ansys project
DESCRIPTION
eTRANSCRIPT
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ANALYSIS OF 2D-ANISOTROPIC PLATE WITH
ELLIPTICAL HOLE
ENGR 6310 Design project
Submitted to
Dr. Jenny Zhou
By
Nadendla Raviteja L20310564
Dinesh Davuluri L20309927
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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
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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
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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
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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.
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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
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RESULTS FROM PROGRAM:
FOR ISOTROPIC:
FOR ORTHOTROPIC (A SINGLE PLY):
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FOR ORTHOTROPIC (0/90)(2D-ANISOTROPIC):
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RESULTS FROM ANSYS:
FOR ISOTROPIC (stress distribution)
FOR ORTHOTROPIC (SINGLE PLY) (stress distribution):
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FOR ORTHOTROPIC (0/90) :
Ply arrangement:
Stress distribution on layer 1:
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Stress distribution on layer 2:
Material properties given to ply1:
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Material properties given to ply2:
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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
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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);
%////////////////////////////////////
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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
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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
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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
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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