UNITED STATES DEPARTMENT OF AGRICULTURE 

Rural Utilities Service 

RUS BULLETIN 1724E-151 

RD-GD-2016-80

SUBJECT: Mechanical Loading on Wooden Distribution Crossarms 

TO: RUS Electric Borrowers and RUS Electric Staff

EFFECTIVE DATE: Date of Approval 

OFFICE OF PRIMARY INTEREST: Distribution Branch, Electric Staff Division 

FILING INSTRUCTIONS: This bulletin replaces RUS Bulletin 1724E-151, "Mechanical 

Loading on Wooden Distribution Crossarrns," issued November 21, 2002. 

AVAILABILITY: This bulletin is available on the Rural Utilities Service website at 

http://www.usda.gov/rus/electric. 

PURPOSE: To furnish design information needed to determine the permitted mechanical 

loading on wood distribution crossaims and crossarm assemblies. 

fi staph r A. McLean, 

Assistant Administrator, 

Electric Prograin 

bate 

Disclaimer: The contents of this guidance document does not have the force and effect of law and is not 

meant to bind the public in any way. This document is intended only to provide clarity to the public regarding 

existing requirements under the law or agency policies.
Bulletin 1724E-151

Page 2

TABLE OF CONTENTS

Page

1 PURPOSE................................................................................................................................3

2 GENERAL INFORMATION REGARDING CROSSARMS ................................................3

3 CROSSARM MOMENT CAPACITIES.................................................................................5

4 VERTICAL LOADING ON CROSSARM ASSEMBLIES....................................................6

5 LONGITUDINAL LOADING ON CROSSARM ASSEMBLIES .......................................11

Exhibits

Exhibit A: NESC District Ice and Wind Loads for Conductors

Exhibit B: Contributors

Exhibit C: Metric Conversion Factors

ABBREVIATIONS

ACSR Aluminum conductor, steel reinforced

ANSI American National Standards Institute

AWG American Wire Gage

CFR Code of Federal Regulations

IEEE Institute of Electrical and Electronics Engineers

NESC National Electrical Safety Code 

RUS Rural Utilities Service
Bulletin 1724E-151

Page 3

1 PURPOSE

This bulletin presents equations, data and other information needed to determine the 

vertical and unbalanced longitudinal loads on wood distribution crossarms and 

assemblies. Sample solved problems that calculate the permitted loading on Rural 

Utilities Service (RUS) standard deadend crossarm assemblies are presented to help the 

reader understand and apply the equations and information presented within this bulletin. 

Exhibit A on page 15 of this bulletin is a table of ice and wind loading data for the most 

commonly used distribution line conductors.

National Electrical Safety Code: This bulletin references rules and presents selected data 

contained in the 2012 Edition of the National Electrical Safety Code (NESC). At the 

time this bulletin was written, the 2012 Edition was the latest edition of the NESC. 

Periodically the NESC is updated and revised. Users of this bulletin should use the rules 

and data, as may be revised and renumbered, from the most recent edition of the NESC. 

Copies of the NESC may be obtained from the Institute of Electrical and Electronics 

Engineers, Inc., (IEEE) at the following address:

IEEE Customer Service

445 Hoes Lane, P.O. Box 1331

Piscataway, NJ 08855-1331

Telephone: 1-800-678-IEEE (1-800-678-4333)

Throughout this bulletin, references to the NESC pertain specifically to the 2012 Edition 

of the National Electrical Safety Code.

2 GENERAL INFORMATION REGARDING CROSSARMS

a References to RUS Standard Crossarm Assemblies: Within the text of this 

bulletin are references to standard RUS distribution construction assemblies such 

as “VC5.21.” These assemblies are contained in RUS Bulletin 1728F-803 (D803), “Specifications and Drawings for 24.9/14.4 kV Line Construction.” 

Crossarm dimensions, attachment spacing, and strength calculations pertaining to 

the example 24.9/14.4 kV assemblies are equally applicable to the corresponding 

RUS standard 12.47/7.2 kV crossarm assemblies.

b Crossarm Dimensions: A standard RUS wood distribution crossarm is 4-5/8 

inches in the vertical (v) direction and 3-5/8 inches in the horizontal (h) direction. 

