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Bar element

The bar element only bears axial strains with a multi-linear axial stiffness. A point P of the structure in the global reference is located by its position on the mean line :

$$\overrightarrow {OP} = \vec {x_o}(s)$$

Then comes the axial strain, $\varepsilon = \left \lVert\frac {\partial \vec {x_o}}{\partial s} \right \rVert -1$, and the axial force, $\vec N_x = EA \varepsilon \vec n$ , with $\vec n = \frac {\vec {x_0}}{\lVert \vec{x_o} \rVert}$.

Finally, the internal efforts contribution of a bar element is given by :

$$G_{internal}^{bar}(\vec x, \delta\vec{x}) = <\vec{N_x}.\delta\vec{x_o}>$$

Let $A_{\rho}$ be the lineic mass diagonal matrix. Then the inertia contribution writes :

$$G_{internal}^{bar} = < A_{\rho} \ddot{\vec {x_o}}.\delta\vec{x_o}>$$