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STAVEBNÍ MECHANIKA 3 - SM 3 - Jenin

STAVEBNÍ MECHANIKA 3 - SM 3 - Jenin

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vyjádřit vztahem<br />

ϕy = N0ϕy0(= − f zL3 ) + NLϕyL(=<br />

24EIy<br />

f zL3 ) (130)<br />

24EIy<br />

kde N0 a NL jsou jednotkové bázové funkce znázorněné graficky na obrázku.<br />

Spojením rovnic (129) a (130) dostaneme<br />

<br />

w(x) = −<br />

wmax(x = L<br />

2 ) = f zL 4<br />

96EIy<br />

ϕy dx + C(= 0, z podminky w(0) = 0) = f zL 3 x<br />

24EIy<br />

<br />

1 − x<br />

<br />

L<br />

(131)

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