Coding an Equation to be solved

I would like some help in coding equation 14, in order to solve equation 15 to obtain the value of C. Is this possible to do? I am not able to do it.

Answer 1

[Alan Stevens]

This should do it, assuming you know the values of the other constants:

% Arbitrary data
L = 1;
rho = 1;
A = 1;
sigma = 1;
Mt = 1;
It = 1;

lambda = 1.162;
k = lambda/L;
% define functions without C
phia = @(x) (cos(k*x)-cosh(k*x)+sigma*(sin(k*x)-sinh(k*x)));
phiasq = @(x) phia(x);
dphidxa = @(x) -k*(sin(k*x)+sinh(k*x)-sigma*(cos(k*x)-cosh(k*x)));

S = rho*A*integral(phiasq,0,L) + Mt*phiasq(L) + It*dphidxa(L)^2;
C = sqrt(1/S);
disp(C)

[Arjun Siddharth]

Thank you so much! This was very helpful!

[Arjun Siddharth]

Can you also tell me how to evaluate a function at a point?

For example, how do i substitute x=L in dphidxa equation?

[Alan Stevens]

Just dphiadx(L)

[Alan Stevens]

It should have worked! What was the error message?

[Alan Stevens]

You just need to define beta_alpha1 as

beta_alpha1 = dphidxa(L);

It is just a number - don't make it a function!

[Arjun Siddharth]

thanks! it worked!

in the intial code, could you clarify why you used 'C=sqrt(1/S)'? Is that from the mathematical solving?

[Alan Stevens]

Yes. each of the terms in eqn (15) is implicitly multiplied by C^2. Hence it could be factored out. You must therefore remember to multiply the "bare" functions, phia and dphidxa, by C when you use them.

[Arjun Siddharth]

how do we multiply a value like 'C' to a function? because it says operators like '*' are incompatible with function handles.

[Alan Stevens]

Example:

a = 1:5;

b = C*phia(a);   % not b = C*phia

You multiply C by evaluated values of the function, not the function definition itself.

[Alan Stevens]

Yes!

[Arjun Siddharth]

thanks so much!

[Alan Stevens]

No, sorry! I must have been half-asleep! Use

phiasq = @(x) phia(x).^2;

[Arjun Siddharth]

that's alright! thanks a lot !

[Arjun Siddharth]

Hey Alan, would you have time to take a look at my code? I'm trying to recreate the graph from a research paper. The code is running, but the graph does not match. Would appreciate if you could have a look. Do let me know , so that I may send you the paper as well as the code.

[Alan Stevens]

Upload your code here. Make use of the > button in the CODE section in the menu header rather than posting a picture.

[Alan Stevens]

Your code has frequency in Hz, whereas the equations want omega in radians/sec. Your Voltage Response section is better expressed as

%% Voltage Response 
freq = linspace(20,140,1200)*2*pi;
v = zeros(numel(freq),1);
for count = 1:numel(freq)
    w = freq(count);
    t1=-1j*w*alpha*RL*M;
    t2=M*(1j*2*Deq*wn*w+wn^2-w^2);
    t3=1j*w*Cp*RL+1;
    t4=1j*w*RL*alpha^2;
    v1=t1/(t2*t3+t4);
    v(count)=abs(v1);
end
figure(1)
plot(freq/(2*pi),v,'b','Linewidth',2);
xlabel('Frequency (Hz)');
ylabel('Voltage');

The magnitude of the voltage response is still different. This might be due to differences in data - I haven't checked everything!

Your value for 'a' is 0.8558 which doesn't match the values used in figure 6.

[Alan Stevens]

Your code needed some reordering and addition of some arguments in function calls. However, end result seems to be the same!

