tugas_3.docx

# tugas_3.docx - a Pi jika ditarik 1 arah dan 2 arah mencari...

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a) Pi jika ditarik 1 arah dan 2 arah? mencari ycgc dan I 1. Elastic shortening I 308277158 60 mm4 Pj/tedo n 1500000 N Ac 217000 mm2 emax 350 mm fc' 30 MPa Ec 25742.96 Es 186000 n 7.23 Ap 1866.67 mm2 p 1607.14 MPa L 20 m wDL 5.208 kN/m MDL 260.4 kNm beton 24 kN/m3 qDL 5.208 N/mm

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qDL+SID L 5.99 N/mm digunakan fci’ karena imidiate loss menggunakan kondisi initial Diujung Bentang pada tendon 1 f cp 11 = 0 f cp 12 = Pj A Pj e 2 e 1 I = 1500000 217000 1500000 0 0 30827715860 =− 6.91 MPa pada tendon 2 f cp 21 = 0 f cp 22 = 0 ´ ∆σ CP = 6.91 2 =− 3.46 MPa Ditengah Bentang (ada beban DL short term , maksimum di tengah) pada tendon 1 f cp 11 = 0 f cp 12 = Pj A Pj e 2 e 1 I + M DL e I = 1500000 217000 1500000 350 350 30827715860 + 260.4 350 30827715860 =− 9.92 MPa pada tendon 2 f cp 21 = 0 f cp 22 = 0 ´ ∆σ CP = 9.92 2 =− 4.96 MPa total
´ σ CP total = 3.46 2 ± 4.96 3 =− 3.96 MPa n = Es Ec = 186000 4700 fci' = 7.23 ´ σ P = n ∗− 3.96 = 7.23 ∗− 3.96 =− 28.59 MPa ( 2 loss ) P elastic shortening = -28.59*As= -28.59*217000= - 53.27 kN 2. Friction and Wooble Pj 3000 kN emax 350 mm hc/emax 0.4 hc 140 mm 0.07 radians 0.25 koef wooble 0.002 /m slip pd baji 3.6 mm Ep 186000 MPa Ap 1866.67 mm2 dianggap bahwa eksentrisitas dimulai dari 0.5 *4000 (setengah dari bentang AB) friction and wooble lokasi A B C D E x (m) 0 4 10 16 20 0 0.07 0 0.07 0 0 0.07 0.14 0.21 0.28 1 0.97482 2 0.946485 0.918972 0.895834 Pi (kN) 3000 2924.46 7 2839.455 2756.915 2687.502 arah sebaliknya lokasi E D C B A x(m) 20 16 10 4 0 Pi (kN) 3000 2924.46 7 2839.455 2756.915 2687.502 contoh perhitungan

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mencari nilai Ldi iterasi Ldi Ldi (m)= 4.00 8.14 8.83 8.85 y 2924.4 7 2869.50 2859.00 2858.72 /2 (rad) 18.88 16.04 15.97 Ldi (m)= 8.14 8.83 8.85 galat (%) 50.83 7.83 0.21 (OK) iterasi dimulai dari titik yang diketahui, dipilih titi B dengan x=4, diketahui bahwa y = 2924.47 mencari nilai /2 α 2 = 0.5 δ P di Ldi = 3000 2924.47 4 = 18.88 radian menghitung nilai Ldi yang seharusnya L di = E p . A p . Δ α 2 = 186000 1866.67 3.6 18.88 = 8.14 mm Galat yang terjadi 50.83 % , terlalu besar ulang kembali iterasi dengan menggunakan Ldi baru posisi y diambil dari persamaan garis di regresi linear dari plot titik-titik pada bidang kartesius (P vs x) , y = -15.18x + 2993 dengan cara yang sama menghitung /2 dan Ldi baru, seterusnya sampai galat kecil . dipilih Ldi = 8.85 m, y= 2858.72 kN gambar grafik
0 5 10 15 20 25 2500 2600 2700 2800 2900 3000 3100 f(x) = 15.18x + 2689.82 f(x) = - 15.18x + 2993.52 tarik arah kiri Linear (tarik arah kiri) Linear (tarik arah kiri) tarik arah kanan Linear (tarik arah kanan) friction wooble friction wooble x(m) P (kN) 0 5 10 15 20 25 2500 2600 2700 2800 2900 3000 3100 f(x) = - 15.18x + 2993.52 tarik arah kiri saja

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