Tugas Beton Nando

8/18/2019 Tugas Beton Nando

1/34

Soal :

A. Pembebanan

a. Pembebanan plat Tipe a, c

Bentang arah y = 5 m = mm

1. Data Perencanaan

Tebal Pelat = m = mmfc' = Mpa = Nmm

fy = Mpa = Nmm

Beban Hidup Pelat = = kN/m²

Berat sendiri beton = kg/m³ = kN/m³

2. Pembebanan ( diambil lebar pias plat yang dihitung sebesar : 1 m )

Beban mati :

Berat sendiri = x 1 x =

= kN/m

Beban guna lantai (Hidup) :

Berat beban guna atap = 1 x = kN/m

ql = kN/m

Beban terfaktor qu = 1.2 qd + 1.6 ql = x + x

qu = kN/m

2,501,602,88

2,500

2,500

PERENCANAAN PELAT LANTAI

350

0,12

25

250

0,12 24,00 2,880

2400

120

25

350

2,5

2,880qd

5000

2,50

1,20

7,456

PENYELESAIAN :

24

6 m 5 m 6 m 5 m

6 m

5 m

6 m

a 12

250

13

300

12

250

13

300

13 12 13 12

12 13 12 13

b c d

e f g h

lk ji

300 300

300 300

250 250

250250

y

x

A B C D E

I

H

G

F

Kg/m2


2/34

b. Pembebanan plat Tipe i, k, f, h


1. Data Perencanaan

Tebal Pelat = m = mm

fc' = Mpa = Nmm

fy = Mpa = Nmm


Berat sendiri beton = Kg/m³ = kN/m³


Beban mati :


= kN/m



ql = kN/m


qu = kN/m

c. Pembebanan plat Tipe e, g, j, l


1. Data Perencanaan


fc' = Mpa = Nmm

fy = Mpa = Nmm



2. Pembebanan ( diambil lebar pias plat yang dihitung sebesar : 1 m )Beban mati :


= kN/m



ql = kN/m


qu = kN/m

d. Pembebanan plat Tipe b,dBentang arah y = 5 m = mm

1. Data Perencanaan


fc' = Mpa = Nmm

fy = Mpa = Nmm



3,000

1,20 3,12 1,60 3,00

8,544

2,880

qd 2,880

2,50 2,500

2,500

1,20 2,88 1,60 2,50

7,456

6000

5000

6000

0,13 130

25 25

350 350

300 Kg/m² 3,0

2400 24

0,13 24,00 3,120

qd 3,120

3,00 3,000

350 350

300 Kg/m² 3,0

2400 24

0,12 120

25 25

350 350

250 Kg/m² 2,5

2400 24

0,12 24,00

0,13 130

25 25


3/34


Beban mati :


= kN/m



ql = kN/m


qu = kN/m

B. Perhitungan Momen

Dengan menggunakan koefisien momen, didapat :1 Plat i = Plat k

Diketahui : Panjang bentang = m

qu = kN/m

● Momen Tumpuan Kiri

1 2 1 2

24 24

● Momen Lapangan

1 2 1 2

14 14

● Momen Tumpuan Kanan

1 2 1 210 10

2 Plat j = Plat l


qu = kN/m


1 2 1 2

24 24

● Momen Lapangan

1 2 1 2

14 14


1 2 1 2

10 10

3 Plat e = Plat g


qu = kN/m


1 2 1 2

11 11

● Momen Lapangan

1 2 1 216 16


= x

0,13 24,00 3,120

qd 3,120

3,00

qu x L

kNm= x 7,456 x 6Mu = x qu x L

x 8,544 x 6

= 26,8416

= 11,184

Mu = x qu x L = x 7,456 x 6

x7,456 6

3,000

3,000

1,20 3,12 1,60 3,00

8,544

= 12,816 kNm

Mu = x qu x L = x 8,544 x 6 = 21,97029 kNm

6

7,456

6,00

8,544

Mu = x qu x L =

kNm

= 19,17257 kNm

Mu = x

x 8,544 x

6 = 30,7584 kNm

6,00

8,544

Mu = x qu x L = x 8,544 x 6 = 27,96218 kNm

Mu = x qu x L = x 8,544 x

6 = 19,224 kNmMu = x qu x L =

I H G F

1/24

1/14

1/10

1/16 1/14

1/241/11 1/101/11

6 m 5 m6 m

http://newkidjoy.blogspot.com/2011/03/perencanaan-struktur-beton-pelat-lantai.htmlhttp://newkidjoy.blogspot.com/2011/03/perencanaan-struktur-beton-pelat-lantai.html


4/34

1 2 1 2

11 11

4 Plat f = Plat h


qu = kN/m


1 2 1 2

11 11

● Momen Lapangan

1 2 1 2

16 16


1 2 1 2

11 11

5 Plat a = Plat c


qu = kN/m● Momen Tumpuan Kiri

1 2 1 2

10 10

● Momen Lapangan

1 2 1 2

14 14


1 2 1 2

24 24

6 Plat b = Plat dDiketahui : Panjang bentang = m

qu = kN/m


1 2 1 2

10 10

● Momen Lapangan

1 2 1 2

14 14


1 2 1 2

24 24

6,00

7,456

Mu = x qu x L = x

Mu = x qu x L = x 8,544 x 6 = 27,96218 kNm

7,456 x 6 = 24,40145 kNm

Mu = x qu x L = x 7,456 x 6 = 16,776 kNm

x 7,456 x

6 = 24,40145 kNm

5,00

7,456

Mu = x qu x L = x 7,456 x 5 = 18,64 kNm

Mu = x qu x L = x 7,456 x

5,00

8,544

Mu = x qu x L = x

5 = 13,31429 kNm

Mu = x qu x L = x 7,456 x 5 = 7,766667 kNm

Mu = x qu x L =

8,544 x 5 = 21,36 kNm

Mu = x qu x L = x 8,544 x 5 = 15,25714 kNm

5 = 8,9 kNmMu = x qu x L = x 8,544 x


5/34

Berikut rekapan momen dapat diperhatikan pada tabel berikut :

