Rao

7,744 views

Published on

Published in: Technology, Business
  • Be the first to comment

Rao

  1. 1. Aurel Boreslav Stodola (1859–1942) was a Swiss engineer who joinedthe Swiss Federal Institute of Technology in Zurich in 1892 to occupythe chair of thermal machinery. He worked in several areas includingmachine design, automatic controls, thermodynamics, rotor dynamic,and steam turbines. He published one of the most outstanding books,namely, Die Dampfturbin, at the turn of the century. This book dis-cussed not only the thermodynamic issues involved in turbine design butalso the aspects of fluid flow, vibration, stress analysis of plates, shellsand rotating discs, thermal stresses and stress concentrations at holesand fillets, and was translated into many languages. The approximatemethod he presented for the computation of natural frequencies ofbeams has become known as the Stodola method. (Photo courtesy ofApplied Mechanics Reviews.)85612.1 IntroductionThe finite element method is a numerical method that can be used for the accurate solu-tion of complex mechanical and structural vibration problems [12.1, 12.2]. In this method,the actual structure is replaced by several pieces or elements, each of which is assumed tobehave as a continuous structural member called a finite element. The elements areassumed to be interconnected at certain points known as joints or nodes. Since it is verydifficult to find the exact solution (such as the displacements) of the original structureunder the specified loads, a convenient approximate solution is assumed in each finite ele-ment. The idea is that if the solutions of the various elements are selected properly, theycan be made to converge to the exact solution of the total structure as the element size isreduced. During the solution process, the equilibrium of forces at the joints and the com-patibility of displacements between the elements are satisfied so that the entire structure(assemblage of elements) is made to behave as a single entity.The basic procedure of the finite element method, with application to simple vibrationproblems, is presented in this chapter. The element stiffness and mass matrices, and forcevectors are derived for a bar element, a torsion element, and a beam element. The trans-formation of element matrices and vectors from the local to the global coordinate systemis presented. The equations of motion of the complete system of finite elements and theC H A P T E R 1 2Finite ElementMethodRaoCh12ff.qxd 10.06.08 13:48 Page 856
  2. 2. 12.2 EQUATIONS OF MOTION OF AN ELEMENT 857incorporation of the boundary conditions are discussed. The concepts of consistent andlumped mass matrices are presented along with a numerical example. Finally, a computerprogram for the eigenvalue analysis of stepped beams is presented. Although the tech-niques presented in this chapter can be applied to more complex problems involvingtwo- and three-dimensional finite elements, only the use of one-dimensional elements isconsidered in the numerical treatment.12.2 Equations of Motion of an ElementFor illustration, the finite element model of a plano-milling machine structure (Fig. 12.1a)is shown in Fig. 12.1(b). In this model, the columns and the overarm are represented by tri-angular plate elements and the cross slide and the tool holder are represented by beam ele-ments [12.3]. The elements are assumed to be connected to each other only at the joints. Thedisplacement within an element is expressed in terms of the displacements at the corners orjoints of the element. In Fig. 12.1(b), the transverse displacement within a typical element e isassumed to be w(x, y, t). The values of w, and at joints 1, 2, and 3—namely—are treatedas unknowns and are denoted as The displacement w(x, y, t)can be expressed in terms of the unknown joint displacements in the form(12.1)where is called the shape function corresponding to the joint displacementand n is the number of unknown joint displacements ( in Fig. 12.1b). If a distributedn = 9wi(t)Ni(x, y)w(x, y, t) = ani=1Ni(x, y)wi(t)wi(t)w9(t).w1(t), w2(t), w3(t), Á ,(0w)/(0y)(x3, y3, t)y1, t), (0w)/(0y)(x1, y1, t), Á ,w(x1 y1, t), (0w)/(0x)(x1,(0w)/(0y)(0w)/(0x),BeamelementsColumnOverarm Tool holderColumnCross-slideCutter(a) Plano-milling machine structure (b) Finite element modelxzyBedFyFxFzCutting forcesPlateelementsElement ew3(t)w1(t)w9(t)w7(t)w8(t)w2(t)w4(t)w6(t)w5(t)2w(x, y, t)1f(x, y, t)3eFIGURE 12.1 Finite element modeling.RaoCh12ff.qxd 10.06.08 13:48 Page 857
  3. 3. 858 CHAPTER 12 FINITE ELEMENT METHODload f (x, y, t) acts on the element, it can be converted into equivalent joint forcesIf concentrated forces act at the joints, they can also be added to theappropriate joint force We shall now derive the equations of motion for determiningthe joint displacements under the prescribed joint forces By using Eq. (12.1),the kinetic energy T and the strain energy V of the element can be expressed as(12.2)(12.3)whereand [m] and [k] are the mass and stiffness matrices of the element. By substitutingEqs. (12.2) and (12.3) into Lagrange’s equations, Eq. (6.44), the equations of motion of thefinite element can be obtained as(12.4)where is the vector of joint forces and is the vector of joint accelerations given byNote that the shape of the finite elements and the number of unknown joint displacementsmay differ for different applications. Although the equations of motion of a single element,Eq. (12.4), are not useful directly (as our interest lies in the dynamic response of theassemblage of elements), the mass matrix [m], the stiffness matrix [k], and the joint forcevector of individual elements are necessary for the final solution. We shall derive theelement mass and stiffness matrices and the joint force vectors for some simple one-dimensional elements in the next section.12.3 Mass Matrix, Stiffness Matrix, and Force VectorConsider the uniform bar element shown in Fig. 12.2. For this one-dimensional element,the two end points form the joints (nodes). When the element is subjected to axial loadsand the axial displacement within the element is assumed to be linear in x as(12.5)u(x, t) = a(t) + b(t)xf2(t),f1(t)12.3.1Bar Elementf:W:$= ew$1w$2...w$nu = ed2w1/dt2d2w2/dt2...d2wn/dt2uW:$f:[m]W:$+ [k]W:= f:W!= ew1(t)w2(t)...wn(t)u, W:#= ew#1(t)w#2(t)...w#n(t)u = edw1/dtdw2/dt...dwn/dtuV =12W:T[k]W:T =12W:#T[m]W:#fi(t).wi(t)fi(t).fi(t)(i = 1, 2, Á , 9).RaoCh12ff.qxd 10.06.08 13:48 Page 858
  4. 4. 12.3 MASS MATRIX, STIFFNESS MATRIX, AND FORCE VECTOR 859u1(t)f1(t)Joint 1Joint 2␳l, E, Au2(t)f2(t)u(x, t)f(x, t) xxFIGURE 12.2 Uniform bar element.When the joint displacements and are treated as unknowns, Eq. (12.5) shouldsatisfy the conditions(12.6)Equations (12.5) and (12.6) lead toand(12.7)Substitution for a(t) and b(t) from Eq. (12.7) into Eq. (12.5) gives(12.8)or(12.9)where(12.10)are the shape functions.The kinetic energy of the bar element can be expressed as(12.11)=12rAl3(u#12+ u#1u#2 + u#22)=12 Ll0rA e a1 -xlbdu1(t)dt+ axlbdu2(t)dtf2dxT(t) =12 Ll0rA e0 u(x, t)0tf2dxN1(x) = a1 -xlb, N2(x) =xlu(x, t) = N1(x)u1(t) + N2(x)u2(t)u(x, t) = a1 -xlb u1(t) +xlu2(t)a(t) + b(t)l = u2(t) or b(t) =u2(t) - u1(t)la(t) = u1(t)u(0, t) = u1(t), u(l, t) = u2(t)u2(t)u1(t)RaoCh12ff.qxd 10.06.08 13:48 Page 859
  5. 5. 860 CHAPTER 12 FINITE ELEMENT METHODwhereis the density of the material and A is the cross-sectional area of the element.By expressing Eq. (12.11) in matrix form,(12.12)whereand the superscript T indicates the transpose, the mass matrix [m] can be identified as(12.13)The strain energy of the element can be written as(12.14)where and E is Young’s modulus. By expressing Eq. (12.14) inmatrix form as(12.15)wherethe stiffness matrix [k] can be identified as(12.16)[k] =EAlc1 -1-1 1du!(t) = eu1(t)u2(t)f and u!(t)T= 5u1(t) u2(t)6V(t) =12u!(t)T[k] u!(t)u1 = u1(t), u2 = u2(t),=12EAl(u12- 2u1u2 + u22)=12 Ll0EA e -1lu1(t) +1lu2(t)f2dxV(t) =12 Ll0EA e0 u(x, t)0xf2dx[m] =rAl6c2 11 2du:#(t) = eu#1(t)u#2(t)fT(t) =12u:#(t)T[m] u:#(t)ru#1 =du1(t)dt, u#2 =du2(t)dt,RaoCh12ff.qxd 10.06.08 13:49 Page 860
  6. 6. 12.3 MASS MATRIX, STIFFNESS MATRIX, AND FORCE VECTOR 861The force vectorcan be derived from the virtual work expression. If the bar is subjected to the distributedforce f(x, t), the virtual work can be expressed as(12.17)By expressing Eq. (12.17) in matrix form as(12.18)the equivalent joint forces can be identified as(12.19)Consider a uniform torsion element with the x axis taken along the centroidal axis, asshown in Fig. 12.3. Let denote the polar moment of inertia about the centroidal axis andIp12.3.2Torsion Elementf1(t) =Ll0f(x, t) a1 -xlb dxf2(t) =Ll0f(x, t) axlb dxtdW(t) = du!(t)Tf:(t) K f1(t) du1(t) + f2(t) du2(t)+ aLl0f(x, t) axlb dxb du2(t)= aLl0f(x, t) a1 -xlb dxb du1(t)=Ll0f(x, t) e a1 -xlb du1(t) + axlb du2(t)f dxdW(t) =Ll0f(x, t) du(x, t) dxdWf:= ef1(t)f2(t)ff1(t)Joint 1 Joint 2f(x, t) f2(t)x1(t) (x, t)␪ 2(t)␪␪␳xl, Ip, G,AFIGURE 12.3 Uniform torsion element.RaoCh12ff.qxd 10.06.08 13:49 Page 861
  7. 7. 862 CHAPTER 12 FINITE ELEMENT METHODGJ represent the torsional stiffness ( for a circular cross section). When the tor-sional displacement (rotation) within the element is assumed to be linear in x as(12.20)and the joint rotations and are treated as unknowns, Eq. (12.20) can be expressed,by proceeding as in the case of a bar element, as(12.21)where and are the same as in Eq. (12.10). The kinetic energy, the strainenergy, and the virtual work for pure torsion are given by(12.22)(12.23)(12.24)where is the mass density and f(x, t) is the distributed torque per unit length. Using theprocedures employed in Section 12.3.1, we can derive the element mass and stiffnessmatrices and the force vector:(12.25)(12.26)(12.27)We now consider a beam element according to the Euler-Bernoulli theory.1Figure 12.4shows a uniform beam element subjected to the transverse force distribution f(x, t). In thiscase, the joints undergo both translational and rotational displacements, so the unknownjoint displacements are labeled as and Thus there will be linearjoint forces and corresponding to the linear joint displacements andand rotational joint forces (bending moments) and corresponding to the rotationalf4(t)f2(t)w3(t)w1(t)f3(t)f1(t)w4(t).w1(t), w2(t), w3(t),12.3.3Beam Elementf:= ef1(t)f2(t)f = dLl0f(x, t) a1 -xlb dxLl0f(x, t) axlb dxt[k] =GJlc1 -1-1 1d[m] =rIpl6c2 11 2drdW(t) =Ll0f(x, t) du (x, t) dxV(t) =12 Ll0GJ e0u(x, t)0xf2dxT(t) =12 Ll0rIp e0u(x, t)0tf2dxN2(x)N1(x)u(x, t) = N1(x)u1(t) + N2(x)u2(t)u2(t)u1(t)u(x, t) = a(t) + b(t)xJ = Ip1The beam element, according to the Timoshenko theory, was considered in Refs. [12.4–12.7].RaoCh12ff.qxd 10.06.08 13:49 Page 862
  8. 8. 12.3 MASS MATRIX, STIFFNESS MATRIX, AND FORCE VECTOR 863Joint 1 Joint 2␳, A, I, Elxxf1(t) f3(t)f2(t) w1(t)w2(t)f4(t)w3(t)w4(t)w(x, t)f(x, t)FIGURE 12.4 Uniform beam element.joint displacements and respectively. The transverse displacement within theelement is assumed to be a cubic equation in x (as in the case of static deflection of abeam):(12.28)The unknown joint displacements must satisfy the conditions(12.29)Equations (12.28) and (12.29) yield(12.30)By substituting Eqs. (12.30) into Eq. (12.28), we can express w(x, t) as(12.31)This equation can be rewritten as(12.32)w(x, t) = a4i=1Ni(x)wi(t)+ a3x2l2- 2x3l3b w3(t) + a -x2l2+x3l3b lw4(t)w(x, t) = a1 - 3x2l2+ 2x3l3b w1(t) + axl- 2x2l2+x3l3b lw2(t)d(t) =1l3[2w1(t) + w2(t)l - 2w3(t) + w4(t)l]c(t) =1l2[-3w1(t) - 2w2(t)l + 3w3(t) - w4(t)l]b(t) = w2(t)a(t) = w1(t)w(0, t) = w1(t),0w0x(0, t) = w2(t)w(l, t) = w3(t),0w0x(l, t) = w4(t)tw(x, t) = a(t) + b(t)x + c(t)x2+ d(t)x3w4(t),w2(t)RaoCh12ff.qxd 10.06.08 13:49 Page 863
