Solution Manual for A First Course in the Finite Element Method, 6th Edition

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Solution Manual for A First Course in the Finite Element Method, 6th Edition - Page 1 preview image

An’s Solutions Manual to AccompanyA FIRST COURSE IN THE FINITE ELEMENTMETHOD, 6THEDITIONDARYLL.LOGAN

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ContentsChapter 1:............................................................................................................................ 1Chapter 2:............................................................................................................................ 3Chapter 3:.......................................................................................................................... 25Chapter 4:........................................................................................................................ 137Chapter 5:........................................................................................................................ 203Chapter 6:........................................................................................................................ 315Chapter 7:........................................................................................................................ 363Chapter 8:........................................................................................................................ 383Chapter 9:........................................................................................................................ 397Chapter 10:...................................................................................................................... 423Chapter 11:...................................................................................................................... 449Chapter 12:...................................................................................................................... 477Chapter 13:...................................................................................................................... 499Chapter 14:...................................................................................................................... 539Chapter 15:...................................................................................................................... 561Chapter 16:...................................................................................................................... 591Appendix A:..................................................................................................................... 629Appendix B: ..................................................................................................................... 635Appendix D:..................................................................................................................... 641

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1Chapter 11.1.A finite element is a small body or unit interconnected to other units to model a largerstructure or system.1.2.Discretization means dividing the body (system) into an equivalent system of finite elementswith associated nodes and elements.1.3.The modern development of the finite element method began in 1941 with the work ofHrennikoff in the field of structural engineering.1.4.The direct stiffness method was introduced in 1941 by Hrennikoff. However, it was notcommonly known as the direct stiffness method until 1956.1.5.A matrix is a rectangular array of quantities arranged in rows and columns that is often usedto aid in expressing and solving a system of algebraic equations.1.6.As computer developed it made possible to solve thousands of equations in a matter ofminutes.1.7.The following are the general steps of the finite element method.Step 1Divide the body into an equivalent system of finite elements with associatednodes and choose the most appropriate element type.Step 2Choose a displacement function within each element.Step 3Relate the stresses to the strains through the stress/strain law—generally calledthe constitutive law.Step 4Derive the element stiffness matrix and equations. Use the direct equilibriummethod, a work or energy method, or a method of weighted residuals to relate thenodal forces to nodal displacements.Step 5Assemble the element equations to obtain the global or total equations andintroduce boundary conditions.Step 6Solve for the unknown degrees of freedom (or generalized displacements).Step 7Solve for the element strains and stresses.Step 8Interpret and analyze the results for use in the design/analysis process.1.8.The displacement method assumes displacements of the nodes as the unknowns of theproblem. The problem is formulated such that a set of simultaneous equations is solved fornodal displacements.1.9.Four common types of elements are: simple line elements, simple two-dimensional elements,simple three-dimensional elements, and simple axisymmetric elements.1.10Three common methods used to derive the element stiffness matrix and equations are(1)direct equilibrium method(2)work or energy methods

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2(3)methods of weighted residuals1.11.The term ‘degrees of freedom’ refers to rotations and displacements that are associated witheach node.1.12.Five typical areas where the finite element is applied are as follows.(1)Structural/stress analysis(2)Heat transfer analysis(3)Fluid flow analysis(4)Electric or magnetic potential distribution analysis(5)Biomechanical engineering1.13.Five advantages of the finite element method are the ability to(1)Model irregularly shaped bodies quite easily(2)Handle general load conditions without difficulty(3)Model bodies composed of several different materials because element equations areevaluated individually(4)Handle unlimited numbers and kinds of boundary conditions(5)Vary the size of the elements to make it possible to use small elements where necessary

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3Chapter 22.1(a)[k(1)]=11110–00000–000000kkkk[k(2)] =22220000000000–00–kkkk[k3(3)] =3333000000–00000–0kkkk[K] = [k(1)]+ [k(2)] + [k(3)][K] =1133112232230–000––0–0––kkkkkkkkkkkk(b)Nodes 1 and 2 are fixed sou1= 0 andu2= 0 and [K] becomes[K] =122223––kkkkkk{F} = [K] {d}34xxFF=122223––kkkkkk34uu0P=122223––kkkkkk34uu{F} = [K] {d}[K]–1{F} = [K]–1[K] {d}

