Polyval function not working in my fatigue failure analysis problem

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Lucas Clark 2020-1-12

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https://ww2.mathworks.cn/matlabcentral/answers/500015-polyval-function-not-working-in-my-fatigue-failure-analysis-problem

评论： Star Strider 2020-1-12

采纳的回答： Star Strider

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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))