longitudinal vibration bars

For educational use at Canterbury and TUHH © S. Ilanko 2005

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4.1 Equations of Motion Using Newton's Second Law

We have seen that for discrete systems, the equations of motion may be obtained by applying Newton's second law to the masses along the directions corresponding to their degrees of freedom. The number of equations is equal to the number of degrees of freedom. The number of natural frequencies (and modes) is also the same. On the other hand continuous systems have distributed elastic properties as well as distributed mass. They have infinite degrees of freedom. For example, a simply supported beam can vibrate freely in any one of the infinite number of modes having any number of half-sine waves. So one may wonder whether there will be an infinite number of equations of motion. Fortunately this is not the case, and for most common engineering problems such as vibration of bars, shafts and thin beams, there is only one equation of motion having an infinite number of possible solutions. There are cases with more than one equation, for example two equations for arches and three equations for shallow shells, but in any case the number of equations of motion is finite. This leads to the question: "How does one obtain an equation of motion for a continuous system that has infinite number of possible solutions?". What is required is a general equation, which represents Newton's second law at any point in the continuous system. This is obtained by applying Newton's second law to an infinitesimal element at an arbitrary location in that system. This results in a partial differential equation. This is the equation of motion for the continuous system. Once an equation of motion for a given system is obtained then applying the boundary conditions to its general solution gives a set of equations. The solution of these simultaneous boundary condition equations results in a determinantal equation involving transcendental functions. This equation is the frequency equation. In some simple cases closed form solutions to the frequency equation is possible and one can get explicit expressions for the natural frequencies. For many practical problems however, the roots of the frequency equation have to be found by a trial and error computational procedure, and in this way the frequencies can be delimited to any desired accuracy. This method is referred to as the exact method because it does give exact results to several sets of standard problems but it must be noted that the derivation of the frequency equation can be very difficult or impossible for many engineering systems. Other approximate numerical techniques are more suitable for such problems but the approximate methods come with a price and that is the uncertainty in the error. For this reason, engineers, researchers and software developers who want to use or develop an approximate method should check the

4. Natural Frequencies and Modes of Continuous Systems Using the “Exact” Method


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accuracy of their numerical procedure by testing it on a related problem for which an exact solution is obtainable. For example one may need to calculate the natural frequencies of a mechanical arm with a complex geometry and support conditions (See Figure 4.1.1). Before using the results of an alternative numerical procedure or software package, one should check if the results obtained using the alternative method agree with results from exact method for a simpler case (Figure 4.1.2). While the actual problem involving a tapered beam with varying axial force is very difficult to solve exactly, any procedure to analyze this system could be tested on a uniform beam carrying an end mass subject to a constant axial end force which could be solved analytically.

Numerical procedures can only tell us information about specific systems and to study the effect of varying system parameters one needs to generate results for a number of sets of input data. The exact method gives us an opportunity to get a deep understanding of the general behavior of various systems. For these reasons, we will devote a considerable amount of space to discuss the exact method and the results of its application to some common continuous systems. We will start by describing the general procedure for deriving an equation of motion for a continuous system. Basic Steps for Deriving Equations of Motion Step 1: An infinitesimal element of the vibrating body is isolated and a free body

diagram is sketched showing all the forces acting on that element. Step 2: The induced elastic actions (forces or moments) are obtained in terms of the co-

ordinate corresponding to the vibratory motion using the following relationships: (a) force (or moment) - stress relationship;

Figure 4.1.1 Figure 4.1.2

P


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(b) constitutive (stress - strain) relationship; (c) strain - displacement relationship. Step 3: The induced actions in the direction of motion are summed to obtain the net

elastic action. Step 4: If there is any imposed dynamic loading (force vibration), its component in the

direction of motion is added to the net elastic action to obtain the net action (force or moment).

Step 5: The net action (force or moment) is related to the corresponding acceleration (linear or angular) using Newton's second law. The resulting equation is the equation of motion. This would be in the form of a partial differential equation.

Free Vibration Analysis For free vibration analysis, step 4 is not necessary as there are no applied dynamic loads. Substituting the prescribed boundary conditions into the general solution of the equation of motion yields a frequency equation, often in a determinantal form. The roots of the frequency equation give the natural frequencies. The corresponding natural modes can also be determined using the general solution and the boundary conditions at each of the natural frequencies. This is best illustrated using the examples in the next section. We will first study the free vibrational behaviour of some simple skeletal elements such as bars, shafts and beams before proceeding to two-dimensional systems. In this chapter we will be concerned only with the calculation of natural frequencies and modes. The determination of the actual displacement of a system for any given set of initial conditions will be discussed in chapter 6.


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4.2 Longitudinal Vibration of Bars Derivation of Equation of Motion: We will start with the free longitudinal vibration of bars which is one of the simplest cases of continuous systems. Figure 4.2.1 shows a finite segment of a bar which has a cross sectional area A (which may vary along the length) and is made of an elastic material having a density ρ and Young's modulus E. The positions of an infinitesimal element of the bar having length δx at rest, and at time t during vibration are also shown in the figure. Let u(x,t) be the longitudinal displacement of the bar at time t, and F(x,t) be the dynamic axial force induced at the same instance. F is taken as positive if it is tensile.

x δx

u δu+δx

Element in Equilibrium Displaced Element

O

Figure 4.2.1 Let us now proceed to derive the equation of motion. Step 1: Sketch the free body diagram of an infinitesimal element:

ρAδx

F F+δF

u

Figure 4.2.2 Step 2:

(a) Force - Stress relationship is: F = Aσ (4.2.1) (b) Stress - Strain relationship is: σ = Eε (4.2.2)

(c) Strain - Displacement relationship: xu

∂∂=ε (4.2.3)


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Using the last three equations, xu

EAF∂∂= (4.2.4)

At any given time t, xxu

EAx

F δδ

∂∂

∂∂= (4.2.5)

Step 3: The net elastic force is: F + δF - F = δF Step 4: Since there are no external forces, this is the net force on the element and is

given by equation (4.2.5) Step 5: Using Newton's second law:

( ) 2

2

tu

xAF∂∂= δρδ

Substituting equation (4.2.5) into the above gives

2

2

tu

Axu

EAx ∂

∂=

∂∂

∂∂ ρ (4.2.6)

This is the partial differential equation governing the longitudinal motion of an elastic bar. For a homogeneous, uniform bar, EA is constant and equation (4.2.6) reduces to

02

2

2

2

=∂∂−

∂∂

tu

xuE

ρ (4.2.7)

This may be solved using the method of separation of variables. Let )()(),( tgxftxu = (4.2.8a)

where f and g are functions of x and t only respectively. To determine the natural frequencies and modes of linear systems, as usual, one can apply the simple harmonic time function. Thus g(t) = sin )( φω +t (4.2.8b)

Substituting equations (4.2.8a,b) into equation (4.2.7) gives the following ordinary differential equation:

022

2

=+

f

dxfdE ω

ρ (4.2.9)

General Solution for Uniform Bars of Finite Length For a uniform bar of length L, the general solution to equation (4.2.9) may be written (see Appendix A for the mathematical derivation)


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( )

+

=Lx

HLx

Hxf λλ cossin 21 (4.2.10)

where H1 and H2 are undetermined coefficients and λ is a dimensionless frequency parameter given by

E

Lρωλ = (4.2.11)

For convenience we will later switch to a non-dimensional axial coordinate ξ which is defined by x/L but until we complete the present case we will retain the axial coordinate x. Equation (4.2.11) may be obtained by substituting equation (4.2.10) into equation (4.2.9). The general solution may be written in other forms also. For example another valid form

is ( ) ( ) ( )[ ]xHxHxf λλ cossin 21 += where Eρωλ = . The natural frequencies do not depend

on the form of general solution as explained later. In this book the general solution for various systems will be formulated to make the frequency parameters non-dimensional. Equation (4.2.7) is a second order partial differential equation. The order of differentiation is two with respect to both x and t. Therefore a complete solution requires two boundary conditions and two initial conditions. However, for the calculation of natural frequencies and modes only the boundary conditions are required. Substituting the general solution into the boundary conditions results in frequency equations the roots of which are the natural frequencies. The natural modes are obtained by substituting the calculated natural frequencies and boundary conditions into the general solution (or its derivative). The following examples illustrate this procedure. Case 1: Both Ends Longitudinally Restrained

x = 0 x = L

Figure 4.2.3 Since both ends are restrained the displacement u is zero at the boundaries. At x=0, u=0.i.e. u(0,t) = 0.


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Since this is true for any t, f(0) = 0 Substituting equation (4.2.10) into this gives: H1 (0.0) + H2 (1.0) = 0 (i) Similarly, f(L) = 0 gives: H1 sin λ + H2 cos λ =0 (ii) From equation (i), H2 =0. Substituting this into equation (ii) gives: H1 sin λ = 0 Therefore either H1 = 0, or sin λ = 0. H1= 0 is a trivial solution as H2 is also zero, which means there is no motion. However if sin λ = 0, H1 is not necessarily zero. This means vibration is possible if and only if sin λ = 0. Hence sin λ = 0 is the frequency equation, the roots of which are λ = 0 or λ = nπ, where n is any integer. It can be seen that λ = 0 also results in a trivial solution as u(x,t) would be zero. Using equation (4.2.11),

ρ

πω EL

n=

As there are infinite possible values of n (1,2,3....∞), there are infinite number of natural frequencies. The lowest (fundamental) natural frequency is given by

ρ

πω EL

=1

The nth natural frequency is ρ

πω EL

nn =

To obtain the modes, H2 = 0 and λ = nπ may be substituted into the general solution for u.

This gives ( )φωπ +

= tL

xnHu sinsin1

The nth mode is

Lxnπ

sin

The first two modes are shown in Figure 4.2.4. It should be noted that this figure is only a representation of the displacement that takes place longitudinally.


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x = 0 x = L

n = 1n = 2

Figure 4.2.4 It is worth noting here that if ( ) ( ) ( )[ ]xHxHxf λλ cossin 21 += had been chosen as the general

solution, the frequency equation would be sin (λL) = 0, giving λL = nπ, but in this case,

sinceEρωλ = , the frequency expression would still be

ρπω EL

nn = .

The above solution may be obtained in a general manner using a matrix approach. Equations (i) and (ii) may be written [B]H =0, (iii)

where

=

λλ cossin10

][B and

=2

1

H

HH

This is an eigen value problem, and for non-trivial solution |B|= 0. (iv) This is the frequency equation in determinantal form. It gives (0.0) cos λ - (1.0) sin λ = 0. ie. sin λ = 0, and as before λ = nπ. Generally, the frequency equations are transcendental functions and sometimes they are only expressed in a determinantal form, as expanding large determinants analytically, could be difficult. The roots may have to be found by a trial and error procedure, by calculating the determinant for various trial values of frequency and searching for a change in the sign of the determinant. This approach is suitable for programming and will be used in some of the examples for which interactive multi-media programs are available on the CD. Case 2: Both Ends Longitudinally Free Let us now consider free boundaries.


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Figure 4.2.5 At a free boundary, the axial strain is zero.

Using the strain - displacement relationship 0=∂∂

xu

. Since this is true for any time, 0=′f

Differentiating the expression for f in equation (4.2.10)

( ) ( )( )λξλξλsincos 21 HH

Lf +

=′ , where ξ = x/L (4.2.12)

0)0( =′f gives: ( ) ( ) 00.00.1 21 =

−

HL

HL

λλ (i)

0)( =′ Lf gives: 0sincos 21 =

−

λλλλH

LH

L (ii)

Once again, we may write this in the form [B]H =0,

[ ]

−=

λλλλλ

sin)/(cos)/(0/LL

LB and

=2

1H

HH

For non-trivial solution of equations (i) and (ii) |B|= 0 gives 0sin2

=

− λλL

This is possible if either λ = 0, or sin λ = 0. Let us investigate the two possibilities. If λ = 0, unlike for case 1 the solution is not trivial. From equation (4.2.8), u = H2 sin(ωt+φ). But ω is zero since λ is zero. Therefore, u = H2 sin φ. The displacement is independent of time. As u′ is zero, the bar remains free of any dynamic axial strain. The relative positions of all points in the bar will remain unchanged. This corresponds to a rigid body motion of the bar with zero frequency. Such systems that have at least one zero natural frequency associated with one or more modes of rigid body motion are classified as semi-definite systems.

