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.”

## 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 Tension | 25% RTS | Newton |

Initial Conductor Temperature | 15 | deg C |

The following is the loading conditions:

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 |

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 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 |

The following the loading conditions:

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 |

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 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