NumericalAlgorithmsforCalculatingTemperature,Layered Stress ... · 2019. 12. 29. · sag and stress (submodule 2 in Figure 3). e two sub-modules needto berepeatedly calledinthe calculation

Research ArticleNumerical Algorithms for Calculating Temperature LayeredStress and Critical Current of Overhead Conductors

Yongdou Liu12 Zhiwei Chen 2 and Quan Gu 12

1Key Laboratory of Earthquake Engineering and Engineering Vibration Institute of Engineering MechanicsChina Earthquake Administration Harbin 150080 China2Xiamen University School of Architecture and Civil Engineering Xiamen 361005 China

Correspondence should be addressed to Quan Gu quanguxmueducn

Received 29 December 2019 Accepted 26 February 2020 Published 30 April 2020

Academic Editor Arkadiusz Zak

Copyright copy 2020 Yongdou Liu et al -is is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Accurate calculation of temperature stress sag and critical current (corresponding to critical temperature) of operationaloverhead conductors is important for ensuring the strength and sag safety of overhead lines Based on 2D steady-state heat transferequations this article studies the temperature fields of the cross section of typical electrified conductors and establishes numericalsimulation methods for calculating the layered stress sag and critical temperature Using the algorithm the relationship betweenthe critical temperature and characteristics of conductors (eg the sag and tensile force) is studied -e results are verified by acomparison with the test results for heat-resistant aluminum alloy conductors JNRLH1G1A-40065 and JNRLH1G1A-63055Finally the paper studies the relationship between the critical temperature of the conductor and its most sensitive factors

1 Introduction

During peak times it is often necessary to dynamicallyincrease the current of overhead lines to meet the powersupply demands In these cases the stress and sag of con-ductors must be accurately calculated to avoid breakage dueto excessive stress multicircuit wire mixture [1] or non-enough safety distance to the ground caused by large sag Inaddition the stress at different layers of strands (calledlayered stress) and the sag are generally affected by thetemperature fields on the conductor cross sections Onlywhen the temperature fields are simulated accurately can thelayered stress and sag be calculated actually Moreover dueto the uneven thermal expansion in different layers duringoperation the mechanical properties of the conductorrsquoscross section will greatly change at definite temperaturewhich is defined as the critical temperature (CT) [2 3]However to the authorrsquos knowledge there is limited re-search on the safety assessment of overhead conductors atthe CT -erefore calculation of the temperature fieldslayered stress sag and tensile force and the analysis of thesafety of overhead lines at CT are of great significance

Many experimental studies [4ndash12] have shown that theradial temperature of conductors is not evenly distributedIEEE specifications [13] proposed that there is radial tem-perature gradient among each layer In recent years somemethods [14ndash16] based on the finite element method wereproposed to simulate the conductor radial temperature fieldWith these methods models better conforming to the actualsectional structure of conductors can be established to avoidcalculation errors due to improper values for effectivethermal conductivity However these studies failed to fullyconsider the impact of cross-section gap distribution inter-strand contact and convection conditions on the radialtemperature field Furthermore there is little research thatsystematically studied the temperature field layered stresssag and critical temperature of conductors -is paperproposes a complete set of numerical methods to calculatethe temperature field layered stress and critical current (ortemperature) of overhead conductors -e temperature fieldwith voids is calculated by using the finite element methodand the layered stress and sag are calculated by using thetemperature results -e CT and the corresponding criticalcurrent (CC) are calculated through the temperature and

HindawiMathematical Problems in EngineeringVolume 2020 Article ID 6019493 14 pageshttpsdoiorg10115520206019493

stress calculative method iteratively By comparing the re-sults of calculation with those of the experiments for a typicalACSR (aluminum-conductor steel reinforced) the algo-rithms are verified In order to facilitate the dynamic as-sessment of safety of the conductors during the operationthis paper studies the relationship between the criticaltemperature of the conductor and its sensitive factors withthe heat-resistant aluminum alloy conductor JNRLH1G1A-40065 as an example

2 Calculation of Conductor Temperature Field

-e thermal balance equation of the entire cross sectionΩ ofthe conductor can be established [14]

k Txx + Tyy1113872 1113873 + s 0 (x y) isin Ω (1)

And the boundary condition on the surface of theconductor Γ can also be given [14]

qn minusk Txnx + Tyny1113872 1113873 α T minus Ta( 1113857 l(x y) isin Γ (2)

where Txx Tyy Tx and Ty indicate the second-orderderivatives and first-order derivatives of the 2D temperaturefield T in the conductor on the x and y coordinates re-spectively qn indicates the heat loss rate along the normaldirection n on the outer surface of the conductor nx and ny

are the components of n in the direction of x and y α in-dicates the composite heat loss coefficient Ta indicates theambient temperature k is the thermal conductivity repre-sented by ka in the area of the aluminum strand ks in thearea of the steel core and kair in the area of the air and sindicates the rate of heat per unit volume represented by sa

in the area of the aluminum strand ss in the area of the steelcore and 0 in the area of the air Herein ka and ks areirrelevant to temperature field while kair is relevant to it andthe relationship is as follows

kair 242 times 10minus2+ 72 times 10minus 5

Tav (3)

where Tav indicates the average temperature of the crosssection -e voltage of steel core per unit length is the sameas that of aluminum strand -e current is inversely pro-portional to the resistance which is inversely proportional tothe conductor cross section and proportional to the resis-tivity [14] -e ratio between the current Is of steel core andthat Ia of aluminum strand is

λ Is

Ia

As

Aa

ρa

ρs

(4)

where As and ρs are the cross-sectional area and electricalresistivity of steel core respectively and Aa and ρa are thearea and electrical resistivity of aluminum strand re-spectively -e rate of heat of steel core per unit volume ss

is

ss Pλ(λ + 1)

As

(5)

External aluminum strands should consider solar energy[17] so the rate of heat per unit volume sa is

sa P(λ + 1)

Aa

+csSDAa

(6)

In equation (6) the first term on the right end of theequation is caused by the current and the second term iscaused by the solar radiation P indicates its total Joule heatgain [17] which is the function of the current averagetemperature and integrated resistance rate of the conductorS indicates the solar intensity cs indicates the solar ab-sorption of the conductor with a value of 023ndash09 and isusually valued at 05 and D indicates the conductordiameter

An aluminum-conductor steel-reinforced cable (ACSR)mainly loses heat by means of convective cooling Pc andradiative cooling Pr -e calculation of heat loss is providedin the literature [17] According to the principle of constantheat loss per unit volume the composite heat transfer co-efficient of conductor surface in equation (2) can bedetermined

α Pc + Pr

Tsur minus Ta( 1113857A (7)

where A indicates the lateral surface area of conductor perunit length (that is the interface of the conductor and theexternal environment)

3 Calculation of Conductor Stress Field

-is article assumes that conductors can only be resistant totension ignoring their resistance to shear bend and torsion[18] and the tensile force on each strand of the same layer isequal [19] -ree basic conditions are required when per-forming calculations for overhead conductors (1) defor-mation compatibility (2) material constitutive relationshipand (3) balance of internal and external forces

31 Deformation Compatibility Using the condition ofdeformation compatibility the strain of each strand can berepresented by the longitudinal strain For a strand with alength of Si if unfolding it as shown in Figure 1 the de-formation compatibility equation can be obtained as follows[20]

ε0Ls ΔSi

sin αi

+ΔRiφi

tan αi

(8)

where ε0 indicates the axial strain of the overhead conductorLs indicates the length of the overhead conductor corre-sponding to a strand with a length of Si αi indicates the anglebetween the tangent of layer i of the strand and the con-ductor cross section ΔRiφi indicates the lateral deformationof the unfolded strand and φi indicates the ratio betweenprojected length of the unfolded strand on the cross sectionand the outer radius Ri of layer i of the strand as follows

φi Si cos αiRi (9)

From the geometrical relationship equation (10) can beobtained as follows

2 Mathematical Problems in Engineering

sin αi Ls

Si

(10)

Dividing the left and right sides of equation (8) by Si andsimplifying it with equations (9) and (10) the relationshipbetween the strain of layer i of the strand and the conductorε0 is

εi ε0sin2αi minusΔRi

Ri

cos2αi (11)

32 Lateral Deformation From the inside to outside thelateral deformation ΔRi of conductors of layer i includesthree parts (1) the deformation Δri1prime arising from cross-section shrinkage due to the Poisson effect when the strandsof layer i are stretched along their own axis [20] (2) thedeformation Δri2prime due to squeezing between layers and (3)the deformation ΔriT

prime due to thermal expansion (or con-traction) Studies have shown that the lateral deformationΔRi has a nonlinear functional relation with axial strain ε0-e calculation of ε0 with ΔRi requires iteration which leadsto complicated procedures Furthermore study on the lat-eral deformation of ACSR (JLLB1A-30050) found that thelateral deformation of the first two parts is very small (thelateral deformation of the first two parts is less than 1) [21]and can be neglected compared with that of the third part-e equation of lateral deformation can be simplified asfollows

ΔRi minusΔriTprime (12)

-e term ΔriTprime in equation (12) indicates the total de-

formation of layer i due to changes in temperature

ΔriTprime

Δr1T i 1

Δr1T + 21113944iminus1

j2ΔrjT + ΔriT i 2 n

⎧⎪⎪⎨

⎪⎪⎩(13)

-e equation ofΔrjT (strand deformation due to thermalexpansion or contraction) in equation (13) is as follows

ΔrjT αjTrj Tj minus 201113872 1113873 (j 1 2 n) (14)

where αjT and Tj indicate the coefficient of thermal ex-pansion and the average temperature of jth layer respec-tively -e strain εiT due to changes in temperature is asfollows

