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4.5 Flexural Vibration of Beams

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In this section, we will study a beam (or rod or shaft) in flexural vibration. The vibration is in the

transverse or lateral direction, which is accompanied by bending (or flexure) of the member. Hence, the

vibrations are perpendicular to the main axis of the member, as in the case of a cable or string, which we

studied in Section 4.2. However, a beam, unlike a string, can support shear forces and bending moments

at its cross section. In the initial analysis of bending vibration, we will assume that there is no axial force

at the ends of the beam. We will make further simplifying assumptions that will be clear in the

development of the governing equation of motion. The analysis procedure will be quite similar to that we

have followed in the previous sections.

The study of the bending vibration (or lateral or transverse vibration) of beams is very important in a

variety of practical situations. Noteworthy are the vibration analyses of structures like bridges, vehicle

guideways, tall buildings, and space stations; the ride quality and structural integrity analysis of buses,

trains, ships, aircraft and spacecraft; the dynamics and control of rockets, missiles, machine tools and

robots; and the vibration testing, evaluation, and qualification of products with continuous members.

4.5.1 Governing Equation for Thin Beams

Now, we will develop the Bernoulli – Euler equation, which governs the transverse vibration of thin beams.

Consider a beam bending in the x – y plane, with x as the longitudinal axis and y as the transverse axis of

bending deflection, as shown in Figure 4.9. We will develop the required equation by considering the

moment – deflection relation, rotational equilibrium, and transverse dynamics of a beam element.

4-26 Vibration and Shock Handbook

(1) Moment – deflection relation

A small beam element of length dx subjected to bending moment M is shown. Neglect any transverse

deflections due to shear stresses. Consider a lateral area element dA in the cross section A of the beam

element, at a distance w (measure parallel to y) from the neutral axis of bending.

Normal strain (at dA)

1 ¼ ðR þ wÞdu 2 R du

R du

Note that the neutral axis joins the points along the beam when the normal strain and stress are zero.

Hence,

1 ¼

R ð4:116Þ

where R ¼ radius of curvature of the bent element. Normal stress in the axial direction

s ¼ E1 ¼ E

R ð4:117Þ

where E ¼ Young’s modulus (of elasticity). Then, bending moment

M ¼

ws dA ¼

w2 E

dA ¼

w2 dA ¼

where I ¼ second moment of area of the beam cross section about the neutral axis. So, we have

M ¼

R ð4:118Þ

Slope at A ¼ ›v=›x; slope at B ¼ ð›v=›xÞ þ ð›2v=›x2Þdx; where v ¼ lateral deflection of the beam at

element dx: Hence, the change in slope ¼ ð›2v=›x2Þdx ¼ du; where du is the arc angle of bending for the

beam element dx; as shown in Figure 4.9.

δx

A B

δθ

δx

δ A

Neutral Axis

FIGURE 4.9 A thin beam in bending.

Distributed-Parameter Systems 4-27

Also, we have dx ¼ R du: Hence, ð›2v=›x2ÞR du ¼ du: Cancel du: We obtain

R ¼

›2v

›x2 ð4:119Þ

Substitute Equation 4.119 into Equation 4.118. We obtain

M ¼ EI

›2v

›x2 ð4:120Þ

(2) Rotatory dynamics (equilibrium)

Again, consider the beam element dx; as shown in Figure 4.10, where forces and moments acting on

the element are indicated. Here, f ðx; tÞ ¼ excitation force per unit length acting on the beam, in the

transverse direction, at location x: Disregard the rotatory inertia of the beam element.

Hence, the equation of angular motion is given by the equilibrium condition of moments:

M þ Q dx 2 M þ

›M

›x

􀀏 􀀐

¼ 0

Q ¼

›M

›x ¼

›

›x

›2v

›x2

􀁻 !

ð4:121Þ

where the previously obtained result (Equation 4.120) for M has been substituted. Note that we have not

assumed a uniform beam and hence I ¼ IðxÞ will be variable along the beam length.

(3) Transverse dynamics

The equation of transverse motion (Newton’s Second Law) for element dx is

ðrA dxÞ

›2v

›t2 ¼ f ðx; tÞdx þ Q 2 Q þ

›Q

›x

􀀏 􀀐

Here, r ¼ mass density of the beam material. So, we obtain

›2v

›t2 þ

›Q

›x ¼ f ðx; tÞ

or, in view of Equation 4.121, we have the governing equation of forced transverse motion for the beam as

›2v

›t2 þ

›2

›x2 EI

›2v

›x2

􀁻 !

¼ f ðx; tÞ ð4:122Þ

f (x,t)δx

M δx ∂x

∂M

M +

Q δx

∂x

∂Q

Q +

x + δx

FIGURE 4.10 Dynamics of a beam element in bending.

4-28 Vibration and Shock Handbook

4.5.2 Modal Analysis

The solution to the flexural vibration problem given by Equation 4.122 may be obtained exactly as for other

continuous members. Specifically, we first obtain the natural frequencies and mode shapes and express the

general solution as a summation of the modal responses. The approach is similar for both free and forced

problems, but the associated generalized coordinates will be different. This approach is followed here.

