5. Linearization

In balance:

\(W(u,\delta u)=0\) [7]

If, within one tolerance, the previous equation is not satisfied, we look for a \(\Delta u\) correction of \(u\) such that:

\(W(u,\delta u)+\text{DW}(u,\delta u)\text{.}\Delta u=0\)

\(\text{DW}(u,\delta u)\text{.}\Delta u\) being the directional derivative of \(\text{W}(u,\delta u)\) in the \(\Delta u\) [bib] and [bib] direction. According to [eq], we obviously have:

\(\text{DN}\text{.}\Delta u={E}_{a}A\mathrm{Dg}\text{.}\Delta u={E}_{a}AB\text{.}\Delta u\)

According to [eq]:

\(DB\text{.}\Delta u={(\left[\frac{d}{{\text{ds}}_{o}}1\right]\Delta u)}^{T}\left[\frac{d}{{\text{ds}}_{o}}1\right]\).

So:

\(\begin{array}{c}{\mathrm{DW}}_{\text{int}}(u,\delta u)\text{.}\Delta u={\int }_{{s}_{1}}^{{s}_{2}}\left\{{(B\delta u)}^{T}{E}_{a}AB\Delta u\right\}\text{ds}\\ +{\int }_{{s}_{1}}^{{s}_{2}}\left\{{(\left[\frac{d}{{\text{ds}}_{o}}1\right]\delta u)}^{T}N\left[\frac{d}{{\text{ds}}_{o}}1\right]\Delta u\right\}\text{ds}\end{array}\) [8]

According to [eq]:

\({\mathrm{DW}}_{\text{iner}}(\ddot{u},\delta u)\text{.}\Delta u=-{\int }_{{s}_{1}}^{{s}_{2}}(\delta {u}^{T}\rho A\Delta \ddot{u})\text{ds}\) [9]