Shigley’s Mechanical Engineering Design, Eighth Edition By Budynas and Nisbet
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Solution Manual for All Chapters Shigley’s MED, 10th edition Chapter 15 Solutions , Page 1/20 Chapter 1 Problems 1-1 through 1-6 are for student research. No standard solutions are provided. 1-7 From Fig. 1 -2, cost of grinding to 0.0005 in is 270%. Cost of turning to 0.003 in is 60%. Relative cost of grinding vs. turning = 270/60 = 4.5 times Ans. ______________________________________________________________________________ 1-8 CA = CB, 10 + 0.8 P = 60 + 0.8 P 0.005 P 2 P 2 = 50/0.005 P = 100 parts Ans. ______________________________________________________________________________ 1-9 Max. load = 1.10 P Min. area = (0.95)2A Min. strength = 0.85 S To offset the absolute uncertainties, the design factor, from Eq. (1 -1) should be 21.101.43 .
0.85 0.95dn Ans ______________________________________________________________________________ 1-10 (a) X1 + X2: 1 2 1 1 2 2
1 2 1 2
12error .x x X e X e
e x x X X
e e Ans
(b) X1 X2:
1 2 1 1 2 2
1 2 1 2 1 2 .x x X e X e
e x x X X e e Ans
(c) X1 X2: 1 2 1 1 2 2
1 2 1 2 1 2 2 1 1 2
12
1 2 2 1 1 2
12 .x x X e X e
e x x X X X e X e e e
eeX e X e X X AnsXX
Solution Manual for All Chapters Shigley’s MED, 10th edition Chapter 15 Solutions , Page 2/20 (d) X1/X2: 1 1 1 1 1 1
2 2 2 2 2 2
1
2 2 1 1 1 2 1 2
2 2 2 2 1 2 1 2
1 1 1 1 2
2 2 2 1 21
1
11 1 then 1 1 11
Thus, .x X e X e X
x X e X e X
e e e X e e e e
X X e X X X X X
x X X e ee Ansx X X X X
______________________________________________________________________________ 1-11 (a) x1 = 7 = 2.645 751 311 1 X1 = 2.64 (3 correct digits) x2 = 8 = 2.828 427 124 7 X2 = 2.82 (3 correct digits) x1 + x2 = 5.474 178 435 8 e1 = x1 X1 = 0.005 751 311 1 e2 = x2 X2 = 0.008 427 124 7 e = e1 + e2 = 0.0 14 178 435 8 Sum = x1 + x2 = X1 + X2 + e = 2.64 + 2.82 + 0.014 178 435 8 = 5.474 178 435 8 Checks (b) X1 = 2.65, X2 = 2.83 (3 digit significant numbers) e1 = x1 X1 = 0.004 248 688 9 e2 = x2 X2 = 0.001 572 875 3 e = e1 + e2 = 0.005 821 564 2 Sum = x1 + x2 = X1 + X2 + e = 2.65 +2.83 0.001 572 875 3 = 5.474 178 435 8 Checks ______________________________________________________________________________ 1-12 3
325 10 32 10001.006 in .2.5dSd Ansnd Table A-17: d = 1
41 in Ans. Factor of safety:
3
325 10
4.79 .32 1000
1.25Sn Ans
______________________________________________________________________________ Solution Manual for All Chapters Shigley’s MED, 10th edition Chapter 15 Solutions , Page 3/20 1-13 (a) Eq. (1-6) 11 8480122.9 kcycles69k
ii
ix f xN Eq. (1-7) 221/ 22
1 1104 600 69(122.9)30.3 kcycles .1 69 1k
ii
i
xf x N x
s AnsN (b) Eq. (1 -5) 115
115115 122.90.2607ˆ 30.3x
xxx x xzs
Interpolating from Table (A -10) 0.2600 0.3974 0.2607 x x = 0.3971 0.2700 0.3936 N(0.2607) = 69 (0.3971) = 27.4 27 Ans. From the data, the number of instances less than 115 kcycles is x f f x f x2 60 2 120 7200 70 1 70 4900 80 3 240 19200 90 5 450 40500 100 8 800 80000 110 12 1320 145200 120 6 720 86400 130 10 1300 169000 140 8 1120 156800 150 5 750 112500 160 2 320 51200 170 3 510 86700 180 2 360 64800 190 1 190 36100 200 0 0 0 210 1 210 44100 69 8480 1 104 600 Solution Manual for All Chapters Shigley’s MED, 10th edition Chapter 15 Solutions , Page 4/20 2 + 1 + 3 + 5 + 8 + 12 = 31 (the data is not perfectly normal) ____________________________________________________________________________ 1-14 x f f x f x2 174 6 1044 181656 182 9 1638 298116 190 44 8360 1588400 198 67 13266 2626668 206 53 10918 2249108 214 12 2568 549552 222 6 1332 295704 197 39126 7789204 Eq. (1-6) 11 39 126198.61 kpsi197k
ii
ix f xN Eq. (1 -7) 22122
1 7 789 204 197(198.61)9.68 kpsi .1 197 1k
ii
i
xf x N x
s AnsN ______________________________________________________________________________ 1-15 122.9 kcycles and 30.3 kcyclesL Ls Eq. (1 -5) 10 10
10122.9
ˆ 30.3x
Lx x L xzs
Thus, x10 = 122.9 + 30.3 z10 = L10 From Table A-10, for 10 percent failure, z10 = 1.282 . Thus, L10 = 122.9 + 30.3( 1.282) = 84.1 kcycles Ans. ___________________________________________________________________________
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