Numerical and Analytical Study of Fatigue and Degradation in Multilayer Composite Plates

The model discussed in this paper is based on the proposed model of Ladeveze for multilayers [14]. This model has been used to predict the degradation response of composites under different conditions. This theory is called the Mesoscale meteorology composite degradation theory because it is based on the assumption of uniform degradation in the thickness of each layer of the composite. Mesoscale meteorology for micro scale analysis at fiber level and is the matrix (and the scale of each layer) [15, 16]. This theory is based on the average value of stress in each layer and the state of degradation at it is a layer. In the Mesoscale meteorology composite degradation theory, it has been hypothesized that the degradation response of each layer at any time point in time (load steps) can be expressed in terms of the modulus loss and inelastic strain due to the degradation or plasticity matrix. Shown in terms of the destruction parameter Which are a function of the thermodynamic force applied to the degradation evolution parameters [17]. The theory contains a condition for the coupling degradation evolution for the multi-axial loading mode that typically occurs in the multilayers and also the difference between the degradation evolution was in the normal tensile and compressive loading conditions [18]. The general form of the law of evolution of destruction varies according to the type of material, which reflects the microstructural dependence of the degradation mechanism [19]. The size of the fibers, the micro-structure and the strength, the matrix strength. The mechanism of degradation at the micro level has not been clearly elucidated. In the Mesoscale meteorology degradation model, degradation evolution is based on experimental observations and on the response of the degraded layers. This is a phenomenological theory. This theory does not allow the law of evolution of destruction during loading to allow new destruction mechanisms to occur [20].  

2.1 Effective tension

The geometry and default dimensions of the problem are shown in Figure 1. The starting point of applying a continuous mechanical model to a multilayer composite is to examine a multilayer layer in plane stress state and determine the effective stress.  

$\left\{\overline{\sigma\}}=\left\{\frac{\left\langle\sigma_{11}\right\rangle+}{\left(1-\mathrm{d}_{1}\right)}+\left\langle\sigma_{11}\right\rangle-\frac{\left\langle\sigma_{22}\right\rangle+}{\left(1-\mathrm{d}_{2}\right)}+\left\langle\sigma_{22}\right\rangle-\frac{\sigma_{12}}{\left(1-\mathrm{d}_{6}\right)}\right\}\right.$     (1)

The effective stress is the applied stress on the degraded surface that effectively resists forces. Parameters $\mathrm{d}_{1} \cdot \mathrm{d}_{2} \cdot \mathrm{d}_{6}$ specifies the degradation status for the three types of stress loading, $\mathrm{d}_{\mathrm{i}}$ between zero (is the healthy) and one (degradable material) [15].

1.png

Figure 1. Dimensions and geometry of the problem

2.2 Effective response

The linear elastic compound equation for the degraded material is written in accordance with the principle of equivalent strain [16]. The elastic composite equations for an orthotropic degraded material are as follows [15]:

$\varepsilon_{11}^{E}=\frac{\left\langle\sigma_{11}\right\rangle+}{E_{1}^{0}\left(1-\mathrm{d}_{1}\right)}+\frac{\left\langle\sigma_{11}\right\rangle-}{E_{1}^{0}}-\frac{v_{12}^{0}}{E_{1}^{0}} \sigma_{22}$

$\varepsilon_{22}^{E}=\frac{\left\langle\sigma_{22}\right\rangle+}{E_{2}^{0}\left(1-\mathrm{d}_{2}\right)}+\frac{\left\langle\sigma_{22}\right\rangle-}{E_{2}^{0}}-\frac{v_{12}^{0}}{E_{1}^{0}} \sigma_{11}$

$\varepsilon_{12}^{E}=\frac{\left\langle\sigma_{12}\right\rangle}{2 G_{12}^{0}\left(1-\mathrm{d}_{6}\right)}$     (2)

Based on the laboratory results for some materials, as well as the values $\frac{v_{12}}{E_{1}}=\frac{v_{12}^{0}}{E_{1}^{0}}=\frac{v_{21}^{0}}{E_{2}^{0}}$ in order to remain constant throughout evolution it has been destroyed. In Eq. (2) answers in modulus $G_{12} \cdot E_{2} \cdot E_{1}$ from the degraded layer it is shown that in terms of parameters $d_{6} . d_{2} . d_{1}$ and $G_{12}^{0} . E_{2}^{0} . E_{1}^{0}$ modules are unmodified material [11]

