Mekanika Teknik I: Tumpuan Jepit Balok Gabungan [PDF]

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Mekanika Teknik I Tumpuan Jepit Balok Gabungan



Penyelesaian : 1. Perhitungan Reaksi Perletakan ΣΣV = 0 Va – 2t – 1t – 2t – 1t = 0 Va = +6t ΣM = 0 Ma + 2t . 2 m + 1t . 3,5 m + 2t . 6 m + 1t . 8 m = 0 Ma + 4 tm + 3,5 tm + 12 tm + 8 tm = 0 Ma + 27,5 tm = 0 Ma = 27,5 tm



Garis Lintang ( Bidang D ) DA = +Va = +6t DB = +Va – Pb = +6t – 2t = +4t DC = +Va – Pb – Pc = +6t – 2t – 1t = +3t DD = +Va – Pb – Pc - Pd = +6t – 2t – 1t – 2t = +1t DE = +Va – Pb – Pc – Pd - Pe = +6t – 2t – 1t – 2t – 1t = 0



Garis Momen ( Bidang M ) Ma =  27,5 tm + Va . 0 m = -27,5 m Mb =  27,5 tm + Va . 2 m – Pb . 0 m =  27,5 tm + 6t . 2m – 0 tm =  -15,5 tm Mc =  27,5 tm + Va . 3,5 m – Pb . 1,5 m – Pc . 0 m = 27,5 tm + 6t . 3,5m – 2t . 1,5m - 0 tm = -9,5 tm Md =  27,5 tm + Va . 6 m – Pb . 4 m – Pc . 2,5 m – Pd . 0 m =  27,5 tm + 6t . 6m – 2t . 4m – 1t . 2,5 m - 0 tm =  -2 tm Me =  27,5 tm + Va . 8 m – Pb . 6 m – Pc . 4,5 m – Pd . 2 m – Pe . 0 m =  27,5 tm + 6t . 8m – 2t . 6m – 1t . 4,5 m – 2t . 2 m - 0 tm = 0 tm



Diagram



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