Problem 2.9. Laminated composite

Problem a)

Elements of the stiffness matrix of the laminate, Q = A/h, where h =4t the thickness of the laminate

Q₁₁ [GPa]=

Q₁₂ [GPa]=

Q₂₂ [GPa]=

Q₃₃ [GPa]=

Problem b)

Elements of the stiffness matrix

Q₁₁ [GPa]=

Q₁₂ [GPa]=

Q₂₂ [GPa]=

Q₃₃ [GPa]=

Problem a)

Step 1. Determine the stiffness matrix of individual plies.

Check individual stiffness matrixHide individual stiffness matrix

Solution is similar to that of Example 2.5.

Netting theory assumes that all loads are carried by the fibers, and plies carry the load in the fiber direction only. The stiffness matrix of an individual ply in a coordinate system attached to the fiber direction is

${\mathit{Q}}_{\mathbf{p}\mathbf{,}\mathbf{0}}=\left[\begin{array}{ccc}0.5\times E& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]=\left[\begin{array}{ccc}0.5\times 260& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\times {10}^9\text{Pa}$

where 0.5 is the volume fraction of the fibers.

Step 2. Transform the matrices of the individual plies from their local coordinate systems to the global x-y coordinate system.

Check transformationHide transformation matrix

Transformation of the stiffness matrix is given by Eqs.(2-9, 2-30, 2-67)

${\mathit{Q}}_{\mathbf{p}}={\mathit{T}}_{\mathit{\sigma }}^{–1}{\mathit{Q}}_{\mathbf{p}\mathbf{,}\mathbf{0}}{\mathit{T}}_{\mathbf{\epsilon }}={\left[\begin{array}{ccc}{\cos }^2\beta & {\sin }^2\beta & 2\cos \beta \sin \beta \\ {\sin }^2\beta & {\cos }^2\beta & –2\cos \beta \sin \beta \\ –\cos \beta \sin \beta & \cos \beta \sin \beta & {\cos }^2\beta –{\sin }^2\beta \end{array}\right]}^{–1}\left[\begin{array}{ccc}0.5\times E& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{ccc}{\cos }^2\beta & {\sin }^2\beta & \cos \beta \sin \beta \\ {\sin }^2\beta & {\cos }^2\beta & –\cos \beta \sin \beta \\ –2\cos \beta \sin \beta & 2\cos \beta \sin \beta & {\cos }^2\beta –{\sin }^2\beta \end{array}\right]$

Performing the above transformation the coordinate systems of the plies are rotated by 0, 45, 90 and 135 degrees, respectively. The transformations results in

$$\begin{gathered} {\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{45}}={\mathbf{T}}_{\mathbf{\sigma }\mathbf{,}\mathbf{45}}^{–1}{\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{0}}{\mathbf{T}}_{\mathbf{\epsilon }\mathbf{,}\mathbf{45}}={\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& 1\\ \frac{1}{2}& \frac{1}{2}& –1\\ –\frac{1}{2}& \frac{1}{2}& 0\end{array}\right]}^{–1}\left[\begin{array}{ccc}130& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}& –\frac{1}{2}\\ –1& 1& 0\end{array}\right]\times {10}^9\mathrm{Pa}=\left[\begin{array}{ccc}32.5& 32.5& 32.5\\ 32.5& 32.5& 32.5\\ 32.5& 32.5& 32.5\end{array}\right]\times {10}^9\mathrm{Pa} \\ {\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{90}}={\mathbf{T}}_{\mathbf{\sigma }\mathbf{,}\mathbf{90}}^{–1}{\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{0}}{\mathbf{T}}_{\mathbf{\epsilon }\mathbf{,}\mathbf{90}}={\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& –1\end{array}\right]}^{–1}\left[\begin{array}{ccc}130& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& –1\end{array}\right]\times {10}^9\mathrm{Pa}=\left[\begin{array}{ccc}0& 0& 0\\ 0& 130& 0\\ 0& 0& 0\end{array}\right]\times {10}^9\mathrm{Pa} \\ {\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{135}}={\mathbf{T}}_{\mathbf{\sigma }\mathbf{,}\mathbf{135}}^{–1}{\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{0}}{\mathbf{T}}_{\mathbf{\epsilon }\mathbf{,}\mathbf{135}}={\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& –1\\ \frac{1}{2}& \frac{1}{2}& 1\\ \frac{1}{2}& –\frac{1}{2}& 0\end{array}\right]}^{–1}\left[\begin{array}{ccc}130& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& –\frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}& \frac{1}{2}\\ 1& –1& 0\end{array}\right]\times {10}^9\mathrm{Pa}=\left[\begin{array}{ccc}32.5& 32.5& –32.5\\ 32.5& 32.5& –32.5\\ –32.5& –32.5& 32.5\end{array}\right]\times {10}^9\mathrm{Pa} \end{gathered}$$

