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Tugas A khir
Perencanaan Struktur Gedung Factory Outlet 2 Lantai
Bab 5 Perencanaan Plat
Tinggi efektif
Gambar 5.6. Perencanaan Tinggi Efektif
dx = h – d’ - ½ Ø
= 120 – 20 – 5 = 95 mm dy
= h – d’ – Ø - ½ Ø = 120 – 20 - 10 - ½ . 10 = 85 mm
untuk plat digunakan ρb
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ + fy
fy fc
600 600
. .
. 85
, β
=
⎟ ⎠
⎞ ⎜
⎝ ⎛
+ 240 600
600 .
85 ,
. 240
25 .
85 ,
= 0,0538
ρ
max
= 0,75 . ρb
= 0,0403
ρ
min
= 0,0025 untuk pelat
5.2.3. Penulangan lapangan arah x
Mu =304,55 kgm = 3,04.10
6
Nmm Mn =
φ Mu
=
6 6
80 .
3 8
, 10
. 04
, 3
= Nmm
Rn =
=
2
.d b
Mn =
2 6
95 .
1000 80
. 3
0,42 Nmm
2
m =
29 ,
11 25
. 85
, 240
. 85
, =
= c
f fy
h d y
d x d
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Tugas A khir
Perencanaan Struktur Gedung Factory Outlet 2 Lantai
Bab 5 Perencanaan Plat
ρ
perlu
= ⎟
⎟ ⎠
⎞ ⎜
⎜ ⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− 240
42 ,
. 29
, 11
. 2
1 1
. 29
, 11
1 = 0,0017
ρ ρ
max
ρ ρ
min
, di pakai ρ
min
= 0,0025
As = ρ
min
. b . d = 0,0025. 1000 . 95
= 237,5 mm
2
Digunakan tulangan D 10 = ¼ .
π . 10
2
= 78,5 mm
2
Jumlah tulangan =
02 ,
3 5
, 78
5 ,
237 =
~ 4 buah. Jarak tulangan dalam 1 m
1
= 250
4 1000 =
mm ~ 240 mm Jarak maksimum
= 2 x h = 2 x 120 = 240 mm As yang timbul
= 4. ¼ . π.10
2
= 314 237,5 As …OK Dipakai tulangan
∅ 10 – 240 mm
5.2.4. Penulangan lapangan arah y
Mu = 184,15 kgm = 1,8415.10
6
Nmm
Mn = φ
Mu =
6 6
10 .
301875 ,
2 8
, 10
. 8415
, 1
= Nmm
Rn =
=
2
.d b
Mn =
2 6
85 .
1000 10
. 301875
, 2
0,318 Nmm
2
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Tugas A khir
Perencanaan Struktur Gedung Factory Outlet 2 Lantai
Bab 5 Perencanaan Plat
m = 29
, 11
25 .
85 ,
240 .
85 ,
= =
c f
fy
i
ρ
perlu
=
⎟ ⎟
⎠ ⎞
⎜ ⎜
⎝ ⎛
− −
× fy
Rn m
m .
. 2
1 1
1
= .
29 ,
11 1
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
− −
240 318
, .
29 ,
11 .
2 1
1 = 0,0013
ρ ρ
max
ρ ρ
min
, di pakai ρ
min
= 0,0025 As =
ρ
min
b . d = 0,0025 . 1000 . 85
= 212,51 mm
2
Digunakan tulangan ∅ 10
= ¼ . π . 10
2
= 78,5 mm
2
Jumlah tulangan =
71 ,
2 5
, 78
5 ,
212 =
~ 4 buah. Jarak tulangan dalam 1 m
1
= 250
4 1000 =
mm ~ 240 mm. Jarak maksimum
= 2 x h = 2 x 120 = 240 mm As yang timbul
= 4. ¼. π.10
2
= 314 212,51 As….OK Dipakai tulangan
∅ 10 – 240 mm
5.2.5. Penulangan tumpuan arah x
Mu = 729,5 kgm = 7,29.10
6
Nmm
Mn = φ
Mu =
= 8
, 10
. 29
, 7
6
9,1125.10
6
Nmm Rn
= =
2
.d b
Mn =
2 6
85 .
1000 10
. 1125
, 9
1,26 Nmm
2
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Tugas A khir
Perencanaan Struktur Gedung Factory Outlet 2 Lantai
Bab 5 Perencanaan Plat
m = 29
, 11
25 .
85 ,
240 .
85 ,
= =
c f
fy
ρ
perlu
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= .
29 ,
11 1
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
− −
240 26
, 1
. 29
, 11
. 2
1 1
= 0,0054 ρ ρ
max
ρ ρ
min
, di pakai ρ
perlu
= 0,0054 As
= ρ
perlu
. b . d = 0,0054 . 1000 . 85
= 459 mm
2
Digunakan tulangan D 10 = ¼ .
π . 10
2
= 78,5 mm
2
Jumlah tulangan =
84 ,
5 5
, 78
459 = ~ 6 buah.
Jarak tulangan dalam 1 m
1
= 66
, 166
6 1000 =
mm ~ 120 mm Jarak maksimum
= 2 x h = 2 x 120 = 240 mm As yang timbul
= 6. ¼. π.10
2
= 392,5 459 As ….OK Dipakai tulangan
∅ 10 – 120 mm
5.2.6. Penulangan tumpuan arah y
