Returning to a function and changing variables if ans is false?

If someone can link me to a similar answer, I am sorry I could not find it.

I am calculating required beam depths to hold a certain load and I have an equation that tests a certain depth, and it must be less than the max strain allowed.

If my calculated value (Fs) is greater than what is allowed (Fs_max), how do I loop back to the equation and add a factor of 10 or so to the intital variable (dRound) so that it can run tests until it is <= max allowed?

Thanks.

if dRound > d 
        
        Fs_max = input('Enter max shearing force (Fs max) in MPa ');
        
        V_max = W/2;
        
        Fs = (3/2)*((V_max*10^3)/(b*dRound));
        
        fprintf('Shearing force is = %f MPa\n', Fs);
    end
    
end

if Fs > Fs_max
    
    return ????

Answer 1

[Star Strider]

I would do something like this, and treat it as an optimization (specifically root-finding) problem, returning the value of ‘dRound’ such that ‘Fs’ approximately equals ‘F_Max’:

Fs = @(dRound,W,b,F_max) (3/2)*((W/2*1E3)/(b*dRound)) - F_max;

dRound0 = 10;
F_max = 1000;                                                               % I Have No Idea What This Values Should Be
W = pi;                                                                     % I Have No Idea What This Values Should Be
b = 42;                                                                     % I Have No Idea What This Values Should Be

[dRound_est, Fs_val] = fsolve(@(dRound)Fs(dRound,W,b,F_max), dRound0)

The fzero function might also work here, however when I tried it with these values, it returned NaN. (It would be a direct substitution for fsolve in this code.)

[Tim Kim]

Hey thanks for that. I didn't even think of solving for dRound. I appreciate it heaps.

[Star Strider]

My pleasure!