Introduction to FEM-2

8/22/2019 Introduction to FEM-2

1/31

Transient Thermal Analysis

= neglecting radiation and q

= Applying Galerkin weighted residual method

= 0

x

x = 0

Rearranging and removing ^ for convenience

=

x

x = 0 (1)


2/31

Transient Thermal Analysis

=

= = 6 2 11 2

=

= = kA

1 1 11 1

6 2 11 2

=

=


3/31

Transient Thermal Analysis (thermal) ) + = 1 1 +

Determine the temperatures

Heat removal rate and efficiency

Of the pin (Home Work)

If the density c are 8900/And 375 respectively find theTemperatures at the middle and end

With a time step of 0.1 s and

=


4/31

Transient Thermal Analysis (thermal)

clear all;

clc;

clf;

stiff_mat;

fun;

T0=25*ones(1,nnode)';

dt=0.05;

k1=C+alpha*dt*K;

k2=C-(1-alpha)*K*dt;

f1=k1*T0;

k11=k1;

k11(1,1)=1;

f11=f1;

f11(1)=85;

f11(2)=f11(2)-k11(2,1)*f11(1);k11(1,2)=0;

k11(2,1)=0;

T=k11\f11;

T1(1)=T(1);

T2(1)=T(2);

T3(1)=T(3);

F';

for i=1:60;

k1=C+alpha*dt*K;

k2=C-(1-alpha)*K*dt;

f1=k2*T+F'*dt;

T=k1\f1;

T=[85

T(2:nnode)];

T1(1+i)=T(1);

T2(1+i)=T(2);

T3(1+i)=T(3);t=dt+i*dt;

hold on;

plot(t,T(1),'+ -', t,T(2),'+ -',t,T(3),'+ -')

end

hold off;


5/31

Transient Thermal Analysis-(stiff_mat)

a=0;b=0.02;nelem=2;

nnode=nelem+1;

h=b/nelem;

gnode=zeros(1,2);elenode=zeros(2,2);

elenode1=zeros(2,2);

k_mat=zeros(nnode,nnode);

m_mat=zeros(nnode,nnode);

global hx kx rho c alpha p A;

hx=150;kx=400;

rho=8900;

c=375;

alpha=2/3;

A=pi*.004^2/4;

p=pi*.004;count=1;

for i=1:nelem

count=count+1;

z=i;

gnode(1)=z;

gnode(2)=gnode(1)+1;for j=1:2

for k=1:2

elenode(j,k)=dN(j,k,0,h);

elenode1(j,k)=N(j,k,0,h);

end

endfor j=1:2

for k=1:2

kk=elenode(j,k);

mm=elenode1(j,k);

m=gnode(j);n=gnode(k);

k_mat(m,n)=k_mat(m,n)+kk;

m_mat(m,n)=m_mat(m,n)+mm;

end

end

end

K=k_mat*kx*A+hx*p*m_mat;

C=rho*c*A*m_mat;


6/31

Transient Thermal Analysis-Time discretization

= = lim ()

()

Substituting for in = ( = )

() K a ( ) = f

+ = + = a f

Forward difference scheme


7/31


( ) Substituting for in = ( = ) ( )

K a ( ) = f

Backward difference scheme

( ) = ()Multiplying with and re-arranging

= f

It can be re-written as

= f

+ = f+


8/31


Central difference Scheme

= ()2 = ()2

() However instead of evaluating the other terms at time , Use the average values

We have

() K ()2 = ()2 Multiplying through by isolating the terms containing ()

2 =

2 ()[ ]

2


9/31


In terms of subscript notation

2 + = 2 + 2 The three recurrence formulae can be combined into

) + = 1 1 +

When = 0 or 1 the above equation reduces to forward difference formulaCentral difference formula or backward difference formula


