Computational handbook for power line engineers

IN THE FIELD OF TECHNOLOGYDEGREE PROJECT VEHICLE ENGINEERINGAND THE MAIN FIELD OF STUDYMECHANICAL ENGINEERING,SECOND CYCLE, 30 CREDITS

, STOCKHOLM SWEDEN 2017

Computational handbook for power line engineersMethodology for design of non-standard transmission line supports

LINNEA SJÖHOLM

KTH ROYAL INSTITUTE OF TECHNOLOGYSCHOOL OF ENGINEERING SCIENCES

Abstract

The high voltage power network in Sweden consists mainly of overhead power lines, thatis conductors suspended in the air by supports. Ideally when projecting a power line, allsupports should be placed equidistantly and the conductors should be suspended at equalheight. When this is not applied, the support placement may have consequences, such asuplifting forces in a support, which should be avoided. The objective of this thesis was toinvestigate both what in�uences the support placement, and what consequences that maycome out of it.

This was investigated using mainly analytical calculations on poles and conductors, butwas also implemented on a speci�c case. When solving the case, support placement andFEM software were also used and compared with the analytical calculations.

It was found that the support placement is in�uenced by both environmental factors; suchas terrain, obstacles and solidity of the ground, as well as how long spans that are possibleto construct. The span length is the distance between two nearby supports and is limitedby sag, that is the de�ection of the conductors, and the strength of the supports.

The sag is dependent on the tension in the conductor, which is dependent on wind andice loads, temperature and creep; a permanent elongation that for certain materials occurover time even if the load is constant. Since the sag will increase over time, and especiallyat high temperatures, the distance between the conductors and the ground will decrease.This extra de�ection must be accounted for when designing the power line and determiningthe span length.

When it comes to support designing, both bending and buckling should be accounted for.The greatest loads the supports are in�uenced by are transferred from the conductors,and therefore are dependent of the span length. An analysis of buckling and bending asfunction of span length was therefore conducted on non-guyed timber pole supports. It wasconcluded that bending stresses; due to wind loads on the support and especially on theconductors, are usually the critical aspect when designing standard power line supports.

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Sammanfattning

Sveriges högspänningskraftledningar består till största delen av luftledning, det vill sägalinor upphängda i stolpar. Vid projektering av en kraftledning är det fördelaktigt att plac-era alla stolpar med lika avstånd mellan varandra och hänga upp alla linor på samma höjd.När detta inte kan tillämpas kan stolpplaceringen få konsekvenser, till exempel lyftkrafteri en stolpe, vilka bör undvikas. Syftet med detta examensarbete var att undersöka vadsom påverkar stolpplaceringen samt vilka konsekvenser den kan ge upphov till.

Detta undersöktes först med analytiska beräkningar på stolpar och linor, och implementer-ades sedan på ett verkligt exempelfall. Vid implementeringen användes även stolpplacer-ingsprogram och FEM, vilket jämfördes med de analytiska beräkningarna.

Stolpplaceringen beror till stor del på omgivningen; bland annat terräng, hinder ochmarkens bärighet, men också på hur långa spann som är möjliga att bygga. Spannlängdenär avståndet mellan två kraftledningsstolpar och är beroende av linans nedhängning samtstolparnas hållfasthet.

Nedhängningen är beroende av dragkraften i linan, vilken i sin tur är beroende av is- ochvindlast, temperatur och krypning, vilket är en permanent töjning som för vissa materialuppstås under en längre tid trots att lasten inte förändras. Eftersom linans nedhängningkommer att öka med tiden, speciellt vid högre temperatur, kommer avståndet mellan linanoch marken att minska. Denna minskning av avstånd måste tas hänsyn till vid projekteringav kraftledningar, och speciellt vid bestämning av spannlängd.

Vid dimensionering av stolpar behöver hänsyn tas till både böjning och knäckning. Destörsta lasterna som verkar på stolparna är lasterna som överförs från linorna, och ärdärför beroende av spannlängden. Med anledning av detta gjordes en analys av böjningoch knäckning av ostagade trästolpar som funktion av spannlängd. Den slutsats somkunde dras från detta var att böjspänningar uppkomna av vindlaster på stolpar och linorär nästan alltid den kritiska aspekten vid dimensionering av vanliga trästolpar.

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Acknowledgements

I would like to express my very great appreciation to Lisa Svanholm, my supervisor atNektab, for always asking the right questions and providing helpful feedback. I would alsolike to thank Stefan Thiel, my second supervisor at Nektab, for sharing of his technicalexpertise and experience with power line projects. My grateful thanks are also extendedto Sören Östlund, my supervisor at KTH, for providing me support and guidance duringthis master thesis work.

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Contents

Nomenclature vi

1 Introduction 1

1.1 Objective . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2 Theoretical background 3

2.1 Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.2 Sag . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2.1 Ruling span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2.2 Slack . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.3 Height di�erence consequences . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.3.1 Uplifting forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.3.2 Insulator swing-out . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.4 Strength of timber poles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.4.1 Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.4.2 Buckling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.4.3 Safety factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3 Implementation: Power line crossing the Göta Canal 15

3.1 Identifying limitations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.2 Support placement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.3 Evaluation of support placement . . . . . . . . . . . . . . . . . . . . . . . . 19

3.4 Uplift solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.4.1 Solution alternative 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 20

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3.5 Support designing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.5.1 Solution alternative 2 - suspension supports with counter weights . . 23

3.5.2 Solution alternative 3 - tension supports . . . . . . . . . . . . . . . . 26

4 Discussion 29

5 Conclusions 31

6 Further work 32

7 References 33

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Nomenclature

Clearance A safety distance to the ground or another area or object, whichshould be kept for all load cases.

Conductors The wires which transfer the current.

Distribution network The local power network.

Guy A wire attached to the ground and the support in order tostrengthen the support.

Insulator An object, usually a chain of plates made of glass, porcelainor composite, which isolates the conductors from the supportstructure.

Overhead lines Power lines in the air, not buried in the ground.

Projection The process in a power line project where computations onconductors, supports, foundations and magnetic �elds are con-ducted.

Sag The de�ection of the conductor.

Span The distance between two transmission line supports.

Strut mount Where the guy wires are attached to the pole.

Supports The structures that the overhead power lines are attached to.

