Behavior of Mindlin-Reissner Plates Under Static Loads
Mindlin-Reissner plate theory (aka. first-order shear deformation theory or FSDT) [6, 7, 8] is based on the following core assumptions:
Strains and displacements are small compared to the thickness of the plate.
The material exhibits linear elastic behavior.
Normals to the mid-surface remain straight but not perpendicular after deformation.
This is the defining kinematic assumption.
Normals remaining straight has the consequence of no transverse normal warping.
Releasing the perpendicularity assumption of Kirchhoff-Love plates enables shear-deformations to be accounted for.
Transverse shear strains are constant through the thickness.
On top of these fundamental assumptions, we make the following simplifications:
Equilibrium Equations
The equilibrium equations for an infinitesimally small section of the plate are:
(56)\[\begin{split}\begin{align*}
\sum F_z:& \quad p_{z}(x,y) + \left. \frac{\partial v_x}{\partial x}\right|_{(x,y)} + \left.\frac{\partial v_y}{\partial y}\right|_{(x,y)} &= 0, \\
\sum M_x:& \quad p_{xx}(x,y) - \left.\frac{\partial m_y}{\partial y}\right|_{(x,y)} - \left.\frac{\partial m_{xy}}{\partial x}\right|_{(x,y)} + v_y(x,y) &= 0, \\
\sum M_y:& \quad p_{yy}(x,y) + \left.\frac{\partial m_x}{\partial x}\right|_{(x,y)} + \left.\frac{\partial m_{xy}}{\partial y}\right|_{(x,y)} - v_x(x,y) &= 0,
\end{align*}\end{split}\]
where the internal forces for a plate of constant thickness $t$ are defined as:
(57)\[\begin{split}\begin{align*}
m_x(x,y) &= \int_{-t/2}^{t/2} z \, \sigma_x(x,y,z) dz, \\
m_y(x,y) &= \int_{-t/2}^{t/2} z \, \sigma_y(x,y,z) dz, \\
m_{xy}(x,y) &= \int_{-t/2}^{t/2} z \, \tau_{xy}(x,y,z) dz, \\
v_x(x,y) &= \int_{-t/2}^{t/2} \tau_{xz}(x,y,z) dz, \\
v_y(x,y) &= \int_{-t/2}^{t/2} \tau_{yz}(x,y,z) dz.
\end{align*}\end{split}\]
These equations can be summarized using the following matrix notation:
(58)\[\begin{split}\underline{m} =
\begin{bmatrix}
m_x \\
m_y \\
m_z
\end{bmatrix}
=
\int_{-t/2}^{t/2} z \, \underline{\sigma} \, dz
\qquad \text{and} \qquad
\underline{v} =
\begin{bmatrix}
v_x \\
v_y
\end{bmatrix}
=
\int_{-t/2}^{t/2} \underline{\tau} \, dz\end{split}\]
with
(59)\[\underline{\sigma} =
\begin{bmatrix}
\sigma_x & \sigma_y & \tau_{xy}
\end{bmatrix} ^ T
\qquad \text{and} \qquad
\underline{\tau} =
\begin{bmatrix}
\tau_{xz} & \tau_{yz}
\end{bmatrix} ^ T.\]
Note
In the Kirchhoff-Love plate theory, equations (56) are typically combined into a single equation. Here, however, we retain them as separate equations.
Geometric Equations
The displacement field is described by:
(60)\[\begin{split}\begin{align*}
u(x,y,z) &= z \, \vartheta_y(x,y), \\
v(x,y,z) &= -z \, \vartheta_x(x,y), \\
w(x,y,z) &= w(x,y),
\end{align*}\end{split}\]
where $w(x,y)$, $\vartheta_y(x,y)$, and $\vartheta_x(x,y)$ are the reduced displacement functions. The relationship between the components of the small-strain tensor and the reduced displacement field is:
(61)\[\begin{split}\begin{align*}
\varepsilon_x(x,y,z) &= \left. \frac{\partial u}{\partial x} \right|_{(x,y)} = z \, \left. \frac{\partial \vartheta_y}{\partial x} \right|_{(x,y)} = z \, \kappa_{x}(x,y), \\
\varepsilon_y(x,y,z) &= \left. \frac{\partial v}{\partial y} \right|_{(x,y)} = -z \, \left. \frac{\partial \vartheta_x}{\partial y} \right|_{(x,y)} = z \, \kappa_{y}(x,y), \\
\gamma_{xy}(x,y,z) &= \left. \frac{\partial u}{\partial y} \right|_{(x,y)} + \left. \frac{\partial v}{\partial x} \right|_{(x,y)} = z \, \left. \left( \frac{\partial \vartheta_y}{\partial y} - \frac{\partial \vartheta_x}{\partial x} \right) \right|_{(x,y)} = z \, \kappa_{xy}(x,y), \\
\gamma_{xz}(x,y,z) &= \left. \frac{\partial u}{\partial z} \right|_{(x,y)} + \left. \frac{\partial w}{\partial x} \right|_{(x,y)} = \vartheta_y(x,y) + \left. \frac{\partial w}{\partial x} \right|_{(x,y)}, \\
\gamma_{yz}(x,y,z) &= \left. \frac{\partial v}{\partial z} \right|_{(x,y)} + \left. \frac{\partial w}{\partial x} \right|_{(x,y)} = -\vartheta_x(x,y) + \left. \frac{\partial w}{\partial y} \right|_{(x,y)}.
