Home > BEAMS MODELED AS SINGLE-DEGREE-OF-FREEDOM SYSTEMS
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) |
| Cantilever | f1 = f2 = 6.268 f1 f3 = 17.456 f1 |
|
Cantilever with
End Mass m |
f1 = |
| Simply-Supported
at both Ends (Pinned-Pinned) |
fn = , n=1, 2, 3, …. |
| Free-Free | f1 = f2 = 2.757 f1 f3 = 5.404 f1 |
| Fixed-Fixed | Same as Free-Free |
| Fixed - Pinned | f1 = |
where
E is the modulus of elasticity
I is the area moment of inertia
L 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 | Both beam mass and the end mass are significant | Approximate |
| D | Cantilever Beam IV | Beam mass only. | Eigenvalue |
| E | Beam Simply-Supported at Both Ends I | Center mass. Beam mass is negligible. | Approximate |
| F | Beam Simply-Supported at Both Ends II | Beam mass only | Eigenvalue |
| G | Free-Free Beam | Beam mass only | Eigenvalue |
| H | Steel Pipe example, Simply Supported and Fixed-Fixed Cases | Beam mass only | Approximate |
| I | Rocket Vehicle Example, Free-free Beam | Beam mass only | Approximate |
| J | Fixed-Fixed Beam | Beam mass only | Eigenvalue |
Reference
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.
m
EI
g
L
Figure A-1.
E is the modulus of elasticity.
I is the area moment of inertia.
L is the length.
g is gravity.
m is the mass.
The free-body diagram of the
system is
mg
R
MR
L
Figure A-2.
R is the reaction force.
MR is the reaction bending moment.
Apply Newton’s law for static
equilibrium.
(A-1)
R - mg = 0
(A-2)
R = mg
(A-3)
At the left boundary,
(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
R
MR
M
x
y
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.
(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
(A-10)
(A-11)
(A-12)
(A-13)
(A-14)
Integrating,
(A-15)
Note that “a” is an integration
constant.
Integrating again,
(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
(zero slope) (A-19)
Applying the boundary condition
to equation (A-16) yields,
a = 0
(A-20)
The resulting deflection equation
is
(A-21)
The deflection at the right
end is
(A-22)
(A-23)
Recall Hooke’s law for a
linear spring,
F = k y
(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
(A-26)
(A-27)
The formula for the natural
frequency fn of a single-degree-of-freedom system is
(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).
(A-29)
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
(B-1)
The boundary conditions at
the fixed end x = 0 are
y(0) = 0
(zero displacement)
(B-2)
(zero slope)
(B-3)
The boundary conditions at
the free end x = L are
(zero bending moment)
(B-4)
(zero shear force)
(B-5)
Propose a quarter cosine wave
solution.
(B-6)
(B-7)
(B-8)
(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
(B-10)
By substitution,
(B-11)
(B-12)
(B-13)
(B-14)
(B-15)
(B-16)
The total kinetic energy T
is
(B-17)
(B-18)
(B-19)
(B-20)
(B-21)
(B-22)
(B-23)
(B-24)
(B-25)
Now equate the potential and
the kinetic energy terms.
(B-26)
(B-27)
(B-28)
(B-29)
(B-30)
(B-31)
(B-32)
(B-33)
Recall that the stiffness at
the free of the cantilever beam is
(B-34)
The effective mass meff at the end of the beam is thus
(B-35)
(B-36)
(B-37)
(B-38)
APPENDIX C
Cantilever Beam III
Consider a cantilever beam
where both the beam mass and the end-mass are significant.
m
g
EI,
L
Figure C-1.
The total mass mt
can be calculated using equation (B-38).
(C-1)
Again, the stiffness at the
free of the cantilever beam is
(C-2)
The natural frequency is thus
(C-3)
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
(D-1)
Note that this equation neglects
shear deformation and rotary inertia.
Separate the dependent variable.
(D-2)
(D-3)
(D-4)
(D-5)
Let c be a constant
(D-6)
Separate the time variable.
(D-7)
(D-8)
Separate the spatial variable.
(D-9)
(D-10)
(D-11)
(D-12)
(D-13)
(D-14)
(D-15)
Substitute (D-15) and (D-11)
into (D-10).
(D-16)
(D-17)
The equation is satisfied if
(D-18)
(D-19)
The boundary conditions at
the fixed end x = 0 are
Y(0) = 0
(zero displacement)
(D-20)
(zero slope)
(D-21)
The boundary conditions at
the free end x = L are
(zero bending moment)
(D-22)
(zero shear force)
(D-23)
Apply equation (D-20) to (D-11).
