ecce 1202 013 wind induced vibration tall steel chimney

8122019 ECCE 1202 013 Wind Induced Vibration Tall Steel Chimney

httpslidepdfcomreaderfullecce-1202-013-wind-induced-vibration-tall-steel-chimney 19

Canadian Journal on Environmental Construction and Civil Engineering Vol 3 No 2 February 2012

71

Wind Induced Vibration of a Tall Steel Chimney

Sule S Nwofor T C

CivilEnvironmental Engineering

University of Port Harcourt

E-mail samvictoryaheadyahoocom

Abstract

In this paper the vortex induced vibration of a 50m steel chimney under wind excitation is discussed A

steel chimney of length 50m is modeled as a cantilever structure subjected to two degrees of freedom

Lumped parameter approach was employed to estimate the natural frequencies of vibration It was shown

among other findings that the intensity of wind loading varies with chimney height with maximum value

at chimney top and minimum value at the base and the frequency of vortex shedding due to wind

excitation was found to be greater in value than the fundamental frequency of vibration showing that the

chimney may go into resonance leading to a very large and severe deflection and damage to the steel

chimney in the form of fatique during the expected lifetime of the structure

Keywords Wind induced vibration steel chimney degrees of freedom lumped mass wind excitation

10 Introduction

A chimney is used to emit exhaust gases higher up in the atmosphere to facilitate diffusion of gases Asteel chimney is ideally suited for process work where a short heat up period and low thermal capacity are

required The effect of wind excited vibration on tall steel chimney is a matter of great concern to both

structural and design engineers as this may lead to undesirable physical phenomenon called resonance if

one of the natural frequencies of vibration is excited[1-10]

The effect is large and severe deflection and

damage to the structure Wind induced vibration of a tall steel chimney is due to vortex shedding process[11] The vortices shed off from a bluff body as the flow region is separated inducing a fluctuating force on

the chimney leading to chimney vibration The amplitude of vibration of the chimney is dependent on the

intensity of this fluctuating force

In this paper a 50m steel chimney is modeled as a cantilever structure subjected to two-degree of freedom

to determine the fundamental frequency of vibration due to wind excitation The value was compared

with the frequency of the vortex shedding due to wind excitation to investigate the possibility of the

chimney going into resonance

20 Formulation of Mathematical Model

Consider a flexural beam with distributed masses 21 mand m at nodal points 1 and 2 respectively as

shown in Figure 1 For self-excited vibration the forces of inertia due to masses 21 mand m are the onlyexternal source of excitation The generalized displacement equation of motion for masses 21 mand m at

individual nodal points is given by

sum=

minus=

2

1

)(i

j jij j xmt x ampampδ (1)

or




72

0)( 221211111 =++ xm xmt x ampampampamp δ δ (2)


Figure 1 A tall steel chimney modeled as a Two-Degree of Freedom system

where

ijδ = deflection coefficient at point j due to unit load at point i

j j xm ampamp = inertia force of the accelerating mass jm

=)(t x jampamp

time-dependent displacement of the mass jm from their respective equilibrium positions

j xampamp = Acceleration of the masses jm

For self-excited vibration the solution of displacement equation is given by

t At x j j ω sin)( = (5)

Equation (5) shows that the inertia force due to accelerating masses 21 mand m has a frequencyω

Differentiating equation (5) wrt ω gives

t At x i j ω ω cos)(= (6)

t At x j j ω ω sin)( 2= (7)

Substitution of equations (6) and (7) into equations (1) gives the generalized displacement equation for

masses 21 mand m as

0sinsin2

2

1

=minussum=

t Amt A j

i

jij j ω ω δ ω (8)

The force of inertia generated by mass jm is given by

t Am xm j j j j ω ω sin 2=minus ampamp (9)

Let the amplitude of force of inertia due to jm be jY

Therefore2ω j j j AmY = (10)

2ω j

i

j m

Y

A =

rArr (11)

Equation (11) represents unknown amplitude of displacement at i th nodal point expressed in terms of

amplitude of force of inertia

From equation (8)

0sin2

12

=

minussum

=

t Y m

Y

i

jij

j

jω δ

ω (12)

1 x 2 x

1m 2m

11 xm ampamp 22 xm ampamp




73

0sin =t (13)

or

02

12

=minussum=i

jij

j

jY

m

Y δ

ω (14)

