Depending upon the location of the tank the tanks can be named as overhead, on ground or underground. The tanks can be made in different shapes usually circular and rectangular shapes are mostly used. The tanks can be made of RCC or even of steel. The overhead tanks are usually elevated from the roof top through column. In the other hand the underground tanks are rested on the foundation. Different types of tanks and their design procedure is discussed in subsequent portion if this chapter. The water tanks in this chapter are designed on the basis of no crack theory. The concrete used are made impervious. Basing on the location of the tank in a building s tanks can be classified into three categories. Those are • Underground tanks • Tank resting on grounds • Overhead tanks In most cases the underground and on ground tanks are circular or rectangular is shape but the shape of the overhead tanks are influenced by the aesthetical view of the surroundings and as well as the design of the construction. Steel tanks are also used specially in railway yards. Basing on the shape the tanks can be circular, rectangular, square, polygonal, spherical and conical. A special type of tank named Intze tank is used for storing large amount of water for an area. The overhead tanks are supported by the column which acts as stages. This column can be braced for increasing strength and as well as to improve the aesthetic views. One of the vital considerations for design of tanks is that the structure has adequate resistance to cracking and has adequate strength. For achieving these following assumptions are made • Concrete is capable of resisting limited tensile stresses the full section of concrete including cover and reinforcement is taken into account in this assumption. • To guard against structural failure in strength calculation the tensile strength of concrete is ignored. • Reduced values of permissible stresses in steel are adopted in steel are adopted in design. If the tank is resting directly over ground, floor may be constructed of concrete with nominal percentage of reinforcement provided that it is certain that the ground will carry the load without appreciable subsidence  in any part and that the concrete floor is cast in panels with sides not more than with contraction or expansion joints between. In such cases a screed or concrete layer less than 75mm thick shall first be placed on the ground and covered with a sliding layer of bitumen paper or other suitable material to destroy the bond between the screed and floor concrete. In normal circumstances the screed layer shall be of grade not weaker than M 10,where injurious soils or aggressive water are expected, the screed layer shall be of grade not weaker than M 15 and if necessary a sulphate resisting or other special cement should be used. Detail Penulis Penerbit Sipilpedia Bahasa Arabic Durasi Format MP4 Ukuran  Mb DOWNLOAD Konten berikutnya khusus bagi buyer yang telah membeli Premium Membership Daftar disini

Recreationalvehicles typically come with three tanks, located on the underside of the RV, to keep things running smoothly. 1. Fresh Water Tank. There are usually three tanks on an RV: fresh water, grey and black. As the name suggests, this tank is for fresh water. This is the water that comes out of your taps. 2.

