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BENDING FREQUENCIES OF BEAMS, RODS, AND PIPES Revision P
By Tom IrvineEmail: [email protected]
April 19, 2011
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 =
12π [3 . 5156
L2 ]√ EIρ
f2 = 6.268 f1
f3 = 17.456 f1
Cantilever with End Mass m f1 =
12π √ 3 EI
( 0 . 2235 ρ L+m ) L3
Simply-Supported at both Ends(Pinned-Pinned) fn =
12π [ n π
L ]2
√ EIρ , n=1, 2, 3, ….
Free-Free
f1 =
12π [22. 373
L2 ]√ EIρ
f2 = 2.757 f1
f3 = 5.404 f1
Fixed-Fixed Same as Free-Free
Fixed - Pinnedf1 =
12π [15. 418
L2 ]√ EIρ
1
where
E is the modulus of elasticityI is the area moment of inertiaL is the lengthρ 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
1. T. Irvine, Application of the Newton-Raphson Method to Vibration Problems, Vibrationdata Publications, 1999.
2
mEI
g
L
mgRMR
L
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.
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
Figure A-2.
R is the reaction force.MR is the reaction bending moment.Apply Newton’s law for static equilibrium.
(A-1)
3
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.
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
V
RMR M
x
y
4
(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)
5
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)
6
(A-27)
The formula for the natural frequency fn of a single-degree-of-freedom system is
fn= 12 π √ k
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= 12 π √3 EI
mL3 (A-29)
7
L
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.
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.
EI,
8
(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)
9
(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)
10
(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)
11
(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)
12
13
L
APPENDIX C
Cantilever Beam III
Consider a cantilever beam where both the beam mass and the end-mass are significant.
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)
m
EI, g
14
L
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.
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)
EI,
15
(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)
A solution for equation (D-10) is
(D-11)
16
(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)
17
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).
18
(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)
19
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. RootsIndex 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)
20
The natural frequency term n is thus
(D-45)
Substitute the value for the fundamental frequency from Table D-1.
(D-46)
f 1=1
2 π [ 3 . 5156L2 ]√ EI
ρ (D-47)
Substitute the value for the second root from Table D-1.
ω2=[ 4 . 69409L2 ]
2
√ EIρ (D-48)
f 2=1
2π [22.034L2 ]√ EI
ρ (D-49)
f 2=6 . 268 f 1 (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
(D-51)
21
(D-52)
(D-53)
(SDOF Approximation) (D-54)
Eigenvalues
n βn 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
∫0
LρY
n2( x ) dx=1 (D-55)
22
The calculation steps are omitted for brevity. The resulting normalized eigenvectors are
Y 1 ( x )={ 1√ ρL }{ [cosh (β1 x )−cos (β1 x ) ]−0 . 73410 [sinh (β1 x )−sin (β1 x ) ] }
(D-56)
Y 2 ( x )={ 1√ ρL }{ [cosh (β2 x )−cos (β2 x ) ]−1 . 01847 [sinh ( β2 x )−sin ( β2 x )] }
(D-57)
Y 3 ( x )={ 1√ ρL }{ [cosh (β3 x )−cos ( β3 x )]−0 . 99922 [ sinh (β3 x )−sin (β3 x ) ] }
(D-58)
Y 4( x )={ 1√ ρL }{ [cosh ( β4 x)−cos ( β4 x) ]−1 . 00003 [sinh (β4 x )−sin (β4 x )] }
(D-59)
The normalized mode shapes can be represented as
Y i( x )={ 1√ ρL }{ [cosh ( β i x)−cos ( β i x) ]−Di [ sinh ( β i x)−sin (β i x ) ] }
(D-60)
where
Di=cos (β i L )+cosh (β i L )sin (β i L )+sinh (β i L ) (D-61)
Participation Factors
The participation factors for constant mass density are
Γn= ρ∫0
LY n ( x ) dx
(D-62)
23
The participation factors from a numerical calculation are
Γ1=0 .7830 √ ρL (D-63)
Γ2=0 . 4339 √ ρL (D-64)
Γ3=0.2544 √ ρL (D-65)
Γ 4=0 .1818 √ ρL (D-66)
The participation factors are non-dimensional.
