BO15-003

1
BO15-003.xmcd
ABB AB
Rapport
Report
R BO15-003
Från - From
Reg
Datum - Date
Sida - Page
1/26
Författare - Author
PPHB/BOD
Ulf Åkesson
Godkännare - Approv ed by
Ordernr - Ref.No.
2015-02-09
X
Utredning, teoretisk
undersökning -Analysis,
theoretical investigation
X
Slutrapport
Final report
Tom my Borg
Antal textsidor - No of text pages
LTB145D1
Load combinations
24
Antal bilagor - No of supplements
2
Breaker type:
LTB72,5D1 3-pole FSA
Dim ension print: 1HSBDG00001-329
Zone:
0.5 g according to IEEE 693 2005
________________________________________________________________________________
Sammanfattning - Summary
Load com binations calculation is carried out according to IEC 62155 and IEEE 693 2005.
TThe circuit breaker is equipped with porcelain insulators and ηsupport fram e.
LC1:
Design pressure 100%, Mass 100%, Normal wind 30%, Static term inal load 100%
and Impact at operation 100%.
Extrem e load cases:
LC2:
Design Pressure 100%, Mass 100%, Ice load 100%, Norm al wind 110%, Static terminal load 100%
and Impact at operation 100%.
LC3:
Design Pressure 100%, Mass 100%, High wind 100%, Static terminal load 100%
and Impact at operation 100%.
LC4:
Design Pressure 100%, Mass 100%, Short circuit load 100%, Normal wind 100%, Static term inal
load 100% and Im pact at operation 100%.
LC5:
Design Pressure 100%, Mass 100%, Earthquake load 100%, Norm al wind 10%, Static terminal
load 100 % and Impact at operation 100 %.
Basic loads:
Horizontal earthquake acceleration 0.5 g
Static term inal loads according to IEC
Short circuit current
Ice
40 kA
20 mm
Norm al wind
High wind
34 m/s
46 m/s
Maximum bending moment on the post insulator occurs under short circuit condition and
maximum bending moment on the support frame occurs under earthquake condition:
Calculated
Minim um
Safety factor Allowed IEEE 693 2005
bending moment
failing load
Calculated
For com bined loads
Nm
Nm
M
1
3
3
Breaking chamber
M5x = 3.8 ´ 10
M = 11.0 ´ 10
= 2.9
SE = 2.0
1
1
1
M5x
2732 2118-BL
1
M
2
3
3
= 2
Post insulator
M4y = 10.0 ´ 10
M = 20.0 ´ 10
SS = 2.0
M4y
2
2
2
2
1HSB422732-G
M
3
3
3
Support fram e
M5x = 39.2 ´ 10
M = 100.0 ´ 10
= 2.6
SE = 1.5
3
3
3
M5x
1HSB425422-VP
3
Maximum foundation load occurs under earthquake condition:
Bending mom ent
Nm
Foundation
Transversal forces
Direction x N
3
M5x = 39.2 ´ 10
3
3
T5x = 12.8 ´ 10
3
Vertical f orce
N
Direction y N
3
T5y = 12.1 ´ 10
Result:
The breaker is well suited to withstand above specified combined loads.
3
3
T5z = 21.3 ´ 10
3
2
BO15-003.xmcd
Content:
1 Introduction
2 Specification of equipment
3 Individual mechanical loads
3.1 Dead weight
3.2 Dead weight from ice load on conductor
3.3 Dead weight from ice load on breaker
3.4 Design pressure according to IEC 62155
3.5 Normal wind load on equipm ent
3.6 High wind load on equipm ent
3.7 Normal wind load on conductor without ice
3.8 Normal wind load on ice loaded conductor
3.9 High wind load on conductor
3.10 Static terminal load
3.11 Im pact load due to operation
3.12 Short circuit load transversal to conductor
3.13 Earthquake load
4 Load combinations
5 Summary of calculated loads
6 Calculation of safety factors for insulators and support frame
7 Foundation load
8 Foundation bolts
Appendix 1 - Dimension print 1HSBDG00001-329
Appendix 2 - General dimensions used in calculation
3
BO15-003.xmcd
1 Introduction
Load com binations calculation is carried out according to:
IEC 62155 and IEEE 693 2005.
The different m echanical loads on the circuit-breaker are individually calculated and after that combined into
actual load combinations.
The following load combinations are considered.
Norm al load case:
LC1:
Design pressure 100%, Mass 100%, Normal wind 30%, Static term inal load 100%
and Impact at operation 100%.
Extrem e load cases:
LC2:
Design Pressure 100%, Mass 100%, Ice load 100%, Norm al wind 110%, Static terminal load 100%
and Impact at operation 100%.
LC3:
Design Pressure 100%, Mass 100%, High wind 100%, Static terminal load 100%
and Impact at operation 100%.
LC4:
Design Pressure 100%, Mass 100%, Short circuit load 100%, Normal wind 100%, Static term inal
load 50% and Im pact at operation 100%.
