User:Tmhoang81/Curling of paper sheet using shell theory

= Small transverse deflections of thin plates =

Consider a load-free plate in which the $$\displaystyle xy$$ plane coincides with plate's midplane and the $$\displaystyle z$$ coordinate is perpendicular to it and is directed downwards. The fundamental assumptions of the linear, elastic, small deflection theory of bending for thin plates may be stated as follows  :

(i) The material of the plate is elastic, homogeneous, and isotropic.

(ii) The plate is initially flat.

(iii) The deflection (the normal component of the displacement vector) of the midplane is small compared with the thickness of the plate. The slope of the deflected surface is therefore very small and the square of the slope is a negligible quantity in comparison with unity.

(iv) The straight lines, initially normal to the middle plane before bending, remain straight and normal to the middle surface during the deformation, and the length of such elements is not altered. This means that the vertical shear strain $$\displaystyle \gamma _{xz}$$ and $$\displaystyle \gamma _{yz}$$ are negligible and the normal strain $$\displaystyle \varepsilon _z$$ may also be omitted. This assumption is referred to as the "hypothesis of straight normals."

(v) The stress normal to the middle plane, $$ \sigma_z$$ is small compared with the other stress components and may be neglected in the stress-strain relations

(vi) Since the displacements of a plate are small, it is assumed that the middle surface remains unstrained after bending.

Strain-curvature relations (Kinematic equations)
Let $$\displaystyle u$$,$$\displaystyle v$$ and $$\displaystyle w$$ be components of the displacement vector of points in the middle surface of the plate occuring in the $$\displaystyle x$$,$$\displaystyle y$$ and $$\displaystyle z$$ directions, respectively. As it follows from the assumption (iv) above


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$$  \displaystyle \begin{array}{l} \varepsilon _z = 0\,\,,\,\,\gamma _{xz}  = 0\,\,,\,\,\gamma _{yz}  = 0 \\ \end{array} $$     (1.1)
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Integrating the expressions (1.1) and taking into account assumptions (vi) we have the following form in the context of Kirchhoff's theory :


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$$   \begin{array}{l} \displaystyle w = w\left( {x,y} \right)\,\,,\,\,u = - z\frac\,\,,\,\,v =  - z\frac \\ \end{array} $$     (1.2)
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From the definitions, strains are given by


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$$  \displaystyle \begin{array}{l} \displaystyle \varepsilon _x = \frac =  - z\frac = z\kappa _x  \\ \displaystyle \varepsilon _y = \frac =  - z\frac = z\kappa _y  \\ \displaystyle \gamma _{xy} = \frac + \frac =  - 2z\frac = z\kappa _{xy}  \\ \end{array} $$     (1.3) where curvatures are
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$$  \displaystyle \begin{array}{l} \displaystyle \kappa _x =  - \frac \\ \displaystyle \kappa _y =  - \frac \\ \displaystyle \kappa _{xy} =  - 2\frac \\ \end{array} $$     (1.4)
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Stresses, stress resultants, and stress couples
Stress-strain relations are related by the Hooke's law :


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$$  \displaystyle \begin{array}{l} \displaystyle \sigma _x = \frac{E}\left( {\varepsilon _x  + \nu \varepsilon _y } \right) =  - \frac\left( {\frac + \nu \frac} \right) \\ \displaystyle \sigma _y = \frac{E}\left( {\varepsilon _y  + \nu \varepsilon _x } \right) =  - \frac\left( {\frac + \nu \frac} \right) \\ \displaystyle \tau _{xy} = G\gamma _{xy}  =  - \frac\frac \\ \end{array} $$     (1.5) where $$\displaystyle G = \frac{E}$$
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The resultant forces and moments are


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$$   \begin{array}{l} \displaystyle \left\{ {\begin{array}{*{20}c} {Q_x } \\ \displaystyle {Q_y } \\ \end{array}} \right\} = \displaystyle \int\limits_{ - h/2}^{h/2} {\left\{ \begin{array}{l} \displaystyle \tau _{xz} \\ \displaystyle \tau _{yz} \\ \end{array} \right\}dz} \\ \end{array} $$     (1.6)
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$$  \displaystyle \begin{array}{l} \left\{ {\begin{array}{*{20}c} {M_x } \\ \displaystyle {M_y } \\ \displaystyle {M_{xy} } \\ \displaystyle \end{array}} \right\} = \displaystyle \int\limits_{ - h/2}^{h/2} {\left\{ {\begin{array}{*{20}c} \displaystyle {\sigma _x } \\ \displaystyle {\sigma _y } \\ \displaystyle {\tau _{xy} } \\ \displaystyle \end{array}} \right\}zdz} \\ \end{array} $$     (1.7)
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Substituting eqs (1.5) into eqs (1.7) one gets


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$$  \displaystyle \begin{array}{l} \displaystyle M_x =  - D\left( {\frac + \nu \frac} \right) \\ \displaystyle M_y =  - D\left( {\frac + \nu \frac} \right) \\ \displaystyle M_{xy} =  - D\left( {1 - \nu } \right)\frac \\ \end{array} $$     (1.8)
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where D is the flexural rigidity of the plate as below
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$$  \displaystyle \begin{array}{l} \displaystyle D = \frac \\ \end{array} $$     (1.9)
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Governing equation for deflection of plates in cartesian coordinates
To get the governing equation for deflection of plates, we imagine to cut out an infinite element and consider its conditions of equilibrium. The force summation in the z axis gives


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$$  \displaystyle \begin{array}{l} \displaystyle \frac + \frac + p\left( {x,y} \right) = 0 \\ \end{array} $$     (1.10)
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The moment summation about the x and y axises lead to
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$$  \displaystyle \begin{array}{l} \displaystyle Q_x = \frac + \frac \\ \displaystyle Q_y = \frac + \frac \\ \end{array} $$     (1.11)
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Substituting eqs (1.11) into (1.10), one finds the following:
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$$  \displaystyle \begin{array}{l} \displaystyle \frac + 2\frac + \frac = - p\left( {x,y} \right) \\ \end{array} $$     (1.12)
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Finally, introduction of the expressions from eqs (1.8) into eqs (1.12) yields


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$$  \displaystyle \begin{array}{l} \displaystyle \frac + 2\frac + \frac = \frac{p}{D} \\ \end{array} $$     (1.13)
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This is the governing differential equation for the deflections for thin plate bending analysis based on Kirchhoff's assumptions. Equation (1.13) can be rewritten as follows :


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$$  \displaystyle \begin{array}{l} \displaystyle \nabla ^4 w = \frac{p}{D} \\ \end{array} $$     (1.14) where
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$$  \displaystyle \begin{array}{l} \displaystyle \nabla ^4 \equiv \frac + 2\frac + \frac \\ \end{array} $$     (1.15)
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Expressions for the vertical forces easily obtain by substituting eqs (1.8) into eqs (1.11)


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$$  \displaystyle \begin{array}{l} \displaystyle Q_x =  - D\frac{\partial }\left( {\frac + \frac} \right) = - D\left( {\frac + \frac} \right) = - D\frac{\partial }\left( {\nabla ^2 w} \right) \\ \displaystyle Q_y =  - D\frac{\partial }\left( {\frac + \frac} \right) =   - D\left( \frac+ {\frac} \right)= - D\frac{\partial }\left( {\nabla ^2 w} \right) \\ \end{array} $$     (1.16)
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Note that the bending moments $$\displaystyle M_{x},M_{y}$$ and twist $$\displaystyle M_{xy}$$ depend only on $$\displaystyle w$$, not on the in-plane displacements.