Effective Modal Mass
The effective modal mass is
meff , n=[∫0
Lm( x ) Y n ( x )dx ]
2
∫0
Lm( x ) [Y n ( x )]2 dx
(D-67)
The eigenvectors are already normalized such that
∫0
Lm( x ) [Y n ( x )]2 dx=1
(D-68)
Thus,
meff , n= [Γn ]2=[∫0
Lm( x ) Y n ( x )dx ]
2
(D-69)
The effective modal mass values are obtained numerically.
meff ,1 =0.6131 ρL (D-70)
meff ,2 =0.1883 ρL (D-71)
meff ,3 =0 .06474 ρL (D-72)
24
m
meff ,4 =0 .03306 ρL (D-73)
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.
Figure E-1.
The free-body diagram of the system is
Figure E-2.
Apply Newton’s law for static equilibrium.
(E-1)
Ra + Rb - mg = 0 (E-2)
EI
g
L
L1 L1
mgRa
L1 L1
Rb
L
25
V
Ra M
L1
y
mg
x
Ra = mg - Rb (E-3)
At the left boundary,
(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)
Sum the moments at the right side of the segment.
(E-9)
- Ra x + mg <x-L1 > - M = 0 (E-10)
26
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)
EI y ' '=[ - (1/2) x +¿ x-L1>][ mg ] (E-15)
y ' '=[ - (1/2) x +¿ x-L1>][mgEI ]
(E-16)
y '=[ - 14
x2+12<x-L1>
2 ] [mgEI ]+a
(E-17)
y ( x )=¿¿ (E-18)
The boundary condition at the left side is
y(0) = 0 (E-19a)
This requires b = 0 (E-19b)
Thus
y ( x )=¿¿ (E-20)
The boundary condition on the right side is
y(L) = 0 (E-21)
27
¿ x−L1> = { 0 , for x<L1
x−L1 , for x≥L1
¿¿ (E-22)
[- 112
L3 +1
48L3 ][mg
EI ]+aL=0 (E-23)
[- 448
L3 +1
48L3] [mg
EI ]+aL=0 (E-24)
[- 348
L3 ] [mgEI ]+aL=0
(E-25)
[- 116
L3 ][mgEI ]+aL=0
(E-26)
aL= [ 116
L3 ][mgEI ]
(E-27)
a= [ 116
L2 ][mgEI ]
(E-28)
Now substitute the constant into the displacement function
y ( x )=¿¿ (E-29)
y ( x )=¿¿ (E-30)
The displacement at the center is
y ( L2 )=¿¿ (E-31)
28
y ( L2 )=[- 196
+ 132 ][mgL3
EI ] (E-32)
y ( L2 )=[- 196
+ 396 ][mgL3
EI ] (E-33)
y ( L2 )=[ 296 ][mgL3
EI ] (E-34)
y ( L2 )=[ 148 ][mgL3
EI ] (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
k = { mg
[ mgL3
48 EI ] } (E-38)
k=48 EI
L3 (E-39)
The formula for the natural frequency fn of a single-degree-of-freedom system is
29
(E-40)
The mass term m is simply the mass at the center of the beam.
fn=( 12π )√48 EI
mL3 (E-41)
fn=( 12π ) (6 . 928 )√ EI
mL3 (E-42)
30
L
APPENDIX F
Beam Simply-Supported at Both Ends II
Consider a simply-supported beam as shown in Figure F-1.
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)
31
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)
32
33
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)
34
(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)
35
(F-35)
(F-36)
The effective mass at the center of the beam for the first mode is
(SDOF Approximation) (F-37)
Normalized Eigenvectors
The eigenvector and its second derivative at this point are
(F-38)
d2 Y ( x )dx2 =a1 β
2 sinh (β x )−a3 β2 sin (β x )
(F-39)
The eigenvector derivation requires some creativity. Recall
Y(L) = 0 (zero displacement) (F-40)
(zero bending moment) (F-41)
Thus,
d2Ydx 2 +Y=0
for x=L and βn L=nπ , n=1,2,3, … (F-42)
(1−( nπL )2) a1sinh (nπ )+(1−(nπL )
2) a3 sin ( nπ )=0 , n=1,2,3, …
(F-43)
36
The sin(n) term is always zero. Thus a1= 0.