LC5:
Design Pressure 100%, Mass 100%, Earthquake load 100%, Norm al wind 10%, Static terminal
load 70% and im pact at operation 100%.
Specified mechanical loads:
v = 34
m /s
Norm al wind
vh = 46
m /s
High wind
t = 20
mm
Ice thickness
Fthy = 750
N
Static horizontal terminal load transversal to line
Fthx = 1250
N
Static horizontal terminal load along line
Ftz = 1000
N
Static vertical terminal load
Fyimp = 1000
N
Horizontal load at operation
Fzimp = 12000
N
Vertical load at operation
Isc = 40
kA
Short circuit current
ahr = 16.00
m /s2
Horizontal earthquake acceleration
Based on 0.5 g horizontal acceleration and damping
of critical dam ping ζ=0.02.
2 Specification of equipment.
Circuit-breaker of type:
Dim ension print:
LTB72,5D1 3-pole operated with FSA mechanism
1HSBDG00001-329 See appendix 1 and 2.
4
BO15-003.xmcd
3 Individual mechanical loads
3.1 Dead weight
m = 138
kgs
1 Breaking unit
m = 112
kgs
1 Post insulator
m = 177
kgs
Link gear
m = 170
kgs
Operating mechanism
m = 160
kgs
1 Support frame
1
2
3
4
5
Total dead weight affecting one support frame
2
mtot :=
å
i=1
æ mi × 3 ö
ç
÷+
è 2 ø
2
mtot = 7.1 ´ 10
4
m
å
i
i=3
kgs
+m
5
2
Total dead weight
Centre of gravities according to appendix 2.
h = 4.499
m
Height to centre of gravity of breaking unit
h = 3.117
m
Height to centre of gravity of post isolator
h = 2.298
m
Height to centre of gravity of link gear
h = 1.702
m
Height to centre of gravity of operating mechanism
h = 1.103
m
Height to centre of gravity of support fram e
1
2
3
4
5
3.2 Dead weight from ice load on conductor according to IEC 62271-1
Lc = 1.3
m
Length of conductor bound to equipment
Dc = 100
mm
Diameter of conductor
t = 20
mm
Thickness of ice
N/m
Ice load
Qicez := QI× Lc× 2
N
Ice load on terminals
Qicez = 646.9
N
Vertical load from ice on term inal of one breaker pole
QI := π× éë( Dc + t) - Dc ùû × 10
2
2
-6
× 900× t
5
BO15-003.xmcd
3.3 Dead weight from ice load on breaker according to IEC 62271-1
Extra weight in kgs due to ice:
mice =
i =
i
1
23.2
Ice on one breaking unit
2
24.1
Ice on one post insulator
3
33.4
Ice on pole beam
4
50.5
5
67.7
Ice on operating m echanism
Ice on one support fram e
3.4 Design Pressure according to IEC 62155
5
Pdes = 7 ´ 10
Pa
Design pressure
p =
1
2
Breaking cham ber insulator
Post insulator
Ds =
p
m
Sealing diam eter
m
Outer diam eter
m
Inner diameter
0.200
0.160
Do =
p
0.225
0.190
Di =
p
0.175
0.130
Mdes := Pdes ×
p
π
32
( p)
× Ds
(× Dop)2 + (Dip)2
2
Do
p
Mdes = 993
Nm
Equivalent bending moment on breaking chamber insulator due to design
pressure
Mdes = 491
Nm
Equivalent bending moment on post insulator due to design pressure
1
2
6
BO15-003.xmcd
3.5 Normal wind load on equipment
v = 34
m /s
cc := 0.9
Drag factor, cyli ndrical surface
cp := 1.3
Drag factor, flat surf ace
A = 0.40
m2
Frontal area of breaking unit
A = 0.33
m2
Frontal area of post insulator
A = 0.77
m2
Frontal area of link gear
A = 0.58
m2
Frontal area of operating mechanism
A = 0.37
m2
Frontal area of support fram e
B = 0.41
m2
Side area of breaking unit
B = 0.33
m2
Side area of post insulator
B = 0.10
m2
Side area of link gear
B = 0.48
m2
Side area of operating mechanism
B = 0.49
m2
Side area of support fram e
1
2
3
4
5
1
2
3
4
5
Norm al wind speed
Load due to normal wind [X direction]:
2
Qwx := 0.625× v × A × cc
1
1
2
Qwx := 0.625× v × A × cc
2
2
2
Qwx := 0.625× v × A × cp
3
3
2
Qwx := 0.625× v × A × cp
4
4
2
Qwx := 0.625× v × A × cp
5
5
Qwx = 262
N
Load in X direction on one breaking unit due to normal wind
Qwx = 217
N
Load in X direction on one post insulator due to normal wind
Qwx = 723
N
Load in X direction on link gear due to normal wind
Qwx = 545
N
Load in X direction on operating mechanism due to
normal wind
Qwx = 348
N
Load in X direction on one support fram e due to normal wind
1
2
3
4
5
7
BO15-003.xmcd
Load due to normal wind [Y direction]:
2
Qwy := 0.625× v × B × cc
1
1
2
Qwy := 0.625× v × B × cc
2
2
2
Qwy := 0.625× v × B × cp
3
3
2
Qwy := 0.625× v × B × cp
4
4
2