$\begin{aligned} E_{1} &=E_{1}^{0}\left(1-d_{1}\right) \\ E_{2} &=E_{2}^{0}\left(1-d_{2}\right) \\ E_{12} &=E_{12}^{0}\left(1-d_{6}\right) \end{aligned}$     (3)

2.3 Thermodynamic forces

Thermodynamic forces $Y_{6} . Y_{2} . Y_{1}$ Which give the variables of internal degradation and elastic strain energy density $E_{D}$ dependent in the current state of stress and degradation; the partial derivative can be shown below [15]:

$Y_{1}=\frac{\partial E_{D}}{d_{1}}\left|\bar{\sigma} \cdot d_{2} \cdot d_{6}: \operatorname{cons}\right|$

$Y_{2}=\frac{\partial E_{D}}{d_{2}}\left|\bar{\sigma} \cdot d_{1} \cdot d_{6}: \operatorname{cons}\right|$

$Y_{6}=\frac{\partial E_{D}}{d_{6}}\left|\bar{\sigma} \cdot d_{1} \cdot d_{2}: \operatorname{cons}\right|$     (4)

The average value of the strain energy density of the degraded layers can be written in terms of effective stresses, which is the difference between normal tensile stress and normal compressive stress by brackets.

$E_{D}=\frac{1}{2}\left[\frac{\left\langle\sigma_{11}\right\rangle^{2}+}{E_{1}^{0}\left(1-\mathrm{d}_{1}\right)}+\frac{\left\langle\sigma_{11}\right\rangle^{2}-}{E_{1}^{0}}-2 \frac{v_{12}^{0}}{E_{1}^{0}} \sigma_{11} \sigma_{22}+\frac{\left\langle\sigma_{22}\right\rangle^{2}+}{E_{2}^{0}\left(1-\mathrm{d}_{2}\right)}+\frac{\left\langle\sigma_{22}\right\rangle^{2}-}{E_{2}^{0}}+\frac{\sigma_{12}^{2}}{G_{12}^{0}\left(1-\mathrm{d}_{6}\right)}\right]$     (5)

By placing Eq. (3) in Eq. (4) thermodynamic forces are obtained in terms of stress components and degradation variables:

$Y_{1}=\frac{\partial E_{D}}{d_{1}}\left|\bar{\sigma} \cdot d_{2} \cdot d_{6}: \operatorname{cons}\right|=\frac{\left\langle\sigma_{11}\right\rangle^{2}+}{E_{1}^{0}\left(1-\mathrm{d}_{1}\right)^{2}}$

$Y_{2}=\frac{\partial E_{D}}{d_{2}}\left|\bar{\sigma} \cdot d_{1} \cdot d_{6}: \operatorname{cons}\right|=\frac{\left\langle\sigma_{22}\right\rangle^{2}+}{E_{2}^{0}\left(1-\mathrm{d}_{2}\right)^{2}}$

$Y_{6}=\frac{\partial E_{D}}{d_{6}}\left|\bar{\sigma} \cdot d_{1} \cdot d_{2}: \operatorname{cons}\right|=\frac{\left\langle\sigma_{12}\right\rangle^{2}+}{G_{12}^{0}\left(1-\mathrm{d}_{6}\right)}$     (6)

2.4 Evolution of destruction

A particular form of the law of evolution of degradation is in the matter of Ladeveze theory of matter. This model has been proven for highly orthotropic composites such as carbon/epoxy. Initially, a test for carbon/epoxy was established, which is mostly for continuous fiber composites. Damages other than fiber breakage do not cause modulus loss in the fiber direction, so in the absence of fiber breakage $d_{1}$ is zero and at plate stresses only $Y_{2}$ and $Y_{6}$ thermodynamic forces are present. $Y_{2}$ and $Y_{6}$ The maximum amount of thermodynamic force obtained outside the loading path $\tau \leq t$ for each $\tau$ Bigger than the current time [15]:

$Y_{2}(t)=\max \left\{Y_{2}(\tau)\right\} \quad T \leq t$     (7) 

$Y_{6}(t)=\max \left\{Y_{6}(\tau)\right\} \quad \tau \leq t$     (8)

Then the linear combination of $Y_{2}$ and $Y_{6}$:

$Y=\left(Y_{6}+b Y_{2}\right)$     (9)

b, it is a matter-dependent constant that indicates the coupling effect between transverse tensile and shear. $Y,$ a simple linear combination of effective thermodynamic forces. Eventually $Y$ maximum $Y$ for each $\tau$ larger than currently introduced.