Since the manufacturing tolerances of crossarms are plus or minus 1/8 inch, the 

dimensions used for strength calculations in this bulletin are 4 1⁄2 inches by 3 1⁄2 

inches. For further information reference RUS construction specifications 

(1728F-803 and 1728F-802) for drawing W1.2G which depicts standard 

dimensions and drilling guides for wood crossarms.
Bulletin 1724E-151

Page 4

c Crossarm Installations: All RUS standard crossarms and crossarm assemblies are 

attached to the pole at the center of the crossarm or crossarm assembly. RUS 

standard crossarm assemblies are symmetrical on each side of the pole.

d Loads Applied to Crossarms: Loading on crossarms is the sum of the following 

forces when applicable: conductor tensions, weight of conductors and ice, and the 

force of the wind on conductors and the crossarm. NESC Rule 261D2a(1) states: 

“Crossarms and braces shall be designed to withstand the loads in Rule 252, 

multiplied by the load factors in Table 253-1 without exceeding their permitted 

stress.” In order to determine the permitted stress, NESC Rule 261D2a(2) states: 

“The permitted stress level of solid sawn or laminated wood crossarms and braces 

shall be determined by multiplying their ultimate fiber stress by the strength 

factors in Table 261-1”. 

e Unbalanced Longitudinal Loading: Unbalanced longitudinal crossarm loading is 

the difference in loading from the front to the back of a crossarm. For example, 

as shown in Diagram 2.5, unbalanced loading is the difference between the loads 

L1-in and L1-out. Unbalanced loading pertains to “double deadend” crossarm 

assemblies as shown in Diagram 2.5 and to “single deadend” assemblies as would 

be the case if L1-out and L2-out (and L3-out) were omitted in Diagram 2.5. 

Conductors attached to the center of a crossarm assembly, (e.g., L3-out) do not 

contribute to longitudinal or vertical loading on the crossarm(s) but do contribute 

to loading on the pole. RUS assumes that in the design and construction of 

distribution lines that the side-to-side loading on crossarm assemblies is balanced, 

for example in Diagram 2.5, L1-in minus L1-out equals L2-in minus L2-out. (Sides 

mean right or left of the pole when looking down the line parallel to the 

conductors.)

Unbalanced loading can occur when the tensions in the conductors that are 

attached “into” and “out from” the crossarm assembly are unequal because of: (1) 

a change in conductor size or type; (2) different installation tensions; (3) unequal 

ice or wind loading; or, (3) different ruling spans.

L1-out L2-out

L3-out

L1-in L3-in L2-in

Diagram 2.5
Bulletin 1724E-151

Page 5

f Grades of Construction: In Subpart E of 7 CFR Part 1724, RUS requires 

overhead distribution line components (poles, guys, etc.,) to be constructed to no 

less than NESC Grade C strength requirements. NESC Section 24 including 

Table 242-1 specifies where Grade B must be used in place of Grade C and 

includes where distribution lines cross over or overhang a railroad track, the 

traveled way of a limited access highway, or navigable waterways requiring 

waterway crossing permits. RUS recommends that crossarm assemblies used at 

deadends on Grade C constructed lines be constructed to meet NESC Grade B 

strength requirements.

(1) Distribution crossarms on transmission structures may be designed for NESC 

Grade C, except at angles where they are required to be designed to Grade B. 

See RUS Bulletin 1724E-200, Section 16 for a more complete discussion of 

distribution underbuild on transmission structures. Note that for distribution 

underbuild on transmission structures, the distribution loads are applied to the 

transmission structures using NESC Grade B load criteria.

3 CROSSARM MOMENT CAPACITIES

a Moment Capacities of Crossarms: The moment capacity of a crossarm, Mu, can 

be expressed as:

M (F ) (X ) u = b × ft-lbs EQ-3.1a

Where:

Mu = the moment capacity of a crossarm

Fb = the designated modulus of rupture for wood crossarms

X = the section modulus of the crossarm

b Section Moduli: The section moduli for resisting longitudinal and vertical loads, 

Xh and Xv respectively, are:

( ) ( )

6

2 d a b X h

− × = in3 EQ-3.2a

( ) ( )

d

d a b Xv 6

3 3 − × = in3 EQ-3.2b

Where:

b = width (horizontal dimension) of the crossarm (= 3 1⁄2 inches)

d = depth (vertical dimension) of the crossarm (= 4 1⁄2 inches)

a = diameter of the crossarm mounting bolt hole (= 11/16 inches) 

Substituting the values given for d, a, and b into equations EQ-3.2a and EQ-3.2b 

yields the following values for the longitudinal and vertical section moduli for a 

RUS standard wood crossarm:

