This article is a primer to Example 2: Pole Loading Analysis of the Angled Structure based on NESC 2017. Also, this is a supplement of the previous post: Loading Analysis of Transmission and Distribution Structures based on NESC 2017.
Angled Structures are greatly affected by the transverse load imposed by the 1) wind on the structure itself, 2) initial conductor tension, and 3) wind load on the conductor. However, as a Transmission Line Designer, you must take into account also the possible increase in conductor tension due to wind with/without ice loads. In order to do that, you must be knowledgeable in performing sag-and-tension calculations.
A note on NESC 2017 pp 481 states the following:
“Rule 252B3 details the requirement at angles in the line. As written, it requires the vector sum of the transverse wind load and the resultant load imposed by the wires due to their change in direction. The latter is required to include the effects of wind on the wire tension.”
Related: You may review my extensive articles regarding sag and tension here.
Formulas
Transmission lines typically contain a majority of tangent (no line angle) structures. However, it is common for the installed locations of tangent structures to vary from design and thus have a small line angle. This can be caused by construction considerations, subsoil issues, right-of-way limitations, etc. Consequently, it is common to design some small angle loading into most tangent structures to provide a bit of excess capacity to allow for these circumstances.
The drawing above shows an angled structure with line angle θ. By trigonometry, we can solve the resultant force due to wire tension as the following:
Example with Increase in Conductor Tension
We will use the data in the previous example to illustrate the effect of wind on the conductor tension. Let us assume that we have a line angle of 15 degrees and consider only the Phase A conductor.
The initial conditions are the following:
| Initial Conditions: | ||
| Initial Horizontal Tension | 25% RTS | Newton |
| Initial Conductor Temperature | 15 | deg C |
The Phase A conductor has the following properties.
| Code Word | Condor |
| Size | 795 MCM ACSR |
| Stranding | 54/7 |
| Rated Tensile Strength (RTS) | 12954.5 kg |
| Unit Weight | 1.527 kg/m |
| Cross-sectional Area | 0.0004548 m^2 |
| Diameter | 0.0278 m |
| Final Modulus of Elasticity | 59 GPa |
| Coefficient of Thermal Expansion | 0.0000193 / deg C |
For Rule250B (Light Loading District)
It must withstand the following loads:
| Rule 250B (Light Loading District) | ||
| Radial Ice Thickness (Table 250-1) | 0 mm | 0 inches |
| Wind Pressure (Table 250-1) | 430 Pa | 9 lb/ft^2 |
| Temperature (Table 250-1) | -1 deg C | 30 deg F |
The result of the sag-tension calculation showed that the horizontal conductor tension was significantly increased by 8,987.5 N. See the full result here.
| Sag and Tension Result for Rule 250B (Light Loading District) | ||
| Initial Condition | Final Condition | |
| Horizontal Tension | 31,771 N | 40,758.5 N |
| Resultant Transverse Load | 8,294 N | 10,640 N |
For Rule 250C (Extreme Wind)
For Rule 250C, we need to calculate first the wind pressure while taking into account the Kz, GRF, Cf, and Importance Factor. These parameters were already found in the previous post. Hence,
| Calculation of Wind Pressure for Phase A Conductor | |||||||
| Height Above GL | Kz | GRF | I | Cf | V | Wind Pressure (Pa) | |
| ft | m/s | =0.613*Kz*GRF*I*Cf*V^2 | |||||
| Phase A | 65.5 | 1.2 | 0.75 | 1.0 | 1.0 | 63 | 2,189.7 |
Hence, the final loading for this load case is the following:
| Rule 250C (Extreme Wind) | ||
| Radial Ice Thickness (Table 250-1) | 0 mm | 0 inches |
| Wind Pressure (Calculated Above) | 2,189.7 Pa | 45.7 lb/ft^2 |
| Temperature (Table 250-1) | 15 deg C | 60 deg F |
The result of the sag-tension calculation showed that the horizontal conductor tension was significantly increased by 32,496.5 N. See the full result here.
| Sag and Tension Result for Rule 250C (Extreme Wind) | ||
| Initial Condition | Final Condition | |
| Horizontal Tension | 31,771 N | 64,267.5 N |
| Resultant Transverse Load | 8,294 N | 16,777 N |
For Rule 250D (Extreme Ice with Concurrent Wind)
Again, using the loading found in the previous post we have the following:
| Rule 250D (Extreme Ice with Concurrent Wind) | ||
| Radial Ice Thickness (Figure 250-3) | 0 mm | 0 inches |
| Wind Pressure (Figure 250-3) | 110 Pa | 2.3 lb/ft^2 |
| Temperature (Table 250-1) | -10 deg C | 15 deg F |
The result of the sag-tension calculation showed that the horizontal conductor tension was significantly increased by 9,223.8 N. See the full result here.
| Sag and Tension Result for Rule 250D (Extreme Ice with Concurrent Wind) | ||
| Initial Condition | Final Condition | |
| Horizontal Tension | 31,771 N | 40,994.8 N |
| Resultant Transverse Load | 8,294 N | 10,701.8 N |
In Conclusion,
The following table summarizes the effect on conductor tension with respect to the four load cases. Extreme wind load case has the highest effect on the resultant transverse load due to the increase in conductor tension.Hence, this highest value must be used for the loading analysis of the structure.
| Summary of Different Load Cases Effect on Conductor Tension | ||||
| Initial | Rule 250B | Rule 250C | Rule 250D | |
| Horizontal Tension | 31,771 N | 40,758.5 N | 64,267.5 N | 40,994.8 N |
| Resultant Transverse Load | 8,294 N | 10,640 N | 16,777 N | 10,701.8 N |
In this article we had laid out the foundation for the next example.
HELLO,Sir dear
We know that the initial tension can be calculated by the cubic equation,how can we calculate tension after loading & after creep ?
note: i have IEEE Std 80- 2013
Hi Sir, including plastic deformations (after load/creep) is quite complex because you need to factor in the conductor stress-strain-creep curves. I haven’t figured it out yet. Rest assured, I will share it on this blog once I have the spreadsheet.