Blogs

Categories

The actual design of a transmission structure not only involves checking the strength and suitability of various components, but also ensuring that all materials and components meet the requirements of the governing standards or codes. Important items such as grounding are also a part of final structure design configuration and are briefly reviewed.

**1-) Strength Checks**

The following sections cover code-mandated design checks for wood poles, steel poles, lattice towers, concrete and composite poles. Formulae given in this section refer mostly to U.S. codes. Comparable equations from other codes are listed in Appendix 15.

** 1-a) Wood Poles**

The behavior of wood poles used as transmission structures is more complex than steel poles. The basic difference is in the material: wood is orthotropic with low flexural strength. Wood poles are generally sized for normal stresses due to bending and axial loads. A variety of other factors including moisture content, effect of bolt holes, defects and environmental deterioration etc. are also often considered.

The equations for computing the pole section moment capacity at a given elevation are:

*Wood pole grounding *

The grounding of a wood pole begins at the pole top where the copper grounding wire is attached to the overhead ground or shield wire with a clamp. From that location, the grounding wire continues down the length of the pole, with down lead clamps every 12 in. (30.5 cm) intervals, and to a ground rod that is installed at a given distance from the pole. The grounding wire is run 18 in. (45.7 cm) below the ground and clamped to the rod. RUS Specifications 810 and 811 (1998) provides another way of grounding wood poles by using a butt wrap where the grounding wire from the pole top travels all the way down the pole and is wrapped in 3 or 4 rounds near the butt of the pole. For H-Frame type wood structures, the same grounding process is applied at both the poles. For additional information on several other methods of H-Frame grounding, the reader is referred to the above mentioned RUS Bulletins.

The grounding of a guyed wood pole depends on whether the shield wire is guyed or unguyed. If it is unguyed, then the procedure for a regular wood pole is adopted. If the shield wire is guyed, the grounding wire is clamped to the shield wire, and then bonded to the guy wire. The guy at the shield wire location then becomes the ground wire and the anchor acts as a ground rod. For additional information on other methods of guyed pole grounding, the reader is referred to the above mentioned RUS Bulletins.

The detail drawings given at the end of this Chapter shows several grounding techniques for wood poles. Examples 3.3, 3.4, 3.5, 3.6, 3.7 and 3.8 given below illustrate various concepts associated with wood pole design, namely, selection of pole class, buckling, selection of H-Frames and allowable spans for H-Frames and single poles.

**Example 3.3** A 55 ft. (16.76 m) Class 1 Douglas Fir (DF) wood pole shown below is subject to transverse loads from a 4-wire distribution circuit (3 phases and 1 neutral). Each load is 500 lbs. (2.23 kN). Assume 18 psf. (Extreme Wind) and neglect moments due to vertical loads. Assume P-Delta effects of 10% and that center of gravity of pole at half the height above ground. RUS Standards apply. Is Class 1 adequate?

**Solution: **

The pole stands 47.5 ft. (14.48 m) above the ground and embedded 7.5 ft. (2.29 m) into the ground.

Properties of Class 1 DF pole from pole tables:

dt = 8.6 in. (21.6 cm) (Table A2.4)

dGL = 14.6 in. (37.1 cm)

Moment due to wire loads = (4) (500)*(47.5 − 5)/1000 = 85 kip-ft. (115.26 kN-m)

Moment due to wind on pole = (18) (47.5) [(8.6 + 14.6)/(2)(12)](47.5/2)/1000 =

19.63 kip-ft. (26.6 kN-m)

(with pole CG assumed approximately at half pole height above ground. Exact value can be computed using a trapezium shape for the pole). Total Moment applied at GL = 85 + 19.63 = 104.63 kip-ft. (141.88 kN-m) Add 10% second order effects: 104.63 + 10.46 = 115.1 kip-ft. (156.1 kN-m) Ultimate capacity of a Class 1 DF pole can be calculated using pole classes,

Appendix 2, Table A2.1 as: (47.5−2) (4500)/1000 = 204.75 kip-ft. (277.6 kN-m) Strength Reduction factor = 0.75 (for Extreme Wind; Table 2.15a and Load Factor of 1.0)

Available Capacity = (0.75) (204.75) = 153.6 kip-ft. (208.2 kN-m) > 115.1 kip-ft. Therefore Pole is adequate.

Note: Extreme Wind load case is not required for poles shorter than 60 ft. (18.3 m) above ground per NESC. However, RUS requires all poles be checked for this load case regardless of height.

**Example 3.4 **Consider the following pole situation. Assume an Extreme Wind pressure of 21 psf. (1.0 kPa) and 8% second order effects. Determine the most suitable wood pole class. Note the incline (gain base) at the horizontal post insulators. Use Strength Factor for High Wind = 0.75 and Load Factor of 1.0.

