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Tugas Akhir Perencanaan Struktur Kafe dan Resto 2 Lantai
BAB 6 Balok Anak
ρ
perlu
= ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ ×
× −
− 360
97 ,
1 17
2 1
1 .
17 1
= 0,006 ρ
ρ
max
ρ ρ
min
, di pakai ρ
perlu
= 0,006 As =
ρ
perlu
. b . d = 0,006. 250 . 294
= 441 mm
2
Digunakan tulangan D 16 = ¼ .
π . 16
2
= 200,96 mm
2
Jumlah tulangan =
19 ,
2 96
, 200
441 =
~ 3 buah.
Dipakai tulangan 3 D 16 mm
As ada = 3 . ¼ .
π . 16
2
= 602,88 mm
2
As ……… aman a =
250 25
85 ,
360 88
, 602
b c
f 0,85
fy ada
As ×
× ×
= ×
× ×
= 40,85 Mn ada = As ada × fy
⎟ ⎠
⎞ ⎜
⎝ ⎛ −
2 a
d
= 602,88 × 360
⎟ ⎠
⎞ ⎜
⎝ ⎛
− 2
85 ,
40 294
= 5,9375 . 10
7
Nmm Mn ada Mn ......... aman
Kontrol Spasi : S
=
1 -
n sengkang
2 -
tulangan n
- 2p
- b
φ φ
=
1 3
8 .
2 -
16 .
3 -
40 .
2 -
50 2
−
= 53 25 mm......oke
Jadi dipakai tulangan D 16 mm
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Tugas Akhir Perencanaan Struktur Kafe dan Resto 2 Lantai
BAB 6 Balok Anak Daerah Lapangan
Dari perhitungan SAP 2000 diperoleh :
Mu = 2033.83
kgm = 2,033 . 10
7
Nmm Mn =
φ Mu
=
7 7
10 .
54 ,
2 8
, 10
. 033
, 2
=
Nmm
Rn =
=
2
.d b
Mn
= ×
2 7
294 250
10 .
54 ,
2 1,17 Nmm
2
m = =
= 0,85.25
360 c
0,85.f fy
17 ρ
perlu
= ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ ×
× −
− 360
17 ,
1 17
2 1
1 .
17 1
= 0,0033 ρ
ρ
max
ρ ρ
min
, di pakai ρ
min
= 0,0038 As =
ρ. b . d = 0,0038. 250 . 294
= 279,3 mm
2
Digunakan tulangan D 16 = ¼ .
π . 16
2
= 200,96 mm
2
Jumlah tulangan =
38 ,
1 96
, 200
3 ,
279 =
~ 2 buah.
Dipakai tulangan 2 D 16 mm
As ada = 2 . ¼ .
π . 16
2
= 401,92 mm
2
As ……… aman a =
250 25
85 ,
360 92
, 401
b c
f 0,85
fy ada
As ×
× ×
= ×
× ×
= 27,23
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Tugas Akhir Perencanaan Struktur Kafe dan Resto 2 Lantai
BAB 6 Balok Anak
Mn ada = As ada × fy d -
2 a
= 401,92 × 360
⎟ ⎠
⎞ ⎜
⎝ ⎛
− 2
23 ,
27 294
= 4,0569 . 10
7
Nmm Mn ada Mn ......... aman
Kontrol Spasi : S
=
1 -
n sengkang
2 -
tulangan n
- 2p
- b
φ φ
=
1 2
8 .
2 -
16 .
2 -
40 .
2 -
50 2
−
= 122 25 mm......oke
Jadi dipakai tulangan D 16 mm
Tulangan Geser
Dari perhitungan SAP 2000 diperoleh : Vu =
3994.36 kg = 39943,6 N
f’c = 25 Mpa
fy = 360 Mpa
d = h – p – ½ Ø = 350 – 40 – ½ 12 = 304 mm
Vc = 1 6 .
c f .b .d
= 1 6 . 25 . 250 . 304 =
63333,33 N
Ø Vc = 0,75 . 63333,33 N = 47500 N
3 Ø Vc = 3 . 47500 = 142500 N
Vu Ø Vc 3 Ø Vc 39943,6 N 47500 N 142500 N
Jadi tidak perlu tulangan geser S max = d2 =
2 304
= 152 mm
Jadi dipakai sengkang dengan tulangan Ø 8 – 150 mm
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Tugas Akhir Perencanaan Struktur Kafe dan Resto 2 Lantai
BAB 6 Balok Anak 6.4.Pembebanan Balok Anak as B’
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