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Bab 6 Perencanaan Balok A nak
A B
C D
C
5 00 5 00
5 00
1 2
1 2
3 4
5 1 0
Vs ada Vs perlu
59202,82 N 17283,33N ...... aman
Jadi dipakai sengkang dengan tulangan Ø 8 – 100 mm
6.4. Perhitungan Balok Anak as 1’ A-D
6.4.1. Pembebanan Balok
Gambar 6.8. Lebar Equivalen Balok Anak as 1’ A-D
a Beban Mati qD
Pembebanan balok as 1’ A-B = 1’ B-C • Beban Plat
= 0,947 + 1,32 x 404 kgm
2
= 915,868 kgm • Berat dinding = 0,15 x 4,25-0,4 x 1700 kgm
2
= 981,75 kgm qD
1
=1897,618 kgm Pembebanan balok as 1’ C-C”
• Beban Plat = 0,852 + 0,5 + 0,5x 404 kgm
2
= 748,208 kgm • Berat dinding = 0,15 x 4,25-0,4 x 1700 kgm
2
= 981,75 kgm qD
2
=1729,958 kgm Pembebanan balok as 4 C’-D
• Beban Plat = 2 x 0,667 x 404 kgm
2
= 538,936 kgm qD
3
= 538,936 kgm b
Beban hidup qL Beban hidup digunakan 250 kgm
2
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Bab 6 Perencanaan Balok A nak
qL
1
= 0,947 + 1,32 x 250 kgm
2
= 566,75 kgm qL
2
= 0,852 + 0,5 + 0,5 x 250 kgm
2
= 463 kgm qL
3
=0,667 x 2 x 250 kgm
2
= 333,5 kgm
Bidang momen:
Gambar 6.9. Bidang momen Balok Anak as 1’ A-D
Bidang geser:
Gambar 6.10. Bidang geser Balok Anak as 1’ A-D 6.4.2
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 D
t
- Ø
s
fy = 400 Mpa = 400 – 40 – ½ . 16– 8
f’c = 20 MPa = 344 mm
ρb = ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ + fy
600 600
fy c.
β 0,85.f
= ⎟
⎠ ⎞
⎜ ⎝
⎛ + 400
600 600
85 ,
400 20
. 85
, = 0,022
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Bab 6 Perencanaan Balok A nak
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 = 8294,21 kgm = 8,294×10
7
Nmm Mn
= φ
Mu =
8 ,
10 294
, 8
7
× = 10,37×10
7
Nmm Rn =
4 ,
4 44
3 00
2 10
37 ,
10 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
= 013
, 400
4 ,
4 53
, 23
2 1
1 53
, 23
1 =
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
× ×
− −
ρ ρ
min
ρ ρ
max
Pakai tulangan tunggal Digunakan
ρ = 0,013 As perlu =
ρ . b . d = 0,013 × 200 × 344
= 894,4 mm
2
n =
2
.16 π
. 4
1 perlu
As
= tulangan
5 5
, 4
96 ,
200 4
, 894
≈ =
As ada = n . ¼ . π . d
2
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Bab 6 Perencanaan Balok A nak
5D 16
2D 16 ∅8−150
40
= 5 . ¼ . π . 16
2
= 1004,8 As perlu → Aman..
a =
= b
c f
fy Asada
. .
85 ,
. 21
, 118
200 20
85 ,
400 8
, 1004
= ×
× ×
Mn ada = As ada . fy d – a2 = 1004,8. 400 344 – 118,212 = 11,45×10
7
Nmm Mn ada Mn → Aman..
Jadi dipakai tulangan 5 D 16 mm
Kontrol Spasi : S =
1 -
n sengkang
2 -
tulangan n
- 2p
- b
φ φ
= 1
5 8
. 2
- 16
5. -
40 .
2 -
200 −
= 6 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
= 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 +
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Bab 6 Perencanaan Balok A nak
= 5
298.2 344.3
+ = 325,6 mm
T = As
ada
. fy = 1004,8. 400
= 401920 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
, 401920
= 118,21 ØMn
= Ø . T d – a2 = 0,85 . 401920 325,6 – 118,212
= 9,10 × 10
7
Nmm ØMn Mu
→ Aman.. 9,10 × 10
7
Nmm 8,294×10
7
Nmm
Jadi dipakai tulangan 5 D 16 mm
¾ Daerah Lapangan : Dari Perhitungan SAP 2000 diperoleh :
Mu = 6236,71 kgm = 6,236×10
7
Nmm Mn
= φ
Mu =
8 ,
10 236
, 6
7
× = 7,80×10
7
Nmm Rn =
3 ,
3 344
00 2
10 80
, 7
d .
b Mn
2 7
2
= ×
× =
m = 53
, 23
20 .
85 ,
400 .