Xh = 7.784 in3

Xv = 11.770 in3
Bulletin 1724E-151

Page 6

c Longitudinal and Vertical Crossarm Moment Capacities: The following 

longitudinal and vertical moment capacities, Mh and Mv, for single distribution 

crossarms are calculated by multiplying the section moduli, Xh and Xv, by the 

designated modulus of rupture for wood. The modulus of rupture for wood 

needs to be determined for the species of wood utilized in the crossarm 

construction. For the purpose of this bulletin, the modulus of rupture for wood is 

assumed to be 7,800 lbs/in2. Thus,

Mh = (7.784 in3) x (7,800 lbs/in2) = 60,715 in-lbs = 5,060 ft-lbs

Mv = (11.770 in3) x (7,800 lbs/in2) = 91,806 in-lbs = 7,650 ft-lbs

4 VERTICAL LOADING ON CROSSARM ASSEMBLIES

a Assumptions: Rule 261D5b of the NESC specifies that “crossarms shall be 

supported by bracing, if necessary, to support expected loads, including line 

personnel working on them.” RUS assumes that a lineworker and equipment may 

apply a weight of 250 pounds, 2 feet from the center of the pole. (Note: Standard 

construction practices and RUS discourage lineworkers from standing on 

crossarms.) For calculation purposes, RUS applies a load factor (FLF) of 2.0 to 

the weight that might be caused by a lineworker. Thus, the vertical load moment 

due to the lineworker multiplied by the assumed load factor is 1,000 foot-pounds 

(2 feet x 250 pounds x 2.0).

(1) Crossarm braces are primarily installed to stabilize the horizontal position 

of crossarms. RUS does not factor in any additional strength due to 

crossarm braces in its calculations. Furthermore, RUS ignores the weight 

of crossarms, pins, insulators and connectors in its calculations.

(2) RUS assumes that a conductor’s longitudinal (tension) load does not 

contribute any vertical load to a crossarm assembly. This assumption is 

correct only if the conductor is perfectly horizontal. However, because of 

sag, nearly horizontal conductor tensions do exert small vertical loads on 

crossarm assemblies at the points of attachment. Since this additional 

vertical load cannot be precisely determined and is small compared with 

the vertical loading of the conductors and ice, it is ignored.

(3) Within this bulletin, RUS assumes that the distribution line sections 

adjacent to the assembly under consideration have level spans (i.e., the 

conductor attachments are at the same altitude).

b Applied Vertical Loads Equated to Crossarm Assembly Capacity: The NESC 

requires that the sum of each vertical load attached to a crossarm assembly, 

multiplied by the load’s appropriate NESC vertical load factor and the load’s 

distance to the center of the crossarm assembly, not exceed the permitted vertical 
Bulletin 1724E-151

Page 7

moment capacity of the crossarm assembly. Mathematically, this statement is 

written:

∑(Si ×Wi ×D j×F LF )+ 000 ≤ N × Mv ×F s 1, (ft-lbs) EQ-4.2a

Where:

Si = 1⁄2 of the span length of the conductor “into” plus 1⁄2 of the span 

length of the conductor “out from” the crossarm assembly (ft) (“i”, 

the number of conductors, ranges from 1 to 4 for either side [half] 

of the crossarm) (ft)

Wi = unit weight of the conductor (or unit weight of conductor plus 

wind and ice loading, if applicable) “into” or “out from” the 

crossarms (lb/ft)

Dj = distance of load “Li” from center of the crossarm(s) (ft) (“j”, the 

number

 of load attachment points, is 1 to 2 for either side [half] of the 

crossarm) (ft)

FLF = NESC vertical load factor (Table 253-1 of the NESC) 

= 1.90 for Grade C construction or 1.50 for Grade B construction

1,000 = vertical load moment times load factor attributed to the weight of

 a lineworker and equipment (ft-lbs) (See Paragraph 4.1 of this 

bulletin.)

N = number of crossarms (N equals 1, 2 or 3) 

Mv = vertical moment capacity (= 7,650 ft-lbs) (See Paragraph 3.3 of 

this bulletin.)

Fs = NESC strength factor (Table 261-1 of the NESC)

= 0.85 for Grade C construction or 0.65 for Grade B construction

Wi for conductors with applicable NESC wind and ice loading, for each NESC 

loading district, can be found in Exhibit A on page 15 of this bulletin.

(1) All of the conductor spans “into” a crossarm assembly have the same 

length and all of the conductor spans “out of” a crossarm assembly have 

the same length. For purposes of this bulletin, when performing vertical 

loading calculations, the total length of 1⁄2 of the conductor span length 

“into” the assembly (Sin), plus 1⁄2 of the conductor span length “out from” 

the assembly (Sout) is called a “weight span” because the assembly bears 

the conductor weight of these two half spans.

