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127 Perencanaan Struktur Gedung Sekolah 2 Lantai dan RAB
Bab 5 Perencanaan Plat Lantai dan Atap
5.5. Penulangan tumpuan arah y
Mu = 1152,27 kgm = 11,5227.10
6
Nmm
Mn =
f Mu
= =
8 ,
10 .
11,5227
6
14,403.10
6
Nmm
Rn =
=
2
.dx b
Mn
=
2 6
94 .
1000 10
. 14,403
1,63 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 63
, 1
. 12
, 14
. 2
1 1
= 0,00715 r
r
max
r r
min
, di pakai r
perlu
= 0,00715 As
perlu
= r
perlu
. b . dx = 0,00715 . 1000 . 94
= 672,405 mm
2
Digunakan tulangan Æ 12 As = ¼ . p . 12
2
= 113,097 mm
2
S =
perlu
As b
As .
= 405
, 672
1000 .
097 ,
113
= 168,197 ~ 150 mm n =
s b
=
150 1000
= 6,67 ~ 7
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128 Perencanaan Struktur Gedung Sekolah 2 Lantai dan RAB
Bab 5 Perencanaan Plat Lantai dan Atap As ada
=7. ¼ . p . 12
2
= 791,679 mm
2
As
perlu
672,405 mm
2
…..… OK
Dipakai tulangan Æ 12 – 150 mm
Cek kapasitas lentur : a =
b c
f fy
As
ada
. .
85 ,
. =
1000 .
20 .
85 ,
240 .
679 ,
791 = 7,11,176
M
n
= As
ada
.fy.d-a2 = 16,798.10
6
Nmm Mn ada Mn ® 16,798.10
6
Nmm 14,403.10
6
Nmm OK..
5.6. Penulangan lapangan arah x
Mu = 591,77 kgm = 5,9177.10
6
Nmm Mn
= f
Mu =
6 6
10 .
397 ,
7 8
, 10
. 5,9177
= Nmm
Rn =
=
2
.dx b
Mn
=
2 6
94 .
1000 10
. 397
, 7
0,837 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
= ÷
÷ ø
ö ç
ç è
æ -
- 240
837 ,
. 12
, 14
. 2
1 1
. 12
, 14
1
= 0,003578 r r
max
r r
min
, di pakai r
ada
= 0,003578 As
= r
ada
. b . dx = 0,003578. 1000 . 94
= 336,385 mm
2
Digunakan tulangan Æ 12
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129 Perencanaan Struktur Gedung Sekolah 2 Lantai dan RAB
Bab 5 Perencanaan Plat Lantai dan Atap As
= ¼ . p . 12
2
= 113,097 mm
2
S =
perlu
As b
As .
= 235
1000 .
097 ,
113
= 336,21 mm Jarak maksimum = 2 x h
= 2 x 120 = 240 ~ 225 mm
n =
s b
=
225 1000
= 4,44 ~ 5 As ada
= 5. ¼ . p . 12
2
= 565,487 mm
2
As 336,385 mm
2
…...............OK
Dipakai tulangan Æ 12 – 225 mm
Cek kapasitas lentur : a =
b c
f fy
As
ada
. .
85 ,
. =
1000 .
20 .
85 ,
240 .
487 ,
565 = 7,983 mm
M
n
= As
ada
.fy.d-a2 = 12,2157.10
6
Nmm Mn ada Mn = 12,2157.10
6
6
10 .
397 ,
7 Nmm ...... OK
5.7. Penulangan lapangan arah y
Mu = 435,99 kgm = 4,35.10
6
Nmm Mn
= f
Mu =
6 6
10 .
4498 ,
5 8
, 10
. 4,35
= Nmm
Rn =
=
2
.dx b
Mn
=
2 6
94 .
1000 10
. 4498
, 5
0,617 Nmm
2
m =
12 ,
14 20
. 85
, 240
. 85
, =
= c
f fy
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130 Perencanaan Struktur Gedung Sekolah 2 Lantai dan RAB
Bab 5 Perencanaan Plat Lantai dan Atap r
perlu
= ÷
÷ ø
ö ç
ç è
æ -
- fy
Rn .
m 2
1 1
. m
1
= ÷
÷ ø
ö ç
ç è
æ -
- 240
617 ,
. 12
, 14
. 2
1 1
. 12
, 14
1
= 0,0026 r r
max
r r
min
, di pakai r
perlu
= 0,0025 As
= r
erlu
. b . dx = 0,0026. 1000 . 94
= 246,122 mm
2
Digunakan tulangan Æ 12 As
= ¼ . p . 12
2
= 113,097 mm
2
S =
perlu
As b
As .
= 122
, 246
1000 .
097 ,
113
= 459,55 mm Jarak maksimum = 2 x h
= 2 x 120 = 240 ~ 225 mm
n =
s b
=
225 1000
= 4,44 ~ 5 As ada
= 5. ¼ . p . 12
2
= 565,487 mm
2
As 246,122 mm
2
…...............OK
Dipakai tulangan Æ 12 – 225 mm
Cek kapasitas lentur : a =
b c
f fy
As
ada
. .
85 ,
. =
1000 .
20 .
85 ,
240 .
487 ,
565 = 7,983 mm
commit to user Tugas Akhir
131 Perencanaan Struktur Gedung Sekolah 2 Lantai dan RAB
Bab 5 Perencanaan Plat Lantai dan Atap M
n
= As
ada
.fy.d-a2 = 12,2157.10
6
Nmm Mn ada Mn = 12,2157.10
6
6
10 .
4498 ,
5 Nmm...... OK
5.8. Rekapitulasi Tulangan
