# Shigleys ch20

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Chapter 20
20-1 (a)
12 10 8 6 4 2 0

60

70

80

90 100 110 120 130 140 150 160 170 180 190 200 210

(b) f/(N x) = f/(69 · 10) = f/690 x 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 f 2 1 3 5 8 12 6 10 8 5 2 3 2 1 0 1 69 fx 120 70 240 450 800 1320 720 1300 1120 750 320 510 360 190 0 210 8480 f x2 7200 4900 19 200 40 500 80 000 145 200 86 400 169 000 156 800 112 500 51 200 86 700 64800 36 100 0 44 100 1 104 600 f/(N x) 0.0029 0.0015 0.0043 0.0072 0.0116 0.0174 0.0087 0.0145 0.0116 0.0174 0.0029 0.0043 0.0029 0.0015 0 0.0015

Eq. (20-9) Eq. (20-10)

x= ¯

8480 = 122.9 kcycles 69
1/2

1 104 600 − 84802 /69 sx = 69 − 1 = 30.3 kcycles Ans.

2

Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design

20-2

Data represents a 7-class histogramwith N = 197. x 174 182 190 198 206 214 220 f 6 9 44 67 53 12 6 197 x= ¯ fx 1044 1638 8360 13 266 10 918 2568 1320 39 114 Ans.
1/2

f x2 181 656 298 116 1 588 400 2 626 688 2 249 108 549 552 290 400 7 789 900

39 114 = 198.55 kpsi 197

7 783 900 − 39 1142 /197 sx = 197 − 1 = 9.55 kpsi 20-3 Form a table: x 64 68 72 76 80 84 88 92 f 2 6 6 9 19 10 4 2 58 x= ¯ sx = From Eq. (20-14) 1 1 f (x) = √ exp −2 6.57 2π fx 128 408 432 684 1520 840 352 184 4548 Ans.

f x2 8192 27 744 31 104 51 984 121 600 70 560 30 976 16 928 359 088

4548 = 78.4 kpsi 58 359 088 − 45482 /58 58 − 1
1/2

= 6.57 kpsi

x − 78.4 6.57

2

Chapter 20

3

20-4 (a) y 5.625 5.875 6.125 6.375 6.625 6.875 7.125 7.375 7.625 7.875 8.125 f 1 0 0 3 3 6 14 15 10 2 1 55 fy 5.625 0 0 19.125 19.875 41.25 99.75 110.625 76.25 15.75 8.125396.375 f y2 31.640 63 0 0 121.9219 131.6719 283.5938 710.7188 815.8594 581.4063 124.0313 66.015 63 2866.859 y 5.625 5.875 6.125 6.375 6.625 6.875 7.125 7.375 7.625 7.875 8.125 f/(Nw) 0.072 727 0 0 0.218 182 0.218 182 0.436 364 1.018 182 1.090 909 0.727 273 0.145 455 0.072 727 f (y) 0.001 262 0.008 586 0.042 038 0.148 106 0.375 493 0.685 057 0.899 389 0.849 697 0.577 665 0.282 608 0.099 492 g(y)0.000 295 0.004 088 0.031 194 0.140 262 0.393 667 0.725 002 0.915 128 0.822 462 0.544 251 0.273 138 0.106 72

For a normal distribution, y = 396.375/55 = 7.207, ¯ f ( y) = sy = 1 2866.859 − (396.3752 /55) 55 − 1 x − 7.207 0.4358
2

1/2

= 0.4358

1 √ exp − 2 0.4358 2π

For a lognormal distribution, √ x = ln 7.206 818 − ln 1 + 0.060 4742 = 1.9732, ¯ g( y) = (b) Histogram
f 1.2 1 0.8 0.6 0.4 0.2 0 5.63Data N LN

√ sx = ln 1 + 0.060 4742 = 0.0604 ln x − 1.9732 0.0604
2

1 exp − √ 2 x(0.0604)( 2π)

1

5.88

6.13

6.38

6.63

6.88 7.13 log N

7.38

7.63

7.88

8.13

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) Eq. (20-23) µx= σx = a+b 0.5000 + 0.5008 = = 0.5004 2 2 b−a 0.5008 − 0.5000 = 0.000 231 √ = √ 2 3 2 3

(b) PDF from Eq. (20-20) f (x) = (c) CDF from Eq. (20-21)  x < 0.5000 0 F(x) = (x − 0.5)/0.0008 0.5000 ≤ x ≤ 0.5008  x > 0.5008 1 If all smaller diameters are removed by inspection, a = 0.5002, b = 0.5008 µx = σx = ˆ f (x) = 0.5002 + 0.5008 = 0.5005 in 2 0.5008 − 0.5002 = 0.000 173 in √ 2 3 1666.7 0.5002 ≤x ≤ 0.5008 0 otherwise 1250 0.5000 ≤ x ≤ 0.5008 in 0 otherwise

 x < 0.5002 0 F(x) = 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 distribution is uniform. From Eqs. (20-22) and (20-23), √ √ a = µx − 3s = 0.6241 − 3(0.000 581) = 0.6231 in √ √ b = µx + 3s = 0.6241 + 3(0.000 581) = 0.6251 in We suspect thedimension was 0.623 in 0.625 Ans.

Chapter 20

5

20-7

F(x) = 0.555x − 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 = 0.555b − 33 ∴ b = 61.26 mm. Therefore,  x < 59.46 mm 0 F(x) = 0.555x − 33 59.46 ≤ x ≤ 61.26 mm  1 x > 61.26 mm The PDF is d F/dx , thus the range numbers are: f (x) = From the...