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