LAPORAN TUGAS DESAIN BANGUNAN BETON

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LAPORAN TUGAS

DESAIN BANGUNAN BETON

Disusun Oleh :

1. Rifaldi Rahmansyah 01.2019.1.05909

2. Aji Priatmojo 01.2019.1.05843

FAKULTAS TEKNIK SIPIL DAN PERENCANAAN INSTITUT TEKNOLOGI ADHI TAMA SURABAYA

2021

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LEMBAR PENGESAHAN

LAPORAN INI DIBUAT UNTUK MENDUKUNG NILAI DESAIN BANGUNAN BETON

Disusun Oleh:

1. Rifaldi Rahmansyah 01.2019.1.05909

2. Aji Priatmojo 01.2019.1.05843

Mengetahui :

Dosen Pengampu

Eka Susanti, ST., MT NIP : 991106

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DAFTAR ISI

COVER ... i

LEMBAR PENGESAHAN ... ii

DAFTAR ISI ...iii

BAB I PENDAHULUAN ... 1

1.1 Data Perencanan ... 1

BAB II PERENCANAAN PLAT... 3

2.1 Data Plat Lantai & Atap A1 ... 3

2.1.1 Preliminary Tebal Plat Lantai A1 ... 3

2.1.2 Pembebanan Plat Lantai A1... 3

2.1.3 Momen Plat Lantai A1 ... 4

2.1.4 Penulangan Plat Lantai A1 ... 4

2.1.5 Preliminary Tebal Plat Atap A1 ... 6

2.1.6 Pembebanan Plat Atap A1 ... 6

2.1.7 Momen Plat Atap A1 ... 6

2.1.8 Penulangan Plat Atap A1 ... 7

2.2 Data Plat Lantai & Atap A2 ... 9

2.2.1 Preliminary Tebal Plat Lantai A2 ... 9

2.2.2 Pembebanan Plat Lantai A2... 9

2.2.3 Momen Plat Lantai A2 ... 10

2.2.4 Penulangan Plat Lantai A2 ... 10

2.2.5 Preliminary Tebal Plat Atap A2 ... 12

2.2.6 Pembebanan Plat Atap A2 ... 12

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2.2.7 Momen Plat Atap A2 ... 13

2.2.8 Penulangan Plat Atap A2 ... 13

BAB III PERENCANAAN BALOK ANAK ... 18

3.1 Data Balok B-1 Atap ... 18

3.1.1 Pembebanan Balok - B1 Atap ... 19

3.1.2 Momen Balok - B1 Atap ... 20

3.1.3 Penulangan Balok – B1 Atap ... 20

3.1.4 Penulangan Tumpuan ... 21

3.1.5 Penulangan Lapangan ... 22

3.1.6 Penulangan Geser Balok Anak B – 1 Atap ... 24

3.2.1 Pembebanan Balok – B2 Atap ... 26

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3.6 Data Balok B-1 Lantai ... 58

3.6.1 Pembebanan Balok - B1 Lantai ... 58

3.6.2 Momen Balok – B1 Lantai ... 59

3.6.3 Penulangan Balok – B1 Lantai ... 60

3.6.6 Penulangan Geser Balok Anak B – 1 Lantai ... 64

3.7.1 Pembebanan Balok – B2 Lantai ... 67

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BAB IV PERENCANAAN BALOK INDUK ... 99

4.1 Data Balok Induk Melintang B6 – A Atap ... 99

4.1.1 Pembebanan Balok Induk Melintang B6 – A Atap ... 99

4.1.1.1 Beban Terpusat (pada ujung portal) P1 ... 100

4.1.1.2 Beban Terpusat (pada tengah portal) P2 ... 100

4.1.1.6 Beban Merata Q ... 101

4.1.2 Momen – Momen Balok Induk Melintang B6-A Atap (Hasil Output SAP2000 2D Portal) ... 101

4.1.3 Penulangan Balok Induk Melintang B6 – A Atap ... 102

4.1.4 Penulangan Tumpuan 1 ... 102

4.1.7 Penulangan Geser (pada ujung kiri balok induk atap) Hasil Output SAP2000 2D Portal ... 106

4.1.8 Penulangan Geser (pada ujung kanan balok induk atap) Hasil Output SAP2000 2D Portal ... 108

4.2 Data Balok Induk Melintang B6 – A Lantai ... 110

4.2.1 Pembebanan Balok Induk Melintang B6 – A Lantai ... 111

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4.2.1.6 Beban Merata Q ... 112

4.2.2 Momen – Momen Balok Induk Melintang B6-A Lantai (Hasil Output SAP2000 2D Portal) ... 113

4.2.3 Penulangan Balok Induk Melintang B6 – A Lantai ... 113

4.2.7 Penulangan Geser (pada ujung kiri balok induk atap) Hasil Output SAP2000 2D Portal ... 118

4.2.8 Penulangan Geser (pada ujung kanan balok induk atap) Hasil Output SAP2000 2D Portal ... 119

BAB V PERENCANAAN KOLOM ... 122

5.1 Input Data – Data Kolom... 122

5.2 Analisa Perencanaan Kolom dengan Software PCACol ... 122

5.3 Hasil Output PCACol ... 126

DAFTAR PUSTAKA ... 130

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BAB 1 PENDAHULUAN 1.1 Data Perencanaan

● Jarak Portal : 5 m

● Perletakan : A = Jepit B = Jepit C = Jepit

● Jarak Kolom : L1 = 6 m L2 = 7 m

● Tinggi Kolom : h1 = 5,5 m h2 = 4,5 m

● Mutu Bahan : Beton : f”c = 30 Mpa Baja : fy =240 Mpa

● Jumlah Portal : 8 portal

● Beban hidup, dimisalkan lantai gedung untuk : Kantor

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BAB 2

PERENCANAAN PLAT 2.1 Data Plat Lantai & Atap A1

● Fc : 30 Mpa

● Fy : 240 Mpa

● Ly : 5000 mm

● Lx : 3500 mm

● Penentuan jenis plat 𝛽 =𝑙𝑛𝑦

𝑙𝑛𝑥=5000

3500= 1.43 < 2.5, 𝑚𝑎𝑘𝑎 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑏𝑎ℎ𝑤𝑎 𝑝𝑙𝑎𝑡 2 𝑎𝑟𝑎ℎ

2.1.1 Preliminary Tebal Plat Lantai A1

~ ℎ𝑚𝑎𝑥 = ^{𝑙𝑛×[0.8+}

𝑓𝑦 1500]

36+5×𝛽[𝛼𝑚−0.2] = ^5000×[0.8+

240 1500]

36+5×1.43[0.6−0.2]= 123,53 𝑚𝑚

~ℎ 𝑚𝑖𝑛 =^{𝑙𝑛×[0.8+}

𝑓𝑦 100]

36+9×𝛽 =^5000×[0.8+

240 1500]

36+9×1.43 = 98,25 𝑚𝑚

Tebal direncanakan harus memenuhi syarat h min < h < h max, maka dipilih tebal plat sebesar 100 mm

2.1.2 Pembebanan Plat Lantai A1

Beban Hidup : Ruang Kantor = 2,4 Kn/m² = 244,73 Kg/m² Beban Mati : Beban Sendiri = 0.23 Kn/m² = 234,53 Kg/m²

Beban Keramik & Spesi = 78,5169 Kg/m² Beban Plafond = 48,9456 Kg/m² + Total Beban Mati = 361,99 Kg/m²

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Beban Ultimate (qu)

𝑞𝑢 = (1.2 × 𝐷𝐿) + (1.6 × 𝐿𝐿) = (1.2 × 361,99) + (1.6 × 244.73) = 825,92 𝐾𝑔/𝑚²

Jadi beban ultímate (qu) yang diterima oleh plat lantai A1 sebesar 825,92 Kg/m² 2.1.3 Momen Plat Lantai A1

Berdasarkan tabel Marcus 13.3.2.II Plat 2 arah, terdapat perhitungan momen Plat Terjepit Elastis

