PORTAL TIGA SENDI - burhan

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NFD (Normal force diagram)

� Gaya normal

Batang AC

N AC = A V = 1,25 ton (+)

Batang CD

N CD = B H = q.l-A H = 1 . 5 – 3,75 = 1,25 t (-)

Batang DB

N DB = B V = 1,25 t (-)

SFD ( Shear force diagram) � Bidang lintang M mak � di D = 0 D A = + 3,75 t D C = 3,75 – q . 5 = -1,25 t Batang CD dipandang yang bekerja A V = 1,25 t ( ke bawah), sehingga : D C = -1,25 t � sampai titik D Batang DB, dipandang dari B B H = 1,25 t (arah ke dalam/kiri) D B = + 1,25 t D D = + 1,25 t

NFD (Normal force diagram) � Gaya normal Batang AC N AC = A V = 1,25 ton (+) Batang CD N CD = B H = q.l-A H = 1 . 5 – 3,75 = 1,25 t (-) Batang DB N DB = B V = 1,25 t (-)

Page 1 and 2: PORTAL TIGA SENDI Soal 1 Ditanyakan
Page 3 and 4: C � Bidang momen BMD � Batang A
Page 5 and 6: Soal 2 : Portal 3 sendi Diketahui P
Page 7: Gaya-gaya dalam � Momen Pada bata

PORTAL TIGA SENDI - burhan

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