03 Vibration of string

Dynamics of Continuous Structures
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Vibration of Continuous
Structures

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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 Cable/
String
• Estimate the Natural Frequencies
• 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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Objectives
• Derive the equation of motion for
Cable/String
response

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

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Examine a small element of
string
xtxf
t
txw
xFy


),(sinsin
),(
2211
2
2
tt



• 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
t1
x1 x2 = x1 +x
w(x,t)
f (x,t)

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

















)(
:about/ofseriesTaylortheRecall
2
1
112
xO
x
w
x
x
x
w
x
w
xxw
xxx


t




t


t
t
x
t
txw
xtxf
x
txw
x
txw
xx












2
2
),(
),(
),(),(
12





t


t
2
2
),(
),(
),(
t
txw
txf
x
txw
x 




t








x
t
txw
xtxfx
x
txw
x x






2
2
),(
),(
),(
1





t



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0,0),(),0(
0at)()0,(),()0,(
,
),(),(
00
2
2
22
2



ttwtw
txwxwxwxw
c
x
txw
tc
txw
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

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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 twxx(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
Subscript denotes differentiation w.r.t. to that parameter

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Modes and Natural Frequencies
   
2
2
2
2
2
2
2
2
2
)(
)(
)(
)(
0
)(
)(
,
)(
)(
)(
)(
=and=where)()()()(
)()(),(










 






tTc
tT
xX
xX
xX
xX
dx
d
tTc
tT
xX
xX
dt
d
dx
d
tTxXtTxXc
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 









equationsticcharacteri
1
2
2121
2
0sin
0sin)(
0)0(
,0)(,0)0(
0)()(,0)()0(
nintegratioofconstantsareand,cossin)(
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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The temporal solution








1
22
)sin()cos()sin()sin(),(
)sin()cos()sin()sin(
sincossinsin),(
)conditionsinitialfrom(getnintegratioofconstantsare,
cossin)(
3,2,1,0)()(
n
nn
nn
nnnnnnn
nn
nnnnn
nnn
x
n
ct
n
dx
n
ct
n
ctxw
x
n
ct
n
dx
n
ct
n
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,(
:conditionsinitialtheFrom
2,0
,
)sin()sin(
dxx
m
x
n
ddxx
m
xw
x
n
dxwxw
mn
mn
dxx
m
x
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











ndxx
n
xw
cn
c
x
n
cxw
ndxx
n
xwd
nm
mdxx
m
xwd
n
n
nn
n
m







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Objectives
• Derive the equation of motion for
Cables/Strings
response

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A mode shape












t
c
xtxw
d
ndxx
n
xd
ncxw
nxxw
Assume
n
n









cos)sin(),(
1
3,2,0)sin()sin(
2
,0,0)(
1)=(ioneigenfunctfirsttheiswhich,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
sin
n
2
x




nodes

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String mode shapes Video 1
String mode shapes Video 2

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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 
c
l


1.4
t



1.4
11.1104
N
0.0786 kg/m
 2666.69 rad/s or 424 Hz

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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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