In this example calculation of tangent 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. Then we will identify the appropriate standard class steel or wood pole in the succeeding posts.
Since this is a tangent structure with 0 degrees line angle, the wire tension has no effect on the transverse load. Hence, sag-tension calculation will not be required here. For an angled structure that will require sag-tension calculation will be tackled in Example 2.
An important note here to the reader is that NESC 2017 does not require Rule 250C and Rule 250D if no portion of a structure or its supported facilities exceeds 18 m (60 ft) above ground or water level. However, RUS has no 60 ft exemption.
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A. General Data:
| Location | Florida, USA |
| Ruling Span | 200 meters |
| Weight Span ( or Vertical Span ) | Equal to RS = 200 meters |
| Wind Span ( or Horizontal Span ) | Assume Equal to Weight Span = 200 meters |
| Line Angle | 0 degrees ( Tangent Structure ) |
| RUS Structure Designation | TU-1AA ( Steel Upswept Arms ) |
| Voltage Level | 69kV |
| Ground Clearance | 14.85 meters ( 49 ft ) |
| Grade of Construction | Grade B |
Conductor Data:
| Code Word | Condor |
| Size | 795 MCM ACSR |
| Stranding | 54/7 |
| Rated Tensile Strength (RTS) | 12,954.5 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 |
Overhead Ground Wire Data:
| Code Word | OPGW |
| Rated Tensile Strength (RTS) | 4,936 kg |
| Unit Weight | 0.328 kg/m |
| Cross-sectional Area | 0.00007967 m^2 |
| Diameter | 0.012 m |
| Final Modulus of Elasticity | 95.5 GPa |
| Coefficient of Thermal Expansion | 0.0000193 / deg C |
B. Pole Framing
1. Standard Construction
All standard construction drawing here is taken from RUS Bulletin 1728F-810. TU-1AA pole construction is selected as shown below. For 69kV voltage level, the minimum dimensions are:
| A = | 4 ft |
| B = | 5 ft |
| C = | 7 ft |
| D = | 4.5 ft |
2. Insulator Assembly
TM-1A is taken as the insulator assembly to be used in this example. Based on Table 1 below, the length of this insulator assembly is 2.5 ft.
3. Burial Depth
We will use the rule of thumb burial depth of 10%*L + 2ft, where L= pole length. Hence for 80 ft pole, the burial depth would be:
Burial Depth = 10% (80 ft) + 2 ft = 10 ft
4. Ground Clearance
Hence, using an 80 ft pole would yield a ground clearance of:
Ground Clearance = Pole Length – (Burial Depth) – D – C – C – (Insulator Assembly Length)
Ground Clearance = 80′ – 10′ – 4.5′ – 7′ -7′ – 2.5′ = 49 ft.
For the purpose of discussion, let us assume that this ground clearance is enough. However, on actual application follow the required minimum ground clearance.
C. Pole Loading Analysis
We had determined now the required length of the pole. In the next steps, we can determine the forces acting on the conductor due to wind and/or tension. As stated above, the structure is more than 60 ft above the ground level thus Rule 250C and Rule 250D are required by the code.
Step 1: Evaluate the applicable Grade of Construction
Refer to NESC Rule 240 for the detailed determination of required grade of construction. For this example, we will use the highest grade which is Grade B.
Step 2: Apply NESC Rule 250B
Using Figure 250-1, Florida is located within the light loading district. Hence, it must withstand the following minimum loads from Table 250-1:
| Light Loading District (from Table 250-1) | ||
| Radial Ice Thickness | 0 mm | 0 inches |
| Wind Pressure | 430 Pa | 9 lb/ft^2 |
| Temperature | -1 deg C | 30 deg F |
Step 2.1: Calculate the transverse, vertical and longitudinal load
Step 2.1.1 Vertical Load
For this example, we will neglect the weight of the insulators and hardware. For wires, the vertical load is the weight span times the conductor unit weight including ice.
