V - Theory of Compressive Stress in Aluminum of ACSR

7/29/2019 V - Theory of Compressive Stress in Aluminum of ACSR

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Appendix V Theory of Compressive Stress in A lum inum of ACSR

THEORY OF COMPRESSIVE STRESS IN ALUM INUM OF ACSRNomenclatureA a = total aluminum areaA i = area of strand in rth layerd = strand diameterD = outside diameter of strand layerE = Young's modulusI = length ofwire in one lay lengthrii = number of strand s in z'th layerP = Total tension or compression load in aluminum part ofACSRR - radius of strand helix (radius rom conductor axis to strand axes)T = tension (or compres sion) force in one strand of th layerY = interlayer pressure per unit length of stranda = lay angle of strand helix at strand axisAa = Radial expansion of aluminum portion as a unite = conductor axial strainen = conductor strain during unloading after aluminum goes slackk = curvature of strand due to helicity = sin2a/i?A = lay length

IntroductionConsider an ACSR that has been subjected to some initial loading , following which

the load is reduced . The unloading w ill follow the final stress-strain curve, and atsome point the tension in the aluminum portion w ill go to zero. Since the aluminumwill have experienced some plastic deformation, while the steel will have experiencedlittle if any, the steel will still be under tension when the aluminum "goes slack."As the conductor tension is reduced further, the alum inum layers will tend to

expand away from the core and become loose. The rates at which the individua laluminum layers expand will usually be different, due to differences in the layerdiameters and the lay angles of the strands . A shallow lay angle or large lay ratio w illcause more rapid radia l expansion than a large lay angle or short lay ratio . Because ofthis, there are conditions where an inner layer will expand radially more rapidly than

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the layer above it, and there wi ll be interference between them . When that happens,the layers must press against and strain against each other, and one effect of this is anet compressive stress developing in the pair of layers taken as a unit . If there aremore than two aluminum layers, then three or more layers may lock together in thismanner, depending upon their relative rates of radial expansion following unloading ofthe conductor past the "slack aluminum" point .The effect of this net compressive stres s is to extend the inal stress-strain curve for

the aluminum into the negative stres s range , rather than simply having it stop at zerostress as is generally assumed . In that range, the effective aluminum modulus isdif ferent from that in the positive stress range, being determined by the degree towhich the variou s layers interfere with each others' expansion . We can calculate thisnegative-stress leg of the fina l aluminum curve by analyzing that interference.AnalysisWe will focus on what happens as conductor elongation is further reduced, ater

having reached the po int where the aluminum goes slack. Let us assume that allaluminum layers go slack simultaneously, although this is probably only approximatelytrue . We will begin the analysis by determining how rapidly each layer wou ld expandradially, during this further reduction in elongation, in the absence of interference.Then we will determine how much each layer must be deflected radially from its reeexpansion position in order to eliminate the interference, and yet leave the locked-together set of interfeing layers in mechanica l equ ilibrium . These individualdeflection s will require strains in the strands of the layers in question, and those strainswill determine the layer tensions. The sum of these tensions is the tension in thealum inum , and that may be used to calculate an overall average alum inum stress. Theratio of this stress to the conductor strain that provoked the interference betweenlayers is the aluminum modulus in the negative stress range.Let the decreasing conductor strain that follows the point where the aluminum goes

slack be en . Now the length of strand in one lay length is ,. ' Z = VA2+47r2.i?2 , so fl= p *If there is no interference between layers when en occurs , the strand will experiencenegligible stress, so I will be constant. Lay length A will change , however, because ofen , and the rate at which R varies w ith A is,

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1R _ 2 -AdX 47rV /2-A2 27r,tatl a

But the conductor strain is en = y , so dA = A . en , and the radial expansion ofthe layer is,

Note that is negative , so Ai? is positive. Now , if the outside diameter of the layeris D and strand diameter is d, then

But in a well packed strand layer, D - d = , so , A = and ,

Ifwe identify the ind ividual layers by the subscipt, i = 1,2,3, thenA* = - e e . *

These Ai2i are generally different, even though all layers experience the sameconductor strain, en . The next step in our analysis is to force all interfering layers toshare the same radial deflection, Aa. This w ill require add itional de flections , 6Ri ,such that,

ARi -h 8Ri = A Q (2)for all interfering layers. In reality, these additional de flections take placesimultaneously with the ARi , but it is conveniferit for us to treat them as though theytook place sequentially. Thus , we have treated the AR i as though they took placew ithout change in strand tension, letting that remain zero. The layers expandedw ithout restraining each other . Now we will treat the 8Ri as occurring in the absenceof additional conductor strain, as we force the layers to the position where they are incontact, but do no t interpenetrate.If the conductor elongation is indeed held constant during this second step, then so

is lay length. A, so the rate of change of helix radius w ith respect to w ire length is,dR