(D-24)
(D-25)
Apply equation (D-21) to (D-12).
(D-26)
(D-27)
Apply equation (D-22) to (D-13).
(D-28)
Apply equation (D-23) to (D-14).
(D-29)
Apply (D-25) and (D-27) to
(D-28).
(D-30)
(D-31)
Apply (D-25) and (D-27) to
(D-29).
(D-32)
(D-33)
Form (D-31) and (D-33) into
a matrix format.
(D-34)
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.
(D-35)
(D-36)
(D-37)
(D-38)
(D-39)
(D-40)
There are multiple roots which satisfy equation (D-40). Thus, a subscript should be added as shown in equation (D-41).
(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 D-1, 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
(D-42)
Substitute (D-42) into (D-8).
(D-43)
Equation (D-43) is satisfied
by
(D-44)
The natural frequency term
n
is thus
(D-45)
Substitute the value for the
fundamental frequency from Table D-1.
(D-46)
(D-47)
Substitute the value for the
second root from Table D-1.
(D-48)
(D-49)
(D-50)
Compare equation (D-47) with
the approximate equation (B-33).
SDOF
Model Approximation
The effective mass m
eff
at the end of the beam for
the fundamental mode is thus
(D-51)
(D-52)
(D-53)
(SDOF Approximation)
(D-54)
Eigenvalues
|
n |
|
| 1 | 1.875104 |
| 2 | |
| 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
(D-55)
The calculation steps are omitted
for brevity. The resulting normalized eigenvectors are
(D-56)
(D-57)
(D-58)
(D-59)
Participation Factors
The participation factors for
constant mass density are
(D-60)
The participation factors from
a numerical calculation are
(D-61)
(D-62)
(D-63)
(D-64)
The participation factors are
non-dimensional.
Effective Modal Mass
The effective modal mass is
(D-65)
The eigenvectors are already
normalized such that
(D-66)
Thus,
(D-67)
The effective modal mass values
are obtained numerically.
(D-68)
(D-69)
(D-70)
(D-71)
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
L
1
L
1
L
Figure E-1.
The free-body diagram of the
system is
L
Ra
L
1
L
1
Rb
mg
Figure E-2.
Apply Newton’s law for static
equilibrium.
(E-1)
Ra + Rb - mg = 0
(E-2)
Ra = mg - Rb
(E-3)
At the left boundary,
(E-4)
Rb L - mg L
1
= 0
(E-5)
Rb = mg ( L
1
/ 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)
V
Ra
M
L1
y
mg
x
Sum the moments at the right side of the segment.
(E-9)
- R
a
x + mg <x-L
1
> - M = 0
(E-10)
Note that < x-L1> denotes a step function as follows
(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)
(E-15)
(E-16)
(E-17)
(E-18)
y(0) = 0
(E-19a)
b = 0 (E-19b)
(E-20)
y(L) = 0
(E-21)
(E-22)
(E-23)
(E-24)
(E-25)
(E-26)
(E-27)
(E-28)
(E-29)
(E-30)
The displacement at the center
is
(E-31)
(E-32)
(E-33)
(E-34)
(E-35)
Recall Hooke’s law for a
linear spring,
F = k y
(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
(E-38)
(E-39)
The formula for the natural
frequency fn of a single-degree-of-freedom system is
(E-40)
The mass term m is simply the
mass at the center of the beam.
(E-41)
(E-42)
APPENDIX F
Beam Simply-Supported
at Both Ends II
Consider a simply-supported
beam as shown in Figure F-1.
L
Figure F-1.
Recall that the governing differential
equation is
(F-1)
The spatial solution from section
D is
(F-2)
(F-3)
The boundary conditions at
the left end x = 0 are
Y(0) = 0
(zero displacement)
(F-4)
(zero bending moment)
(F-5)
The boundary conditions at
the free end x = L are
Y(L) = 0
(zero displacement)
(F-6)
(zero bending moment)
(F-7)
Apply boundary condition (F-4)
to (F-2).
(F-8)
(F-9)
Apply boundary condition (F-5)
to (F-3).
(F-10)
(F-11)
Equations (F-8) and (F-10)
can only be satisfied if
(F-12)
and
(F-13)
The spatial equations thus
simplify to
(F-14)
(F-15)
Apply boundary condition (F-6)
to (F-14).
(F-16)
Apply boundary condition (F-7)
to (F-15).
