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  • 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 12.4 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Using one beam element and one spring element, find the natural frequencies of the systemshown in Fig. Using two beam elements, find the natural frequencies and mode shapes of the uniform fixed-fixed beam shown in Fig.* 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. 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. 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. 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. 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. 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. 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 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 Use PROGRAM17.F to solve Problem 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. 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. 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. 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. 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. 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$+ F1x#2 + kx = 0FIGURE 13.4 Variable mass system.RaoCh13ff.qxd 10.06.08 13:57 Page 904
  • 50. 13.3 EXACT METHODS 90513.3 Exact MethodsAn exact solution is possible only for a relatively few nonlinear systems whose motion isgoverned by specific types of second-order nonlinear differential equations. The solutionsare exact in the sense that they are given either in closed form or in the form of an expres-sion that can be numerically evaluated to any degree of accuracy. In this section, we shallconsider a simple nonlinear system for which the exact solution is available. For a singledegree of freedom system with a general restoring (spring) force F(x), the free vibrationequation can be expressed as(13.9)where is a constant. Equation (13.9) can be rewritten as(13.10)Assuming the initial displacement as and the velocity as zero at Eq. (13.10) canbe integrated to obtain(13.11)where is the integration variable. Equation (13.11), when integrated again, gives(13.12)where is the new integration variable and corresponds to the time whenEquation (13.12) thus gives the exact solution of Eq. (13.9) in all those situations wherethe integrals of Eq. (13.12) can be evaluated in closed form. After evaluating the integralsof Eq. (13.12), one can invert the result and obtain the displacement-time relation. If F(x)is an odd function,(13.13)By considering Eq. (13.12) from zero displacement to maximum displacement, the periodof vibration can be obtained:(13.14)For illustration, let In this case Eqs. (13.12) and (13.14) become(13.15)t - t0 =1a An + 12 Lx00dj1x0n+1- jn+121/2F1x2 = xn.t =412a Lx00djeLx0jF1h2 dhf1/2tF1-x2 = -F1x2x = 0.t0jt - t0 =112a Lx0djeLx0jF1h2dhf1/2hx# 2= 2a2Lx0xF1h2dh or ƒx#ƒ = 12aeLx0xF1h2dhf1/2t = t0,x0ddx1x# 22 + 2a2F1x2 = 0a2x$+ a2F1x2 = 0RaoCh13ff.qxd 10.06.08 13:57 Page 905
  • 51. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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 =vn1v21vn2RaoCh13ff.qxd 10.06.08 13:57 Page 917
  • 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
  • 64. 13.5 SUBHARMONIC AND SUPERHARMONIC OSCILLATIONS 919The condition to avoid a secular term in the solution is that the coefficient of inEq. (13.79) must be zero. Since in order to have a subharmonic response,(13.80)Equations (13.80) and (13.75) give(13.81)Substituting Eq. (13.81) into Eq. (13.69) and rearranging the terms, we obtain the equationto be satisfied by A and in order to have subharmonic oscillation as(13.82)Equation (13.82) can be seen to be a cubic equation in and a quadratic in A. The rela-tionship between the amplitude (A) and the subharmonic frequency given byEq. (13.82), is shown graphically in Fig. 13.7. It has been observed that the curve PQ,where the slope is positive, represents stable solutions while the curve QR, where the slopeis negative, denotes unstable solutions [13.4, 13.6]. The minimum value of amplitude for the1v2,v2v6- v02v4-3a256164A2v4- 8AFv2+ 2F22 = 0vv1 =34aA2-AF8v2+2F264v4bv1 = 34 1A2+ AC + 2C22A Z 0cos vtFIGURE 13.7 Subharmonic oscillations.RaoCh13ff.qxd 10.06.08 13:57 Page 919
  • 65. existence of stable subharmonic oscillations can be found by setting as4Consider the undamped Duffing’s equation(13.83)The solution of this equation is assumed as(13.84)where the amplitudes of the harmonic and superharmonic components, A and C, are to bedetermined. The substitution of Eq. (13.84) into Eq. (13.83) gives, with the use of thetrigonometric relations of Eq. (13.78),(13.85)Neglecting the terms involving and and equating the coeffi-cients of and on both sides of Eq. (13.85), we obtain(13.86)(13.87)Equations (13.86) and (13.87) represent a set of simultaneous nonlinear equations that canbe solved numerically for A and C.As a particular case, if C is assumed to be small compared to A, the terms involvingand can be neglected and Eq. (13.87) gives(13.88)and Eq. (13.86) gives(13.89)C LF - v02A + v2A - 34 aA334 aA2C L- 14 aA3132 aA2+ v02- 9v22C3C2v02C - 9v2C + 14 aA3+ 34 aC3+ 32 aA2C = 0v02A - v2A + 34 aA3+ 34 aA2C + 32 aAC2= Fcos 3vtcos vtcos 9vt,cos 5vt, cos 7vt,+ cos 9vt C14 aC3D = F cos vt+ cos 5vt C34 aA2C + 34 aAC2D + cos 7vt C34 aAC2D+ cos 3vt C - 9v2C + v02C + 14 aA3+ 34 aC3+ 32 aA2CDcos vt C - v2A + v02A + 34 aA3+ 34 aA2C + 32 aAC2Dx1t2 = A cos vt + C cos 3vtx$+ v02x + ax3= F cos vtA = 1F/16v22.dv2/dA = 0920 CHAPTER 13 NONLINEAR VIBRATION4Equation (13.82) can be rewritten aswhich, upon differentiation, givesBy setting we obtain A = 1F/16v22.dv2/dA = 0,31v222dv2- 2v02v2dv2-3a412A dA21v222-3a2A2v2dv2+3aF32v2dA +3aF32A dv2= 01v223- v021v222-3a4A21v222+3aF32A1v22 -3aF2128= 013.5.2SuperharmonicOscillationsRaoCh13ff.qxd 10.06.08 13:57 Page 920
  • 66. 13.6 SYSTEMS WITH TIME-DEPENDENT COEFFICIENTS (MATHIEU EQUATION) 921Equating C from Eqs. (13.88) and (13.89) leads to(13.90)which can be rewritten as(13.91)Equation (13.88), in conjunction with Eq. (13.91), gives the relationship between theamplitude of superharmonic oscillations (C) and the corresponding frequency13.6 Systems with Time-Dependent Coefficients (Mathieu Equation)Consider the simple pendulum shown in Fig. 13.8(a). The pivot point of the pendulum ismade to vibrate in the vertical direction as(13.92)where Y is the amplitude and is the frequency of oscillation. Since the entire pendulumaccelerates in the vertical direction, the net acceleration is given byThe equation of motion of the pendulum can be derived as(13.93)For small deflections near and Eq. (13.93) reduces to(13.94)u$+ agl-v2Ylcos vtbu = 0u = 0, sin u L uml2u$+ m1g - y$2 l sin u = 0g - v2Y cos vt.g - y$1t2 =vy1t2 = Y cos vt13v2.+ A110 v2v02- 9v4- v042 + 1v02F - 9v2F2 = 0-A511516 a22 + A31334 av2- 94 av022 + A2132 aF2* 1F - v02A + v2A - 34 aA321- 14 aA32134 aA22 = 132 aA2+ v02- 9v22Pivoty(t)0mgm␪y ϭ Y cos t␻(a)Pivoty(t)0mgm␪y ϭ Y cos t␻(b)FIGURE 13.8 Simple pendulum with oscillations ofpivot.RaoCh13ff.qxd 10.06.08 13:57 Page 921
  • 67. 922 CHAPTER 13 NONLINEAR VIBRATIONIf the pendulum is inverted as shown in Fig. 13.8(b), the equation of motion becomesor(13.95)where is the angle measured from the vertical (unstable equilibrium) point. If the pivotpoint 0 vibrates as the equation of motion becomes(13.96)For small angular displacements around Eq. (13.96) reduces to(13.97)Equations (13.94) and (13.97) are particular forms of an equation called the Mathieu equa-tion for which the coefficient of in the differential equation varies with time to form anonautonomous equation. We shall study the periodic solutions and their stability charac-teristics of the system for small values of Y in this section.Periodic Solutions Using Lindstedt’s Perturbation Method [13.7]. Consider the Mathieuequation in the form(13.98)where is assumed to be small. We approximate the solution of Eq. (13.98) as(13.99)(13.100)where are constants. Since the periodic coefficient cos t in Eq. (13.98) varieswith a period of it was found that only two types of solutions exist—one with periodand the other with period [13.7, 13.28]. Thus we seek the functionsin Eq. (13.99) in such a way that y(t) is a solution of Eq. (13.98) withperiod or Substituting Eqs. (13.99) and (13.100) into Eq. (13.98) results in(13.101)+ a0y2) + Á = 0+ P2(y$2 + a2y0 + a1y1 + y1 cos t1y$0 + a0y02 + P1y$1 + a1y0 + y0 cos t + a0y124p.2py01t2, y11t2, Á4p2p2p,a0, a1, Áa = a0 + Pa1 + P2a2 + Áy1t2 = y01t2 + Py11t2 + P2y21t2 + ÁPd2ydt2+ 1a + P cos t2y = 0uu$+ a -gl+v2Ylcos vtbu = 0u = 0,u$+ a -gl+v2Ylcos vtb sin u = 0y1t2 = Y cos vt,uu$-glsin u = 0ml2u$- mgl sin u = 0RaoCh13ff.qxd 10.06.08 13:57 Page 922
  • 68. 13.6 SYSTEMS WITH TIME-DEPENDENT COEFFICIENTS (MATHIEU EQUATION) 923where Setting the coefficients of various powers of inEq. (13.101) equal to zero, we obtain(13.102)(13.103)(13.104)where each of the functions is required to have a period of or The solution ofEq. (13.102) can be expressed as(13.105)andNow, we consider the following specific values of n.When Equation (13.105) gives and and Eq. (13.103) yields(13.106)In order to have as a periodic function, must be zero. When Eq. (13.106) is integratedtwice, the resulting periodic solution can be expressed as(13.107)where is a constant. With the known values of andEq. (13.104) can be rewritten asor(13.108)In order to have as a periodic function, must be zero (i.e., ). Thus,for Eq. (13.100) gives(13.109)a = - 12 P2+ Án = 0,a2 = - 121- 12 - a22y2y$2 = - 12 - a2 - a cos t - 12 cos 2ty$2 + a2 + 1cos t + a2cos t = 0cos t + a,y1 =a0 = 0, a1 = 0, y0 = 1ay11t2 = cos t + aa1y1y$1 + a1 + cos t = 0 or y$1 = - a1 - cos ty0 = 1a0 = 0n = 0:a0 =n24, n = 0, 1, 2, Áy01t2 = dcos 2a0tsin 2a0tK dcosn2tsinn2t, n = 0, 1, 2, Á4p.2pyi...P2: y$2 + a0y2 + a2y0 + a1y1 + y1 cos t = 0P1: y$1 + a0y1 + a1y0 + y0 cos t = 0P0: y$0 + a0y0 = 0Py$i = d2yi/dt2, i = 0, 1, 2, ÁRaoCh13ff.qxd 10.06.08 13:57 Page 923
  • 69. 924 CHAPTER 13 NONLINEAR VIBRATIONWhen For this case, Eq. (13.105) gives and or sin(t/2).With Eq. (13.103) gives(13.110)The homogeneous solution of Eq. (13.110) is given bywhere and are constants of integration. Since the term involving cos(t/2) appears inthe homogeneous solution as well as the forcing function, the particular solution will con-tain a term of the form t cos(t/2), which is not periodic. Thus the coefficient of cos(t/2)—namely must be zero in the forcing function to ensure periodicity ofThis gives and Eq. (13.110) becomes(13.111)By substituting the particular solution into Eq. (13.111), we obtainand hence Using andEq. (13.104) can be expressed as(13.112)Again, since the homogeneous solution of Eq. (13.112) contains the term cos(t/2), thecoefficient of the term cos(t/2) on the right-hand side of Eq. (13.112) must be zero. Thisleads to and hence Eq. (13.100) becomes(13.113a)Similarly, by starting with the solution we obtain the relation (seeProblem 13.17)(13.113b)When Equation (13.105) gives and or sin t. With andEq. (13.103) can be written as(13.114)Since cos t is a solution of the homogeneous equation, the term involving cos t inEq. (13.114) gives rise to t cos t in the solution of Thus, to impose periodicity of wey1,y1.y$1 + y1 = -a1 cos t -12-12cos 2ty0 = cos t,a0 = 1y0 = cos ta0 = 1n = 2:a =14+P2-P28+ Áy0 = sin1t/22,a =14-P2-P28+ Áa2 = - 18= a - a2 -18b cost2+18cos3t2-18cos5t2y$2 +14y2 = - a2 cost2+12a14cos3t2b - a14cos3t2b cos ty1 = 14 cos13t/22,a0 = 14, a1 = - 12y11t2 = 14 cos13t/22.A2 = 14,y11t2 = A2 cos13t/22y$1 +14y1 = -12cos3t2a1 = -1/2,y11t2.1- a1 - 1/22,A2A1y11t2 = A1 cost2+ A2 sint2y$1 +14y1 = a - a1 -12b cost2-12cos3t2y0 = cos1t/22,y0 = cos1t/22a0 = 14n = 1:RaoCh13ff.qxd 10.06.08 13:57 Page 924
  • 70. 13.6 SYSTEMS WITH TIME-DEPENDENT COEFFICIENTS (MATHIEU EQUATION) 925set With this, the particular solution of can be assumed asWhen this solution is substituted into Eq. (13.114), we obtain andThus Eq. (13.104) becomesor(13.115)For periodicity of we set the coefficient of cos t equal to zero in the forcing functionof Eq. (13.115). This gives and hence(13.116a)Similarly, by proceeding with we obtain (see Problem 13.17)(13.116b)To observe the stability characteristics of the system, Eqs. (13.109), (13.113), and (13.116)are plotted in the plane as indicated in Fig. 13.9. These equations represent curves1a, P2a = 1 - 112 P2+ Áy0 = sin t,a = 1 + 512 P2+ Áa2 = 512,y21t2,= cos t1- a1 + 12 - 1122 + 12 cos 3ty$2 + y2 = - a2 cos t - 1- 12 + 16 cos 2t2cos ty$2 + y2 + a2 cos t + y1 cos t = 0B3 = 16.A3 = - 12+ B3 cos 2t.y11t2 = A3y11t2a1 = 0.FIGURE 13.9 Stability of periodic solutions.RaoCh13ff.qxd 10.06.08 13:57 Page 925
  • 71. that are known as the boundary or transition curves that divide the plane intoregions of stability and instability. These boundary curves are such that a point belongingto any one curve represents a periodic solution of Eq. (13.98). The stability of these peri-odic solutions can be investigated [13.7, 13.25, 13.28]. In Fig. 13.9, the points inside theshaded region denote unstable motion. It can be noticed from this figure that stability isalso possible for negative values of a, which correspond to the equilibrium positionThus with the right choice of the parameters, the pendulum can be made to bestable in the upright position by moving its support harmonically.13.7 Graphical MethodsGraphical methods can be used to obtain qualitative information about the behavior of thenonlinear system and also to integrate the equations of motion. We shall first consider abasic concept known as the phase plane. For a single degree of freedom system, twoparameters are needed to describe the state of motion completely. These parameters areusually taken as the displacement and velocity of the system. When the parameters areused as coordinate axes, the resulting graphical representation of the motion is called thephase plane representation. Thus each point in the phase plane represents a possible stateof the system. As time changes, the state of the system changes. A typical or representa-tive point in the phase plane (such as the point representing the state of the system at time) moves and traces a curve known as the trajectory. The trajectory shows how thesolution of the system varies with time.Trajectories of a Simple Harmonic OscillatorFind the trajectories of a simple harmonic oscillator.Solution: The equation of motion of an undamped linear system is given by(E.1)By setting Eq. (E.1) can be written as(E.2)from which we can obtain(E.3)Integration of Eq. (E.3) leads to(E.4)y2+ vn2x2= c2dydx= -vn2xydxdt= ydydt= - vn2xy = x#,x$+ vn2x = 0E X A M P L E 1 3 . 2t = 013.7.1Phase PlaneRepresentationu = 180°.1a, P2926 CHAPTER 13 NONLINEAR VIBRATIONRaoCh13ff.qxd 10.06.08 13:57 Page 926
  • 72. 13.7 GRAPHICAL METHODS 927xy ϭ xиFIGURE 13.10 Trajectories of a sim-ple harmonic oscillator.where c is a constant. The value of c is determined by the initial conditions of the system. Equation(E.4) shows that the trajectories of the system in the phase plane (x-y plane) are a family of ellipses,as shown in Fig. 13.10. It can be observed that the point is surrounded by closedtrajectories. Such a point is called a center. The direction of motion of the trajectories can be deter-mined from Eq. (E.2). For instance, if and Eq. (E.2) implies that andtherefore, the motion is clockwise.■Phase-plane of an Undamped PendulumFind the trajectories of an undamped pendulum.Solution: The equation of motion is given by Eq. (13.1):(E.1)where Introducing and Eq. (E.1) can be rewritten asoror(E.2)y dy = - v02sin x dxdydx= -v02sin xydxdt= y,dydt= - v02sin xy = x#= u#,x = uv02= g/l.u$= - v02sin uE X A M P L E 1 3 . 3dy/dt 6 0;dx/dt 7 0y 7 0,x 7 01x = 0, y = 02RaoCh13ff.qxd 10.06.08 13:57 Page 927
  • 73. 928 CHAPTER 13 NONLINEAR VIBRATION2␲3␲Ϫ3␲ ␲␲Ϫxиz ϭ y/␻0 ␪ ␻0ϭ /0Ϫ2␲FIGURE 13.11 Trajectories of an undamped pendulum.Integrating Eq. (E.2) and using the condition that when (at the end of the swing), weobtain(E.3)Introducing Eq. (E.3) can be expressed as(E.4)The trajectories given by Eq. (E.4) are shown in Fig. 13.11.■Phase-plane of an Undamped Nonlinear SystemFind the trajectories of a nonlinear spring-mass system governed by the equation(E.1)Solution: The nonlinear pendulum equation can be considered as a special case of Eq. (E.1). To seethis, we use the approximation in the neighborhood of in Eq. (E.1) ofExample 13.3 to obtainwhich can be seen to be a special case of Eq. (E.1). Equation (E.1) can be rewritten asdxdt= y,dydt= - v021x - 2ax32u$+ v02au -u36b = 0u = 0sin u M u - u3/6x$+ v021x - 2ax32 = 0E X A M P L E 1 3 . 4z2= 21cos x - cos x02z = y/v0y2= 2v021cos x - cos x02x = x0x#= 0RaoCh13ff.qxd 10.06.08 13:57 Page 928
  • 74. 13.7 GRAPHICAL METHODS 929oror(E.2)Integration of Eq. (E.2), with the condition when (at the end of the swing in the caseof a pendulum), gives(E.3)where and is a constant. The trajectories, in the phase-plane, given byEq. (E.3), are shown in Fig. 13.12 for several values ofIt can be observed that for Eq. (E.3) denotes a circle of radius A and corresponds to asimple harmonic motion. When Eq. (E.3) represents ovals within the circle given byand the ovals touch the circle at the points When Eq. (E.3) becomes(E.4)Equation (E.4) indicates that the trajectories are given by the parabolas(E.5)y = Ϯ aA -x22Aby2+ x2-x44A2- A2= cy - aA -x22Ab d cy + aA -x22Ab d = 0a = 11>42A2,10, ϮA2.a = 0a 6 0,a = 0,a.A2= x0211 - ax022z = y/v0z2+ x2- ax4= A2x = x0x#= 0y dy = - v021x - 2ax32 dxdydx= -v021x - 2ax32yFIGURE 13.12 Trajectories of a nonlinear system.RaoCh13ff.qxd 10.06.08 13:57 Page 929
  • 75. 930 CHAPTER 13 NONLINEAR VIBRATIONThese two parabolas intersect at points which correspond to the points ofunstable equilibrium.When the trajectories given by Eq. (E.3) will be closed ovals lying betweenthe circle given by and the two parabolas given by Since these trajectories areclosed curves, they represent periodic vibrations. When the trajectories given byEq. (E.3) lie outside the region between the parabolas and extend to infinity. These trajectories cor-respond to the conditions that permit the body to escape from the center of force.■To see some of the characteristics of trajectories, consider a single degree of freedomnonlinear oscillatory system whose governing equation is of the form(13.117)By defining(13.118)and(13.119)we obtain(13.120)Thus there is a unique slope of the trajectory at every point (x, y) in the phase plane, pro-vided that is not indeterminate. If and (that is, if the point lies on thex axis), the slope of the trajectory is infinite. This means that all trajectories must cross thex axis at right angles. If and the point is called a singular point, and the slopeis indeterminate at such points. A singular point corresponds to a state of equilibrium ofthe system—the velocity and the force are zero at a singular point. Furtherinvestigation is necessary to establish whether the equilibrium represented by a singularpoint is stable or unstable.The velocity with which a representative point moves along a trajectory is called thephase velocity. The components of phase velocity parallel to the x and y axes are(13.121)and the magnitude of is given by(13.122)ƒv!ƒ = 4vx2+ vy2=Cadxdtb2+ adydtb2v!vx = x#, vy = y#v!x$= -fy = x#f = 0,y = 0f Z 0y = 0f1x, y2dydx=dy/dtdx/dt= -f1x, y2y= f1x, y2, say.dydt= y#= -f1x, y2dxdt= x#= yx$+ f1x, x#2 = 0a 7 11>42A2,a = 11>42A2.a = 011>42A2Ú a Ú 0,1x = Ϯ12A, y = 02,13.7.2Phase VelocityRaoCh13ff.qxd 10.06.08 13:57 Page 930
