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Tugas Akhir
Perencanaan St rukt ur Gedung Sekolah 2 L ant ai
BAB 5 Plat Lant ai 128
5.5. Penulangan Plat Lantai
Tabel 5.1. Perhitungan Plat Lantai
Tipe Plat
LyLx m
Mlx kgm
Mly kgm
Mtx kgm
Mty kgm
4,003,00 = 1,33
361,746 209,009
787,803 618,988
3,002,50 = 1,2
212,135 156,310 474,513 413,105 3,002,50
= 1,2 173,058 156,310 413,105 385,193
4,002,50 = 1,6
256,795 139,563 552,668 429,853 4,002,50
= 1,6 206,553 89,320 441,018 318,203
3,253,00 = 1,08
209,009 217,048 522,522 522,522 3,253,00
= 1,08 200,970 168,815 474,289 434,095
4,003,25 = 1,23
292,485 179,265 651,015 537,795 4,003,25
= 1,23 330,225 169,830 698,190 537,795
2,002,00 = 1,0
75,029 75,029 185,786 185,786
4,001,25 = 3,2
58,603 11,162 118,602 79,534 3,003,00
= 1,0 168,815 209,009 442,134 482,328
3,003,00 = 1,0
168,534 168,534 418,018 418,018
A1 A
A2 A3
B1 B2
B3
B4
B5
B6
C1 C2
A4
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Tugas Akhir
Perencanaan St rukt ur Gedung Sekolah 2 L ant ai
BAB 5 Plat Lant ai 129
4,003,00 = 1,33
273,319 144,698 586,832 458,212 3,253,00
= 1,08 265,280
225,086 618,988 578,794
3,253,00 = 1,08
233,125 160,775 530,561 458,212 4,003,25
= 1,23 339,660 188,700 773,670 679,32
2,002,00 = 1,0
75,029 92,893 196,504 214,368
4,002,50 = 1,6
173,138 157,748 -
380,903
Dari perhitungan momen diambil momen terbesar yaitu: Mlx
= 361,746 kgm Mly
= 225,086 kgm Mtx
= 787,803 kgm Mty
= 679,320 kgm Data – data plat :
Tebal plat h = 12 cm
= 120 mm Diameter tulangan
∅ = 10 mm fy
= 240 MPa f’c
= 25 MPa b =
1000 mm
p =
20 mm
Tebal penutup d’ = p + ½
∅ tul = 20 + 5
= 25
mm
C3
D1
D2
D3 D4
P.atp
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Perencanaan St rukt ur Gedung Sekolah 2 L ant ai
BAB 5 Plat Lant ai 130
Tinggi Efektif d = h - d’
= 120 – 25 =
95 mm
Tingi efektif
Gambar 5.4 Perencanaan Tinggi Efektif dx = h – p - ½Ø
= 120 – 20 – 5 = 95 mm dy = h – d’ – Ø - ½ Ø
= 120 – 20 - 10 - ½ . 10 = 85 mM ρb =
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
+ fy fy
fc 600
600 .
. .
85 ,
β
= ⎟
⎠ ⎞
⎜ ⎝
⎛ + 240
600 600
. 85
, .
240 25
. 85
, =
0,0538 ρ
max
= 0,75 . ρb
= 0,75 . 0,0538 =
0,04035 ρ
min
= 0,0025
h dy
dx d
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Tugas Akhir
Perencanaan St rukt ur Gedung Sekolah 2 L ant ai
BAB 5 Plat Lant ai 131
5.6. Penulangan tumpuan arah x
Mu = 787,803 kgm = 7,878x10
6
Nmm
Mn = φ
Mu =
= 8
, 7,878x106
9,8475x10
6
Nmm Rn
= =
2
.dx b
Mn =
2 6
95 .
1000 10
8475 ,
9 x
1,091 Nmm
2
m = 294
, 11
25 .
85 ,
240 .
85 ,
= =
c f
fy
ρ
perlu
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= . 294
, 11
1 ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− 240
091 ,
1 .
294 ,
11 .
2 1
1 = 0,00467
ρ ρ
max
ρ ρ
min
, di pakai ρ
perlu
= 0,00467 As
perlu
= ρ
perlu
. b . dx = 0,00467. 1000 . 95
= 443,65mm
2
Digunakan tulangan ∅ 10
As = ¼ . π . 10
2
= 78,5
mm
2
Jumlah tulangan dalam 1m
1
= 5
. 78
443.65 = 5,65 ~ 6 buah
Jarak tulangan dalam 1m
1
= 6
1000 = 166,67mm ~ 160mm
Jarak maksimum tulangan = 2 x h = 2 x 120 = 240mm
As yang timbul = 6. ¼ .
π.10
2
= 471 mm
2
As
perlu
…ok Dipakai tulangan
∅ 10 – 160 mm
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Perencanaan St rukt ur Gedung Sekolah 2 L ant ai
BAB 5 Plat Lant ai 132
5.7. Penulangan tumpuan arah y