17_ch 12 Mechanical Design budynas_SM_ch12

# 17_ch 12 Mechanical Design budynas_SM_ch12 -...

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320 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design In this problem we will take journal diameter as the nominal value and the bushing bore as a variable. In the next problem, we will take the bushing bore as nominal and the jour- nal diameter as free. To determine where the constraints are, we will set t b = t d = 0, and thereby shrink the design window to a point. We set d = 2 . 000 in b = d + 2 c min = d + 2 c n d = 2 (This makes Trumpler’s n d 2 tight) and construct a table. cb d ¯ T f * T max h o P st T max n fom 0.0010 2.0020 2 215.50 312.0 × ± × ± 5.74 0.0011 2.0022 2 206.75 293.0 × ±± ± 6.06 0.0012 2.0024 2 198.50 277.0 × ± 6.37 0.0013 2.0026 2 191.40 262.8 × ± 6.66 0.0014 2.0028 2 185.23 250.4 × ± 6.94 0.0015 2.0030 2 179.80 239.6 × ± 7.20 0.0016 2.0032 2 175.00 230.1 × ± 7.45 0.0017 2.0034 2 171.13 220.3 × ± 7.65 0.0018 2.0036 2 166.92 213.9 ±±±± 7.91 0.0019 2.0038 2 163.50 206.9 8.12 0.0020 2.0040 2 160.40 200.6 8.32 *Sample calculation for the ﬁrst entry of this column. Iteration yields: ¯ T f = 215 . 5°F With ¯ T f = 215 . 5°F, from Table 12-1 µ = 0 . 0136(10 6 )exp[1271 . 6 / (215 . 5 + 95)] = 0 . 817(10
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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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