Yes, same results! then, would this be a bug? How they can be different?
Also,
Y:invert(M).K;
eigenvectors(%);
should do the same and would give w^2 eigenvalues and the resulting
eigenvectors.
Eigenvalues ok,
But eigenvectors?
[[[-((64*sqrt(34)-512)*E*I)/(135*m*L^4),((64*sqrt(34)+512)*E*I)/(135*m*L^4),(24*E*I)/(m*L^4)],[1,1,1]],[0,1,(sqrt(2)*sqrt(17)-2)/6],[0,1,-(sqrt(2)*sqrt(17)+2)/6],[1,0,0]]
which is
[[[(1.028289472073339*E*I)/(m*L^4),(6.556895713111847*E*I)/(m*L^4),(24*E*I)/(m*L^4)],[1,1,1]],[0,1,0.63849198247422],[0,1,-1.305158649140884],[
1,0,0]]
from the previous result,
for lamda 2 be 1
lamda 3 is -0.17232165733319 or 49.56815440289505
or lamda 3 be 1
lamda 2 is -2.718367346938776
means,
no -1.305158649140884 or 0.63849198247422
On Fri, May 30, 2008 at 1:02 AM, Raymond Toy (RT/EUS) <
raymond.toy at ericsson.com> wrote:
> ahmet alper parker wrote:
>
>>
>>
>> On Wed, May 28, 2008 at 9:24 PM, ahmet alper parker <aaparker at gmail.com<mailto:
>> aaparker at gmail.com>> wrote:
>>
>> I am trying to do a modal analysis with K the stiffness matrix and M
>> the mass matrix and trying to find K*lamda=w^2*M*lamda. Looking at
>> the matlabs help files, I discovered that I can use (M^-1.K) to
>> calculate the eigenvalues and eigenvectors with the standard
>> functions. But when I tried to find (according to my knowledge) the
>> eigenvectors, The only method I know did not worked for the
>> eigenvectors.
>> For example
>> writing w^2 in the equation and solving for lamda, the lamda is a
>> vector and as far as I know I have to enter 1 to at least one of
>> them, since they are not independent. Then I solve the others from
>> the matrix equation.
>>
>>
>>
>> K:matrix([(12*E*I)/L^3,0,0],[0,(3.555555555555555*E*I)/L^3,-(3.555555555555555*E*I)/L^3],[0,-(3.555555555555555*E*I)/L^3,(7.111111111111111*E*I)/L^3]);
>> M:matrix([(m*L)/2,0,0],[0,(5*m*L)/4,0],[0,0,(3*m*L)/2]);
>> J:K-w^2*M;
>> solve([determinant(J)=0;], [w]);
>>
>> I got eigenvalues correctly.
>> subst((8*sqrt(sqrt(34)+8)*sqrt((E*I)/m))/(3*sqrt(15)*L^2), w, J);
>> t:%;
>>
>
> Ok, I follow you up to this point.
>
> [lamda1=0,lamda2=(48*%r1^2-2046*%r1)/(325*%r1+410),lamda3=%r1];
>>
>
> I assume this is what you get from solving the equation J with the value of
> w given above.
>
> It seems this might be a bug in linsolve. (Well, I used linsolve to get
> the same expression as you did above.)
>
> However, note that K can be simplified to be a be
>
> K1 : matrix([12*c,0,0],[0,3.555555555555555*c,-3.555555555555555*c],
> [0,-3.555555555555555*c,7.111111111111111*c]);
>
> where c = E*I/L^3 and M can be simplified to be
>
> M1 : matrix([ml,0,0],[0,5*ml/2,0],[0,0,3*ml]);
>
> where ml = m*L. Then
>
> J1 : K1-w^2*M1 ->
>
> matrix([12*c-ml*w^2,0,0],
> [0,3.555555555555555*c-5*ml*w^2/2,-3.555555555555555*c],
> [0,-3.555555555555555*c,7.111111111111111*c-3*ml*w^2]);
>
> If we then solve this for w, and substitute a value back into J1, and ask
> linsolve for the solution, we get:
>
> [f1 = 0,f2 = -(sqrt(34)*%r4-2*%r4)/5,f3 = %r4]
>
> so we get nice linear relationship between the eigenvector [f1,f2,f3] and
> the arbitrary parameter %r4.
>
> Ray
>