Vibration of Continuous Structures

Aero631
Dr. Eng. Mohammad Tawfik
Vibration of Continuous
Structures

Aero631
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
txw
xFy


),(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/ofseriesTaylortheRecall
2
1
112
xO
x
w
x
x
x
w
x
w
xxw
xxx












x
t
txw
xtxf
x
txw
x
txw
xx












2
2
),(
),(
),(),(
12









2
2
),(
),(
),(
t
txw
txf
x
txw
x 













x
t
txw
xtxfx
x
txw
x x






2
2
),(
),(
),(
1









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0,0),(),0(
0at)()0,(),()0,(
,
),(),(
00
2
2
22
2



ttwtw
txwxwxwxw
c
x
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
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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Also an eigenvalue problem
0)()0(,0)(,)(
ofvalueindexedtheofbecauseresultsindextheHere
sin)(,1,2,3=For
2
2











nnnnnn
nn
XXxXXX
x
n
x
n
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
happenalsowillexpansionmodal
sfrequencieseigenvalueshapes,modebecomerseigenvecto
rolesametheplays
gnormalizinofconditiontheandresultsalsoityorthoganal
reigenvectoandalso)(
ioneigenfunctreigenvecto),(
operatormatrix,conditionsboundaryplus,2
2
xX
xX
x
A
n
ni






u

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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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A mode shape












t
c
xtxw
d
ndxx
n
xd
ncxw
nxxw
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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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+dFF
Equilibrium
position
Infinitesimal
element
0 l

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0
),(
:endfreeAt the
,0),0(:endclampedAt the
),(),(
constant)(
),(
)(
),(
)(
),(
)(
),(
)(
),(
)(
2
2
2
2
2
2
2
2

















xx
txw
EA
tw
t
txw
x
txwE
xA
t
txw
xA
x
txw
xEA
x
dx
x
txw
xEA
x
dF
x
txw
xEAF
t
txw
dxxAFdFF























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

Aero631
Bending vibrations of a beam
2
2
),(
)(),(
aboutinertia
ofmomentareasect.-cross)(
modulusYoungs
)(stiffnessbending
x
txw
xEItxM
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

Aero631
Summing forces and moments
0)(
2
),(),(
),(
),(
0
2
),(
),(
),(),(
),(
),(
),(
)(),(),(
),(
),(
2
2
2


































dx
txf
x
txV
dxtxVdx
x
txM
dx
dxtxf
dxdx
x
txV
txVtxMdx
x
txM
txM
t
txw
dxxAdxtxftxVdx
x
txV
txV













0

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A
EI
c
x
txw
c
t
txw
txf
x
txw
xEI
xt
txw
xA
t
txw
dxxAdxtxfdx
x
txM
x
txM
txV




























,0
),(),(
),(
),(
)(
),(
)(
),(
)(),(
),(
),(
),(
4
4
2
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)
0forceshear
0momentbending
endFree
2
2
2
2








x
w
EI
x
x
w
EI






0slope
0deflection
endfixed)(orClamped


x
w
w


0momentbending
0deflection
endsupported)simply(orPinned
2
2


x
w
EI
w


0forceshear
0slope
endSliding
2
2








x
w
EI
x
x
w







Aero631
Solution of the time equation:
)()0,(),()0,(
:conditionsinitialTwo
cossin)(
0)()(
)(
)(
)(
)(
00
2
22
xwxwxwxw
tBtAtT
tTtT
tT
tT
xX
xX
c
t












Aero631
Spatial equation (BVP)
xaxaxaxaxX
AexX
EI
A
c
xX
c
xX
x





coshsinhcossin)(
:getto)(Let
Define
.0)()(
4321
22
4
2
















Apply boundary conditions to get 3
constants and the characteristic equation

Aero631
Example: compute the mode shapes and
natural frequencies for a clamped-pinned
beam.
0)coshsinhcossin(
0)(
0coshsinhcossin
0)(
andend,pinnedAt the
0)(0)0(
00)0(
and0endfixedAt
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
nn
nn
nn
nn
nnnn












coscosh)sin(sinh
sinsinh
coscosh
)()(
sinsinh
coscosh
:yieldsSolving
equationfourth)(orthirdthefrom
0)cos(cosh)sinsinh(
equationsecondthefrom
equationfirstthefrom
:4ththeofin termsconstants3yields
4
43
43
42
31
0a

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Mode shapes
X ,n x .cosh n cos n
sinh n sin n
sinh .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 2
Mode 3
Note zero slope
Non zero slope

Aero631
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

Aero631
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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Vibration of Continuous Structures