Solution:

The pole is 47.5 ft. above ground.

Moment due to vertical loads MVL = [(300) (½) + (3)(1000) (3)]/1000 = 9.15 kip-ft. (12.41 kN-m)

Moment due to horizontal loads MHL = (500) (47.5/1000) + (2000)([(41.5 + 1) + (33.5 + 1) + (25.5 + 1)])/1000 = 230.75 kip-ft. (312.9 kN-m)

Pole size not given; so assume Class H2 Douglas Fir (DF) pole

Pole diameter at top = 9.9 in. (25.1 cm)

Pole diameter at GL = 16.35 in. (41.5 cm)

Average diameter from GL to top = (9.9 + 16.35)/2 = 13.1 in. (33.3 cm)

Moment due to wind on pole MWP = (21) (47.5) (13.1/12) (47.5/2) (1/1000) = 25.9 kip-ft. (35.1 kN-m)

(with pole CG assumed approximately at half pole height above ground. Exact value can be computed using a trapezium shape for the pole).

Total moment due to wire and wind loads = 9.15 + 230.75 + 25.9 = 265.8 kip-ft. (360.4 kN-m)

P-Delta secondary effects = 8% of total moment = (0.08) (265.8) = 21.3 kip-ft. (28.8 kN-m)

Total applied moment = 265.8 + 21.3 = 287.1 kip-ft. (389.3 kN-m)

Strength Factor = 0.75

Required lateral load capacity for determining pole class = (287.1) (1000/[(47.5-2) (0.75)] = 8,413 lbs. (37.4 kN)

Lateral Load rating for Class H4 = 8,700 lbs. (Appendix 2, Table A2.1) USE 55 ft. Class H4.

(The student should re-check pole strength with the revised pole diameter of a H4 pole). Note: Extreme Wind load case is not required for poles shorter than 60 ft. (18.3 m) above ground per NESC. However, RUS requires all poles be checked for this load case regardless of height.

**Example 3.5** Determine the ultimate buckling capacity of the multi-guyed wood pole system for the 90 deg. DDE (double deadend) guying configuration. Use the Gere-Carter formula. The pole is guyed in both planes and all guys are inclined to the pole at 45◦ angle. Other data is as follows:

E = 1800 ksi (12.4 GPa)

Spacing between phase wires = 10 ft. (3.05 m)

Distance from lowest phase to ground L = 46 ft. (14 m)

Pole top diameter = 8.6 in. (21.84 cm)

Pole diameter at ground line = 16.8 in. (42.9 cm)

**Solution: **

The largest unsupported column is the 46 ft. segment between the ground and the lowest guy. Therefore, using the Gere-Carter formula (Equation 3.5) for this segment and assuming fixed-pinned end conditions:

Pcr = 2π2EIa/L2 ∗ (dg/da) 2

Pole taper = (16.8 − 8.6)/(46) = 8.2/46 = 0.187 in./ft.

Pole diameter at lowest guy = 8.6 + (30) (0.187) = 14.21 in. (36.1 cm) = da

IA = moment of inertia at the location of the lowest guy = (3.1416) (14.214/64) = 2001.5 in4 (83307.1 cm4)

Pcr = [(2) (3.14162) (1800) (2001.5)/[(46) (12)]2] (16.8/14.21)2 = 326.2 kips (1451.7 kN)

Use a factor of safety of 3.0 since this is a DDE.

Design buckling capacity = 326.2/3 = 108.7 kips (483.9 kN)

**Example 3.6** Discuss the various situations where you would recommend the below H-Frame systems.

**Solution: **

(a) Preferred at small to moderate spans on flat terrains. Lateral soil pressure more important for foundation checks given absence of X-bracing.

(b) Used at moderate to large horizontal (wind) spans where transverse loads due to wind control design. Bearing and uplift are more dominant than lateral soil pressure.

(c) This design is used for larger vertical (weight) spans on uneven and hilly terrains.

(d) Preferred choice for large wind and weight spans and higher voltages. A double cross arm can be adopted for large weight spans; an extra X-Brace can help reduce frame bending.

**Example 3.7** For the 69 kV braced wood H-Frame shown below, with three (3) phase conductors and two (2) overhead ground wires, determine the maximum allowable wind and weight spans, for Extreme Wind load case, based on the strength of X-Braces. Use 10% reduction in span to account for P-Delta effects and a vertical span to horizontal span ratio of 1.15 to account for uneven terrain.