Step 3. Determine the stiffness matrix of the laminate.

Check stiffness matrixHide stiffness matrix

Step 4. Check isotropy.

Check isotropyHide checking

The engineering constants are determined from the stiffness matrix:

$$\begin{gathered} {\nu }_{xy}={\nu }_{yx}=\nu =\frac{Q_{12}}{Q_{11}}=\frac{16.25}{48.75}=0.333 \\ {E}_x=E_y=E=Q_{11}(1–{\nu }^2)=48.75(1–0.{333}^2)=43.333\times {10}^9\mathrm{Pa} \\ G={\mathrm{Q}}_{33}=16.25\times {10}^9\mathrm{Pa} \end{gathered}$$

The material is isotropic when the following relation between the engineering constants is true

Eq.(2-62)

$G=\frac{E}{2(1+\nu )}$

Here

$G=16.25\times {10}^9\mathrm{Pa}=\frac{E}{2\left(1+\nu \right)}=\frac{43.333\times {10}^9\mathrm{Pa}}{2(1+0.333)}=16.25\times {10}^9\mathrm{Pa}$

thus the material is isotropic.

Problem b)

Solution is similar to that of Example 2.7.

Step 1. Determine the stiffness matrix of individual plies.

Check individual stiffness matrixHide individual stiffness matrix

The compliance matrix of an individual ply is determined by the engineering constants given in the initial data, the stiffness matrix is the inverse of the compliance matrix:

Eq.(2-63)

${\mathit{Q}}_{\mathbf{p}\mathbf{,}\mathbf{0}}={\mathit{C}}_{\mathit{p}\mathbf{,}\mathbf{0}}^{–1}={\left[\begin{array}{ccc}\frac{1}{E_1}& –\frac{{\nu }_{21}}{E_2}& 0\\ –\frac{{\nu }_{12}}{E1}& \frac{1}{E_2}& 0\\ 0& 0& \frac{1}{G}\end{array}\right]}^{–1}={\left[\begin{array}{ccc}\frac{1}{148.4}& –\frac{0.3}{148.4}& 0\\ –\frac{0.3}{148.4}& \frac{1}{8.91}& 0\\ 0& 0& \frac{1}{4.5}\end{array}\right]}^{–1}\times {10}^9\text{Pa=}\left[\begin{array}{ccc}149.2& 2.69& 0\\ 2.69& 8.96& 0\\ 0& 0& 4.5\end{array}\right]\times {10}^9\mathrm{Pa}$

Step 2. Transform the matrices of the individual plies from their local coordinate systems to the global x-y coordinate system.

Check transformationHide transformation

The same transformations are performed for the individual layers as in Problem a)

Transformation of the stiffness matrix is given by Eqs.(2-9, 2-30, 2-67)

