Chapter 6 Distributed Parameter
Systems
Extending the first 5
chapters (particularly
Chapter 4) to systems
with distributed mass
and stiffness properties:
Strings, rods and beams
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Tacoma Narrows Bridge
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The string/cable equation
f(x,t)
t
y
x
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• 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
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Examine a small element of the string
t2
f (x,t)
2
1
w(x,t)
t1
x1
 2 w(x,t)
 Fy = x  t 2 =
− t 1 sin 1 + t 2 sin  2 + f (x,t)x
x2 = x1 +x
• Force balance on an infinitesimal element
• Now linearize the sine with the small angle
approximate sinx = tanx =slope of the string
w
• Writing the slope as the derivative:
tan x =
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x
Substitute into F=ma and use a Taylor expansion:
  w(x,t) 

 t
x 
x2
  w(x,t) 
− t


x 
x1
 2 w(x,t)
x
+ f (x,t)x = 
2
t
Recall the Taylor series of t w/ x about x1 :
 w 

 t
x 
x2
 w 
= t
  x 
   w(x,t) 

 t
x
x
x1
x1
  w 
+ x  t

x  x 
+ O(x 2 ) + K
x1
 2 w(x,t)
x 
x + f (x,t)x = 
2
t
 2 w(x,t)
   w(x,t) 
 + f (x,t) = 
 t
t 2
x
x
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(6.7)
Since t is constant, and for no external force, the equation
of motion becomes:
 w( x, t )  w( x, t )
c
=
,
2
2
x
t
2
t
c=

2
2
(6.8)
, wave speed
Second order in time and second order in space, therefore
4 constants of integration. Two from initial conditions:
w( x, 0) = w0 ( x), wt ( x, 0) = w0 ( x) at t = 0
And two from boundary conditions (e.g. a fixed-fixed string):
w(0, t ) = w( , t ) = 0,
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t 0
(6.9)
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Each of these terms has a
physical interpretation:
•
•
•
•
•
Deflection is w(x,t) in the y-direction
The slop 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
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There are Two Solution Types
for Two Situations
• This is called the wave equation and if there
are no boundaries, or they are sufficiently
far away it is solved as a wave phenomena
– Disturbance results in propagating waves*
• If the boundaries are finite, relatively close
together, then we solve it as a vibration
phenomena (focus of Vibrations)
– Disturbance results in vibration
*focus of courses in Acoustics and in Wave Propagation
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Solution of the Wave Equation:
• Interpret w(x,t) as a stress, particle velocity,
or displacement to examine the propagation
of waves in elastic media
– Called wave propagation
• Interpret w(x,t) as a pressure to examine the
propagation of sound in a fluid
– Called acoustics
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Solution of the “string equation” as a Wave
A solution is of the form:
w(x,t) = w1(x − ct) + w2 (x + ct)
This describes on wave traveling forward and one wave
traveling backwards (called traveling waves as the form
of the wave moves along the media)
crest
trough
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The wave speed is c
Think of waves in a
pool of water
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Example 6.1.1 what are the boundary
conditions for this system?
A force balance at l yields
F
x=l
= t sin  + kw(l ,t) = 0
y
w(x,t)
t
= −k w(x,t) x = l
x x = l
At x = 0, the BC is w(x,t) x = 0 = 0
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Fig 6.2
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6.2 Modes and Natural Frequencies
Solve by the method of separation of variables:
w( x, t ) = X ( x)T (t ) 
Substitute into the equation of motion to get:
(6.10)
2
2
d
d
c 2 X ( x)T (t ) = X ( x)T (t ) where = 2 and = 2
dx
dt
X ( x) T (t )
d  X ( x) 
X ( x)
2

= 2
, 
=
0

=
−


X ( x) c T (t )
dx  X ( x) 
X ( x)
(6.14)
Results in two second order equations
T (t )
coupled only by a constant:
 2
= − 2
c T (t )
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(6.15)
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Solving the spatial equation results
in a boundary value problem:
X (x) +  2 X(x) = 0 
(6.16)
A second order equation with solution of the form:
X(x) = a1 sin  x + a2 cos x, a1 and a2 constants of integration
Next apply the boundary conditions:
X(0)T(t) = 0,
X( )T(t) = 0  Since T(t) is not zero
X(0) = 0,
X( ) = 0,
 X(0) = a2 = 0