Multiplying equation (14) by -1 gives

02

12

=+minus

sum=i

jij

j

jY

m

Y δ

ω (15)

At node 1 1= j

02121112

1

1=++

minusrArr Y Y

m

Y δ δ

ω (16)

At node 2 2= j

02221212

2

2=++

minusrArr Y Y

m

Y δ δ

ω (17)

Factorizing equations (16) through (17) gives

01

21212

1

11 =+

minus Y Y

mδ

ω δ (18)

01

22

2

22121 =

minus+ Y

mY

ω δ δ (19)

Equations (18) and (19) give the frequency equations for a tall steel chimney modeled as a two degree of

freedom system

Putting equations (18) and (19) in matrix form we have

01

1

2

1

2

2

2221

222

1

11

=

minus

minus

Y

Y

m

m

ω δ δ

δ

ω

δ

(20)

For non-zero solution the determinant of frequency equation must be zero

Therefore

01

1

2

2

2221

122

1

11

=

minus

minus

ω δ δ

δ ω

δ

m

m (21)

Let

211

2 == j

m jω β (22)

Equation (21) now transforms to

02221

1211=

minus

minus

β δ δ

δ β δ (23)

Wind Induced Vibration




74

Wind induced vibration is due to vortex shedding frequency exceeding the fundamental frequency ofvibration The vortex shedding causes harmonically varying lift forces on the chimney at right angle to

the velocity of wind The frequency of vortex shedding expressed as strouhal number is approximately

equal to 021 [12] for a circular cylinder

210=Ω

=

V

DS (24)

V

D210=Ω (25)

where

S = strouhal number

V = wind velocity

D = cross- sectional dimension at right angle to wind excitation

Ω = frequency of vortex shedding

Figure 2 Wind-induced resonant vibration past a steel chimney

Wind Pressure on Steel ChimneyThe intensity of wind load on the steel chimney is a function of height

The design wind speed at ith height is given by

321 S S S V V hi = (26)

where

=hiV wind velocity at ith height chimney height

=V wind velocity at chimney location

=1S risk coefficient = 10

=2S height terrain and structure size factor

=3S topography factor for flat terrain

260 hihi V =σ (27)

Using a shape factor of 07 the wind load at ith height is given by

( ) 70 ihihi h DP ∆= σ (28)

where=hiP wind load at i

th chimney height

=hiσ wind pressure at ith chimney height

=∆ ih ith chimney height segment

983140bull




75

30 An example for numerical studyA steel chimney has a height of 50m an inner diameter 075m and an outer diameter 080m Investigate

the possibility of the chimney going in to resonance when the velocity of airflow is in the chimney

location is 30ms

Figure 3 (a) A 50m steel chimney for numerical study Figure 4 Division of steel chimney

(b) Dynamic model into segments for wind

pressure estimation

Material and Geometrical properties

Unit weight =33 10576 m N x

Modulus of elasticity ( ) 29 10207 m N x E =

External diameter m800= Internal diameter m750=

Wind speed at chimney location sm 30=

Chimney area ( ) ( )2222 75080044

minus=minus== π π

d D A

206090 m=

Chimney volume = chimney area x length304535006090 m x ==

Moment of inertia of chimney ( ) ( )4444 7508006464

minus=minus== π π

d D A

40045750 m=

Flexural rigidity289 10479004575010207 m N x x x EI ===

Mass of chimney =819

volume xWeight Unit

Kg x x

4123745819

045310576 3

==

983093 983088 983149

2m

1m

X X

m750

m800

983123983141983139983156983145983151983150 X X minus

983088983086983096983088983149bull

983089

983090

983091

983092

983093

1 X

2 X

3 X

4 X

5 X

1 X

2 X

3 X

4 X

5 X

983089 983088 983149

bull

bull

bull

bull

bull

bull bull

bull

983093 983088 983149

bull bull 983088983086983096983088983149

983089 983088 983149

983089 983088 983149

983089 983088 983149

983089 983088 983149

bull




76

Mass intensity of chimney mKg 47550

4123745=== ρ

From equations (27) and (28) wind pressure and wind loading are determined at various height segments

