normalised mode shapes
prashanth
Gordon
> what is the exact meaning of "normalised mode shapes".
A mode shape is an eigenvector. An eigenvector is unique only up to an
arbitrary nonzero scale factor. A normalized mode shape scales the
eigenvector according to some desired criterion, e.g., largest component
equal to unity, or generalized modal mass equal unity, or whatever.
matthias
eigenvector('s) are normalized to the mass matrix (in FEA).
Therefore the whole system of coupled ODE's gets ,,uncoupled"...
Jeff Finlayson
> Gordon schrieb:
>> prashanth wrote:
>>
>>> what is the exact meaning of "normalised mode shapes".
>>
>> A mode shape is an eigenvector. An eigenvector is unique only up to
>> an arbitrary nonzero scale factor. A normalized mode shape scales the
>> eigenvector according to some desired criterion, e.g., largest
>> component equal to unity, or generalized modal mass equal unity, or
>> whatever.
>
> If you want to perform something like a modal analysis the
> eigenvector('s) are normalized to the mass matrix (in FEA).
Eigenvectors can be normalized to mass or the max value to 1.
Greg Locock
news:120sa4j...@corp.supernews.com:
I've seen at least 3 different ways of normalising mode shapes,
basically any nut can invent a new one.
The point is, the scaling (or normalisation factor) is essentially
arbitrary.
Cheers
Greg Locock
matthias
yes it may seem, that the normalization factor (scaling) is arbitrary.
The scaling factor really depend on the application. For modal analysis
in FEA it is recommendend to scale with the mass matrix (or let me say
it's done by the software).
So for further explaination let p be the vector of (unknown) nodal
displacements. Let M be the system's mass matrix, and though K the
system's stiffness matrix. Then you can write:
M.p''+K.p=R ; where here p'' is the second derivative of nodal
displacements, R are the nodal loads
What is done now? Simply an eigenvalue problem has to be solved. This
leads to:
(M-lambda_i.K).p_i=0 ; where lambda_i are the eigenvalues, and p_i are
the corresponding eigenvectors
If one constructs a modal matrix with P=(p_1 p_2 .... p_n), one can
easily show, that Transpose(P).M.P=I , where I is the identity matrix
and Transpose(P).K.P=diag(lambda_i). If one constructs modal
displacements such that: P^-1.p=y, then you can easily insert these
three ,,equations" in our first one, which leads to the ,,uncoupled"
version of our system of coupled ODE's:
I.y'' + diag(lambda_i).y=Transpose(P).R
So as you can now easily see, it really depends on the application how
to scale vectors. For animating purposes it's useful to scale the
eigenvectors to unity.....
Regards
Matthias
P.S.: one can show, that if the general eigenvalue problem shown above
is symmetric, all roots of the characteristic polynom are reals
Jeff Finlayson
> Jeff Finlayson wrote:
>>matthias wrote:
>>>Gordon schrieb:
>>>>prashanth wrote:
>>>>
>>>>>what is the exact meaning of "normalised mode shapes".
>>>>
>>>>A mode shape is an eigenvector. An eigenvector is unique only up to
>>>>an arbitrary nonzero scale factor. A normalized mode shape scales
>>>>the eigenvector according to some desired criterion, e.g., largest
>>>>component equal to unity, or generalized modal mass equal unity, or
>>>>whatever.
>>>
>>>If you want to perform something like a modal analysis the
>>>eigenvector('s) are normalized to the mass matrix (in FEA).
>>
>>Eigenvectors can be normalized to mass or the max value to 1.
> I've seen at least 3 different ways of normalising mode shapes,
> basically any nut can invent a new one.
>
> The point is, the scaling (or normalisation factor) is essentially
> arbitrary.
Thanks, that's what I was getting at.
prashanth
thanx a lot for clearing my doubt.. in modal analysis the physical
meaning of normalising the mode shapes is that we are finding the
individual contributions of the various natural frequencies a system
possesses, to the nodal displacements.. am i correct? i am trying to
develop a code for finding out the natural frequencies and mode shapes
of a model.
how do i apply the boundary conditions to find out the eigen vectors
which represent the mode shapes. all i know is that the model is fixed
at some positions.
in the eigen value problem what do the eigen vectors represent without
the boundary conditions?
matthias
I think you're not completely correct...
Yes the system oscillates in a composition of its natural frequencies...
This fact let us know, wheather we hear a violin or a piano. Just listen
to the tone "a" in Mathematica (440 Hz)... sounds not like a violin!!
Transforming the REAL coordinates into modal coordinates means, that
every single DOF works just like a single harmonic mass-spring
oscillator in modal coordinates. That's the trick!! Every DOF just
performes a sine wave in modal coordinates!! (the terrible 440 Hz
Mathematica-"a"...)
For modal analysis (meaning to solve the eigenvalue problem) b.c.'s are
not really necessary! Why? Because the system isn't solved like a static
one, which means that the system's stiffness matrix isn't formally
inverted....
