04 Vibration of bars

Dynamics of Continuous Structures
Maged Mostafa
#WikiCourses
http://WikiCourses.WikiSpaces.com
Vibration of Continuous
Structures

Maged Mostafa
#WikiCourses
Bar Vibration

Maged Mostafa
#WikiCourses
Course Contents
 SDOF
 M-DOF
 Cables/String
• Bars
• Shafts
• Vibration Attenuation
• Beams
• FEM for Vibration
• Plates
• Aeroelasticity

Maged Mostafa
#WikiCourses
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

Maged Mostafa
#WikiCourses
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

Maged Mostafa
#WikiCourses
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!

Maged Mostafa
#WikiCourses
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+dFF
Equilibrium
position
Infinitesimal
element
0 l

Maged Mostafa
#WikiCourses
Remember!
x
txu
xEAF
x
txu
E
xσAF
From
),(
)(
),(
)(
LawsHawk'











Maged Mostafa
#WikiCourses
0
),(
:endfreeAt the
,0),0(:endclampedAt the
),(),(
constant)(
),(
)(
),(
)(
),(
)(
),(
)(
),(
)(
2
2
2
2
2
2
2
2

















xx
txu
EA
tw
t
txu
x
txuE
xA
t
txu
xA
x
txu
xEA
x
dx
x
txu
xEA
x
dF
x
txu
xEAF
t
txu
dxxAFdFF























Force balance:
Constitutive relation:

Maged Mostafa
#WikiCourses
Apply the boundary conditions to the
spatial solution to get
asin(0)  bcos(0)  0
acos(l )  bsin(l )  0
 b  0 and det
0 1
cos(l ) sin(l )





  0

cosl  0  n 
2n 1
2l
, n  1,2,3,L
Xn (x)  an sin(
(2n 1)x
2l
), n  1,2,3,L







Maged Mostafa
#WikiCourses
time response equation :
&&Tn (t)  c2 2n 1
2l





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)
2l
E

, n  1,2,3L (6.63)
Thus the solution implies oscillation with Frequencies:

Maged Mostafa
#WikiCourses
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

Maged Mostafa
#WikiCourses
Given v0(l)=3 cm/s, =8x103 kg/m3 and
E=20x1010 N/m2, compute the response.
w(x,t)  (cn sinnct
n1

  dn cosnct)sin
(2n 1)
2l
x
dn 
2
l
w0 (x)sin
0
l

(2n 1)
2l
xdx  0 
wt (x,0)  0.03(x  l )  cnnccos(0)
n1

 sin
(2n 1)
2l
x
0
Multiply by the mode shape indexed m and integrate:

Maged Mostafa
#WikiCourses
 0.03 (sin
0
l

(2m 1)
2l
x)(x  l )dx
 cn
0
l
 ncsin
n1


(2m 1)
2l
xsin
(2n 1)
2l
xdx
 0.03sin
(2m 1)
2
 
cml
2
cm  cm 
1


E
0.06(1)m1
(2m 1)
cn 
8 103
210 109
0.12(1)n1
(2n 1)
 7.455 10-6 (1)n1
(2n 1)
m
w(x,t)  7.455 10-6 (1)n1
(2n 1)
sin
2n 1
10
x




n1

 sin 512.348(2n 1)t  m

Maged Mostafa
#WikiCourses
Assignment
• 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

04 Vibration of bars

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Viewers also liked

Viewers also liked (20)

Similar to 04 Vibration of bars

Similar to 04 Vibration of bars (20)

Recently uploaded

Recently uploaded (20)

04 Vibration of bars