In den folgenden Beíspielen wird die Lasteinzugsfläche eines deckengleichen Unterzuges berechnet.
b E = l e f f 2 ∗ t a n ( 60 ) = 2 , 1 m 2 ∗ t a n ( 60 ) = 1 , 82 m _ _ {\displaystyle b_{\mathrm {E} }={\cfrac {l_{\mathrm {eff} }}{2}}*tan(60)={\cfrac {2,1m}{2}}*tan(60)={\underline {\underline {1,82\,\mathrm {m} }}}} l E o = l P , o 2 = 6 , 0 m 2 = 3 , 0 m _ _ > 1 , 82 m ( b E i s t m a s s g e b e n d ) {\displaystyle l_{\mathrm {Eo} }={\cfrac {l_{\mathrm {P,o} }}{2}}={\cfrac {6,0m}{2}}={\underline {\underline {3,0\,\mathrm {m} }}}>1,82m~(b_{\mathrm {E} }~ist~massgebend)} l E u = l P , u 2 = 4 , 0 m 2 = 2 , 0 m _ _ > 1 , 82 m ( b E i s t m a s s g e b e n d ) {\displaystyle l_{\mathrm {Eu} }={\cfrac {l_{\mathrm {P,u} }}{2}}={\cfrac {4,0m}{2}}={\underline {\underline {2,0\,\mathrm {m} }}}>1,82m~(b_{\mathrm {E} }~ist~massgebend)}
b E = l e f f 2 ∗ t a n ( 60 ) = 2 , 1 m 2 ∗ t a n ( 60 ) = 1 , 82 m _ _ {\displaystyle b_{\mathrm {E} }={\cfrac {l_{\mathrm {eff} }}{2}}*tan(60)={\cfrac {2,1m}{2}}*tan(60)={\underline {\underline {1,82\,\mathrm {m} }}}} l E o = l P , o 2 = 3 , 0 m 2 = 1 , 5 m _ _ < 1 , 82 m ( l E o i s t m a s s g e b e n d ) {\displaystyle l_{\mathrm {Eo} }={\cfrac {l_{\mathrm {P,o} }}{2}}={\cfrac {3,0m}{2}}={\underline {\underline {1,5\,\mathrm {m} }}}<1,82m~(l_{\mathrm {Eo} }~ist~massgebend)} l E u = l P , u 2 = 4 , 0 m 2 = 2 , 0 m _ _ > 1 , 82 m ( b E i s t m a s s g e b e n d ) {\displaystyle l_{\mathrm {Eu} }={\cfrac {l_{\mathrm {P,u} }}{2}}={\cfrac {4,0m}{2}}={\underline {\underline {2,0\,\mathrm {m} }}}>1,82m~(b_{\mathrm {E} }~ist~massgebend)}
