Complete Solution Manual Shigleys Mechanical Engineering Design 11th Edition Budynas Questions & Answers with rationales (Chapter 1-20)

Chapter 1 Solutions, Page 1/12 Shigleys Mechanical Engineering Design 11th Edition Budynas Solutions Manual 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 nd 1.10 0.850.952  1.43 Ans. 1-10 (a) X1 + X2: x1  x2  X1  e1  X 2  e2 error  e   x1  x2    X1  X 2 
 e1  e2 Ans. (b) X1  X2: (c) X1 X2: x1  x2  X1  e1   X 2  e2 
e   x1  x2    X1  X 2   e1  e2 x1x2   X1  e1  X 2  e2 
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Ans. e  x1 x2  X1 X 2  X1e2  X 2e1  e1e2  X e  X e  X X  e1  e2 
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Ans. 1 2 2 1 1 2  X X 
 1 2  Chapter 1 Solutions, Page 2/12 4   4.79 d (d) X1/X2: x1  X1  e1  X1  1 e1 X1 
x X  e X  1 e X 
2 2 2 2  2 2 
 e 1 e  1 e X  e  e  e e 1 2  1  2 then  1 1   1 1 1 2   1 1  2  X 2  X 2  1 e2 X 2  X1 X 2  X1 X 2 Thus, e  x1  X1 
 X1  e1  e2 
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Ans. x X X  X X 
2 2 2  1 2 
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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 = 0.014 178 435 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 S 32100025103 
1-12     
   d  1.006 in Ans. n d 3 2.5 Table A-17: d = 11 in Ans. Factor of safet y: n  S 25103 
321000
 1.253 Ans. Chapter 1 Solutions, Page 3/12  i i k f x  N x N  1 2 2 i 1 1-13 (a) Eq. (1-6) 1 k x N i 1 fixi  8480 69  122.9 kcycles Eq. (1-7) 1104 600  69(122.9)2 1/ 2 sx    
 30.3 kcycles Ans.  69  1 
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(b) Eq. (1-5) z115 x  x ˆ  x115  x s  115  122.9 30.3  0.2607 x x 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 Chapter 1 Solutions, Page 4/12 
i1 k f x  N x 2 2 i i N 1 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) 1 k x N i 1 fixi  39 126 197  198.61 kpsi Eq. (1-7)  7 789 204 197(198.61)2 1 2 s    9.68 kpsi Ans. x  197 1 
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1-15 L  122.9 kcycles and sL  30.3 kcycles Eq. (1-5) z  x  x  x10  L  x10 122.9 10 ˆ s 30.3 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. 
L