Dari Perhitungan di atas, diperoleh volume yang harus ditampung ground reservoir di mana diambil volume yang terbesar m 3 jam jam 6 pagi + m 3 jam jam 8 malam = m 3 ≈ 390 m 3 Kapasitas Ground Reservoir Kecamatan Gunem Volume yang dibutuhkan 390 m 3 Direncanakan tinggi ground reservoir 3 m dan lantai dasar ground reservoir persegi P = L Maka dimensi ground reservoir yang lain V = P x L x t 390 m 3 = P x L x 3 m P x L = 130 m 2 P = 13 L = 10 m Jadi dimensi reservoir P = 13 m ; L = 10 m ; t = 3,5 m. 0,5 Freeboard. c. Rencana Desain Bangunan Ground Reservoir 1. Panjang bangunan = 13 m Lebar bangunan = 10 m Tinggi MA dari dasar = 3 m Tinggi jagaan = m Tinggi total bangunan = m 2. Tebal dinding beton = m 3. Tebal lantai beton = m 4. Plat atap beton = m 5. Mutu beton fc = 25 Mpa Mutu baja fy = 400 Mpa 6. Perhitungan struktur menggunakan program SAP dengan acuan buku ”Dasar – dasar Perencanaan Beton Bertulang ” dan ” Grafik dan Tabel Perhitungan Beton Bertulang ” berdasarkan SKSNI T 15 – 1991 – 03. d. Perhitungan Struktur Ground Reservoir Ground Reservoir direncanakan menggunakan struktur beton bertulang. Sebelumnya perlu dilakukan perhitungan terhadap pembebanan ground reservoir . Perhitungan pembebanan ground reservoir sebagai berikut ini Perhitungan Pelat Dasar Tebal plat h = 25 cm = 250 mm Lebar b = 1000 mm Penutup beton p = 40 mm Diameter tulangan utama direncanakan = ø 10 mm Dimeter tulangan bagi direncanakan = ø 8 mm Tinggi efektif adalah Arah x d x = h – p – ½ øD = 250 – 40 – ½ 10 = 205 mm Arah y d y = h – ρ – øD - ½ øS = 250 – 40 – 10 - ½ 8 = 196 mm Dengan spesifikasi - Mutu beton fc = 25 Mpa - Mutu baja fy = 400 Mpa Maka digunakan - ρ min = - ρ max = Dari perhitungan SAP didapat Momen Tumpuan - x = Momen Lapangan - x = Gambar Momen M11 Plat Dasar Arah x Momen Tumpuan - y = -6 Momen Lapangan - y = Gambar Momen M22 Plat Dasar Arah y Momen Tumpuan arah – x 2 .d b Mu = 2 205 . . 1 4 . 6 = kNm 2 ρ min = ρ max = ρ = diinterpolasi ρ min ρ ρ max dipakai ρ min = As = x = x 1000 x x 10 6 = 369 mm 2 digunakan tulangan ф 10 – 200 As terpasang 393 mm 2 Momen Lapangan arah – x 2 .d b Mu = 2 205 . . 1 67 . = kNm 2 ρ min = ρ max = ρ = diinterpolasi ρ min ρ ρ max dipakai ρ min = As = x = x 1000 x x 10 6 = 369 mm 2 digunakan tulangan ф 10 – 200 As terpasang 393 mm 2 Momen Tumpuan arah – y 2 .d b Mu = 2 196 . . 1 6 = kNm 2 ρ min = ρ max = ρ = diinterpolasi ρ min ρ ρ max dipakai ρ min = As = y = x 1000 x x 10 6 = mm 2 digunakan tulangan ф 8 – 125 As terpasang 402 mm 2 Momen Lapangan arah - y 2 .d b Mu = 2 196 . . 1 5 . = kNm 2 ρ min = ρ max = ρ = diinterpolasi ρ min ρ ρ max dipakai ρ min = As = y = x 1000 x x 10 6 = mm 2 digunakan tulangan ф 8 – 125 As terpasang 402 mm 2 Perhitungan Atap Tebal plat h = 20 cm = 200 mm Lebar b = 1000 mm Penutup beton p = 40 mm Diameter tulangan utama direncanakan = ø 10 mm Diameter tulangan bagi direncanakan = ø 10 mm Tinggi efektif adalah Arah x d x = h – p – ½ ø D = 200 – 40 – ½ 10 = 155 mm Arah y d y = h – p – øD - ½ øS = 200 – 40 – 10 - ½ 8 = 146 mm Dengan spesifikasi - Mutu beton fc = 25 Mpa - Mutu baja fy = 400 Mpa Maka digunakan - ρ min = - ρ max = Dari perhitungan SAP didapat Momen Tumpuan - x = -36 Momen Lapangan - x = 23 Gambar Momen M22 Plat Atap Arah x Momen Tumpuan - y = Momen Lapangan - y = Gambar Momen M22 Plat Atap Arah y Momen Tumpuan arah – x 2 .d b Mu = 2 155 . . 1 36 = kNm 2 ρ min = ρ max = ρ = diinterpolasi ρ min ρ ρ max dipakai ρ = As = x = x 1000 x x 10 6 = mm 2 digunakan tulangan ф 10 – 75 As terpasang 1047 mm 2 Momen Lapangan arah – x 2 .d b Mu = 2 155 . . 1 23 = kNm 2 ρ min = ρ max = ρ = diinterpolasi ρ min ρ ρ max dipakai ρ = As = x = x 1000 x x 10 6 = mm 2 digunakan tulangan ф 10 – 150 As terpasang 524 mm 2 Momen Tumpuan arah – y 2 .d b Mu = 2 146 . . 1 5 . 31 = kNm 2 ρ min = ρ max = ρ = diinterpolasi ρ min ρ ρ max dipakai ρ = As = y = x 1000 x x 10 6 = 730 mm 2 digunakan tulangan ф 8 – 50 As terpasang 1005 mm 2 Momen Lapangan arah - y 2 .d b Mu = 2 146 . . 1 2 . 