Bab 4 Perhitungan Plat

GORDING

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BAB IV

PELAT LANTAIIV.1. PELAT LANTAI

Diketahui data-data sebagai berikut: fc = 30 MPa ; fy = 240 MPa Tebal pelat = 120 mm (selimut beton = 20 mm)

Digunakan tulangan 10 mm

dx = 120 20 0,5 x 10= 95 mm

dy = 120 20 10 0,5 x 10= 85 mm

IV.2.Idealisasi struktur

Gambar 4.1 Denah Struktur Pelat Lantai 2 & 3

Gambar 4.2 Denah Struktur Pelat AtapIV.3.Pembebanana) Pembebanan Lantai 2-41. Beban mati (qD)Tebal pelat = 120 mmBerat sendiri=288 kg/m2

Berat penutup lantai=16,8 kg/m2

Berat spesi 3 cm=63 kg/m2

Berat ducting=25 kg/m2

Berat plafon + penggantung=18 kg/m2

TOTAL=410,8 kg/m2

2. Beban hidup (qL)

qL = 250 kg/m2 ........................(Berdasarkan PPIUG 1983 Tabel 3.1 hal. 17)3.Beban ultimate (qu)qu = 1,2 qD + 1,6 qL

= 1,2 x 410,8 + 1,6x 250

= 892,96 kg/m2

= 8,93 kN/m2b) Pembebanan Lantai atap

1. Beban mati (qD)Tebal pelat = 100 mmBerat sendiri=240 kg/m2

Berat penutup lantai=16,8 kg/m2

Berat spesi 3 cm=63 kg/m2

Berat ducting=25 kg/m2

Berat plafon + penggantung=18 kg/m2

TOTAL=362,8 kg/m2

2. Beban hidup (qL)

qL = 100 kg/m2 ........................(Berdasarkan PPIUG 1983 Tabel 3.1 hal. 17)

3.Beban ultimate (qu)qu = 1,2 qD + 1,6 qL

= 1,2 x 362,8 + 1,6x 100

= 595,36 kg/m2

= 5,95 kN/m2IV.4.Perhitungan Tulangan Lantai 2 & 3a. Penulangan Pelat D (Skema III CUR IV) (ly/lx=1,0) Mlx = 0,001x 8,93x 62 x 30 = 6,698 kN.m

DMly = 0,001x 8,93x 62 x 30 = 6,698 kN.m

Mtx = -0,001x 8,93x 62 x 68 = -15,181 kN.m

Mty = -0,001x 8,93x 62 x 68 = -15,181 kN.mMtix = mlx = x 6,698 = 3,349 kN.m

Mtiy = mly = x 6,698 = 3,349 kN.m

Mu (kN.m)Mu/bd2As (mm2)tul. terpasang

Mlx6,698742,1050,0039370,5 10 200

Mly6,698926,9900,0049416,5 10 175

Mtx-15,181-1682,1050,0092874 10 75

Mty-15,181-2101,1760,0117994,5 10 75

Mtix3,349371,0530,0019180,5 10 250

Mtiy3,349463,4950,0024204 10 250

Mu/bdx2 = 6,698/ (1 x 0,0952) = 742,105 ( diperoleh 0,0039Asx = x b x dx = 0,0039 x 1000 x 95 = 370,5 mm2

As min = 0,25% x b x t = 0,25% x 1000 x 120 = 300 mm2 Asx > As minTulangan terpasang = 10 200 (Tabel CUR IV hal 15)b. Penulangan Pelat C (Skema VIIB CUR IV) (ly/lx=1,0) Mlx = 0,001x 8,93x 62 x 28 = 6,251 kN.m

CMly = 0,001x 8,93x 62 x 25 = 5,581 kN.m

Mtx = -0,001x 8,93x 62 x 60 = -13,395 kN.m

Mty = -0,001x 8,93x 62 x 54 = -12,056 kN.mMtiy = mlx = x 6,251 = 3,126 kN.m

Mu (kN.m)Mu/bd2As (mm2)tul. terpasang

Mlx6,251692,6320.0036342 10 225

Mly5,581772,4910.0041348,5 10 225

Mtx-13,3951484,2110.0081769,5 10 100

Mty-12,0561668,5810.0091773,5 10 - 100

Mtiy3,126346,3160.0018171 10 - 250

c. Penulangan Pelat E (Skema VIA CUR IV) (ly/lx=1,71) Mlx = 0,001x 8,93x 3,52 x 63,1 = 5,071 kN.m

