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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition - Document preview page 1

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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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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition - Page 1 preview imageShigley’s MED, 10theditionChapter 1 Solutions,Page1/12Chapter 1Problems1-1through1-6are for student research. No standard solutions are provided.1-7From Fig. 1-2, cost of grinding to0.0005 in is 270%. Cost of turning to0.003 in is60%.Relative cost of grinding vs. turning =270/60 =4.5 timesAns.______________________________________________________________________________1-8CA=CB,10 +0.8P= 60 + 0.8P−0.005P2P2= 50/0.005P= 100partsAns.______________________________________________________________________________1-9Max. load = 1.10PMin. area= (0.95)2AMin. strength = 0.85STo offset the absolute uncertainties, the design factor, from Eq. (1-1)should be()21.101.43.0.85 0.95dnAns==______________________________________________________________________________1-10(a)X1+X2:()()121122121212error.xxXeXeexxXXeeAns+=+++==+−+=+(b)X1−X2:()()()121122121212.xxXeXeexxXXeeAns−=+−+=−−−=−(c)X1X2:()()12112212121221121212211212.x xXeXeex xX XX eX ee eeeX eX eX XAnsXX=++=−=+++=+
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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition - Page 2 preview image
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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition - Page 3 preview imageShigley’s MED, 10theditionChapter 1 Solutions,Page2/12(d)X1/X2:11111122222212211121222221212111122221211111then1111Thus,.xXeXeXxXeXeXeeeXeeeeXXeXXXXXxXXeeeAnsxXXXX−++==++++−+−+−+=−−______________________________________________________________________________1-11(a)x1=7= 2.645 751 311 1X1= 2.64(3correct digits)x2=8= 2.828 427 124 7X2= 2.82(3 correct digits)x1+x2= 5.474 178 435 8e1=x1−X1= 0.005 751 311 1e2=x2−X2= 0.008 427 124 7e=e1+e2= 0.014 178 435 8Sum =x1+x2=X1+X2+e= 2.64 + 2.82 + 0.014 178 435 8 = 5.474 178 435 8Checks(b)X1= 2.65,X2= 2.83(3 digit significant numbers)e1=x1−X1=−0.004 248 688 9e2=x2−X2=−0.001 572 875 3e=e1+e2=−0.005 821 564 2Sum =x1+x2=X1+X2+e= 2.65 +2.83−0.001 572 875 3 = 5.474 178 435 8Checks______________________________________________________________________________1-12()()3325 1032 10001.006 in.2.5dSdAnsnd===TableA-17:d=141inAns.Factor of safety:()()()3325 104.79.32 10001.25SnAns===______________________________________________________________________________
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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition - Page 4 preview imageShigley’s MED, 10theditionChapter 1 Solutions,Page3/121-13(a)Eq.(1-6)118480122.9 kcycles69kiiixf xN====Eq. (1-7)221/ 2211 104 60069(122.9)30.3 kcycles.1691kiiixf xN xsAnsN=−−===−−(b)Eq. (1-5)115115115122.90.2607ˆ30.3xxxxxxzs−−−==== −Interpolating from Table (A-10)0.26000.39740.2607xx= 0.39710.27000.3936N(−0.2607) = 69 (0.3971) = 27.427Ans.From the data, the number of instances less than 115 kcycles isxff xf x2602120720070170490080324019200905450405001008800800001101213201452001206720864001301013001690001408112015680015057501125001602320512001703510867001802360648001901190361002000002101210441006984801 104 600
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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition - Page 5 preview imageShigley’s MED, 10theditionChapter 1 Solutions,Page4/122 + 1 + 3 + 5 + 8 + 12 = 31 (the data is not perfectly normal)____________________________________________________________________________1-14xff xf x217461044181656182916382981161904483601588400198671326626266682065310918224910821412256854955222261332295704197391267789204Eq.(1-6)1139 126198.61 kpsi197kiiixf xN====Eq. (1-7)221 2217 789 204197(198.61)9.68 kpsi.11971kiiixf xN xsAnsN=−−===−−______________________________________________________________________________1-15122.9 kcycles and30.3 kcyclesLLs==Eq. (1-5)101010122.9ˆ30.3xLxxLxzs−−−===Thus,x10= 122.9 + 30.3z10=L10From TableA-10, for 10 percent failure,z10=−1.282. Thus,L10= 122.9 + 30.3(−1.282) =84.1 kcyclesAns.___________________________________________________________________________
