Solution Manual for Introduction to Finite Elements in Engineering, 4th Edition

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Introduction toFinite Elementsin EngineeringTirupathi R. ChandrupatlaRowan UniversityGlassboro, New JerseyAshok D. BelegunduThePennsylvania StateUniversityUniversity Park, PennsylvaniaSolutions ManualPrentice Hall, Upper Saddle River, New Jersey 07458

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CONTENTSPrefaceChapter 1Fundamental Concepts1Chapter 2Matrix Algebra and Gaussian Elimination18Chapter3One-Dimensional Problems26Chapter 4Trusses61Chapter 5Beams and Frames86Chapter 6Two-Dimensional Problems Using Constant103Strain TrianglesChapter 7Axisymmetric Solids Subjected to151Axisymmetric LoadingChapter 8Two-Dimensional Isoparametric Elements181and Numerical IntegrationChapter 9Three-Dimensional Problems in Stress207AnalysisChapter10Scalar Field Problems218Chapter11Dynamic Considerations264Chapter 12Preprocessing and Postprocessing282

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PREFACEThis solutions manual serves as an aid to professors in teaching from the bookIntroduction toFinite Elements in Engineering, 4thEdition.The problems in the book fall into the followingcategories:1. Simple problems to understand the concepts2. Derivations and direct solutions3. Solutions requiring computer runs4. Solutions requiring program modificationsOur basic philosophy in the development of this manual is to provide a complete guidance to theteacher in formulating, modeling, and solving the problems. Complete solutions are given forproblems in all categories stated. For some larger problems such as those in three dimensionalstress analysis, complete formulation and modeling aspects are discussed. The students shouldbe able to proceed from the guidelines provided.For problems involving distributed and other types of loading, the nodal loads are to becalculated for the input data. The programs do not generate the loads. This calculation and theboundary condition decisions enable the student to develop a physical sense for the problems.The students may be encouraged to modify the programs to calculate the loads automatically.The students should be introduced to the programs in Chapter 12 right from the point of solvingproblems in Chapter 6. This will enable the students to solve larger problems with ease. Theinput data file for each program has been provided. Data for a problem should follow thisformat. The best strategy is to copy the example file and edit it for the problem underconsideration. The data from program MESHGEN will need some editing to complete theinformation on boundary conditions, loads, and material properties.We thank you for your enthusiastic response to our first threeeditions of the book. We lookforward to receive your feedback of your experiences, comments, and suggestions for makingimprovements to the book and this manual.Tirupathi R. Chandrupatla P.E., CMfgEDepartment of Mechanical EngineeringRowan University, Glassboro, NJ 08028e-mail:chandrupatla@rowan.eduAshok D. BelegunduDepartment of Mechanical and Nuclear EngineeringThe Pennsylvania State UniversityUniversity Park, PA 16802e-mail:adb3@psu.edu

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1CHAPTER 1FUNDAMENTAL CONCEPTS1.1We use the first three steps of Eq. 1.11EEEEEEEEEzyxzzyxyzyxxAdding the above, we getzyxzyxE21Adding and subtractingExfrom the first equation,zyxxxEE1Similar expressions can be obtained fory, andz.From the relationship foryzand Eq. 1.12,yzyzE12etc.Above relations can be written in the form=DwhereDis the material property matrix defined in Eq. 1.15.1.2Note thatu2(x) satisfies the zero slope boundary condition at the support.

