Fundamental Concepts of Sag-Tension Relationship in Transmission and Distribution Lines

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When a conductor is suspended between two supports and carrying its own uniform weight, the curve formed is called a catenary. For simplified analysis, some of the design engineers assume that the curve is parabolic.

The parabolic approximation is only true if the sag is small as compared with the span length which happens mostly on the distribution lines. Although the formula is quite complex, it is always recommended to follow catenary calculations.

Importance of Sag-Tension Calculations

It is important to predict the sag and tension of the conductor to determine if:

  1. It will pass the required electrical clearances to ground, right-of-way (blowout) and uplift
  2. It will not exceed the tension limits of the conductor at varying loading conditions
  3. It will pass the required levels of electric and magnetic field, aeolian vibration and ice galloping
  4. the structure is sufficient to withstand the conductor tension.

The sag of the conductor depends on:

  1. Conductor unit weight
  2. Span Length
  3. Tension in the conductor
  4. Weather Conditions (effect of wind, ice)
  5. Temperature
  6. Elasticity of the conductor materials
  7. Plastic Deformation (creep, strand settlement, etc.)

The Catenary Curve

If the lowest point of the curve above will be taken as the origin, then it can be modeled by the following equation:

\begin{align} y = C (cosh{\frac{x}{C}}-1) \end{align}


  • x = horizontal distance from lowest point in span (m)
  • y = vertical distance from lowest point of span
  • C = Catenary constant

Under the no-wind condition, the catenary constant is essentially the ratio of the horizontal tension in the conductor to the unit weight.

\begin{align} C = \frac{H}{W} \\ W = m*g \end{align}


  • H  = Horizontal component of tension (N)
  • W = distributed load on conductor (N/m)
  • m = unit mass of conductor (kg/m)
  • g = gravitational constant of 9.81 m/s2

Some important notes about this catenary curve:

  • The tension at any point acts tangentially through the conductor.
  • The horizontal component of  Tension, H, is constant all throughout the conductor.
  • For simplification purposes, some calculations assumed that Tension is equal to the horizontal Tension, T = H. 
  • Also, some calculations assumed that the vertical component of tension Ty = WS/2.
\begin{align} H &= T_{x} \\ T_{y} &= \frac{w*L}{2}\\ T_{at \ supports} &= \sqrt{T_{x}^{2}+T_{y}^{2}}\\ T_{avg\ at \ supports}&= H + w*D \end{align}
Forces Acting on the Half Span

True Conductor Length

The true conductor length, Lc , is only marginally greater than the span length, S. Applying calculus, the conductor length from the lowest point to any other point along the conductor is given by the following equation:

\begin{align} L_{c} (x) = \frac{H}{W} sinh \left( \frac{Wx}{H} \right) = x \left(1 +\frac{x^{2}W^{2}}{6H^{2}} \right) \end{align}

For a level span, the distance from the lowest point to one conductor support is x = S /2. Thus the total conductor length is:

\begin{align} L_{c} = \frac{2H}{W} sinh \left( \frac{WS}{2H} \right) = S \left( 1 +\frac{S^{2}W^{2}}{24H^{2}} \right)= S +\frac{S^{3}W^{2}}{24H^{2}} \end{align}

Analyzing the equation above, the slack of the line is the difference between true conductor length and span length.

\begin{align} Slack = L_{c} – S =\frac{S^{3}W^{2}}{24H^{2}} \end{align}

Changes in Conductor Length

The actual length of the conductor is affected by three factors:

  1. Linear Elastic Strain
  2. Linear Thermal Elongation
  3. Plastic Elongations

1. Linear Elastic Strain

It is a reversible linear elongation due to tension change.

The conductor materials will follow the Hooke’s law at some point, wherein changes in the tension would also change the total conductor length. This elastic elongation is “spring-like”. Ideally, when the tension is removed the conductor will return to its original length.

It is a function of the following:

  • Horizontal Tension, H
  • modulus of elasticity (Young’s Modulus) , E
  • The cross-sectional area of the conductor, A

But note that most overhead conductors are made of aluminum reinforced of steel wires (ACSR). Thus the modulus of elasticity of composite conductor is:

\begin{align} E_{conductor} = E_{al} \left( \frac{A_{al}}{A_{total}} \right) +E_{st} \left( \frac{A_{st}}{A_{total}} \right) \end{align}

The modulus of elasticity is the ratio of the stress to strain, as follows: 

E = Stress / Strain                or          Strain = Stress / E 

Consequently, the change in conductor length due to strain-stress, is as follows:

\begin{align} \frac{L_{C-final} -L_{C-initial} }{L_{C-initial} } = \frac{H}{E_{cond}A_{cond}} \end{align}

2. Linear Thermal Elongation

It is a reversible linear elongation due to temperature changes.

Conductors used in power lines will expand or contract with changes in the temperature. The rate of expansion or contraction is dependent on the conductor materials and the degree of temperature change.