If λ ≠ 0, then λ = nπ, and the natural frequencies are given byρ

πω EL

nn = as for case 1.

x = 0 x = L


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From equation (i), for λ ≠ 0, H1 = 0 giving ( ) ( )φωπξ += tnHu sincos2

The nth mode is: ( )πξncos

The natural modes for this system are shown in Figure 4.2.6.

x = 0 x = L

n = 0 (Rigid Body Motion)

n = 2

n = 1

Figure 4.2.6 Case 3: One End Restrained, Other End Free Now let us consider a combination of fixed and free boundaries.

x = 0 x = L

Figure 4.2.7 At x=0, the condition is the same as that for case 1 resulting in the following equation: H1 (0.0) + H2 (1.0) = 0 (i) At x=L, the condition is the same as that for case 2.

ie. [ ] 0sincos 21 =−

λλλHH

L (ii)

For non-trivial solution of equations (i) and (ii), λ cos λ = 0. (Note we have missed some obvious steps including the definition of the elements of [B] and the statement |B| = 0. From here onwards these steps will not be included except for some complicated systems). There are two possible solutions. Either λ = 0 or cos λ = 0. From equation (i), H2 = 0. If λ = 0 then ( )λξsin would also be zero. This means u(x,t) would

be zero, which is a trivial solution.


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Therefore for non-trivial solution, cos λ = 0 The roots are: λ = (n-½)π, for n = 1,2....

Using equation (4.2.11), ρ

πω EL

n2

)12( −=

Substituting H2 = 0 and the roots for λ into the general solution, The nth mode is: ( )( )πξ21sin −n

Figure 4.2.8 illustrates the modes for this case.

x = 0 x = L

n = 1

n = 2

Figure 4.2.8 It may be noted that whether a system is semi-definite or not could be easily checked by considering the question whether the system can be displaced freely without inducing any internal strains. In the case of a free-free bar, such a rigid body displacement is possible, but for Cases 1 and 3 it is not so. Case 4: One End Restrained, Other End Carrying A Concentrated Mass We have seen cases where either the displacement or the force is zero at a boundary. There are many situations where neither of these is zero. One such example is a bar carrying a concentrated mass (let us say a particle possessing mass or a completely rigid body) at one end as shown in Figure 4.2.9.

Figure 4.2.9 At x = 0, as for Case 1, f(x) = 0. ie. H1 (0.0) + H2 (1.0) = 0 (i) At x = L, the boundary condition we need is not so simple.

x = 0 x = L

m0


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When dealing with a boundary where neither the displacement nor the force is zero, one has to look for a relationship between the two. In the case of a connected mass, this relationship is obtained by applying Newton's second law to the mass. In order to do this for a finite mass m0, it is best to sketch a free-body diagram (see Figure 4.2.10).

F(L,t)

),( tLu

m 0

Figure 4.2.10 - F(L,t) = m0 ü(L,t) i.e. -EA u′(L,t) - m0 ü(L,t) = 0 But in determining natural modes, the motion is simple harmonic giving ü = -ω2 u. Therefore -EA u′(L,t) + m0 ω2 u(L,t) = 0 Since this is true for any t, -EA f ′(L) + m0 ω2 f(L) = 0 Substituting equations (4.2.10) and (4.2.12) into the above gives:

[ ] [ ] 0cossinsincos 212

021 =+−−

λλωλλλHHmHH

LEA

Rearranging we get

0cossin

sincos 2

022

01 =

+−

− λωλλλωλλm

LEA

HmL

EAH (ii)

For non-trivial solution of equations (i) and (ii) |B|= 0. This gives the frequency equation:

0sincos 2

0 =− λωλλm

LEA

Using equation (4.2.11) this may be reduced to λ tan λ - η = 0, where η is the ratio of the total

mass of the bar to the tip mass m0 given by 0m

ALρη = .

Once the frequencies are found, the modes may be determined. Mode calculation means finding the relationship between the constants H1, H2 by using all but one of the boundary conditions. In the present case since H2 is zero, the mode is given by the term associated with H1. That is cos (λx/L).


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Special cases: If m0 → 0, η → ∞. Therefore from the frequency equation tan λ → ∞. The roots are given by λ = (n-½)π which agrees with the result for case 3 as to be expected. If m0 → ∞, η → 0. This means either λ or tan λ must be zero. This agrees with the roots of the frequency equation for case 1. This demonstrates the fact that an infinite mass corresponds to a fully restrained end. However, it should be noted that there would be another mode corresponding to a low frequency vibration of the end mass in which the bar provides the elastic

restraint only. This frequency may be estimated by the formula 0m

k=ω where L

EAk = .

The solution to Case 4 is available as an interactive tutorial modes951.exe. By changing the magnitude of the end mass to a very large value one can observe that the first few natural modes except for the lowest one, have very little translation at the end carrying the mass. The lowest frequency will monotonically decrease with mass, and approach zero as the mass approaches infinity. It is easy to miss this frequency if one looks for the roots of the transcendental frequency equation, but it may be estimated using the simple formula given above. When dealing with non-standard boundary conditions, it is good to check the results by putting limiting values (for example, in the case of an attached mass setting its magnitude to zero and a very large value) and see if the solution for the special cases agree with what is readily available. Case 5: Both Ends Partially Restrained Another end condition where one needs to obtain a relationship between the force and displacement is a bar that is connected to a support that is not fully rigid (immovable). The flexibility of a support may be modelled by an elastic spring of finite stiffness. Consider the vibration of the bar shown in Figure 4.2.11, which is partially axially restrained by linear elastic springs of stiffnesses k1 and k 2, at x = 0 and at x = L respectively.


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x = 0 x = L

k 1 k 2

Figure 4.2.11 At x = 0, the force in the spring = the force induced in the bar (Newton's third law). ie. k1 u(0,t) = EA u′(0,t). Substituting equations (4.2.8) and (4.2.10) into the above gives:

( ) ( )[ ] ( ) ( )[ ]0.00.10.10.0 21211 HHL

EAHHk −

=+ λ

ie. 0211 =+

− HkHL

EAλ

(i)

Similarly at x = L, - k2 u(L,t) = EA u′(L,t) This is true for any t. Therefore - k2 f(L) = EA f′(L) (The sign of spring force is negative because a positive f(L) causes compression in the spring) Substituting equations (4.2.10) and (4.2.12) into the above equation yields:

0sin

coscos

sin 2212 =

−−

+− HL

EAkH

LEA

kλλλλλλ (ii)

For non-trivial solution of equations (i) and (ii) |B|= 0. This results in the following frequency equation:

( )

( )[ ] 0tan 2221

21 =−

++λλλ

EALkkLEAkk

Special Case: If k1 = k2 then the system is symmetrical. This means the system can vibrate either in a symmetrical mode or anti-symmetrical mode. The above frequency equation reduces to:

( )[ ]222

2tan

λλλ

EALkkLEA−

+ (iii)

If a system is structurally symmetrical, its frequencies and modes may be calculated by considering only one half of the structure and by using conditions of symmetry or anti-symmetry at the centre as explained below. Finding modes:


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Using the first boundary condition, 11

2 HLkEA

H

= λ

f(x) =

+

Lx

LkEA

Lx

Hλλλ

cossin1

1

Mode is

+

Lx

LkEA

Lx λλλ

cossin1

When calculating modes, the ratios of the constants is calculated by taking one of them as unity. If the term that is assumed to be unity were zero there would be arithmetic overflow as a division by zero is involved. Therefore if any denominator term has a very small magnitude then a different term should be selected as unity. For example, in the current problem, if k1 is zero the above expression for the mode would become infinity. In such a case H1 should be set to zero and the mode would then be cos (λx/L)

Case 6: Symmetrical and Anti-Symmetrical Vibration of the System in Case 5 In a symmetrical vibration, the mode on one side of the centre line (x = L/2) is a mirror image of that on the other side. This means u(0,t) = -u(L,t) and for 0<x<L, u(x,t) = -u(L-x,t).

At

−=

= tL

utL

uL

x ,2

,2

,2

This is possible only if 0,2

=

tL

u (a)

Equation(a) forms a boundary condition for symmetrical vibration. In an anti-symmetrical vibration, the displacement of both sides of the centreline will be in phase. u(x,t) = u(L-x,t)

As ∆ → 0,

∆−=

∆+ tL

utL

u ,2

,2

This means at 2L

x = , δu = 0


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ie. 0,2

=

′ tL

u (b)

Equation (b) is a boundary condition applicable for the anti-symmetrical vibration of a symmetrical system. This means the strain, stress and the force at the centre are zero. It is safer to consider the anti-symmetrical boundary condition for longitudinal system in terms of force being zero as this would hold even in cases where there is a concentrated mass at the centre (see Case 9). To obtain the natural frequencies and modes corresponding to symmetrical vibration, equation (a) is used in addition to the boundary condition at any one of the ends. Similarly for analysing the vibration corresponding to anti-symmetrical modes, equation (b) is used in conjunction with any one of the end conditions. To illustrate this procedure, let us consider the special case of k1 = k2 = k in case 5. As the system is symmetrical let us consider the left half of it. At x = 0, the spring force is equal to the force induced in the bar. This yields:

021 =+

− kHHL

EAλ

(i)

For symmetrical vibration, substituting equation(4.2.8) into equation (a) gives:

02

cos2

sin 21 =

+

λλHH (ii)

For non-trivial solution of equations (i) and (ii),

02

sin2

cos =

−

− λλλk

LEA

ie. ( ) 02

tan =+

kLEA

λλ (iii)

This is the frequency equation for the symmetrical vibration. For anti-symmetrical vibration, substituting equation (4.2.10) into equation (b) gives:

02

sin2

cos 21 =

−

λλHH (iv)

For non-trivial solution of equations (i) and (iv),

02

cos2

sin =

−

λλλk

LEA


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ie. 02

tan =−

λλ

EAkL

(v)

This is the frequency equation for the anti-symmetrical vibration. Note: It can be shown that the left hand side of equations (iii) and (v) are the factors of

the left hand side of equation (iii) for case 5. This illustrates the fact that the vibration modes of a symmetrical system consist of symmetrical modes and anti-symmetrical modes. This is schematically illustrated in Figure 4.2.12.

x = 0 x = LAxis of Symmetry

Anti-Symmetrical Mode

Symmetrical Mode

Figure 4.2.12 If a symmetrical system has a mass or spring element at the centre, then the condition of anti-symmetry is that the force at the centre of the system is zero, even if that centre is in the middle of a mass or spring.