%Code to validate Improved Lumped Parameter Model 
%% Parameters
rp=7800; %Density of Piezo Plate
rs=9000; %Density of Substrate Plate 
Ep=66*10^9; 
Es=105*10^9;
d31=-190*10^-12;
e31=-11.5;
e0=8.854*10^-12;
e33=1500*e0;
L=50.8*10^-3;
b=31.8*10^-3;
hp=0.26*10^-3;
hs=0.14*10^-3;
DrB=0.002;
Deq=0.027;
Mt=0.012;
RL=1000;

g = 9.81;  % You hadn't defined g (though result seems independent of g).
%% Computed Parameters

I1=2*b*(((hp+(hs/2))^3)-((hs/2)^3));
I2=I1/3;
I3=(b*(hs^3))/12;
I=I2+I3;
E1=Es*((b*(hs^3))/12);
E2=Ep*2*b*(((hp+(hs/2))^3)-((hs/2)^3));
E3=E2/3;
E=(E1+E3)/I;
Cp=(b*L*e33)/(2*hp);
rho=((2*rp*hp)+(rs*hs))/((2*hp)+hs);
A=b*(hs+(2*hp));
It=Mt* (L^2);
%% Bending Mode Shape
lambda = 1.162;
sigma=sigma_calc(lambda,Mt);
k = lambda/L;
%Define functions without 'C'
phia = @(x) ((cos(k*x)-cosh(k*x))+(sigma*(sin(k*x)-sinh(k*x))));
phiasq = @(x) phia(x).^2;
dphidxa = @(x) -k*( sin(k*x)+sinh(k*x) - sigma*(cos(k*x)-cosh(k*x)) );
S = (rho*A*(integral(phiasq,0,L))) + (Mt*phia(L).^2) + (It*((dphidxa(L)).^2));
C = sqrt((1/S));
beta_alpha=(-L)*C*dphidxa(L);
disp(beta_alpha);
disp(C);
alpha=(beta_alpha*b*(hs+hp)*e31)/(2*L);

%% Correction Factors Computation
mew_x=@(x) (Mt*g*x.^2.*(3*L-x))./(6*E*I*C*phia(x));  % need to call phia with x as argument
mew_barx=@(x) ((-2.7*10^-5)/(99*L))*(x-(0.01*L))-(1.207*10^-5);
func= @(x)mew_barx(x).*phia(x)*C;  % need to call mew_bar and phia with x as argument
func_sq=@(x) func(x).^2;
BM1=integral(func_sq,0,L);
BM2 = func_sq(L)*L;
Beta_M=BM1/BM2;
BK1=L^3*Mt*g;
BK2=E*I*mew_x(L)*C*phia(L);
Beta_K=BK1/BK2;


M_beam=(Beta_M*rho*A*L);
M=(Beta_M*rho*A*L)+Mt;
K=(Beta_K*E*I)/(L^3);
wn=sqrt((K/M));


a=Mt/M;


%% Voltage Response 
freq = linspace(20,140,1200)*2*pi;
v = zeros(numel(freq),1);
for count = 1:numel(freq)
    w = freq(count);
    t1=-1j*w*alpha*RL*M;
    t2=M*(1j*2*Deq*wn*w+wn^2-w^2);
    t3=1j*w*Cp*RL+1;
    t4=1j*w*RL*alpha^2;
    v1=t1/(t2*t3+t4);
    v(count)=abs(v1);
end
figure(1)
plot(freq/(2*pi),v,'b','Linewidth',2);
xlabel('Frequency (Hz)');
ylabel('Voltage');

%% Function to calculate Sigma Value
function [sig]=sigma_calc(lambda,Mt)
mL=0.02;
t2=lambda*Mt;
n1=sin(lambda)-sinh(lambda);
n2=cos(lambda)-cosh(lambda);
d1=cos(lambda)+cosh(lambda);
d2=sin(lambda)-sinh(lambda);
b1=(mL*n1)+(t2*n2);
b2=(mL*d1)-(t2*d2);
sig=b1/b2;
end

[Alan Stevens]

You can get a = 10 by setting Mt = 0.0165. However, it only leads to a small increase in the peak; nothing like enough to match that in the Wang paper.

