plat 2 arah baru
DESCRIPTION
Penulangan Plat 2 arah baruTRANSCRIPT
![Page 1: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/1.jpg)
![Page 2: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/2.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
PENULANGAN PLAT DUA ARAH (Plat A)
Diketahui data perencanaan :
h : 12 cm h : 12 cm h 12 cm h : 12 cm
250 Kg/m² 250 Kg/m² 250 Kg/m² 250 Kg/m²
a b c d
h : 12 cm h : 12 cm h 12 cm h : 12 cm
250 Kg/m² 250 Kg/m² 250 Kg/m² 250 Kg/m²
e f g h
h : 12 cm h : 12 cm h 12 cm h : 12 cm
250 Kg/m² 250 Kg/m² 250 Kg/m² 250 Kg/m²
i j k l
4.0 5.0 3.0 4.0
● Kuat tekan beton : 25 MPa
● Kuat leleh baja tulangan : 350 MPa
● : 120 mm = 0.12 m
● : 250 Kg/m² = 2.5 kN/m²
● Berat jenis beton : 2400 Kg/m³ = 24 kN/m³
● : 10 mm
● : 10 mm
● Diameter tul Bagi : 6 mm
ql : ql: ql ql:
ql : ql: ql ql:
ql : ql: ql ql:
Tebal plat (h)
Beban hidup (ql)
Ø tul utama arah x
Ø tul utama arah y
![Page 3: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/3.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
● Tebal selimut Beton : 20 mm
A Pembebanan yang bekerja pada plat
- Beban mati :
Berat sendiri = 0.12 x 1 x 1 x 24.0 = 2.880
= 2.880
- Beban hidup :
Berat beban hidup 1 x 1 x 2.5 = 2.50
= 2.50
- Beban terfaktor
= 1.20 x 2.88 + 1.60 x 2.50
= 7.4560 kN/m
B Perhitungan Momen
●
M = ± 0.001 .
Ly = 4 m
Dimana :Ly
=4
= 1Lx = 4.0 Lx 4
Ly : Bentang terpanjang
Lx : Bentang terpendek
Dengan mengetahui nilai Ly/Lx maka dapat ditentukan nilai C sesuai tabel 13.3.2 Buku Teknik
: 46
: 46
: 38
: 38
Sehingga dapat ditentukan nilai momen sebagai berikut :
Mtx = - 0.001 x qu x x
= - 0.001 x 7.456 x 4 ² x 46
= - 5.487616 kNm
= - 5487616 Nmm
qd
ql
qu = 1.2 q
d + 1.6 q
l
Plat a = Plat d
qu . lx² . C
Sipil Hal 355, sebagai berikut :
Ctx
Clx
Cty
Cly
lx² Ctx
![Page 4: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/4.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
0.001 x qu x x
= 0.001 x 7.456 x 4 ² x 46
= 5.487616 kNm
= 5487616.0 Nmm
Mty = - 0.001 x qu x x
= - 0.001 x 7.456 x 4 ² x 38
= - 4.533248 kNm
= - 4533248 Nmm
0.001 x qu x x
= 0.001 x 7.456 x 4 ² x 38
= 4.533248 kNm
= 4533248.0 Nmm
●
M = ± 0.001 .
Ly = 5.0 m
Dimana :Ly
=5
= 1.25Lx = 4.0 Lx 4
Ly : Bentang terpanjang
Lx : Bentang terpendek
Dengan mengetahui nilai Ly/Lx maka dapat ditentukan nilai C sesuai tabel 13.3.2 Buku Teknik
: 46
: 46
: 38
: 38
Sehingga dapat ditentukan nilai momen sebagai berikut :
Mtx = - 0.001 x qu x x
= - 0.001 x 7.456 x 4 ² x 46
= - 5.487616 kNm
= - 5487616 Nmm
0.001 x qu x x
= 0.001 x 7.456 x 4 ² x 46
Mlx = lx² Clx
lx² Cty
Mly = lx² Cly
Plat b = Plat i = Plat l
qu . lx² . C
Sipil Hal 355, sebagai berikut :
Ctx
Clx
Cty
Cly
lx² Ctx
Mlx = lx² Clx
![Page 5: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/5.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= 5.487616 kNm
= 5487616.0 Nmm
Mty = - 0.001 x qu x x
= - 0.001 x 7.456 x 4 ² x 38
= - 4.533248 kNm
= - 4533248 Nmm
0.001 x qu x x
= 0.001 x 7.456 x 4 ² x 38
= 4.533248 kNm
= 4533248.0 Nmm
●
M = ± 0.001 .
Lx = 3 m
Dimana :Ly
=4
= 1.33Ly = 4
Lx 3Ly : Bentang terpanjang
Lx : Bentang terpendek
Dengan mengetahui nilai Ly/Lx maka dapat ditentukan nilai C sesuai tabel 13.3.2 Buku Teknik
: 46
: 46
: 38
: 38
Sehingga dapat ditentukan nilai momen sebagai berikut :
Mtx = - 0.001 x qu x x
= - 0.001 x 7.456 x 3 ² x 46
= - 3.086784 kNm
= - 3086784 Nmm
0.001 x qu x x
= 0.001 x 7.456 x 3 ² x 46
= 3.086784 kNm
lx² Cty
Mly = lx² Cly
Plat c
qu . lx² . C
Sipil Hal 355, sebagai berikut :
Ctx
Clx
Cty
Cly
lx² Ctx
Mlx = lx² Clx
![Page 6: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/6.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= 3086784.0 Nmm
Mty = - 0.001 x qu x x
= - 0.001 x 7.456 x 3 ² x 38
= - 2.549952 kNm
= - 2549952 Nmm
0.001 x qu x x
= 0.001 x 7.456 x 3 ² x 38
= 2.549952 kNm
= 2549952.0 Nmm
●
M = ± 0.001 .
Lx = 4 m
Dimana :Ly
=6
= 1.5Lx 4
Ly = 6.0 Ly : Bentang terpanjangLx : Bentang terpendek
Dengan mengetahui nilai Ly/Lx maka dapat ditentukan nilai C sesuai tabel 13.3.2 Buku Teknik
: 46
: 46
: 38
: 38
Sehingga dapat ditentukan nilai momen sebagai berikut :
Mtx = - 0.001 x qu x x
= - 0.001 x 7.456 x 4 ² x 46
= - 5.487616 kNm
= - 5487616 Nmm
0.001 x qu x x
= 0.001 x 7.456 x 4 ² x 46
= 5.487616 kNm
= 5487616.0 Nmm
lx² Cty
Mly = lx² Cly
Plat e = Plat h
qu . lx² . C
Sipil Hal 355, sebagai berikut :
Ctx
Clx
Cty
Cly
lx² Ctx
Mlx = lx² Clx
![Page 7: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/7.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Mty = - 0.001 x qu x x
= - 0.001 x 7.456 x 4 ² x 38
= - 4.533248 kNm
= - 4533248 Nmm
0.001 x qu x x
= 0.001 x 7.456 x 4 ² x 38
= 4.533248 kNm
= 4533248.0 Nmm
●
M = ± 0.001 .
