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Tugas Akhir Perencanaan Struktur Gedung Swalayan 2 Lantai RAB
BAB 6 Balok Anak
b. Beban hidup q
L
Beban hidup digunakan 250 kgm
2
qL = 2 x 1,1458 x 250 kgm
2
= 572,9 kgm c. Beban berfaktor q
U
qU1 = 1,2. qD + 1,6. qL = 1,2 . 2140 + 1,6 . 572,9
= 3484,64 kgm
6.3. Perhitungan Tulangan Balok Anak
6.3.1. Perhitungan Tulangan Balok Anak As A- A’
1. Tulangan Lentur Balok Anak
Data Perencanaan : h = 450 mm
Ø
t
= 16 mm b = 225 mm
Ø
s
= 8 mm p = 40 mm
d = h - p - 12 Ø
t
- Ø
s
fy = 320 Mpa = 450
– 40 - 12.16 - 8 f’c = 25 MPa
= 394
Daerah Tumpuan
b =
fy 600
600 β
fy c
0,85.f
=
320 600
600 85
, 320
25 .
85 ,
= 0,037
max
= 0,75 . b = 0,75 . 0,037
= 0,02775
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Tugas Akhir Perencanaan Struktur Gedung Swalayan 2 Lantai RAB
BAB 6 Balok Anak
min
=
004375 ,
320 4
, 1
4 ,
1 fy
Dari perhitungan SAP 2000 diperoleh :
Mu = 5944,67 kgm = 5,9447 . 10
7
Nmm
Mn =
Mu
=
8 ,
10 .
9447 ,
5
7
= 7,431 . 10
7
Nmm
Rn =
2
.d b
Mn
2 7
394 225
10 .
431 ,
7 2,128 mm
2
m =
0,85.25 320
c 0,85.f
fy
15,06
perlu
= fy
Rn .
m 2
1 1
. m
1
= 320
128 ,
2 06
, 15
2 1
1 .
06 ,
15 1
= 0,007
max min
, di pakai = 0,007 As = . b . d
= 0,007. 225 . 394 = 620,55
Digunakan tulangan D 16 = ¼ . . 16
2
= 200,96 mm
2
Jumlah tulangan =
08 ,
3 96
, 200
55 ,
620
~ 4 buah. As ada = 4 . ¼ . . 16
2
= 803,84 mm
2
As ……… aman a =
225 25
85 ,
320 84
, 803
b c
f 0,85
fy ada
As
= 53,79 Mn ada= As ada × fy d
– a2 = 803,84 × 320 394
– 53,792 = 94,328 . 10
7
Nmm
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Tugas Akhir Perencanaan Struktur Gedung Swalayan 2 Lantai RAB
BAB 6 Balok Anak
Mn ada Mn ......... aman
Kontrol Spasi : S =
1 -
n sengkang
2 -
tulangan n
- 2p
- b
=
1 4
8 .
2 -
16 4.
- 40
. 2
- 225
= 21,66 mm 25 mm dipakai tulangan 2 lapis
Karena jarak antar tulangan 25 mm maka kita rancang dengan tulangan berlapis dengan cara mencari d yang baru.
d
1
= h - p - 12 D
t
- Ø
s
= 450 – 40 – ½ . 16– 8
= 394 mm d
2
= h - p - Ø
t
- 12 D
t
– s - Ø
s
= 450 – 40 – 16 – ½ 16 – 30 - 8
= 348 mm d =
n .
d
2 1
n d
n
=
4 348.2
394.2
= 371 mm T = As
ada
. fy = 803,84. 320
= 257228,8 Mpa C
= 0,85 . f’c . a . b T = C
As . fy = 0,85 . f’c . a . b
a =
b c
f fy
As .
. 85
, .
=
225 .
25 .
85 ,
8 ,
257228
= 53,79
O 2 D
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Tugas Akhir Perencanaan Struktur Gedung Swalayan 2 Lantai RAB
BAB 6 Balok Anak
ØMn = Ø . T d
– a2 = 0,85 . 257228,8 371
– 53,792 = 7,081 × 10
7
Nmm ØMn Mu
Aman.. 7,081 × 10
7
Nmm 5,9447 × 10
7
Nmm
Jadi dipakai tulangan 4 D 16 mm Daerah Lapangan :
Dari perhitungan SAP 2000 diperoleh :
Mu = 4404,05 kgm = 4,40405 . 10
7
Nmm
Mn =
Mu
=
8 ,
10 .
40405 ,
4
7
= 5,505 . 10
7
Nmm
Rn =
2
.d b
Mn
2 7
394 225
10 .
505 ,
5 1,576 mm
2
m =
0,85.25 320
c 0,85.f
fy
15,06
perlu
= fy
Rn .
m 2
1 1
. m
1
= 320
576 ,
1 06
, 15
2 1
1 .
06 ,
15 1
= 0,005
max min
, di pakai = 0,005 As = . b . d
= 0,005. 225 . 394 = 443,25
Digunakan tulangan D 16 = ¼ . . 16
2
= 200,96 mm
2
Jumlah tulangan =
21 ,
2 96
, 200
25 ,
443
~ 3 buah. As ada = 3 . ¼ . . 16
2
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Tugas Akhir Perencanaan Struktur Gedung Swalayan 2 Lantai RAB
BAB 6 Balok Anak
= 602,88 mm
2
As ……… aman a =
225 25
85 ,
320 88
, 602
b c
f 0,85
fy ada
As
= 40,35 Mn ada = As ada × fy d
– a2 = 602,88 × 320 394
– 40,352 = 7,212 . 10
7
Nmm Mn ada Mn ......... aman
Kontrol Spasi : S =
1 -
n sengkang
2 -
tulangan n
- 2p
- b
=
1 3
8 .
2 -
16 3.
- 40
. 2
- 225
= 40,5 mm 25 mm dipakai tulangan 1 lapis
Jadi dipakai tulangan 3 D 16 mm
O 2 D 10
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Tugas Akhir Perencanaan Struktur Gedung Swalayan 2 Lantai RAB
BAB 6 Balok Anak
2. Tulangan Geser