For composite conductors like ACSR, the coefficient of thermal expansion (COE) of aluminum strands is different from that of steel. At knee-point temperature, the steel core will carry all the tension while in the aluminum strands will be zero.

Also the effective linear thermal coefficient of expansion of composite conductor is:

\begin{align} \alpha _{conductor} = \alpha_{al} \left(\frac{E_{al}}{E_{conductor}} \right)\left(\frac{A_{al}}{A_{total}} \right) +\alpha_{st} \left(\frac{E_{st}}{E_{conductor}} \right)\left(\frac{A_{al}}{A_{total}} \right) \end{align}

The change in conductor length due to thermal expansion, is as follows:

\begin{align} L_{C-final} = L_{C-initial} (1 +\alpha _{cond} (t_{i}-t_{f})) \end{align}

Conductor State Change Equation

As temperature increases, the conductor expands in length and therefore its tension decreases, while the sag increases. With the reduction in tension, there is also a reduction in strain in the conductor. Thus, there is a complicated relationship between temperature and tension.

If we will only consider the effects of 1) elastic strain and 2) thermal elongation (ignoring the plastic elongations), we will arrive with the following general equation:

Final Conductor Length = Initial Length + Effect of Elastic Strain + Effect of Thermal Elongation

The equation above is a linear elastic model for calculating sag and tension. There are three models to compute sag-tension as described in this post.

By substitution, we can derive the conductor state change equation that follows the linear elastic model:

\begin{align} H_2^{3} +H_2^{2} \left( \frac{(W_{1}S)^{2}AE}{24H_{1}^{2}} – H_{1} + (t_{2}-t_{1}) \alpha AE\right) – \frac{(W_{2}S)^{2}AE}{24} = 0 \end{align}


  • H1 = initial conductor tension at initial temperature (N)
  • W1 = unit weight of the conductor at initial temperature (N/m)
  • Δt =t2 –  t1 = temperature change (°C)
  • E = final modulus of elasticity (Pa)
  • A = cross-sectional area conductor (m2)
  • ∝ = coefficient of linear thermal expansion (/°C)
  • S=length of ruling span ( simply span length for a single span) (m)

Conductor tension, H2 ,at any given temperature can be calculated by iteratively solving the roots of this equation. This can be easily carried out in excel or other spreadsheet that has “goal seek” or solver plug-in.

Sag-Tension Relationship

The basic relationship between sag and span length, for a level single span, is as follows:

\begin{align} D = C\left( cosh \frac{S}{2C} – 1\right) \end{align}

This can be approximated as:

\begin{align} D = \frac{WS^{2}}{8H} \rightarrow H = \frac{WS^{2}}{8D} \end{align}

Effects of Ice and Wind

When the conductor will be covered with ice and/or subjected to wind, the total unit weight of the conductor increases and changes the shape of the catenary curve. Line design engineers must consider expected extreme loading in the area in considering the sag-tension calculation.

Conductor and supports may fail under these extreme loading conditions. Also, it may exceed the required electrical clearances. 

There are different physical forms of ice that may form in the conductor. These are:

  • glaze ice
  • rime ice
  • wet snow

In calculations,  glaze ice formation is usually considered with density of 57 lb/ft3.

Effect of Ice on Conductor Weight

The volume of ice is:

\begin{align} V_{ice} = \frac{\pi }{4}\left( (D + 2t)^{2} – D^{2} + \right) \rightarrow V_{ice} = \pi t (D+t) \end{align}

The weight of ice per unit length:

\begin{align} W_{ice} =Volume\, x\, density = \rho_{ice} \pi t (D +t) \end{align}


  • Vice = Volume of ice per unit length (m3 / m)
  • D = outside diameter of the conductor (m)
  • t = thickness of the ice (m)
  • ρice = ice density (kg/m3)

Effect of Wind on Conductor Weight

The wind pressure is usually in lb/ft2 (Pa) and assumes to act horizontally on the projected area of the conductor.  Weight due to wind per unit length (and/or radial ice included):

\begin{align} W_{wind} = P_{wind} \ (D + 2t) \end{align}

Total Conductor Unit Weight

The total conductor weight is given by:

\begin{align} W_{total} = \sqrt{(W + W_{ice})^{2}+(W_{wind})^{2}} \end{align}

Thus the total sag of the conductor is given by the following equations:

\begin{align} D = \frac{H}{W_{total}}\left( cosh\left( \frac{S}{2 H/W_{total}} \right) – 1\right) \end{align}

But, take note that this sag makes an angle θ with the vertical. So the vertical sag of the conductor is, as follows:

\begin{align} D_{v} = D\; \cos \, \theta \end{align}

where θ = blowout angle. Some textbooks suggest that the effect of wind is essential only if you are concerned in horizontal clearance because of the conductor blowout.


  1. NS220 Overhead Design Manual
  2. RUS Bulletin 1724E-200
  3. Overhead Power Line Design – F. Kiessling , P. Nefzger, J.F. Nolasco and U. Kaintzyk
  4. Principles of Power System – K. Mehta