The following data applies to the problem:

Poles 55 ft. Class 1 DF; Embedment = 7.5 ft. (2.29 m)

Height of pole above ground = h = 55 − 7.5 = 47.5 ft. (14.48 m)

Pole spacing = b = 10.5 ft. (3.20 m)

Pole diameter at top = dt = 8.6 in. (21.84 cm) or 0.716 ft. (See Table A2.4)

Pole diameter at ground line G = dG = 14.6 in. (37.1 cm) or 1.216 ft.

Pole diameter at B = dB = 9.48 in. (24.1 cm) or 0.790 ft.

Pole diameter at E = dE = 10.37 in. (26.3 cm) or 0.864 ft.

Pole diameter at D = dD = 11.7 in. (29.7 cm) or 0.975 ft.

Pole diameter at C = dC = 12.93 in. (32.8 cm) or 1.078 ft. Moment capacity at ground line = MG = 204.75 kip-ft. (277.6 kN-m) (See Example E3.3)

Distance of OHGW from pole top = g = 6 in. (0.15 m)

Distance of Cross Arm from pole top = y1 = 7 ft. (2.13 m)

Distance of X-Brace from pole top = y = 14 ft. (4.27 m) x = h − y − b = 47.5 − 14 − 10.5 = 23 ft. (7.01 m)

Load Factor LF = 1.00 (Extreme Wind)

Wind pressure q = 21 psf. (1 kPa) (Extreme Wind)

Weight of bare conductor per unit length = wc = 1.10 plf. (16.06 N/m)

Weight of bare ground wire per unit length = wg = 0.40 plf. (5.84 N/m)

Wind load per unit length on conductor = pc = 1.75 plf. (25.55 N/m)

Wind load per unit length on ground wire = pg = 0.70 plf. (10.22 N/m) (Wind loads refer to Extreme Wind case, bare wire with no ice).

XBS = Strength of X-Braces = 28, 300 lbs. (125.94 kN)

**Solution:**

Determine location of pt:

pt = total horizontal force per unit length = (2)(pg) + (3)(pc) = (2)(0.70) + (3) (1.75) = 6.65 plf. (97.1 N/m)

To determine location of pt, use moment equilibrium about pole top.

(pt)(distance of pt from pole top, z) = (2)(pg)(g) + (3)(pc)(y1)

From which, distance of pt from pole top, z = [(2)(pg)(6/12) + (3)(pc)(7)]/pt = [(2) (0.70)(0.5) + (3)(1.75)(7)]/6.65 = 5.63 ft. (1.72 m)

h1 = h − z = 47.5 − 5.63 = 41.9 ft. (12.76 m)

Determine xo:

dD = 11.7 in. → CD = πdD = 36.76 in. (93.36 cm) or 3.063 ft.

dG = 14.6 in. → CG = πdG = 45.87 in. (116.51 cm) or 3.822 ft.

From Equation 3.3a:

x0/x = CG(2CG + CD)/2(C2 G + CGCD + C2 D) = 0.573

xo = (0.573)(x) = (0.573)(23) = 13.18 ft. (4.02 m)

h2 = h1 − xo = 41.9 − 13.18 = 28.72 ft. (8.75 m)

Strength Factor ϕ = 0.75 (Extreme Wind)

Maximum Horizontal Span (HS) based on X-Brace strength is given by Equation 3.3f.

HSX = {[ϕ ∗ XBS ∗ b] − [2 ∗ LF ∗ q ∗ (h − x0) 2 ∗ (2dt + dC)/6}/(LF ∗ pt ∗ h2) = {[(0.75)(28300)(10.5)] − [(2)(1.0)(21)(47.5 − 13.18)2((2)(0.716) + 1.078)]}/6 / [(1.0)(6.65)(28.72)] = (222, 862.5 − 20964.2)/190.988 = 1, 058.5 ft. (322.64 m)

Reduce 10% due to P-Delta effects.

HSX = (0.90) (1058.5) = 952.65 ft. (290.4 m)

Maximum Vertical Span (VS) for VS/HS ratio 1.15:

VS = (1.15) (952.65) = 1,095.55 ft. (333.9 m)

**Example 3.8**

(a) Determine the allowable horizontal span for the tangent pole with post insulators shown below for the case of Extreme Wind at 21 psf (1005 Pa). The pole is a 75 ft (22.86 m) Class 1 Douglas-Fir wood pole with a ground line bending moment capacity of 283.5 kip-ft (384.4 kN-m). Other data is as follows:

Load Factor for Vertical Load, LFv = 1.0

Load Factor for Wind Load, LFt = 1.0

Strength Factor for Wood Pole SF = 0.75 for Extreme Wind The overhead ground wire is 3/8’’ EHS and conductors are 477 ACSR 18/1, Pelican. Insulators are horizontal posts as shown. Assume level terrain and ignore deflection effects. Also assume wind span = weight span.

