Konstruksi Kayu
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![Page 1: Konstruksi Kayu](https://reader037.fdocument.pub/reader037/viewer/2022100304/5571f88149795991698d9185/html5/thumbnails/1.jpg)
Cek ketidaksamaan struktur
m = 2 j – 3
17 = 2(10) – 3
17 = 17 ( ok )
Jumlah reaksi tumpuan
Hitung reaksi tumpuan ∑Ma = 0P1 . 3 + P2 . 4 + P3 . 5 + P4 . 6 + P5 . 7 – Rbv . 10 = 01 . 3 + 2 . 4 + 3 . 5 + 2 . 6 + 2 . 7 – Rbv . 10 = 03 + 8 + 15 + 12 + 14 – 10 . Rbv = 0
52 – 10 . Rbv = 0 Rbv = 5,2 ton
∑Mb = 0- P5 . 3 - P4 . 4 - P3 . 5 - P2 . 6 – P1 . 7 + Rav . 10 = 0-2 . 3 + 2 . 4 + 3 . 5 + 2 . 6 + 1 . 7 + Rav . 10 = 03 + 8 + 15 + 12 + 14 + 10 . Rav = 0
-48 + 10 . Rav = 0 Rav = 4,8 ton
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∑V = 0Rav + Rbv – P1 – P2 – P3 – P4 – P5 = 05,2 + 4,8 – 1 – 2 – 3 – 2 – 2 = 0
0 = 0 ( ok )
Potongan I – I dari tinjauan kiri
∑V = 0Rav – S1 . sinα = 04,8 – S1 . sin 33,7 = 0
S1 = 8,73 ton∑Mh = 0Rav . 3 – S2.2 = 04,8 . 3 – S2.2 = 0
S2 = 7,2 tonTan α = 2/3 = 0,667 α = 33,7
Potongan II – II dari tinjauan kiri
∑V = 0Rav – P1 – S3 = 04,8 – 1 – S3 = 0
S3 = 3,8 ton
∑Mh = 0
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Rav . 3 – S4.2 = 04,8 . 3 – S4.2 = 0
S4 = 7,2 ton
Potongan III – III dari tinjauan kiri
∑V = 0Rav – P1 + S5 . cosβ = 04,8 – 1 + S5 . cos 26,56 = 0
S5 = - 4,27 ton∑Md = 0Rav . 4 – P1 . 1 – S6 . 2 = 04,8 . 4 – 1 . 1 – S6 . 2 = 0
S6 = 9,1 tonTan β = 1/2 = 0,5 β = 26,56
Potongan IV – IV dari tinjauan kiri
∑V = 0Rav – P1 – P2 – S7 . sin θ = 04,8 – 1 – 2 – S7 . sin 63,44 = 0
S7 = 2 ton∑Mi = 0
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Rav . 5 – P1 . 2 – P2 . 1 + S8 . 2 = 04,8 . 5 – 1 . 2 – 2 . 1 + S8 . 2 = 0
S8 = - 1,2 tonθ = 90˚ - 26,56˚ = 63,44
Potongan V – V dari tinjauan kiri
∑V = 0Rav – P1 – P2 – P3 – S9 = 04,8 – 1 – 2 – 3 – S9 = 0
S9 = - 1,2 ton∑Mi = 0Rav . 5 – P1 . 2 – P2 . 1 + S10 . 2 = 04,8 . 5 – 1 . 2 – 2 . 1 + S10 . 2 = 0
S10 = - 10 ton
Potongan VI – VI dari tinjauan kiri
∑V = 0Rav – P1 – P2 – P3 + S11 . cos β = 04,8 – 1 – 2 – 3 + S11 . cos 26,56 = 0
S11 = 1,35 ton
∑Mf = 0Rav . 6 – P1 . 3 – P2 . 2 – P3 . 1 – S12 . 2 = 0
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4,8 . 6 – 1 . 3 – 2 . 2 – 3 . 1 – S12 . 2 = 0 S12 = 9,4 ton
Potongan VII – VII dari tinjauan kiri
∑V = 0Rav – P1 – P2 – P3 – P4 – S13 . sin θ = 04,8 – 1 – 2 – 3 – 2 – S13 . sin 63,44 = 0
S13 = 3,59 ton∑Mj = 0Rav . 7 – P1 . 4 – P2 . 3 – P3 . 2 – P4 . 1 + S14 . 2 = 04,8 . 7 – 1 . 4 – 2 . 3 – 3 . 2 – 2 . 1 + S14 . 2 = 0
S14 = - 7,8 ton
Potongan VIII – VIII dari tinjauan kiri
∑V = 0Rav – P1 – P2 – P3 – P4 – P5 – S15 = 04,8 – 1 – 2 – 3 – 2 – 2 – S15 = 0
S15 = - 5,2 ton∑Mj = 0Rav . 7 – P1 . 4 – P2 . 3 – P3 . 2 – P4 . 1 + S16 . 2 = 04,8 . 7 – 1 . 4 – 2 . 3 – 3 . 2 – 2 . 1 + S16 . 2 = 0
