MATLAB Solve function gives [ empty sym ]

Hi,

on the last line of the code I want to use the solve function, but it gives me:

    x=
    [ empty sym ]

Any ideas?

close all
clear all
clc

E = 206900; % Young's Modulus [MPa]
K_ = 3328; % Cyclic Hardening Coefficient
n_ = 0.147; % Cyclic Hardening Exponent
sigma_f = 2619; % Fatigue Strenght Coefficient [MPa]
epsilon_f = 0.145; % Fatigue Ductility Coefficient
b = -0.9280; % Fatigue Strenght Exponent
c = -05596; % Fatigue Ductility Exponent
sigmaE = 233; % Stress - MAX in component[MPa]
epsilon = sigmaE/E + (sigmaE/K_)^(1/n_);
sigma = sigmaE^2/E*epsilon;
Ne=1000000;
sigmaA = sigma_f*(2*Ne)^b;
epsilonA=sigma_f*(2*Ne)^b+epsilon_f*(2*Ne)^c;
M = 0.7; % Midrange Stress Sensitivity
k = M*(M+2); % We use this factor in order to express the 
          % midrange stress sensitivity (mainly for R=0),
          % but keep on using Local Strain Concept 
          % for R = -1 
sigmaM = 0;
PB1=sqrt((sigmaA + k*sigmaM)*epsilonA*E);
PB2=sqrt((sigma + k*sigmaM)*epsilon*E);
sym N;
x = solve('PB2=sqrt(((sigma_f*(2*N)^b)+k*sigmaM)*((sigma_f/E)*(2*Ne)^b+epsilon_f*(2*N)^c)*E)','N')

Cheers!!!

Answer 1

[Thomas]

I assume the equation does not have an explicit solution.

Does matlab throw out an error that explicit solution cannot be found?

does substitution numerical values solve the equation?

You could also be getting into a 0==0 situation or something.

Answer 2

[Roger Stafford]

Squaring both sides of your equation and coalescing given parameters produces a simplified equation of the form

 N^(b+c) + k1*N^b + k2 = 0

where k1 and k2 are known. (The N^c term that would have been there has disappeared since you set sigmaM to zero.) If 'solve' cannot find an explicit solution for this, you will have to do it numerically, say, with 'fzero'.

By the way, your value for exponent 'c' looks like a suspiciously large negative quantity. Don't you mean

 c = -0.5596; % Fatigue Ductility Exponent ?