free vibration of a simply supported beam considering
TRANSCRIPT
Free vibration of a simply supported beam considering nonlinear
effects due to mid-plane stretching
AME-522 Nonlinear Dynamics
Soohan Kim
1
Introduction
Euler-Bernoulli, Rayleigh and Timoshenko beam theories constitute some of the widely
accepted and well established beam theories. These approximate methods are considered linear
due to the absence of non-linear terms in the derivation of the beam’s equation of motion and
yield sufficiently accurate results for a wide range of arrangements and loading conditions.
For configurations in which nonlinear terms necessarily exist within the partial differential
equation, linear beam theories do not suffice and another method is needed. Take for instance the
case of static or dynamic buckling where the coupling between the transverse motion and axial
motion of the beam is important or when the rotation of the beam’s axis is sufficiently large.
Other examples include nonlinear behavior of materials, geometric nonlinearity and nonlinearity
due to damping.
Thus, a very wide range of real problems require the incorporation of nonlinear terms. In
contrast with linear problems, nonlinear problems cannot be broken into smaller problems so
solving them is much more challenging.
In this paper, the free vibration of a geometrically nonlinear simply supported beam with
sinusoidal loading is investigated. The equation of motion is derived for a general loading using
a combination of the Newton’s law and mode shapes of linear case while assuming that first
mode is a good approximation, thus reducing the problem to a single degree of freedom system.
Stability analysis is used to deduce qualitative behavior of the nonlinear system and to
investigate the difference between the nonlinear and its corresponding linear system.
1 Statement of the problem
Vibration of a simply supported beam shown in Figure 1 is considered. Effect of mid-plane
stretching is included in deriving equations of motion. Shear deformations, rotatory inertia and
acceleration in direction are assumed to be negligible. As in general linear beam theory, effect
of rotatory inertia is neglected relative to those of linear inertia, thereby treating the mass
distribution as if it is concentrated on the neutral axis of the beam. Also, deformations due to
transverse shear are can be ignored if the beam is thin and effective wave lengths of the
individual modes are sufficiently large compared to the thickness of the beam [1] . The
aforementioned effects will be neglected in this study.
2
2 Model
Figure 1-Schematic of the problem and free body diagram (redrawn from [2])
Based on the FBD provided above, the equations of motion are re-derived for our problem
with the aid of studies in the literature like Nishawala [3], Zhang et al. [2] and Kovacic and
Brennan [4].
2.1 Dynamic equilibrium equations
To derive equations of motion, dynamic equilibrium equations are written for the element
shown in Figure 1. Dynamic equilibrium in vertical direction is as shown in equation (1).
∑ (
) ( ) ( )
(1)
Where is the shear force, ( ) is the transverse load, ( ) is the axial load, is
transverse deflection, is mass per unit volume of the beam, is elastic modulus, is second
moment of area of beam cross section and is cross sectional area. It is assumed that
are constant throughout the length of the beam. In writing equation (1), it is also
assumed that rotation (
) is small, and thus (
)
(
) .
Moment equilibrium about point is shown in equation (2), where is bending moment,
is axial deformation and is rotation of the element. Since effect of rotatory inertia is neglected,
3
the right hand side of this equation is set equal to zero. Neglecting rotatory inertia essentially
treats the mass distribution as if it is concentrated along the neutral axis of the beam. Neglecting
second order terms and assuming
(where is the axial deformation), equation (2) can
be simplified as shown in equation (3).
∑
(
) ( ) (
) (
)
( )( (
)) (
)
(2)
(3)
Similarly, equilibrium in horizontal direction is as shown in equation (4). It is assumed that
acceleration in longitudinal direction is insignificant compared to the acceleration in vertical
direction, thus the right hand side of equation (4) is set equal to zero. Also, as in equation (1)
small rotation is assumed such that (
) .
∑
( )
( ) (4)
Substituting equation (3) into equation (1) will result in equation (5). Equation (5) can be
simplified using equation (4) as shown in equation (6).
( ) ( )
(5)
( )
(6)
The relationship between bending moment and curvature is shown in equation (7). This
expression can be derived based on the hypothesis that planes remain plane after deformation
and a line perpendicular to the neutral axis of the beam remains perpendicular to the axis after
deformation. Assuming
expression (7) can be approximated by expression (8). If the
4
assumption
is relaxed, equation (7) can be approximated using higher order terms. In
that case another source of nonlinearity will be introduced into the equation of motion. This
source of nonlinearity is neglected here.
