26_ch 06 Mechanical Design budynas_SM_ch06

26_ch 06 Mechanical Design budynas_SM_ch06 - K ts = 1 40...

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172 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design From Table A-10, p f = 0 . 008 89 R = 1 0 . 008 89 = 0 . 991 Ans . Note: The correlation method uses only the mean of S ut ; its variability is already included in the 0.138. When a deterministic load, in this case M , is used in a reliability estimate, en- gineers state, “For a Design Load of M , the reliability is 0.991.” They are in fact referring to a Deterministic Design Load. 6-35 For completely reversed torsion, k a and k b of Prob. 6-34 apply, but k c must also be con- sidered. Eq. 6-74: k c = 0 . 328(110) 0 . 125 LN (1, 0 . 125) = 0 . 590 LN (1, 0 . 125) Note 0.590 is close to 0.577. S Se = k a k b k c S ± e = 0 . 768[ LN (1, 0 . 058)](0 . 878)[0 . 590 LN (1, 0 . 125)][55 . 7 LN (1, 0 . 138)] ¯ S Se = 0 . 768(0 . 878)(0 . 590)(55 . 7) = 22 . 2 kpsi C Se = (0 . 058 2 + 0 . 125 2 + 0 . 138 2 ) 1 / 2 = 0 . 195 S Se = 22 . 2 LN (1, 0 . 195) kpsi Fig. A-15-15: D / d = 1 . 25, r / d = 0 . 125, then K ts = 1 . 40. From Eqs. (6-78), (6-79) and Table 6-15
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Unformatted text preview: K ts = 1 . 40 LN (1, 0 . 15) 1 + ( 2 / √ . 125 ) [(1 . 4 − 1) / 1 . 4](3 / 110) = 1 . 34 LN (1, 0 . 15) τ = K ts 16 T π d 3 τ = 1 . 34[ LN (1, 0 . 15)] ± 16(1 . 4) π (1) 3 ² = 9 . 55 LN (1, 0 . 15) kpsi From Eq. (5-43), p. 242: z = − ln ³ (22 . 2 / 9 . 55) ´ (1 + . 15 2 ) / (1 + . 195 2 ) µ ´ ln [(1 + . 195 2 )(1 + . 15 2 )] = − 3 . 43 From Table A-10, p f = . 0003 R = 1 − p f = 1 − . 0003 = . 9997 Ans . For a design with completely-reversed torsion of 1400 lbf · in, the reliability is 0.9997. The improvement comes from a smaller stress-concentration factor in torsion. See the note at the end of the solution of Prob. 6-34 for the reason for the phraseology....
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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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