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STAVEBNÍ MECHANIKA 3 - SM 3 - Jenin

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– Výpočet natočení průřezů ϑ1, ϑ2 od momentu RML<br />

vzhledem k obrázku dostaneme<br />

(δM1|0 = 1) · ϑ21 =<br />

(δM2|L = 1) · ϑ22 =<br />

φ|RML L<br />

RML<br />

δM|δM <br />

1<br />

<br />

x<br />

(−1) · (1 −<br />

EIy L<br />

x<br />

) dx = −RMLL<br />

L 6EIy<br />

0<br />

L<br />

RML<br />

0<br />

EIy<br />

x x<br />

L L<br />

dx = RMLL<br />

3EIy<br />

(120)

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