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Case 7: Stepped Bar The system shown in Figure 4.2.12 has a geometrical discontinuity. The bar is made of a material having density ρ and Young's modulus E. The first segment of the bar is of length L1 and cross sectional area A1. The second segment has length L2 and cross sectional area A2. as the system contains a discontinuity, it is convenient to use two functions u1(x1,t) and u2(x2,t) to represent the longitudinal displacement in the two segments. x1,x2 are the longitudinal co-ordinates measured from the left hand end of the segments.

x1 = 0 x2 = L2 x1 = L1

x2 = 0

A1 A2

Figure 4.2.13 The dynamic displacements and their derivatives may be expressed in the following form:

( )φωλλ+

+

= t

Lx

HLx

Hu sincossin1

112

1

1111 , (4.2.13)

where 2

1

11

=E

Lρωλ (4.2.14)

( )φωλλ+

+

= t

Lx

HL

xHu sincossin

2

224

2

2232 , (4.2.15)

where 2

1

22

=E

Lρωλ (4.2.16)

It should be noted that the time function sin(ωt+φ) is the same for both segments, as the entire system vibrates together at the same frequency. Boundary Conditions: At x1 = 0, u1 = 0. Using equation (4.2.13) we get H2 = 0 (i)


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At x2 = L2, strain = 02

2 =∂∂

xu

. Differentiating equation (4.2.15) and substituting into this

condition gives: [ ] 0sincos 24232

2 =−

λλλ

HHL

. (ii)

Two more equations are required as we have four undetermined coefficients H1-4. These are obtained using continuity conditions at x1 = L1 (or x2 = 0). Continuity Conditions: Since the displacement is continuous, u1(L1,t) = u2(0,t). Substituting equations (4.2.13) and (4.2.15) into the above gives: H1 sin λ1 + H2 cos λ1 - H4 = 0. (iii) Since the force is continuous (or by Newton's third law), F1(L1,t) = F2(0,t) ie. EA1u1′(L1,t) = EA2u2′(0,t) Differentiating equations (4.2.13) and (4.2.15) and substituting into the above gives:

0sincos

2

223

1

1112

1

1111 =

−

−

LEAH

LEA

HL

EAH

λλλλλ (iv)

Equations (i) to (iv) may be expressed in matrix form [B]H=0 (v) For non-trivial solution |B|= 0. This yields the following frequency equation:

0coscossinsin 2121211

122 =−

λλλλλλ

LALA

.

Case 8: Restrained Bar Carrying A Concentrated Mass at A Mid-Point Consider the longitudinal vibration of the system shown in Figure 4.2.14. It consists of a uniform bar of length L, cross sectional area A, density ρ and Young's modulus E, and a concentrated mass m0 attached to the bar at distance a from the left hand end as shown in the Figure. Both ends of the bar are fully axially restrained. The frequency equation of this system is required.


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x1 = 0

m0

x2 = L-a x1 = a x2 = 0

Figure 4.2.14 The presence of m0 introduces a discontinuity. It is not possible to apply a single governing differential equation for the whole bar. As in case 7, let us use two functions to represent the dynamic axial displacements in the segments on either side of the mass.

For 0<x1<a, ( ) ( )φωλλ +

+

= tax

Hax

Htxu sincossin, 112

11111 (4.2.17)

where E

aρωλ =1 (4.2.18)

For 0<x2<L-a, ( ) ( ) ( ) ( )φωλλ +

−+

−= t

aLx

HaL

xHtxu sincossin, 2

242

2322 (4.2.19)

where E

aLρωλ )(2 −= (4.2.20)

Boundary Conditions: At x1 = 0, u1 = 0 gives H2 = 0 (i) At x2 = L-a, u2 = 0 gives H3 sin λ2 + H4 cos λ2 = 0 (ii) Continuity Conditions: Continuity of displacement at x1 =a is u1(a,t) = u2(0,t). This yields H1 sin λ1 + H2 cos λ1 - H4 = 0 (iii) Continuity of force at x1 = a is ensured by applying Newton's second law to the mass. Consider the freebody shown in Figure 4.2.15,


21

),(1 tau

m0 F2(0,t) F1(a,t)

Figure 4.2.15 The net longitudinal force = F2(0,t) - F1(a,t) = EA[u2′(0,t)-u1′(a,t)] The acceleration is ü1(a,t) = - ω2u1(a,t) Therefore EA[u2′(0,t)-u1′(a,t)] = -m0ω2u2(0,t) Differentiating equations (4.2.17) and (4.2.19) into the above yields:

0sincos2

04

231

121

11 =

−

−−

−

EAm

HaL

Ha

Ha

Hωλλλλλ

(iv)

For non-trivial solution of equations(i)-(iv), |B|= 0 This gives the following frequency equation:

0cotsincos2

02

211

1 =

−

−+

EAm

aLaωλλλλλ

In this case, the origin of the coordinate for the second bar (x2) could have been set at the right end. This would have made the solution slightly simpler because H4 would have been zero. The general approach used in the treatment of non-standard boundary conditions in Cases 4, 5 and 8 for the longitudinal vibration holds for other types of vibrations too. That is, finite masses would require the application of Newton’s second law of motion while elastic elements such as springs would require the use of constitutive equations (F = k e). Case 9: Symmetrical and Anti-Symmetrical Vibration of a Bar with a Mass at the Centre


22

This is a special case of Case 8, where the mass is located at the centre. Let us consider the symmetrical case first and take the left half of the structure. At the centre the displacement is zero. Since we assume that the mass is rigid, the displacement at the right end of the left bar is also zero. This means the system could be treated as a fixed-fixed bar with a length of L/2. Using the results for Case 1, we get an expression for the nth

natural frequency in symmetrical vibration as ρ

πω ELn

n

2=

The anti-symmetrical case is not so straightforward. If one were to apply the condition that 0),2/( =′ tLu as for Case 6, we would get the wrong answer, because this condition only holds

at the centre of the mass, where the displacement form u will not be the same as that for the bar. Since we are not interested in the straining within the mass (which for the purpose of analysis has been taken as rigid) one has to resort to the condition associated with the force. That is we can say the force at the centre of the mass is zero, and then apply Newton’s second law of motion to half the mass to find a boundary condition for the bar. This means treating the problem as that of Case 4, but with the magnitude of mass changed to m0/2 and the length changed to L/2.

m0

L/2 L/2 x= 0 x= L/2

Figure 4.2.16


23

4.3 Torsional Vibration of Shafts Derivation of Equation of Motion: The partial differential equation governing the torsional vibration of a circular shaft is of the same form as the equation of motion for longitudinal vibration of bars. It will be shown that for many simple boundary conditions the natural frequencies and modes of torsional vibration can be deduced from the solution for equivalent longitudinal vibration problems. Figure 4.3.1 shows a segment of a shaft having circular cross section with the following properties: Polar second moment of area J; Shear modulus G; Density ρ. Let θ(x,t) be the angle of twist at time t, and T(x,t) be the dynamic torque induced at the same instance. Figure 4.3.1 also shows the sign convention for T and θ. Let us now derive the equation of motion following the basic steps.

θ

T(x1 ,t) T(x2,t)

x2

x

x1

Figure 4.3.1 Step 1: Sketch the free body diagram of an infinitesimal element:

δx

T T+δT

θ

Figure 4.3.2 Step 2: From Mechanics of Materials, the torque-twist relationship is:


24

GJT

x=

∂∂θ

(4.3.1)

At any time t, δT = xx

GJx

δθ

∂∂

∂∂

(4.3.2)

Step 3: The net elastic action is T + δT - T = δT. Step 4: Since there are no external dynamic torques, this is the net torque on the element and is given by equation 4.3.2. Step 5: Using Newton's second law in the rotational sense:

δT =2

2

)(t

xJ∂∂ θδρ

Substituting equation 4.3.2 into the above gives:

2

2

=xx

t

JGJ∂∂

∂∂

∂∂ θρθ

(4.3.3)

This is the partial differential equation governing the torsional vibration of an elastic shaft of circular cross section. For a homogeneous, uniform shaft GJ is constant and equation (4.3.3) reduces to

0x 2

2

2

2

=∂∂−

∂∂

tG θθρ

. (4.3.4)

Again a wave equation is obtained. It may be noted that changing E and u in equation (4.2.7) to G and θ results in equation (4.3.4). General Solution for Uniform Shafts of Finite Length The general solution takes the form ( ) ( ) ( )[ ] ( )φωλξλξθ ++= tHHtx sincossin, 21 (4.3.5)

where H1 and H2 are undetermined coefficients, ξ is x/L, and the non-dimensional parameter λ is defined by

G

Lρωλ = (4.3.6)


25

The natural frequencies and modes of torsional vibration may be obtained by using the general solution and the end conditions as illustrated in the case of longitudinal vibration. The mathematical similarity between the equations of motion for longitudinal and torsional vibration may be used to deduce the results for one case if the corresponding results for the other case are known. This is illustrated in the following example. Case 1: Both Ends Torsionally Restrained

x = 0 x = L Figure 4.3.3 At x=0, θ=0 for any t. Hence θ(0,t) = 0. Substituting equation(4.3.5) into this gives: H1(0.0) + H2(1.0) = 0 (i) Similarly θ(L,t) = 0 gives: H1 sin λ + H2 cos λ = 0 (ii) For non-trivial solution of equations (i) and (ii), sin λ = 0. This is the frequency equation the roots of which are given by λ = nπ where n is any integer. Using equation (4.3.6) the nth natural frequency is obtained.

ρ

πω GL

nn =

This result could have been deduced from the result for the natural frequency of longitudinal vibration of a bar having restrained ends, by replacing the elastic modulus E with shear modulus G. Substituting condition (i) and the expression for the natural frequency into the general

solution results in the expression for the nth mode

Lxnπ

sin .


26

Similarly the results for other simple boundary conditions may be deduced from the corresponding results for the longitudinal vibration of a bar. It should be noted that for systems with rigid end masses, the mass terms in the frequency equations for longitudinal systems should be replaced by polar moment of inertia terms if corresponding results for torsional systems are required. For example, results from Case 4 of the longitudinal system could be used to find the natural frequencies of a shaft carrying a mass with polar moment of inertia I0, provided the ratio η is defined as JLρ/I0. The same roots of the frequency parameter as that for Case 4 of the longitudinal system could be used to find the natural frequencies of a corresponding shaft this time using equation (4.3.6). Therefore the torsional vibration systems for other common boundary conditions will not be discussed here. We will however consider the torsional oscillations of geared system since we have not dealt with a corresponding longitudinal vibration problem. Case 2: A Geared System

I1

r2

Shaft 2

I2

I3

I4

Shaft 1

r3

x1= 0 x1= L1

x2= 0 x2= L2

),( 11 txθ

),( 22 txθ

Figure 4.3.4 Consider the vibration of the geared system shown in Figure 4.3.4. The polar moments of inertia of the rotors (which are assumed rigid) are I1, I2, I3 and I4. The properties of the two shafts are: G1, J1, ρ1 and L1 for the first shaft, and G2, J2, ρ2 and L2 for the second shaft. The radii of gears I2 and I3 are r2 and r3 respectively. The dynamic rotations of the two shafts may be expressed in the following form: ( ) ( )[ ] ( )φωξλξλθ ++= tHH sincossin 1121111 , (4.3.7)

where 111 / Lx=ξ and λ1 is defined by


27

1

111 G

Lρωλ = and (4.3.8)

( ) ( )[ ] ( )φωξλξλθ ++= tHH sincossin 2242232 , (4.3.9)

where 222 / Lx=ξ and λ2 is defined by

2

222 G

Lρωλ = and (4.3.10)

Differentiating yields:

( ) ( )[ ] ( )φωξλξλλθ++

=

∂∂

tHHLx

sinsincos 1121111

11 (4.3.11)

( ) ( )( )[ ] ( )φωξλξλλθ++

=

∂∂

tHHLx

sinsincos 2242232

22 (4.3.11)

Boundary Conditions: As there are four constants H1 ... H4, four conditions are required. First consider the vibration of the first rotor (see Figure 4.3.5).

