Perencanaan Gelagar Prategang

8/22/2019 Perencanaan Gelagar Prategang

1/18

NIM akhir 4

Diketahui:

A. Material

= 20 m

Beton

f'c = 40.4 MPa

f'ci = 30.4 MPa

Eci = 30000 MPa 3 hariEc = 40000 MPa 28 hari

Baja

fpu = 1860 MPa

fpy = 1450 MPa

Eps = 200000 MPa

dps = 12.7 mm

n strand = 7 (kabel seven wire strand)

1- = 0.15

= 0.85

B. Pembebanan

=

=

Ditanya:

Bentuk & Dimensi Penampang

Gaya Prategang & n kabel perlu

Posisi Kabel

Tulangan Geser Perlu

Penyelesaian:

1. Perhitungan Tegangan Izin

Saat Transfer:

fci = -0,6. f'ci

= -0,6 x 30.4

= -18.24 MPa

fti =

= 0,25 x ( 30.4^0,5)

= 1.378404875 MPa (di tengah bentang)

fti =

= 0,5 x (30.4^0,5)

= 2.75680975 MPa (di tumpuan)

Saat Layan:

fc = -0,45. f'c

= -0,45 x 40.4

= -18.18 MPa

ft =

= 0,5 x (40.4^0,5)

= 3.178049716 MPa

Area = 178064.1600

Centroid x = 203.2000 mm

Centroid y = 319.7700 mm

Moments Inertia x = 9.4668E+09 mm^4

Moments Inertia y = 1.3953E+09 mm^4

2

Ct = 391.4300 mm

Cb = 319.7700 mm

h = 711.2000 mm

b1 = 304.8000 mm

b2 = 406.4000 mm

a. Momen Lentur

qdl = Ac x beton

= 0.1781 x 2.4

= 0.4274 t/m

qsidl = 0.2 t/m

qLL = 0.34 t/m

=

= 1/8 x 0.4274 x 20

= 21.3677 t.m

=

= 1/8 x 0.2 x 20

= 10.0000 t.m

=

= 1/8 x 0.34 x 20

= 17.0000 t.m

Hasil:

Md = 21.3677 t.m

= 213676992.0000 N.mm

Msidl + MLL = 27.0000 t.m

= 270000000.0000 N.mm

3. Perhitungan St dan Sb perlu

a. St dan Sb perlu (akibat pembebanan dan sifat material)

St 15,608,573

St 14,717,503 mm (di tumpuan)

Sb 16,168,009 mm

b. St dan Sb tersedia dari penampang

St = Ic / Ct

= 9466821164.642 / 391.43

= 24,185,221 mm

Sb = Ic / Cb

= 9466821164.642 / 319.77

= 29,605,095 mm

Hasil:

= 24,185,221 > St perlu = ######## (di tengah bentang) OK

= 24,185,221 > St perlu = ######## (di tumpuan) OK

= 29,605,095 > Sb perlu = ######## OK

4. Gaya Prategang Awal

= 1.3785 - ((391.43 / 711.2) x (1.3785-18.24))

= -9.4192 MPa (di tengah bentang)

= 2.7569 - ((391.43 / 711.2) x (2.7569-18.24))

= -8.7994 MPa (di tumpuan)

Pi =

= 178064. 16 x 9.41916573826548

= 1,677,215.8351 N (di tengah bentang)

Pi =

= 178064. 16 x 8.79940684211027

= 1,566,858.9878 N (di tumpuan)

5. Eksentrisitas Maksimum

e

e ((1.3785 - -9.4192) x (24185221.2775/1677215.8351)) + (213676992/1677215.8351)

e 283.0992986 mm (di tengah bentang)

e

e ((2.7569--8.7995) x (24185221.2775/1566858.9879)) + (213676992/1677215.8351)

e 305.7755732 mm (di tumpuan)

*dipakai e = 200

6. Hitung Batas Pi

Di tengah Bentang

Saat Awal

Pi 3,848,953.6312

Pi 2,057,751.9138

Saat akhir

Pi -3,110,653

Pi 565,054

Di tumpuan

Saat Awal

Pi 1,197,655

Pi 1,474,351

Saat akhir

Pi 1,409,006

Pi 1,728,825

Rekap batas

Tengah Bentang

Saat Awal Saat Akhir

Tumpuan


7. Luas Baja Prategang

Pi = 2,057,751.9138 N

Pe = 0.85 x 2057751.9138

= 1,749,089.1267 N

= 2057751.9138 / 1450

= 1419.1393

n kabel =

= 1.599760173

2

dp = h - e= 711.2 - 200

= 511.2000 mm

a = n x Aps x fps

0,85 x f'c x b

= 3299993.4

12211.304

= 270.2408686 mm

Mn = n x Aps x fps x (dp- a/2)

= (((0,25 x (22/7) x (12.7^2))) x 7) x 2 x 1860 x (511.2-(270.2409/2))

= 1.2411E+09

0,8. Mn > +

> 213676992 + 270000000

> 4.8368 x 10^8 N.mm

2 bu ah tendo n deng an P i = 205 7751 .9138 N , menimbulkan Mn = 9 .9285 x 10^8 N.mm > Mu = MD + MT = 4.8368 x 10^8 N .mm, OK!