EMly = 0,001x 8,93x 3,52 x 21,3 = 1,712 kN.m

Mtx = -0,001x 8,93x 3,52 x 108,45 = -8,716 kN.m

Mty = -0,001x 8,93x 3,52 x 77 = -6,188 kN.mMtix = mlx = x 5,071 = 2,536 kN.m

Mtiy = mly = x 1,712 = 0,856 kN.m

Mu (kN.m)Mu/bd2As (mm2)tul. terpasang

Mlx5,071561,9220,0029275,5 10 250

Mly1,712236,9390,0012102 10 250

Mtx-8,716-965,7760,0052494 10 150

Mty-6,188-856,5380,0045382,5 10 200

Mtix2,536280,9610,0014133 10 250

Mtiy0,856118,4690.000651 10 250

d. Penulangan Pelat F (Skema VIIB CUR IV) (ly/lx=1,2)Mlx = 0,001x 8,93x 32 x 28= 1,563 kN.m

Mly = 0,001x 8,93x 32 x 25= 1,395 kN.m

FMtx = -0,001x 8,93x 32 x 60= -3,349 kN.m

Mty = -0,001x 8,93x 32 x 54= -3,014 kN.mMtiy = mlx = x 1,563 = 0,7815 kN.m

Mu (kN.m)Mu/bd2As (mm2)tul. terpasang

Mlx1,563173.1580.000876 10 250

Mly1,395193.1230.000976.5 10 250

Mtx-3,349371.0530.0019180.5 10 250

Mty-3,014417.1450.0023195.5 10 250

Mtiy0,781586,5790.000547.5 10 250

e. Penulangan Pelat Tipe A (Skema VIA CUR IV) (ly/lx=1,2) Mlx = 0,001x 8,93x 32 x 25 = 1,395 kN.m

Mly = 0,001x 8,93x 32 x 28 = 1,563 kN.m

AMtx = -0,001x 8,93x 32 x 54 = -3, 014 kN.m

Mty = -0,001x 8,93x 32 x 60 = -3, 349 kN.mMtiy = mlx = x 1,395= 0,698 kN.m

Mu (kN.m)Mu/bd2As (mm2)tul. terpasang

Mlx1,395193,1230.000985.5 10 250

Mly1,563173,1580.000868 10 250

Mtx-3,014-417,1450.0023218.5 10 250

Mty-3,349-371,0530.0019161.5 10 250

Mtiy0,69877,3030.000438 10 250

f. Penulangan Pelat B (Skema VIIB CUR IV) (ly/lx=2,4)Mlx = 0,001x 8,93x 32 x 58= 3,237 kN.m

BMly = 0,001x 8,93x 32 x 17= 0,949 kN.m

Mtx = -0,001x 8,93x 32 x 83= -4,632 kN.m

Mty = -0,001x 8,93x 32 x 53= -2,958 kN.m

Mtiy = mlx = x 3,237= 1,619 kN.m

Mu (kN.m)Mu/bd2As (mm2)tul. terpasang

Mlx3,237358.6840,0021199.5 10 250

Mly0,949131.3240,000759.5 10 250

Mtx-4,632513.2890,0029275.5 10 250

Mty-2,958409.4200,0024204 10 250

Mtiy1,619179.3420,0011104.5 10 250

IV.5.Perhitungan Tulangan Lantai atap

a. Penulangan Pelat A (Skema III CUR IV) (ly/lx=1,0) Mlx = 0,001x 5,95x 62 x 30 = 4,463 kN.m

