Solution Manual for Shigleys Mechanical Engineering Design 11th Edition Budynas / All Chapters 1 - 20 / Full Complete 2025

Solution Manual for Shigleys Mechanical Engineering Design 11th Edition Budynas / All Chapters 1 - 20 / Full Complete 2025 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  2 1.10 1.43 . 0.850.95 d n Ans   ______________________________________________________________________________ 1-10 (a) X1 + X2:     1 2 1 1 2 2 1 2 1 2 1 2 error . 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 1 2 1 2 2 1 1 2 1 2 . xx X e X e e xx XX Xe Xe ee e e Xe Xe XX Ans X X                   Shigley’s MED, 11th edition Chapter 1 Solutions, Page 1/12 (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 2 1 1 1 1 1 then 1 1 1 1 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 e e Ans x X X X X                                                      ______________________________________________________________________________ 1-11 (a) x1 = 7 = 2. X1 = 2.64 (3 correct digits) x2 = 8 = 2. X2 = 2.82 (3 correct digits) x1 + x2 = 5. e1 = x1  X1 = 0. e2 = x2  X2 = 0. e = e1 + e2 = 0. Sum = x1 + x2 = X1 + X2 + e = 2.64 + 2.82 + 0. = 5. Checks (b) X1 = 2.65, X2 = 2.83 (3 digit significant numbers) e1 = x1  X1 =  0. e2 = x2  X2 =  0. e = e1 + e2 =  0. Sum = x1 + x2 = X1 + X2 + e = 2.65 +2.83  0. = 5. Checks ______________________________________________________________________________ 1-12     3 3 .006 in . 2.5 d S d Ans n d        Table A-17: d = 1 4 1 in Ans. Factor of safety:       3 3 2510 4.79 . 321000 1.25 S n Ans      ______________________________________________________________________________ Shigley’s MED, 11th edition Chapter 1 Solutions, Page 2/12 1-13 (a) Eq. (1-6) 1 .9kcycles 69 k i i i x fx N      Eq. (1-7) 2 2 1/2 2 1 (122.9) 30.3kcycles . 1 69 1 k i i i x fx Nx s Ans N                (b) Eq. (1-5) 115 115 115 122.9 0.2607 ˆ 30.3 x x x x x x z s          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. Shigley’s MED, 11th edition Chapter 1 Solutions, Page 3/12 x f f x f x2 200 0 0 0  From the data, the number of instances less than 115 kcycles is 2 + 1 + 3 + 5 + 8 + 12 = 31 (the data is not perfectly normal) ____________________________________________________________________________ 1-14 x f f x f x2  Eq. (1-6) 1 .61kpsi 197 k i i i x fx N      Eq. (1-7) 2 2 12 2 1 (198.61) 9.68 kpsi . 1 197 1 k i i i x fx Nx s Ans N                ______________________________________________________________________________ 1-15 122.9kcycles and 30.3kcycles L L s   Eq. (1-5) 10 10 10 122.9 ˆ 30.3 x L x x L x z s         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. ___________________________________________________________________________ Shigley’s MED, 11th edition Chapter 1 Solutions, Page 4/12 1-16 x f fx fx2  Eq. (1-6) 1 1 13 364/136 98.26471 =98.26 kpsi k i i i x fx N      Eq. (1-7) 2 2 12 2 1 (98.26471) 4.30 kpsi 1 136 1 k i i i x fx Nx s N                Note, for accuracy in the calculation given above,xneeds to be of more significant figures than the rounded value. For a normal distribution, from Eq. (1-5), and a yield strength exceeded by 99 percent (R = 0.99, pf = 0.01), 0.01 0.01 0.01 98.26 ˆ 4.30 x x x x x x x z s         Solving for the yield strength gives x0.01 = 98.26 + 4.30 z0.01 From Table A-10, z0.01 =  2.326. Thus x0.01 = 98.26 + 4.30( 2.326) = 88.3 kpsi Ans. ______________________________________________________________________________ 1-17 Eq. (1-9): R = 1 n i i R   = 0.98(0.96)0.94 = 0.88 Overall reliability = 88 percent Ans. Shigley’s MED, 11th edition Chapter 1 Solutions, Page 5/12 ______________________________________________________________________________ 1-18 Obtain the coefficients of variance for strength and stress ˆ 23.5 0. sy S S sy C S     ˆ ˆ 145 0. T C T          For R = 0.99, from Table A-10, z =  2.326. Eq. (1-12):                2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 2.326 0.07532 1 2.326 0.09667 1.3229 1.32 . 