Dari tabel 13.3.2.II didapat nilai x dari Ly/Lx = 1.43, yaitu sebagai berikut :

▪ MLx (+) = MTx (-) = 50

▪ MLy (+) = 38

▪ MTy (-) = 38

▪ 𝑀𝑙𝑥 = +0.001𝑞𝑙𝑥²𝑥 = +0.0001 × 825,92 × 3,5² × 50 = −505,89 𝐾𝑔. 𝑚

▪ 𝑀𝑙𝑦 = +0.001𝑞𝑙𝑥²𝑥 = +0.0001 × 825,92 × 3,5² × 38 = 384,48 𝐾𝑔. 𝑚

▪ 𝑀𝑡𝑦 = −0.001𝑞𝑙𝑥²𝑥 = −0.0001 × 825,92 × 3,5² × 38 = −384,48 𝐾𝑔. 𝑚

2.1.4 Penulangan Plat Lantai A1

● Untuk f’c < 30 Mpa, maka digunakan β1 = 0.85 𝜌 𝑚𝑖𝑛 = 1,4

𝑓𝑦 = 1,4

240= 0.0058 𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 =0.85𝑓′𝑐

𝑓𝑦 𝛽1 600

600 + 𝑓𝑦=0.85 × 30

240 0.85 600

600 + 240= 0,065 𝜌 𝑚𝑎𝑥 = 0.75 × 𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 = 0,75 × 0,065 = 0,048

● Penulangan Rencana : Tul. Arah X =ᴓ25 mm & Y = ᴓ10 mm 𝑑𝑥 = ℎ − 𝑠 −1

2∅𝑥 = 100 − 20 −1

2× 25 = 67,5 𝑚𝑚 𝑑𝑦 = ℎ − 𝑠 − ∅𝑥 −1

2∅𝑦 = 100 − 20 − 25 −1

2× 10 = 50 𝑚𝑚

● Penulangan Plat Arah x (untuk daerah Tumpuan & Lapangan) MTx = MLx = 5058986,62 Nmm²

𝑚 = ^𝑓𝑦

0.85×𝑓𝑐 = ²⁴⁰

0.85×30= 9,4

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𝑅𝑛𝑥 = ^𝑀𝑢𝑥

0.8×𝑏×𝑑²= ^5058986,62

0.8×1000×67,5²= 1,38 𝜌 𝑝𝑒𝑟𝑙𝑢 = ¹

𝑚[√1 −^{(2×𝑚×𝑅𝑛𝑥)}

𝑓𝑦 ] = ¹

9,4[√1 −(2×9,4×1,38)

240 ] = 0.0059

Karena hasil ρ min = 0.0058 <ρ perlu = 0.0059< ρ max = 0.048, maka digunakan ρ pakai = 0.0059

Sehingga luas tulangan yang diperlukan adalah :

𝐴𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌 𝑝𝑎𝑘𝑎𝑖 × 𝑏 × 𝑑 = 0.0059 × 1000 × 67,5 = 401,59 𝑚𝑚² Σ𝑡𝑢𝑙𝑎𝑛𝑔𝑎𝑛 =_{0.25×𝜋×𝑑²}^𝐴𝑠 =0.25×3.14×25²^401,59 = 0,8, 𝑑𝑖𝑏𝑢𝑙𝑎𝑡𝑘𝑎𝑛 𝑗𝑎𝑑𝑖 1 𝑏𝑎𝑡𝑎𝑛𝑔 𝑠 =^{1000×𝐴𝑑}

𝐴𝑠−𝑥 =^1000×319,9

401,59 = 1221,7 𝑚𝑚 − − − 1000 𝑚𝑚

𝑠1 = 2 × ℎ = 2 × 100 = 200 𝑚𝑚 (SNI 2847-2019,pasal 9.12.5.4) 𝑠2 = 450 𝑚𝑚 (SNI 2847-2019,pasal 9.12.5.4)

Jarak dipilih yang terkecil, s pakai = 200 mm.

Jadi pada penulangan plat arah X dipasang Tul. Lentur ᴓ25-200 mm, dengan,

𝐴𝑠 𝑝𝑎𝑘𝑎𝑖 =1000 × 𝐴𝑑

𝑠 𝑝𝑎𝑘𝑎𝑖 =1000 × 490,62

200 = 2453,1 𝑚𝑚² > 𝐴𝑠 𝑝𝑒𝑟𝑙𝑢 = 419,92 𝑚𝑚²

● Cek Momen Nomina 𝑎 =𝐴𝑠 𝑝𝑎𝑘𝑎𝑖×𝑓𝑦

0,85×𝑓′𝑐×𝑏 = ^2453,1^×240

0,85×30×1000= 23,1 𝑚𝑚

𝑀𝑛 = (𝐴𝑠 × 𝑓𝑦) × (𝑑 −^𝑎

2) = (2453,1 × 240) × (67,5 −^23,1

2 ) = 32943759,2 𝑁𝑚𝑚²

∅𝑀𝑛 = 0,9 ×2943759,2= 29649383,3 𝑁𝑚𝑚² > 𝑀𝑢 = 5058986,62 𝑁𝑚𝑚²

● Penulangan Plat Arah y (untuk daerah Tumpuan & Lapangan) MTy = MLy = 3844829,83 Nmm²

𝑚 = ^𝑓𝑦

0.85×𝑓𝑐 = ²⁴⁰

0.85×30= 9,4 𝑅𝑛𝑥 = ^𝑀𝑢𝑥

0.8×𝑏×𝑑²= ^4106847.87

0.8×1000×53²= 1,83 𝜌 𝑝𝑒𝑟𝑙𝑢 = ¹

𝑚[√1 −^{(2×𝑚×𝑅𝑛𝑥)}

𝑓𝑦 ] = ¹

9,4[√1 −(2×9,4×1,83)

240 ] = 0.0079

𝐴𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌 𝑝𝑎𝑘𝑎𝑖 × 𝑏 × 𝑑 = 0.0079 × 1000 × 53 = 419,18 𝑚𝑚²

𝛴𝑡𝑢𝑙𝑎𝑛𝑔𝑎𝑛 = 𝐴𝑠

0.25 × 𝜋 × 𝑑²= 419,18

0.25 × 3.14 × 10²= 5,34, 𝑑𝑖𝑏𝑢𝑙𝑎𝑡𝑘𝑎𝑛 𝑗𝑎𝑑𝑖 6 𝑏𝑎𝑡𝑎𝑛𝑔

𝑠 =^{1000×𝐴𝑑}_{𝐴𝑠−𝑥} =^1000×78,5_419,18 = 187,3 𝑚𝑚 − − − 150 𝑚𝑚

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Jadi pada penulangan plat arah Y dipasang Tul. Lentur ᴓ10-150 mm, dengan,

𝑠 𝑝𝑎𝑘𝑎𝑖 =1000 × 78,5

150 = 523,3 𝑚𝑚² > 𝐴𝑠 𝑝𝑒𝑟𝑙𝑢 = 419,18 𝑚𝑚²

0,85×𝑓′𝑐×𝑏 =0,85×25×1000^523,2×240 = 4,93 𝑚𝑚 𝑀𝑛 = (𝐴𝑠 × 𝑓𝑦) × (𝑑 −^𝑎

2) = (523,3 × 240) × (53 −^4,93

2 ) = 6347479,2𝑁𝑚𝑚²

∅𝑀𝑛 = 0,9 × 6347479,2 = 5712731,3 𝑁𝑚𝑚² > 𝑀𝑢 = 4106847.87 𝑁𝑚𝑚²

2.1.5 Preliminary Tebal Plat Atap A1

~ ℎ𝑚𝑎𝑥 = ^{𝑙𝑛×[0.8+}

𝑓𝑦 1500]

36+5×𝛽[𝛼𝑚−0.2] = ^5000×[0.8+

240 1500]

