Annexe ====== Présentation ------------ On considère le portique ci-contre, soumis à diverses charges. +----------------------------------------------------------------+---------------------------------------------------------------------------------+ | |On considère le portique ci-contre, soumis à diverses charges. | | .. image:: images/10000000000001D2000001672BDC41EDF32CABE4.png | | | :width: 3.5626in | Hyperstaticité de degré 1. | | :height: 2.5654in | Inconnue hyperstatique : X : | | | moment en C. | | | | | .. _RefImage_10000000000001D2000001672BDC41EDF32CABE4.png: | Chargement vertical réparti :math:`p` sur :math:`{C}_{1}C` . | | | Deux forces :math:`{F}_{1}` , :math:`{F}_{2}` et un couple en :math:`{C}_{1}` . | | | | +----------------------------------------------------------------+---------------------------------------------------------------------------------+ Hyperstaticité de degré 1. Inconnue hyperstatique : :math:`X` Charges appliquées : * moment en :math:`C` , * chargement vertical réparti :math:`p` sur :math:`{C}_{1}{C}_{2}` , * force :math:`\mathrm{F1}` , :math:`\mathrm{F2}` appliquées en :math:`{C}_{1}` , * couple :math:`\Gamma` appliqué en :math:`{C}_{1}` .. image:: images/100000000000012D000000BA3A22FD6020CAD4B5.png :width: 1.6972in :height: 1.1543in .. _RefImage_100000000000012D000000BA3A22FD6020CAD4B5.png: :math:`\tan(\alpha )=\frac{\mathrm{2a}}{l}=0.4(\Rightarrow {(\cos(\alpha ))}^{-1}=\sqrt{1.16}=1.077033)` :math:`\tan(\beta )=\frac{l}{2(a+h)}=\frac{1}{1.2}` :math:`b=\frac{l}{\mathrm{2cos}(\alpha )};\sin(\alpha )=\frac{a}{b}` Sollicitations isostatiques sous charge réelle répartie :math:`p`sur :math:`{C}_{1}C` ------------------------------------------------------------------------------------- Réactions d'appuis isostatiques ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ :math:`{H}_{A}+{H}_{B}=0` :math:`{V}_{A}+{V}_{B}=\frac{\mathrm{pl}}{\mathrm{2cos}(\alpha )}` :math:`{\mathrm{lV}}_{B}=\frac{{\mathrm{pl}}^{2}}{\mathrm{8cos}(\alpha )}` La partie :math:`\mathrm{CB}` est articulée et chargée seulement à ses extrémités :math:`(\begin{array}{}{H}_{B}\\ {V}_{B}\end{array})\wedge \mathrm{BC}=0\iff {H}_{B}=-{V}_{B}\tan(\beta )` D'où les réactions isostatiques :math:`{H}_{A}=\frac{\mathrm{pl}}{\mathrm{8cos}(\alpha )}\tan(\beta )` ; :math:`{V}_{A}=\frac{3\mathrm{pl}}{\mathrm{8cos}(\alpha )}` ; :math:`{H}_{B}=\frac{-\mathrm{pl}}{\mathrm{8cos}(\alpha )}\tan(\beta )` ; :math:`{V}_{B}=\frac{\mathrm{pl}}{\mathrm{8cos}(\alpha )}` **Remarque:** :math:`\frac{l\tan(\beta )}{8\cos(\alpha )}=\frac{\mathrm{bl}}{8(a+h)}` ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Sollicitations ~~~~~~~~~~~~~~ **Poutre** :math:`{\mathrm{AC}}_{1}` +----------------------------------------------------------------+---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+ | |:math:`\begin{array}{}{N}_{\mathrm{iso}}=\frac{-3\mathrm{pl}}{\mathrm{8cos}(\alpha )}\\ {V}_{\mathrm{iso}}=\frac{\mathrm{pl}}{\mathrm{8cos}(\alpha )}\tan(\beta )\\ {M}_{\mathrm{iso}}=\frac{-\mathrm{pl}}{\mathrm{8cos}(\alpha )}\mathrm{y.tan}(\beta )\end{array}` | | .. image:: images/100000000000007B00000080327158A9E207E182.png | | | :width: 1.2811in | | | :height: 1.2535in | | | | | | | | | .. _RefImage_100000000000007B00000080327158A9E207E182.png: | | | | | | | | +----------------------------------------------------------------+---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+ **Poutre** :math:`{C}_{2}B` +----------------------------------------------------------------+---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+ | |:math:`\begin{array}{}{N}_{\mathrm{iso}}=\frac{-\mathrm{pl}}{\mathrm{8cos}(\alpha )}\\ {V}_{\mathrm{iso}}=\frac{-\mathrm{pl}}{\mathrm{8cos}(\alpha )}\tan(\beta )\\ {M}_{\mathrm{iso}}=\frac{-\mathrm{pl}}{\mathrm{8cos}(\alpha )}\mathrm{y.tan}(\beta )\end{array}` | | | | | .. image:: images/1000000000000070000000809E493CF23E17DE52.png | | | :width: 1.1665in | | | :height: 1.1819in | | | | | | | | | .. _RefImage_1000000000000070000000809E493CF23E17DE52.png: | | | | | | | | +----------------------------------------------------------------+---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+ **Poutre** :math:`{C}_{1}C` +----------------------------------------------------------------+----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+ | |:math:`\begin{array}{cc}{N}_{\mathrm{iso}}& =-{H}_{A}\cos(\alpha )-{V}_{A}\sin(\alpha )+\frac{\mathrm{px}}{\cos(\alpha )}\sin(\alpha )\\ & =-\frac{\mathrm{pl}}{8}(\tan(\beta )+3\tan(\alpha )-8\tan(\alpha )\frac{x}{l})\end{array}` :math:`\begin{array}{cc}{V}_{\mathrm{iso}}& ={H}_{A}\sin(\alpha )-{V}_{A}\cos(\alpha )+\frac{\mathrm{px}}{\cos(\alpha )}\cos(\alpha )\\ & =\frac{\mathrm{pl}}{8}(\tan(\beta )\tan(\alpha )-3+8\frac{x}{l})\end{array}` :math:`\begin{array}{cc}{M}_{\mathrm{iso}}& =-\frac{{\mathrm{px}}^{2}}{\mathrm{2cos}(\alpha )}+{V}_{A}x-{H}_{A}y\\ & =\frac{p}{\cos(\alpha )}(\frac{-{x}^{2}}{2}+\frac{3\mathrm{lx}}{8}-\frac{\mathrm{ly}\tan(\beta )}{8})\end{array}` | | .. image:: images/1000000000000106000000F97237DECA6E26F809.png | avec :math:`{M}_{\mathrm{iso}}=0` en :math:`C` | | :width: 2.4028in | | | :height: 2.1717in | | | | | | | | | .. _RefImage_1000000000000106000000F97237DECA6E26F809.png: | | | | | | | | +----------------------------------------------------------------+----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+ :math:`{M}_{\mathrm{iso}}=\frac{-\mathrm{pl}}{8(a+h)}(2{s}^{2}(\frac{a+h}{b})-s(\mathrm{2a}+\mathrm{3h})+\mathrm{bh})` avec :math:`s=\frac{x}{\cos(\alpha )}\in [0,b]` **Poutre** :math:`{\mathrm{CC}}_{2}` +----------------------------------------------------------------+---------------------------------------------------------------------------------------------------------------------------------------------------------------+ | |:math:`\begin{array}{cc}{N}_{\mathrm{iso}}& ={H}_{B}\cos(\alpha )-{V}_{B}\sin(\alpha )\\ & =-\frac{\mathrm{pl}}{8}(\tan(\beta )+\tan(\alpha ))\end{array}` | | | :math:`\begin{array}{cc}{V}_{\mathrm{iso}}& ={H}_{B}\sin(\alpha )+{V}_{B}\cos(\alpha )\\ & =-\frac{\mathrm{pl}}{8}(\tan(\beta )\tan(\alpha )-1)\end{array}` | | .. image:: images/100000000000013B000000C629026E018D1EA36B.png | :math:`\begin{array}{cc}{M}_{\mathrm{iso}}& ={H}_{B}y-{V}_{B}(l-x)\\ & =-\frac{\mathrm{pl}}{\mathrm{8cos}(\alpha )}(\mathrm{y.tan}(\beta )-(l-x))\end{array}` | | :width: 2.889in | avec :math:`{M}_{\mathrm{iso}}=0` en :math:`C` | | :height: 1.8118in | | | | | | | | | .. _RefImage_100000000000013B000000C629026E018D1EA36B.png: | | | | | | | | +----------------------------------------------------------------+---------------------------------------------------------------------------------------------------------------------------------------------------------------+ Diagrammes ~~~~~~~~~~ :math:`(b=\frac{l}{\mathrm{2cos}(\alpha )})` .. image:: images/1000000000000308000000FF05AA8138C4D73811.png :width: 6.4575in :height: 2.1138in .. _RefImage_1000000000000308000000FF05AA8138C4D73811.png: Sollicitations sous force concentrée :math:`{F}_{1}`(vers le bas) ----------------------------------------------------------------- Réactions d'appui ~~~~~~~~~~~~~~~~~ +--------------------------------------------------------------------------------------------------------------------------------------------+----------------------------------------------------------------+ | | | | :math:`{H}_{A}+{H}_{B}=0;` | .. image:: images/100000000000012D000000E9E1660AAA6525BEA3.png | | :math:`{V}_{A}+{V}_{B}={F}_{1};` | :width: 2.2854in | | :math:`(\begin{array}{}{H}_{A}\\ {V}_{A}\end{array})\wedge \mathrm{AC}=0=(\begin{array}{}{H}_{B}\\ {V}_{C}\end{array})\wedge \mathrm{BC};` | :height: 1.6874in | | | | | | | | | .. _RefImage_100000000000012D000000E9E1660AAA6525BEA3.png: | | | | | | | +--------------------------------------------------------------------------------------------------------------------------------------------+----------------------------------------------------------------+ D'où: :math:`{H}_{A}=\frac{1}{2}{F}_{1}\tan(\beta )` ; :math:`{V}_{A}=\frac{1}{2}{F}_{1}` ; :math:`{H}_{B}=\frac{-1}{2}{F}_{1}\tan(\beta )` ; :math:`{V}_{B}=\frac{1}{2}{F}_{1}` Sollicitations ~~~~~~~~~~~~~~ .. csv-table:: "**Poutre** :math:`{\mathrm{AC}}_{1}` **:**", ":math:`{N}_{\mathrm{iso}}=\frac{-1}{2}{F}_{1}` :math:`{V}_{\mathrm{iso}}=\frac{1}{2}{F}_{1}\tan(\beta )` :math:`{M}_{\mathrm{iso}}=\frac{-1}{2}{F}_{1}y\tan(\beta )`" "**Poutre** :math:`{C}_{2}B` **:**", ":math:`{N}_{\mathrm{iso}}=\frac{-1}{2}{F}_{1}` :math:`{V}_{\mathrm{iso}}=\frac{-1}{2}{F}_{1}\tan(\beta )` :math:`{M}_{\mathrm{iso}}=\frac{-1}{2}{F}_{1}y\tan(\beta )`" "**Poutre** :math:`{C}_{1}C` **:**", ":math:`{N}_{\mathrm{iso}}=\frac{-1}{2}{F}_{1}(\tan(\beta )\cos(\alpha )+\sin(\alpha ))` :math:`{V}_{\mathrm{iso}}=\frac{1}{2}{F}_{1}(\tan(\beta )\sin(\alpha )-\cos(\alpha ))` :math:`{M}_{\mathrm{iso}}=\frac{-1}{2}{F}_{1}(y\tan(\beta )-x)`" "**Poutre** :math:`{\mathrm{CC}}_{2}` **:**", ":math:`{N}_{\mathrm{iso}}=\frac{-1}{2}{F}_{1}(\tan(\beta )\cos(\alpha )+\sin(\alpha ))` :math:`{V}_{\mathrm{iso}}=\frac{1}{2}{F}_{1}(\tan(\beta )\sin(\alpha )-\cos(\alpha ))` :math:`{M}_{\mathrm{iso}}=\frac{-1}{2}{F}_{1}(y\tan(\beta )-(l-x))`" Diagrammes (:math:`{F}_{1}`vers le bas) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ .. image:: images/10000000000002A8000000DEB33AFAF001209DD4.png :width: 5.7528in :height: 1.7854in .. _RefImage_10000000000002A8000000DEB33AFAF001209DD4.png: Sollicitations sous la force concentrée :math:`{F}_{2}`(vers la gauche) ----------------------------------------------------------------------- Réactions d'appui ~~~~~~~~~~~~~~~~~ +-------------------------------------------------------------------------------------------------------------------------+----------------------------------------------------------------+ |:math:`\begin{array}{}{H}_{A}+{H}_{B}={F}_{2};\\ {V}_{A}+{V}_{B}=0;\\ {\mathrm{lV}}_{B}+{\mathrm{hF}}_{2}=0;\end{array}` | | | :math:`(\begin{array}{}{H}_{B}\\ {V}_{B}\end{array})\mathrm{ans}\mathrm{BC}=0` | .. image:: images/1000000000000132000000C2C7933BFEBA320EC3.png | | | :width: 2.4335in | | | :height: 1.5728in | | | | | | | | | .. _RefImage_1000000000000132000000C2C7933BFEBA320EC3.png: | | | | | | | +-------------------------------------------------------------------------------------------------------------------------+----------------------------------------------------------------+ D'où: :math:`{H}_{A}={F}_{1}(1-\frac{h}{l}\tan(\beta ))` ; :math:`{V}_{A}={F}_{2}\frac{h}{l}` ; :math:`{H}_{B}={F}_{1}\frac{h}{l}\tan(\beta )` ; :math:`{V}_{B}=-\mathrm{F2}\frac{h}{l}` ; **Remarque:** :math:`\frac{h}{l}\tan(\beta )=\frac{h}{2(a+h)}` :math:`(1-\frac{h}{l}\tan(\beta ))=\frac{\mathrm{2a}+h}{2(a+h)}` :math:`\tan(\beta )\sin(\alpha )–\cos(\alpha )=\frac{-\mathrm{hl}}{\mathrm{2b}(a+h)}` , :math:`\tan(\beta )\cos(\alpha )–\sin(\alpha )=\frac{{l}^{2}–4({a}^{2}+\mathrm{ah})}{\mathrm{4b}(a+h)}` Sollicitations ~~~~~~~~~~~~~~ .. csv-table:: "**Poutre** :math:`{\mathrm{AC}}_{1}` :", ":math:`{N}_{\mathrm{iso}}=-{F}_{2}\frac{h}{l}` :math:`{V}_{\mathrm{iso}}={F}_{2}(1-\frac{h}{l}\tan(\beta ))` :math:`{M}_{\mathrm{iso}}=-{F}_{2}y(1-\frac{h}{l}\tan(\beta ))`" "**Poutre** :math:`{C}_{2}B` :", ":math:`{N}_{\mathrm{iso}}={F}_{2}\frac{h}{l}` :math:`{V}_{\mathrm{iso}}={F}_{2}\frac{h}{l}\tan(\beta )` :math:`{M}_{\mathrm{iso}}=-{F}_{2}y\frac{h}{l}y\tan(\beta )`" "**Poutre** :math:`{C}_{1}C` :", ":math:`{N}_{\mathrm{iso}}={F}_{2}((1-\frac{h}{l}\tan(\beta ))\cos(\alpha )–\frac{h}{l}\cos(\alpha ))` :math:`{V}_{\mathrm{iso}}={F}_{2}((1–\frac{h}{l}\tan(\beta ))–\frac{h}{l}\cos(\alpha ))` :math:`{M}_{\mathrm{iso}}={F}_{2}(\frac{h}{l}x-(1–\frac{h}{l}\tan(\beta ))y)`" "**Poutre** :math:`C{C}_{2}` **:**", ":math:`{N}_{\mathrm{iso}}={F}_{2}\frac{h}{l}(\tan(\beta )\cos(\alpha )+\sin(\alpha ))` :math:`{V}_{\mathrm{iso}}={F}_{2}\frac{h}{l}(\tan(\beta )\sin(\alpha )-\cos(\alpha ))` :math:`{M}_{\mathrm{iso}}={F}_{2}\frac{h}{l}(y\tan(\beta )-(l-x))`" Diagrammes ~~~~~~~~~~ .. image:: images/100000000000029C000000EDCA1A24F3881A8B90.png :width: 6.098in :height: 2.0827in .. _RefImage_100000000000029C000000EDCA1A24F3881A8B90.png: Sollicitations sous le couple concentré:math:`\Gamma`(positif) -------------------------------------------------------------- Réactions d'appui ~~~~~~~~~~~~~~~~~ +---------------------------------------------------------------------------------------------------------+----------------------------------------------------------------+ |:math:`\begin{array}{}{H}_{A}+{H}_{B}=0;\\ {V}_{A}+{V}_{B}=0;\\ {\mathrm{lV}}_{B}+\Gamma =0;\end{array}` | | | :math:`(\begin{array}{}{H}_{B}\\ {V}_{B}\end{array})\wedge \mathrm{BC}=0;` | .. image:: images/100000000000012D000000BA53EF896D4CDEBDCA.png | | | :width: 2.4492in | | | :height: 1.4827in | | | | | | | | | .. _RefImage_100000000000012D000000BA53EF896D4CDEBDCA.png: | | | | | | | +---------------------------------------------------------------------------------------------------------+----------------------------------------------------------------+ D'où: :math:`{H}_{A}=-\Gamma \tan\frac{\beta}{l}` , :math:`{V}_{A}=\frac{\Gamma}{l}` , :math:`{H}_{B}=\Gamma \tan\frac{\beta}{l}` , :math:`{V}_{A}=\frac{-\Gamma }{l}` **Remarque:** :math:`\frac{\tan(\beta )}{l}=\frac{1}{2(a+h)}` Sollicitations ~~~~~~~~~~~~~~ .. csv-table:: "**Poutre** :math:`A{C}_{1}` :", ":math:`{N}_{\mathrm{iso}}=\frac{-\Gamma }{l}` :math:`{V}_{\mathrm{iso}}=\frac{-\Gamma \tan(\beta )}{l}` :math:`{M}_{\mathrm{iso}}=\frac{\Gamma y\tan(\beta )}{l}`" "**Poutre** :math:`{C}_{2}B` :", ":math:`{N}_{\mathrm{iso}}=\frac{-\Gamma }{l}` :math:`{V}_{\mathrm{iso}}=\frac{\Gamma \tan(\beta )}{l}` :math:`{M}_{\mathrm{iso}}=\frac{\Gamma y\tan(\beta )}{l}`" "**Poutre** :math:`{C}_{1}C` :", ":math:`{N}_{\mathrm{iso}}=\frac{\Gamma}{l}(\tan(\beta )\cos(\alpha )–\sin(\alpha ))` :math:`{N}_{\mathrm{iso}}=\frac{-\Gamma }{l}(\tan(\beta )\sin(\alpha )+\cos(\alpha ))` :math:`{N}_{\mathrm{iso}}=\frac{\Gamma}{l}(x+y\tan(\beta )–l)`" "**Poutre** :math:`C{C}_{2}` **:**", ":math:`{N}_{\mathrm{iso}}=\frac{\Gamma}{l}(\tan(\beta )\cos(\alpha )+\sin(\alpha ))` :math:`{N}_{\mathrm{iso}}=\frac{\Gamma}{l}(\tan(\beta )\sin(\alpha )-\cos(\alpha ))` :math:`{N}_{\mathrm{iso}}=\frac{\Gamma}{l}(y\tan(\beta )–(l-x))`" Diagrammes (:math:`\Gamma`positif) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ .. image:: images/1000000000000343000001154B3444319539B639.png :width: 5.9835in :height: 1.9827in .. _RefImage_1000000000000343000001154B3444319539B639.png: Sollicitations sous le moment:math:`X`hyperstatique --------------------------------------------------- Réactions d'appui ~~~~~~~~~~~~~~~~~ +------------------------------+----------------------------------------------------------------+ |:math:`{H}_{A}+{H}_{B}=0;` | | | :math:`{V}_{A}+{V}_{B}=0;` | .. image:: images/1000000000000151000000B7847C7CFB1B3CBF5C.png | | :math:`{\mathrm{lV}}_{B}=0;` | :width: 3.0228in | | :math:`{H}_{B}(a+h)–X=0;` | :height: 1.5965in | | | | | | | | | .. _RefImage_1000000000000151000000B7847C7CFB1B3CBF5C.png: | | | | | | | +------------------------------+----------------------------------------------------------------+ D'où les réactions: :math:`{H}_{a}=\frac{-X}{a+h}` , :math:`{V}_{A}=0` , :math:`{H}_{B}=\frac{X}{a+h}` , :math:`{V}_{B}=0` Sollicitations ~~~~~~~~~~~~~~ .. csv-table:: "**Poutre** :math:`{\mathrm{AC}}_{1}` :", ":math:`{N}_{X}=0` :math:`{V}_{X}=\frac{-X}{a+h}` :math:`{M}_{X}=\frac{X}{a+h}y`" "**Poutre** :math:`{C}_{2}B` :", ":math:`{N}_{X}=0` :math:`{V}_{X}=\frac{X}{a+h}` :math:`{M}_{X}=\frac{X}{a+h}y`" "**Poutre** :math:`{C}_{1}C` :", ":math:`{N}_{X}=\frac{X}{a+h}\cos(\alpha )` :math:`{V}_{X}=\frac{X}{a+h}\sin(\alpha )` :math:`{M}_{X}=\frac{X}{a+h}y=\frac{X}{a+h}(h+x\tan(\alpha ))`" "**Poutre** :math:`{\mathrm{CC}}_{2}` **:**", ":math:`{N}_{X}=\frac{X}{a+h}\cos(\alpha )` :math:`{V}_{X}=\frac{X}{a+h}\sin(\alpha )` :math:`{M}_{X}=\frac{X}{a+h}y`" Diagrammes ~~~~~~~~~~ .. image:: images/10000000000003010000015D27D5B04219FF9F97.png :width: 6.4035in :height: 2.9043in .. _RefImage_10000000000003010000015D27D5B04219FF9F97.png: Sollicitations sous charges fictives ponctuelles en:math:`C` ------------------------------------------------------------ Afin de calculer les déplacement en :math:`C` , à l’aide du Principe des travaux Virtuels (:math:`\mathrm{cf.}` le paragraphe :math:`[\S 8]` ), il est nécessaire d’établir les diagrammes de sollicitations sous l’action de deux forces “fictives” :math:`f` et :math:`g` appliquées en :math:`C` . Réactions d'appui ~~~~~~~~~~~~~~~~~ +-------------------------------------------------------------------------------------------------------------------------------------------+----------------------------------------------------------------+ |:math:`{H}_{A}+{H}_{B}=-f;` | | | :math:`{V}_{A}+{V}_{B}=-g;` | .. image:: images/100000000000012D000000E266E8AC0680ABB2F0.png | | :math:`(\begin{array}{}{H}_{A}\\ {V}_{A}\end{array})\wedge \mathrm{AC}=0=(\begin{array}{}{H}_{B}\\ {V}_{B}\end{array})\wedge \mathrm{BC}` | :width: 2.0854in | | | :height: 1.6311in | | | | | | | | | .. _RefImage_100000000000012D000000E266E8AC0680ABB2F0.png: | | | | | | | +-------------------------------------------------------------------------------------------------------------------------------------------+----------------------------------------------------------------+ D'où: :math:`{H}_{A}=\frac{-1}{2}(f+g\tan(\beta ))` , :math:`{V}_{A}=\frac{-1}{2}(g+f\mathrm{cot}(\beta ))` :math:`{H}_{B}=\frac{-1}{2}(f-g\tan(\beta ))` , :math:`{V}_{B}=\frac{-1}{2}(g-f\mathrm{cot}(\beta ))` Sollicitations ~~~~~~~~~~~~~~ .. csv-table:: "**Poutre** :math:`{\mathrm{AC}}_{1}` **:**", ":math:`n=\frac{1}{2}(g+f\mathrm{cot}(\beta ))` :math:`v=\frac{-1}{2}(f+g\tan(\beta ))` :math:`m=\frac{1}{2}(f+g\tan(\beta ))`" "**Poutre** :math:`{C}_{2}B` :", ":math:`n=\frac{1}{2}(g-f\mathrm{cot}(\beta ))` :math:`v=\frac{-1}{2}(f-g\tan(\beta ))` :math:`m=\frac{-1}{2}(f-g\tan(\beta ))y`" "**Poutre** :math:`{C}_{1}C` :", ":math:`n=\frac{1}{2}(f+g\tan(\beta ))\cos(\alpha )+\frac{1}{2}(g+f\mathrm{cot}(\beta ))\sin(\alpha )` :math:`v=\frac{-1}{2}(f+g\tan(\beta ))\sin(\alpha )+\frac{1}{2}(g+f\mathrm{cot}(\beta ))\cos(\alpha )` :math:`m=\frac{1}{2}(f+g\tan(\beta ))y-\frac{1}{2}(g+f\mathrm{cot}(\beta ))x`" "**Poutre** :math:`{\mathrm{CC}}_{2}` **:**", ":math:`n=\frac{-1}{2}(f-g\tan(\beta ))\cos(\alpha )+\frac{1}{2}(g-f\mathrm{cot}(\beta ))\sin(\alpha )` :math:`v=\frac{-1}{2}(f-g\tan(\beta ))\sin(\alpha )-\frac{1}{2}(g-f\mathrm{cot}(\beta ))\cos(\alpha )` :math:`m=\frac{-1}{2}(f-g\tan(\beta ))y-\frac{1}{2}(g-f\mathrm{cot}(\beta ))(l-x)`" Diagrammes ~~~~~~~~~~ Voici les diagrammes de sollicitations sous l’action des deux forces “fictives” :math:`f` et :math:`g` . On considère ici: :math:`f\ge 0,g\ge \mathrm{fcot}(\beta )` **.