IV- 1
BAB IV PERHITUNGAN STRUKTUR
BAB IV PERHITUNGAN STRUKTUR 4.1
Analisis Arsitektur
Luas lapangan sepakbol sepakbola a adalah 102 x 68 mdan m dan luas lapangan lapangan rumput adalah 120 x 80 m, sedangkan panjang tribun tribun dari pusat lapangan lapangan adalah 104.5 4.5 m dan 107 m, dengan demikian ikian desain memenuhi syarat sebagai stadion stadion internasional. internasional.
m 5 . 4 0 1
m 8 6
102m
106m
Gbr.4.1.Denah dan ukuran stadion
Analisis Analisis Konfigu Konfigurasi Struktur Atap Stadion Stadion didesain didesain berbentuk cangkang, selai selain n optimal dalamm dalammendukung beban juga memiliki iliki nilai nilai estetik estetik yang tinggi. Bagian tengah stadion stadion dirancang terbuka tepat di atas lapangan berfungsi untuk menerim enerima cahaya matahari secara secara memadai dan memperlancar perlancar sir sirkulas kulasii udara. udara. Atap stadion stadion mengekspos keempat busur utamanya sebagai unsur artistiknya artistiknya sekaligus sekaligus menjadi ciri ciri khas orisi orisinal nal stadion stadion tersebut. Bahan atap stadion adalah polycarbonate yang dirancang dirancang untuk memantulkan panas yang berlebihan berlebihan sehingga penggunaanya cocok untuk daerah tropis seperti di Indonesia. Indonesia. Selai elain n mampu memantulkan antulkan panas, panas, polycarbonate ju juga memiliki be berat yang relatih ringan dan pemasangannyarelati relatiff mudah. Struktur tribun di didesain desain mengitari lapangan dengan bagian tengah lapang lapangan sebagai pusat pandangannya. K Kem emiringan ringan tribun tribun menyesuaikan enyesuaikan kenyamanan pandangan penonton kearah lapang
Tugas Akhir Desain Stadion Internasional
Arbo r Reseda/ Reseda/L2A0010 L2A001020 20 Titi Puji Astu ti/L2 ti/L2A0011 A001150 50
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IV- 2
BAB IV PERHITUNGAN STRUKTUR
2 u n b i r r t m 5 . 6 3
1 u n i r b t r
Lapangan sepakbola
6 8m
6m
63.5m
Gbr.4.2.Sudut pandang penonton
Untuk akses penonton, perencanaan pintu, koridor, koridor, dan tangga harus didesain didesain agar arus penonton menjadi enjadi nyaman namun tidak terlal terlalu u lama. Berdas Berdasarkan arkan data arsitek, arsitek, kecepatan orang untuk arus penuh sebagai berikut berikut : Ta Tabel 4.1.Data desain arus dan kapasitas penonton V(m/s) /s)
Kapasitas Kapasitas (org/m (org/mnt/m)
-
100
Koridor
1.0
84
Ta Tangga
0.6
60
Pintu
Sebagai contoh contoh kita bisa bisa menganalisi nalisis s tribun tribun 2 (Gb. 4.3) dim dimana kapasitas pasitas penontonnya adalah yaitu9536penonton. Perhitungan Lebar pintu : Semua penonton harus keluar dalam dalamw waktu 5 menit, dengan kapasitas penuh sebesar sebesar 100orang tiap tiap menit per meter lebar, lebar, makalebar pintu pintu : penonton 100 × 5menit
=
9536 500
= 19.072 m
J ika pintu pada tribun 1 dibagi 6 bagian, maka masing – masing pintu lebarnya 19 6
→
≈ 3. 2 m .
Perhitungantangga Waktu tempuh tempat akses keluar keluar yang paling ling ekstrimadalah ekstrimadalah akses 1 ke akses 4. Akses 1 berupa ja jalur data sepanjang 48msedangkan akses 4adalah jalur tangga sepanjang 60m.(gambar 4.2) Perhitunganwaktuakses 1=
jarak
+ waktukelua r
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IV- 3
BAB IV PERHITUNGAN STRUKTUR
= 80.4 0.6 + 300 = 434dt Waktu tempuh paling paling lama adalah akses akses 4 sel selam ama 434 detik.