8. Kontrol Tegangan Izin

8.a) Saat transfer

Tengah Bentang

ft = -3.374655756 MPa < 1.378405

fb = -18.24 MPa < -18.24

Tumpuan

ft = 5.460367482 MPa > 2.75681

fb = -25.457575 MPa < -18.24

8.b) Saat layan

Tengah Bentang

ft = -6.522530092 MPa < 3.17805

fb = -12.51888688 MPa > -18.18

Tumpuan

ft = 4.641312359 MPa > 3.17805

fb = -21.63893875 MPa < -18.18

9. Perhitungan Batas Aman Tendon

9.a) Nilai Momen Per Pias Bentang

Akibat MD Akibat MT

Ra = Rb = 4.2735 t Ra = Rb = 5.4000 t

qdl = 0.4274 t/mqsidl = 0.2 t/m

qLL = 0.34 t/m

kt =2

= -(53165.2252 / 319.77)

= -166.2608 mm

kb =2

= (53165.2252 / 391.43)

= 135.8231 mm

Pi = 2,057,751.9138 N

Pe = 1,749,089.1267 N

Di atas Tumpuan 0.0000 m

Batas bawah

=

e5 = 135.8231 mm

0 8 x 1.2411 x 10^9

.9285 x 10^8 N.mm

=

=

akibat LL

Momen tengah bentang

akibat DL


akibat SIDL


t/m0.54

0.15 x 213676992 + 270000000

0.85 x 1.37840487520902 - -18.18

St

+

0.85 x 2.75680975041804 - -18.18

St

0.15 x 213676992 + 270000000

3.17804971641414 - 0.85 x -18.24

Sb

cti

LSDDt

ff

MMMS

)1(

cit

LSDDb

ff

MMMS

)1(

cif'.25,0

cif'.5,0

cf'.5,0

101.

600

76.

200

304.800

711.

200

127.

000

127.

000

319.

770

391.

430

)( fcciftcih

cftci

t

fcci

fcci

fcci

)( fcciftcih

cftci

t

fcci

fcci

fcci

I

D

I

t

p

M

p

Sxfcciftci

I

D

I

t

p

M

p

Sxfcciftci

)/.1(

)/(

2rce

ASMfP

tc

tDti

i

)/.1(

)/(

2rce

ASMfP

b

cbDcii

)/.1(

)/(

2rce

ASMfP

tc

tTc

i

)/.1(

)/(

2rce

ASMfP

b

cbTtsi

)/.1(

)(2rce

AfP

t

ctii

)/.1(

)(2

rce

AfP

b

cci

i

t

Dt

c

i

S

M

r

ce

A

Ptf )1(2

.

b

Db

c

i

S

M

r

ce

A

Pbf )1( 2

.

)1(2

.

r

ce

A

Pt t

c

if

)1( 2.

r

ce

A

P

bb

c

if

tTt

c

e

S

M

r

ce

A

Ptf )1(2

.

b

Tb

c

e

S

M

r

ce

A

Pbf )1( 2

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)1( 2.

r

ce

A

Pt t

c

ef

)1( 2.

r

ce

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P

bb

c

ef

)/.1(

)(

2rce

AfP

tcts

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)/.1(

)(

2rce

AfP

b

cci


2/18

Batas atas

=

e5 = -166.2608 mm

1/8 Bentang 2.5000 m

=

= 9.3484 t.m

Batas bawah

=

= 93483684 / 2057751.9138

= 45.4300 mm

e4 =

= 135.8231 + 45.4301

= 181.2531 mm

+ = ############## N.mm

= 11.8125 t.m

Batas atas

=

= 118125000 / 1749089.12673

= 67.5352 mm

e4 = -

= 67.5352 - -166.2609

= 233.7960 mm


=

= 16.0258 t.m

Batas bawah

=

= 160257744 / 2057751.9138

= 77.8800 mm

e3 =

= 135.8231 + 77.8801

= 213.7031 mm

+ = ############## N.mm

= 20.2500 t.m

Batas atas

=

= 202500000 / 1749089.12673

= 115.7745 mm

e3 = -

= 115.7746 - -166.2609

= 282.0354 mm


=

= 20.0322 t.m

Batas bawah

=

= 200322180 / 2057751.9138

= 97.3500 mm

e2 =

= 135.8231 + 97.3501

= 233.1731 mm

= ############## N.mm

= 25.3125 t.m

Batas atas

=

= 253125000 / 1749089.12673

= 144.7182 mm

e2 = -

= 144.7182 - -166.2609

= 310.9790 mm

1/2 Bentang 10.0000 m

=

= 21.3677 t.m

Batas bawah

=

= 213676992 / 2057751.9138

= 103.8400 mm

e1 =

= 135.8231 + 103.8401

= 239.6631 mm

= ##############

= 27.0000 t.m

Batas atas

=

= 270000000 / 1749089.12673

= 154.3661 mm

e1 = -

= 154.3661 - -166.2609

= 320.6269 mm

10. Perencanaan Balok Prategang

10.a) Perhitungan Tulangan Utama

qdl = 0.4274 t/m

L = 4.0000 m *(1 segmen 4 m)

Mu = 0.8547 t.m

= 8,547,079.6800 N.mm

= 13 mm

= 10 mm

= 30 mm

fy = 400 MPa

f'c = 40.4 MPa

h = 711.2000 mm

b1 = 304.8000 mm

b2 = 406.4000 mm

d = h p - sengkang tul. Utama

= 711.2 30 - 10 ( .13)

= 664.7 mm

Ac' = 153883.36

'

a = 186.2970297

b = -320

c = 0.0836

dengan rumus abc diperoleh akar2 persamaan:

x1 = 1.718

x2 = -0.0002612

= 0.0002612

min = 1,4 / fy

= 1,4 / 400

= 0.0035

= 0.0035000

As = . b. D

= 0.0035 x ((304.8 + 406.4) / 2) x 664.7

= 827.28562

10.b) Perhitungan Tulangan Geser

*Pengaruh Beban Mati

= 0.6274 t/m

L = 20 m

= 20000 mm

Vu = 1/2. q. L

= 0,5 x 0.627353984 x 20

= 6.27353984 t

= 62,735.3984 N

d = 664.7000 mm

Vc = gaya lintang yang ditahan oleh beton

Untuk perhitungan Vc ini, harus dilihat dari dua hal yaitu

-retak akibat geseran pada badan penampang (Vcw)

