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Perencanaan Struktur Factory Outlet dan Resto 2 Lantai
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m = 17
25 0,85
360 c
0,85.f fy
= ×
=
ρ = ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− fy
2.m.Rn 1
1 m
1
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ ×
× −
− 360
85 ,
2 17
2 1
1 17
1 = 0,0085
ρ ρ
min
ρ ρ
max
→ dipakai tulangan tunggal Digunakan
ρ = 0,0085 As perlu =
ρ. b . d = 0,0085 × 300 × 440,5
= 1123,27 mm
2
Digunakan tulangan D 19 n
= 385
, 283
27 ,
1123 19
. 4
1 perlu
As
2
= π
= 3,964
≈ 4 tulangan As’ =
2
19 .
4 1 π
=
2
19 .
14 ,
3 4
1 = 283,385 mm
As = 4 × 283,385 = 1133,54 mm
2
As’ As………………….aman Ok
Dipakai tulangan 4 D 19
Kontrol Spasi : S =
1 -
n sengkang
2 -
tulangan n
- 2p
- b
φ φ
= 1
4 10
. 2
- 19
. 4
- 40
. 2
- 00
3 −
= 120 25 mm......oke
Jadi dipakai tulangan D 19 mm
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BAB 7 Portal Daerah Lapangan
Dari Perhitungan SAP 2000 diperoleh momen terbesar pada batang nomor 278.
Mu = 11375,55 = 11,375 × 10
7
Nmm Mn
= φ
Mu =
8 ,
10 375
, 11
7
× = 14,22 × 10
7
Nmm Rn =
44 ,
2 440,5
300 10
14,22 d
. b
Mn
2 7
2
= ×
× =
m = 17
25 0,85
360 c
0,85.f fy
= ×
= ρ =
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
− −
fy 2.m.Rn
1 1
m 1
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ ×
× −
− 360
44 ,
2 17
2 1
1 17
1 = 0,0072
ρ ρ
min
ρ ρ
max
→ dipakai tulangan tunggal Digunakan
ρ
perlu
= 0,0072 As perlu =
ρ. b . d = 0,0072 × 300 × 440,5
= 951,48 mm
2
Digunakan tulangan D 19 n
= 385
, 283
48 ,
951 19
. 4
1 perlu
As
2
= π
= 3,357 ≈ 4 tulangan
As’ =
2
19 .
4 1 π
=
2
19 .
14 ,
3 4
1 = 283,385 mm
As ada = 4 × 283,385 = 1133,54 mm
2
As’ As………………….aman Ok
Dipakai tulangan 4 D 19
commit to user
Tugas Akhir
Perencanaan Struktur Factory Outlet dan Resto 2 Lantai
BAB 7 Portal
Kontrol Spasi : S =
1 -
n sengkang
2 -
tulangan n
- 2p
- b
φ φ
= 1
4 10
. 2
- 19
. 4
- 40
. 2
- 00
3 −
= 120 25 mm......oke
Jadi dipakai tulangan D 19 mm 7.6.4.
Perhitungan Tulangan Geser Balok Portal Melintang Dari Perhitungan SAP 2000 diperoleh momen terbesar pada batang nomor 278:
Vu = 13607,10
kg = 136071,0 N
f’c = 25 Mpa
fy = 360 Mpa
d = 440,5
Vc = 1 6 .
c f .b .d
= 1 6 . 25 .300.440,5
= 110125 N φ Vc
= 0,75 . 110125 = 82593,75 N ½ Ø Vc
= 0,5 . 82593,75 = 41296,87 N 3
φ Vc = 3 . 82593,75 = 247781,25 N Syarat tulangan geser : Ø Vc Vu 3Ø Vc
: 82593,75 N 136071,0 N 247781,25 N Jadi diperlukan tulangan geser
Ø Vs = Vu – Ø Vc = 136071,0 - 82593,75
= 53477,25 N Vs perlu =
75 ,
53477,25 75
, =
Vs φ
= 71303 N Av
= 2 . ¼ π 10
2
= 2 . ¼ . 3,14 . 100 = 157 mm
2
S = 17
, 349
71303 5
, 440
. 360
. 157
perlu Vs
d .
fy .
Av =
= mm
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Tugas Akhir
Perencanaan Struktur Factory Outlet dan Resto 2 Lantai
BAB 7 Portal
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