homework 6 solution

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clc clear all close all %% geometry Lx=0.1; %m Ly=0.1; %m L=sqrt(Lx^2+Ly^2); A=0.001; %m2 alpha=pi/4; E=210e9; Ex=[0 Lx]; Ey=[0 Ly]; % Bouandary and load condition pres=[4]; fix=[1 2 3]; % Time stepping acc. to Newton Raphson nstep=100; tol=1e-6; maxiter=100; % Max number iterations ndof=4; % initial values x=zeros(ndof,1); % history variables x_hist=zeros(ndof,nstep); f_hist=zeros(ndof,nstep); u=zeros(4,1); for i=1:nstep+1 x=[0 0 0.1 -Ly+2*Ly*(i-1)/nstep]; du=2*Ly/nstep; alpha=atan(-x(4)/x

TRANSCRIPT

Page 1: Homework 6 solution
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clc

clear all

close all

%% geometry

Lx=0.1; %m

Ly=0.1; %m

L=sqrt(Lx^2+Ly^2);

A=0.001; %m2

alpha=pi/4;

E=210e9;

Ex=[0 Lx];

Ey=[0 Ly];

% Bouandary and load condition

pres=[4];

fix=[1 2 3];

% Time stepping acc. to Newton Raphson

nstep=100;

tol=1e-6;

maxiter=100; % Max number iterations

ndof=4;

% initial values

x=zeros(ndof,1);

% history variables

x_hist=zeros(ndof,nstep);

f_hist=zeros(ndof,nstep);

u=zeros(4,1);

for i=1:nstep+1

x=[0 0 0.1 -Ly+2*Ly*(i-1)/nstep];

du=2*Ly/nstep;

alpha=atan(-x(4)/x(3));

K=(A*E/L)*[1 -1;-1 1];

R=[cos(alpha) sin(alpha) 0 0;

0 0 cos(alpha) sin(alpha)];

K_glob=R’*K*R;

f=K_glob*x’;

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Page 7: Homework 6 solution

x_hist(:,i)=x’;

f_hist(:,i)=f;

end

plot(x_hist(4,:)/L,f_hist(4,:)/(E*A))

grid on

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Page 8: Homework 6 solution

Continuum mechanics modelclc

clear all

close all

%% Material parameters

global E

E=210e9;

alfa=pi/4;

b=0.2; % the base

%% ------------------ Part a

A1=0.001;

A2=A1;

A=[A1 A2];

nen=2;

% Positions

x=[-b/2 0 b/2];

h=b/(2*tan(alfa));

y=[0 h 0];

z=[0 0 0];

Coord=[x’ y’ z’];

Dof=[1 2 3;4 5 6;7 8 9];

Edof=[1 1 2 3 4 5 6; 2 4 5 6 7 8 9];

[ex,ey,ez]=coordxtr(Edof,Coord,Dof,nen);

nelem=2;

figure(3)

plotpar=[1 3 1];

eldraw2(ex,ey,plotpar)

% Bouandary and load condition

free=[4];

pres=[1 2 3 5 6 7 8 9];

% Time stepping acc. to Newton Raphson

nstep=50;

tol=1e-6;

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Page 9: Homework 6 solution

maxiter=100; % Max number iterations

ndof=max(max(Edof));

% initial values

x=zeros(ndof,1);

x(free)=[0];

% history variables

x_hist=zeros(ndof,nstep);

T_hist=zeros(ndof,nstep);

tau_hist=zeros(nelem,nstep);

eps_hist=zeros(nelem,nstep);

for i=1:nstep

x(pres)=[-b/2 0 0 h-2*h*(i-1)/nstep 0 b/2 0 0];

for k=1:maxiter

K=zeros(ndof);

T=zeros(ndof,1);

for j=1:nelem

Xa=[ex(j,1) ey(j,1) ez(j,1)];

Xb=[ex(j,2) ey(j,2) ez(j,2)];

Xe=[Xa Xb]’;

xa=x(Edof(j,2:4))’;

xb=x(Edof(j,5:7))’;

xe=[xa xb]’;

[Te,Ke,tau(j),eps(j)]=bar_element(Xa,Xb,xa,xb,A(j));

[K,T]=assem(Edof(j,:),K,Ke,T,Te);

end

g=T(free);

if norm(g)<tol

disp([’Timestep ’ num2str(i) ’, convergence at ’ num2str(k)

’ iterations. Residual: ’ num2str(norm(g))])

break

elseif k==maxiter

error([’Divergence at ’ num2str(maxiter) ’ iterations’])

end

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Page 10: Homework 6 solution

x(free)=x(free)-inv(K(free,free))*T(free);

end

Ex=[x(1) x(4);x(4) x(7)];

Ey=[x(2) x(5);x(5) x(8)];

Ez=[x(3) x(6);x(6) x(9)];

figure(4)

plotpar=[1 3 1];

eldraw2(Ex,Ey,plotpar); hold all;

function [Te,Ke,tau,eps]=bar_element(Xa,Xb,xa,xb,A)

global E

L=norm(Xb-Xa);

l=norm(xb-xa);

n=(xb-xa)/norm(xb-xa);

lambda=l/L;

eps=log(lambda);

tau=E*eps;

tau_prim=E;

Tb=tau*exp(-eps)*A*n;

Ta=-Tb;

Te=[Ta’; Tb’];

dTbdxb=(tau_prim-2*tau)*exp(-2*eps)*A/L*n’*n+tau*exp(-2*eps)*A/L*eye(3);

dTadxa=dTbdxb;

dTbdxa=-dTbdxb;

dTadxb=-dTbdxb;

Ke=[dTadxa dTadxb;

dTbdxa dTbdxb];

end

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Page 11: Homework 6 solution

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force-displacement in symetrical case

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