Modal Transient Vibration Response of a Cantilever Beam Subjected to Base Excitation

By Tom Irvine Email: [email protected] January 29, 2013 _______________________________________________________________________________ Normal Modes Consider the cantilever beam in Figure 1. EI, 

L

Figure 1.

The governing differential equation for the displacement y(x,t) is

4 y 2 y  EI  x 4 t 2

(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 this equation neglects shear deformation and rotary inertia.

1

Separate the dependent variable.

y( x, t )  Y( x)T(t )

(2)

 4 Y( x)T( t )  2 Y( x)T( t )  EI  x 4 t 2

(3)

     d4   d2   EI T(t )  Y( x)    Y( x)  T(t )  4 2      dx   dt 

(4)

 d 4   4 Y( x )     EI   dx    Y( x )   

 d 2   2 T(t )   dt  T(t )

 d 4  Y( x )      EI   dx 4    Y( x )   

 d 2   2 T(t )   dt    c2 T(t )

(5)

Let c be a constant

(6)

Separate the time variable.  d 2   2 T(t )   dt    c2 T(t )

(7)

d2 T(t )  c 2 T(t )  0 2 dt

2

(8)

Separate the spatial variable.

 d 4   4 Y( x )     EI   dx   c2   Y( x )   

(9)

d4  Y( x)  c 2  Y( x)  0  EI  dx 4

(10)

A solution for equation (10) is

Y( x)  a1 sinhx  a 2 coshx  a 3 sinx  a 4 cosx

(11)

dY( x)  a1 coshx  a 2 sinh x  a 3 cos x  a 4 sinx dx

(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

(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 4 Y( x)  a1 4 sinh x  a 2 4 cosh x  a 3 4 sin x  a 4 4 cos x 4 dx

(14)

(15)

Substitute (15) and (11) into (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 





(16)

3





 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 





(17) The equation is satisfied if  4  c2    EI 

(18)

 1/4    c 2   EI 

(19)

The boundary conditions at the fixed end x = 0 are

Y(0) = 0

(zero displacement)

dY  0 dx x 0

(zero slope)

(20) (21)

The boundary conditions at the free end x = L are d 2Y dx 2 d 3Y dx 3

 0

(zero bending moment)

(22)

 0

(zero shear force)

(23)

x L

x L

Apply equation (20) to (11). a2  a4  0

(24)

a4   a2

(25)

4

Apply equation (21) to (12). a1  a 3  0

(26)

a 3  a1

(27)

Apply equation (22) to (13).

a1 sinhL  a 2 coshL  a 3 sinL  a 4 cosL  0

(28)

Apply equation (23) to (14).

a1 coshL  a 2 sinhL  a 3 cosL  a 4 sinL  0

(29)

Apply (25) and (27) to (28).

a1 sinhL  a 2 coshL  a1 sinL  a 2 cosL  0









a1 sinL  sinhL  a 2 cosL  coshL  0

(30) (31)

Apply (25) and (27) to (29).

a1 coshL  a 2 sinhL  a1 cosL  a 2 sinL  0









a1 cosL  coshL  a 2  sinL  sinhL  0

(32) (33)

Form (31) and (33) into a matrix format.  sin L  sinh L   cos L  cosh L 

cos L  cosh L   a1  0           sin L  sinh L  a 2  0

(34)

By inspection, equation (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

5

(35)

 sin2 L  sinh2 L  cos2 L  2 cosL coshL  cosh2 L  0 (36)

 sin 2 L  sinh 2 L  cos2 L  2 cosL coshL  cosh 2 L  0 (37)

 2  2 cosL cosh L  0

(38)

1  cosL coshL  0

(39)

cosL coshL  1

(40)

There are multiple roots which satisfy equation (40). Thus, a subscript should be added as shown in equation (41). (41) cos n L cosh n L  1

The subscript is an integer index. The roots can be determined through a combination of graphing and numerical methods. The Newton-Raphson method is an example of an appropriate numerical method. The roots of equation (41) are summarized in Table 1, as taken from Reference 1.