 X( ) = a1 sin  = 0
characteristic equation
sin  = 0
 n =
n
(6.21)
an infinite number of values of 
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© D. J. Inman
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A more systematic way to generate the
characteristic equation is write the boundary
conditions (6.20) in matrix form
a1 sin  = 0 and a2 = 0
0   a1  0 
sin 

= 



1   a2  0 
 0
0
sin 
 det( 
) = 0  sin  = 0

1
 0
This seemingly longer approach works in
general and will be used to compute the
characteristic equation in more complicated situations
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The the solution of the boundary value
problem results in an eigenvalue
problem
The spatial solution becomes:
 n 
(6.22)
, X n ( x) = an sin 
x


Here the index n results because of the indexed value of 
For n=1,2,3
The spatial problem also can be written as:
 2
− 2 ( X n ) = n X n ,
x
X n ( x)  0, X n (0) = X n ( ) = 0
(6.23)
Which is also an eigenvalue, eigenfunction problem where
 =2 is the eigenvalue and Xn is the eigenfunction.
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This is Analogous to Matrix
Eigenvalue Problem
2
A → −t 2 , plus boundary conditions, matrix → operator
x
ui → Xn (x), eigenvector → eigenfunction
 Xn (x) also and eigenvector
orthoganality also results and the condition of normalizing
plays the same role
eigenvectors become mode shapes, eigenvalues frequencies
modal expansion will also happen
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Next consider the temporal
solution resulting in oscillation.
Tn (t ) + c 2 n2Tn (t ) = 0, n = 1, 2,3
(6.24)
Again a second order ode with solution of the form:
Tn (t ) = An sin  n ct + Bn cos  n ct
(6.25)
An , Bn are constants of integration (get from initial conditions)
Substitution back into the separated form X(x)T(t) yields:
wn ( x, t ) = cn sin  n ct sin  n x + d n cos  n ct sin  n x
= cn sin(
n
ct ) sin(
n
The total solution becomes:

w( x, t ) =  cn sin(
n =1
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n
ct ) sin(
x) + d n cos(
n
n
x) + d n cos(
ct ) sin(
n
n
ct ) sin(
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x)
n
x)
(6.27)
Orthogonality is used to evaluate the
remaining constants from the initial
conditions
 2, n = m
0 sin( x) sin( x)dx =  0, n  m = 2  nm (6.28)
(proving this looks a lot like homework!)
From the initial poistion:
n
m

w( x, 0) = w0 ( x) =  d n sin(
n
x) cos(0) 
n =1
 w ( x) sin(
0
0
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m

x)dx =  d n  sin(
n =1
n
x) sin(
0
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m
x)dx 
dm =
2
 w ( x)sin(
m
0
x)dx, m = 1, 2,3
(6.31)
0
m→n
dn =
2
 w ( x)sin(
n
0
x)dx, n = 1, 2,3
0

w0 ( x) =  cn n csin(
n
x) cos(0)
n =1
cn =
2
w ( x)sin(

n c
0
n
x)dx, n = 1, 2,3
0
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(6.32)
(6.33)
A Eigenfunctions become
the vibration mode shapes
w0 ( x) = sin

x, which is the first eigenfunction (n =1)
w0 ( x) = 0,  cn = 0, n
dn =
2
 sin(

x) sin(
n
x)dx = 0, n = 2,3
0
d1 = 1 

c 
w( x, t ) = sin( x) cos  t 


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Causes vibration in the first
mode shape
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Plots of mode shapes (fig 6.3)
 n 
sin 
x
2 
1
0.5
X 1, x
X 2, x
0
X 3, x
0.5
1
1.5
2
nodes
0.5
1
x
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Example 6.2.2: 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
c

t
 11.1  10 N
1 =
=
=
l
1.4  1.4 0.0786 kg/m
4
= 2666.69 rad/s or 424 Hz
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Example 6.2.3: Compute the mode
shapes and natural frequencies
for the following system:
0
x
w(x,t)
A cable hanging from
the top and attached
to a spring of tension t
and density 
k
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In this case the characteristic equation
must be solved numerically
F
y x=
= 0  t sin  + kw( , t ) = 0 
 w( x, t )
= −kw( , t )
 x x=
X ( x) = a1 sin  x + a2 cos  x,
t
X (0) = 0  a2 = 0 and X ( x) = a1 sin  x
 t cos  = −k sin 
 tan  = −
t
k
The characteristic equation must be
solved numerically for n
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Fig 6.4
The Mode Shapes and
Natural Frequencies are
k = 1000,t = 10, = 2
 n = 1.497, 2.996, 4.501
6.013
(2n − 1)
2
X n = an sin( n x)
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• The values of n must be
found numerically
• The eigenfunctions are
again sinusoids
• The value shown for
n is for large n
• See Window 6.3 for a
summary of method
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Example: Compute the response of the piano wire
to :
initial conditions: w0 ( x) = sin

