commit to user
Tugas Akhir
Perencanaan St rukt ur Gedung Sekolah 2 L ant ai
BAB 5 Plat Lant ai 132
5.7. Penulangan tumpuan arah y
Mu = 679,320 kgm = 6,7932x10
6
Nmm Mn =
φ Mu
=
6 6
10 492
, 8
8 ,
10 7932
, 6
x x
= Nmm
Rn =
=
2
.dy b
Mn 175
, 1
85 .
1000 10
492 ,
8
2 6
= x
Nmm
2
m = 294
, 11
25 .
85 ,
240 .
85 ,
= =
c f
fy
ρ
perlu
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− 240
175 ,
1 .
294 ,
11 .
2 1
1 .
294 ,
11 1
= 0,00504 ρ
ρ
max
ρ ρ
min
, di pakai ρ
perlu
= 0,00504 As
perlu
= ρ
perlu
. b . d = 0,00504 . 1000 . 85
= 428,5 mm
2
Digunakan tulangan ∅ 10
As = ¼ . π . 10
2
= 78,5
mm
2
Jumlah tulangan dalam 1m
1
= 5
. 78
428,5 = 5,458 ~ 6 buah
Jarak tulangan dalam 1m
1
= 6
1000 = 166,67mm ~ 160mm
Jarak maksimum tulangan = 2 x h = 2 x 120 = 240mm
As yang timbul = 6. ¼ .
π.10
2
= 471 mm
2
As
perlu
…ok Dipakai tulangan
∅ 10 – 160 mm
commit to user
Tugas Akhir
Perencanaan St rukt ur Gedung Sekolah 2 L ant ai
BAB 5 Plat Lant ai 133
5.8. Penulangan lapangan arah x
Mu = 361,746 kgm = 3,6175x10
6
Nmm Mn =
φ Mu
=
6 6
10 523
, 4
8 ,
10 6175
, 3
x x
= Nmm
Rn =
=
2
.dx b
Mn =
2 6
95 .
1000 10
523 ,
4 x
0,501 Nmm
2
m = 294
, 11
25 .
85 ,
240 .
85 ,
= =
c f
fy
ρ
perlu
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− 240
501 ,
. 294
, 11
. 2
1 1
. 294
, 11
1
= 0,00211
ρ ρ
max
ρ ρ
min
, di pakai ρ
min
= 0,0025 As =
ρ
min
. b . dx = 0,0025. 1000 . 95
= 237,5
mm
2
Digunakan tulangan ∅ 10
As = ¼ . π . 10
2
= 78,5
mm
2
Jumlah tulangan dalam 1m
1
= 5
. 78
237,5 = 3,025 ~ 4 buah
Jarak tulangan dalam 1m
1
= 4
1000 = 250mm
Jarak maksimum tulangan = 2 x h = 2 x 120 = 240mm
As yang timbul = 4. ¼ .
π.10
2
= 314 mm
2
As
perlu
…ok Dipakai tulangan
∅ 10 – 240 mm
commit to user
Tugas Akhir
Perencanaan St rukt ur Gedung Sekolah 2 L ant ai
BAB 5 Plat Lant ai 134
5.9. Penulangan lapangan arah y
Mu = 225,086 kgm = 2,251x10
6
Nmm
Mn = φ
Mu =
6 6
10 769
, 2
8 ,
10 251
, 2
x x
= Nmm
Rn =
=
2
.dy b
Mn =
2 6
85 .
1000 10
769 ,
2 x
0,383 Nmm
2
m = 294
, 11
25 .
85 ,
240 .
85 ,
= =
c f
fy
i
ρ
perlu
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− fy
Rn m
m .
. 2
1 1
. 1
= . 294
, 11
1 ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− 240
383 ,
. 294
, 11
. 2
1 1
= 0,00161 ρ
ρ
max
ρ ρ
min
, di pakai ρ
min
= 0,0025 As =
ρ
min
b . d = 0,0025 . 1000 . 85
= 212,5 mm
2
Digunakan tulangan ∅ 10
As = ¼ . π . 10
2
= 78,5
mm
2
Jumlah tulangan dalam 1m
1
= 5
. 78
212,5 = 2,70 ~ 3 buah
Jarak tulangan dalam 1m
1
= 3
1000 = 333,333mm
Jarak maksimum tulangan = 2 x h = 2 x 120 = 240mm
As yang timbul = 3. ¼ .
π.10
2
= 235,5mm
2
As
perlu
…ok Dipakai tulangan
∅ 10 – 240 mm
commit to user
Tugas Akhir
Perencanaan St rukt ur Gedung Sekolah 2 L ant ai
BAB 5 Plat Lant ai 135
5.10. Rekapitulasi Tulangan
