BENDING FREQUENCIES OF BEAMS, RODS, AND PIPES
Revision M
By Tom Irvine
Email: tomirvine@aol.com
September 13, 2010
Introduction
The fundamental frequencies for typical beam configurations are given in Table 1.
Higher frequencies are given for selected configurations.
Table 1. Fundamental Bending Frequencies
Configuration
Frequency (Hz)
f1 =
Cantilever
1  3.5156  EI


2  L2  
f2 = 6.268 f1
f3 = 17.456 f1
Cantilever with
End Mass m
Simply-Supported
at both Ends
(Pinned-Pinned)
Free-Free
f1 = 1
2
3EI
 0.2235  L  m  L3
2
fn =
1 n 
2  L 
f1 =
1  22.373  EI


2  L2  
EI
, n=1, 2, 3, ….

f2 = 2.757 f1
f3 = 5.404 f1
Fixed-Fixed
Same as Free-Free
Fixed - Pinned
f1 =
1 15.418  EI


2  L2  
1
where
E
I
L

is the modulus of elasticity
is the area moment of inertia
is the length
is the mass density (mass/length)
Note that the free-free and fixed-fixed have the same formula.
The derivations and examples are given in the appendices per Table 2.
Table 2. Table of Contents
Appendix
Title
Mass
Solution
A
Cantilever Beam I
End mass. Beam mass
is negligible
Approximate
B
Cantilever Beam II
Beam mass only.
Approximate
C
Cantilever Beam III
Approximate
D
Cantilever Beam IV
Both beam mass and
the end mass are
significant
Beam mass only.
E
Beam SimplySupported at Both
Ends I
Beam SimplySupported at Both
Ends II
Free-Free Beam
Center mass. Beam
mass is negligible.
Approximate
Beam mass only
Eigenvalue
Beam mass only
Eigenvalue
Steel Pipe example,
Simply Supported
and Fixed-Fixed
Cases
Rocket Vehicle
Example, Free-free
Beam
Fixed-Fixed Beam
Beam mass only
Approximate
Beam mass only
Approximate
Beam mass only
Eigenvalue
F
G
H
I
J
Eigenvalue
Reference
1. T. Irvine, Application of the Newton-Raphson Method to Vibration Problems,
Vibrationdata Publications, 1999.
2
APPENDIX A
Cantilever Beam I
Consider a mass mounted on the end of a cantilever beam. Assume that the end-mass is
much greater than the mass of the beam.
g
EI
m
L
Figure A-1.
E
I
L
g
m
is the modulus of elasticity.
is the area moment of inertia.
is the length.
is gravity.
is the mass.
The free-body diagram of the system is
L
MR
mg
R
Figure A-2.
R is the reaction force.
MR is the reaction bending moment.
Apply Newton’s law for static equilibrium.

 forces  0
(A-1)
3
R - mg = 0
(A-2)
R = mg
(A-3)
At the left boundary,

 moments  0
(A-4)
MR - mg L = 0
(A-5)
MR = mg L
(A-6)
Now consider a segment of the beam, starting from the left boundary.
V
y
x
MR
M
R
Figure A-3.
V is the shear force.
M is the bending moment.
y is the deflection at position x.
Sum the moments at the right side of the segment.

 moments  0
(A-7)
MR - R x - M = 0
(A-8)
M = MR - R x
(A-9)
The moment M and the deflection y are related by the equation
4
M  EI y 
(A-10)
EI y   M R  R x
(A-11)
EI y   mgL  mg x
(A-12)
EI y   mg  L  x
(A-13)
 mg 
 L  x
y   
 EI 
(A-14)
Integrating,

 x2  
 mg  
y     Lx      a
 EI  
 2  

(A-15)
Note that “a” is an integration constant.
Integrating again,
2
3 

 mg    x   x  
y( x)    L      ax  b
 EI    2   6  


(A-16)
A boundary condition at the left end is
y(0) = 0
(zero displacement)
(A-17)
Thus
b=0
(A-18)
Another boundary condition is
y'  0  0
(zero slope)
(A-19)
5
Applying the boundary condition to equation (A-16) yields,
a=0
(A-20)
The resulting deflection equation is
2
3 

 mg    x   x  
y( x)    L    
 EI    2   6  


(A-21)
The deflection at the right end is
2
3 

 mg    L   L  
y( L)    L    
 EI    2   6  


(A-22)
 mgL3 

y( L)  
 3EI 
(A-23)
Recall Hooke’s law for a linear spring,
F=ky
(A-24)
F is the force.
k is the stiffness.
The stiffness is thus
k=F/y
(A-25)
The force at the end of the beam is mg. The stiffness at the end of the beam is