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21.3Plane strain condition implies that0yxzzEEEWhich gives()zxy On substitution of the given valuesz= 0.3 (3000–15000) = 4500 psi1.4u= (x2+4y2–16xy)×10-424(58 )10vyxy410ux(2x–16y)4(816 )10uyxy4510uy  uy=(2y+8) ×10-4uxuyuuyx Atx=1,y=04281021xyxy

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31.5From the figure P 1.5, it is revealed that the displacementsuandvare given byu(x) = 0.2y+ 6v(y) = 0u1= 0.2× 0 + 6 = 6v1= 0u2=u1+ [0.2×0+6] = 12v2= 0u4= 0.2 × 15 + 6 =9u3=u4+ [0.2×15+6]= 15The strain values are computed as:εx=ux= 0εy=vy= 0xyuvxy= 0.21.6u=2+2x+4x2+3xy2v=xy=8x2a)Expressions forεx,εyandγxyεx=ux= 2 + 8x+ 3y2εy=vy=xxyuvxy=6xy+y–16x

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4(b)In order to draw the contours of the strain field using MATLAB, we need to create ascriptfile, which may be edited as a text file and save with “.m” extension. The filefor plottingεxεyandγxyis given belowb) MATLAB code forεx, εy, γxy:[X,Y] = meshgrid(-1:.1:1,-1:.1:1);Z = 2.+8.*X + 3.*Y.^2;k = X.;L = 6.*X.*Y+Y.-16.*X;[C,h]=contour(X,Y,Z);clabel(C,h);On running the program, the contour map is shown as follows:

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5c)The maximum value ofεxis 12 in the location (0.9, 0.9) of the square region.1.7a)0.20.201uyuyvb)000.2xyxyuvuvxyyx(x,y)(u,v)

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61.8x= 120 MPa,y=55MPa,z=–85MPayz=33 MPa,xz= 75 MPa andxy=–55MPa.n=111,,222TFrom Eq. 1.8 we getTx=xxn+xyyxzznn=85.50 MPaTy=xyxyyyzznnn=23.33 MPaTZ=xyxyzyzznnn=–6.10 MPaThe normal stressnxxyyzzT nT nT nn= 50 MPa1.9From the derivation made in P1.1, we have  yzyzvxxzyxxEEE12and21211formin thewrittenbecanwhich1211Lame’s constantsandare defined in the expressions

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712211,inspectionOn2EEyzyzxvxis same as the shear modulusG.1.105210T=45°CE=123GPa616.210/°CFrom Eq. 1.20 and 1.21,47.2910oT=E(ε–εo)=–87.2MPa1.11The strain at any pointεx= 5+4x2=dudxδ=3453LLOOdudxxxdx2453LL1.12With reference to Eq.1.29, the global stiffness matrix can be formed as;

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8123025505045505030qqSolving105q1+50q2=45–(i)–50q1+50q2=30–(ii)Adding (i) and (ii)55q1=75q1= 1.363 mm and on substitution ofq1,q2= 1.963 mm1.13When the wall is smooth,0x.Tis the temperature rise.a)When the block is thinin thezdirection, it corresponds to plane stress condition. Therigid walls in theydirection require0y. The generalized Hooke’s law yields theequationsyxyyTETE From the second equation, setting0y, we getyET .xis then calculatedusing the first equationas1T.b)When the block is very thick in thezdirection, plain strain condition prevails. Now wehave0z, in addition to0y.zis not zero.

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900yzxyzyyzzTEETEETEE  From the last two equations, we get1211yzETET xis now obtained from the first equation.1.14For thin block, it is plane stress condition. Treating the nominal size as 1, we may set theinitial strain00.11Tin part (a) of problem 1.13.Thus0.1yE .1.15The potential energy π is written as21.. .2LLooduEAdxu g A dxdx -1Let us consider a polynomial eqn. for the displacementu=a(–2x+x2)dudx= 2a(–1 +x)Substituting the expression in the first term ofEqn 1 and integrating:2201[2 (1)]2EAa xdx 114103aThe second term of the eqn(1)becomes;22200. .9.81( 2)u g Adxaxxdx 

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1023209.813xax 4(9.81)3a Eqn(1)becomes:11244109.8133aa Then114109.813aa Minimizing the potential energy yields:na= 0 =43[ 2a× 1011+ 9.81] = 0a=–4.90 ×10-11The displacement of the mid node is;ux=1=–4.90 × 10-11(–2 + 1)ux=1= 4.90 × 10-11mWe use the displacement field byu=a0+a1x+a2x2u=0 atx=0a0= 0u= 0 atx=1a1+a2= 0a2=–a,We haveu=a1x(1–x) &dudx=a1(1–x)2110012duEdxf u dxdx 11112110021012(1)2axxx ax x dx21111210112360aa0dda 