36+5×1.43[0.6−0.2]= 123,53 𝑚𝑚

~ℎ 𝑚𝑖𝑛 =𝑙𝑛 × [0.8 + ^𝑓𝑦

100]

36 + 9 × 𝛽 =5000 × [0.8 + ²⁴⁰

1500]

36 + 9 × 1.43 = 98,25 𝑚𝑚

2.1.6 Pembebanan Plat Atap A1

Beban Hidup : atap datar, berbubung, dan lengkung= 0.96 Kn/m² = 97,89 Kg/m² Beban Mati : Beban Sendiri = 0,23 Kn/m² = 234,53 Kg/m²

Beban Waterproofing = 0,1 Kn/m² = 10,197 Kg/m² Beban Plafond = 0,48 Kn/m² = 48,9456 Kg/m² + Total Beban Mati = 293,67Kg/m² Beban Ultimate (qu)

𝑞𝑢 = (1.2 × 𝐷𝐿) + (1.6 × 𝐿𝐿) = (1.2 × 293,67) + (1.6 × 97,89) = 565,32 𝐾𝑔/𝑚²

Jadi beban ultímate (qu) yang diterima oleh plat lantai A1 sebesar 565,32 Kg/m² 2.1.7 Momen Plat Atap A1

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▪ MLx (+) = MTx (-) = 50

▪ MLy (+) = 38

▪ MTy (-) = 38

▪ 𝑀𝑙𝑥 = +0.001𝑞𝑙𝑥²𝑥 = +0.0001 × 565,32 × 3,5² × 50 = −346,26 𝐾𝑔. 𝑚

▪ 𝑀𝑙𝑦 = +0.001𝑞𝑙𝑥²𝑥 = +0.0001 × 565,32 × 3,5² × 38 = 263,16 𝐾𝑔. 𝑚

▪ 𝑀𝑡𝑦 = −0.001𝑞𝑙𝑥²𝑥 = −0.0001 × 565,32 × 3,5² × 38 = −263,16𝐾𝑔. 𝑚

2.1.8 Penulangan Plat Atap A1

𝑓𝑦 = 1,4

240= 0.0058 𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 =0.85𝑓′𝑐

𝑓𝑦 𝛽1 600

600 + 𝑓𝑦=0.85 × 30

240 0.85 600

600 + 240= 0,065 𝜌 𝑚𝑎𝑥 = 0.75 × 𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 = 0,75 × 0,065 = 0,048

● Penulangan Rencana : Tul. Arah X & Y = ᴓ8 mm 𝑑𝑥 = ℎ − 𝑠 −1

2∅𝑥 = 100 − 40 −1

2× 8 = 58 𝑚𝑚 𝑑𝑦 = ℎ − 𝑠 − ∅𝑥 −1

2∅𝑦 = 100 − 40 − 80 −1

2× 8 = 48 𝑚𝑚

𝑚 = ^𝑓𝑦

0.85×𝑓𝑐 = ²⁴⁰

0.85×30= 9,4 𝑅𝑛𝑥 = ^𝑀𝑢𝑥

0.8×𝑏×𝑑²= ^3462595,3

0.8×1000×56²= 1,38 𝜌 𝑝𝑒𝑟𝑙𝑢 = ¹

𝑚[√1 −^{(2×𝑚×𝑅𝑛𝑥)}

𝑓𝑦 ] = ¹

9,4[√1 −(2×9,4×1,38)

240 ] = 0.0059

𝐴𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌 𝑝𝑎𝑘𝑎𝑖 × 𝑏 × 𝑑 = 0.0059 × 1000 × 56 = 331,26 𝑚𝑚²

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8 𝛴𝑡𝑢𝑙𝑎𝑛𝑔𝑎𝑛 = ^𝐴𝑠

0.25×𝜋×𝑑²= ^331,26

0.25×3.14×8²= 6,6, 𝑑𝑖𝑏𝑢𝑙𝑎𝑡𝑘𝑎𝑛 𝑗𝑎𝑑𝑖 7 𝑏𝑎𝑡𝑎𝑛𝑔 𝑠 =^{1000×𝐴𝑑}

𝐴𝑠−𝑥 =^1000×50,24

331,26 = 151,7 𝑚𝑚 − − − 150 𝑚𝑚

𝑠 𝑝𝑎𝑘𝑎𝑖 =1000 × 50,24

150 = 334,9 𝑚𝑚² > 𝐴𝑠 𝑝𝑒𝑟𝑙𝑢 = 331,26 𝑚𝑚²

0,85×𝑓′𝑐×𝑏 =0,85×30×1000^334,9×240 = 3,15𝑚𝑚 𝑀𝑛 = (𝐴𝑠 × 𝑓𝑦) × (𝑑 −^𝑎

2) = (334,9 × 240) × (56 −^3,15

2 ) = 4374806,2 𝑁𝑚𝑚²

∅𝑀𝑛 = 0,9 × 4374806,2 = 3937325,6 𝑁𝑚𝑚² > 𝑀𝑢 = 3462595,3 𝑁𝑚𝑚²

𝑚 = ^𝑓𝑦

0.85×𝑓𝑐 = ²⁴⁰

0.85×30= 9,4 𝑅𝑛𝑥 = ^𝑀𝑢𝑥

0.8×𝑏×𝑑²= ^2631572,42

0.8×1000×48²= 1,43 𝜌 𝑝𝑒𝑟𝑙𝑢 = ¹

𝑚[√1 −^{(2×𝑚×𝑅𝑛𝑥)}

𝑓𝑦 ] = ¹

9,4[√1 −(2×9,4×1,43)

240 ] = 0.0061

𝐴𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌 𝑝𝑎𝑘𝑎𝑖 × 𝑏 × 𝑑 = 0.0061 × 1000 × 48 = 294,02 𝑚𝑚² 𝛴𝑡𝑢𝑙𝑎𝑛𝑔𝑎𝑛 =_{0.25×𝜋×𝑑²}^𝐴𝑠 =0.25×3.14×8²^294,02 = 5,6, 𝑑𝑖𝑏𝑢𝑙𝑎𝑡𝑘𝑎𝑛 𝑗𝑎𝑑𝑖 6 𝑏𝑎𝑡𝑎𝑛𝑔 𝑠 =^{1000×𝐴𝑑}

𝐴𝑠−𝑥 =^1000×50,24

294,02 = 170,87 𝑚𝑚 − − − 150 𝑚𝑚

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9 𝐴𝑠 𝑝𝑎𝑘𝑎𝑖 =1000 × 𝐴𝑑

𝑠 𝑝𝑎𝑘𝑎𝑖 =1000 × 50,24

150 = 334,93 𝑚𝑚² > 𝐴𝑠 𝑝𝑒𝑟𝑙𝑢 = 294,02 𝑚𝑚²

0,85×𝑓′𝑐×𝑏 = 334,93,5×240

0,85×30×1000= 3,15 𝑚𝑚 𝑀𝑛 = (𝐴𝑠 × 𝑓𝑦) × (𝑑 −^𝑎

2) = (334,93 × 240) × (48 −^3,15

2 ) = 3731734,21 𝑁𝑚𝑚²

∅𝑀𝑛 = 0,9 × 3731734,21 = 3358560,79 𝑁𝑚𝑚² > 𝑀𝑢 = 2631572,42 𝑁𝑚𝑚²

2.2 Data Plat Lantai & Atap A2

● Fc : 30 Mpa

● Fy : 240 Mpa

● Ly : 5000 mm

● Lx : 3000 mm

● Penentuan jenis plat 𝛽 =𝑙𝑛𝑦

𝑙𝑛𝑥=5000

3000= 1.67 < 2.5, 𝑚𝑎𝑘𝑎 𝑚𝑒𝑛𝑢𝑛𝑗𝑢𝑘𝑘𝑎𝑛 𝑏𝑎ℎ𝑤𝑎 𝑝𝑙𝑎𝑡 2 𝑎𝑟𝑎ℎ

2.2.1 Preliminary Tebal Plat Lantai A2

~ ℎ𝑚𝑎𝑥 = ^{𝑙𝑛×[0.8+}

𝑓𝑦 1500]

