30_ch 06 Mechanical Design budynas_SM_ch06

# 30_ch 06 Mechanical Design budynas_SM_ch06 - . 138) kpsi S...

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176 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design K t = 2 . 1 ¯ K f = 2 . 1 1 + ( 2 / 0 . 25 ) [(2 . 1 1) / (2 . 1)](4 / 64) = 1 . 857 h = ¯ K f ¯ nF a w 2 ¯ S e = 1 . 857(2 . 02)(10) 2 . 5(25 . 5) = 0 . 667 Ans. This thickness separates ¯ S e and ¯ σ a so as to realize the reliability goal of 0.999 at each shoulder. The design decision is to make t the next available thickness of 1018 CD steel strap from the same heat. This eliminates machining to the desired thickness and the extra cost of thicker work stock will be less than machining the fares. Ask your steel supplier what is available in this heat . 6-39 F a = 1200 lbf S ut = 80 kpsi (a) Strength k a = 2 . 67(80) 0 . 265 LN (1, 0 . 058) = 0 . 836 LN (1, 0 . 058) k b = 1 k c = 1 . 23(80) 0 . 078 LN (1, 0 . 125) = 0 . 874 LN (1, 0 . 125) S ± a = 0 . 506(80) LN (1, 0 . 138) = 40 . 5 LN (1, 0
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Unformatted text preview: . 138) kpsi S e = . 836[ LN (1, 0 . 058)](1)[0 . 874 LN (1, 0 . 125)][40 . 5 LN (1, 0 . 138)] ¯ S e = . 836(1)(0 . 874)(40 . 5) = 29 . 6 kpsi C Se = (0 . 058 2 + . 125 2 + . 138 2 ) 1 / 2 = . 195 Stress: Fig. A-15-1; d /w = . 75 / 1 . 5 = . 5, K t = 2 . 17. From Eqs. (6-78), (6-79) and Table 6-15 K f = 2 . 17 LN (1, 0 . 10) 1 + ( 2 / √ . 375 ) [(2 . 17 − 1) / 2 . 17](5 / 80) = 1 . 95 LN (1, 0 . 10) σ a = K f F a ( w − d ) t , C σ = . 10 ¯ σ a = ¯ K f F a ( w − d ) t = 1 . 95(1 . 2) (1 . 5 − . 75)(0 . 25) = 12 . 48 kpsi 1200 lbf 3 4 " 1 4 " 1 2 1 "...
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## This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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