2. Reference solution: plane stress analytics#
2.1. 2D solution#
Using the Cartesian coordinate system \((x,y)\), the displacement of any point \(M(x,y)\) on the upper plate is written as:
\(u(x,y)\mathrm{=}{u}_{x}(x,y)\overrightarrow{{e}_{x}}+{u}_{y}(x,y)\overrightarrow{{e}_{y}}\) eq 2-1
Remarks:
The top plate and the bottom plate are separated because the interface completely separates the plate in two. Since the lower plate is embedded at its base, the result is that it is completely immobile and that the analytical study is carried out only on the upper plate.
We break down the components of displacement in base \(\mathrm{\{}\mathrm{1,}x,y,{x}^{2},{y}^{2},\mathit{xy},{x}^{2}y,{\mathit{xy}}^{2}\mathrm{\}}\):
\({u}_{x}(x,y)\mathrm{=}{I}_{1}+{I}_{2}x+{I}_{3}y+{I}_{4}{x}^{2}+{I}_{5}{y}^{2}+{I}_{6}\mathit{xy}+{I}_{7}{x}^{2}y+{I}_{8}x{y}^{2}\) eq 2-2
\({u}_{y}(x,y)\mathrm{=}{J}_{1}+{J}_{2}x+{J}_{3}y+{J}_{4}{x}^{2}+{J}_{5}{y}^{2}+{J}_{6}\mathit{xy}+{J}_{7}{x}^{2}y+{J}_{8}x{y}^{2}\) eq 2-3
The problem has geometric and loading symmetry with respect to the y-axis. This involves:
\({I}_{1}\mathrm{=}{I}_{3}\mathrm{=}{I}_{4}\mathrm{=}{I}_{5}\mathrm{=}{I}_{7}\mathrm{=}{J}_{2}\mathrm{=}{J}_{6}\mathrm{=}{J}_{8}\mathrm{=}0\) eq 2-4
The local equilibrium equations expressed in the Cartesian coordinate system give:
\({I}_{8}\mathrm{=}{J}_{7}\mathrm{=}0\) eq 2-5
\({I}_{6}\mathrm{=}\mathrm{-}\frac{2(1\mathrm{-}\nu )}{1+\nu }{J}_{4}\mathrm{-}\frac{4}{1+\nu }{J}_{5}\) eq 2-6
By applying the Dirichlet boundary conditions of the upper face \(d\mathrm{=}{d}_{2}{x}^{2}+{d}_{0}\), we deduce:
\({J}_{4}\mathrm{=}{d}_{2}\mathrm{=}\mathrm{-}\mathrm{2,5}{.10}^{\mathrm{-}6}\) eq 2-7
\({J}_{1}+{J}_{3}{L}_{y}+{J}_{5}{L}_{y}^{2}\mathrm{=}{d}_{0}\) eq 2-8
By applying the Neumann limit conditions of the lateral edges \(P\mathrm{=}{p}_{1}y+{p}_{0}\) to the constraints resulting from the generalized Hooke’s law:
\({J}_{5}\mathrm{=}\mathrm{-}\frac{{d}_{2}}{2+\nu }+\frac{{p}_{1}{(1+\nu )}^{2}}{2E(2+\nu )}\mathrm{=}\mathrm{-}\mathrm{0,5}{.10}^{\mathrm{-}6}\) eq 2-9
\({I}_{6}\mathrm{=}\mathrm{-}\frac{{\mathrm{2d}}_{2}(1\mathrm{-}\nu )}{1+\nu }{p}_{0}\mathrm{-}\frac{{\mathrm{4J}}_{5}}{1+\nu }\mathrm{=}\mathrm{-}\frac{5}{3}{.10}^{\mathrm{-}6}\) eq 2-10
\({I}_{2}\mathrm{=}\mathrm{-}\frac{1\mathrm{-}{\nu }^{2}}{E}{p}_{0}\mathrm{-}\nu {J}_{3}\) eq 2-11
The interface is a free edge, i.e. the stress vector at any point on this surface in the normal direction outside the structure is zero:
\({J}_{3}\mathrm{=}\frac{\nu {p}_{0}}{E}\mathrm{=}{2.10}^{\mathrm{-}6}\) eq 2-12
\({I}_{2}\mathrm{=}\frac{\mathrm{-}{p}_{0}}{E}\mathrm{=}\mathrm{-}{1.10}^{\mathrm{-}6}\) eq 2-13
Therefore, by combining the results and expressions obtained, we get \({J}_{1}\):
\({J}_{1}\mathrm{=}{d}_{0}\mathrm{-}{J}_{3}{L}_{y}\mathrm{-}{J}_{5}{L}_{y}^{2}\mathrm{=}\mathrm{8,02}{.10}^{\mathrm{-}6}\) eq 2-14
The solution obtained is as follows:
\({u}_{x}(x,y)\mathrm{=}\mathrm{-}{1.10}^{\mathrm{-}6}(\mathrm{10x}+\frac{5}{3}\mathit{xy})\) eq 2-15
\({u}_{y}(x,y)\mathrm{=}{1.10}^{\mathrm{-}6}(\mathrm{8,02}+\mathrm{2y}\mathrm{-}\mathrm{2,5}{x}^{2}\mathrm{-}\mathrm{0,5}{y}^{2})\) eq 2-16
On the interface, the following result is given:
\({u}_{x}(x,y\mathrm{=}0)\mathrm{=}\mathrm{-}{1.10}^{\mathrm{-}5}x\) eq 2-17
\({u}_{y}(x,y\mathrm{=}0)\mathrm{=}{1.10}^{\mathrm{-}6}(\mathrm{8,02}\mathrm{-}\mathrm{2,5}{x}^{2})\) eq 2-18
2.2. 3D solution#
According to the hypothesis of plane stresses, the 3D stress field does not vary in the \(z\) direction, which implies that the deformations are also independent of it. The problem can then be brought back to the problem in 2D (plane \(\mathit{ABCD}\)) for the resolution of constraints and deformations.
