18_ch 20 Mechanical Design budynas_SM_ch20

# 18_ch 20 Mechanical Design budynas_SM_ch20 -...

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

18 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design z =− ln 49 . 6 38 . 197 ± 1 + 0 . 170 2 1 + 0 . 076 81 2 ² ln ³ (1 + 0 . 076 81 2 )(1 + 0 . 170 2 ) ´ 1 . 470 From Table A-10 p f = 0 . 0708 R = 1 p f = 0 . 929 Ans. 20-25 xn n x nx 2 93 19 1767 164 311 95 25 2375 225 625 97 38 3685 357 542 99 17 1683 166 617 101 12 1212 122 412 103 10 1030 106 090 105 5 525 55 125 107 4 428 45 796 109 4 436 47 524 111 2 222 24 624 136 13364 1315 704 ¯ x = 13 364 / 136 = 98 . 26 kpsi s x = µ 1 315 704 13 364 2 / 136 135 1 / 2 = 4 . 30 kpsi Under normal hypothesis, z 0 . 01 = ( x 0 . 01 98 . 26) / 4 . 30 x 0 . 01 = 98 . 26 + 4 . 30 z 0 . 01 = 98 . 26 + 4 . 30( 2 . 3267) = 88 . 26 . = 88 . 3 kpsi Ans. 20-26 From Prob. 20-25,
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online