_1

_2 -1 _ _L . 1 _ 1dl 2-k ' aV -A2 ' -Jl-j 27r-sina

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So ,K - 2ir-Sina t

But, sina = ,-SO Z = . Also, D - d = , so,= eS fc ' (f (3)

This gives the radial deflection of the layer as a function of the longitudinal strain ofthe w ire of the strands, dl/l . We w ill take the source of dl/ l to be a change in strandtension, Ti , and this tension arises because the interfering layers press against eachother .

If the area of a strand in the ith layer is Ai , thenT -Cf)., sothat . SRi-fzg& -Ti

The tension required to cause delection 8R i is thus,T i = " . SRi (4 )

Now, strand tension ord inarily results in a bind ing pressure rom the layer inquestion upon the layer below . In the present case, however, the inner layer(s) of theinterfering group of layers will be in compression, so they will exert a pressureoutward to meet the inward pressures from above. Within the group of layers that isexpanding as a unit, the radial forces from the various layers must be in equilibrium ;they must add up to zero , since the group is out of contact with layers below andabove . We need to relate these interlayer forces to the strand tensions, Ti .Let the radial force acting between the layers be Y per unit strand length for eachstrand in the layer . Now it can be shown that Y = k .JT,Jwhere T is the tension (or

compression) in the strand along its axis , and k is the curvature of the strand . For ahelix, k - sm2a/R. The total inward radial force rom a layer, per unit length alongthe conductor is thus,

S = . . T = 71 . sin a . tan a . T = f . sin a . tan a Tos a cos a K nd aThe cos a in the denominator occurs because we are now working per unit length ofconductor, instead of strand . These interlayer radial forces must add up to zero, so

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(5)

Substituing (4) into (5), .g _ sinttj-tana, , e-EjAj-sm2 m _ q

and ,

J2 j - sin j . tan Qj 6Ri = 0 (6 )This gives one equation relating the 8R i as unknowns . All other parameters in (6)are defined by the conductor structure. The equation imposes balance of radial forces

within the interfering group of layers .Equation (1) may be substituted into (2), to eliminate the Ai thus:

providing as many equations as there are interfering layers . These equations, with (6)form a se t of simultaneou s equations that maybe solved for Aa and the i - asunknowns.For compactness , de fine C i = 1 *4 . sin3a; . tan a; (8)

TT E i . 3= . sin a ,- . tan a;

(9)Then these equations can be written in matrix form . For illustration we will assumethree interfeing layers. Thus,

1 0 00 1 00 0 1Ci C2 Cz

-1-1- 10

}=..{} (10)5/1 av


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This may be w ritten even more compactly as , \ 1k$i \ "C-6 = enB (11)

'"1-"'.where C is the square matrix, 6 is the column vector, {6Ri 6R2 6R3 Aa} , and Bis the column vector, {Bi B2 B3 0}T.Then, 5 = n.C-1-S (12)where C-1 is the matix inverse of C. Now, rom (4), the tension per strand in the ithlayer is,

Define

ThenHi = g- W i (13)

Ti Hi-SRi (14)

The component of 7 in the direction of the conductor axis is Ti . cos ai , so thecontribution of that layer to conductor tension is,

riiTi cos Qf = riiH i cos a ; .

Thus, the total tension in the group of interfeing layers is,P = riiHi cos a ,- . SR i (15)

i

D eine the row vector, F = {niHiCos ai n2if2C0s Q2 n Hzcos qj ... 0}Then P is given by a quadratic form :

P = en . F . CT 1 . B (16)Let the total aluminum area be A . Then the effective aluminum modulus in thenegative stress region is ,

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(17)

Note that En is defined on the aluminum area, rather than the total conductor area, soit must be multiplied by Ha oefore being used in sag-tension calculations .

This ana lysis assumes that the magnitude of Aa is small compared to the layer radii.In add ition, it is assumed that the the expansion of the aluminum does not becomelocalized, forming a birdcage. Rather, the looseness of the aluminum is taken to beuniform along the conductor . There is evidence rom Nigo l et al that, at least at highconductor temperatures, the buckling of the alum inum may concentrate into localizedbirdcages. In that case, the elfective value of En is reduced. Finally, it is assumed thatthe normal compliance at interlayer contacts is negligible.