In the 3D case, the solution on \({u}_{x}\) and \({u}_{y}\) therefore has the following form:
\({u}_{x}(x,y,z)\mathrm{=}{1.10}^{\mathrm{-}6}(\mathrm{-}\mathrm{10x}\mathrm{-}\frac{5}{3}\mathit{xy}+{h}_{x}(z))\) eq 2-19
\({u}_{y}(x,y,z)\mathrm{=}{1.10}^{\mathrm{-}6}(\mathrm{8,02}+\mathrm{2y}\mathrm{-}\mathrm{2,5}{x}^{2}+{h}_{y}(z))\) eq 2-20
Additionally, the warp on \({e}_{z}\) is written as:
\({\epsilon }_{\mathit{zz}}(x,y,z)\mathrm{=}\mathrm{-}\frac{\nu }{1\mathrm{-}\nu }(\frac{\mathrm{\partial }{u}_{x}}{\mathrm{\partial }x}+\frac{\mathrm{\partial }{u}_{y}}{\mathrm{\partial }y})\mathrm{=}{1.10}^{\mathrm{-}6}(2+\frac{2}{3}y)\) eq 2-21
With:
\({\epsilon }_{\mathit{zz}}(x,y,z)\mathrm{=}\frac{\mathrm{\partial }{u}_{z}(x,y,z)}{\mathrm{\partial }z}\) eq 2-22
Consequently, by combining the results and expressions obtained, we obtain:
\({u}_{z}(x,y,z)\mathrm{=}{1.10}^{\mathrm{-}6}\mathrm{[}(2+\frac{2}{3}y)z+g(x,y)\mathrm{]}\) eq 2-23
According to \({\epsilon }_{\mathit{xz}}\mathrm{=}0\) and \({\epsilon }_{\mathit{yz}}\mathrm{=}0\), we get:
\({h}_{x}(z)\mathrm{=}{C}_{1}z+{C}_{0}\) eq 2-24
\({h}_{y}(z)\mathrm{=}\mathrm{-}\frac{{z}^{2}}{3}+{C}_{3}z+{C}_{5}\) eq 2-25
\(g(x,y)\mathrm{=}\mathrm{-}{C}_{\mathrm{1x}}\mathrm{-}{C}_{3}y+{C}_{4}\) eq 2-26
Plane \(\mathit{ABCD}\) is stuck in direction \({e}_{z}\), we get:
\({u}_{z}(x,y,z\mathrm{=}0)\mathrm{=}0\) eq 2-27
That means: \({C}_{1}\mathrm{=}{C}_{3}\mathrm{=}{C}_{4}\mathrm{=}0\).
Plane \((x\mathrm{=}0)\) is stuck in direction \({e}_{x}\), we get:
\({u}_{x}(x\mathrm{=}\mathrm{0,}y,z)\mathrm{=}0\) eq 2-28
That means: \({C}_{0}\mathrm{=}0\).
In addition, the displacement imposed on the upper surface leads to: \({C}_{5}\mathrm{=}0\).
Finally, we get:
\({u}_{x}(x,y,z)\mathrm{=}\mathrm{-}(\mathrm{10x}+\frac{5}{3}\mathit{xy}){.10}^{\mathrm{-}6}\) eq 2-29 \({u}_{y}(x,y,z)\mathrm{=}(\mathrm{8,02}+\mathrm{2y}\mathrm{-}\mathrm{2,5}{x}^{2}\mathrm{-}\mathrm{0,5}{y}^{2}\mathrm{-}\frac{1}{3}{z}^{2}){.10}^{\mathrm{-}6}\) eq 2-30
\({u}_{z}(x,y,z)\mathrm{=}(2+\frac{2}{3}y){\mathit{z.10}}^{\mathrm{-}6}\) eq 2-31
On the interface, the following result is given:
\({u}_{x}(x,y\mathrm{=}\mathrm{0,}z)\mathrm{=}\mathrm{-}{1.10}^{\mathrm{-}5}x\) eq 2-32
\({u}_{y}(x,y\mathrm{=}\mathrm{0,}z)\mathrm{=}(\mathrm{8,02}\mathrm{-}\mathrm{2,5}{x}^{2}\mathrm{-}\frac{1}{3}{z}^{2}){.10}^{\mathrm{-}6}\) eq 2-33
\({u}_{z}(x,y\mathrm{=}\mathrm{0,}z)\mathrm{=}{2.10}^{\mathrm{-}6}z\) eq 2-34