36+5×𝛽[𝛼𝑚−0.2] = ^5000×[0.8+

240 1500]

36+5×1,67[0.6−0.2]= 122,03 𝑚𝑚

~ℎ 𝑚𝑖𝑛 =𝑙𝑛 × [0.8 + ^𝑓𝑦

100]

36 + 9 × 𝛽 =5000 × [0.8 + ²⁴⁰

1500]

36 + 9 × 1,67 = 94,11 𝑚𝑚

2.2.2 Pembebanan Plat Lantai A2

Beban Hidup : Ruang Kantor = 2,4 Kn/m² = 244,73 Kg/m²

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Beban Mati : Beban Sendiri = 0.23 Kn/m² = 234,53 Kg/m² Beban Keramik & Spesi = 78,5169 Kg/m²

Beban Plafond = 48,9456 Kg/m² + Total Beban Mati = 361,99 Kg/m² Beban Ultimate (qu)

𝑞𝑢 = (1.2 × 𝐷𝐿) + (1.6 × 𝐿𝐿) = (1.2 × 361,09) + (1.6 × 244.73) = 825.95 𝐾𝑔/𝑚²

Jadi beban ultímate (qu) yang diterima oleh plat lantai A1 sebesar 825,95 Kg/m²

2.2.3 Momen Plat Lantai A2

▪ MLx (+) = MTx (-) = 58

▪ MLy (+) = 36

▪ MTy (-) = 36

▪ 𝑀𝑙𝑥 = +0.001𝑞𝑙𝑥²𝑥 = +0.0001 × 825,95 × 3² × 58 = −460,53 𝐾𝑔. 𝑚

▪ 𝑀𝑙𝑦 = +0.001𝑞𝑙𝑥²𝑥 = +0.0001 × 825,95 × 3² × 36 = 285,85 𝐾𝑔. 𝑚

▪ 𝑀𝑡𝑦 = −0.001𝑞𝑙𝑥²𝑥 = −0.0001 × 825,95 × 3² × 36 = −285,85 𝐾𝑔. 𝑚

2.2.4 Penulangan Plat Lantai A2

𝑓𝑦 = 1,4

240= 0.0058 𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 =0.85𝑓′𝑐

𝑓𝑦 𝛽1 600

600 + 𝑓𝑦=0.85 × 30

240 0.85 600

600 + 240= 0,065 𝜌 𝑚𝑎𝑥 = 0.75 × 𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 = 0,75 × 0,065 = 0,048

● Penulangan Rencana : Tul. Arah X =ᴓ22 mm & Y = ᴓ10 mm 𝑑𝑥 = ℎ − 𝑠 −1

2∅𝑥 = 100 − 25 −1

2× 22 = 64 𝑚𝑚 𝑑𝑦 = ℎ − 𝑠 − ∅𝑥 −1

2∅𝑦 = 100 − 25 − 22 −1

2× 10 = 48 𝑚𝑚

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𝑚 = ^𝑓𝑦

0.85×𝑓𝑐 = ²⁴⁰

0.85×30= 9,4 𝑅𝑛𝑥 = ^𝑀𝑢𝑥

0.8×𝑏×𝑑²= ^4605315,98

0.8×1000×64²= 1,41 𝜌 𝑝𝑒𝑟𝑙𝑢 = ¹

𝑚[√1 −^{(2×𝑚×𝑅𝑛𝑥)}

𝑓𝑦 ] = ¹

9,4[√1 −(2×9,4×1,41)

240 ] = 0.0060

𝐴𝑠−𝑥 =1000×379,94

385,72 = 985,01 𝑚𝑚 − − − 900 𝑚𝑚

𝑠 𝑝𝑎𝑘𝑎𝑖 =1000 × 379,94

200 = 1899,7 𝑚𝑚² > 𝐴𝑠 𝑝𝑒𝑟𝑙𝑢 = 333,33 𝑚𝑚²

0,85×𝑓′𝑐×𝑏 = ^1899,7×240

0,85×30×1000= 17,9 𝑚𝑚 𝑀𝑛 = (𝐴𝑠 × 𝑓𝑦) × (𝑑 −^𝑎

2) = (1899,7 × 240) × (64 −^17,9

2 ) = 25103502,9 𝑁𝑚𝑚²

∅𝑀𝑛 = 0,9 × 25103502,9 = 22593152,7 𝑁𝑚𝑚² > 𝑀𝑢 = 4605315,98 𝑁𝑚𝑚²

● Penulangan Plat Arah y (untuk daerah Tumpuan & Lapangan) MTy = MLy = 2588471,99 Nmm2

𝑚 = ^𝑓𝑦

0.85×𝑓𝑐 = ²⁴⁰

0.85×30= 9,4 𝑅𝑛𝑥 = ^𝑀𝑢𝑥

0.8×𝑏×𝑑²= ^2588471,99

0.8×1000×48²= 1,55 𝜌 𝑝𝑒𝑟𝑙𝑢 = ¹

𝑚[√1 −^{(2×𝑚×𝑅𝑛𝑥)}

𝑓𝑦 ] = ¹

9,4[√1 −(2×9,4×1,55)

240 ] = 0.0066

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𝐴𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌 𝑝𝑎𝑘𝑎𝑖 × 𝑏 × 𝑑 = 0.0066 × 1000 × 48 = 320,2 𝑚𝑚² 𝛴𝑡𝑢𝑙𝑎𝑛𝑔𝑎𝑛 =_{0.25×𝜋×𝑑²}^𝐴𝑠 =0.25×3.14×10²^320,2 = 4,08, 𝑑𝑖𝑏𝑢𝑙𝑎𝑡𝑘𝑎𝑛 𝑗𝑎𝑑𝑖 5𝑏𝑎𝑡𝑎𝑛𝑔 𝑠 =^{1000×𝐴𝑑}

𝐴𝑠−𝑥 =^1000×78,5

320,2 = 245,14 𝑚𝑚 − − − 200 𝑚𝑚

𝑠 𝑝𝑎𝑘𝑎𝑖 =1000 × 78,5

200 = 392,5 𝑚𝑚² > 𝐴𝑠 𝑝𝑒𝑟𝑙𝑢 = 320,2 𝑚𝑚²

0,85×𝑓′𝑐×𝑏 = ^392,5×240

0,85×30×1000= 3,69 𝑚𝑚 𝑀𝑛 = (𝐴𝑠 × 𝑓𝑦) × (𝑑 −^𝑎

2) = (392,5 × 240) × (48 −^3,69

2 ) = 4347607,1 𝑁𝑚𝑚²

∅𝑀𝑛 = 0,9 × 4347607,1 = 3912846,35 𝑁𝑚𝑚² > 𝑀𝑢 = 2858471,99 𝑁𝑚𝑚²

2.2.5 Preliminary Tebal Plat Atap A2

~ ℎ𝑚𝑎𝑥 = ^{𝑙𝑛×[0.8+}

𝑓𝑦 1500]

36+5×𝛽[𝛼𝑚−0.2] = ^5000×[0.8+

240 1500]

36+5×1.67[0.6−0.2]= 122,03 𝑚𝑚

~ℎ 𝑚𝑖𝑛 =^{𝑙𝑛×[0.8+}

𝑓𝑦 100]

36+9×𝛽 =^5000×[0.8+

240 1500]

36+9×1.67 = 94,11 𝑚𝑚

2.2.6 Pembebanan Plat Atap A2

Beban Hidup : atap datar, berbubung, dan lengkung= 0.96 Kn/m² =97,89 Kg/m² Beban Mati : Beban Sendiri = 0,23 Kn/m² = 234,53 Kg/m²

Beban Waterproofing = 0,1 Kn/m² = 10,197 Kg/m² Beban Plafond = 0,48 Kn/m² = 48,9456 Kg/m² +

