Integration along the laminate thickness

II.1.11.4 Integration along the laminate thickness

One considers separately the laminate thermal in-plane conductivity, thermal out-of-plane conductivity and thermal capacity.

II.1.11.4.1 In-plane conductivity

To calculate the laminate in-plane thermal conductivity properties, one assumes that temperature is constant along the laminate thickness. Consequently, the temperature gradient does not depend on $z$ . The thermal flux ${\{q\}}_{x y}$ however depends on $z$ because the thermal conductivity does:

{\{q\}}_{x y} (z) = {[λ^{T}]}_{x y} (z) \cdot {\{\nabla T\}}_{x y} .

The laminate in-plane thermal conductivity is calculated as follows:

\begin{align} {\{Q\}}_{x y} & = \int_{- h ∕ 2}^{- h ∕ 2} {\{q\}}_{x y} (z) d z, \\ = \int_{- h ∕ 2}^{- h ∕ 2} {[λ^{T}]}_{x y} (z) \cdot {\{\nabla T\}}_{x y} d z, \\ = (\int_{- h ∕ 2}^{- h ∕ 2} {[λ^{T}]}_{x y} (z) d z) \cdot {\{\nabla T\}}_{x y}, \\ = (\int_{- h ∕ 2}^{- h ∕ 2} {[S_{+}]}_{(θ)} (z) \cdot {[λ^{T}]}_{12} (z) \cdot {[S_{-}]}_{(θ)} (z) d z) \cdot {\{\nabla T\}}_{x y}, \\ = {[Λ^{T}]}_{x y} \cdot {\{\nabla T\}}_{x y} . \end{align}

In the previous equation, one introduced the laminate in-plane thermal conductivity:

\begin{align} {[Λ^{T}]}_{x y} & = \int_{- h ∕ 2}^{- h ∕ 2} {[S_{+}]}_{(θ)} (z) \cdot {[λ^{T}]}_{12} (z) \cdot {[S_{-}]}_{(θ)} (z) d z, \\ = \sum_{k = 1}^{N} {[S_{+}]}_{(θ_{k})} \cdot {[λ_{k}^{T}]}_{12} \cdot {[S_{-}]}_{(θ_{k})} e_{k}, \end{align}

where $e_{k}$ is the thickness of ply $k$ .

II.1.11.4.2 Out-of-plane conductivity

One assumes that out-of-plane thermal flux is constant across laminate thickness. Then, as thermal conductivity depends on $z$ , so will the out-of-plane gradient of temperature:

\nabla T_{z} = \frac{q_{z}}{λ_{33}^{T}} .

The integration across the thickness gives the difference of temperature between upper and lower laminate surfaces:

\begin{align} T_{sup} - T_{inf} & = \int_{- h ∕ 2}^{- h ∕ 2} \nabla T_{z} d z, \\ = \int_{- h ∕ 2}^{- h ∕ 2} \frac{q_{z}}{λ_{33}^{T}} d z, \\ = (\int_{- h ∕ 2}^{- h ∕ 2} \frac{1}{λ_{33}^{T}} d z) q_{z}, \\ = R_{33}^{T} q_{z} . \end{align}

In previous expression, one introduced the out-of-plane thermal resistance:

\begin{align} R_{33}^{T} & = \int_{- h ∕ 2}^{- h ∕ 2} \frac{1}{λ_{33}^{T}} d z, \\ = \sum_{k = 1}^{N} \frac{1}{{(λ_{33}^{T})}_{k}} e_{k} . \end{align}

II.1.11.4.3 Thermal capacity

To estimate the laminate thermal capacity, one again assumes a temperature constant across the laminate thickness. Then, the heat energy stored per unit of surface is:

\begin{align} U & = \int_{- h ∕ 2}^{- h ∕ 2} ρ C_{p} T d z, \\ = (\int_{- h ∕ 2}^{- h ∕ 2} ρ C_{p} d z) T, \\ = {(ρ C_{p} h)}_{lam} T . \end{align}

In previous expression, one introduced the surfacic thermal capacity

\begin{align} {(ρ C_{p} h)}_{lam} & = \int_{- h ∕ 2}^{- h ∕ 2} ρ C_{p} d z, \\ = \sum_{k = 1}^{N} {(ρ C_{p})}_{k} e_{k} . \end{align}

Note that in Nastran, when a thermal material MAT4 or MAT5 is defined, the density $ρ$ and the heat capacity per unit of mass $C_{p}$ are defined separately. So it is the responsibility of the user to select appropriate values for these two quantities. Also, in Nastran, the thickness is defined separately in the PSHELL property card. (PCOMP or PCOMPG cards do not accept thermal materials.)

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