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Penalty solution, TUBE56C1

Mesh: Ten hexahedra--see appendix B.5.

Material properties: $\alpha=10^{-5}$ 1/K, $E=2\times 10^5$ MPa, $\nu=0.3$.

$\dot\epsilon_c=a_1 +a_2\sigma_e+a_3\sigma_e^2+a_4\sigma_e^3~,~~~a_1=a_2=a_3=0~,~~~a_4=2\times 10^{-28}\,\mbox{1/h}$

Support: Plain strain/axisymmetric conditions.

v=0 nodes: 	 all 	 (plain strain condition) 
v=w=0 nodes: 	 1 to 53 	 (z=0 plane) 
un=0 nodes: 	 55 57 59 ...(symmetry plane) 


${\bf n}(\mbox{symmetry plane}) = \{-\sin\alpha~0~\cos\alpha\}, ~\mbox{where}~\alpha=3.1481^\circ$

Loading: The internal pressure p=100 MPa.

Solution: The task lying ahead is to define an axisymmetric problem as viewed in 3D space. To this end we consider a segment created by the rotation of a certain area about the axis of revolution with the symmetry condition imposed on both meridian faces of the segment.

In this particular example the mesh shown in appendix B.5 has been generated by the clockwise rotation $3.1481^\circ$ about y-axis of the 2D mesh plotted in appendix A.3. Therefore the symmetry condition u_n=0, where u_n is the displacement component normal to the symmetry plane, must be enforced on plane z=0 and the one defined by the unit normal ${\bf n}$.

Since the normal direction on the ${\bf n}$-plane does not coincide with any coordinate axis, the symmetry condition takes on a character of a general linear constraint. There are two common ways to handle such constraints in the PMD system: we can either employ a ``periodicity'' approach, setting parameters KPER, NPER and ALPHA in the I1 input file, or use the penalty method. The first method is only available for the solution of heat transfer and linear elastic problems therefore we have to resort to the latter one.

The penalty method can be thought of as a prescription of very stiff springs acting on nodes in the constraint direction. The spring stiffness should be large enough to enforce the constraint with sufficient accuracy but not excessively large so as not to spoil conditioning of the stiffness matrix. Typically we may opt for numbers exceeding by six or seven orders of magnitude the ``average'' stiffness of the body computed as

K = EL

where E is the Young modulus and L a characteristic dimension. Substituting $E=2\times10^{11}\,\mbox{Pa}\,$, $L=1\,\mbox{m}$ and enhancing stiffness by 10⁷ we arrive at the penalty value $10^{18}\,\mbox{N/m}$ which has been used in the I2 input file.

It should be noted that the displacement field computed with the aid of the penalty method agrees up to three digits with the solution obtained in example VI.4.1. This can be viewed as a good result having in mind the problem is highly nonlinear, involving nearly one percent inelastic strain.

Execute from prompt:
>rmd3 tube56c1.I1
>rpd3 tube56c1.I2
>srh3 tube56c1.I3
>fefs tube56c1.I4
>hpp3 tube56c1.IP
>hpls tube56c1.IL
>str3 tube56c1.I5

• INPUT
• OUTPUT