Chapter_20_Solutions

Chapter_20_Solutions - Chapter 20 20-1 (a) (b) f / (Nx) = f...

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Chapter 20 20-1 ( a ) ( b ) f / ( N x ) = f / [69(10)] = f / 690 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 19200 0.0043 90 5 450 40500 0.0072 100 8 800 80000 0.0116 110 12 1320 145200 0.0174 120 6 720 86400 0.0087 130 10 1300 169000 0.0145 140 8 1120 156800 0.0116 150 5 750 112500 0.0174 160 2 320 51200 0.0029 170 3 510 86700 0.0043 180 2 360 64800 0.0029 190 1 130 36100 0.0015 200 0 0 0 0 210 1 210 44100 0.0015 69 8480 1 104 600 Chapter 20, Page 1/29
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Eq. (20-9): 8480 122.9 kcycles 69 x  Eq. (20-10): 12 2 1 104 600 8480 / 69 30.3 kcycles . 69 1 x sA    n s ______________________________________________________________________________ 20-2 Data represents a 7-class histogram with N = 197. 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 39 114 198.55 kpsi . 197 x Ans 2 7 783 900 39 114 /197 9.55 kpsi . 197 1 n s ______________________________________________________________________________ 20-3 Form a Table: x f fx fx 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 Chapter 20, Page 2/29
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4548 78.4 kpsi . 58 x Ans  12 2 359 088 4548 / 58 6.57 kpsi . 58 1 x sA    n s From Eq. 20-14  2 1 1 78.4 exp . 26 . 5 7 6.57 2 x f xA     n s ______________________________________________________________________________ 20-4 (a) x 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 66.015 63 8.125 0.072 727 0.099 492 0.106 720 55 396.375 2866.859 For a normal distribution,   2 2866.859 396.375 /55 396.375/55 7.207, 0.4358 55 1 y ys  2 1 1 7.207 exp 2 0.4358 0.4358 2 x fy For a lognormal distribution, 22 ln 7.206 818 ln 1 0.060 474 1.9732, ln 1 0.060 474 0.0604 x xs   2 1 1 ln 1.9732 exp 2 0.0604 0.0604 2 x gy x Chapter 20, Page 3/29
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(b) Histogram ______________________________________________________________________________ 20-5 Distribution is uniform in interval 0.5000 to 0.5008 in, range numbers are a = 0.5000 in, b = 0.5008 in. (a) Eq. (20-22) 0.5000 0.5008 0.5004 22 x ab  Eq. (20-23) 0.5008 0.5000 0.000 231 2 23 x ba (b) PDF, Eq. (20-20) 1250 0.5000 x 0.5008 in () 0 otherwise fx  (c) CDF, Eq. (20-21) 0 0.5000 in ( ) ( 0.5) / 0.0008 0.5000 0.5008 in 1 0.5008 in x Fx x x x  If all smaller diameters are removed by inspection, a = 0.5002 in, b = 0.5008 in, 0.5002 0.5008 0.5005 in 2 0.5008 0.5002 ˆ 0.000 173 in x x 1666.7 0.5002 0.5008 in 0 otherwise x Chapter 20, Page 4/29
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0 0.5002 in ( ) 1666.7( 0.5002) 0.5002 0.5008 in 1 0.5008 in x Fx x x x  ______________________________________________________________________________ 20-6 Dimensions produced are due to tool dulling and wear. When parts are mixed, the distribution is uniform. From Eqs. (20-22) and (20-23),    3 0.6241 3 0.000 581 0.6231 in 3 0.6241 3 0.000 581 0.6251 in x x as bs  We suspect the dimension was 0.623 in . 0.625 Ans ______________________________________________________________________________ 20-7 F ( x ) = 0.555 x – 33 mm. (a) Since F ( x ) is linear, 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, 0 59.46 mm ( ) 0.555 33 59.46 61.26 mm 61.26 mm x x x x The PDF is dF/dx , thus the range numbers are: 0.555 59.46 61.26 mm ( ) . 0 otherwise x f xA  n s From the range numbers, 59.46 61.26 60.36 mm . 2 61.26 59.46 ˆ 0.520 mm . 23 x x Ans Ans  Chapter 20, Page 5/29
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( b ) is an uncorrelated quotient 2 3600 lbf, 0.112 in FA  300 3600 0.083 33, 0.001 0.112 0.008 929 CC From Table 20-6, For  1/2 22 2 3600 32 143 psi .
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This note was uploaded on 11/20/2011 for the course MCE 325 taught by Professor Ahm during the Spring '11 term at American Dubai.

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Chapter_20_Solutions - Chapter 20 20-1 (a) (b) f / (Nx) = f...

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