vibration of cables

8/7/2019 vibration of cables

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1

Chapter 9 Mathematical Models of Continuous Systems

• Vibration of Strings(Cables)

Ref: “Anaytical Methods in Vibrations,” L Meirovitch 2ed, 1967,

§5-6( pp. 143-144)

Figure 5.2

2

2

2

2

),()(

),()(),(),(),()()(

t

t x ydx x

x t x yx T dx t x pdx

x t x y

x t x ydx

x x T x T

∂

∂=

∂

∂−+

∂

∂+∂

∂ ∂

∂+

ρ

2

2),(

)(),(),(

)(t

t x yx t x p

x

t x yx T

x ∂

∂=+

∂

∂

∂

∂ρ (5.59)

Partial differential equation

Two boundary conditions and two initial conditions are required

for solution

•Boundary Conditions

end freeforced t x pe

−= 0),(

end fixed t x y e 0),( =



2

end freex

yx T

ex 0)( =

∂

∂

•Initial Conditions

)0,(),0,( x yx y &

Change Notation

),(),( t x vt x y →

• Vibration of Taut Cable with Damper

Non-dimensionalized Equation of Motion

The motion of the taut cable in the linear range is described by the

following partial differential equation

)x x ( t

vc

t

vm

x

v)x ( T c−

∂

∂+

∂

∂=

∂

∂δ

2

2

2

v=lateral displacement from static position.

x=coordinate along the cable chord line,xc = location of the damper , t=time, and

T=initial static tension force in the cable,

m=mass per unit length of cable,

c=damper coefficient,

δ = Dirac’s delta function,

The inherent damping of the cable is assumed to be negligible.

§9.1 Longitudinal Vibration of Bars (Rods, Truss Members)



3

),( t x u

),( t x p

y

x t x p ∆),(

),( t x p

),( t x x p ∆+

)( x x A ∆+)(x A

Figure 9.1. Member undergoing axial deformation. (a) portion of a

member undergoing axial deformation. (b) Freebody diagram

x

u

∂

∂=ε (9.1)

εσ E = (9.2)

σAP = (9.3)

∑ ∆=+x x

amF )( (9.4)

2

2

),(),(t

ux At x P t x x P x p∂

∂∆=−∆++∆ ρ (9.5)

2

2

0

),(),(lim

t

uAp

x

t x P t x x P

x ∂

∂=+

∆

−∆+

→∆

ρ

2

2

t

uAp

x

P

∂

∂=+

∂

∂ρ (9.6)

2

2

t

uAp

x

uAE

x ∂

∂=+

∂

∂

∂

∂ρ (9.7)

x x ∆

x



4


end freeforced t x P e

−= 0),( (9.8a)

end fixed t x ue

0),( = (9.8b)

end freex

uex

0=∂

∂(9.8c)

Example 9.1 Boundary Conditions

Determine the appropriate axial deformation boundary conditions at

x=0 for the two members shown below.

Solution

02

2

),0( =∂

∂=

x t

umt P (1)

0),0( =∂

∂=

x x

uAE t P (2)

02

2

0 == ∂

∂=

∂

∂x x

t

um

x

uAE (3)

k

x

m

m ),0( t P



6

§9.2 Transverse Vibration of Beams (Bernoulli-Euler Theory)

Figure 9.2. Member undergoing transverse vibration

R

y−

=ε (9.9)

R

EI t x M =),( (9.10)

+↑∑ ∆=yy

amF )( (9.11)

∑ ∆=+ α)(G G

I M (9.12)

∑ = 0G

M (9.13)

2

2

),(),(),(t

vx Ax t x pt x x S t x S ∂

∂∆=∆+∆+− ρ (9.14)

2

2

),(t

vAt x p

x

S

∂

∂=+

∂

∂− ρ (9.15)

x

M S

∂

∂= (9.16)

2

2

),(

x

vEI t x M

∂

∂= (9.17)



7

),(2

2

2

2

2

2

t x pt

v

Ax

v

EI x =∂

∂

+

∂

∂

∂

∂

ρ (9.18)


a. Fixed end:

0),( =t x ve

(9.19a)