Weight of OPGW:
- V_opgw = (unit weight with ice) x (weight span)
- V_opgw = 0.328*9.81*200
- V_opgw = 643.536 N
Weight of Conductor:
- V_condA = (unit weight with ice) x (weight span)
- V_condA = 1.527*9.81*200
- V_condA = 2995.974 N, this is the same for Phase B and C
- V_condB = 2995.974 N
- V_condC = 2995.974 N
Step 2.1.2 Transverse Load
For tangent pole structure, the transverse load is due to:
- Wind on Wires
- Wind on Equipment, Insulators and hardware -> will be neglected in this example
- Wind on Structures/Poles
Wind on OPGW:
- W_opgw = (Wind Pressure) x (Projected Area with Ice)
- Projected Area = (diameter with ice) x (Wind Span)
- Hence,
- W_opgw = 430 x 0.012 x 200
- W_opgw = 1032 N
Wind on Conductor:
- W_condA = (Wind Pressure) x (Projected Area with Ice)
- W_condA = 430 x 0.0278 x 200
- W_condA = 2390.8 N, this is the same for Phase B and C
- W_condB = 2390.8 N
- W_condC = 2390.8 N
Wind on Pole:
- W_pole = Wind Pressure x Projected Area above ground
- W_pole = 430A, where A = Area of Pole above GL
We don’t have the dimension of the pole right now to determine the area.
Step 2.1.3 Longitudinal Load
Longitudinal load is zero since it is assumed that forward and ahead spans are equal.
Step 2.2 If angle structure, calculate transverse component of wire tension using sag-tension principles.
This step is not applicable for tangent structure of 0 degrees line angle.
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 Longidutinal = 1.1
Hence,
| Point Loads | Load Factors | Factored Load | |||
| Vertical | Transverse | Longitudinal | Point Load x LF | ||
| V_opgw | 644 | 1.5 | 966 N | ||
| V_condA | 2996 | 1.5 | 4494 N | ||
| V_condB | 2996 | 1.5 | 4494 N | ||
| V_condC | 2996 | 1.5 | 4494 N | ||
| W_opgw | 1032 | 2.5 | 2580 N | ||
| W_condA | 2391 | 2.5 | 5978 N | ||
| W_condB | 2391 | 2.5 | 5978 N | ||
| W_condC | 2391 | 2.5 | 5978 N | ||
| W_pole | 430A | 2.5 | 1075A |
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
The applicable maps for Florida is Figure 250-2(b) as shown below. We will use 140 miles/hr or 63 m/s as the Basic Wind Speed.
Step 3.2 Calculate transverse, vertical and longitudinal load
Step 3.2.1 Vertical Load
There is no change in vertical load. It is the same as the previous calculations.
Step 3.2.2 Transverse Load
The formulas to be used in this step was discussed in the previous post. However, in this example, we will use Table 250-2 and Table 250-3 to find Kz and GRF respectively. You may also use the formula and compare it with the values on the table.
Wind on Wires:
Note that in the code Cf, shape factor, for conductor is always taken as 1.0, unless a wind tunnel test is performed.
| Point Loads | Height above GL | Kz | GRF | I | Cf | Diameter | Wind Span | Wind Load |
| ft | m | m | N | |||||
| W_opgw | 69.5 | 1.2 | 0.75 | 1.0 | 1.0 | 0.012 | 200 | 5255 |
| W_condA | 65.5 | 1.2 | 0.75 | 1.0 | 1.0 | 0.0278 | 200 | 121747 |
| W_condB | 58.5 | 1.2 | 0.75 | 1.0 | 1.0 | 0.0278 | 200 | 121747 |
| W_condC | 51.5 | 1.2 | 0.75 | 1.0 | 1.0 | 0.0278 | 200 | 121747 |
Wind on Structure:
Note that at this point we don’t have yet the pole dimension and shape. The area and shape factor is dependent on that data.
| Point Loads | Height above GL | Kz | GRF | I | Cf | Area | Wind Load |
| ft | m^2 | N | |||||
| W_pole | 70 | 1.1 | 0.93 | 1.0 | varies | varies | 2489*Cf*Area |
Step 3.2.3 Longitudinal Load
Longitudinal load is zero since it is assumed that forward and ahead spans are equal.
