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,
P X
S = -------
.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
P
D³ = ------- X.
.0982 S
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 D³ (1.5)³
------- = --- = ------ = 0.675.
.0982 S X 5
Substituting 0.675 for the quantity (P/0.0982 S) in the formula D³ =
(P/0.0982 S) X gives the formula D³ = 0.675 X, from which the diameters
at all cross sections of a tapered pin above its shoulder, that will
give it a strength just equal to that of a section of 1.5 inches
diameter and 5 inches from the line wire, may be found. To use the
formula for this purpose it is only necessary to substitute any desired
values of X therein and then solve in each case for the corresponding
values of D. Let it be required, for instance, to determine what
diameter a pin should have at a cross section one inch below the line
wire in order that the maximum strain at that cross section may equal
the corresponding strain at a cross section five inches below the line
wire and of 1.5 inch diameter. Substituting one as the value of X, the
last-named formula becomes D³ = 0.675, and from this, D = 0.877, which
shows that the diameter of the pin one inch below the line wire should
be 0.877-inch. A similar calculation will show that if a pin is long
enough so that a cross section above the cross-arm is 12 inches below
the line wire, the diameter of this cross section should be equal to the
cube root of 0.675 × 12 = 8.1, which is 2.008, or practically two
inches. It should be observed that the calculations just made have
nothing to do with the ability of a pin to resist any particular pull of
its line wire. These calculations simply show what diameters a pin
should have at different distances below its line wire in order that the
maximum stress at each of its cross sections may equal that at a cross
section 5 inches below the wire where the diameter is 1.5 inches. In
Vol. xx., A. I. E. E., pp. 415 to 419, specifications are proposed for
standard insulator pins based on calculations like those just made. As a
result of such calculations, the following table for the corresponding
values of X and D, as used in the above formula, are there presented,
each expressed in inches.
+----------+
| X D |
+----------+
| 1 0.877|
| 2 1.106|
| 3 1.263|
| 4 1.395|
| 5 1.500|
| 6 1.592|
| 7 1.678|
| 8 1.754|
| 9 1.825|
|10 1.888|
|11 1.95 |
|13 2.06 |
|15 2.17 |
|17 2.25 |
|19 2.34 |
|21 2.42 |
+----------+
A pin twenty-one inches long between the line wire and the cross-arm
will have a uniform strength to resist the pull of the wire if it has
the diameter given in this table at the corresponding distances below
the line wire. From this it follows that a pin of any length between
wire and cross-arm corresponding to X in the table will be equally
strong to resist a pull of the line wire as a standard 1.5-inch diameter
pin with its wire five inches above the cross-arm. In other words, if a
pin that is twenty-one inches long between the line wire and the
cross-arm has the diameters given in the table at the corresponding
distances below the wire, then a pin of equal strength to resist
bending, and of any shorter length, would correspond in the part above
the cross-arm to an equal length cut from the top end of the longer pin.
Designating that part of a pin that is above the cross-arm as the
“stem,” and that part in the cross-arm as the “shank,” each pin in the
specifications under consideration is named by the length of its stem,
as a 5-, 7- or 11-inch pin. It is proposed that each pin of whatever
length be threaded for a distance of 2.5 inches at the top of its stem
with four threads per inch, the sides of each thread being at an angle
of ninety degrees with each other. Each thread is to cut into the pin
about 3/32 inch, come to a sharp angle at the bottom, and be about 1/16
inch wide on top. At the end of the pin the proposed diameter over the
thread is one inch in all cases, and at the lower end of the threaded
portion the outside diameter is 1.25 inches. Near the end of the pin the
diameter at the bottom of the thread is thus only 13/16 inch, and the
corresponding diameter at the lower end of the threaded portion is about
1-1/16 inches on all pins. Each pin is to have a square shoulder to rest
on the cross-arm, and the diameter of this shoulder is to be 3/8 inch
greater than the nominal diameter of the shank of the pin. The proposed
length of this shoulder on all pins is 1/4 inch before the taper begins.