Section 1 9802 =S m H m Dmh 5080010 1 ===∆

sm x x xV h 42998011301 ==

232

1 519010)429(60 mKN xh == minusσ

KN x x xPh 9127010080051901 ==

Section 2

m D

m H

800

40

2

2

=

=

9302 =S

sm x x xV h 92793011302 ==

232

2 467010)927(60 mKN xh == minusσ

KN x x xPh 622701080046702 ==

Section 3

m D

m H

800

30

2

3

=

=

8502 =S

sm x x xV h 52585011303 ==

232

3 390010)525(60 mKN xh == minusσ

KN x x xPh 182701080039003 ==

Section4

m D

m H

800

20

2

4

=

=

7502 =S

sm x x xV h 52275011304 ==

232

4 304010)522(60 mKN xh == minusσ

KN x x xPh 701701080030404 ==

Section5

m D

m H

800

10

2

5

=

=

6202 =S




77

sm x x xV h 61862011305 ==

232

5 208010)618(60 mKN xh == minusσ

KN x x xPh 171701080020805 ==

Figure 5 Derivation of deflection coefficients

From moment diagram multiplication table

EI

ll xl xl x

EI lac

3

8222

3

1

3

13

11 ===δ

EI

l

EI

l

xll xl

EI

lac

3

3

83

3

322 ===δ

( ) ( ) EI

lll xcbal

6

522

6

12

6

1 3

2112 =+=+==δ δ

The flexibility matrix of the above deflection coefficient is

152

528

3

3

EI

l (29)

Substituting for l and multiplying through by EI transforms equation (24) to

03352088313020

83130206741666=

minus

minus

β

β

EI

EI (30)

Let

β α EI = (31)

03352088313020

83130206741666=

minus

minus

α

α (32)

Evaluation of the above determinant gives

6912361 =α

96404292 =α

From statical consideration

1=P

bull bull 11

1m l 2m l

l2

12δ

bull bull 1=P

l 2m

l

1m 22δ l

21δ




78

ρ ρ ρ

252

25252 =

+=m and

ρ ρ

5122

251 ==m

For 96404291 =α and substituting for ρ and EI

we have

srad x x

xw 9861

4755129640429

10479 21

8

1 =

=

For 6912361 =α

srad x x

xw 0308

47525961236

10479 21

8

2 =

=

From equation (26) the frequency of the vortex shedding is

srad x

8757

800

30210==Ω

40 Discussion of Results

Table 1 Wind pressure at different height of chimney (V=30ms)

Chimney height (m) 10 20 30 40 50

Wind pressure (KNm2) 0208 0304 0390 0467 0519

Wind load (KN) 117 170 218 262 291

Table 2 Comparison of frequency of vortex shedding and natural vibration frequencies

Wind velocity at

chimney location

30ms

Vortex shedding

frequency (rads)

Natural frequencies (rads)

1w 2w

7875 9861 8030

The wind induced vibration of a 50m steel chimney subjected to two-degree of freedom using lumped

mass approach has been presented From Table 1 it can be seen that the intensity of wind loading on the

chimney varies with height with maximum value at the top and minimum value at the chimney baseFrom Table 2 it can be seen that the fundamental frequency (1986rads) is much lower in value than the

frequency of the vortex shedding due to wind excitation showing that resonance may occur leading to a

very large and severe deflection and damage to the steel chimney

50 Conclusion

The dynamic analysis of a 50m steel chimney modeled as a two-degree of freedom structural systemunder wind excitation was carried out using a lumped mass approach The results of the analysis showed

that the fundamental frequency of vibration of the chimney was much lower in value than the frequency

of the vortex shedding showing the possibility of the chimney going into resonance resulting in large

displacement and stresses which may cause fatique failure during the excepted lifetime of the chimney

Guyed cables should be provided between the top of the chimney and the ground to minimize the effect of

flow induced vibration The intensity of wind loading is a function of the height with maximum value at

chimney top and minimum value at chimney base




79

References1 Kolousek B Wind Effects on Civil Engineering Structure Elsevier Amsterdam 1984