0=∂

∂= ex x

x

v(9.19b)

b. Simply supported*

end:

0),( =t x ve

(9.20a)

0),( =t x M e

(9.20b)

02

2

=

∂

∂= ex x

x

v(9.20c)

c. Force-free end:

0),( =t x S e

(9.21a)

0),( =t x M e

(9.21b)

02

2

=

∂

∂

∂

∂= ex x

x

vEI x (9.22a)

02

2

=∂

∂= ex x

x

v(9.22b)



8

Example 9.2 Boundary Conditions

Determine the appropriate boundary conditions if a point mass m is

attached at the end of the beam at x =L

Solution

+↑

∑ =yy maF (1)

Lx t

vmt LS =∂

∂=

2

2

),( (2)

Lx Lx t

vm

x

vEI

x == ∂

∂=

∂

∂

∂

∂2

2

2

2

(3)

∑ = 0G

M (4)

0),( =t LM (5)

02

2

=∂

∂= Lx

t

vm (6)

m

L

),( t LS

),( t LM m



9

Example 9.3 Compressive Load

Determine the equation of motion for a beam that is subjected to a

compressive load N , which remains parallel to the x-axis. Neglect axial

strain. (Note: Coupled axial-bending motion will be considered in

Problem 9.1.)

Solution

+↑∑ ∆= yymaF (1)

+∑ = 0G

M (2)

2

2

),(t

vAt x p

x

S

∂

∂=+

∂

∂− ρ (3)

0),(

)],(),([),(),(

=∆∆+−

−∆++−∆+

x t x x S

t x vt x x vN t x M t x x M (4)



10

S x

vN

x

M =

∂

∂+

∂

∂(5)

),(2

2

2

2

2

2

t x pt vA

x vN

x M =∂∂+∂∂+∂∂

ρ (6)

),(2

2

2

2

2

2

2

2

t x pt

vA

x

vN

x

vEI

x =

∂

∂+

∂

∂+

∂

∂

∂

∂ρ (7)



11

●Torsional Vibration of Bars: Newton’s Laws


§5-9( pp.156-157)

Figure 5.6

x

t x x GJ t x M

T

∂

∂=

),()(),(

θ(5.125)

)(x GJ : torsional stiffness.

2

2),(

)(),(),(),(

),(x

t x dx x I t x M dx t x mdx

x

t x M t x M

T T

T

T ∂

∂=−+

∂

∂+

θρ

(5.126)

2

2),(

)(),(

),(

t

t x

x I t x mx

t x M T

T

∂

∂

=+∂

∂ θ

ρ (5.127)



12

2

2),(

)(),(),(

)(t

t x x I t x m

x

t x x GJ

x T

∂

∂=+

∂

∂

∂

∂ θρ

θ(5.128)


0),0( =t θ Fixed end (5.130)

0),(

)( =∂

∂

=Lx x

t x x GJ

θFree end (5.131)

§9.3 Torsional Vibration of Bars: Hamilton’s Principle:

Figure 9.3. Rod undergoing torsional deformation

∫∫ =+−2

1

2

1

0)(t

t nc

t

t dt W dt V T δδ (9.23)

2

4

2 RdAr J I

p

π=== ∫∫ (9.25)

∫ ′=L

dx GJ V 0

2)(

2

1θ (9.26)

∫=L

pdx I T

0

2)(

2

1θρ & (9.27)



13

t

x

∂∂=

∂

∂=′

• )()(

)()(

),()(),(),(0

t Lt T dx t x t x t W L

L

ncδθδθδ += ∫ (9.28)

[ ] 0),()(),(),(

)(2

1)(

2

1

2

1

2

1

0

0

2

0

2

=++

′−

∫ ∫

∫ ∫∫

dt t Lt T dx t x t x t

dt dx GJ dx I t t

t L

L

t

t

LL

p

δθδθ

θθρδ &

(9.29)

∫ ′′=L

dx GJ V 0

θδθδ (9.30)

∫=L

pdx I T

0θδθρδ && (9.31)

