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Old 07-17-2011, 12:05 PM   #1
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design review

an elongation, the formula and parameters:
1, the average pre-stress formula and the parameters of Tension:

where:
Pp-tendon mean tensile force (N)
P-tendon tension side of the tensile coerce (N)
X-calculated from the tension side of the cross-section of the channel length apt (m)
θ-calculated from the tension end-to-channel cross-section of part of the bend and the tangent of the angle (rad)
k-channel deviation per measure of local influence above Friction coefficient, catching 0.002
μ-tendons and pore walls Friction coefficient, taking 0.14
2,Supra Shoes, tendon of the academic merit of elongation formula and parameters:

where:
Pp-tendon Average Tension (N)
L-tendon length (mm)
Ap-tendon cross-sectional place (mm2), take the 140 mm2
Ep-tendon modulus of bounce ( N / mm2), taking 1.95 × 105 N / mm2

Second, the elongation calculation:
1, N1 beam end elongation:
unattached strand tensioning Zhang Rally
P = 0.75 × 1860 × 140 = 195300N
X = 15.812 / 2 = 7.906m
θ = 11.4 × π/180 = 0.19897rad
kx + μθ = 0.002 × 7.906 +0.14 × 0.19897 = 0.0436678
Pp = 195300 × (1-e-0.0436678) / 0.0436678 = 191097N
ΔL = PpL / (Ap Ep) = 191097 × 7.906 / (140 × 1.95 × 105) = 55.3mm
compared with the design (55.3-57.1) / 57.1 =- 3.15%
2, N2 beam end elongation:
single strand tension Tension
P = 0.75 × 1860 × 140 = 195300N
X = 15.821 / 2 = 7.9105m
θ = 12.8 × π/180 = 0.2234rad
kx + μθ = 0.002 × 7.9105 +0.14 × 0.2234 = 0.047097
Pp = 195300 × ( 1-e-0.047097) / 0.047097 = 190772N
ΔL = PpL / (Ap Ep) = 190772 × 7.9105 / (140 × 1.95 × 105) = 55.27mm
compared with the design (55.27-57.1) / 57.1 = -3.2%


Tension
a theoretical measurement of elongation, the measurement parameters:
1,Supra Cruizer Shoes, K-channel deviation per meter of local shock of Friction coefficient: Take 0.002
2, μ-tendons and the pore wall Friction coefficient: Take 0.14
3, Ap-tendon cross-sectional area measured: 140 mm2
4, Ep-tendon springy modulus amounted : 2.02 × 105 N / mm2
5, beneath the control of anchor stress: σk = 0.75Ryb = 0.75 × 1860 = 1395 N / mm2
6, anchor ring jaws friction loss: 3.3% σk
7, single strand tension side of the tension control: P = 103.3% × σkAp = 201745N
8, jack calculate length: 56cm
9, working anchor length: 7cm
10, limit embark computing Length: 2.5cm
11, calculate the length of the anchor tools: excluding
Second,Supra Skytop III Shoes, the methodology of stress when the elongation calculation:
1,Supra Skylow Shoes, N1 beam end elongation:
X = 15.812 / 2 = 7.906m
L = 7.906 + (0.56 +0.07 +0.025) = 8.561m
θ = 11.4 × π/180 = 0.19897rad
kx + μθ = 0.002 × 7.906 +0.14 × 0.19897 = 0.0436678
Pp = 201745 × (1-e-0.0436678) / 0.0436678 = 197404N
ΔL = PpL / (Ap Ep) = 197404 × 8.561 / (140 × 2.02 × 105) = 59.8mm
2, N2 beam end of elongation:
X = 15.821 / 2 = 7.9105m
L = 7.9105 + (0.56 +0.07 +0.025) = 8.566m
θ = 12.8 × π/180 = 0.2234rad
kx + μθ = 0.002 × 7.9105 +0.14 × 0.2234 = 0.047097
Pp = 201745 (1-e-0.047097) / 0.047097 = 197068N
ΔL = PpL / (Ap Ep) = 197068 × 8.566 / (140 × 2.02 × 105) = 59.7mm
Third, fuel jack Tension meter readings and the corresponding reckoned
1 strand of the tension control stress:
12 strand bundle: σcon = 103.3σk = 103.3 % × 2343 = 2420.32KN
Second, the 1523 Jack-tensioned, 0050,Supra Trainers, Yau table:
jack regression equation:
P =- 0.35 +0.01035 F
where: P - Oil pressure gauge reading (MPa)
F - Jack pull (KN)
(1), 10% σcon = 242.032 KN time:
P =- 0.35 +0.01035 F =- 0.35 +0.01035 × 242.032 = 2.16MPa
(2), 40% σcon = 968.13KN time:
P =- 0.35 +0.01035 F =- 0.35 +0.01035 × 968.13 = 9.67 MPa
(3), 70% σcon = 1694.22KN When:
P =- 0.35 +0.01035 F =- 0.35 +0.01035 × 1694.22 = 17.19 MPa
(4), 100% σcon = 2420.32KN time:
P =- 0.35 +0.01035 F =- 0.35 +0.01035 × 2420.32 = 24.7 MPa
Third, the tension jack, No. 1524, 0054, Yau table:
jack regression equation:
P = 0.21 +0.01022 F:
formula: P - oil pressure gauge reading (MPa)
F - Jack pull (KN)
(1),Supra Indy NS Shoes, 10% σcon = 242.032KN time:
P = 0.21 +0.01022 F = 0.21 +0.01022 × 242.032 = 2.68 MPa
(2), 40% σcon = 968.13KN time:
P = 0.21 +0.01022 F = 0.21 +0.01022 × 968.13 = 10.10 MPa
(3), 70% σcon = 1694.22KN time:
P = 0.21 +0.01022 F = 0.21 +0.01022 × 1694.22 = 17.52 MPa
(4), 100% σcon = 2420.32KN time:
P = 0.21 +0.01022 F = 0.21 +0.01022 × 2420.32 = 24.95 MPa
Fourth, 1525 jack-tensioned, 0077, Yau table:
jack regression equation: P =- 0.47 +0.01024 F:
formula: P - oil pressure gauge reading (MPa)
F - Jack pull (KN)
(1), 10% σcon = 242.032KN time:
P =- 0.47 +0.01024 F =- 0.47 +0.01024 × 242.032 = 2.0 MPa
(2) , 40% σcon = 968.13KN when
P =- 0.47 +0.01024 F =- 0.47 +0.01024 × 968.13 = 9.44 MPa
(3), 70% σcon = 1694.22KN time:
P =- 0.47 +0.01024 F =- 0.47 +0.01024 × 1694.22 = 16.88 MPa
(4), 100% σcon = 2420.32KN time:
P =- 0.47 +0.01024 F =- 0.47 +0.01024 × 2420.32 = 24.31 MPa
May, 1526 jack-tensioned, 0064, Yau table:
jack regression equation: P =- 0.05 +0.01021 F:
formula: P - oil pressure gauge reading (MPa)
F - Jack drag (KN)
(1), 10% σcon = 242.032KN time:
P =- 0.05 +0.01021 F =- 0.05 +0.01021 × 242.032 = 2.42 MPa
(2), 40 % σcon = 968.13KN when
P =- 0.05 +0.01021 F =- 0.05 +0.01021 × 968.13 = 9.83 MPa
(3), 70% σcon = 1694.22KN time:
P =- 0.05 +0.01021 F =- 0.05 +0.01021 × 1694.22 = 17.24 MPa
(4), 100% σcon = 2420.32KN time:
P =- 0.05 +0.01021 F =- 0.05 +0.01021 × 2420.32 = 24.66 MPa

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