Example 2: Pole Loading Analysis of Angled Structure based on NESC 2017

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In this example calculation of angled pole structure, the objective is to apply NESC 2017 rules to generate the loading trees for different load cases. We will apply the step-by-step process outlined in the previous post.

Since this is an angled structure, the wire tension has an effect on the transverse load. Hence, sag-tension calculation will be required here. We will utilize the same data as the previous post with the addition of line angle equal to 15 degrees.

Conductor blow out during a Bora event in Croatia, January 2003. From CIGRE TB 256 (Photo: F. Jakl)

Theory:

Angled structures are most affected by the increase in wire tension. This wire tension may increase due to additional wind and ice loading.

A commentary on NESC Rule 252B3 states: “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.

In addition, IEC 60826:2003, states: “Wind acting on the conductors will cause an increase in their mechanical tension that can be computed with standard sag-tension methods.”

C. Pole Loading Analysis

Please refer to the Example 1 for the values calculated on each step. We will highlight only here the calculations of transverse load due to wire tension.

Step 1: Evaluate the applicable Grade of Construction

Step 2: Apply NESC Rule 250B

Step 2.1: Calculate the transverse, vertical and longitudinal load

Step 2.1.1 Vertical Load

Step 2.1.2 Transverse Load due to wind

Step 2.1.3 Longitudinal Load

Step 2.2 If angle structure, calculate transverse component of wire tension using sag-tension principles.

Line Angle θ = 15 degrees

Initial Conditions:
Initial Horizontal Tension25% RTSNewton
Initial Conductor Temperature15deg C

The following is the loading conditions:

Rule 250B (Light Loading District)
Radial Ice Thickness (Table 250-1)0 mm0 inches
Wind Pressure (Table 250-1)430 Pa9 lb/ft^2
Temperature (Table 250-1)-1 deg C30 deg F

The result of the sag-tension calculations are the following: (Using my spreadsheet.)

Important Note: Include the NESC “k- Constant” in sag-tension calculations.

Joint LabelFinal Tension for Rule250B
OPGW15, 889.2 N
Cond A40,758.5 N
Cond B40,758.5 N
Cond C40,758.5 N

Hence, the transverse load due to tension are:

Point LoadsLine Angle (deg)Tension (N)Resultant (N)
θT=2*T*sin (θ/2)
T_opgw1515 889.24 148 N
T_condA1540 758.510 640 N
T_condB1540 758.510 640 N
T_condC1540 758.510 640 N

Step 2.3 Apply Load Factors (LF)

From Table 253-1, the load factor for Rule 250B and Grade B Construction:

  • LF Vertical = 1.5
  • LF Transverse (Wind) = 2.5
  • LF Transverse (Wire Tension) = 1.65
  • LF Longitudinal = 1.1

Hence,

Joint LabelPoint LabelLoadLoad FactorsFactored Load
( N )VerticalTransverseLongitudinalPoint Load x LF
OPGWV_opgw6441.5966 N
CondAV_condA29961.54494 N
CondBV_condB29961.54494 N
CondCV_condC29961.54494 N
OPGWW_opgw10322.52580 N
CondAW_condA23912.55978 N
CondBW_condB23912.55978 N
CondCW_condC23912.55978 N
PoleW_pole430*A2.51075*A
OPGWT_opgw4 1481.656 844 N
CondAT_condA10 6401.6517 556 N
CondBT_condB10 6401.6517 556 N
CondCT_condC10 6401.6517 556 N

Now we can draw the loading tree diagram,

Step 3: Apply NESC Rule 250C (Extreme Wind)

Step 3.1 Select Basic Wind Speed in appropriate maps

Step 3.2.1 Vertical Load

Step 3.2.2 Transverse Load

Step 3.2.3 Longitudinal Load

Step 3.3 If angled, calculate the transverse component of wire tension using sag-tension principles.

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
Joint LabelHeight Above GLKzGRFICfVWind Pressure (Pa)
ftm/s=0.613*Kz*GRF*I*Cf*V^2
OPGW69.51.20.751.01.0632,189.7
CondA65.51.20.751.01.0632,189.7
CondB58.51.20.751.01.0632,189.7
CondC51.51.20.751.01.0632,189.7

The following the loading conditions:

Rule 250C (Extreme Wind)
Radial Ice Thickness (Table 250-1)0 mm0 inches
Wind Pressure (Calculated Above)
Temperature (Table 250-1)15 deg C60 deg F

The result of the sag-tension calculations are the following: (Using my spreadsheet.)

