plat 2 arah baru

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


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


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



: 46

: 46

: 38

: 38


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


Ctx

Clx

Cty

Cly

lx² Ctx

Mlx = lx² Clx


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



: 46

: 46

: 38

: 38


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


Ctx

Clx

Cty

Cly

lx² Ctx

Mlx = lx² Clx


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


: 46

: 46

: 38

: 38


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


Ctx

Clx

Cty

Cly

lx² Ctx

Mlx = lx² Clx


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


: 46

: 46

: 38

: 38


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


Ctx

Clx

Cty

Cly

lx² Ctx

Mlx = lx² Clx

lx² Cty


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


: 46

: 46

: 38

: 38


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


Ctx

Clx

Cty

Cly

lx² Ctx

Mlx = lx² Clx


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


: 46

: 46

: 38

: 38


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


Ctx

Clx

Cty

Cly

lx² Ctx

Mlx = lx² Clx

lx² Cty


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


: 46

: 46

: 38

: 38


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


Ctx

Clx

Cty

Cly

lx² Ctx

Mlx = lx² Clx

lx² Cty


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


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


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)





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


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


4 fy fy

=25


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


= 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


yakni :

ρb

ρmaks

ρmax

=


Asperlu


STRUKTUR BETON I

= 380


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





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 .


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


4 fy fy

=25


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


= ρ x b x d

= 0.0040 x 1000 x 95

= 380


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


yakni :

ρb

ρmaks

ρmax

=


Asperlu

mm²

mm2


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






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


STRUKTUR BETON I

=16.470588

1 - 1 -350

= 0.0022838076

=f'c


4 fy fy

=25


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


= ρ x b x d

= 0.004 x 1000 x 85

= 340


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


yakni :

ρb

ρmaks

ρmax

=


Asperlu

mm²

mm2

mm2

mm2

mm2 ≥ mm2


STRUKTUR BETON I

D PENULANGAN PLAT B = PLAT I = PLAT L

a) Tumpuan





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


4 fy fy

Plat Arah X

Tul utama arah x


½ .

ρmin

ρmin


STRUKTUR BETON I

=25


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


= ρ x b x d

= 0.0040 x 1000 x 95

= 380


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


yakni :

ρb

ρmaks

ρmax

=


Asperlu

mm²

mm2

mm2

mm2

mm2 ≥ mm2


STRUKTUR BETON I




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


4 fy fy

=25


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


½ .

Mlx

ρmin

ρmin

ρmin

ρmin


yakni :

ρb


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


= ρ x b x d

= 0.004 x 1000 x 95

= 380


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





Tulangan utama

b = 1000 mm

h = 120dy

ρmaks

ρmax

=


Asperlu

mm²

mm2

mm2

mm2

mm2 ≥ mm2

Plat Arah Y


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


4 fy fy

=25


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


½ .

ρmin

ρmin

ρmin

ρmin


yakni :

ρb

ρmaks

ρmax

=



STRUKTUR BETON I


= ρ x b x d

= 0.0040 x 1000 x 95

= 380


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






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


½ .


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


4 fy fy

=25


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


= ρ x b x d

= 0.004 x 1000 x 85

= 340


As =1

π . ²ɸ =1

x 3.14 x 10² = 78.5

Mly

ρmin

ρmin

ρmin

ρmin


yakni :

ρb

ρmaks

ρmax

=


Asperlu

mm²

mm2


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





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


½ .


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


4 fy fy

=25


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


= ρ x b x d

= 0.0040 x 1000 x 95

= 380


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


yakni :

ρb

ρmaks

ρmax

=


Asperlu

mm²

mm2


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






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


½ .

Mlx


STRUKTUR BETON I

= 0.0012340628

=f'c


4 fy fy

=25


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


= ρ x b x d

= 0.004 x 1000 x 95

= 380


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


yakni :

ρb

ρmaks

ρmax

=


Asperlu

mm²

mm2

mm2

mm2

mm2 ≥ mm2

Plat Arah Y


STRUKTUR BETON I

a) Tumpuan





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


4 fy fy

=25


0.004

Tul utama arah y


½ .

