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181 Perencanaan St rukt ur dan Rencana A nggaran Biaya Gedung Sekolah D asar
2 lant ai
BAB 6 Balok Anak
2. Beban hidup qL Beban hidup 250 kgm
2
qL
1
= 1,267 x 2 x 250 kgm
2
= 633,5 kgm qL
2
= 2 x 0,667 x 250 kgm
2
= 333,5 kgm
6.2.2. Perhitungan Tulangan Balok Anak as E 1-9 a. Tulangan Lentur Balok Anak
Data Perencanaan : h
= 400 mm Ø
t
= 16 mm b
= 200 mm` Ø
s
= 8 mm p
= 40 mm d
= h - p - 12 Ø
t
- Ø
s
fy = 380 Mpa
= 400 – 40 – ½ . 16 – 8 f’c = 30 MPa
= 344 mm
1. Daerah Tumpuan
ρ b
=
+
fy fy
fc 600
600 .
. .
85 ,
β
=
+
380 600
600 .
380 85
, .
30 .
85 ,
= 0,035 ρ
max
= 0,75 . ρ
b = 0,75 . 0,036
= 0,026 ρ
min
= 380
4 ,
1 = 0,0037
Dari Perhitungan SAP 2000 diperoleh :
Mu = 4257,11 kgm = 4,26 .10
7
Nmm Mn
= Mu
= 8
, 10
. 4,26
7
= 5,325 .10
7
Nmm
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182 Perencanaan St rukt ur dan Rencana A nggaran Biaya Gedung Sekolah D asar
2 lant ai
BAB 6 Balok Anak
Rn =
2
.d b
Mn =
2 7
344 .
200 10
. 5,325
= 2,25
m =
c f
fy
,
. 85
, =
30 .
85 ,
380 = 14,9
ρ =
− −
fy 2.m.Rn
1 1
m 1
=
−
− 380
25 ,
2 .
9 ,
14 .
2 1
1 .
9 ,
14 1
= 0,0062 ρ
ρ
min
ρ ρ
max
→ dipakai
ρ Digunakan
ρ
perlu
= 0,0062 As perlu
= ρ
perlu
. b . d = 0,0062. 200 . 344
= 426,56 mm
2
n =
2
16 .
. 14
perlu As
π =
06 ,
201 56
, 426
= 2,12 ~3 tulangan As ada
= 3 . ¼ . π
. 16
2
= 3 . ¼ . 3,14 . 16
2
= 603,19 mm
2
As perlu →
Aman a
= =
b c
f fy
Asada .
, 85
, .
200 .
30 .
85 ,
380 .
603,19 = 44,94
Mn ada = As ada . fy d – a2
= 603,19. 380 344 – 44,942 = 7,4. 10
7
Nmm Mn ada Mn
→ Aman..
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Tugas A khir
183 Perencanaan St rukt ur dan Rencana A nggaran Biaya Gedung Sekolah D asar
2 lant ai
BAB 6 Balok Anak
Kontrol Spasi : S
= 1
- n
sengkang 2
- tulangan
n -
2p -
b φ
φ
= 1
3 8
. 2
16 .
3 40
. 2
200 −
− −
− = 28 25 mm…..ok
Jadi dipakai tulangan 3 D 16 mm
2. Daerah Lapangan
ρ b
=
+
fy fy
fc 600
600 .
. .
85 ,
β
=
+
380 600
600 .
380 85
, .
30 .
85 ,
= 0,035 ρ
max
= 0,75 . ρ
b = 0,75 . 0,036
= 0,026 ρ
min
= 380
4 ,
1 = 0,0037
Dari Perhitungan SAP 2000 diperoleh :
Mu = 3150,04 kgm = 3,2.10
7
Nmm Mn =
φ Mu
= 8
, 3,2.10
7
= 4.10
7
Nmm
Rn =
7 ,
1 344
. 00
2 4.10
d .
b Mn
2 7
2
= =
m =
c f
fy
,
. 85
, =
30 .
85 ,
380 = 14,9
ρ
ada
=
−
− fy
2.m.Rn 1
1 m
1
=
−
− 380
7 ,
1 .
9 ,
14 .
2 1
1 9
, 14
1
= 0,0046
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184 Perencanaan St rukt ur dan Rencana A nggaran Biaya Gedung Sekolah D asar
2 lant ai
BAB 6 Balok Anak
ρ ρ
min
ρ ρ
max
→ dipakai
ρ
ada
Digunakan ρ
ada
= 0,0046 As
perlu
= ρ
ada
. b . d = 0,0046. 200 . 344
= 316,48 mm
2
n =
2
16 .
4 1
perlu As
= tulangan
2 57
, 1
06 ,
201 48
, 316
≈ =
As ada = 2 . ¼ . π
. 16
2
= 2. ¼ . 3,14 . 16
2
= 402,12 mm
2
As perlu →
Aman a =
= b
c f
fy Asada
. ,
85 ,
. 96
, 29
200 30
85 ,
380 12
, 402
= x
x x
Mn ada = As ada . fy d – a2 = 402,12. 380 344 – 29,962
= 5.10
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
200 −
− −
− = 72 25 mm…..ok
Jadi dipakai tulangan 2 D 16 mm
3. Tulangan Geser Balok anak Dari perhitungan SAP 2000 Diperoleh :
