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
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
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
commit to user Tugas Akhir
Perencanaan Struktur Gedung Sekolah 2 Lantai
103
BAB 5 Plat Lantai
= 8
As ada = 8. ¼ . p . 10
2
= 628 mm
2
As
perlu
…..… OK J Cek kapasitas lentur :
a = b
c f
fy As
ada
. .
85 ,
. =
1000 .
25 .
85 ,
240 .
628 = 7,093 mm
M
n
= As
ada
.fy.d-a2 = 13,784.10
6
Nmm Mn ada Mn = 13,784.10
6
13,2355.10
6
® OK J
5.7. Penulangan lapangan arah x
Mu = 435,99 kgm = 4,35599.10
6
Nmm Mn
= f
Mu =
6 6
10 .
44 ,
5 8
, 10
. 4,35599
= Nmm
Rn =
=
2
.dx b
Mn
=
2 6
95 .
1000 10
. 44
, 5
0,602 Nmm
2
m =
294 ,
11 25
. 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
- -
240 602
, .
294 ,
11 .
2 1
1 .
294 ,
11 1
= 0,00255 r r
max
r r
min
, di pakai r
perlu
= 0,00255 As
= r
min
. b . dx = 0,00255. 1000 . 95
= 242,25 mm
2
Digunakan tulangan Æ 10
commit to user Tugas Akhir
Perencanaan Struktur Gedung Sekolah 2 Lantai
103
BAB 5 Plat Lantai
As = ¼ . p . 10
2
= 78,5 mm
2
S =
perlu
As b
As .
= 25
, 242
1000 .
5 ,
78
= 324,045 ~ 330 mm Jarak maksimum = 2 x h
= 2 x 120 = 240 mm
n =
s b
=
240 1000
= 4,2 ~ 5 As ada
= 5. ¼ . p . 10
2
= 392,5 mm
2
As…...............OK J
Dipakai tulangan Æ 10 – 240 mm
Cek kapasitas lentur : a =
b c
f fy
As
ada
. .
85 ,
. =
1000 .
25 .
85 ,
240 .
5 ,
392 = 4,433 mm
M
n
= As
ada
.fy.d-a2 = 8,740.10
6
Nmm Mn ada Mn = 8,740.10
6
6
10 .
44 ,
5 ® OK J
5.8. Penulangan lapangan arah y
Mu = 435,99 kgm = 4,35599.10
6
Nmm Mn
= f
Mu =
6 6
10 .
44 ,
5 8
, 10
. 4,35599
= Nmm
Rn =
=
2
.dx b
Mn
=
2 6
95 .
1000 10
. 44
, 5
0,602 Nmm
2
m =
294 ,
11 25
. 85
, 240
. 85
, =
= c
f fy
commit to user Tugas Akhir
Perencanaan Struktur Gedung Sekolah 2 Lantai
103
BAB 5 Plat Lantai
r
perlu
=
÷ ÷
ø ö
ç ç
è æ
- -
fy Rn
. m
2 1
1 .
m 1
=
÷ ÷
ø ö
ç ç
è æ
- -
240 602
, .
294 ,
11 .
2 1
1 .
294 ,
11 1
= 0,00255 r r
max
r r
min
, di pakai r
perlu
= 0,00255 As
= r
min
. b . dx = 0,00255. 1000 . 95
= 242,25 mm
2
Digunakan tulangan Æ 10 As
= ¼ . p . 10
2
= 78,5 mm
2
S =
perlu
As b
As .
= 25
, 242
1000 .
5 ,
78
= 324,045 ~ 330 mm Jarak maksimum = 2 x h
= 2 x 120 = 240 mm
n =
s b
=
240 1000
= 4,2 ~ 5 As ada
= 5. ¼ . p . 10
2
= 392,5 mm
2
As…................... OK J
Dipakai tulangan Æ 10 – 240 mm
Cek kapasitas lentur : a =
b c
f fy
As
ada
. .
85 ,
. =
1000 .
25 .
85 ,
240 .
5 ,
392 = 4,433 mm
M
n
= As
ada
.fy.d-a2
commit to user Tugas Akhir
Perencanaan Struktur Gedung Sekolah 2 Lantai
103
BAB 5 Plat Lantai
= 8,740.10
6
Nmm Mn ada Mn = 8,740.10
6
6
10 .
44 ,
5 ® OK J
5.9. Rekapitulasi Tulangan
