03 vibration of string

Dynamics of Continuous StructuresMaged Mostafa

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Vibration of Continuous Structures



Course Contents SDOF M-DOF• Cables/String• Bars• Shafts• Vibration Attenuation• Beams• FEM for Vibration• Plates• Aeroelasticity



Objectives

• Derive the equation of motion for Cable/ String

• Estimate the Natural Frequencies• Understand the concept of mode shapes• Apply BC’s and IC’s to obtain structure

response



String and Cables



Objectives

• Derive the equation of motion for Cable/String


response



Strings and Cables

• This type of structures does not bare any bending or compression loads

• It resists deformations only by inducing tension stress

• Examples are the strings of musical instruments, cables of bridges, and elevator suspension cables



The string/cable equation• Start by considering a

uniform string stretched between two fixed boundaries

• Assume constant, axial tension t in string

• Let a distributed force f(x,t) act along the string

f(x,t)

t

xy



Examine a small element of string

xtxfttxwxFy

),(sinsin

),(

2211

2

2

tt

• Where is the mass per unit length of the cable• Force balance on an infinitesimal element• Now linearize the sine with the small angle

approximate sin(x) = tan(x) = slope of the string

1

2

t2

t1x1 x2 = x1 +x

w(x,t)

f (x,t)



)(

:about / of seriesTaylor theRecall

2

1

112

xOxw

xx

xw

xw

xxw

xxx t

t

t

t

xttxwxtxf

xtxw

xtxw

xx

2

2 ),(),(),(),(

12

t

t

2

2 ),(),(),(ttxwtxf

xtxw

x

t

xttxwxtxfx

xtxw

x x

2

2 ),(),(),(

1

t



0 ,0),(),0(

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

,),(),(

00

2

2

22

2

ttwtw

txwxwxwxw

cxtxw

tctxw

t

t

Since t is constant, and for no external force the equation of motion becomes:

Second order in time and second order in space, therefore 4 constants of integration. Two from initial conditions:

And two from boundary conditions:

, wave speed



Physical quantities

• Deflection is w(x,t) in the y-direction• The slope of the string is wx(x,t)• The restoring force is twxx(x,t) • The velocity is wt(x,t)• The acceleration is wtt(x,t) at any point x

along the string at time tNote that the above applies to cables as well as stringsSubscript denotes differentiation w.r.t. to that parameter



Objectives

• Derive the equation of motion for Cable/String


response



Modes and Natural Frequencies

22

22

2

2

2

22

)()(

)()(0

)()(,

)()(

)()(

= and = where)()()()(

)()(),(

tTctT

xXxX

xXxX

dxd

tTctT

xXxX

dtd

dxdtTxXtTxXc

tTxXtxw

Solve by the method of separation of variables:

Substitute into the equation of motion to get:

Results in two second order equations coupled only by a constant:



Solving the spatial equation:

naXaX

XXtTXtTX

aaxaxaxX

xXxX

n

equation sticcharacteri

1

2

2121

2

0sin0sin)(

0)0(

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

nintegratio of constantsare and , cos sin)(

0)()(

Since T(t) is not zero

an infinite number of values of

A second order equation with solution of the form:

Next apply the boundary conditions:



The temporal solution

1

22

)sin()cos()sin()sin(),(

)sin()cos()sin()sin(

sincossinsin),(

)conditions initial from(get n integratio of constants are ,cossin)(

3,2,1 ,0)()(

nnn

nn

nnnnnnn

nn

nnnnn

nnn

xnctndxnctnctxw

xnctndxnctnc

xctdxctctxw

BActBctAtT

ntTctT

Again a second order ode with solution of the form:

Substitution back into the separated form X(x)T(t) yields:

The total solution becomes:



Using orthogonality to evaluate the remaining constants from the initial conditions

0100

10

2

0

)sin()sin()sin()(

)0cos()sin()()0,(

:conditions initial theFrom2 ,0

,)sin()sin(

dxxmxnddxxmxw

xndxwxw

mnmn

dxxmxn

nn

nn

nm



3,2,1 ,)sin()(2

)0cos()sin(c)(

3,2,1 ,)sin()(2

3,2,1 ,)sin()(2

00

10

00

00

ndxxnxwcn

c

xncxw

ndxxnxwd

nm

mdxxmxwd

n

nnn

n

m



Objectives

• Derive the equation of motion for Cables/Strings


response



A mode shape

tcxtxw

d

ndxxnxd

ncxw

nxxw

Assume

n

n

cos)sin(),(

1

3,2 ,0)sin()sin(2

,0,0)(

1)=(ion eigenfunctfirst theis which ,sin)(

1

0

0

0

Causes vibration in the first mode shape



Plots of mode shapes

0 0.5 1 1.5 2

1

0.5

0.5

1

X ,1 x

X ,2 x

X ,3 x

x

sinn2x

nodes



String mode shapes Video 1 String mode shapes Video 2



Example :Piano wire: L=1.4 m, t=11.1x104 N, m=110 g. Compute the first natural frequency.

110 g per 1.4 m = 0.0786 kg/m

1 cl

1.4

t

1.4

11.1104 N0.0786 kg/m

2666.69 rad/s or 424 Hz



Assignment

1. Solve the cable problem with one side fixed and the other supported by a flexible support with stiffness k N/m

2. Solve the cable problem for a cable that is hanging from one end and the tension is changing due to the weight N/m

03 vibration of string

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