2. Benchmark solution#
2.1. Calculation method#
In both modes of stress, the membrane is in a state of uniform stress. It is therefore possible to calculate analytically the solution of this problem in both cases.
Traction solicitation
In the case of a simple tensile load, it is simply shown that, in the global frame of reference:
\(\mathrm{\{}\begin{array}{c}{\varepsilon }_{\mathit{XX}}\mathrm{=}{F}_{X}\frac{{M}_{\mathit{LLLL}}}{({M}_{\mathit{TTTT}}{M}_{\mathit{LLLL}}\mathrm{-}{M}_{\mathit{LLTT}}^{2})}\\ {\varepsilon }_{\mathit{YY}}\mathrm{=}\mathrm{-}{F}_{X}\frac{{M}_{\mathit{LLTT}}}{({M}_{\mathit{TTTT}}{M}_{\mathit{LLLL}}\mathrm{-}{M}_{\mathit{LLTT}}^{2})}\\ {\varepsilon }_{\mathit{XY}}\mathrm{=}0\end{array}\)
Shear load
In the case of shear stress, it is shown that, in the global frame of reference:
\(\mathrm{\{}\begin{array}{c}{\varepsilon }_{\mathit{XX}}\mathrm{=}0\\ {\varepsilon }_{\mathit{YY}}\mathrm{=}0\\ {\varepsilon }_{\mathit{XY}}\mathrm{=}\frac{{\sigma }_{\mathit{XY}}}{{M}_{\mathit{LTLT}}}\end{array}\)
2.2. Reference quantities and results#
Displacement, stress, and deformation are tested at summit \(\mathit{POINT}\). The quantities tested are summarized in the table below, for the two modes of stress.
Size |
Component |
Simple Traction |
Shear |
Tolerance |
|
Displacement |
DX |
3/8 |
3/8 |
1/2 |
1.E-6 |
DY |
-1/8 |
1/2 |
1.E-6 |
||
Membrane deformations (local coordinate system) |
EXX |
-1/8 |
0 |
1.E-6 |
|
EYY |
3/8 |
0 |
1.E-6 |
||
EXY |
0 |
|
1.E-6 |
||
Membrane constraints (local coordinate system) |
NXX |
0 |
0 |
1.E-6 |
|
NYY |
1 |
0 |
1.E-6 |
||
NXY |
0 |
|
1.E-6 |