A. IDEALISASI STRUKTUR RANGKA ATAP (TRUSS)
Perencanaan kuda – kuda dalam bangunan sederhana dengan panjang bentang 10 m. jarak antara kuda – kuda adalah 3 m dan 2 m, jarak mendatar antara kedua gording adalah 3 m dan sudut kemiringan atap adalah 30˚ terhadap batang horizontal. Kayu yang digunakan memiliki Kode Mutu E22 dengan klasifikasi Kelas A. Rencana kuda
– kuda dan atap dapat dilihat pada Gambar 1 dan Gambar 2.
7 BC9
BC8
6 BC6 BC2 2 0 °
BC1 1 0 °
3 3 °
° 3 3
2
1 0 , 2
BC11
BC7 BC5
8
BC10
3
BC12
BC13
BC3
4
BC4 ° 0 1
1
6 6 , 0
° 0 2
5 3,00
2,00
2,00
3,00
10,00
Gambar 1 Rencana Kuda – Kuda
Contoh Tugas Struktur Kayu
1
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Luas Atap Yang Membebani Satu Titik Buhul
0 0 , 3
Bumbungan
Rangka Atap
0 0 , 3
0 0 , 3
Gording 0 0 , 3
0 0 , 3
3,00
2,00
2,00
3,00
Gambar 2 Rencana Atap
Berdasarkan gambar rencana kuda - kuda dan gambar rencana atap, beberapa data untuk perencanaan seperti luas atap dan panjang gording yang membebani masingmasing titik buhul kuda - kuda serta panjang batang kuda – kuda dapat di analisis. Luas atap dapat dilihat pada gambar rencana atap (Gambar 2). Panjang elemen kuda -
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Tabel 1 Panjang Bentang Per Elemen Panjang
No
Batang
1
BC1
3,05
2
BC2
2,03
3
BC3
2,03
4
BC4
3,05
5
BC5
3,46
6
BC6
1,20
7
BC7
2,17
8
BC8
2,31
9
BC9
2,31
10
BC10
2,01
11
BC11
2,17
12
BC12
1,20
13
BC13
3,46
(m)
Sumber : Perhitungan Berdasarkan Gambar Rencana
B. PERENCANAAN GORDING
1. Data – data:
Beban pada gording Ukuran gording
=
60/120 mm
Luas gording
=
0,06 0,12 0,0072 m²
Berat per satuan volume
=
600 Kg/m³ = 6000 N/m³
Berat sendiri gording
=
Luas gording x Bj
=
0,06 0,12 6000 = 43,20 N/m
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Beban Vertikal (Penutup Atap) Berat atap seng gelombang (BWG 24) = 10 Kg/m² = 100 N/m² Berat sendiri atap seng
=
Jarak gording x Bj
=
3,46 100 100 = 346 N/m
Beban Hidup (L) Beban hidup (L) = 70 Kg = 700 N
Beban Angin 2
2
Beban angin = 45 Kg/m = 450 N/m
2. Pembebanan
Beban Mati (D) Berat penutup atap
=
346 N/m
Berat gording
=
43,20 N/m +
Total
=
389,20 N/m
Dx
Dy
=
D sin α
=
389,20 sin 30˚
=
194,60 N/m
=
D cos α
=
389,20 cos 30˚
=
337,06 N/m
Beban hidup (L) L
=
700 N
Lx
=
L sin α
=
700 sin 30˚
=
350 N
=
L cos α
=
700 cos 30˚
=
606,22 N
Ly
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=
=
Koefisien angin isap (C2)
0,2
=
(PMI Pasal 4.3 Ayat 1.b, Hal 20) Wtekan
=
Wtekan
=
= 311,40 N/m Wisap
=
Wisap
=
= - 622,80 N/m
Perhitungan momen untuk beban angin Mx tekan
=
Mx tekan
=
= 350,30 Nm My tekan
=
My tekan
=
= 0 Nm Mx isap
=
Mx isap
=
= - 700,65 Nm My isap
=
My isap
=
= 0 Nm
3. Perhitungan momen
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=
218,92 Nm
Mx (W)
=
L
=
Mx tekan
=
3m
= 350,30 Nm Mx isap
= =
=
= 3m
Mx (L) =
- 700,65 Nm
Mx (L) L
= 262,50 Nm
Beban pada sumbu lemah:
My (D)
=
L
=
My (D) = =
3m
379,19 Nm
My (W)
=
L
=
My tekan
=
3m
= 0 Nm My isap
= =
My (L)
L
=
0 Nm
3
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Momen terfaktor:
Kombinasi pembebanan sementara (DL + LL + Wtekan)
Nm [SNI 03 – xxxx – 2000 butir 4.2.2]
Nm [SNI 03 – xxxx – 2000 butir 4.2.2]
Kombinasi pembebanan sementara (DL + LL + Wisap)
Nm [SNI 03 – xxxx – 2000 butir 4.2.2]
Nm [SNI 03 – xxxx – 2000 butir 4.2.2]
Tegangan acuan kayu: Ew = 21000 MPa dan F b = 54 MPa
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3
mm
3
mm