S16 = - 7,8 ton Tinjauan dari kanan
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∑V = 0Rbv – S17 . sin α = 05,2 – S17 . sin 33,7 = 0
S17 = 9,45 ton∑Mj = 0- Rbv . 3 – S16 . 2 = 0- 5,2 . 3 – S16 . 2 = 0
S16 = - 7,8 ton
Tabel gaya – gaya batang
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No. Besar gaya batang Jenis batang1 8,73 ton Batang tarik 2 7,2 ton Batang tarik3 3,8 ton Batang tarik4 7,2 ton Batang tarik5 -4,3 ton Batang tekan6 9,1 ton Batang tarik7 2 ton Batang tarik8 -10 ton Batang tekan9 -1,2 ton Batang tekan10 -10 ton Batang tekan11 1,35 ton Batang tarik12 9,4 ton Batang tarik13 3,59 ton Batang tarik14 -7,8 ton Batang tekan15 -5,2 ton Batang tekan16 -7,8 ton Batang tekan17 9,45 ton Batang tarik
Batang horizontal atas
No. Besar gaya batang Jenis batang2 7,2 ton Batang tarik4 7,2 ton Batang tarik8 -10 ton Batang tekan10 -10 ton Batang tekan14 -7,8 ton Batang tekan16 -7,8 ton Batang tekan
Batang horizontal bawah
No. Besar gaya batang Jenis batang6 9,1 ton Batang tarik12 9,4 ton Batang tarik
Batang vertical
No. Besar gaya batang Jenis batang3 3,8 ton Batang tarik9 -1,2 ton Batang tekan15 -5,2 ton Batang tekan
Batang diagonal
No. Besar gaya batang Jenis batang1 8,73 ton Batang tarik5 -4,3 ton Batang tekan7 2 ton Batang tarik11 1,35 ton Batang tarik13 3,59 ton Batang tarik17 9,45 ton Batang tarik
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Rancangan sebuah jembatan terbuat dari kayu ulin mutu A. Konstruksi tidak terlindungi dengan muatan tetap dan muatan tidak tetap
PERHITUNGAN DIMENSI Mendimensikan batang horizontal bawah dengan jenis batang tarik dan menggunakan
sambungan baut.
o Kayu ulin
g = 1,04 kg/cm³ P = 9,4 ton = 9400 kg
γ = 5/6 perlemahan baut 20 %
β = 5/4
BATANG TARIK
σ trk = 150 . γ . β . g
= 150 . 5/6 . 5/4 . 1,04
= 162,5
Fnt = = = 57,85
С = 100% - perlemahan = 100% - 20% = 80 %
Fbr = = = 72,31
Dimensi => b = 8
h = 10
Fbr > 72,31
Fbr = b . h = 8 . 10 = 80 > 72,31
Fnt = Fbr . С = 80 . 0,8 = 64
σ trk < 162,5
σ trk = = = 146,87 < 162,5 ( diizinkan )
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Mendimensikan batang horizontal atas dengan jenis batang tekan dan tarik, menggunakan sambungan baut.
o Kayu ulin
g = 1,04 kg/cm³ P = 7,2 TON (TARIK) = 7200Kg
γ = 5/6 P= -10 TON (TEKAN)= -10000Kg
β = 5/4 perlemahan baut 20 %
BATANG TARIK
σ trk = 150 . γ . β . g
= 150 . 5/6 .5/4 . 1,04
= 162,5
Fnt = = = 44,3
C = 100% - PERLEMAHAN = 100% -20% = 80%
Fbr = = = 55,4
DIMENSI => b = 8
h = 8 Fbr > 72,31 Fbr = b x h = 8x8 =64 > 55,4 Fnt = Fbr x C = 64 x 0,8 = 51,2
σ trk < 162,5
σ trk = = = 140,6 < 162,5 ( diizinkan )
BATANG TEKAN
σ tkn = 150 . γ . β . g
= 150 . 5/6 .5/4 . 1,04
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= 162,5
Imin = 40 . Ptk .
= 40 . 10 .