${\mathit{Q}}_{\mathbf{p}\mathbf{,}\mathbf{45}}={\mathit{T}}_{\mathit{\sigma }\mathbf{,}\mathbf{45}}^{–1}{\mathit{Q}}_{\mathbf{p}\mathbf{,}\mathbf{0}}{\mathit{T}}_{\mathbf{\epsilon }\mathbf{,}\mathbf{45}}={\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& 1\\ \frac{1}{2}& \frac{1}{2}& –1\\ –\frac{1}{2}& \frac{1}{2}& 0\end{array}\right]}^{–1}\left[\begin{array}{ccc}149.2& 2.69& 0\\ 2.69& 8.96& 0\\ 0& 0& 4.5\end{array}\right]\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}& –\frac{1}{2}\\ –1& 1& 0\end{array}\right]\times {10}^9\mathrm{Pa}=\left[\begin{array}{ccc}45.385& 36.385& 35.062\\ 36.385& 45.385& 35.062\\ 35.062& 35.062& 38.197\end{array}\right]\times {10}^9\mathrm{Pa}$${\mathit{Q}}_{\mathbf{p}\mathbf{,}\mathbf{90}}={\mathit{T}}_{\mathit{\sigma }\mathbf{,}\mathbf{90}}^{–1}{\mathit{Q}}_{\mathbf{p}\mathbf{,}\mathbf{0}}{\mathit{T}}_{\mathbf{\epsilon }\mathbf{,}\mathbf{90}}={\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& –1\end{array}\right]}^{–1}\left[\begin{array}{ccc}149.2& 2.69& 0\\ 2.69& 8.96& 0\\ 0& 0& 4.5\end{array}\right]\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& –1\end{array}\right]\times {10}^9\mathrm{Pa}=\left[\begin{array}{ccc}8.96& 2.69& 0\\ 2.69& 149.2& 0\\ 0& 0& 4.5\end{array}\right]\times {10}^9\mathrm{Pa}$

${\mathit{Q}}_{\mathbf{p}\mathbf{,}\mathbf{135}}={\mathit{T}}_{\mathit{\sigma }\mathbf{,}\mathbf{135}}^{–1}{\mathit{Q}}_{\mathbf{p}\mathbf{,}\mathbf{0}}{\mathit{T}}_{\mathbf{\epsilon }\mathbf{,}\mathbf{135}}={\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& –1\\ \frac{1}{2}& \frac{1}{2}& 1\\ \frac{1}{2}& –\frac{1}{2}& 0\end{array}\right]}^{–1}\left[\begin{array}{ccc}149.206& 2.69& 0\\ 2.69& 8.96& 0\\ 0& 0& 4.5\end{array}\right]\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& –\frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}& \frac{1}{2}\\ 1& –1& 0\end{array}\right]\times {10}^9\mathrm{Pa}=\left[\begin{array}{ccc}45.385& 36.385& –35.062\\ 36.385& 45.385& –35.062\\ –35.062& –35.062& 38.197\end{array}\right]\times {10}^9\mathrm{Pa}$

 

Step 3. Determine the stiffness matrix of the laminate.

Check stiffness matrixHide stiffness matrix

The stiffness matrix of the laminate is

$\mathbf{A}=t{\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{0}}+t{\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{45}}+t{\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{90}}+t{\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{135}}$

The replacement stiffness matrix of the laminate’s material is 

$$\begin{gathered} \mathbf{Q}=\frac{\mathbf{A}}{4t}=\frac{0.1}{4\times 0.1}\times \left(\left[\begin{array}{ccc}149.2& 2.69& 0\\ 2.69& 8.96& 0\\ 0& 0& 4.5\end{array}\right]+\left[\begin{array}{ccc}45.385& 36.385& 35.062\\ 36.385& 45.385& 35.062\\ 35.062& 35.062& 38.197\end{array}\right]+\left[\begin{array}{ccc}8.96& 2.69& 0\\ 2.69& 149.2& 0\\ 0& 0& 4.5\end{array}\right]+\left[\begin{array}{ccc}45.385& 36.385& –35.062\\ 36.385& 45.385& –35.062\\ –35.062& –35.062& 38.197\end{array}\right]\right)\times {10}^9\mathrm{Pa} \\ \mathbf{A}\mathbf{=}\left[\begin{array}{ccc}\mathbf{62}\mathbf{.}\mathbf{23}& \mathbf{19}\mathbf{.}\mathbf{55}& \mathbf{0}\\ \mathbf{19}\mathbf{.}\mathbf{55}& \mathbf{62}\mathbf{.}\mathbf{23}& \mathbf{0}\\ \mathbf{0}& \mathbf{0}& \mathbf{21}\mathbf{.}\mathbf{35}\end{array}\right]\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{Pa} \end{gathered}$$

Step 4. Check isotropy.