Qwy := 0.625× v × B × cp
5
5
Qwy = 267
N
Load in Y direction on one breaking unit due to normal wind
Qwy = 217
N
Load in Y direction on one post insulator due to normal wind
Qwy = 94
N
Load in Y direction on link gear due to norm al wind
Qwy = 451
N
Load in Y direction on operating mechanism due to
normal wind
Qwy = 460
N
Load in Y direction on one support fram e due to norm al wind
1
2
3
4
5
Heights to locations where loads on the equipm ent will be calculated
L = 5.204
m
Height to upper terminal
L = 3.837
m
Height to bottom of breaking unit
L = 2.417
m
Height to bottom of post insulator
0
1
2
8
BO15-003.xmcd
Bending mom ent due to normal wind [X direction]:
( 1 1)
Mwx := Qwx × (h - L ) + Qwx × ( h - L )
2
1 1
2
2 2
2
Mwx := Qwx × h - L
1
1
2
Mwx :=
3
å
i=1
æ Qwx × h × 3 ö +
ç
i i 2÷
è
ø
4
å
i=3
æ Qwxi × hi ö
ç
÷ + Qwx5× h5
è 2 ø
Mwx = 173
Nm
Bending mom ent in X direction on one breaking unit due to
normal wind
Mwx = 697
Nm
Bending mom ent in X direction on one post insulator due to
normal wind
Mwx = 4460
Nm
Bending mom ent in X direction on one support fram e due
to normal wind
1
2
3
Bending mom ent due to normal wind [Y direction]:
( 1 1)
Mwy := Qwy × (h - L ) + Qwy × ( h - L )
2
1 1
2
2 2
2
Mwy := Qwy × h - L
1
1
2
Mwy :=
3
å
i=1
æ Qwy × h × 3 ö +
ç
i i 2÷
è
ø
4
å
i=3
æ Qwyi × hi ö
ç
÷ + Qwy5× h5
è 2 ø
Mwy = 177
Nm
Bending mom ent in Y direction on one breaking unit due to
normal wind
Mwy = 707
Nm
Bending mom ent in Y direction on one post insulator due to
normal wind
Mwy = 3813
Nm
Bending mom ent in Y direction on one support f ram e due
to normal wind
1
2
3
9
BO15-003.xmcd
3.6 High wind load on equipment
Load due to high wind [X direction]:
2
Qwhx := 0.625× vh × A × cc
1
1
2
Qwhx := 0.625× vh × A × cc
2
2
2
Qwhx := 0.625× vh × A × cp
3
3
2
Qwhx := 0.625× vh × A × cp
4
4
2
Qwhx := 0.625× vh × A × cp
5
5
Qwhx = 479
N
Load in X direction on one breaking unit due to high wind
Qwhx = 397
N
Load in X direction on one post insulator due to
high wind
Qwhx = 1324
N
Load in X direction on link gear due to high wind
Qwhx = 997
N
Load in X direction on operating mechanism due to
high wind
Qwhx = 636
N
Load in X direction on one support fram e due to high wind
1
2
3
4
5
Load due to high wind [Y direction]:
2
Qwhy := 0.625× vh × B × cc
1
1
2
Qwhy := 0.625× vh × B × cc
2
2
2
Qwhy := 0.625× vh × B × cc
3
3
2
Qwhy := 0.625× vh × B × cp
4
4
2
Qwhy := 0.625× vh × B × cp
5
5
Qwhy = 488
N
Load in Y direction on one breaking unit due to high wind
Qwhy = 397
N
Load in Y direction on one post insulator due to
high wind
Qwhy = 119
N
Load in Y direction on link gear due to high wind
Qwhy = 825
N
Load in Y direction on operating mechanism due to
high wind
Qwhy = 842
N
Load in Y direction on one support fram e due to high wind
1
2
3
4
5
10
BO15-003.xmcd
Heights to locations where loads on the equipm ent will be calculated
L = 5.204
m
Height to upper terminal
L = 3.837
m
Height to bottom of breaking unit
L = 2.417
m
Height to bottom of post insulator
0
1
2
Bending mom ent due to high wind [X direction]:
( 1 1)
Mwhx := Qwhx × ( h - L ) + Qwhx × ( h - L )
2
1 1
2
2 2
2
Mwhx := Qwhx × h - L
1
1
2
Mwhx :=
3
å
i=1
æ Qwhx × h × 3 ö +
ç
1 i 2÷
è
ø
4
å
i=3
æ Qwhxi × hi ö
ç
÷ + Qwhx5× h5
è 2
ø
Mwhx = 317
Nm
Bending mom ent in X direction on one breaking unit due to
high wind
Mwhx = 1276
Nm
Bending mom ent in X direction on one post insulator due to
high wind
Mwhx = 8547
Nm
Bending mom ent in X direction on one support fram e due
to high wind
1
2
3
Bending mom ent due to high wind [Y direction]:
( 1 1)
Mwhy := Qwhy × ( h - L ) + Qwhy × (h - L )
2
1 1
2
2 2
2
Mwhy := Qwhy × h - L
1
1
2
Mwhy :=
3
å
i=1
æ Qwhy × h × 3 ö +
ç
i i 2÷
è
ø
4
å
i=3
æ Qwhyi × hi ö
ç
÷ + Qwhy5× h5
è 2 ø
Mwhy = 323.1
Nm
Bending mom ent in Y direction on one breaking unit due to
high wind
Mwhy = 1294
Nm
Bending mom ent in Y direction on one post insulator due to
high wind
Mwhy = 6920
Nm
Bending mom ent in Y direction on one support f ram e due
to high wind
1