  • 76. 13.7 GRAPHICAL METHODS 931We can note that if the system has a periodic motion, its trajectory in the phase plane is aclosed curve. This follows from the fact that the representative point, having started itsmotion along a closed trajectory at an arbitrary point (x, y), will return to the same pointafter one period. The time required to go around the closed trajectory (the period of oscil-lation of the system) is finite because the phase velocity is bounded away from zero at allpoints of the trajectory.We shall now consider a method known as the method of isoclines for constructing the tra-jectories of a dynamical system with one degree of freedom. By writing the equations ofmotion of the system as(13.123)where and are nonlinear functions of x and the equation for the integral curvescan be obtained as(13.124)The curve(13.125)for a fixed value of c is called an isocline. An isocline can be defined as the locus of pointsat which the trajectories passing through them have the constant slope c. In the method ofisoclines we fix the slope (dy)/(dx) by giving it a definite number and solve Eq. (13.125)for the trajectory. The curve thus represents an isocline in the phaseplane. We plot several isoclines by giving different values to the slopeLet denote these isoclines in Fig. 13.13(a). Suppose that weh1, h2, Áf = 1dy2/1dx2.c1, c2, Áf1x, y2 - c1 = 0c1f1x, y2 = cdydx=f21x, y2f11x, y2= f1x, y2, say.y = x#,f2f1dydt= f21x, y2dxdt= f11x, y213.7.3Method ofConstructingTrajectoriesR1R2R3R4h1h2h3h4R2ЉR3ЉR4ЉR2ЈR3ЈR4Јyx(a)yxO(x0, y0)(b)FIGURE 13.13 Method of isoclines.RaoCh13ff.qxd 10.06.08 13:57 Page 931
  • 77. are interested in constructing the trajectory passing through the point on the isoclineWe draw two straight line segments from one with a slope meeting at andthe other with a slope meeting at The middle point between and lying onis denoted as Starting at this construction is repeated, and the point is deter-mined on This procedure is continued until the polygonal trajectory with sidesis taken as an approximation to the actual trajectory passingthrough the point Obviously, the larger the number of isoclines, the better is theapproximation obtained by this graphical method. A typical final trajectory looks like theone shown in Fig. 13.13(b).Trajectories Using the Method of IsoclinesConstruct the trajectories of a simple harmonic oscillator by the method of isoclines.Solution: The differential equation defining the trajectories of a simple harmonic oscillator is givenby Eq. (E.3) of Example 13.2. Hence the family of isoclines is given by(E.1)This equation represents a family of straight lines passing through the origin, with c representing theslope of the trajectories on each isocline. The isoclines given by Eq. (E.1) are shown in Fig. 13.14.Once the isoclines are known, the trajectory can be plotted as indicated above.■The trajectory plotted in the phase plane is a plot of as a function of x, and time (t) doesnot appear explicitly in the plot. For the qualitative analysis of the system, the trajectoriesare enough, but in some cases we may need the variation of x with time t. In such cases, it ispossible to obtain the time solution x(t) from the phase plane diagram, although the originalx#c = -vn2xyor y =- vn2cxR1.R1R2, R2R3, R3R4, Áh3.R3R2,R2.h2Rfl2Rœ2Rfl2.h2c2Rœ2,h2c1,R1:h1.R1932 CHAPTER 13 NONLINEAR VIBRATIONE X A M P L E 1 3 . 513.7.4Obtaining TimeSolution fromPhase PlaneTrajectoriesc ϭ 1c ϭ ϱc ϭ Ϫ1c ϭ 0c ϭ Ϫ14c ϭ14xyTrajectoryOFIGURE 13.14 Isoclines of a simpleharmonic oscillator.RaoCh13ff.qxd 10.06.08 13:57 Page 932
  • 78. 13.8 STABILITY OF EQUILIBRIUM STATES 933differential equation cannot be solved for x and as functions of time. The method ofobtaining a time solution is essentially a step-by-step procedure; several different schemesmay be used for this purpose. In this section, we shall present a method based on the rela-tionFor small increments of displacement and time ( and ), the average velocity canbe taken as so that(13.126)In the phase plane trajectory shown in Fig. 13.15, the incremental time needed for the rep-resentative point to traverse the incremental displacement is shown as Ifdenotes the average velocity during we have Similarly,etc. Once are known, the time solution x(t) can beplotted easily, as shown in Fig. 13.15(b). It is evident that for good accuracy, the incre-mental displacements must be chosen small enough that the correspon-ding incremental changes in and t are reasonably small. Note that need not beconstant; it can be changed depending on the nature of the trajectories.13.8 Stability of Equilibrium StatesConsider a single degree of freedom nonlinear vibratory system described by two first-order differential equations(13.127)dydt= f21x, y2dxdt= f11x, y213.8.1StabilityAnalysis¢xx#¢xAB, ¢xBC, Á¢tAB, ¢tBC, Á¢tBC = ¢xBC/x#BC,¢tAB = ¢xAB/x#AB.¢tAB,x#AB¢tAB.¢xAB¢t =¢xx#avx#av = 1¢x2/1¢t2,¢t¢xx#= 1¢x2/1¢t2.x#FIGURE 13.15 Obtaining time solution from a phase plane plot.RaoCh13ff.qxd 10.06.08 13:57 Page 933
  • 79. 934 CHAPTER 13 NONLINEAR VIBRATIONwhere and are nonlinear functions of x and For this system, the slopeof the trajectories in the phase plane is given by(13.128)Let be a singular point or an equilibrium point so that (dy)/(dx) has the form 0/0:(13.129)A study of Eqs. (13.127) in the neighborhood of the singular point provides us withanswers as to the stability of equilibrium. We first note that there is no loss of generality ifwe assume that the singular point is located at the origin (0, 0). This is because the slope(dy)/(dx) of the trajectories does not vary with a translation of the coordinate axes x and yto and(13.130)Thus we assume that is a singular point, so thatIf and are expanded in terms of Taylor’s series about the singular point (0, 0), weobtain(13.131)whereIn the neighborhood of (0, 0), x and y are small; and can be approximated by linearterms only, so that Eqs. (13.131) can be written as(13.132)The solutions of Eq. (13.132) are expected to be geometrically similar to those ofEq. (13.127). We assume the solution of Eq. (13.132) in the form(13.133)exyf = eXYf eltex#y# f = ca11 a12a21 a22d exyff2f1a11 =0f10x`10, 02, a12 =0f10y`10, 02, a21 =0f20x`10, 02, a22 =0f20y`10, 02y#= f21x, y2 = a21x + a22y + Higher order termsx#= f11x, y2 = a11x + a12y + Higher order termsf2f1f110, 02 = f210, 02 = 01x = 0, y = 02dydx=dy¿dx¿y¿ = y - y0x¿ = x - x0y¿:x¿f11x0, y02 = f21x0, y02 = 01x0, y02dydx=y#x# =f21x, y2f11x, y2y = x#= dx/dt.f2f1RaoCh13ff.qxd 10.06.08 13:57 Page 934
  • 80. 13.8 STABILITY OF EQUILIBRIUM STATES 935where X, Y, and are constants. Substitution of Eq. (13.133) into Eq. (13.132) leads to theeigenvalue problem(13.134)The eigenvalues and can be found by solving the characteristic equationas(13.135)where and Ifdenote the eigenvectors corresponding to and respectively, the general solution ofEqs. (13.127) can be expressed as (assuming and ):(13.136)where and are arbitrary constants. We can note the following:If the motion is oscillatory.If the motion is aperiodic.If the system is unstable.If the system is stable.If we use the transformationwhere [T] is the modal matrix and and are the generalized coordinates, Eqs. (13.132)will be uncoupled:(13.137)The solution of Eqs. (13.137) can be expressed as(13.138)b1t2 = el2ta1t2 = el1tea#b# f = cl1 00 l2d eabf ora#= l1ab#= l2bbaexyf = cX1 X2Y1 Y2d eabf = [T]eabfp 6 0,p 7 0,1p2- 4q2 7 0,1p2- 4q2 6 0,C2C1exyf = C1eX1Y1fel1t+ C2eX2Y2fel2tl1 Z l2l1 Z 0, l2 Z 0,l2,l1eX1Y1f and eX2Y2fq = a11a22 - a12a21.p = a11 + a22l1, l2 = 121p Ϯ 4p2- 4q2`a11 - l a12a21 a22 - l` = 0l2l1ca11 - l a12a21 a22 - ld eXYf = e00flRaoCh13ff.qxd 10.06.08 13:57 Page 935
  • 81. 936 CHAPTER 13 NONLINEAR VIBRATIONDepending on the values of and in Eq. (13.135), the singular or equilibrium pointscan be classified as follows [13.20, 13.23].Case (i)— and Are Real and Distinct In this case, Eq. (13.138) gives(13.139)where and are the initial values of and respectively. The type of motion dependson whether and are of the same sign or of opposite signs. If and have the samesign the equilibrium point is called a node. The phase plane diagram for the case(when and are real and negative or ) is shown in Fig. 13.16(a).In this case, Eq. (13.139) shows that all the trajectories tend to the origin as andhence the origin is called a stable node. On the other hand, if thearrow heads change in direction and the origin is called an unstable node (see Fig. 13.16b).If and are real but of opposite signs ( irrespective of the sign of p), one solu-tion tends to the origin while the other tends to infinity. In this case the origin is called asaddle point and it corresponds to unstable equilibrium (see Fig. 13.16d).q 6 0l2l1l2 7 l1 7 0 1p 7 02,t : qp 6 0l2l1l2 6 l1 6 01q 7 02,l2l1l2l1b,ab0a0a1t2 = a0el1tand b1t2 = b0el2t(p2>4q).L2L1l2l113.8.2Classification ofSingular Pointsx xy ϭ xи(a) Stable nodexy(b) Unstable nodey(c) Stable nodexy(d) Saddle pointxy(e) Stable focusxy(f) Unstable focusxy(g) CenterFIGURE 13.16 Types of equilibrium points.RaoCh13ff.qxd 10.06.08 13:57 Page 936
  • 82. 13.9 LIMIT CYCLES 937Case (ii)— and Are Real and Equal In this case Eq. (13.138) gives(13.140)The trajectories will be straight lines passing through the origin and the equilibrium point(origin) will be a stable node if (see Fig. 13.16c) and an unstable node ifCase (iii)— and Are Complex Conjugates Let andwhere and are real. Then Eq. (13.137) gives(13.141)This shows that and must also be complex conjugates. We can rewrite Eq. (13.138) as(13.142)These equations represent logarithmic spirals. In this case the equilibrium point (origin) iscalled a focus or a spiral point. Since the factor in represents a vector of unit mag-nitude rotating with angular velocity in the complex plane, the magnitude of the com-plex vector and hence the stability of motion, is determined by If themotion will be asymptotically stable and the focal point will be stable ( and ).If the focal point will be unstable ( and ). The sign of merely givesthe direction of rotation of the complex vector, counterclockwise for and clockwiseforIf the magnitude of the complex radius vector will be constantand the trajectories reduce to circles with the center as the equilibrium point (origin). Themotion will be periodic and hence will be stable. The equilibrium point in this case iscalled a center or vertex point and the motion is simply stable and not asymptotically sta-ble. The stable focus, unstable focus, and center are shown in Figs. 13.16(e) to (g).13.9 Limit CyclesIn certain vibration problems involving nonlinear damping, the trajectories, starting eithervery close to the origin or far away from the origin, tend to a single closed curve, whichcorresponds to a periodic solution of the system. This means that every solution of the sys-tem tends to a periodic solution as The closed curve to which all the solutionsapproach is called a limit cycle.For illustration, we consider the following equation, known as the van der Pol equation:(13.143)This equation exhibits, mathematically, the essential features of some vibratory systems likecertain electrical feedback circuits controlled by valves where there is a source of power thatincreases with the amplitude of vibration. Van der Pol invented Eq. (13.143) by introducinga type of damping that is negative for small amplitudes but becomes positive for largeamplitudes. In this equation, he assumed the damping term to be a multiple ofin order to make the magnitude of the damping term independent of the sign of x.-11 - x22x#x$- a11 - x22x#+ x = 0, a 7 0t : q.a1t2u1 = 0 1p = 02,u2 6 0.u2 7 0u2q 7 0p 7 0u1 6 0,q 7 0p 6 0u1 6 0,eu1t.a1t2,u2a1t2eiu2ta1t2 = 1a0eu1t2eiu2t, b1t2 = 1b0eu1t2e-iu2tbaa#= 1u1 + iu22a and b#= 1u1 - iu22bu2u1l2 = u1 - iu2,l1 = u1 + iu2(p2<4q).L2L1l1 7 0.l1 6 0a1t2 = a0el1tand b1t2 = b0el1t(p2‫؍‬ 4q).L2L1RaoCh13ff.qxd 10.06.08 13:57 Page 937
  • 83. 938 CHAPTER 13 NONLINEAR VIBRATIONThe qualitative nature of the solution depends upon the value of the parameterAlthough the analytical solution of the equation is not known, it can be represented usingthe method of isoclines, in the phase plane. Equation (13.143) can be rewritten as(13.144)(13.145)Thus the isocline corresponding to a specified value of the slope is given byor(13.146)By drawing the curves represented by Eq. (13.146) for a set of values of c, as shown inFig. 13.17, the trajectories can be sketched in with fair accuracy, as shown in the same fig-ure. The isoclines will be curves since the equation, Eq. (13.146), is nonlinear. An infinityof isoclines pass through the origin, which is a singularity.An interesting property of the solution can be observed from Fig. 13.17. All the tra-jectories, irrespective of the initial conditions, approach asymptotically a particular closedy + cx- a11 - x22 + cd = 0dydx=dy/dtdx/dt=a11 - x22y - xy= cdy/dx = cy#=dydt= a11 - x22y - xy = x#=dxdta.0 00.510.5c ϭ Ϫ2c ϭ ϱϪ20 00.510.5c ϭ Ϫ2 Ϫ2Ϫ2Ϫ3 311Ϫ1Ϫ32xyϭ 1␣0Ϫ1 2Ϫ23FIGURE 13.17 Trajectories and limit cycle forvan der Pol’s equation.RaoCh13ff.qxd 10.06.08 13:57 Page 938
  • 84. 13.10 CHAOS 939curve, known as the limit cycle, which represents a steady-state periodic (but not har-monic) oscillation. This is a phenomenon that can be observed only with certain nonlinearvibration problems and not in any linear problem. If the initial point is inside the limitcycle, the ensuing solution curve spirals outward. On the other hand, if the initial pointfalls outside the limit cycle, the solution curve spirals inward. As stated above, the limitcycle in both the cases is attained finally. An important characteristic of the limit cycle isthat the maximum value of x is always close to 2 irrespective of the value of This resultcan be seen by solving Eq. (13.143) using the perturbation method (see Problem 13.34).13.10 ChaosChaos represents the behavior of a system that is inherently unpredictable. In other words,chaos refers to the dynamic behavior of a system whose response, although described bya deterministic equation, becomes unpredictable because the nonlinearities in the equationenormously amplify the errors in the initial conditions of the system.Attractor. Consider a pendulum whose amplitude of oscillation decreases gradually dueto friction, which means that the system loses part of its energy in each cycle and eventu-ally comes to a rest position. This is indicated in the phase plane shown in Fig. 13.18(a).The rest position is called an attractor. Thus the pendulum has just one attractor. If the pen-dulum is given a push at the end of each swing to replenish the energy lost due to friction,the resulting motion can be indicated as a closed loop in the phase plane (see Fig. 13.18b).In general, for a dynamic system, an attractor is a point (or object) toward which all nearbysolutions move as time progresses.Poincaré Section. A pendulum is said to have two degrees of freedom—namely, x andIn general, a phase space of a system can be defined with as many axes as there aredegrees of freedom. Thus, for a system with three degrees of freedom, the phase spacemight appear (as a spiral converging in the z-direction) as shown in Fig. 13.19(a). Sincethe points are displaced from one another and never coincide in Fig. 13.19(a), the systemdoes not have a periodic motion. The intersection of the phase space with the yz planex#.a.Amplitude (x)(a)Amplitude (x)(b)Velocity (x)· Velocity (x)·FIGURE 13.18 Attractor.RaoCh13ff.qxd 10.06.08 13:57 Page 939
  • 85. 940 CHAPTER 13 NONLINEAR VIBRATIONxzyzy(a)(b)16537247564231FIGURE 13.19 Phase space for a three degreeof freedom system.appears as shown in Fig. 13.19(b). This diagram is known as the Poincaré section orPoincaré map, and it represents points that occur at equal intervals of time,where T denotes the fundamental period of the forcing function. Notethat, if the system is periodic, all the dots would be at the same location in the Poincarésection.Consider the sequence of numbers generated by the following equation:(13.147)For any two initial values of which differ by a small amount, the values of con-verge to 1. For example, with and the sequences of numbers givenby Eq. (13.147) are shown below:1.0001 : 1.0000: 1.0090 : 1.0045 : 1.0023 : 1.0011 : 1.0006 : 1.0003 : 1.0001 :10.0 : 3.1623 : 1.7783 : 1.3335 : 1.1548 : 1.0746 : 1.0366 : 1.0182x1 = 10.2,x1 = 10.0xn+1x1,xn+1 = 2xn; n = 1, 2, Á13.10.1Functions withStable OrbitsnT1n = 1, 2, Á 2,RaoCh13ff.qxd 10.06.08 13:57 Page 940
  • 86. 13.10 CHAOS 941andNote that the influence of a change in the initial value of (by 0.2) is lost very soon andthe pattern converges to 1. Any starting value between 0 and 1 also iterates to 1. Thus thefunctional relation, Eq. (13.147), is said to have a stable orbit atConsider the sequence of numbers generated by the following equation:(13.148)where a is a constant. Equation (13.148) has been used as a simple model for populationgrowth with no predators such as fish and fowl. In such cases, the constant a denotes therate of growth of the population, represents the population in generation n, and the fac-tor acts as a stabilizing factor. It can be seen that Eq. (13.148) has the followinglimitations [13.31]:1. has to lie between 0 and 1. If exceeds 1, the iterative process diverges toimplying that the population becomes extinct.2. attains a maximum value of at This implies that3. If converges to zero.4. Thus, for a nontrivial dynamic behavior (to avoid population from becomingextinct), a has to satisfy the relationThe system will have an equilibrium condition if the birth rate replenishes the loss due todeath or migration. The population can be seen to stabilize or reach a definite limitingvalue (predictable) for some values of a—such as For some other values of a—such as with —the species can be seen to disappear after only two gen-erations, as indicated below:However, for with the population count can be seen to be completelyrandom, as indicated below:This indicates that the system is a chaotic one; even small changes in the deterministicequation, Eq. (13.148), can lead to unpredictable results. Physically, this implies that thesystem has become chaotic with population varying wildly from year to year. In fact, aswill be shown in the following paragraph, the system, Eq. (13.148), has unstable orbits.0.466 : 0.996 : 0.018 : Á0.358 : 0.919 : 0.298 : 0.837 : 0.547 : 0.991 : 0.035 : 0.135 :0.4 : 0.96 : 0.154 : 0.520 : 0.998 : 0.006 : 0.025 : 0.099 :Equation 113.1482 with a = 4.0 and x1 = 0.4:x1 = 0.4,a = 4.00.5 : 1.0 : 0.0 : 0.0 : 0.0 : ÁEquation 113.1482 with a = 4.0 and x1 = 0.5:x1 = 0.5a = 4.0a = 3.0.1 … a … 4.a 6 1, xn+1a 6 4.xn = 12.a4xn+1- q,x1x111 - xn2xnxn+1 = axn11 - xn2; n = 1, 2, Á13.10.2Functions withUnstable Orbitsx = 1.x11.0001 : 1.0000: 1.0091 : 1.0045 : 1.0023 : 1.0011 : 1.0006 : 1.0003 : 1.0001 :10.2 : 3.1937 : 1.7871 : 1.3368 : 1.1562 : 1.0753 : 1.0370 : 1.0183RaoCh13ff.qxd 10.06.08 13:57 Page 941