  9. 9. 864 CHAPTER 12 FINITE ELEMENT METHODwhere are the shape functions given by(12.33)(12.34)(12.35)(12.36)The kinetic energy, bending strain energy, and virtual work of the element can beexpressed as(12.37)(12.38)(12.39)where is the density of the beam, E is Young’s modulus, I is the moment of inertia of thecross section, A is the area of cross section, andBy substituting Eq. (12.31) into Eqs. (12.37) to (12.39) and carrying out the necessary inte-grations, we obtain(12.40)[m] =rAl420≥156 22l 54 -13l22l 4l213l -3l254 13l 156 -22l-13l -312-22l 4l2¥dw!(t) = μdw1(t)dw2(t)dw3(t)dw4(t)∂, f:(t) = μf1(t)f2(t)f3(t)f4(t)∂w!(t) = μw1(t)w2(t)w3(t)w4(t)∂, w:#(t) = μdw1/dtdw2/dtdw3/dtdw4/dt∂rdW(t) =Ll0f(x, t) dw(x, t) dx K dw!(t)Tf:(t)V(t) =12 Ll0EIe02w(x, t)0x2f2dx K12w!(t)T[k]w!(t)T(t) =12 Ll0rAe0w(x, t)0tf2dx K12w:#(t)T[m]w:#(t)N4(x) = - laxlb2+ laxlb3N3(x) = 3axlb2- 2axlb3N2(x) = x - 2laxlb2+ laxlb3N1(x) = 1 - 3axlb2+ 2axlb3Ni(x)RaoCh12ff.qxd 10.06.08 13:49 Page 864
  10. 10. 12.4 TRANSFORMATION OF ELEMENT MATRICES AND VECTORS 865(12.41)(12.42)12.4 Transformation of Element Matrices and VectorsAs stated earlier, the finite element method considers the given dynamical system as anassemblage of elements. The joint displacements of an individual element are selected ina convenient direction, depending on the nature of the element. For example, for the barelement shown in Fig. 12.2, the joint displacements and are chosen along theaxial direction of the element. However, other bar elements can have different orientationsin an assemblage, as shown in Fig. 12.5. Here x denotes the axial direction of an individ-ual element and is called a local coordinate axis. If we use and to denote the jointdisplacements of different bar elements, there will be one joint displacement at joint 1, threeat joint 2, two at joint 3, and 2 at joint 4. However, the displacements of joints can be spec-ified more conveniently using reference or global coordinate axes X and Y. Then the dis-placement components of joints parallel to the X and Y axes can be used as the jointdisplacements in the global coordinate system. These are shown asin Fig. 12.5. The joint displacements in the local and the global coordinate system for aUi(t), i = 1, 2, Á , 8u2(t)u1(t)u2(t)u1(t)fi(t) =Ll0f(x, t) Ni(x) dx, i = 1, 2, 3, 4[k] =EIl3≥12 6l -12 6l6l 4l2-6l 2l2-12 -6l 12 -6l6l 2l2-6l 4l2¥211 34243U4(t)U7(t)U8(t)YXU2(t)U1(t) U5(t)U6(t)u1(t)u2(t)u1(t) u2(t) u1(t)u2(t)Loadu2(t)U3(t)u1(t)xxxxFIGURE 12.5 A dynamical system (truss) idealizedas an assemblage of four bar elements.RaoCh12ff.qxd 10.06.08 13:49 Page 865
  11. 11. 866 CHAPTER 12 FINITE ELEMENT METHODtypical bar element e are shown in Fig. 12.6. The two sets of joint displacements are relatedas follows:(12.43)These can be rewritten as(12.44)where is the coordinate transformation matrix given by(12.45)and and are the vectors of joint displacements in the local and the global coordi-nate system, respectively, and are given byIt is useful to express the mass matrix, stiffness matrix, and joint force vector of an ele-ment in terms of the global coordinate system while finding the dynamical response of theu!(t) = eu1(t)u2(t)f, U!(t) = μU2i-1(t)U2i(t)U2j-1(t)U2j(t)∂U!(t)u!(t)[l] = ccos u sin u 0 00 0 cos u sin ud[l]u!(t) = [l]U!(t)u2(t) = U2j-1(t) cos u + U2j(t) sin uu1(t) = U2i-1(t) cos u + U2i(t) sin uU2i(t)U2j(t)u1(t)u2(t)xU2iϪ1(t)U2jϪ1(t)YX␪ijex ϭ local coordinate axisX,Y ϭ global coordinate axesu1(t), u2(t) ϭ local joint displacementsU2iϪ1(t), ... , U2j(t) ϭ global joint displacementsFIGURE 12.6 Local and global joint displacementsof element e.RaoCh12ff.qxd 10.06.08 13:49 Page 866
  12. 12. 12.4 TRANSFORMATION OF ELEMENT MATRICES AND VECTORS 867complete system. Since the kinetic and strain energies of the element must be independentof the coordinate system, we have(12.46)(12.47)where and denote the element mass and stiffness matrices, respectively, in theglobal coordinate system and is the vector of joint velocities in the global coordinatesystem, related to as in Eq. (12.44):(12.48)By inserting Eqs. (12.44) and (12.48) into (12.46) and (12.47), we obtain(12.49)(12.50)Equations (12.49) and (12.50) yield(12.51)(12.52)Similarly, by equating the virtual work in the two coordinate systems,(12.53)we find the vector of element joint forces in the global coordinate system(12.54)Equations (12.51), (12.52), and (12.54) can be used to obtain the equations of motion of asingle finite element in the global coordinate system:(12.55)Although this equation is not of much use, since our interest lies in the equations of motionof an assemblage of elements, the matrices and and the vector are useful inderiving the equations of motion of the complete system, as indicated in the followingsection.f:[k][m][m] U:$(t) + [k]U!(t) = f:(t)f:(t) = [l]Tf:(t)f:(t):dW(t) = du!(t)Tf:(t) = dU!(t)Tf:(t)[k] = [l]T[k][l][m] = [l]T[m][l]V(t) =12U:(t)T[l]T[k][l]U:(t) K12U:(t)T[k]U:(t)T(t) =12U:#(t)T[l]T[m][l]U:#(t) K12U:#(t)T[m] U:#(t)u:#(t) = [l]U:#(t)u:#(t)U:#(t)[k][m]V(t) =12u!(t)T[k]u!(t) =12U!(t)T[k]U!(t)T(t) =12u:#(t)T[m] u:#(t) =12U:#(t)T[m]U:#(t)RaoCh12ff.qxd 10.06.08 13:49 Page 867
  13. 13. 868 CHAPTER 12 FINITE ELEMENT METHOD12.5 Equations of Motion of the Complete System of Finite ElementsSince the complete structure is considered an assemblage of several finite elements, weshall now extend the equations of motion obtained for single finite elements in the globalsystem to the complete structure. We shall denote the joint displacements of the completestructure in the global coordinate system as or, equivalently, as acolumn vector:For convenience, we shall denote the quantities pertaining to an element e in the assem-blage by the superscript e. Since the joint displacements of any element e can be identifiedin the vector of joint displacements of the complete structure, the vectors andare related(12.56)where is a rectangular matrix composed of zeros and ones. For example, for element1 in Fig. 12.5, Eq. (12.56) becomes(12.57)The kinetic energy of the complete structure can be obtained by adding the kinetic ener-gies of individual elements(12.58)where E denotes the number of finite elements in the assemblage. By differentiatingEq. (12.56), the relation between the velocity vectors can be derived:(12.59)Substitution of Eq. (12.59) into (12.58) leads to(12.60)T =12 aEe=1U!# T[A(e)]T[m(e)] [A(e)] U!#U:#(e)(t) = [A(e)]U!(t)#T = aEe=112U:#(e)T[m]U:#(e)U!(1)(t) K μU1(t)U2(t)U3(t)U4(t)∂ = ≥1 0 0 0 0 0 0 00 1 0 0 0 0 0 00 0 1 0 0 0 0 00 0 0 1 0 0 0 0¥ fU1(t)U2(t)...U8(t)v[A(e)]U!(e)(t) = [A(e)]U! (t)U! (t)U!(e)(t)U! (t) = fU1(t)U2(t)...UM(t)vU1(t), U2(t), Á , UM(t)RaoCh12ff.qxd 10.06.08 13:49 Page 868
  14. 14. 12.6 INCORPORATION OF BOUNDARY CONDITIONS 869The kinetic energy of the complete structure can also be expressed in terms of joint veloc-ities of the complete structure(12.61)where is called the mass matrix of the complete structure. A comparison of Eqs. (12.60)and (12.61) gives the relation2(12.62)Similarly, by considering strain energy, the stiffness matrix of the complete structure,can be expressed as(12.63)Finally the consideration of virtual work yields the vector of joint forces of the completestructure,(12.64)Once the mass and stiffness matrices and the force vector are known, Lagrange’s equationsof motion for the complete structure can be expressed as(12.65)Note that the joint force vector in Eq. (12.65) was generated by considering onlythe distributed loads acting on the various elements. If there is any concentrated load act-ing along the joint displacement it must be added to the ith component of12.6 Incorporation of Boundary ConditionsIn the preceding derivation, no joint was assumed to be fixed. Thus the complete structureis capable of undergoing rigid body motion under the joint forces. This means thatis a singular matrix (see Section 6.12). Usually the structure is supported such that the[K ]F! .Ui(t),F![M] U!$+ [K] U:= F!F! = aEe=1[A(e)]Tf:(e)F! :[K ] = aEe=1[A(e)]T[k(e)] [A(e)][K ],[M ] = aEe=1[A(e)]T[m(e)][A(e)][M ]T =12U!# (e)T[M]U!#U!#2An alternative procedure can be used for the assembly of element matrices. In this procedure, each of the rowsand columns of the element (mass or stiffness) matrix is identified by the corresponding degree of freedom in theassembled structure. Then the various entries of the element matrix can be placed at their proper locations in theoverall (mass or stiffness) matrix of the assembled system. For example, the entry belonging to the ith row (iden-tified by the degree of freedom p) and the jth column (identified by the degree of freedom q) of the element matrixis to be placed in the pth row and qth column of the overall matrix. This procedure is illustrated in Example 12.3.RaoCh12ff.qxd 10.06.08 13:49 Page 869
  15. 15. 870 CHAPTER 12 FINITE ELEMENT METHODdisplacements are zero at a number of joints, to avoid rigid body motion of the structure.A simple method of incorporating the zero displacement conditions is to eliminate the cor-responding rows and columns from the matrices and and the vector The finalequations of motion of the restrained structure can be expressed as(12.66)where N denotes the number of free joint displacements of the structure.Note the following points concerning finite element analysis:1. The approach used in the above presentation is called the displacement method offinite element analysis because it is the displacements of elements that are directlyapproximated. Other methods, such as the force method, the mixed method, andhybrid methods, are also available [12.8, 12.9].2. The stiffness matrix, mass matrix, and force vector for other finite elements, includ-ing two-dimensional and three-dimensional elements, can be derived in a similarmanner, provided the shape functions are known [12.1, 12.2].3. In the Rayleigh-Ritz method discussed in Section 8.8, the displacement of the con-tinuous system is approximated by a sum of assumed functions, where each functiondenotes a deflection shape of the entire structure. In the finite element method, anapproximation using shape functions (similar to the assumed functions) is also usedfor a finite element instead of the entire structure. Thus the finite element procedurecan also be considered a Rayleigh-Ritz method.4. Error analysis of the finite element method can also be conducted [12.10].Analysis of a BarConsider a uniform bar, of length 0.5 m, area of cross section Young’s modulus 200 GPa,and density which is fixed at the left end, as shown in Fig. 12.7.a. Find the stress induced in the bar under an axial static load of 1000 N applied at joint 2 alongb. Find the natural frequency of vibration of the bar.Use a one-element idealization.u2.7850 kg/m3,5 * 10-4m2,EXAMPLE 12.1[M]N*NU:$N*1+ [K]N*NU:N*1= F:N*1F! .[K ][M ]21 u1 u20.5 mFIGURE 12.7 Uniform bar with two degrees of freedom.RaoCh12ff.qxd 10.06.08 13:49 Page 870
