24_ch 07 Mechanical Design budynas_SM_ch07

24_ch 07 Mechanical Design budynas_SM_ch07 - Hole: Eq....

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Chapter 7 201 K ± ts has the following attributes: • It exhibits a minimum; • It changes little over a wide range; • Its minimum is a stationary point minimum at a / D . = 0 . 100 ; • Our knowledge of the minima location is 0 . 075 ( a / D ) 0 . 125 We can form a design rule: in torsion, the pin diameter should be about 1 / 10 of the shaft diameter, for greatest shaft capacity. However, it is not catastrophic if one forgets the rule. 7-20 Choose 15 mm as basic size, D , d . Table 7-9: fit is designated as 15H7/h6. From Table A-11, the tolerance grades are ± D = 0 . 018 mm and ± d = 0 . 011 mm. Hole: Eq. (7-36) D max = D + ± D = 15 + 0 . 018 = 15 . 018 mm Ans. D min = D = 15 . 000 mm Ans. Shaft: From Table A-12, fundamental deviation δ F = 0. From Eq. (2-39) d max = d + δ F = 15 . 000 + 0 = 15 . 000 mm Ans. d min = d + δ R ± d = 15 . 000 + 0 0 . 011 = 14 . 989 mm Ans. 7-21 Choose 45 mm as basic size. Table 7-9 designates fit as 45H7/s6. From Table A-11, the tolerance grades are ± D = 0 . 025 mm and ± d = 0 . 016 mm
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Unformatted text preview: Hole: Eq. (7-36) D max = D + D = 45 . 000 + . 025 = 45 . 025 mm Ans. D min = D = 45 . 000 mm Ans. Shaft: From Table A-12, fundamental deviation F = + . 043 mm. From Eq. (7-38) d min = d + F = 45 . 000 + . 043 = 45 . 043 mm Ans. d max = d + F + d = 45 . 000 + . 043 + . 016 = 45 . 059 mm Ans. 7-22 Choose 50 mm as basic size. From Table 7-9 t is 50H7/g6. From Table A-11, the tolerance grades are D = . 025 mm and d = . 016 mm. Hole: D max = D + D = 50 + . 025 = 50 . 025 mm Ans. D min = D = 50 . 000 mm Ans. Shaft: From Table A-12 fundamental deviation = 0.009 mm d max = d + F = 50 . 000 + ( . 009) = 49 . 991 mm Ans. d min = d + F d = 50 . 000 + ( . 009) . 016 = 49 . 975 mm...
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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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