ANALYSIS OF 2D-ANISOTROPIC PLATE WITH

ELLIPTICAL HOLE

ENGR 6310 Design project

Submitted to

Dr. Jenny Zhou

By

Nadendla Raviteja L20310564

Dinesh Davuluri L20309927

2

INDEX

1. Problem Definition 3

2. Flow Chart 4

3. Discretisation Details 5

4. Results from Program 7

5. Results from Ansys 9

6. Mat lab Code 13

7. Comparison of Results 17

8. References 18

3

PROBLEM DEFINITION

An elliptical hole is cut from a rectangular shaped 2 dimensional plate. A uniformly distributed load is applied at the ends. Due to the elliptical cut out some stress concentration

will be there at the elliptical cut out. Our problem is to find out the stress concentration at the

elliptical cut out. A code for the above mentioned problem has to be written and the results

have to be compared with a standard software package. A dimensioned sketch of the 2

dimensional plate with loading is given in fig 1.

Fig 1: 2D plate with elliptical hole

4

FLOW CHART:

Start

Element connectivity table.

(Element no. + Respective

nodes + constraints)

Giving material properties

such as [D] matrix, youngs

modulus, Poissons ratio

Applying the boundary conditions

using constraints on [k].

For each element find [B], [k] and

globalising the stiffness matrix

simultaneously

For each element calling the co-

ordinates of each node.

Create force matrix by lumping the

pressure as a force at the nodes of

line on the plate.

Applying the boundary conditions

to the forces.

Finding the displacements of the

individual nodes.

Again for each element use [B] matrix

and respective displacements and

evaluate the corresponding stress

matrix for the element.

Find stress on each element and

combine it into a single matrix.

Find the maximum stress in the

applied direction.

Evaluate the SCF

5

Discretisation Details:

The plate is symmetric about two mutually perpendicular axes. So we analysed a single

quadrant of the plate to get the results. The meshing of the quadrant is done as per the figure

given below.

6

SOFTWARE USED:

ANSYS Mechanical APDL (v14.0) Standard package used for checking the results.

MATLAB (R2011a) Software used for coding of the problem.

Material properties:

Isotropic material: (Aluminium)

Ex = 70 GPa

= 0.33

Orthotropic material: (T300/5208)

Ex = 181.8 GPa

Ey = 10.3 GPa

Ez = stiffened in z-direction

xy = 0.3

yz = 0.016

zx = 0.016

Gxy = 7.17 GPa

Gyz = 7.17 GPa

Gzx = 7.17 GPa

2D-Anisotropic Material:

Q0 layer =

GPa Q90 layer =

GPa

7

RESULTS FROM PROGRAM:

FOR ISOTROPIC:

FOR ORTHOTROPIC (A SINGLE PLY):

8

FOR ORTHOTROPIC (0/90)(2D-ANISOTROPIC):

9

RESULTS FROM ANSYS:

FOR ISOTROPIC (stress distribution)

FOR ORTHOTROPIC (SINGLE PLY) (stress distribution):

10

FOR ORTHOTROPIC (0/90) :

Ply arrangement:

Stress distribution on layer 1:

11

Stress distribution on layer 2:

Material properties given to ply1:

12

Material properties given to ply2:

13

Program using matlab:

//////// %//////////// COURSE PROJECT CODE

/////////// %///////////* FINITE AND BOUNDARY ELEMENT METHOD (ENGR 6310)

**********/////////////////////////// %%////////////////////////////////////

%%//////////////////////////////////// e=load('newelements.dat'); %.................loading

elements and their respective nodes from ansys file=load('newcoord.dat');

%..............................loading nodal coordinates from ansys cons=load('constraints.txt');

%..............................loading constraints from ansys ncons=size(cons(:,1)); sigapp=10;

%......................applied stress E=[e(:,1) e(:,7) e(:,8) e(:,9)]; nele=size(e(:,1)); file=[file(:,1) file(:,2) file(:,3)]; nnode=size(file(:,1)); ym=70*10^3; nu=0.33; t=1; q11_0=181.8*10^3;

%//////////////////////////////////// q22_0=10.3*10^3;

%//////////////////////////////////// q12_0=2.90*10^3;

%//////////////////////////////////// q66_0=7.17*10^3;

%//////////////////////////////////// q16_0=0;q26_0=0;

%//////////////////////////////////// q11_90=10.3*10^9; %.........[D] MATRIX for generally

orthotropic (2-D ANISOTROPIC) q22_90=181.8*10^9; q12_90=2.90*10^9;

%//////////////////////////////////// q66_90=7.17*10^9;

%//////////////////////////////////// q16_90=0;q26_90=0; qij0=[q11_0 q12_0 q16_0;q12_0 q22_0 q26_0;q16_0 q26_0 q66_0];

%//////////////////////////////////// qij90=[q11_90 q12_90 q16_90;q12_90 q22_90 q26_90;q16_90 q26_90 q66_90]; aij=t*[qij0+qij90]; bij=(t^2/2)*[qij0-qij90]; dij=(t^3/3)*[qij0+qij90];

k=[aij bij;bij dij]; D=[k(1,1) k(1,2) k(1,6);k(1,2) k(2,2) k(2,6);k(1,6) k(2,6) k(6,6)];

%D=(ym/(1-nu^2))*[1 nu 0; nu 1 0; 0 0 ((1-nu)/2)];.. %////// for ISOTROPIC %D=[q11_0 q12_0 q16_0;q12_0 q22_0 q26_0;q16_0 q26_0q66_0];

..%%////////////for ORTHOTROPIC SINGLE PLY K=zeros(2*nnode,2*nnode);