 mg 
k 

  mgL3  
  3EI  
 
 
(A-26)
6
k
3EI
L3
(A-27)
The formula for the natural frequency fn of a single-degree-of-freedom system is
fn 
1 k
2 m
(A-28)
The mass term m is simply the mass at the end of the beam. The natural frequency of the
cantilever beam with the end-mass is found by substituting equation (A-27) into (A-28).
fn 
1 3EI
2 mL3
(A-29)
7
APPENDIX B
Cantilever Beam II
Consider a cantilever beam with mass per length . Assume that the beam has a uniform
cross section. Determine the natural frequency. Also find the effective mass, where the
distributed mass is represented by a discrete, end-mass.
EI, 
L
Figure B-1.
The governing differential equation is
 EI
4 y
2 y

x 4
t 2
(B-1)
The boundary conditions at the fixed end x = 0 are
y(0) = 0
(zero displacement)
dy
 0
dx x 0
(zero slope)
(B-2)
(B-3)
The boundary conditions at the free end x = L are
d2y
dx 2
 0
(zero bending moment)
(B-4)
x L
d 3y
dx 3
 0
(zero shear force)
x L
Propose a quarter cosine wave solution.
8
(B-5)

 x  
y( x)  y o 1  cos  
 2L  

(B-6)
dy
    x 
 y o   sin 
 2L  2L 
dx
(B-7)
d2y
  2
 x 
 y o   cos 
 2L 
 2L 
dx 2
(B-8)
d 3y
 x  3  x 
  y o   sin 
 2L 
 2L 
dx 3
(B-9)
The proposed solution meets all of the boundary conditions expect for the zero shear
force at the right end. The proposed solution is accepted as an approximate solution for
the deflection shape, despite one deficiency.
The Rayleigh method is used to find the natural frequency. The total potential energy and
the total kinetic energy must be determined.
The total potential energy P in the beam is
2
EI L  d 2 y 
P
dx
2 0  dx 2 
(B-10)
By substitution,
EI L
P
2 0
2
   2
 x  
y o   cos   dx
 2L  
  2 L 

EI     2 
y   
P
2  o  2 L  
EI     2 
y   
P
2  o  2 L  
2
2
(B-11)
  x   2
cos 2 L   dx


(B-12)
L  1 
 x  
 2  1  cos L  dx


(B-13)
L
0
0
9
2
EI     2   1    L   x  
y   
P
x    sin 
2  o  2 L    2       L  
L
(B-14)
0
 4 
EI
2  
P
y
L
2 o  32 L4 


(B-15)
1 4  EI 
2
   yo
64
 L3 
(B-16)
 
P
 
The total kinetic energy T is
T 
1
L 2
  2n   y dx
0
2
T 
1
L
  2n 
0
2
2
 
 x   
y
1

cos
dx


 o
 2 L   
 

(B-18)
T 
1
 x 
 x  
2 L
  2n y o  1  2 cos   cos2   dx
 2L 
 2L  
0 
2
(B-19)
T 
1
 x 
 x  
2 L
  2n y o  1  2 cos   cos2    dx


 2L  
0 
2
2L
(B-20)
T 
1
 x  1 1  x  
2 L
  2n y o  1  2 cos    cos  dx
 2L  2 2  L  
0 
2
(B-21)
(B-17)
 
 
 
1
 x 
 x  
2 L 3
  2n y o    2 cos   cos  dx
 2L
 L 
0 2
2
(B-22)
L
1
 4 L   x   L   x  
2 3
2
T    n y o  x    sin     sin  
    2L      L  
2
2
0
(B-23)
T 
 
 
T 
1
 4L 
2 3
  2n y o  L    
  
2
2
 
(B-24)
10
T 
1
2   8
  2n y o L 3    
4
   
 
(B-25)
Now equate the potential and the kinetic energy terms.
1
2   8   1 4  EI 
2
  2n y o L 3     
   yo
3


4
  64

L 
 
 
(B-26)
 EI 
  8 1
  2n L 3       4  
     16
 L3 
(B-27)

 EI  
 4   

 L3  
 2n  

16L 3   8   
    




(B-28)
1/2

 EI  
4
    

 L3  
n  

16L 3   8   
    




(B-29)
1/2

 EI  
4
    
 1 
 L4  
f n   

  8
 2  
16 3    

     

(B-30)
1/2

 EI  
4
    
 1 
 L4  
f n   

  8
 2  
16 3    

     

(B-31)
11

1/2

2 
EI

 1   
f n   



 2   4 L2    8   
 3  
      
 1  3.664  EI
f n   

 2   L2  
(B-32)
(B-33)
Recall that the stiffness at the free of the cantilever beam is
k
3EI
L3
(B-34)
The effective mass meff at the end of the beam is thus
m eff 
meff 
k
(B-35)
2 fn2
3EI
  1   3.664  EI  2
3
L 2    


