- Email: [email protected]
Free vibration of a free-free beam with rotary inertia effect — A flexibility matrix approach
Journal
of Sound and Vibration
(1988)
125(3),
565-569
LETTERS TO THE EDITOR FREE
VIBRATION
OF -A
A
FREE-FREE FLEXIBILITY
BEAM WITH ROTARY MATRIX APPROACH
INERTIA
EFFECT
1. INTRODUCTION
Ships and aircraft are treated as a beam having varying cross-sectional properties for the determination of their strength and vibration characteristics. Their ends, however, are unconstrained as a result of which rigid body deflections and rotations are introduced into the equation. In the displacement approach, special techniques are needed to remove the singularity of the stiffness matrix for the solution of the eigenvalue problem. A method is presented in what follows for evaluating the natural frequencies and mode shapes for these beams. It is based on the flexibility matrix approach and is an extension of the work of Scanlan and Rosenbaum [l] applied to the free-free beam by including the shear deformation and rotary inertia effects. 2.
THE
FLEXIBILITY
MATRIX
METHOD
Consider a beam having varying cross-section and variable mass loading. The left hand station is denoted as 0 and succeeding stations to the right as 1,2,. . . , n respectively. The deflection (positive upward) and rotation (positive counter clockwise) at the zero station are y, and 8, respectively (a list of nomenclature is given in the Appendix). The flexibility coefficients ai, can be calculated relative to the zero station by assuming the beam to be cantilevered at that station. The flexibility coefficients include both the bending deflection and the shearing deflection. Let y, be the total deflection of station i from the equilibrium position and the horizontal distance from the zero station to station i (see Figure 1). In order to consider the rotary inertia effects, additional inertial elements which rotate the beam are to be taken into account. Then the equations of motion for the vibrating beam are
.+%a,.y,) +t(pl,b,,8,+pl2b2,82+. ~+d,h,,&,)l,
.vl-Y,-x,~,~=P2[(“,a,,Y,+mza,,Yz+~.
.+w@2&)
y2-Y,-~2~,=P2~~~,~21Y,+~2~22Y2+~~
+(pl,b,2e,+plzb22e2+.
.
...
yn -~,-x,B”=P~[(m,a,ty,+
m2an2y2+.
~+p~,,b,z~,)l,
. .+
mnannyn)
. ~+PM,A)I, ..* 8,-~o=~2[(m,b,,y,+m2b,2y2+~ ~~+~LJL) +(pl,b,,6,+pl2b,282+.
+(pz,c,,~,+pz,c,2~2+~
.
(14
~+PLc,.~“)l,
b- %=p2[(m,b2,y, + m2b22y2+.. *+ dw+,) +(P~,c*,~,+P~2c*2~2+. ...
. ~+P~“C2n&)l,
f(pZ,c,,~,+pZ,c,2~2+.
. .-tPLC”,&)l.
(lb)
Here a# is the transverse deflection at i due to a unit transverse force at j and includes both the flexural and the shear deformation, cii is the rotation at i due to a unit couple 565 0022-460X/88/
180565 + 05 $03.00/O
@j 1988 Academic
Press Limited
566
LEITERS
TO THE
EDITOR
Equilbrium position
Figure 1. A free-free beam.
applied at j, b, is the rotation at i due to a unit transverse force at j, and transverse deflection at i due to a unit couple at j. It can be shown that
dy is the
b, = dji.
(2)
The relation (2) has been directly used in equation (la). Let the first of equations (la) be multiplied by m, , the second results when added then yield
by m2 and so on. The
Cmi~i-_Cmiy~-_mmixieo=p~[(A,y,+A2y2+...A,y,)+(C,y,+C2y2+...+C,y”)l,
(3) where Ai = C mimjaji and Ci = C plimjbji. The following equations of equilibrium
can be written:
C rn$; = 0, C ( mixfli + pIiei) = 0. Combining
equation
(4) with equation
(435)
(3) yields
-MYo-S80=p2
C (Aai + C,ei). j
(6)
Similarly, multiplying the first of equations (la) by m,x, ,the second by m2x2 and so on, and the first of equations (lb) by pl,, the second by p12 and so on, and then adding up the results yields C (m,xfl,+pI$,)-(2
mixi)yo-C
(mixf+pZi)eo=p2C
(&J~+ZI,~~),
where Bi = Cj mi( mjxjaji + pl,bji) and 0, = xi pl,( mjxjbq + pljcji). Combining and (7) yields -sY,Solving
equations
Yl,Y2,...,Yn,e1,e2,..., in the equations
(6)
Te,=$c
(Biyi+oie,).