%...initialiation of K matrix for co=1:nele

14

nodecoord=file(E(co,2),2:3); % Xi=nodecoord; %

////////////////// nodecoord2=file(E(co,3),2:3); %////////// Xj=nodecoord2; %///////////calling

each element respective nodal coordinates nodecoord3=file(E(co,4),2:3); %

///////////////// Xk=nodecoord3; % areaele =abs((Xi(1)*(Xj(2)-Xk(2)) + Xj(1)*(Xk(2)-Xi(2)) + Xk(1)*(Xi(2)-

Xj(2)))/2); bi = Xj(2)-Xk(2);

%%//////////////////////////////////// bj = Xk(2)-Xi(2);

%%//////////////////////////////////// bk = Xi(2)-Xj(2); %

%//////////////////////////////////// ci = Xk(1)-Xj(1); %...elements of B matrix cj = Xi(1)-Xk(1);

%//////////////////////////////////// ck = Xj(1)-Xi(1);

%////////////////////////////////////

B=(([bi 0 bj 0 bk 0 ;0 ci 0 cj 0 ck ; ci bi cj bj ck

bk])/(2*areaele));%........defining B matrix k=t*areaele*B'*D*B;

%......element stiffness matrix i=E(co,2); j=E(co,3); z=E(co,4); ecm=[2*i-1 2*i 2*j-1 2*j 2*z-1 2*z]; for x=1:6 for y=1:6 K(ecm(x),ecm(y))=K(ecm(x),ecm(y))+k(x,y);

%...............commnad to create global stiffness matrix end end end rem=0;

size(K); for i=1:ncons

%//////////////////////////////////// if (cons(i,2)==0)

%////////////////////////////////////// K(2*cons(i,1)-1-rem,:)=[]; % /////////////// K(:,2*cons(i,1)-1-rem)=[]; %

////////////////// end %applying boundary

conditions (constraints) if (cons(i,2)==1) % K(2*cons(i,1)-rem,:)=[]; %

/////////////////// K(:,2*cons(i,1)-rem)=[]; %////////////////// end rem=rem+1;

%///////////////////////////////////// end size(K);

%////////////////////////////////////

15

F=zeros(2*nnode,1);

%//////////////////////////////////// array1=[2,4,12,14,16,18,20]; % ....nodes on a line in which

pressure is applied i=1; for i=1:150

%//////////////////////////////////// i=array1(:);

%//////////////////////////////////// F(i)=35.714; %,,,,,,,,,,,,,,,applying forces by

lumping on nodes end

rem=0; for i=1:ncons if (cons(i,2)==0) F(2*cons(i,1)-1-rem)=[]; end if (cons(i,2)==1) F(2*cons(i,1)-rem)=[]; end rem=rem+1; end dis=K\F; %%,,,,,,,,,,,,,,,finding

displacements for i=1:ncons if (cons(i,2)==0)

%//////////////////////////////////// dis=[dis(1:(2*cons(i,1)-2));0;dis((2*cons(i,1)-

1):end)];%,,,,,,,,,,,,,,, end

%including constarints in displacement matrix to find out stresses if (cons(i,2)==1)

%//////////////////////////////////// dis=[dis(1:(2*cons(i,1)-

1));0;dis((2*cons(i,1)):end)];%//////////////////////////////////// end end stress=zeros(nele,3); for i=1:nele nodecoord=file(E(co,2),2:3); Xi=nodecoord;

%//////////////////////////////////// nodecoord2=file(E(co,3),2:3);

%//////////////////////////////////// Xj=nodecoord2; %......calling

nodal coordinates of each element to find out the stresses nodecoord3=file(E(co,4),2:3); Xk=nodecoord3; areaele = ((Xi(1)*(Xj(2)-Xk(2)) + Xj(1)*(Xk(2)-Xi(2)) + Xk(1)*(Xi(2)-

Xj(2)))/2);

n1=E(i,2);n2=E(i,3);n3=E(i,4); q=[dis(2*n1-1);dis(2*n1);dis(2*n2-1);dis(2*n2);dis(2*n3-1);dis(2*n3)];

bi = Xj(2)-Xk(2);

%//////////////////////////////////// bj = Xk(2)-Xi(2);

%//////////////////////////////////// bk = Xi(2)-Xj(2); %.....elements of B matrix

16

ci = Xk(1)-Xj(1);

%//////////////////////////////////// cj = Xi(1)-Xk(1);

%//////////////////////////////////// ck = Xj(1)-Xi(1);

B=(([bi 0 bj 0 bk 0 ;0 ci 0 cj 0 ck ; ci bi cj bj ck bk])/(2*areaele)); stress(i,:)=D*B*q; end sigy=max(stress(:,2)) %////////////////////////////////////

maximum value of stess in applied pressure direction scf=sigy/sigapp

%////////////////////////////////////stress concentration factor

17

COMPARISION OF RESULTS: Stress concentration

factor

Isotropic Orthotropic Orthotropic(0/90) (2d-anisotropic)

by MATLAB 5.9077 3.6798 4.9947

by ANSYS 5.7151 4.335 4.167

18

References:

http://www.emeraldinsight.com/journals.htm?articleid=878141

Kelly, D.W., Tosh, M.W. (2000), "Interpreting load paths and stress

trajectories in elasticity", Engineering Computations, Vol. 17 No.2,

pp.11735.

http://ijeit.com/Vol%203/Issue%203/IJEIT1412201309_61.pdf

http://www.iosrjournals.org/iosr-jmce/papers/vol2-issue2/D0222832.pdf

http://www.iaa.ncku.edu.tw/~chwu/PDFfile/PDF_Journal/(7)GREEN'S%

20FUNCTIONS%20OF....pdf

http://www.ewp.rpi.edu/hartford/~ernesto/Su2012/EP/MaterialsforStuden

ts/Aiello/Roark-Ch06.pdf