CHAPTER XXII.
DESIGN OF INSULATOR PINS FOR TRANSMISSION LINES.
Bending strains due to the weights, degree of tension, and the directions of line wires, plus those resulting from wind-pressure, are the chief causes that lead to the mechanical failure of insulator pins.
Considering the unbalanced component of these forces at right angles to the axis of the pin, which alone produce bending, each pin may be considered as a beam of circular cross section secured at one end and loaded at the other.
For this purpose the secured end of the beam is to be taken as the point where the pin enters its cross-arm, and the loaded end of the beam is the point where the line wire is attached to the insulator. The distance between these two points is the length of the beam. The maximum strain in the outside fibres of a pin measured in pounds per square inch of its cross section, represented by S, may be found from the formula,
S = P X .0982 D3
where P is the pull of the wire in pounds, D is the diameter of the pin at any point, and X is the distance in inches of that point from the wire. Inspection of this formula shows that S, the maximum strain at any point in the fibres of a pin, when the pull of the line-wire, P, is constant, increases directly with the distance, X, from the wire to the point where the strain, S, takes place. This strain, S, with a constant pull of the line wire, decreases as the cube of the diameter, D, at the point on the pin where S occurs increases. That cross section of a pin just at the top of its hole in the cross-arm is thus subject to the greatest strain, if the pin is of uniform diameter, because this cross section is more distant from the line wire than any other that is exposed to the bending strain. For this reason it is not necessary to give a pin a uniform diameter above its cross-arm, and in practice it is always tapered toward its top. Notwithstanding this taper, the weakest point in pins as usually made is just at the top of the cross-arm, and it is at this cross section where pins usually break. This break comes just below the shoulder that is turned on each pin to prevent its slipping down through the hole in its cross-arm. If the shoulder on a pin made a tight fit all around down onto the cross-arm, the strength of the pin to resist bending would be thereby increased, but it is hard to be sure of making such fits, and they should not be relied on to increase the strength of pins. By giving a pin a suitable taper from its shoulder at the cross-arm to its top, the strain per square inch, S, in the outside fibres of the pin may be made constant for every cross section throughout its length above the cross-arm, whatever that length may be. The formula above given may be used to determine the diameters of a pin at various cross sections that will make the maximum stress, S, at each of these cross sections constant. By transposition the formula becomes
D3 = P .0982 S X.
Where the pin is tapered so that S is constant for all cross sections, then for any pull, P, of the line wire on the pin the quantity (P.0982 S) must be constant at every diameter, D, distant any number of inches, X, from the point where the wire is attached. If the constant, (P.0982 S) is found for any one cross section of a pin, therefore, the diameter at each other cross section with the same maximum stress, S, may be readily found by substituting the value of this constant in the formula. The so-called “standard” wooden pin that has been very generally used for ordinary distribution lines, and to some extent even on high-voltage transmission lines, has a diameter of nearly 1.5 inches just below the shoulder. The distance of the line wire above this shoulder varies between about 4.5 and 6 inches, according to the type of insulator used, and to whether the wire is tied at the side or top of the insulator. If the line wire is tied to the insulator 5 inches above the shoulder of one of the standard pins, then X becomes 5, and D becomes 1.5 in the formula last given. From that formula by transposition and substitution
P.0982 S = D3X = (1.5)35 = 0.675.