Static analysis of plates March 25, 2026
1 Deflection of a plate using 3D elasticity ¶ Figure 1: Plate with a point load at the centre.
Using three-dimensional (3D) elasticity, the strain-displacement equations including van Kármán non-linear terms can be written as:
ε x x = u , x + 1 2 w , x 2 ε y y = v , y + 1 2 w , y 2 ε z z = w , z + 1 2 w , z 2 γ x y = u , y + v , x + w , x w , y γ x z = u , z + w , x + w , x w , z γ y z = v , z + w , y + w , y w , z \begin{aligned}
\varepsilon_{xx} &= u_{,x} + \frac{1}{2}w_{,x}^2 \\
\varepsilon_{yy} &= v_{,y} + \frac{1}{2}w_{,y}^2 \\
\varepsilon_{zz} &= w_{,z} + \frac{1}{2}w_{,z}^2 \\
\gamma_{xy} &= u_{,y} + v_{,x} + w_{,x} w_{,y} \\
\gamma_{xz} &= u_{,z} + w_{,x} + w_{,x} w_{,z} \\
\gamma_{yz} &= v_{,z} + w_{,y} + w_{,y} w_{,z}
\end{aligned} ε xx ε yy ε zz γ x y γ x z γ yz = u , x + 2 1 w , x 2 = v , y + 2 1 w , y 2 = w , z + 2 1 w , z 2 = u , y + v , x + w , x w , y = u , z + w , x + w , x w , z = v , z + w , y + w , y w , z The displacement field consists of:
u ( x , y , z ) v ( x , y , z ) w ( x , y , z ) \begin{aligned}
u(x,y,z) \\
v(x,y,z) \\
w(x,y,z) \\
\end{aligned} u ( x , y , z ) v ( x , y , z ) w ( x , y , z ) Using the Ritz method:
u ( x , y , z ) = S u u ˉ v ( x , y , z ) = S v u ˉ w ( x , y , z ) = S w u ˉ \begin{aligned}
u(x,y,z) = \boldsymbol{S}^u \bar{\boldsymbol{u}} \\
v(x,y,z) = \boldsymbol{S}^v \bar{\boldsymbol{u}} \\
w(x,y,z) = \boldsymbol{S}^w \bar{\boldsymbol{u}}
\end{aligned} u ( x , y , z ) = S u u ˉ v ( x , y , z ) = S v u ˉ w ( x , y , z ) = S w u ˉ The linear strains become:
ε x x = S , x u u ˉ ε y y = S , y v u ˉ γ x y = S , y u + S , x v u ˉ \begin{aligned}
\varepsilon_{xx} &= \boldsymbol{S}^u_{,x} \bar{\boldsymbol{u}} \\
\varepsilon_{yy} &= \boldsymbol{S}^v_{,y} \bar{\boldsymbol{u}} \\
\gamma_{xy} &= \boldsymbol{S}^u_{,y} + \boldsymbol{S}^v_{,x} \bar{\boldsymbol{u}}
\end{aligned} ε xx ε yy γ x y = S , x u u ˉ = S , y v u ˉ = S , y u + S , x v u ˉ and:
ε = B u ˉ \boldsymbol{\varepsilon} = \boldsymbol{B} \bar{\boldsymbol{u}} \nonumber ε = B u ˉ The stiffness matrix becomes, using the 3D constitutive matrix C \boldsymbol{C} C
K = ∭ x , y , z B ⊤ C B d x d y d z \boldsymbol{K} = \iiint_{x,y,z} \boldsymbol{B}^\top \boldsymbol{C} \boldsymbol{B} dx dy dz \nonumber K = ∭ x , y , z B ⊤ C B d x d y d z The external force vector in this example includes only a point load P P P
F e x t = P S w ∣ x = a / 2 y = b / 2 z = + h / 2 \boldsymbol{F}_{ext} = P \boldsymbol{S}^w \Big|_{\begin{matrix}x=a/2 \\ y=b/2 \\ z=+h/2\end{matrix}} \nonumber F e x t = P S w ∣ ∣ x = a /2 y = b /2 z = + h /2 Solution for the unknown Ritz coefficients:
u ˉ = K − 1 F e x t \bar{\boldsymbol{u}} = \boldsymbol{K}^{-1} \boldsymbol{F}_{ext} \nonumber u ˉ = K − 1 F e x t And the strain recovery becomes:
ε = B u ˉ \boldsymbol{\varepsilon} = \boldsymbol{B} \bar{\boldsymbol{u}} \nonumber ε = B u ˉ An example of the 3D deflection of a plate can be seen in this notebook
This example is also available through this documentation .
