4_ch 20 Mechanical Design budynas_SM_ch20

# 4_ch 20 Mechanical Design budynas_SM_ch20 -...

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4 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 20-5 Distribution is uniform in interval 0.5000 to 0.5008 in, range numbers are a = 0.5000, b = 0.5008 in. (a) Eq. (20-22) µ x = a + b 2 = 0 . 5000 + 0 . 5008 2 = 0 . 5004 Eq. (20-23) σ x = b a 2 3 = 0 . 5008 0 . 5000 2 3 = 0 . 000 231 (b) PDF from Eq. (20-20) f ( x ) = ± 1250 0 . 5000 x 0 . 5008 in 0 otherwise (c) CDF from Eq. (20-21) F ( x ) = 0 x < 0 . 5000 ( x 0 . 5) / 0 . 0008 0 . 5000 x 0 . 5008 1 x > 0 . 5008 If all smaller diameters are removed by inspection, a = 0.5002, b = 0.5008 µ x = 0 . 5002 + 0 . 5008 2 = 0 . 5005 in ˆ σ x = 0 . 5008 0 . 5002 2 3 = 0 . 000 173 in f ( x ) = ± 1666 . 70 . 5002 x 0 . 5008 0 otherwise F ( x ) =
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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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