Ch-07 (523180), страница 3
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This equation has nontrivial solutiononly if the determinant of the coefficient matrix [A]–[I] is equal to zero. In otherwords, the linear algebraic equations represented by Equation 1 are not all independent. The number of equations which are dependent on the other equations, is equalto the multiplicity of the specified . For example, if the matrix [A] is of order Nand if the multiplicity of is M which means M characteristic roots are equal to ,then there are M equations in Equation 1 are dependent on the other N-M equations.© 2001 by CRC Press LLCWhen Gaussian Elimination method is applied for solving {V} from Equation1, the normalization of the last equation cannot be carried out if has a multiplicityequal to 1 even with the pivoting provision in the program. This is because one ofthe N equation is dependent on the other N–1 equations.
But, it suggests that wemay assign the last component of {V} to be equal to an arbitrary constant c andexpress the other components of {V} in terms of c. This concept can be extendedto the case when λ has a multiplicity of M. Since only N-M equations of (1) areindependent, there are M independent solutions of {V}. To obtain the first solution,we assign the last component of {V} a value c1 and the other last M–1 componentsof {V} equal to zero and then proceed to express the first N-M components of {V}in terms of c1. To obtain the second solution, we assign the next to the last componentof {V} a value c2 and the other last M–1 components of {V} equal to zero andexpress the first N-M components of {V} in terms of c2, and so on.
The M solutionof {V} can thus be expressed in terms of ci for i = 1,2,…,M.The program EigenVec is developed from modifying the program Gauss byfollowing the above-explained procedure. This program requires the user to interactively specify the order N of [A], the elements of [A], a specified value of andits multiplicity M. The results produced by the program EigenVec are the M set ofeigenvectors {V}.
Both FORTRAN and QuickBASIC versions of this programs arelisted below along with sample applications.QUICKBASIC VERSION© 2001 by CRC Press LLCSample ApplicationFORTRAN VERSION© 2001 by CRC Press LLCSample Application© 2001 by CRC Press LLCMATLAB APPLICATIONSMATLAB has a file called eig.m which can be applied for finding the eigenvalues and normalized vectors of a specified square matrix. To do so, we firstinteractively specify the elements of a matrix [A} and then ask for the eigenvaluesLambda and normalized eigenvectors EigenVec by entering (such as for the sampleproblem in the FORTRAN and QuickBASIC versions)>> A = [3,0,2;0,5,0;2,03]; [EigenVec,Lambda] = eig(A)It results in a display on the screen:Notice that the eigenvalues are listed in the diagonal of the matrix Lambda andthe corresponding normalized eigenvectors are listed in the matrix EigenVec ascolumns.
To list the eigenvalues in a vector Lambda, we could enter:>> [Lambda] = eig(A)The resulting display is:Lambda =515MATHEMATICA APPLICATIONSMathematica has functions Eigenvalues and Eigenvectors which can beapplied to find the eigenvalues and eigenvectors, respectively, for a specified matrixas illustrated by the following example:© 2001 by CRC Press LLCIn[1]: = a = {{3,0,2},{0,5,0},{2,0,3}}; MatrixForm[a]Out[1]//MatrixForm: =302050203In[2]: = Eigenvalues[a]Out[2]: ={1, 5, 5}In[3]: = Eigenvectors[a]Out[3]: ={–1, 0, 1}, {1, 0, 1}, {0, 1, 0}}Notice that the computed eigenvectors are not normalized.As another example, consider the matrix M generated in the program EigenODEfor the buckling problem when the number of stations is equal to 5. To obtain theeigenvalues, the interactive application of Mathematica goes as:In[4]: = Eigenvalues[M]Out[4]: ={134.354, 108., 72., 36., 9.64617}Notice that the smallest eigenvalue is equal to 9.64617 which predicts the lowestbuckling load.
Since the exact solution is 9.8696, this further indicates that bycontinuously increasing the number of stations the smallest eigenvalue in magnitudewill eventually converge to the expected value.PRINCIPAL STRESSESANDPLANESAs another example of solving the eigenvalues and eigenvectors, consider theproblem of determining the principal stresses at a point within a two-dimensionalbody which is subjected to in-plane loadings. If the normal stresses (x and σy) andshear stresses (xy = yx), Figure 5, at that point are known, it is a common practiceto graphically determine the principal stresses and principal planes, on which theprincipal stresses act by use of Mohr’s circle.4 But, here we demonstrate how theprincipal stresses and principal planes can be solved as the eigenvalues and eigenvectors, respectively, of a matrix [A] constructed using the values of x, y, and xyas follows:© 2001 by CRC Press LLCFIGURE 5.
