commit to user Tugas Akhir
Perencanaan Struktur Gedung Sekolah 2 Lantai
103
BAB 5 Plat Lantai
5.13. Penulangan plat atap
Data – data plat : Tebal plat h
= 10 cm = 100 mm
Diameter tulangan Æ = 8 mm fy
= 240 MPa f’c
= 25 MPa p
= 20 mm Tebal penutup d’
= p + ½Æ tul = 20 + 4
= 24 mm Tinggi Efektif d
= h - d’ = 120 – 24
= 96 mm Tingi efektif
Gambar 5.5 Perencanaan Tinggi Efektif
dx = h – p - ½Ø = 100 – 20 – 4 = 76 mm
dy = h – d’ – Ø - ½ Ø = 100 – 20 - 8 - ½ . 8 = 68 mm
h d y
d x d
commit to user Tugas Akhir
Perencanaan Struktur Gedung Sekolah 2 Lantai
103
BAB 5 Plat Lantai
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
5.14. Penulangan tumpuan arah x
Mu = 225,616 kgm = 2,528.10
6
Nmm
Mn =
f Mu
= =
8 ,
10 .
2,528
6
3,16.10
6
Nmm
Rn =
=
2
.dx b
Mn
=
2 6
76 .
1000 10
. 16
, 3
0,547 Nmm
2
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
- -
240 547
, .
2942 ,
11 .
2 1
1
= 0,002 r
r
max
r r
min
, di pakai r
min
As
perlu
= r
min
. b . dx = 0,0025 . 1000 . 76
= 190 mm
2
Digunakan tulangan Æ 8 As = ¼ . p . 8
2
commit to user Tugas Akhir
Perencanaan Struktur Gedung Sekolah 2 Lantai
103
BAB 5 Plat Lantai
= 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
As ada = 5. ¼ . p . 8
2
= 251,2 mm
2
As
perlu
…..… OK J
Dipakai tulangan Æ8 – 200 mm
5.15. Penulangan tumpuan arah y
Mu = 151,048 kgm =1,6925.10
6
Nmm
Mn =
f Mu
= =
8 ,
10 .
6925 ,
1
6
2,036.10
6
Nmm
Rn =
=
2
.dx b
Mn
=
2 6
76 .
1000 10
. 036
, 2
0,3525 Nmm
2
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 3525
, .
2942 ,
11 .
2 1
1
= 0,00148 r
r
max
r r
min
, di pakai r
min
As
perlu
= r
min
. b . dx = 0,0025 . 1000 . 76
commit to user Tugas Akhir
Perencanaan Struktur Gedung Sekolah 2 Lantai
103
BAB 5 Plat Lantai
= 190 mm
2
Digunakan tulangan Æ 8 As = ¼ . p . 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 As ada
= 5. ¼ . p . 8
2
= 251,2 mm
2
As
perlu
…..… OK J
Dipakai tulangan Æ8 – 200 mm
5.16. Penulangan lapangan arah x