[Arjun Siddharth]

oh right. What might be causing this variation? Some other constant?

[Alan Stevens]

Possibly some other constant.

Possibly, the graphs in the paper were generated with an entirely different set of constants, or there is an error in one or more of their published equations or their non-published coding. Since the authors don't list their coding it's not possible to know!

Perhaps there are other papers you can check against.

[Arjun Siddharth]

alright, thanks a bunch!

[Alan Stevens]

I've found one error: your definition of zbar is missing some brackets. I think it should be

z_bar=(ts^2+tm^2*n+2*ts*tm*n)/(2*(ts+(n*tm)));

As a general comment, you use too many brackets! It makes the code difficult to read. You don't need them around parameters that are multiplied together, for example.

Also, you need to include more comments, explaining what the statements are, or where they connect with the paper - especially if you want help from others! I followed it down to your computed parameters section, but lost the connection with the paper at that point.

[Alan Stevens]

You can fit a straight line as follows:

L = 0.01:0.01:0.05;
mu = [-1.2121, -1.2182, -1.2235, -1.2292, -1.2338];

mc0 = [-1, -1];

mc = fminsearch(@(mc) fn(mc,L,mu), mc0);
m = mc(1); c = mc(2);
disp([m,c])

   -0.5440   -1.2070

x = 0.01:0.001:0.05;
mufit = m*x + c;

plot(L,mu,'o',x,mufit),grid
xlabel('x'), ylabel('\mu')
legend('data','fit')
text(0.015,-1.227,['\mu = ' num2str(m) 'x ' num2str(c)])

function F = fn(mc, L, mu)

         m = mc(1); c = mc(2);
         F = norm(m*L + c - mu);
end

I'll let you manipulate it to look like the expression in the JPG if that's what you want it to look like.

[Alan Stevens]

The first two lines of the listing above are simply the data taken from Table 1 of the Wang paper. The resulting straight line fit for mu is mufit = m*x + c, where m and c are given as -0.54398 and -1.207 respectively, and x represents the values of length. You could define the straight line as a function:

mufit = @(x) -0.54398*x - 1.207;

then call it with the required value of x when you need to, e.g.

y = mufit(0.025);

[Alan Stevens]

You need to recalculate a few values of mu (using equation (16)) over the range of interest of values of L, then do a new curve-fit with those values.

Note: I left out the 10^-5 from the fitted values of mu. The final fit should be multiplied by this.

[Alan Stevens]

x is a dummy variable and represents values if L.

You can see from the graph that the curve fit is pretty good. It doesn't necessarily match exactly at the tabulated points because it is a best-fit line - it's not forced to go through every point. When you say the value is different, what sort of doifference do you get?

Incidentally, I should really have used the function 'polyfit' to get the line; however, as you can see below this doesn't make a significant difference here.

L = 0.01:0.01:0.05;
mu = [-1.2121, -1.2182, -1.2235, -1.2292, -1.2338];
p = polyfit(L,mu,1);
m = p(1); c = p(2);
disp([m,c])

   -0.5440   -1.2070

x = 0.01:0.001:0.05;
mufit = m*x + c;

plot(L,mu,'o',x,mufit),grid
xlabel('L'), ylabel('\mu')
legend('data','fit')
text(0.015,-1.227,['\mu = ' num2str(m) 'x ' num2str(c)])

What I've called 'mufit' is the average value function; your 'mew_bar'. (Again, I've left out the 10^-5 multiplier, which is needed when you actualy use mufit.)

[Alan Stevens]

If you change EI and L you should really recalculate equation 16 for a range of values of L, covering the range of interest. Here, you might manage by simply rescaling the average value of mu function by (EIold/EInew), since mu is inversely proportional to EI.

[Alan Stevens]

Well, strictly, your value of L is outside the range of the values used for the curve fit, but the line is so straight I suspect this won't introduce a significant error here.

[Arjun Siddharth]

oh okay. thanks!