74
Dipakai tulangan φ 10 – 125 mm
As tx
2
= As x n = 78,5 mm
2
x 9 = 706,5 mm
2
As tx
1
= 667 mm
2
Tulangan memenuhi syarat
3. Penulangan plat tangga
Momen Rancang Plat β =
2 75
, 3
= 1,875 cx+ = 57,38
cx- =
81,88 cy+ = 15
cy- =
53,13
Mtx = -cx . 0.001 . Wu . Lx
2
Mtx = -81,88 . 0,001. 9,921 . 2
2
Mtx = -3,25 kNm
Mlx = +cx . 0,001 . Wu . Lx
2
Mlx = +57,38. 0,001. 9,921 . 2
2
Mlx = 2,28 kNm
Mty = -cy . 0,001 . Wu . Lx
2
Mty = -53,13. 0,001. 9,921 . 2
2
Mty = -2,11 kNm 2000
3750
75
Dy
h
Dx dy
dx Mly = +cy . 0,001 . Wu . Lx
2
Mly = +15 . 0,001. 9,921 . 2
2
Mly = 0,6 kNm
Penulangan Plat •
Tebal Plat = 150 mm •
Selimut Beton = p = 20 mm Direncanakan
¾ Diameter tulangan utama arah x =
φ10 mm ¾
Diameter tulangan utama arah y = φ10 mm
Tinggi efektif - Arah x = dx = h – p –Dx2
= 150 – 20 – 102 = 125 mm
- Arah y = dy = h – p – Dx – Dy2 = 150 – 20 – 10 – 102
= 115 mm
76
Penulangan tepi arah x
Ditinjau 1000 mm Mtx = 3,25 kNm
k =
2
xbxd Mu
φ =
2
125 1000
8 ,
3250000 x
x = 0,26
Dari Tabel Beton Apendiks pada bagian tabel A-10 maka
ρ = 0,0058 As tx
1
= xbxd ρ
= 0,0058 x 1000 x 125 = 725 mm
2
Direncanakan tulangan φ 10 mm
As = ¼ x π x D
2
= ¼ x 3,14 x 10
2
= 78,5 mm
2
Jumlah tulangan n =
As Astx
1
n = 5
, 78
725 = 9,24 dibulatkan 10 batang
Spasi =
1 −
n b
=
9 1000mm
= 111,11 mm dibulatkan 110 mm Dipakai tulangan
φ 10 – 110 mm As
tx
2
= As x n = 78,5 mm
2
x 10 = 785 mm
2
As tx
1
= 725 mm
2
Tulangan memenuhi syarat
77
Penulangan lapangan arah x
Ditinjau 1000 mm Mlx = 2,28 kNm
k =
2
xbxd Mu
φ =
2
125 1000
8 ,
2280000 x
x = 0,23
Dari Tabel Beton Apendiks pada bagian tabel A-10 maka
ρ = 0,0058 As tx
1
= xbxd ρ
= 0,0058 x 1000 x 125 = 725 mm
2
Direncanakan tulangan φ 10 mm
As = ¼ x π x D
2
= ¼ x 3,14 x 10
2
= 78,5 mm
2
Jumlah tulangan n =
As Astx
1
n = 5
, 78
725 = 9,24 dibulatkan 10 batang
Spasi =
1 −
n b
=
9 1000mm
= 111,11 mm dibulatkan 110 mm Dipakai tulangan
φ 10 – 110 mm As
tx
2
= As x n = 78,5 mm
2
x 10 = 785 mm
2
As tx
1
= 725 mm
2
Tulangan memenuhi syarat
78
Penulangan tepi arah y
Ditinjau 1000 mm Mty = 2,11 kNm
k =
2
xbxd Mu
φ =
2
115 1000
8 ,
2110000 x
x = 0,2
Dari Tabel Beton Apendiks pada bagian tabel A-10 maka
ρ =0,0058 As tx
1
= xbxd ρ
= 0,0058 x 1000 x 115 = 667 mm
2
Direncanakan tulangan φ 10 mm
As = ¼ x π x D
2
= ¼ x 3,14 x 10
2
= 78,5 mm
2
Jumlah tulangan n =
As Astx
1
n = 5
, 78
667 = 8,5 dibulatkan 9 batang
Spasi =
1 −
n b
=
8 1000mm
= 125 mm Dipakai tulangan
φ 10 – 125 mm As
tx
2
= As x n = 78,5 mm
2
x 9 = 706,5 mm
2
As tx
1
= 667 mm
2
Tulangan memenuhi syarat
79
Penulangan lapangan arah y
Ditinjau 1000 mm Mly = 0,6 kNm
k =
2
xbxd Mu
φ =
2
115 1000
8 ,
6000 x
x = 0,056
Dari Tabel Beton Apendiks pada bagian tabel A-10 maka
ρ =0,0058 As tx
1
= xbxd ρ
= 0,0058 x 1000 x 115 = 667 mm
2
Direncanakan tulangan φ 10 mm
As = ¼ x π x D
2
= ¼ x 3,14 x 10
2
= 78,5 mm
2
Jumlah tulangan n =
As Astx
1
n = 5
, 78
667 = 8,5 dibulatkan 9 batang
Spasi =
1 −
n b
=
8 1000mm
= 125 mm Dipakai tulangan
φ 10 – 125 mm As
tx
2
= As x n = 78,5 mm
2
x 9 = 706,5 mm
2
As tx
1
= 667 mm
2
Tulangan memenuhi syarat
80
4.3 Perencanaan Berat Bangunan
Kategori