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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition

Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition offers a comprehensive guide to solving every question in your textbook, helping you master the material.

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    Shigley’s MED, 10 th edition Chapter 1 Solutions , Page 1 /12 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 C A = C B , 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) 2 A Min. strength = 0.85 S To offset the absolute uncertainties, the design factor, from Eq. (1 - 1) should be ( ) 2 1.10 1.43 . 0.85 0.95 d n Ans = = ______________________________________________________________________________ 1 - 10 ( a ) X 1 + X 2 : ( ) ( ) 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 ) X 1 − X 2 : ( ) ( ) ( ) 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 ) X 1 X 2 : ( )( ) 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 . x x X e X e e x x X X X e X e e e e e X e X e X X Ans X X = + + = − = + +    + = +    

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    Shigley’s MED, 10 th edition Chapter 1 Solutions , Page 2 /12 ( d ) X 1 / X 2 : 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 ) x 1 = 7 = 2.645 751 311 1 X 1 = 2.64 (3 correct digits) x 2 = 8 = 2.828 427 124 7 X 2 = 2.82 (3 correct digits) x 1 + x 2 = 5.474 178 435 8 e 1 = x 1 − X 1 = 0.005 751 311 1 e 2 = x 2 − X 2 = 0.008 427 124 7 e = e 1 + e 2 = 0.0 14 178 435 8 Sum = x 1 + x 2 = X 1 + X 2 + e = 2.64 + 2.82 + 0.014 178 435 8 = 5.474 178 435 8 Checks ( b ) X 1 = 2.65, X 2 = 2.83 (3 digit significant numbers) e 1 = x 1 − X 1 = − 0.004 248 688 9 e 2 = x 2 − X 2 = − 0.001 572 875 3 e = e 1 + e 2 = − 0.005 821 564 2 Sum = x 1 + x 2 = X 1 + X 2 + e = 2.65 +2.83 − 0.001 572 875 3 = 5.474 178 435 8 Checks ______________________________________________________________________________ 1 - 12 ( ) ( ) 3 3 25 10 32 1000 1.006 in . 2.5 d S d Ans n d   =  =  = Table A - 17: d = 1 4 1 in Ans. Factor of safety: ( ) ( ) ( ) 3 3 25 10 4.79 . 32 1000 1.25 S n Ans   = = = ______________________________________________________________________________

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    Shigley’s MED, 10 th edition Chapter 1 Solutions , Page 3 /12 1 - 13 ( a ) Eq . (1 - 6) 1 1 8480 122.9 kcycles 69 k i i i x f x N = = = =  Eq. (1 - 7) 2 2 1/ 2 2 1 1 104 600 69(122.9) 30.3 kcycles . 1 69 1 k i i i x f x N x 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. From the data, the number of instances less than 115 kcycles is x f f x f x 2 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 19 0 36100 200 0 0 0 210 1 210 44100  69 8480 1 104 600

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    Shigley’s MED, 10 th edition Chapter 1 Solutions , Page 4 /12 2 + 1 + 3 + 5 + 8 + 12 = 31 (the data is not perfectly normal) ____________________________________________________________________________ 1 - 14 x f f x f x 2 174 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 1 39 126 198.61 kpsi 197 k i i i x f x N = = = =  Eq. (1 - 7) 2 2 1 2 2 1 7 789 204 197(198.61) 9.68 kpsi . 1 197 1 k i i i x f x N x s Ans N = −   − = = =   − −    ______________________________________________________________________________ 1 - 15 122.9 kcycles and 30.3 kcycles L L s = = Eq. (1 - 5) 10 10 10 122.9 ˆ 30.3 x L x x L x z s   − − − = = = Thus, x 10 = 122.9 + 30.3 z 10 = L 10 From Table A - 10, for 10 percent failure, z 10 = − 1.282 . Thus, L 10 = 122.9 + 30.3( − 1.282) = 84.1 kcycles Ans. ___________________________________________________________________________

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    Shigley’s MED, 10 th edition Chapter 1 Solutions , Page 5 /12 1 - 16 x f fx fx 2 93 19 1767 164331 95 25 2375 225625 97 38 3686 357542 99 17 1683 166617 101 12 1212 122412 103 10 1030 106090 105 5 525 55125 107 4 428 45796 109 4 436 47524 111 2 222 24642  136 13364 1315704 Eq. (1 - 6) 1 1 13 364 / 136 98.26471 = 98.26 kpsi k i i i x f x N = = = =  Eq. (1 - 7) 2 2 1 2 2 1 1 315 704 136(98.26471) 4.30 kpsi 1 136 1 k i i i x f x N x s N = −   − = = =   − −    Note , for accuracy in th e calculation given above , x needs 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 , p f = 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 x 0.01 = 98.26 + 4.30 z 0.01 From Table A - 10, z 0.01 = − 2.326 . Thus x 0.01 = 98.26 + 4.30 ( − 2.326 ) = 88.3 kpsi Ans . ______________________________________________________________________________ 1 - 1 7 Eq. (1 - 9 ): R = 1 n i i R =  = 0.98(0.96)0.94 = 0.88 Overall reliability = 88 percent Ans. ______________________________________________________________________________