  • 87. 942 CHAPTER 13 NONLINEAR VIBRATIONBifurcations. Equation (13.148) exhibits a phenomenon known as bifurcation. To seethis, we start with and a few different values of With this, can be seen toconverge to 0.5. When we start with and a few different values of the processconverges to 0.6. If we use and the process converges to a single value,but while getting there, oscillates between two separate values (namely, and). If we use and then the value of oscillates between thetwo values and At that point the system is said to have aperiodicity of 2. The solution, in this case, moves into a two-pronged fork-type of statewith two equilibrium points. If we use and the system will have aperiod 4—that is, the equilibrium state will oscillate between the four valuesand This implies that thestable behavior of each of the previous two solutions has been broken into further bifurca-tion paths. In fact, the system continues to bifurcate with the range of a needed for eachsuccessive birth of bifurcations becoming smaller as a increases, as shown in Fig. 13.20.Figure 13.20 is known as a bifurcation plot or Feigenbaum diagram [13.31, 13.35]. It canbe observed that the system has reached a chaotic state through a series of bifurcations,with the number of values assumed by Eq. (13.148) doubling at each stage.Strange Attractors. For several years, it was believed that the attractors toward whichphysical systems tend are equilibrium or rest points (as in the case of the rest position ofa pendulum), limit cycles, or repeating configurations. However, in recent years it has beenfound that the attractors associated with chaotic systems are more complex than the restpoints and limit cycles. The geometric points in state space to which chaotic trajectoriesare attracted are called strange attractors.x162= 0.8749.x132= 0.3828, x142= 0.5008, x152= 0.8269,x1 = 0.5,a = 3.5x122= 0.8124.x112= 0.4952xn+1x1 = 0.5,a = 3.250.65 Á0.68 Áx1 = 0.1,a = 3.0x1,a = 2.5xn+1x1.a = 20.02.5 3.0 3.5Chaoticregion4.ϩ1ax(6)x(2)x(1)x(3)x(4)x(5)⌬a2⌬ai ϭ Range of a for period i (i ϭ 2, 4, 8, . . . )⌬a4⌬a8FIGURE 13.20 Bifurcation plot.RaoCh13ff.qxd 10.06.08 13:57 Page 942
  • 88. 13.10 CHAOS 943Consider a single degree of freedom system with a nonlinear spring and a harmonic forc-ing function. The equation of motion (Duffing’s equation) can be expressed as(13.149)First, we consider the free, undamped vibration of the system with(13.150)The static equilibrium positions of this system can be found by setting the spring forceequal to zero, as It can be easily verified that the equilibrium solutionis unstable (saddle point) with respect to infinitesimal disturbances, while the equi-librium solutions and are stable (centers) with respect to infinitesimal disturbances.The stability of the system about the three equilibrium positions can be seen more clearlyfrom the graph of its potential energy. To find the potential energy of the system, we mul-tiply Eq. (13.150) by and integrate the resulting equation to obtain(13.151)where C is a constant. The first term on the left-hand side of Eq. (13.151) represents thekinetic energy and the second and third terms denote the potential energy (P) of the system.Equation (13.151) indicates that the sum of the kinetic and potential energies is a constant(conservative system).A plot of the potential energy, is shown in Fig. 13.21.Next, we consider the free, damped vibration of the system. The governing equation is(13.152)Let the boundary conditions be given by(13.153)It can be expected from Fig. 13.21 that the static equilibria, and areunstable with respect to finite disturbances. Physically, when a finite disturbance is givenx = -1,x = +1x1t = 02 = x0, x#1t = 02 = x#0x$+ mx#- 0.5 x + 0.5 x3= 0P = 18 x4- 14 x2,121x#22+18x4-14x2= Cx#+1-1x = 0x = 0, +1, -1.x$- 0.5 x + 0.5 x3= 0a = b = 0.5:x$+ mx#- ax + bx3= F0 cos v t13.10.3ChaoticBehavior ofDuffing’sEquationWithout theForcing TermϪ1 ϩ10xPStable point(center)Unstable(saddle)pointStable point(center)FIGURE 13.21 Plot of potential energy.RaoCh13ff.qxd 10.06.08 13:57 Page 943
  • 89. 944 CHAPTER 13 NONLINEAR VIBRATIONabout one static equilibrium point, the system could be driven to the other static equilib-rium point. In fact, the steady-state solution can be shown to be extremely sensitive to theinitial conditions, exhibiting a form of chaos. It can be verified easily [13.32] that forand the steady-state solution is Thephase plane trajectory for is shown in Fig. 13.22(a). Note that for allvalues of t. For and the steady-state solution isFigure 13.22(b) shows the phase plane trajectory for indi-cating a single crossing of the axis. For the steady-statesolution is The phase plane trajectory for is shown inFig. 13.22(c), which indicates two crossings of the axis.In fact, by giving a series of values to we can construct several phase plane tra-jectories, from which a composite plot, known as the shell plot, can be constructed asshown in Fig. 13.23 [13.33, 13.34]. Here also, the steady-state solution can be seen to beor depending on the initial conditions, and It can be seen that the var-ious regions are identified by the numbers First, consider the region labeled“0” with If the initial conditions fall in this region, the solution spirals intoas and the solution crosses the axis zero times (similar toFig. 13.22a). Next, consider the region labeled “1.” If the initial conditions fall in thisregion, the solution moves clockwise, crosses the axis once, and settles intoas (similar to Fig. 13.22b). Next, consider the region labeled “2.” If thesystem starts with the initial conditions falling in this region, the phase plane moves clock-wise, crosses the axis continues to move clockwise, crosses the axis again,moves into the region “0” for and spirals into as (similar to Fig.13.22c). This pattern continues with other regions as well, with the labeled region numberindicating the number of crossings of the axis by the phase plane trajectory.Figure 13.23 indicates that if there is sufficient uncertainty in the initial conditionsand the final state of the system, or is unpredictable or uncertain. If damp-ing is reduced further, the width of each region in Fig. 13.23 (except the regions labeled“0”) becomes even smaller and vanishes as Thus the final steady state of the sys-tem is unpredictable as for any finite uncertainty in or both. This means thatthe system exhibits chaos.Consider Duffing’s equation with and including the forcing term.Within the forcing term, small variations in the frequency or the amplitude canlead to chaos, as indicated below.When Is Changed. For a fixed value of the phase plane response of Eq. (13.149)can be periodic or chaotic, depending on the value of For example, Figs. 13.24 and13.25 indicate the two situations that are described by the equations(13.154)(13.155)where has been assumed. Figure 13.24 has been plotted using an approximateanalysis, known as the harmonic balance method. On the other hand, Fig. 13.25 representsF0 = 0.3x$+ 0.2 x#- x + x3= 0.3 cos 1.29 t 1chaotic, Fig. 13.252x$+ 0.2 x#- x + x3= 0.3 cos 1.4 t 1periodic, Fig. 13.242v.F0,V1F021v2a = b = 1m = 0.2,13.10.4ChaoticBehavior ofDuffing’sEquation withthe ForcingTermx0, x#0,m : 0m : 0.-1,x = +1x#0,x0x = 0t : qx = +1x 7 0,x = 0x = 0,t : qx = -1x = 0x = 0t : qx = +1x0 Ú 0.0, 1, 2, 3, Áx#0.x0-1x = +1x#0,x = 0x#0 = 0.56x1t : q2 = +1.0.5572 6 x#0 6 0.5952,x = 0x#0 = 0.54,x1t : q2 = -1.0.521799 6 x#0 6 0.5572,x0 = 1x 7 0x#0 = 0.52x1t : q2 = +1.0 6 x#0 6 0.521799,x0 = 1RaoCh13ff.qxd 10.06.08 13:57 Page 944
  • 90. 13.10 CHAOS 945(a)0.24 0.48 0.72 0.96 1.20 1.44Displacement (x)0Ϫ0.60Ϫ0.48Ϫ0.36Ϫ0.24Ϫ0.Ϫ0.96 Ϫ0.48 0 0.48 0.96 1.44Displacement (x)Ϫ1.44Ϫ0.60Ϫ0.48Ϫ0.36Ϫ0.24Ϫ0. ϭ 1.0, x0 ϭ 0.52.x0 ϭ 1.0, x0 ϭ 0.54.(c)Displacement (x)Ϫ1.5 Ϫ1.2 Ϫ0.9 Ϫ0.6 Ϫ0.3 0.0 0.3 0.6 0.9 1.2 1.5Ϫ0.60Ϫ0.48Ϫ0.36Ϫ0.24Ϫ0. ϭ 1.0, x0 ϭ 0.56.Velocity (x).Velocity (x).Velocity (x).FIGURE 13.22 Phase-plane trajectories for different initial veloci-ties. (From [13.32]; reprinted with permission of Society of Industrial and AppliedMathematics and E. H. Dowell and C. Pierre.)RaoCh13ff.qxd 10.06.08 13:57 Page 945
  • 91. 946 CHAPTER 13 NONLINEAR VIBRATIONϪ1.8Ϫ0.96Ϫ0.72Ϫ0.48Ϫ0.2400.240.480.720.96Ϫ1.2 Ϫ0.6 0DisplacementVelocity0.6 1.2 1.8ϭ 0.0168␮FIGURE 13.23 Shell plot. (From [13.33]; reprinted with per-mission of American Society of Mechanical Engineers.)a Poincaré map, which indicates points that occur at equal intervals of timewhere is the fundamental period of excitation,When Is Changed. Chaos can also be observed when the amplitude of the forcechanges. To illustrate, we consider the following equation [13.33]:(13.156)For definiteness, we assume the initial conditions as and When is suf-ficiently small, the response of the system will be a simple harmonic motion (that is, thephase plane will be an ellipse) about its static equilibrium position, When isincreased, additional harmonics beyond the fundamental are detected and the phase planewill be distorted from a simple ellipse, as shown in Fig. 13.26(a) for Note thatF0 = 0.177.F0x = +1.F0x#0 = 0.x0 = 1x$+ 0.168 x#- 0.5 x + 0.5 x3= F0 sin vt K F0 sin tF0T0 = 2pv = 2p1.29.T0T0, 2T0, 3T0, Á0.40 0.60 0.80 1.00 1.20Ϫ0.80Ϫ0.60Ϫ0.40Ϫ0.200.800.600.400.200.00x1.400.50 0.70 0.90 1.10 1.30x·FIGURE 13.24 Phase plane of Eq. (13.154).(From [13.34]; reprinted with permission of AcademicPress.)RaoCh13ff.qxd 10.06.08 13:57 Page 946
  • 92. 13.10 CHAOS 947Ϫ2.00 Ϫ1.20 Ϫ0.40 0.40 1.20Ϫ1.60 Ϫ0.80 0.00 0.80 1.60Ϫ0.80Ϫ0.60Ϫ0.40Ϫ0.200.800.600.400.200.00xx·FIGURE 13.25 Poincaré map ofEq. (13.155). (From [13.34]; reprinted with permission ofAcademic Press.)Ϫ0.4Ϫ0.8Ϫ1.2Ϫ1.6Ϫ0.6Ϫ0.4Ϫ0.200.2Velocity(b) 2 Period0.40.60.8Ϫ0.8Displacement ϭ 1, F0 ϭ 0.178, ϭ 0.168␻ ␮Ϫ0.4Ϫ0.8Ϫ1.2Ϫ1.6Ϫ0.6Ϫ0.4Ϫ0.200.2Velocity(d)Ϫ0.8DisplacementTrajectory of chaos F0 ϭ 0.21, ϭ 0.168␮Ϫ0.4Ϫ0.8Ϫ1.2Ϫ1.6Ϫ0.6Ϫ0.4Ϫ0.200.2Velocity(c) 4 Period0.40.60.8Ϫ0.8Displacement ϭ 1, F0 ϭ 0.197, ϭ 0.168␻ ␮Ϫ0.4Ϫ0.8Ϫ1.2Ϫ1.6Ϫ0.6Ϫ0.4Ϫ0.200.2Velocity(a) 1 Period0.40.60.8Ϫ0.8Displacement ϭ 1, F0 ϭ 0.177, ϭ 0.168␻ ␮FIGURE 13.26 Distortion of phase plane. (From [13.33]; reprinted with permission of AmericanSociety of Mechanical Engineers.)RaoCh13ff.qxd 10.06.08 13:57 Page 947
  • 93. 948 CHAPTER 13 NONLINEAR VIBRATIONthe boundary of the region labeled “0” in Fig. 13.23 is also shown in Fig. 13.26(a) for acomparison. The response is called a 1 period motion for implying thatthe response oscillates through one period while the force oscillates through one period.For the phase plane is shown in Fig. 13.26(b), which indicates that theresponse is a 2 period motion. Thus, for the response to oscillate through one period, theforce must oscillate through two periods. This change from a 1 to a 2 period response aschanges from 0.177 to 0.178 is called a bifurcation. When the responsewill be a 4 period motion (see Fig. 13.26c). As is increased, period motionsoccur and finally, for chaos can be observed with no apparent periodicity, asindicated in Fig. 13.26(d).13.11 Numerical MethodsMost of the numerical methods described in the earlier chapters can be used for finding theresponse of nonlinear systems. The Runge-Kutta method described in Section 11.4 isdirectly applicable for nonlinear systems and is illustrated in Section 13.12. The centraldifference, Houbolt, Wilson, and Newmark methods considered in Chapter 11 can also beused for solving nonlinear multidegree of freedom vibration problems with slight modifi-cation. Let a multidegree of freedom system be governed by the equation(13.157)where the internal set of forces opposing the displacements are assumed to be nonlinearfunctions of For the linear case, In order to find the displacement vectorthat satisfies the nonlinear equilibrium in Eq. (13.157), it is necessary to perform an equi-librium iteration sequence in each time step. In implicit methods (Houbolt, Wilson, andNewmark methods), the equilibrium conditions are considered at the same time for whichsolution is sought. If the solution is known for time and we wish to find the solution fortime then the following equilibrium equations are considered(13.158)where and is computed as(13.159)where is the linearized or tangent stiffness matrix computed at time Substitution ofEq. (13.159) in Eq. (13.158) gives(13.160)Since the right-hand side of Eq. (13.160) is completely known, it can be solved forusing any of the implicit methods directly. The found is only an approximate vectordue to the linearization process used in Eq. (13.159). To improve the accuracy of the solu-tion and to avoid the development of numerical instabilities, an iterative process has to beused within the current time step [13.21].x!i+1x!i+1$[m] x:i+1 + [c] x:#i+1 + [ki]x!i+1 = F!Ni+1 = F!i+1 - P!i + [ki]x!iti.[ki]P!i+1 = P!i + [ki]¢x!= P!i + [ki]1x!i+1 - x!i2P!i+1F!i+1 = F!1t = ti+12$[m] x:i+1 + [c] x:#i+1 + P!i+1 = F!i+1ti+1,tix!P!= [k]x!.x!.P![m] x:$1t2 + [c] x:#1t2 + P!1x!(t2) = F!1t2F0 Ú 0.205,8, 16, ÁF0F0 = 0.197,F0F0 = 0.178,0 … F0 … 0.177,RaoCh13ff.qxd 10.06.08 13:57 Page 948
  • 94. 13.12 EXAMPLES USING MATLAB 94913.12 Examples Using MATLABSolution of the Pendulum EquationUsing MATLAB, find the solution of the following forms of the pendulum equation with(a) (E.1)(b) (E.2)(c) (E.3)Use the following initial conditions:(E.4)(E.5)(E.6)Solution: Using and each of the Eqs. (E.1) to (E.3) can be rewritten as a system oftwo first-order differential equations as follows:(a)(E.7)(b)(E.8)(c)(E.9)Equations (E.7) to (E.9) are solved using the MATLAB program ode23 for each of the initial con-ditions given by Eqs. (E.4) to (E.6). The solutions given by Eqs. (E.7), (E.8), and (E.9) for aspecific initial conditions, are plotted in the same graph.% Ex13_6.m% This program will use the function dfunc1_a.m, dfunc1_b.m anddfun1_c.m% they should be in the same foldertspan = [0: 1: 250];u1t2x#2 = - v02sin x1 1Nonlinear equation2x#1 = x2x#2 = - v02x1 +16v02x131Nonlinear equation2x#1 = x2x#2 = - v02x1 1Linear equation2x#1 = x2x2 = u#,x1 = u1iii2 u102 =p2, u#102 = 01ii2 u102 =p4, u#102 = 01i2 u102 = 0.1, u#102 = 0u$+ v02sin u = 0u$+ v02u - 16 v02u3= 0u$+ v02u = 0v0 =Agl= 0.09.E X A M P L E 1 3 . 6RaoCh13ff.qxd 10.06.08 13:57 Page 949
  • 95. 950 CHAPTER 13 NONLINEAR VIBRATIONx0 = [0.1; 0.0];x0_1 = [0.7854; 0.0];x0_2 = [1.5708; 0.0];[t, xa] = ode23 (dfunc1_a, tspan, x0);[t, xb] = ode23 (dfunc1_b, tspan, x0);[t, xc] = ode23 (dfunc1_c, tspan, x0);[t, xa1] = ode23 (dfunc1_a, tspan, x0_1);[t, xb1] = ode23 (dfunc1_b, tspan, x0_1);[t, xc1] = ode23 (dfunc1_c, tspan, x0_1);[t, xa2] = ode23 (dfunc1_a, tspan, x0_2);[t, xb2] = ode23 (dfunc1_b, tspan, x0_2);[t, xc2] = ode23 (dfunc1_c, tspan, x0_2);plot (t, xa(:, 1));ylabel (Theta(t));xlabel (t);ylabel (i.c. = [0.1; 0.0]);title. . .(Function a: solid line, Function b: dashed line, Function c: dottedline);hold on;plot (t, xb(:, 1), --);hold on;plot (t, xc(:, 1), :);pause;hold off;plot (t, xa1(:, 1));ylabel (Theta(t));xlabel (t);ylabel (i.c. = [0.7854; 0.0]);title. . .(Function a: solid line, Function b: dashed line, Function c: dottedline);hold on;plot (t, xb1(:, 1), --);hold on;plot (t, xc1(:, 1), :);pause;hold off;plot(t,xa2(:,1));hold on;ylabel(Theta(t));xlabel(t);ylabel(i.c. = [1.5708; 0.0])title. . .(Function a: solid line, Function b: dashed line, Function c: dottedline);plot(t, xb2(:,1),--);hold on;plot(t,xc2(:,1),:);% dfunc1_a.mfunction f = dfunc1_a(t,x);f = zeros(2,1);f(1) = x(2);f(2) = Ϫ0.0081 * x(1);% dfunc1_b.mfunction f = dfunc1_b(t,x);f = zeros(2,1);f(1) = x(2);f(2) = 0.0081 * ((x(1))^3) / 6.0 Ϫ 0.0081 * x(1);% dfunc1_c.mfunction f = dfunc1_c(t,x);f = zeros(2,1);f(1) = x(2);f(2) = Ϫ0.0081 * sin(x(1));RaoCh13ff.qxd 10.06.08 13:57 Page 950
  • 96. 13.12 EXAMPLES USING MATLAB 951RaoCh13ff.qxd 10.06.08 13:57 Page 951
  • 97. 952 CHAPTER 13 NONLINEAR VIBRATION■Solution of Nonlinearly Damped SystemUsing MATLAB, find the solution of a single degree of freedom system with velocity squared damping.Governing equation:(E.1)Data: andAlso find the solution of the system using the equivalent viscous damping constant(E.2)where is given by Eq. (E.4) of Example 3.7 as(E.3)ceq =8 cv X3 pceqm x$+ ceq x#+ k x = F0 sin v t1ceq220, x102 = 0.5, x#102 = 1.0m = 10, c = 0.01, k = 4000, F0 = 200, v = 10m x$+ c1x#22sign1x#2 + k x = F0 sin vtE X A M P L E 1 3 . 720 50i.c.ϭ[1.5708;0.0]100tFunction a: solid line, Function b: dashed line, Function c: dotted line150 200 2501.510.50Ϫ0.5Ϫ1Ϫ1.5Ϫ2RaoCh13ff.qxd 10.06.08 13:57 Page 952
  • 98. 13.12 EXAMPLES USING MATLAB 953Solution: By introducing and Eqs. (E.1) and (E.2) are written as systems of twofirst-order differential equations as(E.4)(E.5)and X, in Eq. (E.3), is taken as the steady-state or static deflection of the system as TheMATLAB solutions given by Eqs. (E.4) and (E.5) are plotted in the same graph for a specificvalue of% Ex13_7.m% This program will use the function dfunc3_a.m, dfunc3_b.m% dfunc3_a1.m, dfunc3_b1.m, they should be in the same foldertspan = [0: 0.005: 10];x0 = [0.5; 1.0];[t,xa] = ode23 (dfunc3_a, tspan, x0);[t,xb] = ode23 (dfunc3_b, tspan, x0);[t,xa1] = ode23 (dfunc3_a1, tspan, x0);[t,xb1] = ode23 (dfunc3_b1, tspan, x0);subplot (211);plot (t,xa (:,1));title (Theta(t): function a (Solid line), function b (Dashedline));ylabel (w = 10 );hold on;plot (t,xb(:,1), --);subplot (212);plot (t,xa1(:,1));ylabel (w = 20 );hold on;plot (t,xb1 (:,1), --);xlabel (t);% dfunc3_a.mfunction f = dfunc3_a (t,x);f0 = 200;m = 10;a = 0.01;k = 4000;w = 10;f = zeros (2,1);f(1) = x(2);f(2) = f0* sin (w*t) /m Ϫ a* x(2)^2 * sign(x(2)) /m Ϫ k*x(1) /m;% dfunc3_a1.mfunction f = dfunc3_a1 (t,x);f0 = 200;m = 10;a = 0.01;k = 4000;w = 20;f = zeros (2,1);v.X =F0k .x#2 =F0msin vt -ceqmx2 -kmx1 1Linear equation21b2 x#1 = x2x#2 =F0msin v t -cmx22sign 1x22 -kmx1 1Nonlinear equation21a2 x#1 = x2x2 = x#,x1 = xRaoCh13ff.qxd 10.06.08 13:57 Page 953
  • 99. 954 CHAPTER 13 NONLINEAR VIBRATIONf(1) = x(2);f(2) = f0* sin (w*t) /m Ϫ a* x(2)^2 * sign (x(2)) /m Ϫ k*x(1) /m;% dfunc3_b.mfunction f = dfunc3_b (t,x);f0 = 200;m = 10;a = 0.01;k = 4000;ceq = sqrt (8*a*f0 / (3*pi));w = 10;f = zeros (2,1);f(1) = x(2);f(2) = f0* sin (w*t) /m Ϫ ceq * x(2) /m Ϫ k*x(1) /m;% dfunc3_b1.mfunction f = dfunc3_b1 (t,x);f0 = 200;m = 10;a = 0.01;k = 4000;ceq = sqrt (8*a*f0 / (3*pi));w = 20;f = zeros (2,1);f(1) = x(2);f(2) = f0* sin (w*t) /m Ϫ ceq * x(2) /m Ϫ k*x(1) /m;■RaoCh13ff.qxd 10.06.08 13:57 Page 954
  • 100. 13.12 EXAMPLES USING MATLAB 955Solution of Nonlinear System Under Pulse LoadingUsing MATLAB, find the solution of a nonlinear single degree of freedom system governed by theequation(E.1)where F(t) is a rectangular pulse load of magnitude applied over Assume the fol-lowing data: Solve Eq. (E.1)for two cases: one with and the other withSolution: Using and Eq. (E.1) is rewritten, as a set of two first-order differentialequations as(E.2)The responses, x(t), found with (linear system) and (nonlinear system) are plottedin the same graph.k2 = 500k2 = 0x#2 =F1t2m-k1mx1 -k2mx13x#1 = x2x2 = x#,x1 = xk2 = 500.k2 = 0m = 10, k1 = 4000, F0 = 1000, t0 = 1.0, x102 = 0.05, x#102 = 5.0 … t … t0.F0m x$+ k1 x + k2 x3= f1t2E X A M P L E 1 3 . 80.50Ϫ0.6Ϫ0.7Ϫ0.5Ϫ0.4Ϫ0.3Ϫ0.2Ϫ0. 1.5 2Solid line: k2 ϭ 500Dashed line: k2 ϭ 02.5tx(t)3 3.5 4 4.5 5RaoCh13ff.qxd 10.06.08 13:57 Page 955
  • 101. 956 CHAPTER 13 NONLINEAR VIBRATION% Ex13_8.m% This program will use the function dfunc13_8_1.m and dfunc13_8_2.m% they should be in the same foldertspan = [0: 0.01: 5];x0 = [0.05; 5];[t,x] = ode23(dfunc13_8_1, tspan, x0);plot(t,x(:,1));xlabel(t);ylabel(x(t));hold on;[t,x] = ode23(dfunc13_8_2, tspan, x0);plot(t,x(:,1),--);gtext(Solid line: k_2 = 500);gtext(Dashed line: k_2 = 0)% dfunc13_8_1.mfunction f = dfunc13_8_1(t,x)f = zeros(2,1);m = 10;k1 = 4000;k2 = 500;F0 = 1000;F = F0 * (stepfun(t, 0) ؊ stepfun(t, 1));f(1) = x(2);f(2) = ؊F/m ؊ k1 * x(1)/m ؊ k2 * (x(1))^3/m;% dfunc13_8_2.mfunction f = dfunc13_8_2(t,x)f = zeros(2,1);m = 10;k1 = 4000;k2 = 0;F0 = 1000;F = F0 * (stepfun(t, 0) ؊ stepfun(t, 1));f(1) = x(2);f(2) = ؊F/m ؊ k1 * x(1)/m ؊ k2 * (x(1))^3/m;■Solution of Nonlinear Differential EquationUsing the fourth-order Runge-Kutta method, develop a general MATLAB program calledProgram18.m to find the solution of a single degree of freedom equation of the form(E.1)Use the program to solve Eq. (E.1) for the following data:Solution: Equation (E.1) is rewritten as(E.2)x#2 = f21x1, x22 = -cmx2 -kmx1 -k*mx13x#1 = f11x1, x22 = x2x102 = 7.5, x#102 = 0.k* = 0.5,m = 0.01, c = 0.1, k = 2.0,m x$+ c x#+ k x + k* x3= 0E X A M P L E 1 3 . 9RaoCh13ff.qxd 10.06.08 13:57 Page 956