4 3
2 1
Gbr.4.3.Akses keluar penonton
Perhitungan lebar tangga :
jumlpenont on waktu ⋅ 1.25
=
8450 434 ⋅ 1.25
= 15.57 m . J ika tangga dibagi menjadi
empat bagian bagian dengan lebar lebar yang sam sama, maka desai desain lebar lebar tangga masing-m asing-masing asing 4 m. Berdasarkan perhitungan di atas, stadion dapat dikosongka dikosongkan dalamw dalamwaktu 434 detik ≈ 7.23menit enit 4.2
Data Perencanaan Perencanaan Struktur Data Stadion
Kapasitas Kapasitas total Stadionl
: 80.000penonton
VIP VIP
: 2.000penonton
Penyandangcacat
:
Luas Area Stadion
: 39.261m²
Luas Area Area Keamanan
: 10.000m²
Luas Area Parkir
: 40.000m²
Fasilitas
: flash light, pencahayaan, pengkondisian kondisian udara, sistem sistemsuara, suara, monitor,
500penonton
air minum, air bersih, bersih, air kotor, drainase, dan lain-lain. lain-lain.
4.3
Analisa Struktur Atap ( rangka ruang baja )
Atap stadion disusun isusun oleh space frame baja diduk didukung oleh 4 struktur struktur lengkun
elalui elalui kabel kabel
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IV- 4
BAB IV PERHITUNGAN STRUKTUR
mengalami perpindahan perpindahan pada arah sumbu x,y,z, x,y,z, maka perencanaan jjenis enis tumpuan yang paling paling tepat adalah tumpuan pegas. Masalah perpindahan pada desain atap ini lebih lebih dominan dari gaya yang terjadi erjadi pada setiap setiap elemen, sehingga sehingga pada desain desain akan banyak ditemukan profilrofil-profi profill yang memiliki iliki kapasitas sitas jauh lebih lebih tingg tinggi dari beban rencana. Hal ini diperlukan diperlukan untuk membentuk struktur yang kaku. Sel Seluruh uruh elemen batang menggunakan profil profil pipa dengan berbagai ukuran, karena karena pertimbangan kestabilan kestabilan struktur, kemudahan pemasangan, dan nilai nilai estetikny estetiknya. Rangka baja dibagi menjadi enambagian enambagian yang terpis terpisah ah namun semuanya tetap ditopang ol oleh eh struktur lengkung, lengkung, sehingga sehingga jika jika salah salah satu bagian rangka baja mengalami kerusakan, kerusakan, rangka itu itu tidak mempengaruhi rangkalainny lainnya.
Di
Di
m 9
L
1
L
6m
6m
Pot. I-I
II
m 0 8
m D i 0 2
2 i D
Pot. II-II
II
Sendi
Sendi
275m
Gbr.4.4.Struktur lengkung utama
I G
I I
i D
D i
m 7
I
G
I
i D
L
G I
G I
G I
H
J
H
J
H
J
H
J
H
J
H
J
H
J
H
J
H J
H
H
J m 6 7
Pot. II-II
K
K
K
K
Sendi 212m
K
K
II Sendi
K
K
K II
H
m 6
Di 6m
6m
Pot. I-I
G
K
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IV- 5
BAB IV PERHITUNGAN STRUKTUR
Menggunakan profil profil pipa pipa dengan sambunganlas.
→
Rangkaruang Menggunakan profil profil pipa pipa dengan sambunganlas.