. .+g,.Y,+h,,~,+. ... Y” = P2C&,Y1 +. . .+g,,Y,+k,o,+.
6 = p2h1Yl+. . .+q,.h+r114+~~ .. . 0, =P2r9,1Y,+*
. .+4,,Y,+r,,e,+.
(5)
(8)
and (8) will yield values of Y, en which on substitution into equations
Y1 = p2[(g1 ,y1+.
equations
(7)
and e. in terms of (la) and (lb) will result
. .+hlnen)i. . .+k,e,)i, ~+hn6Jl,
. .+r,,e,)i.
(9)
0.2
0.4
0.6
324.452 450.579 4.714 4.714 13.682 13.682 20,799 20.799
252.088 374.644 520.284
308.701
457.744
634.609
436.569
647.347
897.473
Inclusion of the effect of rotary inertia 9.427 6.666 5-443 9.427 6.666 5443 15.799 27.365 19.350 19-350 15.799 27.365 41.598 29.414 24.016 41.598 29.414 24.016
t Values within the parentheses are those obtained from “exact”
218.314
132.747
153.283
187.732
265.494
68.063
24.822
0.8
22.202
1.0
beam;
1
solution
equation
3442 3442 9.992 9.992 15.189 15.189
329.241
236.948
159.434
96.945
49.706
18.128
1.5
[2].
2.981 2.981 8.654 8.654 13.154 13.154
285.131
205.203
138.074
83.957
43.047
15.690
2.0
2.434 2-434 7.066 7.066 10.741 10.741
232.809
167.547
2.108 2.108 6.119 6.119 9.302 9.302
g
201.618
8
E
3
2
?
21
145.100
97.633
59.366
68.550 112.737
30.439
11.101
4.0
35.148
12.818
3.0
E = l-0, I = l-0, L,= l-0, A = l-0, numberof stations =
of the differential
4.216 4.216 12.238 12.238 18.603 18.603
(22.368)t 60.877 (61.700) 118.733 (120.637) 195.266 (199.805) 290.198 (298.451) 403.010
of a free-free
78.592
96.255
136.126
Without the effect of rotary inertia 49.644 35.104 28.662
P
E$ect of rotary inertia on the naturalfrequencies
TABLE
568
LEl-l-ERS
TO
THE
EDITOR
Equations (9) in matrix form are { YI = P’[
cl{ Y).
(10)
Equation (10) is a typical eigenvalue problem which can be solved by using any standard procedure. 3.
EXAMPLES
A uniform free-free beam having the cross-sectional and mass properties given in Table 1 has been analyzed by the proposed method by dividing it into 20 divisions. The first six frequencies for varying values of p are shown in Table 1, both with and without the inclusion of rotary inertia effects. As expected the incorporation of rotary inertia has reduced the natural frequencies. For the beam with p = 1, the frequencies compared well with the “exact” value when the rotary inertia effect is not considered [2]. The eigenvalues appear in pairs when the t
1
Without rotary inertio effect
I -/
/’ ,/ “,\x-/’
‘\ ‘\
/’ -\-_____~* I
I
Figure
2. Mode shape
for first frequency.
(b)
I
I
Figure 3. Mode shapes (with p>2) for (a) first, (b) second, mode of pair; - - -, second mode of pair.
and (c) third
paired
frequencies.