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4 K]–1{F} = {d}Using the adjoint method to find [K–1]C11=k2+k3C21=(–1)3(–k2)C12= (–1)1 + 2(–k2) =k2C22=k1+k2[C] =232212kkkkkkandCT=232212kkkkkkdet [K] = | [K] | = (k1+k2) (k2+k3)–(–k2) (–k2)| [K] | = (k1+k2) (k2+k3)–k22[K–1] =[]detTCK[K–1] =232212212232() () –kkkkkkkkkkk=232212121323kkkkkkk kk kkk34uu=2322121213230kkkkkkPk kk kkku3=2121323kPk kk kkku4=12121323()kkPk kk kkk(c)In order to find the reaction forces we go back to the global matrixF= [K]{d}1234xxxxFFFF=1112331122332234000000ukkukkkkkkukkkkuF1x=–k1u3=–k12121323kPk kk kkkF1x=12121323k kPk kk kkkF2x=–k3u4=–k312121323()kkPk kk kkkF2x=312121323()kkkPk kk kkk2.2

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5k1=k2=k3= 1000lbin.(1)(2)(2)(3)[k(1)] =(1)(2)kkkk;[k(2)] =(2)(3)kkkkBy the method of superposition the global stiffness matrix is constructed.(1)(2)(3)[K] =0 (1)(2)0(3)kkkkkkkk[K] =020kkkkkkkNode 1 is fixedu1= 0 andu3={F} = [K] {d}123?0?xxxFFF=020kkkkkkk1230?uuu 30xF=2232022xk ukkkuFk ukkku2=2kk=2=1 in.2u2= 0.5F3x=–k(0.5) +k(1)F3x= (–1000lbin.) (0.5) + (1000lbin.) (1)F3x= 500 lbsInternal forcesElement (1)(1)1(2)2xxff=1200.5ukkkku(1)1xf= (–1000lbin.) (0.5)(1)1xf=–500 lb(1)2xf= (1000lbin.) (0.5)(1)2xf= 500 lbElement (2)

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6(2)2(2)3xxff=22330.50.511uukkkkuukkkk(2)2(2)3– 500 lb500 lbxxff2.3(a)[k(1)] = [k(2)] = [k(3)] = [k(4)] =kkkkBy the method of superposition we construct the global [K] and knowing {F} = [K] {d}we have12345?00?xxxxxFFFPFF=000200020002000kkkkkkkkkkkkk1234500uuuuu(b)00P=22332343442002220202ukkkukukkkuPkukukukkkukuuu2=32u;u4=32uSubstituting in the second equation aboveP=–ku2+ 2ku3–ku4P=–k32u+ 2ku3–k32uP=ku3u3=Pku2=2Pk;u4=2Pk(c)In order to find the reactions at the fixed nodes 1 and 5 we go back to the globalequation {F} = [K] {d}F1x=–ku2=–2PkkF1x=2PF5x=–ku4=–2PkkF5x=2PCheckFx= 0F1x+F5x+P= 0(1)(2)(3)

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72P+2P+P= 00 = 02.4(a)[k(1)] = [k(2)] = [k(3)] = [k(4)] =kkkkBy the method of superposition the global [K] is constructed.Also{F} = [K]{d} andu1= 0 andu5=12345?000?xxxxxFFFFF=000200020002000kkkkkkkkkkkkk123450???uuuuu(b)0 = 2ku2–ku3(1)0 =–ku2+ 2ku3–ku4(2)0 =–ku3+ 2ku4–k(3)From (2)u3= 2u2From (3)u4=222uSubstituting in Equation (2)–k(u2) + 2k(2u2)–k222u –u2+ 4u2–u2–2= 0u2=4u3=2 4u3=2u4=422u4=34(c)Going back to the global equation