Solution: w( x, t ) =  (cn sin
3 x
n
, w0 ( x) = 0
x sin
n c
i =1

w0 ( x) = 0 =  (cn
n c
sin
n
i =1

w( x, t ) =  d n sin
n
x cos
t + d n sin
n
x cos
x cos(0))  cn = 0, n
n c
t , at t = 0, Eq.(6.31) 
i =1
dn =
2
 sin
3
x sin
m
xdx, m = 1, 2,3
0
=0, for all m except m =3, d3 = 1
w( x, t ) = sin
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3
x sin
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3 c
t
n c
t)
Some calculation details:
 w0 ( x) sin
m
xdx
0

=  d n  sin
n =1
= dm
n
x sin
m
xdx
0
w( , t ) = sin 6 sin
2
=0
3 c
t
2
3
3 c
w( , t ) = sin
sin
t
4
4
3 c
= 0.707 sin
t
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= 1.4 m, c = 11.89 
1.4
w( , t ) = 0.707 sin 80.4t
4
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Summary of Separation of Variables
• Substitute w(x,t)=X(x)T(t) into equation
of motion and boundary conditions
• Manipulate all x dependence onto one
side and set equal to a constant (-2)
• Solve this spatial equation which results
in eigenvalues n and eigenfunctions Xn
• Next solve the temporal equation to get
Tn(t) in terms of n and two constants of
integration
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• Re form the product wn(x,t)=Xn(x)Tn(t) in
terms of the 2n constants of integration
• Form the sum

w(x,t) =  Xn (x)(An sin  n ct +Bn cos  n ct)
n=1
• Use the initial displacement, initial
velocity and the orthogonality of Xn(x) to
compute An and Bn.
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6.3 Vibration of Rods and Bars
Equilibrium
position
x
x +dx
dx
Infinitesimal
element
F
F+dF
x
w(x,t)
0
Fig 6.5
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• Consider a small element
of the bar
• Deflection is now along x
(called longitudinal
vibration)
• F= ma on small element
yields the following:
l
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Force balance:
 2 w( x, t )
F + dF − F =  A( x)dx
(6.53)
2
 t
Constitutive relation:
 w( x, t )
 
 w( x, t ) 
F = EA( x)
 dF =
 EA( x)
 dx
 x
 x
 x 
 
 w( x, t ) 
 2 w( x, t )
(6.55)
 EA( x)
 =  A( x)
2
 x
 x 
 t
E  2 w( x, t )  2 w( x, t )
A( x) = constant 
=
(6.56)
2
2
  x
 t
At the clamped end: w(0, t ) = 0,
(6.57)
 w( x, t )
At the free end: EA
=0
 x x=
(6.58)
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Example 6.3.1: compute the mode shapes
and natural frequencies of a cantilevered bar
with uniform cross section.
w( x, t ) = X ( x)T (t ) → c 2 wxx ( x, t ) = wtt ( x, t )
X ( x)
T (t )

= 2
= − 2
X ( x) c T (t )
(6.59)
 X ( x) +  2 X ( x) = 0, X (0) = 0, AEX ( ) = 0 (i)

2 2
T
(
t
)
+
c
 T (t ) = 0, initial conditions (ii)

From equation (i) the form of the spatial solution is
X ( x) = a sin  x + b cos  x
Next apply the boundary conditions in (i) to get the
characteristic equation and the form of the eigenfunction:
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Apply the boundary conditions to the
spatial solution to get
a sin(0) + b cos(0) = 0
a cos( ) − b sin( ) = 0
1
 0

 b = 0 and det 
=0

cos( ) − sin( ) 
2n − 1

cos  = 0   n = 2  ,

 X ( x) = a sin( (2n − 1) x ),
n
 n
2
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n = 1, 2,3,
n = 1, 2,3,
Next consider the time response equation (ii):
 2n − 1 
Tn (t ) + c 
  T (t ) = 0
 2

(2n − 1)c
(2n − 1)c
 Tn (t ) = An sin
t + Bn cos
t
2
2
2
2
Thus the solution implies oscillation with
Frequencies:
(2n − 1)c (2n − 1)
n =
=
2
2
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E

,
n = 1, 2,3
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(6.63)
Example 6.3.2 Given v0(l)=3 cm/s, =8x103
kg/m3 and E=20x1010 N/m2, compute the
response.