Lx = 5 m
Dimana :Ly
=6
= 1.2Ly = 6 Lx 5
Ly : Bentang terpanjangLx : Bentang terpendek
Dengan mengetahui nilai Ly/Lx maka dapat ditentukan nilai C sesuai tabel 13.3.2 Buku Teknik
: 46
: 46
: 38
: 38
Sehingga dapat ditentukan nilai momen sebagai berikut :
Mtx = - 0.001 x qu x x
= - 0.001 x 7.456 x 5 ² x 46
= - 8.5744 kNm
= - 8574400 Nmm
0.001 x qu x x
= 0.001 x 7.456 x 5 ² x 46
= 8.5744 kNm
= 8574400.0 Nmm
Mty = - 0.001 x qu x x
lx² Cty
Mly = lx² Cly
Plat f
qu . lx² . C
Sipil Hal 355, sebagai berikut :
Ctx
Clx
Cty
Cly
lx² Ctx
Mlx = lx² Clx
lx² Cty
![Page 8: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/8.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= - 0.001 x 7.456 x 5 ² x 38
= - 7.0832 kNm
= - 7083200 Nmm
0.001 x qu x x
= 0.001 x 7.456 x 5 ² x 38
= 7.0832 kNm
= 7083200.0 Nmm
●
M = ± 0.001 .
Lx = 3 m
Dimana :Ly
=6
= 2Lx 3
Ly = 6Ly : Bentang terpanjangLx : Bentang terpendek
Dengan mengetahui nilai Ly/Lx maka dapat ditentukan nilai C sesuai tabel 13.3.2 Buku Teknik
: 46
: 46
: 38
: 38
Sehingga dapat ditentukan nilai momen sebagai berikut :
Mtx = - 0.001 x qu x x
= - 0.001 x 7.456 x 3 ² x 46
= - 3.086784 kNm
= - 3086784 Nmm
0.001 x qu x x
= 0.001 x 7.456 x 3 ² x 46
= 3.086784 kNm
= 3086784.0 Nmm
Mly = lx² Cly
Plat g
qu . lx² . C
Sipil Hal 355, sebagai berikut :
Ctx
Clx
Cty
Cly
lx² Ctx
Mlx = lx² Clx
![Page 9: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/9.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Mty = - 0.001 x qu x x
= - 0.001 x 7.456 x 3 ² x 38
= - 2.549952 kNm
= - 2549952 Nmm
0.001 x qu x x
= 0.001 x 7.456 x 3 ² x 38
= 2.549952 kNm
= 2549952.0 Nmm
●
M = ± 0.001 .
Ly = 5 m
Dimana :Ly
=5
= 1Lx 5
Ly = 5 Ly : Bentang terpanjangLx : Bentang terpendek
Dengan mengetahui nilai Ly/Lx maka dapat ditentukan nilai C sesuai tabel 13.3.2 Buku Teknik
: 46
: 46
: 38
: 38
Sehingga dapat ditentukan nilai momen sebagai berikut :
Mtx = - 0.001 x qu x x
= - 0.001 x 7.456 x 5 ² x 46
= - 8.5744 kNm
= - 8574400 Nmm
0.001 x qu x x
= 0.001 x 7.456 x 5 ² x 46
= 8.5744 kNm
= 8574400.0 Nmm
Mty = - 0.001 x qu x x
lx² Cty
Mly = lx² Cly
Plat j
qu . lx² . C
Sipil Hal 355, sebagai berikut :
Ctx
Clx
Cty
Cly
lx² Ctx
Mlx = lx² Clx
lx² Cty
![Page 10: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/10.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= - 0.001 x 7.456 x 5 ² x 38
= - 7.0832 kNm
= - 7083200 Nmm
0.001 x qu x x
= 0.001 x 7.456 x 5 ² x 38
= 7.0832 kNm
= 7083200.0 Nmm
●
M = ± 0.001 .
Lx = 3 m
Dimana :Ly
=5
= 1.6666667Lx 3
Ly = 5 Ly : Bentang terpanjangLx : Bentang terpendek
Dengan mengetahui nilai Ly/Lx maka dapat ditentukan nilai C sesuai tabel 13.3.2 Buku Teknik
: 46
: 46
: 38
: 38
Sehingga dapat ditentukan nilai momen sebagai berikut :
Mtx = - 0.001 x qu x x
= - 0.001 x 7.456 x 3 ² x 46
= - 3.086784 kNm
= - 3086784 Nmm
0.001 x qu x x
= 0.001 x 7.456 x 3 ² x 46
= 3.086784 kNm
= 3086784.0 Nmm
Mty = - 0.001 x qu x x
Mly = lx² Cly
Plat k
qu . lx² . C
Sipil Hal 355, sebagai berikut :
Ctx
Clx
Cty
Cly
lx² Ctx
Mlx = lx² Clx
lx² Cty
![Page 11: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/11.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= - 0.001 x 7.456 x 3 ² x 38
= - 2.549952 kNm
= - 2549952 Nmm
0.001 x qu x x
= 0.001 x 7.456 x 3 ² x 38
= 2.549952 kNm
= 2549952.0 Nmm
Dengan demikian, momen yang diperoleh dapat diperhatikan pada tabel dibawah :
PlatTumpuan Lapangan Tumpuan Lapangan
Plat A -5487616.0 5487616.0 -4533248.0 4533248.0
Plat B -5487616.0 5487616.0 -4533248.0 4533248.0
Plat C -3086784.0 3086784.0 -2549952.0 2549952.0
Plat D -5487616.0 5487616.0 -4533248.0 4533248.0
Plat E -5487616.0 5487616.0 -4533248.0 4533248.0
Plat F -8574400.0 8574400.0 -7083200.0 7083200.0
Plat G -3086784.0 3086784.0 -2549952.0 2549952.0
Plat H -5487616.0 5487616.0 -4533248.0 4533248.0
Plat I -5487616.0 5487616.0 -4533248.0 4533248.0
Plat J -8574400.0 8574400.0 -7083200.0 7083200.0
Plat K -3086784.0 3086784.0 -2549952.0 2549952.0
Plat L -5487616.0 5487616.0 -4533248.0 4533248.0
C PENULANGAN PLAT A = PLAT D
a) Tumpuan
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.85
Tulangan utama
b = 1000 mm
Mly = lx² Cly
Momen Arah x (Nmm) Momen Arah y (Nmm)
Plat Arah X
![Page 12: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/12.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
h = 120dxTul bagi
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn =Mtx
=5487616
= 6859520 NmmØ 0.8
Rn =Mn
=6859520
= 0.760058 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530,85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.760058 . 16.47059
16.470588 350
= 0.0022118838
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
Tul utama arah x
h - tebal selimut beton - ½ diameter tulangan utama arah x
½ .
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
![Page 13: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/13.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= 0.024445
ρ = 0.002 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.0040 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 392.5s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
b) Lapangan (Momen Positif)
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.85
Tulangan utama
b = 1000 mm
h = 120dx
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
mm2
mm2
mm2 ≥ mm2
Tul utama arah y
Tul utama arah x
![Page 14: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/14.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON Ih = 120
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn = =5487616
= 6859520 NmmØ 0.8
Rn =Mn
=6859520.0
= 0.7600576177 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530.85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.760058 . 16.47059
16.470588 350
= 0.0022118838
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.002 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.004 x 1000 x 95
h - tebal selimut beton - ½ diameter tulangan Utama arah x
½ .