Table 1. Roots Index

n L

n=1

1.87510

n=2

4.69409

n=3

7.85476

n=4

10.99554

n>5

(2n-1)/2

Note: the root value formula for n > 5 is approximate.

6

Rearrange equation (19) as follows  EI  c2  n 4   

(42)

Substitute (42) into (8).

 T( t )   n 4 dt 2 

 EI   T( t )  0   

(43)

  EI   EI    t   b 2 cos   n 2  t  0 T( t )  b1 sin   n 2         

(44)

d2

Equation (43) is satisfied by

The natural frequency term n is thus  n   n2

EI 

(45)

Substitute the value for the fundamental frequency from Table 1.

. 187510  2 EI 1     L 

(46)

1  3.5156  EI   2  L2  

(47)

Y( x)  a1 sinhx  a 2 coshx  a 3 sinx  a 4 cosx

(48)

f1 

Find the eigenvectors.

a 4  a 2

(49)

a 3  a1

(50)

Y(x)  a1 sinhx   a 2 coshx   a1 sinx   a 2 cosx 

(51)

Y(x)  a1sinhx   sinx   a 2 coshx   cosx 

(52)

7

a1sinhL  sinL  a 2 coshL  cosL  0

a 2  a 1

a1  a 2

sinhL  sinL coshL  cosL

(53)

(54)

coshL  cosL sinhL  sinL

(55)

 coshL  cosL sinhx   sinx  Y( x )  a 2  coshx   cosx    sinhL  sinL  

(56)

Eigenvectors with arbitrary scale. Y( x )   cosh n L  cos n L sinh  x   sin  x  aˆ n  cosh n x   cos n x   n n  sinh  n L  sin  n L  

(57) Again, the  n terms are given in Table 1. The eigenvectors are normalized such that L

0 Yn

2 ( x ) dx  1

(58)

Normalize with respect to mass. The leading coefficient is 1 for each mode. eigenvectors normalized with respect to mass are

Thus the

  1   Y1 ( x )   cosh1 x   cos1 x   0.73410 sinh 1 x   sin 1 x     L   

(59)

  1   Y2 ( x )   cosh 2 x   cos 2 x   1.01847 sinh  2 x   sin  2 x     L   

(60)

8

  1   Y3 ( x )   cosh3 x   cos3 x   0.99922 sinh 3 x   sin 3 x     L   

(61)

  1   Y4 ( x )   cosh 4 x   cos 4 x   1.00003 sinh  4 x   sin  4 x     L   

(62)

ρ and L are numerical values only. Y is non-dimensional. The units must be consistent, however. Again, the  n values for equations (59) through (62) are given in Table 1.

CANTILEVER BEAM MODE SHAPES 3 n=3 n=2 n=1

DISPLACEMENT * SQRT (  L )

2

1

0

-1

-2

-3

0

0.1

0.2

0.3

0.4

0.5

0.6

DISTANCE (X/L)

Figure 2. The first three mode shapes are plotted in Figure 2.

9

0.7

0.8

0.9

1.0

Participation Factors The participation factors for constant mass density are L 0

n   Yn ( x ) dx

(63)

The participation factors from a numerical calculation are

1  0.7830 L

(64)

2  0.4339 L

(65)

3  0.2544 L

(66)

4  0.1818 L

(67)

The participation factors are non-dimensional. 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

(68)

The eigenvectors are already normalized such that

L

0

m( x ) Yn ( x )2 dx  1

(69)

Thus,

 L  m eff , n  n 2    m( x ) Yn ( x )dx   0 

2 (70)

The effective modal mass values are obtained numerically.

m eff , 1  0.6131 L

10

(71)

m eff , 2  0.1883 L

(72)

m eff , 3  0.06474 L

(73)

m eff , 4  0.03306 L

(74)

The effective modal mass terms have units of mass.