2
3
11
BO15-003.xmcd
3.7 Normal wind load on conductor without ice
v = 34
m /s
cc = 0.9
Drag factor, cyli ndrical surface
Norm al wind
Ac := Lc× Dc× 0.001× 2
Ac = 0.3
m2
Frontal area of conductor
Transversal load due to normal wind on c onductor:
2
Qwno := .625× v × cc × Ac
Qwno = 169.1
N
Norm al wind load on conductor per term inal
Bending mom ent due to normal wind from on conductor:
( 1 1) 2
Mwno := Qwno× (h - L )
2
1
2
Mwno := Qwno× h - L ×
1
1
Mwno := Qwno× h ×
3
3
1 2
Mwno = 56
Nm
Bending mom ent in Y direction on one breaking unit due to
normal wind on conductor
Mwno = 352
Nm
Bending mom ent in Y direction on one post insulator due to
normal wind on conductor
Mwno = 1141
Nm
Bending mom ent in Y direction on one support f ram e due
to normal wind on conductor
1
2
3
12
BO15-003.xmcd
3.8 Normal wind load on ice loaded conductor
v = 34
m /s
cc = 0.9
Drag factor, cyli ndrical surface
Norm al wind
Ac := Lc× ( Dc× 0.001 + t× 0.001) × 2
Ac = 0.3
m2
Frontal area of conductor
Transversal load due to normal wind on iced conductor:
2
Qwice := .625× v × cc× Ac
N
Norm al wind load on iced conductor
Qwice = 202.9
Bending mom ent due to normal wind from iced conductor:
( 1 1) 2
Mwice := Qwice× ( h - L )
2
1
2
Mwice := Qwice× h - L ×
1
1
Mwice := Qwice× h ×
3
3
1 2
Mwice = 67.2
Nm
Bending mom ent in Y direction on one breaking unit due to
normal wind on conductor with ice
Mwice = 422
Nm
Bending mom ent in Y direction on one post insulator due to
normal wind on conductor with ice
Mwice = 1369
Nm
Bending mom ent in Y direction on one support f ram e due
to normal wind on conductor with ice
1
2
3
13
BO15-003.xmcd
3.9 High wind load on conductor
At high wind it is considered that no ice loading exists.
vh = 46
m /s
cc = 0.9
Drag factor, cyli ndrical surface
High wind
Ach := Lc× Dc× 0.001× 2
Ach = 0.3
m2
Frontal area of conductor
Transversal load due to high wind on c onductor:
2
Qwhno := .625× vh × cc × Ach
N
High wind load on conductor
Qwhno = 309.5
Bending mom ent due to high wind on conductor:
( 1 1) 2
Mwhno := Qwhno× ( h - L )
2
1
2
Mwhno := Qwhno× h - L ×
1
1
Mwhno := Qwhno× h ×
3
3
1 2
Mwhno = 102
Nm
Bending mom ent in Y direction on one breaking unit due to
high wind on conductor
Mwhno = 644
Nm
Bending mom ent in Y direction on one post insulator due to
high wind on conductor
Mwhno = 2088
Nm
Bending mom ent in Y direction on one support fram e due
to high wind on conductor
1
2
3
14
BO15-003.xmcd
3.10 Static terminal load according to IEC 62271-100
The static terminal load due to weight and type of connection will be added to all load cases.
Static term inal load due to wind and ice are separately calculated in other paragraphs.
Fthy = 750
N
Horizontal load transversal to line
Fthx = 1250
N
Horizontal load along line
Ftz = 1000
N
Vertical load
Lb = 0.255
m
Distance from centre line to end of terminal flange
Bending mom ent along line due to static term inal load:
( 0 1) + Ftz× Lb
MFthx := Fthx× (L - L ) + Ftz× Lb
2
0
2
3
MFthx := ( Fthx× L + Ftz× Lb) ×
3
0
2
MFthx := Fthx× L - L
1
MFthx = 1964
Nm
Bending mom ent in X direction on one breaking unit due to
static terminal load
MFthx = 3739
Nm
Bending mom ent in X direction on one post insulator due to
static terminal load
Nm
Bending mom ent in X direction on one support fram e due to
static terminal load
1
2
3
MFthx = 10.140 ´ 10
3
Bending mom ent transversal to line due to static terminal load:
( 0 1)
MFthy := Fthy× ( L - L )
2
0
2
MFthy := Fthy× L - L
1
MFthy := Fthy× L ×
3
3
0 2
MFthy = 1025
Nm
Bending mom ent in Y direction on one breaking unit due to
static terminal load
MFthy = 2090
Nm
Bending mom ent in Y direction on one post insulator due to
static terminal load
MFthy = 5855
Nm
Bending mom ent in Y direction on one support fram e due to
static terminal load
1
2
3
15
BO15-003.xmcd
3.11 Im pact load due to operation
The bending mom ent at the support frame and bottom of post insulator due to operational impact
is added to all load com binations.