εiT αiT Ti minus 20( 1113857 (15)

33 Balance Equation of Internal and External ForceLinear elastic material is supposed in this article -e re-sultant axial internal force of the conductor section can beobtained by summing up all axial components of internalforces of strands on the cross section -e internal force ofeach strand of layer i is as follows

pi Ei εi minus εiT( 1113857Ai (16)

whereEi indicates themodulus of layer i of the conductorAi

indicates the cross-sectional area of one strand in layer iWhat needs to be mentioned is that the temperature strainεiT (see equation (15)) of each layer of strands may differ dueto the difference of the thermal expansion coefficient As thetemperature arises aluminum strands with a higher thermalexpansion coefficient may get zero stress and even some ofthem get negative stress As a result all the external force isresisted by the steel strands -is may cause changes to thecomposite property of the overhead conductors and will bestudied in detail in Section 42 later

-e equation of the axial internal forces on the crosssection can be obtained as follows

P ε0( 1113857 1113944n

i1pini sin αi (17)

where ni indicates the number of layer i strands Figure 2 isthe diagrammatic sketch of single span overhead conductorin which the letters ldquoArdquo ldquoBrdquo and ldquoOrdquo represent the sus-pension points on the left and right and the lowest pointrespectively -e span of the overhead conductor is l and theheight difference is h (when suspension point B on the rightis higher than A on the left h is positive otherwise it isnegative)-e ratio between h and l is denoted as β-e axialtensile external force N(x) at the distance x from the leftsuspension point A (as shown in Figure 2) can be calculatedaccording to the axial tensile stress at the lowest point O σ0[18] In equation (18) c denotes the gravity of conductor perunit length

N(x) σ0A0

1 + tan β minusc(1 minus 2x)

2σ0 cos β1113890 1113891

2

11139741113972

(18)

-us the force balance equation of the conductor sectionat distance of x from the origin can be obtained as follows

f ε0( 1113857 N(x) minus P ε0( 1113857 0 (19)

Brsquorsquo Brsquo

B

ΔSi

αiαiA

B

Riφi

Riφi ΔRiφiA

SiLs

Lsε0

Figure 1 Unfolding view of the strand

Mathematical Problems in Engineering 3

4 Calculation of Sag and Critical Current(or Temperature)

41 Sag Sag refers to the vertical distance between the lo-cation of actual overhead lines and the line connecting theleft and right suspension points such as the sag fM at themidpoint of the conductor (shown in Figure 2) Studiesshowed that the sag of overhead conductors can be ap-proximately represented by a parabola [18] If suspensionpoint A on the left is regarded as the origin the sag fx at thelocation x is as follows

fx fM middot 4x(l minus x)l2 (20)

-e relation between the sag fM in the midpoint of theconductor and its arc length L is as follows [18]

fM

L minusl

cos β1113888 1113889 middot

3l

8 cos3 β

1113971

(21)

If the deformation (due to the gravity and other ex-ternal loads) is neglected (that is the axial rigidity of theconductor is assumed to be infinite) its initial arc length L0is as follows

L0

2σo

csh

cl

2σo

1113888 1113889

2

+ h2

11139741113972

(22)

If the sag fM at the midpoint is known the horizontaltensile stress can be calculated with the followingequation

σ0 cl2

8fM cos β (23)

If the sag fM at the midpoint is unknown but the heightdifference fAO between the left suspension point and thelowest point is known the horizontal tensile stress can beobtained by solving the following equation

σ0c

1 minus chcl

2σ01113888 1113889

1 +h

Lh01113888 1113889

2

11139741113972

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ +h

2+ fAO 0 (24)

In the operation the arc length of the overhead line willchange with the effect of current and load If the average strainof the whole overhead conductor is ε the arc length L is

L L0(1 + ε) (25)

42 Critical Temperature and Critical Current Overheadconductors are generally twisted by multilayers of strands-e deformation of each strand is caused by stress or due tochanges in temperature -e deformation due to changes intemperature will not cause internal force in the strands Asthe coefficients of thermal expansion and temperature ofeach layer of strands are different their temperature de-formations are also different With the increase of tem-perature the distribution of stresses in the strands of eachlayer will vary If the resultant external force on the con-ductor is constant when the temperature rises the tensilestresses of strands with greater thermal deformation (alu-minum strands with higher thermal expansion coefficients)will decrease while the stress of other strands (generally steelcore with lower coefficients) will increase When the tem-perature reaches a certain critical value the resultant axialforces on all aluminum strands are zero At this point theaverage temperature of the cross section and the corre-sponding current are defined as critical temperature (CT)and critical current (CC) respectively Note that the CT isthe same concept as the knee-point temperature [22ndash24]-e CC Ic and CT Tc of the conductor meet the followingconditions

Pa(I)1113868111386811138681113868IIc

0 (26)

Pa(I) 1113944

na

i1σi(I)miAi sin αi (27)

Tc T Ic( 1113857 (28)

where Pa(I) indicates the resultant force on the aluminumstrands when the current is I mi indicates the number ofaluminum strands of layer I αi indicates the angle betweenthe layer i aluminum strands and the cross section Ai in-dicates the cross-sectional area of a single strand of the layer ialuminum strand σi indicates the stress of layer i aluminumstrands and na indicates the total layers of aluminumstrands CT and CC are related to span height differenceexternal load (including its own weight) ambient temper-ature wind speed and solar heating

43 Calculation Process -e CT temperature field sag andlayered stress proposed above can be calculated according tothe flowcharts shown in Figure 3 -e main calculativemodule indicates the process for calculating CTand CC-esubmodules involve the calculation process of temperaturefield (submodule 1 in Figure 3) and the calculation process ofsag and stress (submodule 2 in Figure 3) -e two sub-modules need to be repeatedly called in the calculation ofcritical temperature -e calculation steps of the mainmodule are as follows

Step 1 set the initial values of current step size dInumber of iterations i 0 and allowable error tolCalculate the upper and lower limits of current Per-form steps 11ndash13

A

B

x

y

l

hfM

O

fx

l2

xfAO

β

Figure 2 Diagrammatic sketch of the overhead conductors


Step 11 call the submodule 1 to calculate thetemperature field of the conductor cross section with agiven current I i middot dI

Step 12 call the submodule 2 to calculate the layeredstress of the conductor and the resultant force Pa onaluminum strands (see equation (27))

Step 13 if Pa ge 0 i i + 1 go to Step 11 Oth-erwise the upper and lower limits of CCI(2) i middot dI

and I(1) (i minus 1) middot dI and the corresponding resul-tant force P(2)

a and P(1)a on the aluminum strands are

obtained

Step 2 calculate the CT and CC with the false positionmethod Perform steps 21sim23

Step 21 update the CC I I(2) minus P(2)a (I(2) minus I(1))

(P(2)a minus P(1)

a ) according to its upper and lower limits

Step 22 if |I minus I(2)|ge tol perform step 221simstep 223

Step 221 call the submodules 1 and 2 to calculatethe resultant force Pa on the aluminum strands whenthe current is I

Step 222 if Pa middot P(2)a lt 0 update the lower limit

ie I(1) I(2) P(1)a P(2)

a

Given values for paremeters ks ka I Ta and vwSet initial values Itmax tol i = 0T(i) = Ta

Import the mesh of the cross section

||T(i)ndashT(indash1)|| ge tolYes

No

Solve T(i) using finite element method

i = i + 1If i le Itmax continue

Else exit and prompt an error

Calculate Tav Tsur kair ss sa α

Output T(i)

Given dI let the initial value be i = 0

I = imiddotdI solve the temperature field

Pa ge 0

Determine the upper and lower limits

I(1) = (i minus 1) middot dI Pa(1) = Pa|I=I

(1)

Pa(2) = Pa|I=I

(2)I(2) = i middot dI

If (Pa middot Pa(2) lt 0) then I(1) = I(2) Pa

(1) = Pa(2)

|I minus I(2)| ge tol Solve temperature field calculate Pa

i = i + 1Yes

No

Output Ic and Tc etc

Yes

No

Enter submodule 1

Enter submodule 2

Enter submodule 1 Enter submodule 2

Solve the stress field obtain Pa

I = I(2) ndash Pa(2)(I(2) ndash I(1))(Pa

(2) ndash Pa(1))

Sub-module1 the calculationprocess of temperature fields

The main module

Read in h l γ etcLet initial values i = 0

Solve σ0(using equation 24)calculate fm(0) (using equation

22 and 21)

for each node j = 1~nperform

1solve N (x)using equation 182calculate Pa

(j) solving equation19

Output ε(j) = ε Pa(j)

layered stress and strain

Given ε(1) and ε(2) Read in parametersof ith layer Ei Ai αiT αi ni Ti

i = i + 1If i le Itmax continueElse exit and promt

an error

Output fm(i) min (Pa(j) j = 1~n) etc

||fm(i) minus fm(indash1)|| ge tol

Calculate f (ε(1) f (ε(2)) and ∆Ri

Calculate f (ε)

If f (ε)f (ε)(2) lt 0 thenε(1) = ε(2) f (ε(1)) = f (ε(2))

Yes

No

Yes

No

Submodule 2 the calculationprocess of sag and stress

Solve the stress andstrain for each cross

section

I(2) = I Pa(2) = Pa

ε = ε(2) minus f (ε(2))(ε(2)minus ε(1))[f (ε(2)) minus f (ε(1))]

|ε minus ε(2)| ge tol

ε(2) = ε f (ε(2)) = f (ε)

Calculate εndash =nj=1 ε(j)n L(i) and fm

(i)

Figure 3 -e calculation methods for critical temperature temperature field sag and layered stress