Plat a

Plat b

Plat c

Plat d

Plat e

Plat f

Plat g

Plat h

Plat i

Plat j

Plat k

Plat l

C. Penulangan Plat1. Penulangan Plat a dan c

a) Tumpuan Kiri

Tebal pelat (h) = mm

Tebal penutup (p) = mm

Ditentukan diameter tulangan f p = mm

Tinggi efektif d = h – p – ½ f p

= - - ½ = mm

fc' = Mpa = , untuk fc' < 30 Mpa

fy = Mpaβ1 fc'

+ fy

x x

+

ρ maks = x

= x

=

ρ min =

Mu = kNm = x Nmm

x

x 94 ²

fc x 25

1 2 m Rnm

=0,85 0,85

0,85=

fy

600

600

0,0244

= 0,0326

0,75

0,75 0,0326

25

350x

600

600

350

x

25

350

0,85

12

120 20 12 94

Rn =Mn

= =1000b . d²

18,640 18,64

Mn =Mu

ɸ=

18,64

0,8

ρ = 1 -fy

=m0,85

fy=

0,85

350

Tumpuan Kiri Lapangan Tumpuan Kanan

MOMEN (kNm)

19,173 26,842

21,970 30,758

19,173

18,640

21,360

18,640

21,360

27,962

24,401

27,962

24,401

11,184

ρ b

fy

1,4=

1,4

350=

26,842

21,970 30,758

PLAT

12,816

11,184

12,816

13,314 7,767

15,257 8,900

13,314 7,767

15,257 8,900

19,224 27,962

16,776 24,401

19,224 27,962

16,776 24,401

0,004

10 ⁶

10 ⁶23300000,000 Nmm

2,636939792 N/mm²23300000,00

120

20

16,47059=

=

M terbesar 30,75821,97027,962

1

ρ


6/34

2 x x

=

ρ min = < ρ = < ρ maks=

ρ =

Asperlu = ρ b d = x x = mm

Diperlukan tulangan Ø 12 -

1 22 2 2

4 7

As ada > As perlu OK

Maka, Dipakai tulangan Ø 12 -

Menentukan tulangan bagi tumpuan kiri


As bagi = 20% As perlu 20% x = mm2

As bagi = 0,002 . b . h = x x = mm2

→ Dipilih yang terbesar, As bagi mm2

s =

= x x2

x

= mm

s ≤ 5 h = 5 x =

s ≤ mm

→ Dipilih yang terkeci ≈ mm

As ada =

= x x2

x

=mm

2

> As bagiAs ada > As perlu OK

→ dipakai tulangan 8 - mm200

200

1/4 . π .Ф2 .b

s

0,25 3,142 8 1000

200

251,327

758,627 151,725

1000 240120

8

0,002

240

1/4 . π . Ф2 .b

= ρ1

16,47059-

As ada x x 12 x1000

120

10000,008071 758,6273

120

= = 942,9 mm

116,47059 2,63694

1350

0,008071

0,008071dipakai

0,004 0,0080705 0,0244

94

120

As bagi

0,25 3,142 8 1000

240

209,44

120 600

450

209,440


7/34

b) Lapangan





= - - ½ = mm


fy = Mpa

β1 fc'

+ fy

x x

+

ρ maks = x

= x

=

ρ min =

Mu = kNm = x Nmm

x

x 94 ²

fc x 25

1 2 m Rn

m

2 x x

=


ρ =



1 22 2 2

4 7

As ada > As perluOK


530,5197

200

As ada

=0,85 0,85 25

x600

350 600 350

0,75

0,75 0,0326

0,0244

1,4

16,47059 1,883528

= 0,0326

=1,4

=

350

0,005644

0,004 0,0056438 0,0244

dipakai 0,005644

0,005644 1000 94

120

20

12

0,004fy 350

13,314 13,314286 10 ⁶

Mn =Mu

=13,314 10 ⁶

= 16642857,143

120 20 12 94

25 0,85

350

ρ b =0,85

x600

fy 600

m =fy

=350

= 16,470590,85 0,85

Nmmɸ 0,8

Rn =Mn

=16642857,14

= 1,883528423 N/mm²b . d² 1000

12 x1000

= 565,7 mm200

ρ = 1 -fy

ρ =1

1 1 -16,47059

200

ρb

= x x

1


8/34

Menentukan tulangan bagi Lapangan



As bagi = 0,002 . b . h = x x = mm2


s =

= x x2

x

= mm

s ≤ 5 h = 5 x =

s ≤ mm


As ada =

= x x2

x

= mm2

> As bagi


→ dipakai tulangan 8 - mm

c) Tumpuan Kanan


Tebal penutup (p) = mmDitentukan diameter tulangan f p = mm


= - - ½ = mm


fy = Mpa

β1 fc'

