14
Perencanaan Struktur Hotel 2 lantai
BAB I Pendahuluan d
= h – p – ½ Ø
= 250 – 40 – ½ 8 = 206 mm
Vc = 1 6 .
c f .b .d = 1 6 .
25
.170.206 = 29183,333 N
Ø Vc = 0,6 . 29183,33 N = 17510 N
3 Ø Vc = 3 . 17510 = 52530 N Ø Vc Vu 3 Ø Vc, diperlukan tulangan geser
Ø Vs = Vu - Ø Vc
= 25850 – 17510 = 8340
Vs Perlu = 13900
6 ,
8340 6
,
s
V
Av = 2. ¼
8
2
= 2. ¼ . 3,14. 64 = 100,48 mm
2
S =
12 ,
791 13900
456 .
240 .
48 ,
100 .
.
Vsper lu
d fy
Av
mm
S max = h2 =
125 2
250
mm
Jadi dipakai sengkang dengan tulangan Ø 8 – 125
6.3.3 Perhitungan Tulangan Balok Anak as C-
C’
Data Perencanaan : h
= 500 mm Ø
t
= 12 mm b
= 340 mm` Ø
s
= 8 mm p
= 40 mm d
= h - p - 12 Ø
t
- Ø
s
fy = 380 Mpa = 500
– 40 – ½ .12 – 8 = 446 mm f’c = 25 MPa
Daerah Lapangan
15
Perencanaan Struktur Hotel 2 lantai
BAB I Pendahuluan b
=
fy
600 600
fy c.
β 0,85.f
=
380 600
600 85
, 380
25 .
85 ,
= 0,0291
max
= 0,75 . b
= 0,75 . 0,0291 = 0,0219
min
= 00368
, 380
4 ,
1 4
, 1
fy
Dari Perhitungan SAP 2000 diperoleh :
Mu = 4225,6 kgm= 4,226.10
7
Nmm Mn =
φ M u
= 8
, 10
. 226
, 4
7
= 5,283.10
7
Nmm Rn =
781 ,
446 .
340 10
5,283. d
. b
Mn
2 7
2
m =
8824 ,
17 25
. 85
, 380
. 85
,
c f
fy
=
fy 2.m.Rn
1 1
m 1
=
00174 ,
380 781
, .
8824 ,
17 .
2 1
1 8824
, 17
1
min
max
dipakai tulangan tunggal Digunakan
min
= 0,00368 As perlu =
. b . d = 0,00368. 340 .446
= 558,035 mm
2
n =
2
12 4
1 perlu
As
16
Perencanaan Struktur Hotel 2 lantai
BAB I Pendahuluan =
tulangan 5
937 ,
4 04
, 113
035 ,
558
Dipakai tulangan 5 D 10 mm
As ada = 5 . ¼ . . 12
2
= 5 . ¼ . 3,14 . 12
2
= 565,2mm
2
As perlu Aman..
a =
b c
f fy
Asada
. ,
85 ,
. 727
, 29
340 .
25 .
85 ,
380 .
2 ,
565
Mn ada = As ada . fy d – a2
= 565,2 .380 446 – 29,7272
= 9,26.10
7
Nmm Mn ada Mn
Aman..
Jadi dipakai tulangan 5 D 12 mm
Daerah Tumpuan Dari Perhitungan SAP 2000 diperoleh :
Mu = 8451,2 kgm= 8,451.10
7
Nmm Mn = φ
M u =
8 ,
10 .
451 ,
8
7
= 10,564.10
7
Nmm
Rn = 562
, 1
446 .
340 10
10,564. d
. b
Mn
2 7
2
m =
8824 ,
17 25
. 85
, 380
. 85
,
c f
fy
=
fy 2.m.Rn
1 1
m 1
=
00428 ,
380 562
, 1
. 8824
, 17
. 2
1 1
8824 ,
17 1
min
max
dipakai tulangan tunggal Digunakan
= 0,00428
17
Perencanaan Struktur Hotel 2 lantai
BAB I Pendahuluan As perlu =
. b . d = 0,00428. 340 . 446
= 649,019 mm
2
n =
2
10 .
4 1
perlu As
= tulangan
6 74
, 5
04 ,
113 019
, 649
Dipakai tulangan 6 D 12 mm As ada = 6 . ¼ .
. 12
2
= 6 . ¼ . 3,14 . 12
2
= 678,24 mm
2
As perlu Aman..
a =
b c
f fy
Asada
. ,
85 ,
. 135
, 34
340 .
25 .
85 ,
380 .
24 ,
678
Mn ada = As ada . fy d – a2
= 678,24 . 380 446 – 34,1352
= 11,055.10
7
Nmm Mn ada Mn
Aman..
Jadi dipakai tulangan 6 D 12 mm
4. Tulangan Geser Balok anak