Suspension support A support where the conductor is not �xed, only suspended.

Tension support A support where the conductor is �xed and pre-tensioned.

Transmission network The high voltage networks; the national grid and the regionalpower network.

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1 Introduction

Sweden's power line industry has its roots in the 19th century, and has since then developedinto a nation wide power network with 555,000 km of power lines with connections abroad.Most of the transmission supports in use today were built many decades ago and areapproaching their end of service date. Renewing the existing supports and expandingthe power network will cost several billion Swedish kronor the upcoming years, and thisraises a demand for �nding a more e�ective method when designing the supports. Ane�ective method will save costs both by minimizing the work hours and avoiding oversizedconstructions without compromising safety.

The power network can be divided into three di�erent levels; the national grid, the regionalnetwork and the local network. The national grid, which is the backbone of the Swedishpower network (Figure 1), is owned by the Swedish state through the company SvenskaKraftnät. The grid consists of power lines with voltages of 220 kV and 400 kV respectively,and is used to transport power to the regional network all over Sweden. The regionalnetwork has a voltage of 40-130 kV and is mostly owned by three major power companies;E.ON Elnät Sverige AB, Vattenfall Eldistribution AB and Ellevio AB. It transports powerto larger industries and local networks, which distributes electricity to smaller industriesand all households in Sweden. The local network has voltages up to 20 kV.

Figure 1: Map of the national grid.[1]

Only the national grid and the regional network will be considered in this thesis since theprojection of those kinds of power lines has more mechanical and less electrical aspects

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than projection of distribution power lines. Overhead power lines is the alternative mostlyused when dealing with these high voltage alternating current networks. One of the majorreasons why ground cables usually are not used is that substantial phase shifts betweencurrent and voltage quickly arises when the cable is buried in the ground.[2] A consequenceof this is that great amounts of electricity become useless after short distances. Otheradvantages with overhead lines are that they have twice as long life length before theyneed to be replaced, they are much easier to troubleshoot and repair in case an error arisesand they are much cheaper to build.[2]

All high voltage power line spans need an individual investigation, but some are moredi�cult to �nd a solution to than others; these are called non-standard spans. Non-standard spans usually means that there is some kind of obstacle which makes it necessaryto make the span longer or higher than usual, or that the power line can not go straightand one or several angle points have to be introduced. For this thesis, the de�nition of anon-standard span has been chosen as when a support either has to be modi�ed or replacedby another support type, as a result of non-standard complications. Even if there are onlya small amount of non-standard spans in a power line project, these spans usually requiresa majority of the projection time since their design is complex.

Nektab is a consulting company in the power line business. Since the projection of thesenon-standard spans are complex, Nektab requested a computational handbook in orderto �nd a good projection method and speed up the projection process, and this will beinvestigated in this master thesis report.

1.1 Objective

The objective of this master thesis is to investigate what in�uences the support placementof transmission lines, as well as what consequences non-standard spans can have and howto identify them.

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2 Theoretical background

When projecting a power line, some parameters are speci�ed by the client. This usuallyincludes the start and end point, the path of the power line and the type of supports andconductor. A lot of data such as the material properties of the conductor are thereforealready given, what is to be decided is mostly the placement of the supports and supportmodi�cations, if needed, due to loads transferred from the conductors.

There are two length dimensions used when calculating the loads on the conductors. Thedimensions are wind span, ah, and weight span, av. The wind span, or horizontal loadinglength, is the dimension used for horizontal loads such as wind. The wind load on theconductors is assumed to be transferred equally to the adjacent supports, and thereforethe wind span is de�ned as the mean value of the span length before and after the support,see Figure 2.[9]

Figure 2: De�nition of wind span, ah.

The weight span, or vertically loading length, is the dimension used when calculatingvertical loads such as dead weight of the conductor and ice loads. A support is assumed tobe in�uenced by vertical loads acting on the section of adjacent spans which weigh it down,and therefore is the weight span de�ned as the distance between the lowest line point inthe span before the support and the lowest line point in the span after the support, seeFigure 3.[9] Since the sag, which is the de�ection of the conductor, changes for di�erentload cases, so does the weight span.

Figure 3: De�nition of weight span, av.

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2.1 Loads

The highest loads acting on the conductors are dead weight, ice load, wind load andtemperature load. The dead weight transferred to the support, Gc [N], can be calculatedaccording to

Gc = g ge av, (1)

where g [m/s2] is the gravitational constant, ge [kg/m] is the self weight of the conductorper meter and the weight span av is given in meters. If the conductor is covered in ice, thevertical load will increase. The thickness of the ice layer might vary for di�erent regions,but the ice load transferred to the support at normal wind conditions, QI [N], can usuallybe calculated as

QI = I av, (2)

where the weight span av is given in meters and the ice load per meter conductor, I [N/m],is given by[3]

I = 9.2 + 5.1 10−4 d, (3)

where d [m] is the diameter of the conductor.

The wind load on the conductor transferred to the support, Qwc [N], can be calculated as

Qwc = qh Gq Gc Cc d ah cos2φ (4)

where qh [N/m2] is the dynamic wind pressure, d [m] is the diameter of the conductor, ah[m] is the wind span and φ is the angle of incidence of the wind. The rest of the parametersare design factors which, according to the Swedish standard, should be chosen as Gq = 1.0,Gc = 0.5 and Cc = 1.0 for normal circumstances.[3] When calculating the wind load withice coating the diameter d should be replaced by

dice = d+ 0.036. (5)

The dynamic wind pressure is given by[3] where h is the conductor's height above the

For normal wind: h ≤ 25 m qh = 500 N/m2

h > 25 m qh = 500 + 6 (h− 25) N/m2

For high wind: h ≤ 10 m qh = 800 N/m2

h > 10 m qh = 800 + 6 (h− 10) N/m2,

ground. The wind load per meter on timber poles, Qwpol [N/m2], should be calculatedas[3]

Qwpol = 0.16 qh. (6)

When calculating the total load transferred to the support, Ed [N], the following equationshould be used

Ed =∑

γG GK + Ψ∑

γQ QnK , (7)

where GK denotes permanent loads, such as dead weight, and QnK denotes variable loads,such as ice and wind loads. The permanent loads are usually multiplied with a unit factorsince the dead weight of supports and equipment are assumed not to change. The variable