The eigenvector for all n modes is
Y n( x )=ansin (nπ x /L ) (F-44)
Mass normalize the eigenvectors as follows
∫0
LρY
n2( x ) dx=1 (F-45)
ρan2∫0
Lsin2 (nπ x /L ) dx=1
(F-46)
ρan2
2 ∫0
L[1−cos (2n π x /L)]=1
(F-47)
ρan2
2 [ x− 12 βn
sin(2nπ x /L)]|0L =1 (F-48)
ρan2 L2
=1 (F-49)
an2 =
2ρL (F-50)
an =√ 2ρL (F-51)
Y n( x ) = √ 2ρL
sin (nπ x /L ) (F-52)
37
38
Participation Factors
The participation factors for constant mass density are
Γn= ρ∫0
LY n ( x ) dx
(F-53)
Γn= ρ∫0
L √ 2ρL
sin (nπ x /L ) dx (F-53)
Γn=√ 2 ρL ∫0
Lsin (nπ x /L ) dx
(F-54)
Γn=− √ 2 ρL [ L
nπ ] cos (nπ x /L ) |0L
(F-55)
Γn=− √2 ρL [ 1nπ ] [cos (nπ )−1 ]
, n=1, 2, 3, …. (F-56)
Effective Modal Mass
The effective modal mass is
meff , n=[∫0
Lm( x ) Y n ( x )dx ]
2
∫0
Lm( x ) [Y n ( x )]2 dx
(F-57)
The eigenvectors are already normalized such that
∫0
Lm( x ) [Y n ( x )]2 dx=1
(F-58)
39
Thus,
meff , n= [Γn ]2=[∫0
Lm( x ) Y n ( x )dx ]
2
(F-59)
meff , n=[− √2 ρL [ 1nπ ] [cos (nπ )−1 ] ]
2
(F-60)
meff , n=2 ρL 1(nπ )2
[ cos (nπ )−1 ]2 , n=1, 2, 3, …. (F-61)
40
L
APPENDIX G
Free-Free Beam
Consider a uniform beam with free-free boundary conditions.
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)
EI,
41
d3 Y ( x )dx3 =a1 β
3 cosh (βx )+a2 β3 sinh ( βx )−a3 β
3 cos (βx )+a4 β3 sin (βx )
(G-6)Apply the boundary conditions.
d2Ydx 2 |x=0= 0
(zero bending moment) (G-7)
a2−a4=0 (G-8)
a4=a2 (G-9)
d3Ydx3 |x=0= 0
(zero shear force) (G-10)
a1−a3=0 (G-11)
a3=a1 (G-12)
d2 Y ( x )dx2 =a1 β
2 [sinh ( βx )−sin ( βx ) ]+a2 β2 [cosh ( βx )−cos (βx ) ]
(G-13)
d3 Y ( x )dx3 =a1 β
3 [cosh (βx )−cos (βx ) ]+a2 β3 [sinh (βx )+sin (βx ) ]
(G-14)
(zero bending moment) (G-15)
42
a1 [sinh ( βL )−sin (βL ) ]+a2 [cosh (βL )−cos (βL ) ]=0 (G-16)
(zero shear force) (G-17)
a1 [cosh ( βL )−cos ( βL ) ]+a2 [sinh (βL )+sin (βL ) ]=0 (G-18)
Equation (G-16) and (G-18) can be arranged in matrix form.
[sinh (βL )−sin ( βL ) cosh ( βL )−cos (βL )
cosh (βL )−cos (βL ) sinh ( βL )+sin (βL ) ] [a1
a2]=[00 ]
(G-19)
Set the determinant equal to zero.
[sinh (βL )−sin (βL ) ] [sinh (βL )+sin (βL ) ]−[cosh (βL )−cos (βL ) ]2=0 (G-20)
sinh2 (βL )−sin2 (βL )−cosh2 ( βL )+2cosh (βL )cos ( βL )−cos2 ( βL )=0 (G-21)
+2 cosh (βL ) cos ( βL )−2=0 (G-22)
cosh (βL )cos ( βL )−1=0 (G-23)
The roots can be found via the Newton-Raphson method, Reference 1. The first root is
βL=4 . 73004 (G-24)
43
(G-25)
ω 1=[ 4 .73004L ]
2
√ EIρ (G-26)
ω 1=[22 .373L2 ]√ EI
ρ (G-27)
The second root is
βL=7. 85320 (G-28)
ω n=βn2√ EI
ρ (G-29)
ω 2=[ 7 .85320L ]
2
√ EIρ (G-30)
ω 2=[61 . 673L2 ]√ EI
ρ (G-31)
ω 2= 2 .757 ω 1 (G-32)
The third root is
βL=10. 9956 (G-33)
ω n=βn
2√ EIρ (G-34)
ω 3=[10 .9956L ]
2
√ EIρ (G-35)
44
ω 3=[120 . 903L2 ]√ EI
ρ (G-36)