+ fy

x x

+

ρ maks = x

= x

=

ρ min =

Mu = kNm = x Nmm

x

20 100

450

209,440 200

1/4 . π .Ф2 .b

s

0,25 3,142 8 1000

200

251,327

200

8

530,520 106,104

0,002 1000 120 240

240

1/4 . π . Ф2 .b

As bagi

0,25 3,142 8 1000

240

209,44

7,767 7,7666667 10 ⁶

0,0244

1,4=

1,4= 0,004

fy 350

0,85

ρ b =0,85

x600

fy 600

=0,85 25

x600

94

25 0,85

350

Mn =Mu

=7,767 10 ⁶

= 9708333,333 Nmmɸ 0,8

= 0,0326350 600 350

0,75 ρb

0,75 0,0326

120

20

12

120 20 12

1


9/34

x 94 ²

fc x 25

1 2 m Rn

m

2 x x

=


ρ =



1 22 2 2

4 7



Menentukan tulangan bagi tumpuan kanan



As bagi = 0,002 . b . h = x x = mm2


s =

= x x2

x

= mm

s ≤ 5 h = 5 x =

s ≤ mm


As ada =

= x x2

x

= mm2

> As bagi



0,002 1000 120 240

240

1/4 . π . Ф2 .b

As bagi

0,25 3,142 8 1000

240

209,44

20 100

450

209,440 200

8

303,137 60,627

350

1000

16,47059 1,098725

m =fy

=350

= 16,470590,85 0,85

Rn =Mn

=9708333,33

= 1,098724913 N/mm²b . d²

200

As ada = x x 12 x1000

=

0,003225

0,004 0,0032249 0,0244

dipakai 0,003225

0,003225 1000 94 303,1367

ρ = 1 -fy

ρ =1

1

565,7 mm200

200

1 -16,47059

0,25

1/4 . π .Ф2 .b

s

3,142 8 1000

200

251,327

200


10/34

2. Penulangan Plat b dan d

a) Tumpuan Kiri





= - - ½ = mm


fy = Mpa

β1 fc'

+ fy

x x

+

ρ maks = x

= x

= ρ min =

Mu = = x Nmm

x

x 104 ²

fc x 25

1 2 m Rn

m

2 x x

=


ρ =



1 22 2 2

4 7



As ada x x 12 x1000

= 942,9 mm120

120

1 -16,47059 2,468565

16,47059 350

0,007519

0,004 0,0075186 0,0244

dipakai 0,007519

0,007519 1000 104 781,9319

120

10 ⁶

Mn =Mu

=21,36 10 ⁶

= 26700000,000 Nmmɸ 0,8

Rn =Mn

=26700000,00

= 2,468565089 N/mm²b . d² 1000

kNm

x600

fy 600

=0,85 0,85 25

x600

= 0,0326350 600 350

0,75

0,75 0,0326

ρb

130

20

12

130 20 12 104

25 0,85

350

ρ b =0,85

0,02441,4

=1,4

= 0,004fy 350

21,360 21,36

m =fy

=350

=

= 16,470590,85 0,85

ρ = 1 -fy

ρ =1

1

1


11/34




As bagi = 0,002 . b . h = x x = mm2


s =

= x x2

x

= mm

s ≤ 5 h = 5 x =

s ≤ mm


As ada =

= x x2

x

= mm2

> As bagi



b) Lapangan




= - - ½ = mm


fy = Mpa

β1 fc'

+ fy

x x

+

ρ maks = x

= x

=

ρ min =

Mu = kNm = x Nmm

x

x 104 ²

fc x 25

Nmmɸ 0,8

Rn =Mn

=19071428,57

= 1,763260778 N/mm²

b . d² 1000

m =fy

=350

= 16,470590,85

0,75 ρb

0,75 0,03260,0244

1,4= = 0,004

fy

15,257 15,257143 10 ⁶

Mn =Mu

=15,257 10 ⁶

= 19071428,571

=0,85

x600

fy 600

=0,85 0,85 25

x600

= 0,0326350 600 350

193,329 180

1/4 . π .Ф2 .b

s

0,25 3,142 8 1000

180

279,253

180

130

20

12

130 20 12

8

781,932 156,386

0,002 1000 130 260

260

1/4 . π . Ф2 .b

As bagi

0,25 3,142 8 1000

260

193,33

130 650

450

1,4

350

104

25 0,85

350

ρ b

0,85

1


12/34

1 2 m Rn

m

2 x x

=


ρ =



1 22 2 2

4 7






As bagi = 0,002 . b . h = x x = mm2


s =

= x x2

x

= mm

s ≤ 5 h = 5 x =

s ≤ mm


As ada =

= x x2

x

= mm2

> As bagi



260

1/4 . π . Ф2 .b

As bagi

0,25 3,142 8 1000

260

193,33

20 100

450

193,329 180

1/4 . π .Ф2 .b

s

0,25 3,142 8

0,005266 1000 104 547,6935

200

As ada = x x 12 x1000

= 565,7 mm200

ρ =1

1

- ρ = 1fy

1 -16,47059 1,763261

35016,47059

0,005266

0,004 0,0052663 0,0244

dipakai 0,005266

200

8

547,693 109,539

0,002 1000 130 260

1000

180

279,253

180


13/34

c) Tumpuan Kanan





= - - ½ = mm


fy = Mpa

β1 fc'