(b) Determine the effects of using LFv = LFt = 1.10

(c) Determine the allowable span if bundled conductors are used (2 wires per insulator).

Solution:

(a) Let the allowable horizontal span be H.

From wire tables in RUS Bulletin 200, the following data on the ground wire and conductor are obtained.

Diameter ground wire = 0.360 in. (9.1 mm)

conductor = 0.814 in. (20.7mm)

Vertical wgw = 0.273 plf (3.98 N/m) wcon = 0.518 plf (7.56 N/m)

Compute unit loads due to wind on wires.

Wind pgw = (21)(0.36)(1/12) = 0.630 plf (9.19 N/m)

pcon = (21)(0.814)(1/12) = 1.425 plf (20.8 N/m)

Compute vertical and transverse (wind) loads on the pole for a span of H.

Vertical Loads

Due to ground wire weight = (wgw)(H)(LFv)

Due to conductor weight = (wcon)(H)(LFv)

Wind Loads

Due to wind on ground wire = (pgw)(H)(LFt)

Due to wind on conductor = (pcon)(H)(LFt)

Compute bending moment at ground line due to above loads for LFv = LFt = 1.0.

Moment due to vertical load at ground wire = wgw H LFv (6/12) = 0.5 wgw H

Moment due to vertical load at conductor = wcon H LFv(2)(3 wires) = 6 wcon H

Moment due to wind load at ground wire = pgw H LTt (75 − 9.5 − 0.5) = 65 pgw H

Moment due to wind load at conductor = pcon H WLF (49.5 + 55.5 + 61.5) = 166.5 pcon H

Pole Data pole top diameter = 8.6 in. 21.8 cm) ground line diameter = 16.3 in. (41.4 cm) average diameter = (8.6 + 16.3)/2 = 12.45 in. (31.6 cm) pole height above ground = 65.5 ft. (19.96 m)

Moment due to wind on pole = (21) (12.45/12)(0.5)(65.52)/1000 = 46.74 kip-ft. (63.4 kN-m)

Total Applied Moment = MA = 0.5 wgw H + 6 wcon H + 65 pgw H + 166.5 pcon H + (46.74) (1000) = [0.5 wgw + 6 wcon + 65 pgw + 166.5 pcon] H + 46,740 = [(0.5) (0.273) + (6) (0.518) + (65) (0.63) + (166.5) (1.425)] (H) + 46, 740 = (281.46) (H) + 46, 740 lb-ft

MA ≤ (SF) (Mcap) ≤ (0.75) (283.5) (1000) lb-ft

281.46 H ≤ 212,625 − 46,740 = 165,885 lb-ft

H = 589.4 ft. (179.6 m)

(b) For load factors of 1.10, the allowable horizontal span will be H = 589.4/1.10 = 535.8 ft. (163.3 m) That is, increasing the load factors to 1.10 resulted in a 9% reduction in allowable span.

(c) For bundled conductors, the parameters affected will be the weight and wind load at conductor points.

wcon = (2) (0.518) = 1.036 plf (15.12 N/m)

pcon = (2) (1.425) = 2.850 plf (41.6 N/m)

MA = [(0.5) (0.273) + (6) (1.036) + (65) (0.63) + (166.5) (2.85)] (H) + 46, 470 ≤ (0.75)(283,500) 521.83 H ≤ 165,885

H = 317.9 ft. (96.9 m)

That is, bundling the conductors resulted in almost half the allowable span. (Note: In a 2-wire horizontal bundle, one conductor shields the other; but this shielding is ignored for conservative purposes).

** 1-b) Steel Poles**

Steel poles are sized for bending, axial and shear stresses, supplemented by local buckling checks given the width-to-thickness ratios and stress interaction. The equations for computing the pole section moment capacity at a given elevation are the same as wood except that geometrical properties are a function of shaft thickness and diameter and are given in Appendix A3. The overall usage of a tubular steel pole is determined with reference to the most highly stressed quadrant of the cross section. This is given by the ASCE interaction equation shown in PLS-POLETM:

Section properties for various steel pole cross sections are given in Appendix 3. Other design checks per ASCE 48-11 are as follows:

Permissible Compressive Stress: Rectangular, Hexagonal and Octagonal Members

Note: If the axial stress is greater than 1 ksi (6.9 MPa), Equations 3.9 of Dodecagonal members shall be used for rectangular members.