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Total Beban Mati = 293,67 Kg/m²

Beban Ultimate (qu)

𝑞𝑢 = (1.2 × 𝐷𝐿) + (1.6 × 𝐿𝐿) = (1.2 × 293,67) + (1.6 × 97,89) = 565,32 𝐾𝑔/𝑚²

Jadi beban ultímate (qu) yang diterima oleh plat lantai A2 sebesar 565,32 Kg/m²

2.2.7 Momen Plat Atap A2

▪ MLx (+) = MTx (-) = 58

▪ MLy (+) = 36

▪ MTy (-) = 36

▪ 𝑀𝑙𝑥 = +0.001𝑞𝑙𝑥²𝑥 = +0.0001 × 565,32 × 3² × 58 = −295,10 𝐾𝑔. 𝑚

▪ 𝑀𝑙𝑦 = +0.001𝑞𝑙𝑥²𝑥 = +0.0001 × 565,32 × 3² × 36 = 183,16 𝐾𝑔. 𝑚

▪ 𝑀𝑡𝑦 = −0.001𝑞𝑙𝑥²𝑥 = −0.0001 × 565,32 × 3² × 36 = −183,16𝐾𝑔. 𝑚

2.2.8 Penulangan Plat Atap A2

𝑓𝑦 = 1,4

240= 0.0058 𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 =0.85𝑓′𝑐

𝑓𝑦 𝛽1 600

600 + 𝑓𝑦=0.85 × 30

240 0.85 600

600 + 240= 0,065 𝜌 𝑚𝑎𝑥 = 0.75 × 𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 = 0,75 × 0,065 = 0,048

● Penulangan Rencana : Tul. Arah X = ᴓ16 mm & Y = ᴓ8 mm 𝑑𝑥 = ℎ − 𝑠 −1

2∅𝑥 = 100 − 40 −1

2× 16 = 52 𝑚𝑚 𝑑𝑦 = ℎ − 𝑠 − ∅𝑥 −1

2∅𝑦 = 100 − 40 − 16 −1

2× 8 = 40 𝑚𝑚

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𝑚 = ^𝑓𝑦

0.85×𝑓𝑐 = ²⁴⁰

0.85×30= 9,4 𝑅𝑛𝑥 = ^𝑀𝑢𝑥

0.8×𝑏×𝑑²= ^2950979,17

0.8×1000×52²= 1,36 𝜌 𝑝𝑒𝑟𝑙𝑢 = ¹

𝑚[√1 −^{(2×𝑚×𝑅𝑛𝑥)}

𝑓𝑦 ] = ¹

9,4[√1 −(2×9,4×1,36)

240 ] = 0.00584

𝐴𝑠−𝑥 =1000×200,96

303,93 = 661,2 𝑚𝑚 − − − 600 𝑚𝑚

𝑠 𝑝𝑎𝑘𝑎𝑖 =1000 × 200,96

200 = 1004,8 𝑚𝑚² > 𝐴𝑠 𝑝𝑒𝑟𝑙𝑢 = 303,93 𝑚𝑚²

0,85×𝑓′𝑐×𝑏 =0,85×30×1000^1004,8×240 = 9,46 𝑚𝑚 𝑀𝑛 = (𝐴𝑠 × 𝑓𝑦) × (𝑑 −^𝑎

2) = (1004,8 × 240) × (52 −^9,46

2 ) = 11399623,9 𝑁𝑚𝑚²

∅𝑀𝑛 = 0,9 × 11399623,9 = 10259661,5 𝑁𝑚𝑚² > 𝑀𝑢 = 2950979,2 𝑁𝑚𝑚²

𝑚 = ^𝑓𝑦

0.85×𝑓𝑐 = ²⁴⁰

0.85×30= 9,4 𝑅𝑛𝑥 = ^𝑀𝑢𝑥

0.8×𝑏×𝑑²= 1831642,243

0.8×1000×40²= 1,43 𝜌 𝑝𝑒𝑟𝑙𝑢 = ¹

𝑚[√1 −^{(2×𝑚×𝑅𝑛𝑥)}

𝑓𝑦 ] = ¹

9,4[√1 −(2×9,4×1,43)

240 ] = 0.0061

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𝐴𝑠−𝑥 =^1000×50,24

245,59 = 204,57 𝑚𝑚 − − − 200 𝑚𝑚

𝑠 𝑝𝑎𝑘𝑎𝑖 =1000 × 50,24

200 = 251,2 𝑚𝑚² > 𝐴𝑠 𝑝𝑒𝑟𝑙𝑢 = 245,59 𝑚𝑚²

0,85×𝑓′𝑐×𝑏 = ^251,2×240

0,85×30×1000= 2,36 𝑚𝑚 𝑀𝑛 = (𝐴𝑠 × 𝑓𝑦) × (𝑑 −^𝑎

2) = (251,2 × 240) × (40 −^2,36

2 ) = 2340252,49 𝑁𝑚𝑚²

∅𝑀𝑛 = 0,9 × 2340252,49 = 2106227,24 𝑁𝑚𝑚² > 𝑀𝑢 = 1831642,24 𝑁𝑚𝑚²

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18 BAB 3

PERENCANAAN BALOK ANAK

3.1 Data Balok – B1 Atap

● Lx : 3 m

● Ly : 5 m

● Tebal plat : 10cm

● Mutu Beton (Fc) : 30 Mpa

● Mutu Baja (Fy) : 240 Mpa

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3.1.1 Pembebanan Balok – B1 Atap

● qD Plat = 293,67 Kg/m² (sesuai pembebanan plat SNI 2847-2020)

● qL Plat = 97,8912 Kg/m² (sesuai pembebanan plat SNI 2847-2020)

● Preliminary h balok

h = L/21 = 500/21 = 23,809 cm ---- 22 cm (SNI 2847-2019) b = 2/3*h = 2/3*22 = 14,67 cm --- 15 cm

Jadi pakai dimensi balok 15x22 cm

● Beban Mati

Beban sendri balok = 0,23 Kn/m3 x (bxh)

= (0,23 x 101,97) x (0,15 x 0,22) = 0,77 Kg/m² Beban mati plat (ekuivalen plat atap A2)

qD ekuivalen plat = ¹

2× 𝑞 × (1 − (

1 2𝑙𝑥² 1 3𝑙𝑦²))

1

2× 293,674 × (1 − (

1 23²

1

35²)) = 387,65 𝐾𝑔/𝑚²

Total beban mati QD = 0,77 + 387,65 = 388,42 Kg/m² Beban hidup plat (ekuivalen plat atap A2)

qL ekuivalen plat = ¹

2× 𝑞 × (1 − (

1 2𝑙𝑥² 1 3𝑙𝑦²))

1

2× 97,89 × (1 − (

1 23²

1

35²)) = 129,22 𝐾𝑔/𝑚²

Beban Ultimate : Qu = 1,2QD + 1,6qL ekuivalen plat

= 1,2 (388,42) + 1,6 (129,22)

= 672,85 Kg/m²

Jadi beban ultímate (qu) yang diterima oleh balok B1 Atap sebesar 672,85 Kg/ m²

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3.1.2 Momen Balok – B1 Atap

▪ 𝑀𝑡 = ¹

10× 𝑊𝑢 × 𝑙𝑛²= ¹

10× 672,85 × 5²= 1682,134 ^𝐾𝑔

𝑚² = 16821348 𝑁𝑚𝑚²

▪ 𝑀𝑙 = ¹

16× 𝑊𝑢 × 𝑙𝑛²= ¹

16× 672,85 × 5²= 1051,334 ^𝐾𝑔

𝑚² = 10513342 𝑁𝑚𝑚²

3.1.3 Penulangan Balok – B1 Atap

● Tinggi balok : 220 mm

● Tebal penutup beton : 40 mm

● Tulangan Pokok : 13 mm

● Tulangan Sengkang : 10 mm Sehingga didapatkan tinggi efektif ;