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loads are usually multiplied with factors greater than one since they vary with weatherand might be both higher and lower than the values used when calculating wind and iceloads. The factors Ψ = 1.0, γG = 1.0 and γQ = 1.3 should be used for ultimate limitstate calculations. Ultimate limit state is when the overhead power line no longer satis�esthe design performance and is usually associated with structural failure due to excessivedeformation, loss of stability, buckling, etcetera.[3]

2.2 Sag

The span length, which is the distance between the supports, is determined by the maxi-mum allowed sag and the strength of the supports. The vertical de�ection of a conductor,y [m], can be estimated according to[7]

y(z) = −Tw

[cosh

(wS

2T

)− cosh

w

T

(S

2− z

)], (8)

where T [N] is the horizontal tension component in the conductor, which is assumed to beuniform along the conductor. The weight of the conductor per unit length is denoted byw [N/m], S [m] is the span length and z [m] is the position along the conductor where thesag is calculated. For highly tensioned wires, which a conductor is assumed to be since itis pre-tensioned, Equation (8) can be simpli�ed as

y(z) ≈ w

2T

(z2 − zS

). (9)

The sag at the middle of the span can then be estimated as

D = y

(S

2

)≈ −wS

2

8T. (10)

A comparison between Equations 8 and 9 is plotted in Figure 4. The comparison wasmade with a standard conductor for the regional network with span a length of 150 m.The conductor type was a 593 mm2 Al59 conductor, that is a aluminum alloy conductorwith a cross sectional area of 593 mm2. No wind or ice loads where added, only pre-tensionat 0◦C. The pre-tension, U [N/mm2], was varied between 1-50 N/mm2. The de�ection de-noted by y (continuous line) was calculated using the cosine hyperbolic expression whereasthe de�ection denoted D (noncontinuous line) was calculated using the parabolic approx-imation. The pre-tension in the conductor is denoted by U . The di�erence between thetwo solutions in Figure 4 is plotted in percent in Figure 5.

From the graphs can be concluded that, depending on the accuracy of the calculations,a 593 mm2 AL59 conductor pre-tensioned to 10 N/mm2 or more can be considered to bea highly tensioned wire. Since conductors usually are pre-tensioned to 45 N/mm2, theapproximation is applicable.

5

Figure 4: Vertical de�ection using Equation (8) versus simple parabolic approximation. A 593 mm2 Al59conductor with span length 150 m was used, with no loading at 0◦C only di�erent pre-tension.

2.2.1 Ruling span

The length dimension used when calculating sag is ruling span, which is de�ned as

an =

√∑S3k∑Sk, (11)

where Sk [m] is the span length for span k = 1, 2, 3, .., n. The ruling span does not includeall spans for one power line, but all spans between two tension supports, that is supportswith tension insulators. A power line consists mostly of suspension supports, that issupports with suspension insulators, in which the conductors are not �xed only suspended,see Figure 6 for visual explanation.

2.2.2 Slack

The elongation of the conductor depends on the tension in the conductor, which varies withwind, ice and temperature loads, as well as creep. Creep is a permanent elongation which,for certain materials, can arise over time even if the load is constant. When mounting thepower line, sag calculations can be made assuming that the only tension in the conductoris the pre-tension. This is however not valid otherwise, various loads and permanentelongation will soon a�ect the sag. The elongation of the conductor is mainly in�uencedby three strain components; thermal strain, elastic strain and long-time creep strain, asshown in Figure 7. The thermal strain varies with temperature changes in the air, but itis also caused by the rise in temperature due to the current in the conductor. The elasticstrain varies with wind and ice loads. The long-time creep strain is a permanent elongation

6

Figure 5: Di�erences between the de�ections plotted in Figure 4.

which will only get larger with time, and occurs due to self weight of the conductor andice loads.

This elongation, or the length di�erence compared to the original conductor length, iscalled slack. Slack does not a�ect the span length, S [m], only the line length, L [m]. Slackis de�ned as

Slack = L− S. (12)

Assuming that all sag components come from slack, the sag calculation can be simpli�edas[5]

Dslack =

√3S(L− S)

8. (13)

2.3 Height di�erence consequences

When placing supports, it is favorable to suspend the conductors at as equal height aspossible. Sometimes however, for example when crossing a railway or river, one supporthas to be higher than the others. In some cases, height di�erences in the power line comesfrom height di�erences in the terrain. Regardless of the reason, the height di�erence maycause problems with nearby supports.

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Figure 6: Example of tension support (strain structure) to the left and suspension suspension (suspension

structure) to the right.[4]

2.3.1 Uplifting forces

One of the consequences can be uplifting forces. Uplift means that the conductors pushthe suspension insulators upwards instead of hanging in the insulator hooks with all forcesacting downwards or horizontally, as they should, see Figure 8.

The uplifting force, LF [kg/phase], can be calculated according to[9]

LF = qe av−40, (14)

where qe [kg/m] is the dead weight of the conductor and av−40 [m] is the weight span at-40◦C. The weight span at minimum temperature is used since that is when the conductorde�ects the least. The weight span at -40◦C can be calculated according to

av−40 = ah +bmax

b−40(av − ah), (15)

where bmax [m] is the sag at maximum temperature after �nal stage of creep and b−40 [m]is the sag at -40◦C. A negative LF means that there are uplifting forces. Whether upliftingforces occur or not is decided by the quota av

ah. See Figure 9 for uplifting forces for normal

ruling spans for regional power lines. Negative uplift forces means that uplift occurs. Whatcan be observed in the graph is that the ratio when uplift occurs is dependent on the rulingspan, but how big the uplift is, is determined by the wind span. From the graph can it beconcluded that uplift can arise when av

ah≤ 0.8.

2.3.2 Insulator swing-out

Another consequence of height di�erences between supports is that the swing-out angle ofthe insulators on the low supports may become too large, see Figure 10. This occurs since

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Figure 7: Elongation of the conductor causing slack.[4]

Figure 8: Normal suspension insulator to the left, suspension insulator subjected to uplift to the right.

the weight of the conductors does not hold them down. The swing-out angle, α [◦], can beestimated as[6]

α = arctanqH ah + Qins

2

qV av +Wins, (16)

where qH [N/m] is horizontal loads (wind load) per meter conductor, Qins [N] is the windload on the insulator set, qV [N/m] is vertical loads (dead weight and ice load) per meterconductor and Wins [N] is the weight of the insulator set including counter weights if any.Which loads and av that should be used is determined by which load cases to investigate.The swing-out angle should preferable not be greater than 30◦, and de�nitely not largerthan 45◦.