Permissible Compressive Stress: Dodecagonal Members

Permissible Compressive Stress: Hexdecagonal Members

*Steel davit arms *

The most common structures with davit arms are tangent (suspension) and light angle poles. Longer davit arms facilitate greater phase separation without increase in structure height. Upswept davit arms are often used for aesthetic reasons with the upsweep ranging from 6 in. (15.2 cm) to 18 in. (45.7 cm). The length and vertical spacing of davit arms depend on required phase separation for that particular voltage, clearances for insulator swing, galloping conditions etc. Arms are also employed for shield wires to reduce the shielding angle and improve lightning protection. Such arm lengths are determined solely by the required shielding angle.

Other than some nominal guidance offered in ASCE 48-11, there is no standardized design method for steel davit arms. Designs are generally based on empirical approaches, finite element analysis as well as inferences from full-scale testing. Davit arms with suspension or light angle insulators are usually designed as a cantilever subject to vertical and transverse loads. Stress checks for steel davit arms are similar to those of steel pole shafts described in the previous section. If PLS-POLETM program can be used for design, then axial, shear, bending and torsional stresses produced by each load case are checked.

Connections to the steel pole usually involve brackets with pin-type bolts transferring flexure and shear effects. For double circuit structures, where davit arms are needed on both sides of the pole, the connection involves brackets with through plates. In a majority of cases, the pole maker or fabricator designs, fabricates and details the davit arms. Wind-induced vibration of unloaded davit arms is a big concern during construction; weights are often suspended from the arm tips to provide some damping. Another means of reducing vibration effects is to tie the arm tips with a cable to the pole.

Items such as vangs welded to steel poles for installing running angle or strain insulators are designed for full tension effects. Vangs are fabricated and welded to the pole shaft prior to galvanizing.

**Base plates **

Currently only one design guide is available – AISC (2006) – to provide guidance for analysis and design of base plates for tubular steel poles supported by concrete piers.

Pole fabricators also have in-house and proprietary design processes which vary widely from one manufacturer to another. However, ASCE 48-11 outlines a procedure based on effective bend lines and a 45◦ bend-line limitation. The manual also gives equations to calculate effective anchor bolt loads, base plate stresses and plate thickness.

*Steel pole grounding *

Steel transmission poles are usually provided with a metal grounding pad welded to the side of the pole about 12 in. (30.5 cm) to 18 in. (45.7 cm) above the ground or base plate. The shield wire is bonded to the steel pole near the pole top with a copper (grounding) wire and a stainless steel nut. Grounding rods, if used, are usually installed about 3 ft. (0.9 m) from the pole connected to the ground pad. Examples 3.9, 3.10 and 3.11 given below illustrate some concepts associated with steel pole design, namely, thickness requirement and determination of number of anchor bolts for a pole transferring moment to the base plate and concrete pier.

**Example 3.9 **

An 80 ft. (24.4 m) steel pole has been rated for a GL moment capacity of 280 kip-ft. (379.7 kN-m). The pole diameters are 8.7 in. (22.1 cm) at the top and 20.4 in. (51.8 cm) at the butt. These are mean diameters measured to the midpoint of the thickness across the flats. Assume 12-sided cross section, uni-axial bending about X-axis, a material yield stress of 65 ksi (448.2 MPa) and standard embedment of 10% + 2 ft. Neglect axial, shear and torsional stresses. Determine the approximate pole shaft thickness required. Verify if thickness satisfies local buckling criteria.

Solution:

Therefore four assumption about width-to-thickness ratio is valid. (See Table A3.8 to verify this analysis).

**Example 3.10**

For the dual use pole system shown below, determine the required pole height. The transmission circuit is 3-phase 161 kV and the under-build distribution is 34.5 kV. Assume sag of the transmission and distribution conductors as 8.0 ft. (2.44 m) and 6.0 ft. (1.83 m), respectively. Use Tables 2.6a-1, 2.6b-1 and 2.7. Assume a buffer of 2.0 ft. (0.61 m) for ground clearance.

**Solution:**

The process basically involves determining the various wire spacing associated with the system. From Table 2.6a-1: Required ground clearance for 34.5 kV = 18.7 ft. (5.7 m) From Table 2.7: S = shield wire to phase separation = 4.3 ft. (1.31 m) P = phase to phase clearance = 6.7 ft. (2.04 m) Sag of transmission conductor = 8.0 ft. (2.44 m) C1 = clearance from sagged 161 kV wire to 34.5 kV wire = 7.0 ft. (2.13 m) (Table 2.6b-1)

Sag of distribution conductor = 6.0 ft. (1.83 m)

Therefore required height of the pole above ground is:

H = 4.3 + 6.7 + 6.7 + 8.0 + 7.0 + 6.0 + 18.7 + 2.0 = 59.4 ft. (18.11 m) Consider a 70 ft. (21.34 m) pole with 9 ft. (2.74 m) embedment De giving 61 ft. (18.6 m) above ground. 70 ft.

