%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
% 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
if MomentCols==1 %just sent a dummy
[d sl]=displace(x,Moment,['place' 'place'],[0 L],E,I);
coefs=[2*L^2 -L^2;-L^2 2*L^2]/(6*E*I*L);
SubSca=[2*L^2 -L^2;-L^2 2*L^2];
coefs=[SubSca SubRow; SubCol SubMat]/(6*E*I*L);