-,
First
569
LETTERS TO THE EDITOR
rotary inertia is included. If r, is the ratio of the nth frequency of a beam without rotary inertia to the nth paired frequency for the beam with rotary inertia, then it is interesting to note that for all values of p one has r, = 5.266, r2 = 4.97 and r3 = 6.38. The mode shapes for the first frequency for the beam without rotary inertia and for beams with rotary inertia for p > 2 and p < 2 are plotted in Figure 2. When rotary inertia is considered, the mode shapes for all beams having p < 2 are identical for all modes and for those having p > 2 are identical to each other in the different modes. Mode shapes for both the modes corresponding to the first, second and third paired frequencies for and (e) respectively. Inclusion of the beam with p > 2 are plotted in Figures 3(a), (b) rotary inertia tends to make the mode shape rather flat. It is also interesting to note that whereas the number of nodes increases with the increase of modes for beams when rotary inertia is not considered, the same observation cannot be made when the rotary inertia is incorporated in the analysis. For the first paired frequencies, the beam has two nodes, for the second and third paired frequencies, it has one node only.
4. CONCLUSIONS A flexibility matrix method has been applied for determining free vibration characteristics of free-free beams with rotary inertia effects. A uniform beam having various mass densities has been studied. The incorporation of rotary inertia has resulted in paired natural frequencies and the mode shapes tend to flatten. Department of Naval Architecture, Indian Znstitute of Technology, Kharagpur 721302, West Bengal, India (Received
M. MUKHOPAHYAY
3 May 1988) REFERENCES
1. R. H. SCANLAN
and R. ROSENBAUM
1968 Aircraft
Publications Inc. 2. G. L. ROGERS 1959 Dynamics of FramedStructures. APPENDIX: A a, b, cu dij E I 2 P S T Xi Y, P 4
Vibration and Flutter.
New York: Dover
New York: John Wiley and Sons Inc.
NOMENCLATURE
cross-sectional area of the beam transverse deflection at i due to a unit force at j rotation at i due to a unit transverse force at j rotation at i due to a unit couple at j transverse deflection at i due to a unit couple at j modulus of elasticity second moment of the area mass at station i total mass of the beam natural frequency of the beam static moment of the masses about zero station second moment of area of the masses about zero station horizontal distance of station i from zero station the transverse deflection at station i mass density angular rotation at station i
of Sound and Vibration
(1988)
125(3),
565-569
LETTERS TO THE EDITOR FREE
VIBRATION
OF -A
A
FREE-FREE FLEXIBILITY
BEAM WITH ROTARY MATRIX APPROACH
INERTIA
EFFECT
1. INTRODUCTION
Ships and aircraft are treated as a beam having varying cross-sectional properties for the determination of their strength and vibration characteristics. Their ends, however, are unconstrained as a result of which rigid body deflections and rotations are introduced into the equation. In the displacement approach, special techniques are needed to remove the singularity of the stiffness matrix for the solution of the eigenvalue problem. A method is presented in what follows for evaluating the natural frequencies and mode shapes for these beams. It is based on the flexibility matrix approach and is an extension of the work of Scanlan and Rosenbaum [l] applied to the free-free beam by including the shear deformation and rotary inertia effects. 2.
THE
FLEXIBILITY
MATRIX
METHOD
Consider a beam having varying cross-section and variable mass loading. The left hand station is denoted as 0 and succeeding stations to the right as 1,2,. . . , n respectively. The deflection (positive upward) and rotation (positive counter clockwise) at the zero station are y, and 8, respectively (a list of nomenclature is given in the Appendix). The flexibility coefficients ai, can be calculated relative to the zero station by assuming the beam to be cantilevered at that station. The flexibility coefficients include both the bending deflection and the shearing deflection. Let y, be the total deflection of station i from the equilibrium position and the horizontal distance from the zero station to station i (see Figure 1). In order to consider the rotary inertia effects, additional inertial elements which rotate the beam are to be taken into account. Then the equations of motion for the vibrating beam are
.+%a,.y,) +t(pl,b,,8,+pl2b2,82+. ~+d,h,,&,)l,
.vl-Y,-x,~,~=P2[(“,a,,Y,+mza,,Yz+~.
.+w@2&)
y2-Y,-~2~,=P2~~~,~21Y,+~2~22Y2+~~
+(pl,b,2e,+plzb22e2+.
