commit to user
ρ ρ
min
, di pakai ρ
min
As
perlu
= ρ
min
. b . dx = 0,0025 . 1000 . 76
= 190 mm
2
Digunakan tulangan ∅ 8
As = ¼ . π . 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 . π . 8
2
= 251,2
mm
2
As
perlu
…..… OK ☺
Dipakai tulangan
∅8 – 200 mm
5.15. Penulangan tumpuan arah y
Mu = 151,048 kgm =1,6925.10
6
Nmm
Mn = φ
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
ρ
perlu
= ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
commit to user
T ugas Ak hir
Perencanaan Struktur Gedung Sekolah 2 Lantai
66
BAB 5 Plat Lantai ATAP
= . 2942
, 11
1 ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ −
− 240
3525 ,
. 2942
, 11
. 2
1 1
= 0,00148 ρ
ρ
max
ρ ρ
min
, di pakai ρ
min
As
perlu
= ρ
min
. b . dx = 0,0025 . 1000 . 76
= 190 mm
2
Digunakan tulangan ∅ 8
As = ¼ . π . 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 . π . 8
2
= 251,2
mm
2
As
perlu
…..… OK ☺
Dipakai tulangan
∅8 – 200 mm
5.16. Penulangan lapangan arah x
Mu = 110,896 kgm = 1,2426.10
6
Nmm Mn =
φ Mu
=
6 6
10 .
55 ,
1 8
, 10
. 2426
, 1
= Nmm
Rn =
=
2
.dx b
Mn =
2 6
76 .
1000 10
. 55
, 1
0,268 Nmm
2
m = 294
, 11
25 .
85 ,
240 .
85 ,
= =
c f
fy
commit to user
ρ
perlu
= ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ −
− 240
268 ,
. 294
, 11
. 2
1 1
. 294
, 11
1 =
0,001123 ρ
ρ
max
ρ ρ
min
, di pakai ρ
min
As
perlu
= ρ
min
. b . dx = 0,0025 . 1000 . 76
= 190 mm
2
Digunakan tulangan ∅ 8
As = ¼ . π . 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 . π . 8
2
= 251,2
mm
2
As
perlu
…..… OK ☺
Dipakai tulangan
∅8 – 200 mm
5.17. Penulangan lapangan arah y
Mu = 40,152 kgm = 0,4499.10
6
Nmm Mn =
φ Mu
=
5 6
10 .
623 ,
5 8
, 10
. 4499
, =
Nmm
commit to user
T ugas Ak hir
Perencanaan Struktur Gedung Sekolah 2 Lantai
66
BAB 5 Plat Lantai ATAP
Rn =
=
2
.dx b
Mn =
2 5
95 .
1000 10
. 623
, 5
0,097 Nmm
2
m = 294
, 11
25 .
85 ,
240 .
85 ,
= =
c f
fy
ρ
perlu
= ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ −
− 240
097 ,
. 294
, 11
. 2
1 1
. 294
, 11
1 =
0,000405 ρ
ρ
max
ρ ρ
min
, di pakai ρ
min
As
perlu
= ρ
min
. b . dx = 0,0025 . 1000 . 76
= 190 mm
2
Digunakan tulangan ∅ 8
As = ¼ . π . 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 . π . 8
2
= 251,2
mm
2
As
perlu
…..… OK ☺
Dipakai tulangan
∅8 – 200 mm
5.18. Rekapitulasi Tulangan