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Figure 9: Uplifting forces as function of the quota avah. Negative uplift forces means that uplift occurs.

2.4 Strength of timber poles

The span length depends both on the sag and the strength of the supports. Supportsfor the national grid are usually made of steel frameworks, which can be manufacturedto withstand the stresses they are subjected to. Supports for the regional network areoften made of two timber poles and a beam, which makes the supports limited by thewood properties. Pinewood (Pinus Sylvetris) is the wood used for timber poles in Sweden.According to Swedish standard, these are the design values of resistance in ultimate limitstate:[3] The modulus of elasticity is the material property which should be used when

Maximum bending stress: 27.3 MPaMaximum shearing stress: 2.4 MPa

Maximum compression without risk of bucklingperpendicular to �bers: 3.6 MPaalong �bers: 13.2 MPa

Modulus of elasticity in bending: 10000 MPaModulus of elasticity for Euler-buckling: 4760 MPa

designing timber poles. Which modulus of elasticity that should be used depends on if itis bending or buckling calculations.

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Figure 10: Insulator swing-out.

2.4.1 Bending

The most critical stress that the poles are subjected to is bending stress, but also Eulerbuckling has to be accounted for. The maximum bending stress, |σ(x)|max [Pa], can beestimated according to

|σ(x)|max =|M(x)|Wb(x)

, (17)

whereM(x) [Nm] is the bending moment in the pole and Wb(x) [m3] is the elastic bendingresistance, which is given by

Wb(x) =πd(x)3

32, (18)

where d(x) [m] is the diameter of the pole where the moment is calculated. The greatestmoment forces in the pole occurs at the ground surface for non-guyed structures and at thestrut mount for guyed structures, see marked areas in Figure 11. The bending arise from

Figure 11: Two-legged suspension supports; non-guyed to the left and guyed to the right. The circlesmark where the greatest moment forces occur.

wind loads on the conductors as well as wind loads on the support. The bending moment

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at the ground surface, Mj [Nm], for a non-guyed two-legged structure can be estimated as

Mj =1

2Qwpol l

2 dj + dt4

γQ +Qwci lcγQ, (19)

where Qwpol [N/m2] is the wind load on the pole, l [m] is the length of the pole aboveground in accordance with Figure 11, dj [m] is the pole diameter at the ground surface, dt[m] is the pole diameter at the top, Qwci [N] is the wind load on ice coated conductors, lc[m] is the distance between the ground and the beam where the forces from the conductorsare applied and γQ = 1.3 is the design factor for variable loads.

General bending calculations of non-guyed timber pole supports are shown in Figure 12.The graph shows how bending stress in the poles varies with wind span for di�erent polethicknesses. The graph also shows maximum allowed bending stress, which makes it possi-ble to observe maximum span length for di�erent pole thicknesses, when using non-guyedtimber supports.

Figure 12: Bending stress in 16 m long timber poles as function of wind span.

From the graph can it be observed that the span length between non-guyed timber polesupports is limited to 100-180 m, depending on pole thickness, due to high bending stresses.

2.4.2 Buckling

The buckling load, Pk [N], can be estimated according to Euler's buckling cases, see Figure13. Euler 1 is used for non-guyed structures and Euler 3 is used for guyed structures. The

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Figure 13: Euler buckling, case 1-3.

buckling load according to Euler 1 is given by

Pk,1 =π2EI

4l2, (20)

where E [Pa] is the wood's elasticity modulus along the �bers, I [m4] is the area momentof inertia and l [m] is the length of the pole from the ground to where the load is applied.The area moment of inertia can be calculated according to

I =πdt

4

4, (21)

where dt [m] is the diameter at the top of the pole, which is the smallest dimension of thepole since it is cone shaped. The buckling load for Euler 3 is given by

Pk,3 =2.05π2EI

l2, (22)

where l [m] is the length of the pole from the ground to the strut mount.

General buckling calculations of non-guyed timber pole supports are shown in Figure 14.The graph shows the maximum vertical loads di�erent pole thicknesses can withstand.The solid line in the graph shows an approximation of the load they would be in�uencedby from conductors and equipment, as function of weight span.

From the graph can it be observed that the span length between non-guyed timber polesupports is not limited by buckling for span lengths shorter than 240 m, except for thethinnest poles.

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Figure 14: Buckling in 16 m long timber poles.

2.4.3 Safety factor

All equations above assumes a perfectly straight pole, but timber usually has an initialcrookedness. According to Swedish standard, this initial crookedness does not have to beaccounted for as long as a straight line can be drawn from the bottom to the top of thepole and always be inside the pole. As compensation, a safety factor of 1.5 have beenused against bending and a safety factor of 2.1 against buckling. These safety factors werechosen since 1.5 is the design factor for timber poles according to Swedish standard, and2.1 is the safety factor against buckling used by Svensk energi in their timber pole designcourses.[3],[9]

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3 Implementation: Power line crossing the Göta Canal

A 130 kV power line reaching from Gullspång to Käckestad shall be replaced. The powerline crosses the Göta Canal, which makes the crossing span a non-standard span, andtherefore the crossing and nearby spans will be investigated. The following data are given:

Minimum temperature: -40◦CMaximum temperature: +50◦CHighest system voltage: 145 kVConductor: 3x593 mm2 Al59Support type: Double poleSupport material: Timber (preferred, but only up to 18 m), steel

When designing the new power line there are three �xed points; the canal tower, the supportbefore the section and the support after the section. The last two can be denominated aspoint A and point B, see Figure 15. The �xed points can be summarized as: Between

Point Distance from point A [m]

Point A 0Canal tower 613Point B 1156

Figure 15: Assembled map model displaying the �xed points and the path of the power line.

these points a suitable number of supports should be chosen and placed. Loads actingon the supports and e�ects from height di�erences etcetera should be accounted for. The

15

problem does not include calculations on the canal tower or all line equipment, only thetimber poles. According to the problem description, both steel and composite supports arepossible alternatives. In order to limit this thesis, the support type is delimited to onlytimber pole supports, except for the canal tower.