Pole is adequate.

**Example 3.11 **

Determine the approximate number of anchor bolts required for the following situation:

Pole GL diameter = 48 in. (121.9 cm)

Moment transmitted from pole loads = 4800 kip-ft. (6509 kN-m)

Use ASTM A615 #14 anchor bolts (1.75 in. diameter) with a ultimate tensile strength of 100 ksi. (689 MPa). Assume 5 threads per inch. Consider bending effects only.

**Solution: **

(The aim of this problem is to illustrate a quick approximate method for determining the preliminary number of anchor bolts needed for a steel pole with a base plate. The exact number must be determined either by a detailed analysis per ASCE 48-11 or via the PLS-POLETM program.)

The relationship between pole shaft, base plate, anchor bolt circle and pier is shown below. For the 48 in. diameter pole shaft, the approximate size of the anchor bolt circle is 54 in (see Figure 5.9 in Chapter 5). Assume all anchor bolts arranged in a circular fashion.

Distance to the nearest anchor bolt from shaft center = 54/2 = 27 in. (76.2 cm) From ASCE 48-11, cross sectional stress area of an anchor bolt A = (π/4) [d − (0.9743/n)]2

Where ‘n’ is the number of threads per unit length. A = (3.1416/4) [1.75 − (0.9743/5)]2 = 1.90 in2 (1225.5 mm2) (Area reduction due to bolt threads is often considered during such calculations). Permitted tensile stress in anchor bolt = 0.75 Fu Capacity of a single anchor bolt = (1.90) (100)(0.75) = 142.5 kips (634.1 kN) Number of anchor bolts required = Moment/(bolt capacity)(lever arm) = (4800)(12)/(142.5)(27) = 14.9 or 15 per half circle Use 30 anchor bolts for the whole pole, arranged in a circular fashion.

**1-c) Lattice Towers**

Design checks given below for angle members in steel lattice towers refer to ASCE 10-15 (2015) and include allowable slenderness ratios and associated tensile and compressive stresses in steel angles. TOWER™ program also determines the usage level of each member group based on ASCE. Connection checks include bolt shear, bearing, block shear capacity and tensile rupture.

Compression Capacity is the minimum of:

(a) Member compressive strength based on slenderness ratio, kL/r

(b) Connection shear capacity

(c) Connection bearing capacity

Tension Capacity is the minimum of:

(a) Member tensile strength based on net section

(b) Connection rupture

(c) Connection shear capacity

(d) Connection bearing capacity

Then the revised design compressive stress (Fa) is obtained by replacing Fy in Equation 3.17a, and in Cc, with Fcr given below.

In all equations above, ψ = 1 for stress in ksi and 2.62 if stress is in MPa.

Compressive capacity of the angle member is given by the design compressive stress from above equations times the gross cross sectional area.

Effective Lengths of Angle Members

For leg members bolted in both faces:

kL/r = L/r for 0 ≤ L/r ≤ 150 (3.20a)

Unsupported panels

For other compression members with concentric load at both ends:

kL/r = L/r for 0 ≤ L/r ≤ 120 (3.20b)

For other compression members with concentric load at one end and normal framing eccentricity (NFE) at the other:

kL/r = 30 + 0.75L/r for 0 ≤ L/r ≤ 120 (3.20c)

For other compression members with NFE at both ends:

kL/r = 60 + 0.50L/r for 0 ≤ L/r ≤ 120 (3.20d)

For other compression members unrestrained against rotation at both ends:

kL/r = L/r for 120 ≤ L/r ≤ 200 (3.20e)

For other compression members partially restrained against rotation at one end:

kL/r = 28.6 + 0.762L/r for 120 ≤ L/r ≤ 225 (3.20f)

For other compression members partially restrained against rotation at both ends:

kL/r = 46.2 + 0.615L/r for 120 ≤ L/r ≤ 250 (3.20g)

NFE is the Normal Framing Eccentricity which is defined as the condition when the centroid of the bolt pattern is located between the heel of the angle and the centerline of the connected leg. NFE may cause a reduction of up to 20% in the axial capacity of short, stocky, single angle struts.

Angles in Tension:

For angle members connected by one leg:

Design tensile stress on net cross-sectional area = Ft = 0.90 Fy (3.21a)

Net Section Capacity Ncap = AnetFt (3.21b)

(For angle members connected in both legs, the 0.90 factor is replaced by 1.0 (i.e.) Ft = Fy).