1994 May 30 C. B . RawlinsCorrected 1996 October 5

Cautions

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Addendum

The equation between (4) and (5) is not complete . The radial pressure Y isinfluenced not only by T but also by the bending moment M and torque H inthe strand . The complete relation is, from Love , 254 , Eqs (10) and (11) asapp lied to a uniform helix,

Y = Tk-# kt + Mt 2 (Al)We consider the helix to be in equilibrium before being delected by 6R , so weare concerned w ith increments in T , H and M , w ith and r sensibly constant.Thus,

6Y dT dH o dM- = k kt h -r (A2)6R dR dR dR y 1Now ,

dT dT dedR de dRdH dH drdR dr dR

dM dM dKdR dn dR

ldl_IdR

_ dTdR

sin a cos a 2~\

Then ,dl At R = 2 /TsmQdR VA2+47r 9-i22

and1 dl 2ir sin2 aIdR = R

(A3)

(A4)

A lso ,

l = y/\2+4iiR* sina=-7=SL= cosa = / (A6)\/A2+47r 2 VA2+ 47r2 JR2sin2Q 47rit:K _ _ _ = _ _ (,47

(A9)

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Furthermore,dn sin2 adR R? (1-2 sin2 a)

and ,dr sin a cos a= -2 -dR

Collecting all parts,dY = EA sin4 a -2G* sin6 a . co s2 a

K

+ EI

(A10)

(All )

sin4 a . cos2 adR - R? RA R4

Since A = 7r


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' 1 0 0 -10 1 0 -10 0 1 -1c l C2 0

1 0 0 0 -1 -0 1 0 0 -10 0 1 0 -10 0 0 1 -1Cl 2 C3 C4 0

(C3 + C2+Cl)-C-c-C

-c 2C3 + C-C o-c

-c .-cCj+Cj 1

-C , 1

(c4+ C -j -i-C ->-t-c

-c 2 -C3 -C4C4+C3+C! -C3 -C4

"Cl -c 2 C4+C2+C! -C4"Cl -c 2 -C3 C3-C2+C"Cl -c 2 -C3 -C4

A

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Compression Modulus of a Free Strand LayerWhen the aluminum layers of an ACSR go slack, following a tension loading thatleaves permanent set in them, they are able to sustain some compressive load because they

act as helical compression springs . The stif fness, or spring constant for a helical strandcan be calculated from Love1 . This equation actually gives the axial force and torque thatresult from axial deflection 6h and torsional delection 5%. Since we are not interested inthe change in torque , and are assuming that the conductor does not twist, we need onlythe part of (42) that gives,

R = (C cos2d + B sin20) 6h (1)where C is the torsional rigidity of the strand or wire and B is its lexural rigidity. (As 0approaches zero, this equation approaches the formula for the constant of a coiledcompression spring as given byMarks.2) For round wire,

7rd4 E nd4B = El = E ---, and C = G$ = - -.- (2)64 2(1+ */) 32 wThus,

R 1 / COS20 . o \ _ TTd46h l r2 \ l + v + sin20 E-- (3)

Now, per unit length of conductor, I ~ l/sin0 . Furthermore , in a well packedstrand layer, r = nd/6 . Thus, the spring constant for a strand becomes,R 36 sin0 nd4 { cos26 .

9Trd2E smd16 TV2

{ 2d . 2 \h snr0+ u )The spring constant per unit area of strand, that is, its apparent Young 'sModulus, is.

4 R 9 /cos20. 2 \ E .

Note that Love's 6 = tv/2 - a, where a is the angle between the helix axis and thestrand axis. The factor,

lT heMathematicalTheory of Elasticity, by A . E. H . Love , Dover, 1944, 271, Eq .(42) .2Marks'Engineers 'Handbook , 4th Edition, page 486, f 15 .

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COS" + sin20 .sin4 +is nearly constant for practical values ofa . Taking v = 1/3,

a 5 10 15Factor 2 .237 2.199 2.137

Thus, to a good approximation,(5)

There is actually som e additional compliance of the strand that results rom thecompressive stress on it . From Progress Report 16-P-77 , Eq (10a),de = 4 cos3 a

The effective modulus then becom es,

1 S2 .2-4 Sco a 2 .2 T co q

(6)

(7)

(8)

For a 6 strand layer with a - 15 this changes Ee// rom 611,111 psi to 572,304, areduction of 6.35% ." For a 7 strand layer, the change is 4.74%; for 8 strand , 3 .00% ; 9strand, 2.92%; 10 strand, 2.38% .C . B . Raw lins1996 October 9M assena, NY

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V - Theory of Compressive Stress in Aluminum of ACSR

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