Solution Manual for Finite Element Analysis: Theory and Application with ANSYS, 5th Edition

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Solution Manual for Finite Element Analysis: Theory and Application with ANSYS, 5th Edition - Page 1 preview image

Chapter 11.1(a)Using two elements( )0.250.0125A yy210.25 inA220.1875 inA230.125 inA1()2iieqAAEK61(0.250.1875)(10.410 )lb455,0002(5)inK62(0.18750.125) (10.410 )lb325,0002(5)inK3455101455045512031230455325325010003253250,0.002197 in0.005274 inuuuuuu    (b)210.25 inA220.234375 inA230.21875 inA240.203125 inA250.1875 inA260.171875 inA270.15625 inA280.140625 inA290.125 inA

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1.1 Cont.1lb2015000 inK2lb1885000 inK3lb1755000 inK4lb1625000 inK5lb1495000 inK6lb1365000 inK7lb1235000 inK8lb1105000 inK20150001201500002015000020150001885000188500018850001885000175500017550001755000175500016250001625000162500016250001495000149500014950001495000136500013650001365000136500012350001235000123500012350001105012345678900000000100011050000011050001105000uuuuuuuuu            10,u20.00049628,u30.0010268,u40.0015965,u50.0022119,u60.0028808,u70.0036134,u80.0044232,u90.0053281in.uy(in)exact defl.Two-elementeight element00001.250.000496450.000496282.50.00102720.00102683.750.00159720.00159655.00.00221290.0021970.00221196.250.00288220.00288087.50.00361540.00361348.750.00442590.004423210.00.00533190.0052740.0053281

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1.2avg.AEK61510.512(3)(3.2710 )26KK15lb18,393,750 inKK62410.59(3)(3.2710 )lb215,941,2506inKK63(9)(3)(3.2710 )lb22, 072,5004inK1839375011839375001839375010234561839375015941250159412501594125015941250220725002207250022072500220725001594125015941250159412501594125018393750183937501839375018393750uuuuuu 00000500 5554402.71831105.8548310{ }in8.12009101.12566101.3974910u 556(3)( 8.12009105.85483)10(3.2710 )4(3)18.5 Psi C

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1.2Cont.as a check:(3)50018.5 Psi(9)(3)FA1()iiuuE56(1)( 2.71831100)(3.2710 )14.8 Psi C646(5)( 1.397491.12566)10(3.2710 )14.8 Psi C6as a check:(1)(5)50014.8 Psi C10.512(3)2FA5625.854832.7183110(3.2710 )17.1 Psi C6456(4)( 1.12566108.1200910) (3.2710 )17.1 Psi C6as a check:(2)(4)50017.1 Psi10.59(3)2FA

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1.3AEK913(0.08)(0.006)(68.910 )0.025KK913N1.3228810mKK92(0.02)(0.006)(68.910 )0.1K92N0.0826810mK91.322881011.3228801.3228810234001.322880.082680.0826800.082680.082681.322881.3228818001.322881.32288uuuu     23946552341.405560.08268000.082681.405561.32288001.322881.322881800101.36066810m2.31313510m2.44920210muuuuuu   6(1)921(1.360668100)(68.910 )37500003.75 MPa0.025uuEas a check:(1)1800 N3.75 MPa(0.08)(0.006)FA56(2)932(2.313135101.36066810)(68.910 )15 MPa0.1uuEas a check:(2)1800 N15 MPa(0.02)(0.006)5(3)943(2.4492022.313135)10(68.910 )3.75 MPa0.025uuE

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1.4616(4)(0.125)(2810 )2KK616lb710inKK62345(0.625)(0.125)(2810 )8KKKK2lb273438 inK2newlb()(4)(273438)1093750 inK6710167100671010263664660071010937501093750010937501093750710710500710710uuuu    263664544234809375010937500010937508093750710050007107107.14285710in5.28571410in610inuuuuuu   5(1)62127.14285710in1b281010002inuuEas a check:(1)2500lb1000(4)(0.125)inFA45(2)63225.285714107.14285710lb281016008inuuEas a check:(2)2500lb1600(2.5)(0.125)in44(3)64326105.28571410lb281010002inuuE