2 Deflection of a plate using CLPT ¶ The kinematic (strain-displacement) equations using the clasical laminated plate theory (CLPT), including van Kármán non-linear are:
ε x x = u , x − z w , x x + 1 2 w , x 2 ε y y = v , y − z w , y y + 1 2 w , y 2 γ x y = u , y + v , x − 2 z w , x y + w , x w , y \begin{aligned}
\varepsilon_{xx} &= u_{,x} - z w_{,xx} + \frac{1}{2}w_{,x}^2 \\
\varepsilon_{yy} &= v_{,y} - z w_{,yy} + \frac{1}{2}w_{,y}^2 \\
\gamma_{xy} &= u_{,y} + v_{,x} - 2z w_{,xy} + w_{,x} w_{,y}
\end{aligned} ε xx ε yy γ x y = u , x − z w , xx + 2 1 w , x 2 = v , y − z w , yy + 2 1 w , y 2 = u , y + v , x − 2 z w , x y + w , x w , y In the CLPT, the rotation of the plate is assumed constant through the thickness and field approximation (CLPT kinematics):
u ( x , y , z ) = u 0 ( x , y ) − z w , x ( x , y ) v ( x , y , z ) = v 0 ( x , y ) − z w , y ( x , y ) w ( x , y , z ) = w 0 ( x , y ) \begin{aligned}
u(x,y,z) &= u_0(x,y) - z w_{,x}(x,y) \\
v(x,y,z) &= v_0(x,y) - z w_{,y}(x,y) \\
w(x,y,z) &= w_0(x,y)
\end{aligned} u ( x , y , z ) v ( x , y , z ) w ( x , y , z ) = u 0 ( x , y ) − z w , x ( x , y ) = v 0 ( x , y ) − z w , y ( x , y ) = w 0 ( x , y ) Using the Ritz method:
u ( x , y , z ) = ( S u − z S , x w ) u ˉ v ( x , y , z ) = ( S v − z S , y w ) u ˉ w ( x , y ) = S w u ˉ \begin{aligned}
u(x,y,z) &= (\boldsymbol{S}^u - z\boldsymbol{S}^w_{,x})\bar{\boldsymbol{u}} \\
v(x,y,z) &= (\boldsymbol{S}^v - z\boldsymbol{S}^w_{,y})\bar{\boldsymbol{u}} \\
w(x,y) &= \boldsymbol{S}^w \bar{\boldsymbol{u}}
\end{aligned} u ( x , y , z ) v ( x , y , z ) w ( x , y ) = ( S u − z S , x w ) u ˉ = ( S v − z S , y w ) u ˉ = S w u ˉ The linear strains then become:
ε x x = ( S , x u − z S , x x w ) u ˉ ε y y = ( S , y v − z S , y y w ) u ˉ γ x y = ( S , y u + S , x v − 2 z S , x y w ) u ˉ \begin{aligned}
\varepsilon_{xx} &= (\boldsymbol{S}^u_{,x} - z\boldsymbol{S}^w_{,xx})\bar{\boldsymbol{u}} \\
\varepsilon_{yy} &= (\boldsymbol{S}^v_{,y} - z\boldsymbol{S}^w_{,yy})\bar{\boldsymbol{u}} \\
\gamma_{xy} &= (\boldsymbol{S}^u_{,y} + \boldsymbol{S}^v_{,x} - 2z\boldsymbol{S}^w_{,xy})\bar{\boldsymbol{u}}
\end{aligned} ε xx ε yy γ x y = ( S , x u − z S , xx w ) u ˉ = ( S , y v − z S , yy w ) u ˉ = ( S , y u + S , x v − 2 z S , x y w ) u ˉ Matrix separation into membrane (B m \boldsymbol{B}_m B m B b \boldsymbol{B}_b B b
ε = { ε x x ε y y γ x y } = ( [ S , x u S , y v S , y u + S , x v ] + z [ − S , x x w − S , y y w − 2 S , x y w ] ) u ˉ \boldsymbol{\varepsilon} = \begin{Bmatrix} \varepsilon_{xx} \\ \varepsilon_{yy} \\ \gamma_{xy} \end{Bmatrix} = \left( \begin{bmatrix} \boldsymbol{S}^u_{,x} \\ \boldsymbol{S}^v_{,y} \\ \boldsymbol{S}^u_{,y} + \boldsymbol{S}^v_{,x} \end{bmatrix} + z \begin{bmatrix} -\boldsymbol{S}^w_{,xx} \\ -\boldsymbol{S}^w_{,yy} \\ -2\boldsymbol{S}^w_{,xy} \end{bmatrix} \right) \bar{\boldsymbol{u}} \nonumber ε = ⎩ ⎨ ⎧ ε xx ε yy γ x y ⎭ ⎬ ⎫ = ⎝ ⎛ ⎣ ⎡ S , x u S , y v S , y u + S , x v ⎦ ⎤ + z ⎣ ⎡ − S , xx w − S , yy w − 2 S , x y w ⎦ ⎤ ⎠ ⎞ u ˉ ε = ( B m + z B b ) u ˉ \boldsymbol{\varepsilon} = (\boldsymbol{B}_m + z\boldsymbol{B}_b)\bar{\boldsymbol{u}} \nonumber ε = ( B m + z B b ) u ˉ The stiffness matrix becomes, using the laminate constitutive matrices A , B , D \boldsymbol{A}, \boldsymbol{B}, \boldsymbol{D} A , B , D
K = ∬ x , y ( B m ⊤ A B m ) + ( B b ⊤ D B b ) + ( B m ⊤ B B b ) + ( B b ⊤ B B m ) d x d y \boldsymbol{K} = \iint_{x,y} \left( \boldsymbol{B}_m^\top \boldsymbol{A} \boldsymbol{B}_m \right) + \left( \boldsymbol{B}_b^\top \boldsymbol{D} \boldsymbol{B}_b \right) + \left( \boldsymbol{B}_m^\top \boldsymbol{B} \boldsymbol{B}_b \right) + \left( \boldsymbol{B}_b^\top \boldsymbol{B} \boldsymbol{B}_m \right) dx dy \nonumber K = ∬ x , y ( B m ⊤ A B m ) + ( B b ⊤ D B b ) + ( B m ⊤ B B b ) + ( B b ⊤ B B m ) d x d y External force vector (point load P P P
F e x t = P S w ∣ x = a / 2 y = b / 2 \boldsymbol{F}_{ext} = P \boldsymbol{S}^w \Big|_{\begin{matrix}x=a/2 \\ y=b/2\end{matrix}} \nonumber F e x t = P S w ∣ ∣ x = a /2 y = b /2 Solution for unknown coefficients:
u ˉ = K − 1 F e x t \bar{\boldsymbol{u}} = \boldsymbol{K}^{-1} \boldsymbol{F}_{ext} \nonumber u ˉ = K − 1 F e x t An example of the deflection of a plate using the CLPT can be seen in this notebook
This example is also available through this documentation .