$Y(t)=\max \left\{Y_{6}(\tau)+b Y_{2}(\tau)\right\} \quad T \leq t$     (10)

Laboratory results for highly orthotropic composites such as carbon / epoxy under quasi-static shear loading have shown that the degradation evolution law for $d_{6}$  :

$d_{6}=\frac{\langle\sqrt{Y-\sqrt{Y_{0}}\rangle}+}{\sqrt{Y_{c}}}$     (11)

Laboratory results for carbon/epoxy also show the degradation evolution law for transverse tensile loading as follows [15]:

$d_{2}=\frac{\langle\sqrt{Y-\sqrt{\tilde{Y}_{0}}}\rangle+}{\sqrt{\tilde{Y}_{c}}}$     (12)

Parameters $\tilde{Y}_{0} \cdot \tilde{Y}_{c} \cdot Y_{c} \cdot Y_{0}$ and b constants are the material through the testing for the composite IM6/914 the properties of which are given in Table 2 specified, and the Table 3 specified. The thickness of each layer is 0.125 mm.

Table 2. IM6/914 composite the properties [15]

$$\begin{array}{|c|c|c|c|c|}

\hline \mathbf{E}_{1}(\mathbf{G P a}) & \mathbf{E}_{2}(\mathbf{G P a}) & \mathbf{E}_{12}(\mathbf{G P a}) & \mathbf{V}_{12} & \mathbf{V}_{13} \\

\hline 170 & 10.8 & 5.8 & 0.34 & 0.4 \\

\hline

\end{array}$$

Table 3. Constant material parameters [15]

$$\begin{array}{|c|c|c|c|c|}

\hline \tilde{\mathbf{Y}}_{\mathbf{0}}(\mathbf{M P a}) & \tilde{\mathbf{Y}}_{\mathbf{c}}(\mathbf{M P a}) & \mathbf{Y}_{\mathbf{0}}(\mathbf{M P a}) & \mathbf{Y}_{\mathbf{c}}(\mathbf{M P a}) & \mathbf{b} \\

\hline 0.0576 & 14.29 & 0.0225 & 7.673 & 2.5 \\

\hline

\end{array}$$

2.5 Criterion of destruction

The criterion for destruction of transverse traction in each layer can be illustrated as follows:

$\left\{Y_{2} \geq Y_{2}^{c} \quad d_{2}=1\right.$     (13)

And otherwise $d_{2}$ Eq. (9) it is calculated; and can be written for damage in the shear direction

$\left\{Y_{2} \geq Y_{2}^{c} \quad d_{6}=1\right.$     (14)

And otherwise $d_{6}$ Eq. (8) it will be counted. Fragility and degradation thresholds are consistent with the failure of the interface between $Y_{2}^{c}$ and the matrix is due to transverse tensile loading.

In solving problems with finite element method, it is necessary to examine the independence of the mesh. To check the independence of the mesh, the mesh size is reduced so that the answer is not much different from the previous answer (according to the required accuracy of the problem), the previous mesh is selected as the optimal mesh. In this paper, the maximum stress is selected for comparison. The results are compared for different sizes of mesh in the Table 3. The dimensions used are as shown in Table 4.

Table 4. Investigating the independence of mesh

$$\begin{array}{|c|c|}

\hline \text { Maximum stress (Mpa) } & \text { Mesh size (mm) } \\

\hline 511.62 & 4 \\

\hline 497.71 & 3 \\

\hline 478.22 & 2 \\

\hline 474.23 & 1 \\

\hline 473.83 & 0.5 \\

\hline

\end{array}$$

Therefore, Structural mesh with size of 1 mm can be the optimal size for the problem. But with this mesh size for the whole geometry, the solution time is extended, Therefore, by using the geometry section tool, this mesh size is used only around the hole. The problem mesh is shown in the Figure 2.

