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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
6.5.1. Pembebanan
Gambar 6.5 Lebar Equivalen Balok Anak as 6
Perencanaan Dimensi Balok h = 112 . L
= 112 . 3500 = 291,67 mm = 300 mm
b = 23 . h = 23 . 300
= 200 mm h dipakai = 300 mm, b = 250 mm 1. Beban Mati q
D
Pembebanan balok B 5
Berat sendiri = 0,20x0,30 0,12 x 2400 kgm
3
= 86,4 kgm Beban plat
= 2 x 1,04 x 411 kgm
2
= 854,88 kgm qD = 941,28 kgm
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2. Beban hidup q
L
Beban hidup digunakan 250 kgm
2
qL = 2 x 1,04 x 250 kgm
2
= 520 kgm 3. Beban berfaktor q
U
qU = 1,2. qD + 1,6. qL
= 1,2 . 941,28 + 1,6.520 = 1961,54 kgm
6.5.2. 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 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
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Daerah Tumpuan Dari perhitungan SAP 2000 diperoleh :
Mu = 2990,21
kgm = 2,99021 . 10
7
Nmm Mn
= Mu
=
7 7
10 .
74 ,
3 8
, 10
. 2,99021
Nmm
Rn =
2
.d b
M n
2 7
244 250
10 .
74 ,
3
2,51 Nmm
2
m =
0,85.25 390
c 0,85.f
fy 18,35
perlu
= fy
Rn .
m 2
1 1
. m
1
= 390
51 ,
2 35
, 18
2 1
1 .
35 ,
18 1
= 0,0069
max min
, di pakai
perlu
= 0,0069 As
= . b . d = 0,0069 . 250 . 244
= 420,9 mm
2
Digunakan tulangan D 16 = ¼ . . 16
2
= 200,96 mm
2
Jumlah tulangan =
09 ,
2 96
, 200
9 ,
420 ~3 buah.
Dipakai 2 D 16 mm
As ada = 3 . ¼ . . 16
2
= 602,88 mm
2
Aman
a =
250 25
85 ,
390 88
, 602
b c
f 0,85
fy ada
As = 44,26
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Mn ada = As ada × fy d -
2 a
= 602,88× 390 244 -
2 26
, 44
= 5,22 . 10
7
Nmm Mn ada Mn ......... Aman
Jadi dipakai tulangan 3 D 16 Daerah Lapangan
Dari perhitungan SAP 2000 diperoleh :
Mu = 1711
kgm = 1,711. 10
7
Nmm Mn
= Mu
=
7 7
10 .
14 ,
2 8
, 10
. 1,711
Nmm
Rn =
2
.d b
M n
2 7
244 250
10 .
14 ,
2
1,44 Nmm
2
m =
0,85.25 390
c 0,85.f
fy 18,35
perlu
= fy
Rn .
m 2
1 1
. m
1
= 390
44 ,
1 35
, 18
2 1
1 .
35 ,
18 1
= 0,0038
max min
, di pakai
perlu
= 0,0038 As
=
min
. b . d = 0,0038 . 250 . 244
= 231,8 mm
2
Digunakan tulangan D 16 = ¼ . . 16
2
= 200,96 mm
2
Jumlah tulangan =
15 ,
1 96
, 200
8 ,
231 ~ 2 buah.
Dipakai 2 D 16 mm
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As ada = 2 . ¼ . . 16
2
= 401,91 mm
2
Aman a
= 250
25 85
, 390
91 ,
401 b
c f
0,85 fy
ada As
= 29,51 Mn ada = As ada × fy d -
2 a
= 401,91 × 390 244 -
2 29,51
= 3,6 . 10
7
Nmm Mn ada Mn ......... Aman
Jadi dipakai tulangan 2 D 16 Tulangan Geser
Dari perhitungan SAP 2000 diperoleh : Vu
= 4345,62 kg = 43456,2 N
= 25 Mpa fy
= 240 Mpa d
= h p ½ Ø Ø
s
= 300 40 ½ 16 8 = 244 mm Vc
= 1 6 . c
f .b .d = 1 6 . 25 . 250 . 244
= 50833,34 N Ø Vc = 0,75 . 50833,34 N
= 38125,00 N 3 Ø Vc = 3 . N
= 114375,00 N ½ Ø Vc Vu Ø Vc
19062,5 43456,2 N 114375,00 N Jadi di perlukan tulangan geser minimum,
Ø Vs = 13 b.d
= 0,75.13.250.244 = 15250 N
Vs perlu = 75
, Vs
= 75
, 15250
= 20333,34 N
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Av = 2 . ¼ 8
2
= 2 . ¼ . 3,14 . 64 = 100,48 mm
2
S =
3823 ,
289 20333,34
244 240
48 ,
100 perlu
Vs d
. fy
. Av
mm
S max = d2 = 2
244 = 122 ~ 150 mm 600 mm
Jadi dipakai sengkang dengan tulangan Ø 8 150 mm
6.6. Pembebanan Balok Anak as A 4 - 6