-retak miring akibat lentur (Vci)

-nilai Vc adalah nilai terkecil dari Vcw dan Vci

Vcw = '0,5

Vp = komponen vertikal dari gaya prategang

Vp = 27,949.0187 N

Bw = 250 mm *Separuh Elastomer

fpc = Pe / Ac

= 1749089.1268 / 178064.16

= 9.822802785

Vcw = '0,5

= (((0,29 x (40.4^0,5)) + (0,3 x 9.8229)) x 250 x 664.7) + 27949.0187

= 823,945.4933 N

Mcr = ' '0,5

= (9466821164.642 / 391.43) x ((0,5 x (40.4^0,5)) + 9.8229)

= ############## N.mm

Tegangan terbesar terdapat pada jarak x atau 0,25.L dari tumpuan, maka:

x = 0,25 x L

= 0,25 x 20 m

= 5000 mm

Mmax / Vt =

= ((20000. 5000) - 5000) / (20000 - 2.5000)

= 7500 mm

Vci = '0,5

= (0,05 x 250 x 664.7 x (40.4^0,5)) + (314428494.5389 / 7500)

= 52,811.2412 + 41923.79927

= 94,735.0404 N

dipakai = 94,735.0404 N

= 104,558.9973 N

Vs =

= (62735.3984 / 0, 6) - 94735.0405

= 9,823.9569 N

0,5.Vc = 0,5 x 94735.0404343639

= 47,367.5202 N

Dibutuhkan Tulangan Geser, (Vu / ) > 0,5.Vc

Av / S = Vs / (fy.d)

Av / S = 9823.95689896942 / (400 x 664.7)

Av / S = 0.036948837

Av = 78.57142857

0.03695 = 78.5714285714286 / S

S = Av / (Av / S)

= 78.5715 / 0.037

= 2126.492578 mm

= 212.6492578 cm

S = 0,75 x h

= 533.4 mm

= 53.34 cm

S = 24 inch

= 60.96 cm

S = 53.34 *(Dipakai yg terkecil)

40 *Sesuaikan dengan bentang

Mu / Ac' x d

Mu / Ac' x d

0.0836

tulan an utama

tulan an sen kan

Mu / Ac' x d

0.0836


3/18

angan Beton Prategang

NIM akhir 4

iketahui:

. MaterialPanjang bentang :

L = 20 m

Beton

Mutu beton pada umur 28 hari :

f'c = 40.4 MPa

Mutu beton pada umur 3 hari :

f'ci = 30.4 MPa

Modulus elastisitas beton pada umur 3 hari :

Eci = 30000 MPa

Modulus elastisitas beton pada umur 28 hari :

Ec = 40000 MPa

Berat jenis beton :

beton = 2.4 t/m

Baja

Mutu baja prategang ultimit :

fpu = 1860 MPa

Mutu baja prategang leleh :

fpy = 1450 MPa

Modulus elastisitas baja prategang :

Eps = 200000 MPa

Diameter baja prategang :dps = 12.70 mm (1 strand)

Jumlah Strand (dalam 1 stendon):


Kehilangan prategang :

1- = 0.15 (prategang residual)

= 0.85

. Pembebanan

Beban berat sendiri balok :

qdl = - t/m

Beban mati :

qsidl = 0.20 t/m

Beban hidup :

qLL = 0.34 t/m

itanya:



Posisi Kabel


BAB 3

PERHITUNGAN

Anggen


4/18


enyelesaian:

. Perhitungan Tegangan Izin

Saat Transfer:

Tegangan tekan ijin saat awal :

fci = -0,6. f'ci

= -0,6 x 30.4

= -18.24 MPa

Tegangan tarik ijin saat awal di tengah bentang :

fti =

= 0,25 x ( 30.4^0,5)

= 1.3784 MPa

Tegangan tarik ijin saat awal di tumpuan :

fti =

= 0,5 x (30.4^0,5)

= 2.7568 MPa

Saat Layan:

Tegangan tekan ijin saat akhir :

fc = -0,45. f'c

= -0,45 x 40.4

= -18.1800 MPa

Tegangan akhir ijin saat akhir :

ft =

= 0,5 x (40.4^0,5)

= 3.1780 MPa

. Perhitungan Momen Yang Terjadi

Anggen


5/18


Data Penampang :

Luas = 178064.1600 mm



R2

= 53165.2252 mm

Ct = 391.4300 mm

Cb = 319.7700 mm

h = 711.2000 mm

b1 = 304.8000 mm

b2 = 406.4000 mm

. Momen Lentur

Beban berat sendiri balok :

qdl = Ac x beton

= 0.1781 x 2.4

= 0.4274 t/m

Beban mati :

qsidl = 0.20 t/m

Beban hidup :

qLL = 0.34 t/m

= 1/8.qdl.l

= 1/8 x 0.4274 x 20

= 21.3677 t.m

= 1/8.qsidl.l

= 1/8 x 0.2 x 20

= 10.0000 t.m

= 1/8.qLL.l

= 1/8 x 0.34 x 20

= 17.0000 t.m

Hasil perhitungan momen :

Md = 21.3677 t.m

= 2.1368E+08 N.mm


= 2.7000E+08 N.mm

. Perhitungan Modulus Penampang Serat Atas (St) dan Modulus Penampang Serat Bawah (Sb) perlu

. Modulus Penampang St dan Modulus Sb perlu (akibat pembebanan dan sifat material)

Perhitungan Modulus Penampang Serat Atas (St) perlu di tengah bentang :

