General Order 95
Appendix F
Typical Problems
Part II
Deadend Problem
It is the object of this problem to indicate the construction
requirements for a typical deadend structure, since the longitudinal stresses
imposed upon such a structure differ substantially from those on a pole on
which the conductors supported are normally balanced. The deadend structure
considered herein is assumed to support an 11,000 volt circuit, a 4,000 volt
circuit and two secondary circuits. It is also assumed that the deadend pole
takes Grade “A” construction by virtue of its location.
The deadend structure diagram and dimensions are shown on
Page F–22
. The primary data chosen for this structure are as follows:
Data For Deadend Structure
Supply Conductors
11 kV circuit 
3 #0 AWG Stranded, hard–drawn copper 
4 kV circuit 
4 #2 AWG Stranded, hard–drawn copper 
120/240 volt circuit 
3 #4 AWG Solid, hard–drawn copper 
120/240 volt circuit 
3 #2 AWG Stranded, hard–drawn copper 
Insulators  Strain Type (to conform to
Rule 49.5
).
Conductor fastenings (to meet the safety
factor of Table 4
, Rule 44.1
)
Crossarms:
11 kV circuit 
Douglas fir 4–3/4” x 5–3/4” x 8’–0” 
4 kV circuit 
Douglas fir 4–3/4” x 4–3/4” x 7’–8” 
Secondary circuits 
Douglas fir 4–3/4” x 4–3/4” x 7’–0” 
Pole  western red cedar.
Pole dimensions: 55’
in length; 25” top circumference; 50” ground line circumference (ground line
diameter 15.9”).
Construction Requirements
1.
Conductor Tensions
It is assumed
that the conductors are strung with the minimum sags specified in sag curves
of Appendix C
, hence the tension values at 60_F and no wind (normal tensions) are 35%
of the ultimate tensions shown in Table 18
. These tensions for each of the conductor sizes and corresponding tensions
at maximum loading (25°F and wind of 8 pounds) are as follows, where span
length is 250 feet:

Tension–Pounds 


35% of Ultimate 
At Maximum Loading 
#O AWG Stranded, hard–drawn copper 
1,664 
2,125 
#2 AWG Stranded, hard–drawn copper 
1,065 
1,360 
#4 AWG Solid, hard–drawn copper 
690 
890 
2.
Crossarms
Spacings
assumed are shown on the pole framing diagram on
Page F–22
. Double crossarms of Douglas fir, dense, are employed for each of the four
different circuits.
Computations of the fiber stresses imposed upon the various crossarms by
the unbalanced wire loads of conductors in the physical configuration shown
on the diagram are made in accordance with the method outlined in
Part 1
to show these stresses under the conditions of long–time loading and maximum
loading. Furthermore, double crossarm construction of this type with separation
maintained by space bolts is assumed to have a horizontal strength equivalent
to 130% of the sum of the strengths of two single crossarms acting independently.
The stresses computed in this manner are:

Fiber Stress  lbs per Sq. In. 


Long–Time 
Maximum 

Loading 
Loading 
Top crossarms 
1,412 
1,804 
Second Crossarms 
1,598 
2,040 
Third crossarms 
932 
1,202 
Fourth crossarms 
1,438 
1,811 
Since a factor of safety of 2 permits a maximum stress of modulus of rupture
in bending of 1,732 lbs per square inch.
under the conditions of longtime loading (60o F and no Wind) and 3150 lbs per square inch at maximum loading see Table 5 the crossarm chosen are satisfactory.
3. Pole (See
Page F–22
)
Rule 44
provides that poles supporting unbalanced longitudinal loads in Grade “A”
construction shall have a safety factor of 4 against such loads.
Rule 47.3
specifies that guys used to support unbalanced longitudinal loads shall
have a safety factor of 2 for all grades of construction (Where guys are
used they must take the entire load with the designated safety factor, the
pole being considered merely as a strut).
Using the values given above for tensions at maximum loading, the following
moments due to dead ending the conductors are obtained:
3 x 2,125 x 47.3 
= 
301,500 pound–feet 
4 x 1,360 x 38.3 
= 
208,400 pound–feet 
3 x 890 x 30.3 
= 
80,900 pound–feet 
3 x 1,360 x 25.3 
= 
103,200 pound–feet 
Total Moments 
= 
694,000 pound–feet 
The total deadend stress, using the tension values for maximum loading given
above, will be:
3 x 2,125 
= 
6,380 pounds 
4 x 1,360 
= 
5,440 pounds 
3 x 890 
= 
2,670 pounds 
3 x 1,360 
= 
4,080 pounds 
Total 
= 
18,570 pounds 
The tension of a single guy with a lead to height ratio of 1 to 1 (assumed)
and a safety factor of 2 would be:
A stranded guy attached at the center of load could be used provided the
allowable fiber stress of the pole is not exceeded. The stress due to guying
at this point would be as follows:
The center of load (37.4’ above ground) would be
9.9 ft. (118.8”) below the top conductors (11 kV) and
0.9 ft. ( 10.8”) below the second crossarm (4 kV)
The fiber stress in the pole at the center of load due to the tension in
the conductors above the center of load is computed as follows:
Bending moment 
3 x 2,125 x 118.8 
= 
757,400 pound–inches 

4 x 1,360 x 10.8 
= 
58,800 pound–inches 
Total moment 

= 
816,200 pound–inches 
The section modulus of a solid circular section is
The diameter of the pole at the center of load is d = 9.7inches
Then, E = 0.0982 x (9.7)^{3} = 89.6” ^{3}
Since a pole in Grade “A” construction must have a safety factor of 4, the
allowable value of fiber stress would be 5,600/4 = 1,400 pounds per square
inch; therefore, the pole cannot be guyed by a single guy but can be guyed
as illustrated on
Page F–22.