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Gambar 5.7 Plat tipe B3 2,0
2,0 4,0
Lx Ly
= =
Mlx = 0,001.qu . Lx
2
. x = 0.001. 973,2. 2,0
2
.55 = 214,104 kg m
Mly = 0,001.qu . Lx
2
. x = 0.001. 973,2. 2,0
2
.21 = 81,75 kg m
Mtx = - 0,001.qu . Lx
2
. x = - 0.001 .973,2. 2,0
2
.114 = - 443,78 kg m Mty = - 0,001.qu . Lx
2
. x = - 0.001 . 973,2. 2,0
2
.78 = - 303,64 kgm g.Tipe pelat C3
C3
Lx Ly
Gambar 5.8 Plat tipe C3 2,0
2,0 4,0
Lx Ly
= =
Mlx = 0,001.qu . Lx
2
. x = 0.001. 973,2. 2,0
2
.85 = 330,88 kg m
Mly = 0,001.qu . Lx
2
. x = 0.001. 973,2. 2,0
2
.39 = 151,82 kg m
Mty = - 0,001.qu . Lx
2
. x = - 0.001 . 973,2. 2,0
2
.119 = - 463,24 kgm
5.4. Penulangan Plat Lantai
Tabel 5.1. Perhitungan Plat Lantai Tipe Plat LyLx m
Mlx kgm Mly kgm
Mtx kgm Mty kgm A
4,04,0 = 1 323,98
386,69 -
-877,9 B1
4,04,0 = 1 435,99
435,99 - 1058,84
- 1058,84 B2
4,04,0 = 1 404,85
327 - 937,28
- 856,42 B3
4,04,0 = 1 327
327 - 809,71
- 809,71 C1
4,02,0= 2 205,78
73,96 - 459,75
- 307,53
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Perencanaan Struktur Gedung Sekolah 2 Lantai
66
BAB 5 Plat Lantai ATAP
C2 4,02,0= 2
214,104 81,75
- 443,78 - 303,64
C3 4,02,0= 2
330,88 151,82
- - 463,24
Dari perhitungan momen diambil momen terbesar yaitu: Mlx
= 435,99 kgm Mly
= 435,99 kgm Mtx
= - 1058,84 kgm Mty
= - 1058,84 kgm Data – data plat :
Tebal plat h = 12 cm
= 120 mm Diameter tulangan
∅ = 10 mm fy
= 240 MPa f’c
= 25 MPa b =
1000 mm
p =
20 mm
Tebal penutup d’ = p + ½
∅ tul = 20 + 5
= 25
mm Tinggi Efektif d
= h - d’ = 120 – 25
= 95
mm Tingi efektif
Gambar 5.9 Perencanaan Tinggi Efektif dx = h – p - ½Ø
h d y
d x d
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= 120 – 20 – 5 = 95 mm dy = h – d’ – Ø - ½ Ø
= 120 – 20 - 10 - ½ . 10 = 85 mm
ρb = ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ + fy
fy fc
600 600
. .
. 85
, β
= ⎟
⎠ ⎞
⎜ ⎝
⎛ + 240
600 600
. 85
, .
240 25
. 85
, =
0,05376 ρ
max
= 0,75 . ρb
= 0,75 . 0,05376 =
0,04032 ρ
min
= 0,0025 5.5.
Penulangan tumpuan arah x
Mu = 1058,84 kgm = 10,5884.10
6
Nmm
Mn = φ
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
ρ
perlu
= ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= . 2942
, 11
1 ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ −
− 240
466 ,
1 .
2942 ,
11 .
2 1
1 = 0,00633
ρ ρ
max
ρ ρ
min
, di pakai ρ
perlu
= 0,00633 As
perlu
= ρ
perlu
. b . dx = 0,00633 . 1000 . 95
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Perencanaan Struktur Gedung Sekolah 2 Lantai
66
BAB 5 Plat Lantai ATAP
= 598,5 mm
2
Digunakan tulangan ∅ 10
As = ¼ . π . 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 = 8
As ada = 8. ¼ .
π . 10
2
= 628
mm
2
As
perlu
…..…OK ☺
Dipakai tulangan
∅ 10 – 120 mm
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
→ OK ☺
5.6. Penulangan tumpuan arah y
Mu = 1058,84 kgm = 10,5884.10
6
Nmm
Mn = φ
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
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ρ
perlu
= ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ −
− fy
Rn .
m 2
1 1
. m
1
= . 2942
, 11
1 ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ −
− 240
466 ,
1 .
2942 ,
11 .
2 1
1 = 0,00633
ρ ρ
max
ρ ρ
min
, di pakai ρ
perlu
= 0,00633 As
perlu
= ρ
perlu
. b . dx = 0,00633 . 1000 . 95
= 598,5 mm
2
Digunakan tulangan ∅ 10
As = ¼ . π . 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 = 8
As ada = 8. ¼ .
π . 10
2
= 628
mm
2
As
perlu
…..… OK ☺
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 ☺
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Perencanaan Struktur Gedung Sekolah 2 Lantai
66
BAB 5 Plat Lantai ATAP
5.7. Penulangan lapangan arah x