  • 102. 13.12 EXAMPLES USING MATLAB 957Program18.m is developed to accept the values of m, c, k, and as YM, YC, YK, and YKS, respec-tively. The time step and the number of time steps (NSTEP) are specified as 0.0025 and 400,respectively. A subprogram, called fun; is to be given to define and The pro-gram gives the values of and NSTEP as output. The program also plotsandSolution of nonlinear vibration problemby fourth order Runge-Kutta methodData:ym = 1.000000eϪ002yc = 1.000000eϪ001yk = 2.00000000e+000yks = 5.00000000eϪ001Resultsi time(i) x(i,1) x(i,2)1 2.500000eϪ003 7.430295e+000 Ϫ5.528573e+0016 1.500000eϪ002 5.405670e+000 Ϫ2.363166e+00211 2.750000eϪ002 2.226943e+000 Ϫ2.554475e+00216 4.000000eϪ002 Ϫ8.046611eϪ001 Ϫ2.280328e+00221 5.250000eϪ002 Ϫ3.430513e+000 Ϫ1.877713e+00226 6.500000eϪ002 Ϫ5.296623e+000 Ϫ1.002752e+002...x#1t2 = x21t2.x1t2 = x11t2x#1ti2, i = 1, 2, Á ,ti, x1ti2f21x1, x22.f11x1, x221¢t2k*X(2)0.1 0.2 0.3 0.4 0.5Time0.6 0.7 0.8 0.9 10Ϫ200Ϫ1000100200Ϫ300X(1)0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10Ϫ50510Ϫ10RaoCh13ff.qxd 10.06.08 13:57 Page 957
  • 103. 958 CHAPTER 13 NONLINEAR VIBRATION371 9.275000eϪ001 1.219287eϪ001 7.673075eϪ002376 9.400000eϪ001 1.209954eϪ001 Ϫ2.194914eϪ001381 9.525000eϪ001 1.166138eϪ001 Ϫ4.744062eϪ001386 9.650000eϪ001 1.093188eϪ001 Ϫ6.853283eϪ001391 9.775000eϪ001 9.966846eϪ002 Ϫ8.512093eϪ001396 9.900000eϪ001 8.822462eϪ002 Ϫ9.724752eϪ001■13.13 C++ ProgramAn interactive C++ program called Program18.cpp is given for finding the solution of asingle degree of freedom system with a nonlinear spring using the fourth-order Runge-Kutta method. The input and output of the program are similar to those of the MATLABprogram, Program18.m, given in Example 13.9.Solution of Nonlinear Spring-Mass-Damper SystemUse Program18.cpp to solve the problem described in Example 13.9.Solution: The input data are to be entered interactively. The input and output of the program aregiven below.SOLUTION OF NONLINEAR VIBRATION PROBLEMBY FOURTH ORDER RUNGE-KUTTA METHODDATA:YM = 0.010000YC = 0.100000YK = 2.000000YKS = 0.500000RESULTS:I TIME(I) X(I,1) X(I,2)1 0.00250000 7.43029541 Ϫ55.285731342 0.00500000 7.22711455 Ϫ106.350200863 0.00750000 6.90397627 Ϫ150.943618274 0.01000000 6.47900312 Ϫ187.661521315 0.01250000 5.97267903 Ϫ216.01444338...395 0.98750000 0.09063023 Ϫ0.95172604396 0.99000000 0.08822462 Ϫ0.97247522397 0.99250000 0.08576929 Ϫ0.99150470398 0.99500000 0.08326851 Ϫ1.00883370399 0.99750000 0.08072652 Ϫ1.02448309400 1.00000000 0.07814748 Ϫ1.03847532■EXAMPLE 13.10RaoCh13ff.qxd 10.06.08 13:57 Page 958
  • 104. REFERENCES 95913.14 Fortran ProgramA Fortran program, called PROGRAM18.F, is given for the numerical solution of the non-linear free vibration problem of a single degree of freedom system using the fourth-orderRunge-Kutta method. The input and output of the program are similar to those of theMATLAB program, Program18.m, given in Example 13.9.Solution of Nonlinear Vibration ProblemFind the solution of the problem described in Example 13.9 using PROGRAM18.F.Solution: The output of the program is given below.SOLUTION OF NONLINEAR VIBRATION PROBLEMBY FOURTH ORDER RUNGE-KUTTA METHODDATA:YM = 0.100000E؊01YC = 0.100000E؉00YK = 0.200000E؉01YKS = 0.500000E؉00RESULTS:I TIME(I) X(I,1) X(I,2)1 0.002500 0.743030E+01 Ϫ0.552857E+022 0.005000 0.722711E+01 Ϫ0.106350E+033 0.007500 0.690398E+01 Ϫ0.150944E+034 0.010000 0.647900E+01 Ϫ0.187662E+035 0.012500 0.597268E+01 Ϫ0.216014E+03...395 0.987511 0.906302EϪ01 Ϫ0.951725E+00396 0.990011 0.882246EϪ01 Ϫ0.972474E+00397 0.992511 0.857693EϪ01 Ϫ0.991504E+00398 0.995011 0.832685EϪ01 Ϫ0.100883E+01399 0.997511 0.807265EϪ01 Ϫ0.102448E+01400 1.000011 0.781475EϪ01 Ϫ0.103847E+01■REFERENCES13.1 C. Hayashi, Nonlinear Oscillations in Physical Systems, McGraw-Hill, New York, 1964.13.2 A. A. Andronow and C. E. Chaikin, Theory of Oscillations (English language edition),Princeton University Press, Princeton, N.J., 1949.13.3 N. V. Butenin, Elements of the Theory of Nonlinear Oscillations, Blaisdell Publishing, NewYork, 1965.13.4 A. Blaquiere, Nonlinear System Analysis, Academic Press, New York, 1966.13.5 Y. H. Ku, Analysis and Control of Nonlinear Systems, Ronald Press, New York, 1958.13.6 J. N. J. Cunningham, Introduction to Nonlinear Analysis, McGraw-Hill, New York, 1958.13.7 J. J. Stoker, Nonlinear Vibrations in Mechanical and Electrical Systems, IntersciencePublishers, New York, 1950.EXAMPLE 13.11RaoCh13ff.qxd 10.06.08 13:57 Page 959
  • 105. 960 CHAPTER 13 NONLINEAR VIBRATION13.8 J. P. Den Hartog, Mechanical Vibrations (4th ed.), McGraw-Hill, New York, 1956.13.9 N. Minorsky, Nonlinear Oscillations, D. Van Nostrand, Princeton, N.J., 1962.13.10 R. E. Mickens, “Perturbation solution of a highly nonlinear oscillation equation,” Journal ofSound and Vibration, Vol. 68, 1980, pp. 153–155.13.11 B. V. Dasarathy and P. Srinivasan, “Study of a class of nonlinear systems reducible to equiv-alent linear systems,” Journal of Sound and Vibration, Vol. 7, 1968, pp. 27–30.13.12 G. L. Anderson, “A modified perturbation method for treating nonlinear oscillation problems,”Journal of Sound and Vibration, Vol. 38, 1975, pp. 451–464.13.13 B. L. Ly, “A note on the free vibration of a nonlinear oscillator,” Journal of Sound andVibration, Vol. 68, 1980, pp. 307–309.13.14 V. A. Bapat and P. Srinivasan, “Free vibrations of nonlinear cubic spring mass systems in thepresence of Coulomb damping,” Journal of Sound and Vibration, Vol. 11, 1970, pp. 121–137.13.15 H. R. Srirangarajan, P. Srinivasan, and B. V. Dasarathy, “Analysis of two degrees of freedomsystems through weighted mean square linearization approach,” Journal of Sound andVibration, Vol. 36, 1974, pp. 119–131.13.16 S. R. Woodall, “On the large amplitude oscillations of a thin elastic beam,” InternationalJournal of Nonlinear Mechanics, Vol. 1, 1966, pp. 217–238.13.17 D. A. Evenson, “Nonlinear vibrations of beams with various boundary conditions,” AIAAJournal, Vol. 6, 1968, pp. 370–372.13.18 M. E. Beshai and M. A. Dokainish, “The transient response of a forced nonlinear system,”Journal of Sound and Vibration, Vol. 41, 1975, pp. 53–62.13.19 V. A. Bapat and P. Srinivasan, “Response of undamped nonlinear spring mass systems sub-jected to constant force excitation,” Journal of Sound and Vibration, Vol. 9, 1969, Part I:pp. 53–58 and Part II: pp. 438–446.13.20 W. E. Boyce and R. C. DiPrima, Elementary Differential Equations and Boundary ValueProblems (4th ed.), Wiley, New York, 1986.13.21 D. R. J. Owen, “Implicit finite element methods for the dynamic transient analysis of solidswith particular reference to nonlinear situations,” in Advanced Structural Dynamics, J. Donéa(ed.), Applied Science Publishers, London, 1980, pp. 123–152.13.22 B. van der Pol, “Relaxation oscillations,” Philosophical Magazine, Vol. 2, pp. 978–992, 1926.13.23 L. A. Pipes and L. R. Harvill, Applied Mathematics for Engineers and Physicists (3rd ed.),McGraw-Hill, New York, 1970.13.24 N. N. Bogoliubov and Y. A. Mitropolsky, Asymptotic Methods in the Theory of NonlinearOscillations, Hindustan Publishing, Delhi, 1961.13.25 A. H. Nayfeh and D. T. Mook, Nonlinear Oscillations, Wiley, New York, 1979.13.26 G. Duffing, “Erzwungene Schwingungen bei veranderlicher Eigenfrequenz und ihre technis-che Bedeutung,” Ph.D. thesis (Sammlung Vieweg, Braunschweig, 1918).13.27 C. A. Ludeke, “An experimental investigation of forced vibrations in a mechanical systemhaving a nonlinear restoring force,” Journal of Applied Physics, Vol. 17, pp. 603–609, 1946.13.28 D. W. Jordan and P. Smith, Nonlinear Ordinary Differential Equations (2nd ed.), ClarendonPress, Oxford, 1987.13.29 R. E. Mickens, An Introduction to Nonlinear Oscillations, Cambridge University Press,Cambridge, 1981.RaoCh13ff.qxd 10.06.08 13:57 Page 960
  • 106. REVIEW QUESTIONS 96113.30 S. H. Crandall, “Nonlinearities in structural dynamics,” The Shock and Vibration Digest, Vol. 6,No. 8, August 1974, pp. 2–14.13.31 R. M. May, “Simple mathematical models with very complicated dynamics,” Nature, Vol. 261,June 1976, pp. 459–467.13.32 E. H. Dowell and C. Pierre, “Chaotic oscillations in mechanical systems,” in Chaos inNonlinear Dynamical Systems, J. Chandra (ed.), SIAM, Philadelphia, 1984, pp. 176–191.13.33 E. H. Dowell and C. Pezeshki, “On the understanding of chaos in Duffing’s equation includ-ing a comparison with experiment,” ASME Journal of Applied Mechanics, Vol. 53, March1986, pp. 5–9.13.34 B. H. Tongue, “Existence of chaos on a one-degree-of-freedom system,” Journal of Sound andVibration, Vol. 110, No. 1, October, 1986, pp. 69–78.13.35 M. Cartmell, Introduction to Linear, Parametric, and Nonlinear Vibrations, Chapman andHall, London, 1990.REVIEW QUESTIONS13.1 Give brief answers to the following:1. How do you recognize a nonlinear vibration problem?2. What are the various sources of nonlinearity in a vibration problem?3. What is the source of nonlinearity in Duffing’s equation?4. How is the frequency of the solution of Duffing’s equation affected by the nature of thespring?5. What are subharmonic oscillations?6. Explain the jump phenomenon.7. What principle is used in the Ritz-Galerkin method?8. Define these terms: phase plane, trajectory, singular point, phase velocity.9. What is the method of isoclines?10. What is the difference between a hard spring and a soft spring?11. Explain the difference between subharmonic and superharmonic oscillations.12. What is a secular term?13. Give an example of a system that leads to an equation of motion with time-dependentcoefficients.14. Explain the significance of the following: stable node, unstable node, saddle point, focus,and center.15. What is a limit cycle?16. Give two examples of physical phenomena that can be represented by van der Pol’sequation.13.2 Indicate whether each of the following statements is true or false:1. Nonlinearity can be introduced into the governing differential equation through mass,spring and/or dampers.2. Nonlinear analysis of a system can reveal several unexpected phenomena.3. The Mathieu equation is an autonomous equation.4. A singular point corresponds to a state of equilibrium of the system.5. Jump phenomenon is exhibited by both linear and nonlinear systems.RaoCh13ff.qxd 10.06.08 13:57 Page 961
  • 107. 962 CHAPTER 13 NONLINEAR VIBRATION6. The Ritz-Galerkin method finds the approximate solution by satisfying the nonlinearequation in the average.7. Dry friction can introduce nonlinearity in the system.8. Poincaré’s solution of nonlinear equations is in the form of a series.9. The secular term appears in the solution of free Duffing’s equation.10. According to Lindstedt’s perturbation method, the angular frequency is assumed to be afunction of the amplitude.11. An isocline is the locus of points at which the trajectories passing through them have aconstant slope.12. Time does not appear explicitly in a trajectory plotted in the phase plane.13. The time variation of the solution can be found from the phase plane trajectories.14. A limit cycle denotes a steady-state periodic oscillation.15. Approximate solutions of nonlinear vibration problems can be found using numericalmethods such as Houbolt, Wilson, and Neumark method.13.3 Fill in each of the following blanks with the appropriate words:1. When finite amplitudes of motion are involved, _____ analysis becomes necessary.2. _____ principle is not applicable in nonlinear analysis.3. _____ equation involves time-dependent coefficients.4. The governing equation of a simple pendulum whose pivot is subjected to vertical vibra-tion is called _____ equation.5. The representation of the motion of a system in the displacement-velocity plane is knownas _____ plane representation.6. The curve traced by a typical point in the phase plane is called a _____.7. The velocity with which a representative point moves along a trajectory is called the_____ velocity.8. The phenomenon of realizing two amplitudes for the same frequency is known as _____phenomenon.9. The forced vibration solution of Duffing’s equation has _____ values of the frequencyfor any given amplitude10. The Ritz-Galerkin method involves the solution of _____ equations.11. Mechanical chatter is a _____ vibration.12. If time does not appear explicitly in the governing equation, the corresponding system issaid to be _____.13. The method of _____ can be used to construct the trajectories of a one degree of free-dom dynamical system.14. Van der Pol’s equation exhibits _____ cycles.13.4 Select the most appropriate answer out of the choices given:1. Each term in the equation of motion of a linear system involves displacement, velocity,and acceleration of the(a) first degree (b) second degree (c) zero degree2. A nonlinear stress-strain curve can lead to nonlinearity of the(a) mass (b) spring (c) damper3. If the rate of change of force with respect to displacement, is an increasing func-tion of x, the spring is called a(a) soft spring (b) hard spring (c) linear springdf/dx,ƒAƒ.vRaoCh13ff.qxd 10.06.08 13:57 Page 962
  • 108. REVIEW QUESTIONS 9634. If the rate of change of force with respect to displacement, is a decreasing func-tion of k, the spring is called a(a) soft spring (b) hard spring (c) linear spring5. The point surrounded by closed trajectories is called a(a) center (b) mid-point (c) focal point6. For a system with periodic motion, the trajectory in the phase plane is a(a) closed curve (b) open curve (c) point7. In subharmonic oscillations, the natural frequency and the forcing frequencyare related as(a)(b)(c)8. In superharmonic oscillations, the natural frequency and the forcing frequencyare related as(a)(b)(c)9. If time appears explicitly in the governing equation, the corresponding system iscalled(a) an autonomous system(b) a nonautonomous system(c) a linear system.10. Duffing’s equation is given by(a)(b)(c)11. Lindstedt’s perturbation method gives(a) periodic and nonperiodic solutions(b) periodic solutions only(c) nonperiodic solutions only13.5 Match the items in the two columns below for the nature of equilibrium points in the con-text of the stability analysis of equilibrium states with and as eigenvalues:(1) and with same sign (a) Unstable node( real and distinct)(2) and (b) Saddle point( real and distinct)(3) and (c) Node( real and distinct)(4) and real with opposite signs (d) Focus or spiral point(5) and complex conjugates (e) Stable nodel2:l1l2:l1l1, l2:l2 7 0l1l1, l2:l2 6 0l1l1, l2:l2l1l2l1x$+ a x3= 0x$+ v02x = 0x$+ v02x + a x3= 0vn =vn; n = 2, 3, 4, Ávn = nv; n = 2, 3, 4, Ávn = v1v21vn2vn =vn; n = 2, 3, 4, Ávn = nv; n = 2, 3, 4, Ávn = v1v21vn2df/dx,RaoCh13ff.qxd 10.06.08 13:57 Page 963
  • 109. 964 CHAPTER 13 NONLINEAR VIBRATION13.6 Match the items in the two columns below:(1) (a) Nonlinearity in mass(2) (b) Nonlinearity in damping(3) (c) Linear equation(4) (d) Nonlinearity in springPROBLEMSThe problem assignments are organized as follows:Problems Section Covered Topic Covered13.1 13.1 Introduction13.2–13.8 13.2 Examples of nonlinear vibration13.9–13.11 13.3 Exact methods of solution13.12, 13.13 13.4 Approximate analytical methods13.14–13.16 13.5 Subharmonic and superharmonicoscillations13.17 13.6 Mathieu equation13.18–13.22 13.7 Graphical methods13.23–13.33 13.8 Stability of equilibrium states13.34 13.9 Limit cycles13.35, 13.36 13.10 Chaos13.37–13.42 13.12 MATLAB programs13.43–13.45 13.13 C++ program13.46–13.49 13.14 Fortran program13.50–13.51 — Design projects13.1 The equation of motion of a simple pendulum, subjected to a constant torque, isgiven by(E.1)If is replaced by its two-term approximation, the equation of motionbecomes(E.2)u$+ v02u = f +v026u3u - 1u3/62,sin uu$+ v02sin u = fMt = ml2f,x$+ cx#+ kx = ax3axx$+ kx = 0x$+ v02ax -x36b = 0x$+ fx#ƒx#ƒ+ vn2x = 0RaoCh13ff.qxd 10.06.08 13:57 Page 964
  • 110. PROBLEMS 965Let the solution of the linearized equation(E.3)be denoted as and the solution of the equation(E.4)be denoted as Discuss the validity of the total solution. given byfor Eq. (E.2).13.2 Two springs, having different stiffnesses and with are placed on either side ofa mass m, as shown in Fig. 13.27. When the mass is in its equilibrium position, no spring isin contact with the mass. However, when the mass is displaced from its equilibrium position,only one spring will be compressed. If the mass is given an initial velocity at deter-mine (a) the maximum deflection and (b) the period of vibration of the mass.t = 0,x#0k2 7 k1,k2k1u11t2 + u21t2,u1t2 =u1t2,u21t2.u$+ v02u =v026u3u11t2,u$+ v02u = fmk1x(t)k2 Ͼ k1FIGURE 13.27mx0k kkx(t)␻P sin tx0kFIGURE 13.28m1 m2l1 l2 l3FIGURE 13.2913.4 Two masses and are attached to a stretched wire, as shown in Fig. 13.29. If the initialtension in the wire is P, derive the equations of motion for large transverse displacements ofthe masses.m2m113.3 Find the equation of motion of the mass shown in Fig. 13.28. Draw the spring force versus xdiagram.RaoCh13ff.qxd 10.06.08 13:57 Page 965
  • 111. 966 CHAPTER 13 NONLINEAR VIBRATIONkF(t)␪2l3l3FIGURE 13.31k1k2x(t)F(t)mlhFIGURE 13.3213.5 A mass m, connected to an elastic rubber band of unstretched length l and stiffness k, is per-mitted to swing as a pendulum bob, as shown in Fig. 13.30. Derive the nonlinear equations ofmotion of the system using x and as coordinates. Linearize the equations of motion anddetermine the natural frequencies of vibration of the system.u13.7 Derive the nonlinear equation of motion of the spring-mass system shown in Fig. 13.32Rubber band,stiffness kml ϩ x␪FIGURE 13.3013.6 A uniform bar of length l and mass m is hinged at one end supported by a spring atand acted by a force at as shown in Fig. 13.31. Derive the nonlinear equationof motion of the system.x = l,x = 2l3 ,1x = 02,RaoCh13ff.qxd 10.06.08 13:57 Page 966
  • 112. PROBLEMS 96713.8 Derive the nonlinear equations of motion of the system shown in Fig. 13.33. Also, find the lin-earized equations of motion for small displacements, x(t) and u1t2.13.9 Find the natural time period of oscillation of the pendulum shown in Fig. 13.1(a) when itoscillates between the limits and using Eqs. (13.1) and (13.12).13.10 A simple pendulum of length 30 in. is released from the initial position of 80° from the ver-tical. How long does it take to reach the position13.11 Find the exact solution of the nonlinear pendulum equationwith when where denotes the maximum angular displacement.13.12 Find the solution of Example 13.1 using the following two-term approximation for x(t):13.13 Using a three-term expansion in Lindstedt’s perturbation method [Eq. (13.30)], find the solu-tion of the pendulum equation, Eq. (13.20).13.14 The equation of motion for the forced vibration of a single degree of freedom nonlinear sys-tem can be expressed asDerive the conditions for the existence of subharmonics of order 3 for this system.13.15 The equation of motion of a nonlinear system is given byInvestigate the subharmonic solution of order 2 for this system.13.16 Prove that, for the system considered in Section 13.5.1, the minimum value of for whichthe amplitude of subharmonic oscillations A will have a real value is given byvmin = v0 +212048F2v05v2x$+ cx#+ k1x + k2x2= a cos 2vtx$+ cx#+ k1x + k2x3= a1 cos 3vt - a2 sin 3vtx1t2 = A0 sin vt + A3 sin 3 vtu0u = u0,u#= 0u$+ v02au -u36b = 0u = 0°?u = p/2,u = -p/2Rigid bar, length l(massless)mmkF(t)x(t)␪(t)Mt(t)FIGURE 13.33RaoCh13ff.qxd 10.06.08 13:57 Page 967