(2n − 1)
w( x, t ) =  (cn sin  n ct + d n cos  n ct ) sin
x
2
n =1
dn =
2
0
(2n − 1)
0 w0 ( x) sin 2  xdx = 0 
(2n − 1)
wt ( x, 0) = 0.03 ( x − ) =  cn n c cos(0)sin
x
2
n =1

Multiply by the mode shape indexed m and integrate:
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 0.03 (sin
0
(2m − 1)
 x) ( x − )dx
2
(2m − 1)
(2n − 1)
=   cn n c sin
 x sin
 xdx
2
2
n =1 0

c
(2m − 1)
1
 0.03sin
 = m cm  cm =
2
2

 0.06(−1) m +1
E
(2m − 1)
n +1
8 103 0.12(−1) n +1
-6 ( −1)
cn =
= 7.455 10
m
9
210 10  (2n − 1)
(2n − 1)
(−1) n +1
 2n − 1 
w( x, t ) = 7.455 10 
sin 
 x  sin ( 512.348(2n − 1)t ) m
 10

n =1 (2n − 1)

-6
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Various Examples Follow From
Other Boundary Conditions
• Table 6.1 page 485 gives a variety of different
boundary conditions. The resulting frequencies
and mode shapes are given in Table 6.2 page 486.
• Various problems consist of computing these
values
• Once modes/frequencies are determined, use mode
summation to compute the response from IC
• See the book by Blevins:Mode Shapes and
Natural Frequencies for more boundary
conditions
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6.4 Torsional Vibrations

(x,t)
t
x
(x,t)+d
t+dt
x+dx
t
dx, from calculus
x
 (x,t)
t = GJ
, from solid mechanics
x
dt =
G=shear modulus
J =polar moment of area cross section
Summing moments on the element dx Combining these expressions yields;
t
 2 (x,t)
t + dx − t =  J
dx
2
x
t
Where  is the shaft's mass density
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   (x,t) 
 2 (x,t)
,GJ constant 
 GJ
 = J
x 
x 
t 2
 2 (x,t) G  2 (x,t)
=
(6.66)
t 2
  x2
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The initial and boundary
conditions for torsional vibration
problems are:
• Two spatial conditions (boundary conditions)
• Two time conditions (initial conditions)
• See Table 6.4 for a list of conditions and Equation (6.67)
and Table 6.3 for odd cross section
• Clamped-free rod:
 (0, t ) = 0 Clamped boundary (0 deflection)
G x ( , t ) = 0 Free boundary (0 torque)
 (x,0)= 0 ( x) and t (x,0)= 0 ( x)
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Example 6.4.1: grinding
shaft vibrations
• Top end of shaft is connected to pulley (x = 0)
• J1 includes collective inertia of drive belt, pulley
and motor
J1
Drive pulley, collective inertia
x
Grinding head inertia
(x,t)
Shaft of stiffness GJ, length l
J2
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Use torque balance at top and bottom
to get the Boundary Conditions:
 (x,t)
 (x,t)
GJ
= J1
at top
2
x x= 0
t
x= 0
2
 (x,t)
 (x,t)
GJ
= −J 2
at bottom
2
x x=
t
x=
2
The minus sign follows from right hand rule.
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Again use separation of variables
to attempt a solution
 ( x, t ) = ( x)T (t ) 
( x)    T (t )
= 
= − 2
( x)  G  T (t )
G
c = 

( x) +  2( x) = 0, T (t ) +  2T (t ) = 0
2
 =c =
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G

Mechanical Engineering at Virginia Tech
The next step is to use the boundary
conditions:
Boundary Condition at x = 0 
GJ (0)T (t ) = J1(0)T (t ) 
GJ (0) T (t )
=
= −c 2 2 
J1(0) T (t )
 2 J1
(0) = −
(0)
J
Similarly the boundary condition at l yields:
 2 J1
( ) =
( )
J
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The Boundary Conditions reveal the
Characteristic Equation
( x) = a1 sin  x + a2 cos  x  (0) = a2
( x) = a1 cos  x − a2 sin  x  (0) = a1
x=0
 2 J1
 J1