( (
) )
(7)
(8)
Substituting equation (8) into equation (6) will lead to equation (9) which is the governing
equation of forced vibration of an Euler-Bernoulli beam considering the effects of mid-plane
stretching.
( )
(9)
When the beam is under large deflections, axial characteristics of the beam cannot be ignored.
In this case ( ) would be a function of ( ) and equation (9) is nonlinear.
2.2 Internal axial force due to mid-plane stretching
In contrast to linear beam theory where the length of neutral axis of the beam doesn’t change
during the deflection in the model derived in section 2.1, neutral axis of the beam begins to
stretch when ( ) Based on equation (4), if there is no external axial force (i.e. ( )
) then
and so ( ) ( ) is only a function of time and is constant through the
beam length. In this case, ( ) can be written as a function of axial strain ( ). ( ) is a function
of differential length of the deformed element and can be approximated by Taylor series
expansion as shown in equation (11). Details of this approximation are shown in Appendix A.
( ) ( ) (10)
( )
√( ) ( )
(
)
(11)
5
Substituting equation (11) in (10) will result in the following expression for internal axial
force.
( ) (
(
)
) (12)
Integrating both sides of equation (12) with respect to will result in equation of ( ) as
shown by equation (13).
( ) * ( ) ( )
∫ (
)
+ (13)
Substituting equation (13) in (9), the equation of motion can be written as follows.
* ( ) ( )
∫ (
)
+ (
) ( )
(14)
2.3 Satisfying Boundary conditions
Boundary conditions (B.C’s) assumed to solve equation (9) are two immovable hinges at the
ends of the beam. In this case the following expressions are true.
( ) (15)
( ) (16)
( ) (17)
( ) (18)
( )
(19)
( )
(20)
Substituting equations (15) and (16) in equation (14) will result in (21), a nonlinear
differential equation in terms of derivatives of ( ). Note that substituting equations (15) and
(16) in equation (13) will result in equation of axial force as shown by equation (22). One may
6
notice that we already assumed
to simplify equation (7) so we can expect that for the
range of
, our previous assumption is true and axial force will also be small.
(∫ (
)
)(
) ( ) (21)
( )
∫ (
)
(22)
To solve the partial differential equation (21), separation of variables is used as shown in
equation (23), where ( ) is the mode shape and ( ) is the time function of the mth
mode.
Since mode shapes are an orthogonal set of functions, solution can be expanded in terms of mode
shapes. In case of a simply supported beam, the mode shape can be defined by a sine functions as
shown by equation (24). It can be easily shown that this mode shape satisfies B.C.’s in equations
(17) to (20). External force ( ) can be also expanded in terms of mode shapes as shown in
equation (25). Multiplying both sides of equation (25) and integrating over the distance between
supports ( ) results in equation (26). Note that due to orthogonality of mode shapes, there is
only one non-zero integral in the summation on the right hand side.
( ) ∑ ( ) ( )
(23)
( ) (
) (24)
( ) ∑ (
) ( )
(25)
( ) ∫ ( )
(
)
∫ ( (
))
∫ ( )
(
) (26)
Using equations (23), (24) and (25), equation (21) can be written as follows.
7
( (
)) [ ∑ ( )
∑ ( ) (
)
(∫ (∑∑(
)
( ) (
)
(
) ( ) (
))
)(∑ ( )
(
)
)]
∑ (
) ( )
(27)
Using orthogonally property for cosine functions inside the integral, equation (27) can be
simplified as follows.
( (
)) [ ∑ ( )
∑ ( ) (
)
(∑(
)
( ( ))
)(∑ ( )
(
)
)]
∑ (
) ( )
(28)
So equation on mth
mode of vibration is as follows.
( ) (
)
( )
(∑(
)
( ( ))
) (
)
( ( )) ( ) (29)
As you can see in equation (28), equation of mth
mode is highly coupled to other modes. Due
to the nonlinear nature of our equations, mode shapes of linear case cannot decouple our
equations. Even if first n modes are used as an approximation of the exact solution, there is still a
highly coupled system of n equations and n unknowns [3]. If only the first mode is used as an
approximation of the response, there will be no coupling terms as shown in equation (30), which
is in the form of a Duffing equation with cubic nonlinearity. Note that due to the difficulty that
arises in computing nonlinear frequency and spatial distribution of the multiple-degree-of-
freedom system, many existing references treat the study of the nonlinear vibration of beams
assuming a single mode shape [5]. Here, the problem is treated in a similar manner.
