# Step-by-Step Sag and Tension Computation using Ruling Span Theory

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In the previous post “What is Ruling Span?”, we understood that the sag characteristics of the ruling span sets the sag characteristics of every span in the section.

In this post, you learn how to apply the formulas and equation associated with Ruling Span Theory. However, the following paragraphs will explain when to use and when this theory is not applicable.

WHEN NOT TO USE RULING SPAN THEORY?

In actual construction, the stringing section is not a single dead-end span but it consist of series of unequal spans between the rigid dead-end supports. During stringing, the conductors can freely move between spans because it is  temporarily supported by free-wheeling rollers. Under this conditions, the conductor behaves according to the “Ruling Span Theory”.

However, when the conductor are tied or fixed at the rigid support insulators (pin insulator/post insulator) the conductor can no longer move freely between spans. The spans, in a sense, become dead-end spans or mechanically independent spans. Then, the future behavior of the conductor under various loading conditions will not follow the “Ruling Span Theory”. Its behavior can be determined based on calculation procedures used for single dead-end spans. The difference in horizontal tension between span will then cause longitudinal movement or flexing in the supporting structures or insulators.

On the other hand, when the conductor is fixed at suspension insulator or strings, the difference in horizontal tension between spans will be compensated by the longitudinal or transverse movement or swing of the strings. Hence, it is safe to assume that the conductor will behave according to ruling span theory.

THEORY:

Since the tension in all of the suspension spans is equal (or nearly so),  the sag in any of the corresponding suspension spans can be calculated using the following formula:

$\fn_phv&space;D_{n}&space;=&space;D_{RS}&space;\left&space;(&space;\frac&space;{S_{n}}{S_{RS}}&space;\right&space;)^{2}$

where:

Dn = Sag of the nth span

DRS = Sag using the Ruling Span Length

Sn = Length of the nth span

SRS = Ruling Span Length

According to CIGRE Brochure 324, sags errors from using ruling span method may be significant where there is relatively large variations of the span length and where conductors are operated at temperatures above 100oC.

A more accurate sag calculation of the sag can be done with numerical analysis. There are several computer programs that can do this, two of which are the Sagsec and PLS-CADD. These programs lets the user model the full line section and calculates, using finite element analysis, the forces on each insulator and the resulting sags of the line section can be determined.

EXAMPLE CALCULATION OF RULING SPAN:

A tensioning section of a power line consist of six spans with the length 350, 200, 450, 500 and 325 meters. The conductor 304-AL1/49-ST1A ( ACSR 300/50) is installed with initial horizontal tension of 17685 N at 10oC .  Compute the initial and final sag of every span.

STEP 1: Compute the Ruling Span Length

Formula:                    $\fn_phv&space;S_{R}&space;=&space;\sqrt{\frac{\sum&space;S^{3}}{\sum&space;S}}&space;=&space;\sqrt{\frac{S_{1}^{3}+S_{2}^{3}+...&space;S_{n}^{3}}{S_{1}+S_{2}+...&space;S_{n}}}$

$\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_phv&space;S_{RS}=&space;\sqrt{\frac{350^{3}&space;+&space;200^{3}+&space;450^{3}+275^{3}+500^{3}+325^{3}}{350+200+450+275+500+325}}&space;=&space;\sqrt{\frac{322.125x10^{6}}{2100}}&space;=&space;391.65\;&space;meters$

STEP 2: Compute initial sag of the Ruling Span

Formula:              $\fn_cm&space;\small&space;\fn_phv&space;D_{RS}&space;=&space;\frac{H}{W}\left&space;(&space;cosh&space;\frac{S_{RS}W}{2H}&space;-&space;1\right&space;)$

$\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_phv&space;D_{RS}&space;=&space;\frac{17685}{12.115}\left&space;(&space;cosh&space;\frac{391.65\;x\;12.115}{2\;x\;17685}&space;-&space;1\right&space;)&space;=&space;13.155\;meters$

STEP 3: Compute initial sag of all other spans

Formula:                                      $\fn_phv&space;D_{n}&space;=&space;D_{RS}&space;\left&space;(&space;\frac&space;{S_{n}}{S_{RS}}&space;\right&space;)^{2}$

$\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_phv&space;D_{@350}&space;=&space;13.155\left&space;(&space;\frac&space;{350}{391.65}&space;\right&space;)^{2}&space;=&space;10.5&space;\;meters$$\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_phv&space;D_{@200}&space;=&space;13.155\left&space;(&space;\frac&space;{200}{391.65}&space;\right&space;)^{2}&space;=&space;3.43&space;\;meters$$\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_phv&space;D_{@450}&space;=&space;13.155\left&space;(&space;\frac&space;{450}{391.65}&space;\right&space;)^{2}&space;=&space;17.37&space;\;meters$$\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_phv&space;D_{@275}&space;=&space;13.155\left&space;(&space;\frac&space;{275}{391.65}&space;\right&space;)^{2}&space;=&space;6.49&space;\;meters$$\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_phv&space;D_{@500}&space;=&space;13.155\left&space;(&space;\frac&space;{500}{391.65}&space;\right&space;)^{2}&space;=&space;21.44&space;\;meters$$\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_phv&space;D_{@325}&space;=&space;13.155\left&space;(&space;\frac&space;{325}{391.65}&space;\right&space;)^{2}&space;=&space;9.06&space;\;meters$

You can observe in the data above that spans shorter than the ruling span have less sag while spans longer than the ruling span have greater sag. It is important to determine the actual sag of the longer spans for calculations of the clearances.