ρmin

ρmin

ρmin

ρmin


yakni :


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


= ρ x b x d

= 0.0040 x 1000 x 95

= 380


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






ρb

ρmaks

ρmax

=


Asperlu

mm²

mm2

mm2

mm2

mm2 ≥ mm2


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


4 fy fy

=25


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


½ .

Mly

ρmin

ρmin

ρmin

ρmin


yakni :

ρb

ρmaks


STRUKTUR BETON I

= 0.024445

ρ = 0.001 < ρmin = 0.0040 < 0.024445

0.0040


= ρ x b x d

= 0.004 x 1000 x 85

= 340


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





Tulangan utama

b = 1000 mm

h = 120dx

ρmax

=


Asperlu

mm²

mm2

mm2

mm2

mm2 ≥ mm2

Plat Arah X

Tul utama arah x


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


4 fy fy

=25


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


½ .

ρmin

ρmin

ρmin

ρmin


yakni :

ρb

ρmaks

ρmax

=



STRUKTUR BETON I


= ρ x b x d

= 0.0040 x 1000 x 95

= 380


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






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


½ .


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


4 fy fy

=25


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


= ρ x b x d

= 0.004 x 1000 x 95

= 380


As =1

π . ²ɸ =1

x 3.14 x 10² = 78.54 4

Mlx

ρmin

ρmin

ρmin

ρmin


yakni :

ρb

ρmaks

ρmax

=


Asperlu

mm²

mm2


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





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


½ .


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


4 fy fy

=25


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


= ρ x b x d

= 0.0040 x 1000 x 95

= 380


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


yakni :

ρb

ρmaks

ρmax

=


Asperlu

mm²

mm2

mm2


STRUKTUR BETON I

As ada =s

=200

= 392.5

As perlu = 380.00

As ada ≥ As perlu

393 380.00 ………………….. OK






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


4 fy fy

mm2

mm2

mm2 ≥ mm2

Tul Utama arah y

Tul Utama arah x


½ .

Mly

ρmin

ρmin


STRUKTUR BETON I

=25


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


= ρ x b x d

= 0.004 x 1000 x 85

= 340


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


yakni :

ρb

ρmaks

ρmax

=


Asperlu

mm²

mm2

mm2

mm2

mm2 ≥ mm2

Plat Arah X


STRUKTUR BETON I





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


4 fy fy

=25


0.004

= 0.85 βf'c

x600

Tul utama arah x


½ .

ρmin

ρmin

ρmin

ρmin


yakni :

ρb


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


= ρ x b x d

= 0.0040 x 1000 x 95

= 380


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






Tulangan utama

ρb

ρmaks

ρmax

=


Asperlu

mm²

mm2

mm2

mm2

mm2 ≥ mm2


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


4 fy fy

=25


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


½ .

Mlx

ρmin

ρmin

ρmin

ρmin


yakni :

ρb

ρmaks


STRUKTUR BETON I

ρ = 0.003 < ρmin = 0.0040 < 0.024445

0.0040


= ρ x b x d

= 0.004 x 1000 x 95

= 380


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





Tulangan utama

b = 1000 mm

h = 120dyTul bagi

ρmax

=


Asperlu

mm²

mm2

mm2

mm2

mm2 ≥ mm2

Plat Arah Y

Tul utama arah y


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


4 fy fy

=25


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


= ρ x b x d

= 0.0040 x 1000 x 95


½ .

ρmin

ρmin

ρmin

ρmin


yakni :

ρb

ρmaks

ρmax

=


Asperlu


STRUKTUR BETON I

= 380


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






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


½ .

