Armaduras
Enviado por luk_su • 7 de Octubre de 2014 • 239 Palabras (1 Páginas) • 142 Visitas
METODO DE SECCIONES
∑ Mf = -3.28 ( 2.31) – 3.28 ( 4.62) – 1.64 ( 6.93) + 9.84 ( 6.93) + EG ( 2.31 ) = 0
EG = (34.0956)/(2.31) = 14.76
∑ Ma = 3.28 ( 2.31 ) + 3.28 ( 4.62 ) – EF ( sen 45º ) (4.62) = 0
EF = (22,73)/(sen 45º ( 4.62 )) = 6.96
∑ Me = - 1.64 ( 4.62) – 3.28 ( 2.31 ) + 9.84 ( 4.62 ) + DF ( cos 18.43º )( 1.54) = 0
DF = (30.31)/(cos 18.43 ( 1.54 )) = 20.74
∑ Md = - 1.64 ( 4.62) – 3.28 ( 2.31 ) + 9.84 ( 4.62 ) + CE ( 1.54) = 0
CE = (30.31)/( 1.54) = 19.68
∑ Mc = - 1.64 ( 2.31) + 9.84 ( 2.31 ) + BD ( cos 18.43º )( .77) = 0
BD = (18.94)/(cos 18.43 ( .77 )) = 25.93
∑ Ma = – 3.28 ( 2.31 ) + CD ( sen 33.69 )( 2.31 ) = 0
CD = (-7.58)/(sen 33.69 ( 2.31 )) = - 5.91
∑ Mc = - 1.64 ( 2.31 ) + 9.84 ( 2.31 ) + AB ( cos 18.43º )( .77) = 0
AB = (18.94)/(cos 18.43 ( .77 )) = 25.93
∑ Mb = - 1.64 ( 2.31) + 9.84 ( 2.31 ) +AC ( .77) = 0
DF = (18.94)/(( .77 )) = 24.60
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