  2    L2   
m eff 
3EI
(B-36)
(B-37)
 EI 
L3
13.425  

L4
meff  0.2235  L
(B-38)
12
APPENDIX C
Cantilever Beam III
Consider a cantilever beam where both the beam mass and the end-mass are significant.
g
EI, 
m
L
Figure C-1.
The total mass mt can be calculated using equation (B-38).
m t  0.2235L  m
(C-1)
Again, the stiffness at the free of the cantilever beam is
k
3EI
L3
(C-2)
The natural frequency is thus
fn 
1
2
3EI
(C-3)
 0.2235L  m  L3
13
APPENDIX D
Cantilever Beam IV
This is a repeat of part II except that an exact solution is found for the differential
equation. The differential equation itself is only an approximation of reality, however.
EI, 
L
Figure D-1.
The governing differential equation is
4 y
2 y
 EI

x 4
t 2
(D-1)
Note that this equation neglects shear deformation and rotary inertia.
Separate the dependent variable.
y( x, t )  Y( x)T(t )
 EI
(D-2)
 4 Y( x)T( t )
 2 Y( x)T( t )

x 4
t 2




 d4

 d2

 EI T(t ) 
Y ( x )    Y( x ) 
T(t ) 
4
2




 dx

 dt

14
(D-3)
(D-4)
 d 4

 4 Y( x ) 

  EI   dx



Y( x )
  
 d 2

 2 T(t ) 
 dt

T(t )
(D-5)
Let c be a constant
 d 4

 4 Y( x) 

  EI   dx



Y( x )
  
 d 2

 2 T(t ) 
 dt

  c2
T(t )
(D-6)
Separate the time variable.
 d 2

T(t
)
 2

 dt

  c2
T(t )
(D-7)
d2
T(t )  c 2 T(t )  0
2
dt
(D-8)
Separate the spatial variable.
 d 4

Y( x ) 

  EI   dx 4

  c2



Y
(
x
)


(D-9)
d4

Y( x)  c 2  Y( x)  0
 EI 
dx 4
(D-10)
A solution for equation (D-10) is
Y( x)  a1 sinhx  a 2 coshx  a 3 sinx  a 4 cosx
15
(D-11)
dY( x)
 a1 cosh x  a 2  sinh x  a 3 cos x  a 4  sin x
dx
(D-12)
d 2 Y( x)
 a1 2 sinh x  a 2 2 cosh x  a 3 2 sin x  a 4 2 cos x
2
dx
(D-13)
d 3 Y( x)
 a1 3 cosh x  a 2 3 sinh x  a 3 3 cos x  a 4 3 sin x
3
dx
(D-14)
d 4 Y( x)
 a1 4 sinh x  a 2 4 cosh x  a 3 4 sin x  a 4 4 cos x
4
dx
(D-15)
Substitute (D-15) and (D-11) into (D-10).
a14 sinhx  a 24 coshx  a 34 sinx  a 44 cosx

 c 2   a1 sinh x  a 2 cosh x  a 3 sin x  a 4 cos x  0
 EI 


(D-16)


 4 a1 sinh x  a 2 cosh x  a 3 sin x  a 4 cos x

 c 2   a1 sinh x  a 2 cosh x  a 3 sin x  a 4 cos x  0
 EI 


(D-17)
The equation is satisfied if

4  c2  
 EI 
(D-18)
 2  1/4
  c

 EI 
(D-19)
16
The boundary conditions at the fixed end x = 0 are
Y(0) = 0
(zero displacement)
dY
 0
dx x 0
(zero slope)
(D-20)
(D-21)
The boundary conditions at the free end x = L are
d 2Y
dx 2
d 3Y
dx 3
 0
(zero bending moment)
(D-22)
 0
(zero shear force)
(D-23)
x L
x L
Apply equation (D-20) to (D-11).
a2  a4  0
(D-24)
a4   a2
(D-25)
Apply equation (D-21) to (D-12).
a1  a 3  0
(D-26)
a 3   a1
(D-27)
Apply equation (D-22) to (D-13).
a1 sinh L  a 2 coshL  a 3 sinL  a 4 cosL  0
(D-28)
Apply equation (D-23) to (D-14).
a1 cosh L  a 2 sinh L  a 3 cos L  a 4 sinL  0
(D-29)
Apply (D-25) and (D-27) to (D-28).
a1 sinhL  a 2 coshL  a1 sin L  a 2 cosL  0
17
(D-30)




a1 sinL  sinhL  a 2 cosL  coshL  0
(D-31)
Apply (D-25) and (D-27) to (D-29).
a1 cosh L  a 2 sinhL  a1 cos L  a 2 sin L  0



(D-32)

(D-33)
0
 
 
0
(D-34)
a1 cosL  coshL  a 2  sinL  sinhL  0
Form (D-31) and (D-33) into a matrix format.
 sin L  sinh L


cos L  cosh L

cos L  cosh L   a1 
 
  
 sin L  sinh L  a 2 
By inspection, equation (D-34) can only be satisfied if a1 = 0 and a2 = 0. Set the
determinant to zero in order to obtain a nontrivial solution.
 sin2 L  sinh2 L