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1.5121()0RKuu121232()()0KuuKuu232343()()0KuuKuu`343()0KuuPIn matrix form:111112222233333400KKuRKKKKuKKKKuKKuP   

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1.6Size of the global matrix: 5×5 1[]5555K①②①②2[]8888K②③②③ 3[]5555K②③②③4[]20202020K②④②④ 5[]10101010K③④③④6[]20202020K④⑤④⑤(6)555585208520[]858510102010201020202020KapplyingB.Csandloads;150uu210 lb,F4101bF234234381320101323100201050100.962 in0.874 in0.760 inuuuuuu   15(0.9620)4.8 lbR520(0.7600)15.2 lbR

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1.6 Cont.45()K uunote also as a check,15externalRRF4.8 + 15.2 = 10 +10

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1.75.8815011505.8805.885.882.272.272.272.27101010100.5810.5810.5810.5810.7810.7810.7810.7812.222.222.222.221.471.471.4701.471234567811501000000068TTTTTTTT           1012.0417.3118.51{ }39.1254.4659.F8568T qUA TBtu(1.47)(150)(6859.85)1800 hrqor as another example:Btu(0.781)(150)(54.4639.12)1800 hrqorinout1(150)(6810)Btu()1800Resistance0.170.440.11.721.280.450.68hrqA TT

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1.85.8815011505.8805.885.881.231.231.231.230.760.760.760.760.0530.0530.0530.0532.222.222.222.221.471.471.4701.4712345671150200000068TTTTTTT         2020.3722.12{ }24.9565.5766.F5468T qUA Tas an example:Btu(1.47)(150)(6866.54)320 hrqorinout1(150)(6820)Btu()320Resistance0.170.811.32190.450.68hrqA TT

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1.9with the help of a psychometric chart,usinga dry bulbTemp.of68 Fand40%, we identify condensation temperature to be42 F.Thus condensation will occur betweensurfaces 4 and 5.

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1.101.471000110001.4701.471.470.0530.0530.0530.0532.222.222.222.221.471.471.4701.4712345110001500070TTTTT        1516.81{ }F66.9968.1970TqUA Tas an example:Btu(1.47)(1000)(7068.19)2660 hrqorBtu(1.47)(1000)(16.8115)2660 hrq

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1.115.8822.5122.55.8805.885.882.562.562.562.561.471.471.4701.471234122.5200070TTTT     2026.85{ }F42.5970T1Btu(5.88)(22.5)(26.8520)910 hrqU A Tor as another example:2Btu(2.56)(22.5)(42.5926.85)910 hrqU A Torinout11()(22.5)(7020)Resistance(0.170.390.68)qA TTBtu910 hrq

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1.1222151210.5()()(3)(3)(0.5)33.161 in24ccAA2224(10.59)()()(3)3(0.5)28.661 in24ccAA223()(9)(3)3(0.5)26.411 in4cA615(33.161)(3.2710 )lb()()18,072,7456inccKK624(28.661)(3.2710 )1b()()15,620,2456inccKK63(26.411)(3.2710 )lb()21,590,9924incK261245(3)(0.5) (2910 )lb4()()()()2,847,0686inssssKKKK263(3)(0.5) (2910 )lb4()4,270,6024insKThe Combined stiffnesses:1518,072,7452,847,06820,919,813KK2415,620,2452,847,06818,467,313KK321,590,9924,270,60225,861,594K20,919,813120,919,813020,919,813020,919,81318, 467,31318, 467,31318, 467,31318, 467,31325,861,59425,861,59425,861,59425,861,59418, 467,31318, 467,31318, 467,31318, 467,31320,919,81320,919,81320,919,81320,919,813123456000001000uuuuuu        5412344445604.7801610in1.0195110in1.40618101.9476810in2.4257010inuuuuuu     

