commit to user
114
Tugas Akhir
Perencanaan Struktur Bangunan Sekolahan 2 Lantai
BAB 5 Plat Lantai 5.5. Penulangan lapangan arah x
Mu = 345,67 kgm = 3,46.10
6
Nmm Mn =
φ Mu
=
6 6
10 .
33 ,
4 8
, 10
. 46
, 3
= Nmm
Rn =
=
2
.d b
Mn =
2 6
95 .
1000 10
. 33
, 4
0,48 Nmm
2
m =
29 ,
11 25
. 85
, 240
. 85
, =
= c
f fy
ρ
perlu
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− 240
48 ,
. 29
, 11
. 2
1 1
. 29
, 11
1 = 0,00202
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 ∅ 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 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 – 250 mm
commit to user
115
Tugas Akhir
Perencanaan Struktur Bangunan Sekolahan 2 Lantai
BAB 5 Plat Lantai 5.6. Penulangan lapangan arah y
Mu = 217,05 kgm = 2,17.10
6
Nmm
Mn = φ
Mu =
6 6
10 .
71 ,
2 8
, 10
. 17
, 2
= Nmm
Rn =
=
2
.d b
Mn =
2 6
85 .
1000 10
. 71
, 2
0,375 Nmm
2
m = 29
, 11
25 .
85 ,
240 .
85 ,
= =
c f
fy
i
ρ
perlu
=
⎟ ⎟
⎠ ⎞
⎜ ⎜
⎝ ⎛
− −
× fy
Rn m
m .
. 2
1 1
1
= .
29 ,
11 1
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
− −
240 375
, .
29 ,
11 .
2 1
1 = 0,00158
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 =
~ 3 buah. Jarak tulangan dalam 1 m
1
= 333
3 1000 =
Jarak maksimum = 2 x h = 2 x 120 = 240 mm
As yang timbul = 3. ¼.
π.10
2
= 235,5 212,51 As ….ok
Dipakai tulangan
∅ 10 – 333 mm
commit to user
116
Tugas Akhir
Perencanaan Struktur Bangunan Sekolahan 2 Lantai
BAB 5 Plat Lantai 5.7. Penulangan tumpuan arah x
Mu = 755,65
kgm = 7,55.10
6
Nmm
Mn = φ
Mu =
= 8
, 10
. 55
, 7
6
9,44.10
6
Nmm Rn
= =
2
.d b
Mn =
2 6
85 .
1000 10
. 44
, 9
1,31 Nmm
2
m = 29
, 11
25 .
85 ,
240 .
85 ,
= =
c f
fy
ρ
perlu
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= .
29 ,
11 1
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
− −
240 31
, 1
. 29
, 11
. 2
1 1
= 0,00564 ρ ρ
max
ρ ρ
min
, di pakai ρ
perlu
= 0,00564 As
= ρ
perlu
. b . d = 0,00564 . 1000 . 85
= 479,4 mm
2
Digunakan tulangan ∅ 10
= ¼ . π . 10
2
= 78,5 mm
2
Jumlah tulangan =
11 ,
6 5
, 78
4 ,
479 =
~ 7 buah. Jarak tulangan dalam 1 m
1
= 143
7 1000 =
mm Jarak maksimum
= 2 x h = 2 x 120 = 240 mm As yang timbul
= 7. ¼. π.10
2
= 549,5 479,4 As ….ok Dipakai tulangan
∅ 10 – 143 mm
commit to user
117
Tugas Akhir
Perencanaan Struktur Bangunan Sekolahan 2 Lantai
BAB 5 Plat Lantai 5.8. Penulangan tumpuan arah y