Fyimp = 1000
N
Horizontal force at operation
Fzimp = 12000
N
Vertical f orce at operat ion
Mimp = 1500
Nm
Bending mom ent on structure at operation
3.12 Short circuit load transversal to conductor
The electromagnetic force between interrupters is calculated by the following formula:
K := 2.95
Engineering factor
Isc = 40
kArms Max. three phase fault current
Lph = 1.500
m
Spacing between phases
a := 1
Support factor
(
Lsc := L - L
0
1
) + Lb×2 + Lc×2 m
Interrupter length and part of conductor length
2
Fscy := 0.2 × K×
Isc
Lph
× a× Lsc
N
Electromagnetic force on equipm ent
Fscy = 2818
Bending mom ent due to short circuit:
hsc = 4.5
m
Height to centre of conductor under short circuit
(
1)
Mscy := Fscy× (hsc - L )
2
2
Mscy := Fscy× hsc - L
1
Mscy = 1865
Nm
Bending mom ent in X direction on one breaking unit due to
short circuit load
Mscy = 5866
Nm
Bending mom ent in X direction on one post insulator due to
short circuit load
1
2
16
BO15-003.xmcd
3.13 Earthquake load
fnatural = 2.8
Hz
Fundamental m ode
ζ = 0.02
Damping of critical dam ping
ah = 5.0
m /s2
Horizontal acceleration
av = 2.5
m /s2
Vertical acceleration
ahr = 16.00
m /s2
Horizontal acceleration response
avr = 8.00
m /s2
Vertical acceleration response
Load on each mass due to horizontal earthquake acceleration:
Feh := ahr × m
i
i
Load on each mass due to vertical earthquake acceleration:
Fev := avr × m
i
i
Bending mom ent due to earthquake acceleration:
(1
1
(1
2
Meh := Feh × h - L
1
1
)
Meh := Feh × h - L
) + Feh2×(h2 - L2)
2
4
2
Meh :=
3
1
å
i=1
æ Feh × h × 3 ö +
ç i i 2÷
è
ø
å
i=3
æ Fehi × hi ö
ç
÷ + Feh5× h5
è 2 ø
Meh = 1462
Nm
Bending mom ent on one breaking unit due to
horizontal earthquake acceleration
Meh = 5851
Nm
Bending mom ent in on one post insulator due to
horizontal earthquake acceleration
Nm
Bending mom ent in on one support fram e due to
horizontal earthquake acceleration
1
2
3
Meh = 31.672 ´ 10
3
17
BO15-003.xmcd
4 Load combinations
LC1 (see definition clause1):
Horizontal loads [X direction]:
T1x := Qwx × 0.3 + Fthx
1
1
T1x := T1x + Qwx × 0.3
2
1
2
Qwx + Qwx
ö
3 æ
3
4
T1x := T1x × + ç
+ Qwx ÷ × 0.3
3
2 2
5
2
è
ø
Horizontal loads [Y direction]:
(
)
T1y := Qwy + Qwno × 0.3 + Fthy
1
1
T1y := T1y + Qwy × 0.3
2
1
T1y := T1y ×
3
2
3
2 2
+
æ Qwy + Qwy
3
ç
è
4
ö
+ Qwy ÷ × 0.3 + Fyimp
ø
5
2
Bending moment [X direction]:
M1x := Mdes + Mwx × 0.3 + MFthx
1
1
1
1
M1x := Mdes + Mwx × 0.3 + MFthx
2
2
2
2
M1x := Mwx × 0.3 + MFthx
3
3
3
Bending moment [Y direction]:
(
1
(
2
)
1
)
2
M1y := Mdes + Mwy + Mwno × 0.3 + MFthy
1
1
1
M1y := Mdes + Mwy + Mwno × 0.3 + MFthy + Mimp
2
2
(
2
)
M1y := Mwy + Mwno × 0.3 + MFthy + Mimp
3
3
3
3
Vertical loads [Z direction]:
T1z := m × 9.81 + Ftz
1
1
T1z := T1z + m × 9.81
2
1
T1z := T1z ×
3
2
3
2 2
4
+
å
i=3
æ mi × 9.81 ö
ç
÷ + m5× 9.81 + Fzimp
è 2 ø
18
BO15-003.xmcd
LC2 (see definition clause1):
Horizontal loads [X direction]:
T2x := Qwx × 1.1 + Fthx