Step 223 update the upper limit ieI(2) I P(2)

a Pa and go to step 21

Step 23 otherwise output the critical current Iccritical temperature Tc and corresponding layered stress

-e calculation of temperature field is performed withfixed point iteration and the calculative process is as follows

Step 1 set the initial values of allowed maximumnumber of iterations Itmax allowed error tol coefficientof thermal conductivity ks for steel core and ka foraluminum strands current I ambient temperature Tcand wind speed vw respectively Import the cross-sectional finite element mesh

Step 2 update the number of iterations i i + 1 Ifile Itmax continue otherwise the calculation fails exitand prompt an error

Step 3 calculate the average temperature Tav of theconductor cross section and the average temperatureTsur of the conductor surface Calculate kair ss sa and αwith equations (3)ndash(7)Step 4 solve the temperature field of the conductorcross section by using the finite element program de-rived from equations (1) and (2)Step 5 if the temperature field T(i) of the iteration isvery different from the result T(iminus 1) of the previousiteration (ie T(i) minus T(iminus 1)ge tol) perform step 2ndash5Step 6 otherwise converge the calculation and save theresults

-e iterative process for calculating the sag and stress isas follows

Step 1 import the geometric information of weight perunit length c span l height difference h and initial sagfAO Set the initial value of the iteration number i 0and the maximum number of iterations ItmaxStep 2 solve equation (24) to obtain the initial value ofhorizontal tensile stress σ0 Calculate the initial iterativevalues f(0)

m of sag at the midpoint according to equa-tions (22) and (21) respectivelyStep 3 equally divide the whole span of conductors inton minus 1 sections (totally n nodes) For each nodej isin [1 n] calculate the strain and layered stress and theresultant force P

(j)a of aluminum strands Perform the

following stepsStep 31 calculate the resultant external force N(x)

at the node j of x with equation (18)Step 32 perform step 321simstep 324 (false position

method) to solve equation (19)Step 321 given the upper and lower limits of the axial

strain ε(1) and ε(2) read in the cross-sectional parametersof ith layer includingEi (see equation (16))Ai (in equation(16)) αiT (in equation (15)) αi (in equation (8)) ni (inequation (17)) and Ti (in equation (14))

Step 322 calculate the lateral deformation ΔRi ofeach layer with equations (12)ndash(14) Calculate f(ε(1))

and f(ε(2)) combined with equations (11) and(15)ndash(19)

Step 323 update the axial strainε ε(2) minus f(ε(2))(ε(2) minus ε(1))[f(ε(2)) minus f(ε(1))]

according to the upper and lower limits of axial strainStep 324 if |ε minus ε(2)|ge tol perform steps 3241sim

3243Step 3241 calculate f(ε) combined with

equations (11) and (15)ndash(19)Step 3242 if f(ε)f(ε(2))lt 0 update the lower

limit of strain ε(1) ε(2) f(ε(1)) f(ε(2))Step 3243 update the upper limit of strain

ε(2) ε f(ε(2)) f(ε) go to step 322Step 325 otherwise (ie |ε minus ε(2)|lt tol) output

the axial strain ε(j) ε resultant force P(j)a on alumi-

num strands and layered stress and strainStep 4 calculate the average ε 1113936

nj1ε(j)n of strain at n

nodes Calculate the arc length L(i) and sag f(i)m at it-

eration i with equations (25) and (21)Step 5 if the sag f(i)

m at the midpoint in the iteration isvery different from the result f(iminus1)

m of the previousiteration (ie f(i)

m minus f(iminus1)m ge tol) execute the steps

51ndash53Step 51 update the iteration number i i + 1Step 52 if the iteration number ile Itmax go to step 3Step 53 else the calculation fails exit and prompt an

errorStep 6 otherwise (ie f(i)

m minus f(iminus1)m lt tol) output the

sag f(i)m at the midpoint the resultant force

min(P(j)a j 1 sim n) on the aluminum strands at the

lowest point in the span etc

5 Comparison of Experimental andNumerical Results

-is section provides comparative analyses of the resultsfrom experiments and numerical simulation includingthose for (1) cross-section temperature of ACSR JLG1A-63055 [25] (2) layered stress of ACSR JLG1A-40035 and(3) temperature-sag and temperature-tensile force charac-teristics of heat-resistant aluminum alloy strands JNRLH1G1A-40065 and JNRLH1G1A-63055 with a span of 60m

51 Layered Temperature With reference to the experi-mental study [26] of conductor JLG1A-63055 in the dif-ferent current and wind speed conditions this articleconducts numerical simulation of the conductor and sets upa finite element model as shown in Figure 4 -e model isperformed with meshing by using triangular elements with amaximum size of 24mm and with 8080 elements and 4184finite element nodes -e results of numerical simulationand experiments are compared in Table 1 In the mea-surement of the conductorrsquos temperature the surfacetemperature of the conductors is recorded However in thenumerical simulation the temperature field on the whole


cross section of the conductor can be calculated and then theaverage temperature of strands of each layer can be obtained

Table 1 and Figure 5 show that (1) the surface tem-perature simulation results of the conductor at each windspeed fit well with the experimental values apart from in-dividual data when the wind speed is 1ms and the current is922A -e simulation results are different from the ex-perimental ones by 991 and the experimental data at thispoint are obviously unreasonable Under the currents of1106A and 1290A when the wind speed increases from1ms to 2ms the surface temperature decreases by 9degC and10degC respectively compared to a 1degC reduction when thecurrent is 922A-erefore the experimental value when thewind speed is 1ms and the current is 922A is lower thanexpected (2) -e difference of the cross-sectional temper-ature is small (about 1degC) and the temperature of the internalstrands is slightly higher than that of external strands-erefore the temperature gradients can be ignored in thecalculation of layered stress (3)With the increase of current

the cross-section temperature of the conductor rises whilewith the increase of wind speed it drops

52 Layered Stress -e layered stress experiment on JLG1A-40035 ACSRs is performed through stripping that isstretching the conductor to a certain strain and successivelystripping the conductors by layer to calculate the differenceof axial force so the layered stress of the conductor can becalculated under the strain Only the layered stresses of thealuminum strands are studied as the axial force is mainlyresisted by the aluminum strands (more than 3 times thansteel core)-e experiment is performed at temperature 20degCwithout applying current

-e calculation and test results are compared in Figure 6where the legend layer 1 2 and 3 represent the externalmiddle and internal layers of aluminum strands respec-tively As shown in the figure (1) the numerical simulationresults in the elastic phase (with strain of less than 02) and

AluminumSteelAir

41232

3432

Figure 4 -e finite element mesh of ACSR JLG1A-63055mm

Table 1 -e experimental and numerical results of ACSR JLG1A-63055

Windspeed (ms) Current (A) Ambient

temperature (degC)

Experimental value (degC) Numerical result (degC)

Surface temperature Steelcore

Aluminum strandof inner layer

Aluminum strandof outer layer

0922

3170 6834 6834 6801

1106 84 8287 8286 82381290 98 10016 10015 9948

1922

33542 4690 4690 4662

1106 54 5397 5396 53561290 61 6273 6272 6215

2922

33541 4150 4150 4125

1106 45 4597 4596 45601290 51 5148 5147 5097

4922

3338 3756 3755 3733

1106 40 4011 4011 39801290 45 4322 4321 4278


the experimental and numerical results agree each other withan acceptable accuracy (2) the tensile stresses of the internalaluminum strands are slightly greater than external ones-ere are still some differences between the two results (withmaximum error of 176) which may be because the dis-placement deformation of strands under tension is randomas strands of each layer cannot be closely arranged withoutany gap in the ACSR construction It should be noted thatalthough layered stress does not perform well the resultanttensile stresses fit well with the experimental results whichwill be discussed later

53 Temperature-Sag Characteristics -e sag-temperaturecurves of heat-resistant aluminum alloy conductor JNRLH1G1A-40065 (with initial tension of 388 kN) and JNRLH1G1A-63055 (with initial tension of 485 kN) are comparedin Figures 7 and 8-e curves reflect the change law of sag (atthe midpoint) with the cross-sectional temperature -e

experiment is performed as follows fix the overhead lines(with a height difference of 0) on the equipment with a spanof 60m -en gradually increase the temperature of theconductors by charging with electricity and measure the sagat the midpoint

Figures 7 and 8 show that the numerical and experi-mental results of the two examples fit well (the maximumdifference is 737 and 449 respectively) With the in-crease of temperature the sag increases with a rate thatinitially increase and then decrease When the temperatureincreases to a certain value (ie Point A in Figures 7 and 8)the resultant tensile force on the aluminum strands reducesto zero and the sag gradually increases According toequation (26) at point A the temperature reaches CT -eCTand CC of JNRLH1G1A-40065 are 875degC and 8162 Arespectively and those of JNRLH1G1A-63055 are 1054degC

vw (ms)0 05 1 15 2 25 3 35 4

Tem

pera

ture

(degC)

30

40

50

60

70

80

90

100

Exp(922A)Num(922A)Exp(1106A)

Num(1106A)Exp(1290A)Num(1290A)

Figure 5 -e experimental and numerical results of ACSR JLG1A-63055

Axial strain times10ndash3

0 05 1 15 2 25

Laye

red

stres

s (M

Pa)

0

20

40

60

80

100

120

Exp values of layer 1Sim values of layer 1Exp values of layer 2

Sim values of layer 2Exp values of layer 3Sim values of layer 3

Figure 6 -e comparison of layered stress-strain relationship ofACSR JLG1A-40035

Temperature (degC)20 40 60 80 100 120

Sag

(mm

)

200

400

600

800

1000

1200

A

Num resultsExp resuts

Figure 7 -e temperature and sag characteristics of conductorJNRLH1G1A-40065


Temperature (degC)20 40 60 80 100 120 140

Sag

(mm

)

