TUGAS 3 BETON R.2

PERHITUNGAN DIMENSI & TULANGAN RANGKAP PADA BALOK0.85 Fc'

Sc

As' - Aub Cc

50 43

Cub

As +

7 ST

20

9.28 Tm} 0.8

7.6

Fc' = 300

Fy = 4000

Persamaan Deformasi/Kompatibilitas :

1.=

Cub

d - Cub0.003

=Cub 0.129 - 0.003Cub = 0.002Cub

0.0020 43-Cub 0.129 = 0.002 Cub +0.003 CubCub = 25.8 cmaub = 0.85 x 25.8 = 21.93 cm

Persamaan statika : bajaSc = As'.Fs' = 0.8 As . Fy = 3275.9 AsCc =0.85 Fc'. Aub . B = 111843St = As. Fy = 4000 As

1. Sc + Cc = ST3275.9 As + 111843= 4000 As

111843 = 724.1 As

Asb = 154.46

As'b = 126.50

Jadi , Sc = 126.5 x 4000 = 506000.0 kgCc =0.85 x 300 x 21.93 x 20 = 111843.0 kgSt = 154.46 x 4000 = 617840.0 kg

2. Sc (d - d') + Cc (d - 0.5au) = MnbMnb = (506000 x (43 - 7)) + (111843 x (43 - (0.5 x21.93)))Mnb = 7588501.51 kgcm

Mub = 0.8 x 7588501.51 kgcm= 60.70801 Tm > Mu, dimensi balok kuat

DesignCu = 24.8 cmAu = 0.85Cu = 0.85 x 24.8= 21.08 cm


1.=

Cu=

24.8

Cu - d' 24.8 - 70.003

=24.8

17.8

0.053 = 24.8

0.0022 > 0.002

c' = 0.003s'

s

dia dia

M - =

M + =

kg/cm2

kg/cm2

c'

s

cm2

cm2

c' c'

s' s'

s'

s'

s' =

Fs' = 4306.5

tulangan tekan sudah leleh,maka Fs' = Fy = 4000

kg/cm2

kg/cm2

Persamaan statika : bajaSc = As'.Fs' = 0.8 As x 4000 = 3276 AsCc = 0.85 Fc'. Au . B = 107508St = As. Fy = 4000 As

1. Sc + Cc = ST3276 As + 107508 = 4000 As

107508 = 724 As

As = 148.49

As' = 121.61

Jadi , Sc = 121.61 x 4000 = 486440.0 kgCc = 0.85 x 300 x 21.08 x 20 = 107508.0 kgSt = 148.49 x 4000 = 593960.0 kg


Mub = 0.8 x 21001549.68 kgcm= 168.01 Tm > Mu, dimensi balok kuat

Keadaan Under Reinforce = 0.7 x Asb (max).

diambil As = 0.2 Asb = 30.9 OK, As max < As < As min

As' = 0.2 As'b = 25.3

PERSAMAAN :1. Sc + Cc = ST

Sc = 25.3 x 4000 = 101200.00 kgCc = 0.85 x 300 x 21.08 x 20= 5100.00 kgSt = 30.9 x 4000 = 123568.00 kg

2. Sc (d - d') + Cc (d - 0.5au) = MnMn = 3808746.00 kg cmMu = 3046996.80 kg cm

= 30.47 Tm > M = 9.28 TmOK, dimensi & tulangan cukup

Tulangan diameter 24 , Luas = 4.5216Tulangan tarik 7 buahTulangan tekan 6 buah

0.0035 = As min = 3.01

= As max = 108.1

Dimensi & tulangan terpasang

6d24 6d24

50 50

7d24 6d24

20 20

Tumpuan Lapangan

cm2

cm2

cm2

cm2

Mu = Mn

cm2

tulangan min = cm2

tulangan max = 0.7 x Asb cm2

PERHITUNGAN DIMENSI & TULANGAN RANGKAP PADA BALOK0.85 Fc'

Sc

As' - aub Cc

35 30

Cub

As +

5 ST

20

Mu = 6.4 Tm} 0.5

Fc' = 250

Fy = 4000

Keadaan Balance


1.=

Cub

d - Cub0.003

=Cub 0.09 - 0.003Cub = 0.002Cub

0.0020 30-Cub 0.09 = 0.002 Cub + 0.003 CubCub = 18 cmaub = 0.85 x 18 = ### cm

Persamaan statika : bajaSc = As'.Fs' = 0.5 As . Fy = 2000 AsCc =0.85 Fc'. Aub . B = 65025St = As. Fy = 4000 As