),0(1 tθI1

T1 (0,t)

View from positive x direction

T1 (0,t)

),0(1 tθ

Figure 4.3.5 Applying Newton's second law to I1 gives:

( ) ( )tt

ItT ,0,021

2

11 ∂∂= θ

=-ω2I1θ1(0,t)

Substituting the torque-twist relationship x

GJT∂∂= θ

into the above equation gives:


28

G1J1θ1′(0,t) + ω2I1θ1(0,t) = 0 Substituting the general solution for θ1 and its derivative [equations (4.3.7) and (4.3.11)] into the above equation yields:

0212

11

111 =+

HIH

LJG ωλ

(i)

At x1=L1 (or x2=0) the gear constraint equation gives:

( ) ( )tLrr

t ,,0 113

22 θθ

−=

Substituting equations (4.3.7) and (4.3.9) into the above gives:

0cossin 4123

211

3

2 =+

+

HH

rr

Hrr λλ (ii)

The third condition is obtained by applying Newton's second law to the rotors 2 and 3, and eliminating the contact force between these rotors from the two equations. Figure 4.3.6 gives an exploded view of these rotors.


29

),0(2 tθ

I2

T1(L1,t)

View from positive x direction

(P)(r2 )

P

P

),( 11 tLθ

),( 11 tLθT1(L1,t)

(P)(r2 ) I3

),0(2 tθ

T2(0,t) T2(0,t)

Figure 4.3.6 Applying Newton's second law to I2 gives:

( ) ( )tLIPrtLT ,, 12211 θ=+− (iiia)

Similarly for I3 we get

T2(0,t) + P r3 = -ω2I3θ2(0,t) (iiib) Eliminating P from equations (iiia) and (iiib) gives:

( ) ( ) ( ) ( ) 0,0,,0, 233

2112

22

3

211 =

−−+ tI

rr

tLItTrr

tLT θθω

Substituting the gear constraint condition and the torque-twist relationships gives

( ) ( ) ( ) 0,,0, 113

2

3

22

2222

3

21111 =

+−′+′ tLI

rr

ItJGrr

tLJG θωθθ

Substituting equations (4.3.7),(4.3.11) and (4.3.12) into the above yields: B31 H1 + B32 H2 + B33 H3 + B34 H4= 0 (iii)


30

where

+−= 13

2

3

22

21

1

11131 sincos λωλλ

Irr

IL

JGB (iiic)

++−= 13

2

3

22

21

1

11132 cossin λωλλ

Irr

IL

JGB (iiid)

=

3

2

2

22233 r

rL

JGBλ

(iiie)

B34 = 0 (iiif) The last equation may be obtained by applying Newton's second law to I4. Consider the free body shown in Figure 4.3.7:

T2(L2,t) ),( 22 tLθ

I4

),( 22 tLθ T2(L2,t)

Figure 4.3.7 From Newton's second law,

( ) ( )tLItLT ,, 2242

22 θω−=−

Substituting the torque-twist relationship and equations (4.3.9) and(4.3.12) into the above gives: B41 H1 + B42 H2 + B43 H3 + B44 H4 = 0 (iv) where B41 = 0 (iva) B42 = 0 (ivb)

−= 24

22

2

22243 sincos λωλλ

IL

JGB (ivc)

+−= 24

22

2

22244 cossin λωλλ

IL

JGB (ivd)


31

Equations (i),(ii),(iii) and (iv) may be written in matrix form [B]H = 0. For the non-trivial solution of H, |B| = 0. This is the frequency equation. The coefficients of [B] in the third and fourth rows are explicitly given in equations (iiic) ...(ivd). Other coefficients can easily be obtained from equations (i) and (ii).


32

4.4 Lateral Vibration of Strings

Derivation of Equation of Motion: Another interesting category of vibration that is governed by the one dimensional wave equation is the small amplitude lateral vibration of strings. Again the natural frequencies and modes of strings may be deduced from the corresponding solution for the longitudinal vibration of a bar or the torsional vibration of a shaft. However, some physical boundary conditions are not common to all these three systems. There are some similarities between the lateral vibration of strings and the more complicated lateral vibration of beams (which is more commonly encountered in engineering). For these reasons some examples of vibration of strings is discussed following the derivation of the equation of motion. A string is a structural element that can only sustain axial tensile straining. It has no flexural rigidity. Figure 4.4.1 shows a segment of a vibrating string subject to a uniform static tension T0. The elastic restoring action is provided primarily by the component of the tension in the direction of motion. For small amplitude vibration, the fluctuation in the axial tension F may be neglected as it is small compared to the static tension T0. Let m be the mass of the string per unit length, and v(x,t) be the lateral dynamic displacement.

x1

x T0

T0+ F1 v v(x2,t)

o

T0+ F2

x2

T0

v(x1,t)

Figure 4.4.1 Step 1: Sketch the free body diagram of an infinitesimal element, Figure 4.4.2:


33

Figure 4.4.2

Step 2: As the change in tension during vibration is small (it would be of the order of the dynamic strain), this step is not necessary. Step 3: The net elastic action in the direction of motion is given be (T0) sin(θ+δθ) - (T0) sin θ As the amplitude of vibration is small, sin θ→θ. Therefore the net elastic action is T0 (δθ) + F (δθ) + (δF)θ which may be written as T0 δ(∂v/∂x), (4.4.1) since θ→∂v/∂x as θ→0 Step 4: Since there are no external dynamic forces, this is the net force on the element Step 5: Using Newton's Second law in the v direction:

( ) 2

2

0 tv

xmtv

T∂∂=

∂∂ δδ

i.e. 2

2

0

)/(tv

mx

xvT

∂∂=∂∂

δδ

as δx → 0, this becomes

02

2

2

20 =

∂∂−

∂∂

tv

xv

mT

(4.4.2)

For a uniform string (if m is a constant) this may be solved using the same form of general solution as those used in the previous cases.

x δx

x T0+ F

T0 + F+δF

v

v+δv v

o

θ+δθ

θ


34

General Solution for Uniform Strings of Finite Length The general solution is ( ) ( ) ( )[ ] ( )φωλξλξ ++= tHHtxv sincossin, 21 (4.4.3)

where H1 and H2 are undetermined coefficients, x is given by Lx /=ξ and λ is given by:

0T

mLωλ = (4.4.4)

Substituting the above equations into the boundary conditions yield frequency equations and modes as illustrated in the following examples. Case 1: String Fixed at Both Ends

x = 0 x = LSecond Mode

First Mode

Figure 4.4.3 (a) at x=0, v=0. Substituting this into the general solution gives: H1 (0.0) + H2 (1.0) = 0.0 i.e. H2 =0.0 (i) (b) Similarly, v=0 at x=L gives: H1 sin λ + H2 cos λ = 0. (ii) Substituting equation (i) into (ii) gives H1 sin λ = 0 For non-trivial solution sin λ = 0, which means λ = nπ


35

From this and equation (4.4.4.) the nth natural frequency is obtained as

mT

Ln

n0πω =

The corresponding mode is:

Lx

nπsin

It is worth noting that the above expressions for the natural frequency and mode was obtained using only the transverse boundary conditions. The results are valid even for the lateral vibration of the systems shown in Figure 4.4.4, which have different longitudinal end conditions. The longitudinal motion of the end masses are secondary for small amplitude lateral vibration. The effect of the end springs in Figure 4.4.3a is also negligible. The springs and/or masses in these cases can only cause a dynamic fluctuation in the axial tension which is small compared to the static tension T0.

x = 0 x = L (a)

x = 0 x = L

m1

m2

(b)

k1

k

T0 T0

k1

Figure 4.4.4


36

Case 2: Two Identical Strings Connected to a Solid Sphere Consider the lateral vibration of the system shown in Figure 4.4.5, which consists of a uniform solid sphere of radius R and mass m0 connected to two strings symmetrically. The far ends of the strings are laterally restrained. The tension in the strings is T0.

x = 0

2R m0

x = L

Figure 4.4.5 The discontinuity at the centre means two functions v1 and v2 should be used and four boundary conditions would be required. However, the symmetrical and anti-symmetrical vibrations can be analysed separately as shown here. Taking the left side of the centre line, the boundary condition at x1 = 0 (which is common to both symmetrical and anti-symmetrical vibrations) is: (a) v1(0,t) = 0. Substituting the general solution in terms of non-dimensional coordinate Lx /11 =ξ

( ) ( )[ ] ( )φωξλξλ ++= tHHv sincossin 1121111 (i)

gives: H2 = 0 (ii) Now consider the symmetrical vibration. A sketch of a symmetrical mode is shown in Figure 4.4.6. The sphere translates laterally without any rotation.


37

x = 0 x = L

2R

m0

T0 θ(L,t)

Free Body Diagram for m0

T0 θ(L,t)

Figure 4.4.6 As can be seen from the free body diagram in the Figure, there is no net moment on the sphere from the string reactions. The lateral force components from the strings are equal and their sum is given by

∂∂↓

1

10 sin2

xv

T at x = L

As v1 is very small this may be written as 2T0 v1′(L,t) Hence, using Newton's second law: ↑ -2T0 v1′(L,t) = m0 ),( 11 tLv

i.e. 2T0 v1′(L,t) - m0 ω2 v1(L,t)= 0 (iiia) Substituting the general solution for v1 and equation (ii) into the above yields: 2(λ1/L)T0 cos λ1 - m0 ω2 sin λ1 = 0

i.e. 00

20

1 2cot

λωλ

TLm=

Substituting equation (4.4.4) into the above gives:

( )strings both of mass2cot 00

2

1 mmLm ==

λλ

(iv)


38

This is the frequency equation. Special Cases: (1) If m0 → 0, cot λ1 → 0.

i.e. ( )

212

1πλ −= n

This corresponds to the case of a fixed-free string of length L (a difficult condition to achieve physically) or the symmetrical modes of a fixed-fixed string of length 2L. (2) If m0 → ∞, cot λ1 → ∞. i.e. λ1 = nπ. This corresponds to the case of a fixed-fixed string of length L or the anti-symmetrical modes of a fixed-fixed string of length 2L. The fundamental modes corresponding to the two special cases of symmetrical vibration are shown in Figure 4.4.7.

x = 0 x = L 2R

(b) Large Mass

x = 0 x = L

2R

m0→0

(a) Negligible Mass

m0→∞

Figure 4.4.7


39

Let us now consider the anti-symmetrical modes. In anti-symmetrical vibration, the sphere undergoes a pure rotation as shown in Figure 4.4.8.

x = 0 x = L

2R

m0

T0 θ(L,t)

Free Body Diagram for m0

γ

T0

θ(L,t)

Figure 4.4.8 Let θ1(L,t) = Θ The net moment on the sphere is 2[T0 sin Θ (R cos γ) + T0 cos Θ (R sin γ)] (v) As Θ and γ are small, sin Θ → Θ , sin γ → γ, cos Θ = cos γ → 1 Substituting these into equation (v) gives ΣM = 2T0R( Θ +γ) But from Newton's second law: ΣM = I0 γ

Therefore 2T0R(γ+ Θ ) = -I0 γ

Substituting the compatibility condition, Rγ=v1(L,t) into the above equation gives:

( ) ( )[ ] ( )R

tLvItLvtLRvT

,,,2 10

2

10

ω=+

Using equations (i) and (ii) this becomes:


40

02sincos2 02

0111

0 =

−+

RI

TL

RTωλλλ

This gives:

RTLI

L

R

0

201

1

2

tanωλ

λ

−−= (vi)

Special Cases: (a) If I0 → 0, then ΣM = 0 giving Θ = -γ

LR−=

1

1tanλ

λ

Note that Θ = -γ means the cable forces on the mass would be acting radially. This is to be expected since otherwise there will be a net moment.