Boundary conditions
Considering boundary condition at one edge, for example, edge $$\displaystyle x=a$$. We have three common boundary conditions :

(A) Plates are simple supported at the edge:

The deflection and bending moment are both zero:
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$$  \displaystyle \begin{array}{l} \displaystyle w = M_x = 0|_{x = a}  \\ \end{array} $$     (1.17)
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(B) Plates are clamped or built in, or fixed at this edge

The deflection and slope are zero:
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$$  \displaystyle \begin{array}{l} \displaystyle w = \frac = 0|_{x = a} \\ \end{array} $$     (1.18)
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(C) Plates are free at this edge

All forces and moments are zero:


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$$  \displaystyle \begin{array}{l} \displaystyle Q_x = M_x  = M_{xy}  = 0|_{x = a}  \\ \end{array} $$     (1.19)
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However, three boundary conditions are too many for the governing differential equation. Kirchhoff suggested that the three boundary conditions at the free edge be combined into two by equating to zero the bending moment and the so-called effective shear force.


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$$  \displaystyle \begin{array}{l} \displaystyle M_x = 0|_{x = a} \,\,,\,\,V_x  = Q_x  + \frac = 0|_{x = a}  \\ \end{array} $$     (1.20)
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= Large transverse deflections of thin plates =

Large deformation theory


To calculate deformations of a elastic body, we consider a chance in length of an infinitesimal line PQ before and after deformation. The square of length PQ in the original configuration is given by :


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$$  \displaystyle dS^2 = dX_1^2 + dX_2^2  + dX_3^2  = \delta _{ij} dX_i dX_j  = dX_k dX_k $$     (2.1)
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Due to deformation, the square of length pq in the deformed configuration is:


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\displaystyle ds^2 = dx_1^2  + dx_2^2  + dx_3^2  = \delta _{ij} dx_i dx_j  = dx_k dx_k $$     (2.2)
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The differentials $$\displaystyle dx_i$$ can be transformed into coordinate systems $$\displaystyle X_i$$


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\displaystyle dx_k = \fracdX_i $$     (2.3)
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Hence

\displaystyle ds^2 = \fracdX_i \fracdX_j  = \frac\fracdX_i dX_j $$

By defitinion, Green's strain tensor $$\varepsilon _{ij}$$ is obtained as following:


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\displaystyle ds^2 - dS^2  = \left( {\frac\frac - \delta _{ij} } \right)dX_i dX_j  = 2\varepsilon _{ij} dX_i dX_j $$     (2.4)
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or


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\displaystyle \varepsilon _{ij} = \frac{1}{2}\left( {\frac\frac - \delta _{ij} } \right) $$     (2.5)
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From the defition of displacements:


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\displaystyle x_i - X_i  = u_i $$     (2.6)
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we have:

\displaystyle \frac = \frac + \frac = \frac + \delta _{ki} $$

Substituting above equation into (2.5):

\displaystyle \varepsilon _{ij} = \frac{1}{2}\left( {\frac\frac - \delta _{ij} } \right) = \frac{1}{2}\left( {\left( {\frac + \delta _{ki} } \right)\left( {\frac + \delta _{kj} } \right) - \delta _{ij} } \right) = \frac{1}{2}\left( {\frac\frac + \frac\delta _{kj}  + \frac\delta _{ki} } \right) $$ or


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$$  \displaystyle \varepsilon _{ij} = \frac{1}{2}\left( {\frac + \frac + \frac\frac} \right) $$     (2.7)
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In unabridged notations, in-plane strains, for example, are given by:

\displaystyle \varepsilon _{11} = \frac + \frac{1}{2}\left[ {\left( {\frac} \right)^2  + \left( {\frac} \right)^2  + \left( {\frac} \right)^2 } \right] $$



\displaystyle \varepsilon _{22} = \frac + \frac{1}{2}\left[ {\left( {\frac} \right)^2  + \left( {\frac} \right)^2  + \left( {\frac} \right)^2 } \right] $$

\displaystyle \gamma _{12} = 2\varepsilon _{12}  = \frac + \frac + \left( {\frac\frac + \frac\frac + \frac\frac} \right) $$

Strain-displacement relations
The linear theory of thin plate ignores strains of middle surface of the plate and the corresponding in-plane stresses are neglected. However, if the magnitude of the lateral deflections are large compared to the thickness of plate, these deflections are accompanied by stretching of the middle surface. As a result, the membrane forces are included. The large deflection plate bending theory uses two assumptions:

- The hypothesis of the straight normals: the straight lines, initially normal to the middle plane before deformation remain straight and normal to the middle surface during the deformation therefore $$\displaystyle \gamma _{xz}  = \gamma _{yz}  = \varepsilon _z  = 0$$

- The stress normal to the middle surface, $$\displaystyle \sigma _z $$, is small compared with the other stress components and may be neglected in the stress-strain relations.



The above figure shows a section of the plate normal to the y-axis before and after deformation. The x-y plane is equidistant from the top and bottom surfaces of the plate and is called the midplane of reference plane.

The reference plane displacements $$u_0$$ and $$v_0$$ in the x and y directions and the out of plane displacment w in the z direction are functions of x and y only:


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$$  \displaystyle \begin{array}{l} u_0 = u_0 \left( {x,y} \right) \\ \displaystyle v_0 = v_0 \left( {x,y} \right) \\ w = w\left( {x,y} \right) \\ \end{array} $$     (2.8)
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The in-plane displacements components of a point of coordinate z are:


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\displaystyle \begin{array}{l} \displaystyle u = u_0 - z\frac \\ \displaystyle v = v_0 - z\frac \\ \end{array} $$     (2.9) where z is the through-thickness coordinate of a general point of the cross section.
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Due to large deformations, strain-displacement relations (2.7) must be used. However, we assume that displacements u and v are small, one can neglect $$\left( {\partial u/\partial x} \right)^2 ,\,\left( {\partial u/\partial y} \right)^2 ,\,\left( {\partial v/\partial x} \right)^2 ,\,\left( {\partial v/\partial y} \right)^2 ,\,\left( {\partial u/\partial x} \right)\left( {\partial u/\partial y} \right),\,\left( {\partial v/\partial x} \right)\left( {\partial v/\partial y} \right)$$

Thus strain-displacement relations for large deformation of thin plate are:


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\displaystyle \begin{array}{l} \displaystyle \varepsilon _x = \frac + \frac{1}{2}\left( {\frac} \right)^2  \\ \displaystyle \varepsilon _y = \frac + \frac{1}{2}\left( {\frac} \right)^2  \\ \displaystyle \gamma _{xy} = 2\varepsilon _{xy}  = \frac + \frac + \frac\frac   \\ \end{array} $$     (2.10) Substituting (2.9) into (2.10) we get
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\displaystyle \begin{array}{l} \displaystyle \varepsilon _x = \frac + \frac{1}{2}\left( {\frac} \right)^2  - z\frac = \varepsilon _x^0  + z\kappa _x  \\ \displaystyle \varepsilon _y = \frac + \frac{1}{2}\left( {\frac} \right)^2  - z\frac = \varepsilon _x^0  + z\kappa _y  \\ \displaystyle \gamma _{xy} = \frac + \frac + \frac\frac - 2z\frac = \gamma _{xy}^0  + z\kappa _{xy}  \\ \end{array} $$     (2.11)
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where