1
1
T2x := T2x + Qwx × 1.1
2
1
2
T2x := T2x ×
3
3
+
2 2
æ Qwx + Qwx
3
ç
è
ö
4
+ Qwx ÷ × 1.1
ø
5
2
Horizontal loads [Y direction]:
(
)
T2y := Qwy + Qwice × 1.1 + Fthy
1
1
T2y := T2y + Qwy × 1.1
2
1
2
Qwy + Qwy
ö
3 æ
3
4
T2y := T2y × + ç
+ Qwy ÷ × 1.1 + Fyimp
3
2 2
5
2
è
ø
Bending moment [X direction]:
M2x := Mdes + Mwx × 1.1 + MFthx
1
1
1
1
M2x := Mdes + Mwx × 1.1 + MFthx
2
2
2
2
M2x := Mwx × 1.1 + MFthx
3
3
3
Bending moment [Y direction]:
(
1
(
2
)
1
)
2
M2y := Mdes + Mwy + Mwice × 1.1 + MFthy
1
1
1
M2y := Mdes + Mwy + Mwice × 1.1 + MFthy + Mimp
2
2
(
2
)
M2y := Mwy + Mwice × 1.1 + MFthy + Mimp
3
3
3
3
Vertical loads [Z direction]:
(
)
T2z := m + mice × 9.81 + Qicez + Ftz
1
1
1
(
)
T2z := T2z + m + mice × 9.81
2
1
T2z := T2z ×
3
3
2 2
2
2
4
+
å
i=3
é ( mi + micei ) × 9.81ù
ê
ú + ( m5 + mice5) × 9.81 + Fzimp
2
ë
û
19
BO15-003.xmcd
LC3 (see definition clause1):
Horizontal loads [X direction]:
T3x := Qwhx + Fthx
1
1
T3x := T3x + Qwhx
2
1
T3x := T3x ×
3
3
2 2
2
Qwhx + Qwhx
3
+
4
+ Qwhx
2
5
Horizontal loads [Y direction]:
T3y := Qwhy + Qwhno + Fthy
1
1
T3y := T3y + Qwhy
2
1
T3y := T3y ×
3
2
Qwhy + Qwhy
3
3
+
2 2
4
+ Qwhy + Fyimp
5
2
Bending moment [X direction]:
M3x := Mdes + Mwhx + MFthx
1
1
1
M3x := Mdes + Mwhx + MFthx
2
2
2
1
2
M3x := Mwhx + MFthx
3
3
3
Bending moment [Y direction]:
M3y := Mdes + Mwhy + Mwhno + MFthy
1
1
1
1
1
M3y := Mdes + Mwhy + Mwhno + MFthy + Mimp
2
2
2
2
2
M3y := Mwhy + Mwhno + MFthy + Mimp
3
3
3
3
Vertical loads [Z direction]:
T3z := m × 9.81 + Ftz
1
1
T3z := T3z + m × 9.81
2
1
T3z := T3z ×
3
2
3
2 2
4
+
å
i=3
æ mi × 9.81 ö
ç
÷ + m5× 9.81 + Fzimp
è 2 ø
20
BO15-003.xmcd
LC4 (see definition clause1):
Horizontal loads [X direction]:
T4x := Qwx + Fthx× 0.5
1
1
T4x := T4x + Qwx
2
1
2
T4x := T4x ×
3
3
2 2
Qwx + Qwx
3
+
4
+ Qwx
5
2
Horizontal loads [Y direction]:
T4y := Qwy + Qwno + Fthy× 0.5 + Fscy
1
1
T4y := T4y + Qwy
2
1
T4y := T4y ×
3
2
3
2 2
Qwy + Qwy
3
+
4
2
+ Qwy + Fyimp
5
Bending moment [X direction]:
M4x := Mdes + Mwx + MFthx × 0.5
1
1
1
1
M4x := Mdes + Mwx + MFthx × 0.5
2
2
2
2
M4x := Mwx + MFthx × 0.5
3
3
3
Bending moment [Y direction]:
M4y := Mdes + Mwy + Mwno + MFthy × 0.5 + Mscy
1
1
1
1
1
1
M4y := Mdes + Mwy + Mwno + MFthy × 0.5 + Mscy + Mimp
2
2
2
2
2
M4y := Mwy + Mwno + MFthy × 0.5 + Mimp
3
3
3
3
Vertical loads [Z direction]:
T4z := m × 9.81 + Ftz× 0.5
1
1
T4z := T4z + m × 9.81
2
1
T4z := T4z ×
3
2
3
2 2
4
+
å
i=3
æ mi × 9.81 ö
ç
÷ + m5× 9.81 + Fzimp
è 2 ø
2
21
BO15-003.xmcd
LC5 (see definition clause1):
Horizontal loads [X direction]:
T5x := Qwx × 0.1 + Fthx× 0.7 + Feh
1
1
1
T5x := T5x + Qwx × 0.1 + Feh
2
1
2
T5x := T5x ×
3
3
2 2
2
æ Qwxi × 0.1 ö
ç
÷ + Qwx5× 0.1 +
è 2
ø
4
å
+
i=3
4
å
i=3
Feh
i
+ Feh
5
2
Horizontal loads [Y direction]:
(
)
T5y := Qwy + Qwno × 0.1 + Fthy× 0.7 + Feh
1
1
1
T5y := T5y + Qwy × 0.1 + Feh
2
1
2
2
4
Qwy
é 2 æ Qwyi × 3 ö
ù
i
ê
ç
÷
T5y :=
+
+ Qwy + Qwnoú × 0.1 + Fthy× 0.7 ...