14 = kNm 2 ρ min = ρ max = ρ = diinterpolasi ρ min ρ ρ max dipakai ρ = As = y = x 1000 x x 10 6 = mm 2 digunakan tulangan ф 8 – 150 As terpasang 335 mm 2 Perhitungan Dinding Tebal plat = 20 cm = 200 mm Penutup beton p = 40 mm Diameter tulangan utama direncanakan = ø 10 mm Dimeter tulangan bagi direncanakan = ø 8 mm Tinggi efektif adalah Arah x d x = h – p – ½ øD = 200 – 40 – ½ 10 = 155 mm Arah y d y = h – p – øD - ½ øS = 200 – 40 – 10 - ½ 8 = 146 mm Dengan spesifikasi - Mutu beton fc = 25 Mpa - Mutu baja fy = 400 Mpa Maka digunakan - ρ min = - ρ max = Dinding arah xz Dari perhitungan SAP didapat Momen Tumpuan - x = -7,5 Momen Lapangan - x = 5 Gambar Momen M22 Plat dinding arah x Pu Tumpuan - x = - 40 Pu Lapangan - x = 25 Gambar Gaya Aksial F22 Plat dinding arah x Momen Tumpuan arah – x e 1 = Pu Mu = 40 5 , 7 = m = 187,5 mm h e 1 = 1000 5 , 187 = 0,1875 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ . 85 , . c gr u f A P φ . ⎥⎦ ⎤ ⎢⎣ ⎡ h e 1 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 25 . 85 , . 200 . 1000 . 65 , 40000 . 0,1875 = 0,0027 Dari grafik tulangan kolom Grafik dan Tabel Perhitungan Beton Bertulang Didapat r = 0,0020 ; β = 1,0 ρ = r . β = 0,0020 . 1,0 = 0,0020 Tulangan Utama As tot = ρ . = 0,0020 . 200 . 1000 = 400 mm 2 digunakan tulangan ф 10 – 175 As terpasang 449 mm 2 Momen Lapangan arah – x e 1 = Pu Mu = 25 5 = m = 200 mm h e 1 = 1000 200 = 0,2 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ . 85 , . c gr u f A P φ . ⎥⎦ ⎤ ⎢⎣ ⎡ h e 1 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 25 . 85 , . 200 . 1000 . 65 , 5000 . 0,2 = 0,0004 Dari grafik tulangan kolom Grafik dan Tabel Perhitungan Beton Bertulang Didapat r = 0,00155 ; β = 1,0 ρ = r . β = 0,00155 . 1,0 = 0,00155 Tulangan Utama As tot = ρ . = 0,00155 . 200 . 1000 = 310 mm 2 digunakan tulangan ф 10 – 250 As terpasang 314 mm 2 Tulangan bagi diambil 20 .As Tumpuan = 20 . 400 mm 2 = 80 mm 2 digunakan tulangan ф 8 – 250 As terpasang 201 mm 2 Lapangan = 20 . 310 mm 2 = 62 mm 2 digunakan tulangan ф 8 – 250 As terpasang 201 mm 2 Dinding arah yz Dari perhitungan SAP didapat Momen Tumpuan - y = -19 Momen Lapangan - y = 3,8 Gambar Momen M22 plat dinding arah y Gaya Aksial Pu Tumpuan - y = -44 Gaya Aksial Pu lapangan - y = 27,5 Gambar Gaya Aksial F22 plat dinding arah y Momen Tumpuan arah – y e 1 = Pu Mu = 44 19 = 0,432m = 432 mm h e 1 = 1000 432 = 0,432 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ . 85 , . c gr u f A P φ . ⎥⎦ ⎤ ⎢⎣ ⎡ h e 1 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 25 . 85 , . 200 . 1000 . 65 , 44000 . 0,432 = 0,00688 Dari grafik tulangan kolom Grafik dan Tabel Perhitungan Beton Bertulang Didapat r = 0,0025 ; β = 1,0 ρ = r . β = 0,0025 . 1,0 = 0,0025 Tulangan Utama As tot = ρ . b. h = 0,0025 . 200 . 1000 = 500 mm 2 digunakan tulangan ф 10 – 150 As terpasang 524 mm 2 Momen Lapangan arah - y e 1 = Pu Mu = 5 , 27 8 , 3 = 0,1382 m = 138,2 mm h e 1 = 1000 2 , 138 = 0,1382 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ . 85 , . c gr u f A P φ . ⎥⎦ ⎤ ⎢⎣ ⎡ h e 1 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 25 . 85 , . 200 . 1000 . 65 , 40000 . 0,1382 = 0,0014 Dari grafik tulangan kolom Grafik dan Tabel Perhitungan Beton Bertulang Didapat r = 0,0017 ; β = 1,0 ρ = r . β = 0,0017 . 1 = 0,0017 Tulangan Utama As tot = ρ . = 0,0017 . 200 . 1000 = 340 mm 2 digunakan tulangan ф 10 – 225 As terpasang 349 mm 2 Tulangan bagi diambil 20 .As tumpuan = 20 . 500 mm 2 = 100 mm 2 digunakan tulangan ф 8 – 250 As terpasang 201 mm 2 Lapangan = 20 . 340 mm 2 = 68 mm 2 digunakan tulangan ф 8 – 250 As terpasang 201 mm 2 Tabel Rangkuman Penulangan Ground Reservoir Komponen Struktur Ukuran Penulangan - Pelat Atas Tebal 200 mm Tumpuan arah – x P10 - 75 Lapangan arah – x P10 - 150 Lapangan arah – y P8 - 50 Lapangan arah – y P8 - 150 - Pelat Dinding Tebal 200 mm Tumpuan arah – xz P10 -175 Lapangan arah – xz P10 - 250 Tulangan bagi – xz P8 - 250 Tumpuan arah – yz P10 -150 Lapangan arah – yz P10 - 225 Tulangan bagi – yz P8 – 250 - Pelat Dasar Tebal 250 mm Tumpuan arah – x P10 - 200 Lapangan arah – x P10 - 200 Lapangan arah – y P8 - 125 Lapangan arah – y P8 - 125 Sumber Hasil Perhitungan, 2008 Gambar Pemodelan Ground Reservoir pada program SAP PERENCANAAN TEKNIS PIPA TRANSMISI · In these cases, having a buffer tank to replenish the hot water just used, with even 100ish F water, will keep the finish tank from working as hard to recover. But in situations where the total daily hot water usage is low (10-20 gal) and distributed through out the day, having a large buffer of even 120F water available is not so obvious .

