TMM4175 Polymer Composites

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Laminate deformation¶

The Kirchhoff assumption implies that the displacements at a position $z$ through the thickness are related to the midplane displacements and rotations expressed by:

\begin{equation} \begin{aligned} &u(x,y,z) = u_0(x,y)-z \frac{\partial w_0}{\partial x} \\ &v(x,y,z) = v_0(x,y)-z \frac{\partial w_0}{\partial y} \\ &w(x,y)=w_0(x,y) \end{aligned} \tag{1} \end{equation}

where $u$, $v$ and $w$ are the displacements in $x$, $y$ and $z$ directions respectively, and $u_0$, $v_0$ and $w_0$ are the displacements of the reference plane.

Figure-1: Kirchoff assumption

From the strain-displacement equations;

\begin{equation} \begin{aligned} &\varepsilon_x = \frac{\partial u}{\partial x} =\frac{\partial u_0}{\partial x} - z \frac{\partial^2 w_0}{\partial x^2} = \varepsilon_x^0 + z \kappa_x \\ &\varepsilon_y = \frac{\partial v}{\partial y} =\frac{\partial v_0}{\partial y} - z \frac{\partial^2 w_0}{\partial y^2} = \varepsilon_y^0 + z \kappa_y \\ &\gamma_{xy} = \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} =\frac{\partial u_0}{\partial y} + \frac{\partial v_0}{\partial x} - 2z \frac{\partial^2 w_0}{\partial x \partial y} = \gamma_{xy}^0 + z \kappa_{xy} \end{aligned} \tag{2} \end{equation}

where $\varepsilon_x^0$, $\varepsilon_y^0$ and $\gamma_{xy}^0$ are mid-plane strains and $\kappa_x$, $\kappa_y$ and $\kappa_{xy}$ are curvatures given by

\begin{equation} \kappa_x=-\frac{\partial^2 w}{\partial x^2}, \quad \kappa_y=-\frac{\partial^2 w}{\partial y^2}, \quad \kappa_{xy}=-2\frac{\partial^2 w}{\partial x \partial y} \tag{3} \end{equation}

The relations can now be summarized using matrix form as

\begin{equation} \begin{bmatrix} \varepsilon_x \\ \varepsilon_y \\ \gamma_{xy} \end{bmatrix}= \begin{bmatrix} \varepsilon_x^0 \\ \varepsilon_y^0 \\ \gamma_{xy}^0 \end{bmatrix}+z \begin{bmatrix} \kappa_x \\ \kappa_y \\ \kappa_{xy} \end{bmatrix} \tag{4} \end{equation}

or the short notation

\begin{equation} \boldsymbol{\varepsilon}' = \boldsymbol{\varepsilon}^0 + z \boldsymbol{\kappa} \tag{5} \end{equation}

Illustrating the curvatures:

The curvatures can be visualized by superposing the resulting out-of-plane displacement $w$ given by the contributions from $\kappa_x$, $\kappa_y$ and $\kappa_{xy}$:

$$ \frac{\partial^2 w}{\partial x^2} = -\kappa_x \Rightarrow \frac{\partial w}{\partial x} = -\kappa_x x + c_1 \Rightarrow w=-\frac{1}{2}\kappa_x x^2 + c_1 x + c_2 $$

The boundary conditions $w(x=0) = 0$ and $\frac{\partial}{\partial x}w(x=0)$ leads to

$$ w=-\frac{1}{2}\kappa_x x^2 $$

Correspondingly, the contribution from the curvature $\kappa_y$ when we assume that $\frac{\partial}{\partial y}w(x=0)$ is

$$ w=-\frac{1}{2}\kappa_y y^2 $$

For $\kappa_{xy}$:

$$ \frac{\partial^2 w}{\partial x \partial y} = -\frac{1}{2}\kappa_{xy} \Rightarrow \frac{\partial w}{\partial y} = -\frac{1}{2} \kappa_{xy} x + c_5 \Rightarrow w=-\frac{1}{2}\kappa_{xy} x y + c_5 y + c_4 $$

The boundary conditions $w(x=0) = 0$, $\frac{\partial}{\partial x}w(x=0)$ and $\frac{\partial}{\partial y}w(x=0)$ gives

$$ w=-\frac{1}{2}\kappa_{xy} x y $$

Finally;

\begin{equation} w(x,y) = -\frac{1}{2}\kappa_x x^2 -\frac{1}{2}\kappa_y y^2 -\frac{1}{2} \kappa_{xy} x y \tag{6} \end{equation}

Implementation and examples:

def illustrateCurvatures(Kx,Ky,Kxy):
    from mpl_toolkits.mplot3d import Axes3D
    import matplotlib.pyplot as plt
    from matplotlib import cm
    from matplotlib.ticker import LinearLocator, FormatStrFormatter
    import numpy as np
    fig = plt.figure(figsize=(6,4))
    ax = fig.gca(projection='3d')
    X = np.arange(-0.5, 0.6, 0.1)
    Y = np.arange(-0.5, 0.6, 0.1)
    X, Y = np.meshgrid(X, Y)
    Z1 =  (- Kx*X**2 - Ky*Y**2 - Kxy*X*Y)/2
    surf1 = ax.plot_surface(X, Y, Z1, cmap=cm.coolwarm, linewidth=0, antialiased=False)

%matplotlib inline
illustrateCurvatures(Kx=0.5, Ky=0.0, Kxy=0.0)

%matplotlib inline
illustrateCurvatures(Kx=-0.5, Ky=0.5, Kxy=0.0)

%matplotlib inline
illustrateCurvatures(Kx=0.0, Ky=0.0, Kxy=-0.25)

%matplotlib inline
illustrateCurvatures(Kx=0.0, Ky=0.25, Kxy=0.5)

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