ω 3=5 . 404 ω 1 (G-37)
Equation (G-18) can be expressed as
a2=a1 [−cosh ( βL )+cos ( βL )sinh (βL )+sin ( βL ) ] (G-38)
Recall
a4=a2 (G-39)
a3=a1 (G-40)
The displacement mode shape is thus
Y ( x )=a1 [sinh (βx )+sin ( βx ) ]+a2 [cosh (βx )+cos ( βx ) ] (G-41)
Y ( x )=a1 {[sinh (βx )+sin (βx ) ]+[−cosh ( βL )+cos (βL )sinh ( βL )+sin ( βL ) ] [cosh (βx )+cos ( βx ) ]}
(G-42)
The first derivative is
dydx
=a1 β { [cosh (βx )+cos (βx ) ]+[−cosh (βL )+cos (βL )sinh (βL )+sin ( βL ) ] [sinh (βx )−sin (βx ) ]}
(G-43)
45
The second derivative is
d2 ydx2 =a1 β
2 { [sinh (βx )−sin (βx ) ]+[−cosh (βL )+cos (βL )sinh (βL )+sin (βL ) ] [cosh (βx )−cos (βx ) ]}
(G-44)
46
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= π64 [Do4−D
i4 ] (H-1)
Do=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= π32
[2 .24−1 .04 ] in4
(H-6)
I=1 . 101 in4 (H-7)
The elastic modulus is
E=30 (106 ) lbfin2
(H-8)
The mass density is
ρ=mass per unit length. (H-9)
ρ=[0 . 282 lbmin3 ] [ π4 [2. 22−1 . 02 ] in2 ]
(H-10)
47
ρ=0 .850 lbmin (H-11)
√ EIρ
=√30 (106 ) lbfin2 1 .101 in4 (1 slug ft /sec2
1 lbf ) (12 in1 ft )
0 . 850 lbmin ( 1 slug
32 .2 lbm )(H-12)
√ EIρ
=1 . 225 (105 ) in2
sec (H-13)
The natural frequency for the simply-supported case is
(H-14)
f 1=[ 12π ] [ π
(20 ft )(12 in1 ft )]
2
1. 225 ( 105) in2
sec
(H-15)
f 1=3 .34 Hz (simply-supported) (H-16)
The natural frequency for the fixed-fixed case is
f 1=[ 12π ] [22 . 37
L2 ]√ EIρ (H-17)
48
f 1=[ 12π ] [22 .37
[ (20 ft )(12 in1 ft )]
2 ] 1 .225 (105 ) in2
sec
(H-18)
f 1= 7 .58 Hz (fixed-fixed) (H-19)
49
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.
ρ=14078. 9 lbm372. 0 inches
ρ=37 . 847 lbmin
The average stiffness is
EI = 63034 (106) lbf in^2
The vehicle behaves as a free-free beam in flight. Thus
f 1=1
2 π [22. 37L2 ]√ EI
ρ (I-1)
f 1=1
2 π [22. 37(372 in)2 ]√ [63034e+06 lbf in2] [ slug ft /sec2
lbf ] [12 inft ] [32 . 2 lbm
slugs ]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.
50
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
∂4
∂ x4 Y ( x )−c2{ ρEI } Y ( x )=0
(J-2)
The boundary conditions for the fixed-fixed beam are:
Y(0) = 0 (J-3)
dY ( x )dx
|x=0=0 (J-4)
Y(L)=0 (J-5)
dY ( x )dx
|x=L=0 (J-6)
The eigenvector has the form
(J-7)
(J-8)
51
(J-9)
Y(0) = 0 (J-10)
a 2 + a 4 = 0 (J-11)
- a 2 = a 4 (J-12)
dY ( x )dx
|x=0=0 (J-13)
a1 + a3 = 0 (J-14)
a1 + a3 = 0 (J-15)
- a1 = a3 (J-16)
Y ( x )=a1 [sinh (β x )−sin (β x ) ]+a2 [cosh ( β x )−cos ( β x ) ] (J-17)
dY ( x )dx
=a1 β [cosh (β x )−cos ( β x ) ]+a2 β [sinh (β x )+sin ( β x ) ] (J-18)
Y(L) = 0 (J-19)
a1 [sinh (β L )−sin (β L ) ]+a2 [cosh ( β L )−cos (β L ) ]=0 (J-20)
dY ( x )dx
|x=L=0 (J-21)
a1 β [cosh (β L )−cos (β L ) ]+a2 β [sinh (β L )+sin ( β L ) ]=0 (J-22)
