Why are the natural frequencies of a clamped-clamped beam smaller than usual?

  • Thread starter Thread starter jeffziggy
  • Start date Start date
  • Tags Tags
    Fem Weird
AI Thread Summary
The discussion revolves around the calculation of natural frequencies for a clamped-clamped beam using Matlab. The user, Jeff, notes that the frequencies obtained are significantly smaller than expected, specifically by a factor of N^2, where N is the number of elements used in the model. A contributor explains that this discrepancy arises because the global mass matrix is proportional to the beam length, while the global stiffness matrix is inversely proportional to the cube of the length, leading to the frequencies being proportional to 1/l^2. This clarification helps Jeff understand the relationship between the number of elements and the calculated frequencies. The conversation concludes with Jeff expressing gratitude for the insight.
jeffziggy
Messages
3
Reaction score
0
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

Code:
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)
 
Engineering news on Phys.org
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 l, the global mass is proportional to l and the global stiffness proportional to 1/l^3, so the frequences would be proportional to \sqrt{k/m} = 1/l^2.
 
AlephZero said:
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 l, the global mass is proportional to l and the global stiffness proportional to 1/l^3, so the frequences would be proportional to \sqrt{k/m} = 1/l^2.

That was exactly it! Thank you kind sir!

Jeff
 
Thread 'Turbocharging carbureted petrol 2 stroke engines'
Hi everyone, online I ve seen some images about 2 stroke carbureted turbo (motorcycle derivation engine). Now.. In the past in this forum some members spoke about turbocharging 2 stroke but not in sufficient detail. The intake and the exhaust are open at the same time and there are no valves like a 4 stroke. But if you search online you can find carbureted 2stroke turbo sled or the Am6 turbo. The question is: Is really possible turbocharge a 2 stroke carburated(NOT EFI)petrol engine and...
I need some assistance with calculating hp requirements for moving a load. - The 4000lb load is resting on ball bearing rails so friction is effectively zero and will be covered by my added power contingencies. Load: 4000lbs Distance to travel: 10 meters. Time to Travel: 7.5 seconds Need to accelerate the load from a stop to a nominal speed then decelerate coming to a stop. My power delivery method will be a gearmotor driving a gear rack. - I suspect the pinion gear to be about 3-4in in...
Back
Top