Check isotropyHide checking

The engineering constants from the stiffness matrix are

$$\begin{gathered} {\nu }_{xy}={\nu }_{yx}=\nu =\frac{Q_{12}}{Q_{11}}=\frac{19.536}{62.24}=0.314 \\ {E}_{xy}=E_{yx}=E=Q_{11}(1–{\nu }^2)=62.24\times {10}^8(1–0.3142)=56.101\times {10}^9\mathrm{Pa} \\ \mathrm{G}={\mathrm{Q}}_{33}=21.349\times {10}^9\mathrm{Pa} \end{gathered}$$

The material is isotropic when the following relation between the engineering constants is true

Eq.(2-62)

$G=\frac{E}{2(1+\nu )}$

Here

$G=21.349\times {10}^9\mathrm{Pa}=\frac{E}{2\left(1+\nu \right)}=\frac{56.101\times {10}^9\mathrm{Pa}}{2(1+0.314)}=21.349\times {10}^9\mathrm{Pa}$

thus the material is isotropic.

Problem a)

Netting theory assumes that all loads are carried by the fibers, and plies carry the load in the fiber direction only. 

Solution is similar to that of Example 2.5.

The stiffness matrix of an individual ply in a coordinate system attached to the fiber direction is

${\mathit{Q}}_{\mathbf{p}\mathbf{,}\mathbf{0}}=\left[\begin{array}{ccc}0.5\times E& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]=\left[\begin{array}{ccc}0.5\times 260& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\times {10}^9\text{Pa}$

where 0.5 is the volume fraction of the fibers.

Stiffness matrices of the individual plies must be transformed from their local coordinate systems to the global x-y coordinate system.

Transformation of the stiffness matrix is given by Eqs.(2-9, 2-30, 2-67)

Performing the above transformation the coordinate systems of the plies are rotated by 0, 45, 90 and 135 degrees, respectively. The transformations results in 

$$\begin{gathered} {\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{45}}={\mathbf{T}}_{\mathbf{\sigma }\mathbf{,}\mathbf{45}}^{–1}{\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{0}}{\mathbf{T}}_{\mathbf{\epsilon }\mathbf{,}\mathbf{45}}={\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& 1\\ \frac{1}{2}& \frac{1}{2}& –1\\ –\frac{1}{2}& \frac{1}{2}& 0\end{array}\right]}^{–1}\left[\begin{array}{ccc}130& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}& –\frac{1}{2}\\ –1& 1& 0\end{array}\right]\times {10}^9\mathrm{Pa}=\left[\begin{array}{ccc}32.5& 32.5& 32.5\\ 32.5& 32.5& 32.5\\ 32.5& 32.5& 32.5\end{array}\right]\times {10}^9\mathrm{Pa} \\ {\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{90}}={\mathbf{T}}_{\mathbf{\sigma }\mathbf{,}\mathbf{90}}^{–1}{\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{0}}{\mathbf{T}}_{\mathbf{\epsilon }\mathbf{,}\mathbf{90}}={\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& –1\end{array}\right]}^{–1}\left[\begin{array}{ccc}130& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& –1\end{array}\right]\times {10}^9\mathrm{Pa}=\left[\begin{array}{ccc}0& 0& 0\\ 0& 130& 0\\ 0& 0& 0\end{array}\right]\times {10}^9\mathrm{Pa} \\ {\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{135}}={\mathbf{T}}_{\mathbf{\sigma }\mathbf{,}\mathbf{135}}^{–1}{\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{0}}{\mathbf{T}}_{\mathbf{\epsilon }\mathbf{,}\mathbf{135}}={\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& –1\\ \frac{1}{2}& \frac{1}{2}& 1\\ \frac{1}{2}& –\frac{1}{2}& 0\end{array}\right]}^{–1}\left[\begin{array}{ccc}130& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& –\frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}& \frac{1}{2}\\ 1& –1& 0\end{array}\right]\times {10}^9\mathrm{Pa}=\left[\begin{array}{ccc}32.5& 32.5& –32.5\\ 32.5& 32.5& –32.5\\ –32.5& –32.5& 32.5\end{array}\right]\times {10}^9\mathrm{Pa} \end{gathered}$$The stiffness matrix of the laminate is

$\mathbf{A}=t{\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{0}}+t{\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{45}}+t{\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{90}}+t{\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{135}}$