3 Deflection of a plate using FSDT ¶ For the first-order shear deformation theory (FSDT), the rotations of the displacement field approximation are decoupled from the gradients of w w w ϕ x \phi_x ϕ x ϕ y \phi_y ϕ y
u ( x , y , z ) = u 0 ( x , y ) + z ϕ x ( x , y ) v ( x , y , z ) = v 0 ( x , y ) + z ϕ y ( x , y ) w ( x , y , z ) = w ( x , y ) \begin{aligned}
u(x,y,z) &= u_0(x,y) + z \phi_x(x,y) \\
v(x,y,z) &= v_0(x,y) + z \phi_y(x,y) \\
w(x,y,z) &= w(x,y)
\end{aligned} u ( x , y , z ) v ( x , y , z ) w ( x , y , z ) = u 0 ( x , y ) + z ϕ x ( x , y ) = v 0 ( x , y ) + z ϕ y ( x , y ) = w ( x , y ) Using the Ritz method:
u ( x , y , z ) = ( S u + z S ϕ x ) u ˉ v ( x , y , z ) = ( S v + z S ϕ y ) u ˉ w ( x , y ) = S w u ˉ \begin{aligned}
u(x,y,z) &= (\boldsymbol{S}^u + z\boldsymbol{S}^{\phi_x})\bar{\boldsymbol{u}} \\
v(x,y,z) &= (\boldsymbol{S}^v + z\boldsymbol{S}^{\phi_y})\bar{\boldsymbol{u}} \\
w(x,y) &= \boldsymbol{S}^w \bar{\boldsymbol{u}}
\end{aligned} u ( x , y , z ) v ( x , y , z ) w ( x , y ) = ( S u + z S ϕ x ) u ˉ = ( S v + z S ϕ y ) u ˉ = S w u ˉ The kinematic (strain-displacement) equations for the FSDT, including van Kármán non-linear are:
ε x x = u , x + z ϕ x , x + 1 2 w , x 2 ε y y = v , y + z ϕ y , y + 1 2 w , y 2 γ x y = u , y + v , x + z ϕ x , y + z ϕ y , x + w , x w , y γ x z = ϕ x + w , x γ y z = ϕ y + w , y \begin{aligned}
\varepsilon_{xx} &= u_{,x} + z \phi_{x,x} + \frac{1}{2}w_{,x}^2 \\
\varepsilon_{yy} &= v_{,y} + z \phi_{y,y} + \frac{1}{2}w_{,y}^2 \\
\gamma_{xy} &= u_{,y} + v_{,x} + z\phi_{x,y} + z\phi_{y,x} + w_{,x} w_{,y} \\
\gamma_{xz} &= \phi_x + w_{,x} \\
\gamma_{yz} &= \phi_y + w_{,y}
\end{aligned} ε xx ε yy γ x y γ x z γ yz = u , x + z ϕ x , x + 2 1 w , x 2 = v , y + z ϕ y , y + 2 1 w , y 2 = u , y + v , x + z ϕ x , y + z ϕ y , x + w , x w , y = ϕ x + w , x = ϕ y + w , y The linear strains terms are:
ε x x = ( S , x u + z S , x ϕ x ) u ˉ ε y y = ( S , y v + z S , y ϕ y ) u ˉ γ x y = ( S , y u + S , x v + z S , y ϕ x + z S , x ϕ y ) u ˉ γ y z = ( S ϕ y + S , y w ) u ˉ γ x z = ( S ϕ x + S , x w ) u ˉ \begin{aligned}
\varepsilon_{xx} &= (\boldsymbol{S}^u_{,x} + z\boldsymbol{S}^{\phi_x}_{,x})\bar{\boldsymbol{u}} \\
\varepsilon_{yy} &= (\boldsymbol{S}^v_{,y} + z\boldsymbol{S}^{\phi_y}_{,y})\bar{\boldsymbol{u}} \\
\gamma_{xy} &= (\boldsymbol{S}^u_{,y} + \boldsymbol{S}^v_{,x} + z\boldsymbol{S}^{\phi_x}_{,y} + z\boldsymbol{S}^{\phi_y}_{,x})\bar{\boldsymbol{u}} \\
\gamma_{yz} &= (\boldsymbol{S}^{\phi_y} + \boldsymbol{S}^w_{,y})\bar{\boldsymbol{u}} \\
\gamma_{xz} &= (\boldsymbol{S}^{\phi_x} + \boldsymbol{S}^w_{,x})\bar{\boldsymbol{u}}
\end{aligned} ε xx ε yy γ x y γ yz γ x z = ( S , x u + z S , x ϕ x ) u ˉ = ( S , y v + z S , y ϕ y ) u ˉ = ( S , y u + S , x v + z S , y ϕ x + z S , x ϕ y ) u ˉ = ( S ϕ y + S , y w ) u ˉ = ( S ϕ x + S , x w ) u ˉ Stress resultant operator definitions:
B N = A { B ε x x ( 0 ) B ε y y ( 0 ) B γ x y ( 0 ) } + B { B ε x x ( 1 ) B ε y y ( 1 ) B γ x y ( 1 ) } B M = B { B ε x x ( 0 ) B ε y y ( 0 ) B γ x y ( 0 ) } + D { B ε x x ( 1 ) B ε y y ( 1 ) B γ x y ( 1 ) } B Q = A { B γ y z ( 0 ) B γ x z ( 0 ) } \begin{aligned}
\boldsymbol{B}^N &= \boldsymbol{A} \begin{Bmatrix} \boldsymbol{B}^{\varepsilon_{xx}^{(0)}} \\ \boldsymbol{B}^{\varepsilon_{yy}^{(0)}} \\ \boldsymbol{B}^{\gamma_{xy}^{(0)}} \end{Bmatrix} + \boldsymbol{B} \begin{Bmatrix} \boldsymbol{B}^{\varepsilon_{xx}^{(1)}} \\ \boldsymbol{B}^{\varepsilon_{yy}^{(1)}} \\ \boldsymbol{B}^{\gamma_{xy}^{(1)}} \end{Bmatrix} \\
