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Perencanaan Struktur dan Rencana Anggaran Biaya Gedung Kuliah 2 lantai
BAB 6 Balok Anak
7 6
6 3
7 6
D
5
200 200
250 40
250 40
8 - 160 O
8 - 170 O
2 D 19
2 D 19 2 D 19
3 D 19
POT. TUMPUAN POT. LAPANGAN
= 2 . ¼ . 3,14 . 64 = 100,48 mm
2
s = =
= 51804
356 .
240 .
48 ,
100 perlu
Vs d
. fy
. Av
165,72 mm ~ 160 mm s
max
= d2 = 2
356 = 178 mm ~ 180 mm
Jadi dipakai sengkang dengan tulangan Ø 8 – 160 mm
6.5. Perhitungan Balok Anak As D’ 5 – 6 = As 2’ E – F , K – L
6.5.1 Pembebanan
Gambar 6.5 lebar equivalen pembebanan balok anak As D’5-6
a. Dimensi Balok h = 110 . L
b = 12 . h = 110 . 400
= 12 . 40 = 40 cm
= 20 – 25 cm dipakai 25 cm
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Perencanaan Struktur dan Rencana Anggaran Biaya Gedung Kuliah 2 lantai
BAB 6 Balok Anak
6 3
5
200 200
qD = 1705,12 kgm qL = 320 kgm
b. Pembebanan Setiap Elemen ¾
Beban Mati qD Berat sendiri balok = 0,25 x 0,4 – 0,12 x 2400 = 168 kgm’
Berat plat = 0,61 x 0,67 x 404
= 517,12 kgm’ Berat dinding
= 0,15 x 4 x 1700 = 1020 kgm’
qd = 1705,12 kgm
’
¾ Beban Hidup qL
Beban hidup digunakan 250 kgm
2
qL = 250 x 0,61 + 0,67 = 320 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
400
250
d
d
6.5.2 Perhitungan Tulangan
a Tulangan Lentur Balok Anak
Data Perencanaan: b = 250 mm
Ø
t
= 16
mm h = 400 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 = 400 – 56 = 344 mm
m =
9412 ,
16 25
. 85
, 360
. 85
, =
= fc
fy
ρb =
⎟⎟⎠ ⎞
⎜⎜⎝ ⎛
+ fy fy
fc 600
600 .
. .
85 ,
β
= ⎟
⎠ ⎞
⎜ ⎝
⎛ + 360
600 600
. 85
, .
360 25
. 85
, =
0,03136 ρ
max
= 0,75 . ρb
= 0,0235
ρ
min
= 0039
, 360
4 ,
1 4
, 1
= =
fy ¾
Daerah Lapangan
Mu = 657,44 kgm = 0,657.10
7
Nmm Perhitungan SAP Mn
= φ
Mu =
8 ,
10 .
567 ,
7
= 0,709.10
7
Nmm Rn =
2 7
2
344 .
50 2
10 0,709.
d .
b Mn =
= 0,240 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
0,240 .
16,94 .
2 1
1 16,94
1
= 0,00067
ρ
ada
ρ
min
ρ
max
Digunakan ρ
min
= 0,0039 As perlu =
ρ
min
. b . d = 0,0039. 250. 344
= 335,4 mm
2
n =
2
16 .
π .
4 1
perlu As
= 96
, 200
4 ,
335 = 1,669 ~ 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 335,4 → 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 = 1243,27 kgm = 1,243.10
7
Nmm Perhitungan SAP Mn
= φ
Mu =
8 ,
10 .
243 ,
1
7
= 1,554.10
7
Nmm Rn =
2 7
2
344 .
50 2
10 1,554.
d .
b Mn =
= 0,525 m =
25 .
85 ,
360 .
85 ,
= c
f fy
= 16,94 ρ =
⎟⎟⎠ ⎞
⎜⎜⎝ ⎛
− −
fy 2.m.Rn
1 1
m 1
= ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ −
− 360
0,525 .
16,94 .
2 1
1 16,94
1
= 0,0015
ρ
ada
ρ
min
ρ
max
Digunakan ρ
min
= 0,0039 As perlu =
ρ
min
. b . d = 0,0056 . 250. 344
= 335,4 mm
2
n =
2
16 .
π .
4 1
perlu As
= 96
, 200
4 ,
335 = 1,669 ~ 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 Mn ada = As ada . fy d – a2
= 401,92. 360 244 – 34,042
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Perencanaan Struktur dan Rencana Anggaran Biaya Gedung Kuliah 2 lantai
BAB 6 Balok Anak
250 250
40 40
2 D 16 2 D 16
8 - 170 O
8 - 170 O
2 D 16 2 D 16
POT. TUMPUAN POT. LAPANGAN
= 3,284.10
7
Nmm Mn ada Mn
→ Aman..
Jadi dipakai tulangan 2 D 16 mm b Tulangan Geser Balok anak
Vu = 3179,78 kg = 31797,8 N Perhitungan SAP
f’c = 25 Mpa
fy = 240 Mpa
d = h – p – ½ Ø
= 400 – 40 – ½ 8 = 356 mm Vc
= 1 6 . c
f .b . d = 1 6 . 25 . 250. 356
= 74166,667 N Ø Vc
= 0,6 . 74166,667 N = 44500 N
3 Ø Vc = 3 . 44500 = 133500 N
Vu Ø Vc 3 Ø Vc
31797,8 N 44500 N 133500 N → tidak perlu tulangan geser
s
max
= d2 = 2
356 = 178 mm ~ 170 mm
Jadi dipakai sengkang dengan tulangan Ø 8 – 170 mm
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Perencanaan Struktur dan Rencana Anggaran Biaya Gedung Kuliah 2 lantai
BAB 6 Balok Anak
4
5
E D
D
4
5
3
P
1
150 250
6.6. Perhitungan Balok Anak As 3’’ D – E =As E’ 2 – 3 , As K’ 2 – 3
