What is a continuous structure? How to analyse the vibration of string, bars and shafts? How to analyse the vibration of beams? #WikiCourses https://wikicourses.wikispaces.com/Topic+Vibration+of+Continuous+Structures https://eau-esa.wikispaces.com/Vibration+of+structures

1. Aero631
Dr. Eng. Mohammad Tawfik
Vibration of Continuous
Structures

2. Aero631
Dr. Eng. Mohammad Tawfik
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

3. Aero631
Dr. Eng. Mohammad Tawfik
String and Cables

4. Aero631
Dr. Eng. Mohammad Tawfik
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

5. Aero631
Dr. Eng. Mohammad Tawfik
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

6. Aero631
Dr. Eng. Mohammad Tawfik
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)

7. Aero631
Dr. Eng. Mohammad Tawfik


















)(
: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









8. Aero631
Dr. Eng. Mohammad Tawfik
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

9. Aero631
Dr. Eng. Mohammad Tawfik
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

10. Aero631
Dr. Eng. Mohammad Tawfik
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:

11. Aero631
Dr. Eng. Mohammad Tawfik
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:

12. Aero631
Dr. Eng. Mohammad Tawfik
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.

13. Aero631
Dr. Eng. Mohammad Tawfik
Analogy to Matrix Eigenvalue
Problem
happenalsowillexpansionmodal
sfrequencieseigenvalueshapes,modebecomerseigenvecto
rolesametheplays
gnormalizinofconditiontheandresultsalsoityorthoganal
reigenvectoandalso)(
ioneigenfunctreigenvecto),(
operatormatrix,conditionsboundaryplus,2
2
xX
xX
x
A
n
ni






u

14. Aero631
Dr. Eng. Mohammad Tawfik
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:

15. Aero631
Dr. Eng. Mohammad Tawfik
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





16. Aero631
Dr. Eng. Mohammad Tawfik












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







17. Aero631
Dr. Eng. Mohammad Tawfik
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

18. Aero631
Dr. Eng. Mohammad Tawfik
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

19. Aero631
Dr. Eng. Mohammad Tawfik
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

20. Aero631
Dr. Eng. Mohammad Tawfik
Bar Vibration

21. Aero631
Dr. Eng. Mohammad Tawfik
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!

22. Aero631
Dr. Eng. Mohammad Tawfik
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

23. Aero631
Dr. Eng. Mohammad Tawfik
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:

24. Aero631
Dr. Eng. Mohammad Tawfik
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

25. Aero631
Dr. Eng. Mohammad Tawfik
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

26. Aero631
Dr. Eng. Mohammad Tawfik
Beam Vibration

27. Aero631
Dr. Eng. Mohammad Tawfik
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

28. Aero631
Dr. Eng. Mohammad Tawfik
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

29. Aero631
Dr. Eng. Mohammad Tawfik
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

30. Aero631
Dr. Eng. Mohammad Tawfik
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:

31. Aero631
Dr. Eng. Mohammad Tawfik
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







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

33. Aero631
Dr. Eng. Mohammad Tawfik
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

34. Aero631
Dr. Eng. Mohammad Tawfik
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

35. Aero631
Dr. Eng. Mohammad Tawfik


  






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

36. Aero631
Dr. Eng. Mohammad Tawfik
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:

37. Aero631
Dr. Eng. Mohammad Tawfik
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

38. Aero631
Dr. Eng. Mohammad Tawfik
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

39. Aero631
Dr. Eng. Mohammad Tawfik
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

40. Aero631
Dr. Eng. Mohammad Tawfik
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