+ fy

x x

+

ρ maks = x

= x

=

ρ min =

Mu = kNm = x Nmm

x

x 104 ²

fc x 25

1 2 m Rn

m

2 x x

=


ρ =



1 22 2 2

4 7

As ada > As perluOK


1,028569

16,47059 350

0,003014

0,004 0,0030136 0,0244

dipakai 0,003014

0,003014 1000 104 313,4099

200

As ada = x x 12 x1000

= 565,7 mm200

Rn =Mn

=11125000,00

= 1,028568787 N/mm²b . d² 1000

m =fy

=350

= 16,470590,85 0,85

0,0326350 600 350

0,75 ρb

0,75 0,0326

0,0244

1,4 1,4= 0,004

fy 350

8,9 10 ⁶

20

12

130 12 104

25 0,85

350

ρ b = x600

fy 600

=25

x600

130

0,85

20

=

0,85 0,85=

ɸ

8,900

Mn =Mu

=8,900 10 ⁶

= 11125000,000 Nmm0,8

ρ = -1fy

ρ =1

1 1 -16,47059

200

1


14/34




As bagi = 0,002 . b . h = x x = mm2


s =

= x x2

x

= mm

s ≤ 5 h = 5 x =

s ≤ mm


As ada =

= x x 2 x

= mm2

> As bagi



3. Penulangan Plat e dan g

a) Tumpuan Kiri




= - - ½ = mm


fy = Mpa

β1 fc'

+ fy

x x

+

ρ maks = x

= x

=

ρ min =

Mu = = x Nmm

x

x 104 ²

fc x 25

= 34952727,273 Nmmɸ 0,8

Rn =Mn

=34952727,27

= 3,231576116 N/mm²

b . d² 1000

m =fy

=350

= 16,470590,85

350

0,75 ρb

0,75 0,03260,0244

1,4=

1,4= 0,004

fy 350

27,962 kNm 27,962182 10 ⁶

130

20

12

130 20 12 104

25 0,85

350

ρ b =0,85

x600

fy 600

1000 130 260

260

1/4 . π . Ф2 .b

As bagi

0,25 3,142 8 1000

260

193,33

20 100

450

193,329 180

8

313,410 62,682

0,002

1/4 . π .Ф2 .b

s

0,25 3,142 8 1000

180

279,253

180

=0,85 0,85 25

x600

= 0,0326350 600

0,85

10 ⁶Mn =

Mu=

27,96218

1


15/34

1 2 m Rn

m

2 x x

=


ρ =



1 22 2 2

4 7






As bagi = 0,002 . b . h = x x = mm2


s =

= x x2

x

= mm

s ≤ 5 h = 5 x =

s ≤ mm


As ada =

= x x2

x

= mm2

> As bagi



180

279,253

180

2601/4 . π . Ф

2 .b

As bagi

0,25 3,142 8 1000

260

193,33

130 650

450

193,329 180

1/4 . π .Ф2

.b

s

0,25 3,142 8

As ada = x x 12 x1000

= 1131 mm

100

8

1047,05 209,410

0,002 1000 130 260

ρ = 1 -fy

ρ =

1

1 1 -

16,47059 3,231576

16,47059 350

0,010068

0,004 0,0100678 0,0244

dipakai 0,010068

100

0,010068 1000 104 1047,052

100

1000


16/34

b) Lapangan





= - - ½ = mm


fy = Mpa

β1 fc'

+ fy

x x

+

ρ maks = x

= x

=

ρ min =

Mu = kNm = x Nmm

x

x 104 ²

fc x 25

1 2 m Rn

m

2 x x

=


ρ =



1 22 2 2

4 7

As ada > As perluOK

Maka, Dipakai tulangan Ø 12 - 140

0,00672

0,004 0,0067196 0,0244

dipakai 0,00672

0,00672 1000 104 698,8369

140

As ada = x x 12 x1000

= 808,2 mm140

24030000,000 Nmmɸ 0,8

Rn =Mn

=24030000,00

= 2,22170858 N/mm²b . d² 1000

m =fy

=350

= 16,470590,85 0,85

= 0,0326350 600 350

0,75 ρb

0,75 0,0326

0,0244

1,4=

1,4= 0,004

fy 350

130

20

130 20 12 104

25 0,85

350

ρ b =0,85

x600

fy 600

12

0,85=

0,85 25x

600

19,224 19,224 10 ⁶

Mn =Mu

=19,224 10 ⁶

=

ρ = 1 -fy

ρ =1

1 1 -16,47059 2,221709

16,47059 350

1


17/34




As bagi = 0,002 . b . h = x x = mm2


s =

= x x2

x

= mm

s ≤ 5 h = 5 x =

s ≤ mm


As ada =

= x x 2 x

= mm2

> As bagi



c) Tumpuan Kanan



Ditentukan diameter tulangan f p = mmTinggi efektif d = h – p – ½ f p

= - - ½ = mm


fy = Mpa

β1 fc'