The actual diameter of the shank of each pin just below its shoulder is
to be 1/32 inch less than the nominal diameter, and the actual diameter
of the lower end of each shank is to be 1/16 inch less than the nominal
diameter. With these explanations the proposed sizes of pins have
dimensions as follows in inches:
+---------+---------+-----------+
| | | Nominal |
|Length of|Length of|Diameter of|
| Stem. | Shank. | Shank. |
+---------+---------+-----------+
| 5 | 4-1/4 | 1-1/2 |
| 7 | 4-1/4 | 1-3/4 |
| 9 | 4-1/4 | 1-7/8 |
| 11 | 4-3/4 | 2 |
| 13 | 4-3/4 | 2-1/8 |
| 15 | 4-3/4 | 2-1/4 |
| 17 | 5-3/4 | 2-3/8 |
| 19 | 5-3/4 | 2-1/2 |
+---------+---------+-----------+
In order rightly to appreciate the utility of this table of proposed
standard pins, it is necessary to have in mind the fact that all the
dimensions are based on the assumption that a wooden pin with a shank of
one and one-half inches diameter, and with its line wire attached five
inches above the cross-arm, is strong enough for general use on
transmission lines. Such an assumption covers a wide range of practice,
but its truth may well be doubted for many cases. That this assumption
does form the basis of the entire table is clearly shown by the fact
that the calculated diameter at the shank of each pin is made to depend
on a uniform pull, _P_, of the line wire, giving a uniform maximum
stress, _S_, in the outer fibres of the wood just where the shank joins
the stem. In other words, every pin in the table is designed to break
with a uniform pull of the line wire, provided that the point on the
insulator where the wire is attached is just on a level with the top of
its pin in each case. It will at once occur to practical men that while
a five-inch pin with one and one-half inch shank, or a larger pin of
equal ability to resist the pull of a line wire, may be strong enough
for the conductors of some transmission lines, this same pin may be
entirely too weak for the longer spans, sharper angles, and heavier
conductors of other lines.
Thus, on the sixty-five-mile line between Cañon Ferry and Butte, Mont.,
each conductor is of copper and has a cross section of 106,500 cm.,
while on the older line between Niagara Falls and Buffalo each copper
conductor has a cross section of 350,000 cm. Evidently with equal
conditions as to length of span, amount of sag, and sharpness of angles
on these two lines, pins ample in strength for the smaller wire might be
much too weak for the larger wire.
A little consideration will show that it is neither rational nor
desirable to adopt pins of uniform strength for all transmission lines,
but that several degrees of strength are necessary to correspond with
the range in sizes of conductors in regular use. The size of pins for
use on any transmission line, when the maximum bending strain exerted by
the conductors has been determined, should be found by calculation and
experiment, or by experiment alone. According to Trautwine, the average
compressive strength of yellow locust is 9,800 pounds, of hickory 8,000
pounds, and of white oak 7,000 pounds per square inch in the direction
of the grain. These compressive strengths are less than the tensile
strengths of the same woods, and should therefore be employed in
calculation, since the fibres on one side of a bending pin are
compressed while the fibres on the other side are elongated.
Substituting 1,000 for the value of S in the formula, S =
(P X/(.0982 D³)), and also 5 for the value of X, and 1-1/2 for the value
of D, the resulting value of P is found to be 736.5 pounds. This result
shows that with a locust pin of 1-1/2 inches diameter at the shank, and
with its line wire attached five inches above the shoulder, the
unbalanced side pull of the wire that will break the pin by bending is
736 pounds, provided that the wood of the pin has a strength of 1,000
pounds per square inch in compression. As all of the proposed standard
pins in the above table are designed for uniform strength to resist the
same pull of a line wire attached on a level with the top of the pin in
each case, it follows that the pull of 736 pounds by the wire will break
any one of these pins under the conditions stated.
The calculation just made takes no account of the fact that the actual
diameter of the shank of each pin just below the shoulder is 1/32 inch
less than the nominal diameter, but this of course reduces the strength
somewhat. Trautwine states that the figures above given for the
compressive strengths of wood are only averages and are subject to much
variation. Of course no pin should be knowingly loaded in regular
practice to the breaking point, and to provide against variations in the
strength of wood, and for unexpected strains, a liberal factor of
safety, say four, should be adopted in fixing the maximum strains on
insulator pins. Applying this factor to the calculations just made, it
appears that the maximum pull of the line wire at the top of any one of
the above proposed standard pins should not exceed 736 ÷ 4 = 184 pounds
in regular work. A little calculation will readily show that the side
pull of some of the larger conductors now in use on transmission lines
will greatly exceed 184 pounds under conditions, as to sag, angles and
wind pressure, that are frequently met in practice.
On page 448, Vol. xx., A. I. E. E., some tests are reported on six
locust wood pins with shank diameters of 1-7/16 to 1-1/2 inches. Each of
these pins was tested by inserting its shank in a hole of 1-1/2 inches
diameter in a block of hard wood, and then applying a strain at about
right angles to the pin and about 4-1/2 inches from the block by means
of a Seller’s machine. The pull on each pin was applied gradually, and
in most of the pins the fibres of the wood began to part when the side
pull reached 700 to 750 pounds, though the maximum loads sustained were
about ten per cent above these figures. The average calculated value of
S, the compressive strength of the wood in these pins, was 11,130 pounds
per square inch on the basis of the loads at which the fibres of the
wood began to break, and 13,623 pounds per square inch for the loads at
which the pins gave way. On pages 650 to 653 of the volume last cited,
results are reported of tests on twenty-two pins of eucalyptus wood,
which is generally used for this purpose in California. Twelve of these
pins were of a size much used in California on lines where the voltage
is not above 30,000. Each of the twelve pins was 6-7/8 inches long in
the stem, 4-5/8 inches long in the shank, 1-1/2 inches in diameter at
the shank, 2 inches in diameter at the square shoulder where the shank
joins the stem, and 1-3/8 inches in diameter at the top of the thread.
