beton prategang tugas
TRANSCRIPT
BETON PRATEGANG
“GIRDER”
Prediksi jumlah selongsongLO => 25 ~ 35 = 3 ~ 4H prediksi = 0.07 x 2900
= 203Digunakan => h = 190
Ac = (0,26 x 1,3) + (0,6 x 0,34)
= 0,542 m2
b1 ≤ 8 x 26 = 208b2 ≤ 8 x 26 = 208
91 ≤ 12x154 = 77
91 ≥ 77b yang digunakan = 77b = b1 + b2 + bwb = 72 + 72 + 26 = 180
qDl => aspal = 0,05 x 2 x 2,2 = 0,22lantai = 0,26 x 2 x 2,4 = 1,248girder = 0,542 x 2,4 = 1,3008 +
2,72 tm
qLL = 0,9 x 2= 1,8
pLL = 4,9 x 1,4 x 2= 13,72 ton
Mu beban (14
x 13,72 x 29 x 1,8 ) = 179,046
(18
x 1,8 x 292 x 1,8 ) = 340,605
(18
x 2,77 x 292 x 1,25 ) = 363,995 +
883,6461
Mu kap > Mu beban
Mu kap = 0,071 x fc’ x bd2
= 0,071 x 410 x 180 x 1622
= 1375 tm > 883,646 (oke!)
bfd
= 26162
= 0,16
Mu
fc ' . bd2 = 0,071 (oke!)
bwb
= 26172
= 0,15
DAERAH AMAN KABEL
1. penampang tengah
Kondisi awal
PENAMPANG KONDISI AWALA1 = 3380 cm²A2 = 2040 cm² ∑ 5420 cm² ∑Sx = 369300 cm³ Yb = 68.13653137 Cm = 68 Cm Ya = 96 Cm Ka = 36.59 Cm = 37 CmKb = 26.00 Cm = 26 Cm
Pot A (cm²) Y A x Y Io Ip
I 3380 96 324018.5
3248180 4760167
II 2040 68 138998. 5306040 196520
5∑ 5420 463017 13510907
Yb = ƩAc . yƩAc
= 68 cm
Ya = 164 – 68 = 96 cm.Ix = ƩI0 + ƩIg = 13510907 cm3cm
Wa = Ixya
= 140939 cm3
Wb = Ixyb
= 198291 cm3
Ka = IxA . yb
= 37 cm
Kb = IxA . ya
= 26 cm
Kondisi Akhir
PENAMPANG KONDISI AKHIRA1 = 4680 cm²A2 = 3380 cm²A3 = 2040 cm² ∑ 10100 cm² ∑Sx = 1197660 cm³ Yb = 118.580198 Cm = 119 Cm Ya = 71 Cm Ka = 36.354 Cm = 36 CmKb = 60.360 Cm = 60 Cm
POT A Y A x Y Io IpI 4680 177 828360 15743520 263640
II 3380 99 334620 1352000 4760167
III 2040 17 34680 21224160 196520∑ 10100 1197660 43540006.67
Yb = ƩAc . yƩAc
= 119 cm
Ya = 190 – 119 = 71 cmIx = ƩI0 + ƩIg = 43540006.67 cm3cm
Wa = Ixya
= 609634.9 cm3
Wb = Ixyb
= 367177.7 cm3
Ka = IxA . yb
= 36 cm
Kb = IxA . ya
= 60 cm
2. Penampang Ujung
PENAMPANG KONDISI AWALA = 9840 cm²
∑Sx = 806880 cm³ Yb = 82 Cm Ya = 82 Cm I = 22054720 cm³cm
Ka = 27.33 Cm = 27 CmKb = 27.33 Cm = 27 Cm
Kondisi awal
Yb = ƩAc . yƩAc
= 82 cm
Ya = 164 – 80 = 82 cmIx = ƩI0 + ƩIg = 22054720 cm4
Wa = Ixya
= 22054720
82= 268960 cm3
Wb = Ixyb
= 22054720
82= 268960 cm3
Ka = IxA . yb
= 220547209840x 82
= 27cm
Kb = IxA . ya
= 220547209840x 82
= 27cm
Kondisi Akhir
A1 = 4680 cm²A2 = 9840 cm² ∑ 14520 cm² ∑Sx = 1635240 cm³
Yb = 112.6198347
Cm
= 112 Cm Ya = 78 Cm I₁ = 20036640 cm³cmI₂ = 30910720 cm³cmItotal = 50947360 cm³cm Ka = 31.328 Cm = 32 CmKb = 44.984 Cm = 45 Cm
DLMDL
M
Wa
Pi
kaya
cgc
Pi
Pi
Ac
piMpi
M
Wa
e1
ekb
yb
DLMDL
M
Wb
Pi
Pi
Ac pi
Mpi
M
Wb
,
0,6 fc
,
0,25 fc
Yb = ƩAc . yƩAc
= 112 cm
Ya = 180 – 107 = 78 cmIx = ƩI0 + ƩIg = 50947360cm3cm
Wa = Ixya
= 50947360
78= 653171 cm3
Wb = Ixyb
= 50947360
112= 454887 cm3
Ka = IxA . yb
= 50947360
14520 x 112= 31 cm
Kb = IxA . ya
= 5094736014520x 78
= 45 cm
5.Penentuan Gaya Prategang dan Diameter Kabel
e = 68 – 28 = 40 cm.