Ncap is the strength based on tearing of a member across its net area

Anet which is defined as: Anet = Ag − (d)(t)(nh) (3.21c)

where:

Ag = gross cross-sectional area of the angle section

d = bolt hole diameter (generally 1/16 in. or 1.6 mm more than bolt diameter)

nh = number of holes

t = thickness of angle

If there is a chain of holes in a zigzag fashion in an equal leg angle, then the net width of an element hn and the net area Anet must be determined as follows:

hn = [2h − (nh)(d) + ng(s 2/4g)] (3.22a) Anet = (hn)(t) (3.22b)

where:

h = width of angle leg

nh = number of bolt holes in the chain

ng = number of gauge spaces in the chain

s = bolt spacing or pitch along the line of force

g = gauge length or transverse spacing of the bolts

If the centroid of the bolt pattern on the connected leg is outside the center of gravity of the angle, then all connections must be checked for block shear or rupture using Equation 3.23:

RBSH = 0.60AvFu + AtFy (3.23)

where:

Av = minimum net area in shear along a line of transmitted force for a single angle = (t) {a + (nb − 1)(b)}

t = angle thickness

a = effective end distance = e − d/2 (See Figure 3.20)

b = s − d (See Figure 3.20)

s = bolt spacing, center to center

Fu = specified minimum tensile strength of the angle steel

Fy = specified minimum yield stress of the angle steel

At = (t)(c)

c = effective edge distance = f − d/2 (See Figure 3.20)

d = bolt hole diameter

e, f = as shown in Figure 3.20

nb = number of bolt holes

Connection rupture often occurs due to insufficient edge and end distances as well as bolt spacing. ASCE 10-15 therefore specifies the following minimum values for the parameters.

*Minimum end and edge distances*

The minimum end distance ‘e’ (inches) shall be the largest value of:

e = 1.2 P/tFu or

= 1.3d or

= t + (d/2)

P = force transmitted by bolt

d = nominal bolt hole diameter

Minimum edge distance ‘f’ (inches) shall not be less than:

(a) 0.85 emin for a rolled edge

(b) 0.85 emin + 0.0625 in. for a sheared or mechanically-cut edge where emin is the largest value determined from Equations 3.24a, 3.24b and 3.24c.

*Bolt spacing *

The center-to-center distance between bolt holes shall not be less than:

smin = 1.2 P/tFu + 0.6 d (3.24d)

Tower grounding The process discussed for steel poles is also applicable to steel towers, except that grounding must be facilitated at a minimum of 2 tower legs. In situations where ground resistance is not optimum, all 4 legs can be grounded.

Examples 3.12, 3.13, 3.14 and 3.15 given below illustrate various concepts associated with tower design, namely, crossing diagonals, allowable compressive stress, net section areas and block shear determination.

**Example 3.12** Determine the effective buckling lengths of the crossing diagonals of the tower panel shown below.

b = 11.7 ft. (3.57 m)

v = 13.25 ft. (4.04 m)

k = 1.483 ft. (0.452 m)

Solution:

The AISC Steel Manual’s cross bracing equations were used to estimate the lengths of the main diagonals. The diagonals usually have a bolt at the meeting point which helps with reducing effective buckling lengths.

Example 3.13 For the 10 ft. (3.05 m) tower angle member shown, determine the design compressive stress Fa. Assume Fy = 50 ksi (344.8 MPa) and E = 29,000 ksi (200 GPa).

Solution:

For the 3’’ × 3’’ × ¼’’ angle:

Cross Sectional Area = 1.44 in2 (929 mm2)

Slenderness Ratios:

rx = 0.93 in. (23.6 mm)

ry = 0.93 in. (23.6 mm)

rz = 0.59 in. (15 mm)

The presence of two bolts at each end and the member framing into other members at the joints provides for partial restraint. Assume normal framing eccentricity at both ends. Using Equation 3.20g:

kL/r = 46.2 + 0.615 L/r = 46.2 + (0.615) ((10)(12)/r) = 46.2 + 73.8/r About X- and Y-axes: kL/rx = kL/ry = 46.2 + 73.8/0.93 = 125.6 About Z-axis: kL/rz = 46.2 + 73.8/0.59 = 171.3 − controls The width-to-thickness ratio of the member is w/t = 3/0.25 = 12

Limiting w/t ratios:

(w/t)lim1 = 80/ Fy = 80/7.071 = 11.3

(w/t)lim2 = 144/Fy = 144/7.071 = 20.4

11.3 < 12 < 20.4

(w/t)lim1< (w/t) < (w/t)lim2

Therefore, from Equation 3.19a:

Fcr = [1.677 − (0.677) (12/11.3)] (50) = 47.93 ksi (330.24 MPa)

Cc = π (2)(29000)/47.93 = (3.1416) (34.78) = 109.3

kL/r > Cc

Design compressive stress using Equation 3.17b:

Fa = π2E/(kL/r) 2 = (3.14162)(29000)/171.32 = 9.8 ksi (67.6 kPa)

Corresponding design compressive force = (1.44) (9.8) = 14.1 kips (62.8 kN)

Note: The above value is only from slenderness point of view. The connection’s shear and bearing capacity must also be evaluated to determine which one controls.