1. Sc + Cc = ST2000 As + 65025 = 4000 As

65025 = 2000 As

Asb = 32.51

As'b = 16.26

Jadi , Sc = 16.26 x 4000 = 65040.0 kgCc =0.85 x 250 x 15.3 x 20 = 65025.0 kgSt = 32.51 x 4000 = 130040.0 kg


Mub = 0.8 x 1778658.75 kgcm= 14.229 Tm > ... Mu, dimensi balok kuat

c' = 0.003

s'

s

dia dia

kg/cm2

kg/cm2

c'

s

H = 0

cm2

cm2

M = 0

Design Tulangan (Under-reinforced)Cu = 10 cmAu = 0.85Cu = 0.85 x 10 = 8.5 cm


1.=

Cu=

10

Cu - d' 10 - 50.003

=10

5

0.015 = 10

0.0015 < 0.002

karena tulangan tekan belum leleh,maka Fs'= 3000

Persamaan statika : bajaSc = As'.Fs' = 0.5 As x 3000 = 1500 AsCc = 0.85 Fc'. Au . B = 36125St = As. Fy = 4000 As

1. Sc + Cc = ST1500 As + 36125 = 4000 As

36125 = 2500 As

As = 14.45

As' = 7.23

Jadi , Sc = 7.23 x 3000 = 21690.0 kgCc = 0.85 x 250 x 8.5 x 20 = 36125.0 kgSt = 14.45 x 4000 = 57800.0 kg

2. Sc (d - d') + Cc (d - 0.5au) = MnbMn = (21690 x (30 - 5)) + (36125 x (30 - (0.5 x8.5)))Mn = 1472468.75 kgcm

Mu = 0.8 x 1472468.75 kgcm= 11.78 Tm > Mu, dimensi balok kuat

Keadaan Under Reinforce = 0.7 x Asb (max).

diambil As = 14.45 Asb = 469.8 OK, As max < As < As min

As' = 14.45 As'b = 235.0Perhitungan tulangan di bawah ini tidak terpakai.

Sc = 235 x 3000 = 704871.00 kgCc = 0.85 x 250 x 8.5 x 20 = 36125.00 kgSt = 469.8 x 4000 = 1879078.00 kg

PERSAMAAN :1. Sc (d - d') + Cc (d - 0.5au) = Mn

Mn = 18551993.75 kg cmMu = 14841595.00 kg cm

= 148.42 Tm > M = 6.4 TmOK, dimensi & tulangan cukup

Tulangan diameter 24 , Luas = 4.52Tulangan tarik 104 buahTulangan tekan 52 buah

0.0035 = As min = 2.10

= As max = 22.76

Dimensi & tulangan terpasang

52d24 52d24

35 35

104d24 52d24

20 20

c' c'

s' s'

s'

s'

s' =

kg/cm2

kg/cm2

H = 0

cm2

cm2

M = 0

cm2

cm2

M = 0

Mu = Mn

cm2

tulangan min = cm2

tulangan max = 0.7 x Asb cm2

Tumpuan Lapangan

PERHITUNGAN PEMERIKSAAN TULANGAN KOLOM DUA SISI

0.5 As

5055

PuMu

0.5 As5

30

Fc' = 300 0.85

Fy = 2400

Es = 2000000Pu = 188 Ton

}Beban Luar

Mu = 20.8 Ton meter

Tulangan diambil/perkirakan = 2 x 5 D-16Ast = 20 cm2

Dipasang tulangan 10 16 0.5As= 10.050 .

1. Keadaan normal murni :

0.003Sc Fy

= 0.00120.5 As Es

- Cc Pn0 tulangan tekan sudah leleh makaFs = Fy = 2400 kg/cm2

0.5 AsSc

Sc = 0.5As x Fy = 10.05 x 2400 = 24120.000 kgCc = b x h x 0.85 Fc' = 30 x 55 x 0.85 x 300 = 420750.000 kg

Persamaan statika :Sc + Sc + Cc - Pn0 = 0

Pn0 = 24120 + 24120 + 420750 = 468990.000 kgPu0 = 0.65 Pn0 = 0.65 x 468990 = 304843.500 kg

