%----------------------------------------------------------------------------
%                                 ENGR311
%
%                            Method of Joints:
%                    Simple Space Truss Examples- Tripod
%
%                           Dr. Chris L. Mullen
%                                  1-25-99
%
%----------------------------------------------------------------------------
  

%  Position Vectors for Joints

ra=[3,-4,0]
rb=[3,2,0]
rc=[-2,2,0]
rd=[0,0,8]

%  Relative Position Vectors Defining Member Orientation

rab=rb-ra
rbc=rc-rb
rca=ra-rc
rda=ra-rd
rdb=rb-rd
rdc=rc-rd

%  Member Lengths

lab=norm(rab)
lbc=norm(rbc)
lca=norm(rca)
lda=norm(rda)
ldb=norm(rdb)
ldc=norm(rdc)

%  Unit Vectors Oriented Along Member Axes

uab=rab/lab
ubc=rbc/lbc
uca=rca/lca
uda=rda/lda
udb=rdb/ldb
udc=rdc/ldc

%  Global Equilibrium to Find Vertical Support Reactions
%  {FAy, FCy, FAz, FBz, FCz}

%  Sum {Fy, Fz, Mx, My, Mz} = 0   ; Sum Fx = 0 is trivial here.

A= [ 1,  1,  0,  0,  0;
     0,  0,  1,  1,  1;
     0,  0, -4,  2,  2;
     0,  0, -3, -3,  2;
     3, -2,  0,  0,  0 ]         

b=-[500; 0; -500*8; 0; 0]

FR=A\b


% Joint D equilibrium to Find Bar Forces in Members {DA, DB, DC}.
%   Sum {Fx, Fy, Fz} = 0   
%
%   Note: Solution here does not require knowledge of reactions

A= [uda(1),udb(1),udc(1);
    uda(2),udb(2),udc(2);
    uda(3),udb(3),udc(3)]

b=-[0;500;0]

FD=A\b

%Joint A Equilibrium to Find Bar Forces in Members {AB,CA,DA}.

A=[uab(1),-uca(1),-uda(1);
   uab(2),-uca(2),-uda(2);
   uab(3),-uca(3),-uda(3)]

b=-[0;FR(1);FR(3)]

FA=A\b

%Joint B Equilibrium to Find Bar Forces in Member {BC}.

A=[-uab(1),ubc(1),-udb(1);
   -uab(2),ubc(2),-udb(2);
   -uab(3),ubc(3),-udb(3)]

b=-[0;0;FR(4)]

FB=A\b

%Joint C Equilibrium

A=[-ubc(1),uca(1),-udc(1);
   -ubc(2),uca(2),-udc(2);
   -ubc(3),uca(3),-udc(3)]

b=-[0;FR(2);FR(5)]

FC=A\b