𝑡𝑖𝑛𝑔𝑔𝑖 𝑒𝑓𝑒𝑘𝑡𝑖𝑓 = 𝑡𝑖𝑛𝑔𝑔𝑖 𝑏𝑎𝑙𝑜𝑘 − 𝑡𝑒𝑏𝑎𝑙 𝑝𝑒𝑛𝑢𝑡𝑢𝑝 − ∅𝑡𝑢𝑙. 𝑠𝑒𝑛𝑔𝑘𝑎𝑛𝑔 −1

2𝑡𝑢𝑙. 𝑝𝑜𝑘𝑜𝑘

= 220 − 40 − 10 −1

213 = 163,5 𝑚𝑚

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3.1.4 Penulangan Tumpuan

● Untuk f’c < 30 Mpa, maka digunakan β1 = 0.85 𝑚 = ^𝑓𝑦

0.85×𝑓𝑐 = ²⁴⁰

0.85×30= 9,4 𝑅𝑛 = ^{𝑀𝑢 𝑡}

0.8×𝑏×𝑑²= ^16821348

0.8×150×163,5²= 5,24

● Rasio tulangan yang dibutuhkan 𝜌 𝑚𝑖𝑛 = 1,4

𝑓𝑦 = 1,4

240= 0.0058 𝜌 𝑝𝑒𝑟𝑙𝑢 = 1

𝑚(1 − √1 −2𝑚 ∙ 𝑅𝑛 𝑓𝑦 )

= 1

9,4(1 − √1 −2(9,4) ∙ 5,24 240 )

= 0,024

Dipakai : ρ perlu = 0.024>ρ min = 0,0058 (OK) 𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 =0.85 ∙ 𝑓^′𝑐

𝑓𝑦 × 𝛽1 × 600 600 + 𝑓𝑦 𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 =0.85 ∙ 30

𝑓𝑦240 × 0,85 × 600 600 + 240 𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 = 0,06451

𝜌 𝑚𝑎𝑥 = 0.75 × 𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 = 0,75 × 0,065 = 0,048

● Luas tulangan yang diperlukan 𝐴𝑠 = 𝜌 𝑝𝑎𝑘𝑎𝑖 × 𝑏 × 𝑑 𝐴𝑠 = 0,024 × 150 × 163,5 𝐴𝑠 = 606,4 𝑚𝑚²

𝑛 = 𝐴𝑠

0.25× 𝜋 × ∅ 𝑡𝑢𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑜𝑘𝑜𝑘² 𝑛 = 606,4

0.25 × 𝜋 × 13²

𝑛 = 4,57 dibulatkan menjadi 5

𝐴𝑠 𝑝𝑎𝑘𝑎𝑖 = 𝑛 × 0,25 × 𝜋 × ∅ 𝑡𝑢𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑜𝑘𝑜𝑘² 𝐴𝑠 𝑝𝑎𝑘𝑎𝑖 = 5 × 0,25 × 𝜋 × 13²

𝐴𝑠 𝑝𝑎𝑘𝑎𝑖 = 663,32 𝑚𝑚² dipakai tulangan 5D13

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𝐴𝑠 𝑝𝑎𝑘𝑎𝑖 > 𝐴𝑠 = 663,32 𝑚𝑚² > 606,4 𝑚𝑚² (𝑂𝐾)

● Cek momen nomina 𝑎 =𝐴𝑠 𝑝𝑎𝑘𝑎𝑖 × 𝑓𝑦

0,85 × 𝑓′𝑐 × 𝑏 = 663,32 × 240

0,85 × 30 × 150 = 41,6 𝑀𝑛 = (𝐴𝑠 × 𝑓𝑦) × (𝑑 −𝑎

2) = (663,32 × 240) × (163,5 −41,6 2 ) 𝑀𝑛 = 22715931,4 𝑁𝑚𝑚²

∅𝑀𝑛 = 0,9 × 22715931,4 = 20444338,2 𝑁𝑚𝑚²

∅𝑀𝑛 > 𝑀𝑡1 = 20444338,2 𝑁𝑚𝑚² > 16821348 𝑁𝑚𝑚² (𝑂𝐾)

3.1.5 Penulangan Lapangan

● Untuk f’c < 30 Mpa, maka digunakan β1 = 0.85

𝑚 = ^𝑓𝑦

0.85×𝑓𝑐= ²⁴⁰

0.85×30= 9,4

𝑅𝑛 = ^{𝑀𝑢 𝑙}

0.8×𝑏×𝑑²= ^10513342

0.8×150×163,5²= 3,27

𝑓𝑦 = 1,4

240= 0.0058 𝜌 𝑝𝑒𝑟𝑙𝑢 = 1

𝑚(1 − √1 −2𝑚 ∙ 𝑅𝑛 𝑓𝑦 )

𝜌 𝑝𝑒𝑟𝑙𝑢 = 1

9,4(1 − √1 −2(9,4).3,27

240 ) = 0.014

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23

𝑓𝑦 × 𝛽1 × 600 600 + 𝑓𝑦 𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 =0.85 ∙ 30

𝑓𝑦240 × 0,85 × 600

600 + 240= 0,065 𝜌 𝑚𝑎𝑥 = 0.75 × 𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 = 0,75 × 0,065 = 0,048

𝑛 = 𝐴𝑠

0.25× 𝜋 × ∅ 𝑡𝑢𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑜𝑘𝑜𝑘² 𝑛 =359,73

0.25 × 𝜋 × 13²

𝐴𝑠 𝑝𝑎𝑘𝑎𝑖 = 397,99 𝑚𝑚²dipakai tulangan 3D13 𝐴𝑠 𝑝𝑎𝑘𝑎𝑖 > 𝐴𝑠 =397,99 𝑚𝑚² > 359,73 𝑚𝑚² (𝑂𝐾)

0,85 × 𝑓′𝑐 × 𝑏 = 397,99 × 240

0,85 × 30 × 150 = 24,97 𝑀𝑛 = (𝐴𝑠 × 𝑓𝑦) × (𝑑 −𝑎

2) = (397,99 × 240) × (163,5 −24,97 2 ) 𝑀𝑛 = 14424664,8 𝑁𝑚𝑚²

∅𝑀𝑛 = 0,9 × 14424664,8 = 12982198 𝑁𝑚𝑚²

∅𝑀𝑛 > 𝑀𝑙 = 12982198 𝑁𝑚𝑚² > 10513342 𝑁𝑚𝑚² (𝑂𝐾)

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24

3.1.6 Penulangan Geser Balok Anak B-1 Atap

 Vu = ½.wu.ln 𝑉𝑢 = ¹

2× 6728,5 × 5 = 16821,34 N = 16,82 kN

 Kondisi 1

𝑉𝑐 =1

6× √𝑓𝑐 × 𝑏𝑤 × 𝑑 =1

6× √30 × 150 × 163,5 = 22388,15 N = 22,38 kN 𝑉𝑢 ≤ 0,5∅𝑉𝑐 = 16,82 𝑘𝑁 ≤ 0,5 × 0,75 × 22,38 = 8,39 kN … . . (not ok)

 Kondisi 2

0,5∅𝑉𝑐 ≤ 𝑉𝑢 ≤ ∅𝑉𝑐

8,39 𝑘𝑁 ≤ 16,82 𝑘𝑁 ≤ 0,75 × 22,66 = 16,79 kN … … (not ok)

 Kondisi 3 𝑉𝑠 𝑚𝑖𝑛 =1

3× 𝑏𝑤 × 𝑑 =1

3× 150 × 163,5 = 8175 N = 8,2 kN

∅𝑉𝑐 ≤ 𝑉𝑢 ≤ ∅(𝑉𝑐 + 𝑉𝑠min)

16,79 𝑘𝑁 ≤ 16,82 𝑘𝑁 ≤ 0.75 × (22,38 + 8,2) = 22,92 𝑘𝑁 … … . (𝑜𝑘)