200

400

600

800

1000

1200

1400

A



and 14206A respectively -erefore the slope of the sag-temperature curves of the conductors near the criticaltemperature is in the transition phase ie the slope of thesag-temperature curves increases below and decreases abovethe critical temperature but does not change greatly near thecritical temperature When the conductor temperature isclose to the critical temperature the steel core in the con-ductor starts to bear all the loads However around CT thesag reaches a relatively large value in which the axial forcealong the conductor becomes relatively lower and the rate ofsag increase becomes smaller

54 Temperature-Tension Force Characteristics -e experi-mental and numerical temperature-tensile force character-istics of heat-resistant aluminum alloy conductor JNRLH1G1A-40065 (with initial tension of 34 kN) and JNRLH1G1A-63055 (with initial tension of 42 kN) are compared inFigures 9 and 10 -e experiment is performed using thesame process in Section 53

Figures 9 and 10 show that the numerical and experi-mental results fit well (the maximum difference is 519 and563 respectively) -e tensile force of the overheadconductors decreases gradually with the increase of tem-perature -e CT and CC are respectively (8356degC82764A) and (10096degC 139186A) for the two examplesBelow the CT the tensile force rapidly decreases as thetemperature increases and gradually decreases until thecritical temperature It should be noted that the critical pointcannot be directly observed according to the curve oftemperature vs tensile force but needs be calculated

6 Study of Critical Temperature andCritical Current

61 Layered Stress and Sag at the Critical Point In thissection the experiment in Section 54 is simulated and therelationship between the current and the layered stress (andsag) at the midpoint of overhead conductors is studied AtCT (or CC) the characteristics of the layered stress and sagare discussed Figures 11 and 12 show the law of the currentwith the layered stress on the cross section at the midpoint ofJNRLH1G1A-40065 and JNRLH1G1A-63055 respec-tively -e legend layer 1sim2 (in Figures 11 and 12) representthe internal and external layer of steel core respectively thelegend layer 3sim4 (in Figure 11) and layer 3sim5 (in Figure 12)indicate the internal and external layers of aluminumstrands respectively Figures 13 and 14 show relationshipsbetween their sags at the midpoint and the current re-spectively -e ldquoArdquo point in Figures 11ndash14 indicates thecritical current

Figures 11 and 12 show that the stresses of the steel coreof the internal layers are basically the same and the stresses ofthe aluminum strands of the external layers are basically thesame too With the increase of current the stresses of theexternal aluminum strands gradually decrease and those ofthe external steel core decrease first and then increase Whenthe current is greater than the critical current the stresses ofthe steel core increase rapidly In Figure 11 the stress of the

internal steel core at critical current is less than the initialstress (no current) However its stress is far greater than theinitial stress in Figure 12 It is shown (in Figures 13 and 14)that the sag (at the midpoint) gradually increases with theincrease of current and that under the critical current theyare 38 and 47 times more than the initial sag respectively-is is because when the current is far below the CC with theincrease of current (the temperature rises) the length andsag gradually increase and the tensile force on it decreasesleading to gradual reduction in the stress of the internal steelcore When the current (or temperature) is higher than theCC (or CT) the strands (such as aluminum strands) with alarger thermal expansion coefficient will be under pressurerather than tension -e strands (steel core) with a smallerthermal expansion coefficient will resist the tensile forcecaused by the aluminum strands under pressure but not thatof external load which leads to a rapid increase in the stress


Tens

ile fo

rce (

kN)

5

10

15

20

25

30

35


Figure 9 -e temperature and tensile force characteristics ofconductor JNRLH1G1A-40065



Tens

ile fo

rce (

kN)

5

15

25

35

45



of the steel core and linear increase in the sag It is obviousthat this stress state is not conducive to the synergy of eachlayer of strands In actual operation the current (or tem-perature) of the conductor should be controlled to be lowerthan the critical current (temperature) as much as possible

62=e Sensitive Factors of Critical Temperature andCurrentAs discussed in Section 51 the operational current andtemperature should be less than the critical values As thecritical temperature and current are related to layered stressand temperature field of conductors the following factors

0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

100

125

150

175

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

ndash50

ndash25

0

25

50

75

100

A

Layer 3Layer 4

(b)

Figure 11 Relationship between layered stress of conductor JNRLH1G1A-40065 and the current (a) stress of steel core at each layer and(b) stress of aluminum strand at each layer

0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

120

160

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

ndash50ndash25

0255075

100

A

Layer 3Layer 4Layer 5

(b)

Figure 12 Relationship between layered stress of conductor JNRLH1G1A-63055 and current (a) stress of steel core at each layer and (b)stress of aluminum strand at each layer

Current (A)0 200 400 600 800 1000 1200

Sag

(mm

)

0

300

600

900

1200

1500

A

Figure 13 Relationship between the sag at the midpoint ofconductor JNRLH1G1A-40065 and the current

Current (A)0 400 800 1200 1600

Sag

(mm

)

0

300

600

900

1200

1500

A



F (kN)0 10 20 30 40 50

I c (A

)

550

600

650

700

750

800

850

900

Ic vs FFitted curve

(a)

I c (A

)

Ic vs vw

Fitted curve

vw (ms)0 1 2 3 4 5 6 7

800

1200

1600

2000

2400

(b)

I c (A

)

Ic vs as

Fitted curve

as

02 03 04 05 06 07 08 09680

700

720

740

760

780

800

(c)

I c (A

)

Ic vs Ta

Fitted curve

Ta (degC)ndash20 ndash10 0 10 20 30 40 50

600

800

1000

1200

(d)

I c (A

)

Ic vs SFitted curve

S (Wm2)0 200 400 600 800 1000

740

760

780

800

820

(e)

Figure 15 Relationship between critical current and its factors (a) Relationship between critical current and horizontal tensile force (b)Relationship between critical current and wind speed (c) Relationship between critical current and solar absorption (d) Relationshipbetween critical current and ambient temperature (e) Relationship between critical current and solar intensity


are assumed to be sensitive (1) solar intensity S with a unit ofWm2 and range of 0sim1000 (2) solar absorption αs di-mensionless with a range of 023sim09 reflecting the age ofthe conductors new bright conductors have an absorptionrate of 023 and old gray conductors have an absorption rateof 09 and conductors in service period have an absorptionrate between them (3) ambient temperature Ta with a unitof degC and range of minus20sim50 (4) wind speed vw with a unit ofms and range of 0sim7 (5) initial tensile force F with a unit ofkN and range of 0sim03Fmax where Fmax indicates the ulti-mate bearing capacity In other words critical temperatureand critical current have a functional relation with thesevariables Tc(αs Ta vw F S) and Ic(αs Ta vw F S)

-is section takes the heat-resistant aluminum alloyconductor JNRLH1G1A-40065 as an example First thefitting of the functional relationships (eg linearquadraticexponential functionstrigonometric functions) betweencritical current and the abovementioned five variables isconducted -e relationships between critical current andeach of the five factors are respectively shown inFigures 15(a)ndash15(e) -en the relationship between criticalcurrent and its five factors can be postulated as equation (29)according to the fittings of the individual factor -e fittingcoefficients ai (i 1 sim 11) with a confidence level of 95can be obtained with the multifactor method (see Table 2 for

specific data) -e relationship between critical current andits influencing factors is shown in Figure 16 -e two dottedlines in the figure represent the confidence interval ofpredicted values (with a corresponding confidence level of95)

I a1Fa2 + a3 sin a4vw + a5( 1113857 + a6αs + a7 sin a8Ta + a9( 1113857 + a10S + a11

(29)

In the same way the change law of critical temperaturewith each influencing factor can be obtained (as shown inFigure 17)-e results show that the critical temperature hasa great correlation with the initial tensile force and littlecorrelation with other factors (ie they can be ignored) Forfunction relationship fitting the following equation can beobtained

Tc aFb

+ c (30)

where a 1354 b 04206 and c 2472Figures 15 and 16 show that (1) the sensitive factors of

critical current are successively wind speed ambient tem-perature horizontal tensile force solar intensity and solarabsorption according to their significance level (2) Hori-zontal tensile force and wind speed are positively correlatedwith critical current Solar absorption solar intensity and

Table 2 -e values of the fitting coefficient ai (i 1sim11)

Coefficients Value Confidence boundsa1 2034001 minus671671 1078472a2 0300959 minus038589 098781a3 7833395 minus866757 1023425a4 0091287 minus050256 0685134a5 0998746 minus253519 4532683a6 minus39569 minus110065 3092676a7 2386236 minus440629 4883541a8 minus000761 minus009472 0079501a9 1132626 minus266625 4931498a10 minus010697 minus014894 ndash0065a11 minus830829 minus107465 9084825

20 40400

800

1200

1600

2000

2400

0 2 4 0 02 04 06 08 ndash20 0 20 40 0 500 1000F (kN) vw (ms) αs Ta (degC) S (Wm2)

I c (deg

C)

Figure 16 Fitting of critical current and its sensitive factors


ambient temperature are negatively correlated with itFigure 17 shows that the critical temperature is positivelycorrelated with horizontal tensile force and has little cor-relation with other factors

7 Conclusion

-is article proposes a set of numerical methods for cal-culating the temperature field layered stress and sag ofoverhead conductors based on which the critical temper-ature and critical current are studied -e calculationmethods and procedures are verified through experimentalresults-e authors also studied the relationship between thefive most sensitive factors and the critical temperature (orcurrent) -e following conclusions are made

(1) -e results of numerical calculation and experimentsfor conductor temperature agree well -e temper-ature gradient on the cross section is relatively smalltherefore its impact on the layered stress can beignored in the calculation Meanwhile the averagetemperature of cross sections rises with the increaseof current and drops with the increase of wind speed