The pins were tested by mounting each of them in a cross-arm, securing
the cross-arm in a testing machine so that the pin was horizontal,
placing an insulator on the pin, and exerting the strain on a cable
wrapped around the side groove of the insulator. This cable varied a
little from right angles to the axis of each pin, but the component of
the strain at right angles to this axis was calculated and the breaking
load here mentioned is that component. Nearly all of these twelve pins
broke square off at the cross-arm.
For a single pin, the lowest breaking strain was 705 pounds, the largest
1,360 pounds, and the average for the twelve pins was 1,085 pounds.
Unfortunately, the exact distance of the cable from the cross-arm is not
stated, but as the cable was wound about the side groove of the
insulator it was probably either in line with or a little below the top
of the pin. It seems probable also that the diameter of these pins at
the shoulder--that is, two inches--may have increased the breaking
strain somewhat by giving the shoulder a good bearing on the cross-arm.
The ten other pins were of the size in use on the 60,000-volt line
between Colgate power-house and Oakland, Cal. Each of these pins had a
length of 5-3/8 inches and a maximum diameter of 2-1/8 inches in the
shank, and a length of 10-3/8 inches in the stem, with a diameter of
2-1/2 inches at the shoulder. This shoulder was not square, but its
surface formed an angle of forty-five degrees with the axis of the pin,
and this bevel shoulder took up 1/4 inch of the length just given for
the stem of the pin. At 2-1/2 inches from its threaded end the stem of
the pin had a diameter of 1-15/16 inches, and the diameter slopes to
1-3/8 inches at two inches from the end. The two inches of length at the
top of the stem has the uniform diameter of 1-3/8 inches, and is
threaded with four threads per inch for the insulator. Each of these ten
pins was tested, as already described, until it broke, but the break in
this case started as a split at the lower end of the threaded portion
and ran down the stem to the shoulder in a line nearly parallel with the
axis of the pin. The pull on the cable at right angles to the axis of
each pin had a maximum value of 1,475 pounds in one case, and a
corresponding value of 3,190 pounds in another, while the average
breaking strain for the ten pins was 2,310 pounds. Unfortunately, the
report of this test above named does not distinctly state just how far
the testing cable was attached above the shank of each of these large
pins; but it seems probable that the same insulator was used with the
larger as with the smaller pins, and if this was so the testing cable
was attached near the end of each pin, as this cable was wound about the
side groove of the insulator used on the smaller pins. With the types of
insulator in actual use on the Colgate and Oakland line the wire is
carried at the top groove and its centre is about two and a half inches
above the top of the pin. It is therefore probable that these pins would
not withstand as great strains on the lines as they did in these tests.
The bevel shoulder on each of these larger pins no doubt increases its
ability to resist a bending strain, because the bevel surface fits
tightly down into a counterbore in the cross-arm. Where the pin has a
shoulder at right angles with the axis, as is more usually the case, and
the top of the cross-arm is a little rounding, the square shoulder does
not have a firm seat and is of slight importance as far as the strength
of the pin to resist a bending strain is concerned. Evidently the
weakest point in the ten larger pins of this test was at the lower end
of the threaded portion, since in each case the break was in the form of
a long split starting where the thread ended. There seems to be no
sufficient reason for the reduction of the diameter of a pin intended
for a heavy line wire to a diameter as small as one inch at the threaded
end, or for limiting the length of the threaded portion to 2.5 inches,
as proposed in the specifications for standard pins. It is certain that
the cost of the pin would be no more if its diameter at the threaded end
were 1-1/4 or 1-3/8 inches with a uniform taper from the end of the pin
down to the shoulder and with the thread cut down the stem for three or
four inches. Furthermore, any increase in the cost of insulators for
these larger threaded ends of pins would no doubt be a small matter.
Some excess of strength in the stem of a pin over that of its shank is
to be desired, for the stem is more exposed to the weather and to
charring by leakage currents over the surface of the insulator. On
high-voltage lines, this charring is usually worse at that part of each
pin just below its thread, and the commonest breaks of pins on these
lines leave the insulators with the threaded portions of their pins
hanging on the wire, while the remainder of each pin remains on the
cross-arm. From the tests just noted it is evidently poor design to give
the threaded portion of a pin a short length of uniform diameter, and
then to increase the diameter at once by a shoulder, as was done with
the pins on the Colgate and Oakland line. This design evidently leads to
failure of pins by splitting from the lower end of the threads. The
better design is the more common one which gives the stem of the pin a
uniform taper from the shoulder to the top. Where the line wire is
secured to the top of its insulator, anywhere from one to three inches
above the top of the pin, there is a strong tendency for the insulator
to tip on its pin, and this tendency is more effectively met the longer
the joint between the pin and insulator.