3
5
2
ê
ú
è 2 ø
i=3
ëi = 1
û
å
å
é 2
+ ê
ê
ëi = 1
å
æ Fehi × 3 ö
ç
÷+
è 2 ø
2
ù
2
+ Feh ú + Fyimp
5ú
2
û
4
Feh
å
i
i= 3
Bending moment [X direction]:
M5x := Mdes + Mwx × 0.1 + MFthx × 0.7 + Meh
1
1
1
1
M5x := Mdes + Mwx × 0.1 + MFthx × 0.7 + Meh
2
2
2
2
M5x := Mwx × 0.1 + MFthx × 0.7 + Meh
3
3
3
1
2
3
Bending moment [Y direction]:
(
1
(
2
)
1
)
2
M5y := Mdes + Mwy + Mwno × 0.1 + MFthy × 0.7 + Meh
1
1
1
M5y := Mdes + Mwy + Mwno × 0.1 + MFthy × 0.7 +
2
2
(
2
)
M5y := Mwy + Mwno × 0.1 + MFthy × 0.7 +
3
3
3
3
1
(Meh2)2 + Mimp2
(Meh3)2 + Mimp2
Vertical loads [Z direction]:
T5z := m × 9.81 + Ftz× 0.7 + Fev
1
1
1
T5z := T5z + m × 9.81 + Fev
2
T5z :=
3
1
é 2
ê
ê
ëi = 1
å
2
æ mi × 3 ö
ç
÷+
è 2 ø
2
4
å
i=3
ù
æ mi ö
ç ÷ + m5ú × 9.81 + Ftz× 3 × 0.7 ...
2
ú
è2 ø
û
2
4
Fev
é 2 æ Fevi × 3 ö
ù
i
2
ê
ç
÷
+
+
+ Fev ú + Fzimp
5
2
ê
ú
è 2 ø
i=3
ëi = 1
û
å
å
22
BO15-003.xmcd
5 Summary of calculated loads
Horizontal loads [N]:
X
Norm al
wind
T1x =
k=
k
k := 1 , 2 .. 3
Ice
T2x =
k
High
wind
Short
circuit
Earthquake
T3x =
T4x =
T5x =
k
k
k
1
1.3·103
1.5·103
1.7·103
886.9
3.1·103
2
1.4·103
1.8·103
2.1·103
1.1·103
4.9·103
3
2.4·103
3.7·103
5.0·103
2.6·103
12.8·103
Ice
High
wind
Short
circuit
Earthquake
T3y =
T4y =
T5y =
Y
Norm al
wind
T1y =
k=
k
T2y =
k
k
k
k
1
880.7
1.3·103
1.5·103
3.6·103
2.8·103
2
945.8
1.5·103
1.9·103
3.8·103
4.6·103
3
2.6·103
4.1·103
5.2·103
7.5·103
12.1·103
Ice
High
wind
Short
circuit
Earthquake
M3x =
M4x =
M5x =
Bending moment [Nm]:
X
Norm al
wind
M1x =
k=
k
M2x =
k
k
k
k
1
3.0·103
3.1·103
3.3·103
2.1·103
3.8·103
2
4.4·103
5.0·103
5.5·103
3.1·103
9.0·103
3
11.5·103
15.0·103
18.7·103
9.5·103
39.2·103
Ice
High
wind
Short
circuit
Earthquake
M3y =
M4y =
M5y =
Y
Norm al
wind
M1y =
k=
k
M2y =
k
k
k
k
1
2.1·103
2.3·103
2.4·103
3.6·103
3.2·103
2
4.4·103
5.3·103
6.0·103
10.0·103
8.1·103
3
8.8·103
13.1·103
16.4·103
9.4·103
36.3·103
Ice
High
wind
Short
circuit
Earthquake
T3z =
T4z =
T5z =
Vertical loads [N]:
Z
Norm al
wind
T1z =
k=
k
T2z =
k
k
k
k
1
2.4·103
3.2·103
2.4·103
1.9·103
3.2·103
2
3.5·103
4.6·103
3.5·103
3.0·103
5.2·103
3
20.5·103
23.2·103
20.5·103
19.7·103
21.3·103
23
BO15-003.xmcd
6 Calculation of safety factors for insulators and support frame
Minim um failing loads for insulators and support fram e:
3
M = 11.0 ´ 10
1
3
M = 20.0 ´ 10
2
3
M = 100.0 ´ 10
3
Nm
Minim um value of breaking cham ber insulator and top of post insulator
Nm
Post insulator 1HSB422732-G
Nm
Support fram e 1HSB425422-VP
Allowed safety factors according to IEC 62155