Mlx
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
![Page 15: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/15.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 393s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
a) Tumpuan
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dyTul bagi
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn =Mty
=4533248
= 5666560 NmmØ 0.8
mm²
mm2
mm2
mm2
mm2 ≥ mm2
Plat Arah Y
Tul utama arah y
h - tebal selimut beton - ½ diameter tulangan utama y
½ .
![Page 16: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/16.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Rn =Mn
=5666560
= 0.627874 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530,85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.627874 . 16.47059
16.470588 350
= 0.0018212406
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.002 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.0040 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
![Page 17: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/17.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
s =As perlu
=380.00
= 206.5789 mm ≈ 200
Kontrol :
As ada =As . b
=78.5 x 1000
= 392.5s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
b) Lapangan (Momen Positif)
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dy
d =
= 120.00 - 20 - 10 - 10
= 85.00 mm
Mn = =4533248
= 5666560 NmmØ 0.8
Rn =Mn
=5666560.0
= 0.7842989619 N/mm²b . d² 1000 x 85 ²
m =fy
=350
= 16.47058823530.85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.784299 . 16.47059
mm2
mm2
mm2 ≥ mm2
Tul Utama arah y
Tul Utama arah x
h - tebal selimut beton - Ø tul utama arah x - ½ Ø tul Utama arah y
½ .
Mly
![Page 18: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/18.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
=16.470588
1 - 1 -350
= 0.0022838076
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.002 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.004 x 1000 x 85
= 340
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 230.8824 mm ≈ 230As perlu 340.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 341s 230
As perlu = 340.00
As ada ≥ As perlu
341 340.00 ………………….. OK
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
mm2
mm2
mm2 ≥ mm2
![Page 19: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/19.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
D PENULANGAN PLAT B = PLAT I = PLAT L
a) Tumpuan
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dxTul bagi
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn =Mtx
=5487616
= 6859520 NmmØ 0.8
Rn =Mn
=6859520
= 0.760058 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530,85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.760058 . 16.47059
16.470588 350
= 0.0022118838
=f'c
tidak lebih kecil dari =1.4
4 fy fy
Plat Arah X
Tul utama arah x
h - tebal selimut beton - ½ diameter tulangan utama arah x
½ .
ρmin
ρmin
![Page 20: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/20.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.002 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.0040 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 392.5s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
b) Lapangan (Momen Positif)
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
mm2
mm2
mm2 ≥ mm2
![Page 21: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/21.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dx
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn = =5487616
= 6859520 NmmØ 0.8
Rn =Mn
=6859520.0
= 0.7600576177 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530.85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.760058 . 16.47059
16.470588 350
= 0.0022118838
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
Tul utama arah y
Tul utama arah x
h - tebal selimut beton - ½ diameter tulangan Utama arah x
½ .
Mlx
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
![Page 22: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/22.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.002 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.004 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 393s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
a) Tumpuan
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dy
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
mm2
mm2
mm2 ≥ mm2
Plat Arah Y
![Page 23: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/23.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
h = 120dyTul bagi
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn =Mty
=4533248
= 5666560 NmmØ 0.8
Rn =Mn
=5666560
= 0.627874 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530,85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.627874 . 16.47059
16.470588 350
= 0.0018212406
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.002 < ρmin = 0.0040 < 0.024445
0.0040
Tul utama arah y
h - tebal selimut beton - ½ diameter tulangan utama y
½ .
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
![Page 24: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/24.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.0040 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 392.5s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
b) Lapangan (Momen Positif)
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dy
d =
= 120.00 - 20 - 10 - 10
Asperlu
mm²
mm2
mm2
mm2
mm2 ≥ mm2
Tul Utama arah y
Tul Utama arah x
h - tebal selimut beton - Ø tul utama arah x - ½ Ø tul Utama arah y
½ .
![Page 25: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/25.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= 85.00 mm
Mn = =4533248
= 5666560 NmmØ 0.8
Rn =Mn
=5666560.0
= 0.7842989619 N/mm²b . d² 1000 x 85 ²
m =fy
=350
= 16.47058823530.85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.784299 . 16.47059
16.470588 350
= 0.0022838076
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.002 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.004 x 1000 x 85
= 340
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.5
Mly
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
![Page 26: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/26.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
As =4
π . ²ɸ =4
x 3.14 x 10² = 78.5
s =As . b
=78.5 x 1000
= 230.8824 mm ≈ 230As perlu 340.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 341s 230
As perlu = 340.00
As ada ≥ As perlu
341 340.00 ………………….. OK
E PENULANGAN PLAT C
a) Tumpuan
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dxTul bagi
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn =Mtx
=3086784
= 3858480 NmmØ 0.8
Rn =Mn
=3858480
= 0.427532 N/mm²
mm2
mm2
mm2
mm2 ≥ mm2
Plat Arah X
Tul utama arah x
h - tebal selimut beton - ½ diameter tulangan utama arah x
½ .
![Page 27: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/27.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Rn =b . d²
=1000 x 95 ²
= 0.427532 N/mm²
m =fy
=350
= 16.47058823530,85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.427532 . 16.47059
16.470588 350
= 0.0012340628
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.001 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.0040 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
![Page 28: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/28.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Kontrol :
As ada =As . b
=78.5 x 1000
= 392.5s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
b) Lapangan (Momen Positif)
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dx
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn = =3086784
= 3858480 NmmØ 0.8
Rn =Mn
=3858480.0
= 0.42753241 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530.85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.427532 . 16.47059
16.470588 350
mm2
mm2
mm2 ≥ mm2
Tul utama arah y
Tul utama arah x
h - tebal selimut beton - ½ diameter tulangan Utama arah x
½ .
Mlx
![Page 29: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/29.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= 0.0012340628
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.001 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.004 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 393s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
mm2
mm2
mm2 ≥ mm2
Plat Arah Y
![Page 30: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/30.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
a) Tumpuan
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dyTul bagi
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn =Mty
=2549952
= 3187440 NmmØ 0.8
Rn =Mn
=3187440
= 0.353179 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530,85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.353179 . 16.47059
16.470588 350
= 0.0010176106
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
Tul utama arah y
h - tebal selimut beton - ½ diameter tulangan utama y
½ .
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
![Page 31: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/31.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.001 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.0040 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 392.5s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
b) Lapangan (Momen Positif)
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
mm2
mm2
mm2 ≥ mm2
![Page 32: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/32.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Tulangan utama
b = 1000 mm
h = 120dy
d =
= 120.00 - 20 - 10 - 10
= 85.00 mm
Mn = =2549952
= 3187440 NmmØ 0.8
Rn =Mn
=3187440.0
= 0.4411681661 N/mm²b . d² 1000 x 85 ²
m =fy
=350
= 16.47058823530.85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.441168 . 16.47059
16.470588 350
= 0.0012738437
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
Tul Utama arah y
Tul Utama arah x
h - tebal selimut beton - Ø tul utama arah x - ½ Ø tul Utama arah y
½ .