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\displaystyle \begin{array}{l} \displaystyle \varepsilon _x^0 = \frac + \frac{1}{2}\left( {\frac} \right)^2  \\ \displaystyle \varepsilon _y^0 = \frac + \frac{1}{2}\left( {\frac} \right)^2  \\ \displaystyle \gamma _{xy}^0 = 2\varepsilon _{xy}^0  = \frac + \frac + \frac\frac \\ \displaystyle \kappa _x =  - \frac \\ \displaystyle \kappa _y =  - \frac \\ \displaystyle \kappa _{xy} =  - 2\frac \\ \end{array} $$     (2.12)
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or in matrix form
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$$  \displaystyle \begin{array}{l} \displaystyle \left[ {\begin{array}{*{20}c} {\varepsilon _x } \\ \displaystyle {\varepsilon _x } \\ \displaystyle {\gamma _{xy} } \\ \end{array}} \right] = \left[ {\begin{array}{*{20}c} \displaystyle {\varepsilon _x^0 } \\ \displaystyle {\varepsilon _y^0 } \\ \displaystyle {\gamma _{xy}^0 } \\ \end{array}} \right] + z\left[ {\begin{array}{*{20}c} \displaystyle {\kappa _x } \\ \displaystyle {\kappa _y } \\ \displaystyle {\kappa _{xy} } \\ \end{array}} \right] \\ \end{array} $$     (2.13)
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From Hooke's Law, one obtains:


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\displaystyle \left\{ {\begin{array}{*{20}c} {\sigma _x } \\ \displaystyle {\sigma _y } \\ \displaystyle {\sigma _{xy} } \\ \end{array}} \right\} = \frac{E}\left[ {\begin{array}{*{20}c} \displaystyle 1 & \nu & 0  \\ \displaystyle \nu & 1 & 0  \\ \displaystyle 0 & 0 & \displaystyle {\frac{2}} \\ \end{array}} \right]\left\{ {\begin{array}{*{20}c} \displaystyle {\varepsilon _x } \\ \displaystyle {\varepsilon _y } \\ \displaystyle {\tau _{xy} } \\ \end{array}} \right\} = \displaystyle \frac{E}\left[ {\begin{array}{*{20}c} \displaystyle 1 & \nu & 0  \\ \displaystyle \nu & 1 & 0  \\ \displaystyle 0 & 0 & \displaystyle {\frac{2}} \\ \end{array}} \right]\left\{ {\begin{array}{*{20}c} \displaystyle {\varepsilon _x^0 } \\ \displaystyle {\varepsilon _y^0 } \\ \displaystyle {\tau _{xy}^0 } \\ \end{array}} \right\} + \displaystyle \frac\left[ {\begin{array}{*{20}c} \displaystyle 1 & \nu & 0  \\ \displaystyle \nu & 1 & 0  \\ \displaystyle 0 & 0 & \displaystyle {\frac{2}} \\ \end{array}} \right]\left\{ {\begin{array}{*{20}c} \displaystyle {\kappa _x } \\ \displaystyle {\kappa _y } \\ \displaystyle {\kappa _{xy} } \\ \end{array}} \right\} $$
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Relation (2.14), however, is valid for isotropic materials except for anisotropic materials.

Force resultants and moment resultants


Similar to the small transverse deflection of thin plates, shear forces and moments are the same in (1.6) (1.7):


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$$   \displaystyle \begin{array}{l} \displaystyle \left\{ {\begin{array}{*{20}c} {Q_x } \\ \displaystyle {Q_y } \\ \end{array}} \right\} = \displaystyle \int\limits_{ - h/2}^{h/2} {\left\{ \begin{array}{l} \displaystyle \tau _{xz} \\ \displaystyle \tau _{yz} \\ \end{array} \right\}dz} \\ \end{array} $$     (1.6)
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$$  \displaystyle \begin{array}{l} \left\{ {\begin{array}{*{20}c} {M_x } \\ \displaystyle {M_y } \\ \displaystyle {M_{xy} } \\ \displaystyle \end{array}} \right\} = \displaystyle \int\limits_{ - h/2}^{h/2} {\left\{ {\begin{array}{*{20}c} \displaystyle {\sigma _x } \\ \displaystyle {\sigma _y } \\ \displaystyle {\tau _{xy} } \\ \displaystyle \end{array}} \right\}zdz} \\ \end{array} $$     (1.7)
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The linear theory ignores straining of the middle surface of the plate and the corresponding in-plane stresses are neglected. However, in case of large deflection of thin plates, we have additional in-plane forces as below:


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$$ \displaystyle \left\{ {\begin{array}{*{20}{c}} \\    \\     \\ \end{array}} \right\} = \displaystyle  \int\limits_{ - h/2}^{h/2} {\left\{ {\begin{array}{*{20}{c}} \\    \\     \\ \end{array}} \right\}dz  }$$ (2.15)
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Using (2.14), Eqs (1.7) and (2.15) are rewritten as followings:


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$$   \displaystyle \left\{ {\begin{array}{*{20}{c}} \displaystyle \\ \displaystyle \\ \displaystyle \\ \end{array}} \right\} = D\left[ {\begin{array}{*{20}{c}} \displaystyle  1 & \nu  & 0  \\ \displaystyle  \nu  & 1 & 0  \\ \displaystyle  0 & 0 & \displaystyle{\frac{2}}  \\ \end{array}} \right]\left\{ {\begin{array}{*{20}{c}} \displaystyle    \\ \displaystyle \\ \displaystyle \\ \end{array}} \right\} $$   (2.16)
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Or
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$$  \displaystyle \begin{array}{l} \displaystyle M_x =  - D\left( {\frac + \nu \frac} \right) \\ \displaystyle M_y =  - D\left( {\frac + \nu \frac} \right) \\ \displaystyle M_{xy} =  - D\left( {1 - \nu } \right)\frac \\ \end{array} $$     (1.8)
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And
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$$   \displaystyle \left\{ {\begin{array}{*{20}{c}} \\    \\     \\ \end{array}} \right\} =  A\left[ {\begin{array}{*{20}{c}} 1 & \nu & 0  \\ \nu & 1 & 0  \\ 0 & 0 & \displaystyle {\frac{2}} \\ \end{array}} \right]\left\{ {\begin{array}{*{20}{c}} {\varepsilon _x^0} \\ {\varepsilon _y^0} \\ {\gamma _{xy}^0} \\ \end{array}} \right\} $$   (2.17)
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Or
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$$   \displaystyle \begin{array}{l} {N_x} = A\left( {\varepsilon _x^0 + \nu \varepsilon _y^0} \right) \\ {N_y} = A\left( {\varepsilon _y^0 + \nu \varepsilon _x^0} \right) \\ {N_{xy}} = Gh\gamma _{xy}^0 \\ \end{array}$$ (2.18)
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where $$\displaystyle A = \displaystyle \frac\,\,,\,\,G = \displaystyle \frac{E}\,\,,\,\,D = \displaystyle \frac $$

The equilibrium of thin plate with large deflections
From six equilibrium equations:

$$\displaystyle \sum {{F_x} = 0\,,\,\,\,} \sum {{F_y} = 0\,,\,\,} \sum {{F_z} = 0\,,\,\,} \sum {{M_x} = 0\,,\,\,} \sum {{M_y} = 0\,,\,\,} \sum {{M_z} = 0\,\,\,}$$

One gets:


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$$ \displaystyle \frac + \frac + p\left( {x,y} \right) = 0$$ (2.19)
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$$ \displaystyle \begin{array}{l} \displaystyle {Q_x} = \frac + \frac + {N_x}\frac + {N_{xy}}\frac \\ \displaystyle {Q_y} = \frac + \frac + {N_{xy}}\frac + {N_y}\frac \\ \end{array}$$ (2.20)
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$$ \displaystyle \begin{array}{l} \displaystyle \frac + \frac = 0 \\ \displaystyle \frac + \frac = 0 \\ \end{array}$$ (2.21)
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Substituting (2.20) into (2.19) to obtain the following:


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$$ \displaystyle \frac + 2\frac + \frac = - \left[ {p\left( {x,y} \right) + {N_x}\frac + 2{N_{xy}}\frac + {N_y}\frac} \right]$$ (2.22)
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Or


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$$ \displaystyle \frac + 2\frac + \frac = \frac{1}{D}\left[ {p\left( {x,y} \right) + {N_x}\frac + 2{N_{xy}}\frac + {N_y}\frac} \right]$$ (2.23)
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Thus we have 4 unknown functions $$\displaystyle {N_x},{N_y},{N_{xy}}$$ and $$\displaystyle w$$ with three equations (2.21) and (2.23). The rest equation obtained by eliminating the tangential components of displacements $$\displaystyle u_{0}$$ and $$\displaystyle v_{0}$$ from Eqs (2.11) to get the following equation of the compatibility of deformations in the middle surface of the plate:


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$$ \displaystyle \frac + \frac - \frac = {\left( {\frac} \right)^2} - \frac\frac$$ (2.24)
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The strain-membrane force equations, according to Hooke's law (2.18) are of the form:


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$$ \displaystyle \begin{array}{l} \displaystyle {{\varepsilon }_{x}}=\frac{1}{Eh}\left( {{N}_{x}}-\nu {{N}_{y}} \right) \\ \displaystyle {{\varepsilon }_{y}}=\frac{1}{Eh}\left( {{N}_{y}}-\nu {{N}_{x}} \right) \\ \displaystyle {{\gamma }_{xy}}=\frac{Gh} \\ \end{array}$$ (2.25)
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Upon substitution of Eqs (2.25) into Eqs (2.24), we derive the following:


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$$ \displaystyle \frac{1}{Eh}\left[ \frac{\partial {{y}^{2}}}\left( {{N}_{x}}-\nu {{N}_{y}} \right)+\frac{\partial {{x}^{2}}}\left( {{N}_{y}}-\nu {{N}_{x}} \right)-2\left( 1+\nu \right)\frac{\partial x\partial y} \right]={{\left( \frac{{{\partial }^{2}}w}{\partial x\partial y} \right)}^{2}}-\frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}}$$ (2.26)
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Introducing a stress function $$\displaystyle\phi$$ for in-plane stress components, as follows:


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$$ \displaystyle {{N}_{x}}=\frac{{{\partial }^{2}}\Phi }{\partial {{y}^{2}}},\,\,{{N}_{y}}=\frac{{{\partial }^{2}}\Phi }{\partial {{x}^{2}}},\,\,\,{{N}_{xy}}=-\frac{{{\partial }^{2}}\Phi }{\partial x\partial y}$$ (2.27)
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Where $$\displaystyle\Phi=\phi h$$

It is easy to find that by doing so, Eqs (2.21) are identically satisfied. Substituting for $$\displaystyle \Phi$$ from (2.27) into Eqs (2.23) we obtain:


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$$ \displaystyle \begin{array}{l} \displaystyle {{\nabla }^{4}}\Phi \equiv \frac{{{\partial }^{4}}\Phi }{\partial {{x}^{4}}}+2\frac{{{\partial }^{4}}\Phi }{\partial {{x}^{2}}\partial {{y}^{2}}}+\frac{{{\partial }^{4}}\Phi }{\partial {{y}^{4}}}=Eh\left[ {{\left( \frac{{{\partial }^{2}}w}{\partial x\partial y} \right)}^{2}}-\frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}} \right] \\ \displaystyle {{\nabla }^{4}}w\equiv \frac{{{\partial }^{4}}w}{\partial {{x}^{4}}}+2\frac{{{\partial }^{4}}w}{\partial {{x}^{2}}\partial {{y}^{2}}}+\frac{{{\partial }^{4}}w}{\partial {{y}^{4}}}=\frac{1}{D}\left( p\left( x,y \right)+\frac{{{\partial }^{2}}\Phi }{\partial {{x}^{2}}}\frac{{{\partial }^{2}}w}{\partial {{x}^{2}}}-2\frac{{{\partial }^{2}}\Phi }{\partial x\partial y}\frac{{{\partial }^{2}}w}{\partial x\partial y}+\frac{{{\partial }^{2}}\Phi }{\partial {{y}^{2}}}\frac{{{\partial }^{2}}w}{\partial {{y}^{2}}} \right) \\ \end{array}$$ (2.28)
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These are the governing differential equations for large deflections of thin plates. Equations (2.28) are coupled, nonlinear, partial differential equations, each of fourth order. The first equation can be interpreted as a compatibility equation in terms of the stress function, while the second one is an equilibrium equation. They are often called VON KARMAN the system of equations.

For small deflection, the above P.D.Es become uncoupled and we can use displacement- based method to find solutions as following. And strain-displacement relations of middle surface obtained from (2.12) when ignoring $$\displaystyle \left( \frac{\partial w}{\partial x} \right)^{2},\left( \frac{\partial w}{\partial y} \right)^{2},\frac{\partial w}{\partial x}\frac{\partial w}{\partial y}$$:
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$$ \displaystyle \begin{array}{l} \displaystyle \varepsilon _{x}^{0}=\frac{\partial u_{0}}{\partial x} \\ \displaystyle \varepsilon _{y}^{0}=\frac{\partial v_{0}}{\partial y} \\ \displaystyle \gamma _{xy}^{0}=2\varepsilon _{xy}^{0}=\frac{\partial u_{0}}{\partial x}+\frac{\partial v_{0}}{\partial y} \\ \end{array}$$ (2.29)
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Substituting Eqs (2.29) into (2.18) yields:


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$$ \displaystyle \begin{array}{l} \displaystyle N_{x}=A\left[ \frac{\partial u_{0}}{\partial x}+\nu \frac{\partial v_{0}}{\partial y} \right] \\ \displaystyle N_{y}=A\left[ \frac{\partial v_{0}}{\partial y}+\nu \frac{\partial u_{0}}{\partial x} \right] \\ \displaystyle N_{xy}=N_{yx}=Gt\left[ \frac{\partial u_{0}}{\partial y}+\frac{\partial v_{0}}{\partial x} \right] \\ \end{array}$$ (2.30)
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Substituting Eqs (2.30) into Eqs (2.21):


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$$ \displaystyle \begin{array}{l} \displaystyle A\frac{\partial ^{2}u_{0}}{\partial x^{2}}+\nu A\frac{\partial ^{2}v_{0}}{\partial x\partial y}+Gt\frac{\partial ^{2}u_{0}}{\partial y^{2}}+Gt\frac{\partial ^{2}v_{0}}{\partial x\partial y}=0 \\ \displaystyle A\frac{\partial ^{2}v_{0}}{\partial y^{2}}+\nu A\frac{\partial ^{2}u_{0}}{\partial x\partial y}+Gt\frac{\partial ^{2}u_{0}}{\partial x\partial y}+Gt\frac{\partial ^{2}v_{0}}{\partial x^{2}}=0 \\ \end{array}$$ (2.31)
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The above Eqs can be reduced to:


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$$ \displaystyle \begin{array}{l} \displaystyle \nabla ^{4}u_{0}=0 \\ \displaystyle \nabla ^{4}v_{0}=0 \\ \end{array}$$ (2.32)
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The boundary conditions
According to