concrete(1) Concrete (2nd Edition) by Sidney Mindess, J. Francis Young, David Darwin. Concrete - Micro-structure Properties and Materials (2006) Concrete - Neville's Insights and Issues. Concrete - Nevilles Insights and Issues. Concrete - The Sustainable Infrastructure Material for 21st Century. Concrete Admixtures Handbook.

Italso requires clean water capacity of 75.2 m3 / day perencanaan struktur bangunan, arsitektur bangunan, suatu gedung juga membutuhkan Kapasitas ground tank : Air pada ground tank digunakan untuk kebutuhan selama 2 hari kapasitas ground tank=(2 hari x jumlah total kebutuhan air bersih)

a Sistem plumbing untuk air bersih. Salah satu sistem plumbing pada gedung yaitu digunakan untuk penyediaan sumber air bersih dan distribusi air bersih ke berbagai bagian gedung. Biasanya sistem penyediaan air bersih pada sebuah gedung dinamakan dengan WTP (water treatment plant).

Hidraulikair tanah, hukum Darcy, persamaan energi aliran (ground water head), jenis aquifer, (unconfined, confined, semi confined aquifer). Teknik explorasi dan exploitasi serta pengelolaan air tanah, parameter aquifer S & T aliran transsient dan steady flow. Persamaan ground head dengan debit, Penyelesaian numerik dan finite element. Luaran: a.

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perencanaan struktur ground water tank