  16. 16. 12.6 INCORPORATION OF BOUNDARY CONDITIONS 871Solutiona. Using the stiffness matrix of a bar element, Eq. (12.16), the equilibrium equations can be writ-ten as(E.1)With Eq. (E.1) becomes(E.2)where is the displacement and is the unknown reaction at joint 1. To incorporate theboundary condition we delete the first scalar equation (first row) and substitutein the resulting Eq. (E.2). This gives(E.3)The stress - strain relation gives(E.4)where denotes the change in length of the element and indicates the strain.Equation (E.4) yields(E.5)b. Using the stiffness and mass matrices of the bar element, Eqs. (12.16) and (12.13), the eigen-value problem can be expressed as(E.6)where is the natural frequency and and are the amplitudes of vibration of the bar atjoints 1 and 2, respectively. To incorporate the boundary condition we delete the firstrow and first column in each of the matrices and vectors and write the resulting equation asor(E.7)■v =B3Erl2=B3(2 * 1011)7850 (0.5)2= 17,485.2076 rad/sAElU2 = v2r Al6(2)U2U1 = 0,U2U1vAElc1 -1-1 1d eU1U2f = v2r Al6c2 11 2d eU1U2fs = 2 * 1011a500 * 10-8- 00.5b = 2 * 106Pa¢ll¢l = u2 - u1s = E e = E¢ll= E au2 - u1lb(e)(s)2 * 108u2 = 1000 or u2 = 500 * 10-8mu1 = 0u1 = 0,f1u12 * 108c1 -1-1 1d eu1u2f = ef11000fA = 5 * 10-4, E = 2 * 1011, l = 0.5, f2 = 1000,AElc1 -1-1 1d eu1u2f = ef1f2fRaoCh12ff.qxd 10.06.08 13:49 Page 871
  17. 17. 872 CHAPTER 12 FINITE ELEMENT METHODNatural Frequencies of a Simply Supported BeamFind the natural frequencies of the simply supported beam shown in Fig. 12.8(a) using one finiteelement.Solution: Since the beam is idealized using only one element, the element joint displacements arethe same in both local and global systems, as indicated in Fig. 12.8(b). The stiffness and massmatrices of the beam are given by(E.1)(E.2)and the vector of joint displacements by(E.3)The boundary conditions corresponding to the simply supported ends ( and ) can beincorporated3by deleting the rows and columns corresponding to and in Eqs. (E.1) and (E.2).W3W1W3 = 0W1 = 0W! = μW1W2W3W4∂ K μw1(1)w2(1)w3(1)w4(1)∂[M ] = [M(1)] =rAl420≥156 22l 54 -13l22l 4l213l -3l254 13l 156 -22l-13l -3l2-22l 4l2¥[K ] = [K(1)] =EIl3≥12 6l -12 6l6l 4l2-6l 2l2-12 -6l 12 -6l6l 2l2-6l 4l2¥EXAMPLE 12.2l(1)ϭ lw(x, t)w4(1) ϭ W4w2(1) ϭ W2w3(1) ϭ W3w1(1) ϭ W1l(b)(a)1 221 xFIGURE 12.8 Simply supported beam.3The bending moment cannot be set equal to zero at the simply supported ends explicitly, since there is no degreeof freedom (joint displacement) involving the second derivative of the displacement w.RaoCh12ff.qxd 10.06.08 13:49 Page 872
  18. 18. 12.6 INCORPORATION OF BOUNDARY CONDITIONS 873This leads to the overall matrices(E.4)(E.5)and the eigenvalue problem can be written as(E.6)By multiplying throughout by Eq. (E.6) can be expressed as(E.7)where(E.8)By setting the determinant of the coefficient matrix in Eq. (E.7) equal to zero, we obtain the fre-quency equation(E.9)The roots of Eq. (E.9) give the natural frequencies of the beam as(E.10)(E.11)These results can be compared with the exact values (see Fig. 8.15):(E.12)■v1 = a97.41EIrAl4b1/2, v2 = a1558.56EIrAl4b1/2l2 = 3 or v2 = a2520EIrAl4b1/2l1 =17or v1 = a120EIrAl4b1/2`2 - 4l 1 + 3l1 + 3l 2 - 4l` = (2 - 4l)2- (1 + 3l)2= 0l =rAl4v2840EIc2 - 4l 1 + 3l1 + 3l 2 - 4ld eW2W4f = e00fl/(2EI),c2EIlc2 11 2d -rAl3v2420c4 -3-3 4d d eW2W4f = e00f[M] =rAl3420c4 -3-3 4d[K] =2EIlc2 11 2dRaoCh12ff.qxd 10.06.08 13:49 Page 873
  19. 19. 874 CHAPTER 12 FINITE ELEMENT METHODStresses in a Two-Bar TrussFind the stresses developed in the two members of the truss shown in Fig. 12.9(a), under a verticalload of 200 lb at joint 3. The areas of cross section are for member 1 and for member 2,and the Young’s modulus is 30 * 106psi.2 in.21 in.2EXAMPLE 12.310 in.10 in.(a)(b)5 in.200 lbX311232Element 2Element 1YU2U1U4U3U6U5XXxx1␪F 3 ϭ Ϫ200 lb␥2␪FIGURE 12.9 Two bar truss.RaoCh12ff.qxd 10.06.08 13:49 Page 874
  20. 20. 12.6 INCORPORATION OF BOUNDARY CONDITIONS 875SolutionApproach: Derive the static equilibrium equations and solve them to find the joint displacements.Use the elasticity relations to find the element stresses. Each member is to be treated as a bar ele-ment. From Fig. 12.9(a), the coordinates of the joints can be found asThe modeling of the truss as an assemblage of two bar elements and the displacement degreesof freedom of the joints are shown in Fig. 12.9(b). The lengths of the elements can be computed fromthe coordinates of the ends (joints) as(E.1)The element stiffness matrices in the local coordinate system can be obtained as(E.2)The angle between the local x-coordinate and the global X-coordinate is given by(E.3)(E.4)cos u2 =X3 - X2l(2)=10 - 011.1803= 0.8944sin u2 =Y3 - Y2l(2)=5 - 011.1803= 0.4472t for element 2cos u1 =X3 - X1l(1)=10 - 011.1803= 0.8944sin u1 =Y3 - Y1l(1)=5 - 1011.1803= - 0.4472t for element 1= 5.3666 * 106c1 -1-1 1d[k(2)] =A(2)E(2)l(2)c1 -1-1 1d =(2)(30 * 106)11.1803c1 -1-1 1d= 2.6833 * 106c1 -1-1 1d[k(1)] =A(1)E(1)l(1)c1 -1-1 1d =(1)(30 * 106)11.1803c1 -1-1 1d= 11.1803 in.l(2)= 5(X3 - X2)2+ (Y3 - Y2)261/2= 5(10 - 0)2+ (5 - 0)261/2= 11.1803 in.l(1)= 5(X3 - X1)2+ (Y3 - Y1)261/2= 5(10 - 0)2+ (5 - 10)261/2(X1, Y1) = (0, 10) in.; (X2, Y2) = (0, 0) in.; (X3, Y3) = (10, 5) in.RaoCh12ff.qxd 10.06.08 13:49 Page 875
  21. 21. 876 CHAPTER 12 FINITE ELEMENT METHODThe stiffness matrices of the elements in the global (X, Y) coordinate system can be derived as(E.5)(E.6)where(E.7)(E.8)Note that the top and right-hand sides of Eqs. (E.5) and (E.6) denote the global degrees of freedomcorresponding to the rows and columns of the respective stiffness matrices. The assembled stiffnessmatrix of the system, can be obtained, by placing the elements of and at their properplaces in as(E.9)1 2 3 4 5 6[K] = 2.6833 * 106H0.8 - 0.4 - 0.8 0.4- 0.4 0.2 0.4 - 0.21.6 0.8 - 1.6 - 0.80.8 0.4 - 0.8 - 0.4- 0.8 0.4 - 1.6 - 0.8 (0.8 (- 0.4+1.6) +0.8)0.4 - 0.2 - 0.8 - 0.4 (- 0.4 (0.2+0.8) +0.4)X123456[K ],[k(2)][k(1)][K ]= c0.8944 0.4472 0 00 0 0.8944 0.4472d[l(2)] = ccos u2 sin u2 0 00 0 cos u2 sin u2d= c0.8944 - 0.4472 0 00 0 0.8944 - 0.4472d[l(1)] = ccos u1 sin u1 0 00 0 cos u1 sin u1d= 5.3666 * 106E3 4 5 60.8 0.4 - 0.8 - 0.40.4 0.2 - 0.4 - 0.2- 0.8 - 0.4 0.8 0.4- 0.4 - 0.2 0.4 0.2U3456[k(2)] = [l(2)]T[k(2)][l(2)]= 2.6833 * 106E1 2 5 60.8 - 0.4 - 0.8 0.4- 0.4 0.2 0.4 - 0.2- 0.8 0.4 0.8 - 0.40.4 - 0.2 - 0.4 0.2U1256[k(1)] = [l(1)]T[k(1)][l(1)]RaoCh12ff.qxd 10.06.08 13:49 Page 876
  22. 22. 12.6 INCORPORATION OF BOUNDARY CONDITIONS 877The assembled force vector can be written as(E.10)where, in general, denote the forces applied at joint i along (X, Y) directions. Specifically,and represent the reactions at joints 1 and 2, while lbshows the external forces applied at joint 3. By applying the boundary conditions(i.e., by deleting the rows and columns 1, 2, 3, and 4 in Eqs. E.9 andE.10), we get the final assembled stiffness matrix and the force vector as(E.11)(E.12)The equilibrium equations of the system can be written as(E.13)where The solution of Eq. (E.13) can be found as(E.14)The axial displacements of elements 1 and 2 can be found as(E.15)= e083.3301 * 10-6 f in.= c0.8944 - 0.4472 0 00 0 0.8944 - 0.4472d μ0023.2922 * 10-6-139.7532 * 10-6∂eu1u2f(1)= [l(1)]dU1U2U5U6tU5 = 23.2922 * 10-6in., U6 = -139.7532 * 10-6in.U!= eU5U6f.[K]U!= F!F!= e0-200f565 6[K] = 2.6833 * 106c2.4 0.40.4 0.6d56U1 = U2 = U3 = U4 = 0(FX3, FY3) = (0, -200)(FX2, FY2)(FX1, FY1)(FXi, FYi)F! = fFX1FY1FX2FY2FX3FY3vRaoCh12ff.qxd 10.06.08 13:49 Page 877
  23. 23. 878 CHAPTER 12 FINITE ELEMENT METHOD(E.16)The stresses in elements 1 and 2 can be determined as(E.17)(E.18)where denotes the stress, represents the strain, and indicates the change in length of ele-ment■12.7 Consistent and Lumped Mass MatricesThe mass matrices derived in Section 12.3 are called consistent mass matrices. They areconsistent because the same displacement model that is used for deriving the element stiff-ness matrix is used for the derivation of mass matrix. It is of interest to note that severaldynamic problems have been solved with simpler forms of mass matrices. The simplestform of the mass matrix, known as the lumped mass matrix, can be obtained by placingpoint (concentrated) masses at node points i in the directions of the assumed displace-ment degrees of freedom. The concentrated masses refer to translational and rotationalinertia of the element and are calculated by assuming that the material within the meanlocations on either side of the particular displacement behaves like a rigid body while theremainder of the element does not participate in the motion. Thus this assumption excludesthe dynamic coupling that exists between the element displacements and hence the result-ing element mass matrix is purely diagonal [12.11].mii (i = 1, 2).¢l(i)e(i)s(i)=(30 * 106)(- 41.6651 * 10-6)11.1803= -111.7996 psis(2)= E(2)P(2)=E(2)¢l(2)l(2)=E(2)(u2 - u1)(2)l(2)=(30 * 106)(83.3301 * 10-6)11.1803= 223.5989 psis(1)= E(1)P(1)= E(1)¢l(1)l(1)=E(1)(u2 - u1)(1)l(1)= e0- 41.6651 * 10-6 f in.= c0.8944 0.4472 0 00 0 0.8944 0.4472d μ0023.2922 * 10-6-139.7532 * 10-6∂eu1u2f(2)= [l(2)]μU3U4U5U6∂RaoCh12ff.qxd 10.06.08 13:49 Page 878
  24. 24. 12.7 CONSISTENT AND LUMPED MASS MATRICES 879By dividing the total mass of the element equally between the two nodes, the lumped massmatrix of a uniform bar element can be obtained as(12.67)In Fig. 12.4, by lumping one half of the total beam mass at each of the two nodes, alongthe translational degrees of freedom, we obtain the lumped mass matrix of the beam ele-ment as(12.68)Note that the inertia effect associated with the rotational degrees of freedom has beenassumed to be zero in Eq. (12.68). If the inertia effect is to be included, we compute themass moment of inertia of half of the beam segment about each end and include it at thediagonal locations corresponding to the rotational degrees of freedom. Thus, for a uniformbeam, we have(12.69)and hence the lumped mass matrix of the beam element becomes(12.70)It is not obvious whether the lumped mass matrices or consistent mass matrices yield moreaccurate results for a general dynamic response problem. The lumped mass matrices areapproximate in the sense that they do not consider the dynamic coupling present betweenthe various displacement degrees of freedom of the element. However, since the lumpedmass matrices are diagonal, they require less storage space during computation. On theother hand, the consistent mass matrices are not diagonal and hence require more storagespace. They too are approximate in the sense that the shape functions, which are derivedusing static displacement patterns, are used even for the solution of dynamics problems.12.7.3Lumped MassVersusConsistent MassMatrices[m] =rAl2F1 0 0 00 al212b 0 00 0 1 00 0 0 al212bVI =13arAl2b al2b2=rAl324[m] =rAl2≥1 0 0 00 0 0 00 0 1 00 0 0 0¥12.7.2Lumped MassMatrix for aBeam Element[m] =rAl2c1 00 1d12.7.1Lumped MassMatrix for a BarElementRaoCh12ff.qxd 10.06.08 13:49 Page 879
