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Tugas Akhir Perencanaan St rukt ur Rencana Anggararn Biaya Gedung K uliah 2 L ant ai
BAB 6 Balok Anak
7 7
7 7
7 7
7 7
7 7
7 7
7 7
7 7
6 6
6 6
8 9
8 9
7 7
4 4
6.7 Pembebanan Balok Anak as 3 A-H 6.7.1 Pembebanan
Gambar 6.7. Lebar Equivalen Balok Anak as 3 A – H
Perencanaan Dimensi Balok : h = 112. Ly
= 112. 6000 = 500 mmh dipakai = 500 mm b = 23 . h
= 23 . 500 = 333,33 mm
≈ 350h dipakai = 500 mm, b = 350 mm
1. Beban Mati q
D
Pembebanan balok as 3 A – B = 3’ B – C = 3’ F – G = 3’ G – H Beban Reaksi = R
A’
= R
B’
= R
F’
= R
F’
= 10082,71 kg Berat sendiri = 0,35 x 0,50 – 0,12 x 2400 kgm
3
= 319,2 kgm Beban Plat = 2 2 x Leq7 x 411 kgm
2
= 2 2 x 1 x 411 kgm
2
= 1644 kgm
qD
1
= 1963,2 kgm Pembebanan balok as 3 C – D = 3 E – F
Beban Reaksi R
C’
= R
E”’
= 5856,94 kg R
C’’’
= R
E’
= 15586,00 kg Berat sendiri = 0,35 x 0,50 – 0,12 x 2400 kgm
3
= 319,2 kgm Beban Plat = Leq8+Leq9+ Leq4+22Leq6
= 0,5+0,83+1,33+220,67x411 kgm
2
= 2194,74 kgm qD
2
= 2513,94 kgm
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BAB 6 Balok Anak
Pembebanan balok as 3 D – E Beban Reaksi
= R
D’
= 3619,12 kg Beban Reaksi Tangga = 7106,53 kg
Berat sendiri = 0,35 x 0,50 – 0,12 x 2400 kgm
3
= 319,2 kgm Beban Plat = 2 x Leq7
= 2 x 1 x 411 kgm
2
= 822 kgm qD
3
= 1141,2 kgm 2. Beban hidup q
L
Beban hidup digunakan 250 kgm
2
qL
1
= 2 2 x Leq7 x 250 kgm
2
= 2 2 x 1 x 250 kgm
2
= 1000 kgm qL
2
= Leq8+Leq9+ Leq4+22Leq6 = 0,5+0,83+1,33+220,67x250 kgm
2
= 1335 kgm qL
3
= 2 x Leq7 = 2 x 1 x 250 kgm
2
= 500 kgm 3. Beban berfaktor q
U
qU
1
= 1,2. q
D
+ 1,6. q
L
= 1,2 x 1963,2 + 1,6 x 1000 = 3955,84 kgm
qU
2
= 1,2. q
D
+ 1,6. q
L
= 1,2 x 2513,94 + 1,6 x 1335 = 5152,73 kgm
qU
3
= 1,2. q
D
+ 1,6. q
L
= 1,2 x 1141,2 + 1,6 x 500 = 2169,44 kgm
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BAB 6 Balok Anak
6.7.2. Perhitungan Tulangan
Tulangan Lentur Balok Anak Data Perencanaan :
h = 500 mm Ø
t
= 22
mm b = 350 mm
Ø
s
= 10 mm p = 40 mm
d = h - p - 12 Ø
t
- Ø
s
fy = 360 MPa = 500 – 40 - 12.22 - 10
f’c = 30 MPa = 439 mm
Tulangan Lentur Daerah Lapangan
ρb = ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ + fy
600 600
fy c.
β 0,85.f
= ⎟
⎠ ⎞
⎜ ⎝
⎛ + 360
600 600
85 ,
360 30
. 85
, = 0,038
ρ
max
= 0,75 . ρb
= 0,75 . 0,038 =
0,0285 ρ
min
= 00389
, 360
4 ,
1 4
, 1
= =
fy
Daerah Tumpuan Dari perhitungan SAP 2000 diperoleh :
Mu = 27422.88
kgm 27,42 . 10
7
Nmm
Mn = φ
Mu =
7 7
10 .
27 ,
34 8
, 10
. 42
, 27
= Nmm
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BAB 6 Balok Anak
Rn =
=
2
.d b
Mn =
×
2 7
439 350
10 .
27 ,
34 5,08Nmm
2
m = =
= 0,85.30
360 c
0,85.f fy
14,12 ρ
perlu
= ⎟
⎟ ⎠
⎞ ⎜
⎜ ⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ ×
× −
− 360
08 ,
5 12
, 14
2 1
1 .
12 ,
14 1
= 0,0158 ρ ρ
max
ρ ρ
min
, di pakai ρ
perlu
= 0,0158 As =
ρ. b . d = 0,0158. 350 . 439
= 2427,67 mm
2
Digunakan tulangan D 22 = ¼ .
π . 22
2
= 379,94 mm
2
Jumlah tulangan =
38 ,
6 94
, 379
2427,67 = ~ 7 buah.
Kontrol:
As ada = 7 . ¼ .
