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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...
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