(If R=0, that is for a point mass, this would correspond to the case of a fixed-fixed string of length L) (b) If I0 → ∞, then tan λ1 = 0 corresponding to the vibration of a fixed- fixed string of length L. The modes for these special cases are shown in Figure 4.4.9

x = 0 x = L

2R

x = L 2R

γ I0→0

(a) Negligible Inertia

(b) Large Inertia

I0→∞

x = 0

Figure 4.4.9


41

4.5 Lateral Vibration of Beams Derivation of Equation of Motion All the systems considered so far had equations of motion in the form of a wave equation. The lateral vibration of beams is governed by a fourth order partial differential equation. The derivations presented in this section are based on the following assumptions: Assumptions: 1. The beam is made of a material having linear, elastic properties; 2. The displacements are small; 3. The stresses induced are within the limit of proportionality; 4. The cross section of the beam has at least one axis of symmetry; 5. The transverse cross sections of the beam remain plane during bending; 6. The cross sectional dimensions of the beam are small compared to the span so that: (a). The displacement due to shear strain is small, and (b). The rotary inertia of the beam is negligible. A beam satisfying the above assumptions is called an "Euler-Bernoulli Beam", and is often referred to as a thin beam. It should be noted here that even if the beam is slender (cross sectional dimensions are small compared to its span) and the amplitude of displacements remain small, for higher modes of vibration, the slope of the beam could be large and the rotary inertia may not be negligible. The effect of neglecting rotary inertia will be discussed later.

x1

x2

x

v

o

M(x2,t) Sf(x2,t)

Sf(x1,t)

M(x1,t)

v(x2,t) v(x1,t)

Figure 4.5.1


42

Figure 4.5.1 shows a segment of a laterally vibrating beam. The induced actions [dynamic shearing force Sf(x,t) and bending moment M(x,t)] are shown acting in their positive directions. Sagging moments and anticlockwise shear forces are positive. The properties of the beam are: Young's modulus E; Second moment of area about the neutral axis I; Mass per unit length (density × area) m. Let v(x,t) be the lateral dynamic displacement of the beam. Let us now follow the basic steps in deriving the equation of motion. Step 1: Sketch the free body diagram of an infinitesimal element:

x δx x

v(x,t)

v+δv

o

M M+δΜ

Sf v

Neutral Surface

Sf+δSf

v x

Figure 4.5.2 It is convenient to replace the transverse shearing force Sf(x,t) with its

components, S(x,t) in the lateral direction and F(x,t) in the longitudinal direction as shown in Figure 4.5.3.

S

S+δS

x δx x

v+δv

o

M M+δM

v

Neutral Surface

F+δF

v(x,t)

F

Figure 4.5.3 Step 2: From Mechanics if Materials the moment curvature relationship for a thin beam is


43

2

2

xv

EIM∂∂= (4.5.1)

Neglecting the rotary inertia of the beam, summing the moments on the

infinitesimal element and equating to zero gives: (M + δM) - M + S (δx) - F (δv) = 0 i.e. δM + S (δx) - F (δv) = 0 But F (δv) being a product of two small dynamic terms is negligible

As δx is infinitesimal, this gives S = -x

M∂

∂ (4.5.2)

Step 3: The net elastic action in the positive lateral direction is -S + (S + δS) = δS

But from equation (4.5.2) δS = δ

∂∂−

xM

= - xxM δ2

2

∂∂

= xxv

EIx

δ

∂∂

∂∂− 2

2

2

2

(4.5.3)

Step 4: Since there are no external dynamic forces this is the net lateral force. Step 5: Applying Newton's second law in the lateral direction we get

δS = (m δx) 2

2

tv

∂∂

Using equation (4.5.3) this becomes:

xxv

EIx

δ

∂∂

∂∂− 2

2

2

2

= (m δx) 2

2

tv

∂∂

This gives:

0=2

2

2

2

2

2

tv

+mxv

EI x ∂

∂

∂∂

∂∂ (4.5.4)

This is the partial differential equation governing the lateral vibration of an Euler-Bernoulli beam. For a homogeneous, uniform beam this reduces to the following form:


44

02

2

4

4

=∂∂+

∂∂

tv

mxv

EI (4.5.5)


45

General Solution for Uniform Beams of Finite Length The general solution to equation (4.5.5) may be written in the following form: v(x,t) = f(x) g(t), (4.5.6) where, in terms of the non-dimensional axial coordinate ξ = x/L,

( ) ( ) ( ) ( ) ( )λξλξλξλξ sincossinhcosh 4321 HHHHxf +++= (4.5.6a)

and, g(t) = sin(ωt+φ) (4.5.6b) Substituting equations (4.5.6) into equation (4.5.5) gives

42

EIm

Lωλ = (4.5.7)

The above equations are valid for any boundary conditions. The natural frequencies and modes can be determined as illustrated in the following examples. Case 1: Simply Supported at Both Ends

x = 0 x = L

n = 1n = 2

Figure 4.5.4 (a) At x=0, v=0. Substituting the general solution into this equation gives: H1 (1.0) + H2 (0.0) + H3 (1.0) + H4 (0.0) = 0 (i) (b) At x=0, M=0.

i.e. at x = 0, 02

2

=∂∂xv

EI .

Substituting equation (4.5.6) into this yields:


46

(λ/L)2 [H1 (1.0) + H2 (0.0) - H3 (1.0) - H4 (0.0)] = 0 (ii) (c) Similarly, at x = L, v = 0 leads to: H1 cosh λ + H2 sinh λ + H3 cos λ + H4 sin λ = 0 (iii) (d) Also at x = L, M= 0 results in the following equation:

( ) ( ) ( ) ( )[ ] 0sincossinhcosh 4321

2

=−−+

λλλλλHHHH

L (iv)

Equations (i) to (iv) may be written in matrix form as [B] H = 0,

where [ ]

−−

−=

λλλλλλλλλλλλ

λλ

sincossinhcoshsincossinhcosh

000101

2222

22

B and

=

4

3

2

1

H

H

H

H

H

The characteristic equation (frequency equation) is |B| = 0 which gives 0sinsinh4 =λλλ For this particular case the frequency equation may be obtained as follows: From equation (i) H3 = -H1 (ia)

From equation (ii) ( ) ( ) 0/ 312 =− HHLλ

Substituting equation (ia) into the above gives: either λ = 0, or H1 = H3 =0 (iia) Let us consider the possibility that λ = 0. Substituting this into the general solution for v(x,t) gives v(x,t) = [H1 + 0 + H3 + 0] sin(ωt+φ) = 0 [using equation (ia)] This is a trivial solution. Therefore, for non-trivial solution H1 = H3 = 0. Substituting this into equation (iii) and (iv) gives: H2 sinh λ + H4 sin λ = 0 (iiia)

( ) ( ) 0sinsinh2

4

2

2 =

−

λλλλL

HL

H (iva)

For non-trivial solution of equations (iiia) and (iva) sin λ = 0


47

Hence λ = nπ and mEI

Ln

2

= πω

The nth mode is: sin (nπx/L) Case 2: Both Ends Clamped

x = 0 x = L

n=1

n=2

Figure 4.5.5 Boundary Conditions: (a) At x=0, v=0. As for case 1, this gives: H1 (1.0) + H2 (0.0) + H3 (1.0) + H4 (0.0) = 0 (i) (b) At x=0, ∂v/∂x=0. Substituting the general solution for v into this equation gives

( ) ( ) ( ) ( )[ ] 00.10.00.10.0 4321 =+−+

HHHHLλ

(ii)

(c) At x=L, v=0 yields: H1 cosh λ + H2 sinh λ + H3 cos λ + H4 sin λ = 0 (iii) (d) At x=L, ∂v/∂x=0 gives:

( ) ( ) ( ) ( )[ ] 0cossincoshsinh 4321 =+−+

λλλλλHHHH

L (iv)

This may be written in matrix form as [B]H=0 For non-trivial solution of this equation |B| = 0. It can be shown that this determinantal equation reduces to the following frequency equation: λ2(1-cosh λ cos λ) = 0.


48

While λ=0 satisfies the above equation, substituting λ=0 into the general solution gives v = (H1+H3) g(t). But H1 + H3 = 0 from equation (i). This means λ=0 is a trivial solution. The first non-trivial root of this equation is λ = 4.73... which gives the fundamental frequency

mEI

L

2

173.4

=ω rad/s

The expression for mode is:

( ) ( )

+

−

−

Lx

rLx

Lx

rLx λλλλ sincossinhcosh ,

where λλλλ

sinsinhcoscosh

−−=r .

Case 3: Free - Free Beam

x = 0 x = L

Rigid Body Translation(n=0)n=1

n=2

Figure 4.5.6 This is an example of a semi-definite system. The boundary conditions are: (a) At x=0,M=0.

i.e 02

2

=∂∂xv

EI .

Using the general solution we get:

( ) ( ) ( ) ( )[ ] 00.00.10.00.1 4321

2

=+−+

HHHHLλ

(i)


49

(b) Similarly at x =L, M=0 gives 02

2

=∂∂xv

resulting in the following equation:

( ) ( ) ( ) ( )[ ] 0sincossinhcosh 4321

2

=−−+

λλλλλHHHH

L (i)

These are the same as equations (ii) and (iv) for case 1. The remaining boundary conditions are obtained by setting the shearing force to zero at both ends.

(c) At x=0, S=0 gives: 0=∂

∂−x

M

For a uniform beam this means 03

3

=∂∂xv

.

Again substituting the general solution [equation (4.5.6)] into the above yields:

( ) ( ) ( ) ( )[ ] 00.10.00.10.0 4321

3

=−++

HHHHLλ

(iii)

(d) Similarly at x=L, S=0 gives:

( ) ( ) ( ) ( )[ ] 0cossincoshsinh 4321

3

=−++

λλλλλHHHH

L (iii)

For non-trivial solution of equations (i)-(iv) the determinantal equation can be shown to be λ10(1-cosh λ cos λ) = 0. (v) For this case λ=0 is not a trivial solution as illustrated below: Substituting λ=0 into the general solution for v gives: v = [H1 (1) + H2 (0) + H3 (1) + H4 (0)] sin(0t+φ) = (H1+H3) sin(φ) = constant. This corresponds to a rigid body translation. [There is another semi-definite mode corresponding to a rigid body rotation.] The non-zero roots of equation (v) are identical to the roots for the clamped - clamped beam. This means the non-zero natural frequencies of free-free beams and clamped-clamped beams are the same. However, the modes are different.


50

For free-free beams the expression for the modes is:

( ) ( )

−

+

−

Lx

rLx

Lx

rLx λλλλ sincossinhcosh ,

where λλλλ

sinsinhcoscosh

−−=r .


51

Case 4: Beams Carrying Masses In finding the whirling speed of shafts (see section 4.7) it is often necessary to consider the flexural vibration of shafts carrying masses. Let us consider the general case of a shaft system carrying a rotor at x1=L1 and x2 =-L2 as shown in Figure 4.5.7. Both ends are supported on short bearings, which may be treated as simple supports. Let the mass of the rotor be m0 and its moment of inertia about its centroidal axis parallel to the neutral axis of the cross section be I0. Let the centroid G be located at distances e1 and e2 from the ends of the two shaft segments to which it is connected. Let the flexural rigidity length and mass per unit length of the first shaft segment be EI1, L1 and m1 respectively, and let the corresponding parameters for the second shaft be EI2, L2 and m2.

Figure 4.5.7 Let the displacements of the two shaft/beam segments be v1(x1,t) and v2(x2,t) where v1(x1,t) = f1(x1).g(t), and v2(x2,t) = f2(x2).g(t) (i) in which, in terms of the non-dimensional axial coordinates ξ1 (=x1/L1) and ξ2 (=x2/L2),

( ) ( ) ( ) ( ) ( )1141131121111 sincossinhcosh ξλξλξλξλ HHHHxf +++= (ii)

g(t) = sin(ωt+φ) (iii)

and 4

1

21

11 EIm

Lωλ = (iv)

Also ( ) ( ) ( ) ( ) ( )2282272262252 sincossinhcosh ξλξλξλξλ HHHHxf +++= (v)

in which 4

2

22

22 EIm

Lωλ = (vi)

e1

e2

L1 L2

EI1, m1 EI2, m2 G


52

There are eight constants H1…8 but in this case by setting the origins of the co-ordinates x1 and x2 at the supports, we can eliminate four of the constants, simplifying the problem to that of a four by four matrix equation. At x1=0, v1=0 and 1v ′′ = 0 give:

H1+H3=0 and (λ1)2 (H1-H3)=0 giving H1=H3=0 as in case 1. Similarly at the right hand end x2=0, v2=0 and 2v ′′ = 0 will yield: H5=H7=0

Substituting these into equations (ii) and (v) results in the following simplified expressions for the functions f1 and f2.