I
M N
i D
D i
N
m 7
L M N N
m 3
6m
Det. I-I
Jepit
212m
Gbr.4.6.Struktur space frame
Gbr.4.7.P ermodelan struktur struktur atap
4.3.1
Pembebanan pada atap
n o t e b m o l o K
Jepit
m 2 . 6 4
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IV- 6
BAB IV PERHITUNGAN STRUKTUR
BebanHidup(LL) (LL) Beban Manusia Manusia
= 100kg
BebanHujan Beban terbagi merata per m2 bidang datar berasal dari beban air hujan sebesar (40 (40 – 0.8α ) kg/m2
≤ 20kg/m2 α =sudutkemiringan iringanatap(°) (°)
Beban Angin (WL) p
=
V 2 16
(kg/m2)
V asum asumsi =60 Km/jam /jam V =Kecepatan =Kecepatan angin (m (m/s) /s) BerdasarkanPBI 1983, koefis koefisien ien angin seba sebagai berikut Atap lengkung lengkung dengan sudut pangkal β : β
< 22 ° : untuk bidang lengkung dipihak ipihak angin
→ Pada 1 4 busurpertama Pada 1 busur Kedua 4
=-0.6 =-0.6 =-0.7
untuk bidang lengkung dibelakang dibelakang angin → Pada 1 4 busur pertama =-0.5 =-0.5 Pada 1 4 busurKedua
β
≥ 22 ° : untukbidanglengkungdipihak dipihak angin → Pada 1 4 busur pertama Pada 1 busur Kedua 4
=-0.2 =-0.5 =-0.5 =-0.6 =-0.6
untuk bidang lengkung dibelakang dibelakang angin → Pada 1 4 busur pertama =-0.4 =-0.4 Pada 1 4 busurKedua o
Kombinasi Beban Analisis lisis Statik Bebantetap
=1, 2 DL + 1,6 LL
Beban sementara 1
=1.2 DL + 1.6 LL + 0.8W 1
Beban sementara 2
=1.2 Dl + 1.6 LL + 0.8W 2
KetentuanMaterial
=-0.2
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IV- 7
BAB IV PERHITUNGAN STRUKTUR
• TeganganLeleh Leleh
(f y)
=
2400kg/cm2
• TeganganBatas
(f U)
=
3700kg/cm2
Analisi Analisis sS Struktu truktur menggunakan sistem sistem space frame 3 dimensi yang terdiri dari 10104 elemen batang, 3016 joi joint, nt, 24 tumpuan pegas, dan 44 tumpuan sendi. Berat total rangka baja dan sambungannyaadalah adalah1851.25ton. 4.3.2
Perhitungan Baja
:
Perencanaan Gording
Batang atas space frame berfungsi sebag sebagai gording, sehingga dalamanali dalamanalisi sis s struktu struktur batang atas dianalogi dianalogikan kan sebagai elem elemen lentur lentur yang menahan momen lentur lentur dan gaya geser karena batang atas menderita beban merata secara langsung. Profil rofil yang digunakan adalah : D=190.7 mm
I =1330cm4
i =6.56 =6.56
G =24.2 kg/m
S =139cm3
F =30.87cm2
L =6m
t =5.3mm
Pembebanan:
1 ⋅ 6 ⋅12 + 24.2 =60.2kg/m 2
WD =
1 ⋅ 6 ⋅ 20 2
WL =
=60 kg/m
WU =1.2 WD +1.6 WL
=168.24 .24 kg/m
=0.16 .168 t/m
A
A
Diagram Momen M=18qL²
Diagram Geser
5.3mm
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IV- 8
BAB IV PERHITUNGAN STRUKTUR
akibat beban = Mux =Momen Lentur akibat
1 8
1