2  0
 cos L  cosh L
(D-35)
 sin2 L  sinh2 L  cos2 L  2 cosL coshL  cosh2 L  0
(D-36)
 sin 2 L  sinh 2 L  cos2 L  2 cosL coshL  cosh 2 L  0
(D-37)
 2  2 cosL coshL  0
(D-38)
1  cosL coshL  0
(D-39)
cosL coshL  1
(D-40)
There are multiple roots which satisfy equation (D-40). Thus, a subscript should be
added as shown in equation (D-41).
18
cos n L cosh n L  1
(D-41)
The subscript is an integer index. The roots can be determined through a combination of
graphing and numerical methods. The Newton-Rhapson method is an example of an
appropriate numerical method. The roots of equation (D-41) are summarized in Table D1, as taken from Reference 1.
Table D-1. Roots
Index
n L
n=1
1.87510
n=2
4.69409
n>3
(2n-1)/2
Note: the root value formula for n > 3 is approximate.
Rearrange equation (D-19) as follows
 EI 
c2   n4  

(D-42)
Substitute (D-42) into (D-8).

 EI  
d2
T(t )   n 4    T(t )  0
  

dt 2
(D-43)
Equation (D-43) is satisfied by


EI  
EI  
 t   b 2 cos  n 2
 t
T(t )  b1 sin   n 2
  
  


19
(D-44)
The natural frequency term n is thus
EI

 n   n2
(D-45)
Substitute the value for the fundamental frequency from Table D-1.
.
187510
 2 EI
1  

 L 
f1 
(D-46)
1  3.5156  EI


2  L2  
(D-47)
Substitute the value for the second root from Table D-1.
 4.69409 
2  

 L2 
f2 
2
EI

(D-48)
1  22.034  EI


2  L2  
(D-49)
f 2  6.268 f1
(D-50)
Compare equation (D-47) with the approximate equation (B-33).
SDOF Model Approximation
The effective mass meff at the end of the beam for the fundamental mode is thus
m eff 
k

2  fn

(D-51)
2
20
m eff 
3EI
2
 EI 
.
  1   35156
3
L 2    


  2    L2   
m eff 
3EI
3
 EI 
L
12.3596  

L4
meff  0.2427  L
(D-52)
(D-53)
(SDOF Approximation)
(D-54)
Eigenvalues
n
n L
1
1.875104
2
4.69409
3
7.85476
4
10.99554
5
(2n-1)/2
Note that the root value formula for n > 5 is approximate.
Normalized Eigenvectors
Mass normalize the eigenvectors as follows
L
0 Yn
2 ( x ) dx  1
(D-55)
21
The calculation steps are omitted for brevity. The resulting normalized eigenvectors are
 1 
Y1 ( x )  
 cosh 1 x   cos1 x   0.73410 sinh 1 x   sin 1 x  
 L 
(D-56)
 1 
Y2 ( x )  
cosh  2 x   cos 2 x   1.01847 sinh  2 x   sin  2 x  
 L 
(D-57)
 1 
Y3 ( x )  
cosh  3 x   cos 3 x   0.99922 sinh  3 x   sin  3 x  
 L 
(D-58)
 1 
Y4 ( x )  
cosh  4 x   cos 4 x   1.00003 sinh  4 x   sin  4 x  
 L 
(D-59)
Participation Factors
The participation factors for constant mass density are
L
0
n   Yn (x) dx
(D-60)
The participation factors from a numerical calculation are
1  0.7830 L
(D-61)
2  0.4339 L
(D-62)
3  0.2544 L
(D-63)
4  0.1818 L
(D-64)
The participation factors are non-dimensional.
22
Effective Modal Mass
The effective modal mass is
2
 L

m
(
x
)
Y
(
x
)
dx
n
 0


m eff , n  
L
2
0 m(x) Yn (x) dx
(D-65)
The eigenvectors are already normalized such that
m(x) Yn (x)2 dx  1
(D-66)
2
 L

2
m eff , n  n     m( x ) Yn ( x )dx 
 0

(D-67)
L
0
Thus,
The effective modal mass values are obtained numerically.
m eff , 1  0.6131 L
(D-68)
m eff , 2  0.1883 L
(D-69)
m eff , 3  0.06474 L
(D-70)
m eff , 4  0.03306 L
(D-71)
23
APPENDIX E
Beam Simply-Supported at Both Ends I
Consider a simply-supported beam with a discrete mass located at the middle. Assume
that the mass of the beam itself is negligible.
g
EI
m
L1
L1
L
Figure E-1.
The free-body diagram of the system is
L
L1
Ra
L1
Rb
mg
Figure E-2.
Apply Newton’s law for static equilibrium.

 forces  0
(E-1)
Ra + Rb - mg = 0
(E-2)
Ra = mg - Rb
(E-3)
24
At the left boundary,
 moments  0
(E-4)
Rb L - mg L1 = 0
(E-5)
Rb = mg ( L1 / L )
(E-6a)

Rb = (1/2) mg
(E-6b)
Substitute equation (E-6) into (E-3).
Ra = mg – (1/2)mg
(E-7)
Ra = (1/2)mg
(E-8)
x
V
y
L1
Ra
mg
M
Sum the moments at the right side of the segment.