  25. 25. 880 CHAPTER 12 FINITE ELEMENT METHODThe following example illustrates the application of lumped and consistent mass matricesin a simple vibration problem.Consistent and Lumped Mass Matrices of a BarFind the natural frequencies of the fixed-fixed uniform bar shown in Fig. 12.10 using consistent andlumped mass matrices. Use two bar elements for modeling.Solution: The stiffness and mass matrices of a bar element are(E.1)(E.2)(E.3)where the subscripts c and l to the mass matrices denote the consistent and lumped matrices, respec-tively. Since the bar is modeled by two elements, the assembled stiffness and mass matrices aregiven by(E.4)(E.5)(E.6)1 2 3[M]l =rAl2C1 0 00 1 +1 00 0 1S123=rAl2C1 0 00 2 00 0 1S1 2 3[M]c =rAl6C2 1 01 2 +2 10 1 2S123=rAl6C2 1 01 4 10 1 2S1 2 3[K] =AElC1 -1 0-1 1 +1 -10 -1 1S123=AElC1 -1 0-1 2 -10 -1 1S[m]l =rAl2c1 00 1d[m]c =rAl6c2 11 2d[k] =AElc1 -1-1 1dEXAMPLE 12.4U1 U2 U3Element 1 Element 2l lLFIGURE 12.10 Fixed-fixed uniform bar.RaoCh12ff.qxd 10.06.08 13:49 Page 880
  26. 26. 12.8 EXAMPLES USING MATLAB 881The dashed boxes in Eqs. (E.4) through (E.6) enclose the contributions of elements 1 and 2. Thedegrees of freedom corresponding to the columns and rows of the matrices are indicated at the topand the right-hand side of the matrices. The eigenvalue problem, after applying the boundary condi-tions becomes(E.7)The eigenvalue can be determined by solving the equation(E.8)which, for the present case, becomes(E.9)and(E.10)Equations (E.9) and (E.10) can be solved to obtain(E.11)(E.12)These values can be compared with the exact value (see Fig. 8.7)(E.13)■12.8 Examples Using MATLABFinite Element Analysis of a Stepped BarConsider the stepped bar shown in Fig. 12.11 with the following data:Write a MATLAB program to determine the following:a. Displacements and under loadb. Natural frequencies and mode shapes of barp3 = 1000 Nu3u1, u2,l3 = 0.25 m.l2 = 0.5 m,i = 1, 2, 3, l1 = 1 m,ri = 7.8 * 103kg/m3,i = 1, 2, 3,Ei = 20 * 1010Pa,A2 = 9 * 10-4m2, A3 = 4 * 10-4m2,A1 = 16 * 10-4m2,EXAMPLE 12.5v1 = pAErL2vl =A2Erl2= 2.8284AErL2vc =A3Erl2= 3.4641AErL2`AEl[2] - v2rAl2[2] ` = 0 with lumped mass matrices`AEl[2] - v2rAl6[4] ` = 0 with consistent mass matricesƒ[K] - v2[M]ƒ = 0v2[[K] - v2[M]] 5U26 = 506U1 = U3 = 0,RaoCh12ff.qxd 10.06.08 13:49 Page 881
  27. 27. 882 CHAPTER 12 FINITE ELEMENT METHODSolution: The assembled stiffness and mass matrices of the stepped bar are given by(E.1)(E.2)The system matrices [K] and [M] can be obtained by incorporating the boundary condition —that is, by deleting the first row and first column in Eqs. (E.1) and (E.2).a. The equilibrium equations under the load are given by(E.3)where(E.4)[K] = GA1E1l1+A2E2l2-A2E2l20-A2E2l2A2E2l2+A3E3l3-A3E3l30-A3E3l3A3E3l3W[K]U!= P!p3 = 1000 Nu0 = 0[M ] =16D2r1A1l1 r1A1l1 0 0r1A1l1 2r1A1l1 + 2r2A2l2 r2A2l2 00 r2A2l2 2r2A2l2 + 2r3A3l3 r3A3l30 0 r3A3l3 2r3A3l3T[K] = HA1E1l1-A1E1l10 0-A1E1l1A1E1l1+A2E2l2-A2E2l200-A2E2l2A2E2l2+A3E3l3-A3E3l30 0-A3E3l3A3E3l3Xu0 u1 u2u3 p3l3l2l1A1, E1, 1␳A2, E2, 2␳A3, E3, 3␳FIGURE 12.11 Stepped bar.RaoCh12ff.qxd 10.06.08 13:49 Page 882
  28. 28. 12.8 EXAMPLES USING MATLAB 883b. The eigenvalue problem can be expressed as(E.5)where [K] is given by Eq. (E.4) and [M] by(E.6)The MATLAB solution of Eqs. (E.3) and (E.5) is given below.%------ Program Ex12_5.m%------Initialization of values-------------------------A1 = 16eϪ4 ;A2 = 9eϪ4 ;A3 = 4eϪ4 ;E1 = 20e10 ;E2 = E1 ;E3 = E1 ;R1 = 7.8e3 ;R2 = R1 ;R3 = R1 ;L1 = 1 ;L2 = 0.5 ;L3 = 0.25 ;%------Definition of [K]---------------------------------K11 = A1*E1/L1+A2*E2/L2 ;K12 = ϪA2*E2/L2 ;K13 = 0 ;K21 = K12 ;K22 = A2*E2/L2+A3*E3/L3 ;K23 = ϪA3*E3/L3 ;K31 = K13 ;K32 = K23 ;K33 = A3*E3/L3 ;K = [ K11 K12 K13; K21 K22 K23; K31 K32 K33 ]%-------- Calculation of matrixP = [ 0 0 1000]’U = inv(K)*P[M] =16C2r1A1l1 + 2r2A2l2 r2A2l2 0r2A2l2 2r2A2l2 + 2r3A3l3 r3A3l30 r3A3l3 2r3A3l3Sc[K] - v2[M]dU!= 0!U!= cu1u2u3s, P!= c001000sRaoCh12ff.qxd 10.06.08 13:49 Page 883
  29. 29. 884 CHAPTER 12 FINITE ELEMENT METHOD%------- Definition of [M] -------------------------M11 = (2*R1*A1*L1+2*R2*A2*L2) / 6;M12 = (R2*A2*L2) / 6;M13 = 0;M21 = M12;M22 = (2*R2*A2*L2+2*R3*A3*L3) / 6;M23 = R3*A3*L3;M31 = M13;M32 = M23;M33 = 2*M23;M= [M11 M12 M13; M21 M22 M23; M31 M32 M33 ]MI = inv (M)KM = MI*K%-------------Calculation of eigenvector and eigenvalue--------------[L, V] = eig (KM)>> Ex12_5K =680000000 Ϫ360000000 0Ϫ360000000 680000000 Ϫ3200000000 Ϫ320000000 320000000P =001000U =1.0eϪ005 *0.31250.59030.9028M =5.3300 0.5850 00.5850 1.4300 0.78000 0.7800 1.5600MI =0.2000 Ϫ0.1125 0.0562Ϫ0.1125 1.0248 Ϫ0.51240.0562 Ϫ0.5124 0.8972KM =1.0e+008 *1.7647 Ϫ1.6647 0.5399Ϫ4.4542 9.0133 Ϫ4.91912.2271 Ϫ6.5579 4.5108L =Ϫ0.1384 0.6016 0.39460.7858 Ϫ0.1561 0.5929Ϫ0.6028 Ϫ0.7834 0.7020RaoCh12ff.qxd 10.06.08 13:49 Page 884
  30. 30. 12.8 EXAMPLES USING MATLAB 885V =1.0e+009 *1.3571 0 00 0.1494 00 0 0.0224>>■Program for Eigenvalue Analysis of a Stepped BeamDevelop a MATLAB program called Program17.m for the eigenvalue analysis of a fixed-fixedstepped beam of the type shown in Fig. 12.12.Solution: Program17.m is developed to accept the following input data:global degree of freedom number corresponding to the local jth degree of freedom ofelement iThe program gives the natural frequencies and mode shapes of the beam as output.Natural frequencies of the stepped beams1.6008e+002 6.1746e+002 2.2520e+003 7.1266e+003Mode shapes1 1.0333eϪ002 1.8915eϪ004 1.4163eϪ002 4.4518eϪ0052 Ϫ3.7660eϪ003 2.0297eϪ004 4.7109eϪ003 2.5950eϪ0043 1.6816eϪ004 Ϫ1.8168eϪ004 1.3570eϪ003 2.0758eϪ0044 1.8324eϪ004 6.0740eϪ005 3.7453eϪ004 1.6386eϪ004rho = mass densitye = Young’s modulusbj(i, j) =a(i) = area of cross section of element ixi(i) = moment of inertia of element ixl(i) = length of element (step) iEXAMPLE 12.6x, XW1W2W3W4W5W6W7W82Љ ϫ 2Љl1 ϭ 40Љ l2 ϭ 32Љ l3 ϭ 24Љ1Љ ϫ 1Љ3Љ ϫ 3ЉE ϭ 30 ϫ 106psi, ϭ 0.283 lb/in3␳FIGURE 12.12 Stepped beam.■RaoCh12ff.qxd 10.06.08 13:49 Page 885
  31. 31. 886 CHAPTER 12 FINITE ELEMENT METHOD12.9 C++ ProgramAn interactive C++ program, called Program17.cpp, is given for the eigenvalue solu-tion of a stepped beam. The input and output of the program are similar to those ofProgram17.m.Eigenvalue Solution of a Stepped BeamFind the natural frequencies and mode shapes of the stepped beam shown in Fig. 12.12 usingProgram17.cpp.Solution: The input data are to be entered interactively. The output of the program is shown below.NATURAL FREQUENCIES OF THE STEPPED BEAM160.083080 617.459700 2251.975785 7126.595358MODE SHAPES1 0.0103332469 0.0001891485 0.0141625494 0.00004451372 Ϫ0.0037659939 0.0002029733 0.0047108916 0.00025949713 0.0001681571 Ϫ0.0001816828 0.0013569926 0.00020758374 0.0001832390 0.0000607403 0.0003745272 0.0001638648■12.10 Fortran ProgramA Fortran program called PROGRAM17.F is given for the eigenvalue analysis of a steppedbeam. The input and output of the program are similar to those of Program17.m.Eigenvalue Analysis of a Stepped BeamFind the natural frequencies and mode shapes of the stepped beam shown in Fig. 12.12 usingPROGRAM17.F.Solution: The output of the program is given below.NATURAL FREQUENCIES OF THE STEPPED BEAM0.160083E+03 0.617460E+03 0.225198E+04 0.712653E+04MODE SHAPES1 0.103333EϪ01 0.189147EϪ03 0.141626EϪ01 0.445258EϪ042 Ϫ0.376593EϪ02 0.202976EϪ03 0.471097EϪ02 0.259495EϪ033 0.167902EϪ03 Ϫ0.181687EϪ03 0.135672EϪ02 0.207580EϪ034 0.182648EϪ03 0.607247EϪ04 0.373775EϪ03 0.163869EϪ03■EXAMPLE 12.8EXAMPLE 12.7RaoCh12ff.qxd 10.06.08 13:49 Page 886
  32. 32. REVIEW QUESTIONS 887REFERENCES12.1 O. C. Zienkiewicz, The Finite Element Method (4th ed.), McGraw-Hill, London, 1987.12.2 S. S. Rao, The Finite Element Method in Engineering (3rd ed.), Butterworth-Heinemann,Boston, 1999.12.3 G. V. Ramana and S. S. Rao, “Optimum design of plano-milling machine structure using finiteelement analysis,” Computers and Structures, Vol. 18, 1984, pp. 247–253.12.4 R. Davis, R. D. Henshell, and G. B. Warburton, “A Timoshenko beam element,” Journal ofSound and Vibration, Vol. 22, 1972, pp. 475–487.12.5 D. L. Thomas, J. M. Wilson, and R. R. Wilson, “Timoshenko beam finite elements,” Journalof Sound and Vibration, Vol. 31, 1973, pp. 315–330.12.6 J. Thomas and B. A. H. Abbas, “Finite element model for dynamic analysis of Timoshenkobeams,” Journal of Sound and Vibration, Vol. 41, 1975, pp. 291–299.12.7 R. S. Gupta and S. S. Rao, “Finite element eigenvalue analysis of tapered and twistedTimoshenko beams,” Journal of Sound and Vibration, Vol. 56, 1978, pp. 187–200.12.8 T. H. H. Pian, “Derivation of element stiffness matrices by assumed stress distribution,” AIAAJournal, Vol. 2, 1964, pp. 1333–1336.12.9 H. Alaylioglu and R. Ali, “Analysis of an automotive structure using hybrid stress finite ele-ments,” Computers and Structures, Vol. 8, 1978, pp. 237–242.12.10 I. Fried, “Accuracy of finite element eigenproblems,” Journal of Sound and Vibration, Vol. 18,1971, pp. 289–295.12.11 P. Tong, T. H. H. Pian, and L. L. Bucciarelli, “Mode shapes and frequencies by the finite ele-ment method using consistent and lumped matrices,” Computers and Structures, Vol. 1, 1971,pp. 623–638.REVIEW QUESTIONS12.1 Give brief answers to the following:1. What is the basic idea behind the finite element method?2. What is a shape function?3. What is the role of transformation matrices in the finite element method?4. What is the basis for the derivation of transformation matrices?5. How are fixed boundary conditions incorporated in the finite element equations?6. How do you solve a finite element problem having symmetry in geometry and loadingby modeling only half of the problem?7. Why is the finite element approach presented in this chapter called the displacementmethod?8. What is a consistent mass matrix?9. What is a lumped mass matrix?10. What is the difference between the finite element method and the Rayleigh-Ritz method?11. How is the distributed load converted into equivalent joint force vector in the finite ele-ment method?12.2 Indicate whether each of the following statements is true or false:1. For a bar element of length l with two nodes, the shape function corresponding to node2 is given by x/l.RaoCh12ff.qxd 10.06.08 13:49 Page 887
  33. 33. 2. The element stiffness matrices are always singular.3. The element mass matrices are always singular.4. The system stiffness matrix is always singular unless the boundary conditions areincorporated.5. The system mass matrix is always singular unless the boundary conditions are incorporated.6. The lumped mass matrices are always diagonal.7. The coordinate transformation of element matrices is required for all systems.8. The element stiffness matrix in the global coordinate system, can be expressed interms of the local matrix [k] and the coordinate transformation matrix as9. The derivation of system matrices involves the assembly of element matrices.10. Boundary conditions are to be imposed to avoid rigid body motion of the system.12.3 Fill in each of the following blanks with appropriate word:1. In the finite element method, the solution domain is replaced by several _____.2. In the finite element method, the elements are assumed to be interconnected at certainpoints known as _____.3. In the finite element method, an _____ solution is assumed within each element.4. The displacement within a finite element is expressed in terms of _____ functions.5. For a thin beam element, _____ degrees of freedom are considered at each node.6. For a thin beam element, the shape functions are assumed to be polynomials of degree_____.7. In the displacement method, the _____ of elements is directly approximated.8. If the displacement model used in the derivation of the element stiffness matrices is alsoused to derive the element mass matrices, the resulting mass matrix is called _____ massmatrix.9. If the mass matrix is derived by assuming point masses at node points, the resulting massmatrix is called _____ mass.10. The lumped mass matrices do not consider the _____ coupling between the various dis-placement degrees of freedom of the element.11. Different orientations of finite elements require _____ of element matrices.12.4 Select the most appropriate answer out of the choices given:1. For a bar element of length l with two nodes, the shape function corresponding to node 1is given by(a) (b) (c)2. The simplest form of mass matrix is known as(a) lumped mass matrix(b) consistent mass matrix(c) global mass matrix3. The finite element method is(a) an approximate analytical method(b) a numerical method(c) an exact analytical methoda1 +xlbxla1 -xlb[l]T[k][l].[l][k],888 CHAPTER 12 FINITE ELEMENT METHODRaoCh12ff.qxd 10.06.08 13:49 Page 888