Mly
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
![Page 33: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/33.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= 0.024445
ρ = 0.001 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.004 x 1000 x 85
= 340
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 230.8824 mm ≈ 230As perlu 340.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 341s 230
As perlu = 340.00
As ada ≥ As perlu
341 340.00 ………………….. OK
F PENULANGAN PLAT E = PLAT H
a) Tumpuan
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dx
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
mm2
mm2
mm2 ≥ mm2
Plat Arah X
Tul utama arah x
![Page 34: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/34.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
h = 120Tul bagi dx
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn =Mtx
=5487616
= 6859520 NmmØ 0.8
Rn =Mn
=6859520
= 0.760058 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530,85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.760058 . 16.47059
16.470588 350
= 0.0022118838
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.002 < ρmin = 0.0040 < 0.024445
0.0040
h - tebal selimut beton - ½ diameter tulangan utama arah x
½ .
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
![Page 35: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/35.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.0040 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 392.5s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
b) Lapangan (Momen Positif)
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dx
d =
= 120.00 - 20 - 10
= 95.00 mm
Asperlu
mm²
mm2
mm2
mm2
mm2 ≥ mm2
Tul utama arah y
Tul utama arah x
h - tebal selimut beton - ½ diameter tulangan Utama arah x
½ .
![Page 36: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/36.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Mn = =5487616
= 6859520 NmmØ 0.8
Rn =Mn
=6859520.0
= 0.7600576177 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530.85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.760058 . 16.47059
16.470588 350
= 0.0022118838
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.002 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.004 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
Mlx
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
![Page 37: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/37.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 393s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
a) Tumpuan
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dyTul bagi
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn =Mty
=4533248
= 5666560 NmmØ 0.8
Rn =Mn
=5666560
= 0.627874 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530,85xf'c 0.85 x 25
mm2
mm2
mm2 ≥ mm2
Plat Arah Y
Tul utama arah y
h - tebal selimut beton - ½ diameter tulangan utama y
½ .
![Page 38: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/38.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.627874 . 16.47059
16.470588 350
= 0.0018212406
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.002 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.0040 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 392.5
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
mm2
![Page 39: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/39.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
As ada =s
=200
= 392.5
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
b) Lapangan (Momen Positif)
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dy
d =
= 120.00 - 20 - 10 - 10
= 85.00 mm
Mn = =4533248
= 5666560 NmmØ 0.8
Rn =Mn
=5666560.0
= 0.7842989619 N/mm²b . d² 1000 x 85 ²
m =fy
=350
= 16.47058823530.85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.784299 . 16.47059
16.470588 350
= 0.0022838076
=f'c
tidak lebih kecil dari =1.4
4 fy fy
mm2
mm2
mm2 ≥ mm2
Tul Utama arah y
Tul Utama arah x
h - tebal selimut beton - Ø tul utama arah x - ½ Ø tul Utama arah y
½ .
Mly
ρmin
ρmin
![Page 40: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/40.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.002 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.004 x 1000 x 85
= 340
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 230.8824 mm ≈ 230As perlu 340.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 341s 230
As perlu = 340.00
As ada ≥ As perlu
341 340.00 ………………….. OK
G PENULANGAN PLAT F
a) Tumpuan
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
mm2
mm2
mm2 ≥ mm2
Plat Arah X
![Page 41: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/41.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dxTul bagi
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn =Mtx
=8574400
= 10718000 NmmØ 0.8
Rn =Mn
=10718000
= 1.187590 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530,85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 1.18759 . 16.47059
16.470588 350
= 0.0034936298
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
Tul utama arah x
h - tebal selimut beton - ½ diameter tulangan utama arah x
½ .
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
![Page 42: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/42.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= 0.85 βfy
x600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.003 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.0040 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 392.5s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
b) Lapangan (Momen Positif)
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
mm2
mm2
mm2 ≥ mm2
![Page 43: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/43.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
b = 1000 mm
h = 120dx
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn = =8574400
= 10718000 NmmØ 0.8
Rn =Mn
=10718000.0
= 1.1875900277 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530.85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 1.18759 . 16.47059
16.470588 350
= 0.0034936298
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
Tul utama arah y
Tul utama arah x
h - tebal selimut beton - ½ diameter tulangan Utama arah x
½ .
Mlx
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
![Page 44: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/44.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
ρ = 0.003 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.004 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 393s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
a) Tumpuan
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dyTul bagi
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
mm2
mm2
mm2 ≥ mm2
Plat Arah Y
Tul utama arah y
![Page 45: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/45.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn =Mty
=7083200
= 8854000 NmmØ 0.8
Rn =Mn
=8854000
= 0.981053 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530,85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.981053 . 16.47059
16.470588 350
= 0.0028708825
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.003 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.0040 x 1000 x 95
h - tebal selimut beton - ½ diameter tulangan utama y
½ .
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
![Page 46: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/46.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 392.5s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
b) Lapangan (Momen Positif)
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dy
d =
= 120.00 - 20 - 10 - 10
= 85.00 mm
Mn = =7083200
= 8854000 NmmØ 0.8
Rn =Mn
=8854000.0
= 1.225467128 N/mm²
mm²
mm2
mm2
mm2
mm2 ≥ mm2
Tul Utama arah y
Tul Utama arah x
h - tebal selimut beton - Ø tul utama arah x - ½ Ø tul Utama arah y
½ .
Mly
![Page 47: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/47.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Rn =b . d²
=1000 x 85 ²
= 1.225467128 N/mm²
m =fy
=350
= 16.47058823530.85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 1.225467 . 16.47059
16.470588 350
= 0.003608573
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.004 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.004 x 1000 x 85
= 340
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 230.8824 mm ≈ 230As perlu 340.00
Kontrol :
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
![Page 48: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/48.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
As ada =As . b
=78.5 x 1000
= 341s 230
As perlu = 340.00
As ada ≥ As perlu
341 340.00 ………………….. OK
H PENULANGAN PLAT G
a) Tumpuan
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dxTul bagi
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn =Mtx
=3086784
= 3858480 NmmØ 0.8
Rn =Mn
=3858480
= 0.427532 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530,85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
mm2
mm2
mm2 ≥ mm2
Plat Arah X
Tul utama arah x
h - tebal selimut beton - ½ diameter tulangan utama arah x
½ .
![Page 49: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/49.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
ρ =m
1 - 1 -fy
=1
1 - 1 -2 . 0.427532 . 16.47059
16.470588 350
= 0.0012340628
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.001 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.0040 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 392.5s 200
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
mm2
![Page 50: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/50.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
b) Lapangan (Momen Positif)
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dx
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn = =3086784
= 3858480 NmmØ 0.8
Rn =Mn
=3858480.0
= 0.42753241 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530.85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.427532 . 16.47059
16.470588 350
= 0.0012340628
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =
mm2
mm2 ≥ mm2
Tul utama arah y
Tul utama arah x
h - tebal selimut beton - ½ diameter tulangan Utama arah x
½ .
Mlx
ρmin
ρmin
ρmin
ρmin
![Page 51: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/51.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
=4 x 350
= 0.0036 tidak lebih kecil dari =
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.001 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.004 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 393s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
a) Tumpuan
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
mm2
mm2
mm2 ≥ mm2
Plat Arah Y
![Page 52: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/52.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dyTul bagi
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn =Mty
=2549952
= 3187440 NmmØ 0.8
Rn =Mn
=3187440
= 0.353179 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530,85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.353179 . 16.47059
16.470588 350
= 0.0010176106
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
Tul utama arah y
h - tebal selimut beton - ½ diameter tulangan utama y
½ .
