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01 rozÅ¡Ã­Å�enÃ½ prÅ¯vodce

# 36 o hly 59 kde t1x td th hx vka pr ezu u nosnku

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Unformatted text preview: + 0,222 X 2 − 7,156 X 3 = −3,333, 0,222 X 1 + 1,444 X 2 − 5,111X 3 = −4,667, − 7,156 X 1 − 5,111X 2 + 53,511X 3 = 38,667 a ešení X 1 = M a = 2,945kNm (po sm m ) , - 30 (50) - Silová metoda X 2 = M b = 0,403kNm (proti sm m ) , X 3 = Rbx = 1,155kN (←). Obr. 6.9: Pr b hy M, V, N na jednoduchém rámu Výpo et zbývajících statických veli in provedeme na základní soustav . Složky reakcí ze superpozi ních vztah Rax = 0 + 0 ⋅ 2,945 + 0 ⋅ 0,403 + 1 ⋅ 1,155 = 1,155kN (→), Raz = 8⋅2 1 1 6−4 + − 2,945 + ⋅ 0,403 + ⋅ 1,155 = 1,971kN (↑), 8 8 8 8 Rbz = 8⋅6 1 1 6−4 + ⋅ 2,945 + − ⋅ 0,403 + − ⋅ 1,155 = 6,029kN (↑); 8 8 8 8 nebo m žeme využít statických podmínek rovnováhy Fix = 0, M ib = 0, M i a = 0. - 31 (50) - Statika I P itom platí kontrola Fiz = 0 : − Raz − Rbz + F = 0. Ohybové momenty ze superpozi ních vztah M ac = M a = 2,945kNm, M bd = M b = 0,403kNm, M ca = 0 + 1 ⋅ 2,945 + 0 ⋅ 0,403 + (−6) ⋅ 1,155 = −3,985kNm = M cd , M dc= 0 + 0...
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