% Recursive algorithm

clear,clc,format compact

FiFEA=[0:15:345];%whole turn of the shaft 360 deg, every 15 deg -> reaction forces 0–345 deg

%example shaft: S-12075-3600-10-8-149-144-C-04, example main journal: #10, FEA reaction forces:

RFEA=[749.87 768.80 775.20 767.37 747.40 720.64 694.26 675.33 668.93 676.77 696.74 723.50 749.87 768.80 775.20 767.37 747.40 720.64 694.26 675.33 668.93 676.77 696.74 723.50];

DeltaFi=1;% calculation step (deg)

R0=mean(RFEA);

disp([‘Mean value of the reaction forces: ‘ num2str(R0) ‘ N’])

CR2=0.5*(max(RFEA)-min(RFEA));

disp([‘Reaction forces amplitude of the 2nd harmonic: ‘ num2str(CR2) ‘ N’])

for j=1:360 % calculation loop

FiR2=(j-1)*DeltaFi;

P1=RFEA-(R0+CR2*sind(FiFEA*2+FiR2));

Q(j)=P1*P1’; % sum of squares of deviations

end

k=length(Q);

[Qmin,L]=min(Q);

[Qmax,H]=max(Q);

FiR2a=L-1;

FiR2b=H+1;

disp([‘Reaction forces phase shift of the 2nd harmonic: ‘ num2str(FiR2a) ‘ deg’])