PORTAL TIGA SENDI - burhan

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� Reaksi perletakan ∑ M B = 0 � A V . 10 + q . 5 . 2,5 = 0 Av = - (1 . 5. 2,5)/10 = - 1,25 t ( arah ke bawah ) ∑ M A = 0 � B V . 10 – q . 5 . 2,5 = 0 B V = (1. 5 . 2,5) / 10 = + 1,25 t ( arah ke atas) ∑ M S = 0 � A V . 5 + A H . 5 + q . 5 . 2,5 = 0 A H = (- 1,25 . 5 – 1 . 5 . 2,5)/5 = -3,75 t ( arah ke kiri) B H . 5 – BV . 5 = 0 B H = 1,25 ton ( arah ke kiri)
Gaya-gaya dalam � Momen Pada batang AC M A = 0 M x = A H . x – ½ q . x 2 M mak � dMx/dx = 0 � x = 3,75 / 1 = 3,75 m M mak = 3,75 . 3,75 – ½.1.3,75 2 = 7,03125 tm M tengah = 3,75 . 2,5 – ½.1.2,5 2 = 6,25 tm M C = M x=5 m = 3,75 . 5 – ½.1.5 2 = 6,25 tm Pada Batang CD M x = A H . 5 – q . 5. 2,5 –A V . x = 3,75 . 5 – 5 . 2,5 – 1,25 . x = 6,25 – 1,25x M x=0 = M C = 6,25 – 1.25.0 = 6,25 tm M x=5 = M S = 6,25 – 1,25 . 5 = 0 tm M x=10 = M D = 6,25 – 1,25 . 10 = -6,25 tm Pada batang BD M x = B H . X M X=0 = o tm M X=5 = M D = -1,25 . 5 = -6,25 tm
Page 1 and 2: PORTAL TIGA SENDI Soal 1 Ditanyakan
Page 3 and 4: C � Bidang momen BMD � Batang A
Page 5: Soal 2 : Portal 3 sendi Diketahui P
Page 9 and 10: NFD (Normal force diagram) � Gaya

PORTAL TIGA SENDI - burhan

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