  • 113. Also, show that the minimum value of the amplitude, for stable subharmonic oscillations, isgiven by13.17 Derive Eqs. (13.113b) and (13.116b) for the Mathieu equation.13.18 The equation of motion of a single degree of freedom system is given bywith initial conditions and (a) Plot the graph x(t) versus t for(b) Plot a trajectory in the phase plane.13.19 Find the equilibrium position and plot the trajectories in the neighborhood of the equilibriumposition corresponding to the following equation:13.20 Obtain the phase trajectories for a system governed by the equationwith the initial conditions and using the method of isoclines.13.21 Plot the phase-plane trajectories for the following system:The initial conditions are13.22 A single degree of freedom system is subjected to Coulomb friction so that the equation ofmotion is given byConstruct the phase plane trajectories of the system using the initial conditionsand13.23 The equation of motion of a simple pendulum subject to viscous damping can be expressed asIf the initial conditions are and show that the origin in the phase planediagram represents (a) a stable focus for and (b) an unstable focus for13.24 The equation of motion of a simple pendulum, subjected to external force, is given byFind the nature of singularity at u = sin-110.82.u$+ 0.5 u#+ sin u = 0.8c 6 0.c 7 0u#102 = 0,u102 = u0u$+ cu#+ sin u = 0x#102 = 0.101f/vn22x102 =x$+ fx#ƒx#ƒ+ vn2x = 0x102 = x#102 = 0.x$+ 0.1x#+ x = 5x#102 = 1x102 = 2x$+ 0.4x#+ 0.8x = 0x$+ 0.11x2- 12x#+ x = 00 … t … 10.x#102 = 2.x102 = -12x$+ 0.8x#+ 1.6x = 0Amin =F16v2968 CHAPTER 13 NONLINEAR VIBRATIONRaoCh13ff.qxd 10.06.08 13:57 Page 968
  • 114. 13.25 The phase plane equation of a single degree of freedom system is given byInvestigate the nature of singularity at for13.26 Identify the singularity and find the nature of solution near the singularity for van der Pol’sequation:13.27 Identify the singularity and investigate the nature of solution near the singularity for anundamped system with a hard spring:13.28 Solve Problem 13.27 for an undamped system with a soft spring:13.29 Solve Problem 13.27 for a simple pendulum:13.30 Determine the eigenvalues and eigenvectors of the following equations:a.b.13.31 Find the trajectories of the system governed by the equations13.32 Find the trajectories of the system governed by the equations13.33 Find the trajectories of the system governed by the equations13.34 Using Lindstedt’s perturbation method, find the solution of the van der Pol’s equation,Eq. (13.143).13.35 Verify that the following equation exhibits chaotic behavior:Hint: Give values of 3.25, 3.5 and 3.75 to k and observe the sequence of values generated withx1 = 0.5.xn+1 = k xn 11 - xn2x#= 2x + y, y#= - 3x - 2yx#= x - y, y#= x + 3yx#= x - 2y, y#= 4x - 5yx#= x + y, y#= 4x + yx#= x - y, y#= x + 3yu$+ vn2sin u = 0x$+ vn211 - k2x22 x = 0x$+ vn211 + k2x22 x = 0x$- a 11 - x22 x#+ x = 0c 7 0.1x, y2 = 10, 02dydx=- cy - 1x - 0.1x32yPROBLEMS 969RaoCh13ff.qxd 10.06.08 13:57 Page 969
  • 115. 13.36 Verify that the following equation exhibits chaotic behavior:Hint: Observe the sequence of values generated using and Using MATLAB, solve the simple pendulum equations, Eqs. (E.1) to (E.3) given in Example13.6, for the following data:a.b.13.38 Using MATLAB, find the solution of the nonlinearly damped system, Eq. (E.1) of Example 13.7,for the following data:13.39 Using MATLAB, find the solution of a nonlinear single degree of freedom system governedby Eq. (E.1) of Example 13.8 under a pulse load for the following data:13.40 Solve the equation of motion using the Runge-Kuttamethod with and Plot the variation of x with t. UseProgram18.m for the solution.13.41 Find the time variation of the angular displacement of a simple pendulum (i.e., the solution ofEq. 13.5) for using the initial conditions and Use the Runge-Kutta method given in Program18.m.13.42 In the static firing test of a rocket, the rocket is anchored to a rigid wall by a nonlinear spring-damper system and fuel is burnt to develop a thrust, as shown in Fig. 13.34. The thrust actingon the rocket during the time period is given by where is the con-stant rate at which fuel is burned and v is the velocity of the jet stream. The initial mass of therocket is M, so that its mass at any time t is given by The datais: spring dampingand (a) Using the Runge-Kutta method ofnumerical integration, derive the equation of motion of the rocket and (b) Find the variationof the displacement of the rocket. Use Program18.m.t0 = 100 s.10 kg/s, v = 2000 m/s, M = 2000 kg,force = 10x#+ 20x# 2N, m0 =force = 8 * 105x + 6 * 103x3N,m = M - m0t, 0 … t … t0.m0F = m0v,0 … t … t0u#0 = 0.u0 = 45°g/l = 0.5,x0 = x#0 = 0.¢t = 0.05, tmax = 5.0,x$+ 0.5x#+ x + 1.2x3= 1.8 cos 0.4t,k2 = 1000, F0 = 1000, t0 = 5, x102 = 0.05, x#102 = 5.m = 10, k1 = 4000,x#102 = 1.0.m = 10, c = 0.1, k = 4000, F0 = 200, v = 20, x102 = 0.5,v0 = 0.1, u102 = 0.01, u#102 = 10v0 = 0.1, u102 = 0.01, u#102 = 0x1 = 1.001, 1.002xn+1 = 2.0 xn 1xn - 12970 CHAPTER 13 NONLINEAR VIBRATIONFvkcx(t)FIGURE 13.3413.43 Using Program18.cpp, solve Problem Using Program18.cpp, solve Problem Using Program18.cpp, solve Problem Using PROGRAM18.F, solve Problem 13.40.RaoCh13ff.qxd 10.06.08 13:57 Page 970
  • 116. 13.47 Using PROGRAM18.F, solve Problem Using PROGRAM18.F, solve Problem Write a computer program for finding the period of vibration corresponding to Eq. (13.14).Use a suitable numerical integration procedure. Using this program, find the solution ofProblem 13.41.DESIGN PROJECTS13.50 In some periodic vibratory systems, external energy is supplied to the system over part of aperiod and dissipated within the system in another part of the period. Such periodic oscilla-tions are known as relaxation oscillations. Van der Pol [13.22] indicated several instances ofoccurrence of relaxation oscillations such as a pneumatic hammer, the scratching noise of aknife on a plate, the squeaking of a door, and the fluctuation of populations of animal species.Many relaxation oscillations are governed by van der Pol’s equation:(E.1)a. Plot the phase plane trajectories for three values of and Usethe initial conditions (i) and (ii) andb. Solve Eq. (E.1) using the fourth-order Runge-Kutta method using the initial conditionsstated in (a) for and13.51 A machine tool is mounted on two nonlinear elastic mounts, as shown in Fig. 13.35. The equa-tions of motion, in terms of the coordinates x(t) and are given by(E.1)+ k22 1x + l2 u23= 0mx$+ k11 1x - l1 u2 + k12 1x - l1 u23+ k21 1x + l2 u2u1t2,a = 10.a = 0.1, a = 1,x#102 = 5.x102 = 0x102 = 0.5, x#102 = 0a = 10.a: a = 0.1, a = 1,x$- a 11 - x22 x#+ x = 0DESIGN PROJECTS 971x (t)x(t)Gk1 k2l1 l2G ␪(t)FIGURE 13.35RaoCh13ff.qxd 10.06.08 13:57 Page 971
  • 117. (E.2)where m is the mass and is the mass moment of inertia about G of the machine tool. Usingthe Runge-Kutta method, find x(t) and for the following data:and k2 = 50x2 + 5x23kN/m.2500 kg-m2, l1 = 1 m, l2 = 1.5 m, k1 = 40x1 + 10x13kN/m,m = 1000 kg, J0 =u1t2J0+ k21 1x + l2 u)l2 + k22 1x + l2 u23l2 = 0J0 u$- k11 1x - l1 u2 l1 - k12 1x - l1 u23l1972 CHAPTER 13 NONLINEAR VIBRATIONRaoCh13ff.qxd 10.06.08 13:57 Page 972
  • 118. Karl Friedrich Gauss (1777–1855) was a German mathematician, astronomer,and physicist. Gauss, Archimedes, and Newton are considered to be in a class bythemselves among the great mathematicians. Although Gauss was born in a poorfamily, his extraordinary intelligence and genius in childhood inspired the Dukeof Brunswick to pay all the expenses of Gauss during his entire education. In1795, Gauss entered the University of Göttingen to study mathematics and beganto keep his scientific diary. It was found after his death that his diary containedtheories which were rediscovered and published by others. He moved to theUniversity of Helmstedt in 1798 from which he received his doctor’s degree in1799. He published his most famous work, Disquisitiones Arithmeticae(Arithmetical Researches), in 1801. After that, Gauss was made the director ofGöttingen Observatory and also broadened his activities to include the mathemat-ical and practical aspects of astronomy, geodesy, and electromagnetism. Theinstrument used for measuring magnetic field strength is called the “Gaussmeter.”He invented the method of least squares and the law of normal distribution(Gaussian distribution) which is widely used in probability and random vibration.(Photo courtesy of Dirk J. Struik, A Concise History of Mathematics, 2nd ed.,Dover Publications, New York, 1948.)C H A P T E R 1 4Random Vibration97314.1 IntroductionIf vibrational response characteristics such as displacement, acceleration, and stress areknown precisely as functions of time, the vibration is known as deterministic vibration.This implies a deterministic system (or structure) and a deterministic loading (or excita-tion); deterministic vibration exists only if there is perfect control over all the variables thatinfluence the structural properties and the loading. In practice, however, there are manyprocesses and phenomena whose parameters cannot be precisely predicted. Such processesare therefore called random processes [14.1–14.4]. An example of a random process ispressure fluctuation at a particular point on the surface of an aircraft flying in air. If sev-eral records of these pressure fluctuations are taken under the same flight speed, altitude,and load factor, they might look as indicated in Fig. 14.1. The records are not identicaleven though the measurements are taken under seemingly identical conditions. Similarly,a building subjected to ground acceleration due to an earthquake, a water tank under windloading, and a car running on a rough road represent random processes. An elementarytreatment of random vibration is presented in this chapter.RaoCh14ff.qxd 10.06.08 14:00 Page 973
  • 119. 974 CHAPTER 14 RANDOM VIBRATIONFIGURE 14.1 Ensemble of a random process( is the ith sample function of the ensemble).x1i21t214.2 Random Variables and Random ProcessesMost phenomena in real life are nondeterministic. For example, the tensile strength of steeland the dimensions of a machined part are nondeterministic. If many samples of steel aretested, their tensile strengths will not be the same—they will fluctuate about a mean oraverage value. Any quantity, like the tensile strength of steel, whose magnitude cannot beprecisely predicted, is known as a random variable or a probabilistic quantity. If experi-ments are conducted to find the value of a random variable x, each experiment will give anumber that is not a function of any parameter. For example, if 20 samples of steel are tested,their tensile strengths might beEach of these outcomes is called a sample point. If n experiments are con-ducted, all the n possible outcomes of the random variable constitute what is known as thesample space of the random variable.There are other types of probabilistic phenomena for which the outcome of an exper-iment is a function of some parameter such as time or a spatial coordinate. Quantities suchas the pressure fluctuations shown in Fig. 14.1 are called random processes. Each outcomeof an experiment, in the case of a random process, is called a sample function. If n exper-iments are conducted, all the n possible outcomes of a random process constitute what isknown as the ensemble of the process [14.5]. Notice that if the parameter t is fixed at a par-ticular value is a random variable whose sample points are given by14.3 Probability DistributionConsider a random variable x such as the tensile strength of steel. If n experimental valuesof x are available as the probability of realizing the value of x smaller thanx1, x2, Á , xn,x1121t12, x1221t12, Á , x1n21t12.t1, x1t12298 N/mm2.x112= 284, x122= 302, x132= 269, Á , x1202=RaoCh14ff.qxd 10.06.08 14:00 Page 974
  • 120. 14.3 PROBABILITY DISTRIBUTION 975FIGURE 14.2 Random time function.some specified value can be found as(14.1)where denotes the number of values which are less than or equal to As the numberof experiments Eq. (14.1) defines the probability distribution function of x, P(x):(14.2)The probability distribution function can also be defined for a random time function. Forthis, we consider the random time function shown in Fig. 14.2. During a fixed time spant, the time intervals for which the value of x(t) is less than are denoted asand Thus the probability of realizing x(t) less than or equal to is given by(14.3)As Eq. (14.3) gives the probability distribution function of x(t):(14.4)If x(t) denotes a physical quantity, the magnitude of x(t) will always be a finite number,so (impossible event), and(certain event). The typical variation of P(x) with x is shown in Fig. 14.3(a).The function P(x) is called the probability distribution function of x. The derivative ofP(x) with respect to x is known as the probability density function and is denoted asp(x). Thus(14.5)p1x2 =dP1x2dx= lim¢x:0P1x + ¢x2 - P1x2¢xP1q2 = 1Prob[x1t2 6 q] =Prob[x1t2 6 - q] = P1- q2 = 0P1x2 = limt:q1t ai¢tit : q,Prob [x1t2 … x] =1t ai¢tix¢4.¢t1, ¢t2, ¢t3,xP1x2 = limn:qnnn : q,x.xinProb1x … x2 =nnxRaoCh14ff.qxd 10.06.08 14:00 Page 975
  • 121. 976 CHAPTER 14 RANDOM VIBRATIONFIGURE 14.3 Probability distributionand density functions.where the quantity denotes the probability of realizing x(t) betweenthe values x and Since p(x) is the derivative of P(x), we have(14.6)As Eq (14.6) gives(14.7)which means that the total area under the curve of p(x) is unity.14.4 Mean Value and Standard DeviationIf f(x) denotes a function of the random variable x, the expected value of f(x), denoted asor E[ f(x)] or is defined as(14.8)If Eq. (14.8) gives the expected value, also known as the mean value of x:(14.9)mx = E[x] = x =Lq- qxp1x2 dxf1x2 = x,mf = E[f1x2] = f1x2 =Lq- qf1x2p1x2 dxf1x2,mfP1q2 =Lq- qp1x¿2 dx¿ = 1P1q2 = 1,P1x2 =Lx- qp1x¿2 dx¿x + ¢x.P1x + ¢x2 - P1x2RaoCh14ff.qxd 10.06.08 14:00 Page 976
  • 122. 14.4 MEAN VALUE AND STANDARD DEVIATION 977Similarly, if we get the mean square value of x:(14.10)The variance of x, denoted as is defined as the mean square value of x about the mean,(14.11)The positive square root of the variance, is called the standard deviation of x.Probabilistic Characteristics of Eccentricity of a RotorThe eccentricity of a rotor (x), due to manufacturing errors, is found to have the following distribution:(E.1)where k is a constant. Find (a) the mean, standard deviation, and the mean square value of the eccen-tricity and (b) the probability of realizing x less than or equal to 2 mm.Solution: The value of k in Eq. (E.1) can be found by normalizing the probability density function:That is,That is,(E.2)(a) The mean value of x is given by Eq. (14.9):(E.3)The standard deviation of x is given by Eq. (14.11):=L501x2+ x2- 2xx2 p1x2 dxsx2=L501x - x22p1x2 dxx =L50p1x2x dx = kax44b05= 3.75 mmk =3125kax33b05= 1Lq- qp1x2 dx =L50kx2dx = 1p1x2 = ekx2, 0 … x … 5 mm0, elsewhereE X A M P L E 1 4 . 1s1x2,sx2= E[1x - x22] =Lq- q1x - x22p1x2 dx = 1x22 - 1x22sx2,mx2 = E[x2] = x2=Lq- qx2p1x2 dxf1x2 = x2,RaoCh14ff.qxd 10.06.08 14:00 Page 977
  • 123. 978 CHAPTER 14 RANDOM VIBRATION(E.4)The mean square value of x is(E.5)(b)(E.6)■14.5 Joint Probability Distribution of Several Random VariablesWhen two or more random variables are being considered simultaneously, their jointbehavior is determined by a joint probability distribution function. For example, while test-ing the tensile strength of steel specimens, we can obtain the values of yield strength andultimate strength in each experiment. If we are interested in knowing the relation betweenthese two random variables, we must know the joint probability density function of yieldstrength and ultimate strength. The probability distributions of single random variables arecalled univariate distributions; the distributions that involve two random variables arecalled bivariate distributions. In general, if a distribution involves more than one randomvariable, it is called a multivariate distribution.The bivariate density function of the random variables and is defined by(14.12)that is, the probability of realizing the value of the first random variable between andand the value of the second random variable between and Thex2 + dx2.x2x1 + dx1x1p1x1, x22 dx1 dx2 = Prob [x1 6 xœ1 6 x1 + dx1, x2 6 xœ2 6 x2 + dx2]x2x1= kax33b02=8125= 0.064Prob [x … 2] =L20p1x2 dx = kL20x2dxx2= ka31255b = 15 mm2‹ sx = 0.9682 mm= ka31255b - 13.7522= 0.9375= kax55b05- 1x22=L50kx4dx - 1x22RaoCh14ff.qxd 10.06.08 14:00 Page 978
  • 124. 14.5 JOINT PROBABILITY DISTRIBUTION OF SEVERAL RANDOM VARIABLES 979joint probability density function has the property that(14.13)The joint distribution function of and is(14.14)The marginal or individual density functions can be obtained from the joint probabilitydensity function as(14.15)(14.16)The variances of x and y can be determined as(14.17)(14.18)The covariance of x and y, is defined as the expected value or average of the productof the deviations from the respective mean values of x and y. It is given by(14.19)= E[xy] - mxmy- mxLq- q Lq- qyp1x, y2 dx dy + mxmyLq- q Lq- qp1x, y2 dx dy=Lq- q Lq- qxyp1x, y2 dx dy - myLq- q Lq- qxp1x, y2 dx dy=Lq- q Lq- q1xy - xmy - ymx + mxmy2p1x, y2 dx dysxy = E[1x - mx21y - my2] =Lq- q Lq- q1x - mx21y - my2p1x, y2 dx dysxy,sy2= E[1y - my22] =Lq- q1y - my22p1y2 dysx2= E[1x - mx22] =Lq- q1x - mx22p1x2 dxp1y2 =Lq- qp1x, y2 dxp1x2 =Lq- qp1x, y2 dy=Lx1- q Lx2- qp1xœ1, xœ22 dxœ1 dxœ2P1x1, x22 = Prob [xœ1 6 x1, xœ2 6 x2]x2x1Lq- q Lq- qp1x1, x22 dx1 dx2 = 1RaoCh14ff.qxd 10.06.08 14:00 Page 979
  • 125. 980 CHAPTER 14 RANDOM VIBRATIONThe correlation coefficient between x and y, is defined as the normalized covariance:(14.20)It can be seen that the correlation coefficient satisfies the relation14.6 Correlation Functions of a Random ProcessIf are fixed values of t, we use the abbreviations to denote the valuesof x(t) at respectively. Since there are several random variables weform the products of the random variables (values of x(t) at different times) andaverage the products over the set of all possibilities to obtain a sequence of functions:(14.21)and so on. These functions describe the statistical connection between the values of x(t) atdifferent times and are called correlation functions [14.6, 14.7].Autocorrelation Function. The mathematical expectation of —the correlationfunction —is also known as the autocorrelation function, designated asThus(14.22)If the joint probability density function of and is known to be the auto-correlation function can be expressed as(14.23)Experimentally, we can find by taking the product of and in theith sample function and averaging over the ensemble:(14.24)where n denotes the number of sample functions in the ensemble (see Fig. 14.4). Ifand are separated by (with and ), we haveE[x1t2x1t + t2].R1t + t2 =t2 = t + tt1 = ttt2t1R1t1, t22 =1n ani=1x1i21t12x1i21t22x1i21t22x1i21t12R1t1, t22R1t1, t22 =Lq- q Lq- qx1x2p1x1, x22 dx1 dx2p1x1, x22,x2x1R1t1, t22 = E[x1x2]R1t1, t22.K1t1, t22x1x2t1, t2, ÁK1t1, t2, t32 = E[x1t12x1t22x1t32] = E[x1x2x3]K1t1, t22 = E[x1t12x1t22] = E[x1x2]x1, x2, Áx1, x2, Á ,t1, t2, Á ,x1, x2, Át1, t2, Á-1 … rxy … 1.rxy =sxysxsyrxy,RaoCh14ff.qxd 10.06.08 14:00 Page 980