(2) -e numerical simulation and experimental resultsfor layered stress in the elastic phase (with strain ofless than 02) agree each other with acceptableaccuracy -e tensile stresses of the internal alumi-num strands are slightly greater than those of theexternal ones

(3) -e calculated sages fit well with the experimentalresults With the increase of temperature the sagincreases with a rate that first increases and thendecreases that is when the conductor temperatureincreases close to the critical temperature the sag ofthe conductors gradually increases -e calculatedtensile forces agree well with experimental onesBelow the critical temperature the tensile forcedecreases rapidly with the increase of temperaturewhile the changing rate decreases close to the critical

temperature -e critical temperature cannot bedirectly observed according to the curve of tem-perature vs tensile force before calculation

(4) With the increase of current the stresses of the steelcore first decrease and then increase while those ofthe aluminum monotonically decrease When thecurrent (or temperature) of overhead conductors ishigher than the critical current (or temperature) thestresses of the steel core rapidly increase and the saglinearly increases with increasing temperature Atthe critical current the sag increases to 3ndash4 times ofits original value In some cases the stresses of steelcores in the conductor are far greater than theirinitial value In actual operation the current (ortemperature) of conductors should be controlled sothat it remains lower than the critical current (ortemperature) as much as possible

(5) -e critical temperature is positively correlated withhorizontal tensile force and has little correlation withother factors According to their relative sensitivitiesthe importance ranking of the influencing factors tocritical current are (from high to low) wind speedambient temperature horizontal tensile force solarintensity and solar absorption Horizontal tensileforce and wind speed are positively correlated withcritical current Solar absorption solar intensity andambient temperature are negatively correlated

Data Availability

All the data used to support the findings of this study areincluded within the article

Disclosure

Any opinions findings conclusions or recommendationsexpressed in this publication are those of the authors and donot necessarily reflect the views of the sponsors

Conflicts of Interest

-e authors declare that there are no conflicts of interest

Acknowledgments

-e authors acknowledge the financial support from theNational Key Research and Development Program of China(Grant no 2016YFC0701106) and the National NaturalScience Foundation of China (Grant nos 51578473 and51978591) -e authors also acknowledge the financialsupport from the Scientific Research Fund of Institute ofEngineering Mechanics China Earthquake Administration(Grant nos 2019EEEVL0501 2019B02 and 2019A02)

References

[1] J Wei Z Wang X Feng et al ldquoStudy on the mechanicalproperties of heatmdashresistant aluminum alloy conductorsrdquopp 35-36+47 Water Conservancy amp Electric Power Ma-chinery Ziyang China 2005 in Chinese

T c (deg

C)

F (kN)0 10 20 30 40 50

50

60

70

80

90

Tc vs FFitted curve

Figure 17 Relationship between critical temperature and factors


[2] G Wang and H Huang ldquoCalculation of stress and sag of theenergy-saving capacity-expanded conductors (in Chinese)rdquoElectric WireampCable pp 11ndash14 2009

[3] K Yang Comparison of Stress and Sag Characteristics ofAugmented Capacity Conductors pp 31ndash33 Yunnan ElectricPower Kunming China 2013 in Chinese

[4] R Bush W Black and T Champion ldquoExperimental verifi-cation of a real-time program for the determination oftemperature and sag of overhead linesrdquo IEEE Transactions onPower Apparatus and Systems vol PAS-102 no 7pp 2284ndash2288 1983

[5] G Clark ldquoSummary report on the effects of high operatingtemperatures on conductors and hardware behaviorrdquo ReportOntario Hydro Research Division Toronto Canada 1977

[6] D A Douglass ldquoRadial and axial temperate gradients in barestranded condctorrdquo IEEE Transactions on Power Deliveryvol 1 no 2 pp 7ndash15 1986

[7] D Douglass L Kirkpatrick and L Rathbun ldquoAC resistance ofACSR-Magnetic and temperature effectsrdquo IEEE Transactionson Power Apparatus and Systems vol PAS-104 no 6pp 1578ndash1584 1985

[8] S D Foss S H Lin and R Carberry ldquoSignificance of theconductor radial temperature gradient within a dynamic linerating methodologyrdquo IEEE Transactions on Power Deliveryvol 2 no 2 pp 502ndash511 1987

[9] J F Hall A K Deb and J Savoullis ldquoWind tunnel studies oftransmission line conductor temperaturesrdquo IEEE Transac-tions on Power Delivery vol 3 no 2 pp 801ndash812 1988

[10] V T Morgan ldquoRating of bare overhead conductors forcontinuous currentsrdquo Proceedings of the Institution of Elec-trical Engineers vol 114 no 10 pp 1473ndash1482 1967

[11] V T Morgan ldquoRating of bare overhead conductors for in-termittent and cyclic currentsrdquo Proceedings of the Institutionof Electrical Engineers vol 116 no 8 pp 1361ndash1376 1969

[12] V Morgan ldquoA theoretical and experimental analysis of thefactors which determine the thermal behaviour of an exposedcylindrical current-carrying conductorrdquo Ph D thesis Uni-versity London London UK 1971

[13] IEEE IEEE Standard for Calculating the Current-TemperatureRelationship of Bare Overhead Conductors IEEE Std 738-2012(Revision of IEEE Std 738-2006mdashIncorporates IEEE Std 738-2012 Cor 1-2013) pp 1ndash72 IEEE Piscataway NJ USA 2013

[14] Z He Z Li and P Liang ldquoANSYS calculation method oftemperature and current-carrying capacity for transmissionlinesrdquo Zhejiang Electric Power vol 29 pp 1ndash5 2010 inChinese

[15] C Zhao L Zheng C Tian et al ldquoNumerical study ontemperature field of 66kV overhead transmission linerdquo PowerSystem Technology pp 34ndash36 2007 in Chinese

[16] M Zunec I Ticar and F Jakl ldquoDetermination of current andtemperature distribution in overhead conductors by usingelectromagnetic-field analysis toolsrdquo IEEE Transactions onPower Delivery vol 21 no 3 pp 1524ndash1529 2006

[17] R Stephen D Douglas GMirosevic et al=ermal Behaviourof Overhead Conductors Cigre Paris France 2002

[18] T Shao Calculation of Electrical Engineering of OverheadTransmission Line China Electric Power Press Beijing China2003 in Chinese

[19] S Byrd H Ni and X Rui ldquoApplication of the layer char-acteristics to the intensity of overhead conductors (in Chi-nese)rdquo Southern Power System Technology vol 3 pp 49ndash512009

[20] L Zheng and Z Sun ldquoStress analysis of aluminium strandedwire and steel-cored aluminium stranded wirerdquo Journal of

Wuhan Uniersity of Hydraulic and Electric Engineeringpp 66ndash69 1997 in Chinese

[21] K Xiao Y Liu P Li et alNumerical Analysis on Temperatureand Stress Filed Coupling of Overhead Conductors pp 97ndash102Guangdong Electric Power Guangzhou China 2015 inChinese

[22] D Loudon D A Douglass R G Stephen and G C SibilantldquoCalculation accuracy of high-temperature sag for ACSR inexisting linesrdquo Cigre Science amp Engineering vol 7 2017

[23] O Nigol and J S Barrett ldquoCharacteristics of ACSR con-ductors at high temperatures and stressesrdquo IEEE Transactionson Power Apparatus and Systems vol PAS-100 no 2pp 485ndash493 1981

[24] C B Rawlins ldquoSome effects of mill practice on the stress strainbehavior of ACSRrdquo IEEE Transactions on Power Deliveryvol 14 no 2 pp 602ndash629 1999

[25] Standardization Administration of China GBT 1179-2008Round Wire Concentric Lay Overhead Electrical StrandedConductors China Standardization Administration of ChinaBeijing China 2008

[26] X Peng and Y Lin Experimental Investigation on Impact ofAmbient Wind Velocity on Ampacity of Overhead Transmis-sion Lines and Temperature Rise pp 20ndash25 GuangdongElectric Power Guangzhou China 2012 in Chinese


stress calculative method iteratively By comparing the re-sults of calculation with those of the experiments for a typicalACSR (aluminum-conductor steel reinforced) the algo-rithms are verified In order to facilitate the dynamic as-sessment of safety of the conductors during the operationthis paper studies the relationship between the criticaltemperature of the conductor and its sensitive factors withthe heat-resistant aluminum alloy conductor JNRLH1G1A-40065 as an example

2 Calculation of Conductor Temperature Field

-e thermal balance equation of the entire cross sectionΩ ofthe conductor can be established [14]

k Txx + Tyy1113872 1113873 + s 0 (x y) isin Ω (1)

And the boundary condition on the surface of theconductor Γ can also be given [14]

qn minusk Txnx + Tyny1113872 1113873 α T minus Ta( 1113857 l(x y) isin Γ (2)

where Txx Tyy Tx and Ty indicate the second-orderderivatives and first-order derivatives of the 2D temperaturefield T in the conductor on the x and y coordinates re-spectively qn indicates the heat loss rate along the normaldirection n on the outer surface of the conductor nx and ny

are the components of n in the direction of x and y α in-dicates the composite heat loss coefficient Ta indicates theambient temperature k is the thermal conductivity repre-sented by ka in the area of the aluminum strand ks in thearea of the steel core and kair in the area of the air and sindicates the rate of heat per unit volume represented by sa

in the area of the aluminum strand ss in the area of the steelcore and 0 in the area of the air Herein ka and ks areirrelevant to temperature field while kair is relevant to it andthe relationship is as follows

kair 242 times 10minus2+ 72 times 10minus 5

Tav (3)

where Tav indicates the average temperature of the crosssection -e voltage of steel core per unit length is the sameas that of aluminum strand -e current is inversely pro-portional to the resistance which is inversely proportional tothe conductor cross section and proportional to the resis-tivity [14] -e ratio between the current Is of steel core andthat Ia of aluminum strand is