3.1 Identifying limitations

The �rst step solving this problem is to identify the limitations; except for the �xed pointsand the timber poles. The geographical limitations can be identi�ed from Figure 16 whereseveral areas of interest are marked, see Table 1 for explanations.

Figure 16: Assembled map model including areas of interest.

The greatest limitation mentioned above is the crossing with the Göta canal, which hasa sail less height of 22 m. According to Swedish standard, a power line with 145 kV asits highest system voltage has to have a minimum clearance of 1.5 m above the sail lessheight. This means that the conductors has to have a minimum clearance of 23.5 m abovethe water during maximum sag.[3]

For a power line of this voltage, the clearance between the conductors and the groundshould be at least 7.63 m during maximum sag, according to Swedish standard.[3]

16

Table 1: Areas of interest from the assembled map model shown in Figure 16.

Area of interest Indication inmap model

Meaning

Shoreline - - - - - - Placement of supports within the shoreline shouldbe avoided.

Underground cable - - - - - - , (K) Supports can not be placed on the crossing cable.

Parallel power line - - - - - - A clearance between two power lines is needed.

Car road �� The power may cross the road but has to keepa certain vertical clearance. Supports can not beplaced on top of the road or right next to it.

Crossing with theGöta canal

�� The power line has to keep certain distance to thewater, and since the Göta canal is a fairway thesail less height has to be accounted for.

River, stream �� Supports can not be placed in rivers or streams,and placement close nearby should be avoidedsince the ground there may be less �rm.

3.2 Support placement

An appropriate distance between the supports should be chosen, the so called span length.In order to avoid using unnecessary many supports, which are expensive, a long span lengthis desirable. However, a long span length means greater loads on the supports and largersag.

As starting point for suitable span length, allowed maximum sag was chosen. The followingbasic assumptions were made, which concluded in an allowed sag of 3.87 m, see Figure 17for the calculation principle.

Pole length 17 mBurial deep 2 mSupport top height 0.5 mInsulator length 2 mAverage terrain variance 1 mMinimum ground clearance 7.63 m

The load case which causes the largest sag is maximum temperature after �nal stage ofcreep, that is how much the conductor is estimated to be elongated due to creep duringits life length of 50 years before being replaced. The maximum temperature is given as

17

Figure 17: Principle for calculation of allowed sag.

50◦C, whereof 15◦C is assumed to be heat from the air and 35◦C is assumed to be heatingfrom the current in the conductor. A 593 mm2 Al59 conductor has properties accordingto Table 2.[3]

Table 2: Material properties of 593 mm2 Al59 conductor.

Property Denotation and value

Modulus of elasticity,Initial (before creep): EiL = 53 GPaFinal (after creep): Ep = 60 GPa

Thermal expansion coe�cient: α = 23 10−6 [1/◦C]Creep elongation: εc = 0.4 10−3

Using Equation (13), sag based on slack as function of span length, the graph in Figure 18is obtained.

Using the �gure, 150 m was found to be a suitable span length. Using this, supports wereplaced with this distance in a pole placement program called ICEtow.[11] ICEtow reads inthe assembled map model and terrain data for the power line path, and creates a groundpro�le. Poles can then be placed at a chosen distance and with a chosen height. ICEtowthen calculates and draws the maximum sag of the conductor, as well as weight span, windspan and other useful parameters.

After the �rst placement, the positions were altered to avoid supports too close to roadsand to take advantage of the height di�erences in the terrain. See Figure 19 for the poleplacement after modi�cations. The upper half of the �gure shows the terrain pro�le, thesupports and the conductor. The lower half of the �gure shows the areas of interest fromabove.

18

Figure 18: Calculated sag for di�erent span lengths.

The placement shown in Figure 19 is summarized in Table 3. The span length betweensupport 5 and point B is 149 m.

3.3 Evaluation of support placement

Since there is a big height di�erence between the canal tower and surrounding supports,supports 2 and 4 are at risk of being uplifted. Equation (14) was used to calculate theuplift. The sag for maximum and minimum temperature was calculated in a program calledHoF.[12] HoF is a simple calculation program which uses ruling span and type of conductoras input and gives tension and sag for di�erent load cases as output. The uplifting forcecould then be calculated to +10 kg/phase for support 2 and -150 kg/phase for support4. This means that there are uplifting forces in support 4 but not in support 2, even ifit is very close to getting uplifted since the uplifting forces is close to 0, that is the limitbetween uplift and no uplift.

Another consequence of the height di�erence between the supports is that the insulatorson support 2 and 4 may get too big swing-out angles since the conductors do not holdthem down, but before looking into that problem, solutions to the uplift-problem shouldbe investigated.

19

Figure 19: Preliminary support placement in ICEtow.

3.4 Uplift solutions

There are three di�erent alternatives when solving an uplift problem. Alternative 1 is tomove the support or make it higher until the uplift disappears. This is the cheapest and bestalternative, if it is possible to solve the problem this way. Alternative 2 is to add counterweights at the end of the insulators. This alternative weighs the insulators down but alsointroduces vibration-sensitive supports into the power line. Alternative 3 is to change thesuspension insulators to tension insulators. This alternative is an expensive option, buta too great swing-out angle is no longer an issue since the insulator is pre-tensioned in ahorizontal position.

3.4.1 Solution alternative 1

The problem was approached by trying to use alternative 1, moving or changing the heightof the supports. Since timber poles usually are not made longer than 18 m, moving asupport was the only option. Support 4 was moved in ICEtow, and since it was placed soclose to support 5 that support was removed. The resulting placement is shown in Figure20. The span between the canal tower and support 4 is around 350 m, which makes thebending stress in the support too big due to wind loads on the conductors.

20

Table 3: Preliminary support table.

