MIT - 16.20
Fall, 2002
Unit 23
Vibration of Continuous Systems
Paul A. Lagace, Ph.D.
Professor of Aeronautics & Astronautics
and Engineering Systems
Paul A. Lagace © 2001
MIT - 16.20
Fall, 2002
The logical extension of discrete mass systems is one of an
infinite number of masses. In the limit, this is a continuous
system.
Take the generalized beam-column as a generic representation:
d 2  d 2w 
d  d w
EI
−
F
 = pz
2 
2 
dx  dx 
dx  dx 
Figure 23.1
(23-1)
Representation of generalized beam-column
dF
= − px
dx
This considers only static loads. Must add the inertial load(s). Since the
concern is in the z-displacement (w):
˙˙
Inertial load unit length = m w
where: m(x) = mass/unit length
Paul A. Lagace © 2001
(23-2)
Unit 23 - 2
MIT - 16.20
Fall, 2002
Use per unit length since entire equation is of this form. Thus:
d 2  d 2
w 
d  d w
˙˙
EI
−
F
 = pz − m w
2 
2 
dx  dx 
dx  dx 
or:
d 2  d 2
w 
d  d w
˙˙
= pz
EI 2  −
F
 + mw
2 
dx  dx 
dx  dx 
(23-3)
Beam Bending Equation
often, F = 0 and this becomes:
d 2  d 2
w 
˙˙ = pz
EI 2  + m w
2 
dx  dx 
--> This is a fourth order differential equation in x
--> Need four boundary conditions
--> This is a second order differential equation in time
--> Need two initial conditions
Paul A. Lagace © 2001
Unit 23 - 3
MIT - 16.20
Fall, 2002
Notes:
• Could also get via simple beam equations. Change occurs in:
dS
˙˙
= pz − m w
dx
• If consider dynamics along x, must include mu̇˙ in px term: ( px − mu̇˙)
Use the same approach as in the discrete spring-mass systems:
Free Vibration
Again assume harmonic motion. In a continuous system, there are an
infinite number of natural frequencies (eigenvalues) and associated
modes (eigenvectors)
so:
w ( x , t) = w (x) e i ω t
separable solution spatially (x) and temporally (t)
Consider the homogeneous case (pz = 0) and let there be no axial forces
(px = 0 ⇒ F = 0)
Paul A. Lagace © 2001
Unit 23 - 4
MIT - 16.20
Fall, 2002
So:
d 2  d 2
w 
˙˙ = 0
EI 2  + m w
2 
dx  dx 
Also assume that EI does not vary with x:
d 4
w
˙˙ = 0
EI 4
+ m w
(23-5)
dx
Placing the assumed mode in the governing equation:
d 4
w i ω t
− m ω 2 w e i ω t = 0
EI 4
e
dx
This gives:
d 4
w
EI 4
− m ω 2 w = 0
(23-6)
dx
which is now an equation solely in the spatial variable (successful
separation of t and x dependencies)
_
Must now find a solution for w(x) which satisfies the differential equations
and the boundary conditions.
Note: the shape and frequency are intimately linked
(through equation 23-6)
Paul A. Lagace © 2001
Unit 23 - 5
MIT - 16.20
Fall, 2002
Can recast equation (23-6) to be:
d 4w
mω 2
(23-7)
−
w = 0
4
dx
EI
The solution to this homogeneous equation is of the form:
w ( x) = e p x
Putting this into (23-7) yields
4
pe
mω 2 p x
−
e
= 0
EI
px
⇒ p
4
=
mω 2
EI
So this is an eigenvalue problem (spatially). The four roots are:
p = + λ, - λ, + iλ, - iλ
where:
 mω 2 
λ = 