.
...
yn -~,-x,B”=P~[(m,a,ty,+
m2an2y2+.
~+p~,,b,z~,)l,
. .+
mnannyn)
. ~+PM,A)I, ..* 8,-~o=~2[(m,b,,y,+m2b,2y2+~ ~~+~LJL) +(pl,b,,6,+pl2b,282+.
+(pz,c,,~,+pz,c,2~2+~
.
(14
~+PLc,.~“)l,
b- %=p2[(m,b2,y, + m2b22y2+.. *+ dw+,) +(P~,c*,~,+P~2c*2~2+. ...
. ~+P~“C2n&)l,
f(pZ,c,,~,+pZ,c,2~2+.
. .-tPLC”,&)l.
(lb)
Here a# is the transverse deflection at i due to a unit transverse force at j and includes both the flexural and the shear deformation, cii is the rotation at i due to a unit couple 565 0022-460X/88/
180565 + 05 $03.00/O
@j 1988 Academic
Press Limited
566
LEITERS
TO THE
EDITOR
Equilbrium position
Figure 1. A free-free beam.
applied at j, b, is the rotation at i due to a unit transverse force at j, and transverse deflection at i due to a unit couple at j. It can be shown that
dy is the
b, = dji.
(2)
The relation (2) has been directly used in equation (la). Let the first of equations (la) be multiplied by m, , the second results when added then yield
by m2 and so on. The
Cmi~i-_Cmiy~-_mmixieo=p~[(A,y,+A2y2+...A,y,)+(C,y,+C2y2+...+C,y”)l,
(3) where Ai = C mimjaji and Ci = C plimjbji. The following equations of equilibrium
can be written:
C rn$; = 0, C ( mixfli + pIiei) = 0. Combining
equation
(4) with equation
(435)
(3) yields
-MYo-S80=p2
C (Aai + C,ei). j
(6)
Similarly, multiplying the first of equations (la) by m,x, ,the second by m2x2 and so on, and the first of equations (lb) by pl,, the second by p12 and so on, and then adding up the results yields C (m,xfl,+pI$,)-(2
mixi)yo-C
(mixf+pZi)eo=p2C
(&J~+ZI,~~),
where Bi = Cj mi( mjxjaji + pl,bji) and 0, = xi pl,( mjxjbq + pljcji). Combining and (7) yields -sY,Solving
equations
Yl,Y2,...,Yn,e1,e2,..., in the equations
(6)
Te,=$c
(Biyi+oie,).
. .+g,.Y,+h,,~,+. ... Y” = P2C&,Y1 +. . .+g,,Y,+k,o,+.
6 = p2h1Yl+. . .+q,.h+r114+~~ .. . 0, =P2r9,1Y,+*
. .+4,,Y,+r,,e,+.
(5)
(8)
and (8) will yield values of Y, en which on substitution into equations
Y1 = p2[(g1 ,y1+.
equations
(7)
and e. in terms of (la) and (lb) will result
. .+hlnen)i. . .+k,e,)i, ~+hn6Jl,
. .+r,,e,)i.
(9)
0.2
0.4
0.6
324.452 450.579 4.714 4.714 13.682 13.682 20,799 20.799
252.088 374.644 520.284
308.701
457.744
634.609
436.569
647.347
897.473
Inclusion of the effect of rotary inertia 9.427 6.666 5-443 9.427 6.666 5443 15.799 27.365 19.350 19-350 15.799 27.365 41.598 29.414 24.016 41.598 29.414 24.016
t Values within the parentheses are those obtained from “exact”
218.314
132.747
153.283
187.732
265.494
68.063
24.822
0.8
22.202
1.0
beam;
1
solution
equation
3442 3442 9.992 9.992 15.189 15.189
329.241
236.948
159.434
96.945
49.706
18.128
1.5
[2].
2.981 2.981 8.654 8.654 13.154 13.154
285.131
205.203
138.074
83.957
43.047
15.690
2.0
2.434 2-434 7.066 7.066 10.741 10.741
232.809
167.547
2.108 2.108 6.119 6.119 9.302 9.302
g
201.618
8
E
3
2
?