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
![Page 53: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/53.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.001 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.0040 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 392.5s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
b) Lapangan (Momen Positif)
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dy
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
mm2
mm2
mm2 ≥ mm2
Tul Utama arah y
![Page 54: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/54.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
h = 120dy
d =
= 120.00 - 20 - 10 - 10
= 85.00 mm
Mn = =2549952
= 3187440 NmmØ 0.8
Rn =Mn
=3187440.0
= 0.4411681661 N/mm²b . d² 1000 x 85 ²
m =fy
=350
= 16.47058823530.85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.441168 . 16.47059
16.470588 350
= 0.0012738437
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.001 < ρmin = 0.0040 < 0.024445
0.0040
Tul Utama arah x
h - tebal selimut beton - Ø tul utama arah x - ½ Ø tul Utama arah y
½ .
Mly
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
![Page 55: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/55.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.004 x 1000 x 85
= 340
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 230.8824 mm ≈ 230As perlu 340.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 341s 230
As perlu = 340.00
As ada ≥ As perlu
341 340.00 ………………….. OK
I PENULANGAN PLAT J
a) Tumpuan
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dxTul bagi
d =
Asperlu
mm²
mm2
mm2
mm2
mm2 ≥ mm2
Plat Arah X
Tul utama arah x
h - tebal selimut beton - ½ diameter tulangan utama arah x
![Page 56: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/56.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= 120.00 - 20 - 10
= 95.00 mm
Mn =Mtx
=8574400
= 10718000 NmmØ 0.8
Rn =Mn
=10718000
= 1.187590 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530,85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 1.18759 . 16.47059
16.470588 350
= 0.0034936298
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.003 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.0040 x 1000 x 95
= 380
½ .
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
![Page 57: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/57.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 392.5s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
b) Lapangan (Momen Positif)
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dx
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn = =8574400
= 10718000 NmmØ 0.8
Rn =Mn
=10718000.0
= 1.1875900277 N/mm²b . d² 1000 x 95 ²
mm2
mm2
mm2
mm2 ≥ mm2
Tul utama arah y
Tul utama arah x
h - tebal selimut beton - ½ diameter tulangan Utama arah x
½ .
Mlx
![Page 58: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/58.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
m =fy
=350
= 16.47058823530.85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 1.18759 . 16.47059
16.470588 350
= 0.0034936298
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.003 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.004 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 393
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
mm2
![Page 59: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/59.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
As ada =s
=200
= 393
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
a) Tumpuan
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dyTul bagi
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn =Mty
=7083200
= 8854000 NmmØ 0.8
Rn =Mn
=8854000
= 0.981053 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530,85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.981053 . 16.47059
16.470588 350
mm2
mm2
mm2 ≥ mm2
Plat Arah Y
Tul utama arah y
h - tebal selimut beton - ½ diameter tulangan utama y
½ .
![Page 60: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/60.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= 0.0028708825
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.003 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.0040 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 392.5s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
mm2
mm2
mm2 ≥ mm2
![Page 61: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/61.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
b) Lapangan (Momen Positif)
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dy
d =
= 120.00 - 20 - 10 - 10
= 85.00 mm
Mn = =7083200
= 8854000 NmmØ 0.8
Rn =Mn
=8854000.0
= 1.225467128 N/mm²b . d² 1000 x 85 ²
m =fy
=350
= 16.47058823530.85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 1.225467 . 16.47059
16.470588 350
= 0.003608573
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
Tul Utama arah y
Tul Utama arah x
h - tebal selimut beton - Ø tul utama arah x - ½ Ø tul Utama arah y
½ .
Mly
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
![Page 62: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/62.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= 0.85 βfy
x600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.004 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.004 x 1000 x 85
= 340
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 230.8824 mm ≈ 230As perlu 340.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 341s 230
As perlu = 340.00
As ada ≥ As perlu
341 340.00 ………………….. OK
J PENULANGAN PLAT K
a) Tumpuan
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
mm2
mm2
mm2 ≥ mm2
Plat Arah X
![Page 63: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/63.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Tulangan utama
b = 1000 mm
h = 120dxTul bagi
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn =Mtx
=3086784
= 3858480 NmmØ 0.8
Rn =Mn
=3858480
= 0.427532 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530,85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.427532 . 16.47059
16.470588 350
= 0.0012340628
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
Tul utama arah x
h - tebal selimut beton - ½ diameter tulangan utama arah x
½ .
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
![Page 64: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/64.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.001 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.0040 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 392.5s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
b) Lapangan (Momen Positif)
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dx
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
mm2
mm2
mm2 ≥ mm2
Tul utama arah y
![Page 65: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/65.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
h = 120dx
d =
= 120.00 - 20 - 10
= 95.00 mm
Mn = =3086784
= 3858480 NmmØ 0.8
Rn =Mn
=3858480.0
= 0.42753241 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530.85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.427532 . 16.47059
16.470588 350
= 0.0012340628
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.001 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
Tul utama arah x
h - tebal selimut beton - ½ diameter tulangan Utama arah x
½ .
Mlx
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
![Page 66: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/66.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= ρ x b x d
= 0.004 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 393s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
a) Tumpuan
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dyTul bagi
d =
= 120.00 - 20 - 10
= 95.00 mm
Asperlu
mm²
mm2
mm2
mm2
mm2 ≥ mm2
Plat Arah Y
Tul utama arah y
h - tebal selimut beton - ½ diameter tulangan utama y
½ .
![Page 67: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/67.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Mn =Mty
=2549952
= 3187440 NmmØ 0.8
Rn =Mn
=3187440
= 0.353179 N/mm²b . d² 1000 x 95 ²
m =fy
=350
= 16.47058823530,85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
m fy
=1
1 - 1 -2 . 0.353179 . 16.47059
16.470588 350
= 0.0010176106
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.001 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.0040 x 1000 x 95
= 380
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.5
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
![Page 68: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/68.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
As =4
π . ²ɸ =4
x 3.14 x 10² = 78.5
s =As . b
=78.5 x 1000
= 206.5789 mm ≈ 200As perlu 380.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 392.5s 200
As perlu = 380.00
As ada ≥ As perlu
393 380.00 ………………….. OK
b) Lapangan (Momen Positif)
Tebal plat (h) = 120 mm Lebar plat (b) = 1000
Dipakai tulanganɸ = 10 mm Mutu beton fc' = 25
tulangan bagiɸ = 6 mm Mutu tulangan fy = 350
Tebal selimut beton = 20 mm β1 = 0.00
Tulangan utama
b = 1000 mm
h = 120dy
d =
= 120.00 - 20 - 10 - 10
= 85.00 mm
Mn = =2549952
= 3187440 NmmØ 0.8
Rn =Mn
=3187440.0
= 0.4411681661 N/mm²b . d² 1000 x 85 ²
m =fy
=350
= 16.47058823530.85xf'c 0.85 x 25
ρ =1
1 - 1 -2 . Rn . m
mm2
mm2
mm2
mm2 ≥ mm2
Tul Utama arah y
Tul Utama arah x
h - tebal selimut beton - Ø tul utama arah x - ½ Ø tul Utama arah y
½ .