** .. image:: images/100000000000037600000129AFAACC0851F87FFF.png :width: 6.3543in :height: 2.1209in .. _RefImage_100000000000037600000129AFAACC0851F87FFF.png: Détermination du moment:math:`X`hyperstatique --------------------------------------------- On se place en élasticité; on ne considère que l'énergie de flexion, les poutres étant élancées. L’état naturel est supposé vierge (pas de précontraintes ni de déplacement d'appui). Le potentiel complémentaire est alors: :math:`F\ast (X)={\int}_{\mathrm{poteaux}}\frac{{({M}_{\mathrm{iso}}+{M}_{1}X)}^{2}}{{\mathrm{EI}}_{1}}+{\int}_{\mathrm{charpentes}}\frac{{({M}_{\mathrm{iso}}+{M}_{1}X)}^{2}}{{\mathrm{EI}}_{2}}` Il est stationnaire à l'équilibre, d’où: :math:`\delta .X=\left[{\int}_{\mathrm{pot}}\frac{{M}_{1}^{2}}{{\mathrm{EI}}_{1}}+{\int}_{\mathrm{charp}}\frac{{M}_{1}^{2}}{{\mathrm{EI}}_{2}}\right].X=-{\int}_{\mathrm{pot}}\frac{{M}_{1}{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{1}}-{\int}_{\mathrm{charp}}\frac{{M}_{1}{M}_{\mathrm{iso}}}{{\mathrm{Ei}}_{2}}=S` Le coefficient de souplesse :math:`\delta` est la somme de: :math:`{\int}_{\mathrm{pot}}\frac{{M}_{1}^{2}}{{\mathrm{EI}}_{1}}=\frac{\mathrm{2h}}{{\mathrm{3EI}}_{1}}{(\frac{h}{a+h})}^{2}` :math:`{\int}_{\mathrm{charp}}\frac{{M}_{1}^{2}}{{\mathrm{EI}}_{2}}=\frac{\mathrm{2b}}{{\mathrm{EI}}_{2}}[{(\frac{h}{a+h})}^{2}+\frac{1}{3}{(\frac{a}{a+h})}^{2}+\frac{\mathrm{ah}}{{(a+h)}^{2}}]` soit: :math:`E.\delta =\frac{2}{{(a+h)}^{2}}[\frac{{h}^{3}}{{\mathrm{3I}}_{1}}+\frac{b({\mathrm{3h}}^{2}+{a}^{2}+\mathrm{3ah})}{{\mathrm{3I}}_{2}}]` **Application numérique:** Dans l'exemple considéré: :math:`{I}_{1}={\mathrm{2I}}_{2}=5.0E-4{m}^{4}` , :math:`h=\mathrm{2a}=8m` , :math:`l=20m` , :math:`b=\frac{l}{2}\sqrt{1.16}` D'où: :math:`\gamma =\frac{2}{E{(a+h)}_{1}^{\mathrm{2I}}}\underset{2353.45347{m}^{3}}{\underset{\underbrace{}}{\frac{{h}^{2}}{3}(h+\frac{\mathrm{19b}}{2})}}` On étudie l'un après l'autre les divers chargements pour calculer les seconds membres :math:`S` . Charge répartie:math:`p`sur:math:`{C}_{1}C` ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Le second membre :math:`S` dû à :math:`f` est: :math:`-{\int}_{\mathrm{pot}}\frac{{M}_{1}{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{1}}=\frac{\mathrm{3h}}{{\mathrm{3EI}}_{1}}(\frac{h}{a+h})(\frac{\mathrm{pblh}}{8(a+h)})=\frac{2}{E{(a+h)}^{2}{I}_{1}}\frac{{\mathrm{ph}}^{3}\mathrm{bl}}{24}` :math:`-{\int}_{{\mathrm{CC}}_{2}}\frac{{M}_{1}{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{2}}=\frac{{\mathrm{pb}}^{2}\mathrm{hl}}{8(a+h){\mathrm{EI}}_{2}}(\frac{1}{2}\frac{h}{a+h})+(\frac{a}{6}\frac{a}{a+h})=\frac{2}{E{(a+h)}^{2}{I}_{2}}\frac{{\mathrm{phb}}^{\mathrm{2l}}(\mathrm{3h}+a)}{48}` :math:`\begin{array}{}-{\int}_{{C}_{1}C}\frac{{M}_{1}{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{2}}=\frac{1}{{\mathrm{EI}}_{2}}\frac{\mathrm{pl}}{8{(a+h)}^{2}}{\int}_{0}^{b}\left[{\mathrm{2s}}^{2}\frac{a+h}{b}–s(\mathrm{2a}+\mathrm{3h})+\mathrm{bh}\right]\left[h+s\frac{a}{b}\right]\mathrm{ds}\\ =\frac{1}{E{(a+h)}^{2}{I}_{2}}\frac{{\mathrm{plb}}^{2}}{48}({h}^{2}+\mathrm{2ah}+{a}^{2})\end{array}` D'où: :math:`S=\frac{2}{E{(a+h)}^{2}}\frac{\mathrm{plb}}{96}\left[\frac{{\mathrm{4h}}^{3}}{{I}_{1}}+\frac{\mathrm{hb}(\mathrm{3h}+a)}{{I}_{2}}+\frac{b({h}^{2}–\mathrm{2ah}-{a}^{2})}{{I}_{2}}\right]` **Application numérique:** :math:`{I}_{1}=2{I}_{2}` **;** :math:`h=\mathrm{2a}` **;** :math:`p=3000{\mathrm{N.m}}^{1}` (vers le bas) :math:`S=\frac{2}{E{(a+h)}^{2}{I}_{1}}\underset{43946021.89{\mathrm{N.m}}^{4}}{\underset{\underbrace{}}{\frac{{\mathrm{plbh}}^{2}}{96}\left[\mathrm{4h}+\frac{13}{2}b\right]}}` D'où: • le moment en :math:`C` : :math:`X=18672994\mathrm{N.m}` • la réaction en :math:`A` : :math:`{H}_{A}=p\frac{\mathrm{bl}}{8(a+h)}-\frac{X}{a+h}=\frac{\frac{\mathrm{pbl}}{8-X}}{a+h}` , :math:`{H}_{A}=5175.37N` :math:`{V}_{A}=\frac{\mathrm{3pb}}{4}-0` , :math:`{V}_{A}=24233.24N` Charge ponctuelle:math:`{F}_{1}`en:math:`C` ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Le second membre s’obtient à l’aide de: :math:`-{\int}_{\mathrm{pot}}\frac{{M}_{1}{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{1}}=\frac{\mathrm{2h}}{{\mathrm{3EI}}_{1}}(\frac{h}{a+h})(\frac{{F}_{1}\mathrm{lh}}{4(a+h)})=\frac{2}{E{(a+h)}^{2}{I}_{1}}\frac{{F}_{1}{\mathrm{lh}}^{3}}{12}` :math:`-{\int}_{\mathrm{charp}}\frac{{M}_{1}{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{2}}=\frac{\mathrm{2b}}{{\mathrm{EI}}_{2}}\frac{{F}_{1}\mathrm{lh}}{4(a+h)}(\frac{1}{2}(\frac{h}{a+h})+\frac{1}{6}(\frac{a}{a+h}))=\frac{2}{E{(a+h)}^{2}{I}_{2}}\frac{{F}_{1}\mathrm{blh}(\mathrm{3h}+a)}{24}` D'où: :math:`S=\frac{2}{E{(a+h)}^{2}}\frac{{F}_{1}\mathrm{lh}}{24}\left[\frac{{\mathrm{2h}}^{2}}{{I}_{1}}+\frac{b(\mathrm{3h}+a)}{{I}_{2}}\right]` **Application numérique:** :math:`{I}_{1}=2{I}_{2}` **;** :math:`h=\mathrm{2a}` **;** :math:`{F}_{1}=20000N` (vers