+ fy

x x

+

ρ maks = x

= x

=

ρ min =

Mu = kNm = x Nmm

x

x 104 ²

fc x 25m =

fy=

350= 16,47059

0,85 0,85

350

0,75 ρb

0,75 0,0326

0,02441,4

=1,4

= 0,004fy 350

27,962 27,962182 10 ⁶

Mn =Mu

=27,962 10 ⁶

= 34952727,273ɸ 0,8

0,25 3,142 8 1000

180

279,253

180

130

20

12

130 20 12 104

25 0,85

350

8

139,767

0,002 130 260

260

1/4 . π . Ф2 .b

As bagi

0,25 3,142 8 1000

260

193,33

20 100

698,837

1000

450

193,329 180

1/4 . π .Ф2 .b

s

ρ b =0,85

x600

fy 600

=0,85 0,85 25

x600

= 0,0326350 600

Mn

Nmm

Rn = =34952727,27

= 3,231576116 N/mm²b . d² 1000

1


18/34

1 2 m Rn

m

2 x x

=


ρ =



1 22 2 2

4 7






As bagi = 0,002 . b . h = x x = mm2


s =

= x x2

x

= mm

s ≤ 5 h = 5 x =

s ≤ mm


As ada =

= x x2

x

= mm2

> As bagi



180

279,253

180

dipakai 0,010068

0,010068 1000 104 1047,052

100

As ada = x x 12 x1000

= 1131 mm100

ρ = 1 -fy

ρ =

1

1 1 -

16,47059

16,47059 350

3,231576

0,010068

0,004 0,0100678 0,0244

100

8

1047,05 209,410

0,002 1000 130 260

2601/4 . π . Ф

2 .b

As bagi

0,25 3,142 8 1000

260

193,33

3,142

20 100

450

193,329 180

1/4 . π .Ф2

.b

s

0,25 8 1000


19/34

4. Penulangan Plat f dan h

a) Tumpuan Kiri





= - - ½ = mm


fy = Mpa

β1 fc'

+ fy

x x

+

ρ maks = x

= x

=

ρ min =

Mu = = x Nmm

x

x 94 ²

fc x 25

1 2 m Rn

m

2 x x

=


ρ =



1 22 2 2

4 7

As ada > As perluOK


As ada = x x 12 x1000

= 1131 mm100

100

m =fy

=350

= 16,470590,85

ρ = 1 -fy

ρ =1

1 1 -16,47059

16,47059 350

24,401 kNm 24,401455 10 ⁶

Mn =Mu

=24,40145 10 ⁶

= 30501818,182 Nmmɸ 0,8

Rn =Mn

=30501818,18

= 3,451993909 N/mm²b . d² 1000

= 0,0326350 600 350

0,75 ρb

0,75 0,0326

0,0244

1,4 = 1,4 = 0,004fy 350

120

20

12

120 20 12 94

25 0,85

350

ρ b =0,85

x600

fy 600

0,85

25=

0,85 0,85x

600

3,451994

0,010828

0,004 0,0108285 0,0244

dipakai 0,010828

0,010828 1000 94 1017,877

100

1


20/34




As bagi = 0,002 . b . h = x x = mm2


s =

= x x2

x

= mm

s ≤ 5 h = 5 x =

s ≤ mm


As ada =

= x x 2 x

= mm2

> As bagi



b) Lapangan




= - - ½ = mm


fy = Mpa

β1 fc'

+ fy

x x

+

ρ maks = x

= x

=

ρ min =

Mu = kNm = x Nmm

x

x 94 ²

fc x 25

600

450

209,440 200

1/4 . π .Ф2 .b

s

0,25 3,142 8 1000

200

251,327

200

8

1017,88 203,575

0,002 1000 120 240

240

1/4 . π . Ф2 .b

As bagi

0,25 3,142 8 1000

240

209,44

120

120

20

12

120 20 12 94

25 0,85

600= 0,0326

350 600 350

0,75 ρb

0,75 0,0326

350

ρ b =0,85

x600

fy 600

=0,85 0,85 25

x

10 ⁶

Mn =Mu

=19,224 10 ⁶

= 24030000,000

0,02441,4

=1,4

= 0,004fy 350

19,224 19,224

m =fy

=350

= 16,470590,85 0,85

Nmmɸ 0,8

Rn =Mn

=24030000,00

= 2,71955636 N/mm²b . d² 1000

1


21/34

1 2 m Rn

m

2 x x

=


ρ =



1 22 2 2

4 7






As bagi = 0,002 . b . h = x x = mm2


s =

= x x2

x

= mm

s ≤ 5 h = 5 x =

s ≤ mm


As ada =

= x x2

x

= mm2

> As bagi



3,142 8 1000

240

209,44

20 100

450

209,440

1/4 . π .Ф2

.b

s

0,25 3,142 8 1000

200

251,327

8

784,284 156,857

0,002 1000 120

16,47059 2,719556

16,47059 350

0,008343

0,004 0,0083434 0,0244

dipakai 0,008343

ρ = 1 -fy

ρ =

1

1 1 -

140

0,008343 1000 94 784,2838

140

As ada = x x 12 x1000

= 808,2 mm140

240

2401/4 . π . Ф

2 .b

As bagi

0,25

200

200


22/34

c) Tumpuan Kanan





= - - ½ = mm


fy = Mpa

β1 fc'