 (0) = −
(0)  a1 = −
a2
J
J
x= 
 2 J1
 2 J1
( ) =
( )  a1 cos  − a2 sin  =
a1 sin  + a2 cos 
J
J
 J ( J1 + J 2 )( )
 tan( ) =
THE CHARACTERISTIC EQUATION
J1 J 2 ( ) 2 − (  J ) 2
(6.82)
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Solving the for the first mode
shape
 J ( J1 + J 2 )( )
tan( ) =
has 0 as its first solution:
2
2
J1 J 2 ( ) − (  J )
Numerically solve for  n , n = 1, 2,3,
, and n =  n
Note for n = 1,  1 = 0  1 = 0  T (t ) = 0 
T (t ) = a + bt the rigid body mode of the shaft turning
 1( x) = 0,  1 ( x) = a1 + b1 x 
x = 0  b1 = 0  1 ( x) = a1 the first mode shape
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G

Solutions of the Characteristic Equation involve
solving a transcendental equation
(bx 2 − a ) tan x = x
J
J1 J 2
x = , a =
, b=
J1 + J 2
( J1 + J 2 )  J
J1 = J 2 = 10 kg  m 2 / rad,  =2700 kg/m 3 ,
J =5 kg  m 2 / rad, =0.25 m
G = 25 109 Pa

f1 = 0 Hz, f 2 = 38, 013 Hz,
f3 = 76, 026 Hz, f 4 = 114, 039 Hz,
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Fig 6.9 Plots of each side of eq (6.82)
to assist in find initial guess for numerical routines used
to compute the roots.
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6.5 Bending vibrations of a beam
w (x,t)
f (x,t)
f(x,t)
h1
x
h2
dx
M(x,t)
w(x,t)
V(x,t)
·Q
V(x,t)+Vx(x,t)dx
A(x)= h1h2
bending stiffness = EI(x)
E = Youngs modulus
I(x) = cross-sect. area moment of
inertia about z
 2 w(x,t)
M (x,t) = EI(x)
 x2
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M(x,t)+Mx(x,t)dx
x
x +dx
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)
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
Summing forces and moments yields:

V (x,t) 
 2 w(x,t)
dx  − V (x,t) + f (x,t)dx = A(x)dx
V (x,t) +


x
t 2


M(x,t) 
V (x,t) 
dx  − M(x,t) + V (x,t) +
dxdx
 M(x,t) +




x
x
dx
+ f (x,t)dx
=0
2
0
M(x,t)

V (x,t) f (x,t) 
2

+ V (x,t)dx + 
+
(dx)
=0

 x

 x
2 
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M(x,t)
 V (x,t) = −
x
Substitute into force balance equation yields:
 2 w(x,t)
 2 M(x,t)
dx + f (x,t)dx = A(x)dx
−
2
t 2
x
Dividing by dx and substituting for M yields
 2 w(x,t) 
 2 w(x,t)  2 
+ 2 EI(x)
A(x)
 = f (x,t)
2
2
x 
x 
t
Assume constant stiffness to get:
EI
 2 w(x,t) 2  4 w(x,t)
= 0, c =
+c
4
2
A
x
t
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The possible boundary
conditions are (choose 4):
Free end
 2w
bending moment = EI 2 = 0
x
   2w 
shear force = EI 2  = 0
x  x 
Pinned (or simply supported) end
deflection = w = 0
 2w
bending moment=EI 2 = 0
x
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Clamped (or fixed) end
deflection = w = 0
w
slope =
=0
x
Sliding end
w
slope =
=0
x
   2w 
shear force = EI 2  = 0
x  x 
Mechanical Engineering at Virginia Tech
Solution of the time equation
yields the oscillatory nature:
X ( x)
T (t )
c
=−
= 2 
X ( x)
T (t )
2
T (t ) +  T (t ) = 0 
T (t ) = A sin t + B cos t
Two initial conditions:
w( x , 0) = w0 ( x ), wt ( x , 0) = w0 ( x )
2
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Spatial equation results in a
boundary value problem (BVP)
2
 
X (x) −   X(x) = 0.
c
2
2



A
4
Define  =   =
c
EI
x
Let X(x) = Ae to get :
X(x) = a1 sin x + a2 cos x + a3 sinh x + a4 cosh x
Apply boundary conditions to get 3
constants and the characteristic equation
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Example 6.5.1: compute the mode shapes
and natural frequencies for a clampedpinned beam.
At fixed end x = 0 and
X(0) = 0  a2 + a4 = 0
X (0) = 0   (a1 + a3 ) = 0
At the pinned end, x = and
X( ) = 0 
a1 sin  + a2 cos  + a3 sinh  + a4 cosh  = 0
EIX ( ) = 0 
 2 (−a1 sin  − a2 cos  + a3 sinh  + a4 cosh  ) = 0
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The 4 boundary conditions in the 4 constants can be
written as the matrix equation:

0



 sin 
 2
− sin 
1
0
cos 
− 2 cos 
0

sinh 
 2 sinh 
1
0
cosh 
 2 cosh 
 a1  0
   
a2  = 0
a3  0
   
a4  0
a
B
Ba = 0,a  0  det(B) = 0 
tan  = tanh 
The characteristic equation
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Solve numerically (fsolve) to obtain solution
to transcendental (characteristic) equation
1 = 3.926602  2 = 7.068583 3 = 10.210176
 4 = 13.351768 5 = 16.493361
n 5
n
(4n + 1)
=
4
Next solve Ba=0 for 3 of the constants:
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With the eigenvalues known, now
solve for the eigenfunctions:
Ba = 0 yields 3 constants in terms of the 4th :
a1 = −a3 from the first equation
a2 = −a4 from the second equation
(sinh  n − sin  n )a3 + (cosh  n − cos  n )a4 = 0
from the third (or fourth) equation
Solving yields :
a3 = −
cosh  n − cos  n
a4
sinh  n − sin  n
cosh  n − cos  n
 X n (x) = (a4 ) n 
(sinh  n x − sin  n x) − cosh  n x + cos  n
 sinh  n − sin  n
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
x

Plot the mode shapes to help
understand the system response
X n, x
cosh n
sinh n
cos n .
sinh n. x
sin n
sin n. x
cosh n. x
cos n. x
1.5
1
Mode 2
Mode 3
Note zero slope
0.5
X 3.926602 , x
0
X 7.068583 , x
X 10.210176 , x
0.2
0.4
0.6
0.8
Non zero slope
0.5
1
1.5
2
Mode 1
x
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Again, the modeshape orthogonality becomes
important and is computed as follows:
Write the eigenvalue problem twice, once for n and once for m.
X n( x) =  n4 X n ( x)
and X m ( x) =  m4 X m ( x)
Multiply by X m ( x) and X n ( x) respectively, integrate
and subtact to get:
4
4


X
(
x
)
X
(
x
)
dx
−
X
(
x
)
X
(
x
)
dx
=
(

−

m
n
n
m )  X n ( x ) X m ( x ) dx
 n
 m
0
0
0
Then integrate the left hand side twice by parts to get:
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Use integration by parts to evaluate the
integrals in the orthogonality condition.
apply  udv =uv-  vdu twice:
X
0
m
( x) X n( x)dx = X m X n −  X n X m dx
u
u
dv
v
0
0
v
du
= X m ( ) X n( ) − X m (0) X n(0) −  X m X ndx
0
0
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u
0
Mechanical Engineering at Virginia Tech
dv
−  X m X ndx = − X m ( x) X n( x) 0 +  X n( x) X m ( x)dx
0
u
0
dv
=-X m ( ) X n( ) + X m (0) X n(0) +  X n( x) X m ( x) dx
0
Thus
0
0
4
4


X
(
x
)
X
(
x
)
dx
−
X
(
x
)
X
(
x
)
dx
=
(

−

m
n
n
m )  X n ( x ) X m ( x ) dx
 n
 m
0
0
0
  X n( x) X m ( x)dx −  X n( x) X m ( x)dx = (  n4 −  m4 )  X n ( x) X m ( x)dx
0
0
0
0
0
  X n ( x) X m ( x)dx = 0, n, m, n  m
0
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The solution can be computed via
modal expansion based on
orthogonality of the modes.
X
n
( x) X m ( x)dx =  nm
0

w( x, t ) =  ( An sin nt + Bn cos nt ) X n ( x)
n =1

w( x, 0) = w0 ( x) =  Bn X n ( x)  Bn =  w0 ( x) X n ( x)dx
n =1
0

wt ( x, 0) = w0 ( x) =  n An X n ( x)  An =
n =1
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1
n
 w ( x) X
0
0
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n
( x)dx
Summary of the Euler-Bournoulli
Beam Assumptions
• 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
• Tables 6.4 and 6.5 give eigensolutions for several
configurations (see Blevins for more)
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