Mly


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


4 fy fy

=25


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


= ρ x b x d

= 0.004 x 1000 x 85

= 340


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


yakni :

ρb

ρmaks

ρmax

=


Asperlu

mm²

mm2


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





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


½ .


STRUKTUR BETON I

ρ =m

1 - 1 -fy

=1

1 - 1 -2 . 0.427532 . 16.47059

16.470588 350

= 0.0012340628

=f'c


4 fy fy

=25


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


= ρ x b x d

= 0.0040 x 1000 x 95

= 380


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


yakni :

ρb

ρmaks

ρmax

=


Asperlu

mm²

mm2

mm2


STRUKTUR BETON I

As perlu = 380.00

As ada ≥ As perlu

393 380.00 ………………….. OK






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


4 fy fy

=25

= 0.0036 tidak lebih kecil dari =

mm2

mm2 ≥ mm2

Tul utama arah y

Tul utama arah x


½ .

Mlx

ρmin

ρmin

ρmin

ρmin


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


= ρ x b x d

= 0.004 x 1000 x 95

= 380


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




ρmin

ρmin


yakni :

ρb

ρmaks

ρmax

=


Asperlu

mm²

mm2

mm2

mm2

mm2 ≥ mm2

Plat Arah Y


STRUKTUR BETON I


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


4 fy fy

=25


0.004

= 0.85 βf'c

x600

fy 600 + fy

= 0.85 x 0.85 x25

x600

350 600 + 350

Tul utama arah y


½ .

ρmin

ρmin

ρmin

ρmin


yakni :

ρb


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


= ρ x b x d

= 0.0040 x 1000 x 95

= 380


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






Tulangan utama

b = 1000 mm

h = 120dy

ρmaks

ρmax

=


Asperlu

mm²

mm2

mm2

mm2

mm2 ≥ mm2

Tul Utama arah y


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


4 fy fy

=25


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


½ .

Mly

ρmin

ρmin

ρmin

ρmin


yakni :

ρb

ρmaks

ρmax

=



STRUKTUR BETON I


= ρ x b x d

= 0.004 x 1000 x 85

= 340


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





Tulangan utama

b = 1000 mm

h = 120dxTul bagi

d =

Asperlu

mm²

mm2

mm2

mm2

mm2 ≥ mm2

Plat Arah X

Tul utama arah x



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


4 fy fy

=25


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


= ρ x b x d

= 0.0040 x 1000 x 95

= 380

½ .

ρmin

ρmin

ρmin

ρmin


yakni :

ρb

ρmaks

ρmax

=


Asperlu

mm²


STRUKTUR BETON I


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






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


½ .

Mlx


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


4 fy fy

=25


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


= ρ x b x d

= 0.004 x 1000 x 95

= 380


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


yakni :

ρb

ρmaks

ρmax

=


Asperlu

mm²

mm2

mm2


STRUKTUR BETON I

As ada =s

=200

= 393

As perlu = 380.00

As ada ≥ As perlu

393 380.00 ………………….. OK

a) Tumpuan





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


½ .


STRUKTUR BETON I

= 0.0028708825

=f'c


4 fy fy

=25


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


= ρ x b x d

= 0.0040 x 1000 x 95

= 380


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


yakni :

ρb

ρmaks

ρmax

=


Asperlu

mm²

mm2

mm2

mm2

mm2 ≥ mm2


STRUKTUR BETON I






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


4 fy fy

=25


0.004

= 0.85 βf'c

x600

Tul Utama arah y

Tul Utama arah x


½ .

Mly

ρmin

ρmin

ρmin

ρmin


yakni :

ρb


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


= ρ x b x d

= 0.004 x 1000 x 85

= 340


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





ρb

ρmaks

ρmax

=


Asperlu

mm²

mm2

mm2

mm2

mm2 ≥ mm2

Plat Arah X


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


4 fy fy

=25


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


½ .

