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Tugas Akhir Perencanaan Struktur Factory Outlet dan Resto 2 Lantai
BAB 6 Balok Anak
ρ
perlu
= ⎟
⎟ ⎠
⎞ ⎜
⎜ ⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ ×
× −
− 360
01 ,
2 17
2 1
1 .
17 1
= 0,006 ρ ρ
max
ρ ρ
min
, di pakai ρ
perlu
= 0,006 As =
ρ
perlu
. b . d = 0,006 . 250 . 294
= 441 mm
2
Digunakan tulangan D 16 = ¼ .
π . 16
2
= 200,96 mm
2
Jumlah tulangan =
19 ,
2 96
, 200
441 =
~ 3 buah.
Dipakai 3 D 16 mm
As ada = 3 . ¼ .
π . 16
2
= 602,88 mm
2
As ……… aman a =
250 25
85 ,
360 88
, 602
b c
f 0,85
fy ada
As ×
× ×
= ×
× ×
= 40,85 Mn ada = As ada × fy
⎟ ⎠
⎞ ⎜
⎝ ⎛ −
2 a
d = 602,88 × 360
⎟ ⎠
⎞ ⎜
⎝ ⎛
− 2
85 ,
40 294
= 5,9375 . 10
7
Nmm Mn ada Mn ......... aman
Kontrol Spasi : S
= 1
- n
sengkang 2
- tulangan
n -
2p -
b φ
φ
= 1
3 8
. 2
- 16
. 3
- 40
. 2
- 50
2 −
= 53 25 mm......oke
Jadi dipakai tulangan D 16 mm
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Tugas Akhir Perencanaan Struktur Factory Outlet dan Resto 2 Lantai
BAB 6 Balok Anak Daerah Lapangan
Dari perhitungan SAP 2000 diperoleh :
Mu = 2577.40
kgm = 2,577 . 10
7
Nmm Mn =
φ Mu
=
7 7
10 .
19 ,
3 8
, 10
. 577
, 2
= Nmm
Rn =
=
2
.d b
Mn =
×
2 7
294 250
10 .
19 ,
3 1,47 Nmm
2
m =
= =
0,85.25 360
c 0,85.f
fy 17
ρ
perlu
= ⎟
⎟ ⎠
⎞ ⎜
⎜ ⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ ×
× −
− 360
47 ,
1 17
2 1
1 .
17 1
= 0,004 ρ ρ
max
ρ ρ
min
, di pakai ρ
perlu
= 0,004 As
= ρ
min
. b . d = 0,004 . 250 . 294
= 294 mm
2
Digunakan tulangan D 16 = ¼ .
π . 16
2
= 200,96 mm
2
Jumlah tulangan =
46 ,
1 96
, 200
294 =
~ 2 buah.
Dipakai 2 D 16 mm
As ada = 2 . ¼ .
π . 16
2
= 401,92 mm
2
As ……… aman
a = 250
25 85
, 360
92 ,
401 b
c f
0,85 fy
ada As
× ×
× =
× ×
× = 27,23
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Tugas Akhir Perencanaan Struktur Factory Outlet dan Resto 2 Lantai
BAB 6 Balok Anak
Mn ada = As ada × fy ⎟
⎠ ⎞
⎜ ⎝
⎛ − 2
a d
= 401,92 × 360 ⎟
⎠ ⎞
⎜ ⎝
⎛ −
2 23
, 27
294 = 4,0568 . 10
7
Nmm Mn ada Mn ......... aman
Kontrol Spasi : S
= 1
- n
sengkang 2
- tulangan
n -
2p -
b φ
φ
= 1
2 8
. 2
- 16
. 2
- 40
. 2
- 50
2 −
= 122 25 mm......oke
Jadi dipakai tulangan D 16 mm Tulangan
Geser
Dari perhitungan SAP 2000 diperoleh : Vu =
4280.92 kg = 42809,2 N
f’c = 25 Mpa
fy = 360 Mpa
d = h – p – ½ Ø = 350 – 40 – ½ 12 = 304 mm
Vc = 1 6 .
c f .b .d
= 1 6 . 25 . 250 . 304
= 63333,33
N Ø Vc = 0,75 . 63333,33 N
= 47500 N 3 Ø Vc = 3 . 47500
= 142500 N Vu Ø Vc 3 Ø Vc
42809,2 N 47500 N 142500 N Jadi tidak perlu tulangan geser
S max = d2 = 2
304 = 152 mm
Jadi dipakai sengkang dengan tulangan Ø 8 – 150 mm
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Tugas Akhir Perencanaan Struktur Factory Outlet dan Resto 2 Lantai
BAB 6 Balok Anak
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