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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition - Page 6 preview imageShigley’s MED, 10theditionChapter 1 Solutions,Page5/121-16xffxfx293191767164331952523752256259738368635754299171683166617101121212122412103101030106090105552555125107442845796109443647524111222224642136133641315704Eq. (1-6)1113 364 / 13698.26471 = 98.26 kpsikiiixf xN====Eq. (1-7)221 2211 315 704136(98.26471)4.30 kpsi11361kiiixf xN xsN=−−===−−Note, for accuracy in thecalculationgiven above,xneeds to be of more significantfiguresthan the rounded value.For anormaldistribution,from Eq. (1-5),anda yield strength exceeded by 99 percent(R= 0.99,pf= 0.01),0.010.010.0198.26ˆ4.30xxxxxxxzs−−−===Solving for the yield strength givesx0.01= 98.26 + 4.30z0.01FromTable A-10,z0.01=−2.326. Thusx0.01= 98.26 + 4.30(−2.326) = 88.3kpsiAns.______________________________________________________________________________1-17Eq. (1-9):R=1niiR== 0.98(0.96)0.94 = 0.88Overallreliability = 88 percentAns.______________________________________________________________________________
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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition - Page 7 preview imageShigley’s MED, 10theditionChapter 1 Solutions,Page6/121-18Obtain the coefficients of variance for strength and stressˆ23.50.07532312sySSsyCS===ˆˆ1450.096671 500TCT====ForR= 0.99, from Table A-10,z=−2.326.Eq. (1-12):()()() ()() ()() ()222222222222111111112.3260.0753212.3260.096671.32291.32.12.3260.07532SSz Cz Cnz CAns+−−−=− +−−−−− ===−−From the given equation for stress,max316sySTnd==Solving fordgives1/ 31/ 361616(1500)1.32290.0319 m31.9 mm.(312)10syT ndAnsS====______________________________________________________________________________1-19Obtain the coefficients of variance for stress and strengthˆˆ50.0923165PCP====ˆˆ6.590.0690195.5ySSSSyCS====(a)1.2n=
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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition - Page 8 preview imageShigley’s MED, 10theditionChapter 1 Solutions,Page7/12Eq.(1-11):()22222211.211.61271.20.069010.09231ddSnzn CC−−= −= −= −++Interpolating Table A-10,1.610.05371.6127= 0.05341.620.0526R= 1−0.0534 = 0.9466Ans.()()224 65 1.241.020 in./ (/ 4)495.5yyyySSd SPnndAnsPdPS======(b)1.5n=()2221.513.6051.50.069010.09231z−= −= −+3.60.0001593.605= 0.000156453.70.000108R= 1−0.00015645 = 0.9998Ans.()()4 65 1.541.140 in.95.5yPndAnsS===______________________________________________________________________________1-20maxmax90383473 MPaab==+=+=From footnote 9, p. 25 of text,()max1/ 22222 1/ 2ˆˆˆ(8.422.3 )23.83 MPaab=+=+=maxmaxmaxmaxmaxˆˆ23.830.0504473C====ˆˆ42.70.0772553yyyySSSSyCS====
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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition - Page 9 preview imageShigley’s MED, 10theditionChapter 1 Solutions,Page8/12max5531.1691.17.473ySnAns====Eq. (1-11):()22222211.16911.6351.1690.07720.0504ddSnzn CC−−= −= −= −++From TableA-10,(−1.635) = 0.05105R= 1−0.05105 = 0.94895 = 94.9 percentAns.______________________________________________________________________________1-21a= 1.5000.001 inb= 2.0000.003 inc= 3.0000.004 ind= 6.5200.010 in(a)dabc=−−−w= 6.520−1.5−2−3 = 0.020 inalltt=w= 0.001 + 0.003 + 0.004 +0.010 = 0.018w= 0.0200.018 inAns.(b) From part (a),wmin= 0.002 in. Thus, must add 0.008 in tod. Therefore,d= 6.520 + 0.008 =6.528 inAns.______________________________________________________________________________1-22V = xyz, andx = aa,y =bb,z =cc,Vabc=()()()Vaabbccabcbcaacbab cab cb cacabab c=   =      Thehigher orderterms inarenegligible. Thus,Vbc aac bab c + +and,.Vbcaacbab cabcabc AnsVabcabcabc++=++=++For the numerical values given,()31.500 1.875 3.0008.4375 inV==
