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JMBCE 7323 COMPUTER ANALYSIS OF
STRUCTURES
DESIGN ASSIGNMENT - MATLAB
NAME : BANDAR K.A.J.M.
REG No : RU/E/2008/014
DATE : 25/04/2012
01.
R = 8.014
15
16
14
13
11
12
10
06
08
09 05
04
07
03
02
01
x- y- z- x- y- z-
NODS coordinates coordinates coordinates NODS coordinates coordinates coordinates
01 -4.007 4.007 0 09 -2.672 2.672 8.000
02 -3.339 3.339 4.000 10 -2.672 -2.672 8.000
03 -4.007 -4.007 0 11 2.672 -2.672 8.000
04 -3.339 -3.339 4.000 12 2.672 2.672 8.000
05 4.007 -4.007 0 13 -2.004 2.004 12.000
06 3.339 -3.339 4.000 14 -2.004 2.004 12.000
07 4.007 4.007 0 15 2.004 -2.004 12.000
08 3.339 3.339 8.000 16 2.004 2.004 12.000
function D=Data
% Definition of Data
% 1. Nodal Coordinates
Coord=[-4.007 4.007 0;-3.339 3.339 4.000;
-4.007 -4.007 0;-3.339 -3.339 4.000;
4.007 -4.007 0;3.339 -3.339 4.000;
4.007 4.007 0;3.339 3.339 4.000;
-2.672 2.672 4.000; -2.672 -2.672 8.000;
2.672 -2.672 8.000;2.672 2.672 8.000;
-2.004 2.004 12.000;-2.004 -2.004 12.000;
2.004 -2.004 12.000;2.004 2.004 12.000];
% 2. Connectivity
Con=[1 2;1 4;1 8;3 2;3 4;3 6;
5 4;5 6;5 8;7 6;7 8;7 2;
2 4;2 8;2 12;2 9;2 10;
4 9;4 10;4 6;4 11;6 10;
6 8;6 12;6 11;8 11;8 12;
8 9;9 13;9 10;9 14;10 13;
10 14;10 11;10 15;11 14;
11 15;11 12;11 16;12 15;
12 16;12 9;12 13;9 16;
13 14;14 15;15 16;16 13];
% 3. Definition of Degree of freedom (free=0 & fixed=1); for 2-D trusses the
last
column is equal to 1
Re=[1 1 1; 0 0 0; 1 1 1;0 0 0;1 1 1;0 0 0;1 1 1;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];
% 4. Definition of Nodal loads
Load=[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 0 0;0 0 0;8014 8014 -8014;0 0 0;0 0 0;0 0 0];
% 5. Definition of Modulus of Elasticity
E=[210000000000 210000000000 210000000000 210000000000 210000000000 210000000000
210000000000 210000000000 210000000000 210000000000 210000000000 210000000000
210000000000 210000000000 210000000000 210000000000 210000000000 210000000000
210000000000 210000000000 210000000000 210000000000 210000000000 210000000000
210000000000 210000000000 210000000000 210000000000 210000000000 210000000000
210000000000 210000000000 210000000000 210000000000 210000000000 210000000000
210000000000 210000000000 210000000000 210000000000 210000000000 210000000000
210000000000 210000000000 210000000000 210000000000 210000000000 210000000000];
% 6. Definition of Area
A=[0.01 0.01 0.01 0.01 0.01 0.01
0.01 0.01 0.01 0.01 0.01 0.01
0.01 0.01 0.01 0.01 0.01 0.01
0.01 0.01 0.01 0.01 0.01 0.01
0.01 0.01 0.01 0.01 0.01 0.01
0.01 0.01 0.01 0.01 0.01 0.01
0.01 0.01 0.01 0.01 0.01 0.01
0.01 0.01 0.01 0.01 0.01 0.01];
% Convert to structure array
D=struct('Coord',Coord','Con',Con','Re',Re','Load',Load','E',E','A',A');
function [F,U,R]=ST(D)
% w:matrix size, S:force vector size,
w=size(D.Re);S=zeros(3*w(2));U=1-D.Re;f=find(U);
for i=1:size(D.Con,2)
H=D.Con(:,i);
C=D.Coord(:,H(2))-D.Coord(:,H(1));
Le=norm(C);
T=C/Le;% calculation the cos value
s=T*T';% lamda gobal stiffness matrix
G=D.E(i)*D.A(i)/Le;
Tj(:,i)=G*T;
e=[3*H(1)-2:3*H(1),3*H(2)-2:3*H(2)];%check the corresponding degree of
freed
S(e,e)=S(e,e)+G*[s -s;-s s];% global matrix
end
U(f)=S(f,f)\D.Load(f);%normal equation f=ku calculate
F=sum(Tj.*(U(:,D.Con(2,:))-U(:,D.Con(1,:))))*0.001;% reation and
displaced calculaed
R=reshape(S*U(:),w);
R(f)=0;
02.
Reaction at Base level
Nods point X- Direction(kN) Y- Direction(kN) Z- Direction(kN)
01 -0.141 -2.338 5.085
03 -2.524 -4.066 -11.074
05 0.128 1.002 1.077
07 -5.477 -2.612 12.926
03.
Deformation at Top level
Nods point X- Direction(mm) Y- Direction(mm) Z-
Direction(mm)
13 -0.0038 0.1854 -0.0801
14 0.0475 0.1814 -0.0033
15 0.0473 0.0471 -0.0018
16 -0.0067 0.0473 -0.0403