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    Shigley’s MED, 10 th edition Chapter 1 Solutions , Page 6 /12 1 - 18 Obtain the coefficients of variance for strength and stress ˆ 23.5 0.07532 312 sy S S sy C S  = = = ˆ ˆ 145 0.09667 1 500 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 z C z C n z C 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 0.0319 m 31.9 mm . (312)10 sy T n 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 =

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    Shigley’s MED, 10 th edition Chapter 1 Solutions , Page 7 /12 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 n C C  − − = − = − = − + + 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 4 65 1.2 4 1.020 in . / ( / 4) 4 95.5 y y y y S S d S 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    = 0.00015645 3.7 0.000108 R = 1 − 0.00015645 = 0.9998 Ans . ( ) ( ) 4 65 1.5 4 1.140 in . 95.5 y Pn d Ans S   = = = ______________________________________________________________________________ 1 - 20 max max 90 383 473 MPa a b      = = + = + = From footnote 9, p. 25 of text, ( ) max 1/ 2 2 2 2 2 1/ 2 ˆ ˆ ˆ (8.4 22.3 ) 23.83 MPa 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    = = = =

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    Shigley’s MED, 10 th edition Chapter 1 Solutions , Page 8 /12 max 553 1.169 1.17 . 473 y S n Ans  = = = = 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 n C 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), w min = 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 term s 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 1.500 1.875 3.000 8.4375 in V = =

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    Shigley’s MED, 10 th edition Chapter 1 Solutions , Page 9 /12 ( ) 3 0.002 0.003 0.004 0.004267 0.004267 8.4375 0.0360 in 1.500 1.875 3.000 V V V   + + =    = V = 8.4375  0.036 0 in 3 Ans. This answer yields 8.4735 8.4015 V  in, whereas, exact is 8.473551.. 8.401551.. V = in ______________________________________________________________________________ 1 - 23 w max = 0.05 in, w min = 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 = t w = all t   0.023 = t a + 0.002 + 0.005  t a = 0.016 in Thus, a = 1.569  0.016 in A ns. ______________________________________________________________________________ 1 - 24 ( ) 2 3.734 2 0.139 4.012 in o i D D d = + = + = ( ) all 0.028 2 0.004 0.036 in o D t t = = + =  D o = 4.012  0.036 in Ans. ______________________________________________________________________________ 1 - 25 From O - Rings, Inc. (oringsusa.com), D i = 9.19  0.13 mm, d = 2.62  0.08 mm ( ) 2 9.19 2 2.62 14.43 mm o i D D d = + = + = ( ) all 0.13 2 0.08 0.29 mm o D t t = = + =  D o = 14.43  0.29 mm Ans. ______________________________________________________________________________

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    Shigley’s MED, 10 th edition Chapter 1 Solutions , Page 10 /12 1 - 26 From O - Rings, Inc. (oringsusa.com), D i = 34.52  0.30 mm, d = 3.53  0.10 mm ( ) 2 34.52 2 3.53 41.58 mm o i D D d = + = + = ( ) all 0.30 2 0.10 0.50 mm o D t t = = + =  D o = 41.58  0.50 mm Ans. ______________________________________________________________________________ 1 - 2 7 From O - Rings, Inc. (oringsusa.com), D i = 5.237  0.035 in, d = 0.103  0.003 in ( ) 2 5.237 2 0.103 5.443 in o i D D d = + = + = ( ) all 0.035 2 0.003 0.041 in o D t t = = + =  D o = 5.443  0.041 in Ans. ______________________________________________________________________________ 1 - 2 8 From O - Rings, Inc. (oringsusa.com) , D i = 1.100  0.012 in, d = 0.210  0.005 in ( ) 2 1.100 2 0.210 1.520 in o i D D d = + = + = ( ) all 0.012 2 0.005 0.022 in o D t t = = + =  D o = 1.520  0.022 in Ans. ______________________________________________________________________________ 1 - 2 9 From Table A - 2, ( a )  = 150/6.89 = 21.8 kpsi Ans. ( b ) F = 2 /4.45 = 0.449 kip = 449 lbf Ans. ( c ) M = 150/0.113 = 1330 lbf  in = 1.33 kip  in Ans. ( d ) A = 1500/ 25.4 2 = 2.33 in 2 Ans. ( e ) I = 750/2.54 4 = 18.0 in 4 Ans. ( f ) E = 145/6.89 = 21.0 Mpsi Ans. ( g ) v = 75/1.61 = 46.6 mi/h Ans.