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11111211100360aa1=1111210603a1= 0.275 × 10-11Displacement =u= 0.275x(1–x)× 10-11StressEdudx = 2×1011×0.275 ×10-11(1–x)=0.55 (1–x)1.16We use the displacement field defined byu=a0+a1x+a2x2.u= 0 atx= 0a0= 0u= 0 atx= 1a1+a2= 0a2=a1We then haveu=a1x(1x), anddu/dx=a1(1x).The potential energy is now written as30661513424121441211212121121121101054122110101322110102aaaadxxxadxxxadxxxaxdxxafudxdxdxduf=x3E= 1A= 1x=0x=1

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1203013011aaThis yields,a1= 0.1Displacemenu= 0.1x(1x)Stress=Edu/dx= 0.1(1x)1.17Letu1be the displacement atx= 300 mm piece wise linear displacement that iscontinuous in the interval 0≤x≤800 is represented as shown in the figure below:For element(1)u= 0 atx= 0;a1=0u=u1atx= 300;a2=1300uThe linear displacement field1300uux 1300ududx For element(2)u= 0 atx=800a3+ 800a4= 0-(i)u=u1atx=800a3+ 300a4=u1-(ii)From (i) and (ii)14500ua 3185au Substitution ofa3anda4in the linear displace at field

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131185500uuux1500ududx22300800121030011.800022brduduEAdxEcu Adxudxdx 221112111300500800023002500brcuuuEAEAu 21211180002300500brcuEAEAuu 121080000300500brcuEAEAuuSubstituting the values ofEbr,Ecu,A1andA2in the above eqation.,u1= 0.015 mm1.18In the Galerkin method, we start from the equilibrium equation0gdxduEAdxdFollowing the steps of Example 1.3, we get2020dxgdxdxddxduEAIntroducing12122and,2xxuxxuwhereu1and1are the values ofuandatx= 1 respectively,022120202211dxxxdxxuOn integrating, we get

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140343811uThis is to besatisfied for every1, which gives the solutionu1=0.51.19We use2at00at0342321xuxuxaxaxaauThis implies that4321184200aaaaand4312422433423xaxadxduxxaxxaua3anda4are considered as independent variables in2043224332431221aadxxaxaon expanding and integrating the terms, we get434324236288.12333.1aaaaaaWe differentiate with respect to the variables and equate to zero.066.258028667.2434433aaaaaaOn solving, we geta3=0.74856 anda4=0.00045.On substituting in the expression foru, atx= 1,u1= 0.749This approximation is close to the value obtained in the example problem.

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151.20(a) udxxTAdxLL00T21dxduEandOn substitution,udxudxxdxdxduudxTudxTdxdxduEA603030060026603030060023001010602121(b)Sinceu= 0 atx= 0 andx= 60, andu=a0+a1x+a2x2, we have6026022xadxduxxauOn substituting and integrating,22210877500010216aaSettingd/da2= 0 gives602935.601003125.262xdxduEaPlots of displacement and stress are given below:

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16010203040506000.20.40.60.811.21.41.61.82x 10-3Displacementu0102030405060-4000-3000-2000-100001000200030004000Stress.1.21y= 20 atx= 60implies that2102101803120yieldswhich,36006020aaaaaaSubstituting fork,h,L, anda0inI, we get76050001070210117104561200045600391835000441080018031202521210241422521216002212222121600221221aaaaaaIaadxaxaxaaIaadxxaaI0107021090612000010117612000912000425124211aadadIaadadI