Response to Base Excitation

y(x, t) w(t)

Figure 3.

The forced response equation for a beam with base motion is taken from Reference 1, page 345. y(x,t) is the relative displacement.

EI

4y x 4



2y t 2

 

2w

(75)

t 2

where w is the base displacement. The term on the right-hand-side is the inertial force per unit length.

11

y( x , t ) 

m

 Yn (x)T n (t )

(76)

n 1

m  2w   Yn ( x )T n ( t )   2 t n 1 

(77)

m  m  4 2 2w EI   T n ( t ) Yn ( x )    Yn ( x ) T n ( t )   x 4 t 2 t 2 n 1  n 1 

(78)

m  m  d4 d2 d2w EI   T n ( t ) Yn ( x )    Yn ( x ) T n ( t )   dx 4 dt 2 dt 2 n 1  n 1 

(79)

EI

 4  m 2 Y ( x ) T ( t )      n n x 4 n 1 t 2 

d4 dx

4

Yn ( x )   n 4 Yn ( x )

m  m  d2 d2w EI    n 4 T n ( t )Yn ( x )    Yn ( x ) T n ( t )   dt 2 dt 2 n 1  n 1 

m  m  d2 d2w EI  n 4   T n ( t )Yn ( x )    Yn ( x ) T n ( t )   dt 2 dt 2 n 1  n 1 

(80)

(81)

(82)

Multiply each term by Yp ( x ) .

m  m  d2 d2w EI  n 4   T n ( t )Yn ( x )Yp ( x )    Yn ( x )Yp ( x ) T n ( t )   Y (x) 2 2 p dt dt n 1  n 1 

12

(83)

Integrate with respect to length. L 

m  m   d2 4 EI  T ( t ) Y ( x ) Y ( x )   Y ( x ) Y ( x ) T ( t )       0  n  n n p   n p dt 2 n  dx n 1  n 1   L d2w Y ( x )dx 0 dt 2 p

  

(84)

  L m L m d2 EI  n 4    T n ( t )Yn ( x )Yp ( x )dx     Yn ( x )Yp ( x ) T n ( t )dx 0 0 dt 2 n 1  n 1  L d2w Y ( x )dx 0 dt 2 p

  

(85) m m  2  L L d EI  n 4  T n ( t )  Yn ( x )Yp ( x ) dx      T n ( t )  Yn ( x )Yp ( x )dx  0 0  n 1 dt 2  n 1 

 

d2w L

 dt 2 0

Yp ( x )dx (86)

m  2  L L EI  n 4 m  d  T ( t )  Y ( x ) Y ( x ) dx  n 0    2 T n ( t ) 0 Yn ( x )Yp ( x )dx   n p  n 1   n 1 dt 



d2w L

 dt 2 0

Yp ( x )dx (87)

13

The eigenvectors are orthogonal such that L

0 Yn (x)Yp (x) dx  0 L

0 Yn (x)Yp (x) dx  1 d2 dt 2

T n (t) 

for n ≠ p

(88)

for n = p

(89)

EI  n 4 d2w L T n (t)    Yp (x)dx  dt 2 0

EI 

(91)

EI n 4 

(92)

n   n 2

n 2 

d2 dt

(90)

T ( t )  n 2 T n ( t )   2 n

d2w L

 dt 2 0

d2

d2w

dt

dt 2

T ( t )  n 2 T n ( t )   n 2 n

Yp ( x )dx

(93)

(94)

Add a damping term.

d2

d2w 2  T n ( t )  2  n n T n ( t )  n T n ( t )   n dt 2 dt 2

(95)

 (t )  2   T (t )   2 T (t )    w T n n n n n n  ( t )

(96)

The damped natural frequency d is d  n 1 -  2

(97)

14

The time variable can be determined from a ramp invariant digital recursive filtering relationship from Reference 2. Note that Δt is the time step.