\boldsymbol{B}^M &= \boldsymbol{B} \begin{Bmatrix} \boldsymbol{B}^{\varepsilon_{xx}^{(0)}} \\ \boldsymbol{B}^{\varepsilon_{yy}^{(0)}} \\ \boldsymbol{B}^{\gamma_{xy}^{(0)}} \end{Bmatrix} + \boldsymbol{D} \begin{Bmatrix} \boldsymbol{B}^{\varepsilon_{xx}^{(1)}} \\ \boldsymbol{B}^{\varepsilon_{yy}^{(1)}} \\ \boldsymbol{B}^{\gamma_{xy}^{(1)}} \end{Bmatrix} \\
\boldsymbol{B}^Q &= \boldsymbol{A} \begin{Bmatrix} \boldsymbol{B}^{\gamma_{yz}^{(0)}} \\ \boldsymbol{B}^{\gamma_{xz}^{(0)}} \end{Bmatrix}
\end{aligned} B N B M B Q = A ⎩ ⎨ ⎧ B ε xx ( 0 ) B ε yy ( 0 ) B γ x y ( 0 ) ⎭ ⎬ ⎫ + B ⎩ ⎨ ⎧ B ε xx ( 1 ) B ε yy ( 1 ) B γ x y ( 1 ) ⎭ ⎬ ⎫ = B ⎩ ⎨ ⎧ B ε xx ( 0 ) B ε yy ( 0 ) B γ x y ( 0 ) ⎭ ⎬ ⎫ + D ⎩ ⎨ ⎧ B ε xx ( 1 ) B ε yy ( 1 ) B γ x y ( 1 ) ⎭ ⎬ ⎫ = A { B γ yz ( 0 ) B γ x z ( 0 ) } For the FSDT, the variation of the strain energy is:
δ U = ∬ x y ( N ⊤ { δ ε x x ( 0 ) δ ε y y ( 0 ) δ γ x y ( 0 ) } + M ⊤ { δ ε x x ( 1 ) δ ε y y ( 1 ) δ γ x y ( 1 ) } + Q ⊤ { δ γ y z ( 0 ) δ γ x z ( 0 ) } ) d x d y \delta U = \iint_{xy} \left( \boldsymbol{N}^\top \begin{Bmatrix} \delta \varepsilon_{xx}^{(0)} \\ \delta \varepsilon_{yy}^{(0)} \\ \delta \gamma_{xy}^{(0)} \end{Bmatrix} + \boldsymbol{M}^\top \begin{Bmatrix} \delta \varepsilon_{xx}^{(1)} \\ \delta \varepsilon_{yy}^{(1)} \\ \delta \gamma_{xy}^{(1)} \end{Bmatrix} + \boldsymbol{Q}^\top \begin{Bmatrix} \delta \gamma_{yz}^{(0)} \\ \delta \gamma_{xz}^{(0)} \end{Bmatrix} \right) dx dy \nonumber δ U = ∬ x y ⎝ ⎛ N ⊤ ⎩ ⎨ ⎧ δ ε xx ( 0 ) δ ε yy ( 0 ) δ γ x y ( 0 ) ⎭ ⎬ ⎫ + M ⊤ ⎩ ⎨ ⎧ δ ε xx ( 1 ) δ ε yy ( 1 ) δ γ x y ( 1 ) ⎭ ⎬ ⎫ + Q ⊤ { δ γ yz ( 0 ) δ γ x z ( 0 ) } ⎠ ⎞ d x d y Such that the stiffness matrix becomes, using the laminate constitutive matrices A , B , D \boldsymbol{A}, \boldsymbol{B}, \boldsymbol{D} A , B , D
K = ∬ x y ( B N ⊤ { B ε x x ( 0 ) B ε y y ( 0 ) B γ x y ( 0 ) } + B M ⊤ { B ε x x ( 1 ) B ε y y ( 1 ) B γ x y ( 1 ) } + B Q ⊤ { B γ y z ( 0 ) B γ x z ( 0 ) } ) d x d y \boldsymbol{K} = \iint_{xy} \left( {\boldsymbol{B}^N}^\top \begin{Bmatrix} \boldsymbol{B}^{\varepsilon_{xx}^{(0)}} \\ \boldsymbol{B}^{\varepsilon_{yy}^{(0)}} \\ \boldsymbol{B}^{\gamma_{xy}^{(0)}} \end{Bmatrix} + {\boldsymbol{B}^M}^\top \begin{Bmatrix} \boldsymbol{B}^{\varepsilon_{xx}^{(1)}} \\ \boldsymbol{B}^{\varepsilon_{yy}^{(1)}} \\ \boldsymbol{B}^{\gamma_{xy}^{(1)}} \end{Bmatrix} + {\boldsymbol{B}^Q}^\top \begin{Bmatrix} \boldsymbol{B}^{\gamma_{yz}^{(0)}} \\ \boldsymbol{B}^{\gamma_{xz}^{(0)}} \end{Bmatrix} \right) dx dy \nonumber K = ∬ x y ⎝ ⎛ B N ⊤ ⎩ ⎨ ⎧ B ε xx ( 0 ) B ε yy ( 0 ) B γ x y ( 0 ) ⎭ ⎬ ⎫ + B M ⊤ ⎩ ⎨ ⎧ B ε xx ( 1 ) B ε yy ( 1 ) B γ x y ( 1 ) ⎭ ⎬ ⎫ + B Q ⊤ { B γ yz ( 0 ) B γ x z ( 0 ) } ⎠ ⎞ d x d y The external force vector is then defined as:
F e x t = P S w ∣ x = a / 2 y = b / 2 \boldsymbol{F}_{ext} = P \boldsymbol{S}^w \Big|_{\begin{matrix}x=a/2 \\ y=b/2\end{matrix}} \nonumber F e x t = P S w ∣ ∣ x = a /2 y = b /2 Which can be solved for the Ritz coefficients with:
u ˉ = K − 1 F e x t \bar{\boldsymbol{u}} = \boldsymbol{K}^{-1} \boldsymbol{F}_{ext} \nonumber u ˉ = K − 1 F e x t An example of the deflection of a plate using the FSDT can be seen in this notebook
This example is also available through this documentation .