= 14400
= 14400
800
h = 20,388
DIMENSI => b= 8
h= 20
Fbr = b x h = 8x20 = 160
imin=
w= 1,25
σ tkn= = = 78,125 < 162,5 (diizinkan)
Mendimensikan batang vertical dengan jenis batang tekan dan tarik serta sambungan baut.
o Kayu ulin
g = 1,04 kg/cm³ P = 3,8 (TARIK) = 3800Kg
γ = 5/6 P= -5,2 TON (TEKAN) = -5200Kg
β = 5/4 perlemahan baut 20 %
BATANG TARIK
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σ trk = 150 . γ . β . g
= 150 . 5/6 .5/4 . 1,04
= 162,5
Fnt = = = 23,38
C = 100% - PERLEMAHAN = 100% -20% = 80%
Fbr = = = 29,225
DIMENSI => b = 4
h = 8 Fbr > 29,225 Fbr = b x h = 4x8 =32 > 29,225 Fnt = Fbr x C = 32 x 0,8 = 25,6
σ trk < 162,5
σ trk =
BATANG TEKAN
σ tkn = 150 . γ . β . g
= 150 . 5/6 .5/4 . 1,04
= 162,5
Imin = 40 . Ptk .
= 40 . 5,2 .
= 832
= 832
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h = 9,99
DIMENSI => b= 8
h= 10
Fbr = b x h = 8x10 = 80
imin=
w= 1,95
σ tkn= = = 92,625 < 162,5 (diizinkan)
Mendimensikan batang diagonal dengan jenis batang tekan dan tarik serta sambungan baut.
o Kayu ulin
g = 1,04 kg/cm³ P = 9,45 (TARIK) = 9450Kg
γ = 5/6 P= -4,3TON (TEKAN) = -4300Kg
β = 5/4 perlemahan baut 20 %
BATANG TARIK
σ trk = 150 . γ . β . g
= 150 . 5/6 .5/4 . 1,04
= 162,5
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Fnt = = = 58,15
C = 100% - PERLEMAHAN = 100% -20% = 80%
Fbr = = = 72,69
DIMENSI => b = 8
h = 10 Fbr > 72,69 Fbr = b x h = 8x10 =80 > 72,69 Fnt = Fbr x C = 80 x 0,8 = 64
σ trk < 162,5
σ trk =
BATANG TEKAN
σ tkn = 150 . γ . β . g
= 150 . 5/6 .5/4 . 1,04
= 162,5
Imin = 40 . Ptk .
= 40 . 4,3 .
= 863
= 863
h = 10,1
DIMENSI => b= 8
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h= 10
Fbr = b x h = 8x10 = 80
imin=
w= 1,68
σ tkn= = = 90,3 < 162,5 (diizinkan)
Tabel dimensi batang
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No. Batang Dimensi Batang Jenis batangb H
1 Horizontal bawah 8 10 Batang tarik
2 Horizontal atas 8 8 Batang tarik8 20 Batang tekan
3 Vertikal 4 8 Batang tarik8 10 Batang tekan
4 Diagonal 8 10 Batang tarik8 10 Batang tekan
Dimensi batang yang digunakan adalah :
No. Batang Dimensi Batangb H
1 Horizontal bawah 8 102 Horizontal atas 8 203 Vertikal 8 104 Diagonal 8 10
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PERHITUNGAN SAMBUNGAN
Merencanakan sambungan bertampang dua kayu ulin kelas 1 mutu A, sambungan dgunakan menggunakan baut berdiameter 20mm dan plat besi setebal 15mm.
Sambungan pada buhul A dan buhul B.
Kayu ulin
g = 1,04 g/cm³ P = 10 ton = 10000 kg
γ = 5/6 d = 20 mm = 2 cm
β = 5/4 b1 = 1,5 cm
α = 33,7˚ b3 = 8 cm
Ѕ = 125 . d . b3 (1 – 0,6 sin α )
= 125 . 2 . 8 ( 1 – 0,6 sin 33,7 ) = 1334,187
Ѕ = 250 . d . b1 (1 – 0,6 sin α )
= 250 . 2 . 1,5 ( 1 – 0,6 sin 33,7 ) = 500,32 terkecil
Ѕ = 125 . d² (1 – 0,35 sin α )
= 125 . 2² ( 1 – 0,35 sin 33,7 ) = 1547,14
Ѕ = 500,32 . 5/6 . 1,25 = 521,167
n = = = 19,12 => 20 baut
o Gambar kerja terlampir
Sambungan pada buhul C, E dan G.