Joint LabelFinal Tension for Rule250C
OPGW24, 201.7 N
Cond A64, 267.5 N
Cond B64, 267.5 N
Cond C64, 267.5 N

Hence, the transverse load due to tension are:

Point LoadsLine Angle (deg)Tension (N)Resultant (N)
θT=2*T*sin (θ/2)
T_opgw1524, 201.76 318 N
T_condA1564, 267.516 777 N
T_condB1564, 267.516 777 N
T_condC1564, 267.516 777 N

Step 3.4 Apply Load Factors

From Table 253-1, the load factors for Rule 250C and Grade B Construction:

  • LF Vertical = 1.0
  • LF Transverse (Wind) = 1.0
  • LF Transverse ( Wire Tension) = 1.0
  • LF Longitudinal = 1.0

Hence,

Joint LabelPoint LabelLoadLoad FactorsFactored Load
( N )VerticalTransverseLongitudinalPoint Load x LF
OPGWV_opgw6441.0644N
CondAV_condA29961.02996 N
CondBV_condB29961.02996 N
CondCV_condC29961.02996 N
OPGWW_opgw52551.05255 N
CondAW_condA12, 17471.0121747 N
CondBW_condB12, 17472.5121747 N
CondCW_condC12, 17472.5121747 N
PoleW_pole2489*Cf*A1.02489*Cf*A
OPGWT_opgw6 3181.06 318 N
CondAT_condA16 7771.016 777 N
CondBT_condB16 7771.016 777 N
CondCT_condC16 7771.016 777 N

Now we can draw the loading tree diagram,

Step 4: Apply NESC Rule 250D

Step 4.1 Select Ice thickness and wind speed in the appropriate maps

Step 4.2 Convert Wind Speed to wind pressure using Table 250-4

Step 4.3 Calculate transverse, vertical and longitudinal load

Step 4.3.1 Vertical Load

Step 4.3.2 Transverse Load

Step 4.3.3 Longitudinal Load

Step 4.1 If angled, calculate the transverse component of wire tension using sag-tension principles.

The following is the loading conditions:

Rule 250B (Light Loading District)
Radial Ice Thickness (Table 250-1)0 mm0 inches
Wind Pressure (Figure 250-3)110 Pa2.3 lb/ft^2
Temperature (Table 250-1)-10 deg C15 deg F

The result of the sag-tension calculations are the following: (Using my spreadsheet.)

Joint LabelFinal Tension for Rule250B
OPGW15, 517.7 N
Cond A40, 994.8 N
Cond B40, 994.8 N
Cond C40, 994.8 N

Hence, the transverse load due to tension are:

Point LoadsLine Angle (deg)Tension (N)Resultant (N)
θT=2*T*sin (θ/2)
T_opgw1515 517.74 051 N
T_condA1540 994.810 702 N
T_condB1540 994.810 702 N
T_condC1540 994.810 702 N

Step 4.1 Apply Load Factors

From Table 253-1, the load factors for Rule 250D and Grade B Construction:

  • LF Vertical = 1.0
  • LF Transverse (Wind) = 1.0
  • LF Transverse ( Wire Tension) = 1.0
  • LF Longitudinal = 1.0
Joint LabelPoint LabelLoadLoad FactorsFactored Load
( N )VerticalTransverseLongitudinalPoint Load x LF
OPGWV_opgw6441.0644N
CondAV_condA29961.02996 N
CondBV_condB29961.02996 N
CondCV_condC29961.02996 N
OPGWW_opgw2641.0264 N
CondAW_condA6121.0612 N
CondBW_condB6121.0612 N
CondCW_condC6121.0612 N
PoleW_pole110*A1.0110*A
OPGWT_opgw4 0511.04 051 N
CondAT_condA10 7021.010 702 N
CondBT_condB10 7021.010 702 N
CondCT_condC10 7021.010 702 N

What’s Next?

In the next post, we will select the appropriate pole that can withstand the loading cases we have created.

References:

  • NESC 2017
  • RUS BULLETIN 1728F-810