( ) ( ) ( )ξλξλ 141121 sinsinh HHxf += (ii.a)

( ) ( ) ( )2282262 sinsinh ξλξλ HHxf += (v.a)

The remaining conditions are found by applying the following conditions at x1 = 0 and x2 = -L2. Continuity of slope gives: )()( 2211 LfLf −′−′ =0

Continuity of displacement gives: 0)()()()( 22112111 =−−′++ LfLfeeLf

Applying Newton’s second law in the rotational sense gives:

M2(-L2,t)-M1(L1,t)+S2(-L2,t) e2 + S1(L1,t) e1 = I0

tLxv

t,1

12

2

1

∂∂

∂∂ =−ω2I0 ),( 11 tLv′

And for the translation, S2(-L2,t)- S1(L1,t)=m0

∂∂+

∂∂

tLxv

etLvt

,1

11112

2

1

),(

= −ω2m0 (v1(L1,t)+ e1 ),( 11 tLv′

Substituting the reduced general solutions in equations (ii.a) and (v.a) and the expressions for the bending moments and shear forces into the above equations yields four equations of the form: [B]H=0. For non-trivial solution |B|=0. An interactive multimedia program modes951.exe (windows95 version) can be used to calculate the natural frequency parameter λ for the above problem. In the program all parameters are input in non-dimensional form as follows: For the mass m0 use m0/m1L1 where m1 is the mass per unit length of the first beam/shaft segment.


53

For the moment of inertia I0 use I0/m1L13.

All lengths are non-dimensionalised with respect to the length of the first beam segment L1. i.e. the eccentricities are e1/L1 and e2/L1. The properties of the second beam-segment are also required as ratios. i.e. m2/m1 and EI2/EI1 are required as inputs. The trial frequency parameters λmin, λmax, dλ are also for the first segment. The results displayed give the roots λ1 for the first beam-segment. Actual frequencies may be calculated using equation (iv). The program may be used to study the behaviour of the system under various conditions. The results may be tested against simple known cases. For example, by setting m0 and I0 to zero, and m2/m1 , L2/L1 and EI2/EI1 to unity, one can calculate the natural frequencies of a simply supported beam of length 2L1. Putting large values for the mass m0 or inertia I0 would give results corresponding to constrained cases (with the exception of some low frequency modes where the system behaves like a discrete system). the program can also be used to calculate the natural frequencies of stepped beams by setting m0 and I0 to zero, and putting values other than unity for m2/m1, L2/L1 and EI2/EI1. It can also be used to study symmetrical and anti-symmetrical modes, but in some cases some roots may be missed. This is because the program finds the roots of the frequency equation by searching for a change in the sign of the determinant, and for some symmetrical cases the determinant takes a stationary value at zero. By using slightly unsymmetrical properties, the coincident roots may be separated and delimited. Symmetrical and Anti-symmetrical Vibration of Beams

The vibration modes of any symmetrical structure will be either symmetrical or anti-symmetrical. Applying special conditions for each of these cases, and analysing only half the structure often results in considerable savings in time. Symmetrical modes are the ones whose actual displacement form on one side of the centre-line coincides with the image (formed by a mirror placed along the axis of symmetry) of its displacement on the other side. Anti-symmetrical modes are symmetrical but opposite in sign. Symmetrical modes In the case of beams, symmetrical vibration modes will normally have a zero slope at the centre (exceptions to this are described later) for the following reason:

L/2 L/2

f((L-s)/2) f((L+s)/2)

s/2 s/2

Figure 4.5.8


54

Consider the displacement of two corresponding points on either side of the centre line, located at equidistant from the centre.

The amplitude of displacement of these two points are

−2

sLf and

+2

sLf respectively.

For symmetry, they must be equal.

This means, 022

=

+−

− sLf

sLf

If we take two points that are very close to the centre (i.e. let s = δx), if the displacement is continuous over the centre,

as δx →0, 022

=

+−

− xLf

xLf

δδ

i.e. δx →0, 0=xf

δδ

i.e. 0=dxdf

.

Hence the slope of a continuous curve representing the lateral displacement should vanish at the centre. This may be used as a boundary condition for half of the structure, if only symmetrical modes are considered. This would be true, even if the beam carries a mass at the centre, provided the beam is continuous, or the mass is rigidly connected. However, a notable exception must be mentioned here. In the case of a connected structure, a discontinuity may occur, if two beam segments were hinged together at the centre, thus allowing a discontinuity in the slope to take place. In such a case, continuity condition for slope, and moment -angular acceleration equation, will be replaced by the conditions that the bending moment will be zero in both beams at the hinge. This should not be confused with a continuous beam on which a mass is hinged/pinned at the centre.

Beams hinged to each other

free to rotate relatively

Figure 4.5.9


55

m0

m0/2 m0/2

S1((L-b)/2),t) S2(0,t)

b

L

),2/)((1 tbLv − ),0(2 tv

zero slope & zero shear force

Figure 4.5.10

In the case of a symmetrical vibration, the shear force at a section through the centre line would be zero. If the beam has no attached mass, and is not restrained by any other structural elements including springs, then the shear force in the beam at the centre will also be zero. However, if the beam carries a mass, or is attached to a spring, then the shear force in the beam will not be zero, even if it is close to the centre-line of the structure where the shear force is zero. A sudden jump in the shearing force would occur within the boundaries of the mass. For example, let us consider a beam carrying a mass at the centre. Making an imaginary cut through the centre of the mass, and the end of a beam, which is connected to the mass, we may obtain one of the free-bodies shown in the following diagram: By applying Newton's second law in the lateral direction, another symmetrical boundary condition may be obtained. Taking the origin of co-ordinate axes as the left most point on the beam segment, for the left-half beam we get:

−=

−− tbL

vm

tbL

S ,22

,2 1

01

Alternatively considering the right-half beam gives:

( ) ( )tvm

tS ,02

,0 20

2 =

If neither a mass nor a spring is attached to the beam, then the shear force at the centre of the beam would be zero. The symmetrical boundary condition of a beam without any attached masses or springs may be conveniently modeled using a sliding boundary condition as shown here. Anti-symmetrical modes Similar arguments may be presented for the anti-symmetrical modes. For anti-symmetrical modes, the corresponding points on either side of the centre must have equal and opposite displacement. For any finite displacement at f = 0 at the centre.

Figure 4.5.11


56

the centre would therefore cause a contradiction, which leads to the conclusion that there can be no displacement at the centre. It may also be shown that the bending moment distribution will also be ant-symmetrical, becoming zero exactly at the centre of the ant-symmetric system. Once again, for a beam carrying a mass with a moment of inertia, the bending moment may change abruptly within the boundaries of the mass, and will not be zero in the beam where it is rigidly connected to the mass. In such a case, application of Newton's second law of motion in the rotational

sense for half the mass (i.e. Moment = θ20I

) will yield a boundary condition for anti-

symmetry. The problems in Section 4.8 include a number of practical non-standard boundary and continuity conditions, including partial elastic linear and rotational restraints, cable restraints, pin connections and rigid connections. The solution manual contains the derivation of the boundary conditions in terms of displacement functions and its derivatives.


57

4.6 Lateral Vibration of Statically Axially Loaded Beams Derivation of Equation of Motion

x1

x2

x

v

o

M(x2,t) Sf(x2,,t)

Sf(x1,,t)

M(x1,t)

v(x2,t) v(x1,t)

P P

Figure 4.6.1 Figure 4.6.1 shows a segment of an axially loaded beam that is vibrating laterally. Comparing this with figure 4.5.1, it can be seen that the only change is the addition of the axial force P. In many practical problems the induced axial force would remain constant. However, it should be noted that the axial force can vary in some common applications such as in the case of rotating beams (the steady state centripetal force varies with distance). The solution procedure described in this section is not applicable for such cases. Let v(x,t) be the lateral dynamic displacement of the beam. Let us now follow the basic steps in deriving the equation of motion. Step 1: Sketch the free body diagram of an infinitesimal element:

x δx x

v(x,t)

v+δv

o

M M+δΜ

Sf v

Neutral Surface

P

Sf+δSf

P

v

Figure 4.6.2


58

As in section 4.5, let us replace the transverse shearing force Sf(x,t) with its components, S(x,t) in the lateral direction and F(x,t) in the longitudinal direction as shown in Figure 4.6.3.

S

S+δS

x δx x

v+δv

o

M M+δM

v

Neutral Surface

P+F+δF

v(x,t)

P+F

Figure 4.6.3 Step 2: From Mechanics if Materials[ ] the moment curvature relationship for a thin beam is

2

2

xv

EIM∂∂= (4.6.1)

Neglecting the rotary inertia of the beam, summing the moments on the infinitesimal element and equating to zero gives: (M + δM) - M + S (δx) - F (δv) –P (δv) = 0 i.e. δM + S (δx) - F (δv) –P (δv) = 0 But F (δv) being a product of two small dynamic terms is negligible

As δx is infinitesimal, this gives S = xv

Px

M∂∂+

∂∂− (4.6.2)

Step 3: The net elastic action in the positive lateral direction is -S + (S + δS) = δS

But from equation (4.6.2) δS = δ

∂∂+

∂∂−

xv

Px

M

= xxv

Px

Mx

δ

∂∂+

∂∂−

∂∂ = x

xv

PxM δ

∂∂+

∂∂− 2

2

2

2

= xxv

Pxv

EIx

δ

∂∂+

∂∂

∂∂− 2

2

2

2

2

2

(4.6.3)


59

Step 4: Since there are no external dynamic forces this is the net lateral force. Step 5: Applying Newton's second law in the lateral direction we get

δS = (m δx) 2

2

tv

∂∂

Using equation (4.6.3) this becomes:

xxv

Pxv

EIx

δ

∂∂+

∂∂

∂∂− 2

2

2

2

2

2

= (mδx) 2

2

tv

∂∂

This gives:

0=2

2

2

2

2

2

2

2

tv

+mxv

-Pxv

EI x ∂

∂∂∂

∂∂

∂∂ (4.6.4)

This is the partial differential equation governing the lateral vibration of an Euler-Bernoulli beam. For a homogeneous, uniform beam this reduces to the following form:

0=+ 2

2

2

2

4

4

tv

mxv

Pxv

EI∂∂

∂∂−

∂∂ (4.6.5)

General Solution for Uniform Beams of Finite Length The general solution to equation (4.6.5) may be written in the following form: v(x,t) = f(x) g(t), (4.6.6) where,

( )

+

+

+

=Lx

HLx

HLx

HLx

Hxf ααββ sincossinhcosh 4321 (4.6.6a)

and, g(t) = sin(ωt+φ) (4.6.6b) Substituting equations (4.6.6) into equation (4.6.5) gives

β=

++

42

244

2 γπλγπ and α=

++

−

42

244

2 γπλγπ, in which (4.6.7a,b)

EIPL

2

2

πγ = , and λ = L 4

2

EImω

(4.6.7c,d)


60

The above equations are valid for any boundary conditions. The natural frequencies and modes can be determined as illustrated in the following examples.


61

Case 1: Simply Supported at Both Ends

x = 0 x = L

n = 1n = 2P P

Figure 4.6.4 (a) At x=0, v=0. Substituting the general solution into this equation gives: H1 (1.0) + H2 (0.0) + H3 (1.0) + H4 (0.0) = 0 (i) (b) At x=0, M=0.

i.e. 02

2

=∂∂xv

EI at x = 0.