le bas) :math:`S=\frac{2}{E{(a+h)}^{2}{I}_{1}}\underset{97485127.76{\mathrm{N.m}}^{4}}{\underset{\underbrace{}}{\frac{{F}_{1}{\mathrm{lh}}^{2}}{24}\left[\mathrm{2h}+7b\right]}}` D'où: • le moment en :math:`C` : :math:`X=41422.161\mathrm{N.m}` • la réaction en :math:`A` : :math:`{H}_{A}=\frac{1}{4}{F}_{1}\frac{l}{a+h}-\frac{X}{a+h}=\frac{\frac{{F}_{1}l}{4-X}}{a+h}` , :math:`{H}_{A}=4881.4866N` :math:`{V}_{A}=\frac{1}{2}{F}_{1}-0` , :math:`{V}_{A}=10000.0N` Charge ponctuelle:math:`{F}_{2}`en:math:`{C}_{1}` ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Le second membre s’obtient à l’aide de: :math:`-{\int}_{{\mathrm{AC}}_{1}}\frac{{M}_{1}{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{1}}=\frac{h}{{\mathrm{3EI}}_{1}}(\frac{h}{a+h})\frac{{F}_{2}h(\mathrm{2a}+h)}{2(a+h)}=\frac{2}{E{(a+h)}^{2}{I}_{1}}\frac{{F}_{2}{h}^{3}(\mathrm{2a}+h)}{12}` :math:`-{\int}_{{C}_{2}B}\frac{{M}_{1}{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{1}}=\frac{h}{{\mathrm{3EI}}_{1}}(\frac{h}{a+h})\frac{(-{F}_{2}{h}^{2})}{2(a+h)}=\frac{2}{E{(a+h)}_{1}^{\mathrm{2I}}}\frac{{F}_{2}{h}^{3}(\mathrm{2a}+h)}{12}` :math:`-{\int}_{{C}_{1}C}\frac{{M}_{1}{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{2}}=\frac{b}{{\mathrm{EI}}_{2}}\frac{{F}_{2}h(\mathrm{2a}+h)}{2(a+h)}\left[\frac{1}{2}(\frac{h}{a+h})+\frac{1}{6}(\frac{a}{a+h})\right]=\frac{2}{E{(a+h)}^{2}{I}_{2}}\frac{{F}_{2}\mathrm{bh}({\mathrm{3h}}^{2}+\mathrm{7ah}+{\mathrm{2a}}^{2})}{24}` :math:`-{\int}_{{\mathrm{CC}}_{2}}\frac{{M}_{1}{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{2}}=\frac{b}{{\mathrm{EI}}_{2}}\frac{-{F}_{2}{h}^{2}}{2(a+h)}\left[\frac{1}{2}(\frac{h}{a+h})+\frac{1}{6}(\frac{a}{a+h})\right]=\frac{2}{E{(a+h)}^{2}{I}_{2}}\frac{-{F}_{2}{\mathrm{bh}}^{2}(\mathrm{3h}+a)}{24}` :math:`S=\frac{2}{E{(a+h)}^{2}}\frac{{F}_{1}\mathrm{lh}}{24}\left[\frac{{\mathrm{2h}}^{2}}{{I}_{1}}+\frac{b(\mathrm{3h}+a)}{{I}_{2}}\right]` **Application numérique:** :math:`{I}_{1}=2{I}_{2}` **;** :math:`h=\mathrm{2a}` **;** :math:`{F}_{2}=10000N` (vers la gauche) :math:`S=\frac{2}{E{(a+h)}^{2}{I}_{1}}\underset{19497025.55{\mathrm{N.m}}^{4}}{\underset{\underbrace{}}{\frac{{F}_{2}{h}^{2}a}{12}\left[\mathrm{2h}+7b\right]}}` D'où: • le moment en :math:`C` : :math:`X=8284.4321\mathrm{N.m}` • la réaction en :math:`A` : :math:`{H}_{A}={F}_{2}\frac{\mathrm{2a}+h}{2(a+h)}-\frac{X}{a+h}=\frac{{F}_{2}(a+\frac{h}{2})-X}{a+h}` , :math:`{H}_{A}=5976.297N` :math:`{V}_{A}=\frac{{F}_{2}h}{l}` , :math:`{V}_{A}=4000.0N` Couple ponctuel:math:`\Gamma`en:math:`{C}_{1}` ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Le second membre s’obtient à l’aide de: :math:`-{\int}_{\mathrm{pot}}\frac{{M}_{1}{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{1}}=\frac{-\mathrm{2h}}{{\mathrm{3EI}}_{1}}(\frac{h}{a+h})\frac{\Gamma h}{2(a+h)}=\frac{2}{E{(a+h)}^{2}{I}_{1}}\frac{-\Gamma {h}^{3}}{6}` :math:`-{\int}_{{C}_{1}C}\frac{b}{{\mathrm{EI}}_{2}}=\frac{\Gamma (h+\mathrm{2a})}{2(a+h)}\frac{b}{{\mathrm{EI}}_{2}}\left[\frac{1}{2}(\frac{h}{a+h}+\frac{1}{6}(\frac{a}{a+h}))\right]=\frac{2}{E{(a+h)}^{2}{I}_{2}}\frac{\Gamma (h+\mathrm{2a})(\mathrm{3h}+a)b}{24}` :math:`-{\int}_{{\mathrm{CC}}_{2}}\frac{{M}_{1}{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{2}}=\frac{b}{{\mathrm{EI}}_{2}}\frac{-\Gamma h}{2(a+h)}\left[\frac{1}{2}(\frac{h}{a+h})+\frac{1}{6}(\frac{a}{a+h})\right]=\frac{2}{E{(a+h)}^{2}{I}_{2}}\frac{-\Gamma \mathrm{hb}(\mathrm{3h}+a)}{24}` D'où: :math:`S=\frac{2}{E{(a+h)}^{2}}\frac{-\Gamma }{12}\left[\frac{{\mathrm{2h}}^{3}}{{I}_{1}}+\frac{\mathrm{ab}(\mathrm{3h}+a)}{{I}_{2}}\right]` **Application numérique:** :math:`{I}_{1}=2{I}_{2}` **;** :math:`h=\mathrm{2a}` **;** :math:`\Gamma =-100000\mathrm{N.m}` (sens aiguilles de montre) :math:`S=\frac{2}{E{(a+h)}^{2}{I}_{1}}\underset{11571281.93{\mathrm{N.m}}^{4}}{\underset{\underbrace{}}{\frac{-\Gamma }{6}\left[{h}^{3}-\mathrm{ab}(\mathrm{3h}+a)\right]}}` D'où: • le moment en :math:`C` : :math:`X=4916.7243\mathrm{N.m}` • la réaction en :math:`A` : :math:`{H}_{A}=\frac{-\Gamma }{2(a+h)}-\frac{X}{a+h}=\frac{\frac{-\Gamma }{2}-X}{a+h}` , :math:`{H}_{A}=4576.394N` :math:`{V}_{A}=\frac{\Gamma}{l}` , :math:`{V}_{A}=5000.0N` Récapitulatif ~~~~~~~~~~~~~ .. csv-table:: "**CAS**", "**Moment en** :math:`C`", "**Réactions en** :math:`A(N)`", "" "", ":math:`(\mathrm{N.m})`", ":math:`{H}_{A}`", ":math:`{V}_{A}`" ":math:`p` sur :math:`{C}_{1}C`", "18672.994", "5175.37", "24233.240" ":math:`{F}_{1}` en :math:`C`", "41422.161", "4881.487", "10000.000" ":math:`{F}_{2}` en :math:`{C}_{1}`", "8284.432", "5976.297", "4000.000" ":math:`\Gamma` en :math:`{C}_{1}`", "4916.724", "4576.394", "5000.000" "TOTAL", "73296.311", "22033.31", "43233.24" **Remarque** Rappel: dans