25_ch 07 Mechanical Design budynas_SM_ch07

# 25_ch 07 Mechanical Design budynas_SM_ch07 - p max = E δ...

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202 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 7-23 Choose the basic size as 1.000 in. From Table 7-9, for 1.0 in, the ﬁt is H8/f7. From Table A-13, the tolerance grades are ± D = 0 . 0013 in and ± d = 0 . 0008 in. Hole: D max = D + ( ± D ) hole = 1 . 000 + 0 . 0013 = 1 . 0013 in Ans. D min = D = 1 . 0000 in Ans. Shaft: From Table A-14: Fundamental deviation = − 0.0008 in d max = d + δ F = 1 . 0000 + ( 0 . 0008) = 0 . 9992 in Ans. d min = d + δ F ± d = 1 . 0000 + ( 0 . 0008) 0 . 0008 = 0 . 9984 in Ans. Alternatively, d min = d max ± d = 0 . 9992 0 . 0008 = 0 . 9984 in. Ans. 7-24 (a) Basic size is D = d = 1 . 5in . Table 7-9: H7/s6 is speciﬁed for medium drive ﬁt. Table A-13: Tolerance grades are ± D = 0 . 001 in and ± d = 0 . 0006 in . Table A-14: Fundamental deviation is δ F = 0 . 0017 in . Eq. (7-36): D max = D + ± D = 1 . 501 in Ans . D min = D = 1 . 500 in Ans . Eq. (7-37): d max = d + δ F + ± d = 1 . 5 + 0 . 0017 + 0 . 0006 = 1 . 5023 in Ans . Eq. (7-38): d min = d + δ F = 1 . 5 + 0 . 0017 + 1 . 5017 in Ans . (b) Eq. (7-42): δ min = d min D max = 1 . 5017 1 . 501 = 0 . 0007 in Eq. (7-43): δ max = d max D min = 1 . 5023 1 . 500 = 0 . 0023 in Eq. (7-40):
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Unformatted text preview: p max = E δ max 2 d 3 ± ( d 2 o − d 2 )( d 2 − d 2 i ) d 2 o − d 2 i ² = (30)(10 6 )(0 . 0023) 2(1 . 5) 3 ³ (2 . 5 2 − 1 . 5 2 )(1 . 5 2 − 0) 2 . 5 2 − ´ = 14 720 psi Ans . p min = E δ min 2 d 3 ± ( d 2 o − d 2 )( d 2 − d 2 i ) d 2 o − d 2 i ² = (30)(10 6 )(0 . 0007) 2(1 . 5) 3 ³ (2 . 5 2 − 1 . 5 2 )(1 . 5 2 − 0) 2 . 5 2 − ´ = 4480 psi Ans . (c) For the shaft: Eq. (7-44): σ t ,shaft = − p = − 14 720 psi Eq. (7-46): σ r ,shaft = − p = − 14 720 psi Eq. (5-13): σ ± = ( σ 2 1 − σ 1 σ 2 + σ 2 2 ) 1 / 2 = [( − 14 720) 2 − ( − 14 720)( − 14 720) + ( − 14 720) 2 ] 1 / 2 = 14 720 psi n = S y /σ ± = 57 000 / 14 720 = 3 . 9 Ans ....
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