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Tugas A khir
Perencanaan Struktur Gedung Factory Outlet 2 Lantai
Bab 5 Perencanaan Plat
Digunakan tulangan ∅ 8
As = ¼ . π . 8
2
= 50,24
mm
2
S =
perlu
As b
As .
= 190
1000 .
24 ,
50 = 264,42 ~ 200 mm Smax = 2h
n = s
b
= 42
, 264
1000 = 3,8 buah ~ 4 buah
As yang timbul = 4. ¼ . π . 8
2
= 200,9
mm
2
As
perlu
…..…ok Dipakai tulangan
∅8 – 200 mm
5.3.5. Penulangan Lapangan Arah Y
Mu = 74,68 kgm = 0,7468.10
6
Nmm Mn =
φ Mu
=
6 6
10 .
9335 ,
8 ,
10 .
7468 ,
= Nmm
Rn =
=
2
.dx b
Mn =
2 6
76 .
1000 10
. 9335
, 0,16 Nmm
2
m = 294
, 11
25 .
85 ,
240 .
85 ,
= =
c f
fy
ρ
perlu
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− 240
16 ,
. 294
, 11
. 2
1 1
. 294
, 11
1
= 0,0006
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Tugas A khir
Perencanaan Struktur Gedung Factory Outlet 2 Lantai
Bab 5 Perencanaan Plat
ρ ρ
max
ρ ρ
min
, di pakai ρ
min
As
perlu
= ρ
min
. b . dx = 0,0025 . 1000 . 76
= 190 mm
2
Digunakan tulangan ∅ 8
As = ¼ . π . 8
2
= 50,24
mm
2
S =
perlu
As b
As .
= 190
1000 .
24 ,
50 = 264,42 ~ 200 mm Smax = 2h
n = s
b
= 42
, 264
1000 = 3,8 buah ~ 4 buah
As yang timbul = 4. ¼ . π . 8
2
= 200,9
mm
2
As
perlu
…..…ok Dipakai tulangan
∅8 – 200 mm
5.3.6. Penulangan Tumpuan Arah X
Mu = 307,71 kgm = 3,0771.10
6
Nmm
Mn = φ
Mu =
= 8
, 10
. 3,0771
6
3,8463.10
6
Nmm Rn
= =
2
.dx b
Mn =
2 6
76 .
1000 10
. 8463
, 3
0,66 Nmm
2
m = 2942
, 11
25 .
85 ,
240 .
85 ,
= =
c f
fy
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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
= . 2942
, 11
1 ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− 240
66 ,
. 2942
, 11
. 2
1 1
= 0,0027 ρ
ρ
max
ρ ρ
min
, di pakai ρ
perlu
As
perlu
= ρ
min
. b . dx = 0,0027 . 1000 . 76
= 205,2 mm
2
Digunakan tulangan ∅ 8
As = ¼ . π . 8
2
= 50,24
mm
2
S =
perlu
As b
As .
= 2
, 205
1000 .
24 ,
50 = 244,83 ~ 200 mm Smax = 2h
n = s
b
= 42
, 244
1000 = 4,09 buah ~ 5 buah
As yang timbul = 5. ¼ . π . 8
2
= 251,2
mm
2
As
perlu
…..…ok Dipakai tulangan
∅8 – 200 mm
5.3.7. Penulangan Tumpuan Arah Y
Mu = 230,03 kgm =2,3003.10
6
Nmm
Mn = φ
Mu =
= 8
, 10
. 3003
, 2
6
2,8753.10
6
Nmm
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134
Tugas A khir
Perencanaan Struktur Gedung Factory Outlet 2 Lantai
Bab 5 Perencanaan Plat
Rn =
=
2
.dx b
Mn =
2 6
76 .
1000 10
. 8753
, 2
0,49 Nmm
2
m = 2942
, 11
25 .
85 ,
240 .
85 ,
= =
c f
fy
ρ
perlu
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= . 2942
, 11
1 ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− 240
49 ,
. 2942
, 11
. 2
1 1
= 0,0020 ρ
ρ
max
ρ ρ
min
, di pakai ρ
min
As
perlu
= ρ
min
. b . dx = 0,0025 . 1000 . 76
= 190 mm
2
Digunakan tulangan ∅ 8
As = ¼ . π . 8
2
= 50,24
mm
2
S =
perlu
As b
As .
= 190
1000 .
24 ,
50 = 264,42 ~ 200 mm Smax = 2h
n = s
b
= 200
1000 = 5 buah
As yang timbul = 5. ¼ . π . 8
2
= 251,2
mm
2
As
perlu
…..…ok Dipakai tulangan
∅8 – 200 mm
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135
Tugas A khir
Perencanaan Struktur Gedung Factory Outlet 2 Lantai
Bab 5 Perencanaan Plat
5.3.8. Rekapitulasi Tulangan
