commit to user
Tugas Akhir Perencanaan Struktur Factory Outlet dan Resto 2 Lantai
BAB 6 Balok Anak
3. Beban berfaktor q
U
qU
1
= 1,2. qD + 1,6. qL = 1,2 . 1009,32 + 1,6.530
= 2059,18 kgm qU
2
= 1,2. qD + 1,6. qL = 1,2 . 1033,98 + 1,6.545
= 2112,77 kgm
6.2.2. Perhitungan Tulangan
Tulangan Lentur Balok Anak Data Perencanaan :
h = 350 mm Ø
t
= 16
mm b = 250 mm
Ø
s
= 8 mm p = 40 mm
d = h - p - 12 Ø
t
- Ø
s
fy = 360 Mpa = 350 – 40 - 12.16 - 8
f’c = 25 Mpa = 294
Tulangan Lentur Daerah Lapangan
ρb = ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ + fy
600 600
fy c.
β 0,85.f
= ⎟
⎠ ⎞
⎜ ⎝
⎛ + 360
600 600
85 ,
360 25
. 85
, = 0,031
ρ
max
= 0,75 . ρb
= 0,75 . 0,031 =
0,0232 ρ
min
= 0038
, 360
4 ,
1 4
, 1
= =
fy
commit to user
Tugas Akhir Perencanaan Struktur Factory Outlet dan Resto 2 Lantai
BAB 6 Balok Anak Daerah Tumpuan
Dari perhitungan SAP 2000 diperoleh
:
Mu = 2800,35
kgm = 2,8003 . 10
7
Nmm Mn =
φ Mu
=
7 7
10 .
5 ,
3 8
, 10
. 8003
, 2
= Nmm
Rn =
=
2
.d b
Mn =
×
2 7
294 250
10 .
5 ,
3 1,62 Nmm
2
m = =
= 0,85.25
360 c
0,85.f fy
17 ρ
perlu
= ⎟
⎟ ⎠
⎞ ⎜
⎜ ⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ ×
× −
− 360
62 ,
1 17
2 1
1 .
17 1
= 0,0046 ρ ρ
max
ρ ρ
min
, di pakai ρ
perlu
= 0,0046 As =
ρ
perlu
. b . d = 0,0046 . 250 . 294
= 338,1 mm
2
Digunakan tulangan D 16 = ¼ .
π . 16
2
= 200,96 mm
2
Jumlah tulangan =
68 ,
1 96
, 200
1 ,
338 =
~ 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
commit to user
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 = 4,0569 . 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 Daerah Lapangan
Dari perhitungan SAP 2000 diperoleh :
Mu = 2789.97 kgm = 2,789 . 10
7
Nmm Mn = φ
Mu =
7 7
10 .
48 ,
3 8
, 10
. 789
, 2
= Nmm
Rn = =
2
.d b
Mn =
×
2 7
294 250
10 .
48 ,
3 1,61 Nmm
2
m = =
= 0,85.25
360 c
0,85.f fy
17 ρ
perlu
= ⎟
⎟ ⎠
⎞ ⎜
⎜ ⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ ×
× −
− 360
61 ,
1 17
2 1
1 .
17 1
= 0,004
ρ ρ
max
ρ ρ
min
, di pakai ρ
perlu
= 0,004
commit to user
Tugas Akhir Perencanaan Struktur Factory Outlet dan Resto 2 Lantai
BAB 6 Balok Anak