Perencanaan Struktur Hotel 2 lantai
BAB I Pendahuluan Gambar 5.5 Plat tipe B2
2,5 2
5 Lx
Ly
Mlx = 0,001.qu .Lx
2
.x = 0.001 x 893,2 x 2
2
x 60 = 214,37 kg m Mly = 0,001.qu .Lx
2
.x = 0.001 x 893,2 x 2
2
x 18 = 64,31 kg m Mtx = 0,001.qu .Lx
2
.x = 0.001 x 893,2 x 2
2
x 120 = 428,74 kg m Mty = 0,001.qu .Lx
2
.x = 0.001 x 893,2 x 2
2
x 79 = 282,25 kg m
5.4. Penulangan Plat Lantai
Tabel 5.1. Perhitungan Plat Lantai Tipe Plat
LyLx m Mlx kgm Mly kgm Mtx kgm Mty kgm
A 52,5 = 2
36,62 10,72
296,54 203,65
B 2,51,5 = 1,7
76,37 28,14
162,79 114,55
C 3,52,5 = 1,4
189,81 100,49
407,52 318,2
D 52 = 2,5
214,37 64,31
428,74 282,25
Dari perhitungan momen diambil momen terbesar yaitu: Mlx
= 214,37 kgm Mly
= 100,49 kgm Mtx
= 428,74 kgm Mty
= 318,2 kgm
Data – data plat :
Perencanaan Struktur Hotel 2 lantai
BAB I Pendahuluan Tebal plat h
= 12 cm = 120 mm Diameter tulangan
= 12 mm fy
= 240 MPa f’c
= 25 MPa b
= 1000 mm tebal penutup d’
= 20 mm
Gambar 5.8 Perencanaan Tinggi Efektif
Tinggi efektif : dx = h
– d’ - ½Ø = 120
– 20 – 6 = 94 mm dy = h
– d’ – Ø - ½ Ø = 120
– 20 - 12 - ½ . 12 = 82 mm
b =
fy fy
fc
600 600
. .
. 85
,
f’c = 25 Mpa 30 Mpa →
= 0,85 =
240 600
600 .
85 ,
. 240
25 .
85 ,
= 0,054
max
= 0,75 . b
= 0,75 . 0,054 = 0,0405
min
= 0,0025 untuk plat 5.5. Penulangan tumpuan arah x
Mu = 428,74 kgm = 4,287.10
6
Nmm
h d
dy dx
Perencanaan Struktur Hotel 2 lantai
BAB I Pendahuluan Mn
= M u
=
8 ,
10 .
287 ,
4
6
5,359.10
6
Nmm
Rn =
2
.
dx b
Mn
2 6
94 .
1000 10
. 359
, 5
0,606 Nmm
2
m =
3 ,
11 25
. 85
, 240
. 85
,
c f
fy
perlu
=
fy Rn
. m
2 1
1 .
m 1
= .
3 ,
11 1
240 606
, .
3 ,
11 .
2 1
1
= 0,00256
perlu
max
perlu
min
, di pakai
perlu
= 0,00256 As
perlu
=
perlu
. b . dx = 0,00256 . 1000 . 94
= 240,64 mm
2
Digunakan tulangan 12
As = ¼ . . 12
2
= 113,04 mm
2
S =
per lu
As b
As
. =
64 ,
240 1000
. 04
, 113
= 469,75 mm S
max
= 2 x h = 240 mm n =
s b
=
240 1000
= 5 buah As yang timbul = 5. ¼ .
. 12
2
= 339,12 mm
2
As
perlu
…..…ok Dipakai tulangan
12 – 240 mm
Perencanaan Struktur Hotel 2 lantai
BAB I Pendahuluan
5.6. Penulangan tumpuan arah y
Mu = 318,2 kgm = 3,182.10
6
Nmm
Mn =
M u =
8
, 10
. 182
, 3
6
3,978.10
6
Nmm
Rn =
2
.dy b
Mn
2 6
82 .
1000 10
. 798
, 3
0,592 Nmm
2
m =
3 ,
11 25
. 85
, 240
. 85
,
c f
fy
perlu
=
fy Rn
. m
2 1
1 .
m 1
= .
3 ,
11 1
240 592
, .
3 ,
11 .
2 1
1
= 0,002502
perlu
max
perlu
min
, di pakai
perlu
= 0,002502 As
perlu
=
perlu
. b . dy = 0,002502. 1000 . 82
= 205,164 mm
2
Digunakan tulangan 12
As = ¼ . . 12
2
= 113,04 mm
2
S =
per lu
As b
As
. =
164 ,
205 1000
. 04
, 113
= 550,97 mm S
max
= 2 x h = 240 mm n =
s b
=
240 1000
Perencanaan Struktur Hotel 2 lantai
BAB I Pendahuluan = 5 buah
As yang timbul = 5. ¼ . . 12
2
= 565,2 mm
2
As
perlu
…..…ok Dipakai tulangan
12 – 240 mm
5.7. Penulangan lapangan arah x
