Page:Mémoires de l’Académie des sciences, Tome 8.djvu/718

$\left.{\begin{aligned}{\frac {du'}{d\eta }}=&-{\frac {dy}{dx}},&{\frac {du'}{d\zeta }}=&-{\frac {dz}{dx}},\\{\frac {dv'}{d\eta }}=&0,&{\frac {dw'}{d\zeta }}=&0,&{\frac {dv'}{d\zeta }}+&{\frac {dw'}{d\eta }}=0,\\{\frac {d^{2}u'}{d\zeta ^{2}}}=&0,&{\frac {d^{2}u'}{d\eta d\zeta }}=&0,&{\frac {d^{2}u'}{d\eta ^{2}}}=&0,\\{\frac {d^{2}v'}{d\eta ^{2}}}=&{\frac {1}{4}}{\frac {d^{2}y}{dx^{2}}},&{\frac {d^{2}v'}{d\eta d\zeta }}=&{\frac {1}{4}}{\frac {d^{2}z}{dx^{2}}},&{\frac {d^{2}v'}{d\zeta ^{2}}}=&-{\frac {1}{4}}{\frac {d^{2}y}{dx^{2}}},\\{\frac {d^{2}w'}{d\eta ^{2}}}=&-{\frac {1}{4}}{\frac {d^{2}z}{dx^{2}}},&{\frac {d^{2}w'}{d\eta d\zeta }}=&{\frac {1}{4}}{\frac {d^{2}y}{dx^{2}}},&{\frac {d^{2}w'}{d\zeta ^{2}}}=&{\frac {1}{4}}{\frac {d^{2}z}{dx^{2}}}.\end{aligned}}\right\}\quad$(11)

D’après cela, nous aurons

${\begin{aligned}{\frac {d\mathrm {P} _{3}}{d\eta }}=&-k\left(3{\frac {d^{2}u'}{dxd\eta }}+{\frac {d^{2}v'}{d\eta ^{2}}}+{\frac {d^{2}w'}{d\eta d\zeta }}\right)={\frac {5k}{2}}{\frac {d^{2}y}{dx^{2}}},\\{\frac {d\mathrm {P} _{3}}{d\zeta }}=&-k\left(3{\frac {d^{2}u'}{dxd\zeta }}+{\frac {d^{2}v'}{d\eta d\zeta }}+{\frac {d^{2}w'}{d\zeta ^{2}}}\right)={\frac {5k}{2}}{\frac {d^{2}z}{dx^{2}}}\,;\end{aligned}}$

et les équations (10) deviendront

$\left.{\begin{aligned}\mathrm {Y} +{\frac {\varepsilon ^{2}\rho }{8}}\left({\frac {d^{2}\mathrm {Y} '}{d\eta ^{2}}}+{\frac {d^{2}\mathrm {Y} '}{d\zeta ^{2}}}+2{\frac {d^{2}\mathrm {X} '}{dxd\eta }}\right)&={\frac {5k\varepsilon ^{2}}{8\rho }}{\frac {d^{4}y}{dx^{4}}},\\\mathrm {Z} +{\frac {\varepsilon ^{2}\rho }{8}}\left({\frac {d^{2}\mathrm {Z} '}{d\eta ^{2}}}+{\frac {d^{2}\mathrm {Z} '}{d\zeta ^{2}}}+2{\frac {d^{2}\mathrm {X} '}{dxd\zeta }}\right)&={\frac {5k\varepsilon ^{2}}{8\rho }}{\frac {d^{4}z}{dx^{4}}}.\end{aligned}}\right\}\quad$(12)

Dans le cas de l’équilibre, les forces $\mathrm {Y,Z,X',Y',Z'} ,$ seront données en fonctions de $x,\eta ,\zeta .$ Dans le cas du mouvement, il faudra les diminuer respectivement des différences partielles secondes de $y,z,u',v',w',$ relatives au temps $t\,;$ et si l’on suppose qu’aucune force donnée n’agisse sur le points de la verge, cela reviendra à faire

$\mathrm {Y} =-{\frac {d^{2}y}{dt^{2}}},\quad \mathrm {Z} =-{\frac {d^{2}z}{dt^{2}}},\quad \mathrm {X} '=-{\frac {d^{2}u'}{dt^{2}}},\quad \mathrm {Y} '=-{\frac {d^{2}v'}{dt^{2}}},\quad \mathrm {Z} '=-{\frac {d^{2}w'}{dt^{2}}}.$

En vertu des équations (11) on aura donc