04 vibration of bars

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Vibration of Continuous Structures

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Bar Vibration

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

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Objectives

• Derive the equation of motion for Bars• Estimate the Natural Frequencies• Understand the concept of mode shapes• Apply BC’s and IC’s to obtain structure

response

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Objectives

• Derive the equation of motion for Bars• Apply BC’s and IC’s to obtain structure

response• Estimate the Natural Frequencies• Understand the concept of mode shapes

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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

u(x,t) x

dx

F+dF F

Equilibrium position

Infinitesimal element

0 l

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Remember!

xtxuxEAF

xtxu

ExσAF

From

),()(

),(

)(Law sHawk'

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0

),( :end free At the

,0),0( :end clamped At the

),(

),( constant)(

),( )(

),()(

),()(

),()(

),( )(

2

2

2

2

2

2

2

2

xxtxuEA

twt

txux

txuExA

ttxuxA

xtxuxEA

x

dxx

txuxEAx

dFx

txuxEAF

ttxudxxAFdFF

Force balance:

Constitutive relation:

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Objectives

• Derive the equation of motion for Bars• Apply BC’s and IC’s to obtain structure

response• Estimate the Natural Frequencies• Understand the concept of mode shapes

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Apply the boundary conditions to the spatial solution to get

asin(0) bcos(0)0acos( l ) bsin( l )0

b 0 and det0 1

cos( l ) sin( l )

0

cos l 0 n

2n 12l , n 1,2,3,L

Xn (x)an sin((2n 1) x

2l ), n 1,2, 3,L

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Objectives

• Derive the equation of motion for Bars• Apply BC’s and IC’s to obtain structure

response• Estimate the Natural Frequencies• Understand the concept of mode shapes

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time response equation :

Tn (t) c2 2n 12l

2

T (t)0

Tn (t)An sin(2n 1)c

2l t Bn cos(2n 1)c

2l t

n (2n 1)c

2l (2n 1)

2lE

, n 1,2,3L (6.63)

Thus the solution implies oscillation with Frequencies:

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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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Given v0(l)=3 cm/s, =8x103 kg/m3 and E=20x1010 N/m2, compute the response.

w(x,t) (cn sin nctn1

dn cos nct)sin(2n 1)

2l x

dn 2l w0 (x)sin

0

l

(2n 1)

2l xdx 0

wt (x,0)0.03 (x l ) cn nccos(0)n1

sin(2n 1)

2l x

0

Multiply by the mode shape indexed m and integrate:

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0.03 (sin0

l

(2m 1)

2l x) (x l )dx

cn0

l

ncsinn1

(2m 1)2l x sin

(2n 1)2l xdx

0.03sin(2m 1)

2

cml2

cm cm 1

E

0.06( 1)m1

(2m 1)

cn 8103

210109

0.12( 1)n1

(2n 1)7.45510-6 ( 1)n1

(2n 1) m

w(x, t)7.45510-6 ( 1)n1

(2n 1)sin

2n 110

x

n1

sin 512.348(2n 1)t m

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Assignment

• Solve the equation of motion of a bar with constant cross-section properties with

1. Fixed-Fixed boundary conditions2. Free-Free boundary conditions

• Compare the natural frequencies for all three cases