MATLAB: How to resolve Error Using max function

error using max and too many input arguments.MATLAB

clc
clear all;
close all;
for kk =1:44
for k=1:50
 % here drift values are in the form of drift ratio
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Roof.txt']);
 Roof_Drift_max(kk,k,:)= max(abs(Roof(:,2)));
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Story1.txt']);
 Story1_Drift_max(kk,k,:)= max(abs(Story1(:,2)));
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Story2.txt']);
 Story2_Drift_max(kk,k,:)= max(abs(Story2(:,2)));
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Story3.txt']);
 Story3_Drift_max(kk,k,:)= max(abs(Story3(:,2)));
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Story4.txt']);
 Story4_Drift_max(kk,k,:)= max(abs(Story4(:,2)));
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Story5.txt']);
 Story5_Drift_max(kk,k,:)= max(abs(Story5(:,2)));
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Story6.txt']);
 Story6_Drift_max(kk,k,:)= max(abs(Story6(:,2)));
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Story7.txt']);
 Story7_Drift_max(kk,k,:)= max(abs(Story7(:,2)));
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Story8.txt']);
 Story8_Drift_max(kk,k,:)= max(abs(Story8(:,2)));
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Story9.txt']);
 Story9_Drift_max(kk,k,:)= max(abs(Story9(:,2)));
% Maximum Interstorey Drift Ratio
 MIDR(kk,k,:) =max(Roof_Drift_max(kk,k,:),Story1_Drift_max(kk,k,:),Story2_Drift_max(kk,k,:),Story3_Drift_max(kk,k,:),Story4_Drift_max(kk,k,:),...
                   Story5_Drift_max(kk,k,:),Story6_Drift_max(kk,k,:),Story7_Drift_max(kk,k,:),Story8_Drift_max(kk,k,:),Story9_Drift_max(kk,k,:));
end
end

Best Answer

I am not certain what you want.
One option is to enclose all the arguments within square brackets [], effectively concatenating them: 
 MIDR(kk,k,:) =max([Roof_Drift_max(kk,k,:),Story1_Drift_max(kk,k,:),Story2_Drift_max(kk,k,:),Story3_Drift_max(kk,k,:),Story4_Drift_max(kk,k,:),...
                   Story5_Drift_max(kk,k,:),Story6_Drift_max(kk,k,:),Story7_Drift_max(kk,k,:),Story8_Drift_max(kk,k,:),Story9_Drift_max(kk,k,:)]);
This will produce an ‘MIDR’ matrix with the same dimensions as the argument matrices.
Experiment with other approaches to get different results.
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