 EI 
1/ 4
This yields:
Paul A. Lagace © 2001
Unit 23 - 6
MIT - 16.20
Fall, 2002
w ( x ) = Ae λ x + Be − λ x + C e i λ x + De − i λ x
or:
w ( x ) = C1 sinh λx + C2 cosh λx + C3 sin λx + C4 cos λx
(23-8)
The constants are found by applying the boundary conditions
(4 constants ⇒ 4 boundary conditions)
Example: Simply-supported beam
Figure 23.2
Representation of simply-supported beam
EI, m = constant with x
d 4w
d 2w
EI 4 + m 2 = pz
dx
dt
Paul A. Lagace © 2001
Unit 23 - 7
MIT - 16.20
Fall, 2002
Boundary conditions:
@x=0
w=0
@x=l
d 2w
M = EI 2 = 0
dx
with:
w ( x ) = C1 sinh λx + C2 cosh λx + C3 sin λx + C4 cos λx
Put the resulting four equations in matrix form
w( 0 ) =
d 2w
0
2 ( )
dx
w (l ) =
d 2w
l
2 ( )
dx
0
= 0
0
= 0
1
0
1 
 0
 0
1
0
−1 


cos λl 
sinh λl cosh λl sin λl
sinh λl cosh λl − sin λl − cos λl 


 C1 
0 
C 
0 
 2
 
=
 
 
C
 3
0 
C4 
0 
Solution of determinant matrix generally yields values of λ which then
yield frequencies and associated modes (as was done for multiple mass
systems in a somewhat similar fashion)
Paul A. Lagace © 2001
Unit 23 - 8
MIT - 16.20
Fall, 2002
In this case, the determinant of the matrix yields:
C3 sin λl = 0
Note: Equations (1 & 2) give C2 = C4 = 0
Equations (3 & 4) give 2 C3 sin λl = 0
⇒ nontrivial: λl = nπ
The nontrivial solution is:
λl = nπ
(eigenvalue problem!)
Recalling that:
2
 mω 
λ = 

 EI 
mω 2
⇒
=
EI
1/ 4
n 4π 4
l4
⇒
Paul A. Lagace © 2001
(change n to r to be consistent with
previous notation)
2
ωr = r π
2
EI
ml 4
<-- natural frequency
Unit 23 - 9
MIT - 16.20
Fall, 2002
As before, find associated mode (eigenvector), by putting this back in the
governing matrix equation.
Here (setting C3 = 1…..one “arbitrary” magnitude):
w ( x ) = φr = sin
rπ x
l
<-- mode shape (normal mode)
for: r = 1, 2, 3,……∞
Note: A continuous system has an infinite number of
modes
So total solution is:
 2 2 EI
rπ x
w ( x, t) = φr sin ω r t = sin
sin  r π
l
ml 4


t

--> Vibration modes and frequencies are:
Paul A. Lagace © 2001
Unit 23 - 10
MIT - 16.20
Fall, 2002
Figure 23.3
Representation of vibration modes of simply-supported beam
2
EI
ml 4
1st mode
ω1 = π
2nd mode
ω 2 = 4π 2
EI
ml 4
3rd mode
ω 3 = 9π 2
EI
ml 4
etc.
Same for other cases
Paul A. Lagace © 2001
Unit 23 - 11
MIT - 16.20
Fall, 2002
Continue to see the similarity in results between continuous and multimass (degree-of-freedom) systems. Multi-mass systems have
predetermined modes since discretization constrains system to deform
accordingly.
The extension is also valid for…
Orthogonality Relations
They take basically the same form except now have continuous functions
integrated spatially over the regime of interest rather than vectors:
∫
l
0
where:
m( x ) φr ( x ) φ s ( x ) dx = Mr δ rs
δ rs
= kronecker delta
Mr =
∫
l
0
(23-9)
= 1 for r = s
= 0 for r ≠ s
m( x ) φr 2 ( x ) dx
generalized mass of the rth mode
Paul A. Lagace © 2001
Unit 23 - 12
MIT - 16.20
Fall, 2002
So:
∫
l
0
∫
l
0
r≠s
m φr φ s dx = 0
m φr φr dx = Mr
Also can show (similar to multi degree-of-freedom case):
∫
l
0
d 2  d 2φr 
2
EI
φ
dx
=
δ
M
ω