Support Previous Support height ah av Burial Notesspan length (m) (m) (m) (m) deep (m)

1 198 16 175 171 2.1

2 151 17 208 134 2.1

3 264 36 + top 256 438 Canal tower

4 248 16 199 93 2.0

5 149 16 149 144 2.0


The second alternative, counterweights, was investigated. Support 2 was not subjected touplift, but since the quota av

ahwas small that it was found appropriate to move the support

further away from the canal tower, see Figure 21. The new placement altered the weightand wind spans, which resulted in uplifting forces of 180 kg/phase for support 4. A counterweight of 200 kg/phase was added to support 4. When investigating the swing-out anglesof the insulators with counter weights, the swing-out angle at minimum temperature andwind was 43◦for support 4. When using counter weights, it is important that the swing-out angle never exceeds 45◦. Since the safety factor in this case is almost non-existent, aheavier counter weight is recommended. Counter weights are manufactured up to 450 kgbut it is usually not recommended to use counter weights heavier than 200 kg/phase dueto the vibration sensitivity they introduce into the system.[10]


The third solution alternative, tension supports, was investigated. Two options were pos-sible; either to change both support 2 and 4 to tension supports or to only change support4. If only support 4 is changed to a tension support, support 2 has to be moved in orderto avoid too big swing-out angles.

3.5 Support designing

The supports have to be designed with respect to both buckling and bending. UsingEquations (20) and (21), the minimum top diameter, with respect to buckling, of the poleson a non-guyed structure can be estimated as

dt =

(16 Pk,1 l

2

π3 E

) 14

. (23)

21

Figure 20: Support placement in ICEtow when trying to solve the uplift-problem by moving support 4.

The minimum top diameter was calculated for all supports. The buckling load, Pk,1 [N],was calculated including dead weight of and ice load on conductors, weight of beam andinsulators, one third of the weight of the timber pole, counter weights if any and the weightof a lines man with a weight of 1000 N. The lines man is accounted for in case the supportshould need maintenance by a person climbing the support. A safety factor of 2.1 againstbuckling was used, but even so the minimum top diameter never exceeded 17 cm. The topdiameter of a 17 m high pole is usually greater than 18 cm, which means that the bucklingload is not the critical load when designing the supports.

The minimum pole diameter at the ground surface, dj [m], can be calculated by transposingEquation (19) according to[9]

dj =

(32 MJ

fmd π

) 13

, (24)

where fmd = 27.3 MPa is the maximum bending resistance and MJ [Nm] is the bendingmoment calculated as

MJ = l HF +l

2HV , (25)

where l [m] is the pole length above ground, HF [N] is the wind load transferred from theconductors and HV [N] is the wind load on the pole. The loads can be calculated accordingto[9]

HF = 1.5 Qwci γQ (26)

22

Figure 21: Support placement in ICEtow when trying to solve the uplift-problem by moving support 2and adding counter weights on support 4.

andHV = Qwpol l γQ, (27)

where Qwci [N] is the wind load on ice covered conductors, Qwpol [N/m] is the wind loadon the pole per meter pole and γQ = 1.3 is the design factor for variable loads.

3.5.1 Solution alternative 2 - suspension supports with counter weights

The minimum pole diameter at the ground surface was calculated according to Equation(24) with a safety factor of 1.5, see Table 4.

Table 4: Minimum allowed pole diameter at the ground surface, safety factor 1.5.

Support 1 2 4 5

Minimum dj [cm] 33.6 35.9 34.2 32.4

Based on this calculation, poles of type S were assumed, that is 36 cm as pole diameter atthe ground surface and 23 cm as top diameter for 17 m long poles and 35 cm respectively

23

22.5 cm diameter for 16 m long poles. The strength of the supports where investigated.A simple analytical calculation was made, where only wind loads on conductors and poleswere considered. A pole was assumed to be in�uenced by the loads from 1.5 conductorsand the wind load on the pole was assumed to be an evenly distributed pressure on a �atsurface as wide as the pole. No consideration to the e�ect of an assembled structure wasmade, that is the support was simpli�ed to be two individual poles with no link in-betweenthem.

The analytical solution was then compared with a FEM analysis using the program AnsysWorkbench.[13] The analysis was conducted in the Static Structural module with a simpleFEM model; two poles with pinewood properties and a beam with three idealized insulatorsbetween them. The beam was mounted to the poles with a bounded connection. The windpressure on the poles and the forces applied at the bottom of the insulators were calculatedaccording to Chapter 2.1. The poles were �xed supported at the bottom, which correspondsto the burial deep of the support. The setup with boundary conditions is shown in Figure22.

Figure 22: FEM model and boundary conditions used in Ansys Workbench.

All loads were calculated according to Chapter 2.1. Assuming non-guyed supports andice coated conductors, the bending stress in the supports at the ground surface wherecalculated according to Table 5. The maximum calculated bending stress in the poles isspeci�ed both in MPa and percentage of maximum allowed bending stress in the table. Thepercentage is calculated with a safety factor of 1.5 with respect to the maximum allowedbending stress of 27.3 MPa.

The stress distribution in the support is shown in Figure 23. The largest stresses in the

24

Table 5: Maximum bending stress in the poles, calculated with a �rst order analytical solution and withFEM.

Support 1 2 4 5

Analytical Bending stress (MPa) 18.5 23.2 21.8 16.3solution Bending stress (%) 102 128 120 90

FEM Bending stress (MPa) 17.7 21.4 21.3 15.6solution Bending stress (%) 97 118 117 86

Di�erence in solutions (%) 4 8 2 4

poles occur where the beam is attached to the poles. Since the beam is connected with abonded contact, unrealistic high shear, normal and bending stresses occurs in the contactregion due to how the support is modeled and where the forces are applied. The contactregions between the beam and the poles are small since not that many elements on thecurved surfaces (the poles) are in contact with the �at surface (the beam). This causeslarge normal stresses since they are dependent on the force in relation to the area wherethe force is applied. How the force is transferred to the poles are not realistic either; inreality the forces will be applied on all sides of the poles and not just on one side of thepoles, which also makes the stresses too large in the contact area.

This local stress maximum can be neglected when investigating the stresses in the pole,and therefore only a relevant section of the pole has been selected for the stress analysis,see the sub�gure to the right in Figure 23.

Figure 23: FEM model with equivalent von Mises stress distribution to the left and maximum principalstress in relevant section of pole to the right.