 moments  0
(E-9)
- Ra x + mg <x-L1 > - M = 0
(E-10)
25
Note that < x-L1> denotes a step function as follows
for x  L1
 0,

 x  L1   
 x  L , for x  L
1
1

(E-11)
M = - Ra x + mg <x-L1 >
(E-12)
M = - (1/2)mg x + mg <x-L1 >
(E-13)
M = [ - (1/2) x + <x-L1 > ][ mg ]
(E-14)
EI y  [ - (1/2) x   x - L1  ][ mg ]
(E-15)
 mg 
y  [ - (1/2) x   x - L1  ]
 EI 
(E-16)
y  [ -
1 2 1
 mg 
x   x - L1  2 ]
a
4
2
 EI 
1
 1
  mg 
y( x )  - x 3   x - L1  3  
 ax  b
6
 12
  EI 
(E-17)
(E-18)
The boundary condition at the left side is
y(0) = 0
(E-19a)
This requires
b=0
(E-19b)
Thus
1
 1
  mg 
y( x )  - x 3   x - L1  3  
 ax
6
 12
  EI 
(E-20)
The boundary condition on the right side is
y(L) = 0
(E-21)
26
 1 3 1
3   mg 
- 12 L  6  L - L1    EI   aL  0
(E-22)
1 3   mg 
 1 3
- 12 L  48 L   EI   aL  0
(E-23)
1 3   mg 
 4 3
- 48 L  48 L   EI   aL  0
(E-24)
 3 3   mg 
- 48 L   EI   aL  0
(E-25)
 1 3   mg 
- 16 L   EI   aL  0
(E-26)
 1 3   mg 
aL 
L 
 16
  EI 
(E-27)
 1 2   mg 
a 
L 
 16
  EI 
(E-28)
Now substitute the constant into the displacement function
1
 1
  mg   1 2   mg 
x 
y( x )  - x 3   x - L1  3  

L
6
 12
  EI  16   EI 
(E-29)
1
1
 1
  mg 
y( x )  - x 3  xL2   x - L1  3  
16
6
 12
  EI 
(E-30)
The displacement at the center is
3 1 L

 mg
1 L
L  1 L
 


y   -     L2     - L1  3  
16  2 
6 2
  EI 
 2   12  2 

27
(E-31)
1   mgL 3 
L  1

y    
 2   96 32   EI 
(E-32)
3   mgL 3 
L  1

y    
 2   96 96   EI 
(E-33)
3
 L   2   mgL 

y     
 2   96   EI 
(E-34)
3
 L   1   mgL 

y     
 2   48   EI 
(E-35)
Recall Hooke’s law for a linear spring,
F=ky
(E-36)
F is the force.
k is the stiffness.
The stiffness is thus
k=F/y
(E-37)
The force at the center of the beam is mg. The stiffness at the center of the beam is




 mg 
k 

  mgL 3  


48
EI

 
 

k
(E-38)
48 EI
L3
(E-39)
The formula for the natural frequency fn of a single-degree-of-freedom system is
28
 1 k
fn   
 2  m
(E-40)
The mass term m is simply the mass at the center of the beam.
 1  48 EI
fn   
 2  mL3
(E-41)
EI
 1 
fn   6.928
 2 
mL3
(E-42)
29
APPENDIX F
Beam Simply-Supported at Both Ends II
Consider a simply-supported beam as shown in Figure F-1.
EI,
L
Figure F-1.
Recall that the governing differential equation is
4 y
2 y
 EI

x 4
t 2
(F-1)
The spatial solution from section D is
Y( x)  a1 sinhx  a 2 coshx  a 3 sinx  a 4 cosx
d 2 Y( x)
 a1 2 sinh x  a 2 2 cosh x  a 3 2 sin x  a 4 2 cos x
2
dx
(F-2)
(F-3)
The boundary conditions at the left end x = 0 are
Y(0) = 0
d 2Y
dx 2 x  0
(zero displacement)
 0
(zero bending moment)
30
(F-4)
(F-5)
The boundary conditions at the free end x = L are
Y(L) = 0
d 2Y
dx 2
(zero displacement)
 0
(zero bending moment)
(F-6)
(F-7)
x L
Apply boundary condition (F-4) to (F-2).
a2  a4  0
(F-8)
a4   a2
(F-9)
Apply boundary condition (F-5) to (F-3).
a2  a4  0
(F-10)
a2  a4
(F-11)
Equations (F-8) and (F-10) can only be satisfied if
and
a2  0
(F-12)
a4  0
(F-13)
The spatial equations thus simplify to
Y( x)  a1 sinhx  a 3 sinx
d 2 Y( x)
 a1 2 sinh x  a 3 2 sin x
2
dx
(F-14)
(F-15)
Apply boundary condition (F-6) to (F-14).
a1 sinhL  a 3 sinL  0
31
(F-16)
Apply boundary condition (F-7) to (F-15).
a1 2 sinhL  a 3 2 sin L  0
(F-17)
a1 sinh L  a 3 sin L  0
(F-18)
sinh L


