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3 3
3 7
3
10 8
7 3
1 1
2 3
4 4
5
200 400
200 300
300 200
6.1.2. Lebar Equivalent Balok Anak
Tabel 6.1. Hitungan Lebar equivalent
No. Ukuran Plat
m
2
Lx m
Ly m
Leq segitiga
Leq trapesium
1. 2 × 5
2 5
- 0,947
2. 3 × 5
3 5
- 1,32
3. 2 × 3
2 3
0,667 0,852
4. 1,5 × 1,5
1,5 1,5
0,5 -
5. 2 × 2
2 2
0,667 -
6. 1 × 1,5
1 1,5
0,333 0,426
7. 3 × 3
3 3
1 -
8. 2 × 4
2 4
- 0,917
9. 3 × 4
3 4
1 1,219
10. 1,5 × 4
1,5 4
0,5 0,715
6.2. Perhitungan Balok Anak as
C” 1-5
6.2.1 Pembebanan
Gambar 6.2. Lebar Equivalen Balok Anak as C’’ 1-5
a. Beban Mati qD
Pembebanan balok elemen as C”1-1’ = C’’4’-5 Beban Plat = 2 x 0,667 x 404 kgm
2
= 538,936 kgm qD
1
=538,936 kgm
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Pembebanan balok as C” 1’-2’ • Beban Plat
= 0,917 + 0,715 x 404 kgm
2
= 659,328 kgm • Berat dinding = 0,15 x 4,25-0,3 x 1700 kgm
2
= 1007,25 kgm qD
2
=1666,578 kgm Pembebanan balok as C’’ 2’-3
• Beban Plat = 2 x 0,667 x 404 kgm2
= 538,936 kgm • Berat dinding = 0,15 x 4,25-0,3 x 1700 kgm2 = 1007,25 kgm
qD
3
= 1546,186 kgm Pembebanan balok as C’’ 3-4’
• Beban Plat = 1 + 0,852 x 404 kgm2
= 748,208 kgm • Berat dinding = 0,15 x 4,25-0,3 x 1700 kgm2 = 1007,25 kgm
qD
4
= 1755,458 kgm Pembebanan balok as C’’ 3-4’
• Beban Plat = 1 + 0,852 x 404 kgm2
= 748,208 kgm qD
5
= 748,208 kgm b.
Beban hidup qL Beban hidup digunakan 250 kgm
2
qL
1
= 2 x 0,667 x 250 kgm = 333,5 kgm
qL
2
= 0,917 + 0,715 x 250 kgm = 408 kgm
qL
3
= 2 x 0,667 x 250 kgm = 333,5 kgm
qL
4
= 1 + 0,852 x 250 kgm = 463 kgm
qL
5
= 1 + 0,852 x 250 kgm = 463 kgm
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Bidang momen:
Gambar 6.3. Bidang Momen Balok Anak as C’’ 1-5
Bidang geser:
Gambar 6.4. Bidang Geser Balok Anak as C’’ 1-5
6.2.2
Perhitungan Tulangan
a
Tulangan Lentur Balok Anak
Data Perencanaan
: h =
300 mm
D
t
= 13
mm b =
150 mm
Ø
s
= 8 mm p
= 40 mm d
= h - p - 12 D
t
- Ø
s
fy = 400 Mpa = 300 – 40 – ½ . 13– 8
f’c = 20 MPa = 245,5 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
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ρ
min
= 0035
, 400
4 ,
1 4
, 1
= =
fy
¾ Daerah Tumpuan : Dari Perhitungan SAP 2000 diperoleh :
Mu = 2789,54 kgm = 2,789 × 10
7
Nmm Mn
= φ
Mu =
8 ,
10 789
, 2
7
× = 3,50 × 10
7
Nmm Rn =
8 ,
3 245,5
50 1
10 50
, 3
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
= 0109
, 400
8 ,
3 53
, 23
2 1
1 53
, 23
1 =
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
× ×
− −
ρ ρ
min
ρ ρ
max
Pakai tulangan tunggal Digunakan
ρ = 0,0109
As perlu = ρ . b . d
= 0,0109 × 150 × 245,5 = 401,39mm
2
n =
2
.13 π
. 4
1 perlu
As
= tulangan
4 02
, 3
665 ,
132 39
, 401
≈ =
As ada = n . ¼ . π . d
2
= 4 . ¼ . π . 13
2
= 530,66 As perlu → Aman..
a =
= b
c f
fy Asada
. .
85 ,
. 24
, 83
150 20
85 ,
400 66
, 530
= ×
× ×
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4D 13
2D 13 ∅8−100
30
15
Mn ada = As ada . fy d – a2 = 530,66 . 400 245,5 – 83,242
= 4,3×10
7
Nmm Mn ada Mn
→ Aman..
Jadi dipakai tulangan 4 D 13 mm
Kontrol Spasi : S
= 1
- n
sengkang 2
- tulangan
n -
2p -
b φ
φ
= 1
4 8
. 2
- 13
4. -
40 .