=Classical laminate theory=

The strain-displacement (geometry) relations remain those of equations (2.9), (2.13), the membrane forces, bending moments and twists and vertical shear forces remain those of equations (1.6), (1.7), (2.15). The P.D.E's equations (2.19), (2.21) involving the vertical and membrane forces, respectively, remain valid. The relationship between transverse shear forces and the moments or also accounting for the effect of membrane forces as expressed by equation (2.20) are valid irrespective of whether the response is isotropic or anisotropic. For general orthotropy : $$\displaystyle \left\{ \sigma \right\}=\left[ Q \right]\left\{ \varepsilon  \right\}$$

Thus, the $$\displaystyle \sigma -\varepsilon $$ in the $$\displaystyle k$$ orthotropic ply or lamina of a stacked laminate is (assuming $$\displaystyle \sigma_{z}=\gamma_{xz}=\gamma_{yz}=0$$)
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$$ \displaystyle {{\left\{ \begin{align} & {{\sigma }_{x}} \\ & {{\sigma }_{y}} \\ & {{\tau }_{xy}} \\ \end{align} \right\}}_{k}}={{\left[ \begin{matrix} {{Q}_{xx}} & {{Q}_{xy}} & {{Q}_{xs}} \\ {{Q}_{yx}} & {{Q}_{yy}} & {{Q}_{ys}} \\ {{Q}_{sx}} & {{Q}_{sy}} & {{Q}_{ss}} \\ \end{matrix} \right]}_{k}}{{\left\{ \begin{align} & {{\varepsilon }_{x}} \\ & {{\varepsilon }_{y}} \\ & {{\gamma }_{xy}} \\ \end{align} \right\}}_{k}}$$ (3.1)
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Where $$\displaystyle x, y$$ are general orthotropic coordinates, $$\displaystyle {{\left[ {{Q}_{ij}} \right]}_{k}}$$ are the transformed reduced stiffness of the $$\displaystyle k$$ ply which can be related to $$\displaystyle E_{11}^{k},E_{22}^{k},\nu _{12}^{k}$$ and $$\displaystyle G_{12}^{k}$$ of this $$\displaystyle k$$ ply and the angle between the 1-direction (fiber direction) of this lamina and the $$\displaystyle x$$-axis Since the strains in the $$\displaystyle k$$ ply are given by equation (2.13):
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$$ \displaystyle {{\left[ \begin{matrix} {{\varepsilon }_{x}} \\ {{\varepsilon }_{x}} \\ {{\gamma }_{xy}} \\ \end{matrix} \right]}_{k}}=\left[ \begin{matrix} \varepsilon _{x}^{0} \\ \varepsilon _{y}^{0} \\ \gamma _{xy}^{0} \\ \end{matrix} \right]+z\left[ \begin{matrix} {{\kappa }_{x}} \\ {{\kappa }_{y}} \\ {{\kappa }_{xy}} \\ \end{matrix} \right]$$ (3.2)
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$$ \displaystyle {{\left\{ \begin{align} & {{\sigma }_{x}} \\ & {{\sigma }_{y}} \\ & {{\tau }_{xy}} \\ \end{align} \right\}}_{k}}={{\left[ \begin{matrix} {{Q}_{xx}} & {{Q}_{xy}} & {{Q}_{xs}} \\ {{Q}_{yx}} & {{Q}_{yy}} & {{Q}_{ys}} \\ {{Q}_{sx}} & {{Q}_{sy}} & {{Q}_{ss}} \\ \end{matrix} \right]}_{k}}\left( \left[ \begin{matrix}  \varepsilon _{x}^{0}  \\   \varepsilon _{y}^{0}  \\   \gamma _{xy}^{0}  \\ \end{matrix} \right]+z\left[ \begin{matrix}   {{\kappa }_{x}}  \\   {{\kappa }_{y}}  \\   {{\kappa }_{xy}}  \\ \end{matrix} \right] \right)$$ (3.3)
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$$ \displaystyle \left\{ \begin{align} & {{N}_{x}} \\ & {{N}_{y}} \\ & {{N}_{xy}} \\ \end{align} \right\}=\int\limits_{\begin{smallmatrix} -h/2 \\ \left( z=-{{z}_{0}} \right) \end{smallmatrix}}^{\begin{smallmatrix} \left( z={{z}_{N}} \right) \\ h/2 \end{smallmatrix}}{\left\{ \begin{align} & {{\sigma }_{x}} \\ & {{\sigma }_{y}} \\ & {{\tau }_{xy}} \\ \end{align} \right\}}dz=\sum\limits_{k=1}^{N}{{{\int\limits_^{\left\{ \begin{align} & {{\sigma }_{x}} \\ & {{\sigma }_{y}} \\ & {{\tau }_{xy}} \\ \end{align} \right\}}}_{k}}dz}=\sum\limits_{k=1}^{N}\left( \int\limits_^{\left[ \begin{matrix}  \varepsilon _{x}^{0}  \\   \varepsilon _{y}^{0}  \\   \gamma _{xy}^{0}  \\ \end{matrix} \right]}dz+\int\limits_^{\left[ \begin{matrix}   {{\kappa }_{x}}  \\   {{\kappa }_{y}}  \\   {{\kappa }_{xy}}  \\ \end{matrix} \right]}zdz \right)$$ (3.4)
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$$ \displaystyle \left\{ \begin{align} & {{N}_{x}} \\ & {{N}_{y}} \\ & {{N}_{xy}} \\ \end{align} \right\}=\left[ \begin{matrix} {{A}_{xx}} & {{A}_{xy}} & {{A}_{xs}} \\ {{A}_{yx}} & {{A}_{yy}} & {{A}_{ys}} \\ {{A}_{sx}} & {{A}_{sy}} & {{A}_{ss}} \\ \end{matrix} \right]\left\{ \begin{align} & \varepsilon _{x}^{0} \\ & \varepsilon _{y}^{0} \\ & \gamma _{xy}^{0} \\ \end{align} \right\}+\left[ \begin{matrix} {{B}_{xx}} & {{B}_{xy}} & {{B}_{xs}} \\ {{B}_{yx}} & {{B}_{yy}} & {{B}_{ys}} \\ {{B}_{sx}} & {{B}_{sy}} & {{B}_{ss}} \\ \end{matrix} \right]\left\{ \begin{align} & {{\kappa }_{x}} \\ & {{\kappa }_{y}} \\ & {{\kappa }_{xy}} \\ \end{align} \right\}$$ (3.5)
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Where
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$$ \displaystyle {{A}_{ij}}=\sum\limits_{k=1}^{N}\int\limits_^{dz}=\sum\limits_{k=1}^{N}\underbrace{\left( {{z}_{k}}-{{z}_{k-1}} \displaystyle \right)}_=\sum\limits_{k=1}^{N}{{t}_{k}}$$ (3.6)
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Also
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$$ \displaystyle \left\{ \begin{align} & {{M}_{x}} \\ & {{M}_{y}} \\ & {{M}_{xy}} \\ \end{align} \right\}=\int\limits_{\begin{smallmatrix} -h/2 \\ \left( z=-{{z}_{0}} \right) \end{smallmatrix}}^{\begin{smallmatrix} \left( z={{z}_{N}} \right) \\ h/2 \end{smallmatrix}}{\left\{ \begin{align} & {{\sigma }_{x}} \\ & {{\sigma }_{y}} \\ & {{\tau }_{xy}} \\ \end{align} \right\}z}dz=\sum\limits_{k=1}^{N}{{{\int\limits_^{\left\{ \begin{align} & {{\sigma }_{x}} \\ & {{\sigma }_{y}} \\ & {{\tau }_{xy}} \\ \end{align} \right\}}}_{k}}zdz}=\sum\limits_{k=1}^{N}\left( \int\limits_^{\left[ \begin{matrix}  \varepsilon _{x}^{0}  \\   \varepsilon _{y}^{0}  \\   \gamma _{xy}^{0}  \\ \end{matrix} \right]z}dz+\int\limits_^{\left[ \begin{matrix}   {{\kappa }_{x}}  \\   {{\kappa }_{y}}  \\   {{\kappa }_{xy}}  \\ \end{matrix} \right]}{{z}^{2}}dz \right)$$ (3.7)
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$$ \displaystyle \left\{ \begin{align} & {{M}_{x}} \\ & {{M}_{y}} \\ & {{M}_{xy}} \\ \end{align} \right\}=\left[ \begin{matrix} {{B}_{xx}} & {{B}_{xy}} & {{B}_{xs}} \\ {{B}_{yx}} & {{B}_{yy}} & {{B}_{ys}} \\ {{B}_{sx}} & {{B}_{sy}} & {{B}_{ss}} \\ \end{matrix} \right]\left\{ \begin{align} & \varepsilon _{x}^{0} \\ & \varepsilon _{y}^{0} \\ & \gamma _{xy}^{0} \\ \end{align} \right\}+\left[ \begin{matrix} {{D}_{xx}} & {{D}_{xy}} & {{D}_{xs}} \\ {{D}_{yx}} & {{D}_{yy}} & {{D}_{ys}} \\ {{D}_{sx}} & {{D}_{sy}} & {{D}_{ss}} \\ \end{matrix} \right]\left\{ \begin{align} & {{\kappa }_{x}} \\ & {{\kappa }_{y}} \\ & {{\kappa }_{xy}} \\ \end{align} \right\}$$ (3.8)
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$$ \displaystyle \begin{align} & {{B}_{ij}}=\sum\limits_{k=1}^{N}\int\limits_^{zdz}=\frac{1}{2}\sum\limits_{k=1}^{N}{{{\left[ {{Q}_{ij}} \right]}_{k}}\left( z_{k}^{2}-z_{k-1}^{2} \right)} \\ & {{D}_{ij}}=\sum\limits_{k=1}^{N}\int\limits_^{{{z}^{2}}dz}=\frac{1}{3}\sum\limits_{k=1}^{N}{{{\left[ {{Q}_{ij}} \right]}_{k}}\left( z_{k}^{3}-z_{k-1}^{3} \right)} \\ \end{align}$$ (3.9)
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Note that unlike the isotropic case, both the in-plane forces and moments depend on membrane displacements $$\displaystyle u_{0},v_{0}$$ and transverse flexure $$\displaystyle w=w_{0}$$, i.e. there is coupling between in-plane and out-of-plane activity. The coupling vanishes if and only if $$\displaystyle B_{ij}=0$$