  • 126. 14.7 STATIONARY RANDOM PROCESS 981FIGURE 14.4 Ensemble of a random process.14.7 Stationary Random ProcessA stationary random process is one for which the probability distributions remain invari-ant under a shift of the time scale; the family of probability density functions applicablenow also applies five hours from now or 500 hours from now. Thus the probability den-sity function becomes a universal density function p(x), independent of time.Similarly, the joint density function to be invariant under a shift of the timescale, becomes a function of but not a function of or individually. Thuscan be written as The expected value of stationary random processx(t) can be written as(14.25)and the autocorrelation function becomes independent of the absolute time t and willdepend only on the separation time(14.26)where We shall use subscripts to R to identify the random process when morethan one random process is involved. For example, we shall use and todenote the autocorrelation functions of the random processes x(t) and y(t), respectively.The autocorrelation function has the following characteristics [14.2, 14.4]:Ry1t2Rx1t2t = t2 - t1.R1t1, t22 = E[x1x2] = E[x1t2x1t + t2] = R1t2 for any tt:E[x1t12] = E[x1t1 + t2] for any tp1t, t + t2.p1x1, x22t2t1t = t2 - t1,p1x1, x22,p1x12RaoCh14ff.qxd 10.06.08 14:00 Page 981
  • 127. 982 CHAPTER 14 RANDOM VIBRATIONFIGURE 14.5 Autocorrelation function.1. If gives the mean square value of x(t):(14.27)2. If the process x(t) has a zero mean and is extremely irregular, as shown inFig. 14.5(a), its autocorrelation function will have small values except atas indicated in Fig. 14.5(b).3. If the autocorrelation function will have a constant value asshown in Fig. 14.6.4. If x(t) is stationary, its mean and standard deviations will be independent of t:(14.28)and(14.29)The correlation coefficient, of x(t) and can be found as(14.30)=R1t2 - m2s2=E[x1t2x1t + t2] - mE[x1t + t2] - mE[x1t2] + m2s2r =E[5x1t2 - m65x1t + t2 - m6]s2x1t + t2r,sx1t2 = sx1t+t2 = sE[x1t2] = E[x1t + t2] = mR1t2x1t2 M x1t + t2,t = 0,R1t2R102 = E[x2]t = 0, R1t2RaoCh14ff.qxd 10.06.08 14:00 Page 982
  • 128. 14.7 STATIONARY RANDOM PROCESS 983FIGURE 14.6 Constant values of function.that is,(14.31)Since Eq. (14.31) shows that(14.32)This shows that the autocorrelation function will not be greater than the mean squarevalue,5. Since depends only on the separation time and not on the absolute time t fora stationary process,(14.33)Thus is an even function of6. When is large there will not be a coherent relationship between the twovalues x(t) and Hence the correlation coefficient and Eq. (14.31)gives(14.34)R1t : q2 : m2r : 0x1t + t2.1t : q2,tt.R1t2R1t2 = E[x1t2x1t + t2] = E[x1t2x1t - t2] = R1-t2tR1t2E[x2] = s2+ m2.-s2+ m2… R1t2 … s2+ m2ƒrƒ … 1,R1t2 = rs2+ m2RaoCh14ff.qxd 10.06.08 14:00 Page 983
  • 129. 984 CHAPTER 14 RANDOM VIBRATIONFIGURE 14.7 Autocorrelation function.A typical autocorrelation function is shown in Fig. 14.7.Ergodic Process. An ergodic process is a stationary random process for which we canobtain all the probability information from a single sample function and assume that it isapplicable to the entire ensemble. If represents a typical sample function of dura-tion T, the averages can be computed by averaging with respect to time along Suchaverages are called temporal averages. By using the notation to represent the tem-poral average of x(t) (the time average of x), we can write(14.35)where has been assumed to be defined from to with(T very large). Similarly,(14.36)and(14.37)14.8 Gaussian Random ProcessThe most commonly used distribution for modeling physical random processes iscalled the Gaussian or normal random process. The Gaussian process has a number ofR1t2 = 8x1t2x1t + t29 = limT:q1T LT/2- T/2x1i21t2x1i21t + t2 dtE[x2] = 8x21t29 = limT:q1T LT/2- T/2[x1i21t2]2dtT : qt = T/2t = - T/2x1i21t2E[x] = 8x1t29 = limT:q1T LT/2- T/2x1i21t2 dt8x1t29x1i21t2.x1i21t2RaoCh14ff.qxd 10.06.08 14:00 Page 984
  • 130. 14.8 GAUSSIAN RANDOM PROCESS 985FIGURE 14.8 Gaussian probabilitydensity function.remarkable properties that permit the computation of the random vibration characteris-tics in a simple manner. The probability density function of a Gaussian process x(t) isgiven by(14.38)where and denote the mean value and standard deviation of x. The mean andstandard deviation of x(t) vary with t for a nonstationary process but are constants(independent of t) for a stationary process. A very important property of the Gaussianprocess is that the forms of its probability distributions are invariant with respect to linearoperations. This means that if the excitation of a linear system is a Gaussian process, theresponse is generally a different random process, but still a normal one. The only changesare that the magnitude of the mean and standard deviations of the response are differentfrom those of the excitation.The graph of a Gaussian probability density function is a bell-shaped curve, symmet-ric about the mean value; its spread is governed by the value of the standard deviation, asshown in Fig. 14.8. By defining a standard normal variable z as(14.39)Eq. (14.38) can be expressed as(14.40)The probability of x(t) lying in the interval and where c is any positive numbercan be found, assuming(14.41)e- 12x2s2dxProb [-cs … x1t2 … cs] =Lcs-cs112psx = 0:+cs-csp1x2 =112pe-12 z2z =x - xsx1sx21x2sxxp1x2 =112psxe- 12 Ax - xqsxB2RaoCh14ff.qxd 10.06.08 14:00 Page 985
  • 131. 986 CHAPTER 14 RANDOM VIBRATIONFIGURE 14.9 Graphical representation ofProb[-cs … x1t2 … cs].The probability of x(t) lying outside the range is one minus the value given byEq. (14.41). This can also be expressed as(14.42)The integrals in Eqs. (14.41) and (14.42) have been evaluated numerically and tabulated[14.5]; some typical values are indicated in the following table (see Fig. 14.9 also).Value of c 1 2 3 40.6827 0.9545 0.9973 0.9999370.3173 0.0455 0.0027 0.00006314.9 Fourier AnalysisWe saw in Chapter 1 that any periodic function x(t), of period can be expressed in theform of a complex Fourier series(14.43)where is the fundamental frequency given by(14.44)v0 =2ptv0x1t2 = aqn= - qcneinv0tt,14.9.1Fourier SeriesProb[ƒx1t2ƒ 7 cs]Prob[-cs … x1t2 … cs]Prob [ƒx1t2ƒ 7 cs] =212ps Lqcse- 12x2s2dxϯcsRaoCh14ff.qxd 10.06.08 14:00 Page 986
  • 132. 14.9 FOURIER ANALYSIS 987and the complex Fourier coefficients can be determined by multiplying both sides ofEq. (14.43) with and integrating over one time period:(14.45)Equation (14.45) can be simplified to obtain (see Problem 14.27)(14.46)Equation (14.43) shows that the function x(t) of period can be expressed as a sum of aninfinite number of harmonics. The harmonics have amplitudes given by Eq. (14.46) andfrequencies which are multiples of the fundamental frequency The difference betweenany two consecutive frequencies is given by(14.47)Thus the larger the period the denser the frequency spectrum becomes. Equation (14.46)shows that the Fourier coefficients are, in general, complex numbers. However, ifx(t) is a real and even function, then will be real. If x(t) is real, the integrand of inEq. (14.46) can also be identified as the complex conjugate of that of Thus(14.48)The mean square value of x(t)—that is, the time average of the square of the functionx(t)—can be determined as(14.49)= c02+ aqn=12ƒcn ƒ2= aqn= - qƒcn ƒ2=1t Lt/2- t/2e aqn=12cncn… + c02f dt=1t Lt/2- t/2e aqn=11cneinv0t+ cn…e-inv0t2 + c0 f2dt=1t Lt/2- t/2a a-1n= - qcneinv0t+ c0 + aqn=1cneinv0tb2dtx21t2 =1t Lt/2- t/2x21t2 dt =1t Lt/2- t/2a aqn= - qcneinv0tb2dtcn = c*- nc- n.cncncnt,vn+1 - vn = 1n + 12v0 - nv0 = ¢v =2pt= v0v0.tcn =1t Lt/2- t/2x1t2e-inv0tdt= aqn= - qcnLt/2- t/2[cos1n - m2v0t + i sin1n - m2v0t] dtLt/2- t/2x1t2e-imv0tdt = aqn= - q Lt/2- t/2cnei1n-m2v0tdte-imv0tcnRaoCh14ff.qxd 10.06.08 14:00 Page 987
  • 133. 988 CHAPTER 14 RANDOM VIBRATIONFIGURE 14.10 Complex Fourier series representation.Thus the mean square value of x(t) is given by the sum of the squares of the absolute val-ues of the Fourier coefficients. Equation (14.49) is known as Parseval’s formula for peri-odic functions [14.1].Complex Fourier Series ExpansionFind the complex Fourier series expansion of the function shown in Fig. 14.10(a).Solution: The given function can be expressed as(E.1)where the period and the fundamental frequency are given by(E.2)t = 2a and v0 =2pt=pa1v021t2x1t2 = μAa1 +tab, -t2… t … 0Aa1 -tab, 0 … t …t2EXAMPLE 14.2RaoCh14ff.qxd 10.06.08 14:00 Page 988
  • 134. 14.9 FOURIER ANALYSIS 989The Fourier coefficients can be determined as(E.3)Using the relation(E.4)can be evaluated as(E.5)This equation can be reduced to(E.6)Noting that(E.7)Eq. (E.6) can be simplified to obtain(E.8)The frequency spectrum is shown in Fig. 14.10(b).■cn = eA2, n = 0a4Aatn2v02b =2An2p2, n = 1, 3, 5, Á0, n = 2, 4, 6, Áeinpor e-inp= c1,-1,1,n = 0n = 1, 3, 5, Án = 2, 4, 6, Á+Aa1n2v021inp2ein p-Aa1n2v021inp2e-inpdcn =1tcAinv0einp+2Aa1n2v02-Ainv0e-inp-Aa1n2v02einp-Aa1n2v02e-inp+A-inv0e-inv0t`0t/2-Aaee-inv0t1-inv022[-inv0t - 1] f `0t/2dcn =1tcA-inv0e-inv0t`- t/20+Aaee-inv0t1-inv022[-inv0t - 1] f `- t/20cnLtektdt =ektk21kt - 12=1tcL0- t/2Aa1 +tabe-inv0tdt +Lt/20Aa1 -tabe-inv0tdt dcn =1t Lt/2- t/2x1t2e-inv0tdtRaoCh14ff.qxd 10.06.08 14:00 Page 989
  • 135. 990 CHAPTER 14 RANDOM VIBRATIONFIGURE 14.11 Nonperiodic function.A nonperiodic function, such as the one shown by the solid curve in Fig. 14.11, can betreated as a periodic function having an infinite period The Fourier seriesexpansion of a periodic function is given by Eqs. (14.43), (14.44) and (14.46):(14.50)with(14.51)and(14.52)As the frequency spectrum becomes continuous and the fundamental frequencybecomes infinitesimal. Since the fundamental frequency is very small, we can denote itas as and rewrite Eq. (14.52) as(14.53)By defining as(14.54)X1v2 = limt:q1tcn2 =Lq- qx1t2e-ivtdtX1v2limt:qtcn = limt:qLt/2- t/2x1t2e-ivtdt =Lq- qx1t2e-ivtdtv,¢v, nv0v0t : q,cn =1t Lt/2- t/2x1t2e-inv0tdtv0 =2ptx1t2 = aqn= - qcneinv0t1t : q2.14.9.2Fourier IntegralRaoCh14ff.qxd 10.06.08 14:00 Page 990
  • 136. 14.9 FOURIER ANALYSIS 991we can express x(t) from Eq. (14.50) as(14.55)This equation indicates the frequency decomposition of the nonperiodic function x(t) in acontinuous frequency domain, similar to Eq. (14.50) for a periodic function in a discretefrequency domain. The equations(14.56)and(14.57)are known as the (integral) Fourier transform pair for a nonperiodic function x(t), similarto Eqs. (14.50) and (14.52) for a periodic function x(t) [14.9, 14.10].The mean square value of a nonperiodic function x(t) can be determined from Eq.(14.49):(14.58)Since and as Eq. (14.58) gives the meansquare value of x(t) as(14.59)Equation (14.59) is known as Parseval’s formula for nonperiodic functions [14.1].x21t2 = limt:q1t Lt/2- t/2x21t2 dt =Lq- qƒX1v2ƒ22ptdvt : q,v0 : dvtcn : X1v2, tcn… : X*1v2,=1t aqn= - q1tcn21cn…t2v02p= aqn= - qcncn…tv0tv0= aqn= - qcncn…tv0ta2ptb1t Lt/2- t/2x21t2dt = aqn= - qƒcn ƒ2X1v2 =Lq- qx1t2e-ivtdtx1t2 =12p Lq- qX1v2eivtdv=12p Lq- qX1v2eivtdv= limt:q aqn= - q1cnt2eivta2ptb12px1t2 = limt:q aqn= - qcneivt 2pt2ptRaoCh14ff.qxd 10.06.08 14:00 Page 991
  • 137. 992 CHAPTER 14 RANDOM VIBRATIONFIGURE 14.12 Fourier transform of a triangular pulse.Fourier Transform of a Triangular PulseFind the Fourier transform of the triangular pulse shown in Fig. 14.12(a).Solution: The triangular pulse can be expressed as(E.1)The Fourier transform of x(t) can be found, using Eq. (14.57), as(E.2)=L0- qAa1 +tabe-ivtdt +Lq0Aa1 -tabe-ivtdtX1v2 =Lq- qAa1 -ƒtƒabe-ivtdtx1t2 =cAa1 -ƒtƒab, ƒtƒ … a0, otherwiseE X A M P L E 1 4 . 3RaoCh14ff.qxd 10.06.08 14:00 Page 992
  • 138. 14.10 POWER SPECTRAL DENSITY 993Since for Eq. (E.2) can be expressed as(E.3)Equation (E.3) can be simplified to obtain(E.4)Equation (E.4) is plotted in Fig. 14.12(b). Notice the similarity of this figure with the discrete Fourierspectrum shown in Fig. 14.10(b).■14.10 Power Spectral DensityThe power spectral density of a stationary random process is defined as the Fouriertransform of(14.60)so that(14.61)Equations (14.60) and (14.61) are known as the Wiener-Khintchine formulas [14.1]. Thepower spectral density is more often used in random vibration analysis than the autocor-relation function. The following properties of power spectral density can be observed:1. From Eqs. (14.27) and (14.61), we obtain(14.62)R102 = E[x2] =Lq- qS1v2 dvR1t2 =Lq- qS1v2eivtdvS1v2 =12p Lq- qR1t2e-ivtdtR1t2/2pS1v2=2Aav211 - cos va2 =4Aav2sin2ava2b=2Aav2-Aav21cos va + i sin va2 -Aav21cos va - i sin va2X1v2 =2Aav2+ eivaa -Aav2b + e-ivaa -Aav2b+ aA-ivbe-ivt`0a-Aaee-ivt1-iv22[-ivt - 1]f `0a= aA-ivbe-ivt`-a0+Aaee-ivt1-iv22[-ivt - 1] f `-a0X1v2 =L0-aAa1 +tabe-ivtdt +La0Aa1 -tabe-ivtdtƒtƒ 7 0,x1t2 = 0RaoCh14ff.qxd 10.06.08 14:00 Page 993
  • 139. 994 CHAPTER 14 RANDOM VIBRATIONFIGURE 14.13 Typical power spectral density function.1When several random processes are involved, a subscript is used to identify the power spectral density function(or simply the spectrum) of a particular random process. Thus denotes the spectrum of x(t).Sx1v2If the mean is zero, the variance of x(t) is given by(14.63)If x(t) denotes the displacement, R(0) represents the average energy. From Eq. (14.62),it is clear that represents the energy density associated with the frequencyThus indicates the spectral distribution of energy in a system. Also, in electricalcircuits, if x(t) denotes random current, then the mean square value indicates thepower of the system (when the resistance is unity). This is the origin of the term powerspectral density.2. Since is an even function of and real, is also an even and real functionof Thus A typical power spectral density function is shown inFig. 14.13.3. From Eq. (14.62), the units of can be identified as those of of angularfrequency. It can be noted that both negative and positive frequencies are counted inEq. (14.62). In experimental work, for convenience, an equivalent one-sided spec-trum is widely used [14.1, 14.2].1The spectrum is defined in terms of linear frequency (i.e., cycles per unittime) and only the positive frequencies are counted. The relationship betweenand can be seen with reference to Fig. 14.14. The differential frequency inFig. 14.14(a) corresponds to the differential frequency in Fig. 14.14(b).Since is the equivalent spectrum defined over positive values of f only, we have(14.64)E[x2] =Lq- qSx1v2 dv KLq0Wx1f2 dfWx1f2df = dv/2pdvWx1f2Sx1v2Wx1f2Wx1f2x2/unitS1v2S1-v2 = S1v2.v.S1v2tR1t2S1v2v.S1v2sx2= R102 =Lq- qS1v2 dvRaoCh14ff.qxd 10.06.08 14:00 Page 994
  • 140. 14.11 WIDE-BAND AND NARROW-BAND PROCESSES 995FIGURE 14.14 Two- and one-sided spectrum.In order to have the contributions of the frequency bands and df to the mean squarevalue to be same, the shaded areas in both Figs. 14.14(a) and (b) must be the same.Thus(14.65)which gives(14.66)14.11 Wide-Band and Narrow-Band ProcessesA wide-band process is a stationary random process whose spectral density functionhas significant values over a range or band of frequencies that is approximately the sameorder of magnitude as the center frequency of the band. An example of a wide-band ran-dom process is shown in Fig. 14.15. The pressure fluctuations on the surface of a rocketdue to acoustically transmitted jet noise or due to supersonic boundary layer turbulence areexamples of physical processes that are typically wide-band. A narrow-band randomprocess is a stationary process whose spectral density function has significant valuesonly in a range or band of frequencies whose width is small compared to the magnitude ofthe center frequency of the process. Figure 14.16 shows the sample function and the cor-responding spectral density and autocorrelation functions of a narrow-band process.A random process whose power spectral density is constant over a frequency range iscalled white noise, an analogy with the white light that spans the visible spectrum more orless uniformly. It is called ideal white noise if the band of frequencies is infinitelywide. Ideal white noise is a physically unrealizable concept, since the mean square valueof such a random process would be infinite because the area under the spectrum would beinfinite. It is called band-limited white noise if the band of frequencies has finite cut-offv2 - v1S1v2S1v2Wx1f2 = 2Sx1v2dvdf= 2Sx1v2dvdv/2p= 4pSx1v22Sx1v2 dv = Wx1f2 dfdvRaoCh14ff.qxd 10.06.08 14:00 Page 995
  • 141. 996 CHAPTER 14 RANDOM VIBRATIONFIGURE 14.15 Wide-band stationaryrandom process.frequencies and [14.8]. The mean square value of a band-limited white noise is givenby the total area under the spectrum—namely, where denotes the con-stant value of the spectral density function.Autocorrelation and Mean Square Value of a Stationary ProcessThe power spectral density of a stationary random process x(t) is shown in Fig. 14.17(a). Find itsautocorrelation function and the mean square value.E X A M P L E 1 4 . 4S02S01v2 - v12,v2v1FIGURE 14.16 Narrow-band stationary random process.RaoCh14ff.qxd 10.06.08 14:00 Page 996
  • 142. 14.11 WIDE-BAND AND NARROW-BAND PROCESSES 997FIGURE 14.17 Autocorrelation function of astationary process.Solution:(a) Since is real and even in Eq. (14.61) can be rewritten asThis function is shown graphically in Fig. 14.17(b).(b) The mean square value of the random process is given by■E[x2] =Lq- qSx1v2 dv = 2S0Lv2v1dv = 2S01v2 - v12=4S0tcosv1 + v22t sinv2 - v12t= 2S0a1tsin vtb `v1v2=2S0t1sin v2t - sin v1t2Rx1t2 = 2Lq0Sx1v2cos vt dv = 2S0Lv2v1cos vt dvv,Sx1v2RaoCh14ff.qxd 10.06.08 14:00 Page 997
  • 143. 998 CHAPTER 14 RANDOM VIBRATIONFIGURE 14.18Single degree offreedom system.2The unit impulse applied at is denoted aswhere is the Dirac delta function with (see Fig. 14.19b)Lq- qd1t - t2 dt = 1 1area under the curve is unity2d1t - t2 = 0 for all t except at t = td1t - t2 : q as t : td1t - t2x1t2 = d1t - t2t = t14.12 Response of a Single Degree of Freedom SystemThe equation of motion for the system shown in Fig. 14.18 is(14.67)whereThe solution of Eq. (14.67) can be obtained by using either the impulse response approachor the frequency response approach.Here we consider the forcing function x(t) to be made up of a series of impulses of vary-ing magnitude, as shown in Fig. 14.19(a) (see Section 4.5.2). Let the impulse applied at timebe denoted as If denotes the response to the unit impulse2excitation it is called the impulse response function. The total response of thesystem at time t can be found by superposing the responses to impulses of magnitudeapplied at different values of The response to the excitation willx1t2 dtt = t.x1t2 dtd1t - t2,y1t2 = h1t - t2x1t2 dt.t14.12.1ImpulseResponseApproachx1t2 =F1t2m, vn =Akm, z =ccc, and cc = 2kmy$+ 2zvny#+ vn2y = x1t2RaoCh14ff.qxd 10.06.08 14:00 Page 998
  • 144. 14.12 RESPONSE OF A SINGLE DEGREE OF FREEDOM SYSTEM 999FIGURE 14.19 Impulse response and the response to the total excitation will be given by the super-position or convolution integral:(14.68)y1t2 =Lt- qx1t2h1t - t2 dt[x1t2 dt]h1t - t2,RaoCh14ff.qxd 10.06.08 14:00 Page 999