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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition - Page 10 preview imageShigley’s MED, 10theditionChapter 1 Solutions,Page9/12()30.0020.0030.0040.0042670.004267 8.43750.0360 in1.5001.8753.000VVV++==V= 8.43750.0360in3Ans.This answer yields8.47358.4015Vin, whereas, exact is8.473551..8.401551..V=in______________________________________________________________________________1-23wmax= 0.05 in,wmin= 0.004 in0.050.0040.027 in2+=w =Thus,w= 0.05−0.027 = 0.023 in, and then,w= 0.0270.023 in.0.0270.0421.51.569 inabcaa−−=−−=w =tw=allt0.023 =ta+ 0.002 + 0.005ta= 0.016 inThus,a= 1.5690.016 inAns.______________________________________________________________________________1-24()23.7342 0.1394.012inoiDDd=+=+=()all0.0282 0.0040.036 inoDtt==+=Do= 4.0120.036inAns.______________________________________________________________________________1-25From O-Rings, Inc. (oringsusa.com),Di= 9.190.13 mm,d= 2.620.08 mm()29.192 2.6214.43mmoiDDd=+=+=()all0.132 0.080.29mmoDtt==+=Do= 14.430.29mmAns.______________________________________________________________________________
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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition - Page 11 preview imageShigley’s MED, 10theditionChapter 1 Solutions,Page10/121-26From O-Rings, Inc. (oringsusa.com),Di= 34.520.30 mm,d= 3.530.10 mm()234.522 3.5341.58mmoiDDd=+=+=()all0.302 0.100.50mmoDtt==+=Do= 41.580.50mmAns.______________________________________________________________________________1-27From O-Rings, Inc. (oringsusa.com),Di= 5.2370.035 in,d= 0.1030.003 in()25.2372 0.1035.443inoiDDd=+=+=()all0.0352 0.0030.041 inoDtt==+=Do= 5.4430.041inAns.______________________________________________________________________________1-28From O-Rings, Inc. (oringsusa.com),Di= 1.1000.012 in,d= 0.2100.005 in()21.1002 0.2101.520inoiDDd=+=+=()all0.0122 0.0050.022 inoDtt==+=Do= 1.5200.022inAns.______________________________________________________________________________1-29FromTableA-2,(a)= 150/6.89 = 21.8 kpsiAns.(b)F=2 /4.45 = 0.449 kip = 449 lbfAns.(c)M= 150/0.113 = 1330 lbfin = 1.33 kipinAns.(d)A= 1500/ 25.42= 2.33 in2Ans.(e)I= 750/2.544= 18.0 in4Ans.(f)E= 145/6.89 = 21.0 MpsiAns.(g)v= 75/1.61 = 46.6 mi/hAns.
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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition - Page 12 preview imageShigley’s MED, 10theditionChapter 1 Solutions,Page11/12(h)V= 1000/946 = 1.06 qtAns.______________________________________________________________________________1-30FromTableA-2,(a)l= 5(0.305) = 1.53 mAns.(b)= 90(6.89) = 620 MPaAns.(c)p= 25(6.89) = 172 kPaAns.(d)Z=12(16.4) = 197 cm3Ans.(e)w= 0.208(175) = 36.4 N/mAns.(f)= 0.001 89(25.4) = 0.0480 mmAns.(g)v= 1200(0.0051) = 6.12 m/sAns.(h)= 0.002 15(1) = 0.002 15mm/mmAns.(i)V= 1830(25.43) = 30.0 (106) mm3Ans.______________________________________________________________________________1-31(a)=M /Z= 1770/0.934 =1895 psi = 1.90 kpsiAns.(b)=F /A= 9440/23.8 = 397 psiAns.(c)y =Fl3/3EI= 270(31.5)3/[3(30)106(0.154)] = 0.609 inAns.(d)= Tl /GJ= 9740(9.85)/[11.3(106)(/32)1.004] = 8.648(10−2) rad = 4.95Ans.______________________________________________________________________________1-32(a)=F /wt= 1000/[25(5)] = 8 MPaAns.(b)I = bh3/12 = 10(25)3/12 = 13.0(103) mm4Ans.(c)I =d4/64 =(25.4)4/64 = 20.4(103) mm4Ans.(d)=16T /d3=16(25)103/[(12.7)3] = 62.2 MPaAns.______________________________________________________________________________1-33
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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition - Page 13 preview imageShigley’s MED, 10theditionChapter 1 Solutions,Page12/12(a)=F /A= 2 700/[(0.750)2/4] =6110 psi = 6.11 kpsiAns.(b)=32Fa/d3= 32(180)31.5/[(1.25)3] = 29 570 psi = 29.6 kpsiAns.(c)Z =(do4−di4)/(32do) =(1.504−1.004)/[32(1.50)] = 0.266 in3Ans.(d)k=(d4G)/(8D3N) = 0.06254(11.3)106/[8(0.760)332] = 1.53 lbf/inAns.______________________________________________________________________________
