| [forces]=fixedfixed(x,s,m,a,L,E,I) |
function [forces]=fixedfixed(x,s,m,a,L,E,I)
%FIXEDFIXED Redundant support moments and forces.
% FIXEDFIXED(X,SHEAR,MOMENT,PLACEMENT,L,E,I) will find the redundant
% moment at the fixed supports and the force supplied by any redundant pin
% supports along the length of the beam.
%
% SHEAR is the shear acting along the beam, this should be created with
% the DIAGRAM routine. It does not have to be summed into a single
% vector for use in the routine.
% MOMENT is the moment acting along the beam, this should be only the
% point moments created with the DIAGRAM routine. It does not have to
% be summed into a single vector for use in the routine. It should not
% include the integral of the shear as created with the routine
% DIAGRAMINTEGRAL.
% PLACEMENT is a vector with the location of every pin support.
% L is the length of the beam.
% E is the Young's modulus.
% I is the area moment of inertia of the beam cross section.
%
% See also DISPLACE, FIXEDPIN, PINPIN.
% Details are to be found in Mastering Mechanics I, Douglas W. Hull,
% Prentice Hall, 1998
% Douglas W. Hull, 1998
% Copyright (c) 1998-99 by Prentice Hall
% Version 1.00
b=L-a;
[ShearRows, ShearCols]=size(s);
[MomentRows, MomentCols]=size(m);
if ShearRows>1
Shear=sum(s);
else
Shear=s;
end
if MomentCols==1 %just sent a dummy
Moment=diagramintegral(x,Shear);
else
m(MomentRows+1,:)=diagramintegral(x,Shear);
Moment=sum(m);
end
[d sl]=displace(x,Moment,['place' 'place'],[0 L],E,I);
Deltas(1)=-interpolate(x,sl,0);
Deltas(2)=-interpolate(x,sl,L);
coefs=[2*L^2 -L^2;-L^2 2*L^2]/(6*E*I*L);
if a~=0
i=1;
for gapli=3:length(a)+2;
Deltas(gapli)=-interpolate(x,d,a(gapli-2));
i=i+1;
end
SubSca=[2*L^2 -L^2;-L^2 2*L^2];
SubCol=[(a.*((a.^2)-(3*L*a)+(2*L^2)))' -((L-a).*(((L-a).^2)-(3*L*(L-a))+(2*L^2)))'];
SubRow=[(a.*b.*(L+b));-(a.*b.*(L+a))];
SubMat=makepins(a,L,0);
coefs=[SubSca SubRow; SubCol SubMat]/(6*E*I*L);
end
forces=inv(coefs)*Deltas';
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