2.png

Figure 2. Problem mesh

Samples review containing stress concentration from the outputs obtained which Table 5 illustrates the dimensions of the sample.

Table 5. The Dimensions Model

$$\begin{array}{|c|c|c|c|}

\hline \begin{array}{c}

\text { The } \\

\text { Length } \\

\text { (mm) }

\end{array} & \begin{array}{c}

\text { The } \\

\text { Width } \\

\text { (mm) }

\end{array} & \begin{array}{c}

\text { The diameter of } \\

\text { the hole } \\

\text { (mm) }

\end{array} & \begin{array}{c}

\text { The thickness of } \\

\text { each layer } \\

\text { (mm) }

\end{array} \\

\hline 250 & 100 & 40 & 0.125 \\

\hline

\end{array}$$

Reaction force and displacement curves is shown in Figure 6.

6.png

Figure 6. Reaction force and displacement curves

7.png

Figure 7. Stress contour x at direction 0° for lay-up technique [0/90]_2s

8.png

Figure 8. Stress contour x at direction 90° for lay-up technique [0/90]_2s

In Figure 7 stress contour x at direction 0° for lay-up technique [0/90]_2s and in Figure 8 at layer 90° under 25 MPa loading for lay-up technique [0/90]_2s. Figure 9 stress contour y at direction 0° for lay-up technique [0/90]_2s and in Figure 10 layer 90° under 25 MPa loading for lay-up technique [0/90]_2s.

9.png

Figure 9. Stress contour y at direction 0° for lay-up technique [0/90]_2s

10.png

Figure 10. Stress contour y at direction 90° for lay-up technique [0/90]_2s

11.png

Figure 11. Stress contour xy at direction 0° for lay-up technique [0/90]_2s

12.png

Figure 12. Stress contour xy at direction 90° under 25 MPa loading for lay-up technique [0/90]_2s

13.png

Figure 13. Destruction contour d₂ y at direction 0° for lay-up technique [0/90]_2s

14.png

Figure 14. Destruction contour d₂ y at direction 90° for lay-up technique [0/90]_2s

15.png

Figure 15. Destruction contour d₆ y at direction 0° for lay-up technique [0/90]_2s

16.png

Figure 16. Destruction contour d₆ y at direction 90° for lay-up technique [0/90]_2s

17.png

Figure 17. Stress to initiate first degradation in the 0° layer for (d₂ = 1 or d₆ = 1)

18.png

Figure 18. Defined line

19.png

Figure 19. Stress in the x direction 0° layer under 15 MPa load

20.png

Figure 20. Stress in the y direction 0° layer under 15 MPa load

21.png

Figure 21. Stress in the x direction 90° layer under 15 MPa load

In Figure 11 stress contour xy at direction 0° for lay-up technique [0/90]_2s and in Figure 12 stress contour xy at direction 90° under 25 MPa loading for lay-up technique [0/90]_2s. Figure 13 destruction contour d₂ y at direction 0° for lay-up technique [0/90]_2s and in Figure 14 destruction contour d₂ y at direction 90° for lay-up technique [0/90]_2s and in Figure 15 destruction contour d₆ y at direction 0° for lay-up technique [0/90]_2s and in Figure 16 destruction contour d₆ y at direction 90° for lay-up technique [0/90]_2s. 

Figure 17 is shown for five different diameters of the stress segment internal openings, in order to initiate the first degradation (d₂ = 1 or d₆ = 1), all occurring in the 0° layer. As can be seen from the figure, as the hole diameter decreases the stress tolerance of the specimen decreases in order to form the first degradation. In Figure 18 is shown defined line. In Figure 19 stress in the x direction for five different diameters and in the 0 ° layer calculated under 15 MPa load and the results are compared with each other. In Figure 20 stress in the y direction for five different diameters, and in the layer 0° is calculated under 15 MPa load and compared with each other. In Figure 21 stresses in the x direction for five different diameters and in the 90 ° layer at 15 MPa load are calculated and compared with each other. increasing the diameter increases the maximum stress in the sample.