vibration of continuous structures

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Dr. Eng. Mohammad Tawfik

Vibration of Continuous

Structures

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Objectives

• Derive the equation of motion for simple

structures

• Understand the concept of mode shapes

• Apply BC’s and IC’s to obtain structure

response

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String and Cables

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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

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The string/cable equation• Start by considering a

uniform string

stretched between two

fixed boundaries

• Assume constant,

axial tension in string

• Let a distributed force

f(x,t) act along the

string

f(x,t)

x

y

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Examine a small element of

string

xtxf

t

txwxFy

),(sinsin

),(

2211

2

2

• Force balance on an infinitesimal element

• Now linearize the sine with the small angle

approximate sinx=tanx=slope of the string

1

2

2

1

x1 x2 = x1 +x

w(x,t)

f (x,t)

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)(

:about / of seriesTaylor theRecall

2

1

112

xOx

w

xx

x

w

x

w

xxw

xxx

xt

txwxtxf

x

txw

x

txw

xx

2

2 ),(),(

),(),(

12

2

2 ),(),(

),(

t

txwtxf

x

txw

x

xt

txwxtxfx

x

txw

xx

2

2 ),(),(

),(

1

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0 ,0),(),0(

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

,),(),(

00

2

2

22

2

ttwtw

txwxwxwxw

cx

txw

tc

txw

t

Since 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

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Physical quantities

• Deflection is w(x,t) in the y-direction

• The slope of the string is wx(x,t)

• The restoring force is wxx(x,t)

• The velocity is wt(x,t)

• The acceleration is wtt(x,t) at any point x

along the string at time t

Note that the above applies to cables as well as strings

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Modes and Natural Frequencies

2

2

2

2

2

2

2

22

)(

)(

)(

)(0

)(

)(,

)(

)(

)(

)(

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

)()(),(

tTc

tT

xX

xX

xX

xX

dx

d

tTc

tT

xX

xX

dt

d

dx

dtTxXtTxXc

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:

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Solving the spatial equation:

n

aX

aX

XX

tTXtTX

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:

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Also an eigenvalue problem

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

of valueindexed theof because results index theHere

sin)( ,1,2,3=For

2

2

nnnnnn

nn

XXxXXXx

n

xn

axXn

The spatial solution becomes:

The spatial problem also can be written as:

Which is also an eigenvalue, eigenfunction problem where

=2 is the eigenvalue and Xn is the eigenfunction.

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Analogy to Matrix Eigenvalue

Problem

happen also willexpansion modal

sfrequencie seigenvalue shapes, mode become rseigenvecto

role same theplays

gnormalizin ofcondition theand results alsoity orthoganal

reigenvecto and also )(

ioneigenfunctreigenvecto ),(

operator matrix ,conditionsboundary plus ,2

2

xX

xX

xA

n

ni

u

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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)()(

n

nn

nn

nnnnnnn

nn

nnnnn

nnn

xn

ctn

dxn

ctn

ctxw

xn

ctn

dxn

ctn

c

xctdxctctxw

BA

ctBctAtT

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:

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Using orthogonality to evaluate the remaining

constants from the initial conditions

010

0

1

0

2

0

)sin()sin()sin()(

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

:conditions initial theFrom

2 ,0

,)sin()sin(

dxxm

xn

ddxxm

xw

xn

dxwxw

mn

mndxx

mx

n

n

n

n

n

nm

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

)0cos()sin(c)(

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

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

0

0

1

0

0

0

0

0

ndxxn

xwcn

c

xn

cxw

ndxxn

xwd

nm

mdxxm

xwd

n

n

nn

n

m

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A mode shape

tc

xtxw

d

ndxxn

xd

ncxw

nxxw

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

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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

sinn

2x

nodes

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Homework #3

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

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Bar Vibration

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Bar Vibration

• The bar is a structural element that bears

compression and tension loads

• It deflects in the axial direction only

• Examples of bars may be the columns of

buildings, car shock absorbers, legs of

chairs and tables, and human legs!

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Vibration of Rods and Bars

• Consider a small

element of the bar

• Deflection is now along

x (called longitudinal

vibration)

• F= ma on small element

yields the following:

x x +dx

w(x,t)

x

dx

F+dF F

Equilibrium

position

Infinitesimal

element

0 l

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0

),( :end free At the

,0),0( :end clamped At the

),(

),( constant)(

),( )(

),()(

),()(

),()(

),( )(

2

2

2

2

2

2

2

2

xx

txwEA

tw

t

txw

x

txwExA

t

txwxA

x

txwxEA

x

dxx

txwxEA

xdF

x

txwxEAF

t

txwdxxAFdFF

Force balance:

Constitutive relation:

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Note

• The equation of motion of the bar is similar

to that of the cable/string the response

should have similar form

• The bar may have different boundary

conditions

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Homework #4

• Solve the equation of motion of a bar

with constant cross-section properties

with

1. Fixed-Fixed boundary conditions

2. Free-Free boundary conditions

• Compare the natural frequencies for all

three cases

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Beam Vibration

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Beam Vibration

• The beam element is the most famous

structural element as it presents a lot of

realistic structural elements

• It bears loads normal to its longitudinal

axis

• It resists deformations by inducing bending

stresses

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Bending vibrations of a beam

2

2 ),()(),(

about inertia

ofmoment area sect.-cross)(

modulus Youngs

)( stiffness bending

x

txwxEItxM

z

xI

E

xEI

Next sum forces in the y - direction (up, down)

Sum moments about the point Q

Use the moment given from

stenght of materials

Assume sides do not bend

(no shear deformation)

f (x,t)

w (x,t)

x

dx A(x)= h1h2

h1

h2

M(x,t)+Mx(x,t)dx

M(x,t)

V(x,t)

V(x,t)+Vx(x,t)dx

f(x,t)

w(x,t)

x x +dx

·Q

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Summing forces and moments

0)(2

),(),(),(

),(

02

),(

),(),(),(

),(),(

),()(),(),(

),(),(

2

2

2

dxtxf

x

txVdxtxVdx

x

txM

dxdxtxf

dxdxx

txVtxVtxMdx

x

txMtxM

t

txwdxxAdxtxftxVdx

x

txVtxV

0

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A

EIc

x

txwc

t

txw

txfx

txwxEI

xt

txwxA

t

txwdxxAdxtxfdx

x

txM

x

txMtxV

,0),(),(

),(),(

)(),(

)(

),()(),(

),(

),(),(

4

42

2

2

2

2

2

2

2

2

2

2

2

2

Substitute into force balance equation yields:

Dividing by dx and substituting for M yields

Assume constant stiffness to get:

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Boundary conditions (4)

0 forceshear

0moment bending

end Free

2

2

2

2

x

wEI

x

x

wEI

0slope

0deflection

end fixed)(or Clamped

x

w

w

0moment bending

0deflection

end supported)simply (or Pinned

2

2

x

wEI

w

0forceshear

0slope

end Sliding

2

2

x

wEI

x

x

w

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Solution of the time equation:

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

:conditions initial Two

cossin)(

0)()(

)(

)(

)(

)(

00

2

22

xwxwxwxw

tBtAtT

tTtT

tT

tT

xX

xXc

t

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Spatial equation (BVP)

xaxaxaxaxX

AexX

EI

A

c

xXc

xX

x

coshsinhcossin)(

:get to)(Let

Define

.0)()(

4321

22

4

2

Apply boundary conditions to get 3

constants and the characteristic equation

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Example: compute the mode shapes and

natural frequencies for a clamped-pinned

beam.

0)coshsinhcossin(

0)(

0coshsinhcossin

0)(

and end, pinned At the

0)(0)0(

00)0(

and 0 end fixedAt

4321

2

4321

31

42

aaaa

XEI

aaaa

X

x

aaX

aaX

x

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tanhtan

0)det(,

0

0

0

0

coshsinhcossin

coshsinhcossin

00

1010

4

3

2

1

2222

BB

a

a

a

a

B

0a0a

a

The 4 boundary conditions in the 4 constants can be

written as the matrix equation:

The characteristic equation

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Solve numerically to obtain solution to

transcendental equation

4

)14(

5

493361.16351768.13

210176.10068583.7926602.3

54

321

n

n

n

Next solve Ba=0 for 3 of the constants:

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Solving for the eigenfunctions:

xxxxaxX

aa

aa

aa

aa

B

nnnn

nn

nnnn

nn

nn

nnnn

coscosh)sin(sinhsinsinh

coscosh)()(

sinsinh

coscosh

:yields Solving

equation fourth)(or third thefrom

0)cos(cosh)sinsinh(

equation second thefrom

equationfirst thefrom

:4th theof in terms constants 3 yields

4

43

43

42

31

0a

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Mode shapesX ,n x .

cosh n cos n

sinh n sin nsinh .n x sin .n x cosh .n x cos .n x

0 0.2 0.4 0.6 0.8 1

2

1.5

1

0.5

0.5

1

1.5

X ,3.926602 x

X ,7.068583 x

X ,10.210176 x

x

Mode 1

Mode 2Mode 3

Note zero slope

Non zero slope

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Summary of the Euler-Bernoulli

Beam

• Uniform along its span and slender

• Linear, homogenous, isotropic elastic

material without axial loads

• Plane sections remain plane

• Plane of symmetry is plane of vibration so

that rotation & translation decoupled

• Rotary inertia and shear deformation

neglected

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Homework #5

• Get an expression for for the cases of a

uniform Euler-Bernoulli beam with BC’s as

follows:

– Clamped-Clamped

– Pinned-Pinned

– Clamped-Free

– Free-Free

vibration of continuous structures

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