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1.12 Cont.(1)6521Concrete2()(3.2710 )( 4.7801610)lb26C6incuuE(1121Steel2()lb231CinsuuEas a check:(1)(1)2Concrete11()()(26)(33.161)(231)(3)(0.5)1000 lb4cssAA645(2)322(3.2710 )( 1.01951104.7801610)1b30C6inccuuE(2)2lb231CinsCheck:(2)(2)222()()(30)(28.661)(231)(3)(0.5)1000 lb4ccssAA644(3)432()(3.2710 )( 1.40618101.0195110)lb32C4inccuuE644(3)432()(2910 )( 1.40618101.0195110)lb280C4inssuuECheck:(3)(3)233()()(32)(26.411)(280)(3)(0.5)1000 lb4ccssAA644(4)542()(3.2710 )( 1.94768101.4061810)lb30C6inccuuE(4)2lb231Cins64(5)652()(3.2710 )( 2.425701.94768)10lb26C6inccuuE(5)2lb231Cinsnote(1)(5)ccand(2)(4)ccas expected.

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1.132avg( )12eiiAE uu(1)(2)(3)(4)(5)total     (1)(5)22111121121()()()()22()()()2ccssiiiiccssAEAEuuuuAEAEuu 626(1)5 2(33.161)(3.2710 )3(0.5) (2910 )4( 4.7801610)0.024 lb in(2)(6)or626(5)4 2(33.161)(3.2710 )3(0.5) (2910 )4[( 2.42571.94768)10]0.024(2)(6)(2)2232626452()()()2(28.661)(3.2710 )3(0.5) (2910 )4[( 1.01951104.7801610)]0.0272(6)ccssAEAEuuor626(4)4 2(28.661)(3.2710 )3(0.5) (2910 )4[( 1.947681.40618)10]0.0272(6)(3)3343()()()2ccssAEAEuu626(3)4 2(26.411)(3.2710 )3(0.5) (2910 )4[( 1.406181.0195110]2(4)0.019 lb intotal(2)(0.024)0.0192(0.027)0.121 lb in

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1.14(a)00M( 6 lb)(5 in)(35 lb)(6 in)(8 in)0sF30 lbsF30 lb600.5 insFKXXX(b)( )110nmeiiiiiiiF uuuu213535(6)0248KXXXXX 35(35)(6)048KX35(35)(6)4860X0.5 inX

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1.15(or)Because of the fact that charge is conserved in acircuit (Kirchhoff’scurrent law), at any time,the algebraicsum of the currents entering any node must be zero. Thus we can write1121 ()IVVR2211 ()IVVRNote120IIIn matrix form:112211111VIVIR 

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1.16(1)111[]115K(2)111[]1110K(3)111[];1115KApply30Vas a boundarycondition110551111115510151015110101501110151231120.01000.110.06VVVVV   120.05 VoltsVV230.06 VoltsVV

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1.1722()()2d YM XwX LXEIEIdX2311223dYwXLwXCdXEI341212612wX LwXYC XCEIApplying Boundary Conditions:2310@000@24YXCwLYXLCEI 323exact(2)24wXYXLXLEI(a)Notethatthe assumed Solutionsatisfies the boundaryconditions.21XXYCLL1221dYXCdXLL21222d YCdXL1222wXLXCEILRWe may force the error function to equalzero at2LX41122220216LLwLCwLCEIEIL2416wLXXYEILL

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1.17Cont.(b)00LR dX1202()02LcwX LXdXEIL3341122022324cwLLwLLcEIEIL2424wLXXYEILLForW24 ×104 Beam43100 in ;I62lb2910inEExact44max642lb1 ft12 in5 500020 ftft12 inft50.20 inlb384(384)29103100 ininwLYEI Collocation44max642lb1 ft12 in500020 ftft12 inft0.24 inlb64(64)2910(3100 in )inwLYEI Subdomain44max642lb1 ft12 in500020 ftft12 inft0.16 inlb96(96)29103100 ininwLYEI 

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1.18SeeSection 1.71.1922d udPdXdy11dudP ycdydX2121()2dP yu yc ycdXapplying B.Cs(0)0uand()0u H221211000and020and2dP Hcc HdXHdPccdX 21( )()2dPu yHyydX note becausePressure drops in the direction of flow,0dPdXin velocity equation.(a)2assumed12sind udPyucdXHdynote the assumed solutionSatisfies the boundaryconditions.11sincosdudyyccdydyHHH22112cossind udyyccdyHHHHdy 21sinydPcHHdX RWe may force the error function to equalzero at2Hy21sin02HdPcHHdX 22122then,( )sinHdPHdPycu ydXdXH  