sinh L

sin L   a 1 
 
  
 sin L  a 3 
0
 
 
0
(F-19)
By inspection, equation (F-19) can only be satisfied if a1 = 0 and a3 = 0. Set the
determinant to zero in order to obtain a nontrivial solution.
 sinL sinhL  sinL sinhL  0
(F-20)
 2 sin L sinh L  0
(F-21)
sin L sinh L  0
(F-22)
Equation (F-22) is satisfied if
 n L  n, n  1, 2, 3,....
(F-23)
n
, n  1, 2, 3,....
L
(F-24)
n 
The natural frequency term n is
EI

(F-25)
2
 n  EI
n   
, n  1, 2, 3,...

L
(F-26)
 n   n2
 1   n 
fn     
 2   L 
2 EI

, n  1, 2, 3,...
32
(F-27)
2
 1   n  EI
fn     
, n  1, 2, 3,...

 2   L 
(F-28)
SDOF Approximation
Now calculate effective mass at the center of the beam for the fundamental frequency.

1   
 L
2 EI
(F-29)

Recall the natural frequency equation for a single-degree-of-freedom system.
k
m
1 
(F-30)
Recall the beam stiffness at the center from equation (E-39).
k
48EI
(F-31)
L3
Substitute equation (F-31) into (F-30).
1 
48EI
(F-32)
mL3
Substitute (F-32) into (F-29).

 
3
L
mL
48EI
2
EI

(F-33)
4
   EI
 
mL3  L  
(F-34)
4
 1
 
mL3  L  
(F-35)
48EI
48
33
1  4 


m  48 L 
(F-36)
The effective mass at the center of the beam for the first mode is
m
48 L
(SDOF Approximation)
4
(F-37)
Normalized Eigenvectors
The eigenvector and its second derivative at this point are
Y( x)  a1 sinhx  a 3 sinx
(F-38)
d 2 Y( x )
 a1 2 sinh x   a 3 2 sin x 
2
dx
(F-39)
The eigenvector derivation requires some creativity. Recall
Y(L) = 0
d 2Y
dx 2
(zero displacement)
 0
(zero bending moment)
(F-40)
(F-41)
x L
Thus,
d 2Y
dx 2
Y0
for x=L and  n L  n, n=1,2,3, …
(F-42)
2
2


1   n   a sinh n  1   n   a sin  n  0 , n=1,2,3, …
  L  1
  L  3




(F-43)
The sin(n) term is always zero. Thus a1 = 0.
34
The eigenvector for all n modes is
Yn ( x )  a n sin nx / L
(F-44)
Mass normalize the eigenvectors as follows
L
0 Yn
2 ( x ) dx  1
(F-45)
a n 2  sin 2 nx / L dx  1
(F-46)
a n 2 L
1  cos( 2 n x / L)  1
2 0
(F-47)
L

a n 2 
1
sin( 2nx / L)  1
x 
2 
2 n
0
(F-48)
L
0
a n 2 L
2
1
(F-49)
an2 
2
L
(F-50)
an 
2
L
(F-51)
Yn ( x ) 
2
sin n x / L 
L
(F-52)
35
Participation Factors
The participation factors for constant mass density are
L
0
n   Yn (x) dx
L
0
n   
(F-53)
2
sin n x / L  dx
L
2 L
sin n x / L dx
L 0
n 
2
L
(F-53)
(F-54)
L
L
 n  cosn x / L  0
(F-55)
1
n   2 L   cosn   1 , n=1, 2, 3, ….
 n 
(F-56)
n  
Effective Modal Mass
The effective modal mass is
2
 L

 0 m( x ) Yn ( x )dx 

m eff , n  
L
2
0 m(x) Yn (x) dx
(F-57)
The eigenvectors are already normalized such that
L
2
0 m(x) Yn (x) dx  1
36
(F-58)
Thus,
2
 L

m eff , n  n 2    m( x ) Yn ( x )dx 
 0



1
m eff , n   2 L   cosn   1
 n 


m eff , n  2 L
1
n2
 cosn   12
37
(F-59)
2
(F-60)
, n=1, 2, 3, ….
(F-61)
APPENDIX G
Free-Free Beam
Consider a uniform beam with free-free boundary conditions.
EI, 
L
Figure G-1.
The governing differential equation is
4 y
2 y
 EI

x 4
t 2
(G-1)
Note that this equation neglects shear deformation and rotary inertia.
The following equation is obtain using the method in Appendix D
d4