⋅Wu ⋅ L2 = ⋅ 0.168 ⋅ 6 2 = 0.757 tm 8
kapasitas as penampang= fy ⋅ S = 24000 ⋅ 1.39 ⋅10 − Mn =Momenkapasit
4
=Fakto =Faktor reduksi
φ
= 3.336 tm
=0.9
Perencanaan geser
≤ φ ⋅ Vn 0.504 ≤ 0.9 ⋅ 26.672 0.504 ≤ 24.0048
Vu
(Aman)
Perencanaan lendutan δ
=
δ ijin
5 384
=
⋅
EI
L 240 δ
•
Wu ⋅ L4
=
6 240
=
5
⋅
0.168 ⋅ 6 4
384 2 ⋅ 10 7 ⋅1.33 ⋅10 −5
= 0.0107 m
= 0.025
< δ ijin ⇒ 0.0107 < 0.025
(Aman)
Komponen Struktur Tarik
Nu(+)
Nu(+) A
Diagram Normal
+
5.3mm
190.7mm
POT. A-A
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IV- 9
BAB IV PERHITUNGAN STRUKTUR
:
Nu=Gayaaksial tarik
=35.96t (batang 293dari hasil SAP SAP))
Nn=Kapasitas =Kapasitas tarik penampang
= F ⋅ fy = 30.87 ⋅ 2400 = 74.088t
φ =Faktor reduksi
=0.9
Kabel
Untuk perencanaan kabel sam sama dengan perencanaan batang tarik tarik dengan nilai nilai E dan fy yang fykabel = 13909.5 kg/cm /cm2
berbeda. Ekabel =16,500,000 =16,500,000Mpa
≤ φ ⋅ Nn 11.13 ≤ 0.9 ⋅ 101.4 11.13 ≤ 91.26
Nu
Nu=Gayaaksial tarik
=17.183t (kabel 4346dari hasil SAP SAP))
Nn =Kapasitas =Kapasitas tarik penampang
=101.4t
φ =Faktor reduksi
=0.9
1 2
1 2 3 4
3
= Baut 10mm = Plat = Angker = Kabel 35mm
4
Gbr.4.10.Elem Gbr.4.10.E lemen en kabel :
Komponen Struktur Tekan
A
Nu(-)
Nu(-)
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IV- 10
BAB IV PERHITUNGAN STRUKTUR
Elemen tekan terjadi pada seluruh batang atas dan bawah struktur lengk lengkung dan pada sebagian batang space frame. Batang diagonal pelengk pelengkung lateral menggunakan profil pipa. Profil rofil yang digunakan adalah : D=267 mm
F =90.87 cm2
L =8.838 m
G =24.2kg/m
t =6.6 mm
r =12.8 cm
Perencanaan elemen tekan
Nu ≤ φ ⋅ Nn 82.91 ≤ 0.85 ⋅ 106.293 82.91 ≤ 90.349 Nu=Gaya aksial aksial tekan
=82.91t (batang4139 dari SAP SAP)
Nn=Kuat =Kuat tekan penampang
= F ⋅ f cr
λc =
1 Lk λ
⋅
r
⋅
fy E
=
0.25 < λ c < 1.2 → ω f cr
=
fy ϖ
=
2400 1.508
1 3.14
=
⋅
883.8 12.2
240
⋅
1.43 1.6 − 0.67 ⋅ λ c
2000000
=
= 120.1 ⋅ 1920 = 106.293t
= 0.76
1.43 1.6 − 0.6 ⋅ 0.76
= 1.25
= 1591 .328 kg/m2
Keterangan: D =Diameter (mm)
F =Luas PenampangProfil rofil (m2)
T =T =Tebal Profil (mm)
r =J ari – jari kelembaman (m)
L =P =Panjang anjangProfil (m)
G =BeratProfil (kg/m)
I =MomenInersia (cm4)
λ c =P =Param arameterkelangsingan kelangsinganbatang
Lk =P =Panjang anjangbatang(m)
ϖ =Faktor langsing
f cr =tegangan Kritis Kritis (t/m (t/m2) Ta Tabel 4.2Faktor reduksi elemen baja struktural
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IV- 11
BAB IV PERHITUNGAN STRUKTUR
2
:
Batang Batangbawahlengkung utama
60- 108
914.4
30
108– 137.5
1000
40