St 15,608,573 mm

(akibat LL )


St ((0.15 x 213676992) + 270000000)

((0.85 x 1.37840487520902) - -18.18)

(akibat DL )


(akibat SIDL )


Anggen


6/18


Perhitungan Modulus Penampang Serat Atas (St) perlu di tumpuan :

St 14,717,503 mm

Perhitungan Modulus Penampang Serat Bawah (Sb) :

Sb 16,168,009 mm

. Modulus Penampang Serat Atas (St) dan Modulus Penampang Serat Bawah (Sb) tersedia dari penamp

Modulus Penampang Serat Atas (St) tersedia :

St = Ic / Ct

= 9466821164.642 / 391.43

= 24,185,221 mm

Modulus Penampang Serat Bawah (Sb) tersedia :

Sb = Ic / Cb

= 9466821164.642 / 319.77

= 29,605,095 mm

Hasil perhitungan Modulus Penampang Serat Atas (St) dan Modulus Penampang Serat Bawah (Sb) :

t tersedia = 24,185,221 > St perlu = 15,608,573 mm OK

t tersedia = 24,185,221 > St perlu = 14,717,503 mm OK

b tersedia 29,605,095 > Sb perlu = 16,168,009 mm OK

. Gaya Prategang Awal

= 1.3785 - ((391.43 / 711.2) x (1.3785-18.24))


= 2.7569 - ((391.43 / 711.2) x (2.7569-18.24))


((0.15 x 213676992) + 270000000)

(3.17804971641414 - (0.85 x -18.24))

=

Modulus Penampang Serat Atas (St) :

Modulus Penampang Serat Bawah (Sb) :

=

St ((0.15 x 213676992) + 270000000)

((0.85 x 2.75680975041804) - -18.18)

Sb

Anggen


7/18


Gaya Prategang Awal di tengah bentang :

Pi = Ac x fcci

= 178064.16 x 9.41916573826548

= 1,677,215.8351 N

Gaya Prategang Awal di tumpuan :

Pi = Ac x fcci

= 178064.16 x 8.79940684211027

= 1,566,858.9878 N

. Eksentrisitas Maksimum

Eksentrisitas Maksimum di tengah bentang :

e

e ((1.3785 - -9.4192) x (24185221.2775/1677215.8351)) + (213676992/1677215.8351)

e 283.0992986 mm

Eksentrisitas Maksimum di tumpuan :

e

e ((2.7569--8.7995) x (24185221.2775/1566858.9879)) + (213676992/1677215.8351)

e 305.7755732 mm

Asumsi eksentrisitas kurang dari eksentrisitas maksimum di tengah bentang, maka dipakai :

e = 200 mm

. Mencari Batas - batas Gaya Prategang

Batas - batas Gaya Prategang Di tengah Bentang

Saat Awal

Pi

Pi 3,848,953.6312 N

Pi

Pi 2,057,751.9138 N

((1.3785 + (213676992 / 24185221.2775)) x 178064.16)

(-1 + ((200 x 391.43) / 53165.2252))

((--18.24 + (213676992 / 29605094.8014)) x 178064.16)

(1 + ((200 x 319.77) / 53165.2252))

Anggen


8/18


Saat akhir

Pi

Pi -3,110,653 N

Pi

Pi 565,054 N

Batas - batas Gaya Prategang Di tumpuan

Saat Awal

Pi

(-1 + ((200 x 391.43) / 53165.2252))

Pi 1,197,655 N

Pi

(1 + (200 x 319.77) / 53165.2252)

Pi 1,474,351 N

Saat akhir

Pi

Pi 1,409,006 N

Pi

Pi 1,728,825 N

(3.1781 x 178064.16)

(--18.24 x 178064.16)

(3.1781 x 178064.16)

((-1+((200 x 391.43) / 53165.2252)) x 0.85)

((-1+((200 x 391.43) / 53165.2252)) x 0.85)

((-3.1781 + (270000000 / 29605094.8014)) x 178064.16)

((1+((200 x 319.77) / 53165.2252)) x 0.85)

((-18.18 + (270000000 / 24185221.2775)) x 178064.16)

(--18.18 x 178064.16)

((1+((200 x 319.77) / 53165.2251898529)) x 0.85)

Anggen


9/18


Rekapitulasi Hasil Batas - batas Gaya Prategang :

Tengah Bentang


Pi 3,848,953.6313 N Pi -3,110,652.8725 N

Pi 2,057,751.9138 N Pi 565,053.8796 N

Tumpuan


Pi 1,197,655.2692 N Pi 1,409,006.1991 N

Pi 1,474,350.7547 N Pi 1,728,824.6079 N

. Luas Baja Prategang

Maka batas prategang yang memenuhi persyaratan ( Pi) adalah :

Pi = 1,197,655.2692 N

Gaya Prategang Efektif (Pe)

Pe = . Pi

= 0.85 x 1197655.2692

= 1,018,006.9788 N

Luas Baja Prategang (Apsperlu)

Apsperlu = Pi / fpsi

= 1197655.2692 / 1450

= 825.9692 mm

Jumlah kabel (n)

n = Apsperlu / Aps

= 825.9692 / (((0,25 x (22/7) x (12.7^2))) x 7)

= 0.93109436

1 Tendon

Gaya Prategang Aktual (Pi)

Pi = Pi . n

= 1197655.2692 x 1

= 1,197,655.2692 N

Gaya Prategang Aktual Efektif (Pe)

Pe = Pe . n

= 1018006.97882 x 1

= 1,018,006.9788 N

Luas Baja Prategang Aktual (Apsaktual)

Apsaktual = Jumlah tendon . Luas satu tendon

= n . 1/4 D2

= 1 x (0,25 x (22/7) x (12.7^2) x 7)