+ fy

x x

+

ρ maks = x

= x

=

ρ min =

Mu = kNm = x Nmm

x

x 94 ²

fc x 25

1 2 m Rn

m

2 x x

=


ρ =



1 22 2 2

4 7

As ada > As perluOK


As ada = x x 12 x1000

= 1616 mm70

70

ρ =1

1 1 -16,47059 3,955718

16,47059 350

0,012612

0,004 0,012612 0,0244

dipakai 0,012612

= 3,955718342 N/mm²b . d² 1000

m =fy

=350

= 16,470590,85 0,85

ρ = 1 -fy

0,004fy 350

27,962 27,962182 10 ⁶

Mn =Mu

=27,962 10 ⁶

= 34952727,273 Nmmɸ 0,8

25 0,85

350

ρ b =0,85

x600

fy 600

=0,85 0,85 25

x600

=350 600 350

120

20

12

120 20 12 94

0,0326

0,75 ρb

0,75 0,0326

0,0244

1,4=

1,4=

Rn =Mn

=34952727,27

0,012612 1000 94 1185,526

70

1


23/34




As bagi = 0,002 . b . h = x x = mm2


s =

= x x2

x

= mm

s ≤ 5 h = 5 x =

s ≤ mm


As ada =

= x x 2 x

= mm2

> As bagi



5. Penulangan Plat i dan k

a) Tumpuan Kiri




= - - ½ = mm


fy = Mpa

β1 fc'

+ fy

x x

+

ρ maks = x

= x

=

ρ min =

Mu = = x Nmm

x

x 94 ²

fc x 25

b . d² 1000

m =fy

=350

= 16,470590,85

0,75

0,75 0,03260,0244

1,4=

1,4= 0,004

fy 350

11,184 kNm 11,184 10 ⁶

Mn =Mu

=11,184 10 ⁶

= 13980000,000ɸ 0,8

0,85

350

ρ b =0,85

x600

fy 600

=0,85 0,85 25

x600

= 0,0326350 600 350

20 100

450

209,440 200

1/4 . π .Ф2 .b

s

0,25 3,142 8 1000

200

251,327

200

120

20

12

8

1185,53 237,105

0,002 1000 120 240

240

1/4 . π . Ф2 .b

As bagi

0,25 3,142 8 1000

240

209,44

ρb

120 20 12 94

25

Nmm

Rn =Mn

=13980000,00

= 1,582163875 N/mm²

0,85

1


24/34

1 2 m Rn

m

2 x x

=


ρ =



1 22 2 2

4 7






As bagi = 0,002 . b . h = x x = mm2


s =

= x x2

x

= mm

s ≤ 5 h = 5 x =

s ≤ mm


As ada =

= x x2

x

= mm2

> As bagi



0,25 8 1000

200

251,327

200

442,043 88,409

0,002 1000 120 240

2401/4 . π . Ф

2 .b

As bagi

0,25 3,142 8 1000

240

209,44

120 600

dipakai 0,004703

0,004703 1000 94 442,0431

150

As ada = x x 12 x1000

= 754,3 mm150

ρ = 1 -fy

ρ =

1

1 1 -

16,47059

35016,47059

1,582164

0,004703

0,004 0,0047026 0,0244

150

8

3,142

450

209,440 200

1/4 . π .Ф2

.b

s


25/34

b) Lapangan





= - - ½ = mm


fy = Mpa

β1 fc'

+ fy

x x

+

ρ maks = x

= x

=

ρ min =

Mu = kNm = x Nmm

x

x 94 ²

fc x 25

1 2 m Rn

m

2 x x

=


ρ =



1 22 2 2

4 7

As ada > As perluOK


As ada = x x 12 x1000

= 808,2 mm140

140

1 -16,47059 2,712281

16,47059 350

0,008319

0,004 0,0083194 0,0244

dipakai 0,008319

0,008319 1000 94 782,0191

140

350

19,173 19,172571 10 ⁶

Mn =Mu

=19,173 10 ⁶

= 23965714,286 Nmmɸ 0,8

Rn =Mn

=23965714,29

= 2,712280929 N/mm²b . d² 1000

120

12

120 20 12 94

25 0,85

350

ρ b =0,85

x600

fy 600

20

=0,85 0,85 25

x600

=350 600 350

0,75 ρb

0,0326

0,75 0,0326

0,0244

1,4=

1,4= 0,004

fy

m =fy

=350

= 16,470590,85 0,85

ρ = 1 -fy

ρ =1

1

1


26/34




As bagi = 0,002 . b . h = x x = mm2


s =

= x x2

x

= mm

s ≤ 5 h = 5 x =

s ≤ mm


As ada =

= x x 2 x

= mm2

> As bagi



c) Tumpuan Kanan




= - - ½ = mm


fy = Mpa

β1 fc'

