Polyval function not working in my fatigue failure analysis problem

This is a fatigue failure analysis of a shaft with an unknown diameter.

The issue being we have a lack of information.

So, we assume a diameter and perform multiple iterations to find the correct factor of safety. The factor of safety is given.

u = symunit; 
%Given information at the beginning of problem 
Sy = 420e6*u.Pa; %Yield strength
Sut = 560e6*u.Pa; %Ultimate strength
TA = 340*u.N*u.m; %Given applied torque
d = .15*u.m ;
theta = 20;
BC = .25*u.m; 
CD = .1*u.m; 
n = 2.5;

F = vpa(TA/(cosd(theta)*d/2),5); %Acting force. 

%Sum of forces and Sum of moments to find the critical point of interest and the max bending moment.
C_z = vpa(F*sind(theta)*(BC+CD)/BC,4);
B_z = vpa(F*sind(theta)-C_z ,4);
M_z = abs(B_z*BC );
C_y = vpa(F*cosd(theta)*(BC+CD)/BC,4);
B_y = vpa(F*cosd(theta)-C_y ,4);
M_y = abs(B_y*BC);
M_max = vpa(sqrt(M_y^2 + M_z^2),4)

%Torque on the whole system.
T = F*cosd(theta)*d/2 + TA;



Se_p = .5*Sut;

a = 3.04; 
b = -0.217; 

%kb, ka, kc constants help determine endurance strength
ka = a*separateUnits(Sut)^(b);

Instead of doing multiplie iterations I just created a range of diameter values where it satisfied a factor of safety.

d = linspace(.04,.15)*u.m; %Here's my range of diametric values. 


kb = 1.24.*separateUnits(d).^(-.107); 
kc = 1; 

Se = Se_p.*ka.*kb.*kc;

% sharp fillet concentration factors
kt = 2.7; 
kts = 2.2;

%inertia and polar moment of inertia.
I = pi.*d.^4./64;
J = pi.*d.^4./32;
%Alternating bending stress
sigma_alt = kt.*M_max.*32./(pi.*d.^3);
%Mean bending stress
sigma_m = 0; 

% Alternating torsion stress
tao_alt = 0; 
%mean torsion stress
tao_m = kts.*16.*T./(pi.*d.^3);

%Von mises alternating stress
sigma_altp = (sigma_alt.^2 + 3*tao_alt.^2).^(1/2);
%Von mises mean stress
sigma_mp = (sigma_m^2 + 3.*tao_m.^2).^(1/2);

A = sqrt(4*(kt*M_max)^2+3*(kts*0).^2); 
B = sqrt(4*(kt*0)^2+3*(kts*T).^2);

%Goodman criteria for solving factor of safety
n = simplify(pi.*d.^3./16.*(A./Se + B./Sut).^(-1))

Now that I have a range of values for the factor of safety, I can use polyfit function to find the coefficients to solve for an exact diameter.

When I apply the polyfit function it gives me coefficients.

x1 = polyfit(separateUnits(d),n,2)

When I apply the polyval so that I can get the precise diameter value, it gives me an error.

polyval(x1,2.5)

Specifically:

Error using filter

Invalid data type. Input

arrays must be numeric

or logical.

Error in polyval (line

56)

y = filter(1,[1

-x],p);

Error in Untitled (line

83)

polyval(x1,2.5)

Why doesn't this work? Thank you.

Answer 1

[Star Strider]

If you are calling polyfit as:

x1 = polyfit(separateUnits(d),n,2)

you need to call polyval as:

v = polyval(x1,separateUnits(d))

since ‘separateUnits(d)’ appears to be your independent variable.

(I do not see that anywhere in your posted code, though.)

[Lucas Clark]

Even with the added changes it still produces the same error.

[Star Strider]

The separateUnits call will produce the numeric result for ‘d’. If it is a vector, there should be no problems. I thought ‘n’ was a vector of the same dimensions, however looking more closely, it appears that:

n = 2.5;

If you specify ‘n’ as a vector iinstead, it works:

d = linspace(.04,.15)*u.m;
n = linspace(0, 1, numel(d))*u.s;                   % Create ‘n’ As A Vector
x1 = polyfit(separateUnits(d),n,2)

producing:

x1 =
[ 0, (100/11)*[s], -(4/11)*[s]]

However while polyfit works with symbolic variables, polyval does not.

It is necesary to use:

v = polyval(double(separateUnits(x1)),double(separateUnits(d)));

to get the desired result.

You will have to provide the appropriate units for ‘v’.

[Lucas Clark]

That did it!

When I used the double() function I stopped receiving the error and received an ouput.

Thank you thank you

[Star Strider]

My pleasure!

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