Pu0 = 304.844 ton

kg/cm2 , kg/cm2

kg/cm2

cm2

c' = 0.003 cu' = y=

H = 0

2. Keadaan Decompresi :

0.85fc'

Sc

-Cu = d = 50 au Cc Pnd

Mnd

St 0

au = 0.85 x Cu = 0.85 x 50 = 42.500 cm

Sc = 0.5As.Fy = 10.05 x 2400 = 24120.000 kgCc = b x au x 0.85Fc = 30 x 42.5 x 0.85 x 300 = 325125.000 kgSt = 0 (Decompresi)

Persamaan statika :

Sc + Cc - St - Pnd = 024120 + 325125 - 0 = PndPnd = 349245.0 kgPud = 0.65 x Pnd = 227009.250 kg

= 227.009 ton

Sc(d - d') +Cc (d-0.5au) - Pnd (0.5h-d') - Mnd = 0Mnd = ( 24120 x 45) + (325125 x 28.75) - (349245 x 22.5)

= 2574731.25 kgcm= 25.747 tm

Mud = 0.65 x Mnd = 0.65 x 25.747= 16.736 tm

c' = 0.003

s'

s = 0dia

(1). H = 0

(2).M = 0

3. Keadaan Balance :

0,85fc'

ScCu - au Cc

Pnb + Mnb

St


=Cu

d - Cu0.003

=Cu 0.15 - 0.003Cu = 0.0012Cu

0.0012 50 - Cu 0.015 = 0.0012 Cu +0.003 CuCub = 35.71 cm

au = 0.85 x 35.71 = 30.354 cm

Sc = 0.5As.Fy = 10.05 x 2400 = 24120.000 kgCc = b x au x 0.85Fc = 30 x 30.3535 x 0.85 x 300 = 232204.300 kgSt = 0.5As.Fy = 10.05 x 2400 = 24120.000 kg

Persamaan statika :

Sc + Cc - St - Pnb = 0Pnb = 24120 + 232204.3 - 24120Pnb = 232204.300 kgPub = 0.65 x 232204.3 = 150932.795 kgPub = 150.933 ton

Sc(d - d') +Cc (d-0.5au) - Pnb (0.5h-d') - Mnb = 0Mnb = ( 24120 x 45) + (232204.3 x 34.82325) - (232204.3 x 22.5)

= 3946911.64 kgcm= 39.469 tm

Mub = 0.65 x Mnb = 0.65 x 39.469= 25.655 tm

c' = 0.003

s'

s

s = y

c's

(1).H = 0

(2). M = 0

4. Keadaan Keruntuhan Tekan :

Coba, Cu = 40 cm > Cub = 35.71cm .au = . 34 cm

0.85 fc'.

Sc - Cu = 40 au Cc

d = 50

PntMnt

+St


=Cu

d - Cu

0.003=

40

50 - 40

0.00075

Fs = 1500

Fy = 2400

Fs < Fy , berati tulangan tarik belum leleh maka, Fs = 1500

Sc = 0.5As.Fy = 10.05 x 2400 = 24120.000 kgCc = b x au x 0.85Fc = 30 x 34 x 0.85 x 300 = 260100.000 kgSt = 0.5As.Fs = 10.05 x 1500 = 15075.000 kg

Persamaan statika :

Sc + Cc - St - Pnt = 0Pnt = 24120 + 260100 - 15075Pnt = 269145.000 kgPut = 0.65 x 269145 = 174944.25 kgPut = 174.944 ton

Sc(d - d') +Cc (d-0.5au) - Pnt (0.5h-d') - Mnt = 0Mnt = ( 24120 x 45) + (260100 x 33) - (269145 x 22.5)

= 3612937.5 kgcm= 36.129 tm

Mut = 0.65 x Mnt = 0.65 x 36.129= 23.484 tm

c' = 0.003

s'

s < y

c'

s

s

s =

s x 2.106 = kg/cm2

kg/cm2

kg/cm2

(1).H = 0

(2). M = 0

5. Keruntuhan Tarik1 :

Coba, Cu = 28 cm < Cub = 35.71cm .

au = 0.85 x 28 = 23.8 cm

Sc - Cu Cc

d = 50Pnr

+ Mnr

St


=Cu

Cu - d'