+ fy

x x

+

ρ maks = x

= x

=

ρ min =

Mu = kNm = x Nmm

x

x 94 ²

fc x 25

33552000,00= 3,7971933 N/mm²

b . d² 1000

m =fy

=350

= 16,470590,85

350

0,75 ρb

0,75 0,0326

0,02441,4

=1,4

= 0,004fy 350

26,842 26,8416 10 ⁶

Mn =Mu

=26,842 10 ⁶

= 33552000,000ɸ 0,8

120

20

12

120 20 12 94

25 0,85

350

ρ b =0,85

x600

fy 600

8

782,019 156,404

0,002 1000 120 240

240

1/4 . π . Ф2 .b

As bagi

0,25 3,142 8 1000

240

20

209,44

100

450

209,440 180

1/4 . π .Ф2 .b

s

0,25 3,142 8 1000

180

279,253

180

=0,85 0,85 25

x600

= 0,0326350 600

0,85

Nmm

Rn =Mn

=

1


27/34

1 2 m Rn

m

2 x x

=


ρ =



1 22 2 2

4 7






As bagi = 0,002 . b . h = x x = mm2


s =

= x x2

x

= mm

s ≤ 5 h = 5 x =

s ≤ mm


As ada =

= x x2

x

= mm2

> As bagi



s

0,25 3,142 8 1000

200

251,327

200

8

1132,10 226,421

0,002 1000 120 240

2401/4 . π . Ф

2 .b

As bagi

0,25 3,142 8 1000

240

209,44

20 100

16,47059 350

0,012044

0,0120436 0,0244

dipakai 0,012044

0,012044 1000 94 1132,103

70

As ada = x x 12 x1000

= 1616 mm70

ρ = 1 -fy

ρ =

1

1 1 -

16,47059 3,797193

0,004

70

450

209,440 200

1/4 . π .Ф2

.b


28/34

6. Penulangan Plat j dan i

a) Tumpuan Kiri





= - - ½ = mm


fy = Mpa

β1 fc'

+ fy

x x

+

ρ maks = x

= x

=

ρ min =

Mu = = x Nmm

x

x 104 ²

fc x 25

1 2 m Rn

m

2 x x

=


ρ =



1 22 2 2

4 7

As ada > As perluOK


150

As ada = x x 12 x1000

= 754,3 mm150

Nmm0,8

16020000,001,481139053 N/mm²

1000

350= 16,470590,85 0,85

ρ = 1 -fy

ρ =1

1 1 -16,47059 1,481139

16,47059 350

600

fy

=0,85 0,85 25

x600

= 0,0326350 600 350

0,75 ρb

0,0326

0,0244

130

20

12

130 20 12 104

25 0,85

600

350

ρ b =0,85

x

1,4 1,4fy

0,75

= = 0,004350

Rn =Mn

= =

ɸ

12,816 kNm 12,816 10 ⁶

Mn =Mu

=12,816 10 ⁶

= 16020000,000

b . d²

m =fy

=

0,004391

0,004 0,0043906 0,0244

dipakai 0,004391

0,004391 1000 104 456,6202

150

1


29/34




As bagi = 0,002 . b . h = x x = mm2


s =

= x x2

x

= mm

s ≤ 5 h = 5 x =

s ≤ mm


As ada =

= x x 2 x

= mm2

> As bagi



b) Lapangan




= - - ½ = mm


fy = Mpa

β1 fc'

+ fy

x x

+

ρ maks = x

= x

=

ρ min =

Mu = kNm = x Nmm

x

x 104 ²

fc x 25

= =27462857,14

= 2,53909552 N/mm²b . d² 1000

m =fy

=350

= 16,470590,85 0,85

0,85 25x

600= 0,0326

350 600 350

0,75

0,75 0,0326

0,02441,4 1,4

= 0,004fy 350

8 1000

260

193,33

650

450

193,329 180

1/4 . π .Ф2 .b

s

0,25 3,142 8 1000

180

279,253

180

8

456,620 91,324

0,002 1000 130 260

130

260

1/4 . π . Ф2 .b

As bagi

0,25 3,142

130

20

12

130 20 12 104

25 0,85

ρb

350

ρ b =0,85

x600

fy 600

=0,85

ɸ

Mn

=

Mn =Mu

=

21,970 21,970286 10 ⁶

21,970 10 ⁶= 27462857,143 Nmm

0,8

Rn

1


30/34

1 2 m Rn

m

2 x x

=


ρ =



1 22 2 2

4 7






As bagi = 0,002 . b . h = x x = mm2


s =

= x x2

x

= mm

s ≤ 5 h = 5 x =

s ≤ mm


As ada =

= x x2

x

= mm2

> As bagi



s

0,25 3,142 8 1000

180

279,253

180

0,007749 1000 104 805,9036

140

As ada = x x 12 x1000

= 808,2 mm140

ρ =

1

1 1 -

16,47059 2,539096

16,47059 350

0,007749

0,004 0,0077491 0,0244

dipakai 0,007749

ρ = 1 -fy

140

8

805,904 161,181

0,002 1000 130 260

2601/4 . π . Ф

2 .b

As bagi

0,25 3,142 8 1000

260

193,33

20 100

450

193,329 180

1/4 . π .Ф2

.b


31/34

c) Tumpuan Kanan





= - - ½ = mm


fy = Mpa

β1 fc'

+ fy

x x

+

ρ maks = x

= x

=

ρ min =

Mu = kNm = x Nmm

x

x 104 ²

fc x 25

Nmmɸ 0,8

Rn =Mn

=38448000,00

= 3,554733728 N/mm²b . d² 1000

m =fy

=350

= 16,470590,85 0,85

0,0244

1,4=

1,4= 0,004

fy 350

30,758 30,7584 10 ⁶

Mn =Mu

=30,758

= 38448000,000

600

fy 600

0,85 0,85 25x

600= 0,0326

350 600 350

0,75 ρb

0,75 0,0326

130

20

12

130 20 12 104

25 0,85

10 ⁶

=

350

ρ b =0,85

x

1


32/34

1 2 m Rn

m

2 x x

=


ρ =



1 22 2 2

4 7






As bagi = 0,002 . b . h = x x = mm2


s =

= x x2

x

= mm

s ≤ 5 h = 5 x =

s ≤ mm


As ada =

= x x2

x

= mm2

> As bagi



180

279,253

180

0,25 3,142 8 1000

260

193,33

20 100

450

193,329 180

1/4 . π .Ф2 .b

s

0,25 3,142 8 1000

70

As ada = x x 12 x1000

= 1616 mm70

70

8

1163,45 232,690

fy ρ = 1 -

0,011187

16,47059 ρ =

11 1 -

16,47059 3,554734

350

0,004 0,011187 0,0244

dipakai 0,011187

0,011187 1000 104 1163,451

0,002 1000 130 260260

1/4 . π . Ф2 .b

As bagi


33/34

Berikut Rekapan Tulangan Pokok masing-masing plat :