qDL = Ac x Bj = 1.3008 t/m
Momen Ultimate (MDL) = 18
x q x l2 = 18
x 1.3008 x 292 = 136.7466 tm = 1367466 kgcm
Mpi = Pi x e = 40 Pi
-97 - 0,000185 pi + 0.00185 pi = 16…….................................. ( 1 )
69 - 0.000185pi – 0,002 pi = -246 .......................................... ( 2 )
Maka agar kondisinya aman, diperoleh nilai Pi = 818182 kg
Pi/tendon = 16,5 x 0,85 = 14,025 ton = 14025 kg
Digunakan kabel Ø12,7, dengan Pu = 17500 kgPi kabel = 0,94 x 0,85 x Pu = 19975 kg
Jumlah tendon = Pi
Pi / tendon =81818214025
= 59
Jumlah Tendon/sel = 59/3 = 20 buah tendon/sel
Type DongkrakPi/angkur = 818182/3 = 272727 kg = 272 tonDigunakan dongkrak type k350
Tekanan compressor = 272727
3065 = 556 bar
qDl => p + a = 0.1 x 1.8 = 0.18lantai = 0.26x1.8x2.4 = 1.123lantai kerja =0.07x1.74x2.4 = 0.292
= 1.596 tm
Mlt = qdl x Lo = 167.72904 tm = 16772904 kgm
Prediksi Loss % Awal = 12%
Pe1 = Pix(1-12%) = 818182 x (1- 12/100) = 720000
MPe1 = Pe1 x e = 720000 x 40 = 28800006 kgcm
Ap = 70.725 cm2
KONTROL KEHILANGAN TEGANGAN
1.Penyusutan Beton
σPi = PiAp
= 81818270.25
= 11568 kg/cm
Ec = 4700√ fc ' = 4700√41 = 300947kg/cm2
Loss = 200x 10−5
log (28+2)x Ec = 0,00135 x 300947 = 407 Kg/cm2.
%Loss = Loss x EcσPi
x 100% = 407
11568 x 100% = 3,52%.
2. Slip Anchor.
.Es = 1.95 x (106) = 1950000
L = 2900 cm
Ϫ = 0,4
%Loss = Es x ∆σPi x L
x100% = 1950000 x3
13111x35000 x100% =1,27%.
3. Gesekan Tendon
Po = Pi
0,85 =
8181820,85
= 962567
σPo = PoAp
= 96256770,725
= 13609.997 = 13.609997 < fypy (ok)
α = 5,36◦ = 0,09 radr = 15527 cm
L = 2x(5,36◦ /360)x 3.14 x2 x 15527 = 2904 cm = 29,04 m
e = 2,7183
μ = 0,2
k = 0,044Px = Poe-(µα+kx) = 962567x2,7183-(0,2x0,09+0,044) = 904462.19
Loss of prestress = Po−PxPo
x 100% = 962567−904462,19
962567 x 100% = 6,04
Total Loss = 11,8 % < 12 % (ok)
Kehilangan Tegangan Pada Kondisi AkhirRangkak BetonEs = 1950000 Kg/cm2
Ec = 4700√41 x 10 = 300947 Kg/cm2
Øcc = 1.109Pi = 818182 Kg.Ap = 70,725 cm2.fc = 11,09 Kg/cm2.
fp1 = PiAp
(1-loss awal) = 81818270,25
(1-0,12) = 10180.278
Kg/cm2.