To explain, let K be the system's stiffness matrix, p the vector of
unknown nodal displacments and q the vector of nodal loads. In (linear)
FEA this leads us to the following linear system of equations:
K.p=q ; which is formally solved: p=K^-1.q
In the dynamic case we have to solve the eigenvalue problem:
det(K-lambda.M)=0 ; where M is the system's mass matrix;
By setting det(A)=0 we force that A isn't inverteble!!!!! To obtain a
solution we have to search for roots of the characteristic polynom.
These roots are our natural frequencies! (omega^2=lambda)
(an error occured in my last message!!, I wrote (M-lambda.K))
So if you have some understanding of machine dynamics you know, that a
natural frequency of zero means a rigid body motion; in the static case
this would be horrible (think of bridges->these are static systems
(mainly), there must not be a rigid body motion)!!
(This fact always helped me to see if a hand-made solution could be
right or not e.g. in some machine dynamic exams...:-))
This I tell you because it brings some more understanding of the
differences between static and dynamic FEA. As I agree, it isn't really
fine to have no b.c.'s.... So what is done in the static case? The
applied b.c.'s mean, that some unknown DOF's are not unknown any
more.... This doesn't differ to the dynamic case.....
What kind of code are you going to develop? What language do you use?
Regards
Matthias
P.S.: In all my explainations above, I just didn't tell you about stress
stiffning... This would have made the things more complicated...:-)
matthias
> that was fast.. i wont find all these in a textbook.. thanx a ton..
> i am going to develop a model specific code because this is my first..
> i used matlab..
> its very simple to solve the eigen value problem in that.. i just found
> out the eigenvalues of the matrix [A]= [K]/[M].. with my bookish
> knowledge i am surmising that these values are the square of the
> circular frequencies.. i got the eigen vectors too.. but i dunno HOW TO
> INTERPRET them!!! and in those eigen vectors i got values ranging from
> e-12 to e7.. does this mean that i havent normalised it?
>
Hello Prashanth,
there are many textbooks where you can find all about this...
Maybe the best one is : Clough: ,,Dynamics of Structures"
For internal usage, I could send you my script of machine dynamics, but
it is in german...
I don't know about the syntax in Matlab, but you won't get the
eigenvalues by dividing K by M, this operation is only defined for
scalars.... As I wrote previosly, you have to get the roots of the
characteristic polynom, which is defined by det(K-lambda.M)=0.
These roots, we identify as the squares of our natural (circular)
frequencies omega_i...
The eigenvectors p_i you get by: (K-lambda_i.M).p_i=0... the
eigenvectors depend linear on each other (as we forced them to be by
setting the determinat to zero!!). The eigenvectors you can ,,normalize"
to some quantity (e.g. to unity), these eigenvectors represent our mode
shapes. The mode shapes show us the ,,eigenform" of our systems as it
oscillates in it's specific natural frequency...
What kind of model do you have? Maybe you can send me a sketch of it, so
I can help you better...
Regards
Matthias
Greg Locock
news:1142126519....@v46g2000cwv.googlegroups.com:
> its very simple to solve the eigen value problem in that.. i just
> found out the eigenvalues of the matrix [A]= [K]/[M].. with my bookish
> knowledge i am surmising that these values are the square of the
> circular frequencies..
Fine so far, the eigenvalues of ([M]^-1)[K] are the omega ^2 terms.
i got the eigen vectors too.. but i dunno HOW TO
> INTERPRET them!!! and in those eigen vectors i got values ranging from
> e-12 to e7.. does this mean that i havent normalised it?
>
You don't have to normalise them.
The eigenvector is the displacement of each degree of freedom, for that
modeshape, and you have as many modes as you have degrees of freedom, so
you have a square matrix [A], in which say rows are the displacement
of each degree of freedom, and columns are mode shapes, or vice versa,
depending on how you set the thing up in the first place, I think. It
should be easy to pick if you have any fixed DOF, there should be a
corresponding row or column of zero displacements.
Cheers
Greg Locock
prashanth
how will i get zero displacements if i dont give the condition first..
i will get zeros only if the modeshape itself is like that, right?.
please bear with me if i am saying somethuing fundamentally wrong..
lets say there are two beams.. one is fixed at one end only.. the other
is fixed at its two ends.. can u tell me how the modeshapes of these
two will differ? wont i get the same eigen vectors for both of them if
i dont apply the boundary conditions?
thanking you
prashanth
Greg Locock
news:1142227334.1...@j52g2000cwj.googlegroups.com:
Mode shapes (and frequencies) are not independent of the boundary
conditions, so you are asking me if two identical beams will have identical
modeshapes.
matthias
did you get my eMail with attached pdf? Hopefully this clears out your
doubts and gives a short introduction in dynamics of stuctures....
Regards
Matthias