s
rs
r
r
dx 2 
dx 2 
(23-10)
This again, leads to the ability to transform the equation based on the
normal modes to get the…
Normal Equations of Motion
Let:
∞
w ( x, t) =
∑ φ (x) ξ (t )
r
r
(23-11)
r =1
normal mode
Paul A. Lagace © 2001
normal coordinates
Unit 23 - 13
MIT - 16.20
Fall, 2002
Place into governing equation:
d 2  d 2w 
d 2w
EI 2  + m 2 = pz ( x )
2 
dt
dx  dx 
multiply by φs and integrate
∞
∑
r =1
∫
l
dx to get:
0
∞
l
˙˙
ξr ∫ m φr φ s dx +
∑ξ ∫
r
0
r =1
l
0
d 2  d 2φr 
φ s 2  EI 2  dx =
dx 
dx 
∫
l
0
φ s f dx
Using orthogonality conditions, this takes on the same forms as
before:
Mr ξ˙˙r + Mr ω r 2ξr = Ξr
(23-12)
r = 1, 2, 3,……∞
with:
Mr =
Ξr =
∫
l
0
∫
l
0
m φr 2 dx
- Generalized mass of rth mode
φr pz ( x, t) dx - Generalized force of rth mode
ξ r (t) = normal coordinates
Paul A. Lagace © 2001
Unit 23 - 14
MIT - 16.20
Fall, 2002
Once again
• each equation can be solved independently
• allows continuous system to be treated as a series of “simple”
one degree-of-freedom systems
• superpose solutions to get total response (Superposition of
Normal Modes)
• often only lowest modes are important
• difference from multi degree-of-freedom system: n --> ∞
--> To find Initial Conditions in normalized coordinates…same as before:
w ( x, 0) =
∑ φ (x) ξ (0)
r
r
r
etc.
Thus:
1
ξr ( 0 ) =
Mr
1
ξ˙r (0) =
Mr
Paul A. Lagace © 2001
∫
l
0
∫
l
0
m φr w0 (x) dx
(23-13)
m φr ẇ0 ( x ) dx
Unit 23 - 15
MIT - 16.20
Fall, 2002
Finally, can add the case of…
Forced Vibration
Again… response is made up of the natural modes
• Break up force into series of spatial impulses
• Use Duhamel’s (convolution) integral to get response for each
normalized mode
1
ξ r (t ) =
Mr ω r
t
∫0 Ξ
r
( τ) sin ω r (t − τ) dτ
(23-14)
• Add up responses (equation 23-11) for all normalized modes
(Linear ⇒ Superposition)
What about the special case of…
--> Sinusoidal Force at point xA
Paul A. Lagace © 2001
Unit 23 - 16
MIT - 16.20
Fall, 2002
Figure 23.4
Representation of force at point xA on simply-supported
beam
F (t ) = Fo sin Ω t
As for single degree-of-freedom system, for each normal mode get:
ξr ( t ) =
φr ( x A ) Fo
sin Ω t
2

Ω 
Mr ω r 2  1 −

ωr2 

for steady state response (Again, initial transient of sin ωrt dies out
due to damping)
Add up all responses…
Paul A. Lagace © 2001
Unit 23 - 17
MIT - 16.20
Fall, 2002
Note:
• Resonance can occur at any ωr
• DMF (Dynamic Magnification Factor) associated with each normal
mode
--> Can apply technique to any system.
• Get governing equation including inertial terms
• Determine Free Vibration Modes and frequencies
• Transform equation to uncoupled single degree-of-freedom system
(normal equations)
• Solve each normal equation separately
• Total response equal to sum of individual responses
Modal superposition is a very powerful technique!
Paul A. Lagace © 2001
Unit 23 - 18