0.003=

28

28 - 5

0.0025

Fs = 4929 > fy --> tul.tekan sdh lleh

Fy = 2400

Fy < Fs', tulangan tarik sudah leleh maka Fs' = Fy = 2400

Sc = 0.5As.Fs' = 10.05 x 2400 = 24120.000 kgCc = b x au x 0.85Fc = 30 x 23.8 x 0.85 x 300 = 182070.000 kgSt = 0.5As.Fy = 10.05 x 2400 = 24120.000 kg

Persamaan statika :


Sc(0.5h - d') +Cc (0.5h-0.5au) + St (0.5h-d') - Mnr = 0Mnr = (24120 x (27.5 - 5)) + (182070 x (27.5 - 11.9)) + (24120 x (27.5-5))

= 3925692.000 kgcm= 39.257 tm

Mut = 0.65 x Mnt = 0.65 x 39.257= 25.517 tm

c' = 0.003

s'

s

s > y

c'

s'

s'

s' =

s' x 2.106 = kg/cm2

kg/cm2

kg/cm2

(1).H = 0

(2).= 0

6. Keadaan Tarik2 :

Coba, Cu = 10 cm < Cub = 35.71cm .

au = 0.85 x 10 = 8.5 cm

Sc - Cu Cc

d = 50Pnr

+ Mnr

St


=Cu

Cu - d'

0.003=

10

10 - 5

0.0015

Fs' = 3000

Fy = 2400

Fy < Fs, tulangan tarik sudah leleh Fs' = Fy = 2400


Persamaan statika :

Sc + Cc - St - Pnr = 0Pnr = 24120 + 65025 - 24120Pnr = 65025.0 kgPur = 0.65 x 65025 = 42266.25 kgPur = 42.266 ton

Sc(0.5h - d') +Cc (0.5h-0.5au) + St (0.5h-d') - Mnr = 0Mnr = (24120 x (27.5 - 5)) + (65025 x (27.5 - 4.25)) + (24120 x (27.5-5))

= 2597231.250 kgcm= 25.972 tm

Mur = 0.65 x Mnr = 0.65 x 25.972= 16.882 tm

c' = 0.003

s'

s

s > y

c'

s'

s'

s =

s' x 2.106 = kg/cm2

kg/cm2

kg/cm2

(1).H = 0

(2).Mditengah kolom

= 0

Ringkasan M & P dalam 5 kondisi

No. Kondisi M P( Tm ) ( T )

1. Normal 0.0 304.82. Decompresi 16.7 227.03. Tekan 23.5 174.94. Balance 25.7 150.95. Tarik1 25.5 118.36. Tarik2 16.9 42.3

Hasil grafik interaksi :Kolom berdimensi 55cm x 30cm dibebani dengan gaya luar :Pu = 188 tonMu = 20.8 ton meterTidak kuat menahan beban kombinasi Mu-Pu,

0 5 10 15 20 25 30 35 400

50

100

150

200

250

300

350

Diagram Interaksi

kekuatan batas gaya luar

momen (ton-meter)

Ak

sia

l (to

n)

PERHITUNGAN PEMERIKSAAN TULANGAN KOLOM DUA SISI

0 5 10 15 20 25 30 35 400

50

100

150

200

250

300

350

Diagram Interaksi


momen (ton-meter)

Ak

sia

l (to

n)

DESAIN KOLOM TULANGAN DUA SISI

Dari perhitungan struktur, didapat :Pu = 264 tonMu = 22.6 ton meter

Rencanakn kolom, dengan data-data sbb :

Fc' = 250 0.85

Fy = 2400

Es = 2000000

Perhitungan Desinassumsi : luas tulangan = 2 % dari luas Beton

As max = 6 % dari luas Beton0.0058 As min =

Pu0 = 2 x Pu = 528 Ton 812.31 Ton

Pn0 = ( Ac x 0.85Fc' ) + ( As x Fy )= ( b x h ) x 0.85Fc' + ( 0.02 x b x h ) x Fy= ( b x h ) x 0.85 x 250 + ( 0.02 x b x h ) x 2400= ( 212.5+48 ) x ( b x h ) = 260.5 x ( b x h )

307.692307692 = 260.5 x (b x h)( b x h ) = 528000 / 260.5

( b x h ) = 2026.87 Jika h= 1.48 b

(b x (1.48b)) = 2026.87

1369.51 b = 1369.51 = 37.01

b = 37.01 cm

h = 54.77 cm

coba dimensi kolom 38 x 56 cm

Luas tulangan ( As ) = 2% Luas Beton ( Ac ) = 0.02 x ( 38 x 56 )

56 = 42.56

. 22 , A = 3.80Jumlah tulangan (n) = 42.56 / 3.8 = 11.20

Dipasang tulangan = 12 22 .