0– 60
355.6
9.5
60– 108
406.4
12.7
108– 137.5
711.2
16
0- 40
355.6
9.5
40– 108
457.2
12.7
0- 40
267.4
6.6
40- 108
267.4
6.6
3
Batangatas lengkunglateral
4
Batang bawah lengkunglateral
5
Kolom Kolomlengkung lengkunglateral
-
457.2
12.7
6
BatangDiagonal
-
190.7
5.3
7
BatangLateral
-
165.2
5
8
Gording
-
190.7
5.3
9
Batang diagonal space frame
-
139.8
4.5
10
Batang Batangbawah space frame
-
114.3
4.5
Sambungan
Pada desain space frame atap stadion, setiap setiap elemen pada struktur llengk engkung dan rangka baja utamadisamb disambungdengansam sambunganlas. las. φ y ⋅ Rnw = 0.9 ⋅ t ⋅ fy (beban dasar) φ y ⋅ Rnw = 0.9 ⋅ t ⋅ fyw (las) Maka setiap setiap elemen lengkung lengkung dan rangka utama harus disam disambung dengan las penetrasi penetrasi penuh agarsam sambungantidak tidakmengurangi kapasitas sitasprofil profil.. :
Plat Dasar elemen tekan
Untuk mengandung elemen batang baja dengan kolom kolombeton beton atau penetrasi penetrasi memerlukan erlukan plat dasar
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IV- 12
BAB IV PERHITUNGAN STRUKTUR
B =C = 0.9 D + 2m = 0.9 ⋅ 457.2 + 2m B ⋅ C = 0.662m2 = (0.9 ⋅ 0.4572 + 2m )
2
→ m=0.2010 m=0.2010m
B =C = 0.9 ⋅ 0.4572 + 2 ⋅ 0.2010 = 0.81363 m 0.84 m q=
P A
=
77.55 0.706
= 109.84 t/m2
Fb = 0.75 fy = 0.75 ⋅ 24000 = 18000 t/m2
1
1
1 3 ⋅ q ⋅ m 2 2 3 ⋅ 109.84 ⋅ 0.2010 2 2 6 M 2 = = 0.027m = t = Fb Fb 18000
J umlah angkur : F’t =0.6 fy = 0.6 x 24000 t A=
P Ft
=.
109.84 14400
m
2
0.032 m
=14400t/m t/m2
= 0.008m 2
Direncanakan Direncanakan menggunakan angkur : 12 ø 25 Panjang angkur : f sc
1
= 0.58 f c = 0.58 ⋅ 2500 = 1450
2 ⋅ n ⋅ π ⋅ D ⋅ L =
P f cs
t
m
2
→ 1 2 ⋅ 0.2010 ⋅ π ⋅ 0.4572 ⋅ L =
Desain Desain panjang angkur 50cm.
77.55 1450
→ L =0.37 =0.37 m 0.50 m.
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IV- 13
BAB IV PERHITUNGAN STRUKTUR
koefisien koefisien pegas ( k =F / ). Perpindahan pada kolombag olombagian atas tidak boleh lebih dari L/ L/50 500. Dari hasil analisi lisis s SAP SAP 2000didapat : Ta Tabel 4.4. Output SAP 2000 pada kolomtung kolomtunggal h(m)
L/500
(m)
k(x+;x-;y+;y-;z)(t/m)
Agr bwh(m²)
Agr ats(m²)
Joint Force x,y,z(t)
K1
53
0.106
0.024
91;91;9;208;1429;1000
4.0 x 2.0
2.5 x 1.3
0.98;9.56;56.88
K2
50
0.1
0.028
110;110;249;1603;1074
4.0 x 2.0
2.4 x 1.2
1.65;1.27;46.06
K3
44
0.088
0.038
161;161;365;2353;1387
4.0 x 2.0
2x1
4.61;6.07;19.82
K4
35
0.07
0.07
197;2427;93;93;1972
3.0x1.5
2x1
16.3;1.778;45.029
K5
32
0.064
0.01
233;2849;108;108;2188
3.0x1.5
2x1
9.17;1.89;45.99
K5
21.2
0.0424
0.01
794;10101;370;370;4950
3.0x1.5
2x1