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8{F} = [K]{d}F1x=–ku2=4kF1x=4kF5x=–ku4+k=–k34+kF5x=4k2.5u1u2u2u4[k(1)] =1111;[k(2)] =2222u2u4u2u4[k(3)] =3333;[k(4)] =4444u4u3[k(5)] =5555Assembling global [K] using direct stiffness method[K] =11001123402340055023452345Simplifying[K] =110011009kip0055in.095142.6Now apply + 3 kip at node 2 in spring assemblage of P 2.5.F2x= 3 kip[K]{d} = {F}

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9[K] from P 2.5110011009005509514123400uuuu=1330FF(A)whereu1= 0,u3= 0 as nodes 1 and 3 are fixed.Using Equations (1) and (3) of (A)24109914uu=30Solvingu2=0.712 in.,u4= 0.458 in.2.7f1x=C,f2x=–Cf=–k=–k(u2–u1)f1x=–k(u2–u1)f2x=–(–k) (u2–u1)12xxff=k–k–kk12uu[K] =k–k–kksame as fortensile element2.8k1= 10001111;k2= 10001111So

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10[K] = 1000110121011{F} = [K] {d}123?0500FFF=11010121000111230??uuu0 = 2000u2–1000u3(1)500 =–1000u2+ 1000u3(2)From (1)u2=10002000u3u2= 0.5u3(3)Substituting (3) into (2)500 =–1000 (0.5u3) + 1000u3500 = 500u3u3= 1 in.u2= (0.5) (1 in.)u2= 0.5 in.Element 1–2(1)1(1)2xxff= 1000(1)1(1)2500lb110in.110.5in.500lbxxff    Element 2–3(2)2(2)3xxff= 1000(2)2(2)3500 lb110.5 in.111 in.500 lbxxff  F1x= 500 [1–1 0]100.5in.500 lb1 in.xF  2.9(1)(2)[k(1)] =5000500050005000(2)(3)

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11[k(2)] =5000500050005000(3)(4)[k(3)] =5000500050005000(1)(2)(3)(4)[K] =5000500000500010000500000500010000500000500050001234?100004000xxxxFFFF=50005000005000100005000005000100005000005000500012340uuuuu1= 0 in.u2= 0.6 in.u3= 1.4 in.u4= 2.2 in.ReactionsF1x= [5000–500000]123400.61.42.2uuuuF1x=–3000 lbElement forcesElement (1)(1)1(1)2xxff=500050000500050000.6  (1)1(1)23000lb3000lbxxffElement (2)(2)2(2)3xxff=500050000.6500050001.4  (2)2(2)34000lb4000lbxxffElement (3)(3)(3)33(3)(3)44xxxxffff=500050001.4500050002.2  (3)3(3)44000lb4000lbxxff2.10

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12[k(1)] =1000100010001000[k(2)] =500500500500[k(3)] =500500500500{F} =[K] {d}1234?– 8000??xxxxFFFF=123401000100000?10002000500500005005000050005000uuuuu2=80002000=–4 in.Reactions1234xxxxFFFF=01000100000100020005005004050050000050005000     1234xxxxFFFF=4000800020002000lbElement (1)(1)1(1)2xxff=10001000010001000– 4  (1)1(1)2xxff=40004000lbElement (2)(2)2(2)3xxff=50050045005000  (2)2(2)3xxff=– 20002000lbElement (3)

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13(3)2(3)4xxff=50050045005000  (3)2(3)4xxff20002000lb2.11[k(1)] =1000100010001000;[k(2)]=3000300030003000{F} = [K] {d}123?0?xxxFFF=1000100001000400030000300030001230?0.02 muuuu2= 0.015 mReactionsF1x= (–1000) (0.015)F1x=–15 NElement (1)12xxff=100010001000100000.01512xxff1515NElement (2)23xxff=3000–3000–300030000.0150.0223xxff=1515N2.12[k(1)]= [k(3)]= 100001111[k(2)]= 100003333{F}= [K] {d}

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