= 887.095 mmJarak dari serat tekan terluar :

dp = h - e'

= 711.2 - (319.77 - 200)

= 591.4300 mm

. Anggen


10/18

cangan Beton Prategang

Tinggi blok tegangan persegi panjang ekivalen :

a = n x Aps x fps

0,85 x f'c x b

=

= 1649996.712211.304

= 135.1204343 mm

Kuat momen nominal :


= (((0,25 x (22/7) x (12.7^2))) x 7) x 1 x 1860 x (591.43-(135.1205/2)

= 8.6438E+08 N.mm

Faktor reduksi dikali Momen nominal

0,8. Mn > MD + MT

> 213676992 + 270000000

> 4.8368 x 10^8 N.mm

1 buah tendon dengan Pi = 1197655.2692 N, menimbulkan Mn = 6.9151 x 10^8 N.mm > Mu = MD + MT = 4.8368 x 10^8 N.mm, OK!

Momen tahanan tendon > momen akibat beban sehingga tulangan non-prategang secara teoritis tidak diperlukan,

namun tetap dipasang sebagai tulangan pembentuk.


8.a) Saat transfer

Tegangan serat di tengah bentang

Tepi atas :

ft = -1,197,655.2692 -213676992.0000178,064.1600 24,185,221

ft = -5.656973521 < 1.378404875 MPa

Tepi bawah :

fb = -1,197,655.2692 213676992.0000

178064.1600 29,605,095

fb = -7.599273736 > -18.24 MPa

Tegangan serat di tumpuanTepi atas :

ft = -1,197,655.2692

178064.1600

ft = 3.178049716 > 2.75680975 MPa

(1 - 1.4726)

(1 + 1.203)

(1 - 1.4726)

0,85 x 40.4 x ((304.8 + 406.4) / 2)

1 x 887.095 x 1860

0,8 x 0.8644 x 10^9

6.9151 x 10^8 N.mm

S. Anggen


11/18


Tepi bawah :

fb = -1,197,655.2692

178064.1600

fb = -14.81684873 > -18.24 MPa

.b) Saat layan

Tegangan serat di tengah bentang

Tepi atas :

ft = -1,018,006.9788 -270000000.0000

178064.1600 24,185,221

ft = -8.462500193 < 3.178049716 MPa

Tepi bawah :

fb = -1,018,006.9788 270000000.0000

178064.1600 29,605,095

fb = -3.474269559 > -18.18 MPa

Tegangan serat di tumpuan

Tepi atas :

ft = -1018006.979

178064.1600

ft = 2.701342259 < 3.178049716 MPa

Tepi bawah :

fb = -1,018,006.9788

178064.1600

fb = -12.59432142 > -18.18 MPa

. Perhitungan Batas Aman Tendon

.a) Nilai Momen Per Pias Bentang

Beban yang Bekerja :

Akibat MD

Ra = Rb = 4.2735 t

Akibat MT

Ra = Rb = 5.40 t

(1 + 1.203)

(1 + 1.203)

(1 - 1.4726)

(1 + 1.203)

(1 - 1.4726)

Anggen


12/18


Beban yang bekerja :

qdl = 0.4274 t/m

qsidl = 0.2 t/m

qLL = 0.34 t/m

Posisi kern atas (diatas garis netral)

kt = -(r2

/ Cb)

= -(53165.2252 / 319.77)= -166.2608 mm

Posisi kern bawah (dibawah garis netral)

kb = (r2

/ Ct)

= (53165.2252 / 391.43)

= 135.8231 mm

Gaya prategang

Pi = 1,197,655.2692 N

Gaya prategang efektif

Pe = 1,018,006.9788 N


Batas bawah

amin = 0 mm

e5 = 135.8231 mm

Batas atas

amax = 0 mm

e5 = -166.2608 mm


MD = 93,483,684.0000 N.mm

= 9.3484 t.m

Batas bawah

amin = MD /Pi

= 93483684 / 1197655.2692

= 78.0556 mm

e4 = kb + amin= 135.8231 + 78.0556

= 213.8787 mm

MSD + MLL = ############## N.mm

= 11.8125 t.m

Batas atas

amax = MT /Pe

= 118125000 / 1018006.97882

= 116.0356 mm

e4 = amax - kt

= 116.0356 - -166.2609

= 282.2964 mm

0.54 t/m

. Anggen


13/18

cangan Beton Prategang


MD = ############## N.mm

= 16.0258 t.m

Batas bawah

amin = MD /Pi= 160257744 / 1197655.2692

= 133.8096 mm

e3 = kb + amin= 135.8231 + 133.8096

= 269.6326 mm

MSD + MLL = ############## N.mm

= 20.2500 t.m

Batas atas

amax = MT /Pe

= 202500000 / 1018006.97882= 198.9181 mm

e3 = amax - kt

= 198.9181 - -166.2609

= 365.1789 mm


MD = ############## N.mm

= 20.0322 t.m

Batas bawah

amin = MD /Pi

= 200322180 / 1197655.2692

= 167.2620 mm

e2 = kb + amin= 135.8231 + 167.262

= 303.0850 mm

MSD + MLL = ############## N.mm

= 25.3125 t.m

Batas atas

amax = MT /Pe

= 253125000 / 1018006.97882

= 248.6476 mm

e2 = amax - kt

= 248.6477 - -166.2609

= 414.9084 mm

S. Anggen


14/18


1/2 Bentang 10.0000 m

MD = ############## N.mm

= 21.3677 t.m

Batas bawah

amin = MD /Pi= 213676992 / 1197655.2692

= 178.4128 mm

e1 = kb + amin= 135.8231 + 178.4128

= 314.2358 mm

MSD + MLL = ##############

= 27.0000 t.m

Batas atas

amax = MT /Pe

= 270000000 / 1018006.97882

= 265.2241 mm

e1 = amax - kt

= 265.2242 - -166.2609

= 431.4849 mm


0.a) Perhitungan Tulangan Geser

Pengaruh Beban Mati dan tambahan beban luar

qdl + qsd = 0.6274 t/m

Panjang bentang (L)

