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Perencanaan Struktur dan Rencana Anggaran Biaya Gedung Kuliah 2 lantai
BAB 6 Balok Anak
1
2 2
A B
C D
1
2 2
2
225 300
300
Beban Plat Lantai
¾ Beban Mati qd
Beban plat sendiri = 0,12. 2400 = 288 kgm
2
Spesi pasangan = 0,02. 2100 = 42 kgm
2
Beban pasir = 0,02. 1600 = 32 kgm
2
Beban keramik = 0,01. 2400 = 24 kgm
2
Plafond + penggantung = 11 + 7
= 18 kgm
2
qd = 404 kgm
2
6.2. Perhitungan Balok Anak As 2”A – D,M – P = As 3’A – D,M – P
6.2.1. Pembebanan
Gambar 6.2 lebar equivalen pembebanan balok anak As 2”A-D
a. Dimensi Balok
h = 110 . L b = 12 . h
= 110 . 300 = 12 x 30
= 30 cm = 15 cm – 20 cm dipakai 20 cm
b. Pembebanan Setiap Elemen
¾ Beban Mati qD
Pembebanan balok elemen A - B Berat sendiri balok = 0,2 x 0,3 – 0,12 x 2400 = 86,4 kgm’
Berat plat = 2 x 0,75 x 404
= 606 kgm’ qd = 692,4 kgm’
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Perencanaan Struktur dan Rencana Anggaran Biaya Gedung Kuliah 2 lantai
BAB 6 Balok Anak
qD = 894,4 kgm qL = 500 kgm
qD = 692,4 kgm qL = 375 kgm
A B
C D
225 300
300 Pembebanan balok elemen B – C = C - D
Berat sendiri balok = 0,2 x 0,3 – 0,12 x 2400 = 86,4 kgm’ Berat plat
= 2 x 1 x 404 = 808 kgm’
qd = 894,4 kgm’ ¾
Beban Hidup qL Beban hidup digunakan 250 kgm
2
Pembebanan balok elemen A – B = 250 x 2 x 0,75 = 375 kgm’ Pembebanan balok elemen B – C = C – D = 250 x 2 x 1
= 500 kgm’
Pembebanan Balok Anak
Bidang Momen
Bidang Geser
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Perencanaan Struktur dan Rencana Anggaran Biaya Gedung Kuliah 2 lantai
BAB 6 Balok Anak
300
200
d d
6.2.2. Perhitungan Tulangan
a Tulangan Lentur Balok Anak
Data Perencanaan: b = 200 mm
Ø
t
= 16
mm h = 300 mm
Ø
s
= 8
mm f’c = 25 Mpa
d’ = 40 + 8 + ½ .16 fy = 360 Mpa ulir
= 56 mm fys = 240 Mpa polos
d = h – d’
p = 40 mm = 300 – 56 = 244 mm
m =
94 ,
16 25
. 85
, 360
. 85
, =
= fc
fy
ρb =
⎟⎟⎠ ⎞
⎜⎜⎝ ⎛
+ fy fy
fc 600
600 .
. .
85 ,
β
= ⎟
⎠ ⎞
⎜ ⎝
⎛ + 360
600 600
. 85
, .
360 25
. 85
, =
0,0313 ρ
max
= 0,75 . ρb
= 0,0235
ρ
min
= 0039
, 360
4 ,
1 4
, 1
= =
fy ¾
Daerah Lapangan
Mu = 1263,03 kgm = 1,263.10
7
Nmm Perhitungan SAP Mn
= φ
Mu =
8 ,
10 .
263 ,
1
7
= 1,579.10
7
Nmm Rn =
2 7
2
244 .
00 2
10 1,579.
d .
b Mn =
= 1,326 m =
25 .
85 ,
360 .
85 ,
= c
f fy
= 16,94
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Perencanaan Struktur dan Rencana Anggaran Biaya Gedung Kuliah 2 lantai
BAB 6 Balok Anak
ρ = ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ −
− fy
2.m.Rn 1
1 m
1
= ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ −
− 360
1,326 .
16,94 .
2 1
1 16,94
1 =
0,0038 ρ
ada
ρ
min
ρ
max
Digunakan ρ
min
= 0,0039 As perlu =
ρ
min
. b . d = 0,0039. 200. 244
= 190,32 mm
2
n =
2
16 .
π .
4 1
perlu As
= 96
, 200
32 ,
190 = 0,947 ~ 2 tulangan
Dipakai tulangan 2 D 16 mm As ada = 2. ¼ .
π . 16
2
= 2 . ¼ . 3,14 . 16
2
= 401,92 mm
2
As perlu 190,32 → Aman..
a = =
b .
c f
. 0,85
fy .
ada As
200 .
25 .
0,85 360
. 401,92
= 34,04 Mn ada = As ada . fy d – a2
= 401,92. 360 244 – 34,042 = 3,284.10
7
Nmm Mn ada Mn
→ Aman..