6.11;0.20;27.72
F(t)
d F / = k
0 K1
K2
K2
K3 (a)
K4
d(m)
K5 (b)
Gbr.4.13.a.Kolom tunggal penopang atap. b. Grafik konstanta pegas
Pada bagian bawah kolom lentur biaksi biaksial al dan gaya tekan dapat dihitun dihitung :
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IV- 14
BAB IV PERHITUNGAN STRUKTUR
lc =Panjang =Panjangtekuk h
b
= My Mx = 579.54 51.94 = 11.158
( b )⋅ (1 − B ) B + Mny
Mo = Mnx h
(1 − 0.65 )
Mo = 51.94 ⋅ 11.158
+ 579.54 = 891.6tm = 8916 kNm 0.65
Pu=772.34t=77 =7723.4 kN Agr = 4.0m ⋅ 2.0m − 2.3m ⋅ 0.5m = 6.85m
2
= 6.85 ⋅ 10 6 mm 2
F’c =2500 t/m2 =25 Mpa
et = Mu et
h
Pn
= 8916 7723.4 = 1.154m = 1154mm
= 1154 4000 = 0.289 Pu '
Agr φ ⋅ 0.85 ⋅ f ' c Pu ' Agr φ ⋅ 0.85 ⋅ f ' c
= ⋅
φ
= 0.65
7723400 N 0.65 ⋅ 6.85 ⋅ 10
et h
=
6
⋅ 0.85 ⋅ 25
= 0.082
7723400 N 0.65 ⋅ 6.85 ⋅ 10
6
⋅ 0.85 ⋅ 25
⋅ 0.289 = 0.024
Dari gambar 6.2.d 6.2.d Grafik Grafik dan Tabel Penulangan Penulangan BetonBertulang didapat r =0.001 =r . =0.001 =0.001. 1=0.001 =0.001
dimana untukbeton 25Mpa=1
Untuk desain desain struktur penahan gempa diambil =0.01. =0.01.
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IV- 15
BAB IV PERHITUNGAN STRUKTUR
K4
K5
K6
4.4
-45.01
16.3
0
0
0
25ø32
ø12-200
35
-149.2
16.3
285.27
-49
-75.385
40ø32
ø12-400
17.5
-293.7
16.3
570.54
-124.67
-191.8
56ø32
ø12-200
0
-45.96
-9.17
0
0
25ø32
ø12-200
32
-141.2
-9.17
-48.35
-74.385
40ø32
ø12-400
16
-273.3
-9.17
0 146.73 293.46
-119.43
-183.74
56ø32
ø12-200
0
-27.71
6.11
0
0
0
25ø32
ø12-200
21.2
-90.8
6.11
64.78
-13.57
-20.877
40ø32
ø12-400
10.6
-178.3
6.11
129.56
-42.2
-64.923
56ø32
ø12-200
0
Struktur Tribun ( portal beton bertulang )
Struktur tribun tribun penonton menggunakan struktur beton bertulang yang terdiri terdiri dari elemen plat lantai, lantai, plat tribun, tribun, tangga, balok balok anak, balok balok induk dan kolom. Konfigurasi Konfigurasi struktur struktur direncana irencana berdasarkan posisi posisi sudut pandang penonton, kapasitas penonton dan fungsi bangunan di dalam struktur tersebut. Secara struktural portal terdiri dari delapan struktur yangterpisah terpisah oleh dilatasi s selebar elebar 20cm. Hal ini ini diperlukan diperlukanuntukmenghindari torsi yangberlebihan erlebihansaatterjadi terjadigempa. Dilatasi 20cm
y57
Dilatasi 20cm
y49.5 y42.75 y36 y27 y18 y9
0
m 2 1
(b)
m 2 1 m 5 . 8 m
m 0
x9
x18
x27
x36
x45
x54
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IV- 16
BAB IV PERHITUNGAN STRUKTUR
Gbr.4.15.Perm Gbr.4.15.P ermodelan odelan portal portal
4.4.1
Kombinasi Kombinasi Pembebanan Portal
:
Bebanmati (DL)