21
145.100
97.633
59.366
68.550 112.737
30.439
11.101
4.0
35.148
12.818
3.0
E = l-0, I = l-0, L,= l-0, A = l-0, numberof stations =
of the differential
4.216 4.216 12.238 12.238 18.603 18.603
(22.368)t 60.877 (61.700) 118.733 (120.637) 195.266 (199.805) 290.198 (298.451) 403.010
of a free-free
78.592
96.255
136.126
Without the effect of rotary inertia 49.644 35.104 28.662
P
E$ect of rotary inertia on the naturalfrequencies
TABLE
568
LEl-l-ERS
TO
THE
EDITOR
Equations (9) in matrix form are { YI = P’[
cl{ Y).
(10)
Equation (10) is a typical eigenvalue problem which can be solved by using any standard procedure. 3.
EXAMPLES
A uniform free-free beam having the cross-sectional and mass properties given in Table 1 has been analyzed by the proposed method by dividing it into 20 divisions. The first six frequencies for varying values of p are shown in Table 1, both with and without the inclusion of rotary inertia effects. As expected the incorporation of rotary inertia has reduced the natural frequencies. For the beam with p = 1, the frequencies compared well with the “exact” value when the rotary inertia effect is not considered [2]. The eigenvalues appear in pairs when the t
1
Without rotary inertio effect
I -/
/’ ,/ “,\x-/’
‘\ ‘\
/’ -\-_____~* I
I
Figure
2. Mode shape
for first frequency.
(b)
I
I
Figure 3. Mode shapes (with p>2) for (a) first, (b) second, mode of pair; - - -, second mode of pair.
and (c) third
paired
frequencies.
-,
First
569
LETTERS TO THE EDITOR
rotary inertia is included. If r, is the ratio of the nth frequency of a beam without rotary inertia to the nth paired frequency for the beam with rotary inertia, then it is interesting to note that for all values of p one has r, = 5.266, r2 = 4.97 and r3 = 6.38. The mode shapes for the first frequency for the beam without rotary inertia and for beams with rotary inertia for p > 2 and p < 2 are plotted in Figure 2. When rotary inertia is considered, the mode shapes for all beams having p < 2 are identical for all modes and for those having p > 2 are identical to each other in the different modes. Mode shapes for both the modes corresponding to the first, second and third paired frequencies for and (e) respectively. Inclusion of the beam with p > 2 are plotted in Figures 3(a), (b) rotary inertia tends to make the mode shape rather flat. It is also interesting to note that whereas the number of nodes increases with the increase of modes for beams when rotary inertia is not considered, the same observation cannot be made when the rotary inertia is incorporated in the analysis. For the first paired frequencies, the beam has two nodes, for the second and third paired frequencies, it has one node only.
4. CONCLUSIONS A flexibility matrix method has been applied for determining free vibration characteristics of free-free beams with rotary inertia effects. A uniform beam having various mass densities has been studied. The incorporation of rotary inertia has resulted in paired natural frequencies and the mode shapes tend to flatten. Department of Naval Architecture, Indian Znstitute of Technology, Kharagpur 721302, West Bengal, India (Received
M. MUKHOPAHYAY
3 May 1988) REFERENCES
1. R. H. SCANLAN
and R. ROSENBAUM
1968 Aircraft
Publications Inc. 2. G. L. ROGERS 1959 Dynamics of FramedStructures. APPENDIX: A a, b, cu dij E I 2 P S T Xi Y, P 4
Vibration and Flutter.
New York: Dover
New York: John Wiley and Sons Inc.
NOMENCLATURE
cross-sectional area of the beam transverse deflection at i due to a unit force at j rotation at i due to a unit transverse force at j rotation at i due to a unit couple at j transverse deflection at i due to a unit couple at j modulus of elasticity second moment of the area mass at station i total mass of the beam natural frequency of the beam static moment of the masses about zero station second moment of area of the masses about zero station horizontal distance of station i from zero station the transverse deflection at station i mass density angular rotation at station i