Y( x)  c 2  Y( x)  0
4
 EI 
dx
(G-2)
The proposed solution is
Y( x)  a1 sinhx  a 2 coshx  a 3 sinx  a 4 cosx
(G-3)
dY( x)
 a1 cosh x  a 2  sinh x  a 3 cos x  a 4  sin x
dx
(G-4)
d 2 Y( x)
 a1 2 sinh x  a 2 2 cosh x  a 3 2 sin x  a 4 2 cos x
2
dx
(G-5)
d 3 Y( x )
 a13 cosh x   a 23 sinh x   a 33 cosx   a 43 sin x 
dx 3
(G-6)
38
Apply the boundary conditions.
d 2Y
dx 2
0
(zero bending moment)
x 0
a2  a4  0
(G-8)
a4  a2
d 3Y
dx 3
(G-7)
0
(G-9)
(zero shear force)
(G-10)
x 0
a1  a 3  0
(G-11)
a 3  a1
(G-12)
d 2 Y( x )
 a1 2 sinh x   sin x   a 2 2 cosh x   cosx 
dx 2
(G-13)
d3 Y( x )
 a13cosh x   cosx   a 23sinh x   sin x 
3
dx
(G-14)
d 2Y
dx 2
 0
(zero bending moment)
x L
39
(G-15)
a1sinh L  sin L  a 2coshL  cosL  0
d 3Y
dx 3
 0
(zero shear force)
(G-16)
(G-17)
x L
a1coshL  cosL  a 2sinh L  sin L  0
(G-18)
Equation (G-16) and (G-18) can be arranged in matrix form.
 sinh L   sin L 


cosh L   cosL 
cosh L   cosL   a1  0
    
   
sinh L   sin L   a 2  0
(G-19)
Set the determinant equal to zero.
sinh L   sin Lsinh L   sin L   cosh L   cosL2  0
(G-20)
sinh 2 L   sin 2 L   cosh 2 L   2 cosh L cosL   cos 2 L   0
(G-21)
 2 coshLcosL  2  0
(G-22)
cosh LcosL  1  0
(G-23)
The roots can be found via the Newton-Raphson method, Reference 1. The first root is
L  4.73004
(G-24)
40
 n   n2
 4.73004 
 1 

 L 
EI

(G-25)
EI

(G-26)
2
 22.373  EI
 1 

 L2  
(G-27)
The second root is
L  7.85320
(G-28)
2 EI
 n  n
(G-29)

 7.85320 
 2 

 L 
2
EI

(G-30)
 61.673  EI
 2 

 L2  
(G-31)
 2  2.757 1
(G-32)
 L  10.9956
(G-33)
The third root is
2 EI
 n  n
(G-34)

10.9956 
 3 

 L 
2
EI

(G-35)
41
120.903  EI
 3 

 L2  
(G-36)
 3 5.404  1
(G-37)
Equation (G-18) can be expressed as
  cosh L   cosL 
a 2  a1 

 sinh L   sin L  
(G-38)
Recall
a4  a2
(G-39)
a 3  a1
(G-40)
The displacement mode shape is thus
Y(x)  a1sinh x   sin x   a 2coshx   cosx 
(G-41)


  cosh L  cosL
cosh x   cosx 
Y( x )  a1 sinh x   sin x   

 sinh L  sin L 


(G-42)
The first derivative is


  cosh L  cosL
dy
sinh x   sin x 
 a1  cosh x   cosx   

dx
 sinh L  sin L 


42
(G-43)
The second derivative is


  cosh L   cosL  

 a1 2  sinh x   sin x   
cosh x   cosx 

 sinh L   sin L  
dx 2


d2y
(G-44)
43
APPENDIX H
Pipe Example
Consider a steel pipe with an outer diameter of 2.2 inches and a wall thickness of 0.60
inches. The length is 20 feet. Find the natural frequency for two boundary condition
cases: simply-supported and fixed-fixed.
The area moment of inertia is
I


Do 4  D i 4
64

(H-1)
D o  2.2 in
(H-2)
D i  2.2  2(0.6) in
(H-3)
D i  2.2  1.2 in
(H-4)
D i  1.0 in
(H-5)
I


2.2 4  1.0 4
32

in 4
(H-6)
I  1.101 in 4
(H-7)
The elastic modulus is
 
E  30 10 6
lbf
(H-8)
in 2
The mass density is
  mass per unit length.

(H-9)


lbm   

  0.282
2.2 2  1.0 2 in 2 
3   4

in 

44
(H-10)
lbm
in
  0.850
(H-11)
 
 1 slug ft / sec 2   12 in 


1.101 in 4 
2

  1 ft 
1
lbf
in




lbm
1 slug


0.850
in  32.2 lbm 
lbf
30 106
EI


(H-12)
 
EI
 1.225 105

in 2
sec
(H-13)
The natural frequency for the simply-supported case is
2
 1   n  EI
fn     
, n  1, 2, 3,...