  • 145. 1000 CHAPTER 14 RANDOM VIBRATIONThe transient function x(t) can be expressed in terms of its Fourier transform as(14.69)Thus x(t) can be considered as the superposition of components of different frequenciesIf we consider the forcing function of unit modulus as(14.70)its response can be denoted as(14.71)where is called the complex frequency response function (see Section 3.5). Sincethe actual excitation is given by the superposition of components of different frequencies(Eq. 14.69), the total response of the system can also be obtained by superposition as(14.72)If denotes the Fourier transform of the response function y(t), we can express y(t) interms of as(14.73)Comparison of Eqs. (14.72) and (14.73) yields(14.74)The following characteristics of the response function can be noted:1. Since when or (i.e., the response before the applicationof the impulse is zero), the upper limit of integration in Eq. (14.68) can be replacedby so that(14.75)y1t2 =Lq- qx1t2h1t - t2 dtqt 7 tt 6 th1t - t2 = 014.12.3Characteristicsof the ResponseFunctionY1v2 = H1v2X1v2y1t2 =12p Lq- qY1v2eivtdvY1v2Y1v2=12p Lq- qH1v2X1v2eivtdvy1t2 = H1v2x1t2 =Lq- qH1v212pX1v2eivtdvH1v2y1t2 = H1v2eivtx1t2 = eivtv.x1t2 =12p Lqv= - qX1v2eivtdvX1v214.12.2FrequencyResponseApproachRaoCh14ff.qxd 10.06.08 14:00 Page 1000
  • 146. 2. By changing the variable from to Eq. (14.75) can be rewritten as(14.76)3. The superposition integral, Eq. (14.68) or (14.75) or (14.76), can be used to find theresponse of the system y(t) for any arbitrary excitation x(t) once the impulse-responsefunction of the system h(t) is known. The Fourier integral, Eq. (14.72), can also beused to find the response of the system once the complex frequency response of thesystem, is known. Although the two approaches appear to be different, they areintimately related to one another. To see their inter-relationship, consider the excita-tion of the system to be a unit impulse in Eq. (14.72). By definition, the responseis h(t) and Eq. (14.72) gives(14.77)where is the Fourier transform of(14.78)since everywhere except at where it has a unit area and atEquations (14.77) and (14.78) give(14.79)which can be recognized as the Fourier integral representation of h(t) in whichis the Fourier transformation of h(t):(14.80)14.13 Response Due to Stationary Random ExcitationsIn the previous section, the relationships between excitation and response were derived forarbitrary known excitations x(t). In this section, we consider similar relationships when theexcitation is a stationary random process. When the excitation is a stationary randomprocess, the response will also be a stationary random process [14.15, 14.16]. We considerthe relation between the excitation and the response using the impulse response (timedomain) as well as the frequency response (frequency domain) approaches.H1v2 =Lq- qh1t2e-ivtdtH1v2h1t2 =12p Lq- qH1v2eivtdvt = 0.e-ivt= 1t = 0d1t2 = 0X1v2 =Lq- qx1t2e-ivtdt =Lq- qd1t2e-ivtdt = 1x1t2 = d1t2:X1v2y1t2 = h1t2 =12p Lq- qX1v2H1v2eivtdvd1t2H1v2,y1t2 =Lq- qx1t - u2h1u2 duu = t - t,t14.13 RESPONSE DUE TO STATIONARY RANDOM EXCITATIONS 1001RaoCh14ff.qxd 10.06.08 14:00 Page 1001
  • 147. 1002 CHAPTER 14 RANDOM VIBRATIONMean Values. The response for any particular sample excitation is given by Eq. (14.76):(14.81)For ensemble average, we write Eq. (14.81) for every (x, y) pair in the ensemble and thentake the average to obtain3(14.82)Since the excitation is assumed to be stationary, is a constant independent ofEq. (14.82) becomes(14.83)The integral in Eq. (14.83) can be obtained by setting in Eq. (14.80) so that(14.84)Thus a knowledge of either the impulse response function h(t) or the frequency responsefunction can be used to find the relationship between the mean values of the exci-tation and the response. It is to be noted that both E[x(t)] and E[y(t)] are independent of t.Autocorrelation. We can use a similar procedure to find the relationship between theautocorrelation functions of the excitation and the response. For this, we first write(14.85)# h1u12h1u22 du1 du2=Lq- q Lq- qx1t - u12x1t + t - u22#Lq- qx1t + t - u22h1u22d u2y1t2y1t + t2 =Lq- qx1t - u12h1u12 du1H1v2H102 =Lq- qh1t2 dtv = 0E[y1t2] = E[x1t2]Lq- qh1u2 dut,E[x1t2]=Lq- qE[x1t - u2]h1u2 duE[y1t2] = EcLq- qx1t - u2h1u2 dudy1t2 =Lq- qx1t - u2h1u2 du14.13.1ImpulseResponseApproach3In deriving Eq. (14.82), the integral is considered as a limiting case of a summation and hence the average of asum is treated to be same as the sum of the averages, that is,E[x1 + x2 + Á ] = E[x1] + E[x2] + ÁRaoCh14ff.qxd 10.06.08 14:00 Page 1002
  • 148. 14.13 RESPONSE DUE TO STATIONARY RANDOM EXCITATIONS 1003where and are used instead of to avoid confusion. The autocorrelation function ofy(t) can be found as(14.86)Power Spectral Density. The response of the system can also be described by its powerspectral density, which by definition, is (see Eq. 14.60)(14.87)Substitution of Eq. (14.86) into Eq. (14.87) gives(14.88)Introduction of(14.89)into Eq. (14.88) results in(14.90)In the third integral on the right-hand side of Eq. (14.90), and are constants and theintroduction of a new variable of integration as(14.91)h = t + u1 - u2hu2u1*12p Lq- qRx1t + u1 - u22e-iv1u1 -u22dtSy1v2 =Lq- qh1u12eivu1du1Lq- qh1u22e-ivu2du2eivu1e-ivu2e-iv1u1 -u22= 1*Lq- q Lq- qRx1t + u1 - u22h1u12h1u22du1 du2Sy1v2 =12p Lq- qe-ivtdtSy1v2 =12p Lq- qRy1t2e-ivtdt14.13.2FrequencyResponseApproach=Lq- q Lq- qRx1t + u1 - u22h1u12h1u22du1 du2=Lq- q Lq- qE[x1t - u12x1t + t - u22]h1u12h1u22du1 du2Ry1t2 = E[y1t2y1t + t2]uu2u1RaoCh14ff.qxd 10.06.08 14:00 Page 1003
  • 149. 1004 CHAPTER 14 RANDOM VIBRATIONleads to(14.92)The first and the second integrals on the right-hand side of Eq. (14.90) can be recognizedas the complex frequency response functions and respectively. Sinceis the complex conjugate of Eq. (14.90) gives(14.93)This equation gives the relationship between the power spectral densities of the excitationand the response.Mean Square Response. The mean square response of the stationary random processy(t) can be determined either from the autocorrelation function or from the powerspectral density(14.94)and(14.95)Note: Equations (14.93) and (14.95) form the basis for the random vibration analysisof single- and multidegree of freedom systems [14.11, 14.12]. The random vibrationanalysis of road vehicles is given in Refs. [14.13, 14.14].Mean Square Value of ResponseA single degree of freedom system (Fig. 14.20a) is subjected to a force whose spectral density is awhite noise Find the following:(a) Complex frequency response function of the system(b) Power spectral density of the response(c) Mean square value of the responseSolution(a) To find the complex frequency response function we substitute the input as and thecorresponding response as in the equation of motionmy$+ cy#+ ky = x1t2y1t2 = H1v2eivteivtH1v2,Sx1v2 = S0.E X A M P L E 1 4 . 5E[y2] =Lq- qSy1v2 dv =Lq- qƒH1v2ƒ2Sx1v2 dvE[y2] = Ry102 =Lq- q Lq- qRx1u1 - u22h1u12h1u22 du1 du2Sy1v2:Ry1t2Sy1v2 = ƒH1v2ƒ2Sx1v2H1v2,H1- v2H1- v2,H1v2=12p Lq- qRx1h2e-ivhdh K Sx1v212p Lq- qRx1t + u1 - u22e-iv1t+u1 -u22dtRaoCh14ff.qxd 10.06.08 14:00 Page 1004
  • 150. 14.13 RESPONSE DUE TO STATIONARY RANDOM EXCITATIONS 1005FIGURE 14.20 Single degree offreedom obtainand(E.1)(b) The power spectral density of the output can be found as(E.2)Sy1v2 = ƒH1v2ƒ2Sx1v2 = S0 `1- mv2+ icv + k`2H1v2 =1- mv2+ icv + k1- mv2+ icv + k2H1v2eivt= eivtRaoCh14ff.qxd 10.06.08 14:00 Page 1005
  • 151. 1006 CHAPTER 14 RANDOM VIBRATIONFIGURE 14.21 Single-story building.(c) The mean square value of the output is given by4(E.3)which can be seen to be independent of the magnitude of the mass m. The functions andare shown graphically in Fig. 14.20(b).■Design of the Columns of a BuildingA single-story building is modeled by four identical columns ofYoung’s modulus E and height h anda rigid floor of weight W, as shown in Fig. 14.21(a). The columns act as cantilevers fixed at theground. The damping in the structure can be approximated by an equivalent viscous damping con-stant c. The ground acceleration due to an earthquake is approximated by a constant spectrum Ifeach column has a tubular cross section with mean diameter d and wall thickness find themean diameter of the columns such that the standard deviation of the displacement of the floor rel-ative to the ground does not exceed a specified valueSolutionApproach: Model the building as a single degree of freedom system. Use the relation between thepower spectral densities of excitation and output.The building can be modeled as a single degree of freedom system as shown in Fig. 14.21(b)with(E.1)m = W/gd.t = d/10,S0.E X A M P L E 1 4 . 6Sy1v2H1v2= S0Lq- q`1- mv2+ k + icv`2dv =pS0kcE[y2] =Lq- qSy1v2 dv4The values of this and other similar integrals have been found in the literature [14.1]. For example, ifH1v2 =ivB1 + B0- v2A2 + ivA1 + A0,Lq- qƒ H1v2ƒ2dv = pe1B02/A02A2 + B12A1A2fRaoCh14ff.qxd 10.06.08 14:00 Page 1006
  • 152. 14.13 RESPONSE DUE TO STATIONARY RANDOM EXCITATIONS 1007and(E.2)since the stiffness of one cantilever beam (column) is equal to where E is the Young’smodulus, h is the height, and I is the moment of inertia of the cross section of the columns given by(see Fig. 14.21(c)):(E.3)Equation (E.3) can be simplified, using and as(E.4)With Eq. (E.4) becomes(E.5)and hence Eq. (E.2) gives(E.6)When the base of the system moves, the equation of motion is given by (see Section 3.6)(E.7)where is the displacement of the mass (floor) relative to the ground. Equation (E.7) canbe rewritten as(E.8)The complex frequency response function can be obtained by making the substitution(E.9)x$= eivtand z1t2 = H1v2eivtH1v2z$+cmz#+kmz = - x$z = y - xmz$+ cz#+ kz = - mx$k =12 E10.03966d42h3=0.47592 Ed4h3I =101p8000d4= 0.03966d4t = d/10,=p8dt1d2+ t22=p64[1d + t22+ 1d - t22][1d + t2 + 1d - t2][1d + t2 - 1d - t2]I =p641d02+ di221d0 + di21d0 - di2di = d - t,d0 = d + tI =p641d04- di4213EI/h32,k = 4a3EIh3bRaoCh14ff.qxd 10.06.08 14:00 Page 1007
  • 153. 1008 CHAPTER 14 RANDOM VIBRATIONso thatwhich gives(E.10)The power spectral density of the response z(t) is given by(E.11)The mean square value of the response z(t) can be determined, using Eq. (E.4) of Example 14.5, as(E.12)Substitution of the relations (E.1) and (E.6) into Eq. (E.12) gives(E.13)Assuming the mean value of z(t) to be zero, the standard deviation of z can be found as(E.14)Since we find thator(E.15)d4ÚpS0W2h30.47592g2cEd2pS0W2h30.47592g2cEd4… d2sz … d,sz = 4E[z2] =BpS0W2h30.47592g2cEd4E[z2] = pS0W2h3g2c10.47592 Ed42= S0apm2kcb= S0Lq- q∞-1a - v2+ ivcm+kmb∞2dvE[z2] =Lq- qSz1v2dvSz1v2 = ƒH1v2ƒ2Sx$1v2 = S0 ∞-1a - v2+ ivcm+kmb∞2H1v2 =-1a - v2+ ivcm+kmbc - v2+ ivcm+kmdH1v2eivt= -eivtRaoCh14ff.qxd 10.06.08 14:00 Page 1008
  • 154. 14.14 RESPONSE OF A MULTIDEGREE OF FREEDOM SYSTEM 1009Thus the required mean diameter of the columns is given by(E.16)■14.14 Response of a Multidegree of Freedom SystemThe equations of motion of a multidegree of freedom system with proportional dampingcan be expressed, using the normal mode approach, as (see Eq. 6.128)(14.96)where n is the number of degrees of freedom, is the ith natural frequency, is theith generalized coordinate, and is the ith generalized force. The physical and gener-alized coordinates are related asor(14.97)where [X] is the modal matrix and is the ith component of jth modal vector. The phys-ical and generalized forces are related asor(14.98)where is the force acting along the coordinate Let the applied forces beexpressed as(14.99)so that Eq. (14.98) becomes(14.100)Qi1t2 = a anj=1Xj1i2fjbt1t2 = Ni t1t2Fj1t2 = fj t1t2xj1t2.Fj1t2Qi1t2 = anj=1Xj1i2Fj1t2Q!1t2 = [X]TF!1t2Xi1j2xi 1t2 = anj=1Xi1j2qj1t2x!1t2 = [X]q!1t2Qi1t2qi1t2viq$i1t2 + 2 zi vi q#i1t2 + vi2qi1t2 = Qi1t2; i = 1, 2, Á , nd Ú epS0W2h30.47592g2cEd2f1/4RaoCh14ff.qxd 10.06.08 14:00 Page 1009
  • 155. 1010 CHAPTER 14 RANDOM VIBRATIONwhere(14.101)By assuming a harmonic force variation(14.102)the solution of Eq. (14.96) can be expressed as(14.103)where denotes the frequency response function(14.104)The mean square value of the physical displacement, can be obtained fromEqs. (14.97) and (14.103) as(14.105)From Eq. (3.56), can be expressed as(14.106)where the magnitude of known as the magnification factor, is given by(14.107)and the phase angle, by(14.108)fr = tan-1μ2zrvvr1 - avvrb2∂fr,ƒHr1v2ƒ = e c1 - avvrb2d2+ a2zrvvrb2f-1/2Hr1v2,Hr1v2 = ƒHr1v2ƒe-ifrHr1v2= anr=1ans=1Xi1r2Xi1s2 Nrvr2Nsvs2limT:qLT- THr1v2Hs1v2 t21t2 dtxi21t2 = limt:q12T LT- Txi21t2 dtxi1t2,Hi1v2 =11 - avvib2+ i 2 zivviHi1v2qi1t2 =Nivi2Hi1v2 t1t2t1t2 = eivtNi = anj=1Xj1i2fjRaoCh14ff.qxd 10.06.08 14:00 Page 1010
  • 156. 14.14 RESPONSE OF A MULTIDEGREE OF FREEDOM SYSTEM 1011By neglecting the phase angles, the integral on the right-hand side of Eq. (14.105) can beexpressed as(14.109)For a stationary random process, the mean square value of can be expressed in termsof its power spectral density function, as(14.110)Combining Eqs. (14.109) and (14.110) gives(14.111)Substitution of Eq. (14.111) into Eq. (14.105) yields the mean square value of as(14.112)The magnification factors, and are shown in Fig. 14.22. It can be seenthat the product, for is often negligible compared to andHence Eq. (14.112) can be rewritten as(14.113)For lightly damped systems, the integral in Eq. (14.113) can be evaluated by approximat-ing the graph of to be flat with as(14.114)Lq- qƒHr1v2ƒ2St1v2 dv L St1vr2Lq- qƒHr1v2ƒ2dv =pvr St1vr22zrSt1v2 = St1vr2St1v2xi21t2 L anr=1aXi1r2b2Nr2vr4Lq- qƒHr1v2ƒ2St1v2 dvƒHs1v2ƒ2.ƒHr1v2ƒ2r Z s,ƒHr1v2ƒ ƒ Hs1v2ƒƒHs1v2ƒ,ƒHr1v2ƒxi21t2 L anr=1ans=1Xi1r2Xi1s2 Nrvr2Nsvs2Lq- qƒHr1v2ƒ ƒ Hs1v2ƒSt1v2 dvxi1t2LLq- qƒHr1v2ƒ ƒ Hs1v2ƒSt1v2 dvlimT:q12T LT- THr1v2Hs1v2 t21t2 dtt21t2 = limT:q12T LT- Tt21t2 dt =Lq- qSt1v2 dvSt1v2,t21t2L limT:q12T LT- TƒHr1v2ƒ ƒ Hs1v2ƒt21t2 dtlimT:q12T LT- THr1v2Hs1v2 t21t2 dtRaoCh14ff.qxd 10.06.08 14:00 Page 1011
  • 157. 1012 CHAPTER 14 RANDOM VIBRATIONEquations (14.113) and (14.114) yield(14.115)The following example illustrates the computational procedure.Response of a Building Frame Under an EarthquakeThe three-story building frame shown in Fig. 14.23 is subjected to an earthquake. The ground accel-eration during the earthquake can be assumed to be a stationary random process with a power spec-tral density Assuming a modal damping ratio of 0.02 in each mode,S1v2 = 0.05(m2/s4)/(rad/s).E X A M P L E 1 4 . 7xi21t2 = anr=11Xi1r222 Nr2vr4apvr St1vr22zrbFIGURE 14.23 Three-storybuilding frame.FIGURE 14.22 Magnification factors.RaoCh14ff.qxd 10.06.08 14:00 Page 1012
  • 158. 14.14 RESPONSE OF A MULTIDEGREE OF FREEDOM SYSTEM 1013determine the mean square values of the responses of the various floors of the building frame underthe earthquake.Solution: The stiffness and mass matrices of the building frame can be found as(E.1)(E.2)From Examples 6.10 and 6.11, the eigenvalues and the eigenvectors (normalized with respect to themass matrix [m]) can be computed using the values N/m and as(E.3)(E.4)(E.5)(E.6)(E.7)(E.8)Note that the notation is used to denote the ith mode shape instead of since the relative dis-placements, will be used instead of the absolute displacements, in the solution.By denoting the ground motion as y(t), the relative displacements of the floors, can beexpressed as The equations of motion can be expressed as(E.9)[m]x!$+ [c]z!#+ [k]z!= 0!zi1t2 = xi1t2 - y1t2, i = 1, 2, 3.zi1t2,xi1t2,zi1t2,X!1i2Z!1i2Z!132=0.59111mμ1.0000- 1.24680.5544∂ = μ0.01869- 0.023300.01036∂Z!122=0.73701mμ1.00000.4450- 0.8020∂ = μ0.023310.01037- 0.01869∂Z!112=0.32801mμ1.00001.80192.2470∂ = μ0.010370.018690.02330∂v3 = 1.8025Akm= 57.0001 rad/sv2 = 1.2471Akm= 39.4368 rad/sv1 = 0.44504Akm= 14.0734 rad/sm = 1000 kg,k = 106[m] = mC1 0 00 1 00 0 1S[k] = kC2 -1 0-1 2 -10 -1 1SRaoCh14ff.qxd 10.06.08 14:00 Page 1013
  • 159. which can be rewritten as(E.10)where By expressing the vector in terms of normal modes, we have(E.11)where [Z] denotes the modal matrix. By substituting Eq. (E.11) into Eq. (E.10) and premultiplyingthe resulting equation by we can derive the uncoupled equations of motion. Assuming adamping ratio in mode i, the uncoupled equations of motion are given by(E.12)where(E.13)and(E.14)with denoting the mass of the floor. By representing as(E.15)we note that(E.16)and(E.17)The mean square values, can be determined from Eq. (14.115):(E.18)Nothing that and are given by Eqs. (E.6), (E.7), and (E.8), respectively, and(E.19)(E.20)N2 = a3j=1Zj122fj = - ma3j=1Zj122= - 1000 10.052372 = -52.37N1 = a3j=1Zj112fj = - ma3j=1Zj112= - 1000 10.052362 = -52.36Z!132Z!112, Z!122,zi21t2 = a3r=11Zi1r222Nr2vr3ap2zrbSt1vr2zi21t2,t1t2 = y$1t2fj = - mj = - mFj1t2 = fj t1t2Fj1t2jthmj = mFj1t2 = - mj y$1t2 = - m y$1t2Qi = anj=1Zj1i2Fj1t2q$i + 2 zi vi + vi2qi = Qi; i = 1, 2, 3zi1zi = 0.022[Z]T,z!= [Z]q!z!y!$= μy$y$y$∂.[m]z!$+ [c]z!#+ [k]z!= - [m]y!$1014 CHAPTER 14 RANDOM VIBRATIONRaoCh14ff.qxd 10.06.08 14:00 Page 1014
  • 160. 14.15 EXAMPLES USING MATLAB 1015(E.21)Eq. (E.18) yields the mean square values of the relative displacements of the various floors of thebuilding frame as(E.22)(E.23)(E.24)■Probability of Relative Displacement Exceeding a Specified ValueFind the probability of the magnitude of the relative displacement of the various floors exceeding 1,2, 3, and 4 standard deviations of the corresponding relative displacement for the building frame ofExample 14.7.SolutionApproach: Assume the ground acceleration to be a normally distributed random process with zeromean and use standard normal tables.Since the ground acceleration, is assumed to be normally distributed with zero meanvalue, the relative displacements of the various floors can also be assumed to be normally distributedwith zero mean values. Thus the standard deviations of the relative displacements of the floors aregiven byThe probability of the absolute value of the relative displacement, exceeding a specified num-ber of standard deviations can be found from standard normal tables as (see Section 14.8):■14.15 Examples Using MATLABPlotting of Autocorrelation FunctionUsing MATLAB, plot the autocorrelation function corresponding to white noise with spectral den-sity for the following cases:(a) band limited white noise with and(b) band limited white noise with and(c) ideal white noisev2 = 4 rad/s, 6 rad/s, 8 rad/sv1 = 2 rad/sv2 = 4 rad/s, 6 rad/s, 8 rad/sv1 = 0S0EXAMPLE 14.9P[ƒzi1t2ƒ 7 pszi] = μ0.31732 for p = 10.04550 for p = 20.00270 for p = 30.00006 for p = 4zi1t2,szi = 3zi21t2; i = 1, 2, 3y$1t2,EXAMPLE 14.8z321t2 = 0.00216455 m2z221t2 = 0.00139957 m2z121t2 = 0.00053132 m2N3 = a3j=1Zj132fj = - ma3j=1Zj132= - 1000 10.052352 = -52.35RaoCh14ff.qxd 10.06.08 14:00 Page 1015
  • 161. 1016 CHAPTER 14 RANDOM VIBRATIONSolutionFor (a) and (b), the autocorrelation function, can be expressed as (from Example 14.4)(E.1)For (c) it can be expressed asFor an ideal white noise, we let and which yields where is theDirac delta function. The MATLAB program to plot Eq. (E.1) is given below.% Ex14_9.mw1 = 0;w2 = 4;for i = 1:101t(i) = Ϫ5 + 10* (iϪ1)/100;R1(i) = 2 * ( sin(w2 *t(i)) Ϫ sin(w1*t(i)) )/t(i);endw1 = 0;w2 = 6;for i = 1:101t(i) = Ϫ5 + 10*(iϪ1)/100;R2(i) = 2 * ( sin(w2 *t(i)) Ϫ sin(w1*t(i)) )/t(i);endw1 = 0;w2 = 8;for i = 1:101t(i) = Ϫ 5 + 10*(iϪ1)/100;R3(i) = 2 * ( sin(w2 *t(i)) Ϫ sin(w1*t(i)) )/t(i);endfor i = 1:101t1(i) = 0.0001 + 4.9999*(iϪ1)/100;R3_1(i) = 2 * ( sin(w2 *t(i)) Ϫ sin(w1*t(i)) )/t(i);endxlabel (t);ylabel(R/S_0);plot(t,R1);hold on;gtext(Solid line: w1 = 0, w2 = 4)gtext(Dashed line: w1 = 0, w2 = 6);plot(t,R2,--);gtext(Dotted line: w1 = 0, w2 = 8);plot(t,R3,:);w1 = 2;w2 = 4;for i = 1:101t(i) = Ϫ5 + 10*(iϪ1)/100;R4(i) = 2 * ( sin(w2 *t(i)) Ϫ sin(w1*t(i)) )/t(i);endw1 = 2;w2 = 6;for i = 1:101t(i) = Ϫ5 + 10*(iϪ1)/100;R5(i) = 2 * ( sin(w2 *t(i)) Ϫ sin(w1*t(i)) )/t(i);endw1 = 2;w2 = 8;for i = 1 : 101t (i) = Ϫ5 + 10* (iϪ1) / 100;R6 (i) = 2 * ( sin (w2 *t (i) ) Ϫ sin (w1 *t (i) ) ) / t (i);endpaused1t2R = 2S0 d1t2v2 : q,v1 = 0R102 = limt:0e2 S0av2 sin v2 tv2 tb - 2 S0av1 sin v1 tv1 tb f = 2 S01v2 - v12t : 0,R1t2S0=2t1sin v2t - sin v1t2R1t2,RaoCh14ff.qxd 10.06.08 14:00 Page 1016