IEEE 693 2005
Safety factors direction x:
Norm al
wind
M
M
k
M1x
k=
Ice
=
k
High
wind
M
k
M2x
=
k
Shortcircuit
M
k
M3x
=
k
Earthquake
M
k
M4x
=
k
k
M5x
=
3.7
3.5
3.4
5.1
2.9
2
4.5
4.0
3.6
6.5
2.2
3
8.7
6.6
5.4
10.5
2.6
Ice
SN :=
2.1
2.1
1.5
k
k
1
Norm al
wind
Safety factors direction y:
High
wind
Short
circuit
Earth
quake
SI :=
SH :=
SS :=
SE :=
1.2
1.2
1.2
1.2
1.2
1.2
2.0
2.0
1.5
2.0
2.0
1.5
k
k
k
k
Allowed safety factors according to IEC 62155
IEEE 693 2005
Norm al
wind
M
M
k
M1y
k=
k
1
Ice
=
High
wind
M
k
M2y
k
=
Shortcircuit
M
k
M3y
k
5.3
4.8
4.5
2
4.5
3.8
3
11.3
7.7
=
Earthquake
M
k
M4y
k
=
k
M5y
k
3.1
3.4
3.3
2.0
2.5
6.1
10.7
2.8
=
Norm al
wind
Ice
SN :=
2.1
2.1
1.5
k
High
wind
Shortcircuit
Earth
quake
SI :=
SH :=
SS :=
SE :=
1.2
1.2
1.2
1.2
1.2
1.2
2.0
2.0
1.5
2.0
2.0
1.5
k
k
k
k
24
BO15-003.xmcd
7 Support frame foundation loads
Vertical
Horizontal
direction z [N]
direction x [N]
3
3
LC1 T1z = 20.5 ´ 10
T1x = 2.4 ´ 10
3
3
3
3
3
3
3
3
3
3
M3y = 16.4 ´ 10
3
3
3
M4x = 9.5 ´ 10
M4y = 9.4 ´ 10
3
3
3
3
T5y = 12.1 ´ 10
3
3
3
3
T5x = 12.8 ´ 10
3
3
M2y = 13.1 ´ 10
M3x = 18.7 ´ 10
T4y = 7.5 ´ 10
3
LC5 T5z = 21.3 ´ 10
3
3
3
3
3
3
T4x = 2.6 ´ 10
3
3
M1y = 8.8 ´ 10
M2x = 15.0 ´ 10
T3y = 5.2 ´ 10
3
LC4 T4z = 19.7 ´ 10
3
3
3
Horizontal
direction y [Nm]
3
3
T3x = 5.0 ´ 10
3
M1x = 11.5 ´ 10
T2y = 4.1 ´ 10
3
LC3 T3z = 20.5 ´ 10
3
3
T2x = 3.7 ´ 10
3
Horizontal
direction x [Nm]
T1y = 2.6 ´ 10
3
LC2 T2z = 23.2 ´ 10
Horizontal
direction y [N]
3
M5x = 39.2 ´ 10
3
M5y = 36.3 ´ 10
3
3
8 Foundation bolts.
Maximum load on support frame foundation bolts:
Tz :=
Tx :=
Ty :=
Mx :=
My :=
T1z
T1x
T1y
M1x
M1y
T2z
T2x
T2y
M2x
T3z
T3x
T3y
M3x
T4z
T4x
T4y
M4x
T5z
T5x
T5y
M5x
i
i
3
i
3
3
3
3
3
3
3
3
3
3
3
i
3
3
3
3
i
3
M2y
3
3
M3y
3
3
M4y
3
3
M5y
3
3
Axial load and stress for M24 bolts; Load case 1 to 4:
Axial stress [N/m m 2]
Axial loads [N]
Tz
Faz :=
i
i
4
(
max Mx , My
+
i
2× db
i
)
Faz
σz :=
i
i
352.490
Shear stress X [N/mm 2]
Tx
τx :=
i
σz =
i
Ty
i
π
4
Faz =
Shear stress Y [N/m m 2]
× 21.185
τy :=
τx =
i
i
2
4
i
55.2
69.8
6.8
10.6
7.5
11.5
28.5·103
16.8·103
80.8
14.1
14.8
47.8
7.5
N
154.2
N/m m 2
36.4
21.3
N/m m 2
Equivalent stress:
σe :=
i
× 21.185
τy =
i
19.5·103
24.6·103
54.3·103
i
π
(σzi)2 + 3×éë(τxi)2 + (τyi )2ùû
match( max( σe) , σe) = ( 5 )
Load case that yields the highest load on foundation bolts
max( σe) = 176.8
N/m m 2
For size M24 foundation bolts a material with yield point larger than se shall be used.
34.2 N/m m 2
2
25
BO15-003.xmcd
26
BO15-003.xmcd