PERHITUNGAN DIMENSI & TULANGAN RANGKAP PADA BALOK 0.85 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 - Cub 0.003 = Cub 0.129 - 0.003Cub = 0.002Cub 0.0020 43-Cub 0.129 = 0.002 Cub +0.003 Cub Cub = 25.8 cm aub = 0.85 x 25.8 = 21.93 cm Persamaan statika : baja Sc = As'.Fs' = 0.8 As . Fy = 3275.9 As Cc =0.85 Fc'. Aub . B = 111843 St = As. Fy = 4000 As 1. Sc + Cc = ST 3275.9 As + 111843= 4000 As 111843 = 724.1 As Asb = 154.46 As'b = 126.50 Jadi , Sc = 126.5 x 4000 = 506000.0 kg Cc =0.85 x 300 x 21.93 x 20 = 111843.0 kg St = 154.46 x 4000 = 617840.0 kg 2. Sc (d - d') + Cc (d - 0.5au) = Mnb Mnb = (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 Design Cu = 24.8 cm Au = 0.85Cu = 0.85 x 24.8= 21.08 cm Persamaan Deformasi/Kompatibilitas : 1. = Cu = 24.8 Cu - d' 24.8 - 7 0.003 = 24.8 17.8 0.053 = 24.8 0.0022 > 0.002 c' = 0.003 s' s dia dia M - = M + = kg/cm 2 kg/cm 2 c' s cm 2 cm 2 c' c' s' s' s' s' s' =
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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
Persamaan Deformasi/Kompatibilitas :
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
2. Sc (d - d') + Cc (d - 0.5au) = MnbMnb = (486440 x (43 - 7)) + (107508 x (43 - (0.5 x21.08)))Mnb = 21001549.68 kgcm
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
Persamaan Deformasi/Kompatibilitas :
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
2. Sc (d - d') + Cc (d - 0.5au) = MnbMnb = (65040 x (30 - 5)) + (65025 x (30 - (0.5 x15.3)))Mnb = 1778658.75 kgcm
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
Persamaan Deformasi/Kompatibilitas :
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
Persamaan Deformasi/Kompatibilitas :
=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)
Sc = 0.5As.Fs' = 10.05 x 2400 = 24120.000 kgCc = b x au x 0.85Fc = 30 x 8.5 x 0.85 x 300 = 65025.000 kgSt = 0.5As.Fy = 10.05 x 2400 = 24120.000 kg
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
kekuatan batas gaya luar
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 )
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
kekuatan batas gaya luar
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