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210 Tugas Akhir
Perencanaan Struktur Gedung Swalayan 2 Lantai
Bab 6 Balok Anak
6.4. Pembebanan Balok Anak as 4 A D
6.4.1. Pembebanan
Gambar 6.4 Lebar Equivalen Balok Anak as
4 A D
Perencanaan Dimensi Balok h = 112 . L
= 112 . 3000 = 250 mm = 300 mm
b = 23 . h = 23 . 300
= 200 mm = 250 mm h dipakai = 300 mm, b = 250 mm 1. Beban Mati q
D
Pembebanan balok 2 9
Berat sendiri = 0,25x0,30 0,12 x 2400 kgm
3
= 108 kgm Beban plat
= 2 x 0,96 x 411 kgm
2
= 789,12 kgm qD =897,12 kgm
2. Beban hidup q
L
Beban hidup digunakan 250 kgm
2
qL = 2 x 0,96 x 250 kgm
2
= 480 kgm
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3. Beban berfaktor q
U
qU = 1,2. qD + 1,6. qL
= 1,2 . 897,12 + 1,6.480 = 1844,55 kgm
6.4.2. Perhitungan Tulangan
a. Tulangan Lentur Balok Anak Data Perencanaan :
h = 300 mm Ø
t
= 16 mm b = 250 mm
Ø
s
= 8 mm p = 40 mm
d = h - p - 12 Ø
t
- Ø
s
fy = 390 Mpa = 300 40 - 12.16 - 8
= 25 Mpa = 244
Tulangan Lentur Daerah Lapangan
b =
fy 600
600 fy
c. 0,85.f
= 390
600 600
85 ,
390 25
. 85
, = 0,028
max
= 0,75 . b = 0,75 . 0,028
= 0,021
min
= 0036
, 390
4 ,
1 4
, 1
fy
Daerah Tumpuan Dari perhitungan SAP 2000 diperoleh :
Mu = 1542,50
kgm = 1,5425. 10
7
Nmm Mn
= Mu
=
7 7
10 .
93 ,
1 8
, 10
. 1,5425
Nmm
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212
Rn =
2
.d b
M n
2 7
244 250
10 .
93 ,
1
1,30 Nmm
2
m =
0,85.25 390
c 0,85.f
fy 18,35
perlu
= fy
Rn .
m 2
1 1
. m
1
= 390
30 ,
1 35
, 18
2 1
1 .
35 ,
18 1
= 0,0035
max min
, di pakai
min
= 0,0036 As
= . b . d = 0,0036 . 250 . 244
= 219,6 mm
2
Digunakan tulangan D 16 = ¼ . . 16
2
= 200,96 mm
2
Jumlah tulangan =
10 ,
1 96
, 200
219,6 ~ 2 buah.
Dipakai 2 D 16 mm
As ada = 2 . ¼ . . 16
2
= 401,92 mm
2
Aman
a =
250 25
85 ,
390 92
, 401
b c
f 0,85
fy ada
As = 29,51
Mn ada = As ada × fy d -
2 a
= 401,92 × 360 294 -
2 51
, 29
= 4,38 . 10
7
Nmm Mn ada Mn ......... Aman
Jadi dipakai tulangan 2 D 16
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213
Daerah Lapangan Dari perhitungan SAP 2000 diperoleh :
Mu = 1383,85
kgm = 1,38385. 10
7
Nmm Mn
= Mu
=
7 7
10 .
73 ,
1 8
, 10
. 1,38385
Nmm
Rn =
2
.d b
M n
2 7
244 250
10 .
73 ,
1
1,16 Nmm
2
m =
0,85.25 390
c 0,85.f
fy 18,35
perlu
= fy
Rn .
m 2
1 1
. m
1
= 390
16 ,
1 35
, 18
2 1
1 .
35 ,
18 1
= 0,0031
max min
, di pakai
min
= 0,0036 As
= . b . d = 0,0036 . 250 . 244
= 219,6 mm
2
Digunakan tulangan D 16 = ¼ . . 16
2
= 200,96 mm
2
Jumlah tulangan =
10 ,
1 96
, 200
219,6 ~ 2 buah.
Dipakai 2 D 16 mm
As ada = 2 . ¼ . . 16
2
= 401,92 mm
2
Aman
a =
250 25
85 ,
390 92
, 401
b c
f 0,85
fy ada
As = 29,51
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214
Mn ada = As ada × fy d -
2 a
= 401,92 × 360 294 -
2 51
, 29
= 4,38 . 10
7
Nmm Mn ada Mn ......... Aman
Jadi dipakai tulangan 2 D 16 Tulangan Geser
Dari perhitungan SAP 2000 diperoleh : Vu
= 3331,20 kg = 33312 N
= 20 Mpa fy
= 240 Mpa d
= h p ½ Ø Ø
s
= 350 40 ½ 16 8 = 294 mm Vc
Vc = 1 6 .
c f .b .d
= 1 6 . 25 . 250 . 294 = 61250 N
Ø Vc = 0,75 . 61250 N = 45937,5 N
3 Ø Vc = 3 . 41087,7525 N = 123263,2575 N
½ Ø Vc Vu Ø Vc 22968,75 33312 N 41087,7525 N
Jadi di perlukan tulangan geser minimum, Ø Vs
= 13 b.d = 0,75.13.250.294
= 18375 N Vs perlu =
75 ,
Vs =
75 ,
18375 = 24500 N
Av = 2 . ¼ 8
2
= 2 . ¼ . 3,14 . 64 = 100,48 mm
2
S =
3824 ,
289 24500
294 240
48 ,
100 perlu
Vs d
. fy
. Av
mm
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215 Tugas Akhir
Perencanaan Struktur Gedung Swalayan 2 Lantai
Bab 6 Balok Anak S max = d2 =
2 294
= 147 ~ 150 mm 600 mm
Jadi dipakai sengkang dengan tulangan Ø 8 150 mm
6.5. Pembebanan Balok Anak as 6
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