 2   L 
(H-14)
2





 1 
 1.225 105
f1   


2



12
in
 
 
 20 ft 

 1ft  
 
f1  3.34 Hz
in 2
sec
(simply-supported)
(H-15)
(H-16)
The natural frequency for the fixed-fixed case is
 1   22.37  EI
f1    

 2   L2  
(H-17)
45






1
22
.
37
 
5
f1    
 1.225 10
2
 2   
 12 in  
 
 20 ft 
 
 1ft  
 
f1  7.58 Hz
in 2
sec
(fixed-fixed)
(H-18)
(H-19)
46
APPENDIX I
Suborbital Rocket Vehicle
Consider a rocket vehicle with the following properties.
mass = 14078.9 lbm (at time = 0 sec)
L = 372.0 inches.

14078.9 lbm
372.0 inches
  37.847
lbm
in
The average stiffness is
6
EI = 63034 (10 ) lbf in^2
The vehicle behaves as a free-free beam in flight. Thus
f1 
f1 
1  22.37  EI


2  L2  
1  22.37 


2  372 in 2 


(I-1)
.2 lbm 
63034e  06 lbf in 2  slug ftlbf/ sec 2  12ftin   32slugs



37.847

lbm
in
(I-2)
f1 = 20.64 Hz
(at time = 0 sec)
(I-3)
Note that the fundamental frequency decreases in flight as the vehicle expels propellant
mass.
47
APPENDIX J
Fixed-Fixed Beam
Consider a fixed-fixed beam with a uniform mass density and a uniform cross-section.
The governing differential equation is
 EI
4 y
2 y

x 4
t 2
(J-1)

Y( x )  c 2   Y( x )  0
 EI 
(J-2)
The spatial equation is
4
x 4
The boundary conditions for the fixed-fixed beam are:
Y(0) = 0
(J-3)
dY( x )
0
dx x  0
(J-4)
Y(L)=0
(J-5)
dY( x )
0
dx x  L
(J-6)
The eigenvector has the form
Y( x)  a1 sinhx  a 2 coshx  a 3 sinx  a 4 cosx
dY( x)
 a1 cosh x  a 2  sinh x  a 3 cos x  a 4  sin x
dx
48
(J-7)
(J-8)
d 2 Y( x)
 a1 2 sinh x  a 2 2 cosh x  a 3 2 sin x  a 4 2 cos x
2
dx
(J-9)
Y(0) = 0
(J-10)
a2+a4=0
(J-11)
-a2=a4
(J-12)
dY( x )
0
dx x  0
(J-13)
a1 + a3  = 0
(J-14)
a1 + a3 = 0
(J-15)
- a1 = a3
(J-16)
Y(x)  a1sinh x   sin x   a 2 coshx   cosx 
(J-17)
dY( x )
 a1cosh x   cosx   a 2sinh x   sin x 
dx
(J-18)
Y(L) = 0
(J-19)
a1sinh L  sin L  a 2 coshL  cosL  0
dY( x )
0
dx x  L
(J-20)
(J-21)
a1coshL  cosL  a 2sinh L  sin L  0
49
(J-22)
a1coshL  cosL  a 2 sinh L  sin L  0
(J-23)
 sinh L  sin L cosh L  cosL  a1  0
cosh L  cosL sinh L  sin L  a   0

 2  
(J-24)
 sinh L   sin L  cosh L   cosL 
det 
 0
cosh L   cosL  sinh L   sin L  
(J-25)
[sinh L  sin L][sinh L  sin L]  [cosh L  cosL]2  0
(J-26)
sinh 2 L  sin 2 L  cosh 2 L  2 cosLcosh L  cos 2 L  0
(J-27)
2 cosLcosh L  2  0
(J-28)
cosLcoshL  1  0
(J-29)
The roots can be found via the Newton-Raphson method, Reference 1. The first root is
 L  4.73004
(J-30)
EI

(J-31)
2
n   n
 4.73004 
1  

 L 
2
EI

(J-32)
 22.373  EI
1  

 L2  
(J-33)
50
f1 
1  22.373  EI


2  L2  
(J-34)
a1coshL  cosL  a 2 sinh L  sin L
Let a 2 = 1
(J-36)
a1coshL  cosL  sinh L  sin L
a1 
(J-35)
 sinh L   sin L 
cosh L   cosL 
(J-37)
(J-38)
  sinh L  sin L
Y( x )  cosh x   cosx   
sinh x   sin x   0
 cosh L  cosL 
(J-39)
 sinh L  sin L 
Y( x )  cosh x   cosx   
sinh x   sin x   0
 cosh L  cosL
(J-40)
The mode shape for a fixed-fixed beam is
Yn (x)  coshn x   cosn x   n sinh n x   sin n x   0
(J-41)
where
 sinh L   sin L 
n  

 cosh L  cosL
(J-42)
51
The eigenvalues are
n
n L
1
4.73004
2
10.9956
3
14.13717
4
17.27876
The first derivative is
d
Yn ( x )   n sinh  n x   sin  n x    n  n cosh  n x   cos n x   0
dx
(J-43)
The second derivative is
d2
dx 2
Yn ( x )   n 2 cosh  n x   cos n x    n  n 2 sinh  n x   sin  n x   0
(J-44)
52