L = 20 m= 20000 mm

Gaya geser terfaktor (Vu)

Vu = 1/2. q. L

= 0,5 x 0.627354 x 20

= 6.2735 t

= 62,735.3984 N

Jarak dari serat tekan terluar ke pusat berat tulangan tarik

d = 664.7000 mm



-retak akibat geseran pada badan penampang (Vcw)-retak miring akibat lentur (Vci)


Kuat geser badan (Vcw)

Vcw = (0,29. (f 'c)0,5

+ 0,3.fpc).bw.d + Vp

Komponen vertikal dari gaya prategang di penampang (Vp)


Vp = 16,266.9218 N

. Anggen


15/18


Lebar badan (Bw)

Bw = 350 mm

Tegangan tekan beton akibat gaya prategang dari beban luar (fpc)

fpc = Pe / Ac

= 1018006.9789 / 178064.16

= 5.717079612 N/mm

Vcw = (0,29. (f 'c)0,5 + 0,3.fpc).bw.d + Vp

= (((0,29 x (40.4^0,5)) + (0,3 x 5.7171)) x 350 x 664.7) + 16266.9219

= 844,109.1960 N

Momen akibat beban hidup yang bekerja menimbulkan retak

Mcr = (Ic' / Yt).((0,5.f'c0,5

) + fpc)

= (9466821164.642 / 391.43) x ((0,5 x (40.4 0,5)) + 5.7171)

= ############## N.mm


x = 0,25 x L

= 0,25 x 20 m

= 5000 mm

Mmax / Vt = ((L. x) - x) / (L - 2.x)= ((20000. 5000) - 5000) / (20000 - 2.5000)

= 7500 mm

Kekuatan lentur geser

Vci = 0,05.bw.d.(f'c)0,5

+ (Vt.Mcr) / Mmax

= (0,05 x 350 x 664.7 x (40.4 0,5)) + (215130671.1071 / 7500)

= 73,935.7376 + 28684.08948

= 102,619.8271 N

ipakai Vc = Vci = 102,619.8271 N

Kekuatan geser nominal

Vu / = 62735.3984 / 0,6

= 104,558.9973 N

Kuat geser nominal tulangan geser (Vs)

Vs = (Vu / ) - Vc

= (62735.3984 / 0,6) - 102619.8272

= 1,939.1702 N

0,5 kuat geser nominal beton

0,5.Vc = 0,5 x 102619.8272

= 51,309.9136 N


Mencari kebutuhan luas sengkang per milimeter panjang


Av / S = 1939.1703 / (400 x 664.7)

Av / S = 0.007293 mm/mm

Luas 1 tulangan sengkang dengan diameter 10 mm

Av = 78.5714 mm tulangan sengkang 10 m

idapatkan 0.0072934 = 78.5715 / S

S = Av / (Av / S)

= 78.5715 / 0.0073

= 10772.9436 mm

= 1077.2944 cm

Anggen


16/18


S = 0,75 x h

= 533.4 mm

= 53.34 cm

S = 24 inch

= 60.96 cm



Anggen


17/18

Wandrianto S. Anggen

I 1111 094

NIM akhir 4

Diketahui:

A. Material

= 20 m

Betonf'c = 40.4 MPa

f'ci = 30.4 MPa

Eci = 30000 MPa 3 hari

Ec = 40000 MPa 28 hari

Baja

fpu = 1860 MPa

fpy = 1450 MPa

Eps = 200000 MPa

dps = 12.7 mm


1- = 0.15

= 0.85

B. Pembebanan

= -

=

Ditanya:



Posisi Kabel


Penyelesaian:

1. Perhitungan Tegangan Izin

Saat Transfer:

fci = -0,6. f'ci

= -0,6 x 30.4

= -18.24 MPa

fti =

= 0,25 x ( 30.4^0,5)

= 1.378404875 MPa (di tengah bentang)

fti =

= 0,5 x (30.4^0,5)= 2.75680975 MPa (di tumpuan)

Saat Layan:

fc = -0,45. f'c

= -0,45 x 40.4

= -18.18 MPa

ft =

= 0,5 x (40.4^0,5)

= 3.178049716 MPa

2. Perhitungan Momen Yang Terjadi

Area = 178064.1600

Centroid x = 203.2000 mm

Centroid y = 319.7700 mm



2

Ct = 391.4300 mm

Cb = 319.7700 mm

h = 711.2000 mm

b1 = 304.8000 mm

b2 = 406.4000 mm

a. Momen Lentur

qdl = Ac x beton

= 0.1781 x 2.4

= 0.4274 t/m

qsidl = 0.2 t/m

qLL = 0.34 t/m

=

= 1/8 x 0.4274 x 20

= 21.3677 t.m

=

= 1/8 x 0.2 x 20= 10.0000 t.m

=

= 1/8 x 0.34 x 20

= 17.0000 t.m

Hasil:

Md = 21.3677 t.m

= 213676992.0000 N.mm


= 270000000.0000 N.mm

3. Perhitungan St dan Sb perlu

a. St dan Sb perlu (akibat pembebanan dan sifat material)

St 15,608,573

St 14,717,503 mm (di tumpuan)