Mly
![Page 69: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/69.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
ρ =m
1 - 1 -fy
=1
1 - 1 -2 . 0.441168 . 16.47059
16.470588 350
= 0.0012738437
=f'c
tidak lebih kecil dari =1.4
4 fy fy
=25
= 0.0036 tidak lebih kecil dari =4 x 350
0.004
= 0.85 βf'c
x600
fy 600 + fy
= 0.85 x 0.85 x25
x600
350 600 + 350
= 0.032593985
= 0.75 x ρb
= 0.75 x 0.032594
= 0.024445
ρ = 0.001 < ρmin = 0.0040 < 0.024445
0.0040
Dengan demikian perhitungan As yakni :
= ρ x b x d
= 0.004 x 1000 x 85
= 340
Direncanakan plat dengan Ø tul = 10
As =1
π . ²ɸ =1
x 3.14 x 10² = 78.54 4
s =As . b
=78.5 x 1000
= 230.8824 mm ≈ 230As perlu 340.00
Kontrol :
As ada =As . b
=78.5 x 1000
= 341s 230
As perlu = 340.00
As ada ≥ As perlu
ρmin
ρmin
ρmin
ρmin
Dengan demikian digunakan ρ min
yakni :
ρb
ρmaks
ρmax
=
Maka digunakan ρ yakni :
Asperlu
mm²
mm2
mm2
mm2
![Page 70: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/70.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
341 340.00 ………………….. OK
Dengan demikian digunakan Tulangan utama plat yang digunakan seperti pada tabel berikut :
Plat ATul. Utama Plat arah x Tul. Utama Plat arah y
Tulangan As ada Tulangan
Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20
Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23
Tumpuan Kanan Ø 10 - 20 392.5 mm² Ø 10 - 20
Plat BTul. Utama Plat arah x Tul. Utama Plat arah y
Tulangan As ada Tulangan
Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20
Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23
Tumpuan Kanan Ø 10 - 20 392.5 mm² Ø 10 - 20
Plat CTul. Utama Plat arah x Tul. Utama Plat arah y
Tulangan As ada Tulangan
Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20
Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23
Tumpuan Kanan Ø 10 - 20 392.5 mm² Ø 10 - 20
Plat DTul. Utama Plat arah x Tul. Utama Plat arah y
Tulangan As ada Tulangan
Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20
Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23
Tumpuan Kanan Ø 10 - 20 392.5 mm² Ø 10 - 20
Plat ETul. Utama Plat arah x Tul. Utama Plat arah y
Tulangan As ada Tulangan
Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20
Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23
Tumpuan Kanan Ø 10 - 20 392.5 mm² Ø 10 - 20
Plat FTul. Utama Plat arah x Tul. Utama Plat arah y
Tulangan As ada Tulangan
Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20
Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23
Tumpuan Kanan Ø 10 - 20 392.5 mm² Ø 10 - 20
Plat GTul. Utama Plat arah x Tul. Utama Plat arah y
mm2 ≥ mm2
mm²
mm²
mm²
mm²
mm²
mm²
![Page 71: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/71.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Plat GTulangan As ada Tulangan
Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20
Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23
Tumpuan Kanan Ø 10 - 20 392.5 mm² Ø 10 - 20
Plat HTul. Utama Plat arah x Tul. Utama Plat arah y
Tulangan As ada Tulangan
Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20
Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23
Tumpuan Kanan Ø 10 - 20 392.5 mm² Ø 10 - 20
Plat ITul. Utama Plat arah x Tul. Utama Plat arah y
Tulangan As ada Tulangan
Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20
Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23
Tumpuan Kanan Ø 10 - 20 392.5 mm² Ø 10 - 20
Plat JTul. Utama Plat arah x Tul. Utama Plat arah y
Tulangan As ada Tulangan
Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20
Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23
Tumpuan Kanan Ø 10 - 20 392.5 mm² Ø 10 - 20
Plat KTul. Utama Plat arah x Tul. Utama Plat arah y
Tulangan As ada Tulangan
Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20
Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23
Tumpuan Kanan Ø 10 - 20 392.5 mm² Ø 10 - 20
Plat LTul. Utama Plat arah x Tul. Utama Plat arah y
Tulangan As ada Tulangan
Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20
Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23
Tumpuan Kanan Ø 10 - 20 392.5 mm² Ø 10 - 20
Tetapi untuk mempermudah pekerjaan, diguanakan tulangan sebagai berikut :
● Plat A
Plat arah xDipakai Tulangan Utama
Tulangan As ada As Perlu As ada > As Perlu
mm²
mm²
mm²
mm²
mm²
mm²
![Page 72: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/72.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Tumpuan Kiri Ø 10 - 20 392.5 380.0 OK
Lapangan Ø 10 - 20 392.5 mm² 380.0 mm² OK
Tumpuan Kanan Ø 10 - 20 392.5 mm² 380.0 mm² OK
Plat arah yDipakai Tulangan Utama
Tulangan As ada As Perlu As ada > As Perlu
Tumpuan Kiri Ø 10 - 20 392.5 380.0 OK
Lapangan Ø 10 - 20 392.5 mm² 340.0 mm² OK
Tumpuan Kanan Ø 10 - 20 392.5 mm² 380.0 mm² OK
● Plat B
Plat arah xDipakai Tulangan Utama
Tulangan As ada As Perlu As ada > As Perlu
Tumpuan Kiri Ø 10 - 20 392.5 380.0 OK
Lapangan Ø 10 - 20 392.5 mm² 380.0 mm² OK
Tumpuan Kanan Ø 10 - 20 392.5 mm² 380.0 mm² OK
Plat arah yDipakai Tulangan Utama
Tulangan As ada As Perlu As ada > As Perlu
Tumpuan Kiri Ø 10 - 20 392.5 380.0 OK
Lapangan Ø 10 - 20 392.5 mm² 340.0 mm² OK
Tumpuan Kanan Ø 10 - 20 392.5 mm² 380.0 mm² OK
● Plat C
Plat arah xDipakai Tulangan Utama
Tulangan As ada As Perlu As ada > As Perlu
Tumpuan Kiri Ø 10 - 20 392.5 380.0 OK
Lapangan Ø 10 - 20 392.5 mm² 380.0 mm² OK
Tumpuan Kanan Ø 10 - 20 392.5 mm² 380.0 mm² OK
Plat arah yDipakai Tulangan Utama
Tulangan As ada As Perlu As ada > As Perlu
Tumpuan Kiri Ø 10 - 20 392.5 380.0 OK
Lapangan Ø 10 - 20 392.5 mm² 340.0 mm² OK
Tumpuan Kanan Ø 10 - 20 392.5 mm² 380.0 mm² OK
● Plat D
mm² mm²
mm² mm²
mm² mm²
mm² mm²
mm² mm²
mm² mm²
![Page 73: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/73.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Plat arah xDipakai Tulangan Utama
Tulangan As ada As Perlu As ada > As Perlu
Tumpuan Kiri Ø 10 - 20 392.5 380.0 OK
Lapangan Ø 10 - 20 392.5 mm² 380.0 mm² OK
Tumpuan Kanan Ø 10 - 20 392.5 mm² 380.0 mm² OK
Plat arah yDipakai Tulangan Utama
Tulangan As ada As Perlu As ada > As Perlu
Tumpuan Kiri Ø 10 - 20 392.5 380.0 OK
Lapangan Ø 10 - 20 392.5 mm² 340.0 mm² OK
Tumpuan Kanan Ø 10 - 20 392.5 mm² 380.0 mm² OK
● Plat E
Plat arah xDipakai Tulangan Utama
Tulangan As ada As Perlu As ada > As Perlu
Tumpuan Kiri Ø 10 - 20 392.5 380.0 OK
Lapangan Ø 10 - 20 392.5 mm² 380.0 mm² OK
Tumpuan Kanan Ø 10 - 20 392.5 mm² 380.0 mm² OK
Plat arah yDipakai Tulangan Utama
Tulangan As ada As Perlu As ada > As Perlu
Tumpuan Kiri Ø 10 - 20 392.5 380.0 OK
Lapangan Ø 0 - 20 0.0 mm² 340.0 mm² ERROR
Tumpuan Kanan Ø 0 - 20 0.0 mm² 380.0 mm² ERROR
Perhitungan Tulangan Bagi
mm² mm²
mm² mm²
mm² mm²
mm² mm²
![Page 74: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/74.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
a) Tulangan bagi plat x
Direncanakan tulangan bagi : Ø 6
Tulangan bagi = 20% x As = 20% x 393 = 78.5
As =1
π . ²ɸ =1
x 3.14 x 6² = 28.264 4
s =As . b
=28.26 x 1000
= 360.0 mmAs perlu 78.50
Kontrol :
As ada =As . b
=28.3 x 1000
= 94.2s 300
As perlu = 78.50
As ada ≥ As perlu
94.20 ≥ 78.50 ………………….. OK
Maka dipakai tulangan bagi : Ø 6 - 30
b) Tulangan bagi Plat y
Direncanakan tulangan bagi : Ø 6
Tulangan bagi = 20% x As = 20% x 393 = 78.5
As =1
π . ²ɸ =1
x 3.14 x 6² = 28.264 4
s =As . b
=28.26 x 1000
= 360.0 mmAs perlu 78.50
Kontrol :
As ada =As . b
=28.3 x 1000
= 94.2s 300
As perlu = 78.50
As ada ≥ As perlu
94.20 ≥ 78.50 ………………….. OK
Maka dipakai tulangan bagi : Ø 6 - 30
mm2
mm2
mm2
mm2
mm2 ≥ mm2
mm2
mm2
mm2
mm2
mm2 ≥ mm2
![Page 75: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/75.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 76: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/76.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
PENULANGAN PLAT DUA ARAH (Plat A)
Catatan 1 :
● 0.85
● 0.89
250 Kg/m²4.0
Catatan 2 :
Tipe Tulangan (Isikan angka 1 pada tipe tulangan yang digunakan)
Tulangan Polos (Ø) 1
250 Kg/m² Tulangan Ulir (D)
6.0
250 Kg/m²
5.0
Untuk nilai β1, sebagai berikut :
Jika f'c < 30 Mpa, maka nilai β1 adalah :
Jika f'c < 30 Mpa, maka nilai β1 adalah :
![Page 77: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/77.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
2.880 kN/m
2.880 kN/m
2.50 kN/m
2.50 kN/m
Dengan mengetahui nilai Ly/Lx maka dapat ditentukan nilai C sesuai tabel 13.3.2 Buku Teknik
![Page 78: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/78.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.25
Dengan mengetahui nilai Ly/Lx maka dapat ditentukan nilai C sesuai tabel 13.3.2 Buku Teknik
![Page 79: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/79.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Dengan mengetahui nilai Ly/Lx maka dapat ditentukan nilai C sesuai tabel 13.3.2 Buku Teknik
![Page 80: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/80.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Dengan mengetahui nilai Ly/Lx maka dapat ditentukan nilai C sesuai tabel 13.3.2 Buku Teknik
![Page 81: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/81.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Dengan mengetahui nilai Ly/Lx maka dapat ditentukan nilai C sesuai tabel 13.3.2 Buku Teknik
![Page 82: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/82.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Dengan mengetahui nilai Ly/Lx maka dapat ditentukan nilai C sesuai tabel 13.3.2 Buku Teknik
![Page 83: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/83.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1
Dengan mengetahui nilai Ly/Lx maka dapat ditentukan nilai C sesuai tabel 13.3.2 Buku Teknik
![Page 84: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/84.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Dengan mengetahui nilai Ly/Lx maka dapat ditentukan nilai C sesuai tabel 13.3.2 Buku Teknik
![Page 85: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/85.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Lapangan
4533248.0
4533248.0
2549952.0
4533248.0
4533248.0
7083200.0
2549952.0
