7/14/2011

1

AS/NZS 7000:2010

Worked Examples

11kV Single Circuit Concrete Poles

9.2m

1.8m

Single CircuitLV or 11 kV

Worked Example 1

11kV Single circuit pole line

Ground Line

9.2m

1.8m

Single Circuit

Maximum change in line direction is 10 degrees

Line INLINE POLE

7/14/2011

2

Line Design ParametersComponent Detail Reference

Location Coastal plain North

Island

Terrain Flat / undulating

Exposure Terrain Cat 2 AS/NZS 1170.2

Design working life 50yrs Table 6.1 AS/NZS 7000

Design security level Level 1 Table 6.1 AS/NZS 7000

Design Wind RP 50yrs Table 6.1 AS/NZS 7000

Design Wind velocity VR

39.0 m/s Region A7 AS/NZS 1170.2 Table

3.1

Lee wind area No Fig B2 AS/NZS7000

Topographic multiplier M t

1.0 Cl. B3 AS/NZS7000

Directional multiplier Md

1.0 Cl. B3 AS/NZS7000

Terrain multiplier Mzcat

1.0 Table 4.1 AS/NZS1170.2

Ice load Nil Appendix EE AS/NZS7000

Selection of design wind return period

Australian Panel B2 –Overhead Lines Seminar –AS/NZS Overhead Line Design Sydney 28 -29 March 2011

7/14/2011

3

Australian Panel B2 –Overhead Lines Seminar –AS/NZS Overhead Line Design Sydney 28 -29 March 2011

Australian Panel B2 –Overhead Lines Seminar –AS/NZS Overhead Line Design Sydney 28 -29 March 2011

AS/NZS 1170.2

7/14/2011

4

Line Design Parameters

Component Detail Reference

Pole type 11.0m Prestressed concrete

Conductor ‘Dog’ ACSR

Earthwire Nil

Wind span 70m

Weight span 70m

Deviation angle 10 degrees

Pole details

Item Assumed Pole Details

Pole type 11m PSC Rectangular I

section

Embedment Depth 1.8m

Conductor attachment Ht 9.2m

Transverse Base width @GL 430mm

Longitudinal Base width @GL 150 mm

Transverse Tip width 160 mm

Longitudinal tip width 150 mm

Pole tip load longitudinal capacity 8.0kN

Pole tip load transverse capacity 22.0kN

Assume max bending moment 200mm below ground level

7/14/2011

5

Conductor details

DOG ACSR VALUE UNIT

Everyday Temperature (EDT): 10 DegC

Minimum Temperature: -5.76 DegC

Maximum Design Temperature: 50 DegC

Stringing Temperature: 15 DegC

Stringing Tension, 10% UTS 3.00 kN

Stringing Tension, (EDT) 3.29 kN

Nominal Overall Diam: 14.2 mm

Cross Sectional Area: 118.8 m2

Initial Mod Of Elasticity: 56.54 x 109 Pa

Ultimate Tensile Strength: 3357 kg

32.93 kN

Self weight: 0.396 kg/m

0.00388 kN/m

Wind Pressures

Unit Design Wind Pressures:Regional site design wind velocity

Unit Design Wind Pressure

Vsit

,β

qz

==

VR

Md

Mz,cat

Ms M

t=39m/s ( Cl B3 AS/NZS 7000)

= 913 Pa

Element Design Wind Pressures:

Conductors (Cd = 1.0): = 913 x 1.00 (Design Wind Pressure x Drag Force Coef)

= 913 Pa

Pole (Cd = 1.6): = 913 x 1.6 (Design Wind Pressure x Drag Force Coef)

= 1460 Pa

Crossarm End (Cd = 1.2): = 913 x 1.2 (Design Wind Pressure x Drag Force Coef)

= 1096 Pa

Insulators (Cd = 1.2): = 913 x 1.2 (Design Wind Pressure x Drag Force Coef)

= 1096 Pa

2 3,0.6 10 (kPa)sitV β

−× ( Cl B5 AS/NZS 7000)