8
( ) (
)
( )
(
)
( ) ( ) (30)
In case of a harmonic load at mid span, ( ) can be shown by equation (31). Now, using
equation (26), ( ) can be found as shown in equation (32). Finally by substituting ( ) from
equation (32), equation (30) can be simplified as shown in equation (33). Notice that the
vibration of first mode is in the form of a Duffing equation with cubic nonlinearity. Since
coefficient of the cubic nonlinear term is positive, this oscillator is experiencing hardening
stiffness.
( ) (
) ( ) (31)
( )
( )∫ (
)
(
)
( ) (32)
( ) (
)
( )
(
)
( )
( ) (33)
3 Free vibration of first mode
Now we investigate the free vibration of the first mode. Understanding the behavior of the
free vibrating system may yield insightful results that pertain to the forced system. In case of free
vibrations, the force term is equal to zero as shown in equation (34). Let ( )
and
where √
(
) is the period of equivalent linear system. To be able to better interpret the
results, equation (34) is written in dimensionless form using and as dimensionless parameters
as shown in equation (39), where is gyration radius of the beam and is defined by (38).
( ) (
)
( )
(
)
( ) (34)
√
(
) (35)
9
(36)
( ) ( ) ( ) ( ) ( )
( ) (37)
√
(38)
( ) ( )
( ) (39)
3.1 Stability analysis
Using (39), substitute the dimensionless linear and nonlinear stiffness coefficients as follows.
( ) ( ) ( ) (40)
Where and
are dimensionless linear and nonlinear stiffness
coefficients. Next, express (40) in terms of state variables ( ) and ( ).
(41)
(42)
From (41) and (42), the fixed points can be easily determined by setting .
( ) (√ ) ( √
) (43)
Recall that both and in equation (40) are positive numbers, thus only one fixed point at
the origin (0,0) persists, as the other two fixed points become imaginary. To check the stability
of the trivial fixed point, consider the linearized system ignoring the higher order terms. The
Jacobian matrix is as follows.
( )
[
]
(
)
[
](
)
(44)
10
The linearized system is as follows.
[ ] [
(
) ](
)
* + (45)
The linearized system (45) can be expanded and written as a 2nd order system.
[ ( ) ] (46)
To determine the stability of the fixed point (0,0), compute the Jacobian at (0,0) as such.
[ ] [
] * + (47)
Observe the trace and determinant . The eigenvalues of the linearized system are
( √ ).
√ √ √ √ (48)
In general, the eigenvalues of the linearized system indicate the flow in phase space and the
corresponding eigenvectors form the stable and unstable subspaces. Pure imaginary eigenvalues
indicate existence of centers in the linear system, and polar transformation indicates that the
imaginary part of the eigenvalue represents the angular velocity.
Figure 2-Phase portrait around a center in a linear system
Hyperbolic functions are defined as linearized functions in which all eigenvalues contain non-
zero real parts. Hatrman-Grobman Theorem states that when a linearized system is hyperbolic,
the stability of the linearized system correctly represents the stability of the nonlinear system,
11
and the topology of the linearized system persists in the nonlinear system with a continuous 1:1
invertible transformation [6]. Note that for the system under investigation, all eigenvalues are
pure imaginary numbers so the real part is equal to zero, thus, the linearized system does not
predict the stability or topography of the nonlinear system [6]. Thus, another method is needed
to determine stability of the centers in the nonlinear system.
Recall the nonlinear system described in (40) to (42), which represents the mechanically
oscillating nonlinear system. We can compute the total energy of the system as the sum of its
potential and kinetic energy as follows. Rewriting (31) yields the following.
( ) ( ) ( ) ( )
( )
(49)
Where the potential energy of the spring is computed as the following.
( )
(50)
Thus the total energy is observed to be C-1 continuous and computed as follows.
( )
(51)
Taking the partial derivative of the total energy with respect to the parameters
( )
(52)
Substituting (52) back into (41), (42) and using the gradient of the potential given in (49), we
can rewrite the system in the transformed coordinates as such.
(53)
(54)
The above equations (53) and (54) are known as Hamilton’s Equations [6], and ( )
( ) is known as the Hamiltonian. Taking the time derivative of the Hamiltonian and
substituting (53) and (54), we can see that the total energy of the system is conserved.