4 Deflection of a plate using the TSDT ¶ The third-order shear deformation theory enforces zero transverse shear stresses and strains at the plate faces, z = − h / 2 z = -h/2 z = − h /2 z = + h / 2 z = +h/2 z = + h /2 Reddy, 2003
u ( x , y , z ) = u 0 ( x , y ) + z ϕ x ( x , y ) − 4 3 h 2 z 3 ( ϕ x ( x , y ) + w , x ( x , y ) ) v ( x , y , z ) = v 0 ( x , y ) + z ϕ y ( x , y ) − 4 3 h 2 z 3 ( ϕ y ( x , y ) + w , y ( x , y ) ) w ( x , y , z ) = w ( x , y ) \begin{aligned}
u(x,y,z) &= u_0(x,y) + z \phi_x(x,y) - \frac{4}{3h^2}z^3 \left( \phi_x(x,y) + w_{,x}(x,y) \right) \\
v(x,y,z) &= v_0(x,y) + z \phi_y(x,y) - \frac{4}{3h^2}z^3 \left( \phi_y(x,y) + w_{,y}(x,y) \right) \\
w(x,y,z) &= w(x,y)
\end{aligned} u ( x , y , z ) v ( x , y , z ) w ( x , y , z ) = u 0 ( x , y ) + z ϕ x ( x , y ) − 3 h 2 4 z 3 ( ϕ x ( x , y ) + w , x ( x , y ) ) = v 0 ( x , y ) + z ϕ y ( x , y ) − 3 h 2 4 z 3 ( ϕ y ( x , y ) + w , y ( x , y ) ) = w ( x , y ) Using the Ritz method:
u ( x , y , z ) = ( S u + z S ϕ x + z 3 ( − 4 3 h 2 ) ( S ϕ x + S , x w ) ) u ˉ v ( x , y , z ) = ( S v + z S ϕ y + z 3 ( − 4 3 h 2 ) ( S ϕ y + S , y w ) ) u ˉ w ( x , y ) = S w u ˉ \begin{aligned}
u(x,y,z) &= \left( \boldsymbol{S}^u + z\boldsymbol{S}^{\phi_x} + z^3\left(-\frac{4}{3h^2}\right) (\boldsymbol{S}^{\phi_x} + \boldsymbol{S}^w_{,x}) \right)\bar{\boldsymbol{u}} \\
v(x,y,z) &= \left( \boldsymbol{S}^v + z\boldsymbol{S}^{\phi_y} + z^3\left(-\frac{4}{3h^2}\right) (\boldsymbol{S}^{\phi_y} + \boldsymbol{S}^w_{,y}) \right)\bar{\boldsymbol{u}} \\
w(x,y) &= \boldsymbol{S}^w \bar{\boldsymbol{u}}
\end{aligned} u ( x , y , z ) v ( x , y , z ) w ( x , y ) = ( S u + z S ϕ x + z 3 ( − 3 h 2 4 ) ( S ϕ x + S , x w ) ) u ˉ = ( S v + z S ϕ y + z 3 ( − 3 h 2 4 ) ( S ϕ y + S , y w ) ) u ˉ = S w u ˉ Strain-displacement equations, including van K'arm\an non-linear terms:
ε x x = u , x + 1 2 w , x 2 + z ϕ x , x + z 3 ( − 4 3 h 2 ) ( ϕ x , x + w , x x ) ε y y = v , y + 1 2 w , y 2 + z ϕ y , y + z 3 ( − 4 3 h 2 ) ( ϕ y , y + w , y y ) γ x y = u , y + v , x + w , x w , y + z ϕ x , y + z ϕ y , x + z 3 ( − 4 3 h 2 ) ( ϕ x , y + ϕ y , x + 2 w , x y ) γ x z = ϕ x + w , x + z 2 ( − 4 h 2 ) ( ϕ x + w , x ) γ y z = ϕ y + w , y + z 2 ( − 4 h 2 ) ( ϕ y + w , y ) \begin{aligned}
\varepsilon_{xx} &= u_{,x} + \frac{1}{2}w_{,x}^2 + z\phi_{x,x} + z^3\left(-\frac{4}{3h^2}\right) (\phi_{x,x} + w_{,xx}) \\
\varepsilon_{yy} &= v_{,y} + \frac{1}{2}w_{,y}^2 + z\phi_{y,y} + z^3\left(-\frac{4}{3h^2}\right) (\phi_{y,y} + w_{,yy}) \\
\gamma_{xy} &= u_{,y} + v_{,x} + w_{,x} w_{,y} + z\phi_{x,y} + z\phi_{y,x} + z^3\left(-\frac{4}{3h^2}\right) (\phi_{x,y} + \phi_{y,x} + 2w_{,xy}) \\
\gamma_{xz} &= \phi_x + w_{,x} + z^2\left(-\frac{4}{h^2}\right) (\phi_x + w_{,x}) \\
\gamma_{yz} &= \phi_y + w_{,y} + z^2\left(-\frac{4}{h^2}\right) (\phi_y + w_{,y})
\end{aligned} ε xx ε yy γ x y γ x z γ yz = u , x + 2 1 w , x 2 + z ϕ x , x + z 3 ( − 3 h 2 4 ) ( ϕ x , x + w , xx ) = v , y + 2 1 w , y 2 + z ϕ y , y + z 3 ( − 3 h 2 4 ) ( ϕ y , y + w , yy ) = u , y + v , x + w , x w , y + z ϕ x , y + z ϕ y , x + z 3 ( − 3 h 2 4 ) ( ϕ x , y + ϕ y , x + 2 w , x y ) = ϕ x + w , x + z 2 ( − h 2 4 ) ( ϕ x + w , x ) = ϕ y + w , y + z 2 ( − h 2 4 ) ( ϕ y + w , y ) The linear strains then become:
ε x x = ( S , x u + z S , x ϕ x + z 3 ( − 4 3 h 2 ) ( S , x ϕ x + S , x x w ) ) u ˉ ε y y = ( S , y v + z S , y ϕ y + z 3 ( − 4 3 h 2 ) ( S , y ϕ y + S , y y w ) ) u ˉ γ x y = ( S , y u + S , x v + z S , y ϕ x + z S , x ϕ y + z 3 ( − 4 3 h 2 ) ( S , y ϕ x + S , x ϕ y + 2 S , x y w ) ) u ˉ γ y z = ( S ϕ y + S , y w + z 2 ( − 4 h 2 ) ( S ϕ y + S , y w ) ) u ˉ γ x z = ( S ϕ x + S , x w + z 2 ( − 4 h 2 ) ( S ϕ x + S , x w ) ) u ˉ \begin{aligned}
\varepsilon_{xx} &= \left( \boldsymbol{S}^u_{,x} + z\boldsymbol{S}^{\phi_x}_{,x} + z^3\left(-\frac{4}{3h^2}\right) (\boldsymbol{S}^{\phi_x}_{,x} + \boldsymbol{S}^w_{,xx}) \right)\bar{\boldsymbol{u}} \\
\varepsilon_{yy} &= \left( \boldsymbol{S}^v_{,y} + z\boldsymbol{S}^{\phi_y}_{,y} + z^3\left(-\frac{4}{3h^2}\right) (\boldsymbol{S}^{\phi_y}_{,y} + \boldsymbol{S}^w_{,yy}) \right)\bar{\boldsymbol{u}} \\
\gamma_{xy} &= \left( \boldsymbol{S}^u_{,y} + \boldsymbol{S}^v_{,x} + z\boldsymbol{S}^{\phi_x}_{,y} + z\boldsymbol{S}^{\phi_y}_{,x} + z^3\left(-\frac{4}{3h^2}\right) (\boldsymbol{S}^{\phi_x}_{,y} + \boldsymbol{S}^{\phi_y}_{,x} + 2\boldsymbol{S}^w_{,xy}) \right)\bar{\boldsymbol{u}} \\
\gamma_{yz} &= \left( \boldsymbol{S}^{\phi_y} + \boldsymbol{S}^w_{,y} + z^2\left(-\frac{4}{h^2}\right) (\boldsymbol{S}^{\phi_y} + \boldsymbol{S}^w_{,y}) \right)\bar{\boldsymbol{u}} \\
\gamma_{xz} &= \left( \boldsymbol{S}^{\phi_x} + \boldsymbol{S}^w_{,x} + z^2\left(-\frac{4}{h^2}\right) (\boldsymbol{S}^{\phi_x} + \boldsymbol{S}^w_{,x}) \right)\bar{\boldsymbol{u}}
\end{aligned} ε xx ε yy γ x y γ yz γ x z = ( S , x u + z S , x ϕ x + z 3 ( − 3 h 2 4 ) ( S , x ϕ x + S , xx w ) ) u ˉ = ( S , y v + z S , y ϕ y + z 3 ( − 3 h 2 4 ) ( S , y ϕ y + S , yy w ) ) u ˉ = ( S , y u + S , x v + z S , y ϕ x + z S , x ϕ y + z 3 ( − 3 h 2 4 ) ( S , y ϕ x + S , x ϕ y + 2 S , x y w ) ) u ˉ = ( S ϕ y + S , y w + z 2 ( − h 2 4 ) ( S ϕ y + S , y w ) ) u ˉ = ( S ϕ x + S , x w + z 2 ( − h 2 4 ) ( S ϕ x + S , x w ) ) u ˉ Stress resultant operator definitions:
B N = A { B ε x x ( 0 ) B ε y y ( 0 ) B γ x y ( 0 ) } + B { B ε x x ( 1 ) B ε y y ( 1 ) B γ x y ( 1 ) } + E { B ε x x ( 3 ) B ε y y ( 3 ) B γ x y ( 3 ) } B M = B { B ε x x ( 0 ) B ε y y ( 0 ) B γ x y ( 0 ) } + D { B ε x x ( 1 ) B ε y y ( 1 ) B γ x y ( 1 ) } + F { B ε x x ( 3 ) B ε y y ( 3 ) B γ x y ( 3 ) } B P = E { B ε x x ( 0 ) B ε y y ( 0 ) B γ x y ( 0 ) } + F { B ε x x ( 1 ) B ε y y ( 1 ) B γ x y ( 1 ) } + H { B ε x x ( 3 ) B ε y y ( 3 ) B γ x y ( 3 ) } B Q = A { B γ y z ( 0 ) B γ x z ( 0 ) } + D { B γ y z ( 2 ) B γ x z ( 2 ) } B R = D { B γ y z ( 0 ) B γ x z ( 0 ) } + F { B γ y z ( 2 ) B γ x z ( 2 ) } \begin{aligned}
