Question

For the 3-D indeterminate (4-member) TRUSS structure shown in Figure 2A. Given that Px 10K (in X-direction); Py none (in Y-direction); E 30,000 ksi; A 0.2 square inches. The nodal coordinates, the earth-quake displacement/settlement, and members' connectivity information are given aS Applied Load! Earth-Quake MEMBER #1 NODE # X node-i node-j 120.00" 160.00"| 80.00"| Px=-10 Kips none Py- none 120.00" 160.00"0.00"none 120.00"0.00" 0.00" none 0.00" 0.00"0.00" none 0.00" 0.00" 80.00" none none 2 none 4 4 none 4 +2.00" (in Y-direction Using the "direct stiffness matrix" method that you have (a) Find the ELEMENT stiffness matrices, and the corresponding ELEMENT load vectors in the GLOBAL reference ?? (b) Find the SYSTEM stiffness matrix, and the corresponding SYSTEM load vector "before" applying the boundary conditions?? (c) Find the "new, transformed" SYSTEM stiffness matrix, and the corresponding "new, transformed" SYSTEM load vector "after" imposing the boundary conditions?? (d) Find ALL the joint (global) displacements ?? (e) Find ALL the support reactions ?? (f Find Member-End-Actions (axial forces, in this case) for every truss members??

Answer 1

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b) system stiffness matrix - k_G =

Columns 1 through 12

19.4444 25.9259 5.7630 0 0 0 0 0 0 -8.6444 -11.5259 -5.7630
25.9259 61.4007 21.1004 0 0 0 0 -26.8328 -13.4164 -11.5259 -15.3679 -7.6839
5.7630 21.1004 85.5502 0 0 -75.0000 0 -13.4164 -6.7082 -5.7630 -7.6839 -3.8420
0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 0 -75.0000 0 0 75.0000 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0
0 -26.8328 -13.4164 0 0 0 0 26.8328 13.4164 0 0 0
0 -13.4164 -6.7082 0 0 0 0 13.4164 6.7082 0 0 0
-8.6444 -11.5259 -5.7630 0 0 0 0 0 0 8.6444 11.5259 5.7630
-11.5259 -15.3679 -7.6839 0 0 0 0 0 0 11.5259 15.3679 7.6839
-5.7630 -7.6839 -3.8420 0 0 0 0 0 0 5.7630 7.6839 3.8420
-10.8000 -14.4000 0 0 0 0 0 0 0 0 0 0
-14.4000 -19.2000 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0

Columns 13 through 15

-10.8000 -14.4000 0
-14.4000 -19.2000 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
10.8000 14.4000 0
14.4000 19.2000 0
0 0 0

attaching a MATLAB code. it will be helpul always.

clear all;
clc;
tic;

E=30000;
A=.2;

Con=[1 2;1 3;1 4;1 5];
node_vec=unique(reshape(Con',1,[]));
node=numel(node_vec);
dof1=3;
dof=[];
for i=1:node
dof=[dof;dof1*i-dof1+1, dof1*i-dof1+2, dof1*i-dof1+3];
end
dof_vec=unique(reshape(dof',1,[]));
ndof_tot=numel(dof_vec);  
dof_R=[4:15];
dof_A=setdiff(dof_vec,dof_R);
del_R=zeros(numel(dof_R),1);
del_R(12)=2;
p=1
del=zeros(ndof_tot,1);
F=zeros(ndof_tot,1);
F(1)=-10;
F_A=F(dof_A);
x=[120 120 120 0 0];
y=[160 160 0 0 0];
z =[80 0 0 0 80];
Coord=[x' y' z'];
nmem=size(Con,1);
areapop=ones(nmem,1)*.2;
nCoord=size(Coord,1);
k_G=zeros(dof1*node);
for i=1:size(Con,1)
% coordinates of first node of element
x1=Coord(Con(i,1),1);
y1=Coord(Con(i,1),2);
z1=Coord(Con(i,1),3);
  
%coordinates of second node of element
x2=Coord(Con(i,2),1);
y2=Coord(Con(i,2),2);
z2=Coord(Con(i,2),3);
  
L=sqrt((x2-x1)^2+(y2-y1)^2+(z2-z1)^2);
cx= (x2-x1)/L;
cy= (y2-y1)/L;
cz= (z2-z1)/L;
  
T=[cx cy cz 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 cx cy cz
0 0 0 0 0 0
0 0 0 0 0 0];  
  
k_loc= E*A/L*[1 0 0 -1 0 0
0 0 0 0 0 0
0 0 0 0 0 0
-1 0 0 1 0 0
0 0 0 0 0 0
0 0 0 0 0 0];
  
k=T'*k_loc*T;   
ni=Con(i,:);  
dii=dof(ni,:);
dj=reshape(dii',1,[]);
k_G(dj,dj)=k_G(dj,dj)+k;
  
end
k_RR=k_G(dof_R,dof_R);
k_RA=k_G(dof_R,dof_A);
k_AR=k_RA';
k_AA=k_G(dof_A,dof_A);
del_A=(k_AA)\(F_A-k_AR*del_R);
F_R=k_RR*del_R+k_RA*del_A;
  
del(dof_R,1)=del_R;
del(dof_A,1)=del_A;
F(dof_R)=F_R;
  
u1=del(1:dof1:end);
u2=del(2:dof1:end);
u3=del(3:dof1:end);
  
x_new=x+u1';
y_new=y+u2';
z_new=z+u3';
U=[u1' u2' u3'];
u(p)=max(U);
p=p+1;
k_G