[ ]dt dx GJ GJ

dxdt GJ Vdt

t

t

LL

t

t

Lt

t

∫ ∫

∫ ∫∫

′′−′=

′′=

2

1

2

1

2

1

00

0

)()( δθθδθθ

θδθδ

(9.32)

[ ]dx dt I I

dtdx I Tdt

L t

t p

t

t p

L t

t p

t

t

∫ ∫

∫ ∫∫

−=

=

0

0

2

1

2

1

2

1

2

1

)( δθθρδθθρ

θδθρδ

&&&

&&

(9.33)

Since ),(1

t x θ and ),(2

t x θ are known, ),(1

t x δθ = ),(2

t x δθ =0

[ ] [ ]{ }

[ ] 0)(

)()(

2

1

2

1

0

0

=+−′′+

−′−′

∫ ∫

∫ ==

dxdt t I GJ

dt T GJ GJ

t

t

L

p

t

t Lx Lx

δθθρθ

δθθδθθ

&& (9.34)

00==x

δθ and 0≠=Lx δθ

)()( t T GJ LLx

=′=θ natural boundary condition (9.35)

θρθ &&

pI t x t GJ =+′′ ),()( Euler equation (9.36)



15

∫∫ +=LL

dx I dx vAT 0

2

0

2)(

2

1)(

2

1αρρ && (9.42)

dx t x vt x pW Lnc ∫= 0

),(),( δδ (9.43)

0

])()([2

1

2

1

2

1

0

0

2222

=+

′−−′−+

∫ ∫

∫ ∫t

t

L

t

t

L

vdx p

dx vGAEI I vA

δ

ακααρρδ &&

(9.44)

0])([])([

)]()([

}])([{

00

0

0

2

1

2

1

2

1

2

1

=′−′−+

′−−′′+−+

+′′−−−

∫∫∫ ∫

∫ ∫

dt EI dt vvGA

dt dx vGAEI I

dt dx vpvGAvA

Lt

t

Lt

t

t

t

L

t

t

L

δααδακ

δαακααρ

δακρ

&&

&&

(9.45)

),(])([ t x pvAvGA =+′′− &&ρακ (9.46a)

0)()( =+′′−′− αραακ &&I EI vGA (9.46b)

0,0)( == x at vGA δβκ (9.47a)

Lx at vGA == ,0)( δβκ (9.47b)

0,0)( ==′ x at EI δαα (9.47c)

Lx at EI ==′ ,0)( δαα (9.47d)

)(1

vApGA

v &&ρκ

α −

+′′=′ (9.48)



16

02

2

2

2

2

2

2

2

22

4

2

2

4

4

=

∂

∂−

∂

∂−

∂

∂−

∂

∂+

∂∂

∂−

∂

∂−−

∂

∂

t

vAp

t GA

I

t

vAp

x GA

EI

t x

vI

t

vAp

x

vEI

ρκ

ρρ

κ

ρρ

(9.49)

●Vibration of Membranes

Ref: “Anaytical Methods in Vibrations,” L Meirovitch

●Vibration of Membranes


§5-11( pp.166-167)

2

2

2

t w pw T

∂∂=+∇ ρ (5.171)

2∇ : Laplace operator

w: displacement

T: uniform tension

p: external pressure

ρ : mass per unit area


( )11

,at0 baw = fixed points (5.173)

( )22

,at0 ban

w T =

∂

∂free (5.174)

Bernoulli-Euler theory Principle rotary inertia term

Principle shear

deformation term

Combined rotary inertia

and shear deformation



17

●Vibration of Plates


§5-12( pp.170-171)

( )2

3

2

2

4

112,

v

EhD

t

w pw D

E E −

=∂

∂=+∇− ρ (5.236)

224 ∇∇=∇ : biharmonic operator

E D : plate flexural rigidity.

h: plate thickness

ν : poisson’s ratio


n and s : the coordinates in the directions normal and tangential to

the boundary

0and0 =∂∂=

n

w w clamped edge (5.238)

which are both geometric boundary conditions.

0and0 ==n

M w simply supported edge (5.239)

nM : bending moment per unit length associated with the cross

section whose normal is n .

0and0 =∂

∂−==s

M QV M ns

nnn free edge (5.240)

vibration of cables

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