38 As total = 45.60

}. As max = 127.68 As max < As total < As min

As min = 12.34

0.5 As = 22.8

Jarak antar tulangan = 2.96 cm ok

Cek, dimensi & tulangan :

1. Keadaan Normal MurniPn0 = ( Ac x 0.85Fc' ) + ( As x Fy )

= ( 38x56 ) x 0.85 x 250 + ( 45.6 x 2400 )= 561640.00 kg = 561.64 ton

Pu0 = 0.65 x Pn0= 365.07 ton

kg/cm2 , kg/cm2

kg/cm2

min = b d

Pno = Pu/ ᶲ

b2 =

cm2

Pakai tulangan = cm2

cm2

cm2

cm2

cm2

2. Keadaan Balance

Sc

Cub Cc56 51 Pnb

Mnb5 St

38


=Cub

d - Cu0.003

=Cub 0.153 - 0.003Cu = 0.0012Cub

0.0012 51 - Cub 0.153 = ( 0.003 + 0.0012 )CubCub = 36.43 cm

aub = 0.85 x 36.43 = 30.966 cm

Sc = 0.5As.Fy = 22.8 x 2400 = 54720.00 kgCc = b x aub x 0.85Fc = 38 x 30.9655 x 0.85 x 250 = 250,046.41 kgSt = 0.5As.Fy = 22.8 x 2400 = 54720.00 kg

Persamaan statika :

Sc + Cc - St - Pnb = 0Pnb = 54720 + 250046.41 - 54720Pnb = 250046.41 kgPub = 0.65 x 250046.41 = 162530.1665 kgPub = 162.53 ton

Sc(d - d') +Cc (d-0.5au) - Pnb (0.5h-d') - Mnb = 0Mnb = ( 54720 x 46) + (250046.41 x 35.51725) - (250046.41 x 23)

= 5647013.43 kgcmMub = 0.65x5647013.43 = 3670558.73 kgMub = 36.71 tm > Mu = 22.6tm ok ukuran kolom cukup.

3. Keadaan Decompresi

0.85fc'

Sc

-Cud = 51 au Cc Pnd

Mnd

St 0

au = 0.85 x Cu = 0.85 x 51 = 43.35 cm

Sc = 0.5As.Fy = 22.8 x 2400 = 54720.00 kgCc = b x au x 0.85Fc = 38 x 43.35 x 0.85 x 250 = 350051.25 kgSt = 0 (Decompresi)

Persamaan statika :

Sc + Cc - St - Pnd = 054720 + 350051.25 - 0 = PndPnd = 404771.3 kgPud = 0.65 x Pnd = 263101.35 kg

= 263.101 ton

c' = 0.003

s'

s = y

c's

(1).H = 0

(2). M = 0

c' = 0.003

s'

s = 0dia

(1). H = 0

Sc(d - d') + Cc (d-0.5au) - Pnd (0.5h-d') - = 0 = ( 54720 x 46) + (350051.25 x 48.5) - (404771.3 x 23)= 3472633.01 kgcm

Mnd= 34.726 tmMud = 0.65 x Mnd = 0.65 x 34.726

= 22.572 tm

4. Keadaan Keruntuhan Tekan :

Coba, Cu = 45 cm > Cub = 36.43cm .au = . 38.25 cm

0.85 fc'.

Sc - Cu = 45 au Cc

d = 51

PntMnt

+St


=Cu

d - Cu

0.003=

45

51 - 45

0.00040

Fs = 800.00

Fy = 2,400

Fs < Fy , berati tulangan tarik belum leleh maka, Fs = 800

Sc = 0.5As.Fy = 22.8 x 2400 = 54720.000 kgCc = b x au x 0.85Fc = 38 x 38.25 x 0.85 x 250 = 308868.750 kgSt = 0.5As.Fs = 22.8 x 800 = 18240.000 kg

Persamaan statika :

Sc + Cc - St - Pnt = 0Pnt = 54720 + 308868.75 - 18240Pnt = 345348.750 kgPut = 0.65 x 345348.75 = 224476.688 kgPut = 224.477 ton

Sc(d - d') +Cc (d-0.5au) - Pnt (0.5h-d') - Mnt = 0Mnt = ( 54720 x 46) + (308868.75 x 31.875) - (345348.75 x 23)