2 Kappos AJ Dynamic Loading and Design of Structures E amp FN Spon Press London 2002

3 Sockel H Wind-excited vibrations of structures Springer-Verlag New York 1994

4 Sachs P Wind Forces in Engineering 2nd edn Pergaman Press Oxford 1978

5 Rao SS Mechanical Vibrations 4th Prentice Hall Inc 2003

6 Balendra T Vibration of Buildings to Wind and Earthquake Loads Springer- Verlag London

1993

7 De Silva CW Vibration Fundamentals and Practice CRC Press 1999

8 Clough RW and Penzien J Dynamics of Structures 2nd ed McGraw-Hill New York 1993

9 Humar JL Dynamics of Structures Prentice Hall Inc 1990

10 Thomson WT Theory of Vibration with Applications 3rd

ed CBS Publishers New Delhi

1988

11 RD Blevins Flow-Induced Vibration (2nd ed) Van Nostrand Reinhold New York 1990

12 RW Fox and AT McDonald Introduction to Fluid Mechanics (4th ed) Wiley New York

1992

13 Osadebe NN An improved MDOF model simulating some system with distributed mass

Journal of the University of Science and Technology Kumasi volume 19 Nos 1 2 amp 3 1999

14 CE Reynolds and JC Steedman Reinforced Concrete Designerrsquos Handbook 10th ed Spon

Press Taylor and Francis Publication London 1988




72



Figure 1 A tall steel chimney modeled as a Two-Degree of Freedom system

where

ijδ = deflection coefficient at point j due to unit load at point i

j j xm ampamp = inertia force of the accelerating mass jm

=)(t x jampamp

time-dependent displacement of the mass jm from their respective equilibrium positions

j xampamp = Acceleration of the masses jm

For self-excited vibration the solution of displacement equation is given by

t At x j j ω sin)( = (5)

Equation (5) shows that the inertia force due to accelerating masses 21 mand m has a frequencyω

Differentiating equation (5) wrt ω gives

t At x i j ω ω cos)(= (6)

t At x j j ω ω sin)( 2= (7)

Substitution of equations (6) and (7) into equations (1) gives the generalized displacement equation for

masses 21 mand m as

0sinsin2

2

1

=minussum=

t Amt A j

i

jij j ω ω δ ω (8)

The force of inertia generated by mass jm is given by

t Am xm j j j j ω ω sin 2=minus ampamp (9)

Let the amplitude of force of inertia due to jm be jY

Therefore2ω j j j AmY = (10)

2ω j

i

j m

Y

A =

rArr (11)

Equation (11) represents unknown amplitude of displacement at i th nodal point expressed in terms of

amplitude of force of inertia

From equation (8)

0sin2

12

=

minussum

=

t Y m

Y

i

jij

j

jω δ

ω (12)

1 x 2 x

1m 2m

11 xm ampamp 22 xm ampamp




73

0sin =t (13)

or

02

12

=minussum=i

jij

j

jY

m

Y δ

ω (14)


02

12

=+minus

sum=i

jij

j

jY

m

Y δ

ω (15)

At node 1 1= j

02121112

1

1=++

minusrArr Y Y

m

Y δ δ

ω (16)

At node 2 2= j

02221212

2

2=++

minusrArr Y Y

m

Y δ δ

ω (17)


01

21212

1

11 =+

minus Y Y

mδ

ω δ (18)

01

22

2

22121 =

minus+ Y

mY

ω δ δ (19)


freedom system


01

1

2

1

2

2

2221

222

1

11

=

minus

minus

Y

Y

m

m

ω δ δ

δ

ω

δ

(20)


Therefore

01

1

2

2

2221

122

1

11

=

minus

minus

ω δ δ

δ ω

δ

m

m (21)

Let

211

2 == j

m jω β (22)


02221

1211=

minus

minus

β δ δ

δ β δ (23)





74




210=Ω

=

V

DS (24)

V

D210=Ω (25)

where

S = strouhal number

V = wind velocity






321 S S S V V hi = (26)

where






260 hihi V =σ (27)


( ) 70 ihihi h DP ∆= σ (28)