1 2.326 0.07532 S S zC zC n zC Ans                         From the given equation for stress, max 3 16 sy S T n d     Solving for d gives 1/3 1/3 6 16 16(1500)1.3229 0.0319m 31.9mm . (312)10 sy Tn d Ans S                     ______________________________________________________________________________ 1-19 Obtain the coefficients of variance for stress and strength ˆ ˆ 5 0.09231 65 P C P           ˆ ˆ 6.59 0.06901 95.5 y S S S S y C S        (a) 1.2 n Eq. (1-11):   2 2 2 2 2 2 1 1.2 1 1.6127 1.2 0.06901 0.09231 d d S n z nC C        Shigley’s MED, 11th edition Chapter 1 Solutions, Page 6/12 Interpolating Table A-10, 1.61 0.0537 1.6127    = 0.0534 1.62 0.0526 R = 1  0.0534 = 0.9466 Ans.     2 2 4651.2 4 1.020 in . / ( /4) 4 95.5 y y y y S S dS Pn n d Ans P d P S             (b) 1.5 n   2 2 2 1.5 1 3.605 1.5 0.06901 0.09231 z     3.6 0.000159 3.605    = 0. 3.7 0.000108 R = 1  0. = 0.9998 Ans.     4651.5 4 1.140 in . 95.5 y Pn d Ans S      ______________________________________________________________________________ 1-20 max max MPa a b            From footnote 9 of text,   max 1/2 2 2 2 2 1/2 ˆ ˆ ˆ (8.4 22.3) 23.83MPa a b            max max max max max ˆ ˆ 23.83 0.0504 473 C             ˆ ˆ 42.7 0.0772 553 y y y y S S S S y C S        max 553 1.169 1.17 . 473 y S n Ans      Shigley’s MED, 11th edition Chapter 1 Solutions, Page 7/12 Eq. (1-11):   2 2 2 2 2 2 1 1.169 1 1.635 1.169 0.0772 0.0504 d d S n z nC C        From Table A-10, ( 1.635) = 0.05105 R = 1  0.05105 = 0.94895 = 94.9 percent Ans. ______________________________________________________________________________ 1-21 a = 1.500  0.001 in b = 2.000  0.003 in c = 3.000  0.004 in d = 6.520  0.010 in (a) d a b c     w = 6.520  1.5  2  3 = 0.020 in all t t  w = 0.001 + 0.003 + 0.004 +0.010 = 0.018 w = 0.020  0.018 in Ans. (b) From part (a), wmin = 0.002 in. Thus, must add 0.008 in to d . Therefore, d= 6.520 + 0.008 = 6.528 in Ans. ______________________________________________________________________________ 1-22 V = xyz, and x = a   a, y = b   b, z = c   c, V abc      V a a b b c c abc bc a ac b ab c a b c b c a c a b a b c          The higher order terms in  are negligible. Thus, V bc a ac b ab c        and, . V bc a ac b ab c a b c a b c Ans V abc a b c a b c                    For the numerical values given,   3 1.5001.8753.000 8.4375in V    3 0.002 0.003 0.004 0.004267 0.0042678.4375 0.0360in 1.500 1.875 3.000 V V V          V = 8.4375  0.0360 in3 Ans. Shigley’s MED, 11th edition Chapter 1 Solutions, Page 8/12 This answer yields 8.4735 8.4015 V in, whereas, exact is 8.473551.. 8.401551.. V in ______________________________________________________________________________ 1-23 wmax = 0.05 in, wmin = 0.004 in 0.05 0.004 0.027 in 2   w= Thus,  w = 0.05  0.027 = 0.023 in, and then, w = 0.027  0.023 in. 0.027 0.042 1.5 1.569 in a b c a a       w= tw = all t   0.023 = ta + 0.002 + 0.005  ta = 0.016 in Thus, a = 1.569  0.016 in Ans. ______________________________________________________________________________ 1-24   2 3.734 20.139 4.012 in o i D D d        all 0.028 20.004 0.036 in o D t t      Do = 4.012  0.036 in Ans