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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition - Page 14 preview imageShigley’s MED, 10theditionChapter 2Solutions,Page1/22Chapter 22-1From Tables A-20, A-21, A-22, and A-24c,(a) UNS G10200 HR:Sut= 380 (55)MPa (kpsi),Syt= 210 (30) MPa (kpsi)Ans.(b) SAE 1050 CD:Sut=690 (100)MPa (kpsi),Syt=580 (84) MPa (kpsi)Ans.(c) AISI 1141 Q&Tat540C(1000F):Sut=896(130)MPa (kpsi),Syt=765(111)MPa (kpsi)Ans.(d) 2024-T4:Sut=446(64.8)MPa (kpsi),Syt=296(43.0) MPa (kpsi)Ans.(e) Ti-6Al-4Vannealed:Sut=900 (130)MPa (kpsi),Syt=830 (120) MPa (kpsi)Ans.______________________________________________________________________________2-2(a)Maximize yield strength:Q&Tat425C (800F)Ans.(b)Maximize elongation:Q&Tat650C (1200F)Ans.______________________________________________________________________________2-3ConversionofkN/m3to kg/m3multiply by 1(103) /9.81 = 102AISI1018 CD steel: Tables A-20 and A-5()()3370 1047.4 kN m/kg.76.5 102ySAns==2011-T6 aluminum: Tables A-22and A-5()()3169 1062.3 kN m/kg.26.6 102ySAns==Ti-6Al-4V titanium:Tables A-24c and A-5()()3830 10187 kN m/kg.43.4 102ySAns==ASTM No. 40 cast iron: Tables A-24aand A-5.Does not have ayield strength. Using theultimate strength in tension()()()342.5 6.891040.7 kN m/kg70.6 102utSAns==______________________________________________________________________________2-4AISI1018 CD steel: Table A-5()()6630.0 10106 10in.0.282EAns==2011-T6 aluminum: TableA-5()()6610.4 10106 10in.0.098EAns==
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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition - Page 15 preview imageShigley’s MED, 10theditionChapter 2Solutions,Page2/22Ti-6Al-6Vtitanium:Table A-5()()6616.5 10103 10in.0.160EAns==No. 40 cast iron: Table A-5()()6614.5 1055.8 10in.0.260EAns==______________________________________________________________________________2-522(1)2EGGvEvG−+==Using values forEandGfromTable A-5,Steel:()()30.02 11.50.304.2 11.5vAns−==The percent difference from the value in Table A-5 is0.3040.2920.04114.11 percent.0.292Ans−==Aluminum:()()10.42 3.900.333.2 3.90vAns−==The percent difference from the value in Table A-5 is 0 percentAns.Beryllium copper:()()18.02 7.00.286.2 7.0vAns−==The percent difference from the value in Table A-5 is0.2860.2850.003510.351 percent.0.285Ans−==Gray cast iron:()()14.52 6.00.208.2 6.0vAns−==The percent difference from the value in Table A-5 is0.2080.2110.01421.42 percent.0.211Ans−= −= −______________________________________________________________________________2-6(a)A0=(0.503)2/4= 0.1987 in2,=Pi/A0
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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition - Page 16 preview imageShigley’s MED, 10theditionChapter 2Solutions,Page3/22Fordatain elastic range,=l / l0=l /2For data in plastic range,0000011llAlllllA−===−=−òOn the next two pages, the data and plots are presented. Figure (a) shows the linear part ofthe curve from data points 1-7.Figure (b) shows datapoints 1-12. Figure (c) shows thecomplete range.Note: The exact value ofA0is used without rounding off.(b) From Fig.(a)the slope of the line from a linear regression isE= 30.5 MpsiAns.From Fig. (b) theequation for the dotted offset line is found to be= 30.5(106)−61 000(1)The equation for the line between data points8 and 9 is= 7.60(105)+ 42 900(2)Solving Eqs. (1) and (2) simultaneously yields= 45.6 kpsi which is the 0.2 percentoffset yield strength. Thus,Sy= 45.6 kpsiAns.The ultimate strength from Figure (c) isSu= 85.6 kpsiAns.The reduction in area is given byEq. (2-12)is()()000.19870.107710010045.8 %.0.1987fAARAnsA−−===Data PointPil, Ai10000210000.00040.000205032320000.00060.0003010065430000.0010.0005015097540000.00130.0006520130670000.00230.0011535227784000.00280.0014042272888000.00360.0018044285992000.00890.00445462981088000.19840.00158442851192000.19780.00461462981291000.19630.012294579513132000.19240.032816642814152000.18750.059807649215170000.15630.271368555116164000.13070.520378253117148000.10770.8450674479
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