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    Shigley’s MED, 10 th edition Chapter 1 Solutions , Page 11 /12 ( h ) V = 1000/946 = 1.06 qt Ans. ______________________________________________________________________________ 1 - 30 From Table A - 2, ( a ) l = 5(0.305) = 1.53 m Ans. ( b )  = 90(6.89) = 620 MPa Ans. ( c ) p = 25(6.89) = 172 kPa Ans. ( d ) Z =12(16.4) = 197 cm 3 Ans. ( e ) w = 0.208(175) = 36.4 N/m Ans. ( f )  = 0.001 89(25.4) = 0.048 0 mm Ans. ( g ) v = 1 200(0.0051) = 6.12 m/s Ans. ( h )  = 0.002 15(1) = 0.002 15 mm/ mm Ans. ( i ) V = 1830(25.4 3 ) = 30.0 (10 6 ) mm 3 Ans. ___________________________________________________________________ ___________ 1 - 31 ( a )  = M /Z = 1770/0.934 = 1895 psi = 1.90 kpsi Ans. ( b )  = F /A = 9440/23.8 = 397 psi Ans. ( c ) y =Fl 3 / 3 EI = 270(31.5) 3 /[3(30)10 6 (0.154)] = 0.609 in Ans. ( d )  = Tl /GJ = 9 740(9.85)/[11.3(10 6 )(  /32)1.00 4 ] = 8 .648(10 − 2 ) rad = 4.95  Ans. ______________________________________________________________________________ 1 - 32 ( a )  =F / w t = 1000/[25(5)] = 8 MPa Ans. ( b ) I = bh 3 /12 = 10(25) 3 /12 = 13.0(10 3 ) mm 4 Ans. ( c ) I =  d 4 /64 =  (25.4) 4 /64 = 20.4(10 3 ) mm 4 Ans. ( d )  = 16 T /  d 3 = 16(25)10 3 /[  (12.7) 3 ] = 62.2 MPa Ans. ______________________________________________________________________________ 1 - 33

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    Shigley’s MED, 10 th edition Chapter 1 Solutions , Page 12 /12 ( a )  =F /A = 2 700/[  (0.750) 2 /4] = 6110 psi = 6.11 kpsi Ans. ( b )  = 32 Fa/  d 3 = 32(180)31.5/[  (1.25) 3 ] = 29 570 psi = 29.6 kpsi Ans. ( c ) Z =  ( d o 4 − d i 4 )/(32 d o ) =  (1.50 4 − 1.00 4 )/[32(1.50)] = 0.266 in 3 Ans. ( d ) k = ( d 4 G )/(8 D 3 N ) = 0.062 5 4 (11.3)10 6 /[8(0.760) 3 32] = 1.53 lbf/in Ans. ______________________________________________________________________________

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    Shigley’s MED, 10 th edition Chapter 2 Solutions , Page 1 /22 Chapter 2 2 - 1 From Tables A - 20 , A - 21, A - 22, and A - 24c, ( a ) UNS G10200 HR: S ut = 380 (55) MPa (kpsi), S yt = 210 (30) MP a (kpsi) Ans. ( b ) SAE 1050 CD: S ut = 69 0 ( 100 ) MPa (kpsi), S yt = 58 0 ( 84 ) MP a (kpsi) Ans. ( c ) AISI 1141 Q&T at 540  C (1000  F) : S ut = 896 ( 130 ) MPa (kpsi), S yt = 765 ( 111 ) MP a (kpsi) Ans. ( d ) 2024 - T4: S ut = 446 ( 64.8 ) MPa (kpsi), S yt = 296 ( 4 3 . 0) MP a (kpsi) Ans. ( e ) Ti - 6Al - 4V annealed : S ut = 90 0 ( 130 ) MPa (kpsi), S yt = 83 0 ( 12 0) MP a (kpsi) Ans. ______________________________________________________________________________ 2 - 2 ( a ) Maximize yield strength: Q&T at 425  C ( 800  F) Ans. ( b ) Maximize elongation: Q&T at 650  C (1200  F) Ans. ______________________________________________________________________________ 2 - 3 Conversion of kN/m 3 to kg/ m 3 multiply by 1 (10 3 ) / 9.81 = 102 AISI 1018 CD steel: Tables A - 20 and A - 5 ( ) ( ) 3 370 10 47.4 kN m/kg . 76.5 102 y S Ans  = =  2011 - T6 aluminum: Tables A - 22 and A - 5 ( ) ( ) 3 169 10 62.3 kN m/kg . 26.6 102 y S Ans  = =  Ti - 6Al - 4 V titanium: Tables A - 24c and A - 5 ( ) ( ) 3 830 10 187 kN m/kg . 43.4 102 y S Ans  = =  ASTM No. 40 cast iron: Tables A - 24 a and A - 5 .Does not have a yield strength. Using the ultimate strength in tension ( ) ( ) ( ) 3 42.5 6.89 10 40.7 kN m/kg 70.6 102 ut S Ans  = =  ______________________________________________________________________________ 2 - 4 AISI 1018 CD steel: Table A - 5 ( ) ( ) 6 6 30.0 10 106 10 in . 0.282 E Ans  = = 2011 - T6 aluminum: Table A - 5 ( ) ( ) 6 6 10.4 10 106 10 in . 0.098 E Ans  = =