(2) Standard RUS distribution crossarm assemblies have a maximum of 2 

conductor attachment locations (Dj, where j = 1 or 2) on either side of the 

pole. Each attachment location has a maximum of 2 attached conductors; 

one “into” the crossarm and the other “out from” the crossarm as shown in 
Bulletin 1724E-151

Page 8

Diagram 4.2 below. Thus, standard RUS distribution crossarm assemblies 

have a maximum of 4 conductors (i = 1, 2 or 4) attached to the crossarm 

on either side of the pole.

(3) The induced vertical moments of a lineworker, conductors, and wind and 

ice loads (when applicable) on one side of a standard RUS distribution 

crossarm assembly equated to its permitted vertical moment is expressed 

as:

∑(Si ×Wi ×D j×F LF )+ 000 ≤ N × Mv ×F s 1, =

[( ) ( )] [( ) ( )] in out LF in out W FLF N M v Fs D1 × S ×W1 + S ×W2 × F + D2 × S ×W3 + S × 4 × +1,000 ≤ × × EQ-4.2b

The right-hand side of the equation (N x Mv x Fs) is defined as an 

assembly’s “permitted vertical load moment.” The units for both sides of 

the equation are “ft-lbs.”

c Vertical Loads on RUS Standard Tangent and Small Angle Crossarm Assemblies: 

The conductors on RUS standard tangent and small angle crossarm assemblies are 

attached to the crossarms with crossarm pins and insulators or post type 

insulators. Each side (half) of the crossarm assembly may support 1 or 2 

conductors that “feed through,” or literally, feed over the assembly. Thus, the 

conductors into and out from the assembly are the same type and size and, as 

shown in Diagram 4.2, W1 = W2 and W3 = W4

(1) After substituting the values for Wi that are equal (i equals 1 through 4) 

and rearranging the terms, equation EQ-4.2b in Paragraph 4.2.3 of this 

bulletin, for “two feedthrough conductors” on each side of the pole, 

(e.g., “VC1.41”) becomes:

Sout x W2 Sout x W4

Sin x W1 Sin x W3

D2

D1 

Diagram 4.2
Bulletin 1724E-151

Page 9

( ) ( ) in out LF in out FLF N Mv Fs D1 ×W1 × S + S ×F +D2 ×W3 × S + S × +1,000 ≤ × × (ft-lbs) EQ-4.3a

Where: N = 1 or 2 crossarms (Only 10-foot crossarms are used on these assemblies) 

 D1 = 4-ft, 8-in (4.666 ft); D2 = 1-ft, 7-in (1.583 ft)

If both of the conductors are the same size and type, then W1 = W3 and equation EQ-4.3a 

simplifies to:

( ) ( ) in out FLF N Mv Fs W1 × D1 + D2 × S + S × +1,000 ≤ × × (ft-lbs) EQ-4.3b

(2) For tangent or small angle assemblies that support “one feedthrough 

conductor” on each side of the pole (e.g., “VC1.11”), D2 = 0 and equation 

EQ-4.3b simplifies to:

( ) in out F LF N Mv Fs D1 ×W1 × S + S × +1,000 ≤ × × (ft-lbs) EQ-4.3c

Where: N = 1 or 2 crossarms (Only 8-foot crossarms are used on these assemblies)

D1 = 3-ft, 8-in (3.666 ft) 

d Vertical Loads on RUS Standard Single and Double Deadend Crossarm 

Assemblies: On RUS standard deadend crossarm assemblies the conductors are 

attached to the crossarms with suspension insulators. Each half of the crossarm 

assembly (on either side of the pole) may support 1, 2 or 4 conductors “into” or 

“out from” the assembly.

(1) On double deadend crossarm assemblies, the conductors are attached 

back-to-back on the crossarms as shown in Diagram 4.2 in Paragraph 4.2.2 

of this bulletin. Following is the equation for applied vertical crossarm 

load moments relative to an assembly’s permitted vertical moment 

capacity for assemblies that support 4 conductors (2 “into” and 2 “out 

from”) on each side of the pole (e.g., “VC6.51”):

[( ) ( )] [( ) ( )] in out LF in out W FLF N M v Fs D1 × S ×W1 + S ×W2 × F + D2 × S ×W3 + S × 4 × +1,000 ≤ × ×

EQ-4.4a

Where: N = 2 or 3 crossarms (Only 10-foot crossarms are used on these assemblies.) 

D1 = 4-ft, 6-in (4.5 ft); D2 = 1-ft, 9-in (1.75 ft)

The units of measure for each side of the equation are “ft-lbs.”