12
[
] [
] (55)
Thus, by (55) the Hamiltonian (total energy) of the system is conserved and always constant
along a trajectory. The total energy is positive definite, C-1 continuous and has a local minimum
at the trivial fixed point. This can also be shown by plotting (51) in Figure 3. Thus, for small
perturbations from the trivial fixed point, centers will persist in the nonlinear system and stable
for all time.
Figure 3- Total Energy
3.2 Period of oscillations of first mode
In this section conservation of energy is used to find an expression for period. From equation
(51) total energy of the system can be shown as follows.
(
)
(56)
Let be the maximum displacement of the oscillator. We know that when kinetic
energy is zero and all of energy is in the form of potential energy. Since energy is conserved and
constant on each trajectory, it can be written as a function of as shown below.
(57)
We also know that it takes a quarter of period for the oscillator to go from to . So
if we solve equation (56) for and integrate from to for
the result is a
quarter of the period.
13
√
(58)
∫
∫
√
(59)
Let’s simplify the expression under the square root. If we substitute from (57) into (59).
(
) ( )
( ) ( ( ))
(60)
Let , using this change of variable, equation (60) will be as follows.
( ) ( ( )) ( ) (
)
( ) ( )(
)
(61)
Now put (61) back into (59).
∫
√ ( ) (
) (
)
√(
)
∫
√( ) (
)
√(
)
∫
√( )( )
(62)
14
Where k is defined as follows.
(63)
Integral in the equation (62) is in the form of a complete elliptic integral.
( ) ∫
√( )( )
(64)
So period is as follows
( )
√(
)
(65)
As it can be seen in equation (65) in contrast with a linear system, period of this nonlinear
system depends not only on value of linear and nonlinear stiffness but also on the amplitude of
vibration.
3.3 Example
In this section, an example is studied to see the effect of the nonlinear term on free vibrations
of the beam. First, we need to make some assumptions. Assume that initial condition is in the
form of an initial displacement and initial velocity is zero. Since energy is conserved this
maximum displacement will be the same in the following cycles. As mentioned before and can
be clearly seen in equation (39) (repeated here from before) our case is a case of hardening
stiffness. It was shown in Section 3.2 that period of such free vibration will depend on amplitude
of the initial displacement. A steel beam of length 15 feet is designed for a concentrated load that
consists of 10 Kips of dead load and 20 Kips of live load at its mid span based on Load factor
design method. W14X30 satisfies design requirements and its corresponding gyration ratio and
height of cross section are equal to and . Based on design code beam cross
section should be chosen in a way that maximum deflection under live load is less than
where
L length of the beam. In terms of our parameters it means ( )
. (Note
that ( ) in equation (39) is deflection normalized with respect to length). It should also be noted
that maximum strain up to which stress-strain relationship for steel is linear is . Since
15
we assumed linear material in our model, initial displacement should be chosen in a way that this
assumption is valid.
( ) ( )
( ) (39)
Assuming that initial displacement is applied very slowly, we can use statics to find the
maximum deflection up to which beam steel behaves linearly. We use static deflections under a
concentrated load. If load is applied at mid span, maximum deflection happens right under the
load and is equal to
. Maximum moment happens at the same point and is equal to
. Note that in Euler-Bernoulli beam theory,
where is the largest
distance from the neutral axis. Substituting and using the fact that maximum stress in
linear range is we have
we can find
the maximum value
of force under which steel behaves linearly. So
and
which
for our cross section would be
0.004=0.4%.
Based on the above explanation, assumption of linear behavior of beam is valid as long
as ( ) . Now that the appropriate range of ( ) is known, equation (39) can be solved
numerically. Doing so ( ) versus time, ( ) and phase portrait are plotted for
different values of ( ) assuming that ( ) . Vertical and horizontal scale of phase portrait
is the same in Figure 4 to Figure 7. The following results can be seen in the figures:
1- It can be clearly seen from phase portrait that fixed point (0,0) is a center and response is
periodic (phase portrait is closed).
2- Shape of closed curves around the origin is stretched along the velocity axis. This can be
explained by equation (56) which is repeated here. This equation states that for each
closed curve around the origin there is a corresponding constant value of energy. At the
point of maximum displacement, velocity is zero so assuming that
is dominant
maximum displacement will be proportional to √
. At the point zero displacement
maximum velocity will be proportional to √
. So maximum velocity has a larger value
and phase portrait is stretched along the velocity axis.