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1.19 Cont.(b)00HR dy210sin0HydPcdyHHdX2101 cos0HydPcHHHdXH1( 2)dPcHHdX221then,( )sin22HdPHdPycu ydXdXH  ForComparison, let’s use:582N SPa0.02;0.01 mm110m;110mmdPHdX evaluating max vel. @2HyExact2285 2max11( 110 )(110)m0.0622224(8)(0.02)sdPHHdP HuHdXdX     Collocation method22528max222(110) ( 110 )m2sin0.05 s(0.02)HHdPHdPudXHdX    Subdomain method22528max(110) ( 110 )m2sin0.0822(2)(0.02)sHHdPHdPudXHdX    

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1.20(a)Galerkin method210sinsin0HyydPCdyHHHdX  22100sinsinHHydPyCdydyHHdXH21001121sincos24HHdPCyyyHHdXHHH211( 2)2H dPCHHdX    2213344ythen,sinHdPHdPCuydXdXH (b)least-squares method1221100sinsinsin0RcRHHRyydPyRdyCdyCHHHdXHH  resultsin:2213344and( )sinHdPHdPyCu ydXdXH  using values from Problem 1.19 forcomparison2528max3344(110) ( 110 )m0.06 s(0.02)HdPudX 

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1.2122dyPXEIdX 212dyp XcdXEI3126pXyc XcEI B.Cs①( )0,y L②0XLdydX②2211022P LP LccEIEI①3232210623LP LPPLCCLEIEI 323exact(32)6PyXL XLEIIn terms ofx: Substitute forXLxandsimplify323exact( ()3()2)6PyLxLLxLEI32exact36PyxL xEI220dyEIP LxdxB.Cs00yand00xdydxLet us assume:23assumed12yc xc xnote, the assumedsolutionsatisfies the boundaryconditions.

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1.21 Cont.21223dyc xc xdx212226dycc xdxthen ResidualRbecomes:12(26)()REIcc xP LxSubdomain method:2212000[(26)()]0LLR dxEIcc xP Lxdx(1)22120[(26)()]0LLLLR dxEIcc xP Lxdx(2)Integrating(1):2212222602222LLLLEIccPLSimplifying:1233224PLcc LEI (3)Integrating (2) andSimplifying:12948PLcc LEI (4)Solving (3) and (4) Simultaneously:12and26PLPccEIEI 32(3)6PyxLxEInote, because the assumedsolution has thesame functional form as the exactsolution, thecoefficients are exact.

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Solution Manual for Finite Element Analysis: Theory and Application with ANSYS, 5th Edition - Page 29 preview image

1.21 Cont.Galerkin method:2212000[(26)()]0LLx R dxxEIccP Lxdx(1)3312000[(26)()]0LLx R dxxEIccP Lxdx(2)Integrating and Simplifying (1) and (2), we get:1212233212162520PLcc LEIPLcc LEI  Solving(3) and (4) Simultaneously12and26PLPccEIEI 32(3)6PyxLxEI(3)(4)

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Solution Manual for Finite Element Analysis: Theory and Application with ANSYS, 5th Edition - Page 30 preview image

1.2241312JJr441311.50.497 in22JJ 442110.0982 in22J 6(1)(3)1111202942202942(0.497)(9.810 )[][]1111202942202942(2)(12)JGKK6(2)116110261102(0.0982)(11.210 )[]116110261102(1.5)(12)K2029421202942020294202029426110261102611026110220294220294202029420202942123411234002400000.002222 rad0.009604 rad0      433(0)2029420.0096041949 lb inRK112(0)2029420.002222451 lb inRKnote theyadd up to 2400lb⋅in

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Solution Manual for Finite Element Analysis: Theory and Application with ANSYS, 5th Edition - Page 31 preview image

1.2323264044611021200611022640441200230.005913 rad141233(0)(0)2029420.0059131200 lb inRRKK

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