Beban merata pada plat llantai, antai, plat tribun, tribun, dan tangga
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IV- 17
BAB IV PERHITUNGAN STRUKTUR
R
4.4.2
=faktor reduksi berdasarkanjenis jenis struktur danpendetailan etailan
Menentukan Pusat Massa Tiap Lantai
Pusat massa assa terletak pada koordinat ( X, X,Y ) kemudian struktur tiap lantai disatukan disatukan dengan menggunakan Constraint Diagfragma yang menyebabkan semua joi joint nt bergerak bersam bersama sebagai satukesatuan kesatuan. J enis Constraint
: Diaphragm(Diafragma)
SumbuConstraint Constraint
: Z axes (sum (sumbuZ)
Nama Constraint
: Lantai 1 Diaph1, Lantai 2 Diaph 2, Lantai 3 Diaph 3, Lantai 4 Diaph 4, Lan Lantai 5 Diaph iaph 5, Lantai 6 → Diaph 6, Lantai Lantai 7
→
Diaph 7, Atap → Diaph 8 4.4.3
Perhitungan Beban Gempa
Perhitungan erhitungan beb
pa untuk stadion stadion ini direncanakan direncanakan dengan ketentuan-ketentuan ketentuan-ketentuan sebagai
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IV- 18
BAB IV PERHITUNGAN STRUKTUR
5
100
500
S=
∑ t = ∑ S t
0,00174 0,00174
0,502 0,502
37.203
0,1639207 0,163920727
749,8145015 3839,126961
=0,14 ,14698 kg/cm2
S =14,698Kpa Kondisi tanah : •
Kedalaman lapisan lapisan keras 20m
•
S =14,698Kpa Te Tergolong dalamje mjenis tanah lunak, dimana termasuk pada klasifikasi kedalaman lapisan keras 20
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IV- 19
BAB IV PERHITUNGAN STRUKTUR
4.4.5
Perhitungan Berat TiapLantai
Direncanakan: :
Beban hidup ( WL ) untukLantai 1 – 4
=250kg/m2
:
Beban mati ( WD )
=Wbalok +Wkolom +W pelat
:
Beban desain esain ( WU )
=1.0 .05WD +1.05 WL
( TataCara CaraPerencanaan PerencanaanPembebananuntukRumahdanGedung/ SNISNI-1727-1989)
4.4.5.1. PEMBEBANAN TIAP LANTAI Lantai 1 ( W t1 )
Beban mati ( WD ) Pelat elat lantai
: [(40,5x12 [(40,5x126)+(16,875x (16,875x40,5)+(9x7.5)+ (9x7.5)+(7,5x18 (7,5x18)]x0,14x0 )]x0,14x0,85x2400
=1806831 kg
Balok Balok anak
:(63x5+ :(63x5+54+9x5+48x5+91+6,75x3+ 6,75x3+40,5)x2(0, 40,5)x2(0,2x0,32x0,3-0, 0,14x0,2) 14x0,2)x9, x9,1x2400 1x2400=105551 kg
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BAB IV PERHITUNGAN STRUKTUR
4.4.6. ANALISA PERHITUNGAN BEBAN GEMPA
Besarnya Besarnya Beban Gempa Nominal pada arah hori horisontal sontal yang diakibatkan diakibatkan oleh gempa, menurut Standar Peraturan Peraturan Perencanaan Tahan Gempa Indone Indonesia sia untuk Gedung SNI 03-1726-2003, dinyatakansebagai berikut: V = W.C.I./ R
Dim Dimana: W
=Kombinasi dari bebanmati danbebanhidup vertikal yangdireduksi. direduksi.