th chimney height



983140bull




75



location is 30ms



pressure estimation






Chimney area ( ) ( )2222 75080044

minus=minus== π π

d D A

206090 m=



minus=minus== π π

d D A

40045750 m=



volume xWeight Unit

Kg x x

4123745819

045310576 3

==

983093 983088 983149

2m

1m

X X

m750

m800

983123983141983139983156983145983151983150 X X minus

983088983086983096983088983149bull

983089

983090

983091

983092

983093

1 X

2 X

3 X

4 X

5 X

1 X

2 X

3 X

4 X

5 X

983089 983088 983149

bull

bull

bull

bull

bull

bull bull

bull

983093 983088 983149

bull bull 983088983086983096983088983149

983089 983088 983149

983089 983088 983149

983089 983088 983149

983089 983088 983149

bull




76


4123745=== ρ


Section 1 9802 =S m H m Dmh 5080010 1 ===∆

sm x x xV h 42998011301 ==

232

1 519010)429(60 mKN xh == minusσ

KN x x xPh 9127010080051901 ==

Section 2

m D

m H

800

40

2

2

=

=

9302 =S

sm x x xV h 92793011302 ==

232

2 467010)927(60 mKN xh == minusσ

KN x x xPh 622701080046702 ==

Section 3

m D

m H

800

30

2

3

=

=

8502 =S

sm x x xV h 52585011303 ==

232

3 390010)525(60 mKN xh == minusσ

KN x x xPh 182701080039003 ==

Section4

m D

m H

800

20

2

4

=

=

7502 =S

sm x x xV h 52275011304 ==

232

4 304010)522(60 mKN xh == minusσ

KN x x xPh 701701080030404 ==

Section5

m D

m H

800

10

2

5

=

=

6202 =S




77

sm x x xV h 61862011305 ==

232

5 208010)618(60 mKN xh == minusσ

KN x x xPh 171701080020805 ==



EI

ll xl xl x

EI lac

3

8222

3

1

3

13

11 ===δ

EI

l

EI

l

xll xl

EI

lac

3

3

83

3

322 ===δ

( ) ( ) EI

lll xcbal

6

522

6

12

6

1 3

2112 =+=+==δ δ


152

528

3

3

EI

l (29)


03352088313020

83130206741666=

minus

minus

β

β

EI

EI (30)

Let

β α EI = (31)

03352088313020

83130206741666=

minus

minus

α

α (32)


6912361 =α

96404292 =α


1=P

bull bull 11

1m l 2m l

l2

12δ

bull bull 1=P

l 2m

l

1m 22δ l

21δ




78

ρ ρ ρ

252

25252 =

+=m and

ρ ρ

5122

251 ==m


we have

srad x x

xw 9861

4755129640429

10479 21

8

1 =

=

For 6912361 =α

srad x x

xw 0308

47525961236

10479 21

8

2 =

=


srad x

8757

800

30210==Ω





Wind load (KN) 117 170 218 262 291


Wind velocity at

chimney location

30ms

Vortex shedding

frequency (rads)


1w 2w

7875 9861 8030






50 Conclusion











79







1993






1988



1992








73

0sin =t (13)

or

02

12

=minussum=i

jij

j

jY

m

Y δ

ω (14)


02

12

=+minus

sum=i

jij

j

jY

m

Y δ

ω (15)

At node 1 1= j

02121112

1

1=++

minusrArr Y Y

m

Y δ δ

ω (16)

At node 2 2= j

02221212

2

2=++

minusrArr Y Y

m

Y δ δ

ω (17)


01

21212

1

11 =+

minus Y Y

mδ

ω δ (18)

01

22

2

22121 =

minus+ Y

mY

ω δ δ (19)


freedom system


01

1

2

1

2

2

2221

222

1

11

=

minus

minus

Y

Y

m

m

ω δ δ

δ

ω

δ

(20)


Therefore

01

1

2

2

2221

122

1

11

=

minus

minus

ω δ δ

δ ω

δ

m

m (21)

Let

211

2 == j

m jω β (22)


02221

1211=

minus

minus

β δ δ

δ β δ (23)





74




210=Ω

=

V

DS (24)

V

D210=Ω (25)

where

S = strouhal number

V = wind velocity






321 S S S V V hi = (26)

where






260 hihi V =σ (27)


( ) 70 ihihi h DP ∆= σ (28)


