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ex_planestrain1.m File Reference

Description

EX_PLANESTRAIN1 Plane strain analysis of a pressure vessel.

[ FEA, OUT ] = EX_PLANESTRAIN1( VARARGIN ) Benchmark example for plain strain approximation of a pressure vessel (annular cross section with symmetry).

Reference. B. J. Mac Donald, Practical Stress Analysis with Finite Elements (2nd Ed), case study E on page 327, 2007.

Accepts the following property/value pairs.

Input       Value/{Default}        Description
-----------------------------------------------------------------------------------
E           scalar {207e9}         Modulus of elasticity
nu          scalar {0.27}          Poissons ratio
sfun        string {sflag1}        Shape function for displacements
iplot       scalar 0/{1}           Plot solution (=1)
                                                                                  .
Output      Value/(Size)           Description
-----------------------------------------------------------------------------------
fea         struct                 Problem definition struct
out         struct                 Output struct

Code listing

 cOptDef = { ...
   'E',        207e9; ...
   'nu',       0.27; ...
   'sfun',     'sflag1'; ...
   'iplot',    1; ...
   'igrid',    1; ...
   'tol',      0.1; ...
   'fid',      1 };
 [got,opt] = parseopt( cOptDef, varargin{:} );
 fid       = opt.fid;


% Geometry and grid.
 fea.sdim = { 'x' 'y' };   % Coordinate names.
 fea.grid = ringgrid( 12, 216, 100e-3, 120e-3 );
 fea.grid = delcells( fea.grid, selcells( fea.grid, '(x<=eps) + (y<=eps)') );
 if( opt.igrid~=1 )
   fea.grid = quad2tri( fea.grid );
 end
 n_bdr = max(fea.grid.b(3,:));   % Number of boundaries.


% Problem definition.
 fea = addphys( fea, @planestrain );
 fea.phys.psn.eqn.coef{1,end} = { opt.nu };
 fea.phys.psn.eqn.coef{2,end} = { opt.E  };
 fea.phys.psn.sfun            = { opt.sfun opt.sfun };


% Boundary conditions.
 bctype = mat2cell( zeros(2,n_bdr), [1 1], ones(1,n_bdr) );
 bctype{1,4} = 1;
 bctype{2,3} = 1;
 fea.phys.psn.bdr.coef{1,5} = bctype;

 bccoef = mat2cell( zeros(2,n_bdr), [1 1], ones(1,n_bdr) );
 bccoef{1,1} = '-nx*1e4';
 bccoef{2,1} = '-ny*1e4';
 fea.phys.psn.bdr.coef{1,end} = bccoef;


% Parse and solve problem.
 fea       = parsephys( fea );
 fea       = parseprob( fea );                          % Check and parse problem struct.
 fea.sol.u = solvestat( fea, 'fid', fid, 'icub', 1+str2num(strrep(opt.sfun,'sflag','')) );   % Call to stationary solver.


% Postprocessing.
 s_disp = fea.phys.psn.eqn.vars{2,end};
 if( opt.iplot>0 )
   figure
   postplot( fea, 'surfexpr', s_disp )
   title( 'Total displacement' )
 end


% Error checking.
 s_sx     = fea.phys.psn.eqn.vars{5,end};
 s_sy     = fea.phys.psn.eqn.vars{6,end};
 s_sxy    = fea.phys.psn.eqn.vars{8,end};
 s_sp1    = fea.phys.psn.eqn.vars{9,end};
 s_sp2    = fea.phys.psn.eqn.vars{10,end};
 s_sp3    = fea.phys.psn.eqn.vars{11,end};
 v_disp   = evalexpr( s_disp, [100e-3 120e-3-2*sqrt(eps);0 0]+sqrt(eps), fea )';
 v_dref   = [2.64e-8 2.41e-8];
 [v_sx(1),v_sx(2)] = minmaxsubd( s_sx, fea );
 v_sxref  = [-10000 55454];
 [v_sy(1),v_sy(2)] = minmaxsubd( s_sy, fea );
 v_syref  = [-10000 55454];
 [v_sxy(1),v_sxy(2)] = minmaxsubd( s_sxy, fea );
 v_sxyref = [-32730 0];
 [v_sp1(1),v_sp1(2)] = minmaxsubd( s_sp1, fea );
 v_sp1ref = [4.5e4 55454];
 [v_sp2(1),v_sp2(2)] = minmaxsubd( s_sp2, fea );
 v_sp2ref = [1.227e4 1.227e4];
 [v_sp3(1),v_sp3(2)] = minmaxsubd( s_sp3, fea );
 v_sp3ref = [-1e4 0];
 out.err  = [ abs([v_dref-v_disp])./v_dref ;
              abs([v_sxref-v_sx])./v_sxref ;
              abs([v_syref-v_sy])./v_syref ;
              abs([v_sxyref(1)-v_sxy(1)])./v_sxyref(1) 0 ;
              abs([v_sp1ref(2)-v_sp1(2)])./v_sp1ref(2) 0 ;
              abs([v_sp2ref-v_sp2])./v_sp2ref      ;
              abs([v_sp3ref(1)-v_sp3(1)])./v_sp3ref(1) 0 ];
 out.pass = all( out.err(:) <= opt.tol );


 if( nargout==0 )
   clear fea out
 end