4533248.0
4533248.0
7083200.0
2549952.0
4533248.0
mm
Mpa
Mpa
Momen Arah y (Nmm)
![Page 86: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/86.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
![Page 87: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/87.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
200 mm
………………….. OK
mm
Mpa
Mpa
![Page 88: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/88.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
![Page 89: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/89.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
200 mm
………………….. OK
mm
Mpa
Mpa
![Page 90: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/90.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
200 mm
![Page 91: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/91.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
200200 mm
………………….. OK
mm
Mpa
Mpa
![Page 92: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/92.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
230 mm
………………….. OK
![Page 93: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/93.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
mm
Mpa
Mpa
![Page 94: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/94.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
200 mm
………………….. OK
mm
![Page 95: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/95.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Mpa
Mpa
1.4= 0.004
350
![Page 96: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/96.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
200 mm
………………….. OK
mm
Mpa
Mpa
![Page 97: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/97.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
![Page 98: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/98.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
200 mm
………………….. OK
mm
Mpa
Mpa
![Page 99: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/99.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
![Page 100: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/100.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
230 mm
………………….. OK
mm
Mpa
Mpa
![Page 101: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/101.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
200 mm
![Page 102: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/102.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
………………….. OK
mm
Mpa
Mpa
![Page 103: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/103.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
200 mm
………………….. OK
![Page 104: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/104.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
mm
Mpa
Mpa
1.4= 0.004
350
![Page 105: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/105.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
200 mm
………………….. OK
mm
Mpa
Mpa
![Page 106: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/106.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
![Page 107: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/107.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
230 mm
………………….. OK
mm
Mpa
Mpa
![Page 108: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/108.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
![Page 109: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/109.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
200 mm
………………….. OK
mm
Mpa
Mpa
![Page 110: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/110.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
![Page 111: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/111.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
200 mm
………………….. OK
mm
Mpa
Mpa
![Page 112: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/112.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
200 mm
![Page 113: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/113.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
………………….. OK
mm
Mpa
Mpa
![Page 114: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/114.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
230 mm
………………….. OK
![Page 115: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/115.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
mm
Mpa
Mpa
1.4= 0.004
350
![Page 116: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/116.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
200 mm
………………….. OK
mm
Mpa
Mpa
![Page 117: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/117.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
![Page 118: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/118.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
200 mm
………………….. OK
mm
Mpa
Mpa
![Page 119: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/119.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
![Page 120: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/120.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
200 mm
………………….. OK
mm
Mpa
Mpa
![Page 121: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/121.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
230 mm
![Page 122: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/122.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
………………….. OK
mm
Mpa
Mpa
![Page 123: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/123.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
200 mm
![Page 124: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/124.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
………………….. OK
mm
Mpa
Mpa
1.4= 0.004
![Page 125: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/125.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
350= 0.004
200 mm
………………….. OK
mm
Mpa
Mpa
![Page 126: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/126.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
![Page 127: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/127.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
200 mm
………………….. OK
mm
Mpa
Mpa
![Page 128: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/128.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
![Page 129: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/129.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
230 mm
………………….. OK
mm
Mpa
Mpa
![Page 130: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/130.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
![Page 131: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/131.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
200 mm
………………….. OK
mm
Mpa
Mpa
![Page 132: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/132.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
200 mm
![Page 133: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/133.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
………………….. OK
mm
Mpa
Mpa
![Page 134: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/134.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
200 mm
………………….. OK
![Page 135: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/135.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
mm
Mpa
Mpa
1.4= 0.004
350
![Page 136: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/136.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
230 mm
………………….. OK
mm
Mpa
Mpa
![Page 137: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/137.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
![Page 138: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/138.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
200 mm
………………….. OK
mm
Mpa
Mpa
![Page 139: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/139.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
![Page 140: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/140.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
200 mm
………………….. OK
mm
Mpa
Mpa
![Page 141: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/141.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
![Page 142: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/142.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
200 mm
………………….. OK
mm
Mpa
Mpa
![Page 143: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/143.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
1.4= 0.004
350
230 mm
![Page 144: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/144.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
………………….. OK
Tul. Utama Plat arah y
As ada
392.5
341.3 mm²
392.5 mm²
Tul. Utama Plat arah y
As ada
392.5
341.3 mm²
392.5 mm²
Tul. Utama Plat arah y
As ada
392.5
341.3 mm²
392.5 mm²
Tul. Utama Plat arah y
As ada
392.5
341.3 mm²
392.5 mm²
Tul. Utama Plat arah y
As ada
392.5
341.3 mm²
392.5 mm²
Tul. Utama Plat arah y
As ada
392.5
341.3 mm²
392.5 mm²
Tul. Utama Plat arah y
mm²
mm²
mm²
mm²
mm²
mm²
![Page 145: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/145.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
As ada
392.5
341.3 mm²
392.5 mm²
Tul. Utama Plat arah y
As ada
392.5
341.3 mm²
392.5 mm²
Tul. Utama Plat arah y
As ada
392.5
341.3 mm²
392.5 mm²
Tul. Utama Plat arah y
As ada
392.5
341.3 mm²
392.5 mm²
Tul. Utama Plat arah y
As ada
392.5
341.3 mm²
392.5 mm²
Tul. Utama Plat arah y
As ada
392.5
341.3 mm²
392.5 mm²
Dipakai Tulangan Utama
As ada > As Perlu
mm²
mm²
mm²
mm²
mm²
mm²
![Page 146: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/146.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
OK
OK
OK
Dipakai Tulangan Utama
As ada > As Perlu
OK
OK
OK
Dipakai Tulangan Utama
As ada > As Perlu
OK
OK
OK
Dipakai Tulangan Utama
As ada > As Perlu
OK
OK
OK
Dipakai Tulangan Utama
As ada > As Perlu
OK
OK
OK
Dipakai Tulangan Utama
As ada > As Perlu
OK
OK
OK
![Page 147: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/147.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Dipakai Tulangan Utama
As ada > As Perlu
OK
OK
OK
Dipakai Tulangan Utama
As ada > As Perlu
OK
OK
OK
Dipakai Tulangan Utama
As ada > As Perlu
OK
OK
OK
Dipakai Tulangan Utama
