1. Determinar la movilidad o nΓΊmero de grados de libertad de los mecanismos presentados. π = 3(π β 1) β 2π½1 β π½2 π = πΓΊ
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1. Determinar la movilidad o nΓΊmero de grados de libertad de los mecanismos presentados.
π = 3(π β 1) β 2π½1 β π½2 π = πΓΊππππ ππ ππ πππππππ = 7 π½1 = π½π’ππ‘ππ ππππππππππ = 7 π½2 = π½π’ππ‘ππ π π’ππππππππ = 1
π = 3(7 β 1) β 2(7) β (1) = 3
π = 3(π β 1) β 2π½1 β π½2 π = πΓΊππππ ππ ππ πππππππ = 10 π½1 = π½π’ππ‘ππ ππππππππππ = 12 π½2 = π½π’ππ‘ππ π π’ππππππππ = 0 π = 3(10 β 1) β 2(12) = 3
π = 3(π β 1) β 2π½1 β π½2 π = πΓΊππππ ππ ππ πππππππ = 12 π½1 = π½π’ππ‘ππ ππππππππππ = 15 π½2 = π½π’ππ‘ππ π π’ππππππππ = 0 π = 3(12 β 1) β 2(15) = 3
π = 3(π β 1) β 2π½1 β π½2 π = πΓΊππππ ππ ππ πππππππ = 6 π½1 = π½π’ππ‘ππ ππππππππππ = 6 π½2 = π½π’ππ‘ππ π π’ππππππππ = 0 π = 3(6 β 1) β 2(6) = 3