# 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.

## 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.”

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 2: Apply NESC Rule 250B

#### Step 2.1: Calculate the transverse, vertical and longitudinal load

Step 2.1.2 Transverse Load due to wind

#### 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 Tension 25% RTS Newton Initial Conductor Temperature 15 deg C

 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 calculations are the following: (Using my spreadsheet.)

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

 Joint Label Final Tension for Rule250B OPGW 15, 889.2 N Cond A 40,758.5 N Cond B 40,758.5 N Cond C 40,758.5 N

Hence, the transverse load due to tension are:

 Point Loads Line Angle (deg) Tension (N) Resultant (N) θ T =2*T*sin (θ/2) T_opgw 15 15 889.2 4 148 N T_condA 15 40 758.5 10 640 N T_condB 15 40 758.5 10 640 N T_condC 15 40 758.5 10 640 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 Label Point Label Load Load Factors Factored Load ( N ) Vertical Transverse Longitudinal Point Load x LF OPGW V_opgw 644 1.5 966 N CondA V_condA 2996 1.5 4494 N CondB V_condB 2996 1.5 4494 N CondC V_condC 2996 1.5 4494 N OPGW W_opgw 1032 2.5 2580 N CondA W_condA 2391 2.5 5978 N CondB W_condB 2391 2.5 5978 N CondC W_condC 2391 2.5 5978 N Pole W_pole 430*A 2.5 1075*A OPGW T_opgw 4 148 1.65 6 844 N CondA T_condA 10 640 1.65 17 556 N CondB T_condB 10 640 1.65 17 556 N CondC T_condC 10 640 1.65 17 556 N

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

#### 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 Label Height Above GL Kz GRF I Cf V Wind Pressure (Pa) ft m/s =0.613*Kz*GRF*I*Cf*V^2 OPGW 69.5 1.2 0.75 1.0 1.0 63 2,189.7 CondA 65.5 1.2 0.75 1.0 1.0 63 2,189.7 CondB 58.5 1.2 0.75 1.0 1.0 63 2,189.7 CondC 51.5 1.2 0.75 1.0 1.0 63 2,189.7

 Rule 250C (Extreme Wind) Radial Ice Thickness (Table 250-1) 0 mm 0 inches Wind Pressure (Calculated Above) Temperature (Table 250-1) 15 deg C 60 deg F

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

 Joint Label Final Tension for Rule250C OPGW 24, 201.7 N Cond A 64, 267.5 N Cond B 64, 267.5 N Cond C 64, 267.5 N

Hence, the transverse load due to tension are:

 Point Loads Line Angle (deg) Tension (N) Resultant (N) θ T =2*T*sin (θ/2) T_opgw 15 24, 201.7 6 318 N T_condA 15 64, 267.5 16 777 N T_condB 15 64, 267.5 16 777 N T_condC 15 64, 267.5 16 777 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 Label Point Label Load Load Factors Factored Load ( N ) Vertical Transverse Longitudinal Point Load x LF OPGW V_opgw 644 1.0 644N CondA V_condA 2996 1.0 2996 N CondB V_condB 2996 1.0 2996 N CondC V_condC 2996 1.0 2996 N OPGW W_opgw 5255 1.0 5255 N CondA W_condA 12, 1747 1.0 121747 N CondB W_condB 12, 1747 2.5 121747 N CondC W_condC 12, 1747 2.5 121747 N Pole W_pole 2489*Cf*A 1.0 2489*Cf*A OPGW T_opgw 6 318 1.0 6 318 N CondA T_condA 16 777 1.0 16 777 N CondB T_condB 16 777 1.0 16 777 N CondC T_condC 16 777 1.0 16 777 N

### Step 4: Apply NESC Rule 250D

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

 Rule 250B (Light Loading District) Radial Ice Thickness (Table 250-1) 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 calculations are the following: (Using my spreadsheet.)

 Joint Label Final Tension for Rule250B OPGW 15, 517.7 N Cond A 40, 994.8 N Cond B 40, 994.8 N Cond C 40, 994.8 N

Hence, the transverse load due to tension are:

 Point Loads Line Angle (deg) Tension (N) Resultant (N) θ T =2*T*sin (θ/2) T_opgw 15 15 517.7 4 051 N T_condA 15 40 994.8 10 702 N T_condB 15 40 994.8 10 702 N T_condC 15 40 994.8 10 702 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 Label Point Label Load Load Factors Factored Load ( N ) Vertical Transverse Longitudinal Point Load x LF OPGW V_opgw 644 1.0 644N CondA V_condA 2996 1.0 2996 N CondB V_condB 2996 1.0 2996 N CondC V_condC 2996 1.0 2996 N OPGW W_opgw 264 1.0 264 N CondA W_condA 612 1.0 612 N CondB W_condB 612 1.0 612 N CondC W_condC 612 1.0 612 N Pole W_pole 110*A 1.0 110*A OPGW T_opgw 4 051 1.0 4 051 N CondA T_condA 10 702 1.0 10 702 N CondB T_condB 10 702 1.0 10 702 N CondC T_condC 10 702 1.0 10 702 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