= 4419290.156 kgcm= 44.193 tm

Mut = 0.65 x Mnt = 0.65 x 44.193= 28.725 tm

5. Keruntuhan Tarik1 :

Coba, Cu = 20 cm < Cub = 36.43cm .

au = 0.85 x 20 = 17 cm

Sc - Cu Cc

d = 51Pnr

+ Mnr

St


=Cu

Cu - d'

(2).M = 0

c' = 0.003

s'

s < y

c'

s

s

s =

s x 2.106 = kg/cm2

kg/cm2

kg/cm2

(1).H = 0

(2). M = 0

c' = 0.003

s'

s

s > y

c'

s'

0.003=

20

20 - 5

0.0023

Fs = 4,500.00 > fy --> tul.tekan sdh leleh

Fy = 2400

Fy < Fs', tulangan tarik sudah leleh maka Fs' = 4500

Sc = 0.5As.Fs' = 22.8 x 4500 = 102600.000 kgCc = b x au x 0.85Fc = 38 x 17 x 0.85 x 250 = 137275.000 kgSt = 0.5As.Fy = 22.8 x 2400 = 54720.000 kg

Persamaan statika :


Sc(0.5h - d') +Cc (0.5h-0.5au) + St (0.5h-d') - Mnr = 0Mnr = (102600 x (28 - 5)) + (137275 x (28 - 8.5)) + (54720 x (28-5))

= 6295222.500 kgcm= 62.952 tm

Mut = 0.65 x Mnt = 0.65 x 62.952= 40.919 tm

6. Keadaan Tarik2 :

Coba, Cu = 15 cm < Cub = 36.43cm .

au = 0.85 x 15 = 12.8 cm

Sc - Cu Cc

d = 51Pnr

+ Mnr

St


=Cu

Cu - d'

0.003=

15

15 - 5

0.0020

Fs' = 4000

Fy = 2400

Fy < Fs', tulangan tarik sudah leleh maka Fs' = 4000


Persamaan statika :

Sc + Cc - St - Pnr = 0Pnr = 91200 + 103360 - 54720Pnr = 139840.0 kgPur = 0.65 x 139840 = 90896 kgPur = 90.896 ton

Sc(0.5h - d') +Cc (0.5h-0.5au) + St (0.5h-d') - Mnr = 0Mnr = (91200 x (28 - 5)) + (103360 x (28 - 6.4)) + (54720 x (28-5))

= 5588736.000 kgcm= 55.887 tm

Mur = 0.65 x Mnr = 0.65 x 55.887= 36.327 tm

Ringkasan M & P dalam 3 kondisi

s'

s' =

s' x 2.106 = kg/cm2

kg/cm2

kg/cm2

(1).H = 0

(2).= 0

c' = 0.003

s'

s

s > y

c'

s'

s'

s =

s' x 2.106 = kg/cm2

kg/cm2

kg/cm2

(1).H = 0

(2).Mditengah kolom

= 0

No. Kondisi M P( Tm ) ( T )

1. Normal 0.0 365.12. Decompresi 22.6 263.14 Tekan 28.7 224.53. Balance 36.7 162.55 Tarik 1 40.9 120.46 Tarik 2 36.3 90.9

Hasil grafik interaksi :Kolom berdimensi 38cm x 56cm dengan tulangan 12 dia22 dibebani dengan gaya luar :Pu = 264 tonMu = 22.6 ton meterMampu menahan gaya luar tsb, karena masih didalam grafik Interaksi.

0 5 10 15 20 25 30 35 40 450

50

100

150

200

250

300

350

400

Diagram Interaksi


Mu

Pu

PERHITUNGAN DIMENSI & TULANGAN TUNGGAL PADA BALOK

0,85fc'

cub - aub Cc

36

As +

4 ST

20

10.24 Tm

Fc' = 250

Fy = 4000 KEADAAN BALANCE

- Persamaan Deformasi/Kompatibilitas :

=Cub

d - Cub0.003

=Cub 0.108 - 0.003Cub = 0.002Cub

0.0020 36-Cub 0.108 = 0.002 Cub +0.003 CubCub = 21.6 cmaub = 0.85 x 21.6 = 18.4 cm

c' = 0.003

s = y

dia dia M - =

kg/cm2

kg/cm2

c'

y

TUGAS 3 BETON R.2

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Transcript of TUGAS 3 BETON R.2