le poteau :math:`{\mathrm{AC}}_{1}` : effort normal= :math:`-{V}_{A}` , effort tranchant= :math:`{H}_{A}` . Calcul du déplacement en:math:`C` --------------------------------- On ne considère aussi que l'énergie élastique de flexion (poutres élancées). En appliquant le Principe des Travaux virtuels sur la structure soumise aux forces fictives du paragraphe :math:`[\S 6]` , travaillant dans les déplacements cherchés, on calcule les nombres :math:`w` et :math:`d` dépendant linéairement de :math:`f` et :math:`g` : :math:`f{u}_{c}+g{v}_{c}={\int}_{\mathrm{pot}}\frac{m({M}_{\mathrm{iso}}+{\mathrm{XM}}_{1})}{{\mathrm{EI}}_{1}}+{\int}_{\mathrm{charp}}\frac{m({M}_{\mathrm{iso}}+{\mathrm{XM}}_{1})}{{\mathrm{EI}}_{2}}=w+\mathrm{Xd},\forall (f,g)` Charge répartie:math:`p`sur:math:`{C}_{1}C` ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ :math:`{\int}_{\mathrm{pot}}\frac{m{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{1}}=\frac{\mathrm{2h}}{{\mathrm{3EI}}_{1}}\frac{\mathrm{ghl}}{4(a+h)}\frac{-\mathrm{pbhl}}{8(a+h)}=\frac{2}{E{(a+h)}^{2}{I}_{1}}\frac{-{\mathrm{gpbh}}^{3}{l}^{2}}{96}` :math:`{\int}_{{C}_{1}C}\frac{m{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{2}}=\frac{2}{E{(a+h)}^{2}{I}_{2}}\frac{-{\mathrm{plhb}}^{2}}{384}(\mathrm{2f}(a+h)+\mathrm{gl})(h-a)` :math:`{\int}_{{\mathrm{CC}}_{2}}\frac{{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{2}}=\frac{b}{{\mathrm{3EI}}_{2}}\frac{\mathrm{pbhl}}{8(a+h)}(\frac{\mathrm{fh}}{2}-\frac{\mathrm{glh}}{4(a+h)})=\frac{2}{E{(a+h)}^{2}{I}_{2}}\frac{-{\mathrm{pb}}^{2}{\mathrm{lh}}^{2}(\mathrm{gl}-\mathrm{2f}(a+h))}{192}` D'où: :math:`w=\frac{2}{E{(a+h)}^{2}}\frac{-\mathrm{pbhl}}{384}(\frac{{\mathrm{4glh}}^{2}}{{I}_{1}}+\frac{\mathrm{glb}(\mathrm{3h}+a)-\mathrm{efb}{(a+h)}^{2}}{{I}_{2}})` **Application numérique:** :math:`{I}_{1}={\mathrm{2I}}_{2}` ; :math:`h=\mathrm{2a}` ; :math:`p=3000{\mathrm{N.m}}^{-1}` (vers le bas) :math:`w=\frac{2}{E{(a+h)}^{2}{I}_{1}}(\mathrm{gl}(\mathrm{2h}+\frac{5}{2}b)-\frac{9}{2}\mathrm{fbh})\underset{-215406.5922{\mathrm{N.m}}^{3}}{\underset{\underbrace{}}{\frac{-{\mathrm{pbh}}^{2}l}{192}}}` Charge ponctuelle :math:`{F}_{1}`en :math:`C` ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ :math:`{\int}_{\mathrm{pot}}\frac{m{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{1}}=\frac{\mathrm{2h}}{{\mathrm{3EI}}_{1}}\frac{\mathrm{ghl}}{4(a+h)}\frac{-{F}_{1}\mathrm{hl}}{4(a+h)}=\frac{2}{E{(a+h)}^{2}{I}_{1}}\frac{-{F}_{1}{\mathrm{gh}}^{3}{l}^{2}}{48}` :math:`{\int}_{\mathrm{charp}}\frac{m{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{2}}=\frac{\mathrm{2b}}{{\mathrm{3EI}}_{2}}\frac{\mathrm{ghl}}{4(a+h)}\frac{-{F}_{1}\mathrm{hl}}{4(a+h)}=\frac{2}{E{(a+h)}^{2}{I}_{2}}\frac{-{F}_{1}{\mathrm{gbh}}^{2}{l}^{2}}{48}` D'où (on constate que :math:`w` ne dépend pas de :math:`f` pour ce chargement): :math:`w=\frac{2}{E{(a+h)}^{2}}\frac{-{F}_{1}{\mathrm{gh}}^{2}{l}^{2}}{48}(\frac{h}{{I}_{1}}+\frac{b}{{I}_{2}})` **Application numérique:** :math:`{I}_{1}={\mathrm{2I}}_{2}` , :math:`h=\mathrm{2a}` , :math:`{F}_{1}=20000N` (vers le bas) :math:`w=\frac{\mathrm{2g}}{E{(a+h)}^{2}{I}_{1}}\frac{-{F}_{1}{h}^{2}{l}^{2}}{48}\underset{-3155100365.0{\mathrm{N.m}}^{5}}{\underset{\underbrace{}}{(h+\mathrm{2b})}}` Charge ponctuelle :math:`{F}_{2}`en :math:`{C}_{1}` ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ :math:`\begin{array}{}{\int}_{\mathrm{pot}}\frac{m{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{1}}=\frac{h}{{\mathrm{3EI}}_{1}}\frac{{F}_{2}h}{2(a+h)}\left[-(\mathrm{2a}+h)(\frac{\mathrm{fh}}{2}+\frac{\mathrm{ghl}}{4(a+h)})+h(\frac{-\mathrm{fh}}{2}+\frac{\mathrm{ghl}}{4(a+h)})\right]\\ =\frac{2}{E{(a+h)}^{2}{I}_{1}}\frac{-{F}_{2}{h}^{3}}{24}(\mathrm{agl}+\mathrm{2f}{(a+h)}^{2})\end{array}` :math:`{\int}_{\mathrm{charp}}\frac{m{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{2}}=\frac{2}{E{(a+h)}^{2}{I}_{2}}\frac{-{F}_{2}{\mathrm{bh}}^{2}}{24}(\mathrm{agl}+\mathrm{2f}{(a+h)}^{2})` D'où: :math:`w=\frac{2}{E{(a+h)}^{2}}\frac{-{F}_{2}{h}^{2}}{24}(\mathrm{agl}+\mathrm{2f}{(a+h)}^{2})(\frac{h}{{I}_{1}}+\frac{b}{{I}_{2}})` **Application numérique:** :math:`{I}_{1}={\mathrm{2I}}_{2}` , :math:`h=\mathrm{2a}` , :math:`{F}_{2}=10000N` (vers la gauche) :math:`w=\frac{2}{E{(a+h)}^{2}{I}_{1}}(\mathrm{gl}+\mathrm{9gh})\underset{-3151003.65{\mathrm{N.m}}^{4}}{\underset{\underbrace{}}{\frac{-{F}_{2}{h}^{3}(h+\mathrm{2b})}{48}}}` Couple ponctuel :math:`\Gamma`en :math:`{C}_{1}` ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ :math:`\begin{array}{}{\int}_{\mathrm{pot}}\frac{m{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{1}}=\frac{h}{{\mathrm{3EI}}_{1}}\frac{\Gamma h}{2(a+h)}\left[(\frac{\mathrm{fh}}{2}+\frac{\mathrm{glh}}{4(a+h)})+(\frac{-\mathrm{fh}}{2}+\frac{\mathrm{glh}}{4(a+h)})\right]\\ =\frac{2}{E{(a+h)}^{2}{I}_{1}}\frac{\Gamma {h}^{3}\mathrm{lg}}{24}\end{array}` :math:`\begin{array}{}{\int}_{\mathrm{charp}}\frac{m{M}_{\mathrm{iso}}}{{\mathrm{EI}}_{2}}=\frac{b}{{\mathrm{3EI}}_{2}}\frac{\Gamma h}{2(a+h)}\left[-(\mathrm{2a}+h)(\frac{\mathrm{fh}}{2}+\frac{\mathrm{glh}}{4(a+h)})+h(\frac{-\mathrm{fh}}{2}+\frac{\mathrm{glh}}{4(a+h)})\right]\\ =\frac{-2}{E{(a+h)}^{2}{I}_{2}}\frac{\Gamma \mathrm{bh}}{24}(\mathrm{agl}+\mathrm{2f}{(a+h)}^{2})\end{array}` **Application numérique:** :math:`{I}_{1}={\mathrm{2I}}_{2}` ; :math:`h=\mathrm{2a}` ; :math:`\Gamma =-100000\mathrm{N.m}` :math:`w=\frac{2}{E{(a+h)}^{2}{I}_{1}}\underset{-3151003.65{\mathrm{N.m}}^{4}}{\underset{\underbrace{}}{\frac{\Gamma {h}^{2}}{24}}}(\mathrm{gl}(h-b)-\mathrm{9fhb})` Calcul de :math:`d=\int\frac{m\cdot {M}_{1}}{\mathrm{EI}}` ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ :math:`{\int}_{\mathrm{pot}}\frac{{\mathrm{mM}}_{1}}{{\mathrm{EI}}_{1}}=\frac{\mathrm{2h}}{{\mathrm{3EI}}_{1}}\frac{\mathrm{glh}}{4(a+h)}\frac{h}{a+h}=\frac{2}{E{(a+h)}^{2}{I}^{1}}\frac{{\mathrm{glh}}^{3}}{12}` :math:`{\int}_{\mathrm{charp}}\frac{m{M}_{1}}{{\mathrm{EI}}_{2}}=\frac{\mathrm{2b}}{{\mathrm{EI}}_{2}}\left[\frac{1}{2}(\frac{h}{a+h})+\frac{1}{6}(\frac{a}{a+h})\right]\frac{\mathrm{glh}}{4(a+h)}=\frac{2}{E{(a+h)}^{2}{I}_{2}}\frac{\mathrm{glbh}(\mathrm{3h}+a)}{24}` D'où (on constate que :math:`d` ne dépend pas de :math:`f` ): :math:`d=\frac{2}{E{(a+h)}^{2}}\frac{\mathrm{glh}}{24}(\frac{{\mathrm{2h}}^{2}}{{I}_{1}}+\frac{b(\mathrm{3h}+a)}{{I}_{2}})` **Application numérique:** :math:`{I}_{1}=2{I}_{2}` , :math:`h=\mathrm{2a}` :math:`d=\frac{2}{E{(a+h)}^{2}{I}_{1}}g\frac{{\mathrm{lh}}^{2}}{24}\underset{-4874.2564{\mathrm{N.m}}^{4}}{\underset{\underbrace{}}{(\mathrm{2h}+\mathrm{7b})}}` Récapitulatif des déplacements :math:`{u}_{c}`et :math:`{v}_{c}` ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ :math:`{I}_{1}=5.0E-4{m}^{4}` :math:`E=210000\mathrm{MPA}` .. csv-table:: "**CAS**", ":math:`X`", ":math:`X\stackrel{ˉ}{d}`", ":math:`{w}_{v}`" "pression sur :math:`{C}_{1}C`", "18672.994", "91016960.3", "–184930109.4" ":math:`{F}_{1}` en :math:`C`", "41422.161", "201902233.4", "–315100365.0" ":math:`{F}_{2}` en :math:`{C}_{1}`", "8284.432", "40380445.6", "–63020073.0" ":math:`\Gamma` en :math:`{C}_{1}`", "4916.724", "23965373.4", "14775091.25" .. csv-table:: "**CAS**", ":math:`{w}_{h}`", ":math:`{u}_{c}(m)`", ":math:`{v}_{c}(m)`" "pression sur :math:`{C}_{1}C`", "83519999.94", "0.0110476", "–0.012422374" ":math:`{F}_{1}` en :math:`C`", "0.00", "0.00", "–0.01497330" ":math:`{F}_{2}` en :math:`{C}_{1}`", "–226872262.8", "–0.03000956", "–0.00299466" ":math:`\Gamma` en :math:`{C}_{1}`", "206790328.5", "0.0273532", "–0.001215646" **Note:** :math:`d=\frac{2}{E{(a+h)}^{2}{I}_{1}}g\stackrel{ˉ}{d}` , avec: :math:`\stackrel{ˉ}{d}=4874.2564{m}^{4}` :math:`w=\frac{2}{E{(a+h)}^{2}{I}_{1}}(g{w}_{v}+f{w}_{h})` voir plus haut :math:`{u}_{c}=\frac{2}{E{(a+h)}^{2}{I}_{1}}{w}_{H}` ; :math:`{v}_{c}=\frac{2}{E{(a+h)}^{2}{I}_{1}}({w}_{v}+X\stackrel{ˉ}{d})` :math:`\frac{2}{E{(a+h)}^{2}{I}_{1}}=1.32275132E-10{N}^{-1}{m}^{-4}` **Comparaison** **Aster- référence analytique (R.)** .. csv-table:: "CAS", "", "**Moment en** :math:`C(\mathrm{N.m})`", "**Réaction** :math:`{H}_{A}(N)`", "**Réaction** :math:`{V}_{A}(N)`", "**Déplacement** :math:`{u}_{c}(m)`", "**Déplacement** :math:`{v}_{c}(m)`" ":math:`P` sur :math:`{C}_{1}C`", "R : *Aster* :", "18672.994 18673.20", "5175.37 5175.36", "24233.24 24233.2", "0.0110476 0.0110472", "–0.012422374 –0.0124233" ":math:`{F}_{1}` en :math:`C`", "R : *Aster* :", "41422.161 41422.40", "4881.487 4881.47", "10000.00 10000.0", "0.00000 0.0000", "–0.01497330 –0.0" ":math:`{F}_{2}` en :math:`{C}_{1}`", "R : *Aster* :", "8284.432 8284.34", "5976.297 5976.31", "4000.00 4000.0", "–0.03000956 –0.0300098", "–0.00299466 –0.00299450" ":math:`\Gamma` en :math:`{C}_{1}`", "R : *Aster* :", "4916.724 4916.62", "4576.394 4576.38", "5000.00 5000.0", "0.0273532 0.0273536", "–0.001215646 –0.00121583" **Nota:** Le calcul *Aster* a été réalisé en prenant des éléments très élancés, de telle sorte que: :math:`{\mathrm{Sl}}^{2}\ll I` . Ainsi, l’énergie de flexion est prédominante. Les valeurs du calcul *Aster* sont issues du cas-test :math:`\mathrm{VPCS}` appelé :math:`\mathrm{SSLL14}` , avec les données suivantes: :math:`{I}_{1}=5.0E-4{m}^{4}` ; :math:`{I}_{2}=2.5E-4{m}^{4}` ; :math:`E=210000\mathrm{MPa}` :math:`h=\mathrm{2a}=8m` ; :math:`l=20m` ; :math:`b=\frac{l}{2}\sqrt{1.16}` :math:`p=3000\mathrm{N.m}` (vers le bas), :math:`{F}_{1}=20000N` (vers le bas), :math:`{F}_{2}=10000N` (vers la gauche), :math:`\Gamma =-100000\mathrm{Nm}` (sens aiguille de montre).