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Proyecto de Estatica
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Informe Proyecto de Estática del Tercer Parcial
UNIVERSIDAD DE LAS FUERZAS ARMADAS ESPE
03/01/2014
Ing. Naranjo Oswaldo
INTEGRANTES:Oviedo Michelle
Rodríguez CatherineZambrano Santiago
Resolver el ejercicio 6.50 mediante el método de las secciones:
∑ Fx=0
Ax=0
∑ Fy=0
Ay+Ly−1−2−2−2−2−2−1=0
Ay+Ly=12
∑M A=0
−2 (2 )−2 (4 )−2 (6 )−2 (8 )−2 (10 )−1 (12 )+12 Ly=0
Ly=6KN
Ay=6KN
θ=tan−1( 16 )=9.46 °
α=tan−1( 2.56 )=22.61°δ= tan−1( 0.66
2)=18.26 °
∑M D=0
(1 )F EG .cos (9.46° )+2 (2 )+4 (5 )=0
FEG=16.22KN
∑MG=0
(1 )F DF .cos (22.61 ° )+2 (2 )+2 (4 )−6 (5 )=0
FDF=13.01KN
∑ Fx=0
−FDF .cos (22.61 ° )−FDG .cos (18.26 ° )+FEG .cos (9.46 ° )=0
FDG=4.21KN