  34. 34. PROBLEMS 8894. The stiffness matrix of a bar element is given by(a) (b) (c)5. The consistent mass matrix of a bar element is given by(a) (b) (c)6. The finite element method is similar to(a) Rayleigh’s method(b) the Rayleigh-Ritz method(c) the Lagrange method7. The lumped mass matrix of a bar element is given by(a) (b) (c)8. The element mass matrix in the global coordinate system, can be expressed in termsof the element mass matrix in local coordinate system [m] and the coordinate transfor-mation matrix as(a) (b) (c)12.5 Match the items in the two columns below. Assume a fixed-fixed bar with one middle node:Element matrices:Steel bar:Aluminum bar:1. Natural frequency of steel bargiven by lumped mass matrices (a) 58,528.5606 rad/sec2. Natural frequency of aluminum bargiven by consistent mass matrices (b) 47,501.0898 rad/sec3. Natural frequency of steel bar givenby consistent mass matrices (c) 58,177.2469 rad/sec4. Natural frequency of aluminum bargiven by lumped mass matrices (d) 47,787.9336 rad/secPROBLEMSThe problem assignments are organized as follows:Problems Section Covered Topic Covered12.1, 12.2, 12.4 12.3 Derivation of element matrices andvectors12.5, 12.7 12.4 Transformation matrix12.6, 12.9 12.5 Assembly of matrices and vectors12.4, 12.8, 12.10–12.30 12.6 Application of boundary conditionsand solution of problemE = 10.3 * 106lb/in.2, r = 0.0002536 lb - sec2/in.4, L = 12 in.E = 30 * 106lb/in.2, r = 0.0007298 lb - sec2/in.4, L = 12 in.[k] =AElc1 -1-1 1d, [m]c =rAl6c2 11 2d, [m]l =rAl2c1 00 1d3m4 = [l]T[m][l]3m4 = [m][l]3m4 = [l]T[m][l],3m4,rAl2c1 00 1drAl6c2 11 2drAlc1 00 1drAl6c1 00 1drAl6c2 -1-1 2drAl6c2 11 2dEAlc1 00 1dEAlc1 -1-1 1dEAlc1 11 1dRaoCh12ff.qxd 10.06.08 13:49 Page 889
  35. 35. 890 CHAPTER 12 FINITE ELEMENT METHODProblems Section Covered Topic Covered12.31, 12.32 12.7 Consistent and lumped massmatrices12.33–12.35 12.8 MATLAB programs12.36 12.9 C++ program12.3, 12.37–12.40 12.10 Fortran programs12.41–12.42 — Design projects12.1 Derive the stiffness matrix and the consistent and lumped mass matrices of the tapered bar ele-ment shown in Fig. 12.13. The diameter of the bar decreases from D to d over its length.lx dD, E␳FIGURE 12.1312.2 Derive the stiffness matrix of the bar element in longitudinal vibration whose cross-sectionalarea varies as where is the area at the root (see Fig. 12.14).A0A(x) = A0e- (x/l),xxlOA0 eϪ(x/l)FIGURE 12.1412.3 Write a computer program for finding the stresses in a planar truss.12.4 The tapered cantilever beam shown in Fig. 12.15 is used as a spring to carry a load P. (a) Derivethe stiffness matrix of the beam element. (b) Use the result of (a) to find the stress induced inthe beam when andUse one beam element for idealization.12.5 Find the global stiffness matrix of each of the four bar elements of the truss shown in Fig. 12.5using the following data:Nodal coordinates:Cross-sectional areas:Young’s modulus of all members: 30 * 106lb/in.2.A1 = A2 = A3 = A4 = 2 in.2.(X4, Y4) = (200, 150) in.(X3, Y3) = (100, 0) in.,(X2, Y2) = (50, 100) in.,Y1) = (0, 0),(X1,P = 1000 N.E = 2.07 * 1011N/m2,l = 2 m,t = 2.5 cm,B = 25 cm, b = 10 cm,RaoCh12ff.qxd 10.06.08 13:49 Page 890
  36. 36. PROBLEMS 89112.6 Using the result of Problem 12.5, find the assembled stiffness matrix of the truss and formu-late the equilibrium equations if the vertical downward load applied at node 4 is 1000 lb.12.7 Derive the stiffness and mass matrices of the planar frame element (general beam element)shown in Fig. 12.16 in the global XY-coordinate system.12.8 A multiple-leaf spring used in automobiles is shown in Fig. 12.17. It consists of five leaves,each of thickness and width Find the deflection of the leaves undera load of Model only a half of the spring for the finite element analysis. TheYoung’s modulus of the material is 30 * 106psi.P = 2000 lb.w = 1.5 in.t = 0.25 in.YXOJoint 2U6U5U4Joint 1U3U2U1FIGURE 12.16 A frame element in global system.Pztxb/2b/2B/2 OB/2lFIGURE 12.15RaoCh12ff.qxd 10.06.08 13:49 Page 891
  37. 37. 892 CHAPTER 12 FINITE ELEMENT METHOD12.9 Derive the assembled stiffness and mass matrices of the multiple-leaf spring of Problem 12.8assuming a specific weight of for the material.12.10 Find the nodal displacements of the crane shown in Fig. 12.18 when a vertically downwardload of 1000 lb is applied at node 4. The Young’s modulus is and the cross-sectional area is for elements 1 and 2 and for elements 3 and 4.1 in.22 in.230 * 106psi0.283 lb/in.312.11 Find the tip deflection of the cantilever beam shown in Fig. 12.19 when a vertical load ofis applied at point Q using (a) a one element approximation and (b) a two-elementapproximation. Assume and12.12 Find the stresses in the stepped beam shown in Fig. 12.20 when a moment of 1000 N-m isapplied at node 2 using a two-element idealization. The beam has a square cross sectionbetween nodes 1 and 2 and between nodes 2 and 3. Assume theYoung’s modulus as 2.1 * 1011Pa.25 * 25 mm50 * 50 mmk = 105N/m.2.07 * 1011Pa,E =b = 50 mm,h = 25 mm,l = 0.25 m,P = 500 N12341 234111122221000 lb100Љ100Љ50Љ50Љ25ЉXYFIGURE 12.185Љ 5Љ 5Љ 5Љ 5ЉP PEyeCenter bolttwFIGURE 12.17 A multiple-leaf spring.RaoCh12ff.qxd 10.06.08 13:49 Page 892
  38. 38. PROBLEMS 89312.13 Find the transverse deflection and slope of node 2 of the beam shown in Fig. 12.21 usinga two element idealization. Compare the solution with that of simple beam theory.12.14 Find the natural frequencies of a cantilever beam of length l, cross-sectional area A, momentof inertia I, Young’s modulus E, and density using one finite element.12.15 Using one beam element, find the natural frequencies of the uniform pinned-free beam shownin Fig. 12.22.r,l, A, I, E␳FIGURE 12.22XXhQklb, A, I, E␳Section X–XFIGURE 12.191000 N-m0.25 m 0.40 m1 2 3FIGURE 12.20P1 2 3l43l4, A, I, E␳FIGURE 12.21RaoCh12ff.qxd 10.06.08 13:49 Page 893
  39. 39. 894 CHAPTER 12 FINITE ELEMENT METHOD12.16 Using one beam element and one spring element, find the natural frequencies of the uniform,spring-supported cantilever beam shown in Fig. 12.19.12.17 Using one beam element and one spring element, find the natural frequencies of the systemshown in Fig. 12.23.12.18 Using two beam elements, find the natural frequencies and mode shapes of the uniform fixed-fixed beam shown in Fig. 12.24.12.19* An electric motor, of mass and operating is fixed at the mid-dle of a clamped-clamped steel beam of rectangular cross section, as shown in Fig. 12.25.Design the beam such that the natural frequency of the system exceeds the operating speed ofthe motor.speed = 1800 rpm,m = 100 kg*An asterisk denotes a problem with no unique answer., A, I, E␳L2L2FIGURE 12.242 m␻FIGURE 12.25mk ϭEll3l, A, I, E␳m ϭ Al␳FIGURE 12.23RaoCh12ff.qxd 10.06.08 13:49 Page 894
  40. 40. PROBLEMS 89512.20 Find the natural frequencies of the beam shown in Fig. 12.26, using three finite elements oflength l each.12.21 Find the natural frequencies of the cantilever beam carrying an end mass M shown in Fig. 12.27,using a one beam element idealization.12.22 Find the natural frequencies of vibration of the beam shown in Fig. 12.28, using two beamelements. Also find the load vector if a uniformly distributed transverse load p is applied toelement 1.12.23 Find the natural frequencies of a beam of length l, which is pin connected at and fixedat using one beam element.x = l,x = 0l, A, I, E␳M ϭ 10 Al␳MFIGURE 12.27l1x, Xy, Yl2, E, A1, I1␳, E, A2, I2␳FIGURE 12.28l l l, A, I, E␳FIGURE 12.26RaoCh12ff.qxd 10.06.08 13:49 Page 895
  41. 41. 896 CHAPTER 12 FINITE ELEMENT METHOD12.24 Find the natural frequencies of torsional vibration of the stepped shaft shown in Fig. 12.29.Assume that andl2 = l.l1 =r1 = r2 = r, G1 = G2 = G, Ip1 = 2Ip2 = 2Ip, J1 = 2J2 = 2J,12.25 Find the dynamic response of the stepped bar shown in Fig. 12.30(a) when its free end is sub-jected to the load given in Fig. 12.30(b).12.26 Find the displacement of node 3 and the stresses in the two members of the truss shown inFig. 12.31. Assume that the Young’s modulus and the cross-sectional areas of the two mem-bers are the same with and A = 1 in.2.E = 30 * 106psitP(t)P(t)llArea ϭ 4A Area ϭ A(a) (b)P0t0OFIGURE 12.3011 lb3225 in.10 in.FIGURE 12.31l2l11, G1, Ip1, J1␳2, G2, Ip2, J2␳FIGURE 12.29RaoCh12ff.qxd 10.06.08 13:49 Page 896
  42. 42. PROBLEMS 89712.27 The simplified model of a radial drilling machine structure is shown in Fig. 12.32. Using twobeam elements for the column and one beam element for the arm, find the natural frequenciesand mode shapes of the machine. Assume the material of the structure as steel.12.28 If a vertical force of 5000 N along the z-direction and a bending moment of 500 N-m in thexz-plane are developed at point A during a metal cutting operation, find the stresses developedin the machine tool structure shown in Fig. 12.32.12.29 The crank in the slider-crank mechanism shown in Fig. 12.33 rotates at a constant clockwiseangular speed of 1000 rpm. Find the stresses in the connecting rod and the crank when theCross section of armCross section of columnColumnArmAzx400 mm415 mm15 mm 550 mm2.4 m2 m0.4 m350 mmFIGURE 12.32 A radial drilling machine structure.2 in.3 in.Section X–XX X ␪XX0.5 in.0.5 in.pFIGURE 12.33 A slider-crank mechanism.RaoCh12ff.qxd 10.06.08 13:49 Page 897
  43. 43. 898 CHAPTER 12 FINITE ELEMENT METHODpressure acting on the piston is 200 psi and The diameter of the piston is 12 in. andthe material of the mechanism is steel. Model the connecting rod and the crank by one beamelement each. The lengths of the crank and connecting rod are 12 in. and 48 in., respectively.12.30 A water tank of weight W is supported by a hollow circular steel column of inner diameter d,wall thickness t, and height l. The wind pressure acting on the column can be assumed to varylinearly from 0 to as shown in Fig. 12.34. Find (a) the bending stress induced in the col-umn under the loads, and (b) the natural frequencies of the water tank using a one beam ele-ment idealization. Data: and pmax = 100 psi.W = 10,000 lb, l = 40 ft, d = 2 ft, t = 1 in.,pmax,u = 30°.12.31 Find the natural frequencies of the stepped bar shown in Fig. 12.35 with the following datausing consistent and lumped mass matrices:and l1 = l2 = 50 in.r = 0.283 lb/in.3,E = 30 * 106psi,A2 = 1 in.2,A1 = 2 in.2,WpmaxColumnWater tanklFIGURE 12.34U3l2l1A1A2U2U1FIGURE 12.3512.32 Find the undamped natural frequencies of longitudinal vibration of the stepped bar shown inFig. 12.36 with the following data using consistent and lumped mass matrices:and103kg/m3.r = 7.8 *l3 = 0.2 m, A1 = 2A2 = 4A3 = 0.4 * 10-3m2, E = 2.1 * 1011N/m2,l1 = l2 =RaoCh12ff.qxd 10.06.08 13:49 Page 898