th chimney height



983140bull




75



location is 30ms



pressure estimation






Chimney area ( ) ( )2222 75080044

minus=minus== π π

d D A

206090 m=



minus=minus== π π

d D A

40045750 m=



volume xWeight Unit

Kg x x

4123745819

045310576 3

==

983093 983088 983149

2m

1m

X X

m750

m800

983123983141983139983156983145983151983150 X X minus

983088983086983096983088983149bull

983089

983090

983091

983092

983093

1 X

2 X

3 X

4 X

5 X

1 X

2 X

3 X

4 X

5 X

983089 983088 983149

bull

bull

bull

bull

bull

bull bull

bull

983093 983088 983149

bull bull 983088983086983096983088983149

983089 983088 983149

983089 983088 983149

983089 983088 983149

983089 983088 983149

bull




76


4123745=== ρ


Section 1 9802 =S m H m Dmh 5080010 1 ===∆

sm x x xV h 42998011301 ==

232

1 519010)429(60 mKN xh == minusσ

KN x x xPh 9127010080051901 ==

Section 2

m D

m H

800

40

2

2

=

=

9302 =S

sm x x xV h 92793011302 ==

232

2 467010)927(60 mKN xh == minusσ

KN x x xPh 622701080046702 ==

Section 3

m D

m H

800

30

2

3

=

=

8502 =S

sm x x xV h 52585011303 ==

232

3 390010)525(60 mKN xh == minusσ

KN x x xPh 182701080039003 ==

Section4

m D

m H

800

20

2

4

=

=

7502 =S

sm x x xV h 52275011304 ==

232

4 304010)522(60 mKN xh == minusσ

KN x x xPh 701701080030404 ==

Section5

m D

m H

800

10

2

5

=

=

6202 =S




77

sm x x xV h 61862011305 ==

232

5 208010)618(60 mKN xh == minusσ

KN x x xPh 171701080020805 ==



EI

ll xl xl x

EI lac

3

8222

3

1

3

13

11 ===δ

EI

l

EI

l

xll xl

EI

lac

3

3

83

3

322 ===δ

( ) ( ) EI

lll xcbal

6

522

6

12

6

1 3

2112 =+=+==δ δ


152

528

3

3

EI

l (29)


03352088313020

83130206741666=

minus

minus

β

β

EI

EI (30)

Let

β α EI = (31)

03352088313020

83130206741666=

minus

minus

α

α (32)


6912361 =α

96404292 =α


1=P

bull bull 11

1m l 2m l

l2

12δ

bull bull 1=P

l 2m

l

1m 22δ l

21δ




78

ρ ρ ρ

252

25252 =

+=m and

ρ ρ

5122

251 ==m


we have

srad x x

xw 9861

4755129640429

10479 21

8

1 =

=

For 6912361 =α

srad x x

xw 0308

47525961236

10479 21

8

2 =

=


srad x

8757

800

30210==Ω





Wind load (KN) 117 170 218 262 291


Wind velocity at

chimney location

30ms

Vortex shedding

frequency (rads)


1w 2w

7875 9861 8030






50 Conclusion











79







1993






1988



1992








74




210=Ω

=

V

DS (24)

V

D210=Ω (25)

where

S = strouhal number

V = wind velocity






321 S S S V V hi = (26)

where






260 hihi V =σ (27)


( ) 70 ihihi h DP ∆= σ (28)


th chimney height



983140bull




75



location is 30ms



pressure estimation






Chimney area ( ) ( )2222 75080044

minus=minus== π π

d D A

206090 m=



minus=minus== π π

d D A

40045750 m=



volume xWeight Unit

Kg x x

4123745819

045310576 3

==

983093 983088 983149

2m

1m

X X

m750

m800

983123983141983139983156983145983151983150 X X minus

983088983086983096983088983149bull

983089

983090

983091

983092

983093

1 X

2 X

3 X

4 X

5 X

1 X

2 X

3 X

4 X

5 X

983089 983088 983149

bull

bull

bull

bull

bull

bull bull

bull

983093 983088 983149

bull bull 983088983086983096983088983149

983089 983088 983149

983089 983088 983149

983089 983088 983149

983089 983088 983149

bull




76


4123745=== ρ


Section 1 9802 =S m H m Dmh 5080010 1 ===∆

sm x x xV h 42998011301 ==

232

1 519010)429(60 mKN xh == minusσ

KN x x xPh 9127010080051901 ==

Section 2

m D

m H

800

40

2

2

=

=

9302 =S

sm x x xV h 92793011302 ==

232

2 467010)927(60 mKN xh == minusσ

KN x x xPh 622701080046702 ==

Section 3

m D

m H

800

30

2

3

=

=

8502 =S

sm x x xV h 52585011303 ==

232

3 390010)525(60 mKN xh == minusσ

KN x x xPh 182701080039003 ==

Section4

m D

m H

800

20

2

4

=

=

7502 =S

sm x x xV h 52275011304 ==

232

4 304010)522(60 mKN xh == minusσ

KN x x xPh 701701080030404 ==

Section5

m D

m H

800

10

2

5

=

=

6202 =S




77

sm x x xV h 61862011305 ==

232

5 208010)618(60 mKN xh == minusσ

KN x x xPh 171701080020805 ==



EI

ll xl xl x

EI lac

3

8222

3

1

3

13

11 ===δ

EI

l

EI

l

xll xl

EI

lac

3

3

83

3

322 ===δ

( ) ( ) EI

lll xcbal

6

522

6

12

6

1 3

2112 =+=+==δ δ


152

528

3

3

EI

l (29)


