commit to user
200 Tugas Akhir
Perencanaan Struktur Gedung Swalayan 2 Lantai
Bab 6 Balok Anak 7
3,75 x 3,50 3,50
3,75 1,24
- 8
3,75 × 3,75 3,75
3,75 -
1,25 9
3,50 x 2,50 2,50
3,50 -
0,83 10
2,50 x 1,875 1,875
2,50 0,76
-
6.2 Pembebanan Balok Anak
as 2 B-D -H
6.2.1. Pembebanan
Gambar 6.2 Lebar Equivalen Balok Anak as 2
B-D -H
Perencanaan Dimensi Balok h = 112 . L
= 112 . 2000
= 166,33 mm ~ 250 mm
b = 23 . h = 23 . 200
= 133,3 mm ~ 200 h dipakai 250 mm, b = 200 mm
1. Beban Mati q
D
Pembebanan balok as 2 B-D -H
Berat sendiri = 0,20x0,25 0,12x1x2400 kgm
3
= 62,4 kgm Beban plat
= 2 x 0,66 x 411 kgm
2
=542,52 kgm Beban dinding
= 0,15 x 2 x 4 x 1700 =2040 kgm +
qD =2644,92 kgm
commit to user
201
2. Beban hidup q
L
Beban hidup digunakan 250 kgm
2
qL = 2 x 0,66 x 250 kgm
2
= 330 kgm 3. Beban berfaktor q
U
qU = 1,2. qD + 1,6. qL
= 1,2. 2644,92 + 1,6.330 = 3701,90 kgm
6.2.2. Perhitungan Tulangan
a. Tulangan Lentur Balok Anak Data Perencanaan :
h = 250 mm Ø
t
= 16 mm b = 200 mm
Ø
s
= 8 mm p = 40 mm
d = h - p - 12 Ø
t
- Ø
s
fy = 390 Mpa = 250 40 - 12.16 - 8
= 25 MPa = 194 mm
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
commit to user
202
Daerah Tumpuan Dari perhitungan SAP 2000 diperoleh :
Mu = 1751,69 kgm = 1,75169 . 10
7
Nmm Mn
= Mu
=
7 7
10 .
20 ,
2 8
, 10
. 1,75169
Nmm Rn
=
2
.d b
M n
2 7
194 200
10 .
20 ,
2 2,92 Nmm
2
m =
0,85.25 390
c 0,85.f
fy 18,35
perlu
= fy
Rn .
m 2
1 1
. m
1
= 390
2,92 35
, 18
2 1
1 .
35 ,
18 1
= 0,0081
max min
, di pakai perlu = 0,024 As
=
perlu
. b . d = 0,0081. 200 . 194 = 314,28 mm
2
Digunakan tulangan D 16 = ¼ . . 16
2
= 200,96 mm
2
Jumlah tulangan =
56 ,
1 96
, 200
314,28 ~ 2 buah.
Dipakai 2 D 16
As ada = 2 . ¼ . . 16
2
= 401,92 mm
2
Aman a
= 200
25 85
, 390
401,92 b
c f
0,85 fy
ada As
= 36,88
commit to user
203
Mn ada = As ada × fy d -
2 a
= 401,92 × 390 194 -
2 36,88
= 2,75 . 10
7
Nmm Mn ada Mn ......... Aman ,
Jadi dipakai tulangan 2 D 16 Daerah Lapangan
Dari perhitungan SAP 2000 diperoleh :
Mu = 991,84 kgm = 0,99184. 10
7
Nmm Mn
= Mu
=
7 7
10 .
24 ,
1 8
, 10
. 0,99184
Nmm Rn
=
2
.d b
M n
2 7
194 200
10 .
24 ,
1 1,65Nmm
2
m =
0,85.25 390
c 0,85.f
fy 18,35
perlu
= fy
Rn .
m 2
1 1
. m
1
= 390
65 ,
1 35
, 18
2 1
1 .
35 ,
18 1
= 0,0044
max min
, di pakai
perlu
= 0,0044 As
=
perlu
. b . d = 0,0044 . 200 . 194
= 170,72 mm
2
Digunakan tulangan D 16 = ¼ . . 16
2
= 200,96 mm
2
Jumlah tulangan =
85 ,
96 ,
200 170,72
~ 2 buah.
Dipakai 2 D 16
commit to user
204
As ada = 2 . ¼ . . 16
2
= 401,91 mm
2
Aman
a =
200 25
85 ,
390 91
, 401
b c
f 0,85
fy ada
As = 36,88
Mn ada = As ada × fy d -
2 a
= 401,91 × 390 194 -
2 36,88
= 2,75 . 10
7
Nmm Mn ada Mn ......... Aman
Jadi dipakai tulangan 2 D 16 Tulangan Geser
Dari perhitungan SAP 2000 diperoleh : Vu
= 4611,23 kg = 46112,3 N
= 25 Mpa fy
= 240 Mpa d
= h p ½ Ø Ø
s
=250 40 ½ 16 8 = 194 mm Vc
= 1 6 . c
f .b .d = 1 6 . 25 . 200 . 194
= 32333,33 N Ø Vc = 0,75 . 32333,33 N
= 24250 N 3 Ø Vc = 3 . 24250 N
= 72750 N Ø Vc Vu 3 Vc
24250 46112,3 N 72750 N Jadi di perlukan tulangan geser
Vs perlu = Vu Vc = 46112,3 32333,33
= 13778,97 N
commit to user
205
Av = 2 . ¼ 8
2
= 2 . ¼ . 3,14 . 64 = 100,48 mm
2
S =
52 ,
339 13778,97
194 240
48 ,
100 perlu
Vs d
. fy
. Av
mm
S max = d2 = 2
194 = 97 ~ 100 mm 600 mm
Jadi dipakai sengkang dengan tulangan Ø 8 100 mm
6.3. Pembebanan Balok Anak