  44. 44. DESIGN PROJECTS 89912.33 Consider the stepped bar shown in Fig. 12.11 with the following data:Pa,Using MATLAB, find the axial dis-placements and under the load N.12.34 Using MATLAB, find the natural frequencies and mode shapes of the stepped bar describedin Problem 12.33.12.35 Use Program17.m to find the natural frequencies of a fixed-fixed stepped beam, similar tothe one shown in Fig. 12.12, with the following data:Cross sections of elements: 1, 2, 3:Lengths of elements:Young’s modulus of all elements:Weight density of all elements:12.36 Use Program17.cpp to solve Problem 12.35.12.37 Use PROGRAM17.F to solve Problem 12.35.12.38 Write a computer program for finding the assembled stiffness matrix of a general planar truss.12.39 Generalize the computer program of Section 12.10 to make it applicable to the solution of anystepped beam having a specified number of steps.12.40 Find the natural frequencies and mode shapes of the beam shown in Fig. 12.12 withand a uniform cross section of throughout the length,using the computer program of Section 12.10. Compare your results with those given inChapter 8. (Hint: Only the data XL, XI, and A need to be changed.)DESIGN PROJECTS12.41 Derive the stiffness and mass matrices of a uniform beam element in transverse vibrationrotating at an angular velocity of rad/sec about a vertical axis as shown in Fig. 12.37(a).Using these matrices, find the natural frequencies of transverse vibration of the rotor blade ofa helicopter (see Fig. 12.37b) rotating at a speed of 300 rpm. Assume a uniform rectangularcross section and a length for the blade. The material of the blade is aluminum.12.42 An electric motor weighing 1000 lb operates on the first floor of a building frame that can bemodeled by a steel girder supported by two reinforced concrete columns, as shown in Fig. 12.38.If the operating speed of the motor is 1500 rpm, design the girder and the columns such thatthe fundamental frequency of vibration of the building frame is greater than the operating speed48–1– * 12–Æ1 in. * 1 in.l1 = l2 = l3 = 10 in.0.1lb/in.3107lb/in.21, 2, 3: 30–, 20–, 10–4– * 4–, 3– * 3–, 2– * 2–p3 = 500u3u1, u2,l3 = 1 m.i = 1, 2, 3, l1 = 3 m, l2 = 2 m,103kg/m3,i = 1, 2, 3, ri = 7.8 *Ei = 2 * 1011A3 = 9 * 10-4m2,A2 = 16 * 10-4m2,A1 = 25 * 10-4m2,l2 l3l1A1OA2 A3x, XFIGURE 12.36RaoCh12ff.qxd 10.06.08 13:49 Page 899
  45. 45. 900 CHAPTER 12 FINITE ELEMENT METHODof the motor. Use two beam and two bar elements for the idealization. Assume the followingdata:Columns: E = 4 * 106psi, r = 2.7 * 10-3lbm/in.3Girder: E = 30 * 106psi, r = 8.8 * 10-3lbm/in.3, h/b = 2R lOBeam element. , A, I, E␳x(a)(b)Rotor blade⍀FIGURE 12.37MotorGirderColumns18 ft9 ft9 fthdbCross section of girderCross section of columnsFIGURE 12.38RaoCh12ff.qxd 10.06.08 13:49 Page 900
  46. 46. 13.1 IntroductionIn the preceding chapters, the equation of motion contained displacement or its derivativesonly to the first degree, and no square or higher powers of displacement or velocity wereinvolved. For this reason, the governing differential equations of motion and the corre-sponding systems were called linear. For convenience of analysis, most systems are mod-eled as linear systems, but real systems are actually more often nonlinear than linear[13.1–13.6]. Whenever finite amplitudes of motion are encountered, nonlinear analysisbecomes necessary. The superposition principle, which is very useful in linear analysis,does not hold true in the case of nonlinear analysis. Since mass, damper, and spring are thebasic components of a vibratory system, nonlinearity into the governing differential equa-tion may be introduced through any of these components. In many cases, linear analysis isinsufficient to describe the behavior of the physical system adequately. One of the mainreasons for modeling a physical system as a nonlinear one is that totally unexpected phe-nomena sometimes occur in nonlinear systems—phenomena that are not predicted or evenhinted at by linear theory. Several methods are available for the solution of nonlinear vibra-tion problems. Some of the exact methods, approximate analytical techniques, graphicalprocedures, and numerical methods are presented in this chapter.Jules Henri Poincaré (1854–1912) was a French mathematician andprofessor of celestial mechanics at the University of Paris and ofmechanics at the Ecole Polytechnique. His contributions to pure andapplied mathematics, particularly to celestial mechanics and electrody-namics, are outstanding. His classification of singular points of nonlin-ear autonomous systems is important in the study of nonlinearvibrations. (Photo courtesy of Dirk J. Struik, A Concise History ofMathematics, 2nd ed. Dover Publications, New York, 1948)C H A P T E R 1 3Nonlinear Vibration901RaoCh13ff.qxd 10.06.08 13:57 Page 901
  47. 47. 902 CHAPTER 13 NONLINEAR VIBRATION13.2 Examples of Nonlinear Vibration ProblemsThe following examples are given to illustrate the nature of nonlinearity in some physicalsystems.Consider a simple pendulum of length l, having a bob of mass m, as shown in Fig. 13.1(a).The differential equation governing the free vibration of the pendulum can be derived fromFig. 13.1(b):(13.1)For small angles, sin may be approximated by and Eq. (13.1) reduces to a linearequation:(13.2)where(13.3)The solution of Eq. (13.2) can be expressed as(13.4)where is the amplitude of oscillation, is the phase angle, and is the angular fre-quency. The values of and are determined by the initial conditions and the angularfrequency is independent of the amplitude Equation (13.4) denotes an approximatesolution of the simple pendulum. A better approximate solution can be obtained by usinga two-term approximation for sin near as in Eq. (13.1)or(13.5)u$+ v021u - 16 u32 = 0ml2u$+ mglau -u36b = 0u - u3/6u = 0uA0.v0fA0v0fA0u1t2 = A0sin1v0t + f2v0 = 1g/l21/2u$+ v02u = 0uuml2u$+ mgl sin u = 013.2.1SimplePendulumOlm␪␪(a)Tmg␪mg sin␪mg cos␪ϩ ␪..,Inertiamomentml2␪..(b)FIGURE 13.1 Simple pendulum.RaoCh13ff.qxd 10.06.08 13:57 Page 902
  48. 48. 13.2 EXAMPLES OF NONLINEAR VIBRATION PROBLEMS 903It can be seen that Eq. (13.5) is nonlinear because of the term involving (due to geo-metric nonlinearity). Equation (13.5) is similar to the equation of motion of a spring-masssystem with a nonlinear spring. If the spring is nonlinear (due to material nonlinearity), therestoring force can be expressed as f(x), where x is the deformation of the spring and theequation of motion of the spring-mass system becomes(13.6)If the spring is linear. If is a strictly increasing functionof x, the spring is called a hard spring, and if is a strictly decreasing function of x,the spring is called a soft spring as shown in Fig. 13.2. Due to the similarity of Eqs. (13.5)and (13.6), a pendulum with large amplitudes is considered, in a loose sense, as a systemwith a nonlinear elastic (spring) component.Nonlinearity may be reflected in the damping term as in the case of Fig. 13.3(a). The sys-tem behaves nonlinearly because of the dry friction between the mass m and the movingbelt. For this system, there are two friction coefficients: the static coefficient of frictioncorresponding to the force required to initiate the motion of the body held by dryfriction; and the kinetic coefficient of friction corresponding to the force required tomaintain the body in motion. In either case, the component of the applied force tangent tothe friction surface (F) is the product of the appropriate friction coefficient and the forcenormal to the surface.The sequence of motion of the system shown in Fig. 13.3(a) is as follows [13.7]. Themass is initially at rest on the belt. Due to the displacement of the mass m along with thebelt, the spring elongates. As the spring extends, the spring force on the mass increasesuntil the static friction force is overcome and the mass begins to slide. It slides rapidlytowards the right, thereby relieving the spring force until the kinetic friction force halts it.The spring then begins to build up the spring force again. The variation of the damping1mk2,1ms2,13.2.2MechanicalChatter, BeltFriction Systemdf/dxdf/dxdf/dx1x2 = k = constant,mx$+ f1x2 = 0u3Oxf(x)(a) Soft springOxf(x)(b) Hard springFIGURE 13.2 Nonlinear spring characteristics.RaoCh13ff.qxd 10.06.08 13:57 Page 903
  49. 49. 904 CHAPTER 13 NONLINEAR VIBRATIONvmϩ ϩkBeltRollerDryfrictionRollerx(a) (b)OvF(x.)Friction force,Velocity ofmass, m(x.)FIGURE 13.3 Dry friction damping.force with the velocity of the mass is shown in Fig. 13.3(b). The equation of motion canbe expressed as(13.7)where the friction force F is a nonlinear function of as shown in Fig. 13.3(b).For large values of the damping force is positive (the curve has a positive slope) andenergy is removed from the system. On the other hand, for small values of the dampingforce is negative (the curve has a negative slope) and energy is put into the system.Although there is no external stimulus, the system can have an oscillatory motion; it cor-responds to a nonlinear self-excited system. This phenomenon of self-excited vibration iscalled mechanical chatter.Nonlinearity may appear in the mass term as in the case of Fig. 13.4 [13.8]. For largedeflections, the mass of the system depends on the displacement x, and so the equation ofmotion becomes(13.8)Note that this is a nonlinear differential equation due to the nonlinearity of the first term.ddt1mx#2 + kx = 03.2.3Variable MassSystemx#,x#,x#,mx$+ F1x#2 + kx = 0FIGURE 13.4 Variable mass system.RaoCh13ff.qxd 10.06.08 13:57 Page 904
  50. 50. 13.3 EXACT METHODS 90513.3 Exact MethodsAn exact solution is possible only for a relatively few nonlinear systems whose motion isgoverned by specific types of second-order nonlinear differential equations. The solutionsare exact in the sense that they are given either in closed form or in the form of an expres-sion that can be numerically evaluated to any degree of accuracy. In this section, we shallconsider a simple nonlinear system for which the exact solution is available. For a singledegree of freedom system with a general restoring (spring) force F(x), the free vibrationequation can be expressed as(13.9)where is a constant. Equation (13.9) can be rewritten as(13.10)Assuming the initial displacement as and the velocity as zero at Eq. (13.10) canbe integrated to obtain(13.11)where is the integration variable. Equation (13.11), when integrated again, gives(13.12)where is the new integration variable and corresponds to the time whenEquation (13.12) thus gives the exact solution of Eq. (13.9) in all those situations wherethe integrals of Eq. (13.12) can be evaluated in closed form. After evaluating the integralsof Eq. (13.12), one can invert the result and obtain the displacement-time relation. If F(x)is an odd function,(13.13)By considering Eq. (13.12) from zero displacement to maximum displacement, the periodof vibration can be obtained:(13.14)For illustration, let In this case Eqs. (13.12) and (13.14) become(13.15)t - t0 =1a An + 12 Lx00dj1x0n+1- jn+121/2F1x2 = xn.t =412a Lx00djeLx0jF1h2 dhf1/2tF1-x2 = -F1x2x = 0.t0jt - t0 =112a Lx0djeLx0jF1h2dhf1/2hx# 2= 2a2Lx0xF1h2dh or ƒx#ƒ = 12aeLx0xF1h2dhf1/2t = t0,x0ddx1x# 22 + 2a2F1x2 = 0a2x$+ a2F1x2 = 0RaoCh13ff.qxd 10.06.08 13:57 Page 905
