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Assalamu'alaikum WR, WB: Nama: Boby S Novansyah NPM: G1B013038

This document discusses the prestressed concrete structure with post-tensioning. It provides the given information such as the diameter of the duct, area of steel, and initial prestressing force. It then asks to calculate the stresses at the center of the span under initial and final conditions. The response shows the calculations of the stresses under initial and final conditions, finding them to be less than the allowable limits.

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Boby Vanshici
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77 views6 pages

Assalamu'alaikum WR, WB: Nama: Boby S Novansyah NPM: G1B013038

This document discusses the prestressed concrete structure with post-tensioning. It provides the given information such as the diameter of the duct, area of steel, and initial prestressing force. It then asks to calculate the stresses at the center of the span under initial and final conditions. The response shows the calculations of the stresses under initial and final conditions, finding them to be less than the allowable limits.

Uploaded by

Boby Vanshici
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PPTX, PDF, TXT or read online on Scribd
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Assalamu’alaikum wr,wb.

NAMA : BOBY S NOVANSYAH


NPM : G1B013038
Presterssed Concrete Structures
Diketahui :
 Struktur prategang dengan cara post-tensioning.
 Diameter lubang (duct) 10 cm,
 Asd = 2000 mm2,
 WADL = 15 kN/m.
 Berat beton 24 kN/m3,
 fc’= 35 Mpa,
 fc’l = 30 MPa,
 n = E/Ec = 7,
 Gaya prategang awal 1400 kN.
 Kehilangan prategang = 20%
Ditanya :
Tegangan – tegangan di tengah bentang yang terjadi pada kedua kondisi
( awal dan akhir) ?

 Penyelesaain :
 WDL = 0,40 x 0,80 x 24 = 7,68 kN/m
1 1
 MDL = 8 x WDL x L2 = 8 x 7,68 x 122 = 138,24 kN/m
1 1
 MADL = 8 x WADL x L2 = 8 x 15 x 122 = 270 kN/m
1 1
 MLL = 8 x WLL x L2 = x 10 x 122 = 180 kN/m
8
 Ms = MDL + MADL + MLL = 138,24 + 270 + 180 = 588,24 kN/m
 Pada kondisi awal ( initial ), baru bekerja berat sendiri balok,
lubang duct belum di grouting ~ grass section.
𝜋
σ𝑛
1=𝑖 𝐴𝑔𝑙.𝛾𝑖
0,4 𝑥 0,8 𝑥 0,4−
4
x0,1 x 0,68
𝑥 0,1 0,123
 𝛾= = 𝜋 = =
σ𝑛
1−𝑖 𝐴𝑔𝑙 0,4 𝑥 0,8− 𝑥 0,1 𝑥 0,1 0,312
4
0,393 𝑚
 ygt = 0,393 m ( dari sisi atas )
 e = 0,407 – 0,120 = 0,287 m
1
 lg = x 0,4 x 0,803 = 0,017067 m4
12
 0,312 ( 0,40 – 0,393)2 = 0,000015 m4 – lg
 = 0,016343 m4 ( nilai menyeseuaikan)
 Tegangan –tegangan awal ( initial stage )
 Sisi atas :
𝜌1 𝜌1.e γtg Mi.𝛾tg 1400 1400 x 0,287 x 0,393
 𝜎tt= - + - = - -
Ag lg lg 0,312 0,016345
0,1383 x 0,393
= 0,471 Mpa < 0,25 fc′
0,016345
= < 1,37 MPa .....OK!!
𝜌1 𝜌1.e γgh Mi.𝛾gh 1400 1400 x 0,287 x 0,407
 𝜎bt= - + - = - -
Ag lg lg 0,312 0,016345
0,1383 x 0,407
= - 11,015 < 0,6 fc′
0,016345
= < 18 MPa ....OK!!
Letak garis netral penampang transformation :

σ 𝐴𝑡.𝑦 0,4 𝑥 0,8 𝑥 0,4 + 7−1 0,002 𝑥 0,68 0,13616


𝛾 = = = 𝛾= = 0,41 m
𝐴𝑡 0,4 𝑥 0,8+ 7−1 𝑥 0,002 0,332
ytt = 0,41 m ( dari sisi atas )
ybt = 0,80 – 0,41 = 0,39 m
e = 0,27 m
1
lt = 12 x 0,4 x 0,83 + 0,4 x 0,8 x (0,01)2 + ( 7-1) 0.002 x 0,272 = 0,01797 m4
Tegangan- tegangan akhir ( sentence) :
 Sisi Atas :
𝜌s 𝜌s.e .γgh Mi.𝛾gh 0,8 x 1,40 0,8 x 1,40 x 0,41 0,1383 x 0,393
𝜎t = - Ag + lg
- lg
=− 0,332
+ 0,01797
- 0,01797
= - 3,373 + 6,899 – 13,412
𝜎t = - 9,895 < 0,45 fc’ = 15,75 MPa ...OK!!
 Sisi Bawah :
𝜌s 𝜌s.e .γbt Mi.𝛾bt 1,12 01,12x 0,27 x 0,39 0,58824 x 0,393
𝜎b = - Ag + - = - 0,332 + -
lg lg 0,01797 0,01797
= - 3,373 + 6,563 – 12,766
𝜎t = + 2,83 < 0,5 fc’ = 2,958 ....OK!!

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