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    Shigley’s MED, 10 th edition Chapter 2 Solutions , Page 2 /22 Ti - 6Al - 6V titanium: Table A - 5 ( ) ( ) 6 6 16.5 10 103 10 in . 0.160 E Ans  = = No. 40 cast iron: Table A - 5 ( ) ( ) 6 6 14.5 10 55.8 10 in . 0.260 E Ans  = = ______________________________________________________________________________ 2 - 5 2 2 (1 ) 2 E G G v E v G − + =  = Using values for E and G f rom Table A - 5, Steel: ( ) ( ) 30.0 2 11.5 0.304 . 2 11.5 v Ans − = = The percent difference from the value in Table A - 5 is 0.304 0.292 0.0411 4.11 percent . 0.292 Ans − = = Aluminum: ( ) ( ) 10.4 2 3.90 0.333 . 2 3.90 v Ans − = = The percent difference from the value in Table A - 5 is 0 percent Ans. Beryllium copper: ( ) ( ) 18.0 2 7.0 0.286 . 2 7.0 v Ans − = = The percent difference from the value in Table A - 5 is 0.286 0.285 0.00351 0.351 percent . 0.285 Ans − = = Gray cast iron: ( ) ( ) 14.5 2 6.0 0.208 . 2 6.0 v Ans − = = The percent difference from the value in Table A - 5 is 0.208 0.211 0.0142 1.42 percent . 0.211 Ans − = − = − ______________________________________________________________________________ 2 - 6 ( a ) A 0 =  (0.503) 2 /4 = 0.1987 in 2 ,  = P i / A 0

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    Shigley’s MED, 10 th edition Chapter 2 Solutions , Page 3 /22 For data in elastic range,  =  l / l 0 =  l / 2 For data in plastic range, 0 0 0 0 0 1 1 l l A l l l l l A −  = = = − = − ò On the next two pages, the data and plots are presented. Figure ( a ) shows the linear part of the curve from data points 1 - 7. Figure ( b ) shows data points 1 - 12. Figure ( c ) shows the complete range . Note : The exact value of A 0 is used without rounding off. ( b ) From Fig. ( a ) t he slope of the line from a linear regression is E = 30.5 Mpsi Ans. From Fig. ( b ) t he equation for the dotted offset line is found to be  = 30.5(10 6 )  − 61 000 (1) The equation for the line between data points 8 and 9 is  = 7.60(10 5 )  + 42 900 (2) Solving Eqs. (1) and (2) simultaneously yields  = 45.6 kpsi which is the 0.2 percent offset yield strength . Thus, S y = 45.6 kpsi Ans. The ultimate strength from Figure ( c ) is S u = 85.6 kpsi Ans. The reduction in area is given by Eq. (2 - 12) is ( ) ( ) 0 0 0.1987 0.1077 100 100 45.8 % . 0.1987 f A A R Ans A − − = = = Data Point P i  l, A i   1 0 0 0 0 2 1000 0.0004 0.00020 5032 3 2000 0.0006 0.00030 10065 4 3000 0.001 0.00050 15097 5 4000 0.0013 0.00065 20130 6 7000 0.0023 0.00115 35227 7 8400 0.0028 0.00140 42272 8 8800 0.0036 0.00180 44285 9 9200 0.0089 0.00445 46298 10 8800 0.1984 0.00158 44285 11 9200 0.1978 0.00461 46298 12 9100 0.1963 0.01229 45795 13 13200 0.1924 0.03281 66428 14 15200 0.1875 0.05980 76492 15 17000 0.1563 0.27136 85551 16 16400 0.1307 0.52037 82531 17 14800 0.1077 0.84506 74479

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