By determining $$\displaystyle \tau_{xz} $$ and $$\displaystyle \tau_{yz}$$ from the equilibrium equations:


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$$ \displaystyle \begin{align} & \frac{\partial {{\sigma }_{x}}}{\partial x}+\frac{\partial {{\tau }_{yx}}}{\partial y}+\frac{\partial {{\tau }_{xz}}}{\partial z}=0 \\ & \frac{\partial {{\tau }_{xy}}}{\partial x}+\frac{\partial {{\sigma }_{y}}}{\partial y}+\frac{\partial {{\tau }_{yz}}}{\partial z}=0 \\ \end{align}$$ (3.10)
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and boundary conditions $$\displaystyle \tau_{xz} = \tau_{yz} = 0 $$ at $$\displaystyle z=\pm {}^{t}\!\!\diagup\!\!{}_{2}\;$$ or from (for small deflections)
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$$ \displaystyle \begin{align} & {{Q}_{x}}=\frac{\partial {{M}_{x}}}{\partial x}+\frac{\partial {{M}_{xy}}}{\partial y} \\ & {{Q}_{y}}=\frac{\partial {{M}_{xy}}}{\partial x}+\frac{\partial {{M}_{y}}}{\partial y} \\ \end{align}$$ (3.11)
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One obtains:
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$$ \displaystyle \begin{align} & {{Q}_{x}}=\int\limits_{-h/2}^{h/2}{{{\tau }_{xz}}dz=\left[ {{B}_{11}}\frac{\partial {{x}^{2}}}+2{{B}_{16}}\frac{\partial x\partial y}+{{B}_{66}}\frac{\partial {{y}^{2}}}+{{B}_{16}}\frac{\partial {{x}^{2}}}+\left( {{B}_{12}}+{{B}_{66}} \right)\frac{\partial x\partial y}+{{B}_{26}}\frac{\partial {{x}^{2}}} \right]} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\left[ {{D}_{11}}\frac{{{\partial }^{3}}w}{\partial {{x}^{3}}}+3{{D}_{16}}\frac{{{\partial }^{3}}w}{{\partial }{x}^{2}\partial y}+\left( {{D}_{12}}+2{{D}_{66}} \right)\frac{{{\partial }^{3}}w}{\partial x\partial {{y}^{2}}}+{{D}_{26}}\frac{{{\partial }^{3}}w}{\partial {{y}^{3}}} \right] \\ & {{Q}_{y}}=\int\limits_{-h/2}^{h/2}{{{\tau }_{yz}}dz=\left[ {{B}_{16}}\frac{\partial {{x}^{2}}}+\left( {{B}_{12}}+{{B}_{66}} \right)\frac{\partial x\partial y}+{{B}_{26}}\frac{\partial {{y}^{2}}}+{{B}_{66}}\frac{\partial {{x}^{2}}}+2{{B}_{26}}\frac{\partial x\partial y}+{{B}_{22}}\frac{\partial {{x}^{2}}} \right]} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\left[ {{D}_{16}}\frac{{{\partial }^{3}}w}{\partial {{x}^{3}}}+\left( {{D}_{12}}+2{{D}_{66}} \right)\frac{{{\partial }^{3}}w}{{\partial }{x}^{2}\partial y}+3{{D}_{26}}\frac{{{\partial }^{3}}w}{\partial x\partial {{y}^{2}}}+{{D}_{22}}\frac{{{\partial }^{3}}w}{\partial {{y}^{3}}} \right] \\ \end{align}$$ (3.12)
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Expressions for membrane forces and moments can be written in a matrix form as follows:


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$$ \displaystyle \left[ \begin{align} & N \\ & M \\ \end{align} \right]=\left[ \begin{matrix} A & B \\ B & D \\ \end{matrix} \right]\left[ \begin{align} & {{\varepsilon }^{0}} \\ & \kappa \\ \end{align} \right]$$ (3.13)
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Governing equation of the classic laminate theory and general solution procedures Solving the problem of the classical laminate theory, one is aiming at finding $$\displaystyle \left\{ N \right\}$$ or $$\displaystyle \left\{ {\varepsilon }^{0} \right\}$$ and $$\displaystyle \left\{ M \right\}$$ or $$\displaystyle \left\{ \kappa \right\}$$ by solving equation (2.19) and (2.21). Using the generalized stress-strain relationship (3.13), all the generalized stresses and strains can be determined. The strains in a lamina at any a given $$\displaystyle z$$ coordinate can be found using Eqs (3.2). The strains found are in the laminate loading coordinate system. The stresses in the loading coordinate system can obtained using (3.1) and can be both transformed to the material principal coordinate system of the lamina, using the coordinate transformation matrix.