25

Since support 1, 2 and 4 exceeds 95% of the maximum bending resistance, these supportsshould be strengthened with guy wires, that is if this solution alternative should be chosen.Suspension supports are usually guyed according to Figure 24 with 30◦ ≤ α ≤ 45◦ andγ ≈ 26.6◦ (slope 1 horizontal, 2 vertical). [10]

Figure 24: Drawing of guyed suspension support, top view to the left and front view to the right. Thesupport has three guy wires per pole; two guys on the outside of the support, directed with an angle αfrom the beam, and one guy wire directed almost straight towards the other pole.

For guyed structures, the bending moment at the guy mount is usually the most critical.Since the bending moment becomes greater the further away from the beam that the guywires are mounted it is desirable to mount the guy wires as close to the beam as possible,but it is also important that the conductors can not touch the guy wires at any load case.According to Swedish standard, a 145 kV conductor has to have a clearance of at least0.6 m to an earthed guy wire.[3] Calculating maximum swing-out of the insulators, theminimum distance y [m] can be estimated according to Figure 25.

The minimum diameter at the stay mount can then be calculated according to[9]

ds =

(32 MS

fmd π

) 13

, (28)

where the bending moment at distance s, MS [Nm], is calculated according to

MS = y

(HF +

HV

2

), (29)

with HF and HV according to Equations (26) and (27). Using these equations, guyedsuspension supports of type S were found to be su�cient for support 1, 2 and 4.

3.5.2 Solution alternative 3 - tension supports

Tension supports need to be able to withstand loading from both sides of the support aswell as one-sided line loads, since the conductors are mounted and pre-tensioned on oneside at a time. Tension supports are usually constructed according to Figure 26.

26

Figure 25: Calculated maximum insulator swing-out with and without safety distance to guy wires.

The minimum thickness of the poles are calculated similarly to the poles in the guyedsuspension support in Chapter 3.5.1, but in this case the design loads are not the windloads but the maximum tensile loads from the conductors. In addition to being able towithstand tensile loads from both one side and both sides of the support, the supportshould also be able to withstand a reduction of conductor tension of 60% in one of theconductors. The minimum diameter at the stay mount can be calculated according toEquation (28), where the bending moment at distance s, MS [Nm], is calculated accordingto[3],[9]

MS = y HF . (30)

The horizontal force, HF [N], can be calculated according to

HF = RA (T0 γG + (Tice − T0) γQ), (31)

where RA = 1.8 is a factor to account for the reduction in conductor tension of 60%, T0[N] is the initial tension in the conductor at 0◦C, Tice [N] is the tension at 0◦C, ice loadand no wind, γG = 1.0 is the design factor for permanent loads and γQ = 1.3 is the designfactor for variable loads. Using these equations, the minimum diameter at the stay mount,ds, was calculated to be 16-23 cm depending on the distance between the guy wires andthe beam, which was assumed to be 0.1-0.3 m. Using poles of type S with a top diameterof 23 cm, this is not a problem.

The supports also needs to be designed against buckling. Euler 3 is used for guyed struc-tures, wherefore the pole diameter is given by

ds =

(2 Pk,3 s

2

π3 E

) 14

. (32)

27

Figure 26: Drawing of tension support, top view to the left and front view to the right.

The buckling load, Pk,3 [N], is given by[9]

Pk,3 = 0.5 RT γG +MT γQ +QT

3γG + LT + VS , (33)

where RT [N] is the weight of the beam, MT = 1000 N is the weight of a linesman workingon the support, QT [N] is the weight of one pole, LT [N] is the load from the conductorsand VS [N] is the vertical tension load from the guy wires. The load from the conductorsis given by

LT = RA av(qe γG + qio γQ), (34)

where RA = 1.8 is a factor to account for the conductor tension reduction of 60%, av [m]is the weight span, qe [N/m] is the dead weight of the conductor per meter and qio [N/m]is the ice load at no wind per meter conductor. The vertical tension load from the guywires is given by

VS =HS

cosα tan γ, (35)

where the angles α [◦] and γ [◦] are in accordance with Figure 26 and the horizontal tensionload from the guy wires, HS [N], is given by

HS =s+ y

sHF . (36)

The lengths s [m] and y [m] are given in Figure 26 and the horizontal tension force HF [N]is according to Equation (31).

Using a safety factor of 2.1 against buckling, a minimum top diameter of 24 cm should bechosen to avoid buckling. This is possible by choosing poles of type S+2.

28

4 Discussion

When comparing the two graphs with general calculations of bending versus buckling fornon-guyed suspension supports, Figures 12 and 14, it can be observed that buckling usuallydo not occur for these type of supports. If buckling occurs, the maximum bending stresswas always reached for a shorter span length than in the buckling case. The conclusionthat can be drawn from this is that bending due to wind loads is the critical load whendesigning non-guyed timber pole supports. That is, if the weight span is not extremelylong in comparison with the wind span.

There was no similar general bending-buckling comparison conducted on guyed suspensionsupports in this thesis. The reason was that the moment forces, and so too the tensionin the guy wires, are dependent on the position of the strut mount, which is dependenton insulator swing outs and internal clearances at the support. No general conclusion cantherefore be drawn about these type of supports, but from the calculations made in theGöta canal case, the conclusion can be drawn, that the moment forces at the strut mountare usually the most critical aspect when designing guyed timber pole suspension supports.It seems very likely that this would be the critical aspect as well, since a greater distancebetween the strut mount and the beam causes higher bending stresses, but this distancecan not be too small since the guy wires may not be too close to the conductors.

There was no general bending-buckling comparison conducted of tension supports either,only the calculations made in the Göta canal case. A di�erence between guyed suspensionsupports and tension supports, regarding the guy wires, is that the strut mount can bea lot closer to the top in a tension support since the insulators are horizontal instead ofvertical, and especially since the conductors are not close to the poles. The guy wires aredesigned to counteract the moment forces at the ground surface, so bending stresses in thatarea is not very likely to be critical. Since the distance between the beam with insulatorsand the strut mount usually is very short, it is not very likely that bending stresses atthe strut mount would be critical. Buckling on the other hand seems more likely to bethe crucial factor when designing tension supports. Since the guy wires are designed tocounteract not just wind loads, but the pre-tension to keep the entire line in tension, untilthe next tension support, the tension in the guy wires are a lot greater than in the onesused in suspension supports. This adds an extra vertical load which can be critical withrespect to buckling.