Substituting equation (4.6.6) into this yields:

( ) ( )[ ] ( ) ( )[ ] 00.00.10.00.1 43

2

21

2

=+

−+

HHL

HHL

αβ (ii)

(c) Similarly, at x=L, v=0 leads to: H1 cosh β + H2 sinh β + H3 cos α + H4 sin α = 0 (iii) (d) Also at x=L, M=0 results in the following equation:

[ ] [ ] 0sincossinhcosh 43

2

21

2

=+

−+

αααβββHH

LHH

L (iv)

Equations (i) to (iv) may be written in matrix form as [B] H = 0, where B11 = 1.0, .... B44 = -(α)2 sin α. The characteristic equation (frequency equation) is |B| = 0. For this particular case the frequency equation may be obtained as follows:


62

From equation (i) H3 = -H1 (ia) From equation (ii) β2H1 - α2H3 = 0 Substituting equation (ia) into the above gives: either α2+β2 = 0, or H1 = H3 =0 (iia) Let us consider the possibility that α2+β2 = 0. This would be true only if both α and β are zero. Substituting α=β=0, into the general solution for v(x,t) gives v(x,t) = [H1 + 0 + H3 + 0] sin(ωt+φ) = 0 [using equation (ia)] This is a trivial solution. Therefore, for non-trivial solution H1 = H3 = 0. Substituting this into equation (iii) and (iv) gives: H2 sinh β + H4 sin α = 0 (iiia)

0sinsinh2

4

2

2 =

−

ααββL

HL

H (iva)

For non-trivial solution of equations (iiia) and (iva) sin α = 0 Hence α = nπ Substituting equation (4.6.7b) into the above yields:

nπ=

++

−

42

244

2 γπλγπ

This, together with equations (4.6.7c,d) results in the following equation:

cnn

n

PP+=

Ω1

2ω

(4.6.8)

where ωn is the nth natural frequency of the axially loaded beam, Ωn is the corresponding natural frequency of the unloaded beam and Pcn is the nth critical load of the same beam which is

given by 2

Ln

EIπ

. The variation of frequency with load is shown graphically in figure 4.6.5

The nth mode is:

Lxnπ

sin


63

ω2

P

Ωn2

-Pcn Figure 4.6.5


64

4.7 Whirling Speed of Shafts Let us now consider the lateral vibration of a shaft that is rotating at a speed of Ω rad/sec. The lateral motion of an infinitesimal element may be regarded as a radial motion of that element, since its position will change due to the rotation. Consider the motion of the free-body in Figure 4.7.1. This is similar to the free-body in Figure 4.5.3, but this time the free-body is vibrating as well as rotating about the original axis of the beam at a speed of Ω rad/sec. Hence its radial acceleration a consists of two components, acceleration due to vibration v and the acceleration due to centripetal acceleration Ω 2v. i.e. a= v - Ω2v

S

S+dS

δx

x

v+dv

o

M M+dM

v

Neutral Surface

F+dF

v(x,t)

F

Ω

a a

x

v

Figure 4.7.1 Following the derivations for a non-rotating beam, the net lateral force in the element is given by:

xx

vEI

xS δ

∂∂

∂∂δ

−= 2

2

2

2

..........................................................(4.7.1)

Using Newton’s 2nd law of motion,

xvvmxamaxmS δδδδ )(..).( 2−===

For principal vibration, vv 2ω−= giving xmvS δωδ )( 22 +−= ........(4.7.2)

Equating the rights hand sides of equations (4.7.1) and (4.7.2) we get,


65

( )vmxv

EIx

222

2

2

2

+=

∂∂

∂∂ ω ......................................(4.7.3)

For a homogeneous, uniform shaft this reduces to the following form:

( )vxv

EI 224

4

m +=∂∂ ω ............(4.7.4)

For a non-rotating shaft, the natural frequencies ωS may be found by solving the above equation for Ω =0. It may be noted here, that the form of the solution will be the same whether or not the

speed of rotation is zero. That is to say that (ω2+Ω2) and 2Sω may be interchanged. One can

deduce the flexural natural frequencies of a rotating shaft from the natural frequencies of non-rotating shafts, by using the equation

(ω2+Ω2) = 2Sω ....................(4.7.5)

If Ω → ωS then, ω → 0. If the speed of rotation (Ω) approaches the natural frequency of the non-rotating shaft (ωS), then the natural frequency of the rotating shaft (ω) would approach zero! That is a state of instability, referred to as whirling. The speeds at which this occurs are called whirling speeds. The whirling speeds are therefore equal to the flexural natural frequencies of the non-rotating shafts. In modelling a shaft for calculating whirling speeds, short bearings that permit lateral rotations may be taken as ‘simple’ supports, and long-bearings that restrain lateral rotations may be taken as ‘clamped’ supports.


66

4.8 Transverse Vibration of Plates (I just commenced this section potential for error is high at this stage)

So far we have considered one-dimensional structural elements. Equations of motion for two and three-dimensional structures can be derived using the same general approach, although the availability of exact solutions is limited, even for systems with simple geometrical shapes such as rectangular plates. Exact natural frequencies and modes are available in the literature for the out-of-plane vibration of rectangular plates, but only for the case where two opposite edges are simply supported. We will present the equation of motion for a plate subject to uniform static in-plane loading, without deriving it, and then proceed to solve it for some specific boundary conditions. Figure 4.8.1 shows the coordinate system and the notation for the plate properties. The out-of-plane displacement coordinate is w(x,y,t) is a function of x, y and t.

Equation of Motion The net elastic flexural resisting force in an infinitesimal plate element is expressed in terms of a parameter D called the plate rigidity. This is given by

)1(12 2

3

ν−= Eh

D , …(4.8.1)

in which E and ν are the elastic modulus and Poisson’s ratio of the plate material, and h is the plate thickness. If the plate is subject to in-plane stresses σx, σy, and τxy, then their components in the out-of-plane direction will affect the equation of motion in the same way the axial force affects the transverse vibration of beams. Traditionally, the in-plane loading is expressed as load per unit length Nx, Ny, and Nxy, which are given by

x y

w

Figure 4.8.1

x = a y = b

x = 0 y = 0


67

=

xy

y

x

xy

y

x

hNN

N

σσσ

The equation of motion is then:

02 2

2

2

22

2

24 =

∂∂+

∂∂+

∂∂∂+

∂∂+∇

tw

hyw

Nyx

wN

xw

NwD yxyx ρ …(4.8.2)

Let us consider a general solution of the form )sin()()(),,( αω += tygxftyxw …(4.8.3)

This assumes that the mode consists of a product of functions in x and y directions. Such a form of solution exits for a plate with two opposite edges simply supported in the absence of any in-plane shear loading (Nxy = 0). Let us assume that the plate we consider is simply supported at the edges y = 0 and y = b. This means the boundary conditions are 0),,(),0,( == tbxwtxw

It follows from equation (4.8.3) that: 0)0( =g …(4.8.4a)

and 0)( =bg …(4.8.4b)

The bending moment intensity (moment per unit length) along the edge y = 0 is given [Ref] by:

∂∂+

∂∂−= 2

2

2

2 ),0,(),0,(),0,(

xtxw

ytxw

DtxM y ν =0

Since this is true for all t, it follows from equation (4.8.3) that )()0()()0( xfgxfg ′′+′′ = 0

We have from equation (4.8.4a), 0)0( =g

Therefore 0)0( =′′g …(4.8.5a)

Similarly we can show that 0)( =′′ bg …(4.8.5b)

The sinusoidal function )/sin()( bynyg π= …(4.8.6)

satisfies boundary conditions (4.8.4a,b) and (4.8.5a,b). By substituting this into the governing differential equation we can show that it also satisfies the equation of motion as follows. Equation (4.9.3) now becomes )sin()/sin()(),,( αωπ += tbynxftyxw

Let )/cos()/sin()/cosh()/sinh()( 4321 axGaxGaxGaxGxf ϕϕϕϕ +++= …(4.8.7)

Substituting equations (4.8.3), (4.8.6) and (4.8.7) into equation (4.8.2) gives:


68

−

−

+

−

Dh

D

N

bn

DN

abn

axf yx

222222

1 )(ωρπϕπϕ

+

−

−

−

+

Dh

D

N

bn

DN

abn

axf yx

222222

2 )(ωρπψπψ

=0

)/cos()/sin()/cosh()/sinh()( 4321 axGaxGaxGaxGxf ϕϕϕϕ +++= …(4.8.8)

where )/cosh()/sinh()( 211 axGaxGxf ϕϕ += and )/cos()/sin()( 432 axGaxGxf ϕϕ +=

Since equation (4.8.8) is true for any x, the terms associated with f1(x) and f2(x) must vanish. i.e…….. ----this section has yet to be completed---- solution will be given for two cases only: (a) all edges simply supported, (b) two opposite edges supported and the other two clamped, and a brief warning on the limitation of the results for large in-plane loadings ( )xN etc for

which there may be a need to consider the effect of initial imperfections. Another 8 pages or so would tidy up this chapter.


69

4.9 Orthogonality of Principal Modes

In Chapter 3, the following equation was derived showing that the natural modes are orthogonal with respect to the mass.

01 ,, = =

n

i kijii qqm for j ≠ k (4.9.1)

For continuous systems we will derive an orthogonality equation of the form:

dmm

kj φφ for j ≠ k (4.9.2)

where φj, φkare the jth and kth modes respectively. For one-dimensional systems the modes would be functions of one spatial coordinate, and the integral would be a single integral. The integral would be a double integral for 2-D systems (for example, with respect to x and y in a Cartesian coordinate system), and triple for 3-D systems (w.r.t x,y and z.) Proof of Orthogonality: Consider the vibration of an infinitesimal element of a continuous system having a mass dm as shown in Figure 4.9.1. Let the net internal induced restoring action corresponding to the jth mode be dFj sin(ωjt+α). If the coordinate of dynamic displacement is q(x,y,..,t) then from Newton's second law of motion, dFj sin(ωjt+α) = dm q . ……….(4.9.3)

If the vibration corresponds to the jth mode, the dynamic displacement would be of the form: q = φj(x,y..) sin(ωjt+α), where φj(x,y..) is the jth mode of vibration, equation (4.9.3) becomes: dFj = -ωj

2 φj dm ……….(4.9.4) From d’Alembert’s principle, if this element is subject to a static force dFj then the resulting displacement would be φj(x,y..).

Figure 4.9.1

dFjsin(ωt+α) q

dm


70

Similarly it may be shown that the net internal action corresponding to the kth mode and the resulting displacement are dFk and φk(x,y..) respectively, where the force dFk is given by:

dFk = -ωk2 φk dm ……….(4.9.5)

Here again we can say that a static force dFk would cause a displacement of φk(x,y..). As for the discrete systems, consider the work done by the internal actions if forces dFj and dFk are applied on the linear elastic system, in the order stated. Noting that φj is caused by dFj, and φk is caused by dFk and that during displacement φk, as a pre-existing force, dFj would do work to its full potential, the net work done on the element is obtained as:

22kk

kjjj . dF

+ . + dF. dF φφφ

Figure 4.9.2 illustrates the force-displacement relationship for the infinitesimal element, in which the two triangular hatched areas represent the work done by the infinitesimal inertial forces while causing corresponding infinitesimal displacements, and the rectangular shaded

area represents the work done by dFj as it moves by φk which is caused by dFk. The total work done includes the integral of the work done on the element and any work done on the boundaries by inertia forces corresponding to rigid masses.

i.e. Net work done = bjkkk

kjjj + W

. dF + . + dF

. dF

22φφ

φ ……….(4.9.6)

where Wbjk is the work done by any attached masses at the boundaries as a result of forces corresponding to the jth mode and kth mode, applied in the order stated. If the order of application of the forces is reversed then the total work done is:

bkjjj

jkkk + W

. dF + . + dF

. dF

22

φφφ

……….(4.9.7)

For linear elastic structures the total work done on the system is independent of the path. Hence equating the r.h.s of equations (4.9.6) and (4.9.7) gives:

dFj

dFk

φj φk

Force

Displacement

Figure 4.9.2


71

bjkkk

kjjj + W

. dF + . + dF

. dF

22φφ

φ= bkj

jjjk

kk + W. dF

+ . + dF. dF

22

φφφ

Canceling the common terms we get,

dFj.φk + Wbjk = dFk.φj + Wbkj ……….(4.9.8) For beams without any attached masses the terms Wbjk and Wbkj would be zero. In such cases the above equation reduces to:

dFk.φj = dFj.φk Substituting equations (4.9.7) and (4.9.8) into the above equation gives

-ωk2 φk.φj dm =-ωj

2 φj.φk dm

i.e. (ωj2-ωk

2) φk.φj dm = 0

If ωk ≠ωj, this results in equation (4.9.2), i.e. φk.φj dm = 0 This is the orthogonality condition. For a beam carrying a concentrated mass m0 at x = a, an extra term mo φk(a).φj(a) should be included.

