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METODO DE LOS NODOS

2.5 Ton. c. 2.5 Ton. 2.5 Ton. 1.50m. d. 0.75 Ton. b. 1.50m. e. a. Rax. f. 2.00m. 2.00m. 2.00m. 2.00m. Ray. METODO DE LOS NODOS. Rby. 1.- OBTENCIÓN DE LAS REACCIONES :

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METODO DE LOS NODOS

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  1. 2.5 Ton c 2.5 Ton 2.5 Ton 1.50m d 0.75 Ton b 1.50m e a Rax f 2.00m. 2.00m. 2.00m. 2.00m. Ray METODO DE LOSNODOS Rby 1.- OBTENCIÓN DE LAS REACCIONES: ΣMa=2.50ton (2.0m) + 2.50ton (4.0m) + 2.50ton (6.0m) – 0.75ton (1.50) – Rby (8.0m) = 0 ; Rby = 3.61 ton ΣMb=- 2.50ton (2.0m) - 2.50ton (4.0m) - 2.50ton (6.0m) – 0.75ton(1. 50) + Ray (8.0m) = 0 ; Ray = 3.89 ton ΣFx= - 0.75ton + Rax = 0 ; Rax = 0.75 ton

  2. Nodo a: ab Rax = 0.75 θ Por lo tanto: ab = 6.48 ton. a af Por lo tanto: af = 4.44 ton. Ray = 3.89 • Representamos resultados en la armadura: 2.50 c 2.50 2.50 d 0.75 b 6.48 e a Rax = 0.75 4.44 f Ray = 3.89 Rby = 3.61

  3. Nodo e: de θ e Por lo tanto: de = 6.02 ton. ef Rby = 3.61 Por lo tanto: ef = 4.81 ton. • Representamos resultados en la armadura: 2.50 c 2.50 2.50 d 0.75 b 6.48 6.02 e a Rax = 0.75 4.44 4.81 f Ray = 3.89 Rby = 3.61

  4. Nodo b: 2.50 bc b θ θ Del sistema de ecuaciones: bc = 4.40 ton. bf = 2.08 ton. θ 6.48 bf • Representamos resultados en la armadura: 2.50 c 2.50 4.40 2.50 d 0.75 b 6.48 2.08 6.02 e a Rax = 0.75 4.44 4.81 f Ray = 3.89 Rby = 3.61

  5. Nodo d: 2.50 cd d θ 0.75 θ Del sistema de ecuaciones: cd = 4.40 ton. df = 2.55 ton. θ df 6.02 • Representamos resultados en la armadura: 2.50 c 2.50 4.40 2.50 4.40 d 0.75 b 2.55 6.48 2.08 6.02 e a Rax = 0.75 4.44 4.81 f Ray = 3.89 Rby = 3.61

  6. Nodo c: 2.50 c Por lo tanto: cf = 2.78 ton. θ θ 4.40 4.40 cf Y finalmente se demuestra el equilibrio en la ecuación x. • Representamos resultados en la armadura: 2.50 c 2.50 4.40 2.50 4.40 d 0.75 b 2.78 2.55 6.48 2.08 6.02 e a Rax = 0.75 4.81 4.44 f Ray = 3.89 Rby = 3.61

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