As ada > As Perlu
OK
ERROR
ERROR
![Page 148: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/148.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
≈ 300 mm
………………….. OK
≈ 300 mm
………………….. OK
E Syarat Minimum Spasi
● Menurut SNI 2847 - 2002 pasal 9.6, jarak bersih antar tulangan sejajar dalam selapis tidak
boleh kurang dari : 25 mm
Jarak Minimum tulangan = Err:504
![Page 149: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/149.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Jarak bersih antar tulangan utama
● Menurut PBI 1971, bab 8 pasal 8.16 (2)a, jarak tul utama tdk boleh melebihi yg terkecil dari:
- 2 . h = 2 x 120
- 200 mm
Jarak bersih antar tulangan utama
● Menurut PBI 1971, bab 8 pasal 8.16 (2)a, jarak tul bagi tidak boleh melebihi :
Jarak antar tulangan bagi
![Page 150: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/150.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
0.85
0.89
Tipe Tulangan (Isikan angka 1 pada tipe tulangan yang digunakan)
![Page 151: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/151.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 152: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/152.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 153: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/153.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 154: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/154.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 155: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/155.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 156: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/156.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 157: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/157.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 158: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/158.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 159: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/159.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 160: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/160.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 161: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/161.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 162: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/162.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 163: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/163.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 164: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/164.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 165: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/165.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 166: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/166.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 167: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/167.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 168: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/168.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 169: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/169.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 170: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/170.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 171: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/171.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 172: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/172.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 173: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/173.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 174: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/174.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 175: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/175.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 176: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/176.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 177: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/177.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 178: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/178.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 179: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/179.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 180: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/180.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 181: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/181.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 182: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/182.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 183: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/183.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 184: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/184.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 185: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/185.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 186: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/186.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 187: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/187.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 188: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/188.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 189: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/189.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 190: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/190.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 191: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/191.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 192: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/192.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 193: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/193.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 194: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/194.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 195: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/195.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 196: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/196.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 197: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/197.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 198: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/198.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 199: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/199.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 200: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/200.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 201: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/201.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 202: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/202.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 203: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/203.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 204: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/204.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 205: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/205.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 206: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/206.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 207: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/207.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 208: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/208.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 209: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/209.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 210: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/210.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 211: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/211.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 212: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/212.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 213: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/213.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 214: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/214.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 215: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/215.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 216: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/216.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 217: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/217.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 218: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/218.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 219: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/219.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 220: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/220.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 221: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/221.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
![Page 222: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/222.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
Menurut SNI 2847 - 2002 pasal 9.6, jarak bersih antar tulangan sejajar dalam selapis tidak
Err:504
![Page 223: Plat 2 arah baru](https://reader034.vdocuments.site/reader034/viewer/2022042509/577c7d3b1a28abe0549deafb/html5/thumbnails/223.jpg)
Sertin. N. M. Mooy1221103
STRUKTUR BETON I
= Jarak - (2 . ½ .
= Err:504 - (2 . ½ . 10 )
= Err:504 mm > 25 mm ###
Menurut PBI 1971, bab 8 pasal 8.16 (2)a, jarak tul utama tdk boleh melebihi yg terkecil dari:
120 = 240 mm
= Err:504 mm < 200 mm ###
Menurut PBI 1971, bab 8 pasal 8.16 (2)a, jarak tul bagi tidak boleh melebihi : 250 mm
= #REF! mm < 250 mm ###
Ø tul utama)