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Weird FEM issue (nat freqs)

by jeffziggy
Tags: freqs, issue, weird
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jeffziggy
#1
Nov8-11, 01:56 AM
P: 3
Hello,

I recently used Matlab to find the natural frequencies of a clamped-clamped beam. It was fairly simple, as I construct the global mass and stiffness matrices. Then it's just a matter of using the eig() function. (For sake of simplicity I put beam properties all to 1.)

When I choose how many elements (lets call it N) to use, it seems to give me natural frequencies which, seem to be smaller than usual. I somehow stumbled upon the fact that these frequencies are the literature values I have divided by N^2.

Anyone got any ideas as why this is so? Here is my code. Thanks!! Edit: For example when N = 10, my first natural frequency is 0.2237 when the exact value is 22.37.

Jeff

syms y;
syms z;
area = 1;
L = 1;
p = 1;
E = 1;
Ig = 1;
n = 10; %number of elements
if n == 0
stop
end
size = 4+((n-1)*2);
%define global matrices
Mg = zeros(size);
Kg = zeros(size);
%beam function
A = [1 0 0 0 ; 0 1 0 0 ; 1 L L^2 L^3 ; 0 1 2*L 3*L^2];
Ainv = inv(A);
Aitrans = transpose(Ainv);
Yh = [1 ; y ; y^2 ; y^3];
Ya = [1 ; y];
Ydotdot = diff(diff(Yh));


%Solving for Mass matrix (kinetic energy)
prod1 = Yh*transpose(Yh);
M = p*area*Aitrans*int(prod1, y, 0, L)*Ainv;
%solving for stiffness matrix (potential energy)
K = (E*Ig)*Aitrans*int(Ydotdot*transpose(Ydotdot), y, 0, L)*Ainv;

%creating the global mass matrix
Mg(1:4, 1:4) = M(1:4, 1:4);
Kg(1:4, 1:4) = K(1:4, 1:4);

if n > 1
i = 1;
j = 1;
    for i=1:n-1
    Mg(j+2:j+5, j+2:j+5) = Mg(j+2:j+5, j+2:j+5) + M(1:4, 1:4);
    Kg(j+2:j+5, j+2:j+5) = Kg(j+2:j+5, j+2:j+5) + K(1:4, 1:4);
    j = j+2;
    end
   
end

%delete rows/columns based on clamped ends (boundary conditions)

Mg(size,:) = [];
Mg(size-1,:) = [];
Mg(:,size) = [];
Mg(:,size-1) = [];
Mg(:,1) = [];
Mg(:,1) = [];
Mg(1,:) = [];
Mg(1,:) = [];


Kg(size,:) = [];
Kg(size-1,:) = [];
Kg(:,size) = [];
Kg(:,size-1) = [];
Kg(:,1) = [];
Kg(:,1) = [];
Kg(1,:) = [];
Kg(1,:) = [];


eigs = eig(Kg,Mg);
sqrt(eigs)
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AlephZero
#2
Nov8-11, 07:23 AM
Engineering
Sci Advisor
HW Helper
Thanks
P: 7,160
I haven't checked every line of the code, but I think you have each element of length 1, so the length of your beam depends on the number of elements in the model.

If the length of the beam is [itex]l[/itex], the global mass is proportional to [itex]l[/itex] and the global stiffness proportional to [itex]1/l^3[/itex], so the frequences would be proportional to [itex]\sqrt{k/m} = 1/l^2[/itex].
jeffziggy
#3
Nov8-11, 10:20 AM
P: 3
Quote Quote by AlephZero View Post
I haven't checked every line of the code, but I think you have each element of length 1, so the length of your beam depends on the number of elements in the model.

If the length of the beam is [itex]l[/itex], the global mass is proportional to [itex]l[/itex] and the global stiffness proportional to [itex]1/l^3[/itex], so the frequences would be proportional to [itex]\sqrt{k/m} = 1/l^2[/itex].
That was exactly it! Thank you kind sir!

Jeff


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