C
=Spektrum Respon Respon Gempa Rencana, yang besarnya tergantung dari jenis jenis tanah dasar
danwaktu getar struktur T ( Gambar 4.13). I
=Faktor =Faktor keutamaanstuktur.
R
=Faktor =Faktor reduksi gempa
IV- 20
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IV- 21
BAB IV PERHITUNGAN STRUKTUR
Kombinasi Beban Analisa Statik Bebantetap
=1.2 =1.2WD+1.6 .6WL
Dari perhitungan SAP 2000 didapat : Mlx=0.57 .57tm
Mtx=-1.17tm
Mty=-1.17tm
Dengan tebal pla platt : h=0.14 .14 m
p=0.02 .02m
φ tul =0.01 m
d = h − p − φ sengkang − 1
= 0.14 − 0.02 − 0.01 − 0.01 ⋅ 1 2 = 0.115m
Untuk Mlx =
Mly=0.57 .57tm
Mlx b ⋅ d
2
=
2 tul .utama 0.57
1 ⋅ 0.115
2
= 43.1t / m 2
Dari grafik grafik dan tabel perencanaan Beton Bertulang ρ =0.0023 ρ min
Alx = ρ ⋅ b ⋅ d = 0.0025 ⋅ 1000 ⋅ 115 = 288mm 2 (Ø 10-250)
= 0.0025
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IV- 22
BAB IV PERHITUNGAN STRUKTUR
Momen Lapangan Lapanganx
0.135
0.56
30.727 0.002 0.0028 378 Ø10 Ø10– 200
Momen Lapangan Lapangany
0.125
0.45
28.8
Momen Tumpuanx
0.135
-0.97
-53.22
0.0028 378
Momen Tumpuany
0.125
-0.81
-51.84
0.0028 350 Ø10 – 200
0.0028 350 Ø10– 200
Ta Tabel 4 11 Penulangan tangga
Ø10– 200
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BAB IV PERHITUNGAN STRUKTUR
25 bw ⋅ d • 350 • 625.5 15 6 = = 182437 N Vc = 2 2.5 ⋅ d ⋅ T n 2.5 • 625.5 • 20000 1 + 1 + • • 350 700 118000 b ⋅ h ⋅ Uu f ' c
φVc = 0.6Vc = 109462.2 N
IV- 23
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IV- 24
BAB IV PERHITUNGAN STRUKTUR
0.050 Kantilever Ujung kiri
0.1
0.6
-90.2
0.0054
1372.7
8ø25
Ujung kanan
0.1
0.6
0.6
0.0089
2262.4
4ø25
4.6.2 4.6.2
Perencanaan Kolo m
23.7 22.3
0.180 0.180
1.1
0.5
0.6
φ12-150
1.0
0.5
0.5
φ12-150
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IV- 25
BAB IV PERHITUNGAN STRUKTUR
d’ =p +φ sengkang + 1 φ utama 2
= 50 + 12 + 1 2 ⋅ 25 = 74.5mm
d =h – d’ =700 – 74.5 =625.5 mm d '
d
= 74.5 625.5 = 0.12 ≈ 0.15
Dari gambar 6.2.d Grafi Grafik k dan Tabel Penulangan Penulangan Beton Bertulang didapat r =0.004 =0.004 β =1 ρ = r β
= 0 004
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IV- 26
BAB IV PERHITUNGAN STRUKTUR
Tabel 4.14. Data Sondir Kedalaman(m)
Conus
Biconus Biconus
Cleef Cleef (kg/cm2)
total friction friction (kg/cm)
0.00
-
-
-
-
0.20
-
-
-
-
0.40
5
10
5
10
0.60
5
10
5
20
0.80
10
15
5
30
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BAB IV PERHITUNGAN STRUKTUR
IV- 27
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BAB IV PERHITUNGAN STRUKTUR
IV- 28
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BAB IV PERHITUNGAN STRUKTUR
IV- 29
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BAB IV PERHITUNGAN STRUKTUR
IV- 30
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BAB IV PERHITUNGAN STRUKTUR
IV- 31
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