Tn , i   2 exp   n t cosd t T n , i 1  exp  2 n t T n , i  2     n   2    i   2exp   n t  cos d t  1  exp   n t  2  1 sin d t    n t   n w 3     m n t   d   



1



  n   2  i 1   2 n t exp   n t  cos d t   21  exp  2 n t   2 2  1 exp   n t sin d t   n w 3   m n t  d



1



     n  2   i  2   2  1 sin d t   2 cos d t   n w  2   n t  exp  2 n t   exp   n t   3    d  m n t    1





(98)

15

T n , i   2 exp   n t cosd t T n , i 1  exp  2 n t T n , i  2      n    i   exp(  n t )  cos(d t )  sin d t   1  n w 2      d m n t     1

      n   i 1      exp  2 n t  2 exp(  n t ) sin d t   1  n w 2     m n t   d   1



     n      i  2      exp(    t ) cos(   t )  sin   t  exp  2   t n d d n   n w 2     d  m n t    1

(99)

 ,   2 exp   t cos t T  ,  T n i n d n i 1  exp  2 n t T n , i  2 

exp  n t sin d t  md t

 n w i  2 w i 1  w i  2  (100)

16

The beam relative displacement is y( x , t ) 

m

 Yn (x)T n (t )

(101)

n 1

The beam relative velocity is

y ( x, t ) 

m

 Y n (x)T n (t )

(102)

n 1

The beam relative acceleration is

y( x, t ) 

m

 Y n (x)T n (t )

(103)

n 1

The beam absolute acceleration A is  ( y)  y(x, t ) A(x, t )  w

(104)

References 1. W. Thomson, Theory of Vibration with Applications, Second Edition, Prentice-Hall, New Jersey, 1981. 2. T. Irvine, Modal Transient Analysis of a System Subjected to an Applied Force via a Ramp Invariant Digital Recursive Filtering Relationship, Revision K, Vibrationdata, 2012. 3. T. Irvine, Steady-State Vibration Response of a Cantilever Beam Subjected to Base Excitation, Revision C, Vibrationdata, 2012.

17

APPENDIX A

Example Consider a beam with the following properties: Cross-Section

Circular

Boundary Conditions

Fixed-Free

Material

Aluminum

Diameter

D

=

0.5 inch

Cross-Section Area

A

=

0.1963 in^2

Length

L

=

24 inch

Area Moment of Inertia

I

=

0.003068 in^4

Elastic Modulus

E

=

1.0e+07 lbf/in^2

Stiffness

EI

=

30680 lbf in^2

Mass per Volume

v

=

0.1 lbm / in^3 ( 0.000259 lbf sec^2/in^4 )

Mass per Length



=

0.01963 lbm/in (5.08e-05 lbf sec^2/in^2)

L

=

0.471 lbm (1.22E-03 lbf sec^2/in)



=

0.05 for all modes

Mass Viscous Damping Ratio

The analysis is performed using Matlab script: continuous_beam_base_accel.m. The normal modes results are given in Table A-1. Again, both the mode shape and participation factor are considered as dimensionless, but they must be consistent with respect to one another. Table A-1. Natural Frequency Results, Fixed-Free Beam Effective Modal Mass ( lbf sec^2/in )

Effective Modal Mass (lbm)

Mode

fn (Hz)

Participation Factor

1

23.86

0.02736

0.00074837

0.289

2

149.52

0.01516

0.00022982

0.089

3

418.67

0.00889

7.9028e-05

0.031

4

820.42

0.00635

4.0361e-05

0.016

18

Now apply a 1 G, 24 Hz sinusoidal base acceleration input to the beam. This case could be solved more properly using Laplace transforms. But it provides a good test of the ramp invariant digital recursive filtering relationship. The relative displacement and absolute acceleration at the free end are shown in Figures A-1 and A-2 respectively.

Figure A-1.

The peak relative displacement is 0.2656 in. This is consistent with the steady-state response in Reference 3.

19

Figure A-2.

The peak acceleration is 15.53 G. Again, this is consistent with the steady-state response in Reference 3.

20