\end{align*}\end{split}\]
Note
In the Kirchhoff-Love theory, the assumption that normals to the mid-surface remain perpendicular to the mid-surface leads to setting the transverse shear-strain components to zero. Since the Mindlin-Reissner theory relaxes this assumption, we do not make this simplification here.
The generalized strain components with matrix notation:
(62)\[\underline{\varepsilon}
=
\begin{bmatrix}
\varepsilon_x & \varepsilon_y & \gamma_{xy}
\end{bmatrix} ^ T
=
z \, \underline{\kappa}
= z \, \begin{bmatrix}
\kappa_x & \kappa_y & \kappa_{xy}
\end{bmatrix} ^ T.\]
and
(63)\[\underline{\gamma} =
\begin{bmatrix}
\gamma_{xy} & \gamma_{xz}
\end{bmatrix} ^ T.\]
Material Equations
The constitutive relationship at the material level is:
(64)\[\underline{\sigma} = \underline{\underline Q} \, \underline{\varepsilon}
= \underline{\underline Q} \, z \, \underline{\kappa}
\qquad \text{and} \qquad
\underline{\tau} = \underline{\underline P} \, \underline{\gamma},\]
Substituting this into (58) gives:
(65)\[\underline{m} = \underline{\underline D} \, \underline{\kappa}
\qquad \text{and} \qquad
\underline{v} = \underline{\underline S} \, \underline{\gamma}\]
with
(66)\[\begin{split}\underline{\underline D} =
\begin{bmatrix}
D_{11} & D_{12} & 0 \\
D_{21} & D_{22} & 0 \\
0 & 0 & D_{66}
\end{bmatrix}
=
\int_{-t/2}^{t/2} z^2 \, \underline{\underline Q} \, dz\end{split}\]
and
(67)\[\begin{split}\underline{\underline S} =
\begin{bmatrix}
S_{55} & 0 \\
0 & S_{44}
\end{bmatrix}
=
\int_{-t/2}^{t/2} \underline{\underline P} \, dz.\end{split}\]
For a linear elastic, isotropic material and a plate of constant thickness $t$, these become:
\[\begin{split}\underline{\underline D} = \frac{E t^3}{12 (1 - \nu^2)}
\begin{bmatrix}
1 & \nu & 0 \\
\nu & 1 & 0 \\
0 & 0 & \frac{1 - \nu}{2}
\end{bmatrix},\end{split}\]
and
\[\begin{split}\underline{\underline S} =
\begin{bmatrix}
k_{x} G t & 0 \\
0 & k_{y} G t
\end{bmatrix},\end{split}\]
where $E$ is Young’s modulus, $\nu$ is Poisson’s ratio, $G$ is the shear modulus, $k_x$ and $k_y$ are the shear correction factors for the $x$ and $y$ directions.
Summary
Combining the equilibrium equations (56) with the geometric and material relationships (61) and (65), we obtain the following partial differential equations:
\begin{align*}
\sum F_z:& \qquad
&S_{44}\, \left( \frac{\partial \vartheta_x}{\partial y}
- \frac{\partial^2 w}{\partial y^2} \right)
- S_{55}\, \left( \frac{\partial \vartheta_y}{\partial x}
+ \frac{\partial^2 w}{\partial x^2} \right) &= p_z, \\
\sum M_x:& \qquad
&D_{12}\,\frac{\partial^2 \vartheta_y}{\partial x\,\partial y}
- D_{22}\,\frac{\partial^2 \vartheta_x}{\partial y^2}
+ D_{66}\, \left( \frac{\partial^2 \vartheta_y}{\partial x\,\partial y} - \frac{\partial^2 \vartheta_x}{\partial x^2} \right) & \\
&& + S_{44}\,\left( \vartheta_x - \frac{\partial w}{\partial y} \right) &= p_{xx}, \\
\sum M_y:& \qquad
&- D_{11}\,\frac{\partial^2 \vartheta_y}{\partial x^2}
+ D_{12}\,\frac{\partial^2 \vartheta_x}{\partial x\,\partial y}
+ D_{66}\,\left( \frac{\partial^2 \vartheta_x}{\partial x\,\partial y} - \frac{\partial^2 \vartheta_y}{\partial y^2} \right) & \\
&& + S_{55}\, \left( \vartheta_y + \frac{\partial w}{\partial x} \right) &= p_{yy}.
\end{align*}
These PDEs, together with the appropriate boundary conditions, defines the boundary-value problem (BVP) for a Mindlin-Reissner plate.