That is: dmm

ji φφ + mo φk(a).φj(a) = 0 for j ≠ k (4.9.9)

It is possible to include such terms as a part of the integral using the delta functions within the integral of equation (4.9.2). It should also be noted that mass should be replaced with an appropriate moment of inertia term, if the displacement coordinate q is a rotational coordinate. For a one-dimensional structure (bars, beams and shafts) having a constant mass per unit length m, the orthogonality relationship reduces to:

φk.φj dx = 0 ………..…..(4.9. 10)

The above orthogonality relations hold only between two different modes (for j ≠ k) and if j=k then the integrals will give non-zero values.

i.e. φj.φj dm = (φj)2 dm ≠ 0. The above integral is used in dynamic response calculations, and it is therefore convenient to normalize the modes such that this integral is unity.

Using such normal modes, (φj)2 dm =1. ………..…..(4.9.11)


72

Similar orthogonal relationship may also be obtained in terms of stiffness of a structure. This will be discussed in Chapter 5. It should be noted here that in response calculations, the actual displacements would be expressed in terms of the modes (whether normalised or not) and weighting coefficients, which are dependent on initial conditions and/or any applied dynamic force. It is also worth noting that the orthogonality relations may be used to check the accuracy of modes. In the case of approximate methods, calculated values of the integral in equation (4.9.2) (which would be zero if the modes were exact) would give an estimate of any error in the mode calculations.


73

4.10. Review Problems in Natural Frequencies and Modes of Continuous Systems

P.4.1. A uniform cable of length L and mass per unit length m is under a static tensile force T0. Its ends are connected to two identical short cantilever beams of negligible mass (see figure P.4.1). The cantilevers may be modelled by elastic springs of lateral stiffness k each. Show that the frequency equation for symmetrical lateral vibration of the cable is

given by: 02

tanTkL=

λλ , where 0T

mLωλ = .

T0

k

Figure P.4.1. Symmetrical Vibration of a Partially Restrained Cable.

k

T0

P.4.2. Derive the boundary condition equations for the torsional vibration of the geared system

shown in Figure P.4.2, in terms of the angles of twist θ1,θ2 and their derivatives ∂θ1/∂x1 etc.


74

8I

r

2r

Shaft 1: G,J1,ρ1

x1 = L1

Figure P.4.2. Geared System

I

x1 = 0

x2 = 0

x2 = L2

16I

Shaft 2: G,J2,ρ2

P.4.3. The drill pipe of an oil well has a length L and polar second moment of area J, and is

made of a material having shear modulus G, and density ρ. The cutting edge has a polar moment of inertia I0. Obtain the frequency equation for the torsional vibration of this system. From the above result, deduce the frequency equations for the torsional vibration of a shaft subject to the following conditions:

(a) Both ends torsionally restrained; (b) One end restrained and the other end free. P.4.4. A solid shaft, and a tube are rigidly connected to a wall, and a concentrated mass m0 at

the other end as shown in Figure P.4.4. Write down the necessary boundary to determine the longitudinal natural modes and frequencies in terms of u1, u2, u1’, u2’ etc. Proceed to obtain a frequency equation, in determinantal form, for the longitudinal vibration of this system in terms of the parameters shown in FigureP.4.4.


75

Figure P.4.4.

L

Tube, E1,A1,ρ1 Shaft, E2,A2,ρ2

mass m0

P.4.5. A shaft carrying three rotors is shown in Figure P.4.5. The end rotors have a polar

moment of inertia I while the central rotor has an inertia of 4I. The shaft is of circular cross section, radius R, length L, density ρ and shear modulus G.

a) Write down the boundary condition equations for the symmetrical and anti-

symmetrical torsional vibration of this system and ,

b) Derive the frequency equation for the anti-symmetrical case only.

c) Deduce an expression for the natural frequencies corresponding the anti-symmetrical vibration if the polar moment of inertia of the rotors are very large compared to the polar moment of inertia of the shaft.

d) Can the expression you deduced in part (c) be applied to determine the first non-

zero natural frequency, and if not write down an approximate formula for it.

4I

I I

L/2 L/2


76

Figure P.4.5

P.4.6. Write down the boundary conditions in terms of constants H1, H2, H3 etc., for the

vibratory system shown in the following diagram:

Figure P.4.6.

String, Tension T0, length , mass per unit length m, vibrates laterally

E,A,ρ,L


77

P.4.7 A 20 mm diameter circular steel rod is clamped at one end and free at the other as

shown in the diagram below. Show that the frequency equation is of the form cosh λ cos λ + 1=0. If 1.875 is the first root of the frequency equation, find the first flexural natural frequency of the rod in Hz. The material properties are: E = 207 GPa and ρ = 7800 kg/m3. Second moment of area of a circle about its diameter is given by: I = (π r4)/4 where r is the radius of the circle.

Figure P.4.7

1.4 m

P.4.8 For the beam shown in Figure P.4.8, write down the boundary conditions for obtaining the frequency equation for symmetrical flexural modes, in terms of the deflection function f1 and or its derivatives and the parameters shown in the diagram.

Spring stiffness k EI,m

L/2 L/2

Figure (P.4.8)

mass m0


78

P.4.9. Write down the boundary conditions that are necessary for obtaining the frequency equations corresponding to the small amplitude lateral vibration of the beams shown in Figure P.4.9 (a)-1(j), in terms of the lateral dynamic displacement v and/or its derivatives. Do not substitute the general solution to the equation of motion.

System (a): This system consists of two beams.

One of the beams is fully restrained against rotation and partially restrained (stiffness k) against translation at one end. The other end of this beam is hinged to the second beam. The joint also provides partial restraint against relative rotation (stiffness Kr) of the two beams.

System (b): This consists of a uniform beam of

length L and flexural rigidity EI, that is fully restrained at one end and is connected to a cable of length d at the other end The tension in the cable and beam is T. The cable may be assumed to be massless.

System (c): This consists of a

uniform beam supported on knife-edges at two points and carrying a mass m0 at the tip. The mass m0 is rigidly connected to the beam. This mass has a moment of inertia IG about the neutral axis at its centroid. The centroid of the mass is at distance e to the right end of the beam.

L2

E,I,m E,I,m Hinge

System (a)

k Kr

L1

System (b)

System (c)

x = 0 1 x =L1

x = 0 2

EI ,m

e

x = L 2 2

G

m0

E,I,m

L

0

Light, Elastic Cable

T

d


79

System (d): In this system, a uniform beam of flexural rigidity EI, length L and mass per unit length m, is restrained against rotation at the left end, and partially restrained against translation at the right end. It also carries a mass m0 which is pinned to the right end.

System (e): This consists of a stepped

cantilever beam carrying a mass m0 at the tip. The flexural rigidity, mass per unit length, and segmental length of each beam segment are shown in the Figure. The mass m0 is pinned to the beam at the neutral axis.

System (f): In this system, a uniform beam of flexural rigidity EI and mass per unit

length m, is simply supported at two intermediate points. It also carries a mass m0, having a moment of inertia I0 about the neutral axis at mid-span. Write down the eight boundary conditions for the symmetrical flexural vibration of this system. Also write down the two boundary conditions for the anti-symmetrical flexural vibration that are different from the symmetrical case.

System (g): This consists of a beam having flexural

rigidity EI, mass per unit length m and length L carrying end masses m1, m2. The moments of inertia of these masses about the neutral axis of the beam are I1 and I2. The beam is not supported at any point.

m0 EI,m,L

k

System (d)

System (f)

L/4

EI,m

L/4 L/4 L/4

m ,I 0 0

E,I,m m1, I1 ,I1

L

m2, I2 ,I1

System (g)

x1 = 0 x1= L1

x2 = 0

EI1 ,m1

System (e)

. m0

EI2 ,m2

x2 = L2

.


80


81

System (h): In this system, the beam is clamped at one end and elastically restrained by three light cables having Young's modulus E and cross-sectional area A. The two lateral cables have length h and the longitudinal cable is of length l. All cables and the beam are under static tension T0.

System (i): This consists of two beams connected

by a smooth hinge which is partially restrained by a spring of stiffness k. The other end of the first beam is permitted to slide transversely but restrained against rotation. The right end of the second beam is simply supported.

System (j): The left end of this beam is simply supported and partially restrained against rotation by a coil spring that has a stiffness k [moment/unit rotation]. A disc of mass m0 is pinned to the beam at the right end. Neglect any friction in the pin.

E,I,m

h

h

L

0

T0

Light, Elastic Cables E,A

d

System (h)

T0

T0

L1

k

E,I,m E,I,m Hinge

System (i)

L2

L

E,I,m m0

k

System (j)

.


82

3.8.10. The partial differential equation governing the small amplitude flexural vibration of a thin simply supported beam of flexural rigidity EI, length L and mass per unit length m, under static tensile axial loading P is given by

0=m+ 2

2

2

2

4

4

t

v

x

vP

xv

EI∂∂

∂∂−

∂∂ .

The general solution of this equation is of the form: v(x,t) = f(x) sin (ωt+φ). Function f(x) is expressed in the following form:

( )

+

+

+

=Lx

GLx

GLx

GLx

Gxf ααββ sincossinhcosh 4321 , where β and α

are given by:

++

=

42

244

2 γπλγπβ and

++

−=

42

244

2 γπλγπα , in

which EI

PL2

2

πγ = and 4

2

EIm

Lωλ =

(a). Using the above, derive an expression for the natural frequencies of this system. State any assumptions made.

(b). Calculate the fundamental natural frequency of a 1.2 m long simply supported steel beam of 50mm x 50 mm square section, subject to a compressive force of 0.2 Pc1 where Pc1 is the first critical load of the beam. The properties of steel are: elastic modulus E = 207 Gpa and density ρ = 7800 kg/m3.

P.4.11 A 20 mm diameter circular steel shaft is supported on long bearings at both ends. It

carries two rotors very close to the bearings which are located at a distance of 1.5 m apart as shown in Figure P.4.11. Show that the frequency equation for the flexural vibration of this beam is of the form (1 - cosh(λ)cos(λ)) = 0. If 4.73 is the first root (λ1) of the frequency equation, find the first critical speed of the shaft in r.p.m. The Material properties are: E = 207 GPa and ρ = 7800 kg/m3. List all assumptions that were made in the theory and in your model, and explain any practical limitations. Comment on whether or not any research is needed to fully understand the vibrational behaviour of the above or Figure P.4.11

1.5 m


83

similar practical systems. P.4.12 An axially loaded simply supported beam is fully restrained at one end, and is

restrained against translation and partially restrained against rotation at the other end as shown in Figure P.4.12. Write down the boundary conditions to form the frequency equation in determinantal form. Substitute the general solution into the boundary conditions and set up the elements of this determinant.

PKrEI,m,L

Figure P.4.12

longitudinal vibration bars

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