λ Is

Ia

As

Aa

ρa

ρs

(4)

where As and ρs are the cross-sectional area and electricalresistivity of steel core respectively and Aa and ρa are thearea and electrical resistivity of aluminum strand re-spectively -e rate of heat of steel core per unit volume ss

is

ss Pλ(λ + 1)

As

(5)

External aluminum strands should consider solar energy[17] so the rate of heat per unit volume sa is

sa P(λ + 1)

Aa

+csSDAa

(6)

In equation (6) the first term on the right end of theequation is caused by the current and the second term iscaused by the solar radiation P indicates its total Joule heatgain [17] which is the function of the current averagetemperature and integrated resistance rate of the conductorS indicates the solar intensity cs indicates the solar ab-sorption of the conductor with a value of 023ndash09 and isusually valued at 05 and D indicates the conductordiameter

An aluminum-conductor steel-reinforced cable (ACSR)mainly loses heat by means of convective cooling Pc andradiative cooling Pr -e calculation of heat loss is providedin the literature [17] According to the principle of constantheat loss per unit volume the composite heat transfer co-efficient of conductor surface in equation (2) can bedetermined

α Pc + Pr

Tsur minus Ta( 1113857A (7)

where A indicates the lateral surface area of conductor perunit length (that is the interface of the conductor and theexternal environment)

3 Calculation of Conductor Stress Field

-is article assumes that conductors can only be resistant totension ignoring their resistance to shear bend and torsion[18] and the tensile force on each strand of the same layer isequal [19] -ree basic conditions are required when per-forming calculations for overhead conductors (1) defor-mation compatibility (2) material constitutive relationshipand (3) balance of internal and external forces

31 Deformation Compatibility Using the condition ofdeformation compatibility the strain of each strand can berepresented by the longitudinal strain For a strand with alength of Si if unfolding it as shown in Figure 1 the de-formation compatibility equation can be obtained as follows[20]

ε0Ls ΔSi

sin αi

+ΔRiφi

tan αi

(8)

where ε0 indicates the axial strain of the overhead conductorLs indicates the length of the overhead conductor corre-sponding to a strand with a length of Si αi indicates the anglebetween the tangent of layer i of the strand and the con-ductor cross section ΔRiφi indicates the lateral deformationof the unfolded strand and φi indicates the ratio betweenprojected length of the unfolded strand on the cross sectionand the outer radius Ri of layer i of the strand as follows

φi Si cos αiRi (9)

From the geometrical relationship equation (10) can beobtained as follows


sin αi Ls

Si

(10)



Ri

cos2αi (11)






ΔriTprime

Δr1T i 1



⎧⎪⎪⎨

⎪⎪⎩(13)




εiT αiT Ti minus 20( 1113857 (15)






P ε0( 1113857 1113944n

i1pini sin αi (17)


N(x) σ0A0


2σ0 cos β1113890 1113891

2

11139741113972

(18)


f ε0( 1113857 N(x) minus P ε0( 1113857 0 (19)

Brsquorsquo Brsquo

B

ΔSi

αiαiA

B

Riφi

Riφi ΔRiφiA

SiLs

Lsε0







fM

L minusl

cos β1113888 1113889 middot

3l

8 cos3 β

1113971

(21)


L0

2σo

csh

cl

2σo

1113888 1113889

2

+ h2

11139741113972

(22)


σ0 cl2

8fM cos β (23)


σ0c

1 minus chcl

2σ01113888 1113889

1 +h

Lh01113888 1113889

2

11139741113972

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ +h

2+ fAO 0 (24)


L L0(1 + ε) (25)


Pa(I)1113868111386811138681113868IIc

0 (26)

Pa(I) 1113944

na


Tc T Ic( 1113857 (28)




A

B

x

y

l

hfM

O

fx

l2

xfAO

β








obtained



(P(2)a minus P(1)





ie I(1) I(2) P(1)a P(2)

a




No





Output T(i)



Pa ge 0



(1)

Pa(2) = Pa|I=I



(1) = Pa(2)


i = i + 1Yes

No


Yes

No

Enter submodule 1

Enter submodule 2




(2) ndash Pa(1))


The main module



22 and 21)








an error




Calculate f (ε)


Yes

No

Yes

No



section

I(2) = I Pa(2) = Pa



ε(2) = ε f (ε(2)) = f (ε)


(i)

















































AluminumSteelAir

41232

3432




temperature (degC)





0922

3170 6834 6834 6801

1106 84 8287 8286 82381290 98 10016 10015 9948

1922

33542 4690 4690 4662

1106 54 5397 5396 53561290 61 6273 6272 6215

2922

33541 4150 4150 4125

1106 45 4597 4596 45601290 51 5148 5147 5097

4922

3338 3756 3755 3733

1106 40 4011 4011 39801290 45 4322 4321 4278






vw (ms)0 05 1 15 2 25 3 35 4

Tem

pera

ture

(degC)

30

40

50

60

70

80

90

100





0 05 1 15 2 25

Laye

red

stres

s (M

Pa)

0

20

40

60

80

100

120





Sag

(mm

)

200

400

600

800

1000

1200

A





Sag

(mm

)

200

400

600

800

1000

1200

1400

A











Tens

ile fo

rce (

kN)

5

10

15

20

25

30

35





Tens

ile fo

rce (

kN)

5

15

25

35

45





0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

100

125

150

175

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

ndash50

ndash25

0

25

50

75

100

A

Layer 3Layer 4

(b)


0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

120

160

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

ndash50ndash25

0255075

100

A


(b)


Current (A)0 200 400 600 800 1000 1200

Sag

(mm

)

0

300

600

900

1200

1500

A


Current (A)0 400 800 1200 1600

Sag

(mm

)

0

300

600

900

1200

1500

A



F (kN)0 10 20 30 40 50

I c (A

)

550

600

650

700

750

800

850

900

Ic vs FFitted curve

(a)

I c (A

)

Ic vs vw

Fitted curve

vw (ms)0 1 2 3 4 5 6 7

800

1200

1600

2000

2400

(b)

I c (A

)

Ic vs as

Fitted curve

as

02 03 04 05 06 07 08 09680

700

720

740

760

780

800

(c)

I c (A

)

Ic vs Ta

Fitted curve


600

800

1000

1200

(d)

I c (A

)

Ic vs SFitted curve

S (Wm2)0 200 400 600 800 1000

740

760

780

800

820

(e)







(29)


Tc aFb

+ c (30)





20 40400

800

1200

1600

2000

2400


I c (deg

C)




7 Conclusion








Data Availability


Disclosure




Acknowledgments


References


T c (deg

C)

F (kN)0 10 20 30 40 50

50

60

70

80

90

Tc vs FFitted curve






























sin αi Ls

Si

(10)



Ri

cos2αi (11)






ΔriTprime

Δr1T i 1



⎧⎪⎪⎨

⎪⎪⎩(13)




εiT αiT Ti minus 20( 1113857 (15)






P ε0( 1113857 1113944n

i1pini sin αi (17)


N(x) σ0A0


2σ0 cos β1113890 1113891

2

11139741113972

(18)


f ε0( 1113857 N(x) minus P ε0( 1113857 0 (19)

Brsquorsquo Brsquo

B

ΔSi

αiαiA

B

Riφi

Riφi ΔRiφiA

SiLs

Lsε0







fM

L minusl

cos β1113888 1113889 middot

3l

8 cos3 β

1113971

(21)


L0

2σo

csh

cl

2σo

1113888 1113889

2

+ h2

11139741113972

(22)


σ0 cl2

8fM cos β (23)


σ0c

1 minus chcl

2σ01113888 1113889

1 +h

Lh01113888 1113889

2

11139741113972

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ +h

2+ fAO 0 (24)


L L0(1 + ε) (25)


Pa(I)1113868111386811138681113868IIc

0 (26)

Pa(I) 1113944

na


Tc T Ic( 1113857 (28)




A

B

x

y

l

hfM

O

fx

l2

xfAO

β








obtained



(P(2)a minus P(1)





ie I(1) I(2) P(1)a P(2)

a




No





Output T(i)



Pa ge 0



(1)

Pa(2) = Pa|I=I



(1) = Pa(2)


i = i + 1Yes

No


Yes

No

Enter submodule 1

Enter submodule 2




(2) ndash Pa(1))


The main module



22 and 21)








an error




Calculate f (ε)


Yes

No

Yes

No



section

I(2) = I Pa(2) = Pa



ε(2) = ε f (ε(2)) = f (ε)


(i)

















































AluminumSteelAir

41232

3432




temperature (degC)





0922

3170 6834 6834 6801

1106 84 8287 8286 82381290 98 10016 10015 9948

1922

33542 4690 4690 4662

1106 54 5397 5396 53561290 61 6273 6272 6215

2922

33541 4150 4150 4125

1106 45 4597 4596 45601290 51 5148 5147 5097

4922

3338 3756 3755 3733

1106 40 4011 4011 39801290 45 4322 4321 4278






vw (ms)0 05 1 15 2 25 3 35 4

Tem

pera

ture

(degC)

30

40

50

60

70

80

90

100





0 05 1 15 2 25

Laye

red

stres

s (M

Pa)

0

20

40

60

80

100

120





Sag

(mm

)

200

400

600

800

1000

1200

A





Sag

(mm

)

200

400

600

800

1000

1200

1400

A











Tens

ile fo

rce (

kN)

5

10

15

20

25

30

35





Tens

ile fo

rce (

kN)