Step 3.3 If angled, calculate the transverse component of wire tension using sag-tension principles.
This step is not applicable for tangent structures of 0 degrees line angle.
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 Longitudinal = 1.0
Hence,
| Point Loads | Load Factors | Factored Load | |||
| Vertical | Transverse | Longitudinal | Point Load x LF | ||
| V_opgw | 644 | 1.0 | 644 N | ||
| V_condA | 2996 | 1.0 | 2996 N | ||
| V_condB | 2996 | 1.0 | 2996 N | ||
| V_condC | 2996 | 1.0 | 2996 N | ||
| W_opgw | 5255 | 1.0 | 5255 N | ||
| W_condA | 121747 | 1.0 | 121747 N | ||
| W_condB | 121747 | 1.0 | 121747 N | ||
| W_condC | 121747 | 1.0 | 121747 N | ||
| W_pole | 2489*Cf*A | 1.0 | 2489*Cf*A |
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
The appropriate map is Figure 250-3(b). From the map below, the ice thickness is 0″ while the concurrent wind is 30 mph.
Step 4.2 Convert Wind Speed to wind pressure using Table 250-4
Using Table 250-4, 30 mph is equivalent to 110 pascals of wind pressure.
Step 4.3 Calculate transverse, vertical and longitudinal load
Step 4.3.1 Vertical Load
There is no change in the vertical load because ice thickness is equal to zero. It is the same in the previous calculation.
Step 4.3.2 Transverse Load
Wind on OPGW:
- W_opgw = (Wind Pressure) x (Projected Area with Ice)
- Projected Area = (diameter with ice) x (Wind Span)
- Hence,
- W_opgw = 110 x 0.012 x 200
- W_opgw = 264 N
Wind on Conductor:
- W_condA = (Wind Pressure) x (Projected Area with Ice)
- W_condA = 110 x 0.0278 x 200
- W_condA = 612 N, this is the same for Phase B and C
- W_condB = 612 N
- W_condC = 612 N
Wind on Pole:
- W_pole = Wind Pressure x Projected Area above ground
- W_pole = 110A, where A = Area of Pole above GL
We don’t have the dimension of the pole right now to determine the area.
Step 4.3.3 Longitudinal Load
Longitudinal load is zero since it is assumed that forward and ahead spans are equal.
Step 4.1 If angled, calculate the transverse component of wire tension using sag-tension principles.
This step is not applicable for tangent structure of 0 degrees line angle.
Step 4.1 Apply Load Factors
From Table 253-1, the load factor for rule 250D and Grade B Construction:
- LF Vertical = 1.0
- LF Transverse (Wind) = 1.0
- LF Longitudinal = 1.0
Hence,
| Point Loads | Load Factors | Factored Load | |||
| Vertical | Transverse | Longitudinal | Point Load x LF | ||
| V_opgw | 644 | 1.0 | 644 N | ||
| V_condA | 2996 | 1.0 | 2996 N | ||
| V_condB | 2996 | 1.0 | 2996 N | ||
| V_condC | 2996 | 1.0 | 2996 N | ||
| W_opgw | 264 | 1.0 | 264 N | ||
| W_condA | 612 | 1.0 | 612 N | ||
| W_condB | 612 | 1.0 | 612 N | ||
| W_condC | 612 | 1.0 | 612 N | ||
| W_pole | 110*Cf*A | 1.0 | 110*Cf*A |
Now, we can draw the loading tree diagram,
What’s Next?
Generating the loading tree is the first step in designing or selecting an appropriate pole or structure. That is what we have done on this post. In the next example, we will consider an angled structure that will employ the sag-tension calculation.
References:
- NESC 2017
- RUS BULLETIN 1728F-810
