11. Description of document versions

Version Aster	Author (s) Organization (s)	Description of changes
3	M. AUFAURE (EDF/IMA/MMN)	Initial text
10.1	J.M. PROIX (EDF -R&D/ AMA)	Modification of GREEN to GROT_GDEP

Stiffness matrix calculation

Here we show how the formula [éq 5-1] applies to the calculation of the first three rows of the matrix \(K\) [éq 5-2], those relating to the force \({F}_{1}\). The other rows are obtained either by permutation of the indices, or by summing the two previous rows.

The first lines are therefore obtained by calculating the derivative:

\(\underset{e\to 0}{\text{lim}}\frac{d}{\mathrm{de}}{F}_{1}(u+\text{ed}u)\) eq Year1-1

and by factoring the vector \(\deltau\).

Based on the relationships [éq 4-1], [éq 3-4], and [éq 3-3], we have:

\({F}_{1}(u+\varepsilon \delta u)=\left\{\text{EA}\left[\frac{{l}_{1}(u+\varepsilon du)+{l}_{2}(u+\varepsilon \delta u)-{l}_{i}}{{l}_{0}}-\alpha (T-{T}_{i})\right]+{N}_{i}\right\}\frac{{l}_{1}(u+\varepsilon \delta u)}{{l}_{1}(u+\varepsilon \delta u)}\)

with, according to [éq 3-1]:

\({l}_{1}(u+\varepsilon \deltau )={x}_{1}+{u}_{1}+\varepsilon \delta {u}_{1}-{x}_{3}-{u}_{3}-\varepsilon \delta {u}_{3}\)

and, according to [éq 3-2]:

\({l}_{1}(u+\varepsilon \deltau )=\sqrt{{l}_{1}^{T}(u+\varepsilon \deltau ){l}_{1}(u+\varepsilon \deltau )}\text{.}\)

Therefore:

\(\underset{\varepsilon \to 0}{\text{lim}}\frac{d}{d\varepsilon }{l}_{1}(u+\varepsilon \deltau )=\delta {u}_{1}-\delta {u}_{3}\)

\(\underset{\varepsilon \to 0}{\text{lim}}\frac{d}{d\varepsilon }{l}_{1}(u+\varepsilon \deltau )=\frac{1}{{l}_{1}(u)}{l}_{1}^{T}(u)(\delta {u}_{1}-\delta {u}_{3})\)

and by permutation of the indices 1 and 2:

\(\underset{\varepsilon \to 0}{\text{lim}}\frac{d}{d\varepsilon }{l}_{2}(u+\varepsilon \deltau )=\frac{1}{{l}_{2}(u)}{l}_{2}^{T}(u)(\delta {u}_{2}-\delta {u}_{3})\)

Finally:

\(\underset{\varepsilon \to 0}{\text{lim}}\frac{d}{d\varepsilon }{l}_{1}(u+\varepsilon \deltau )=\frac{1}{{l}_{1}^{3}(u)}{l}_{1}^{T}(u)(\delta {u}_{1}-\delta {u}_{3})\)

By bringing the previous expressions to [éq An1-1] and factoring the vector \(\deltau\), we easily get the first three lines of \(K\).

Balance figure of a heavy inextensible cable subjected to a given tension at the end

Let’s take a cable [Figure An2-a] whose one end is fixed to point \(O\) and whose other end \(P\) is subject to tension \({N}_{p}\). \(O\) and \(P\) are on the same horizontal and distant from \(s\). The linear weight is \(w\). We’re looking for arrow \(S\).

Figure AN2-a: Cable weighing in balance

In [bib7], p 6, the following well-known formulas are found:

• cable balance figure:

\(z(x)=\frac{H}{w}\left[\text{cosh}\frac{w}{H}(\frac{s}{2}-x)-\text{cosh}\frac{\mathrm{ws}}{\mathrm{2H}}\right];\) eq Year2-1

• voltage:

\(N(x)=H\text{cosh}\frac{w}{H}(\frac{s}{2}-x)\text{.}\) eq An2-2

\(H\) is the horizontal tension, which is constant along the cable since the distributed external force - the weight - is vertical.

\(H\) is calculated by [éq An2-2] written in \(P\):

\(\text{cosh}\frac{\mathrm{ws}}{\mathrm{2H}}-\frac{{\mathrm{2N}}_{p}}{\mathrm{ws}}\frac{\mathrm{ws}}{\mathrm{2H}}=0\text{.}\)

So \(\frac{\mathrm{ws}}{\mathrm{2H}}\) is the root of the transcendent equation:

\(\text{cosh}X=\frac{{\mathrm{2N}}_{p}}{\mathrm{ws}}X\text{.}\)

This equation has two roots [Figure An2-b] if:

\(\frac{{\mathrm{2N}}_{p}}{\mathrm{ws}}>{p}_{0},\)

with:

\({p}_{0}=\text{sinh}{x}_{0}\)

and:

\({x}_{0}=\text{cotanh}{x}_{0}{x}_{0}>0\text{.}\)

Figure AN2-b: Calculating of \(\frac{\mathrm{ws}}{\mathrm{2H}}\)

Only the smallest root, which corresponds to the greatest tension in the cable, is useful. The other root corresponds to a considerable deflection of the cable, of the order of magnitude of its range.

Since \(\frac{\mathrm{ws}}{\mathrm{2H}}\) is calculated, the arrow is derived from [éq An2-1]:

\(S=\frac{s}{2}\frac{\mathrm{2H}}{\mathrm{ws}}(\text{cosh}\frac{\mathrm{ws}}{\mathrm{2H}}-1)\).