There are two typical ways of obtaining the governing equations.

Displacement-based method
If the generalized strains on the right hand side of the generalized stress-strain relationship (3.13) are replaced by displacement derivatives according to their definition in (2.12), all the generalised stresses have been expressed in terms of displacements $$\displaystyle u^{0}, v^{0}$$ and $$\displaystyle w$$. Substituting them into equilibrium equations (2.19) and (2.21), one obtains three partial differentical equations as the governing equations for three displacements $$\displaystyle u^{0}, v^{0}$$ and $$\displaystyle w$$.
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$$ \displaystyle \begin{align} & {{A}_{11}}\frac{\partial {{x}^{2}}}+2{{A}_{16}}\frac{\partial x\partial y}+{{A}_{66}}\frac{\partial {{y}^{2}}}+{{A}_{16}}\frac{\partial {{x}^{2}}}+\left( {{A}_{12}}+{{A}_{66}} \right)\frac{\partial x\partial y}+{{A}_{26}}\frac{\partial {{x}^{2}}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-{{B}_{11}}\frac{{{\partial }^{3}}w}{\partial {{x}^{3}}}-{{B}_{16}}\frac{{{\partial }^{3}}w}{\partial {{x}^{2}}\partial y}-\left( {{B}_{12}}+2{{B}_{66}} \right)\frac{{{\partial }^{3}}w}{\partial x\partial {{y}^{2}}}-{{B}_{26}}\frac{{{\partial }^{3}}w}{\partial {{y}^{3}}}=0 \\ & {{A}_{16}}\frac{\partial {{x}^{2}}}+\left( {{A}_{12}}+{{A}_{66}} \right)\frac{\partial x\partial y}+{{A}_{26}}\frac{\partial {{y}^{2}}}+{{A}_{66}}\frac{\partial {{x}^{2}}}+2{{A}_{26}}\frac{\partial x\partial y}+{{A}_{22}}\frac{\partial {{y}^{2}}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-{{B}_{16}}\frac{{{\partial }^{3}}w}{\partial {{x}^{3}}}-\left( {{B}_{12}}+2{{B}_{66}} \right)\frac{{{\partial }^{3}}w}{\partial {{x}^{2}}\partial y}-{{B}_{26}}\frac{{{\partial }^{3}}w}{\partial x\partial {{y}^{2}}}-{{B}_{22}}\frac{{{\partial }^{3}}w}{\partial {{y}^{3}}}=0 \\ & {{D}_{11}}\frac{{{\partial }^{4}}w}{\partial {{x}^{4}}}+4{{D}_{16}}\frac{{{\partial }^{4}}w}{\partial {{x}^{3}}\partial y}+2\left( {{D}_{12}}+2{{D}_{66}} \right)\frac{{{\partial }^{4}}w}{\partial {{x}^{2}}\partial {{y}^{2}}}+4{{D}_{26}}\frac{{{\partial }^{4}}w}{\partial x\partial {{y}^{3}}}+{{D}_{22}}\frac{{{\partial }^{4}}w}{\partial {{y}^{4}}} \\ & -{{B}_{11}}\frac{\partial {{x}^{3}}}-3{{B}_{16}}\frac{\partial {{x}^{2}}\partial y}-\left( {{B}_{12}}+2{{B}_{66}} \right)\frac{\partial x\partial {{y}^{2}}}-{{B}_{26}}\frac{\partial {{y}^{3}}} \\ & -{{B}_{22}}\frac{\partial {{y}^{3}}}-{{B}_{26}}\frac{\partial x\partial {{y}^{2}}}-\left( {{B}_{12}}+2{{B}_{66}} \right)\frac{\partial {{x}^{2}}\partial y}-{{B}_{16}}\frac{\partial {{x}^{3}}}=p\left( x,y \right) \\ \end{align}$$ (3.14) Solving these governing equations, one obtains displacements $$\displaystyle u^{0}, v^{0}$$ and $$\displaystyle w$$. Generalised strains can then be obtained using (2.29). The procedure as described in the previous paragraph then prevails. Finite element method is a special case following this approach. It uses the energy minimization to satisfy equilibrium instead of solving the partial differential governing equations directly.
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Stress function-based method
This approach is mostly used for analytical study. The generalized stress-strain relationship can be reorganized into the following form:
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$$ \displaystyle \left\{ \begin{align} & {{\varepsilon }^{0}} \\ & M \\ \end{align} \right\}=\left[ \begin{matrix} a & -{{b}^{T}} \\ b & d \\ \end{matrix} \right]\left\{ \begin{align} & N \\ & \kappa \\ \end{align} \right\}$$ (3.15) Where
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$$ \displaystyle \left[ a \right]={{\left[ A \right]}^{-1}},\,\,\left[ b \right]=\left[ B \right]{{\left[ A \right]}^{-1}},\,\,\left[ d \right]=\left[ D \right]-\left[ B \right]{{\left[ A \right]}^{-1}}\left[ B \right]$$ (3.16)
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Introducing the Airy stress function as (2.27) one finds the in-plane equilibrium equations (2.21) are satisfied identically. From the membrane strains as defined in (2.29) (for small deflections), a compatibility equation for these strains can be obtained as (a compatibility equation (2.24) is for large deformation):
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$$ \displaystyle \frac{{{\partial }^{2}}\varepsilon _{x}^{0}}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}\varepsilon _{y}^{0}}{\partial {{x}^{2}}}=\frac{{{\partial }^{2}}\gamma _{xy}^{0}}{\partial x\partial y}$$ (3.17)
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This should be satisfied by the membrane strains corresponding to the membrane forces derived from the Airy stress function. Replacing $$\displaystyle \left\{ N \right\}$$ and $$\displaystyle \left\{ \kappa \right\}$$ by the expression of $$\displaystyle \phi $$ and $$\displaystyle w$$ in the right hand side of (3.15), $$\displaystyle \left\{ {\varepsilon }^{0} \right\}$$  and $$\displaystyle \left\{ M \right\}$$  have been expressed in terms of $$\displaystyle \phi $$ and $$\displaystyle w$$. Then substituting such $$\displaystyle \left\{ {\varepsilon }^{0} \right\}$$ and $$\displaystyle \left\{ M \right\}$$ into equation (3.17) and (1.12), two fourth order partial differential equations are obtained as the governing equations for $$\displaystyle \phi $$ and $$\displaystyle w$$:
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$$ \displaystyle \begin{align} & {{L}_{a}}\phi +{{L}_{b}}w=0 \\ & {{L}_{b}}\phi +{{L}_{d}}w=p \\ \end{align}$$ (3.18) Where
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$$ \displaystyle \begin{align} & {{L}_{a}}={{a}_{22}}\frac{\partial {{x}^{4}}}-2{{a}_{26}}\frac{\partial {{x}^{3}}\partial y}+\left( 2{{a}_{12}}+{{a}_{66}} \right)\frac{\partial {{x}^{2}}\partial {{y}^{2}}}-2{{a}_{16}}\frac{\partial x\partial {{y}^{3}}}+{{a}_{11}}\frac{\partial {{y}^{4}}} \\ & {{L}_{b}}=-{{b}_{12}}\frac{\partial {{x}^{4}}}-\left( 2{{b}_{62}}-{{b}_{16}} \right)\frac{\partial {{x}^{3}}\partial y}-\left( {{b}_{11}}+{{b}_{22}}-2{{b}_{66}} \right)\frac{\partial {{x}^{2}}\partial {{y}^{2}}}-\left( 2{{b}_{61}}-{{b}_{26}} \right)\frac{\partial x\partial {{y}^{3}}}+{{b}_{21}}\frac{\partial {{y}^{4}}} \\ & {{L}_{d}}={{d}_{11}}\frac{\partial {{x}^{4}}}+4{{d}_{16}}\frac{\partial {{x}^{3}}\partial y}+2\left( {{d}_{12}}+2{{d}_{66}} \right)\frac{\partial {{x}^{2}}\partial {{y}^{2}}}+4{{d}_{26}}\frac{\partial x\partial {{y}^{3}}}+{{d}_{22}}\frac{\partial {{y}^{4}}} \\ \end{align}$$ (3.19)
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=Large deformation of paper under thermal loading=

Remember that if a plate is bent by lateral load only and deflections are so small, the plate middle surface was assumed to be unstrained, hence it is considered as a neutral surface. Occasionally, however, direct forces (acting in the plane of the middle surface of the plate) are applied directly at the boundaries or they arise due to temperature (or moisture) changes. The latter will be considered below in detail.