STEP 4: Compute the final horizontal tension of the Ruling Span

In this step, we need to use the conductor state change equation:

$\fn_jvn&space;\small&space;\fn_phv&space;\fn_phv&space;H_2^{3}&space;+H_2^{2}&space;\left&space;(&space;\frac{(W_{1}S)^{2}AE}{24H_{1}^{2}}&space;-&space;H_{1}&space;+&space;(t_{2}-t_{1})&space;\alpha&space;AE\right&space;)&space;-&space;\frac{(W_{2}S)^{2}AE}{24}&space;=&space;0$

The above equation can be manipulated for easy substitution of values as shown below:

$\fn_cm&space;\small&space;H_{2}^{3}&space;+&space;A\;&space;H_{2}&space;=&space;B$

where:

$\fn_cm&space;\small&space;\fn_cm&space;\small&space;A&space;=&space;C_{2}\:(t_{2}-t_{1})&space;\:+\:&space;\left&space;(&space;\frac{W_{1}S\:C_{1}}{H_{1}}&space;\right&space;)^{2}&space;-&space;H_{1}$

$\fn_cm&space;\small&space;B&space;=&space;\left&space;(&space;W_{2}&space;\:S\:C_{1}&space;\right&space;)^{2}$

$\fn_cm&space;\small&space;C_{1}&space;=&space;\sqrt&space;{\frac{EA}{24}}$

$\fn_cm&space;\small&space;C_{2}&space;=&space;\alpha\:EA$

Using the new form of the conductor state equation, we have:

$\fn_cm&space;\small&space;\fn_cm&space;\small&space;C_{1}&space;=&space;\sqrt{\frac{(77,000)(353.7)}{24}}&space;=&space;1,065.264$

$\fn_cm&space;\small&space;C_{2}&space;=&space;(18.9&space;\:x\:&space;10^{-6})\:(77,000)\:(353.7)&space;=&space;514.740$

$\fn_cm&space;\small&space;A&space;=&space;514.740\:(60-10)&space;+\left&space;[&space;\frac{(12.115)(391.65)(1,065.264)}{17,685}&space;\right&space;]^{2}&space;-17,685&space;=&space;89,742.69$

$\fn_cm&space;\small&space;B&space;=&space;\left&space;[&space;(12.115)(391.65)(1,065.264)&space;\right&space;]^{2}&space;=&space;2554.95\:x\:10^{10}$

Then,

$\fn_cm&space;\small&space;\fn_cm&space;\small&space;H_{2}^{3}&space;+&space;89742.69\;&space;H_{2}&space;=&space;2552.95\:x\:10^{10}$

Solving this cubic equation through calculator or any technique, we get H2 = 15,575 Newtons.

STEP 5: Compute the final sag of the Ruling Span

Formula:                                    $\fn_cm&space;\small&space;\fn_phv&space;D_{RS}&space;=&space;\frac{H}{W}\left&space;(&space;cosh&space;\frac{S_{RS}W}{2H}&space;-&space;1\right&space;)$

$\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_phv&space;D_{RS}&space;=&space;\frac{15575}{12.115}\left&space;(&space;cosh&space;\frac{391.65\;x\;12.115}{2\;x\;15575}&space;-&space;1\right&space;)&space;=&space;14.943\;meters$

STEP 6: Compute final sag of all other spans

Formula:                                         $\fn_phv&space;D_{n}&space;=&space;D_{RS}&space;\left&space;(&space;\frac&space;{S_{n}}{S_{RS}}&space;\right&space;)^{2}$

$\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_phv&space;D_{@350}&space;=&space;14.943\left&space;(&space;\frac&space;{350}{391.65}&space;\right&space;)^{2}&space;=&space;11.934&space;\;meters$

$\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_phv&space;D_{@200}&space;=14.943\left&space;(&space;\frac&space;{200}{391.65}&space;\right&space;)^{2}&space;=&space;3.897&space;\;meters$

$\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_phv&space;D_{@450}&space;=&space;14.943\left&space;(&space;\frac&space;{450}{391.65}&space;\right&space;)^{2}&space;=&space;19.727&space;\;meters$

$\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_phv&space;D_{@275}&space;=&space;14.943\left&space;(&space;\frac&space;{275}{391.65}&space;\right&space;)^{2}&space;=&space;7.367&space;\;meters$

$\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_phv&space;D_{@500}&space;=&space;14.943\left&space;(&space;\frac&space;{500}{391.65}&space;\right&space;)^{2}&space;=&space;24.355&space;\;meters$

$\fn_cm&space;\small&space;\fn_cm&space;\small&space;\fn_phv&space;D_{@325}&space;=&space;14.943\left&space;(&space;\frac&space;{325}{391.65}&space;\right&space;)^{2}&space;=&space;10.290&space;\;meters$

SUMMARY OF SOLUTION:

Clearances must be based on the final computed sag of the power line. Take note that if we will not use the ruling span theory, the final sag would be different. For instance, if we apply directly the conductor state change equation to the 500 meters span (S =500), the final tension would be H2 = 16,265 Newtons with a sag of 23.35 meters. This 1.0 meter difference would have cause clearance violation.

References:

1. RUS Bulletin 1724E-200
2. RUS Bulletion 1724-152
3. IEAI  MAGAZINE: The Effects of Ruling Span on Sag and Tension
4.  CIGRE Brochure 342
5. As-NZS-7000:2010
6. Kiessling F., et. al.. Overhead Power Lines Planning and Design Construction