Plat a Ø 12 - Ø 12 - Ø 12 -

Plat b Ø 12 - Ø 12 - Ø 12 -

Plat c Ø 12 - Ø 12 - Ø 12 -

Plat d Ø 12 - Ø 12 - Ø 12 -

Plat e Ø 12 - Ø 12 - Ø 12 -

Plat f Ø 12 - Ø 12 - Ø 12 -

Plat g Ø 12 - Ø 12 - Ø 12 -

Plat h Ø 12 - Ø 12 - Ø 12 -

Plat i Ø 12 - Ø 12 - Ø 12 -

Plat j Ø 12 - Ø 12 - Ø 12 -

Plat k Ø 12 - Ø 12 - Ø 12 -

Plat l Ø 12 - Ø 12 - Ø 12 -

Berikut Rekapan Tulangan bagi masing-masing plat :

Plat a Ø 8 - Ø 8 - Ø 8 -

Plat b Ø 8 - Ø 8 - Ø 8 -

Plat c Ø 8 - Ø 8 - Ø 8 -

Plat d Ø 8 - Ø 8 - Ø 8 -

Plat e Ø 8 - Ø 8 - Ø 8 -

Plat f Ø 8 - Ø 8 - Ø 8 -

Plat g Ø 8 - Ø 8 - Ø 8 -

Plat h Ø 8 - Ø 8 - Ø 8 -

Plat i Ø 8 - Ø 8 - Ø 8 -Plat j Ø 8 - Ø 8 - Ø 8 -

Plat k Ø 8 - Ø 8 - Ø 8 -

Plat l Ø 8 - Ø 8 - Ø 8 -

Dikarenakan jarak tulangan masing-masing pelat berbeda-beda, untuk mempermudah pemasangannya

dilapangan maka semua jarak tulangan yang dipasang berdasarkan portal arah melintang (Y),

kemudian diambil jarak terkecil

Tulangan pokok yang di pasang searah melintang atau sumbu Y:

Plat a

Plat e

Plat i

Plat b

Plat f

Plat j

Plat c

Plat g

Plat k

Plat d

Plat h

Plat l

12 - 70 1616,327

Ø 12 - 140 808,1633

Ø 12 - 140 808,1633

Ø 12 - 140 808,1633

Ø 12 - 70 1616,327 Ø 12 - 140 808,1633

Ø 12 - 70 1616,327

Ø 12 - 70 1616,327

Ø

PlatTumpuan Kiri Lapangan

Tulangan As Tulangan As

180 279,253 180 279,253 180 279,253

200 251,3274 180 279,2527 200 251,3274

180 279,2527 180 279,2527 180 279,2527

180 279,2527 180 279,2527 180 279,2527

200 251,3274 200 251,3274 200 251,3274

200 251,327 180 279,253 200 251,327

180 279,253 180 279,253 180 279,253

180 279,253 180 279,253 180 279,253

200 251,327 200 251,327 200 251,327

200 251,327 200 251,327 200 251,327

180 279,253 180 279,253 180 279,253

200 251,327 200 251,327 200 251,327

150 754,2857 140 808,1633 70 1616,327

150 754,2857 140 808,1633 70 1616,327

PlatTumpuan Kiri Lapangan Tumpuan Kanan

Tulangan As Tulangan As Tulangan As

100 1131,429 140 808,1633 70 1616,327

150 754,2857 140 808,1633 70 1616,327

150 754,2857 140 808,1633 70 1616,327

100 1131,429 140 808,1633 100 1131,429

100 1131,429 140 808,1633 70 1616,327

100 1131,429 140 808,1633 100 1131,429

120 942,8571 200 565,7143 200 565,7143

120 942,8571 200 565,7143 200 565,7143

120 942,8571 200 565,7143 200 565,7143

PlatTumpuan Kiri Lapangan Tumpuan Kanan

Tulangan As Tulangan As Tulangan As

120 942,8571 200 565,7143 200 565,7143


34/34

Tulangan bagi yang di pasang :

Plat a

Plat e

Plat i

Plat b

Plat f

Plat j

Plat c

Plat g

Plat k

Plat d

Plat h

Plat l

Ø 12 - 180 279,253 Ø 12 - 180 279,253

Ø 12 - 180 279,253 Ø 12 - 180 279,253

Ø 12 - 180 279,253 Ø 12 - 180 279,253

PlatTumpuan Kiri Lapangan

Tulangan As Tulangan As

Ø 12 - 180 279,253 Ø 12 - 180 279,253

Tugas Beton Nando

Documents

Transcript of Tugas Beton Nando