Download - Base Plate Pertemuan 25
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Base Plate Pertemuan 25
Mata kuliah : S0844 - Teori Dan Perancangan Struktur BajaTahun : 2010
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Bina Nusantara University
Learning OutcomesPada akhir pertemuan ini, diharapkan mahasiswa akan mampu :• Mahasiswa dapat menghitung desain kolom struktur baja.
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Bina Nusantara University
Outline Materi
• Pengertian base plate• Tebal base plate
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Bina Nusantara University
Pengertian base plate
Pondasi
Kolom baja
Base plate
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Bina Nusantara University
Pengertian Base Plate (samb)
Base plate sebagai desain penghubung antara kolom baja dan pondasi beton.Ada 2 type perletakan :1. Sendi baut angkur tidak memikul momen,
hanya memikul tekan/tarik.2. Jepit sebagian baut angkur memikul tarik
akibat momen dan yang lainnya memikul tekan.
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Bina Nusantara University
T T
NM
Base Plate
Baut Angkur
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Bina Nusantara University
Menentukan tebal (t), lebar (B), panjang (D) basement
0,8 b b
d
m
m
12 120,95 d
1
1
2
2
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Bina Nusantara University
Penampang 1 dijadikan penampang persegi (Ekivalen)Potongan 1-1
M1-1 = ½ g.n2
g = 0,25 f’c f’c = 0,83 σ’bk
Contoh :Mutu beton K 250σ’bk = 300 kg/cm2f’c = 0,83 (300) Untuk lebar pelat 1 cm
σ1-1 = ≤ σ dimana σ = 1600 kg/cm2
= diperoleh t ≥ n
W
11M
)t)(1(6
1
n.g2
1
2
2
g3
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Bina Nusantara University
Potongan 2-2M2-2 = ½ g.m2
Untuk 1 cm lebar pelatσ2-2 = ≤
W
22M σ
)t)(1(61
m.g2
1
2
2
g2
= ≤ σ
t ≥ m
Dipilih t terlebarB dan D ditentukan sendiriB = 0,8 b + 2 mD = 0,95 d + 2 nM atau n minimum 4 di baut
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Bina Nusantara University
Base Plate
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Bina Nusantara University
Learning OutcomesPada akhir pertemuan ini, diharapkan mahasiswa akan mampu :• Mahasiswa dapat menghitung desain kolom struktur baja.
![Page 13: Base Plate Pertemuan 25](https://reader033.vdocuments.pub/reader033/viewer/2022061503/56813ad5550346895da30b77/html5/thumbnails/13.jpg)
Bina Nusantara University
Outline Materi
• Perencanaan Baut Angkur Pada Base Plate
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Bina Nusantara University
Perencanaan Baut Angkur Pada Base Plate
P
M
1 2
0,75 d
d
B
D
0,8b b
1 2
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Bina Nusantara University
Pada kolom bekerja M dan PJadi perletakan adalah jepitAkibat M baut 1 tertarik
baut 2 tertekan
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Bina Nusantara University
P
M
1 2
a
h
g
T (tarik dipikul angkur)
∂hβb. c1
Grs netral
Diagram tegangan
a = jarak titik tangkap gaya tekan terhadap baut yang tertarik
c = ½ g (β.h) B ∂h = h – 1/3. β.h
c = gaya tekan pada beton lengan momen
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Bina Nusantara University
Ec
beton1 2
Diagram regangan
εs
εc
Grs netral
h
βh
h(1- β)
Ec
g
Εc =
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Bina Nusantara University
Es.As
T
Es.As
T
σ beton adalah gDimana g = 0,83 f’cf’c = mutu beton (contoh silinder)As = Luas penampang angkurEs = 2 x 106 kg/cm2T = gaya tarik yang bekerja pada angkur
εs =
εs : εc = h (1- β) : βh
:
Ec
g
= (1- β) : β
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Bina Nusantara University
Σ gaya vertikal = 0C – T = P½ g (β.h) B – T = PΣ Momen terhadap baut = 0c ∂h = P. a + MAkan diperoleh T gaya tarik pada angkur
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Bina Nusantara University
Base Plate
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Bina Nusantara University
Learning OutcomesPada akhir pertemuan ini, diharapkan mahasiswa akan mampu :• Mahasiswa dapat menghitung desain kolom struktur baja.
![Page 22: Base Plate Pertemuan 25](https://reader033.vdocuments.pub/reader033/viewer/2022061503/56813ad5550346895da30b77/html5/thumbnails/22.jpg)
Bina Nusantara University
Outline Materi
• Langkah perencanaan base plate
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Bina Nusantara University
Contoh soalKolom baja WF 400 X 300 X 11 X 18P = 30 ton dan M = 10 ton mMutu beton : f’c = 30 mpaBaut angkur HSS (High Strenght Steel)Rencanakan Base Plate dan baut angkur ?
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Bina Nusantara University
Jawab :Menetukan ukuran base plate baut diameter 20 mm.
d = 400 n
0,8 b = 240
b = 300
0,95 d = 380
m
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Bina Nusantara University
g = 0,25 f’c = 0,25 (300) = 75 kg/cm2t ≥ n ≥ 8 dimana n = 4d baut ≥ 1,73 cm ambil 18 mm = 80 mm
= 8 cmUkuran base plate0,95 d + 2n = 0,95 (400) + 160 = 5400,8 b + 2m = 0,8 (300) + 160 = 400Ambil 550 X 490 X 18
g3
1600
)75(3
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Bina Nusantara University
Menentukan gaya tarik pada baut (T)1.
2. ½ g (βh) B = P + T3. c ∂ h = P a + M
Data-data :a = 23,5 cm g = 75 kg/cm2P = 30000 kg h = 460 mm = 46 cmM = 10 tm = 106 kgcm c = ½ g (βh) B
= ½ (75) (β. 46) 40 = 69000 β kg
Fs.As
T
Ec
g:
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Bina Nusantara University
Grs netral
h = jrk baut ke baut = 460
a = 275 – 40 = 23580 mm
βh
∂h
½ βh
400
500
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Bina Nusantara University
∂h + 1/3 P h = h∂ + 1/3 P = 1∂ = 1 – 1/3 Pc . ∂h = P a + M 69000 β (1- 1/3 β) (46) = 30000 (23,5) + 1000000
β – 1/3 β2 = 0,537 β2 – 3 β + 1,611 = 0 β =
β = 0,7
)1(2
)611,1(493
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Bina Nusantara University
½ g (βh) B – T = P½ (70) (0,7 . 46) 40 – T = 30000T = 18300 kg
Jumlah Baut angkur tarik (diameter 20 mm)Luas 1 baut = ¼ π (2)2
= 3,14 cm2
σ = 1600 kg/cm2
n = = 3,58
Pasang 4 baut untuk tarik40
)1600(14,3
18300