αe = EsEc
= 1950000300947
= 6,48
loss = Øccx fc x αe = 79,69 Kg/cm2.
%Loss = Øcc x fc x αe
fp1x100% =
1,109x 11,09 x6,4812062
x100% = 0,78%
Relaksasi Tendon
fp = PeAp
= 67680070,25
= 9569.4613 Kg/cm2.
fp’ = 0,6 x fpu = 0,6 x 17500 = 10500 Kg/cm2.
K4 = log [5,4 x j1,6] = log [5,4 x 501,6] = 7.5504143
.fpfp '
= 9569 ,10500
= 0.9113773 > 0,85
K5 = 1,7
K6 = T20
= 3320
= 1,65
Rb = 2%Rt = K4 x K5 x K6 x Rb = 7,5 x 1,7 x 1,65 x 0,02 = 0,423
Ϫ fc = loss susut + loss rangkak = 407 + 76,69 = 487.17 Kg/cm2.
%Loss = Rt[Ϫfcfp 1
]x100% = 0,423[487,17
10180,278]x100% =2,03 %.
Jumlah loss akhir = 0,78% + 2,03% = 2,81%
10. Kontrol Geser Tumpuan
Pe = 676800 Kg
α = 5,36o
Pv = Pesinα = 676800 x sin 5,36o = 63213 Kg = 63 ton
qDl => aspal = 0.05 x2 x 2.2 x1.3 = 0.286
lantai = 0.26 x2 x 2.4 x 1.3 = 1.6224
lantai kerja = 0.07 x 1.74 x2.4 x 1.2 = 0.350784
girder = Ac x 2.4 x 1.2 = 1.56096 +
3.82 t/m
qll = 0.87 x2 x1.8 = 3,132 t/m
PLL = 4.9 x 1.4 x 2 x 1.8 = 24,696
R = 12
qDL L 1,2 + 12
qLL L 16,8 + PLL 1,8
= 12
x 3,82 x 29 + 12
x 3,132 x 29 + 24,696
= 125.50 ton = 125500 Kg.
Agc = 0.6 x (1.9-0.14) = 1,06
Vc = R – Pv = 125,500 + 63 = 188,7 ton.
Ph = Pecosα = 676800 cos 5,36o= 673822 Kg
Vc’ = [ 1 + Ph
1,4 x Agc ] [ √ fc '
6 ] bw d
=[ 1 + 673822
1,4 x (14 x1,06) ] [ √506
x 10 ] x 0,6 x 1,9 = 1775294 Kg.
Vn = Vu/0,7 = 188,7/0,7 = 269,59
Vs = Vn - Vc’ = 269590 - 1775294 = -1505700 kg =-1505.70 ton
Vc < Vc’ ............(Tidak perlu Tulangan Geser)
Smax = 1/2d = ½ x 1,9 = 0,95 m
Tulangan Endzone
Pi per CoverPlate = 818182
3 = 272727 Kg.
A CoverPlate = 31,5 x 31,5 = 992,25 cm2.
f = PiAcp
= 818182992,25
= 274,86 Kg/cm2.
fcc = 0,6 fc’ = 0,41 x 41 x 10 = 246 Kg/cm2.
f >fcc............(Perlu tulanganend zone)
P end zone = ( f – fcc ) x Acp = ( 274,86 – 246 ) x 992,25 = 28634 Kg
Akan digunakan tulangan D19 dengan As = 2,835 cm2
fe = 0,6 x fu = 0,6 x 4000 = 2400 Kg/cm2
Ptulangan
= 2,835 x 4000 x 0,85 = 9639 Kg
Jumlah Tulangan (n) = 286349639
= 2,9706 ~ 4 tulangan
Keliling sengkang bagian luar = 60 + 60 + 160 + 160 = 500 cm
Keliling sengkang bagian dalam = 40 + 40 + 140 + 140 = 420 cm
Jarak antar tulangan bagian luar
Atas dan Bawah = 60−(2,9 x8)
6= 5,25 cm > 1,5 d................ok!
Kiri dan Kanan = 160−(2,9 x25)
20 = 4,52 cm > 1,5 d...............ok!
Jarak antar tulangan bagian dalam
Atas dan Bawah = 40−(2,9 x 6)
5 = 4,52 cm > 1,5 d...............ok!
Kiri dan Kanan = 140−(2,9 x11)
12 = 9 cm > 1,5 d...................ok!
LENDUTAN AKIBAT PRESTRESSMpe = 61588813.69