03352088313020

83130206741666=

minus

minus

β

β

EI

EI (30)

Let

β α EI = (31)

03352088313020

83130206741666=

minus

minus

α

α (32)


6912361 =α

96404292 =α


1=P

bull bull 11

1m l 2m l

l2

12δ

bull bull 1=P

l 2m

l

1m 22δ l

21δ




78

ρ ρ ρ

252

25252 =

+=m and

ρ ρ

5122

251 ==m


we have

srad x x

xw 9861

4755129640429

10479 21

8

1 =

=

For 6912361 =α

srad x x

xw 0308

47525961236

10479 21

8

2 =

=


srad x

8757

800

30210==Ω





Wind load (KN) 117 170 218 262 291


Wind velocity at

chimney location

30ms

Vortex shedding

frequency (rads)


1w 2w

7875 9861 8030






50 Conclusion











79







1993






1988



1992








75



location is 30ms



pressure estimation






Chimney area ( ) ( )2222 75080044

minus=minus== π π

d D A

206090 m=



minus=minus== π π

d D A

40045750 m=



volume xWeight Unit

Kg x x

4123745819

045310576 3

==

983093 983088 983149

2m

1m

X X

m750

m800

983123983141983139983156983145983151983150 X X minus

983088983086983096983088983149bull

983089

983090

983091

983092

983093

1 X

2 X

3 X

4 X

5 X

1 X

2 X

3 X

4 X

5 X

983089 983088 983149

bull

bull

bull

bull

bull

bull bull

bull

983093 983088 983149

bull bull 983088983086983096983088983149

983089 983088 983149

983089 983088 983149

983089 983088 983149

983089 983088 983149

bull




76


4123745=== ρ


Section 1 9802 =S m H m Dmh 5080010 1 ===∆

sm x x xV h 42998011301 ==

232

1 519010)429(60 mKN xh == minusσ

KN x x xPh 9127010080051901 ==

Section 2

m D

m H

800

40

2

2

=

=

9302 =S

sm x x xV h 92793011302 ==

232

2 467010)927(60 mKN xh == minusσ

KN x x xPh 622701080046702 ==

Section 3

m D

m H

800

30

2

3

=

=

8502 =S

sm x x xV h 52585011303 ==

232

3 390010)525(60 mKN xh == minusσ

KN x x xPh 182701080039003 ==

Section4

m D

m H

800

20

2

4

=

=

7502 =S

sm x x xV h 52275011304 ==

232

4 304010)522(60 mKN xh == minusσ

KN x x xPh 701701080030404 ==

Section5

m D

m H

800

10

2

5

=

=

6202 =S




77

sm x x xV h 61862011305 ==

232

5 208010)618(60 mKN xh == minusσ

KN x x xPh 171701080020805 ==



EI

ll xl xl x

EI lac

3

8222

3

1

3

13

11 ===δ

EI

l

EI

l

xll xl

EI

lac

3

3

83

3

322 ===δ

( ) ( ) EI

lll xcbal

6

522

6

12

6

1 3

2112 =+=+==δ δ


152

528

3

3

EI

l (29)


03352088313020

83130206741666=

minus

minus

β

β

EI

EI (30)

Let

β α EI = (31)

03352088313020

83130206741666=

minus

minus

α

α (32)


6912361 =α

96404292 =α


1=P

bull bull 11

1m l 2m l

l2

12δ

bull bull 1=P

l 2m

l

1m 22δ l

21δ




78

ρ ρ ρ

252

25252 =

+=m and

ρ ρ

5122

251 ==m


we have

srad x x

xw 9861

4755129640429

10479 21

8

1 =

=

For 6912361 =α

srad x x

xw 0308

47525961236

10479 21

8

2 =

=


srad x

8757

800

30210==Ω





Wind load (KN) 117 170 218 262 291


Wind velocity at

chimney location

30ms

Vortex shedding

frequency (rads)


