>Balok Statis Tertentu – contoh soal

Posted: May 30, 2010 in Uncategorized

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Dibawah ini saya sajikan contoh perhitungan suatu konstruksi dengan beban merata berbentuk segitiga dan beban terpusat dari berbagai arah:

Diketahui suatu gambar struktur denagan tumpuan sendi dan rol sebagai berikut:

MENGHITUNG REAKSI TUMPUAN

ΣMB = 0

-( ½ x 2 x 2 ) x 6.6667 + RAv x 6 – ( 1×6) x 3 + (4×2) – ( 4×1) =0

-13.3334 + 6 RAV – 18 + 8 – 4 =0

6 RAv – 27.3334 =0

RAv = 27.3334 / 6

Kontrol ΣV = 0RAv + RBv = 1/2×2×2 + 1×6 +4

4.5556 + 7.4444 = 12 ton

( O.K )

= 4.5556 ton

ΣMA = 0

-4 x 1 + (4×8) – RBv x 6 + (1×6)x3 + 4×2 – (4 x1) =0

-4 + 32 – 6RBv – 18 – 1.3334 =0

– 6 RBv + 44.6666 =0

RBv = -44.6666 / -6

= 7.4444 ton

MENGHITUNG BIDANG M ( MOMEN )

M 2m = – ( 1/2 x 2 x 2 ) x 0.6667

= – 1.3334 ton

M 5m = – ( 1/2 x 2 x 2 ) x 3.6667 + RAv x 3 – ( 1 x 6 ) x 1.5

= -7.3334 + 13.6668 – 9

= -2.6666 ton

M 8m = – ( 1/2 x 2 x 2 ) x 6.6667 + RAv x 6 – ( 1 x 6 ) x 3

= -13.3334 + 27.3336 – 18

= -3.9998 ton

M 10m = – ( 1/2 x 2 x 2 ) x 8.6667 + RAv x 8 – ( 1 x 6 ) x 5 + 4

= -17.3334 + 36.4448 – 30 + 14.8888 + 4

= 0 ton

MENGHITUNG BIDANG Q ( GAYA LINTANG )

Q 2m = -( 1/2 x 2 x 2 ) +RAv

= – 2 + 4.5556

= 2.5556 ton

Q 8m = -( 1/2 x 2 x 2 ) +RAv – (1 x 6)

= – 2 + 4.5556 – 6

= -3.4444 ton

Q 10m = -( 1/2 x 2 x 2 ) +RAv – (1 x 6) + RBv – 4

= – 2 + 4.5556 – 6 + 7.4444 – 4

= 0 ton

MENGHITUNG BIDANG N ( GAYA NORMAL )

RBh – ( 4 x 1 ) = 0

RBh = 4 ton

tolong di koreksi kalu ada yang salah bisa masukin dibawah…

Dikutip dari www.ilmusipil.com

http://faqihns.wordpress.com/2010/02/18/contoh-hitungan-mekanika-teknik/

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