x ) i = 6 . 000(1 . 75 / 2) . 5369 = 9 . 78 kpsi Torsion: J = 2 I = 1 . 0738 in 4 r o : ( xy ) o = 8 . 000(2 / 2) 1 . 0738 = 7 . 45 kpsi r i : ( xy ) i = 8 . 000(1 . 75 / 2) 1 . 0738 = 6 . 52 kpsi...
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