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Tugas Akhir
Perencanaan Struktur Bangunan Sekolahan 2 Lantai
BAB 6 Balok Anak
Rn =
=
2
.d b
Mn =
×
2 7
244 200
10 .
66 ,
1 1,39 Nmm
2
m = =
= 0,85.25
360 c
0,85.f fy
16,94
ρ
perlu
= ⎟
⎟ ⎠
⎞ ⎜
⎜ ⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ ×
× −
− 360
39 ,
1 94
, 16
2 1
1 .
94 ,
16 1
= 0,004 ρ ρ
max
ρ ρ
min
, di pakai ρ
perlu
= 0,004 As =
ρ. b . d = 0,004 . 200 . 244
= 195,2 mm
2
Digunakan tulangan D 16 = ¼ .
π . 16
2
= 200,96 mm
2
Jumlah tulangan =
97 ,
96 ,
200 2
, 195
= ~ 2 buah.
As ada = 2 . ¼ .
π . 16
2
= 401,92 mm
2
As ……… aman a =
200 25
85 ,
360 92
, 401
b c
f 0,85
fy ada
As ×
× ×
= ×
× ×
= 34,04 Mn ada = As ada × fy d – a2
= 401,92 × 360 244 – 34,042 = 3,2842 . 10
7
Nmm Mn ada Mn ......... aman
Jadi dipakai tulangan 2 D 16 mm
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125
Tugas Akhir
Perencanaan Struktur Bangunan Sekolahan 2 Lantai
BAB 6 Balok Anak Tulangan
Geser
Dari perhitungan SAP 2000 diperoleh : Vu
= 2389,75 kg = 23897,5 N f’c
= 25 Mpa fy
= 360 Mpa d
= h – p – ½ Ø = 300 – 40 – ½ 12 = 254 mm Vc
= 1 6 . c
f .b .d = 1 6 . 25 . 200 . 254
= 42333,33
N Ø Vc = 0,75 . 42333,33 N
= 31750 N 3 Ø Vc = 3 . 31750
= 95250 N Vu Ø Vc 3 Ø Vc
23897,5 N 31750 N 95250 N Jadi tidak di perlukan tulangan geser
S max = d2 = 2
244 = 122 mm
Jadi dipakai sengkang dengan tulangan Ø 8 - 100
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Tugas Akhir
Perencanaan Struktur Bangunan Sekolahan 2 Lantai
BAB 6 Balok Anak
2
6.3. Pembebanan Balok Anak as 1’ A-I
6.3.1. Pembebanan
Gambar 6.3. Lebar Equivalen Balok Anak as 1’
Perencanaan Dimensi Balok : h = 112 . Ly
= 112 . 3750 = 312,5 mm = 400 mm
b = 23 . h = 23 . 400
= 266,66 mm h dipakai = 400 mm, b = 300 mm 1. Beban Mati q
D
Pembebanan balok as 1’ B – C Berat sendiri = 0,3 x 0,4 – 0,12 x 2400 kgm
3
=201,6 kgm Beban Plat
= 0,79 + 1,18 x 411 kgm
2
=809,67 kgm qD =1011,27 kgm
2. Beban hidup q
L
Beban hidup digunakan 250 kgm
2
qL = 0,79 + 1,18 x 250 kgm
2
= 492,5 kgm 3. Beban reaksi
Beban reaksi a = b = 3186,33 kg 4. Beban berfaktor q
U
q
U
= 1,2. q
D
+ 1,6. q
L
= 1,2 x 1011,27 + 1,6 x 492,5 = 2001,52 kgm
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Tugas Akhir
Perencanaan Struktur Bangunan Sekolahan 2 Lantai
BAB 6 Balok Anak
6.3.2. Perhitungan Tulangan Tulangan Lentur Balok Anak
Data Perencanaan : h = 400 mm
Ø
t
= 16
mm b = 300 mm
Ø
s
= 8 mm p = 40 mm
d = h - p - 12 Ø
t
- Ø
s
fy = 360 Mpa = 400 – 40 - 12.16 - 8
f’c = 25 MPa = 344
Tulangan Lentur Daerah Lapangan
ρb = ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ + fy
600 600
fy c.
β 0,85.f
= ⎟
⎠ ⎞
⎜ ⎝
⎛ + 360
600 600
85 ,
360 25
. 85
, = 0,03136
ρ
max
= 0,75 . ρb
= 0,75 . 0,03136 =
0,02352 ρ
min
= 00389
, 360
4 ,
1 4
, 1
= =
fy
Daerah Tumpuan Dari perhitungan SAP 2000 diperoleh :
Mu = 5343,77 kgm = 5,34.10
7
Nmm Mn =
φ Mu
=
7 7
10 .
675 ,
6 8
, 10
. 34
, 5
= Nmm
Rn =
=
2
.d b
Mn =
×
2 7
344 300
10 .
675 ,
6 1,88 Nmm
2
m = =
= 0,85.25
360 c
0,85.f fy
16,94
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Tugas Akhir
Perencanaan Struktur Bangunan Sekolahan 2 Lantai
BAB 6 Balok Anak
ρ
perlu
= ⎟
⎟ ⎠
⎞ ⎜
⎜ ⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ ×
× −
− 360
88 ,
1 94
, 16
2 1
1 .
94 ,
16 1
= 0,00548 ρ ρ
max
ρ ρ
min
, di pakai ρ
perlu
= 0,00548 As =
ρ. b . d = 0,00548. 300 . 344
= 565,54 mm
2
Digunakan tulangan D 16 = ¼ .
π . 16
2
= 200,96 mm
2
Jumlah tulangan =
81 ,
2 96
, 200
54 ,
565 =
~ 3 buah. As ada
= 3 . ¼ . π . 16
2
= 602,88 mm
2
As ……… aman a =
300 25
85 ,
360 88
, 602
b c
f 0,85
fy ada
As ×
× ×
= ×
× ×
= 34,04 Mn ada = As ada × fy d -
2 a
= 602,88 × 360 344 - 2
04 ,
34 = 7,0966692 . 10
7
Nmm Mn ada Mn ......... aman
Jadi dipakai tulangan 3 D 16 mm Daerah Lapangan
Dari perhitungan SAP 2000 diperoleh :
Mu = 4410,97 kgm = 4,41 . 10
7
Nmm Mn =
φ Mu
=
7 7
10 .
512 ,
5 8
, 10
. 41
, 4
= Nmm
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Tugas Akhir
Perencanaan Struktur Bangunan Sekolahan 2 Lantai
BAB 6 Balok Anak
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