 Gambar kondisi gaya geser

 Hitung Jarak

𝑥1 =16,82

2,5 =8,39

𝑥 = 124,7 𝑐𝑚 𝑥2 = 250 𝑐𝑚

*digunakan sengkang 2 kaki Ø10mm, dengan Av = 157 mm2, gunakan s = 80 mm

 Hitung tulangan geser yang dibutuhkan pada tiap kondisi

 Kondisi 3

𝑉𝑠 =^{𝐴𝑣×𝐹𝑦×𝑑}_𝑠 =157×240×163,5

80 = 49,88 𝑘𝑁 ≥ 𝑉𝑠𝑚𝑖𝑛 = 8,2 𝑘𝑁 (𝑜𝑘) 𝐴𝑣 𝑚𝑖𝑛 =^{𝑏𝑤×𝑠}

3×𝑓𝑦=^150×80

3×240 = 16,66 𝑚𝑚2 ≤ 𝐴𝑣 = 100,48 𝑚𝑚2 (𝑜𝑘)

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Kontrol jarak

S = 80 mm < s max = d/2 = 163,5/2 = 81,75 mm (ok) S = 80 mm < s max = 600 mm (ok)

Maka s = 80 mm cukup memadai

 Kondisi 4

𝑉𝑠 𝑝𝑒𝑟𝑙𝑢 = (∅𝑉𝑐 − 0,5 × ∅𝑉𝑐)

∅ =(16,99 − 8,49)

0,75 = 0,04 𝑘𝑁 𝑠 = (𝐴𝑣 × 𝑓𝑦 × 𝑑)

𝑉𝑠 𝑝𝑒𝑟𝑙𝑢 =(157 × 240 × 163,5)

40 = 153,31 𝑚𝑚 === 150 𝑚𝑚

Kontrol jarak

S = 150 mm < s max = d/2 = 163,5/2 = 81,75 mm (not ok) S = 150 mm < s max = 600 mm (ok)

Karena jika menggunanakan s = 150 mm tidak memenuhi persyaratan s

= 150 mm < s max = d/2 = 81,75 mm, maka digunakan s max = d/2 = 81,75 mm atau lebih kecil dari s pada kondisi 3, yaitu s = 60 mm

 Kondisi 5

𝑉𝑠 𝑝𝑒𝑟𝑙𝑢 = (∅𝑉𝑐 − 0,5 × ∅𝑉𝑐)

∅ =(16,99 − 8,49)

0,75 = 0,04 𝑘𝑁 𝑠 = (𝐴𝑣 × 𝑓𝑦 × 𝑑)

𝑉𝑠 𝑝𝑒𝑟𝑙𝑢 =(157 × 240 × 163,5)

0,04 = 153,31 𝑚𝑚 === 150 𝑚𝑚

Kontrol jarak

Karen jika menggunanakan s = 150 mm tidak memenuhi persyaratan s

= 150 mm < s max = d/4 = 40,87 mm, maka digunakan s max = d/4 = 40,87 mm atau dibulatkan jadi s = 40 mm

 Gambar penulangan geser balok anak B-1 atap

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26 3.2 Data Balok – B2 Atap

● Lx : 3 m

● Ly : 5 m

3.2.1 Pembebanan Balok – B2 Atap

● qD Plat = 293,67 Kg/m² (sesuai pembebanan plat SNI 2847-2020)

● qL Plat = 97,8912 Kg/m² (sesuai pembebanan plat SNI 2847-2020)

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● Preliminary h balok

h = L/21 = 500/21 = 17,85 cm ---- 22 cm (SNI 2847-2019) b = 2/3*h = 2/3*22 = 14,67 cm --- 20 cm

Jadi pakai dimensi balok 20x22 cm

● Beban Mati

Beban sendri balok = 0,23 Kn/m3 x (bxh)

= (0,23 x 101,97) x (0,2 x 0,22) = 1,03Kg/m² Beban mati plat (ekuivalen plat atap A2)

qD ekuivalen plat = ¹

2× 𝑞 × (1 − (

1 2𝑙𝑥² 1 2𝑙𝑦²)) 1

2× 293,674 × (1 − (

1 23²

1

25²)) = 361,22 𝐾𝑔/𝑚² Total beban mati QD = 1,03 + 2x361,22 = 723,47Kg/m² Beban hidup plat (ekuivalen plat atap A2)

qL ekuivalen plat = ¹

2× 𝑞 × (1 − (

1 2𝑙𝑥² 1 2𝑙𝑦²)) 1

2× 97,89 × (1 − (

1 23²

1

25²)) = 120,41 𝐾𝑔/𝑚² Beban Ultimate : Qu = 1,2QD + 1,6qL ekuivalen plat = 1,2 (723,47) + 1,6 (120,41) = 1060,81 Kg/m²

Jadi beban ultímate (qu) yang diterima oleh balok B2 Atap sebesar 1060,81 Kg/ m²

3.2.2 Momen Balok – B2 Atap

▪ 𝑀𝑡 = ¹

10× 𝑊𝑢 × 𝑙𝑛²= ¹

10× 1060,81 × 5²= 1682,134 ^𝐾𝑔

𝑚² = 26520316 𝑁𝑚𝑚²

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▪ 𝑀𝑙 = ¹

16× 𝑊𝑢 × 𝑙𝑛²= ¹

16× 1060,81 × 5²= 1051,334 ^𝐾𝑔

𝑚² = 16575198 𝑁𝑚𝑚²

3.2.3 Penulangan Balok – B2 Atap

● Tinggi balok : 220 mm

● Tebal penutup beton : 40 mm

● Tulangan Pokok : 13 mm

● Tulangan Sengkang : 10 mm Sehingga didapatkan tinggi efektif ;

𝑡𝑖𝑛𝑔𝑔𝑖 𝑒𝑓𝑒𝑘𝑡𝑖𝑓 = 𝑡𝑖𝑛𝑔𝑔𝑖 𝑏𝑎𝑙𝑜𝑘 − 𝑡𝑒𝑏𝑎𝑙 𝑝𝑒𝑛𝑢𝑡𝑢𝑝 − ∅𝑡𝑢𝑙. 𝑠𝑒𝑛𝑔𝑘𝑎𝑛𝑔 −1

2𝑡𝑢𝑙. 𝑝𝑜𝑘𝑜𝑘

= 220 − 40 − 10 −1

213 = 163,5 𝑚

3.2.4 Penulangan Tumpuan

0.85×𝑓𝑐 = ²⁴⁰

0.85×30= 9,4

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𝑅𝑛 = ^{𝑀𝑢 𝑡}

0.8×𝑏×𝑑²= ^26520316

0.8×200×165,5²= 6,2

𝑓𝑦 = 1,4

240= 0.0058 𝜌 𝑝𝑒𝑟𝑙𝑢 = 1

𝑚(1 − √1 −2𝑚 ∙ 𝑅𝑛 𝑓𝑦 )

9,4(1 − √1 −2(9,4) ∙ 6,2 240 ) 𝜌 𝑝𝑒𝑟𝑙𝑢 = 0,03

𝑓𝑦 × 𝛽1 × 600 600 + 𝑓𝑦 𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 =0.85 ∙ 30

𝑓𝑦240 × 0,85 × 600 600 + 240 𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 = 0,06451

𝜌 𝑚𝑎𝑥 = 0.75 × 𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 = 0,75 × 0,065 = 0,048

𝑛 = 𝐴𝑠

0.25× 𝜋 × ∅ 𝑡𝑢𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑜𝑘𝑜𝑘² 𝑛 = 984,2

0.25 × 𝜋 × 13²

𝐴𝑠 𝑝𝑎𝑘𝑎𝑖 = 1061,32 𝑚𝑚² dipakai tulangan 8D13 𝐴𝑠 𝑝𝑎𝑘𝑎𝑖 > 𝐴𝑠 = 1061,32 𝑚𝑚² > 984,2 𝑚𝑚² (𝑂𝐾)

