Quiz 2 solutions_Q3 solution corrected

9935 mm d max 25 00065 250065 mm transion t 5 q3

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Unformatted text preview: e) Lower limit of the shaA Upper limit of the shaA f) d min = 25 − 0.0065 = 24.9935 mm d max = 25 + 0.0065 = 25.0065 mm Transi*on fit. 5 Q3 10 R7/h6 + Fundamental deviation - Zero line R7 h6 Fundamental deviation Basic Size Q3 a) R7/h6 è༎ Basic ShaA System b) Basic size D = 10 mm, tolerance on the hole: R7 → - 13 µm - 28 µm 0 µm c) Tolerance on the shaA h6 → - 9 µm 7 Q3 d) Lower limit of the hole Dmin = 10 - 0.028 = 9.972 mm Upper limit of the hole Dmax = 10 − 0.013 = 9.987 mm e) Lower limit of the shaA Upper limit of the shaA f) d min = 10 − 0.09 = 9.991 mm d max = 10 mm Interference fit. 8...
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This document was uploaded on 01/21/2014.

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