commit to user Tugas Akhir
Perencanaan Struktur Gedung Sekolah 2 Lantai
103
BAB 5 Plat Lantai
Mty = - 0,001.qu . Lx
2
. x = - 0.001 . 973,2. 2,0
2
.119 = - 463,24 kgm
5.4. Penulangan Plat Lantai
Tabel 5.1. Perhitungan Plat Lantai Tipe Plat LyLx m
Mlx kgm Mly kgm Mtx kgm Mty kgm A
4,04,0 = 1 323,98
386,69 -
-877,9 B1
4,04,0 = 1 435,99
435,99 - 1058,84
- 1058,84 B2
4,04,0 = 1 404,85
327 - 937,28
- 856,42 B3
4,04,0 = 1 327
327 - 809,71
- 809,71 C1
4,02,0= 2 205,78
73,96 - 459,75
- 307,53 C2
4,02,0= 2 214,104
81,75 - 443,78
- 303,64 C3
4,02,0= 2 330,88
151,82 -
- 463,24
Dari perhitungan momen diambil momen terbesar yaitu: Mlx
= 435,99 kgm Mly
= 435,99 kgm Mtx
= - 1058,84 kgm Mty
= - 1058,84 kgm
Data – data plat : Tebal plat h
= 12 cm = 120 mm
Diameter tulangan Æ = 10 mm fy
= 240 MPa f’c
= 25 MPa b
= 1000 mm p
= 20 mm Tebal penutup d’
= p + ½Æ tul = 20 + 5
= 25 mm Tinggi Efektif d
= h - d’ = 120 – 25
commit to user Tugas Akhir
Perencanaan Struktur Gedung Sekolah 2 Lantai
103
BAB 5 Plat Lantai
= 95 mm Tingi efektif
Gambar 5.9 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
rb = ÷÷
ø ö
çç è
æ + fy
fy fc
600 600
. .
. 85
, b
=
÷ ø
ö ç
è æ
+ 240 600
600 .
85 ,
. 240
25 .
85 ,
= 0,05376 r
max
= 0,75 . rb = 0,75 . 0,05376
= 0,04032 r
min
= 0,0025
5.5. Penulangan tumpuan arah x
Mu = 1058,84 kgm = 10,5884.10
6
Nmm
Mn =
f Mu
= =
8 ,
10 .
10,5884
6
13,2355.10
6
Nmm
Rn =
=
2
.dx b
Mn
=
2 6
95 .
1000 10
. 13,2355
1,466 Nmm
2
h d y
d x d
commit to user Tugas Akhir
Perencanaan Struktur Gedung Sekolah 2 Lantai
103
BAB 5 Plat Lantai
m =
2942 ,
11 25
. 85
, 240
. 85
, =
= c
f fy
r
perlu
=
÷ ÷
ø ö
ç ç
è æ
- -
fy Rn
. m
2 1
1 .
m 1
= .
2942 ,
11 1
÷ ÷
ø ö
ç ç
è æ
- -
240 466
, 1
. 2942
, 11
. 2
1 1
= 0,00633 r
r
max
r r
min
, di pakai r
perlu
= 0,00633 As
perlu
= r
perlu
. b . dx = 0,00633 . 1000 . 95
= 598,5 mm
2
Digunakan tulangan Æ 10 As = ¼ . p . 10
2
= 78,5 mm
2
S =
perlu
As b
As .
= 5
, 598
1000 .
5 ,
78
= 129,72 ~ 125 mm n =
s b
=
125 1000
= 8
As ada = 8. ¼ . p . 10
2
= 628 mm
2
As
perlu
…..…OK J
Dipakai tulangan Æ 10 – 120 mm
Cek kapasitas lentur : a =
b c
f fy
As
ada
. .
85 ,
. =
1000 .
25 .
85 ,
240 .
628 = 7,093 mm
commit to user Tugas Akhir
Perencanaan Struktur Gedung Sekolah 2 Lantai
103
BAB 5 Plat Lantai
M
n
= As
ada
.fy.d-a2 = 13,784.10
6
Nmm Mn ada Mn ® OK J
5.6. Penulangan tumpuan arah y