The safety factors used in this thesis may seem unnecessary high, especially since it canbe assumed that some kind of safety margin already is included in the design values. But,timber poles are in reality not completely straight, they have twigs and other imperfections.Assuming that the material would behave as perfect as a man-made material is a majorsimpli�cation, which makes it necessary to have a high safety factor when calculating thetimber pole supports.

When comparing the �rst order analytical and the FEM solution of bending stress in thepoles, the analytical solution always gave a somewhat larger value. This can be due to thatthe beam connecting the two poles, which makes the support sti�er, was not accountedfor in the analytical solution. Because of this, the analytical solution can be assumed tobe conservative, that is, it will always give a worse result than reality. The di�erence

29

between the two solutions were not that big though, only 2-8%, which indicates that theanalytical solution is not that far from reality, which is good since too over-sized structuresare not desirable. The conclusion that can be drawn from this is that �rst order analyticalcalculations is su�cient for timber pole calculations. This is very useful since Nektabpresently does not use any FEM computation software for computations on poles, theyonly have software for computations on conductors.

Regarding the Göta canal case, both uplift solution alternative 2 and 3 are feasible solu-tions, but solution 3 are in several aspects the best solution. Firstly, the counterweightsintroduce vibration sensitive supports into the system. Secondly, support 2 and 4 can re-sult in very big insulator swing outs if not very heavy counter weights are chosen. And last,but not least, choosing tension supports allows that part of the power line to be mountedseparately. When it comes to construction with crossings, this can be vary favorable sincethe rest of the power line are not depending on when it is possible to build the crossingspan. In this case for instance, it would be preferable to build the crossing span duringwinter when Göta canal is not in use. If suspension supports was used, the rest of thepower line, until the next tension support, would have to be constructed at the same time,but with tension supports this is not a requirement.

30

5 Conclusions

When projecting a power line, one should always aim to place all supports at equal distancesand suspend all conductors at equal height. Signi�cant di�erences in height or lengthbetween spans and supports causes problems.

The risk of uplifting forces arises when the quota avah≤ 0.8. Uplifting forces occur for

smaller quotas for longer spans and for larger quotas for shorter spans, that is the risk ofuplift is greater for power lines with short spans.

Bending stress is usually the critical factor when designing suspension supports. Non-guyed timber pole supports are usually limited to span lengths of 100-180 m, dependingon pole thickness. Buckling is usually not an issue for these type of supports.

For tension supports on the other hand, buckling is more critical. The supports are builtto withstand the bending loads, but this increases the vertical loads which causes buckling.

Regarding the Göta canal case, the best solution to the problem is to change the supportsclosest to the canal tower to tension supports. This solution eliminates both the uplift andthe swing-out problem, as well as simplifying the construction process.

31

6 Further work

This master thesis mostly looked into the strength and limitations of non-guyed timberpole supports. It would also be interesting to do a similar analysis of guyed timber polesuspension supports as well as tension supports. The question is how long span lengthsthat are possible as function of weight span, wind span and insulator type. That is sincethe bending stress at the strut mount is dependent of the distance between the strut mountand the beam, which is dependent of the insulator swing-out.

It would also be interesting to do comparisons of di�erent supports, both support typesand materials. Today a lot of di�erent alternatives are available, but when it comes to theregional network almost all supports that are built in Sweden are made of timber. Timberis the cheapest pole material, and is therefore dominating the regional support market.

Due to this, it would be interesting to investigate the supports also from an economicalaspect, with mechanical aspects as the starting point. Timber is assumed to always be thecheapest material, but the more guy wires, guy wire foundations, counter weights and soon that are added to the support, the more expensive it becomes. The question is, whenwould it be less expensive, or equally expensive, to change to another support type ormaterial, for example composite poles made of �berglass reinforced polyester.

Another aspect is the strength of other support types and materials. How long spanlengths would be possible to have, and how much could the number of supports thereforebe decreased? Since poles made of manufactured materials can be longer than naturallygrown timber, the sag will not limit the span length as much as for timber supports. Thisis another aspect which in�uences how cheap timber poles are compared to other supporttypes. What can be added then are transportation, construction and maintenance costs,in order to look at the full life time perspective.

32

7 References

1. Svenska kraftnät. Stamnätskarta. 2016.http://prod.svk.se/drift-av-stamnatet/stamnatskarta/ (Accessed 2017-03-09)

2. Svenska kraftnät. Technology. 2016.http://www.svk.se/en/grid-development/the-construction-process/technology/ (Ac-cessed 2017-03-09)

3. SEK Svensk Elstandard. Overhead electrical lines exceeding AC 45 kV. 2nd edition.Stockholm: SIS förlag, 2007. (Swedish standard SS-EN 50341)

4. Douglass, Dale and Springer, Paul. Sag-tension Calculations - A CIGRE Tutorial

Based on Technical Brochure 324. Power point presentation. 2013.https://www.slideshare.net/aziramuda/sag-tension-calcsohltutorial (Accessed 2017-08-23)

5. Douglass, D.A and Thrash, Ridley. Sag and Tension of Conductor. 2006.https://www.slideshare.net/iqbal_haqi/sag-andtensionofconductor (Accessed 2017-05-11)

6. Kiessling, F., Nefzger, P., Nolasco, J. F. and Kaintzyk, U. Overhead Power Lines:

Planning, Design, Construction. Springer. 2014.

7. Bowden, Gordon. Stretched Wire Mechanics. Stanford Linear Accelerator Center,USA. 2004.

8. Necks electric. Product catalog. Bollnäs: Bok & Tryck AB, 2014.

9. Svensk energi AB. EBR Mekanisk dimensionering. Course compendium. Stockholm.2006.

10. Thiel, Stefan; Projector at Nektab. Interviews April 4, 21 and June 2, 2017.

11. EFLA. ICEtow, Version V8i. Computer software. Reykjavik, Iceland. 2008.

12. Ingenjörs�rman Leif Andersson. Dataprogram för beräkning av hängkablar och friled-

ningar, Version 2014-01-30. Computer software. Bålsta, Sweden. 2014.

13. ANSYS, Inc. ANSYS Workbench, Release 17.0. Computer software. Canonsburg,USA. 2017.

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Documents

Computational handbook for power line engineers