  51. 51. and(13.16)By setting Eq. (13.16) can be written as(13.17)This expression can be evaluated numerically to any desired level of accuracy.13.4 Approximate Analytical MethodsIn the absence of an exact analytical solution to a nonlinear vibration problem, we wish to findat least an approximate solution. Although both analytical and numerical methods are avail-able for approximate solution of nonlinear vibration problems, the analytical methods aremore desirable [13.6, 13.9]. The reason is that once the analytical solution is obtained, anydesired numerical values can be substituted and the entire possible range of solutions can beinvestigated. We shall now consider four analytical techniques in the following subsections.Let the equations governing the vibration of a nonlinear system be represented by a sys-tem of n first order differential equations1(13.18)where the nonlinear terms are assumed to appear only in and is a small param-eter. In Eq. (13.18)andg!1x!, t2 = fg11x1, x2, Á , xn, t2g21x1, x2, Á , xn, t2...gn1x1, x2, Á , xn, t2vx!= fx1x2...xnv, x:#= fdx1/dtdx2/dt...dxn/dtv, f:1x!, t2 = ff11x1, x2, Á , xn, t2f21x1, x2, Á , xn, t2...fn1x1, x2, Á , xn, t2vag!1x!, t2x:#1t2 = f:1x!, t2 + ag!1x!, t213.4.1BasicPhilosophyt =4a11x0n-121/2 An + 12 L10dy11 - yn+121/2y = j/x0,t =4a An + 12 Lx00dj1x0n+1- jn+121/2906 CHAPTER 13 NONLINEAR VIBRATION1Systems governed by Eq. (13.18), in which the time appears explicitly, are known as nonautonomous systems.On the other hand, systems for which the governing equations are of the typewhere time does not appear explicitly are called autonomous systems.x!#1t2 = f:1x2 + ag!1x2RaoCh13ff.qxd 10.06.08 13:57 Page 906
  52. 52. 13.4 APPROXIMATE ANALYTICAL METHODS 907The solution of differential equations having nonlinear terms associated with a small param-eter was studied by Poincaré [13.6]. Basically, he assumed the solution of Eq. (13.18) inseries form as(13.19)The series solution of Eq. (13.19) has two basic characteristics:1. As Eq. (13.19) reduces to the exact solution of the linear equations2. For small values of the series converges fast so that even the first two or threeterms in the series of Eq. (13.19) yields a reasonably accurate solution.The various approximate analytical methods presented in this section can be considered tobe modifications of the basic idea contained in Eq. (13.19). Although Poincaré’s solution,Eq. (13.19), is valid for only small values of the method can still be applied to systemswith large values of The solution of the pendulum equation, Eq. (13.5), is presented toillustrate the Poincaré’s method.Solution of Pendulum Equations. Equation (13.5) can be rewritten as(13.20)where and Equation (13.20) is known as the freeDuffing’s equation. Assuming weak nonlinearity (i.e., is small), the solution of Eq. (13.20)is expressed as(13.21)where are functions to be determined. By using a two-termapproximation in Eq. (13.21), Eq. (13.20) can be written asthat is,(13.22)If terms involving and are neglected (since is assumed to be small), Eq.(13.22) will be satisfied if the following equations are satisfied:(13.23)(13.24)The solution of Eq. (13.23) can be expressed as(13.25)x01t2 = A0 sin 1v0t + f2x$1 + v02x1 = -x03x$0 + v02x0 = 0aa4a2, a3,+ a313x0x122 + a4x13= 01x$0 + v02x02 + a1x$1 + v02x1 + x032 + a213x02x121x$0 + ax$12 + v021x0 + ax12 + a1x0 + ax123= 0xi1t2, i = 0, 1, 2, Á , n,x1t2 = x01t2 + ax11t2 + a2x21t2 + Á + anxn1t2 + Áaa = - v02/6.x = u, v0 = 1g/l21/2,x$+ v02x + ax3= 0a.a,a,x!#= f:1x!, t2.a : 0,x!1t2 = x!01t2 + ax!11t2 + a2x!21t2 + a3x!31t2 + ÁRaoCh13ff.qxd 10.06.08 13:57 Page 907
  53. 53. 908 CHAPTER 13 NONLINEAR VIBRATIONIn view of Eq. (13.25), Eq. (13.24) becomes(13.26)The particular solution of Eq. (13.26) is (and can be verified by substitution)(13.27)Thus the approximate solution of Eq. (13.20) becomes(13.28)The initial conditions on x(t) can be used to evaluate the constants andNotes1. It can be seen that even a weak nonlinearity (i.e., small value of ) leads to a nonperiodicsolution since Eq. (13.28) is not periodic due to the second term on the right-hand side ofEq. (13.28). In general, the solution given by Eq. (13.21) will not be periodic if we retainonly a finite number of terms.2. In Eq. (13.28), the second term, and hence the total solution, can be seen to approachinfinity as t tends to infinity. However, the exact solution of Eq. (13.20) is known to bebounded for all values of t. The reason for the unboundedness of the solution, Eq. (13.28),is that only two terms are considered in Eq. (13.21). The second term in Eq. (13.28)is called a secular term. The infinite series in Eq. (13.21) leads to a bounded solutionof Eq. (13.20) because the process is a convergent one. To illustrate this point, con-sider the Taylor’s series expansion of the function(13.29)If only two terms are considered on the right-hand side of Eq. (13.29), the solutionapproaches infinity as However, the function itself and hence its infinite seriesexpansion can be seen to be a bounded one.This method assumes that the angular frequency along with the solution varies as a func-tion of the amplitude This method eliminates the secular terms in each step of theapproximation [13.5] by requiring the solution to be periodic in each step. The solutionand the angular frequency are assumed as(13.30)x1t2 = x01t2 + ax11t2 + a2x21t2 + ÁA0.13.4.2Lindstedt’sPerturbationMethodt : q.-a2t22!sin vt -a3t33!cos vt + Ásin1v + a2t = sin vt + at cos vtsin1vt + at2:af.A0= A0 sin1v0t + f2 +3at8v0A03cos1v0t + f2 -A03a32v02sin 31v0t + f2x1t2 = x01t2 + ax11t2x11t2 =38v0tA03cos1v0t + f2 -A0332v02sin 31v0t + f2= -A03C34 sin 1v0t + f2 - 14 sin 31v0t + f2Dx$1 + v02x1 = -A03sin31v0t + f2RaoCh13ff.qxd 10.06.08 13:57 Page 908
  54. 54. 13.4 APPROXIMATE ANALYTICAL METHODS 909(13.31)We consider the solution of the pendulum equation, Eq. (13.20), to illustrate the perturba-tion method. We use only linear terms in in Eqs. (13.30) and (13.31):(13.32)(13.33)Substituting Eqs. (13.32) and (13.33) into Eq. (13.20), we getthat is,(13.34)Setting the coefficients of various powers of to zero and neglecting the terms involvingand in Eq. (13.34), we obtain(13.35)(13.36)Using the solution of Eq. (13.35),(13.37)into Eq. (13.36), we obtain(13.38)It can be seen that the first and the last terms on the right-hand side of Eq. (13.38) lead tosecular terms. They can be eliminated by taking as(13.39)With this, Eq. (13.38) becomes(13.40)x$1 + v2x1 = 14 A03sin 31vt + f2v1 = 34 A02, A0 Z 0v1+ v1A0 sin1vt + f2= - 34 A03sin1vt + f2 + 14 A03sin 31vt + f2x$1 + v2x1 = - [A0 sin1vt + f2]3+ v1[A0 sin1vt + f2]x01t2 = A0 sin1vt + f2x$1 + v2x1 = -x03+ v1x0x$0 + v2x0 = 0a4a2, a3,a+ a213x1x02- v1x12 + a313x12x02 + a41x132 = 0x$0 + v02x0 + a1v2x1 + x03- v1x0 + x$12x$0 + ax$1 + [v2- av11A02][x0 + ax1] + a[x0 + ax1]3= 0v2= v02+ av11A02 or v02= v2- av11A02x1t2 = x01t2 + ax11t2av2= v02+ av11A02 + a2v21A02 + ÁRaoCh13ff.qxd 10.06.08 13:57 Page 909
  55. 55. The solution of Eq. (13.40) is(13.41)Let the initial conditions be and Using Lindstedt’s method,we force the solution given by Eq. (13.37) to satisfy the initial conditions so thatorSince the initial conditions are satisfied by itself, the solution given by Eq. (13.41)must satisfy zero initial conditions.2ThusIn view of the known relations and the above equations yieldThus the total solution of Eq. (13.20) becomes(13.42)with(13.43)For the solution obtained by considering three terms in the expansion of Eq. (13.30), seeProblem 13.13. It is to be noted that the Lindstedt’s method gives only the periodic solu-tions of Eq. (13.20); it cannot give any nonperiodic solutions, even if they exist.In the basic iterative method, first the equation is solved by neglecting certain terms. Theresulting solution is then inserted in the terms that were neglected at first to obtain a sec-ond, improved, solution. We shall illustrate the iterative method to find the solution ofDuffing’s equation, which represents the equation of motion of a damped, harmonicallyexcited, single degree of freedom system with a nonlinear spring. We begin with the solu-tion of the undamped equation.13.4.3Iterative Methodv2= v02+ a34 A02x1t2 = A0 sin1vt + f2 -aA0332v2sin 31vt + f2A1 = - aA332v2b and f1 =p2.f = p/2,A0 = Ax$1102 = 0 = A1v cos f1 -A0332v213v2 cos 3fx1102 = 0 = A1 sin f1 -A0332v2sin 3fx11t2x01t2A0 = A and f =p2x102 = A = A0 sin f, x#102 = 0 = A0v cos fx01t2x#1t = 02 = 0.x1t = 02 = Ax11t2 = A1 sin1vt + f12 -A0332v2sin 31vt + f2910 CHAPTER 13 NONLINEAR VIBRATION2If satisfies the initial conditions, each of the solutions appearing in Eq. (13.30) must sat-isfy zero initial conditions.x11t2, x21t2, Áx01t2RaoCh13ff.qxd 10.06.08 13:57 Page 910
  56. 56. 13.4 APPROXIMATE ANALYTICAL METHODS 911Solution of the Undamped Equation. If damping is disregarded, Duffing’s equationbecomesor(13.44)As a first approximation, we assume the solution to be(13.45)where A is an unknown. By substituting Eq. (13.45) into Eq. (13.44), we obtain the dif-ferential equation for the second approximation:(13.46)By using the identity(13.47)Eq. (13.46) can be expressed as(13.48)By integrating this equation and setting the constants of integration to zero (so as to makethe solution harmonic with period ), we obtain the second approximation:(13.49)Duffing [13.7] reasoned at this point that if and are good approximations to thesolution x(t), the coefficients of in the two equations (13.45) and (13.49) should notbe very different. Thus by equating these coefficients, we obtainor(13.50)For present purposes, we will stop the procedure with the second approximation. It can be ver-ified that this procedure yields the exact solution for the case of a linear spring (with )(13.51)where A denotes the amplitude of the harmonic response of the linear system.A =Fv02- v2a = 0v2= v02Ϯ34A2a -FAA =1v2aAv02Ϯ34A3a - Fbcos vtx21t2x11t2x21t2 =1v21Av02Ϯ 34 A3a - F2cos vt ϮA3a36v2cos 3vtt = 2p/vx$2 = -1Av02Ϯ 34A3a - F2 cos vt ϯ 14A3a cos 3vtcos3vt = 34 cos vt + 14 cos 3vtx$2 = -Av02cos vt ϯ A3a cos3vt + F cos vtx11t2 = A cos vtx$= - v02x ϯ ax3+ F cos vtx$+ v02Ϯ ax3= F cos vtRaoCh13ff.qxd 10.06.08 13:57 Page 911
