Sag and Tensions in Overhead Lines
11-08-2016, 07:04 AM (This post was last modified: 11-08-2016 09:35 AM by Ángel Martin.)
Post: #1 Ángel Martin Senior Member Posts: 1,046 Joined: Dec 2013
Sag and Tensions in Overhead Lines
Sag and Tension in Overhead Lines.[ CAMELA ]
From the author’s Engineering Collection, included in the ETSII5 module (ETI5 on the CL Library)

This program calculates the sag and tensions at the supports of an overhead line cable, with or without equileveled conditions. The conductor adopts a catenary shape in either case, but the different geometric conditions require different methods to resolve the unknowns.

Besides the slope, posts height and span length the input data includes the minimum (perpendicular) distance to the ground, which occurs at the point of maximum deflection of the cable, thus limiting the maximum sag.

Let V = span length; H = posts height; m = tan(theta) = inclination slope; d = minimum distance (safety)

For level spans the maximum sag occurs at its middle point, with a symmetric catenary curve centered there (xf = 0). Thus the coordinates of posts are Xa = -L/2 and Xb = L/2.

The curve equation in that case is:

(H – d) = [1 + ch(-V/2alph.) ]

For unlevel spans, the following two equations are used to calculate the values Xa and Xf, the coordinates of the post at lower slope and the point of maximum sag:

(1) V m = (xf /ash m) { ch (A) ch (B-1) + sh (B) sh (A) }
(2) f = m(xf-xa) + (xf/ash m) [ ch(A) – ch(ash m)]

Where: A = Xa ash m / Xf and: B = V ash m / Xf; and f= H – d /cos(theta) is the maximum sag.

Solving this system for Xa and Xf determines the rest of unknowns, such as Xb = V –|Xa|; The resolution is done numerically using “SLV2”, a built-in routine to solve non-linear systems of two equations.

The program output includes both geometry and stress results. The geometry results are the X-coordinates of each post referred to the point of maximum deflection (x=0), and the alpha parameter of the catenary curve.

The stress results require the unitary weight of the cable (q) , returning the horizontal tension in the supports (Ta, Tb) and maximum sag point (T0), as well as and the total length of the cable (L). The expressions used are derived from the basic catenary, as follows:

T = q alph. ch x/alph. and: L = [sh xb/alph. - sh xa/alph. ]

Example.

Calculate the tensions in the supports for a power line with 100 m span length, with 42 m height posts and a minimum perpendicular distance to ground of 10 m. Do both cases of level span and 20 deg inclined span to compare the results. The unitary weigh is 10 kg/m.

Code:
 Level span      xa = -50 m      xb = 50 m       alpha=43.5470     u = 10 kg/m     TA=755,4702 N   TB=755,4702     T0=435,4702 N    L=123,4667 m Inclined 20 deg XA=-42,8106 m   XF=8,7983 m     alpha = 44.2816     u = 10/kg/m     TA=666,3870 N   TB=866,3870 N   T0=442,8156 N    L=124,2657 m
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