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Off–axis loading of a lamina

Under off-axis loading of a single lamina the applied stress state can be resolved to give the stresses along the laminar principal axes. The stress state of a body can be described by a stress tensor, shown schematically below, which can be related to the strain tensor by the equation:

\[{\sigma _{ij}} = {C_{ijkl}}{e_{kl}}\;\;\;(11)\]

where Cijkl is a fourth rank stiffness tensor containing 81 components.

Equation 11 can be rearranged to give:

\[{e_{ij}} = {S_{ijkl}}{\sigma _{kl}}\;\;\;(12)\]

Equation for stress tensor

where Sijkl is the compliance tensor. If the body is in equilibrium both the stress tensor and the stiffness tensor must be symmetric about the diagonal. Then writing Equation 11 as a matrix equation (See Nye 1985) and taking into account the symmetry of the composite itself, the number of independent terms in Cpq reduces to a reasonably small number. A few examples are shown here:

Matrix maths

Resolving Stresses within a Lamina

For a single lamina it is reasonable to assume that all the stresses acting are in the laminar plane, so that σ3 = τ23 = τ31 = 0. Assuming orthotropic symmetry (likely for a lamina) equation 12 becomes

\[\left[ {\begin{array}{*{20}{c}} {{\varepsilon _1}}\\ {{\varepsilon _2}}\\ {{\gamma _{12}}} \end{array}} \right] = {\left[ S \right]^{}}\left[ {\begin{array}{*{20}{c}} {{\sigma _1}}\\ {{\sigma _2}}\\ {{\tau _{12}}} \end{array}} \right] = {\left[ {\begin{array}{*{20}{c}} {{S_{11}}}&{{S_{12}}}&0\\ {{S_{21}}}&{{S_{22}}}&0\\ 0&0&{{S_{66}}} \end{array}} \right]^{}}\left[ {\begin{array}{*{20}{c}} {{\sigma _1}}\\ {{\sigma _2}}\\ {{\tau _{12}}} \end{array}} \right]\;\;\rm{(13)}\]

when stresses are applied along the principal axes of the lamina.

Clearly, when σ2 = τ12 = 0. \({S_{11}}\) = \(\frac{1}{{{E_1}}}\)

Similar considerations give

\[{S_{22}} = \frac{1}{{{E_2}}},\;\;{S_{66}} = \frac{1}{{{G_{12}}}},\;\;{S_{12}} = \frac{{ - {\nu _{12}}}}{{{E_1}}} = \frac{{ - {\nu _{21}}}}{{{E_2}}}\]


\[{\nu _{21}} = \left[ {{f^{}}{\nu _f} + {{(1 - f)}^{}}{\nu _m}} \right]\frac{{{E_2}}}{{{E_1}}}\]


ν12 =  [ f νf + (1 - f ) νm]

Click here for derivation of Poisson's ratio.

We can now find elastic constants for a lamina whose fibres are at an angle θ to the loading direction by the following resolving procedure:

Note: This animation requires Adobe Flash Player 8 and later, which can be downloaded here.

\[\left[ {\begin{array}{*{20}{c}} {{\varepsilon _x}}\\ {{\varepsilon _y}}\\ {{\gamma _{xy}}} \end{array}} \right] = {\left[ {\overline S } \right]^{}}\left[ {\begin{array}{*{20}{c}} {{\sigma _x}}\\ {{\sigma _y}}\\ {{\tau _{xy}}} \end{array}} \right]\;\;(15)\]

The result is that under an arbitrary planar loading system, the transformed compliance tensor replaces the compliance tensor in equation 13. Similar to before,

\[{E_x} = \frac{1}{{{{\overline S }_{22}}}},\;\;{E_y} = \frac{1}{{{{\overline S }_{22}}}},\;\;{G_{xy}} = \frac{1}{{{{\overline S }_{66}}}},\;\;{\nu _{xy}} = - {E_x}{\overline S _{12}},\;\;{\nu _{yx}} = - {E_y}{\overline S _{12}}\]

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