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17On solving,a2= 0.1699a1=13.969Substituting into the expression fora0, we geta0= 246.538.1.22Sinceu= 0 atx= 0, the displacement satisfying the boundary condition isu=a1x.Alsothe coordinates arex2= 1, andx3= 3.The potential energy for the problem is232233012duEAdxP uPudxWe haveu2=a1,u3= 3a1,E= 1,A= 1,and1duadx. Thus322111110133422adxaaaa .For stationary value, setting10dda , we get3a1–4 = 0, which givesa1= 0.75.The approximate solution isu= 0.75x.1.23UseGalerkin approach with approximation2uabxcxto solve30101duuxxdxu

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18The week form is obtained by multiplying bysatisfying00.1030duuxdxdx We now set21ubxcxsatisfying01uand212a xa x . On introducing theseinto the above integral,122120112223223334120023330332333230a xa xbcxbxcxx dxabxxxbxcxcxdxabxxxbxcxcxdx On integrating, we get121231231331022334344253177131130212612104bccbbccabaabcabcThismust be satisfied for everya1anda2. Thus the equations to be solved are31770212613113012104bcbcThe solution isb=–1.9157,c= 1.2048.Thus211.91571.2048uxx.1.24The deflection and slope atadue toP1are313PaEIand212PaEI. Using this the deflectionand slope atLdue to loadP1are

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1923111211322PaLaPavEIEIPavEI   The deflection and slope due to loadP2are32222232P LvEIP LvEI   We then get1212vvvvvv1.25(a)The displacement ofBis given by (–0.1, 0.1) andA,C, andDremain in their originalposition. Consider a displacement field of the type12341234uaa xa ya xyvbb xb yb xyThe four constants can be evaluated using the known displacementsAtA(0, 0)1100ab(0,0)(1,0)(1,1)(0,1)

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20AtB(1, 0)12120.10.1aabb AtC(1, 1)1234123400aaaabbbbAtD(0, 1)131300aabbThe solution isa1= 0,a2=–0.1,a3= 0,a4= 0.1b1= 0,b2= 0.1,b3= 0,b4This gives0.10.10.10.1uxxyvxxy (b) The shear strain atBis 0.10.10.10.1 10.10.1 00.2Buvxyyx

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18CHAPTER 2MATRIX ALGEBRA AND GAUSSIAN ELIMINATION2.1312,330342028dA(a)836302623312312100010001TddI(b)detA= 8 [ (4)(3)–(–3) (–3) ]–(–2) [(–2)(3)–(0)(–3)]= 12(c)The characteristic equation is det (A-I) = 0, or0330342028detwhich yields012551523Handbooks (e.g., CRC Mathematical Handbook) give explicit solutions to cubicequations. Here equations given in Chapter 9 are used, which give formulas forfinding the eigenvalues of the (3x3) symmetric stress tensor. Referring to Section9.3 in the text, we haveI1= A11+ A22+ A33= 15, I2= 55, I3= 12Thus,a= 20,b= 13,c= 5.164,= 37.4whence1= 0.2325,2= 5.665,3= 9.103Note: Since alli> 0,Aispositive definite.Now, eigenvectoryicorresponding to eigenvalueiis obtained from(a-iI)yi= 0,i= 1,2,3Thus,y1is obtained as0007675.23037675.32027675.7321yyyThus,

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197.7675y1–2y2= 0-2y1+ 3.7675y2–3y3= 0-3y2+ 2.7675y3= 0______________________Only two of the above three equations are independent. We havey1= 0.2575y2y2= 0.922y3Lettingy3= 1, we gety1= [0.237, 0.922, 1]TThe length of the vector is381.1T1yyy. Normalizingy1to be a unitvector yieldsy1= [0.172, 0.668, 0.724]T.Similarly,y2= [0.495, 0.577,-0.650]T,y3= [0.850,-0.470, 0.232]T.(d)Solution to A x = b using Algorithm 1 for general matrix:n= 3First Step(k= 1)i=2 (2ndrow)3334A02028c=a21/a11=-2/8 =-1/4a22(1)= 4–(-1/4)(-2) = 7/2,a23(1)=-3,d2(1)=-1-(-1/4)(2) =-1/2i= 3 (3rdrow)c = 0a32(1)=-3,a33(1)= 3,d3(1)= 3Thus32/12,33032/70028)1()1(dASecond Step(k= 2)i= 3 (3rdrow)c =-6/7,a33(2)= 3–(-6/7)(-3) = 3/7,d3(2)= 3–(-6/7)(-1/2)=18/7Thus7/182/12,7/30032/70028)1()2(dA