The replacement stiffness matrix of the laminate’s material is 

$$\begin{gathered} \mathbf{Q}=\frac{\mathbf{A}}{4t}=\frac{0.1}{4\times 0.1}\times \left(\left[\begin{array}{ccc}130& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]+\left[\begin{array}{ccc}32.5& 32.5& 32.5\\ 32.5& 32.5& 32.5\\ 32.5& 32.5& 32.5\end{array}\right]+\left[\begin{array}{ccc}0& 0& 0\\ 0& 130& 0\\ 0& 0& 0\end{array}\right]+\left[\begin{array}{ccc}32.5& 32.5& –32.5\\ 32.5& 32.5& –32.5\\ –32.5& –32.5& 32.5\end{array}\right]\right)\times {10}^9\mathrm{Pa} \\ \mathbf{Q}\mathbf{=}\left[\begin{array}{ccc}\mathbf{48}\mathbf{.}\mathbf{75}& \mathbf{16}\mathbf{.}\mathbf{25}& \mathbf{0}\\ \mathbf{16}\mathbf{.}\mathbf{25}& \mathbf{48}\mathbf{.}\mathbf{75}& \mathbf{0}\\ \mathbf{0}& \mathbf{0}& \mathbf{16}\mathbf{.}\mathbf{25}\end{array}\right]\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{Pa} \end{gathered}$$

To check isotropy first the engineering constants are determined from the stiffness matrix:

$$\begin{gathered} {\nu }_{xy}={\nu }_{yx}=\nu =\frac{Q_{12}}{Q_{11}}=\frac{16.25}{48.75}=0.333 \\ {E}_x=E_y=E=Q_{11}(1–{\nu }^2)=48.75(1–0.{333}^2)=43.333\times {10}^9\mathrm{Pa} \\ G={\mathrm{Q}}_{33}=16.25\times {10}^9\mathrm{Pa} \end{gathered}$$

The material is isotropic when the following relation between the engineering constants is true

Eq.(2-62)

$G=\frac{E}{2(1+\nu )}$

Here

$G=16.25\times {10}^9\mathrm{Pa}=\frac{E}{2\left(1+\nu \right)}=\frac{43.333\times {10}^9\mathrm{Pa}}{2(1+0.333)}=16.25\times {10}^9\mathrm{Pa}$

thus the material is isotropic.

Problem b)

Solution is similar to that of Example 2.7.

First the stiffness matrices of individual plies are calculated. The compliance matrix of an individual ply is determined by the engineering constants given in the initial data. The stiffness matrix is the inverse of the compliance matrix:

Eq.(2-63)

Now the stiffness matrices of the individual plies are transformed from their local coordinate systems to the global x-y coordinate system.

The same transformations are performed for the individual layers as in Problem a)

Transformation of the stiffness matrix is given by Eqs.(2-9, 2-30, 2-67)

${\mathit{Q}}_{\mathbf{p}\mathbf{,}\mathbf{45}}={\mathit{T}}_{\mathit{\sigma }\mathbf{,}\mathbf{45}}^{–1}{\mathit{Q}}_{\mathbf{p}\mathbf{,}\mathbf{0}}{\mathit{T}}_{\mathbf{\epsilon }\mathbf{,}\mathbf{45}}={\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& 1\\ \frac{1}{2}& \frac{1}{2}& –1\\ –\frac{1}{2}& \frac{1}{2}& 0\end{array}\right]}^{–1}\left[\begin{array}{ccc}149.2& 2.69& 0\\ 2.69& 8.96& 0\\ 0& 0& 4.5\end{array}\right]\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}& –\frac{1}{2}\\ –1& 1& 0\end{array}\right]\times {10}^9\mathrm{Pa}=\left[\begin{array}{ccc}45.385& 36.385& 35.062\\ 36.385& 45.385& 35.062\\ 35.062& 35.062& 38.197\end{array}\right]\times {10}^9\mathrm{Pa}$${\mathit{Q}}_{\mathbf{p}\mathbf{,}\mathbf{90}}={\mathit{T}}_{\mathit{\sigma }\mathbf{,}\mathbf{90}}^{–1}{\mathit{Q}}_{\mathbf{p}\mathbf{,}\mathbf{0}}{\mathit{T}}_{\mathbf{\epsilon }\mathbf{,}\mathbf{90}}={\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& –1\end{array}\right]}^{–1}\left[\begin{array}{ccc}149.2& 2.69& 0\\ 2.69& 8.96& 0\\ 0& 0& 4.5\end{array}\right]\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& –1\end{array}\right]\times {10}^9\mathrm{Pa}=\left[\begin{array}{ccc}8.96& 2.69& 0\\ 2.69& 149.2& 0\\ 0& 0& 4.5\end{array}\right]\times {10}^9\mathrm{Pa}$

