1 / 7

Rumus Dasar

Rumus Dasar. Berdasarkan Kekuatan M M = 1/10. q.l 2 s = W W = 1/6 b.h 2. = tegangan lentur kayu M = Momen lentur yg terjadi W = Momen perlawanan dr penampang. Berdasar kekakuan. 5.q.l 4 f = perletakan di atas dua tumpuan

aolani
Download Presentation

Rumus Dasar

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Rumus Dasar Berdasarkan Kekuatan M M = 1/10. q.l2 s = W W = 1/6 b.h2 • = tegangan lentur kayu M = Momen lentur yg terjadi W = Momen perlawanan dr penampang

  2. Berdasar kekakuan 5.q.l4 f = perletakan di atas dua tumpuan 384.E.I 2,5.q.l4 f = perletakan di atas tiga tumpuan 384.E.I

  3. Contoh perhitungan bekisting Lantai • Suatu lantai beton dengan ketebalan 150 mm, bahan cetakan dari multipleks sedang balok dan tiang memakai kayu kelas II, dengan ketentuan : • Multipleks tebal 15 mm, E = 95.000 kg/cm2 • Kayu Kls II : slt = 10 MPa (100 kg/cm2) ; E = 100.000 kg/cm2 stk = 2,5 MPa (25 kg/cm2) ; t = 12 kg/cm2 • Dimensi balok kayu 6/12 – 400 (cm) • Berat satuan volume beton = 2400 kg/m3 • Lendutan yg diijinkan 1/400 l • Beban pelaksanaan = 1,5 kN/m2

  4. Menentukan beban terbagi rata > berat lantai beton 0,15 x 24 = 3,6 kN/m2 > berat bekisting = 0,2 kN/m2 > Beban pelaksanan = 1,5 kN/m2 q = 5,3 kN/m2 = 530 kg/m2 Beban per cm’ adalah q’ = 530 kg/m2 x 100 cm = 5,3 kg/cm’ • Menentukan jarak l1 W = 1/6.b.h2 = 1/6 x 100 x 1,52 = 37, 5 cm3 M = 0,1.q.l2 = 0,1 x 5,3 x l2 = 0,53 l2 I = 1/12.b.h3 = 1/12 x 100 x 1,53 = 28,125 cm4

  5. 2,5.q.l4 • f ≥ • 384.E.I • 2,5 x 5,3 x l4 • 1/400.l ≥ • 384 x 95.000 x 28,125 • 5300 x l3 ≤ 10,26 x 108 • l3≤ 193584,91 • l ≤ 57,85 cm • Dipakai l yg paling kecil yaitu 55 cm M s ≥ W 0,53 l2 100 ≥ 37, 5 l2 ≤ 3750 : 0,53 l ≤ sqrt (7075,47) l ≤ 84,11 cm

  6. W = 1/6.b.h2 = 1/6 x 6 x 122 = 144 cm3 M = 0,1.q.l2 = 0,1 x 2,97 x l2 = 0,297. l2 I = 1/12.b.h3 = 1/12 x 6 x 123 = 864 cm4 • Menentukan l2 c-t-c 55 cm Beban terbagi rata q’ = 55 cm x 540 kg/m2 = 2,97 kg/cm M s ≥ W 0,297.l2 100 ≥ 144 0,297. l2 ≤ 14400 l2≤ 48484,85 l ≤ 220,2 cm • 2,5.q.l4 • f ≥ • 384.E.I • 2,5 x 2,97 x l4 • 1/400.l ≥ • 384 x 100.000 x 864 • 2970 x l3 ≤ 3,318 x 1010 • l3 ≤ 11170909,09 • l ≤ 223,54 cm

  7. Menentukan l3 c-t-c 200 cm Beban terbagi rata q’ = 200 cm x 550 kg/m2 = 10,8 kg/cm W = 1/6.b.h2 = 1/6 x 6 x 122 = 144 cm3 M = 0,1.q.l2 = 0,1 x 10,8 x l2 = 1,08.l2 I = 1/12.b.h3 = 1/12 x 6 x 123 = 864 cm4 M s ≥ W 1,08.l2 100 ≥ 144 1,08. l2 ≤ 14400 l2≤ 13333,33 l ≤ 115,5 cm • 2,5.q.l4 • f ≥ • 384.E.I • 2,5 x 10,8 x l4 • 1/400.l ≥ • 384 x 100.000 x 864 • 10800 x l3 ≤ 3,318 x 1010 • l3 ≤ 3072000 • l ≤ 145,37 cm

More Related