commit to user
Perencanaan St rukt ur Gedung Sekolah 2 L ant ai
xxxiii
xxxiii 1,14
3,5 4,0
Lx Ly
= =
∼ 1,2 Mlx = 0,001.qu . Lx
2
. x = 0.001. 973,2. 3,5
2
.36 = 429,18 kg m
Mly = 0,001.qu . Lx
2
. x = 0.001. 973,2. 3,5
2
.18 = 214,59 kg m
Mtx = - 0,001.qu . Lx
2
. x = - 0.001 .973,2. 3,5
2
.77 = - 917,97 kg m
5.4. Penulangan Plat Lantai
Tabel 5.1. Perhitungan Plat Lantai Tipe Plat LyLx m
Mlx kgm Mly kgm
Mtx kgm Mty kgm A
4,03,5 = 1,2 453,02
333,81 - 1013,34
- 882,21 B1
3,52,0= 1,8 214,10
85,64 - 439,89
- 303,64 B2
3,52,0= 1,8 198,53
85,64 - 412,64
- 303,64 B3
3,52,0= 1,8 155,71
50,61 - 319,21
- 221,89 C1
4,03,5 = 1,2 369,57
333,81 - 882,21
- 822,60 C2
4,03,5 = 1,2 333,81
238,43 - 762,99
- 667,62 C3
4,03,5 = 1,2 381,49
226,5 - 846,44
- 679,54 C4
4,03,5 = 1,2 453,02
333,81 - 1013,34
- 882,21 C5
4,03,5 = 1,2 429,18
214,59 - 917,97
- Dari perhitungan momen diambil momen terbesar yaitu:
Mlx = 453,02 kgm
Mly = 333,81 kgm
Mtx = - 1013,34 kgm
Mty = - 882,21 kgm
Data – data plat : Tebal plat h
= 12 cm = 120 mm
Diameter tulangan ∅ = 10 mm
fy = 240 MPa
f’c = 30 MPa
commit to user
Perencanaan St rukt ur Gedung Sekolah 2 L ant ai
xxxiv
xxxiv b =
1000 mm
p =
20 mm
Tebal penutup d’ = p + ½
∅ tul = 20 + 5
= 25
mm Tinggi Efektif d
= h - d’ = 120 – 25
= 95
mm Tingi efektif
Gambar 5.10 Perencanaan Tinggi Efektif dx = h – p - ½Ø
= 120 – 20 – 5 = 95 mm dy = h – d’ – Ø - ½ Ø
= 120 – 20 - 10 - ½ . 10 = 85 mm
ρb =
⎟⎟⎠ ⎞
⎜⎜⎝ ⎛
+ fy fy
fc 600
600 .
. .
85 ,
β
=
⎟ ⎠
⎞ ⎜
⎝ ⎛
+ 240 600
600 .
85 ,
. 240
30 .
85 ,
= 0,0645
ρ
max
= 0,75 . ρb
= 0,75 . 0,0645 =
0,0484
ρ
min
= 0,0025
5.5. Penulangan tumpuan arah x
h d y
d x d
commit to user
Perencanaan St rukt ur Gedung Sekolah 2 L ant ai
xxxv
xxxv Mu
= 1013,34 kgm = 10,1334.10
6
Nmm
Mn = φ
Mu =
= 8
, 10
. 1334
, 10
6
12,667.10
6
Nmm Rn
= =
2
.dx b
Mn =
2 6
95 .
1000 10
. 667
, 12
1,404 Nmm
2
m = 412
, 9
30 .
85 ,
240 .
85 ,
= =
c f
fy
ρ
perlu
= ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= . 412
, 9
1 ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ −
− 240
404 ,
1 .
412 ,
9 .
2 1
1 = 0,006024
ρ ρ
max
ρ ρ
min
, di pakai ρ
perlu
= 0,006024 As
perlu
= ρ
perlu
. b . dx = 0,006024 . 1000 . 95
= 572,28 mm
2
Digunakan tulangan ∅ 10
As = ¼ . π . 10
2
= 78,5
mm
2
S =
perlu
As b
As. =
28 ,
572 1000
. 5
, 78
= 137,17 ~ 130 mm n =
s b
= 100
1000 = 10
As yang timbul = 10. ¼ . π . 10
2
= 785
mm
2
As
perlu
…..…ok
commit to user
Perencanaan St rukt ur Gedung Sekolah 2 L ant ai
xxxvi
xxxvi Dipakai tulangan
∅ 10 – 130 mm
5.6. Penulangan tumpuan arah y
