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135 Perencanaan Struktur Gedung Sekolah 2 Lantai dan RAB
Bab 5 Perencanaan Plat Lantai dan Atap Mlx = 0,001.qu . Lx
2
. x = 0.001. 478. 3,0
2
.36 = 154,872 kg m
Mly = 0,001.qu . Lx
2
. x = 0.001. 478. 3,0
2
.28 = 120,456 kg m
Mtx = - 0,001.qu . Lx
2
. x = - 0.001 .478. 3,0
2
.82 = -352,764 kg m
Mty = - 0,001.qu . Lx
2
. x = - 0.001 . 478. 3,0
2
.72= - 309,744 kgm
5.12. Penulangan plat atap
Tabel 5.2. Perhitungan Plat Atap Tipe Plat
LyLx m Mlx kgm Mly kgm Mtx kgm Mty kgm
K 4,03,0 = 1,33
150,57 77,436
-318,348 -245,214
L 4,03,0 = 1,33
133,362 81,738
-296,838 -245,214
M 4,03,0 = 1,33
180,648 120,456
-395,784 -326,952
N 4,03,0 = 1,33
154,872 120,456
-352,764 -309,744
Dari perhitungan momen diambil momen terbesar yaitu: Mlx
= 180,648 kgm Mly
= 120,456 kgm Mtx
= - 395,784 kgm Mty
= - 326,952 kgm Data – data plat :
Tebal plat h = 100 mm
Diameter tulangan Æ = 10 mm fy
= 240 MPa f’c
= 25 MPa p
= 20 mm Tebal penutup d’
= p + ½Æ tul = 20 + 5
= 25 mm Tinggi Efektif d
= h - d’ = 100 – 25
= 75 mm
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136 Perencanaan Struktur Gedung Sekolah 2 Lantai dan RAB
Bab 5 Perencanaan Plat Lantai dan Atap Tingi efektif
Gambar 5.17 Perencanaan Tinggi Efektif
dx = h – p - ½Ø = 100 – 20 – 5 = 75 mm
dy = h – d’ – Ø - ½ Ø = 100 – 25 - 10 - ½ . 10 = 60 mm
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 ,
= 0,043 r
max
= 0,75 . rb = 0,75 . 0,043
= 0,03225 r
min
= 0,0025
h d y
d x d
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137 Perencanaan Struktur Gedung Sekolah 2 Lantai dan RAB
Bab 5 Perencanaan Plat Lantai dan Atap
5.13. Penulangan tumpuan arah x
Mu = 395,784 kgm = 3,96784.10
6
Nmm
Mn =
f Mu
= =
8 ,
10 .
3,95784
6
4,9473.10
6
Nmm
Rn =
=
2
.dx b
Mn
=
2 6
75 .
1000 10
. 9473
, 4
0,8795 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 8795
, .
12 ,
14 .
2 1
1 = 0,003765
r r
max
r r
min
, di pakai r
perlu
As
perlu
= r
perlu
. b . dx = 0,003765 . 1000 . 75
= 282,355 mm
2
Digunakan tulangan Æ 10 As = ¼ . p . 10
2
= 78,539 mm
2
S =
perlu
As b
As .
=
355 ,
282 1000
. 539
, 78
= 278,157 ~ 275 mm Jarak maksimum
= 2 x h = 200 mm
n =
s b
=
200 1000
= 5
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138 Perencanaan Struktur Gedung Sekolah 2 Lantai dan RAB
Bab 5 Perencanaan Plat Lantai dan Atap As ada
= 5. ¼ . p . 10
2
= 392,699 mm
2
As
perlu
282,355 mm
2
…..… OK
Dipakai tulangan Æ10 – 200 mm
5.14. Penulangan tumpuan arah y
Mu = 326,952 kgm =3,26952.10
6
Nmm
Mn =
f Mu
= =
8 ,
10 .
26952 ,
3
6
4,0869.10
6
Nmm
Rn =
=
2
.dx b
Mn
=
2 6
75 .
1000 10
. 0869
, 4
0,7266 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 7266
, .
12 ,
14 .
2 1
1 = 0,003095
r r
max
r r
min
, di pakai r
perlu
As
perlu
= r
perlu
. b . dx = 0,003095 . 1000 . 75
= 232,122 mm
2
Digunakan tulangan Æ 10 As = ¼ . p . 10
2
= 78,539 mm
2
S =
perlu
As b
As .
=
122 ,
232 1000
. 539
, 78
= 338,352 ~ 300 mm
Jarak maksimum = 2 x h
= 200 mm
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139 Perencanaan Struktur Gedung Sekolah 2 Lantai dan RAB
Bab 5 Perencanaan Plat Lantai dan Atap n =
s b
=
200 1000
= 5 As ada
= 5. ¼ . p . 10
2
= 392,699 mm
2
As
perlu
232,122 mm
2
…..… OK
Dipakai tulangan Æ10 – 200 mm
5.15. Penulangan lapangan arah x