**Example 3.14** For the 5’’ × 5’’ × ½’’ steel angle shown, determine:

(a) Net area

(b) Net section capacity in tension

(c) If the end and edge distances shown are adequate. (assume force transmitted as 20 kips). Assume a rolled edge.

All bolts are ¾ in. (19 mm) diameter. Fy = 36 ksi (248 MPa) and Fu = 58 ksi (400 MPa).

**Solution:**

Angle properties: Area Ag = 4.75 in2 (30.65 cm2)

Bolt hole diameter = ¾ + 1/16 = 13/16 in. (2.06 cm)

**Example 3.15 **

Determine the block shear capacity of the above the 3’’ × 3’’ × ¼’’ steel angle. All bolts are ¾ in. (19 mm) diameter. Fy = 36 ksi (248 MPa) and Fu = 58 ksi (400 MPa).

**Solution: **

Bolt hole diameter = ¾ + 1/16 = 13/16 in. (2.06 cm)

From steel tables, the CG of the angle section is at a distance of 0.842 in. (21.4 mm) from the heel. The bolt pattern line is outside the CG. Equation 3.23 applies.

**1-d) Concrete poles**

The behavior of reinforced concrete poles used as transmission structures is more complex than steel given the basic difference in material – while steel is a uniform and isotropic material, concrete is anisotropic, non-linear with low tensile strength (i.e.) susceptible to cracking at even moderate bending. Concrete poles are therefore evaluated not only in terms of the ultimate moment (factored loads), but also the initial cracking moment at service loads and zero tension moment (no cracking). The initial cracking strength will be roughly 40% to 55% of the ultimate strength while zero tension strength is about 70% to 85% of the initial cracking strength. Element usage is determined only as a function of bending moment. Derivations of various equations are available in ASCE Manual 123 (2012).

The behavior of concrete poles subject to axial and flexural loads is an explicit function of f c, the concrete 28-day strength which in turn controls the Modulus of Elasticity, Ec, and thereby, the pole’s bending resistance. Figures 3.26 and 3.27 show the assumed stress distribution used in deriving the relevant moment expressions. The equations for computing pole cross section moment capacity are based on section equilibrium and at a given elevation are as follows:

c = Depth of stress block (NA to extreme compressive fiber) (in. or mm)

ei = di − c (di is the distance of the ‘i’-th strand from extreme compression fiber) (in. or mm)

K = Factor relating centroid of force Cc to Neutral Axis (NA)

Initial Cracking Strength Mic = (fr Ig/yt) + (PIg/Agyt) (3.25b)

where:

fr = Modulus of Rupture of concrete = 7.5 f c (psi) for normal weight concrete

Ag = Gross area of the cross section (in2 or mm2)

Ig = Gross moment of Inertia of the cross section (in4 or mm4)

yt = Distance of extreme tensile fiber from centroidal axis (in or mm)

P = Effective Prestress Force (lbs. or N)

Mic is approximately equal to 40% to 55% of Mu.

Zero Tension Capacity

Mzt = (PIg/Agyt) (3.25c) Mzt is approximately equal to 70% to 85% of Mic or 28% to 47% of Mu.

The usage of concrete poles is determined for each of the above three definitions of bending moments with the appropriate strength factor, as specified for the design. For square concrete poles, the procedures are the same except that the bending moment M is replaced by the larger of the two moments about the principal axes, X and Y.

** 1-e) Composite Poles**

The design of composite poles is governed by both strength (flexural capacity) and stiffness (deflections). The most common way of selecting a composite pole is by using the design charts provided by the manufacturer – in terms of wood pole equivalency – and then check if the pole geometry is adequate for a given limiting deflection.

RS Technologies (2012), for example, provides a design guide with various pole modules and lengths, and lateral load capacities (load applied 2 ft. from pole top) accompanied by tip deflections.

*Composite pole grounding *

Most composite poles can be grounded in the same manner as wood poles, with the ground wire affixed to the outer surface of the pole using wire clips and screws. Another option is to have the ground wire run internally through the pole, exiting to the ground rod through a hole in the base module.

** **

Posted in: Business

2 people like this.