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30

0,85 × 𝑓′𝑐 × 𝑏 = 1061,32 × 240

0,85 × 30 × 200 = 49,9 𝑀𝑛 = (𝐴𝑠 × 𝑓𝑦) × (𝑑 −𝑎

2) = (1061,23 × 240) × (163,5 −49,9 2 ) 𝑀𝑛 = 35285348 𝑁𝑚𝑚²

∅𝑀𝑛 = 0,9 × 35285348 = 31756814 𝑁𝑚𝑚²

∅𝑀𝑛 > 𝑀𝑡1 = 31756814 𝑁𝑚𝑚² > 26520316 𝑁𝑚𝑚² (𝑂𝐾)

3.2.5 Penulangan Lapangan

0.85×𝑓𝑐 = ²⁴⁰

0.85×30= 9,4 𝑅𝑛 = ^{𝑀𝑢 𝑙}

0.8×𝑏×𝑑²= ^16575198

0.8×200×163,5²= 3,87

𝑓𝑦 = 1,4

240= 0.0058 𝜌 𝑝𝑒𝑟𝑙𝑢 = 1

𝑚(1 − √1 −2𝑚 ∙ 𝑅𝑛 𝑓𝑦 )

9,4(1 − √1 −2(9,4).3,87

240 ) = 0.017 Dipakai : ρ perlu = 0.017>ρ min = 0,0058 (OK)

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𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 =0.85 ∙ 𝑓^′𝑐

𝑓𝑦 × 𝛽1 × 600 600 + 𝑓𝑦 𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 =0.85 ∙ 30

𝑓𝑦240 × 0,85 × 600

600 + 240= 0,065 𝜌 𝑚𝑎𝑥 = 0.75 × 𝜌 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 = 0,75 × 0,065 = 0,048

𝑛 = 𝐴𝑠

0.25× 𝜋 × ∅ 𝑡𝑢𝑙𝑎𝑛𝑔𝑎𝑛 𝑝𝑜𝑘𝑜𝑘² 𝑛 =575,7

0.25 × 𝜋 × 13²

𝐴𝑠 𝑝𝑎𝑘𝑎𝑖 = 663,32 𝑚𝑚² dipakai tulangan 6D13 𝐴𝑠 𝑝𝑎𝑘𝑎𝑖 > 𝐴𝑠 =663,32 𝑚𝑚² > 575,7 𝑚𝑚² (𝑂𝐾)

0,85 × 𝑓′𝑐 × 𝑏 = 663,32 × 240

0,85 × 30 × 200 = 31,21 𝑀𝑛 = (𝐴𝑠 × 𝑓𝑦) × (𝑑 −𝑎

2) = (663,32 × 240) × (163,5 −31,21 2 ) 𝑀𝑛 = 23544166,8 𝑁𝑚𝑚²

∅𝑀𝑛 = 0,9 × 23544166,8 = 21189750,1 𝑁𝑚𝑚²

∅𝑀𝑛 > 𝑀𝑙 = 21189750,1 𝑁𝑚𝑚² > 16575198 𝑁𝑚𝑚² (𝑂𝐾)

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32

3.2.6 Penulangan Geser Balok Anak B-2 Atap

 Vu = ½.wu.ln 𝑉𝑢 = ¹

2× 10608,12 × 5 = 26520,31 N = 26,52 kN

 Kondisi 1

𝑉𝑐 =1

6× √𝑓𝑐 × 𝑏𝑤 × 𝑑 =1

6× √30 × 200 × 163,5 = 22388,15 N = 22,38 kN 𝑉𝑢 ≤ 0,5∅𝑉𝑐 = 26,52 𝑘𝑁 ≤ 0,5 × 0,75 × 22,38 = 8,39 kN … . . (not ok)

 Kondisi 2

0,5∅𝑉𝑐 ≤ 𝑉𝑢 ≤ ∅𝑉𝑐

8,39 𝑘𝑁 ≤ 26,52 𝑘𝑁 ≤ 0,75 × 22,38 = 16,79 kN … … (not ok)

 Kondisi 3 𝑉𝑠 𝑚𝑖𝑛 =1

3× 𝑏𝑤 × 𝑑 =1

3× 200 × 163,5 = 8175 N = 8,2 kN

∅𝑉𝑐 ≤ 𝑉𝑢 ≤ ∅(𝑉𝑐 + 𝑉𝑠min)

16,79 𝑘𝑁 ≤ 26,52 𝑘𝑁 ≤ 0.75 × (22,38 + 8,2) = 22,92 𝑘𝑁 … … . (𝑛𝑜𝑡 𝑜𝑘)

 Kondisi 4

∅(𝑉𝑐 + 𝑉𝑠min) ≤ Vu ≤ ∅(Vc +1

3× √fc × bw × d)

22,92 𝑘𝑁 ≤ 26,52 𝑘𝑁 ≤ 0,75 × (22,38 +1

3× √30 × 200 × 163,5 = 33,59 𝑘𝑁 … . (𝑜𝑘)

 Gambar kondisi gaya geser

 Hitung Jarak 𝑥1 = 26,52

2,5 =8,39

𝑥 = 79,09𝑐𝑚 𝑥2 = 26,52

2,5 =16,79

𝑥 = 158,27 𝑐𝑚 𝑥3 = 26,52

2,5 =22,92

𝑥 = 216,06 𝑐𝑚 𝑥4 = 2,5 𝑚𝑒𝑡𝑒𝑟

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33

*digunakan sengkang 2 kaki Ø10mm, dengan Av = 157 mm2, gunakan s = 80 mm

 Hitung tulangan geser yang dibutuhkan pada tiap kondisi

 Kondisi 3

𝑉𝑠 =^{𝐴𝑣×𝐹𝑦×𝑑}

𝑠 =157×240×163,5

80 = 49,88𝑘𝑁 ≥ 𝑉𝑠𝑚𝑖𝑛 = 8,2 𝑘𝑁 (𝑜𝑘) 𝐴𝑣𝑚𝑖𝑛 =^{𝑏𝑤×𝑠}

3×𝑓𝑦=^150×80

3×240 = 16,66 𝑚𝑚2 ≤ 𝐴𝑣 = 100,48 𝑚𝑚2 (𝑜𝑘)

Kontrol jarak

S = 80 mm < s max = d/2 = 163,5/2 = 81,75 mm (ok) S = 80 mm < s max = 600 mm (ok)

Maka s = 80 mm cukup memadai

 Kondisi 4

𝑉𝑠 𝑝𝑒𝑟𝑙𝑢 = (𝑉𝑢 − ∅𝑉𝑐)

∅ =(26,52 − 16,79)

0,75 = 12,97 𝑘𝑁 𝑠 = (𝐴𝑣 × 𝑓𝑦 × 𝑑)

𝑉𝑠 𝑝𝑒𝑟𝑙𝑢 =(157 × 240 × 163,5)

12,97 = 476,36 𝑚𝑚 === 400 𝑚𝑚

Kontrol jarak

= 400 mm < s max = d/2 = 81,75 mm, maka digunakan s max = d/2 = 81,75 mm atau lebih kecil dari s pada kondisi 3, yaitu s = 60 mm

 Kondisi 5

𝑉𝑠 𝑝𝑒𝑟𝑙𝑢 = (𝑉𝑢 − ∅𝑉𝑐)

∅ =(26,52 − 16,79)

0,75 = 12,97 𝑘𝑁 𝑠 = (𝐴𝑣 × 𝑓𝑦 × 𝑑)

𝑉𝑠 𝑝𝑒𝑟𝑙𝑢 =(157 × 240 × 163,5)

12,97 = 304,87 𝑚𝑚 === 300 𝑚𝑚

Kontrol jarak

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34

= 300 mm < s max = d/4 = 40,87 mm, maka digunakan s max = d/4 = 40,87 mm atau dibulatkan jadi s = 40 mm

 Gambar penulangan geser balok anak B-2 Atap

3.3 Data Balok – B3 Atap

● Lx : 3,5 m

● Ly : 5 m