5

15

25

35

45





0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

100

125

150

175

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

ndash50

ndash25

0

25

50

75

100

A

Layer 3Layer 4

(b)


0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

120

160

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

ndash50ndash25

0255075

100

A


(b)


Current (A)0 200 400 600 800 1000 1200

Sag

(mm

)

0

300

600

900

1200

1500

A


Current (A)0 400 800 1200 1600

Sag

(mm

)

0

300

600

900

1200

1500

A



F (kN)0 10 20 30 40 50

I c (A

)

550

600

650

700

750

800

850

900

Ic vs FFitted curve

(a)

I c (A

)

Ic vs vw

Fitted curve

vw (ms)0 1 2 3 4 5 6 7

800

1200

1600

2000

2400

(b)

I c (A

)

Ic vs as

Fitted curve

as

02 03 04 05 06 07 08 09680

700

720

740

760

780

800

(c)

I c (A

)

Ic vs Ta

Fitted curve


600

800

1000

1200

(d)

I c (A

)

Ic vs SFitted curve

S (Wm2)0 200 400 600 800 1000

740

760

780

800

820

(e)







(29)


Tc aFb

+ c (30)





20 40400

800

1200

1600

2000

2400


I c (deg

C)




7 Conclusion








Data Availability


Disclosure




Acknowledgments


References


T c (deg

C)

F (kN)0 10 20 30 40 50

50

60

70

80

90

Tc vs FFitted curve


































fM

L minusl

cos β1113888 1113889 middot

3l

8 cos3 β

1113971

(21)


L0

2σo

csh

cl

2σo

1113888 1113889

2

+ h2

11139741113972

(22)


σ0 cl2

8fM cos β (23)


σ0c

1 minus chcl

2σ01113888 1113889

1 +h

Lh01113888 1113889

2

11139741113972

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ +h

2+ fAO 0 (24)


L L0(1 + ε) (25)


Pa(I)1113868111386811138681113868IIc

0 (26)

Pa(I) 1113944

na


Tc T Ic( 1113857 (28)




A

B

x

y

l

hfM

O

fx

l2

xfAO

β








obtained



(P(2)a minus P(1)





ie I(1) I(2) P(1)a P(2)

a




No





Output T(i)



Pa ge 0



(1)

Pa(2) = Pa|I=I



(1) = Pa(2)


i = i + 1Yes

No


Yes

No

Enter submodule 1

Enter submodule 2




(2) ndash Pa(1))


The main module



22 and 21)








an error




Calculate f (ε)


Yes

No

Yes

No



section

I(2) = I Pa(2) = Pa



ε(2) = ε f (ε(2)) = f (ε)


(i)

















































AluminumSteelAir

41232

3432




temperature (degC)





0922

3170 6834 6834 6801

1106 84 8287 8286 82381290 98 10016 10015 9948

1922

33542 4690 4690 4662

1106 54 5397 5396 53561290 61 6273 6272 6215

2922

33541 4150 4150 4125

1106 45 4597 4596 45601290 51 5148 5147 5097

4922

3338 3756 3755 3733

1106 40 4011 4011 39801290 45 4322 4321 4278






vw (ms)0 05 1 15 2 25 3 35 4

Tem

pera

ture

(degC)

30

40

50

60

70

80

90

100





0 05 1 15 2 25

Laye

red

stres

s (M

Pa)

0

20

40

60

80

100

120





Sag

(mm

)

200

400

600

800

1000

1200

A





Sag

(mm

)

200

400

600

800

1000

1200

1400

A











Tens

ile fo

rce (

kN)

5

10

15

20

25

30

35





Tens

ile fo

rce (

kN)

5

15

25

35

45





0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

100

125

150

175

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

ndash50

ndash25

0

25

50

75

100

A

Layer 3Layer 4

(b)


0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

120

160

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

ndash50ndash25

0255075

100

A


(b)


Current (A)0 200 400 600 800 1000 1200

Sag

(mm

)

0

300

600

900

1200

1500

A


Current (A)0 400 800 1200 1600

Sag

(mm

)

0

300

600

900

1200

1500

A



F (kN)0 10 20 30 40 50

I c (A

)

550

600

650

700

750

800

850

900

Ic vs FFitted curve

(a)

I c (A

)

Ic vs vw

Fitted curve

vw (ms)0 1 2 3 4 5 6 7

800

1200

1600

2000

2400

(b)

I c (A

)

Ic vs as

Fitted curve

as

02 03 04 05 06 07 08 09680

700

720

740

760

780

800

(c)

I c (A

)

Ic vs Ta

Fitted curve


600

800

1000

1200

(d)

I c (A

)

Ic vs SFitted curve

S (Wm2)0 200 400 600 800 1000

740

760

780

800

820

(e)







(29)


Tc aFb

+ c (30)





20 40400

800

1200

1600

2000

2400


I c (deg

C)




7 Conclusion








Data Availability


Disclosure




Acknowledgments


References


T c (deg

C)

F (kN)0 10 20 30 40 50

50

60

70

80

90

Tc vs FFitted curve



































obtained



(P(2)a minus P(1)





ie I(1) I(2) P(1)a P(2)

a




No





Output T(i)



Pa ge 0



(1)

Pa(2) = Pa|I=I



(1) = Pa(2)


i = i + 1Yes

No


Yes

No

Enter submodule 1

Enter submodule 2




(2) ndash Pa(1))


The main module



22 and 21)








an error




Calculate f (ε)


Yes

No

Yes

No



section

I(2) = I Pa(2) = Pa



ε(2) = ε f (ε(2)) = f (ε)


(i)

















































AluminumSteelAir

41232

3432




temperature (degC)





0922

3170 6834 6834 6801

1106 84 8287 8286 82381290 98 10016 10015 9948

1922

33542 4690 4690 4662

1106 54 5397 5396 53561290 61 6273 6272 6215

2922

33541 4150 4150 4125

1106 45 4597 4596 45601290 51 5148 5147 5097

4922

3338 3756 3755 3733

1106 40 4011 4011 39801290 45 4322 4321 4278






vw (ms)0 05 1 15 2 25 3 35 4

Tem

pera

ture

(degC)

30

40

50

60

70

80

90

100





0 05 1 15 2 25

Laye

red

stres

s (M

Pa)

0

20

40

60

80

100

120





Sag

(mm

)

200

400

600

800

1000

1200

A





Sag

(mm

)

200

400

600

800

1000

1200

1400

A











Tens

ile fo

rce (

kN)

5

10

15

20

25

30

35





Tens

ile fo

rce (

kN)

5

15

25

35

45





0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

100

125

150

175

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

ndash50

ndash25

0

25

50

75

100

A

Layer 3Layer 4

(b)


0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

120

160

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

ndash50ndash25

0255075

100

A


(b)


Current (A)0 200 400 600 800 1000 1200

Sag

(mm

)

0

300

600

900

1200

1500

A


Current (A)0 400 800 1200 1600

Sag

(mm

)

0

300

600

900

1200

1500

A



F (kN)0 10 20 30 40 50

I c (A

)

550

600

650

700

750

800

850

900

Ic vs FFitted curve

(a)

I c (A

)

Ic vs vw

Fitted curve

vw (ms)0 1 2 3 4 5 6 7

800

1200

1600

2000

2400

(b)

I c (A

)

Ic vs as

Fitted curve

as

02 03 04 05 06 07 08 09680

700

720

740

760

780

800

(c)

I c (A

)

Ic vs Ta

Fitted curve


600

800

1000

1200

(d)

I c (A

)

Ic vs SFitted curve

S (Wm2)0 200 400 600 800 1000

740

760

780

800

820

(e)







(29)


Tc aFb

+ c (30)





20 40400

800

1200

1600

2000

2400


I c (deg

C)




7 Conclusion








Data Availability


Disclosure




Acknowledgments


References


T c (deg

C)

F (kN)0 10 20 30 40 50

50

60

70

80

90

Tc vs FFitted curve












































































AluminumSteelAir

41232

3432




temperature (degC)





0922

3170 6834 6834 6801

1106 84 8287 8286 82381290 98 10016 10015 9948

1922

33542 4690 4690 4662

1106 54 5397 5396 53561290 61 6273 6272 6215

2922

33541 4150 4150 4125

1106 45 4597 4596 45601290 51 5148 5147 5097

4922

3338 3756 3755 3733

1106 40 4011 4011 39801290 45 4322 4321 4278






vw (ms)0 05 1 15 2 25 3 35 4

Tem

pera

ture

(degC)

30

40

50

60

70

80

90

100





0 05 1 15 2 25

Laye

red

stres

s (M

Pa)

0

20

40

60

80

100

120





Sag

(mm

)

200

400

600

800

1000

1200

A





Sag

(mm

)

200

400

600

800

1000

1200

1400

A











Tens

ile fo

rce (

kN)

5

10

15

20

25

30

35





Tens

ile fo

rce (

kN)

5

15

25

35

45





0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

100

125

150

175

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

ndash50

ndash25

0

25

50

75

100

A

Layer 3Layer 4

(b)


0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

120

160

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

ndash50ndash25

0255075

100

A


(b)


Current (A)0 200 400 600 800 1000 1200

Sag

(mm

)

0

300

600

900

1200

1500

A


Current (A)0 400 800 1200 1600

Sag

(mm

)

0

300

600

900

1200

1500

A



F (kN)0 10 20 30 40 50

I c (A

)

550

600

650

700

750

800

850

900

Ic vs FFitted curve

(a)

I c (A

)

Ic vs vw

Fitted curve

vw (ms)0 1 2 3 4 5 6 7

800

1200

1600

2000

2400

(b)

I c (A

)

