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cable loads

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Simply stated, it is statics, assuming the wire sags in a parabolic form is easier to do manually, or using a catenary curve for a more exact analysis. The loads on the wire, temperature and initial tension all affect the resultant load on the tower.
ALCOA's SAGTEN program is one of several programs that deal with wire sag and tensions. There are also many books addressing the subject...Structural Engineering Handbook, by Gaylord & Gaylord; The Catenary & Overhead Lines, by Sagline Inc. (914-248-8001?); RUS Bulletin 1724E-200 to name a few.
For a level span, the tension at the point of support is equal to the horizontal tension plus the unit weight of the conductor times the sag in the conductor (T = Th + Wc*S).
That's a five minute answer to a week long seminar.
 
Alcoa's SAGTEN is currently named SAG10(tm) and below is a link to an online paper with all the catenary equations of unlevel/level spans.

These are just half the story in transmission lines. The most important part is the cable behavior (stress-strain curves and creep rate). A lot of people assume a linear model of aerial cables which often gives erroneous results. Alcoa has historically used 4th order polynomials for the stress-strain curves.

If you want more information on sag10, or the alcoa graphical method please email me, or visit
SAG10-White papers

John Alexiou
Product Development Engineer - SAG10/Vibrec
Alcoa.
 
My 2 cents on the parabolic/catenary.

"Just an opinion"

A cable must be completley flexable to hang in a true catenary. Line conductors a very stiff and do not hang in a true catenary. A chain would probably be very close to a true catenary. I believe catenary is latin for chain? So, what curve does a cable hang in? My guess is somewhere between a parabola and a catenary. I have not seen any reasearch on this so I ass_u_me that the catenary is close enough.

Measure with a micrometer, mark with chalk, cut with an ax.
Kevin
 
Actually if you consider the shape of a beam under its own weight w, the shape is still a catenary {Hint: Catenary constant is [tt]a=(E I/w)^(1/3)[/tt]}. Now add axial tension T to it, and it is still a catenary. The catenary constant is now whatever "a" satisfies the equation
[tt]E^2*I^2+2*E*I*T*a^2+T^2*a^4-a^6*w^2=0[/tt]. So the hanging cable with flexular rigidity is no different from a beam with axial loading.

Ref. T=tension, w=linear weight, E=modulus of elasticity, I=2nd moment of area,

John Alexiou
Alcoa Design Engineer
MSME Georgia Tech
 
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