Problem 2.10. Lattice grid

Elements of the stiffness matrix, A of the replacement plate in x-y coordinate system

A₁₁ [kN/mm]=

A₁₂ [kN/mm]=

A₂₂ [kN/mm²]=

A₃₃ [kN/mm²]=

Step 1. Replace the truss with unidirectional layers, where each  layer contains the parallel bars in one direction. Determine stiffness matrix of each individual layer in its local coordinate system attached to the direction of the bars.

See the solution of Example 2.6.

Step 2. Transform the local stiffness matrices into the global x-y coordinate system.

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 Coordinate system of the layers must be rotated by 0, 45, 90, 135 degrees, respectively.

Transformation is given by Eq.(2-67)

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Step 3. Add the individual stiffness matrices to calculate the stiffness matrix of the replacement plate.

Check stiffness matrixHide stiffness matrix

Step 4. Check isotropy.

Check isotropyHide checking

The material is isotropic when the following ratio is hold between the elements of the stiffness matrix:

Eq.(2-62)

$$\begin{gathered} A_{11}:A_{33}=\frac{E}{(1–\nu ^2)}:\frac{E}{2(1+\nu )} \\ \mathrm{where} \\ \nu =\frac{A_{12}}{A_{11}}=0.414 \\ \mathrm{and} \\ {A}_{11}:A_{33}=\frac{61.66}{25.54}=2.414 \\ \frac{E}{(1–\nu ^2)}:\frac{E}{2(1+\nu )}=\frac{2(1+\nu )}{(1–\nu ^2)}=\frac{2}{(1–\nu )}=\frac{2}{(1–0.414)}=3.41 \end{gathered}$$

Thus the plate is not isotropic.

The truss is replaced by unidirectional layers, where each  layer contains the parallel bars in one direction. First stiffness matrix of each individual layer is determined in its local coordinate system attached to the direction of the bars.

See the solution of Example 2.6.

Layers are assumed to be orthotropic, their stiffness is EA/t in the direction of the bars, and zero perpendicular to them. t is the perpendicular distance of the bars, t=a or t=a/√2 for the different layers as it is shown in the Figure.

With this assumption the replacement plate consists of four layers, and the solution is similar to the previous problem. Stiffness matrices of the individual layers are

$$\begin{gathered} {\mathit{A}}_{\mathbf{l}\mathbf{,}\mathbf{1}}={\mathit{A}}_{\mathbf{l}\mathbf{,}\mathbf{3}}=\left[\begin{array}{ccc}\frac{210\times {10}^3\times 344}{2000}& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\frac{\mathrm{N}}{\mathrm{mm}}=\left[\begin{array}{ccc}36.12& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\frac{\mathrm{kN}}{\mathrm{mm}} \\ {\mathit{A}}_{\mathbf{l}\mathbf{,}\mathbf{2}}={\mathit{A}}_{\mathbf{l}\mathbf{,}\mathbf{4}}=\left[\begin{array}{ccc}\frac{210\times {10}^3\times 344}{2000/\sqrt{2}}& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\frac{N}{\mathrm{mm}}=\left[\begin{array}{ccc}51.08& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\frac{\mathrm{kN}}{\mathrm{mm}} \end{gathered}$$

Coordinate system of the layers must be rotated by 0, 45, 90, 135 degrees, respectively. Transformation of the local stiffness matrices into the global x-y coordinate system results in

Eq.(2-67)

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The stiffness matrix of the replacement plate is the sum of the individual stiffness matrices of the layers.

$$\begin{gathered} \mathbf{A}={\mathbf{A}}_{\mathbf{1}\mathbf{,}\mathbf{0}}+{\mathbf{A}}_{\mathbf{2}\mathbf{,}\mathbf{45}}+{\mathbf{A}}_{\mathbf{3}\mathbf{,}\mathbf{90}}+{\mathbf{A}}_{\mathbf{4}\mathbf{,}\mathbf{135}}=\left[\begin{array}{ccc}36.12& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]+\left[\begin{array}{ccc}12.77& 12.77& 12.77\\ 12.77& 12.77& 12.77\\ 12.77& 12.77& 12.77\end{array}\right]+\left[\begin{array}{ccc}0& 0& 0\\ 0& 36.12& 0\\ 0& 0& 0\end{array}\right]+\left[\begin{array}{ccc}12.77& 12.77& –12.77\\ 12.77& 12.77& –12.77\\ –12.77& –12.77& 12.77\end{array}\right] \\ \mathit{A}\mathbf{=}\left[\begin{array}{ccc}\mathbf{61}\mathbf{.}\mathbf{66}& \mathbf{25}\mathbf{.}\mathbf{54}& \mathbf{0}\\ \mathbf{25}\mathbf{.}\mathbf{54}& \mathbf{61}\mathbf{.}\mathbf{66}& \mathbf{0}\\ \mathbf{0}& \mathbf{0}& \mathbf{25}\mathbf{.}\mathbf{54}\end{array}\right]\frac{\mathbf{kN}}{\mathbf{mm}} \end{gathered}$$

The material is isotropic when the following ratio is hold between the elements of the stiffness matrix:

Eq.(2-62)

$$\begin{gathered} A_{11}:A_{33}=\frac{E}{(1–\nu ^2)}:\frac{E}{2(1+\nu )} \\ \mathrm{where} \\ \nu =\frac{A_{12}}{A_{11}}=0.414 \\ \mathrm{and} \\ {A}_{11}:A_{33}=\frac{61.66}{25.54}=2.414 \\ \frac{E}{(1–\nu ^2)}:\frac{E}{2(1+\nu )}=\frac{2(1+\nu )}{(1–\nu ^2)}=\frac{2}{(1–\nu )}=\frac{2}{(1–0.414)}=3.41 \end{gathered}$$

Thus the plate is not isotropic.