Sb 16,168,009 mm

b. St dan Sb tersedia dari penampang

St = Ic / Ct

= 9466821164.642 / 391.43

= 24,185,221 mm

Sb = Ic / Cb

= 9466821164.642 / 319.77

= 29,605,095 mm

Hasil:

St tersedia = 24,185,221 > St perlu = ######## (di tengah bentang) OK

St tersedia = 24,185,221 > St perlu = ######## (di tumpuan) OK

Sb tersedia = 29,605,095 > Sb perlu = ######## OK

4. Gaya Prategang Awal

= 1.3785 - ((391.43 / 711.2) x (1.3785-18.24))


= 2.7569 - ((391.43 / 711.2) x (2.7569-18.24))


Pi =

= 178064. 16 x 9. 41916573826548

= 1,677,215.8351 N (di tengah bentang)

Pi =

= 178064. 16 x 8. 79940684211027

= 1,566,858.9878 N (di tumpuan)

5. Eksentrisitas Maksimum

e

e ((1.3785 - -9.4192) x (24185221.2775/1677215.8351)) + (213676992/1677215.8351)

e 283.0992986 mm (di tengah bentang)

e

e ((2.7569--8.7995) x (24185221.2775/1566858.9879)) + (213676992/1677215.8351)

e 305.7755732 mm (di tumpuan)

*dipakai e = 200

6. Hitung Batas Pi

Di tengah Bentang

Saat Awal

Pi

Pi 3,848,953.6312

Pi

Pi 2,057,751.9138

Saat akhir

Pi

Pi -3,110,653

Pi

Pi 565,054

Di tumpuan

Saat Awal

Pi

(-1 + ((200 x 391.43) / 53165.2252))

Pi 1,197,655

Pi

(1 + (200 x 319.77) / 53165.2252)

Pi 1,474,351

Saat akhir

Pi

Pi 1,409,006

Pi

Pi 1,728,825

Rekap batas

Tengah Bentang


Tumpuan


7. Luas Baja Prategang

Pi = 1,197,655.2692 N

Pe = 0.85 x 1197655.2692

= 1,018,006.9788 N

= 1197655.2692 / 1450

= 825.9692

n kabel =

= 0.93109436

1

Pi baru = 1,197,655.2692 N

Pe baru = 1,018,006.9788 N

Aps baru = 887.095 mm

dp = h - e'

= 711.2 - (319.77 - 200)

= 591.4300 mm

a = n x Aps x fps

0,85 x f'c x b

= 1649996.7

12211.304

= 135.1204343 mm


= (((0,25 x (22/7) x (12.7^2))) x 7) x 1 x 1860 x (591.43-(135.1205/2))

= 8.6438E+08

0,8. Mn >

> 213676992 + 270000000

> 4.8368 x 10^8 N.mm

1 buah tendo n deng an Pi = 1197 655 .2692 N , menimbulkan Mn = 6 .9151 x 10^8 N.mm > Mu = MD + MT = 4.8368 x 10^8 N .mm, OK!


8.a) Saat transfer

Tengah Bentang

ft = -5.656973521 MPa < 1.378405

fb = -7.599273736 MPa > -18.24

Tumpuan

ft = 3.178049716 MPa > 2.75681

fb = -14.81684873 MPa > -18.24

8.b) Saat layan

Tengah Bentang

ft = -8.462500193 MPa < 3.17805

fb = -3.474269559 MPa > -18.18

Tumpuan

ft = 2.701342259 MPa < 3.17805

fb = -12.59432142 MPa > -18.18

9. Perhitungan Batas Aman Tendon

9.a) Nilai Momen Per Pias Bentang

Akibat MD Akibat MT

3.1781 x 178064.16

-1+ 200 x 391.43 / 53165.2252 x 0.85

--18.18 x 178064.16

1+ 200 x 319.77 / 53165.2251898529 x 0.85

0 8 x 0.8644 x 10 9

6.9151 x 10^8 N.mm

+

3.17804971641414 - 0.85 x -18.24

=

1.3785 + 213676992 / 24185221.2775 x 178064.16

-1 + 200 x 391.43 / 53165.2252

--18.24 + 213676992 / 29605094.8014 x 178064.16

1 + 200 x 319.77 / 53165.2252

-18.18 + 270000000 / 24185221.2775 x 178064.16

-1+ 200 x 391.43 / 53165.2252 x 0.85

- +

1+ 200 x 319.77 / 53165.2252 x 0.85

3.1781 x 178064.16

--18.24 x 178064.16

0.15 x 213676992 + 270000000

0.85 x 1.37840487520902 - -18.18

St 0.15 x 213676992 + 270000000

0.85 x 2.75680975041804 - -18.18

akibat LL


akibat DL


akibat SIDL


=

St

Sb

cti

LSDDt

ff

MMMS

)1(

cit

LSDDb

ff

MMMS

)1(

cif'.25,0

cif'.5,0

cf'.5,0

)( fcciftcih

cftci

t

fcci

fcci

fcci

)( fcciftcih

cftci

t

fcci

fcci

fcci

I

D

I

t

p

M

p

Sxfcciftci

I

D

I

t

p

M

p

Sxfcciftci

)/.1(

)/(

2rce

ASMfP

tc

tDti

i

)/.1(

)/(

2rce

ASMfP

b

cbDcii

)/.1(

)/(

2rce

ASMf

P tc

tTc

i

)/.1(

)/(

2rce

ASMfP

b

cbTtsi

)/.1(

)(2rce

AfP

t

ctii

)/.1(

)(2

rce

AfP

b

cci

i

t

Dt

c

i

S

M

r

ce

A

Ptf )1(2

.

b

Db

c

i

S

M

r

ce

A

Pbf )1( 2

.