  57. 57. 912 CHAPTER 13 NONLINEAR VIBRATION3The first approximate solution, Eq. (13.45), can be seen to satisfy the initial conditions andx#102 = 0.x102 = AFor a nonlinear system (with ), Eq. (13.50) shows that the frequency is afunction of A, and F. Note that the quantity A, in the case of a nonlinear system, is notthe amplitude of the harmonic response but only the coefficient of the first term of its solu-tion. However, it is commonly taken as the amplitude of the harmonic response of the sys-tem.3For the free vibration of the nonlinear system, and Eq. (13.50) reduces to(13.52)This equation shows that the frequency of the response increases with the amplitude A forthe hardening spring and decreases for the softening spring. The solution, Eq. (13.52), canalso be seen to be same as the one given by Lindstedt’s method, Eq. (13.43).For both linear and nonlinear systems, when (forced vibration), there are twovalues of the frequency for any given amplitude One of these values of is smallerand the other larger than the corresponding frequency of free vibration at that amplitude.For the smaller value of and the harmonic response of the system is in phasewith the external force. For the larger value of and the response is 180° out ofphase with the external force. Note that only the harmonic solutions of Duffing’s equa-tion—that is, solutions for which the frequency is the same as that of the external force—have been considered in the present analysis. It has been observed [13.2] thatoscillations whose frequency is a fraction, such as of that of the applied force,are also possible for Duffing’s equation. Such oscillations, known as subharmonic oscilla-tions, are considered in Section 13.5.Solution of the Damped Equation. If we consider viscous damping, we obtainDuffing’s equation:(13.53)For a damped system, it was observed in earlier chapters that there is a phase differencebetween the applied force and the response or solution. The usual procedure is to prescribethe applied force first and then determine the phase of the solution. In the present case,however, it is more convenient to fix the phase of the solution and keep the phase of theapplied force as a quantity to be determined. We take the differential equation, Eq. (13.53),in the form(13.54)in which the amplitude of the applied force is considered fixed, but theratio is left to be determined. We assume that and are all small,of order As with Eq. (13.44), we assume the first approximation to the solution to be(13.55)x1 = A cos vta.A2c, A1,A1/A2 = tan-1fF = 1A12+ A2221/2= A1 cos vt - A2 sin vtx$+ cx#+ v02x Ϯ ax3= F cos1vt + f2x$+ cx#+ v02x Ϯ ax3= F cos vt12, 13, Á , 1n,F cos vtv, A 6 0v, A 7 0vƒA ƒ.vF Z 0v2= v02Ϯ 34 A2aF = 0a,va Z 0RaoCh13ff.qxd 10.06.08 13:57 Page 912
  58. 58. 13.4 APPROXIMATE ANALYTICAL METHODS 913where A is assumed fixed and to be determined. By substituting Eq. (13.55) into Eq. (13.54)and making use of the relation (13.47), we obtain(13.56)By disregarding the term involving and equating the coefficients of andon both sides of Eq. (13.56), we obtain the following relations:(13.57)The relation between the amplitude of the applied force and the quantities A and can beobtained by squaring and adding the equations (13.57):(13.58)Equation (13.58) can be rewritten as(13.59)where(13.60)It can be seen that for Eq. (13.59) reduces to which is the same asEq. (13.50). The response curves given by Eq. (13.59) are shown in Fig. 13.5.Jump Phenomenon. As mentioned earlier, nonlinear systems exhibit phenomena thatcannot occur in linear systems. For example, the amplitude of vibration of the systemS1v, A2 = F,c = 0,S1v, A2 = 1v02- v22A Ϯ 34 aA3S21v, A2 + c2v2A2= F2C1v02- v22A Ϯ 34 aA3D2+ 1cvA22= A12+ A22= F2vcvA = A21v02- v22A Ϯ 34 aA3= A1sin vtcos vtcos 3vt= A1 cos vt - A2 sin vtc1v02- v22A Ϯ34aA3d cos vt - cvA sin vt ϮaA34cos 3vtvOͿAͿ ͿAͿ ͿAͿF ϭ 0F ϭ 0F ϭ 0F1F1F1F1 F2F2F2F2␻0␻(a) ϭ 0␣O ␻0␻ ␻(b) Ͼ 0␣O ␻0(c) Ͻ 0␣FIGURE 13.5 Response curves of Duffing’s equation.RaoCh13ff.qxd 10.06.08 13:57 Page 913
  59. 59. 914 CHAPTER 13 NONLINEAR VIBRATION2764315 5O O␻ ␻͉A͉ ͉A͉6342 71(a) Ͼ 0 (hard spring)␣ (b) Ͻ 0 (soft spring)␣FIGURE 13.6 Jump phenomenon.described by Eq. (13.54) has been found to increase or decrease suddenly as the excitationfrequency is increased or decreased, as shown in Fig. 13.6. For a constant magnitude ofF, the amplitude of vibration will increase along the points 1, 2, 3, 4, 5 on the curve whenthe excitation frequency is slowly increased. The amplitude of vibration jumps frompoint 3 to 4 on the curve. Similarly, when the forcing frequency is slowly decreased, theamplitude of vibration follows the curve along the points 5, 4, 6, 7, 2, 1 and makes a jumpfrom point 6 to 7. This behavior is known as the jump phenomenon. It is evident that twoamplitudes of vibration exist for a given forcing frequency, as shown in the shaded regionsof the curves of Fig. 13.6. The shaded region can be thought of as unstable in some sense.Thus an understanding of the jump phenomena requires a knowledge of the mathemati-cally involved stability analysis of periodic solutions [13.24, 13.25]. The jump phenomenawere also observed experimentally by several investigators [13.26, 13.27].In the Ritz-Galerkin method, an approximate solution of the problem is found by satisfy-ing the governing nonlinear equation in the average. To see how the method works, let thenonlinear differential equation be represented as(13.61)An approximate solution of Eq. (13.61) is assumed as(13.62)where are prescribed functions of time and areweighting factors to be determined. If Eq. (13.62) is substituted in Eq. (13.61), we get afunction Since is not, in general, the exact solution of Eq. (13.61),will not be zero. However, the value of will serve as a measure ofthe accuracy of the approximation; in fact, as x : x.E [t] : 0E [t]E 1t2 = E[x1t2]x1t2E[x1t2].ana1, a2, Á ,f11t2, f21t2, Á , fn1t2x1t2 = a1f11t2 + a2f21t2 + Á + anfn1t2E[x] = 013.4.4Ritz-GalerkinMethodvvvRaoCh13ff.qxd 10.06.08 13:57 Page 914
  60. 60. 13.4 APPROXIMATE ANALYTICAL METHODS 915The weighting factors are determined by minimizing the integral(13.63)where denotes the period of the motion. The minimization of the function of Eq. (13.63)requires(13.64)Equation (13.64) represents a system of n algebraic equations that can be solved simulta-neously to find the values of The procedure is illustrated with the follow-ing example.Solution of Pendulum Equation Using the Ritz-Galerkin MethodUsing a one-term approximation, find the solution of the pendulum equation(E.1)by the Ritz-Galerkin method.Solution: By using a one-term approximation for x(t) as(E.2)Eqs. (E.1) and (E.2) lead to(E.3)The Ritz-Galerkin method requires the minimization of(E.4)for finding The application of Eq. (13.64) gives* c av02- v2-38v02A02b sin vt +18v02A02sin 3vtd dt = 0Lt0E0E0A0dt =Lt0c av02- v2-18v02A02bA0 sin vt +v0224A03sin 3vtdA0.Lt0E2[x1t2] dt= av02- v2-18v02A02bA0 sin vt +v0224A03sin 3vtE[x1t2] = - v2A0 sin vt + v02cA0 sin vt -16sin3vtdx1t2 = A0 sin vtE[x] = x$+ v02x -v026x3= 0E X A M P L E 1 3 . 1a1, a2, Á , an.i = 1, 2, Á , n00aiaLt0E2[t] dtb = 2Lt0E [t]0E [t]0aidt = 0,tLt0E2[t] dtaiRaoCh13ff.qxd 10.06.08 13:57 Page 915
  61. 61. 916 CHAPTER 13 NONLINEAR VIBRATIONthat is,that is,(E.5)For a nontrivial solution, and Eq. (E.5) leads to(E.6)The roots of the quadratic equation in Eq. (E.5), can be found as(E.7)(E.8)It can be verified that given by Eq. (E.7) minimizes the quantity of (E.4), while the one given byEq. (E.8) maximizes it. Thus the solution of Eq. (E.1) is given by Eq. (E.2) with(E.9)This expression can be compared with Lindstedt’s solution and the iteration methods (Eqs. 13.43and 13.52):(E.10)The solution can be improved by using a two-term approximation for x(t) as(E.11)The application of Eq. (13.64) with the solution of Eq. (E.11) leads to two simultaneous algebraicequations that must be numerically solved for and■A3.A0x1t2 = A0 sin vt + A3 sin 3vtv2= v0211 - 0.125 A022v2= v0211 - 0.147938 A022v2v2= v0211 - 0.352062 A022v2= v0211 - 0.147938 A022v2,v4+ v2v02a12 A02- 2b + v04a1 - 12 A02+ 596 A04b = 0A0 Z 0,A0c av02- v2-18v02A02b av02- v2-38v02A02b +v04A04192d = 0+v04A05192 Lt0sin23vt dt = 0+18v02A02av02- v2-18v02A02bLt0sin vt sin 3vt dt+v02A0324av02- v2-38v02A02bLt0sin vt sin 3vt dtA0av02- v2-18v02A02b av02- v2-38v02A02bLt0sin2vt dtRaoCh13ff.qxd 10.06.08 13:57 Page 916
  62. 62. 13.5 SUBHARMONIC AND SUPERHARMONIC OSCILLATIONS 917Other approximate methods, such as the equivalent linearization scheme and the har-monic balance procedure, are also available for solving nonlinear vibration problems[13.10–13.12]. Specific solutions found using these techniques include the free vibrationresponse of single degree of freedom oscillators [13.13, 13.14], two degree of freedom sys-tems [13.15], and elastic beams [13.16, 13.17], and the transient response of forced sys-tems [13.18, 13.19]. Several nonlinear problems of structural dynamics have beendiscussed by Crandall [13.30].13.5 Subharmonic and Superharmonic OscillationsWe noted in Chapter 3 that for a linear system, when the applied force has a certain fre-quency of oscillation, the steady-state response will have the same frequency of oscilla-tion. However, a nonlinear system will exhibit subharmonic and superharmonicoscillations. Subharmonic response involves oscillations whose frequencies arerelated to the forcing frequency as(13.65)where n is an integer Similarly, superharmonic response involves oscil-lations whose frequencies are related to the forcing frequency as(13.66)whereIn this section, we consider the subharmonic oscillations of order of an undamped pen-dulum whose equation of motion is given by (undamped Duffing’s equation):(13.67)where is assumed to be small. We find the response using the perturbation method [13.4,13.6]. Accordingly, we seek a solution of the form(13.68)(13.69)where denotes the fundamental frequency of the solution (equal to the third subharmonicfrequency of the forcing frequency). Substituting Eqs. (13.68) and (13.69) into Eq. (13.67)gives(13.70)If terms involving and are neglected, Eq. (13.70) reduces to(13.71)x$0 + v2x0 + ax$1 + av2x1 - av1x0 + ax03= F cos 3vta4a2, a3,+ a1x0 + ax123= F cos 3vtx$0 + ax$1 + v2x0 + v2ax1 - av1x0 - a2x1v1vv2= v02+ av1 or v02= v2- av1x1t2 = x01t2 + ax11t2ax$+ v02x + ax3= F cos 3vt1313.5.1SubharmonicOscillationsn = 2, 3, 4, Á .vn = nv1v21vn21n = 2, 3, 4, Á 2.vn =vn1v21vn2RaoCh13ff.qxd 10.06.08 13:57 Page 917
  63. 63. 918 CHAPTER 13 NONLINEAR VIBRATIONWe first consider the linear equation (by setting ):(13.72)The solution of Eq. (13.72) can be expressed as(13.73)If the initial conditions are assumed as and we obtainand so that Eq. (13.73) reduces to(13.74)where C denotes the amplitude of the forced vibration. The value of C can be determinedby substituting Eq. (13.74) into Eq. (13.72) and equating the coefficients of onboth sides of the resulting equation, which yields(13.75)Now we consider the terms involving in Eq. (13.71) and set them equal to zero:or(13.76)The substitution of Eq. (13.74) into Eq. (13.76) results in(13.77)By using the trigonometric relations(13.78)Eq. (13.77) can be expressed as(13.79)-34AC1A + C2 cos 5vt -3AC24cos 7vt -C34cos 9vt+ av1C -A34-34C3-32A2Cb cos 3vtx$1 + v2x1 = Aav1 -34A2-32C2-34ACb cos vtcos2u =cos3u =cos u cos f =12 + 12 cos 2u34 cos u + 14 cos 3u12 cos1u - f2 + 12 cos1u + f2∂- 3AC2cos vt cos23vt- C3cos33vt - 3A2C cos2vt cos 3vtx$1 + v2x1 = v1A cos vt + v1C cos 3vt - A3cos3vtx$1 + v2x1 = v1x0 - x03a1x$1 + v2x1 - v1x0 + x032 = 0aC = -F8v2cos 3vtx01t2 = A cos vt + C cos 3vtB1 = 0A1 = Ax#1t = 02 = 0,x1t = 02 = Ax01t2 = A1 cos vt + B1 sin vt + C cos 3vtx$0 + v2x0 = F cos 3vta = 0RaoCh13ff.qxd 10.06.08 13:57 Page 918

×