2 -
50 1
− = 0,67mm 25 mm tulangan 2 lapis
d
1
= h - p - 12 D
t
- Ø
s
= 300 – 40 – ½ . 13– 8 = 245,5 mm
d
2
= h - p - Ø
t
- 12 D
t
– s - Ø
s
= 300 – 40 – 13 – ½ 13 – 30 - 8 = 202,5 mm
d = n
. d
2 1
n d
n +
= 4
202,5.2 245,5.2
+ = 224 mm
T = As
ada
. fy = 530,66. 400
= 212264 Mpa C = 0,85 . f’c . a . b
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T = C
As . fy = 0,85 . f’c . a . b a =
b c
f fy
As .
. 85
, .
= 150
. 20
. 85
, 212264
= 83,24 ØMn
= Ø . T d – a2 = 0,85 . 212264 245,5 – 83,242
= 3,68 × 10
7
Nmm ØMn Mu
→ Aman.. 3,68 × 10
7
Nmm 2,789 × 10
7
Nmm
Jadi dipakai tulangan 4 D 13 mm
¾ Daerah Lapangan : Dari Perhitungan SAP 2000 diperoleh :
Mu = 2516,84 kgm = 2,516×10
7
Nmm Mn
= φ
Mu =
8 ,
10 516
, 2
7
× = 3,15×10
7
Nmm Rn =
48 ,
3 245,5
50 1
10 15
, 3
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
= 0098
, 400
48 ,
3 53
, 23
2 1
1 53
, 23
1 =
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
× ×
− −
ρ ρ
min
ρ ρ
max
Pakai tulangan tunggal Digunakan
ρ = 0,0098
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4D 13 ∅8−100
30
15 2D 13
As perlu = ρ
. b . d = 0,0098 × 150 × 245
= 360,88 mm
2
n =
2
.13 π
. 4
1 perlu
As
= tulangan
4 7
, 2
665 ,
132 88
, 360
≈ =
As ada = n . ¼ . π . d
2
= 4 . ¼ . π . 13
2
= 530,66 As perlu → Aman..
a =
= b
c f
fy Asada
. .
85 ,
. 24
, 83
150 20
85 ,
400 66
, 530
= ×
× ×
Mn ada = As ada . fy d – a2 = 530,66 . 400 245,5 – 83,242
= 4,3×10
7
Nmm Mn ada Mn
→ Aman..
Jadi dipakai tulangan 4 D 13 mm
Kontrol Spasi : S
= 1
- n
sengkang 2
- tulangan
n -
2p -
b φ
φ
= 1
4 8
. 2
- 13
4. -
40 .
2 -
50 1
− = 0,67mm 25 mm tulangan 2 lapis
d
1
= h - p - 12 D
t
- Ø
s
= 300 – 40 – ½ . 13– 8 = 245,5 mm
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d
2
= h - p - Ø
t
- 12 D
t
– s - Ø
s
= 300 – 40 – 13 – ½ 13 – 30 - 8 = 202,5 mm
d = n
. d
2 1
n d
n +
= 4
202,5.2 245,5.2
+ = 224 mm
T = As
ada
. fy = 530,66. 400
= 212264 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
, .
= 150
. 20
. 85
, 212264
= 83,24 ØMn
= Ø . T d – a2 = 0,85 . 212264 245,5 – 83,242
= 3,68 × 10
7
Nmm ØMn Mu
→ Aman.. 3,68 × 10
7
Nmm 2,516×10
7
Nmm
Jadi dipakai tulangan 4 D 13 mm
b
Tulangan Geser Balok anak
Dari perhitungan
SAP 2000 Diperoleh :
Vu = 5306,41 kgm = 53064,1 N
f’c = 20 Mpa
fy = 240 Mpa
d = 244 mm
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Vc = 16 .
c f .b .d = 16 .
.150 .244 = 27280,03 N Ø Vc = 0,6 . 27280,03 N = 16368,02 N
3 Ø Vc = 3 . 16368,02 = 49104,05 N 5 Ø Vc = 5 . 16368,02 = 81840,10 N
Syarat tulangan geser : 3 Ø Vc Vu 5 Ø Vc : 49104,05 N 53064,1 N 81840,10
N Jadi diperlukan tulangan geser:
ØVs perlu = Vu – Ø Vc = 53064,1 N - 49104,05 N = 3959,6 N
Vs perlu =
6 ,
Vsperlu φ
= 6
, 59,6
9 3
= 6599,33 N Av
= 2 . ¼ π 8
2
= 2 . ¼ . 3,14 . 64 = 100,48 mm
2
S = 62
, 891
33 ,
6599 244
240 48
, 100
perlu Vs
d .
fy .
Av =
× ×
= mm
S max = d2 = 2
244 = 122 mm
Dipakai Ø 8 – 100 mm :
Vs ada = 088
, 58841
100 244
240 48
, 100
S d
. fy
. Av
= ×
× =
N
Vs ada Vs perlu
82594,56 N 6599,33 N...... aman
Jadi dipakai sengkang dengan tulangan Ø 8 – 100 mm
20
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C C
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6.3. Perhitungan Balok Anak As 1’’C-C’