1w 2w

7875 9861 8030






50 Conclusion











79







1993






1988



1992








76


4123745=== ρ


Section 1 9802 =S m H m Dmh 5080010 1 ===∆

sm x x xV h 42998011301 ==

232

1 519010)429(60 mKN xh == minusσ

KN x x xPh 9127010080051901 ==

Section 2

m D

m H

800

40

2

2

=

=

9302 =S

sm x x xV h 92793011302 ==

232

2 467010)927(60 mKN xh == minusσ

KN x x xPh 622701080046702 ==

Section 3

m D

m H

800

30

2

3

=

=

8502 =S

sm x x xV h 52585011303 ==

232

3 390010)525(60 mKN xh == minusσ

KN x x xPh 182701080039003 ==

Section4

m D

m H

800

20

2

4

=

=

7502 =S

sm x x xV h 52275011304 ==

232

4 304010)522(60 mKN xh == minusσ

KN x x xPh 701701080030404 ==

Section5

m D

m H

800

10

2

5

=

=

6202 =S




77

sm x x xV h 61862011305 ==

232

5 208010)618(60 mKN xh == minusσ

KN x x xPh 171701080020805 ==



EI

ll xl xl x

EI lac

3

8222

3

1

3

13

11 ===δ

EI

l

EI

l

xll xl

EI

lac

3

3

83

3

322 ===δ

( ) ( ) EI

lll xcbal

6

522

6

12

6

1 3

2112 =+=+==δ δ


152

528

3

3

EI

l (29)


03352088313020

83130206741666=

minus

minus

β

β

EI

EI (30)

Let

β α EI = (31)

03352088313020

83130206741666=

minus

minus

α

α (32)


6912361 =α

96404292 =α


1=P

bull bull 11

1m l 2m l

l2

12δ

bull bull 1=P

l 2m

l

1m 22δ l

21δ




78

ρ ρ ρ

252

25252 =

+=m and

ρ ρ

5122

251 ==m


we have

srad x x

xw 9861

4755129640429

10479 21

8

1 =

=

For 6912361 =α

srad x x

xw 0308

47525961236

10479 21

8

2 =

=


srad x

8757

800

30210==Ω





Wind load (KN) 117 170 218 262 291


Wind velocity at

chimney location

30ms

Vortex shedding

frequency (rads)


1w 2w

7875 9861 8030






50 Conclusion











79







1993






1988



1992








77

sm x x xV h 61862011305 ==

232

5 208010)618(60 mKN xh == minusσ

KN x x xPh 171701080020805 ==



EI

ll xl xl x

EI lac

3

8222

3

1

3

13

11 ===δ

EI

l

EI

l

xll xl

EI

lac

3

3

83

3

322 ===δ

( ) ( ) EI

lll xcbal

6

522

6

12

6

1 3

2112 =+=+==δ δ


152

528

3

3

EI

l (29)


03352088313020

83130206741666=

minus

minus

β

β

EI

EI (30)

Let

β α EI = (31)

03352088313020

83130206741666=

minus

minus

α

α (32)


6912361 =α

96404292 =α


1=P

bull bull 11

1m l 2m l

l2

12δ

bull bull 1=P

l 2m

l

1m 22δ l

21δ




78

ρ ρ ρ

252

25252 =

+=m and

ρ ρ

5122

251 ==m


we have

srad x x

xw 9861

4755129640429

10479 21

8

1 =

=

For 6912361 =α

srad x x

xw 0308

47525961236

10479 21

8

2 =

=


srad x

8757

800

30210==Ω





Wind load (KN) 117 170 218 262 291


Wind velocity at

chimney location

30ms

Vortex shedding

frequency (rads)


1w 2w

7875 9861 8030






50 Conclusion











79







1993






1988



1992








78

ρ ρ ρ

252

25252 =

+=m and

ρ ρ

5122

251 ==m


we have

srad x x

xw 9861

4755129640429

10479 21

8

1 =

=

For 6912361 =α

srad x x

xw 0308

47525961236

10479 21

8

2 =

=


srad x

8757

800

30210==Ω





Wind load (KN) 117 170 218 262 291


Wind velocity at

chimney location

30ms

Vortex shedding

frequency (rads)


1w 2w

7875 9861 8030






50 Conclusion











79







1993






1988



1992








79







1993






1988



1992





ecce 1202 013 wind induced vibration tall steel chimney

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