Description of document versions
====================================


.. csv-table::

    "**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 [:ref:`éq 5-1 <éq 5-1>`] applies to the calculation of the first three rows of the matrix :math:`K` [:ref:`éq 5-2 <éq 5-2>`], those relating to the force :math:`{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:


.. _RefEquation An1-1:

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

and by factoring the vector :math:`\deltau`.

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

:math:`{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 [:ref:`éq 3-1 <éq 3-1>`]:

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

and, according to [:ref:`éq 3-2 <éq 3-2>`]:

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

Therefore:

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

:math:`\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:

:math:`\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:

:math:`\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 [:ref:`éq An1-1 <éq An1-1>`] and factoring the vector :math:`\deltau`, we easily get the first three lines of :math:`K`.




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

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


.. image:: images/1000084200001E9800000A566CA8B9528D4E8E66.svg
    :width: 394
    :height: 133


.. _RefImage_1000084200001E9800000A566CA8B9528D4E8E66.svg:

**Figure AN2-a: Cable weighing in balance**

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


* cable balance figure:

:math:`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:


.. _RefEquation An2-2:

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

:math:`H` is the horizontal tension, which is constant along the cable since the distributed external force - the weight - is vertical.

:math:`H` is calculated by [:ref:`éq An2-2 <éq An2-2>`] written in :math:`P`:

:math:`\text{cosh}\frac{\mathrm{ws}}{\mathrm{2H}}-\frac{{\mathrm{2N}}_{p}}{\mathrm{ws}}\frac{\mathrm{ws}}{\mathrm{2H}}=0\text{.}`

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

:math:`\text{cosh}X=\frac{{\mathrm{2N}}_{p}}{\mathrm{ws}}X\text{.}`

This equation has two roots [:ref:`Figure An2-b <Figure An2-b>`] if:

:math:`\frac{{\mathrm{2N}}_{p}}{\mathrm{ws}}>{p}_{0},`

with:

:math:`{p}_{0}=\text{sinh}{x}_{0}`

and:

:math:`{x}_{0}=\text{cotanh}{x}_{0}{x}_{0}>0\text{.}`


.. image:: images/1000062E000018B40000124BDD93E2F6E27C84A1.svg
    :width: 394
    :height: 133


.. _RefImage_1000062E000018B40000124BDD93E2F6E27C84A1.svg:

**Figure AN2-b: Calculating of** :math:`\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 :math:`\frac{\mathrm{ws}}{\mathrm{2H}}` is calculated, the arrow is derived from [:ref:`éq An2-1 <éq An2-1>`]:

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