10/31

Time discretization- FEM

The time domain is discretized from

time

= ++Applying Galerkin method with weight function

= 0

= + + =

1 = 0 + = =

= + =

If we take = 1

= 1

; =

1


11/31

Time discretization- FEMIf we take = 1

= 1 1 1

+ 1

+ 1

+

= 0

+ 12 12 + 12 12 + = 0

2 + =

2 + 2

) + = 1 1 + When = 0 or 1 the above equation reduces to forward difference formulaCentral difference formula or backward difference formula


12/31

Vibration- FEM


13/31


14/31


15/31


16/31

Vibration-Example


17/31

Vibration-Example

=

= , = = 32.2 12 /

= 0.28332.212 =7.32410


18/31

Vibration-Example


19/31

Vibration-Example


20/31

Vibration-Example: mode shapes(2

elements)


21/31

Vibration-Example: first three mode shapes(20

elements) of a straight bar


22/31

Matlab code-stiff_mat.m

a=0;b=0.10;nelem=20;

nnode=nelem+1;

h=b/nelem;

gnode=zeros(1,2);

elenode=zeros(2,2);

elenode1=zeros(2,2);k_mat=zeros(nnode,nnode);

m_mat=zeros(nnode,nnode);

global E A rho;

E=200e9;

A=pi*.10^2/4;

rho=7800;

for i=1:nelem

z=i;

gnode(1)=z;

gnode(2)=gnode(1)+1;for j=1:2

for k=1:2

elenode(j,k)=dN(j,k,0,h);

elenode1(j,k)=N(j,k,0,h);

end

end

for j=1:2

for k=1:2

kk=elenode(j,k);

mm=elenode1(j,k);

m=gnode(j);n=gnode(k);

k_mat(m,n)=k_mat(m,n)+kk;

m_mat(m,n)=m_mat(m,n)+mm;

end

end

end

K=k_mat*E*A;

M=rho*A*m_mat;


23/31

Matlab code-vibration.mclear all;

clc;

clf;

stiff_mat;

M1=M(2:nnode,2:nnode);

K1=K(2:nnode,2:nnode);

y1=zeros(nelem,1);

y2=zeros(nelem,1);

y3=zeros(nelem,1);

y4=zeros(nelem,1);

y5=zeros(nelem,1);

[c d]=eig(K1,M1);

for j=1:nelem

y1(j)=c(j,1);

y2(j)=c(j,2);

y3(j)=c(j,3);

endy1=[0 y1']';

y2=[0 y2']';

y3=[0 y3']';

x=linspace(a,b,nnode);

plot(x,y1,x,y2,x,y3)

Other prograns N.m, dN.m and

functions are the same as the

general 1D program


24/31

Higher order elements: 1D quadraticIn formulating the truss element and the one-dimensional heat conduction element,

only line elements having a single degree of freedom at each of two nodes are

considered. While quite appropriate for the problems considered, the linear element is

by no means the only one-dimensional element that can be formulated for a given

problem type. Figure depicts a three-node line element in which node 2 is an interior

node. The interior node is notconnected to any other node in any other element in the

model. Inclusion of the interior node is a mathematical tool to increase the order of

approximation of the field variable. Assuming that we deal with only 1 degree of

freedom at each node, the appropriate polynomial representation of the field variable

is


25/31

Higher order elements: 1D quadratic

Applying the general procedure outlined previously in the context of thebeam element, we apply the nodal (boundary) conditions to obtain

from which the interpolation functions are obtained via the following

sequence


26/31

Higher order elements: 1D quadratic

clear all;

syms Uu1u2u3a0a1a2xL;

U=[u1;u2;u3];

C=[1 0 0; 1 L/2 L^2/4; 1 L L^2];

a=inv(C)*U;

N=[1 x x^2]*a;

N=collect(N,u1);

u2=0;u3=0;p=eval(coeffs(expand(N), u1));

N1=p(1,2);

syms u2;

u1=0;u3=0;

p=eval(coeffs(expand(N), u2));

N2=p(1,2);syms u3;

u1=0;u2=0;


N3=p(1,2);

N=[N1 N2 N3];