16
(
)
(56)
3- As we increase ( ), we are putting more energy into the system so phase portrait will
expand.
4- Note that time was made dimensionless with respect to period of the equivalent linear
oscillator. So if the period of non-dimensional oscillator is one it means that period is
equal to period of a linear oscillator with the same linear stiffness. By looking at plots of
( ) versus one can see that as long as assumption of linear behavior of beam is valid
( ( ) ), changes in period of vibration due to nonlinear term are negligible and
there are five cycles from 0-5 which is the shown interval on the graph. Effect of
nonlinear term is visible in Figure 8 which is plotted for ( ) and ( ) . One
can clearly see that the period is reduced in this case and more than 7 cycles can be seen
from 0-5. ( ) means that initial displacement is one fifth of beam length of course
at such a large displacement assumption of linear material behavior is not true anymore.
Figure 4- plots of ( ), ( ) and phase portrait for ( ) and ( )
17
Figure 5- plots of ( ), ( ) and phase portrait for ( ) and ( )
Figure 6- plots of ( ), ( ) and phase portrait for ( ) and ( )
18
Figure 7- plots of ( ), ( ) and phase portrait for ( ) and ( )
Figure 8- plots of ( ), ( ) and phase portrait for ( ) and ( )
19
4 Conclusion From the example in the previous section can be concluded that choosing the type of
nonlinearity that we want to model is very critical. As it was seen nonlinearity due to mid-plane
stretching does not play an important role if only linear range of materials is considered.
Although forced vibrations under harmonic forces were not covered here, due to changes in
frequency of system with amplitude dynamics will be much more interesting than a linear
system. In terms of steady state motions there can be harmonic motions and chaotic motions
based on input parameters and initial conditions [7]. Multiple resonances would persist even if
frequency of the exciting force and properties of the system are kept constant and the only
parameter changing is the forcing amplitude.
20
5 References
[1] W. J. Bottega, Engineering Vibration. CRC Press, 2006. [2] S. Zhang, Z. Liu, and G. Lu, “Nonlinear Flexural Waves in Large-Deflection
Beams,” Acta Mechanica Solida Sinica, vol. 22, no. 4, pp. 287–294, Aug. 2009.
[3] V. V. Nishawala, “A study of large deflection of beams and plates,” M.S., Rutgers The State University of New Jersey - New Brunswick, United States -- New Jersey, 2011.
[4] I. Kovacic and M. J. Brennan, The Duffing Equation: Nonlinear Oscillators and their Behaviour. John Wiley & Sons, 2011.
[5] A. Y. T. Leung and S. G. Mao, “A symplectic Galerkin method for non-linear vibration of beams and plates,” Journal of Sound and Vibration, vol. 183, no. 3, pp. 475–491, Jun. 1995.
[6] S. H. A. STROGATZ, Nonlinear Dynamics and Chaos: With Applications to Physics, Biology, Chemistry, and Engineering. Westview Press, 1994.
[7] Y. Ueda, “Steady motions exhibited by Duffing’s equation: A picture book of regular and chaotic motions,” in Presented at the Eng. Foundation Conf. on New Approaches to Nonlinear Problems in Dyn., Monterey, Calif., 9-14 Dec. 1979, 1980, vol. -1, pp. 9–14.
21
Appendix A Derivation of an approximate expression for axial strain by Taylor Series expansion
( )
√( ) ( )
√
( ) ( )
√(
)
(
)
(A 1)
Let ( )be defined equations (A 2), (A 3), (A 4).
(A 2)
(A 3)
( ) √( )
(A 4)
Using 1st and 2
nd order terms in the Taylor series expansion about point (0,0), ( )can be
approximated as follows.Here only the first four terms are used.
( ) ( ) (
( )) (
( ))
* (
( )) (
( ))
(
( ))+
(A 5)
( ) (A 6)
( )
√( )
( )
√( )
( )
(A 7)
√( )
√( )
( )
(A 8)
22
√( )
√√( )
( )
( )
[( )
]
( ) (A 9)
Substituting (A 6) to (A 9) into (A 5), ( ) can be written as follows.
( ) ( ) ( ) (
) ( ) (
) (
) ( ) (
)
(
) (
) (
)
(A 10)