52
a1 [cosh (β L )−cos (β L ) ]+a2 [sinh (β L )+sin (β L ) ]=0 (J-23)
[sinh (β L )−sin (β L ) cosh (β L )−cos (β L )cosh (β L )−cos ( β L ) sinh ( β L )+sin (β L ) ] [a1
a2 ]=[00 ] (J-24)
det [sinh (β L )−sin (β L ) cosh (β L )−cos ( β L )cosh ( β L )−cos ( β L ) sinh (β L )+sin (β L ) ] =0
(J-25)
[ sinh (β L )−sin (β L ) ][sinh ( β L )+sin (β L ) ]−[ cosh (β L )−cos (β L ) ]2=0 (J-26)
sinh2 (β L )−sin2 (β L )−cosh2 (β L )+2cos ( β L ) cosh ( β L )−cos2 ( β L )=0 (J-27)
2 cos (β L )cosh (β L )−2=0 (J-28)
cos (β L )cosh (β L )−1=0 (J-29)
The roots can be found via the Newton-Raphson method, Reference 1. The first root is
β L=4 .73004 (J-30)
ωn=βn2√ EI
ρ (J-31)
ω1=[ 4 . 73004L ]
2
√ EIρ (J-32)
53
ω1=[22. 373L2 ]√ EI
ρ (J-33)
f 1=
12 π [22.373
L2 ]√ EIρ (J-34)
a1 [cosh ( β L )−cos (β L ) ]=−a2 [sinh (β L )+sin (β L ) ] (J-35)
Let a2 = 1 (J-36)
a1 [cosh ( β L )−cos ( β L ) ]=− [sinh (β L )+sin ( β L ) ] (J-37)
a1=
−sinh (β L )−sin (β L )cosh (β L )−cos (β L ) (J-38)
Y ( x )=[cosh (β x )−cos (β x ) ]+[−sinh (β L )−sin ( β L )
cosh (β L )−cos (β L ) ] [sinh (β x )+sin ( β x ) ]=0 (J-39)
Y ( x )=[cosh (β x )−cos (β x ) ]−[sinh (β L )+sin ( β L )cosh ( β L )−cos (β L ) ] [sinh (β x )+sin ( β x ) ]=0
(J-40)
The un-normalized mode shape for a fixed-fixed beam is
Y n ( x )= [cosh ( βn x )−cos (βn x ) ]−σn [ sinh ( βn x )−sin (βn x ) ]=0 (J-41)
where
54
σ n=[sinh ( β L )+sin (β L )cosh ( β L )−cos ( β L ) ] (J-42)
55
The eigenvalues are
n βn L
1 4.73004
2 10.9956
3 14.13717
4 17.27876
Normalized Eigenvectors
Mass normalize the eigenvectors as follows
∫0
LρY
n2( x ) dx=1 (J-43)
The mass normalization is satisfied by
Y n ( x )=1
√ ρL { [cosh (βn x )−cos ( βn x )]−σ n [sinh (βn x )−sin ( βn x )] } (J-44)
where
σ n=[sinh ( β L )+sin (β L )cosh ( β L )−cos ( β L ) ] (J-45)
The first derivative is
ddx
Y n ( x )=1
√ ρL { βn [sinh ( βn x )+sin (βn x )]−σn βn [cosh ( βn x )−cos (βn x )] }
(J-46)
56
The second derivative is
d2
dx2 Y n( x )= 1√ρL { β
n2 [cosh (βn x )+cos (βn x )]−σn βn2 [ sinh ( βn x )+sin (βn x )] }
(J-47)
Participation Factors
The participation factors for constant mass density are
Γn= ρ∫0
LY n ( x ) dx
(J-48)
Γn=ρ
√ ρL∫0
L {[ cosh (βn x)−cos ( βn x ) ]−σ n [sinh ( βn x )−sin ( βn x )]} dx (J-49)
Γn=1βn √ ρ
L { [sinh (βn x )−sin (βn x ) ]−σn [ cosh (βn x )+cos (βn x )] } |0L (J-50)
Γn=1βn √ ρ
L { [sinh (βn L )−sin (βn L )]−σ n [cosh ( βn L )+cos (βn L ) ]+2 σn } (J-51)
Γn=1βn √ ρ
L { [sinh (βn L )−sin (βn L )]+σn [2−cosh (βn L )−cos ( βn L )]} (J-52)
The participation factors from a numerical calculation are
Γ1=0 .8309 √ ρL (J-53)
Γ2=0 .3638 √ ρL (J-54)
Γ3= 0 (J-55)
Γ 4=0 . 2315 √ ρL (J-56)
57
The participation factors are non-dimensional
58