ρmin

ρmin

ρmin

ρmin


yakni :

ρb


STRUKTUR BETON I

= 0.75 x ρb

= 0.75 x 0.032594

= 0.024445

ρ = 0.001 < ρmin = 0.0040 < 0.024445

0.0040


= ρ x b x d

= 0.0040 x 1000 x 95

= 380


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






Tulangan utama

b = 1000 mm

h = 120dx

ρmaks

ρmax

=


Asperlu

mm²

mm2

mm2

mm2

mm2 ≥ mm2

Tul utama arah y


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


4 fy fy

=25


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


½ .

Mlx

ρmin

ρmin

ρmin

ρmin


yakni :

ρb

ρmaks

ρmax

=



STRUKTUR BETON I

= ρ x b x d

= 0.004 x 1000 x 95

= 380


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





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


½ .


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


4 fy fy

=25


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


= ρ x b x d

= 0.0040 x 1000 x 95

= 380


As =1

π . ²ɸ =1

x 3.14 x 10² = 78.5

ρmin

ρmin

ρmin

ρmin


yakni :

ρb

ρmaks

ρmax

=


Asperlu

mm²

mm2


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






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


½ .

Mly


STRUKTUR BETON I

ρ =m

1 - 1 -fy

=1

1 - 1 -2 . 0.441168 . 16.47059

16.470588 350

= 0.0012738437

=f'c


4 fy fy

=25


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


= ρ x b x d

= 0.004 x 1000 x 85

= 340


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


yakni :

ρb

ρmaks

ρmax

=


Asperlu

mm²

mm2

mm2

mm2


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


Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20

Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23


Plat CTul. Utama Plat arah x Tul. Utama Plat arah y


Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20

Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23


Plat DTul. Utama Plat arah x Tul. Utama Plat arah y


Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20

Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23


Plat ETul. Utama Plat arah x Tul. Utama Plat arah y


Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20

Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23


Plat FTul. Utama Plat arah x Tul. Utama Plat arah y


Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20

Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23


Plat GTul. Utama Plat arah x Tul. Utama Plat arah y

mm2 ≥ mm2

mm²

mm²

mm²

mm²

mm²

mm²


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


Plat HTul. Utama Plat arah x Tul. Utama Plat arah y


Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20

Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23


Plat ITul. Utama Plat arah x Tul. Utama Plat arah y


Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20

Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23


Plat JTul. Utama Plat arah x Tul. Utama Plat arah y


Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20

Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23


Plat KTul. Utama Plat arah x Tul. Utama Plat arah y


Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20

Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23


Plat LTul. Utama Plat arah x Tul. Utama Plat arah y


Tumpuan Kiri Ø 10 - 20 392.5 Ø 10 - 20

Lapangan Ø 10 - 20 392.5 mm² Ø 10 - 23


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


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


Tumpuan Kiri Ø 10 - 20 392.5 380.0 OK



● Plat B



Tumpuan Kiri Ø 10 - 20 392.5 380.0 OK





Tumpuan Kiri Ø 10 - 20 392.5 380.0 OK



● Plat C



Tumpuan Kiri Ø 10 - 20 392.5 380.0 OK





Tumpuan Kiri Ø 10 - 20 392.5 380.0 OK



● Plat D

mm² mm²

mm² mm²

mm² mm²

mm² mm²

mm² mm²

mm² mm²


STRUKTUR BETON I



Tumpuan Kiri Ø 10 - 20 392.5 380.0 OK





Tumpuan Kiri Ø 10 - 20 392.5 380.0 OK



● Plat E



Tumpuan Kiri Ø 10 - 20 392.5 380.0 OK





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


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


STRUKTUR BETON I


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 :


STRUKTUR BETON I

2.880 kN/m

2.880 kN/m

2.50 kN/m

2.50 kN/m



STRUKTUR BETON I

1.25



STRUKTUR BETON I



STRUKTUR BETON I

1



STRUKTUR BETON I



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)


