10_ch 05 Mechanical Design budynas_SM_ch05

10_ch 05 Mechanical Design budynas_SM_ch05 - . 27 Ans. (d)...

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

View Full Document Right Arrow Icon
124 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (a) σ x = 90 kpsi, σ y =− 50 kpsi, σ z = 0 ± σ A = 90 kpsi and σ B =− 50 kpsi. For the fourth quadrant, from Eq. (5-31b) n = 1 ( σ A / S yt ) ( σ B / S uc ) = 1 (90 / 235) ( 50 / 275) = 1 . 77 Ans. (b) σ x = 120 kpsi, τ xy =− 30 kpsi ccw. σ A , σ B = 127 . 1, 7 . 08 kpsi. For the fourth quadrant n = 1 (127 . 1 / 235) ( 7 . 08 / 275) = 1 . 76 Ans. (c) σ x =− 40 kpsi, σ y =− 90 kpsi, τ xy = 50 kpsi. σ A , σ B =− 9 . 10, 120 . 9 kpsi . Although no solution exists for the third quadrant, use n =− S yc σ y =− 275 120 . 9 = 2
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . 27 Ans. (d) σ x = 110 kpsi, σ y = 40 kpsi, τ xy = 10 kpsi cw. σ A , σ B = 111 . 4, 38 . 6 kpsi. For the first quadrant n = S yt σ A = 235 111 . 4 = 2 . 11 Ans. Graphical Solution: (a) n = OB O A = 1 . 82 1 . 02 = 1 . 78 (b) n = OD OC = 2 . 24 1 . 28 = 1 . 75 (c) n = OF OE = 2 . 75 1 . 24 = 2 . 22 (d) n = OH OG = 2 . 46 1 . 18 = 2 . 08 O ( d ) ( b ) ( a ) ( c ) E F B D G C A H 1 in ± 100 kpsi ± B ± A...
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