  • 162. 14.15 EXAMPLES USING MATLAB 1017hold off;xlabel (t);ylabel (R/S_0);plot (t, R4);hold on;gtext (Solid line: w1 = 2, w2 = 4)gtext (Dashed line: w1 = 2, w2 = 6);plot (t, R5, --);gtext (Dotted line: w1 = 2, w2 = 8);plot (t, R6, :);■RaoCh14ff.qxd 10.06.08 14:00 Page 1017
  • 163. 1018 CHAPTER 14 RANDOM VIBRATIONEvaluation of a Gaussian Probability Distribution FunctionUsing MATLAB, evaluate the following probability for and 3:(E.1)Assume the mean value of x(t) to be zero and standard deviation of x(t) to be one.SolutionEquation (E.1) can be rewritten, for as(E.2)The MATLAB program to evaluate Eq. (E.2) is given below.Ex14_10.m>> q = quad (normp, Ϫ7, 1);>> prob1 = 2 * qprob1 =1.6827>> q = quad (normp, Ϫ7, 2);>> prob2 = 2 * qprob2 =1.9545>> q = quad (normp, Ϫ7, 3);>> prob3 = 2 * qprob3 =1.9973%normp.mfunction pdf = normp(x)pdf = exp(Ϫ0.5*x.^2)/sqrt(2.0 * pi);■REFERENCES14.1 S. H. Crandall and W. D. Mark, Random Vibration in Mechanical Systems, Academic Press,New York, 1963.14.2 D. E. Newland, An Introduction to Random Vibrations and Spectral Analysis, Longman,London, 1975.Prob [ƒx1t2ƒ Ú c] = 2e112p 3c- qe-0.5x2dx fs = 1,Prob 3 ƒx1t2ƒ Ú c s4 =212ps Lqc se- 12x2s2dxc = 1, 2EXAMPLE 14.10RaoCh14ff.qxd 10.06.08 14:00 Page 1018
  • 164. REVIEW QUESTIONS 101914.3 J. D. Robson, An Introduction to Random Vibration, Edinburgh University Press, Edinburgh,1963.14.4 C. Y. Yang, Random Vibration of Structures, Wiley, New York, 1986.14.5 A. Papoulis, Probability, Random Variables and Stochastic Processes, McGraw-Hill, NewYork,1965.14.6 J. S. Bendat and A. G. Piersol, Engineering Applications of Correlation and Spectral Analysis,Wiley, New York, 1980.14.7 P. Z. Peebles, Jr., Probability, Random Variables, and Random Signal Principles, McGraw-Hill,New York, 1980.14.8 J. B. Roberts, “The response of a simple oscillator to band-limited white noise,” Journal ofSound and Vibration, Vol. 3, 1966, pp. 115–126.14.9 M. H. Richardson, “Fundamentals of the discrete Fourier transform,” Sound and Vibration,Vol. 12, March 1978, pp. 40–46.14.10 E. O. Brigham, The Fast Fourier Transform, Prentice Hall, Englewood Cliffs, N.J., 1974.14.11 J. K. Hammond, “On the response of single and multidegree of freedom systems to non-sta-tionary random excitations,” Journal of Sound and Vibration, Vol. 7, 1968, pp. 393–416.14.12 S. H. Crandall, G. R. Khabbaz, and J. E. Manning, “Random vibration of an oscillator withnonlinear damping,” Journal of the Acoustical Society of America, Vol. 36, 1964,pp. 1330–1334.14.13 S. Kaufman, W. Lapinski, and R. C. McCaa, “Response of a single-degree-of-freedom isola-tor to a random disturbance,” Journal of the Acoustical Society of America, Vol. 33, 1961, pp.1108–1112.14.14 C. J. Chisholm, “Random vibration techniques applied to motor vehicle structures,” Journalof Sound and Vibration, Vol. 4, 1966, pp. 129–135.14.15 Y. K. Lin, Probabilistic Theory of Structural Dynamics, McGraw-Hill, New York, 1967.14.16 I. Elishakoff, Probabilistic Methods in the Theory of Structures, Wiley, New York, 1983.14.17 H. W. Liepmann, “On the application of statistical concepts to the buffeting problem,” Journalof the Aeronautical Sciences, Vol. 19, No. 12, 1952, pp. 793–800, 822.14.18 W. C. Hurty and M. F. Rubinstein, Dynamics of Structures, Prentice Hall, Englewood Cliffs,N.J., 1964.REVIEW QUESTIONS14.1 Give brief answers to the following:1. What is the difference between a sample space and an ensemble?2. Define probability density function and probability distribution function.3. How are the mean value and variance of a random variable defined?4. What is a bivariate distribution function?5. What is the covariance between two random variables X and Y?6. Define the correlation coefficient, rXY.RaoCh14ff.qxd 10.06.08 14:00 Page 1019
  • 165. 1020 CHAPTER 14 RANDOM VIBRATION7. What are the bounds on the correlation coefficient?8. What is a marginal density function?9. What is autocorrelation function?10. Explain the difference between a stationary process and a nonstationary process?11. What are the bounds on the autocorrelation function of a stationary random process?12. Define an ergodic process.13. What are temporal averages?14. What is a Gaussian random process? Why is it frequently used in vibration analysis?15. What is Parseval’s formula?16. Define the following terms: power spectral density function, white noise, band-limitedwhite noise, wide-band process, and narrow-band process.17. How are the mean square value, autocorrelation function, and the power spectral densityfunction of a stationary random process related?18. What is an impulse response function?19. Express the response of a single degree of freedom system using the Duhamel integral.20. What is complex frequency response function?21. How are the power spectral density functions of input and output of a single degree offreedom system related?22. What are Wiener-Khintchine formulas?14.2 Indicate whether each of the following statements is true or false:1. A deterministic system requires deterministic system properties and loading.2. Most phenomena in real life are deterministic.3. A random variable is a quantity whose magnitude cannot be predicated precisely.4. The expected value of x, in terms of its probability density function, p(x), is given by5. The correlation coefficient satisfies the relation6. The autocorrelation function is the same as7. The mean square value of x(t) can be determined as8. If x(t) is stationary, its mean will be independent of t.9. The autocorrelation function is an even function of10. The Wiener-Khintchine formulas relate the power spectral density to the autocorrelationfunction.11. The ideal white noise is a physically realizable concept.14.3 Fill in each of the following blanks with the appropriate word:1. When the vibrational response of a system is known precisely, the vibration is called_____ vibration.2. If any parameter of a vibrating system is not known precisely, the resulting vibration iscalled _____ vibration.3. The pressure fluctuation at a point on the surface of an aircraft flying in the air is a _____process.4. In a random process, the outcome of an experiment will be a function of some _____such as time.5. The standard deviation is the positive square root of _____.t.R1t2E[x2] = R102.E[x1t12x1t22].R1t1, t22ƒrXY ƒ … 1.rXY1q- qxp1x2 dx.RaoCh14ff.qxd 10.06.08 14:00 Page 1020
  • 166. REVIEW QUESTIONS 10216. The joint behavior of several random variables is described by the _____ probability dis-tribution function.7. Univariate distributions describe the probability distributions of _____ random variables.8. The distribution of two random variables is known as _____ distribution.9. The distribution of several random variables is called _____ distribution.10. The standard deviation of a stationary random process x(t) will be independent of _____.11. If all the probability information of a stationary random process can be obtained from asingle sample function, the process is said to be _____.12. The Gaussian density function is a symmetric _____-shaped curve about the mean value.13. The standard normal variable has mean of _____ and standard deviation of _____.14. A nonperiodic function can be treated as a periodic function having an _____ period.15. The _____ spectral density function is an even function of16. If has significant values over a wide range of frequencies, the process is called a_____ process.17. If has significant values only over a small range of frequencies, the process iscalled a _____ process.18. The power spectral density of a stationary random process is defined as the _____transform of19. If the band of frequencies has finite cut-off frequencies for a white noise, it is called_____ white noise.14.4 Select the most appropriate answer out of the choices given:1. Each outcome of an experiment for a random variable is called(a) a sample point (b) a random point (c) an observed value2. Each outcome of an experiment, in the case of a random process, is called a(a) sample point (b) sample space (c) sample function3. The probability distribution function, denotes(a)(b)(c)4. The probability density function, denotes(a)(b)(c)5. Normalization of probability distribution implies(a) (b) (c)6. The variance of x is given by(a) (b) (c)7. The marginal density function of x can be determined form the bivariate density functionp(x,y) as(a) p1x2 =Lq- qp1x, y2dy1x221x22 - 1x22x2Lq- qp1x2 = 0Lq- qp1x2 = 1P1q2 = 1P1x… x … x+ ¢x2P1x 7 x2P1x … x2.p1x2,P1x… x … x+ ¢x2P1x 7 x2P1x … x2P1x2,R1t2/2p.S1v2S1v2S1v2v.RaoCh14ff.qxd 10.06.08 14:00 Page 1021
  • 167. 1022 CHAPTER 14 RANDOM VIBRATION(b)(c)8. The correlation coefficient of x and y is given by(a) (b) (c)9. The standard normal variable, z, corresponding to the normal variable x, is defined as(a) (b) (c)10. If the excitation of a linear system is a Gaussian process, the response will be(a) a different random process(b) a Gaussian process(c) an ergodic process11. For a normal probability density function, is(a) 0.6827 (b) 0.999937 (c) 0.997312. The mean square response of a stationary random process can be determined from the:(a) autocorrelation function only(b) power spectral density only(c) autocorrelation function or power spectral density14.5 Match the items in the two columns below:1. All possible outcomes of a random variable (a) Correlation functions in anexperiment2. All possible outcomes of a random process (b) Nonstationary process3. Statistical connections between the (c) Sample spacevalues of x(t) at times4. A random process invariant under a shift (d) White noiseof the time scale5. Mean and standard deviations of x(t) (e) Stationary processvary with t6. Power spectral density is constant over (f) Ensemblea frequency rangePROBLEMSThe problem assignments are organized as follows:Problems Section Covered Topic Covered14.1, 14.10 14.3 Probability distribution14.3, 14.11 14.4 Mean value and standard deviationt1, t2, ÁProb[-3s … x1t2 … 3s]z =xsxz =x - xsxz =xsxsxsysxy/1sxsy2sxyp1x2 =Lq- q Lq- qp1x, y2dx dyp1x2 =Lq- qp1x, y2dxRaoCh14ff.qxd 10.06.08 14:00 Page 1022
  • 168. PROBLEMS 1023Problems Section Covered Topic Covered14.2, 14.4 14.5 Joint probability distribution14.7, 14.9 14.6 Correlation functions14.5, 14.26 14.7 Stationary random process14.12 14.8 Gaussian random process14.8, 14.13–14.16, 14.27 14.9 Fourier analysis14.6, 14.17–14.22 14.10 Power spectral density14.23–14.25, 14.28–14.30 14.12 Response of a single degree of freedom system14.31, 14.32 14.14 Response of a multidegree of freedom system14.33–14.35 14.15 MATLAB programs14.36 — Design project14.1 The strength of the foundation of a reciprocating machine (x) has been found to vary between20 and according to the probability density function:What is the probability of the foundation carrying a load greater than14.2 The joint density function of two random variables X and Y is given by(a) Find the marginal density functions of X and Y. (b) Find the means and standard deviationsof X and Y. (c) Find the correlation coefficient14.3 The probability density function of a random variable x is given byDetermine E[x], and14.4 If x and y are statistically independent, then That is, the expected valueof the product xy is equal to the product of the separate mean values. If where xand y are statistically independent, show that14.5 The autocorrelation function of a random process x(t) is given byFind the mean square value of x(t).Rx1t2 = 20 +51 + 3t2E[z2] = E[x2] + E[y2] + 2E[x]E[y].z = x + y,E[xy] = E[x]E[y].sx.E[x2],p1x2 = c0 for x 6 00.5 for 0 … x … 20 for x 7 2rX,Y.pX,Y1x, y2 = Lxy9, 0 … x … 2, 0 … y … 30, elsewhere28 kips/ft2?p1x2 = Lka1 -x30b, 20 … x … 300, elsewhere30 kips/ft2RaoCh14ff.qxd 10.06.08 14:00 Page 1023
  • 169. 1024 CHAPTER 14 RANDOM VIBRATIONFIGURE 14.2414.6 The autocorrelation function of a random process is given bywhere A and are constants. Find the power spectral density of the random process.14.7 Find the autocorrelation functions of the periodic functions shown in Fig. 14.24.vRx1t2 = A cos vt; -t2v… t …p2v14.8 Find the complex form of the Fourier series for the wave shown in Fig. 14.24(b).14.9 Compute the autocorrelation function of a periodic square wave with zero mean value andcompare this result with that of a sinusoidal wave of the same period. Assume the amplitudesto be the same for both waves.14.10 The life T in hours of a vibration transducer is found to follow exponential distributionwhere is a constant. Find (a) the probability distribution function of T, (b) mean value of T,and (c) standard deviation of T.14.11 Find the temporal mean value and the mean square value of the function14.12 An air compressor of mass 100 kg is mounted on an undamped isolator and operates at anangular speed of 1800 rpm. The stiffness of the isolator is found to be a random variable withmean value and standard deviation followingnormal distribution. Find the probability of the natural frequency of the system exceeding theforcing frequency.14.13–14.16 Find the Fourier transform of the functions shown in Figs. 14.25–14.28 and plot the corre-sponding spectrum.14.17 A periodic function F(t) is shown in Fig. 14.29. Use the values of the function F(t) at tenequally spaced time stations to find (a) the spectrum of F(t) and (b) the mean square valueof F(t).14.18 The autocorrelation function of a stationary random process x(t) is given bywhere a and b are constants. Find the power spectral density of x(t).Rx1t2 = ae-bƒtƒtisk = 0.225 * 106N/mk!= 2.25 * 106N/mx1t2 = x0 sin1pt/22.lpT1t2 = ele-lt, t Ú 00, t 6 0RaoCh14ff.qxd 10.06.08 14:00 Page 1024
  • 170. PROBLEMS 1025FIGURE 14.25FIGURE 14.26FIGURE 14.27FIGURE 14.28RaoCh14ff.qxd 10.06.08 14:00 Page 1025
  • 171. 1026 CHAPTER 14 RANDOM VIBRATIONFIGURE 14.2914.19 Find the autocorrelation function of a random process whose power spectral density is givenby between the frequencies and14.20 The autocorrelation function of a Gaussian random process representing the unevenness of aroad surface is given bywhere is the variance of the random process and v is the velocity of the vehicle. The val-ues of and for different types of road are as follows:Type of roadAsphalt 1.1 0.2 0.4Paved 1.6 0.3 0.6Gravel 1.8 0.5 0.9Compute the spectral density of the road surface for the different types of road.14.21 Compute the autocorrelation function corresponding to the ideal white noise spectral density.14.22 Starting from Eqs. (14.60) and (14.61), derive the relationsS1f2 = 4Lq0R1t2 cos 2pft # dtR1t2 =Lq0S1f2 cos 2pft # dfBASxbsx, a,sx2Rx1t2 = sx2e-aƒvtƒcos bvtv2.v1S1v2 = S0 = constantRaoCh14ff.qxd 10.06.08 14:00 Page 1026
  • 172. PROBLEMS 1027FIGURE 14.3014.23 Write a computer program to find the mean square value of the response of a single degree offreedom system subjected to a random excitation whose power spectral density function isgiven as14.24 A machine, modeled as a single degree of freedom system, has the following parameters:lb/in., and It is subjected to the forceshown in Fig. 14.29. Find the mean square value of the response of the machine (mass).14.25 A mass, connected to a damper as shown in Fig. 14.30, is subjected to a force F(t). Find thefrequency response function for the velocity of the mass.H1v2c = 1200 lb-in./ = 2000 lb, k = 4 * 104Sx1v2.14.26 The spectral density of a random signal is given byFind the standard deviation and the root mean square value of the signal by assuming its meanvalue to be 0.05 m.14.27 Derive Eq. (14.46) from Eq. (14.45).14.28 A simplified model of a motor cycle traveling over a rough road is shown in Fig. 14.31. It isassumed that the wheel is rigid, the wheel does not leave the road surface, and the cycle movesat a constant speed v. The cycle has a mass m and the suspension system has a spring constantk and a damping constant c. If the power spectral density of the rough road surface is taken asfind the mean square value of the vertical displacement of the motor cycle (mass, m).S0,S1f2 = e0.0001 m2/cycle/sec, 10 Hz … f … 1000 Hz0, elsewhereFIGURE 14.31RaoCh14ff.qxd 10.06.08 14:00 Page 1027
  • 173. 14.29 The motion of a lifting surface about the steady flight path due to atmospheric turbulence canbe represented by the equationwhere is the natural frequency, m is the mass, and is the damping coefficient of the sys-tem. The forcing function F(t) denotes the random lift due to the air turbulence and its spec-tral density is given by [14.17]where c is the chord length and v is the forward velocity of the lifting surface and isthe spectral density of the upward velocity of air due to turbulence, given bywhere A is a constant and L is the scale of turbulence (constant). Find the mean square valueof the response x(t) of the lifting surface.14.30 The wing of an airplane flying in gusty wind has been modeled as a spring-mass-dampersystem, as shown in Fig. 14.32. The undamped and damped natural frequencies of the wingare found to be and respectively. The mean square value of the displacement of(i.e., the wing) is observed to be under the action of the random wind force whose powerspectral density is given by Derive expressions for the system parametersand in terms of and S0.v1, v2, d,ceqmeq, keq,S1v2 = S0.dmeqv2,v1ST1v2 = A21 + aLvvb2e1 + aLvvb2f2ST1v2SF1v2 =ST1v2a1 +pvcvbzvnx$1t2 + 2zvnx#1t2 + vn2x1t2 =1mF1t21028 CHAPTER 14 RANDOM VIBRATIONFIGURE 14.3214.31 If the building frame shown in Fig. 14.23 has a structural damping coefficient of 0.01 (insteadof the modal damping ratio 0.02), determine the mean square values of the relative displace-ments of the various floors.RaoCh14ff.qxd 10.06.08 14:00 Page 1028
  • 174. DESIGN PROJECT 102914.32 The building frame shown in Fig. 14.23 is subjected to a ground acceleration whose powerspectral density is given byFind the mean square values of the relative displacements of the various floors of the build-ing frame. Assume a modal damping ratio of 0.02 in each mode.14.33 Using MATLAB, plot the Gaussian probability density functionover14.34 Plot the Fourier transform of a triangular pulse:(See Fig. 14.12.)14.35 The mean square value of the response of a machine, subject to the force shown inFig. 14.29, is given by (see Problem 14.24):wherewith for 1, 2, 3, 4, 5, 6, 7, 8, 9,10; andUsing MATLAB, find the value of withDESIGN PROJECT14.36 The water tank shown in Fig. 14.33 is supported by a hollow circular steel column. Thetank, made of steel, is in the form of a thin-walled pressure vessel and has a capacity of10,000 gallons. Design the column to satisfy the following specifications: (a) Theundamped natural frequency of vibration of the tank, either empty or full, must exceed aN = 10.E[y2]vn = 2 p n.k = 4 * 104, c = 1200, m = 5.1760j = 0,Fj = 0, 20, 40, 60, 80, 100, -80, -60, -40, -20, 0cn =1N aNj=1Fjecos2 p n jN- i sin2 p n jNfE[y2] = aN-1n=0ƒcn ƒ2ak - mvn2b2+ c2vn2E[y2],X1v2 =4 Aa v2sin2 va2, -7 …vap… 7-7 … x … 7.f1x2 =112pe-0.5x2S1v2 =14 + v2RaoCh14ff.qxd 10.06.08 14:00 Page 1029
  • 175. 1030 CHAPTER 14 RANDOM VIBRATIONvalue of 1 Hz. (b) The mean square value of the displacement of the tank, either empty orfull, must not exceed a value of when subjected to an earthquake ground accelerationwhose power spectral density is given byAssume damping to be 10 percent of the critical value.S1v2 = 0.0002m2/s4rad/s16 in2FIGURE 14.33RaoCh14ff.qxd 10.06.08 14:00 Page 1030