function [c ,ceq] = columnConstraints(x)

%pu=3000*1000;

%fck=20;  %20
%fck=25;    %25               %Characteristic strength of concrete
l=3000;
%value= 0.093;  %(p/fck value from chart no 33,37   SP-16)

%fy=415;   %415                %Characteristic strength of steel
%fy=500;    %500
%p = fck*value;
Asc = 5045;  %value taken frm magic equation(excel sheet)

c=[ %pu/((0.4*fck*x(1)*x(2))+(0.67*fy*(.7853*x(3)^2*x(4))))-1;
        
       (0.8*x(1)*x(2)/100)-(.7853*x(3)^2*x(4)); %constraint for min reinf
        
        (.7853*x(3)^2*x(4))-(6*x(1)*x(2)/100);  %constraint for max reinf
        
        
        Asc-(.7853*x(3)^2*x(4)) ;  %constraint for area of main reinforcement
        
        l/x(2)-12;           %slenderness ratio for design as a short column(length/depth<12).
        
       l/x(1)-12;   %  (length/width<12).
       
        x(3)/4 - x(5);    % dia of ties should not be less than 1/4th of largest dia of longitudinal bar
        0.4 - (x(1)/x(2));  %width/depth ratio should be greater than 0.4.
        x(6)-x(1);          %for shear reinf spacing
        x(6)- (16*x(3));  % the shear reinforcement spacing should not be more than 16 times  the dia of small longitudinal bar.
        x(6)-300             %for shear reinf spacing.
        ];
    
           
ceq=[ ];