>> N.'

ans =

(2*x^2)/L^2 - (3*x)/L + 1

(4*x)/L - (4*x^2)/L^2(2*x^2)/L^2 - x/L

Higher order elements: 1D quadratic natural


27/31

Higher order elements: 1D quadratic- natural

coordinates

=

varies from -1 to 1

= 1 = = = 0 = = = 1 = =

= 1 1 11 0 01 1 1

= 1 1 11 0 0

1 1 1

= 1 . N1= r^2/2 - r/2

N2= 1 - r^2

N3= r^2/2 + r/2

clear all;

syms Uu1u2u3a0a1a2xLr;

U=[u1;u2;u3];

C=[1 -1 1; 1 0 0; 1 1 1];

a=inv(C)*U;

N=[1 r r^2]*a;

N=collect(N,u1);

u2=0;u3=0;


N1=p(1,2);

syms u2;

u1=0;u3=0;


N2=p(1,2);

syms u3;

u1=0;u2=0;p=eval(coeffs(expand(N), u3));

N3=p(1,2);

N=[N1 N2 N3];


28/31

Higher order elements: 1D quadratic- natural coordinates

If the end node are numbered 1 and 2 and the middle node as 3 the shape functions

are as shown below

N1= r^2/2 - r/2N2= r^2/2 + r/2

N3= 1 - r^2

= 12 1 = 12 1 = ( 1 ) ( 1 )

Or

= 12 1 = 12 1 = ( 1 ) ( 1 )


29/31



30/31

Higher order elements: 1D quadratic-

clear all;

syms U u1 u2 u3 a0 a1 a2 x L;

hx=30; kx=168;A=.005*.001;p=2*(.005+.001);

U=[u1;u2;u3];

N1=((2*x^2)/L^2 - (3*x)/L + 1);

N2=((2*x^2)/L^2 - x/L);

N3=((4*x)/L - (4*x^2)/L^2);

N=[N1 N2 N3]; dN=diff(N);

K1=kx*A*int(dN.'*dN,0,L)+hx*p*int(N.'*N,0,L);

K2=kx*A*int(dN.'*dN,0,L)+hx*p*int(N.'*N,0,L);

ff1=hx*p*20*int(N.',0,L);

ff2=hx*p*20*int(N.',0,L);

L=0.04;

K1=eval(K1);

K2=eval(K2);f1=eval(ff1);f2=eval(ff2);K11=ex_rows(K1,2,3);K11=ex_cols(K11,2,3);

K22=ex_cols(K2,2,3);K22=ex_rows(K22,2,3);

f11=ex_rows(f1,2,3);f22=ex_rows(f2,2,3);

K11=[K11 zeros(3,2);zeros(2,5)];

K22=[zeros(2,5);zeros(3,2) K22];

Kg=K11+K22;

f11=[f11;zeros(2,1)]; f22=[zeros(2,1);f22];

fg=f11+f22;

Kg(1,1)=1;Kg(1,2)=0;Kg(1,3)=0;fg(1)=100;Kg(5,5)=Kg(5,5)+hx*A;

fg(5)=fg(5)+hx*A*20;

T=Kg\fg;

x=[0 .02 .04 .06 .08];

m=sqrt(hx*p/(kx*A));

T0=20;Tw=100;L=0.08;T1=T0+(Tw-T0)*cosh(m*(L-x))/cosh(m*L);

TT=[T T1'];

display(TT);

plot(x',T,'rd-',x',T1','b+-')


31/31


Exact Quadratic Linear

100 100 100

75.2737 75.2269 75.0387

60.1591 60.0864 59.7901

52.0278 51.8858 51.5633

49.4659 49.2468 48.9064

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.0840

50

60

70

80

90

100

Length

Temperature

quadratic

exact

linear

Date post:	08-Aug-2018
Category:	Documents
Upload:	aswith-r-shenoy
View:	215 times
Download:	0 times

Introduction to FEM-2

Documents