Ic vs as

Fitted curve

as

02 03 04 05 06 07 08 09680

700

720

740

760

780

800

(c)

I c (A

)

Ic vs Ta

Fitted curve


600

800

1000

1200

(d)

I c (A

)

Ic vs SFitted curve

S (Wm2)0 200 400 600 800 1000

740

760

780

800

820

(e)







(29)


Tc aFb

+ c (30)





20 40400

800

1200

1600

2000

2400


I c (deg

C)




7 Conclusion








Data Availability


Disclosure




Acknowledgments


References


T c (deg

C)

F (kN)0 10 20 30 40 50

50

60

70

80

90

Tc vs FFitted curve



































AluminumSteelAir

41232

3432




temperature (degC)





0922

3170 6834 6834 6801

1106 84 8287 8286 82381290 98 10016 10015 9948

1922

33542 4690 4690 4662

1106 54 5397 5396 53561290 61 6273 6272 6215

2922

33541 4150 4150 4125

1106 45 4597 4596 45601290 51 5148 5147 5097

4922

3338 3756 3755 3733

1106 40 4011 4011 39801290 45 4322 4321 4278






vw (ms)0 05 1 15 2 25 3 35 4

Tem

pera

ture

(degC)

30

40

50

60

70

80

90

100





0 05 1 15 2 25

Laye

red

stres

s (M

Pa)

0

20

40

60

80

100

120





Sag

(mm

)

200

400

600

800

1000

1200

A





Sag

(mm

)

200

400

600

800

1000

1200

1400

A











Tens

ile fo

rce (

kN)

5

10

15

20

25

30

35





Tens

ile fo

rce (

kN)

5

15

25

35

45





0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

100

125

150

175

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

ndash50

ndash25

0

25

50

75

100

A

Layer 3Layer 4

(b)


0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

120

160

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

ndash50ndash25

0255075

100

A


(b)


Current (A)0 200 400 600 800 1000 1200

Sag

(mm

)

0

300

600

900

1200

1500

A


Current (A)0 400 800 1200 1600

Sag

(mm

)

0

300

600

900

1200

1500

A



F (kN)0 10 20 30 40 50

I c (A

)

550

600

650

700

750

800

850

900

Ic vs FFitted curve

(a)

I c (A

)

Ic vs vw

Fitted curve

vw (ms)0 1 2 3 4 5 6 7

800

1200

1600

2000

2400

(b)

I c (A

)

Ic vs as

Fitted curve

as

02 03 04 05 06 07 08 09680

700

720

740

760

780

800

(c)

I c (A

)

Ic vs Ta

Fitted curve


600

800

1000

1200

(d)

I c (A

)

Ic vs SFitted curve

S (Wm2)0 200 400 600 800 1000

740

760

780

800

820

(e)







(29)


Tc aFb

+ c (30)





20 40400

800

1200

1600

2000

2400


I c (deg

C)




7 Conclusion








Data Availability


Disclosure




Acknowledgments


References


T c (deg

C)

F (kN)0 10 20 30 40 50

50

60

70

80

90

Tc vs FFitted curve


































vw (ms)0 05 1 15 2 25 3 35 4

Tem

pera

ture

(degC)

30

40

50

60

70

80

90

100





0 05 1 15 2 25

Laye

red

stres

s (M

Pa)

0

20

40

60

80

100

120





Sag

(mm

)

200

400

600

800

1000

1200

A





Sag

(mm

)

200

400

600

800

1000

1200

1400

A











Tens

ile fo

rce (

kN)

5

10

15

20

25

30

35





Tens

ile fo

rce (

kN)

5

15

25

35

45





0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

100

125

150

175

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

ndash50

ndash25

0

25

50

75

100

A

Layer 3Layer 4

(b)


0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

120

160

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

ndash50ndash25

0255075

100

A


(b)


Current (A)0 200 400 600 800 1000 1200

Sag

(mm

)

0

300

600

900

1200

1500

A


Current (A)0 400 800 1200 1600

Sag

(mm

)

0

300

600

900

1200

1500

A



F (kN)0 10 20 30 40 50

I c (A

)

550

600

650

700

750

800

850

900

Ic vs FFitted curve

(a)

I c (A

)

Ic vs vw

Fitted curve

vw (ms)0 1 2 3 4 5 6 7

800

1200

1600

2000

2400

(b)

I c (A

)

Ic vs as

Fitted curve

as

02 03 04 05 06 07 08 09680

700

720

740

760

780

800

(c)

I c (A

)

Ic vs Ta

Fitted curve


600

800

1000

1200

(d)

I c (A

)

Ic vs SFitted curve

S (Wm2)0 200 400 600 800 1000

740

760

780

800

820

(e)







(29)


Tc aFb

+ c (30)





20 40400

800

1200

1600

2000

2400


I c (deg

C)




7 Conclusion








Data Availability


Disclosure




Acknowledgments


References


T c (deg

C)

F (kN)0 10 20 30 40 50

50

60

70

80

90

Tc vs FFitted curve






































Tens

ile fo

rce (

kN)

5

10

15

20

25

30

35





Tens

ile fo

rce (

kN)

5

15

25

35

45





0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

100

125

150

175

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

ndash50

ndash25

0

25

50

75

100

A

Layer 3Layer 4

(b)


0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

120

160

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

ndash50ndash25

0255075

100

A


(b)


Current (A)0 200 400 600 800 1000 1200

Sag

(mm

)

0

300

600

900

1200

1500

A


Current (A)0 400 800 1200 1600

Sag

(mm

)

0

300

600

900

1200

1500

A



F (kN)0 10 20 30 40 50

I c (A

)

550

600

650

700

750

800

850

900

Ic vs FFitted curve

(a)

I c (A

)

Ic vs vw

Fitted curve

vw (ms)0 1 2 3 4 5 6 7

800

1200

1600

2000

2400

(b)

I c (A

)

Ic vs as

Fitted curve

as

02 03 04 05 06 07 08 09680

700

720

740

760

780

800

(c)

I c (A

)

Ic vs Ta

Fitted curve


600

800

1000

1200

(d)

I c (A

)

Ic vs SFitted curve

S (Wm2)0 200 400 600 800 1000

740

760

780

800

820

(e)







(29)


Tc aFb

+ c (30)





20 40400

800

1200

1600

2000

2400


I c (deg

C)




7 Conclusion








Data Availability


Disclosure




Acknowledgments


References


T c (deg

C)

F (kN)0 10 20 30 40 50

50

60

70

80

90

Tc vs FFitted curve
































0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

100

125

150

175

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200

Laye

red

stres

s (M

Pa)

ndash50

ndash25

0

25

50

75

100

A

Layer 3Layer 4

(b)


0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

120

160

200

A

Layer 1Layer 2

Current (A)

(a)

Current (A)0 200 400 600 800 1000 1200 1400 1600

Laye

red

stres

s (M

Pa)

ndash50ndash25

0255075

100

A


(b)


Current (A)0 200 400 600 800 1000 1200

Sag

(mm

)

0

300

600

900

1200

1500

A


Current (A)0 400 800 1200 1600

Sag

(mm

)

0

300

600

900

1200

1500

A



F (kN)0 10 20 30 40 50

I c (A

)

550

600

650

700

750

800

850

900

Ic vs FFitted curve

(a)

I c (A

)

Ic vs vw

Fitted curve

vw (ms)0 1 2 3 4 5 6 7

800

1200

1600

2000

2400

(b)

I c (A

)

Ic vs as

Fitted curve

as

02 03 04 05 06 07 08 09680

700

720

740

760

780

800

(c)

I c (A

)

Ic vs Ta

Fitted curve


600

800

1000

1200

(d)

I c (A

)

Ic vs SFitted curve

S (Wm2)0 200 400 600 800 1000

740

760

780

800

820

(e)







(29)


Tc aFb

+ c (30)





20 40400

800

1200

1600

2000

2400


I c (deg

C)




7 Conclusion








Data Availability


Disclosure




Acknowledgments


References


T c (deg

C)

F (kN)0 10 20 30 40 50

50

60

70

80

90

Tc vs FFitted curve






























F (kN)0 10 20 30 40 50

I c (A

)

550

600

650

700

750

800

850

900

Ic vs FFitted curve

(a)

I c (A

)

Ic vs vw

Fitted curve

vw (ms)0 1 2 3 4 5 6 7

800

1200

1600

2000

2400

(b)

I c (A

)

Ic vs as

Fitted curve

as

02 03 04 05 06 07 08 09680

700

720

740

760

780

800

(c)

I c (A

)

Ic vs Ta

Fitted curve


600

800

1000

1200

(d)

I c (A

)

Ic vs SFitted curve

S (Wm2)0 200 400 600 800 1000

740

760

780

800

820

(e)







(29)


Tc aFb

+ c (30)





20 40400

800

1200

1600

2000

2400


I c (deg

C)




7 Conclusion








Data Availability


Disclosure




Acknowledgments


References


T c (deg

C)

F (kN)0 10 20 30 40 50

50

60

70

80

90

Tc vs FFitted curve


































(29)


Tc aFb

+ c (30)





20 40400

800

1200

1600

2000

2400


I c (deg

C)




7 Conclusion








Data Availability


Disclosure




Acknowledgments


References


T c (deg

C)

F (kN)0 10 20 30 40 50

50

60

70

80

90

Tc vs FFitted curve































7 Conclusion








Data Availability


Disclosure




Acknowledgments


References


T c (deg

C)

F (kN)0 10 20 30 40 50

50

60

70

80

90

Tc vs FFitted curve

























































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NumericalAlgorithmsforCalculatingTemperature,Layered Stress ... · 2019. 12. 29. · sag and stress (submodule 2 in Figure 3). e two sub-modules needto berepeatedly calledinthe calculation