7/14/2011

6

Australian Panel B2 –Overhead Lines Seminar –AS/NZS 7000 :2010 Overhead line Design Sydney 28-29 March 2011

Design Load Combinations

Conductor Applied LoadsLoad Condition Symbol Design Load

Everyday load condition, 10 DegC, no wind Fat 3.27 kN

Sustained load condition, -5.76 DegC, no wind Fat 4.57 kN

Sustained load condition, 00C, no wind Fat 3.99 kNMaximum wind load condition, 0 DegC, Wind = 913 Pa Ftw 8.76 kN

Maximum wind load condition, minimum temperature -5.76 DegC, Wind = 913 Pa Ftw 9.26 kN

Failure Containment, 10 DegC, max. wind x 0.25 Fb 3.97 kN

7/14/2011

7

Check Conductor Design Tensions

For ultimate loading conditions:

Ultimate Tensile Strength of conductor Rn = 32.93 kN

Conductor Capacity ØRn = 23.05 kN (Material Ø Factor = 0.7 Table 6.2 AS/NZS 7000)

Under maximum wind condition:

Conductor tension = 1.25 x Fat

= 1.25 x 8.76

= 10.95 kN < ØRn

Under minimum temperature condition:

Conductor tension = 1.25 x Fat

= 1.25 x4.57

= 5.71 kN < ØRn

Therefore OK for DOG conductor

Check loads on pin insulatorsMaximum transverse wind load from

conductor on insulator Tiw

= Azx Conductor wind pressure

= (Diam x Span x SRF) x Conductor wind pressure

= 14.2/1000 x 70 x 1 x 913/1000

= 0.91 kN < (0.85 x 9.6) kN for 1130W,

Line deviation transverse conductor load

under maximum wind load Tid

= 2 x Fat

x

= 2 x 8.76 x sin(5)

= 1.53 kN

Sum of design ultimate loads at conductor

level on top of insulator: ∑kiTi = k

ix Wind load + 1.25 x deviation load

= 0.75x0.91+1.25x1.53

= 2.59 kN < (7÷2) or 3.5 kN for 1130W,

< (0.85 x 9.6) kN for M20 high tensile pin,

Maximum vertical load from conductor Vu

= Span x weight per m

(No ice or snow) = 70 x 0.00388 x 1.1

(Load Factor 1.1 from Table 7.3 AS/NZS

7000)

= 0.30 kN < (7÷2) kN for 1130W, OK

Use 1130W, Pin M20 High Tensile

2sin

θ

7/14/2011

8

Check Conductor Separation at Midspan

Clause 3.7.3.2 of AS/NZS 7000:2010 provides

Where; X = Projected horizontal distance (m) between conductors midspan

Y = Projected vertical distance (m) between conductors midspan

U = r.m.s. vector difference in potential (kV)

k = Constant, normally equal to 0.4

D = Greater of two conductor sags centre equiv. span, 50 DegC, still air

l = Length in metres of any free swing conductor

In this structure case

Y = 0 m;U = 11 kV; k = 0.4 ; D = 1.14 m ; and l = 0 m

Therefore; X ≥ 11 / 150 + 0.4 x √ (1.142 + 0) ≥ 0.529 m

and X = (X1 + X2)/2 = (300 + 950)/2 for 1400 x 100 x 75 HW crossarm,

Ie X1 = 300, X2 = 950

Distance between conductors midspan X = 0.625 m

Minimum required separation; ≥ 0.529

OK for 1400 x 100 x 75 HW crossarm and electrical separations are OK

2 2(1.2 )150

UX Y k D l+ ≥ + +

HW Crossarm DesignCheck conductors vertical loads Vertical Bending in crossarm:

Vertical Load of conductors = unit weight x weight span x 9.806

= .00388x70 = .27kN

For intermediate suspension structures the conductor vertical loads are very small and

would only govern design where ice loads apply and for design spans > 250m.

Here, span = 70m .

