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