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Tugas Akhir Perencanaan Struktur Kafe dan Resto 2 Lantai
BAB 6 Balok Anak 6.2.2. Perhitungan Tulangan
a. 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
Daerah Tumpuan Dari perhitungan SAP 2000 diperoleh :
Mu = 3485,07
kgm = 3,485 . 10
7
Nmm Mn =
φ Mu
=
7 7
10 .
35 ,
4 8
, 10
. 485
, 3
=
Nmm
Rn =
=
2
.d b
Mn
= ×
2 7
294 250
10 .
35 ,
4
2,01 Nmm
2
m = =
= 0,85.25
360 c
0,85.f fy
17
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Tugas Akhir Perencanaan Struktur Kafe 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 Kafe 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 Kafe dan Resto 2 Lantai
BAB 6 Balok Anak
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