It was assumed in the foregoing discussions that the temperature of an elastic plate remains constant and has the same value at all points of the plate; whence, temperature effects were not taken into consideration. If the temperature of the plate is raised or lowered it expands or contracts, respectively. Within a certain change, such expansion or contraction, for most structural materials, is directly proportional to the change in temperature. When a free plate made of homogenous isotropic material is heated uniformly, there appear normal strains but no thermal stresses. The thermal stresses will occur in the following cases: first, if the plate experiences a nonuniform temperature field; secondly, if the displacements are prevented from occurring freely because of the restrictions placed on the boundary even with a uniform temperature; and thirdly, if the material displays anisotropy even with uniform heating-for example, if a heated plate consists of several layers of different materials.

Curl is defined as the departure form an ideal flat form of a paper caused by ambient humidity or temperature changes. The deformation of an unconstrained sheet may be expressed by three basic curl curvatures $$\displaystyle{{\kappa }_{x}},{{\kappa }_{y}},{{\kappa }_{xy}}$$ where $$\displaystyle{{\kappa }_{x}},{{\kappa }_{y}}$$ represent bending curvatures while $$\displaystyle{{\kappa }_{xy}}$$ is the twist curvature. Classical laminate theory based on linear kinematics has been used to analyze curl of papers. Such analysis generally predicts a double curvature shape of the sheets. Observations of curled papers and asymmetric composite laminates, however, indicate a single curvature, i.e a cylindrical shape. The reason for this phenomenon was examined by Hyer [??] who showed that laminate theory modeling of curvatures of unsymmetric carbon/epoxy laminates requires incorporation of nonlinear kinematics. The analysis of Hyer showed that, at a certain critical temperature, the cylindrical shape becomes the most energetically favorable.

For simulation of curl in paper incorporating effects of nonlinear kinematics, the finite element code ANSYS was used. A two-layer papers of various stacking sequence,i.e, [0/15], [0/30], [0/45] and [0/90] consisting of orthotropic linear elastic plies were considered in the numerical analyses. The paper has properties:

- Dimension: $$ \displaystyle width \times height=100mm \times 89mm $$ and thickness per layer is $$\displaystyle t=0.31mm$$

- Elastic modulus: $$ \displaystyle {{E}_{11}}=3.8GPa,\,\,{{E}_{22}}=1.7GPa,\,\,{{\nu }_{12}}=0.15,\,\,{{G}_{12}}=1GPa $$

- Thermal coefficients (unit is in/in per temperature change): $$ \displaystyle {{\alpha}_{11}}=4e-6 ,\,\,{{\alpha}_{22}}=1e-5 $$

- Moisture coefficients (unit is in/in per %RH change): $$ \displaystyle {{\beta}_{11}}=4e-5,\,\,{{\beta}_{22}}=1e-4 $$

- Density: $$\displaystyle {\rho}=603kg/m^3 $$

A gird 16x16 was used to discretize a plate.



Two different cases are considering here: the first is to increase the moisture and the second is to increase the temperature.

Curling of paper fixed at the top line due to moisture changes
The paper was moisturized from 50% RH to 90% RH after 12 hours. So the increasing of moisture is 40%. These are the results between experiments (done by J. T. Decker, an undergraduate student from University of Wisconsin,Madison) and ANSYS (done by Tuan Hoang, UW Madison)

Curling of paper fixed at the center due to temperature changes
The reference temperature is $$\displaystyle{{T}_{0}}=0$$ and then paper was heated up to $$\displaystyle{{T}_{end}}=300$$ At the critical temperature, the stiffness matrix is zero and the Ansys program crashes, hence we can obtain the value of $$\displaystyle{{T}_{critical}}=150$$.To find the possible stable cylindrical solution, a three-step solution procedure was performed:

- Firstly, four concentrated loads F=0.01N, but no temperature changes, were applied at four corners of the paper to achieve a slight curvature

- Then the temperature was increased incrementally. When the temperature $$\displaystyle{{T}\ge {T}_{critical}}$$, the concentrated loads were released.

- Finally, the temperature was increased until $$\displaystyle{{T}={T}_{end}}$$ without concentrated loads, the shape of the panel will then remain approximately cylindrical.

Curling of [0-90] (cross-ply) laminate
- Step 1: apply four small loads



- Step 2: increase the temperature from T=0 to T=170 while keeping forces



- Step 3: continue increasing the temperature but remove the forces



- Step 4: keep on increasing temperature until T=300



The final shape for layer [0-90] after heating up to 300 Celsius is shown as below:



Curling of [0/45] laminate
- Step 1: apply four small loads



- Step 2: increase the temperature from T=0 to T=170 while keeping forces



- Step 3: continue increasing the temperature but remove the forces



- Step 4: keep on increasing temperature until T=300



The final shape for layer [0-45] after heating up to 300 Celsius is shown as below:



=Nonlinear buckling analysis in ANSYS=


 * Eccentric loading or geometry/imperfection is required
 * Is more accurate then linear buckling solutions
 * Requires gradual loading to find the point of instability
 * can model "post-buckling" response of a structure

ANSYS nonlinear buckling procedure

 * 1) (OPTIONAL) CLEAR the Options of previous analyses on the same model, see Solution > Reset Options
 * 2) Solve the linear, static analysis of a structure
 * 3) Include large deformation effects NLGEOM,ON from Solution > Analysis Options
 * 4) Use Output Controls to insure data other than the Last Step is available for postprocessing (use ALL SUBSTEPS, if possible)
 * 5) Use gradual (ramped) loading and automatic time stepping from Solution > Time/Frequenc > Time and Substeps
 * 6) Control convergence behavior for number of equilibrium iteration, convergence criteria, etc. under Solution > Nonlinear
 * 7) Difficult problems (or analysis of post-buckling) may require use of the "Arc Length" convergence enhancement tool for a successful run

What to do when it will not converge

 * check the solution output and error log file for messages
 * apply the load in more, smaller substeps
 * allow more equilibrium iterations
 * try alternative material models (bilinear vs. multilinear, different hardening rules, etc.)
 * inspect the mesh in high stress regions for poor element shapes
 * re-evaluate you choice of loads (get rid of point loads, overconstrained edges)
 * look at the last good solution for any irregularities
 * re-check element options (and material properties, and real constants, and load values, and dimensions)
 * experiment with convergence enhancement tools in the code
 * deactivate one or more of the nonlinear conditions to find the problem
 * (rarely) loosen the convergence criteria - to find the problem - then tighten it up again

Commands to calculate curl of papers
For sets of command in Ansys refer to = References =