Karena nilai banding penampang d/b (120/60) = 2,00. Maka pada balok tidak diperlukan pengekang lateral [SNI 03 – xxxx – 2000 butir 8.2.2], Cl = 1,00 Nilai Ct diambil dalam kadar air kering dengan suhu T < 38° < °C maka nilai Ct = 1,00. Untuk kayu dengan mutunya ditetapkan secara maksimal, CF = 1,00 [SNI 03 – xxxx – 2000 butir 5.6.2], faktor koreksi pengawetan kayu, nilai C pt = 1,00 [SNI 03
– xxxx – 2000 butir 5.6.1]. Faktor koreksi layan basah, untuk memperhitungkan kadar air masa layan pada balok kayu balok kayu besar 125 mm x 125 mm, F b = 1,00 nilai CM = (F b/CF) = (1,00/1,00 = 1,00) < 8 Mpa maka CM = 1,00 [Bahan Ajar Modul Struktur Kayu, Tabel 3.3, hal III-12]. Faktor reduksi tegangan untuk
batang lentur, φ b = 0,85 dan faktor waktu (λ) pada kombinasi pembebanan
maka λ = 0,80 [Bahan Ajar Modul Struktur Kayu,
Tabel 3.2, hal III-12].
F bx ' C m C t C pt C F F bx M ' S
F
144000 54
54 MPa
7776000 MPa
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3
3
962 962,95 10
1198,48 10
0,8 0,85 7776000
0,8 0,85 3888000
1,00
0,18 0,45 1,00
OK!
0,63 1,00
Kontrol tegangan lentur (DL + LL + Wisap) M ux
b M x
M uy
b M y
1,00
3
3
122,19 10
1198,48 10
0,8 0,85 7776000
0,8 0,85 3888000
1,00
0,02 0,45 1,00
OK!
0,47 1,00
5. Kontrol lendutan balok:
Lendutkan ijin:
∆maks =
L 300
; ∆maks =
3000 300
Lendutkan akibat beban tetap: E ' E w C m C r C pt
= 10 mm
21000 MPa
Lendutkan sumbu kuat: 5 384
4
w x L E ' I x
;
5 384
194,60 3000
4
10
21000 8640000
3
; 1,13 mm
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C. PERHITUNGAN BEBAN – BEBAN PADA KUDA – KUDA
1. Perhitungan beban akibat berat kuda – kuda sendiri pada titik buhul 2
2
Kayu 60/120 mm adalah batang atas dan batang bawah, Kayu 60/120 mm
adalah batang diagonal. Berat sendiri kuda - kuda yang bekerja pada satu titik buhul dengan rumus ; 0,5 x berat sendiri seluruh batang kuda - kuda yang terletak pada satu titik buhul.
Titik buhul B1 Batang BC1
=
Batang BC5
=
1 2 1 2
3,05 0,06 0,12 6000
=
65,88 N
3,46 0,06 0,12 6000
=
74,74 N
Total
+
= 139,62 N
Titik buhul B2 Batang BC1
=
Batang BC2
=
Batang BC6
=
1 2 1 2 1 2
3,05 0,06 0,12 6000
=
65,88 N
2,03 0,06 0,12 6000
=
43,85 N
1,20 0,06 0,12 6000
=
25,92 N +
Total
=
135,65 N
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Batang BC11
=
1 2
2,17 0,06 0,12 6000
=
224,86 N
2,03 0,06 0,12 6000
=
43,85 N
3,05 0,06 0,12 6000
=
65,88 N
1,20 0,06 0,12 6000
=
25,92 N
Titik buhul B4 Batang BC3
=
Batang BC4
=
Batang BC12
=
1 2 1 2 1 2
+ Total
=
135,65
3,05 0,06 0,12 6000
=
65,88 N
3,46 0,06 0,12 6000
=
74,74 N
=
139,62 N
=
74,74 N
Titik buhul B5 Batang BC4
=
Batang BC13
=
1 2 1 2
Total
46,87 N +
Total
=
Titik buhul B6 Batang BC5
=
1 2
3,46 0,06 0,12 6000
+
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Titik buhul B7 Batang BC8
=
Batang BC9
=
Batang BC10
=
1 2 1 2 1 2
2,31 0,06 0,12 6000
=
49,89 N
2,31 0,06 0,12 6000
=
49,89 N
2,01 0,06 0,12 6000
=
43,42 N
=
143,20 N
2,31 0,06 0,12 6000
=
49,89 N
2,17 0,06 0,12 6000
=
46,87 N
1,20 0,06 0,12 6000
=
25,92 N
Total
Titik buhul B8 Batang BC9
=
Batang BC11
=
Batang BC12
=
1 2 1 2 1 2 1
+
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3. Perhitungan beban terhadap atap Berat atap seng gelombang (BWG 24) = 10 1 0 Kg/m² = 100 N/m² Berat atap seng
= =
Jarak gording x Jarak kuda – kuda x Bj
N
4. Perhitungan beban hidup (L) Beban hidup (L) = 70 Kg = 700 N
5. Perhitungan beban akibat angin Beban angin tekan
= 311,40 N/m