π . 22
2
= 2659,58 mm
2
As ada As ≈ 2659,58 mm
2
2427,67 mm
2
……… aman a =
350 30
85 ,
360 2659,58
b c
f 0,85
fy ada
As ×
× ×
= ×
× ×
= 107,23 Mn ada = As ada × fy d -
2 a
= 2659,58 × 360 439 - 2
23 ,
107 = 36,89 . 10
7
Nmm Mn ada Mn
36,89 . 10
7
Nmm 34,27 . 10
7
Nmm ......... aman
Jadi dipakai tulangan 7 D 22 mm
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BAB 6 Balok Anak
Kontrol spasi tulangan :
Cek jarak =
1 -
n t
n -
s 2
- p
2 -
b φ
φ
1 -
7 7.22
- 2.10
- 2.40
- 350
=
= 25 mm 16 mm dipakai tulangan 7 D22 dua lapis Di pakai d
d1 = 439
mm d2
= d1 – s – 2 x ½ Ø = 439 – 30 – 2 x ½.22
= 398
mm d’ x 7
= d1 x 4 + d2 x 3 d
7 3
x 398
4 x
439 +
=
= 421,428
mm Mn ada = As ada . fy d – a2
= 2659,58. 360 421,428 – 107,232 =
35,22.10
7
Nmm Mn ada Mn
≈ 35,22.10
7
Nmm 34,27 . 10
7
Nmm
→
Aman..
Daerah Lapangan Dari perhitungan SAP 2000 diperoleh :
Mu = 25250.43
kgm = 25,25 . 10
7
Nmm
Mn = φ
Mu =
7 7
10 .
56 ,
31 8
, 10
. 25
, 25
= Nmm
Rn =
=
2
.d b
Mn =
×
2 7
439 350
10 .
56 ,
31 4,67 Nmm
2
m = =
= 0,85.30
360 c
0,85.f fy
14,12 ρ
perlu
= ⎟
⎟ ⎠
⎞ ⎜
⎜ ⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ ×
× −
− 360
67 ,
4 12
, 14
2 1
1 .
12 ,
14 1
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BAB 6 Balok Anak
= 0,014 ρ ρ
max
ρ ρ
min
, di pakai ρ
max
= 0,0144 As =
ρ. b . d = 0,0144. 350 . 439
= 2212,56 mm
2
Digunakan tulangan D 22 = ¼ .
π . 22
2
= 379,94 mm
2
Jumlah tulangan =
8 ,
5 94
, 379
2212,56 = ~ 7 buah.
Kontrol:
As ada = 7 . ¼ .
π . 22
2
= 2659,58 mm
2
As ada As ≈ 2659,58 mm
2
2212,56 mm
2
……… aman a =
350 30
85 ,
360 2659,58
b c
f 0,85
fy ada
As ×
× ×
= ×
× ×
= 107,28 Mn ada = As ada × fy d -
2 a
= 2659,58 × 360 439 - 2
28 ,
107 = 36,89. 10
7
Nmm Mn ada Mn
36,89. 10
7
Nmm 31,56 . 10
7
Nmm ......... aman
Jadi dipakai tulangan 7 D 22 mm
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Tugas Akhir Perencanaan St rukt ur Rencana Anggararn Biaya Gedung K uliah 2 L ant ai
BAB 6 Balok Anak
Kontrol spasi tulangan :
Cek jarak =
1 -
n t
n -
s 2
- p
2 -
b φ
φ
1 -
7 7.22
- 2.10
- 2.40
- 350
=
= 25 mm 16 mm dipakai tulangan 7 D22 dua lapis Di pakai d
d1 = 439
mm d2
= d1 – s – 2 x ½ Ø = 439 – 30 – 2 x ½.22
= 398
mm d’ x 7
= d1 x 4 + d2 x 3 d
7 3
x 398
4 x
439 +
=
= 421,43
mm Mn ada = As ada . fy d – a2
= 2659,58. 360 421,43 – 107,282 =
40,35.10
7
Nmm Mn ada Mn
≈ 40,35.10
7
Nmm 31,56 . 10
7
Nmm
→
Aman..
Tulangan Geser
Dari perhitungan SAP 2000 diperoleh : Vu =
29145.13 kg = 291451,3N
f’c = 30 Mpa
fy = 360 Mpa
d = 540,5
mm Vc
= 1 6 . c
f .b .d = 1 6 . 30 . 350 . 440,5
= 140741,87
N Ø Vc = 0,75 . 140741,87 N
= 105556,40 N
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BAB 6 Balok Anak
3 Ø Vc = 3 . 105556,40 N = 316669,22 N
Ø Vc Vu 3 Ø Vc 105556,40 N 291451,3N 316669,22 N
Syarat tulangan geser : Ø Vc Vu 3 Ø Vc Jadi diperlukan tulangan geser
Ø Vs = Vu - Ø Vc
= 291451,3N – 105556,40 N = 185894,9 N Vs perlu =
6 ,
Vs φ
= 6
, 9
, 185894
= 309824,83 N Digunakan sengkang
∅ 10 Av
= 2 . ¼ π 10
2
= 2 . ¼ . 3,14 . 100 = 157 mm
2
s = =
= 83
, 309824
5 ,
440 .
240 .
157 perlu
Vs d
. fy
. Av
53,57 mm ~ 50 mm s
max
= d2 = 2
5 ,
440 = 220,25 mm ~ 220 mm
Jadi dipakai sengkang dengan tulangan Ø 10 – 50 mm
Vs ada =
S d
fy Av
. .
= 50
439 240
157 ×
× = 330830,04 N
Vs ada Vs perlu 330830,04 N 309824,83 N........Aman
Jadi, dipakai sengkang ∅ 10 – 50 mm
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BAB 6 Balok Anak
7 7
7 7
7 7
7 7
7 7
7 7
7 7
7 7
7 4
4 6
6 6
6 4
4 7
7 7
6.8. Pembebanan Balok Anak as 5 A - H
