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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
6.8.1. Pembebanan
Gambar 6.3. Lebar Equivalen Balok Anak as 5 A – H
Perencanaan Dimensi Balok : h = 112 . Ly
= 110 . 6000 =
500 mm
b = 23 . h = 23 . 600
= 333,33 mm ≈ 350 mm h dipakai = 500 mm, b = 350 mm
1. Beban Mati q
D
Pembebanan balok as 5 A – H
Pembebanan balok as 5 A – B = 5 B – C = 5 D– E = 5 F – G = 5 G – H
Beban reaksi R
A’
= R
B’
= R
F’
= R
g’
= 10082,71 kg R
D’
= 9159,63 kg Berat sendiri = 0,35 x 0,50 – 0,12 x 2400 kgm
3
= 319,2 kgm Beban Plat
= 22xLeq 7x 411 kgm
2
= 22x1,0x 411 kgm
2
= 1644 kgm qD
1
= 1963,2 kgm
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BAB 6 Balok Anak
Pembebanan balok as 5 C – D = 5 E– F Beban reaksi R
C
’’’ = 17724,89 kg Berat sendiri = 0,35 x 0,50 – 0,12 x 2400 kgm
3
= 319,2 kgm Beban Plat
= 2 x Leq4+2x Leq6 x 411 =
2x1,33+2x0,67x411kgm
2
=1644 kgm qD
2
= 1963,2 kgm 2. Beban hidup q
L
Beban hidup digunakan 250 kgm
2
qL
1
= 22xLeq 7 x 250 = 22x1,0 x 250 kgm
2
= 1000 kgm qL
2
= 2 x Leq4+2x Leq6 x 250 = 2x1,33+2x0,67x250 kgm
2
= 1000 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 1963,2 + 1,6 x 1000 = 3955,84 kgm.
6.8.2. Perhitungan Tulangan
Tulangan Lentur Balok Anak Data Perencanaan :
h = 500 mm Ø
t
= 19
mm b = 350 mm
Ø
s
= 10 mm p = 40 mm
d = h - p - 12 Ø
t
- Ø
s
fy = 360 MPa = 500 – 40 - 12.19 - 10
f’c = 30 MPa = 440,5
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BAB 6 Balok Anak
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 = 25886.05
kgm = 25,88 . 10
7
Nmm
Mn = φ
Mu =
7 7
10 .
35 ,
32 8
, 10
. 88
, 25
= Nmm
Rn =
=
2
.d b
Mn =
×
2 7
439 350
10 .
35 ,
32 4,79 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
79 ,
4 12
, 14
2 1
1 .
12 ,
14 1
= 0,0148
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BAB 6 Balok Anak
ρ ρ
max
ρ ρ
min
, di pakai ρ
perlu
= 0,0148 As =
ρ. b . d = 0,0148. 350 . 439
= 2274,02 mm
2
Digunakan tulangan D 22 = ¼ .
π . 22
2
= 379,94 mm
2
Jumlah tulangan =
9 ,
5 94
, 379
2274,02 = ~ 7 buah.
Kontrol:
As ada = 7 . ¼ .
π . 22
2
= 2659,58 mm
2
As ada As ≈ 2659,58 mm
2
2274,02 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 32,35 . 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 = 440,5 – 30 – 2 x ½.22
= 398mm
d’ x 7 = d1 x 4 + d2 x 3
d
7 3
x 398
4 x
439 +
=
= 418,72
mm Mn ada = As ada . fy d – a2,
= 2659,58. 360 418,72 – 91,442 =
35,72.10
7
Nmm Mn ada Mn
≈ 35,72.10
7
Nmm 32,35 . 10
7
Nmm
→
Aman..
Daerah Lapangan Dari perhitungan SAP 2000 diperoleh :
Mu = 22909,93 kgm = 22,90 . 10
7
Nmm
Mn = φ
Mu =
7 7
10 .
625 ,
28 8
, 10
. 90
, 22
= Nmm
Rn =
=
2
.d b
Mn =
×
2 7
439 350
10 .
625 ,
28 4,2 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
2 ,
4 12
, 14
2 1
1 .
12 ,
14 1
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BAB 6 Balok Anak
= 0,0128 ρ ρ
max
ρ ρ
min
, di pakai ρ
perlu
= 0,0128 As =
ρ. b . d = 0,0128. 350 . 439
= 1966,72 mm
2
Digunakan tulangan D 22 = ¼ .
π . 22
2
= 379,94 mm
2
Jumlah tulangan =
1 ,
5 94
, 379
72 ,
1966 =
~ 6 buah.
Kontrol :
As ada = 6. ¼ .
π . 22
2
= 2279,64 mm
2
As 1966,72 mm
2
……… aman a =
350 30
85 ,
360 2279,64
b c
f 0,85
fy ada
As ×
× ×
= ×
× ×
= 91,95 Mn ada = As ada × fy d -
2 a
= 2279,64 × 360 439 - 2
95 ,
91 =
= 32,25.10
7
Nmm Mn ada Mn
32,25.10
7
Nmm 28,625.10
7
Nmm......... aman
Jadi dipakai tulangan 6 D 22 mm Kontrol spasi tulangan :
Cek jarak =
1 -
n t
n -
s 2
- p
2 -
b φ
φ
1 -
6 6.22
- 2.10
- 2.40
- 350
=
25 mm23,6 mm dipakai tulangan 8D19 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 8
= d1 x 4 + d2 x 2
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BAB 6 Balok Anak
d
6 2
x 398
4 x
439 +
=
= 425,33
mm Mn ada = As ada . fy d – a2
= 2279,64. 360 425,33 – 91,952 =
31,13.10
7
Nmm Mn ada Mn
≈ 31,13.10
7
Nmm 28,625 . 10
7
Nmm
→
Aman..
Tulangan Geser
Dari perhitungan SAP 2000 diperoleh : Vu
= 26522.25 kg = 265222,5 N
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
= 105556,406
N 3 Ø Vc = 3 . 148022,02 N
= 316669,22 N Ø Vc Vu 3 Ø Vc
105556,406 N 265222,5 N 316669,22 N
Syarat tulangan geser : Ø Vc Vu 3 Ø Vc Jadi diperlukan tulangan geser
Ø Vs = Vu - Ø Vc
= 265222,5 N – 105556,406 N = 159666,09 N Vs perlu =
6 ,
Vs φ
= 6
, 09
, 159666
= 266110,16 N
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BAB 6 Balok Anak
1 1
1 1
1 1
1 1
1 1
1 1
1 1
Digunakan sengkang ∅ 10
Av = 2 . ¼
π 10
2
= 2 . ¼ . 3,14 . 100 = 157 mm
2
s = =
= 16
, 266110
5 ,
440 .
240 .
157 perlu
Vs d
. fy
. Av
62,37 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
5 ,
440 240
157 ×
× = 331960,8 N
Vs ada Vs perlu 331960,8 N 266110,16 N........Aman
Jadi, dipakai sengkang ∅ 10 – 50 mm
6.9. Balok anak as 7 A– H