Beban angin isap
= -622,80 N/m
Jarak kuda – kuda
= 3,00 m
Beban angin untuk setiap titik buhul = beban angin x jarak daerah beban untuk titik buhul, dimana panjang daerah beban = 0,5 x panjang batang
Tabel 2 Perhitungan Beban Angin Tekan
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Tabel 3 Perhitungan Beban Angin Isap Titik
Batang Yang Berseberangan
BC13
5
BC13 BC9
8
BC9
7
Panjang Daerah Beban (m) 3,46
Beban Angin Isap (N) 1077,44
Total 3,46 2,31 Total 2,31 Total
1077,44 1077,44 719,33 1796,78 719,33 719,33
Sumber : Perhitungan Berdasarkan Gambar Rencana
Tabel 4 Rekapitulasi Gaya – Gaya Batang (Tarik dan Tekan) Nomor Batang BC1 BC2 BC3
Besar Gaya - Gaya Yang Bekerja Pada Truss (N) Tarik (+) Tekan (-) 9263,21 0,00 9243,72 0,00 9704,01 0,00
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0,12 m
0,06 m
Jari – jari girasi (r)
4
mm
4
mm 4
min = 17,32 mm r y < r x, maka r min
K e = 1,0 (sendi – sendi) [SNI 03 – xxxx – 2000 butir 7.2.1 (Gambar 7.2.1)] Angka kelangsingan (K eL)/r =
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Kuat sejajar serat:
MPa
3.
MPa
Menghitung faktor beban
MPa
MPa
N
MPa
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E. KONTROL DIMENSI BATANG TARIK
Dimensi batang tarik dari struktur truss seperti Gambar 1, elemen batang terbuat dari kayu ukuran balok 60/120 mm dan panjang 3 m dengan kayu kelas A kode mutu E22 dan balok di beri beban tarik maksimum 9721,76 N. 1.
Menghitung kuat tarik sejajar Faktor tahanan kayu kelas mutu A = 0,80 [Bahan Ajar Modul Struktur Kayu, Tabel 2.4, hal II-7], faktor kuat tarik sejajar serat F t// = 50 MPa.
MPa
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2
mm
Penampang pada daerah sambungan menentukan tegangan yang timbul karena terjadi pengurangan luas tampang akibat terdapat lubang alat sambung. Untuk itu kegiatan perencanaan diperkirakan terjadi pengurangan luas penampang sebesar 25% sehingga luas penampang bruto yang diperlukan adalah
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Struktur Kayu, Tabel 2.1, hal II-8]. Faktor koreksi layan basah, untuk memperhitungkan kadar air masa layan pada balok kayu balok kayu besar 125 mm x 125 mm, F b = 1,00 nilai CM = (F b/CF) = (1,00/1,00 = 1,00) < 8 Mpa maka CM = 1,00 [Bahan Ajar Modul Struktur Kayu, Tabel 3.3, hal III12]. Nilai Ct diambil dalam kadar air kering dengan suhu T < 38° < °C maka nilai Ct = 1,00. Untuk kayu dengan mutunya ditetapkan secara maksimal. 12331,50 N BATANG BC1 BALOK 6/12 BATANG BC5 BALOK 6/12 BEHEL U 40,4
a h l a
m m 2 0
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Berat jenis kayu pada kadar 15 % (G):
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2.
Sambungan pada batang BC1, BC6 dan BC2 a. Menghitung tahanan perlu sambungan
Faktor tahanan sambungan φz = 0,65 [SNI 03 – xxxx – 2000 butir 10.1.2], ρ 3
= 600 kg/m (berat jenis persatuan volume), Fyb = 320 Mpa (kuat lentur baut), diameter (D) yang digunakan 12,701 mm,
sudut sambungan (θ =
80°), tebal kayu BC6, BC1 dan BC2 = 60 mm, banyaknya baut (n = 2). 157,66 N
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e. Menghitung nilai koreksi : Faktor aksi kelompok (Cg) Menurut NDS dari U.S (Tabel hal 72)
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3.
Sambungan pada batang BC5, BC6, BC7 dan BC8 Sambungan 2 irisan antara batang BC5 dan BC6 a.
Menghitung tahanan perlu sambungan
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