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Perencanaan St rukt ur dan Rencana A nggaran Biaya Gedung Puskesmas 2 L ant ai
Bab 6 Perencanaan Balok A nak
F G
H
5 0 0 5 0 0
2 2
2 2
6.9. Perhitungan Balok Anak as
2 F-H
6.9.1. Pembebanan
Gambar 6.23. Lebar Equivalen Balok Anak as 2 F-H
a. Beban Mati qD
Pembebanan balok as 2 F-G = 2 G-H • Beban Plat
= 2 x 1,32 x 404 kgm
2
= 1066,56 kgm qD
1
= 1066,56 kgm b.
Beban hidup qL Beban hidup digunakan 250 kgm
2
qL
1
= 2 x 1,32 x 250 kgm = 660 kgm Bidang momen:
Gambar 6.24. Bidang momen Balok Anak as 2 F-H
Bidang geser:
Gambar 6.25. Bidang geser Balok Anak as 2 F-H
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Bab 6 Perencanaan Balok A nak
6.9.2. Perhitungan Tulangan
a
Tulangan Lentur Balok Anak
Data Perencanaan
: h =
400 mm
D
t
= 16
mm b =
200 mm`
Ø
s
= 8 mm p
= 40 mm d
= h - p -12 Ø
t
- Ø
s
f’c = 20 MPa = 400 – 40 – ½ . 16– 8
fy = 400 Mpa = 344 mm
ρb = ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ + fy
600 600
fy c.
β 0,85.f
= ⎟
⎠ ⎞
⎜ ⎝
⎛ + 400
600 600
85 ,
400 20
. 85
, = 0,022
ρ
max
= 0,75 . ρb
= 0,75 . 0,022 = 0,0165
ρ
min
= 0035
, 400
4 ,
1 4
, 1
= =
fy
¾ Daerah Tumpuan : Dari Perhitungan SAP 2000 diperoleh :
Mu = 7267,76 kgm = 7,267×10
7
Nmm Mn
= φ
Mu =
8 ,
10 267
, 7
7
× = 9,08×10
7
Nmm Rn =
8 ,
3 344
00 2
10 08
, 9
d .
b Mn
2 7
2
= ×
× =
m = 53
, 23
20 .
85 ,
400 .
85 ,
= =
c f
fy
ρ = ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− fy
2.m.Rn 1
1 m
1
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Tugas A khir
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Bab 6 Perencanaan Balok A nak
= 0108
, 400
8 ,
3 53
, 23
2 1
1 53
, 23
1 =
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
× ×
− −
ρ ρ
min
ρ ρ
max
Pakai tulangan tunggal Digunakan
ρ = 0,0108
As perlu = ρ . b . d
= 0,0108 × 200 × 344 = 743,04 mm
2
n =
2
16 π
. 4
1 perlu
As
= tulangan
4 6
, 3
96 ,
200 04
, 743
≈ =
As ada = n . ¼ . π . d
2
= 4 . ¼ . π . 16
2
= 803,84 As perlu → Aman..
a =
= b
c f
fy Asada
. .
85 ,
. 57
, 94
200 20
85 ,
400 84
, 803
= ×
× ×
Mn ada = As ada . fy d – a2 = 803,84 . 400 344 – 94,572
= 9,54×10
7
Nmm Mn ada Mn
→ Aman..
Jadi dipakai tulangan 4 D 16 mm
Kontrol Spasi : S =
1 -
n sengkang
2 -
tulangan n
- 2p
- b
φ φ
= 1
4 8
. 2
- 16
4. -
40 .
2 -
200 −
= 13,3 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.
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Tugas A khir
Perencanaan St rukt ur dan Rencana A nggaran Biaya Gedung Puskesmas 2 L ant ai
Bab 6 Perencanaan Balok A nak 4D 16
∅8−100 40
20 2D 16
d
1
= h - p - 12 D
t
- Ø
s
= 400 – 40 – ½ . 16– 8 = 344 mm
d
2
= h - p - Ø
t
- 12 D
t
– s - Ø
s
= 400 – 40 – 16 – ½ 16 – 30 - 8 = 298 mm
d = n
. d
2 1
n d
n +
= 4
298.2 344.2
+ = 321 mm
T = As
ada
. fy = 803,84. 400
= 321536 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
, .
