commit to user 153
Tugas Akhir
Perencanaan Struktur Gedung Sekolah 2 Lantai RAB
BAB 5 PERENCANAAN PLAT LANTAI
Tingi efektif dx
= h – p – ½Ø = 100 – 20 – 4
= 76 mm dy
= h – d’ – Øtulangan – ½ . Øtulangan = 100 – 20 – 8 – ½ × 8
= 68 mm
Gambar 5.17. Perencanaan Tinggi Efektif
rb = ÷÷
ø ö
çç è
æ + fy
fy fc
600 600
. .
. 85
, b
=
÷ ø
ö ç
è æ
+ 240 600
600 ×
85 ,
× 240
20 ×
85 ,
= 0,0430 r
max
= 0,75 . rb = 0,75 × 0,0430
= 0,0322 r
min
= 0,0025
1. Penulangan Lapangan Arah X
Mu = 180,68 kgm = 1,8068.10
6
Nmm Mn
= f
Mu =
8 ,
10 .
1,8068
6
= 2,2585.10
6
Nmm
commit to user 154
Tugas Akhir
Perencanaan Struktur Gedung Sekolah 2 Lantai RAB
BAB 5 PERENCANAAN PLAT LANTAI
Rn =
=
2
.dx b
Mn
=
2 6
76 ×
1000 10
. 2,2585
0,391 Nmm
2
m =
12 ,
14 20
× 85
, 240
. 85
, =
= c
f fy
r
perlu
=
÷ ÷
ø ö
ç ç
è æ
- -
fy Rn
. m
2 1
1 .
m 1
= ×
12 ,
14 1
÷ ÷
ø ö
ç ç
è æ
- -
240 0,391
× 12
, 14
× 2
1 1
= 0,0016 r
perlu
r
max
r
perlu
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
= 50,24 mm
2
S =
perlu
As b
As .
=
190 1000
× 50,24
= 264,42 ~ 260 mm Jarak maksimum = 2 × h
= 2 × 100 = 200 mm
n =
s b
=
200 1000
= 5 buah As
ada
= 5 . ¼ . p . 8
2
= 251,2 mm
2
As
perlu
..............OK J Cek kapasitas lentur :
a =
= b
c f
fy As
ada
. .
85 ,
. 1000
20 85
, 240
2 ,
51 2
´ ´
´ = 3,55 mm
commit to user 155
Tugas Akhir
Perencanaan Struktur Gedung Sekolah 2 Lantai RAB
BAB 5 PERENCANAAN PLAT LANTAI
Mn
ada
= As
ada
. fy . dx – a2 = 251,2 × 240 × 76 – 3,552
= 4,4749.10
6
Nmm Mn
ada
Mn 4,4749.10
6
Nmm 2,2585.10
6
Nmm ......OK J
Jadi, dipakai tulangan Æ 8 – 200 mm
2. Penulangan Lapangan Arah Y
Mu = 120,46 kgm = 1,2046.10
6
Nmm Mn
= f
Mu =
8 ,
10 .
1,2046
6
= 1,5058.10
6
Nmm
Rn =
=
2
.dy b
Mn
=
2 6
68 ×
1000 10
. 5058
, 1
0,3256 Nmm
2
m =
12 ,
14 20
× 85
, 240
. 85
, =
= c
f fy
r
perlu
=
÷ ÷
ø ö
ç ç
è æ
- -
fy Rn
. m
2 1
1 .
m 1
= ×
12 ,
14 1
÷ ÷
ø ö
ç ç
è æ
- -
240 0,3256
× 12
, 14
× 2
1 1
= 0,0014 r
perlu
r
max
r
perlu
r
min
, di pakai r
min
As
perlu
= r
min
. b . dy = 0,0025 × 1000 × 68
= 170 mm
2
Digunakan tulangan Æ 8 As
= ¼ . p . 8
2
= 50,24 mm
2
commit to user 156
Tugas Akhir
Perencanaan Struktur Gedung Sekolah 2 Lantai RAB
BAB 5 PERENCANAAN PLAT LANTAI
S =
perlu
As b
As .
=
170 1000
× 50,24
= 295,53 ~ 290 mm Jarak maksimum = 2 × h
= 2 × 100 = 200 mm
n =
s b
=
200 1000
= 5 buah As
ada
= 5 . ¼ . p . 8
2
= 251,2 mm
2
As
perlu
..............OK J Cek kapasitas lentur :
a =
= b
c f
fy As
ada
. .
85 ,
. 1000
20 85
, 240
2 ,
51 2
´ ´
´ = 3,55 mm
Mn
ada
= As
ada
. fy . dy – a2 = 251,2 × 240 × 68 – 3,552
= 3,9926.10
6
Nmm Mn
ada
Mn 3,9926.10
6
Nmm 1,5058.10
6
Nmm ......OK J
Jadi, dipakai tulangan Æ 8 – 200 mm
3. Penulangan Tumpuan Arah X
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