)1(2

.

r

ce

A

Pt t

c

if

)1( 2.

r

ce

A

P

bb

c

if

tTt

c

e

S

M

r

ce

A

Ptf )1(2

.

b

Tb

c

e

S

M

r

ce

A

Pbf )1( 2

.

)1( 2.

r

ce

A

Pt t

c

ef

)1( 2.

r

ce

A

P

bb

c

ef

)/.1(

)(

2rce

AfP

tcts

i

)/.1(

)(

2rce

AfP

b

cci


18/18

Ra = Rb = 4.2735 t Ra = Rb = 5.4000 t

qdl = 0.4274 t/m

qsidl = 0.2 t/m

qLL = 0.34 t/m

kt = 2

= -(53165.2252 / 319.77)

= -166.2608 mm

kb =2

= (53165.2252 / 391.43)

= 135.8231 mm

Pi = 1,197,655.2692 N

Pe = 1,018,006.9788 N


Batas bawah

=

e5 = 135.8231 mm

Batas atas

=

e5 = -166.2608 mm


=

= 9.3484 t.m

Batas bawah

=

= 93483684 / 1197655.2692

= 78.0556 mm

e4 =

= 135.8231 + 78.0556

= 213.8787 mm

= ############## N.mm

= 11.8125 t.m

Batas atas

=

= 118125000 / 1018006.97882

= 116.0356 mm

e4 = -

= 116.0356 - -166.2609

= 282.2964 mm


=

= 16.0258 t.m

Batas bawah

=

= 160257744 / 1197655.2692

= 133.8096 mm

e3 =

= 135.8231 + 133.8096

= 269.6326 mm

+ = ############## N.mm

= 20.2500 t.m

Batas atas

=

= 202500000 / 1018006.97882

= 198.9181 mm

e3 = -

= 198.9181 - -166.2609

= 365.1789 mm


=

= 20.0322 t.m

Batas bawah

=

= 200322180 / 1197655.2692

= 167.2620 mm

e2 =

= 135.8231 + 167.262

= 303.0850 mm

+ = ############## N.mm

= 25.3125 t.m

Batas atas

=

= 253125000 / 1018006.97882

= 248.6476 mm

e2 = -

= 248.6477 - -166.2609

= 414.9084 mm

1/2 Bentang 10.0000 m

=

= 21.3677 t.m

Batas bawah

=

= 213676992 / 1197655.2692

= 178.4128 mm

e1 =

= 135.8231 + 178.4128

= 314.2358 mm

+ = ##############

= 27.0000 t.m

Batas atas

=

= 270000000 / 1018006.97882

= 265.2241 mm

e1 = -

= 265.2242 - -166.2609

= 431.4849 mm


10.a) Perhitungan Tulangan Utama

qdl = 0.4274 t/mL = 4.0000 m *(1 segmen 4 m)

Mu = 0.8547 t.m

= 8,547,079.6800 N.mm

= 13 mm

= 10 mm

= 30 mm

fy = 400 MPa

f'c = 40.4 MPa

h = 711.2000 mm

b1 = 304.8000 mmb2 = 406.4000 mm

d = h p - sengkang tul. Utama

= 711.2 30 - 10 ( .13)

= 664.7 mm

Ac' = 153883.36

'

a = 186.2970

b = -320

c = 0.0836

dengan rumus abc diperoleh akar2 persamaan:

x1 = 1.718

x2 = -0.0002612

= 0.0002612

min = 1,4 / fy

= 1,4 / 400

= 0.0035

= 0.0035000

As perlu = . b. D

= 0.0035 x ((304.8 + 406.4) / 2) x 664.7

= 827.28562

= 18 buah

As tersedia = 2390.1429 mm > As perlu = 827.28562 mm

10.b) Perhitungan Tulangan Geser

*Pengaruh Beban Mati

= 0.6274 t/m

L = 20 m

= 20000 mm

Vu = 1/2. q. L

= 0,5 x 0.627353984 x 20

= 6.27353984 t

= 62,735.3984 N

d = 664.7000 mm



-retak akibat geseran pada badan penampang (Vcw)

-retak miring akibat lentur (Vci)


Vcw = '0,5


Vp = 16,266.9218 N

Bw = 350 mm

fpc = Pe / Ac

= 1018006.9789 / 178064.16

= 5.717079612

Vcw = '0,5

= (((0,29 x (40.4^0,5)) + (0,3 x 5.7171)) x 350 x 664.7) + 16266.9219

= 844,109.1960 N

Mcr = ' '0,5

= (9466821164.642 / 391.43) x ((0,5 x (40.4^0,5)) + 5.7171)= ############## N.mm


x = 0,25 x L

= 0,25 x 20 m

= 5000 mm

Mmax / Vt =

= ((20000. 5000) - 5000) / (20000 - 2.5000)

= 7500 mm

Vci = '0,5

= (0,05 x 350 x 664.7 x (40.4^0,5)) + (215130671.1071 / 7500)= 73,935.7376 + 28684.08948

= 102,619.8271 N

dipakai = 102,619.8271 N

= 104,558.9973 N

Vs =

= (62735.3984 / 0, 6) - 102619.8272

= 1,939.1702 N

0,5.Vc = 0,5 x 102619.8272

= 51,309.9136 N



Av / S = 1939.1703 / (400 x 664.7)

Av / S = 0.007293

Av = 78.5714

didapatkan 0.00729 = 78.5715 / S

S = Av / (Av / S)

= 78.5715 / 0.0073

= 10772.9436 mm= 1077.2944 cm

S = 0,75 x h

= 533.4 mm

= 53.34 cm

S = 24 inch

= 60.96 cm



Mu / Ac' x d

Mu / Ac' x d

0.0836

Di akai tulan an okok

t/m

tulan an utama

tulan an sen kan

Mu / Ac' x d

0.54

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