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Femassignment1, Exercises of Accounting

finiteelemtn

Typology: Exercises

2015/2016

Uploaded on 08/08/2016

annshun_kiat
annshun_kiat 🇬🇧

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bg1
E=100e9; %Young Modulus
A0=100e-6; %Cross Sectional Area a x=L
P=1000; %Apply Load
Length=1; %Total length
L=Length/10; %Length in each element =0.1m
x=1; %for loop purpose
n=L; %for loop purpose to calculate length
for i=1:1:10 % FOr Loop to calculate k1 to k10
c=n/2;%centroid of each element
A=A0*(1-c/2/Length); %Cross section area of each element
k(:,x)=(E*A/L); %forming k1 to k10 matrix
n=n+L; %go to next centroid
ie .15,0.25,0.35,0.45,0.55,0.65,0.75,0.85,0.95
x=x+1; %for next column of matrix
end
K=[
k(:,1)+k(:,2) -k(:,2) 0 0 0 0 0 0 0 0;
-k(:,2) k(:,2)+k(:,3) -k(:,3) 0 0 0 0 0 0 0;
0 -k(:,3) k(:,3)+k(:,4) -k(:,4) 0 0 0 0 0 0;
0 0 -k(:,4) k(:,4)+k(:,5) -k(:,5) 0 0 0 0 0;
0 0 0 -k(:,5) k(:,5)+k(:,6) -k(:,6) 0 0 0 0;
0 0 0 0 -k(:,6) k(:,6)+k(:,7) -k(:,7) 0 0 0;
0 0 0 0 0 -k(:,7) k(:,7)+k(:,8) -k(:,8) 0 0;
0 0 0 0 0 0 -k(:,8) k(:,8)+k(:,9) -k(:,9) 0;
0 0 0 0 0 0 0 -k(:,9) k(:,9)+k(:,10) -k(:,10);
0 0 0 0 0 0 0 0 -k(:,10) k(:,10);]; % [K] Structural stiffness
matrix, with eliminated row1 n column 1
disp('[K]=')
disp(K) %show result of [K]
F=[0; 0; 0; 0; 0; 0; 0; 0; 0; P]; % Vector of applied Force with P=1000
U=K\F; %solve the matrix to obtain displacement
%to plot displacement versus x
X=0:0.1:1; %x-axis
Y=[0 U(1,:) U(2,:) U(3,:) U(4,:) U(5,:) U(6,:) U(7,:) U(8,:) U(9,:) U
(10,:)]; % displacement value with u1=0
disp('x=')
disp(X)
disp('displacement=')
disp(Y)
figure
plot(X,Y,'-o')
title('displacement versus x-cordinate')
xlabel('x-cordinate')
ylabel('Displacement/m')

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E=100e9; %Young Modulus A0=100e-6; %Cross Sectional Area a x=L P=1000; %Apply Load Length=1; %Total length L=Length/10; %Length in each element =0.1m x=1; %for loop purpose n=L; %for loop purpose to calculate length

for i=1:1:10 % FOr Loop to calculate k1 to k c=n/2;%centroid of each element A=A0(1-c/2/Length); %Cross section area of each element k(:,x)=(EA/L); %forming k1 to k10 matrix n=n+L; %go to next centroid ie .15,0.25,0.35,0.45,0.55,0.65,0.75,0.85,0. x=x+1; %for next column of matrix end

K=[

k(:,1)+k(:,2) -k(:,2) 0 0 0 0 0 0 0 0; -k(:,2) k(:,2)+k(:,3) -k(:,3) 0 0 0 0 0 0 0; 0 -k(:,3) k(:,3)+k(:,4) -k(:,4) 0 0 0 0 0 0; 0 0 -k(:,4) k(:,4)+k(:,5) -k(:,5) 0 0 0 0 0; 0 0 0 -k(:,5) k(:,5)+k(:,6) -k(:,6) 0 0 0 0; 0 0 0 0 -k(:,6) k(:,6)+k(:,7) -k(:,7) 0 0 0; 0 0 0 0 0 -k(:,7) k(:,7)+k(:,8) -k(:,8) 0 0; 0 0 0 0 0 0 -k(:,8) k(:,8)+k(:,9) -k(:,9) 0; 0 0 0 0 0 0 0 -k(:,9) k(:,9)+k(:,10) -k(:,10); 0 0 0 0 0 0 0 0 -k(:,10) k(:,10);]; % [K] Structural stiffness matrix, with eliminated row1 n column 1 disp('[K]=') disp(K) %show result of [K] F=[0; 0; 0; 0; 0; 0; 0; 0; 0; P]; % Vector of applied Force with P= U=K\F; %solve the matrix to obtain displacement %to plot displacement versus x X=0:0.1:1; %x-axis Y=[0 U(1,:) U(2,:) U(3,:) U(4,:) U(5,:) U(6,:) U(7,:) U(8,:) U(9,:) U (10,:)]; % displacement value with u1= disp('x=') disp(X) disp('displacement=') disp(Y) figure plot(X,Y,'-o') title('displacement versus x-cordinate') xlabel('x-cordinate') ylabel('Displacement/m')