Therefore OK

Check conductors horizontal loads: (Assuming loads are transferred through to pole via

a single bolts in shear and stabilized by standard flat plate braces)

Ultimate horizontal load ∑Wu = (ki x Wind load + 1.25 x deviation load) x 3

= ((0.75 x 0.91) + (1.25 x 1.53))x 3

i.e. N = 7.76 kN

7/14/2011

9

Timber Joint Design (Crossarm to Pole)

Now for a M16 Grade 4.6 bolt

Qskl = 23.4 kN for Joint Group J1 (ironbark) Table 4.9(B) AS 1720.1

where beff = 75 mm (parallel to grain) & single shear

so design capacity; ØNj ≥ N*

ØNj = Ø k1 k16 k17 n Qskl (Clause 4.4.3.2 of AS1720.1 – 1997)

where Ø = 0.8;

k1 = 0.57 (duration of load - say permanent for deviation);

k16 = 1 ( bolt loads not transferred by side plates);

k17 = 1 (from Table 4.11 AS 1720.1);

n = 1 (No. of bolts = 1)

therefore ØNj = 10.67 kN ≥ 7.76 kN

OK for M16 bolt to pole in 75mm wide crossarm,

1400 x 100 x 75 HW crossarm OK

Concrete pole designTransverse design wind loads:

1. Wind on Pole:

Wnpole = Pole Az x Pole wind pressure

= 9.2x(0.24+0.15)/2 x 1460/1000

= 2.62 kN

This force acts at height =((0.240.15)x9.2/2x9.2/3+(0.15x9.2x9.2/2)/(9.2x(0.24+0.15)/2

= 4.25 m

2. Wind on Crossarm:

Wncrossarm = Crossarm Az x Crossarm wind pressure (assume 100 x 75mm)

= (0.1 x 0.075) x 1096/1000

= 0.01 kN This force acts at height of: 9.20 m

7/14/2011

10

Concrete pole design

3. Wind on Conductors:

Wncond = Conductor Az x Conductor wind pressure (3 conductors)

= 3 x (Diam x Span x SRF ) x Conductor wind pressure

= 3 x (14.2/1000 x 70 x 1) x913/1000

= 2.72 kN This force acts at height of 9.20 m

(Ref Fig B5 and B6 of AS/NZS 7000 for SRF )

4. Wind on Insulators:

(assume 150 x 129mm projected area)

Wninsul = Insulator Az x Insulator wind pressure

= 3 x (0.150 x 0.129) x 1367/1000

= 0.08 kN This force acts at height of 9.20 m

Concrete pole design

Line deviation loads:

Line deviation angle =10 degrees

Tid

=2 x Fat

x sin 10/2

=2 x 8.76 x sin(5)

=1.53kN

Determine pole design overturning moments:

Taking moments about ground line for the maximum wind and weight loading

condition

∑(BM) = 1.0 Wn +1.1 Gs +1.25 Gc + 1.25 Ftw

= 1.0 ((2.62 x 4.25) + (0.01 x 9.2) + (2.72 x 9.2) + (0.08 x 9.2)) + 1.1 ( 3x .27 x .75) +

1.25 (3 x 1.53 x 9.2 )

= 90.435 kN-m

Assume 9.28 m effective cantilever ( ie 9.20 O/A above ground +200 - 120 to king bolt)

Pole design transverse tip load = 90.43 / 9.28

= 9.74 kN Pole Capacity

7/14/2011

11

Concrete pole designAS/NZS 7000 :2010 Appendix I Clause I3 provides for

• strength capacity factor (φ) should not be taken as greater than 1.0 when designs

are load tested.

• 0.9 when designed by calculation

Assuming a proprietary design PSC pole based on calculation supported by testing

Then required ØRn must be > 9.74kN

ie Rn transverse = or > 9.74/0.9 = 10.82kN tip load.

Longitudinal strength capacity (tip load ) as general rule should be not < 0.25 x

transverse capacity

ieRn longitudinal = or > 10.82 x .25 = 2.70 kN tip load.

Pole selection :

From Busck manufacturers catalogue

Suggest selection of 11.0 Busck PSC pole

TLC transverse = 22.0kN TLC longitudinal = 8kN

ie >> required Rn

Pole foundation.