(2) For 1 conductor “into” and 1 conductor “out from” the assembly (back to 

back) on each side of the pole (e.g., ”VC6.21”), D2 = 0 and equation EQ4.4a is simplified to:
Bulletin 1724E-151

Page 10

[( ) ( )] in out W FLF N Mv Fs D1 × S ×W1 + S × 2 × +1,000 ≤ × × (ft-lbs) EQ-4.4b

Where: N = 2 or 3 crossarms (Only 8-foot crossarms are used on these assemblies.)

D1 = 3-ft, 6-in (3.5 ft) 

(3) On single deadend crossarm assemblies the conductors deadend in only 

one direction on the crossarm assembly (see “VC5.21”). Thus, the term 

(Sout x W2) is omitted from the crossarm vertical load moment equations.

(4) RUS has no standard crossarm single deadend assemblies that only 

support 2 “in” conductors on each side of the pole. Thus, no equations are 

offered for this arrangement.

(5) For single deadend assemblies that support 1 conductor into the assembly 

on each side of the pole (e.g., “VC5.21”), the equation for permitted 

vertical loading is simply:

( ) in W FLF N Mv Fs D1 × S × 1 × +1,000 ≤ × × (ft-lbs) EQ-4.4c

e Example Problem – Permitted Vertical Loading on Crossarms: Determine the 

maximum “weight span” (Sin + Sout) for a 3-phase, # 4/0 aluminum conductor, 

steel reinforced (ACSR) primary distribution line with a # 1/0 ACSR neutral 

deadending in both directions on the “VC6.51” crossarm assembly shown in 

Diagram 4.2 in Paragraph 4.2.2 of this bulletin. The structure is installed to 

NESC Grade C construction requirements in the NESC Heavy Loading District. 

The following data is found in Exhibit A on page 15, Diagram 4.2 in Paragraph 

4.2.2, and in Section 4.2 of this bulletin.

W1 = 0.9520 lb/ft (# 4/0 ACSR; conductor plus .50 inches of ice)

W3 = 0.7036 lb/ft (# 1/0 ACSR; conductor plus .50 inches of ice)

N = 2 (10-foot crossarms)

D1 = 4.50 ft; D2 = 1.75 ft (deadends on 10-foot crossarms)

FLF = 1.90; Fs = 0.85 (NESC Grade C)

Mv = 7,650 ft-lbs (See Paragraph 3.3 of this bulletin)

The critical vertical loading on this assembly is the half of the assembly that 

supports the two heavier primary conductors. All of the conductors now under 

consideration are identical #4/0 ACSR and thus in equation EQ-4.4a in Paragraph 

4.4.1 of this bulletin, W1 = W2 = W3 = W4 . After setting all of the Wi terms equal 

to W1 and rearranging, the equation can be simplified to:

( ) ( ) D + D ×W × Sin + Sout × FLF + 000 ≤ N × Mv × Fs 1, 1 2 1 (ft-lbs) EQ-4.5a
Bulletin 1724E-151

Page 11

( ) ( )

( ) 11.305

12,005

4.5 1.75 0.9520 1.9

2 7,650 0.85 1,000 = + × ×× × − Sin + Sout = (ft)

(Sin + Sout) = 1,062 ft/ phase

f Remarks Regarding Vertical Loads on Distribution Crossarm Assemblies: For 

the example problem in Paragraph 4.5 of this bulletin assume that 477 kcmil 

(Pelican, 18/1) ACSR primary conductors are to be installed to NESC Grade B 

strength requirements. Thus:

W1 = 1.3350 lb/ft (conductor plus .50 inches of ice)

FLF = 1.50; Fs = 0.65 (NESC Grade B) and 

( ) ( )

( ) 715

12.516

8,945

4.5 1.75 1.3350 1.50

2 7,650 0.65 1,000 = = + × ×× × − Sin + Sout = ft/phase

The sum (Sin + Sout) can be thought of as an average total span length “into” and 

“out from” the assembly. 

5 LONGITUDINAL LOADING ON CROSSARM ASSEMBLIES

a Combined Vertical and Longitudinal Loads on Crossarm Assemblies: In 

Section 4 of this bulletin it was assumed that longitudinal loads did not contribute 

to the vertical loading on crossarm assemblies. (See Paragraph 4.1.2 of this 

bulletin.) However, applied vertical loads do have to be considered when 

determining the permitted longitudinal load of a crossarm assembly. The 

following relationship needs to be satisfied to avoid overstressing the wood fibers 

of crossarms:

1

( ) ( )

+ ≤ ∑ ∑

Permitted Longitudinal Moment Capacity

Applied Longitudinal Load Moments

Permitted Vertical Moment Capacity

Applied Vertical Load Moments EQ-5.1a

For RUS standard deadend crossarm assemblies:

∑ Applied Vertical Load Moments =

[( ) ( )] [( ) ( )] 1,000 D1 × Sin ×W1 + Sout ×W2 × FLF + D2 × Sin×W3 + Sout ×W4 × FLF + EQ-5.1b

Permitted Vertical Moment of Assembly = N x Mv x Fs (ft-lbs)

Permitted Longitudinal Moment of Assembly = N x Mh x Fs (ft-lbs)

See Paragraph 4.2 of this bulletin for an explanation of the above equations and 

definition of the terms. See the various paragraphs of Section 4.4 of this bulletin 

for simplification of the “applied vertical load moments” equation for specific 
Bulletin 1724E-151

Page 12

assembly types. The determination of the “applied longitudinal load moments” is 

discussed in Section 5.2 of this bulletin.

(1) Example Problem of Combined Vertical and Longitudinal Loads on a 

Crossarm Assembly: Assume the same conditions as those given for the 

example problem in Paragraph 4.5 of this bulletin. Also assume a weight 

span (Sin + Sout) that equals 300 feet. Thus, 

 ∑ Applied Vertical Load Moments = (D1 + D2 )×W1 × (Sin + Sout)× FLF +1,000 ft-lbs

= (4.5 +1.75)×0.9520×(300)×1.9 +1,000 ft-lbs

= 4,392 ft-lbs

N x Mv x Fs = 2 x 7,650 x 0.85 = 13,005 ft-lbs

N x Mh x Fs = 2 x 5,060 x 0.85 = 8,602 ft-lbs

The maximum or permitted sum of the “applied longitudinal load moments” (LPL), 

for the specific given and assumed conditions can be calculated as follows:

1

13,005 8,602

4,392

≤ −

− + −

−

ft lbs

L ft lbs

ft lbs

ft lbs PL

∑ Applied Longitudinal Load Moments = LPL ≤ 5,697 ft-lbs

b Applied Longitudinal Loads Equated to Crossarm Assembly Capacity: The sum 

of each longitudinal load attached to a crossarm assembly, multiplied by its 

appropriate NESC longitudinal load factor and by its distance to the center of the 

crossarm assembly, can not exceed the permitted applied longitudinal load 

moment (LPL) as determined in Paragraphs 5.1 and 5.1.1 of this bulletin. 

Mathematically, this statement is expressed as follows for RUS standard double 

deadend crossarm assemblies that support 2 phases on each side of the pole. 

See Diagram 5.2 that follows.

[ ( ) ( )] D1 × L1−in − L1−out + D2 × L2−in − L2−out × FLF ≤ LPL (ft-lbs) EQ-5.2a

Where:

Li = load (conductor tension) “into” or “out from” the crossarm assembly (ft)

 (i = 1, 2 or 4: See Paragraph 4.2 of this bulletin)

Dj = distance of load “Li” from center of the crossarm(s) (ft)

 (“j” = 1 or 2: See Paragraph 4.2 of this bulletin)

FLF = NESC longitudinal overload factor (Table 253-1 of the NESC) 

= 1.30 for Grade C; 1.65 for Grade B: (At Deadends)

LPL = permitted longitudinal moment of all applied loads (Paragraph 5.1.1 of 

this bulletin)
Bulletin 1724E-151

Page 13

(1) If the two conductors “into” the assembly are at the same tension, and the 

two conductors “out from” the assembly are at the same tension, (but the 

“into” and “out” tensions are not the same), then: L1-in = L2-in, L1-out = L2-

out, and (L1-in - L1-out) equals (L2-in - L2-out). Equation EQ-5.2a in 

Paragraph 5.2 of this bulletin can be rewritten and rearranged as:

[( ) ( )] D1 + D2 × L1−in − L1−out × FLF ≤ LPL (ft-lbs) EQ-5.2b

(2) For one conductor “into” and 1 conductor “out from” the assembly (back 

to back), on each side of the pole (e.g., “VC6.21”), D2 = 0 and equation 

EQ-5.2b is simplified to:

[ ( )] D1 × L1−in − L1−out × FLF ≤ LPL (ft-lbs) EQ-5.2c

Where: N = 2 or 3 crossarms (Only 8-foot crossarms are used on these 

assemblies.)

 D1 = 3-ft, 6-in (3.5 ft);

(3) RUS has no standard, single deadend crossarm assemblies that only 

support two “in” conductors on each side of the pole. Thus, no equations 

are offered for this arrangement.