Jadi dipakai tulangan 2 D 16 mm
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Perencanaan Struktur dan Rencana Anggaran Biaya Gedung Kuliah 2 lantai
BAB 6 Balok Anak
¾
Daerah Tumpuan
Mu = 1830,74 kgm = 1,831.10
7
Nmm Perhitungan SAP Mn
= φ
Mu =
8 ,
10 .
831 ,
1
7
= 2,289.10
7
Nmm Rn =
2 7
2
244 .
00 2
10 2,289.
d .
b Mn =
= 1,922 m =
25 .
85 ,
360 .
85 ,
= c
f fy
= 16,94 ρ =
⎟⎟⎠ ⎞
⎜⎜⎝ ⎛
− −
fy 2.m.Rn
1 1
m 1
= ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ −
− 360
1,922 .
16,94 .
2 1
1 16,94
1
= 0,0056
ρ ρ
min
ρ ρ
max
→ dipakai tulangan tunggal Digunakan
ρ = 0,0056 As perlu =
ρ . b . d = 0,0056 . 200. 244
= 273,54 mm
2
n =
2
16 .
π .
4 1
perlu As
= 96
, 200
54 ,
273 = 1,36 ~ 2 tulangan
Dipakai tulangan 2 D 16 mm As ada = 2 . ¼ .
π . 16
2
= 2 . ¼ . 3,14 . 16
2
= 401,92 mm
2
As perlu 273,54 → Aman..
a = =
b .
c f
. 0,85
fy .
ada As
200 .
25 .
0,85 360
. 401,92
= 34,04
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Perencanaan Struktur dan Rencana Anggaran Biaya Gedung Kuliah 2 lantai
BAB 6 Balok Anak
Mn ada = As ada . fy d – a2 = 401,92. 360 244 – 34,042
= 3,284.10
7
Nmm Mn ada Mn
→ Aman..
Jadi dipakai tulangan 2 D 16 mm b Tulangan Geser Balok anak
¾
Daerah Lapangan
Vu = 2483,53 kg = 24835,3 N Perhitungan SAP
f’c = 25 Mpa
fy = 240 Mpa
d = h – p – ½ Ø
= 300 – 40 – ½ 8 = 256 mm Vc
= 1 6 . c
f .b . d = 1 6 . 25 . 200. 256
= 42666,67 N Ø Vc
= 0,6 . 42666,667 N = 25600,002 N
3 Ø Vc = 3 . 25600,002 = 76800,006 N
Ø Vc Vu 3 Ø Vc 24835,3 N 25600,002 N 76800,006 N
Jadi tidak diperlukan tulangan geser s
max
= d2 = 2
256 = 128 mm ~ 125 mm
Jadi dipakai sengkang dengan tulangan Ø 8 – 125 mm
¾
Daerah Tumpuan
Vu = 3064,73 kg = 30647,3 N Perhitungan SAP
f’c = 25 Mpa
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Perencanaan Struktur dan Rencana Anggaran Biaya Gedung Kuliah 2 lantai
BAB 6 Balok Anak
30
200 30
200
2 D 16 2 D 16
2 D 16 2 D 16
8 - 125 O
8 - 125 O
POT. TUMPUAN POT. LAPANGAN
fy = 240 Mpa
d = h – p – ½ Ø
= 300 – 40 – ½ 8 = 256 mm Vc
= 1 6 . c
f .b . d = 1 6 . 25 . 200. 256
= 42666,67 N Ø Vc
= 0,6 . 42666,667 N = 25600,002 N
3 Ø Vc = 3 . 25600,002 = 76800,006 N
Ø Vc Vu 3 Ø Vc 25600,002 N 30647,3 N 76800,006 N
Jadi diperlukan tulangan geser Ø Vs
= Vu - Ø Vc = 30647,3 – 25600,002 = 5047,298 N
Vs perlu = 6
, Vs
φ =
6 ,
298 ,
5047 = 8412,163 N
Digunakan sengkang ∅ 8
Av = 2 . ¼
π 8
2
= 2 . ¼ . 3,14 . 64 = 100,48 mm
2
s = =
= 163
, 8412
256 .
240 .
48 ,
100 perlu
Vs d
. fy
. Av
733,87 mm s
max
= d2 = 2
256 = 128 mm ~ 125 mm
Jadi dipakai sengkang dengan tulangan Ø 8 – 125 mm
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Perencanaan Struktur dan Rencana Anggaran Biaya Gedung Kuliah 2 lantai
BAB 6 Balok Anak
C
3 4
3 2
2
300 300
300 300
P
1 = 6484,89 kg
P
1 = 6484,89 kg
2 2
3 4
2 2
2 2
3 2
2
2 2
300 300
300 300
C
P
1
P
1
6.3. Perhitungan Balok Anak As C2 – 4 = As N2 – 4