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20Back-Substitution7/182/127/30032/70028321xxxThird row gives: 3/7x3= 18/7 whencex3= 62ndrow then givesx2= 5,1strow givesx1= 1.5Thus, solution isx= [1.5, 5, 6 ]T.Solution to A x = b using Algorithm 2 for symmetric, banded matrixAis stored as033428n= 3,nbw= 2First Step(k= 1)nbk= min (3,2) = 22ndrow (i= 2):i1 = 2c =a12/a11=-1/4j1 = 1j2 = 2a21= 4–(-1/4)(-2) = 7/2ThusA(1)=0332/728Second Step(k= 2)nbk= 23rdrow (i= 3):c =-6/7j1 = 1j2 = 2a31= 3/7ThusA(2)=07/332/728reduction of right-hand-side vectordand back-substitution is same as inAlgorithm 1 above, resulting in the the same solutionx= [1.5, 5, 6 ]T.

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212.221N(a)34011dN(b)15/16003/2)1()1()1(2222211TdNN2.3q=x1–6x2+ 3x12+ 5x1x2=2121216105.25.23xxxxxxxTQx+cTxwhere6105.25.23cQand2.4The detailed algorithm is given in the text. This is an excercise in computerprogramming.The solutions are (a) (–2.25,–11.5,–10.5) (b) (1.55, 5.1, 6.1)2.5The minors areM11= det1312=–1M12= det2332=–5M13= det2131=–1M21= 1;M22=–7;M23=–5M31= 3;M32=–3;M33=–3Co factor

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22Cij= (–1)i+jMijThus, the co-factor matrix isAc=151175333The inverse of matrix AA-1=1detAAcTyieldsdet(A) = (2–3)–2 (4–9) + 3(2–3) = 6A-1=161135731532.6Area =12det12217811112= 6= 6 square units.2.7A1=625.155.215.131221det21A2=25.111155.21221det21A3=75.05.131111221det21A = A1+ A2+ A3= 3.625A1/ A = 0.448, A2/ A = 0.345, A3/ A = 0.2072.8Ai,j= Bi,j-i+ 1forji

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23Thus, A11,14corresponds to B11,4andB6,1A6,62.9Full (10x10) matrixBANDEDn= 10,nbw= 10No. of storage locations = (n) (nbw)= 100SKYLINEx x x x ...x x x ...x x ...x ...No. of storage locations = no. of column entries= 1 + 2 + 3 + ... + 10= (10) (10+1) / 2= 552.10Based on Eq. 2.2;K=1 (1strow)l11=11a=4= 2K=2 (2ndrow)l21=2111al=22= 1l22=22221al= 3K=3 (3rdrow)l31=3111al=–3l32=3231 2132al ll= 4l33=223331321allThus

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24L=2001303412.11By expanding the matrix equations, we obtain a set of simultaneous equations for theuandcoefficients, resulting in the solutionL=100210321andU=1230120012.12The quadrilateralABCDis divided in to two triangles with cornersABCandACD.Area of triangleABCA1=12det111172166= 12.5Area of triangleACDA2=12det111166137= 10Area of the quadrilateralABCDA=A1+A2A= 22.5 square unit.2.13Settingcossinsincos T, it is easy to show thatT1001 TT.2.14a)The minors of the given matrix are:M11= 7;M12= 4;M13=–1M21= 11;M22= 17;M23= 7

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