(4) Single deadend crossarm assemblies have no conductors “out from” the 

assemblies and so the term “Lx-out” is set to 0. Equation EQ-5.2b in 

Paragraph 5.2.1 of this bulletin simplifies to:

D1 × L1−in × FLF ≤ LPL (ft-lbs) EQ-5.2d

L1-out L2-out L3-out L4-out

L1-in L2-in L3-in L2-in

D2 = 1'-9"

Diagram 5.2

 D1 = 4'-6"
Bulletin 1724E-151

Page 14

Where: N = 2 or 3 crossarms (Only 8-foot crossarms are used on these 

assemblies.)

 D1 = 3-ft, 6-in (3.5 ft); (D2 = 0)

c Example Problem – Unbalanced Longitudinal Loads on a Single Deadend: 

Determine the permitted unbalanced load of a 280 foot span, 3-phase, #2/0 ACSR 

distribution line that deadends on a “VC5.21” (2 crossarm) assembly in the NESC 

Medium Loading District for NESC Grade C construction. Exhibit A on page 15 

of this bulletin lists unit weight (W1) of this conductor plus the NESC wind and 

ice loading as 0.3998 lbs/ft. By combining equations EQ-4.4c and EQ-5.1a from 

Paragraphs 4.4.2.2 and 5.1 of this bulletin, respectively, the problem can be 

solved as follows:

 ( ) 1 1 1 1,000 ≤

× ×

+

× ×

× × × +

h s

PL

v s

in LF

N M F

L

N M F

D S W F 

 ( ) 1

2 7,650 .85 2 5,060 .85

3.5 140 .3998 1.9 1,000 ≤

× ×

+

× ×

× × × + LPL

 1

13,005 8602

1,372

+ ≤

LPL and thus: LPL = 7,695 ft-lbs 

 D1 × L1−in × FLF ≤ LPL

 3.5× L1−in ×1.30 ≤ 7,695

 1,691 L1−in ≤ lbs

d Example Problem – Unbalanced Multiple Longitudinal Loads: Calculate the 

permitted unbalanced longitudinal loads per phase for the double deadend 

“VC6.51” assembly as shown in Diagram 5.2 in Paragraph 5.2 of this bulletin. 

Assume NESC Heavy Loading District and Grade B construction. The following 

information is given: 

• All of the conductors into the assembly are #4/0 ACSR (W1 = .9520 lb/ft) and 

are installed at the same tension. The span length into the assembly is 240 

feet.

• All of the conductors out from the assembly are #1/0 ACSR (W2 = .7036 lb/ft) 

and are installed at the same tension. The span length out from the assembly 

is 260 feet.

• NESC Grade B construction: FS = 0.65; FLF = 1.50 (vertical loads); FLF = 1.65 

(longitudinal loads at deadends).
Bulletin 1724E-151

Page 15

The sum or the terms for determining all of the applied vertical load moments is 

given in equation EQ-4.2b in Paragraph 4.2.3 of this bulletin as:

[( ) ( )] [( ) ( )] 1,000 D1 × Sin ×W1 + Sout ×W2 × FLF + D2 × Sin×W3 + Sout ×W4 × FLF + (ft-lbs) EQ-5.4a

Since all of the “in” conductors are the same, W1 = W3; and since all of the “out” 

conductors are the same, W2 = W4. After substitution and rearranging, the sum of 

the terms of equation EQ-5.4a can be rewritten as:

( ) [( ) ( )] 1,000 D1 + D2 × S1n ×W1 + Sout ×W2 × FLF + (ft-lbs) EQ-5.4b

Substituting the values given in this example problem yields:

(4.5 +1.75)×[(120×.9520)+ (130×.7036)]×1.50 +1,000 = 2,929 ft-lbs

The maximum or permitted sum of the “applied longitudinal load moments” (LPL), 

for the specific conditions given is as follows: (See Paragraph 3.3 of this bulletin 

for the values of the moment capacities Mv and Mh.)

1 2,929 ≤

× ×

+

× × h s

PL

v s N M F

L

N M F 

 1

2 7,650 .65 2 5,060 .65

2,929 ≤

× ×

+

× ×

LPL

 1

9,945 6,578

2,929

+ ≤

LPL

 LPL ≤ 4,641 ft-lbs 

The permitted longitudinal loads are calculated using equation EQ-5.2b given in 

Paragraph 5.2.1 of this bulletin as follows:

 [( ) ( )] D1 + D2 × L1−in − L1−out × FLF ≤ LPL (ft-lbs)

 ( ) ( ) LF

PL

in out D D F

L L L

+ × − − − ≤

1 2

1 1 lbs

 ( ) (4.5 1.75) 1.65

4,641

1 1

+ ×

L −in − L −out ≤ lbs

(L1-in – L1-out ) ≤ 450 lbs (for each of the 2 pairs of phase conductors)
Bulletin 1724E-151

