Originally, wood poles were designed as per the 1947 Electricity Commissioners’ Regulations, which stipulated that they had to have a factor of safety of 3.5 with a wind pressure of 8 lb/ft2(380 N/m2) and 3/8 in (9.5 mm) radial ice on the conductors at 22◦F (−5.6◦C).
Today, under the 1988 Electricity Supply Regulations, the criterion is that the pole must be ‘fit for purpose’. This gives the design engineer much more latitude in his pole design and has led to the environmental approach detailed in ENATS 43-40 already covered in chapter 4.
Mechanical design of poles, cross-arms and foundations 77 6.2.2.2 Moments of resistance of wood poles
If it is true to state that the strength of a foundation is dependent on the strength of the ground, then it is also true that it is dependent on the strength of the wood pole as well.
If a pole is subject to a bending moment, then the fibres on one side of the pole are compressed and the fibres on the other side are extended. Compressive and tensile stresses are thereby introduced in the pole producing a moment called the moment of resistance; this is equal and opposite to the bending moment.
The second moment of area or the moment of inertia of a circular section (I ) is:
I =π D
4
64 (6.3)
The modulus of section (Z) is:
Z=π D
3
32 (6.4)
Fibre stress2(F ) equals:
F = bending moment
section modulus
Maximum resisting moment equals section modulus× F = (53 300 × πD3)/32, which equates to approximately 5233D3kNm, where D is the critical diameter of the pole that is either at the groundline or at a point 1.5 times the pole top diameter.
The maximum resisting moment of poles can be obtained from tables. These vary from 28 kNm for a 175 mm diameter pole to 224 kNm for a 350 mm diameter pole. To determine the maximum working resistance of a pole this figure should be divided by the factor of safety adopted for any particular location. This can perhaps be best explained by going through a worked example.
Example
A 12 m intermediate pole is used to support a three-phase 175 mm2 Lynx ACSR conductor (diameter 19.53 mm), with a span length of 100 m. The pole is planted 1.8 m deep. The wind loading is taken to act at 250 mm above the pole top (the approximate height of the conductors). A factor of safety of 2.5 is normally included in these calculations. Assume the critical diameter occurs at ground level. The aim is to determine what diameter of pole is needed, i.e. medium, stout or extra stout.
First calculate the wind load on the three-phase conductors: =3× 380 × 100 × [19 + 19.53]
1000 = 4393 N
Wind load on pole assuming average diameter of 250 mm is: 380× 250 × 10.2
1000 = 969 N
Wind load on pin/post insulators, assuming projected area of each insulator = 0.15 m2:
380× 0.15 × 3 = 171 N
Maximum bending moment at ground level equals:
{(4393 × 10.45) + (969 × 10.2/2) + (171 × [10.2 + 0.125])} × 1000 = 52 614 325 Nmm
Maximum allowable fibre stress with a factor of safety of 2.5 = 21.32 N/mm2
Moment of resistance = f × sect mod
(where sect mod= π × D3/32).
From which the minimum pole diameter D is= 292.9 mm.
Diameter at 1.5 m from butt (assuming taper 11 mm/m) = 296.2 mm. This is less than the specified minimum diameter of 305 mm for a stout pole within BS 1990, and hence a stout grade pole is okay.
At an angle or terminal positions the supports are stayed. In addition to the bending due to the wind loading, the pole must also act as a strut due to the vertical component of the stay tension (Figure 6.3).
In Figure 6.3, the conductors are represented by the forces T . These combine vectorially to give a resultant force, P . This has to be balanced by the force in the pole, V , and the stay wire, S.
6.2.2.3 Crippling load of struts (as applied to wood poles)
A strut refers to a member that is long in comparison with its cross-sectional dimen- sions. This describes a wood pole used as a direct support or as a stay. This will fail due to buckling before the compressive stresses reach yield point. The load that will cause buckling is given by Euler’s theory. This can only be approximate in the case of wood poles. Wood poles have a varying diameter or taper, there are imperfections along their length (e.g. knots in the wood) and the actual strength of the wood may vary along the length. The formula used has taken this into account and has an inbuilt factor of safety.
Although a stayed pole does not strictly have both ends pinned, the formula for a beam pinned at both ends is used. However, difficulty arises with the effective length of the beam. Is the top end to be taken as the stay attachment point or the effective point of conductor attachment? The pole diameter is another contentious
Mechanical design of poles, cross-arms and foundations 79
stay tension, S pole
vertical load, V T in each conductor resultant pull, P T F
Figure 6.3 Stay tension
T = conductor tension in N = angle between stay and pole S= tension in stay in N
V = vertical load on pole
P = horizontal pull at pole top and = T for terminal load
then S= P/sin N and V = S cos N
issue. Where should the diameter be taken? If an average is to be used, where and what should be averaged?
There can thus be large discrepancies depending on which parameters are used. The crippling load on a pole is calculated from Euler’s theory where the length l is taken as the distance between the pole top and one and a half metres from the butt and the diameter as the average of the diameters at these points, since these are readily available.
Euler’s theory
The formula commonly used to calculate the crippling load (P ) for poles is:
P =π
2EI
l2 N (6.5)
where l is the length of strut in mm, E is the modulus of elasticity of pole (average value 10 054 N/mm2) and I is the moment of inertia of a circular section (π× D4/64) mm4.