${\mathit{Q}}_{\mathbf{p}\mathbf{,}\mathbf{135}}={\mathit{T}}_{\mathit{\sigma }\mathbf{,}\mathbf{135}}^{–1}{\mathit{Q}}_{\mathbf{p}\mathbf{,}\mathbf{0}}{\mathit{T}}_{\mathbf{\epsilon }\mathbf{,}\mathbf{135}}={\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& –1\\ \frac{1}{2}& \frac{1}{2}& 1\\ \frac{1}{2}& –\frac{1}{2}& 0\end{array}\right]}^{–1}\left[\begin{array}{ccc}149.206& 2.69& 0\\ 2.69& 8.96& 0\\ 0& 0& 4.5\end{array}\right]\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& –\frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}& \frac{1}{2}\\ 1& –1& 0\end{array}\right]\times {10}^9\mathrm{Pa}=\left[\begin{array}{ccc}45.385& 36.385& –35.062\\ 36.385& 45.385& –35.062\\ –35.062& –35.062& 38.197\end{array}\right]\times {10}^9\mathrm{Pa}$

The stiffness matrix of the laminate is

$\mathbf{A}=t{\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{0}}+t{\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{45}}+t{\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{90}}+t{\mathbf{Q}}_{\mathbf{p}\mathbf{,}\mathbf{135}}$

The replacement stiffness matrix of the laminate’s material is

$$\begin{gathered} \mathbf{Q}=\frac{\mathbf{A}}{4t}=\frac{0.1}{4\times 0.1}\times \left(\left[\begin{array}{ccc}149.2& 2.69& 0\\ 2.69& 8.96& 0\\ 0& 0& 4.5\end{array}\right]+\left[\begin{array}{ccc}45.385& 36.385& 35.062\\ 36.385& 45.385& 35.062\\ 35.062& 35.062& 38.197\end{array}\right]+\left[\begin{array}{ccc}8.96& 2.69& 0\\ 2.69& 149.2& 0\\ 0& 0& 4.5\end{array}\right]+\left[\begin{array}{ccc}45.385& 36.385& –35.062\\ 36.385& 45.385& –35.062\\ –35.062& –35.062& 38.197\end{array}\right]\right)\times {10}^9\mathrm{Pa} \\ \mathbf{A}\mathbf{=}\left[\begin{array}{ccc}\mathbf{62}\mathbf{.}\mathbf{23}& \mathbf{19}\mathbf{.}\mathbf{55}& \mathbf{0}\\ \mathbf{19}\mathbf{.}\mathbf{55}& \mathbf{62}\mathbf{.}\mathbf{23}& \mathbf{0}\\ \mathbf{0}& \mathbf{0}& \mathbf{21}\mathbf{.}\mathbf{35}\end{array}\right]\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}{\mathbf{Pa}}_{} \end{gathered}$$

To check isotropy the engineering constants are determined from the stiffness matrix

$$\begin{gathered} {\nu }_{xy}={\nu }_{yx}=\nu =\frac{Q_{12}}{Q_{11}}=\frac{19.536}{62.24}=0.314 \\ {E}_{xy}=E_{yx}=E=Q_{11}(1–{\nu }^2)=62.24\times {10}^8(1–0.3142)=56.10\times {10}^9\mathrm{Pa} \\ \mathrm{G}={\mathrm{Q}}_{33}=21.35\times {10}^9\mathrm{Pa} \end{gathered}$$

The material is isotropic when the following relation between the engineering constants is true

Equation (2-62)

$G=\frac{E}{2(1+\nu )}$

Here

$G=21.35\times {10}^9\mathrm{Pa}=\frac{E}{2\left(1+\nu \right)}=\frac{56.10\times {10}^9\mathrm{Pa}}{2(1+0.314)}=21.35\times {10}^9\mathrm{Pa}$

thus the material is isotropic.