# Die Biegungsmomente der gleichförmig belasteten Platte

• H. Marcus

## Zusammenfassung

Wir haben bereits vorhin die Lastanteile
$$\left. {\begin{array}{*{20}{c}} {{{p}_{x}} = p\frac{{l_{y}^{4}}}{{l_{x}^{4} + l_{y}^{4}}}} \\ {{{p}_{y}} = p\frac{{l_{x}^{4}}}{{l_{x}^{4} + l_{y}^{4}}}} \\ \end{array} } \right\}$$
(12)
ermittelt und erhalten nunmehr, wenn wir die Streifenbreite b = 1 wählen, der Reihe nach:
$$\begin{array}{*{20}{c}} {{{\mathfrak{M}}_{{{{x}_{{\max }}}}}} = p\frac{{l_{x}^{2}}}{8} \cdot \frac{{l_{y}^{4}}}{{l_{x}^{4} + l_{y}^{4}}},} \hfill & {{{\mathfrak{M}}_{{0x}}} = p\frac{{l_{x}^{2}}}{8},} \hfill \\ {{{\mathfrak{M}}_{{{{y}_{{\max }}}}}} = p\frac{{l_{y}^{2}}}{8} \cdot \frac{{l_{x}^{4}}}{{l_{x}^{4} + l_{y}^{4}}},} \hfill & {{{\mathfrak{M}}_{{0y}}} = p\frac{{l_{y}^{2}}}{8},} \hfill \\ {\frac{{{{\mathfrak{M}}_{{{{x}_{{\max }}}}}}}}{{{{\mathfrak{M}}_{{0x}}}}} = \frac{{l_{y}^{4}}}{{l_{x}^{4} + l_{y}^{4}}},} \hfill & {\frac{{{{\mathfrak{M}}_{{{{y}_{{\max }}}}}}}}{{{{\mathfrak{M}}_{{0y}}}}} = \frac{{l_{x}^{4}}}{{l_{x}^{4} + l_{y}^{4}}},} \hfill \\ \end{array}$$
$$\left. {\begin{array}{*{20}{c}} {{{\varphi }_{x}} = {{\varphi }_{y}} = \varphi = \frac{5}{6} \cdot \frac{{l_{x}^{2}l_{y}^{2}}}{{l_{x}^{4} + l_{y}^{4}}}} \\ {{{M}_{{{{x}_{{\max }}}}}} = {{p}_{x}}\frac{{l_{x}^{2}}}{8}{{v}_{a}}} \\ {{{M}_{{{{y}_{{\max }}}}}} = {{p}_{y}}\frac{{l_{y}^{2}}}{8}{{v}_{a}}} \\ \end{array} } \right\}$$
(13)
wobei
$${{v}_{a}} = 1 - \frac{5}{6} \cdot \frac{{l_{x}^{2}l_{y}^{2}}}{{l_{x}^{4} + l_{y}^{4}}}.$$