AMly = 0,001x 5,95x 62 x 30 = 4,463 kN.m

Mtx = -0,001x 5,95x 62 x 68 = -10,115 kN.m

Mty = -0,001x 5,95x 62 x 68 = -10,115 kN.mMtix = mlx = x 4,463 = 2,231 kN.m

Mtiy = mly = x 4,463 = 2,231 kN.m

Mu (kN.m)Mu/bd2As (mm2)tul. terpasang

Mlx4,463494,4600,0025237,5 10 250

Mly4,463617,6470,0032272 10 250

Mtx-10,115-1120,7760,0060570 10 125

Mty-10,115-14000,0076646 10 100

Mtix2,231247,2300,0012114 10 250

Mtiy2,231308,8240,0016136 10 250

Mu/bdx2 = 4,463/ (1 x 0,0952) = 464,460 ( diperoleh 0,0025Asx = x b x dx = 0,0025 x 1000 x 95 = 237,5 mm2

As min = 0,25% x b x t = 0,25% x 1000 x 120 = 300 mm2 Asx > As minTulangan terpasang = 10 250 (Tabel CUR IV hal 15)

b. Penulangan Pelat B (Skema VIIB CUR IV) (ly/lx=1,0) Mlx = 0,001x 5,95x 62 x 28 = 4,165 kN.m

Mly = 0,001x 5,95x 62 x 25 = 3,719 kN.m

BMtx = -0,001x 5,95x 62 x 60 = -8,925 kN.m

Mty = -0,001x 5,95x 62 x 54 = -8,033 kN.mMtiy = mlx = x 4,165 = 2,083 kN.m

Mu (kN.m)Mu/bd2As (mm2)tul. terpasang

Mlx4,165461,4960.0024228 10 250

Mly3,719514,7060.0026221 10 250

Mtx-8,925988,9200.0053503,5 10 150

Mty-8,0331111,7650.0059501,5 10 - 150

Mtiy2,083230,7480.0012114 10 - 250

c. Penulangan Pelat C (Skema VIA CUR IV) (ly/lx=1,71) Mlx = 0,001x 5,95x 3,52 x 63,1 = 3,379 kN.m

Mly = 0,001x 5,95x 3,52 x 21,3 = 1,141 kN.m

CMtx = -0,001x 5,95x 3,52 x 108,45 = -5,807 kN.m

Mty = -0,001x 5,95x 3,52 x 77 = -4,123 kN.mMtix = mlx = x 3,379 = 1,690 kN.m

Mtiy = mly = x 1,141 = 0,570 kN.m

Mu (kN.m)Mu/bd2As (mm2)tul. terpasang

Mlx3,379374,4050,0019180,5 10 250

Mly1,141157,8710,000759,5 10 250

Mtx-5,807-643,4900,0034323 10 225

Mty-4,123-570,7060,0030255 10 250

Mtix1,690187,2020,000985,5 10 250

Mtiy0,57078,9350.0025212,5 10 250

d. Penulangan Pelat D (Skema II CUR IV) (ly/lx=1,0)Mlx = 0,001x 5,95x 62 x 25= 3,719 kN.m

DMly = 0,001x 5,95x 62 x 25= 3,719 kN.m

Mtx = -0,001x 5,95x 62 x 51= -7,586 kN.m

Mty = -0,001x 5,95x 62 x 51= -7,586 kN.m

Besar Mtix dan Mtiy tidak diperhitungkan karena momen tak terduga sebesar 0.5 Mlx memiliki besar yang sangat kecil sehingga disamakan dengan momen tumpuan.

Mu (kN.m)Mu/bd2As (mm2)tul. terpasang

Mlx3,719412,0500,0021199,5 10 250

Mly3,719514,7060,0027229,5 10 250

Mtx-7,586-840,5820,0045427,5 10 175

Mty-7,586-10500,0056476 10 - 150

e. Penulangan Pelat Penutup (Skema I CUR IV) (ly/lx=2,5) Mlx = 0,001x 5,95x 2.52 x 110 = 2,618 kN.m

Mly = 0,001x 5,95x 2.52 x 24 = 0,571 kN.m

Mtix = mlx = x 2,618 = 1,309 kN.m

Mtiy = mly = x 0,571 = 0,285 kN.m

Mu (kN.m)Mu/bd2As (mm2)tul. terpasang

Mlx2,618290,0830.0015142,5 10 250

Mly0,57179,0580.0025212,5 10 250

Mtix1,309145,0410.000766,5 10 250

Mtiy0,28539,5290.0025212,5 10 250

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