Page 16

EXHIBIT A

NESC DISTRICT ICE AND WIND LOADS FOR CONDUCTORS

Vertical Loads – lbs / ft

Conductor NESC Loading District

Name Size* Strand Light Medium Heavy

ACSR CONDUCTORS

Swante 4 7/1 0.0670 0.2247 0.5379

Sparrow 2 6/1 0.0913 0.2673 0.5989

Sparate 2 7/1 0.1067 0.2855 0.6199

Raven 1/0 6/1 0.1452 0.3467 0.7036

Quail 2/0 6/1 0.1831 0.3998 0.7719

Pigeon 3/0 6/1 0.2309 0.4647 0.8539

Penguin 4/0 6/1 0.2911 0.5439 0.9520

Waxwing 266.8 18/1 0.2894 0.5565 0.9789

Partridge 266.8 26/7 0.3673 0.6446 1.0774

Merlin 336.4 18/1 0.3653 0.6557 1.1015

Linnet 336.4 26/7 0.4630 0.7649 1.2222

Pelican 477.0 18/1 0.5180 0.8488 1.3350

Hawk 477.0 26/7 0.6570 1.0015 1.5014

Osprey 556.5 18/1 0.6040 0.9550 1.4614

Dove 556.5 26/7 0.7660 1.1319 1.6533

Kingbird 636.0 18/1 0.6910 1.0610 1.5864

Grosbeak 636.0 26/7 0.8750 1.2605 1.8014

Drake 795.0 26/7 1.0940 1.5162 2.0938

Tern 795.0 45/7 0.8960 1.3042 1.8678

6201 ALUMINUM ALLOY CONDUCTORS

Azusa 123.3 7 0.1157 0.3172 0.6741

Anaheim 155.4 7 0.1459 0.3626 0.7347

Amherst 195.7 7 0.1837 0.4175 0.8067

Alliance 246.9 7 0.2318 0.4846 0.8927

Butte 312.8 19 0.2936 0.5709 1.0037

Canton 394.5 19 0.3703 0.6722 1.1295

Darien 559.5 19 0.5252 0.8697 1.3696

Elgin 652.4 19 0.6124 0.9783 1.4997

Flint 740.8 37 0.6954 1.0812 1.6225

Greely 927.2 37 0.8704 1.2926 1.8702

Light Loading District: 0.00 in. ice; 9 lbs. Wind

Medium Loading District: 0.25 in. ice; 4 lbs. Wind

Heavy Loading District: 0.50 in. ice; 4 lbs. wind

 * Conductor size: AWG or kcmil
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THIS PAGE IS 

INTENTIONALLY BLANK
Bulletin 1724E-151

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EXHIBIT B

CONTRIBUTORS

The members of the Overhead Distribution Lines Subcommittee of the National Rural Electric 

Cooperative Association, Transmission and Distribution Engineering Committee provided 

invaluable assistance in preparing this document:

• Robby Hamlin, CoServ Electric, Corinth, TX

• Chad Kirkpatrick, Fort Loudoun Electric Cooperative, Vonore, TN

• Kevin Mara, GDS Associates, Inc., Marietta, GA

• Ernest Neubauer, Southern Rivers Energy, Barnesville, GA

• Matthew O’Shea, STAR Energy Services, LLC, Alexandria, MN

• Ken Raming, Ozark Electric Cooperative, Mt. Vernon, MO

• Terry Rosenthal, Laclede Electric Cooperative, Lebanon, MO

• Hank Zarandi, Northern Virginia Electric Cooperative, Gainesville, VA

• Robert Harris, NRECA, Arlington, VA

• Rob Hall, Southern Rivers Energy, Barnesville, GA

• Stan Moore, Fisher and Arnold, Inc., Memphis, TN

• Kevin Jordan, Horry Electric Cooperative, Conway, SC

• Wilson Johnson, Rural Utilities Service, USDA, Washington, DC
Bulletin 1724E-151

Page 19

EXHIBIT C

METRIC CONVERSION FACTORS

To Convert From To Multiply by

Foot (ft.) Meter (m) 0.3048

Ft2 M2 0.092903

Inch (in) Centimeter (cm) 2.54

In2 Cm2 6.452

Degrees Fahrenheit

(X°F)

Degrees Celsius

(°C)

5/9 (X° - 32)

Pound (lb) Kilogram (kg) 0.45359

Foot-pound (ft-lb) Newton-meter (n-m) 1.355818