Assumed pole butt details:

Average pole width: = (150 + 240)/2 = 195 mm

Embedment depth: = 1.8 - 0.20 = 1.60m

Load height: = 9.280 m (from crossarm attachment to 200 below GL.)

Assume no blocking

Assumed soil type : Firm sandy clay

Assumed soil properties: bearing strength fbu = 200 kPa (above water table)

Pole Loads: Pole tip load = 10.82kN

7/14/2011

12

Pole foundationEmbedment depth (D) may be determined from the following equation based on

ASCE Method:

Where C = fbu.b for ultimate limit state,

Fbu = the maximum bearing strength of the foundation material (kPa)

b = the effective width of the footing, projected on a plane perpendicular to

the direction of the resultant horizontal force acting on the pole (m)

M = the overturning moment acting on the pole at ground level (kNm)

= HR.hr

HR= the resultant of the horizontal forces acting on the pole tip (kN)

Hr = the height above ground level at which HR acts (m)

C

CMHHD

2

2.1696.126.3 2RR ++

=

TYPICAL BEARING STRENGTH ( fbu )OF SOILS

Class Very soft Soft Firm Very firm Hard

Soil descriptionSilty clays andsands; loose drysands

Wet clays; siltyloams; wet orloose sands

Damp clays; sandyclays; damp sands

Dry clays; clayeysands; coarse sands; compactsands

Gravels; dryclays

Strength (fbu) kPa fbu ≤ 100 100 < fbu ≤ 150 150 < fbu ≤ 250 250 < fbu ≤ 350 350 < fbu

BEARING STRENGTH OF SOILS AT THE SERVICEABILITY LIMIT STATE

7/14/2011

13

Compacted Soil Footing:

Assume standard heel block (490 x 225)

b = 490 mm

h = 225 mm

Average pole width b = ((240 -(240-150)/3) + 240)/2

= 225 mm

Applying values to formula D = 3.44m > 1.60 available

Stabilized backfill footing :

Assume use compacted stabilised backfill or compacted road base backfill in 600 dia

bored hole with standard heel block. ie b=600

Applying assumed soil values D= 2.00m > 1.60 available

Note: Neither of these two alternatives are acceptable. This is due to the high O/T

moment being applied by the deviation loads (52.78kNm out of 90.43kNm )

Alternative stabilized backfill footing :

Assume use compacted stabilised backfill or compacted road base backfill in

900 mm dia. bored hole with standard heel block. ie b=.900m

Applying assumed soil values D= 1.6m c/w 1.6 available and would be OK

7/14/2011

14

Breast Block Alternative:

Once breast logs are utilised the overturning moments are

resisted by reaction mobilised by the bearing blocks and

base block

Assume 1200wide x 360 deep block 200 below GL

Resolving BM’s around pinned base

Rbreast = 10.82 x (9.280 +1.60) / (1.60 - 0.180) = 82.49 kN

Rbase = 82.49 – 10.82 = 72.08 kN

Bearing block area = 1.2 x .360 = .432 m2

Bearing pressure = 82.49/.432=195kPa < 200 kPa assumed

Stay

θ

Rh

RV

Ts

Back stay alternative :

Due to the high O/T moment being applied by the deviation

loads (52.78kNm out of 90.43kNm ) this option may be more

suitable in soft soil sites

OTM on pole = 90.43 kNm

Assume stay attachment point is at crossarm brace bolt at

330 mm below crossarm attachment point

Stay attachment height = 9.28- 0.33 = 8.95m

Therefore Rh = 90.43/8.95 = 10.10 kN

Stay Tension Ts = Rh / Cos45 = 14.28kN

9.5mm dia stay wire has rated capacity of 45kN

Use 9.5mm stay wire and standard dead man anchor

Vertical load component of stay load = Rh x tan45 =14.28kN

This is a relative small vertical load on the pole and is OK

7/14/2011

15

Questions?

Australian Panel B2 –Overhead Lines Seminar –AS/NZS Overhead Line Design Sydney 28 -29 March 2011