Budynas ch20

# Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

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(b) f / ( N x ) = f / (69 · 10) = f / 690 Eq. (20-9) ¯ x = 8480 69 = 122 . 9 kcycles Eq. (20-10) s x = 1 104 600 8480 2 / 69 69 1 1 / 2 = 30.3 kcycles Ans. x f f x f x 2 f / ( N x ) 60 2 120 7200 0.0029 70 1 70 4900 0.0015 80 3 240 19 200 0.0043 90 5 450 40 500 0.0072 100 8 800 80 000 0.0116 110 12 1320 145 200 0.0174 120 6 720 86 400 0.0087 130 10 1300 169 000 0.0145 140 8 1120 156 800 0.0116 150 5 750 112 500 0.0174 160 2 320 51 200 0.0029 170 3 510 86 700 0.0043 180 2 360 64 800 0.0029 190 1 190 36 100 0.0015 200 0 0 0 0 210 1 210 44 100 0.0015 69 8480 1 104 600 Chapter 20 20-1 (a) 0 60 210 190 200 180 170 160 150 140 130 120 110 100 90 80 70 2 4 6 8 10 12

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2 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design 20-2 Data represents a 7-class histogram with N = 197. 20-3 Form a table: ¯ x = 4548 58 = 78 . 4 kpsi s x = 359 088 4548 2 / 58 58 1 1 / 2 = 6 . 57 kpsi From Eq. (20-14) f ( x ) = 1 6 . 57 2 π exp 1 2 x 78 . 4 6 . 57 2 x f f x f x 2 64 2 128 8192 68 6 408 27 744 72 6 432 31 104 76 9 684 51 984 80 19 1520 121 600 84 10 840 70 560 88 4 352 30 976 92 2 184 16 928 58 4548 359 088 x f f x f x 2 174 6 1044 181 656 182 9 1638 298 116 190 44 8360 1 588 400 198 67 13 266 2 626 688 206 53 10 918 2 249 108 214 12 2568 549 552 220 6 1320 290 400 197 39 114 7 789 900 ¯ x = 39 114 197 = 198 . 55 kpsi Ans. s x = 7 783 900 39 114 2 / 197 197 1 1 / 2 = 9 . 55 kpsi Ans.
Chapter 20 3 20-4 (a) y f f y f y 2 y f / ( N w ) f ( y ) g ( y ) 5.625 1 5.625 31.64063 5.625 0.072 727 0.001 262 0.000 295 5.875 0 0 0 5.875 0 0.008 586 0.004 088 6.125 0 0 0 6.125 0 0.042 038 0.031 194 6.375 3 19.125 121.9219 6.375 0.218 182 0.148 106 0.140 262 6.625 3 19.875 131.6719 6.625 0.218 182 0.375 493 0.393 667 6.875 6 41.25 283.5938 6.875 0.436 364 0.685 057 0.725 002 7.125 14 99.75 710.7188 7.125 1.018 182 0.899 389 0.915 128 7.375 15 110.625 815.8594 7.375 1.090 909 0.849 697 0.822 462 7.625 10 76.25 581.4063 7.625 0.727 273 0.577 665 0.544 251 7.875 2 15.75 124.0313 7.875 0.145 455 0.282 608 0.273 138 8.125 1 8.125 66.01563 8.125 0.072 727 0.099 492 0.106 72 55 396.375 2866.859 For a normal distribution, ¯ y = 396 . 375 / 55 = 7 . 207, s y = 2866 . 859 (396 . 375 2 / 55) 55 1 1 / 2 = 0 . 4358 f ( y ) = 1 0 . 4358 2 π exp 1 2 x 7 . 207 0 . 4358 2 For a lognormal distribution, ¯ x = ln 7 . 206 818 ln 1 + 0 . 060 474 2 = 1 . 9732, s x = ln 1 + 0 . 060 474 2 = 0 . 0604 g ( y ) = 1 x (0 . 0604)( 2 π ) exp 1 2 ln x 1 . 9732 0 . 0604 2 (b) Histogram 0 0.2 0.4 0.6 0.8 1 1.2 5.63 5.88 6.13 6.38 6.63 6.88 log N 7.13 7.38 7.63 7.88 8.13 Data N LN f

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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 . 7 0 . 5002 x 0 . 5008 0 otherwise F ( x ) = 0 x < 0 . 5002 1666 . 7 ( x 0 . 5002 ) 0 . 5002 x 0 . 5008 1 x > 0 . 5008 20-6 Dimensions produced are due to tool dulling and wear. When parts are mixed, the distrib- ution is uniform. From Eqs. (20-22) and (20-23), a = µ x 3 s = 0 . 6241 3(0 . 000 581) = 0 . 6231 in b = µ x + 3 s = 0 . 6241 + 3(0 . 000 581) = 0 . 6251 in We suspect the dimension was 0 . 623 0 . 625 in Ans.
Chapter 20 5 20-7 F ( x ) = 0 . 555 x 33 mm (a) Since F ( x ) is linear, the distribution is uniform at x = a F ( a ) = 0 = 0 . 555( a ) 33 a = 59.46 mm. Therefore, at x = b F ( b ) = 1 = 0 . 555 b 33 b = 61.26 mm. Therefore, F ( x ) = 0 x < 59 . 46 mm 0 . 555 x 33 59 . 46 x 61 . 26 mm 1 x > 61 . 26 mm The PDF is dF / dx , thus the range numbers are: f ( x ) = 0 . 555 59 . 46 x 61 . 26 mm 0 otherwise Ans. From the range numbers, µ x = 59 . 46 + 61 . 26 2 = 60 . 36 mm Ans. ˆ σ x = 61 . 26 59 . 46 2 3 = 0 . 520 mm Ans.

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