(F-17)
(F-18)
(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.
(F-20)
(F-21)
(F-22)
Equation (F-22) is satisfied
if
(F-23)
(F-24)
The natural frequency term n is
(F-25)
(F-26)
(F-27)
(F-28)
SDOF Approximation
Now calculate effective mass
at the center of the beam for the fundamental frequency.
(F-29)
Recall the natural frequency
equation for a single-degree-of-freedom system.
(F-30)
Recall the beam stiffness at
the center from equation (E-39).
(F-31)
Substitute equation (F-31)
into (F-30).
(F-32)
Substitute (F-32) into (F-29).
(F-33)
(F-34)
(F-35)
(F-36)
(SDOF Approximation)
(F-37)
Normalized Eigenvectors
The eigenvector and its second
derivative at this point are
(F-38)
(F-39)
The eigenvector derivation
requires some creativity. Recall
Y(L) = 0
(zero displacement)
(F-40)
(zero bending moment)
(F-41)
Thus,
for x=L and n=1,2,3, …
(F-42)
, n=1,2,3, …
(F-43)
The sin(n) term is always zero. Thus = 0.
The eigenvector for all n modes
is
(F-44)
Mass normalize the eigenvectors
as follows
(F-45)
(F-46)
(F-47)
(F-48)
(F-49)
(F-50)
(F-51)
(F-52)
Participation Factors
The participation factors for
constant mass density are
(F-53)
(F-53)
(F-54)
(F-55)
, n=1, 2, 3, ….
(F-56)
Effective Modal Mass
The effective modal mass is
(F-57)
The eigenvectors are already
normalized such that
(F-58)
Thus,
(F-59)
(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
(G-1)
Note that this equation neglects
shear deformation and rotary inertia.
The following equation is obtain
using the method in Appendix D
(G-2)
The proposed solution is
(G-3)
(G-4)
(G-5)
(G-6)
Apply the boundary conditions.
(zero bending moment)
(G-7)
(G-8)
(G-9)
(zero shear force)
(G-10)
(G-11)
(G-12)
(G-13)
(G-14)
(zero bending moment) (G-15)
(G-16)
(zero shear force)
(G-17)
(G-18)
Equation (G-16) and (G-18) can be arranged in matrix form.
Set the determinant equal to
zero.
(G-20)
(G-21)
(G-22)
(G-23)
(G-24)
(G-25)
(G-26)
(G-27)
The second root is
(G-28)
(G-29)
(G-30)
(G-31)
(G-32)
The third root is
(G-33)
(G-34)
(G-35)
(G-36)
(G-37)
(G-39)
(G-40)
(G-44)
(H-1)
(H-2)
(H-3)
(H-4)
(H-5)
(H-6)
(H-7)
(H-8)
mass per unit
length.
(H-9)
(H-10)
(H-11)
(H-12)
(H-13)
(H-14)
(H-15)
(simply-supported)
(H-16)
(H-17)
(H-18)
(fixed-fixed)
(H-19)
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.
The average stiffness is
EI
= 63034 (106) lbf in^2
The vehicle behaves as a free-free
beam in flight. Thus
(I-1)
(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.
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
(J-1)
The spatial equation is
(J-2)
The boundary conditions for
the fixed-fixed beam are:
Y(0) = 0
(J-3)
(J-4)
Y(L)=0
(J-5)
(J-6)
The eigenvector has the form
(J-7)
(J-8)
(J-9)
Y(0) = 0
(J-10)
a 2 + a 4 =
0
(J-11)
- a 2 = a 4
(J-12)
(J-13)
a1 +
a3 = 0
(J-14)
a1 +
a3
= 0
(J-15)
- a1 = a3
(J-16)
(J-17)
(J-18)
Y(L) = 0
(J-19)
(J-20)
(J-21)
(J-22)
(J-23)
(J-24)
(J-25)
(J-26)
(J-27)
(J-28)
(J-29)
The roots can be found via
the Newton-Raphson method, Reference 1. The first root is
(J-30)
(J-31)
(J-32)
(J-33)
(J-34)
(J-35)
Let =
1
(J-36)
(J-37)
(J-38)
(J-39)
(J-40)
The mode shape for a fixed-fixed
beam is
(J-41)
where
(J-42)
The eigenvalues are
| n | |
| 1 | 4.73004 |
| 2 | 10.9956 |
| 3 | 14.13717 |
| 4 | 17.27876 |
The first derivative is
(J-43)
The second derivative is
(J-44)
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