For a finite strain elasticity problem using axisymmetric formulation, from the FEM modelling we have a solution for the displacement [ u_r, u_z ] from which one can compute the 2*2 Green-Lagrange strain tensor E. In addition to the strains in r and z direction how can one account for the strains in circumferential direction (theta_theta)? As an example one can think of an axisymmetric torus thin tube undergoing pure inflation. The strains in r and z direction would be zero but not in the circumferential direction.
Thus, how can one apply a general transformation to map a 2*2 strain tensor (computed as gradient of 2D finite element shape functions): [ e_rr, e_rz; e_zr, e_zz ] to [ e_rr, e_rz, e_r_theta; e_zr, e_zz, e_z_theta; e_theta_r, e_theta_z, e_theta_theta ]?
I found the answer in the book "The Finite Element Method for Solid and Structural Mechanics" by Prof. Olek C Zienkiewicz. For non-torsional deformation in axisymmetric formulation the deformation gradient (3*3 rank 2 tensor) is given by:
[ u_rr, u_rz, 0; u_zr, u_zz, 0; 0, 0, u_r/ R ].
Where u_r, u_z are the displacements in 2D and the gradient is u_rr and so on. 'R' is the radial distance of the node from the axis of symmetry.
The book explains it in more detail.
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