Kayu ulin
g = 1,04 g/cm³ P = 5,2 ton = 5200 kg
γ = 5/6 d = 20 mm = 2 cm
β = 5/4 b1 = 1,5 cm
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α = 90˚ b3 = 8 cm
Ѕ = 125 . d . b3 (1 – 0,6 sin α )
= 125 . 2 . 8 ( 1 – 0,6 sin 90 ) = 2000
Ѕ = 250 . d . b1 (1 – 0,6 sin α )
= 250 . 2 . 1,5 ( 1 – 0,6 sin 90 ) = 750 terkecil
Ѕ = 125 . d² (1 – 0,35 sin α )
= 125 . 2² ( 1 – 0,35 sin 90 ) = 1920
Ѕ = 750 . 5/6 . 1,25 = 781,25
n = = = 6,66 => 8 baut
o Gambar kerja terlampir
Sambungan pada buhul D dan F.
Kayu ulin
g = 1,04 g/cm³ P = 4,3 ton = 4300 kg
γ = 5/6 d = 20 mm = 2 cm
β = 5/4 b1 = 1,5 cm
α = 63,44˚ b3 = 8 cm
Ѕ = 125 . d . b3 (1 – 0,6 sin α )
= 125 . 2 . 8 ( 1 – 0,6 sin 63,44 ) = 926,64
Ѕ = 250 . d . b1 (1 – 0,6 sin α )
= 250 . 2 . 1,5 ( 1 – 0,6 sin 63,44 ) = 347,49 terkecil
Ѕ = 125 . d² (1 – 0,35 sin α )
= 125 . 2² ( 1 – 0,35 sin 63,44 ) = 1318,92
Ѕ = 347,49 . 5/6 . 1,25 = 361,97
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n = = = 11,88 => 12 baut
o Gambar kerja terlampir
Sambungan pada buhul H, I, dan J.
Kayu ulin
g = 1,04 g/cm³ P = 4,3 ton = 4300 kg
γ = 5/6 d = 20 mm = 2 cm
β = 5/4 b1 = 1,5 cm
α = 63,44˚ b3 = 8 cm
Ѕ = 125 . d . b3 (1 – 0,6 sin α )
= 125 . 2 . 8 ( 1 – 0,6 sin 63,44 ) = 926,64
Ѕ = 250 . d . b1 (1 – 0,6 sin α )
= 250 . 2 . 1,5 ( 1 – 0,6 sin 63,44 ) = 347,49 terkecil
Ѕ = 125 . d² (1 – 0,35 sin α )
= 125 . 2² ( 1 – 0,35 sin 63,44 ) = 1318,92
Ѕ = 347,49 . 5/6 . 1,25 = 361,97
n = = = 11,88 => 12 baut
Untuk batang 3, 9, dan 15 bawah
P = 4,3 ton = 4300 kg
γ = 5/6 d = 20 mm = 2 cm
β = 5/4 b1 = 1,5 cm
α = 90˚ b3 = 8 cm
Ѕ = 125 . d . b3 (1 – 0,6 sin α )
= 125 . 2 . 8 ( 1 – 0,6 sin 90 ) = 2000
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Ѕ = 250 . d . b1 (1 – 0,6 sin α )
= 250 . 2 . 1,5 ( 1 – 0,6 sin 90 ) = 750 terkecil
Ѕ = 125 . d² (1 – 0,35 sin α )
= 125 . 2² ( 1 – 0,35 sin 90 ) = 1920
Ѕ = 750 . 5/6 . 1,25 = 781,25
n = = = 6,66 => 8 baut
Untuk batang I dan 17 bawah
P = 9,45 ton = 9450 kg
γ = 5/6 d = 20 mm = 2 cm
β = 5/4 b1 = 1,5 cm
α = 33,7˚ b3 = 8 cm
Ѕ = 125 . d . b3 (1 – 0,6 sin α )
= 125 . 2 . 8 ( 1 – 0,6 sin 33,7 ) = 1334,187
Ѕ = 250 . d . b1 (1 – 0,6 sin α )
= 250 . 2 . 1,5 ( 1 – 0,6 sin 33,7 ) = 500,32 terkecil
Ѕ = 125 . d² (1 – 0,35 sin α )
= 125 . 2² ( 1 – 0,35 sin 33,7 ) = 1547,14
Ѕ = 500 . 5/6 . 1,25 = 521,167
n = = = 18,1 => 18 baut
o Gambar kerja terlampir
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LAMPIRAN
GAMBAR KERJA
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K O N S T R U K S I K A Y U
R A N C A N G A N J E M B A T A N
2 SIPIL 1 PAGI
Oleh :
Gilang aditya. P
Fredy rosendi
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Jaimiland. P
Dikki dharmawan