= 200
. 20
. 85
, 321536
= 94,57 ØMn
= Ø . T d – a2 = 0,85 . 321536 321 – 94,572 = 7,48 × 10
7
Nmm ØMn Mu
→ Aman.. 7,48 × 10
7
Nmm 7,267×10
7
Nmm
Jadi dipakai tulangan 4 D 16 mm
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Bab 6 Perencanaan Balok A nak
¾ Daerah Lapangan : Dari Perhitungan SAP 2000 diperoleh :
Mu = 4102,10 kgm = 4,102×10
7
Nmm Mn
= φ
Mu =
8 ,
10 102
, 4
7
× = 5,13×10
7
Nmm Rn =
17 ,
2 44
3 00
2 10
13 ,
5 d
. b
Mn
2 7
2
= ×
× =
m = 53
, 23
20 .
85 ,
400 .
85 ,
= =
c f
fy
ρ = ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− fy
2.m.Rn 1
1 m
1
= 0058
, 400
17 ,
2 53
, 23
2 1
1 53
, 23
1 =
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
× ×
− −
ρ ρ
min
ρ ρ
max
Pakai tulangan tunggal Digunakan
ρ
ada
= 0,0058 As perlu =
ρ . b . d = 0,0058 × 200 × 344
= 399,04 mm
2
n =
2
.16 π
. 4
1 perlu
As
= tulangan
2 98
, 1
96 ,
200 04
, 399
≈ =
As ada = n . ¼ . π . d
2
= 2 . ¼ . π . 16
2
= 401,92 As perlu → Aman..
a =
= b
c f
fy Asada
. .
85 ,
. 28
, 47
200 20
85 ,
400 92
, 401
= ×
× ×
Mn ada = As ada . fy d – a2 = 401,92. 400 344 – 47,282 = 5,15×10
7
Nmm
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Tugas A khir
Perencanaan St rukt ur dan Rencana A nggaran Biaya Gedung Puskesmas 2 L ant ai
Bab 6 Perencanaan Balok A nak
2D 16 ∅8−100
40
20 2D 16
Mn ada Mn → Aman..
5,15×10
7
Nmm 5,13×10
7
Nmm
Jadi dipakai tulangan 2 D 16 mm
Kontrol Spasi : S =
1 -
n sengkang
2 -
tulangan n
- 2p
- b
φ φ
= 1
2 8
. 2
- 16
2. -
40 .
2 -
00 2
− = 72 mm 25 mmdipakai tulangan 1 lapis
b
Tulangan Geser Balok anak
Dari perhitungan
SAP 2000 Diperoleh : Vu = 7294,55 kgm = 72945,5 N
f’c = 20 Mpa
fy = 240 Mpa
d =
344 mm
Vc = 16 .
c f .b .d = 16 . 20 .200 .344
= 51280,49 N Ø Vc = 0,6 . 51280,49 = 30768,29 N
3 Ø Vc = 3 . 30768,29 = 92304,88 N 5 Ø Vc = 5 . 30768,29 = 153841,45 N
Syarat tulangan geser : ∅Vc Vu 3Ø Vc
: 30768,29 N 80138,9 N 153841,45 N Jadi diperlukan tulangan geser:
∅ Vs = Vu - ∅ Vc
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Tugas A khir
Perencanaan St rukt ur dan Rencana A nggaran Biaya Gedung Puskesmas 2 L ant ai
Bab 6 Perencanaan Balok A nak
500 500
500 400
500 500
500
A B
C D
E F
G H
2 1
2 1
7 3
3 5
9 8
3 5
7 3
2 1
2 1
= 80138,9 – 30768,29 = 49370,61 N Vs
perlu = 6
, Vsp
φ =
6 ,
49370,61 = 82284,35 N
Av = 2 . ¼
π 8
2
= 2 . ¼ . 3,14 . 64 = 100,48 mm
2
S = 82
, 100
35 ,
82284 344
240 48
, 100
perlu Vs
d .
fy .
Av =
× ×
= mm
S max = d2 = 2
344 = 172 mm
Dipakai Ø 8 – 100 mm :
Vs ada = 29
, 82956
100 344
240 48
, 100
S d
. fy
. Av
= ×
× =
N
Vs ada Vs perlu
82956,29 N 82284,35 N...... aman
Jadi dipakai sengkang dengan tulangan Ø 8 – 100 mm
6.10. Perhitungan Balok Anak as 4’ A-H