01 rozÅ¡Ã­Å�enÃ½ prÅ¯vodce

# 4 v n m provedeme cyklickou zm nu index b a c b ma ab

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2 3EI 3 2 v2 1 =− − (v1 + 2v2 ) = δ 3, 2 = −1,893 ⋅ 10− 4 (kN ) −1 , 2 EI 2 6 EI 3 δ 3, 3 = v13 v3 l 2 + 2+ (v12 + v1v2 + v2 ) = 1,982 ⋅ 10− 3 (kN ) −1 m. 3EI1 3EI 2 3EI 3 Vy íslení sou initel δi,0 v etn rozm rových jednotek je: δ1, 0 = Fpp' Fpp' (l + p ' ) = 1,235 ⋅ 10− 4 [l ], δ 2,0 = (l + p ) = 1,728 ⋅ 10− 4 [l ], 6 EI 3l 6 EI 3l δ 3, 0 = − Fpp' [v1 (l + p' ) + v2 (l + p)] = −1,432 ⋅ 10−3 m. 6 EI 3l Pro numerický výpo et byly zavedeny pom rné deformace δ i′,k = 10–3Eδi,k a δ i′, 0 = 10–3Eδi,0 o velikostech δ '1,1 = 10−3 Eδ 1,1 = 10 −3 E 6 8 6 8 =+ = 1,664m −3 , + E ⋅ 0,005 3E ⋅ 0,006 5 3⋅ 6 8 = 0,222m −3 , 6⋅6 62 8 δ '1,3 = 10−3 Eδ 1, 3 = − − (2 ⋅ 6 + 4) = −7,156m −2 , 2⋅5 6⋅6 δ '2 , 2 = 1,4444m −3 , δ '2 , 3 = −5,111m −2 , δ '3, 3 = 53,111m −1 , δ '1, 2 = 10 −3 Eδ 1, 2 = δ '1, 0 = 3,333kNm −2 , δ '2, 0 = 4,667kNm −2 , δ '3, 0 = −38,667kNm −1. Soustava p etvárných rovnic má tvar 1,644 X 1...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online