85 ,
= =
c f
fy
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Bab 6 Perencanaan Balok A nak
ρ = ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− fy
2.m.Rn 1
1 m
1
= 0093
, 400
3 ,
3 53
, 23
2 1
1 53
, 23
1 =
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
× ×
− −
ρ ρ
min
ρ ρ
max
Pakai tulangan tunggal Digunakan
ρ = 0,0093 As perlu =
ρ . b . d = 0,0093 × 200 × 344
= 639,84 mm
2
n =
2
.16 π
. 4
1 perlu
As
= tulangan
4 18
, 3
96 ,
200 84
, 639
≈ =
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
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Bab 6 Perencanaan Balok A nak 4D 16
∅8−150 40
20 2D 16
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
= 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..
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Bab 6 Perencanaan Balok A nak
7,48 × 10
7
Nmm 6,236×10
7
Nmm
Jadi dipakai tulangan 4 D 16 mm b
Tulangan Geser Balok anak
Dari perhitungan
SAP 2000 Diperoleh :
Vu =
9620,84 kgm = 96208,4 N
f’c = 20 Mpa
fy = 240 Mpa
d =
321 mm
Vc = 16 .
c f .b .d = 16 . 20 .200 .321
= 47851,85 N Ø Vc = 0,6 . 47851,85 N = 28711,11 N
3 Ø Vc = 3 . 28711,11 N = 86133,34 N 5 Ø Vc = 5 . 28711,11 N = 143555,55 N
Syarat tulangan geser : 3 Ø Vc Vu 5 Ø Vc : 86133,34 N 96208,4 N 143555,55 N
Jadi diperlukan tulangan geser: Ø Vsperlu = Vu – 3 Ø Vc
= 96208,4 – 86133,34 = 10075,06 N Vs
perlu = 6
, Vsp
φ =
6 ,
06 ,
10075 = 16791,77 N
Av = 2 . ¼
π 8
2
= 2 . ¼ . 3,14 . 64 = 100,48 mm
2
S = 99
, 460
77 ,
16791 321
240 48
, 100
perlu Vs
d .
fy .
Av =
× ×
= mm
S max = d2 = 2
321 = 160,5 mm
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Bab 6 Perencanaan Balok A nak
A B
C D
E F
G H
C E
2 3
9 2
3
500 500
500 400
500 500
500
2
7
2
7
Dipakai Ø 8 – 150 mm :
Vs ada = 53
, 51606
150 321
240 48
, 100
S d
. fy
. Av
= ×
× =
N
Vs ada Vs perlu
51606,53 N 16791,77 N...... aman
Jadi dipakai sengkang dengan tulangan Ø 8 – 150 mm 6.5.
Perhitungan Balok Anak as 4 A-H 6.5.1
Pembebanan Balok
Gambar 6.11. Lebar Equivalen Balok Anak as 4 A-H
a. Beban Mati qD
Pembebanan balok as 4 A-B = 4 B-C = 4 F-G = 4 G-H • Beban Plat
= 1,32 x 2 x 404 kgm
2
=1066,56 kgm • Berat dinding = 0,15 x 4,25-0,4 x 1700 kgm
2
= 981,75 kgm qD
1
= 2048,31 kgm Pembebanan balok as 4 C-C”
• Beban Plat = 1 x 2 x 404 kgm
2
= 808 kgm • Berat dinding = 0,15 x 4,25-0,4 x 1700 kgm
2
= 981,75 kgm qD
2
= 1789,75 kgm Pembebanan balok as 4 E’-F
• Beban Plat = 1 x 2 x 404 kgm
2
= 808 kgm qD
3
= 808 kgm Pembebanan balok as 4 C”-D = 4 E-E’
• Beban Plat = 2 x 0,667 x 404 kgm
2
= 538,936 kgm qD
4
= 538,936 kgm
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Bab 6 Perencanaan Balok A nak
Pembebanan balok as 4 D-E • Beban Plat
= 2 x 1,219 x 404 kgm
2
= 984,952 kgm qD
5
= 984,952 kgm b.
Beban hidup qL Beban hidup digunakan 250 kgm
2
qL
1
=1,32 x 2 x 250 kgm
2
= 660 kgm qL
2
= 1 x 2 x 250 kgm
2
= 500 kgm qL
3
= 1 x 2 x 250 kgm
2
= 500 kgm qL
4
=0,667 x 2 x 250 kgm
2
= 333,5 kgm qL
5
=1,219 x 2 x 250 kgm
2
=609,5kgm
Bidang momen:
Gambar 6.12. Bidang momen Balok Anak as 4 A-H
Bidang geser:
Gambar 6.13. Bidang momen Balok Anak as 4 A-H
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Bab 6 Perencanaan Balok A nak
6.5.2 Perhitungan Tulangan