BVP of Simply-Supported Mindlin-Reissner Plates
The boundary conditions for a simply-supported Mindlin-Reissner plate are:
(69)\[\begin{split}\begin{align*}
w(x, 0) &= w(x, L_y) = 0 \qquad &\forall x \in (0, L_x), \\
w(0, y) &= w(L_x, y) = 0 \qquad &\forall y \in (0, L_y), \\
m_x(0, y) &= m_x(L_x, y) = 0 \qquad &\forall y \in (0, L_y), \\
m_y(x, 0) &= m_y(x, L_y) = 0 \qquad &\forall x \in (0, L_x).
\end{align*}\end{split}\]
Equations (68) and (69) together constitute the BVP for a simply-supported Mindlin-Reissner plate.
Solution of the BVP Using Navier’s Approach
To simplify further expressions, we introduce the following notation:
(70)\[\begin{split}\begin{align*}
&S_i(x) = \sin \left(\frac{\pi \,i \,x}{Lx}\right), \qquad
&S_j(y) = \sin \left(\frac{\pi \,j \,y}{Ly}\right), \\
&C_i(x) = \cos \left(\frac{\pi \,i \,x}{Lx}\right), \qquad
&C_j(y) = \cos \left(\frac{\pi \,j \,y}{Ly}\right).
\end{align*}\end{split}\]
The displacements and loads are approximated as:
(71)\[\begin{split}\begin{align*}
w(x, y) \approx \overline{w}(x, y) &= \sum_{i=1}^m \sum_{j=1}^n w^{(ij)} \, S_i(x) \, S_j(y), \\
\vartheta_x(x, y) \approx \overline{\vartheta}_x(x, y) &= \sum_{i=1}^m \sum_{j=1}^n \vartheta_x^{(ij)} \, S_i(x) \, C_j(y), \\
\vartheta_y(x, y) \approx \overline{\vartheta}_y(x, y) &= \sum_{i=1}^m \sum_{j=1}^n \vartheta_y^{(ij)} \, C_i(x) \, S_j(y), \\
p_z(x, y) \approx \overline{p}_z(x, y) &= \sum_{i=1}^m \sum_{j=1}^n p_z^{(ij)} \, S_i(x) \, S_j(y), \\
p_{xx}(x, y) \approx \overline{p}_{xx}(x, y) &= \sum_{i=1}^m \sum_{j=1}^n p_{xx}^{(ij)} \, S_i(x) \, C_j(y), \\
p_{yy}(x, y) \approx \overline{p}_{yy}(x, y) &= \sum_{i=1}^m \sum_{j=1}^n p_{yy}^{(ij)} \, C_i(x) \, S_j(y).
\end{align*}\end{split}\]
Substituting the approximations from (71) into (68), using trigonometric identities and rearranging, we find that the solution for the unknown coefficients $w^{(ij)}$, $\vartheta_x^{(ij)}$ and $\vartheta_y^{(ij)}$ can be obtained from
(72)\[\begin{bmatrix}
w^{(ij)} & \vartheta_x^{(ij)} & \vartheta_y^{(ij)}
\end{bmatrix} ^ T
=
{\underline{\underline{A}}^{(ij)}}^{-1}
\begin{bmatrix}
p_z^{(ij)} & p_{xx}^{(ij)} & p_{yy}^{(ij)}
\end{bmatrix} ^ T\]
with coefficient matrix
(73)\[\begin{split}\underline{\underline{A}}^{(ij)} =
\left[\begin{matrix}\frac{\pi^{2} S_{44} j^{2}}{L_{y}^{2}} + \frac{\pi^{2} S_{55} i^{2}}{L_{x}^{2}} & - \frac{\pi S_{44} j}{L_{y}} & \frac{\pi S_{55} i}{L_{x}}\\- \frac{\pi S_{44} j}{L_{y}} & \frac{\pi^{2} D_{22} j^{2}}{L_{y}^{2}} + \frac{\pi^{2} D_{66} i^{2}}{L_{x}^{2}} + S_{44} & - \frac{\pi^{2} D_{12} \,i \, j}{L_{x} L_{y}} - \frac{\pi^{2} D_{66} \,i \, j}{L_{x} L_{y}}\\\frac{\pi S_{55} i}{L_{x}} & - \frac{\pi^{2} D_{12} \, i \, j}{L_{x} L_{y}} - \frac{\pi^{2} D_{66} \, i \, j}{L_{x} L_{y}} & \frac{\pi^{2} D_{11} i^{2}}{L_{x}^{2}} + \frac{\pi^{2} D_{66} j^{2}}{L_{y}^{2}} + S_{55}\end{matrix}\right]\end{split}\]
and load coefficients
(74)\[\begin{split}\begin{align*}
p_z^{(ij)} &= \frac{4}{L_x \, L_y} \int_0^{L_y} \int_0^{L_x} p_z(x,y) \, S_i(x) \, S_j(y) \, dxdy, \\
p_{xx}^{(ij)} &= \frac{4}{L_x \, L_y} \int_0^{L_y} \int_0^{L_x} p_{xx}(x,y) \, S_i(x) \, C_j(y) \, dxdy, \\
p_{yy}^{(ij)} &= \frac{4}{L_x \, L_y} \int_0^{L_y} \int_0^{L_x} p_{yy}(x,y) \, C_i(x) \, S_j(y) \, dxdy.
\end{align*}\end{split}\]