If the normal stresses (x and σy) and shear stresses (xy = yx), are known, itis a common practice to graphically determine the principal stresses and principal planes, onwhich the principal stresses act by use of Mohr’s circle. σx[A] = ττ xy σ y yx(2)For the three-dimensional cases, the normal stresses x, y, and z, and shearstresses xy, yz, and zx (yx = xy, zy = yz, and xz = zx) are involved, Figure 6.Again, the Mohr’s circle method can be applied to graphically solve for the principalstresses and the principal planes, on which they act.6 But, as an extension of Equation2, these principal stresses and principal planes can be determined as the eigenvaluesand eigenvectors, respectively, of a matrix constructed using the values of the normaland shear stresses as follows: σx[A] = τ yxτ zxτ xyσyτ zyτ xz τ yz σ z (3)Presented below are MATLAB solutions of two problems: (a) a two-dimensionalcase of x = 50, y = –30, and xy = yx = –20, and (b) a three-dimensional case ofx = 25, y = 36, z = 49, xy = yx = –12, yz = zy = 8, and zx = xz = –9, all in N/cm2.© 2001 by CRC Press LLCFIGURE 6.
For the three-dimensional cases, the normal stresses x, y, and z, and shearstresses xy, yz, and zx (yx = xy, zy = yz, and xz = zx) are involved.Notice that for Problem (a), the result indicates that maximum principal stressequal to 54.7214 N/cm2 is on a plane having an outward normal vector whosedirectional cosines are equal to –0.9732 and 0.2298. That is to say this principalplane has an outward normal vector making an angle of max = 166.7° (cosmax =–0.9732 and cos[90°–max] = 0.2298) measured counterclockwise from the x-axis.The minimum principal stress is found to be equal to –34.7217 N/cm2 which is ona plane having an outward normal vector whose directional cosines are equal to–0.2298 and –0.9732, or at an angle equal to min = –103.3° (cosmin = 0.2298 and© 2001 by CRC Press LLCcos[90°– min] = –0.9732).
The two principal planes are perpendicular to each other.This can also be proven by taking the dot product of the two normalized eigenvectors:(–0.9732i + 0.2298j)•(–0.2298i- 0.9732j) = 0.Similar observation can be made from the results for Problem (b).
The principalstresses are equal to 16.9085, 34.6906, and 58.4009 N/cm2 and they on the planeshaving outward normal vectors n1 = 0.8632i + 0.4921j + 0.1192k, n2 = 0.3277i –0.7218j + 0.6096k, and n3 = –0.3860i + 0.4867j + 0.7837k, respectively. It is easyto prove that these principal planes are indeed orthogonal by showing that n1•n2 =n2•n3 = n1•n3 = 0.QUADRATIC FORMSANDCANONICAL TRANSFORMATIONAnother interesting application of the procedure involved in solving eigenvaluesand eigenvectors of a square matrix is the canonical transformation of quadraticforms6 for Consider a surface described by the equation:25x 2 + 36y 2 + 49z 2 − 24 xy − 18xz + 16yz = 100(4)The left-hand side is called a quadratic form in x, y, and z.
The surface is anellipsoid but it is difficult to make out what are the values of its major and minoraxes, for which a transformation of the coordinate system is necessary for changingthe quadratic form into a canonical one. By canonical form, it means that only thesquared terms should remain. To find what transformation is needed, we first writeEquation 4 in matrix form as:{V}T [A]{V} = 100(5)where:x {V} = y z and 25[A] = −12 −9−12368−98 49 (6,7)It can be shown6 that if we find the eigenvalues and eigenvectors of [A] whichin fact are already available in the answer to the previously discussed Problem (b),the coordinate system x-y-z can be transformed into another set of x-y-z systemby using the so-called normalized modal matrix, [Q], formed with the normalizedeigenvectors as its columns.
That is:x′ .8623V={ ′} y′ = .4921 z′ .1192© 2001 by CRC Press LLC.3277−.7218.6096−.3860 .4867 .7837 x y = Q {V} [ ] z (8)Since [Q] has the property of [Q]–1 = [Q]T, the quadratic form can then be writtenas {V} T [A]{V} = ([Q] –1 {V}) T [A]([Q] –1 {V}) = {V} T [Q][A][Q] T {V} ={V}[D]{V} where [D] is a diagonal matrix having the eigenvalues of [A] (16.9085,34.6906, and 58.4009) along its diagonal. Hence, the resulting canonical form forthe surface defined by Equation 4 can now be expressed as:16.9085x′ 2 + 34.6906y′ 2 + 58.4009z′ 2 = 10 2(9)z-axis and minor axes equal to 10/(16.9085)_ and 10/(34.6906)_ along x- and yaxes, respectively.
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