STRUKTUR BETON I

1.4= 0.004

350


STRUKTUR BETON I

200 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350


STRUKTUR BETON I

200 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350

200 mm


STRUKTUR BETON I

200200 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350

230 mm

………………….. OK


STRUKTUR BETON I

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350

200 mm

………………….. OK

mm


STRUKTUR BETON I

Mpa

Mpa

1.4= 0.004

350


STRUKTUR BETON I

200 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350


STRUKTUR BETON I

200 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350


STRUKTUR BETON I

230 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350

200 mm


STRUKTUR BETON I

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350

200 mm

………………….. OK


STRUKTUR BETON I

mm

Mpa

Mpa

1.4= 0.004

350


STRUKTUR BETON I

200 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350


STRUKTUR BETON I

230 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350


STRUKTUR BETON I

200 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350


STRUKTUR BETON I

200 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350

200 mm


STRUKTUR BETON I

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350

230 mm

………………….. OK


STRUKTUR BETON I

mm

Mpa

Mpa

1.4= 0.004

350


STRUKTUR BETON I

200 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350


STRUKTUR BETON I

200 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350


STRUKTUR BETON I

200 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350

230 mm


STRUKTUR BETON I

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350

200 mm


STRUKTUR BETON I

………………….. OK

mm

Mpa

Mpa

1.4= 0.004


STRUKTUR BETON I

350= 0.004

200 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350


STRUKTUR BETON I

200 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350


STRUKTUR BETON I

230 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350


STRUKTUR BETON I

200 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350

200 mm


STRUKTUR BETON I

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350

200 mm

………………….. OK


STRUKTUR BETON I

mm

Mpa

Mpa

1.4= 0.004

350


STRUKTUR BETON I

230 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350


STRUKTUR BETON I

200 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350


STRUKTUR BETON I

200 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350


STRUKTUR BETON I

200 mm

………………….. OK

mm

Mpa

Mpa


STRUKTUR BETON I

1.4= 0.004

350

230 mm


STRUKTUR BETON I

………………….. OK

Tul. Utama Plat arah y

As ada

392.5

341.3 mm²

392.5 mm²


As ada

392.5

341.3 mm²

392.5 mm²


As ada

392.5

341.3 mm²

392.5 mm²


As ada

392.5

341.3 mm²

392.5 mm²


As ada

392.5

341.3 mm²

392.5 mm²


As ada

392.5

341.3 mm²

392.5 mm²


mm²

mm²

mm²

mm²

mm²

mm²


STRUKTUR BETON I

As ada

392.5

341.3 mm²

392.5 mm²


As ada

392.5

341.3 mm²

392.5 mm²


As ada

392.5

341.3 mm²

392.5 mm²


As ada

392.5

341.3 mm²

392.5 mm²


As ada

392.5

341.3 mm²

392.5 mm²


As ada

392.5

341.3 mm²

392.5 mm²

Dipakai Tulangan Utama

As ada > As Perlu

mm²

mm²

mm²

mm²

mm²

mm²


STRUKTUR BETON I

OK

OK

OK


As ada > As Perlu

OK

OK

OK


As ada > As Perlu

OK

OK

OK


As ada > As Perlu

OK

OK

OK


As ada > As Perlu

OK

OK

OK


As ada > As Perlu

OK

OK

OK


STRUKTUR BETON I


As ada > As Perlu

OK

OK

OK


As ada > As Perlu

OK

OK

OK


As ada > As Perlu

OK

OK

OK


As ada > As Perlu

OK

ERROR

ERROR


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


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


STRUKTUR BETON I

0.85

0.89

Tipe Tulangan (Isikan angka 1 pada tipe tulangan yang digunakan)


STRUKTUR BETON I


STRUKTUR BETON I

Menurut SNI 2847 - 2002 pasal 9.6, jarak bersih antar tulangan sejajar dalam selapis tidak

Err:504


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)

plat 2 arah baru

Documents