\boldsymbol{B}^N &= \boldsymbol{A} \begin{Bmatrix} \boldsymbol{B}^{\varepsilon_{xx}^{(0)}} \\ \boldsymbol{B}^{\varepsilon_{yy}^{(0)}} \\ \boldsymbol{B}^{\gamma_{xy}^{(0)}} \end{Bmatrix} + \boldsymbol{B} \begin{Bmatrix} \boldsymbol{B}^{\varepsilon_{xx}^{(1)}} \\ \boldsymbol{B}^{\varepsilon_{yy}^{(1)}} \\ \boldsymbol{B}^{\gamma_{xy}^{(1)}} \end{Bmatrix} + \boldsymbol{E} \begin{Bmatrix} \boldsymbol{B}^{\varepsilon_{xx}^{(3)}} \\ \boldsymbol{B}^{\varepsilon_{yy}^{(3)}} \\ \boldsymbol{B}^{\gamma_{xy}^{(3)}} \end{Bmatrix}
\\
\boldsymbol{B}^M &= \boldsymbol{B} \begin{Bmatrix} \boldsymbol{B}^{\varepsilon_{xx}^{(0)}} \\ \boldsymbol{B}^{\varepsilon_{yy}^{(0)}} \\ \boldsymbol{B}^{\gamma_{xy}^{(0)}} \end{Bmatrix} + \boldsymbol{D} \begin{Bmatrix} \boldsymbol{B}^{\varepsilon_{xx}^{(1)}} \\ \boldsymbol{B}^{\varepsilon_{yy}^{(1)}} \\ \boldsymbol{B}^{\gamma_{xy}^{(1)}} \end{Bmatrix} + \boldsymbol{F} \begin{Bmatrix} \boldsymbol{B}^{\varepsilon_{xx}^{(3)}} \\ \boldsymbol{B}^{\varepsilon_{yy}^{(3)}} \\ \boldsymbol{B}^{\gamma_{xy}^{(3)}} \end{Bmatrix}
\\
\boldsymbol{B}^P &= \boldsymbol{E} \begin{Bmatrix} \boldsymbol{B}^{\varepsilon_{xx}^{(0)}} \\ \boldsymbol{B}^{\varepsilon_{yy}^{(0)}} \\ \boldsymbol{B}^{\gamma_{xy}^{(0)}} \end{Bmatrix} + \boldsymbol{F} \begin{Bmatrix} \boldsymbol{B}^{\varepsilon_{xx}^{(1)}} \\ \boldsymbol{B}^{\varepsilon_{yy}^{(1)}} \\ \boldsymbol{B}^{\gamma_{xy}^{(1)}} \end{Bmatrix} + \boldsymbol{H} \begin{Bmatrix} \boldsymbol{B}^{\varepsilon_{xx}^{(3)}} \\ \boldsymbol{B}^{\varepsilon_{yy}^{(3)}} \\ \boldsymbol{B}^{\gamma_{xy}^{(3)}} \end{Bmatrix}
\\
\boldsymbol{B}^Q &= \boldsymbol{A} \begin{Bmatrix} \boldsymbol{B}^{\gamma_{yz}^{(0)}} \\ \boldsymbol{B}^{\gamma_{xz}^{(0)}} \end{Bmatrix} + \boldsymbol{D} \begin{Bmatrix} \boldsymbol{B}^{\gamma_{yz}^{(2)}} \\ \boldsymbol{B}^{\gamma_{xz}^{(2)}} \end{Bmatrix}
\\
\boldsymbol{B}^R &= \boldsymbol{D} \begin{Bmatrix} \boldsymbol{B}^{\gamma_{yz}^{(0)}} \\ \boldsymbol{B}^{\gamma_{xz}^{(0)}} \end{Bmatrix} + \boldsymbol{F} \begin{Bmatrix} \boldsymbol{B}^{\gamma_{yz}^{(2)}} \\ \boldsymbol{B}^{\gamma_{xz}^{(2)}} \end{Bmatrix}
\end{aligned} B N B M B P B Q B R = A ⎩ ⎨ ⎧ B ε xx ( 0 ) B ε yy ( 0 ) B γ x y ( 0 ) ⎭ ⎬ ⎫ + B ⎩ ⎨ ⎧ B ε xx ( 1 ) B ε yy ( 1 ) B γ x y ( 1 ) ⎭ ⎬ ⎫ + E ⎩ ⎨ ⎧ B ε xx ( 3 ) B ε yy ( 3 ) B γ x y ( 3 ) ⎭ ⎬ ⎫ = B ⎩ ⎨ ⎧ B ε xx ( 0 ) B ε yy ( 0 ) B γ x y ( 0 ) ⎭ ⎬ ⎫ + D ⎩ ⎨ ⎧ B ε xx ( 1 ) B ε yy ( 1 ) B γ x y ( 1 ) ⎭ ⎬ ⎫ + F ⎩ ⎨ ⎧ B ε xx ( 3 ) B ε yy ( 3 ) B γ x y ( 3 ) ⎭ ⎬ ⎫ = E ⎩ ⎨ ⎧ B ε xx ( 0 ) B ε yy ( 0 ) B γ x y ( 0 ) ⎭ ⎬ ⎫ + F ⎩ ⎨ ⎧ B ε xx ( 1 ) B ε yy ( 1 ) B γ x y ( 1 ) ⎭ ⎬ ⎫ + H ⎩ ⎨ ⎧ B ε xx ( 3 ) B ε yy ( 3 ) B γ x y ( 3 ) ⎭ ⎬ ⎫ = A { B γ yz ( 0 ) B γ x z ( 0 ) } + D { B γ yz ( 2 ) B γ x z ( 2 ) } = D { B γ yz ( 0 ) B γ x z ( 0 ) } + F { B γ yz ( 2 ) B γ x z ( 2 ) } For the TSDT, the variation of the strain energy is:
δ U = ∬ x y ( N ⊤ { δ ε x x ( 0 ) δ ε y y ( 0 ) δ γ x y ( 0 ) } + M ⊤ { δ ε x x ( 1 ) δ ε y y ( 1 ) δ γ x y ( 1 ) } + P ⊤ { δ ε x x ( 3 ) δ ε y y ( 3 ) δ γ x y ( 3 ) } + Q ⊤ { δ γ y z ( 0 ) δ γ x z ( 0 ) } + R ⊤ { δ γ y z ( 2 ) δ γ x z ( 2 ) } ) d x d y \delta U = \iint_{xy} \left( \boldsymbol{N}^\top \begin{Bmatrix} \delta\varepsilon_{xx}^{(0)} \\ \delta\varepsilon_{yy}^{(0)} \\ \delta\gamma_{xy}^{(0)} \end{Bmatrix} + \boldsymbol{M}^\top \begin{Bmatrix} \delta\varepsilon_{xx}^{(1)} \\ \delta\varepsilon_{yy}^{(1)} \\ \delta\gamma_{xy}^{(1)} \end{Bmatrix} + \boldsymbol{P}^\top \begin{Bmatrix} \delta\varepsilon_{xx}^{(3)} \\ \delta\varepsilon_{yy}^{(3)} \\ \delta\gamma_{xy}^{(3)} \end{Bmatrix} + \boldsymbol{Q}^\top \begin{Bmatrix} \delta\gamma_{yz}^{(0)} \\ \delta\gamma_{xz}^{(0)} \end{Bmatrix} + \boldsymbol{R}^\top \begin{Bmatrix} \delta\gamma_{yz}^{(2)} \\ \delta\gamma_{xz}^{(2)} \end{Bmatrix} \right) dx dy \nonumber δ U = ∬ x y ⎝ ⎛ N ⊤ ⎩ ⎨ ⎧ δ ε xx ( 0 ) δ ε yy ( 0 ) δ γ x y ( 0 ) ⎭ ⎬ ⎫ + M ⊤ ⎩ ⎨ ⎧ δ ε xx ( 1 ) δ ε yy ( 1 ) δ γ x y ( 1 ) ⎭ ⎬ ⎫ + P ⊤ ⎩ ⎨ ⎧ δ ε xx ( 3 ) δ ε yy ( 3 ) δ γ x y ( 3 ) ⎭ ⎬ ⎫ + Q ⊤ { δ γ yz ( 0 ) δ γ x z ( 0 ) } + R ⊤ { δ γ yz ( 2 ) δ γ x z ( 2 ) } ⎠ ⎞ d x d y Such that the stiffness matrix becomes, using the laminate constitutive matrices A , B , D , E , F , G \boldsymbol{A}, \boldsymbol{B}, \boldsymbol{D}, \boldsymbol{E}, \boldsymbol{F}, \boldsymbol{G} A , B , D , E , F , G
K = ∬ x y ( B N ⊤ { B ε x x ( 0 ) B ε y y ( 0 ) B γ x y ( 0 ) } + B M ⊤ { B ε x x ( 1 ) B ε y y ( 1 ) B γ x y ( 1 ) } + B P ⊤ { B ε x x ( 3 ) B ε y y ( 3 ) B γ x y ( 3 ) } + B Q ⊤ { B γ y z ( 0 ) B γ x z ( 0 ) } + B R ⊤ { B γ y z ( 2 ) B γ x z ( 2 ) } ) d x d y \boldsymbol{K} = \iint_{xy} \left( {\boldsymbol{B}^N}^\top \begin{Bmatrix} \boldsymbol{B}^{\varepsilon_{xx}^{(0)}} \\ \boldsymbol{B}^{\varepsilon_{yy}^{(0)}} \\ \boldsymbol{B}^{\gamma_{xy}^{(0)}} \end{Bmatrix} + {\boldsymbol{B}^M}^\top \begin{Bmatrix} \boldsymbol{B}^{\varepsilon_{xx}^{(1)}} \\ \boldsymbol{B}^{\varepsilon_{yy}^{(1)}} \\ \boldsymbol{B}^{\gamma_{xy}^{(1)}} \end{Bmatrix} + {\boldsymbol{B}^P}^\top \begin{Bmatrix} \boldsymbol{B}^{\varepsilon_{xx}^{(3)}} \\ \boldsymbol{B}^{\varepsilon_{yy}^{(3)}} \\ \boldsymbol{B}^{\gamma_{xy}^{(3)}} \end{Bmatrix} + {\boldsymbol{B}^Q}^\top \begin{Bmatrix} \boldsymbol{B}^{\gamma_{yz}^{(0)}} \\ \boldsymbol{B}^{\gamma_{xz}^{(0)}} \end{Bmatrix} + {\boldsymbol{B}^R}^\top \begin{Bmatrix} \boldsymbol{B}^{\gamma_{yz}^{(2)}} \\ \boldsymbol{B}^{\gamma_{xz}^{(2)}} \end{Bmatrix} \right) dx dy \nonumber K = ∬ x y ⎝ ⎛ B N ⊤ ⎩ ⎨ ⎧ B ε xx ( 0 ) B ε yy ( 0 ) B γ x y ( 0 ) ⎭ ⎬ ⎫ + B M ⊤ ⎩ ⎨ ⎧ B ε xx ( 1 ) B ε yy ( 1 ) B γ x y ( 1 ) ⎭ ⎬ ⎫ + B P ⊤ ⎩ ⎨ ⎧ B ε xx ( 3 ) B ε yy ( 3 ) B γ x y ( 3 ) ⎭ ⎬ ⎫ + B Q ⊤ { B γ yz ( 0 ) B γ x z ( 0 ) } + B R ⊤ { B γ yz ( 2 ) B γ x z ( 2 ) } ⎠ ⎞ d x d y External force vector:
F e x t = P S w ∣ x = a / 2 y = b / 2 \boldsymbol{F}_{ext} = P \boldsymbol{S}^w \Big|_{\begin{matrix}x=a/2 \\ y=b/2\end{matrix}} \nonumber F e x t = P S w ∣ ∣ x = a /2 y = b /2 Which can be solved for the Ritz coefficients:
u ˉ = K − 1 F e x t \bar{\boldsymbol{u}} = \boldsymbol{K}^{-1} \boldsymbol{F}_{ext} \nonumber u ˉ = K − 1 F e x t An example of the deflection of a plate using the TSDT can be seen in this notebook
This example is also available through this documentation .
Reddy, J. N. (2003). Mechanics of Laminated Composite Plates and Shells . CRC Press. 10.1201/b12409 