For modal (natural) vibration, consider the free motion described by

›2

›x2 EI

›2v

›x2

􀁻 !

þ rA

›2v

›t2 ¼ 0 ð4:123Þ

For a uniform beam, EI will be constant and Equation 4.123 can be expressed as

›2vðx; tÞ

›t2 ¼ 2c2 ›4vðx; tÞ

›x4 ð4:124Þ

where

c ¼

ffiffiffiffiffi

ð4:125Þ

Observe from Equation 4.124 that it is fourth order in x and second order in t; whereas the governing

equations for the transverse vibration of a cable, longitudinal vibration of a rod, and torsional vibration

of a shaft, are all identical in fourth and second order in x: So the behavior of transverse vibrations of

beams will not be exactly identical to that of these other three types of continuous system. In particular,

the traveling wave solution 4.7 will not be satisfied. However, there are also many similarities.

In each mode the system will vibrate in a fixed shape ratio. Hence, the time and space functions will be

separable for a modal motion; we seek a solution of the form

vðx; tÞ ¼ Y ðxÞqðtÞ ð4:126Þ

This separable solution for a modal response has been justified previously. Note that, even in the lumped

parameter case, we make the same assumption, except in that case we have a modal vector

Y ¼

.. .

66666664

77777775

instead of a mode shape function Y ðxÞ: For a given mode of a lumped parameter system, Yi values denote

the relative displacements of various inertia elements mi; as shown in Figure 4.11. Hence, the vector Y

corresponds to the mode shape. Note that Yi can be either positive or negative. Also, qðtÞ is the harmonic

function corresponding to the natural frequency.

It should be clear that Y and qðtÞ are separable in this lumped-parameter case of modal motion.

Then, in the limit, Y ðxÞ and qðtÞ should also be separable for the distributed parameter case.

Substitute Equation 4.124 into Equation 4.123,

and bring the terms containing x to the left-hand

side (LHS) and terms containing t to the righthand

side (RHS).

rAY

dx2 EI

d2Y

dx2

􀁻 !

¼ 2

qðtÞ

d2q

dt2 ¼ v2

ð4:127Þ

Since a function of x cannot be equal to a function

of t in general, unless each function is equal to the

same constant, we have defined v2 as a constant.

Y1q(t) Y2q(t) Y3q(t) Y4q(t)

m1 m2 m3 m4

FIGURE 4.11 Modal motions of a lumped-parameter

system.

Distributed-Parameter Systems 4-29

We have not shown that this separation constant ðv2Þ should be positive. This requirement can be

verified due to the nature of the particular vibration problem; that is, qðtÞ should have an

oscillatory solution in general. It is also clear that the physical interpretation of v is a natural

frequency of the system. Equation 4.127 corresponds to the two ODEs, one in t and the other in

x; as

d2qðtÞ

dt2 þ v2qðtÞ ¼ 0 ð4:128Þ

dx2 EI

d2Y ðxÞ

dx2 2 v2rAY ðxÞ ¼ 0 ð4:129Þ

The solution of these two equations will provide the natural frequencies v and the corresponding

mode shapes Y ðxÞ of the beam.

For further analysis of the modal behavior, assume a uniform beam. Then, EI will be constant and

Equation 4.129 may be expressed as

d4Y ðxÞ

dx4 2 l4Y ðxÞ ¼ 0 ð4:130Þ

where

v ¼ l2c ¼ l2

ffiffiffiffiffi

ð4:131Þ

The positive parameter l is yet to be determined, and will come from the mode shape analysis.

The characteristic equation corresponding to Equation 4.130 is

p4 2 l4 ¼ 0; or; ðp2 2 l2Þðp2 þ l2Þ ¼ 0 ð4:132Þ

The roots are

p ¼ ^l; ^jl ð4:133Þ

Hence, the general solution for a mode shape (eigenfunction) is given by

Y ðxÞ ¼ A1elx þ A2e2lx A3eþjlx þ A4e2jlx ¼ C1 cosh lx þ C2 sinh lx þ C3 cos lx þ C4 sin lx

ð4:134Þ

There are five unknowns (C1; C2; C3; C4; and l) here. The mode shapes can be normalized and one of

the first four unknowns can be incorporated into qðtÞ as usual. The remaining four unknowns

are determined by the end conditions of the beam. So, four BCs will be needed.

Note:

cosh lx ¼

elx þ e2lx

; sinh lx ¼

elx 2 e2lx

cos lx ¼

e jlx þ e2jlx

; sin lx ¼

e jlx 2 e2jlx

cosh lx ¼ l sinh x;

sinh x ¼ l cosh x

4.5.3 Boundary Conditions

The four modal BCs that are needed can be derived in the usual manner, depending on the conditions at

the two ends of the beam. The procedure is to apply the separable (modal) solution Equation 4.126 to the

end relation with the understanding that this relation has to be true for all possible values of qðtÞ: The

relation (Equation 4.128) may be substituted as well, if needed.

4-30 Vibration and Shock Handbook

For example, consider an end x ¼ x0 that is completely free. Then, both bending moment and shear

force have to be zero at this end. From Equation 4.120 and Equation 4.121, we have

›2vðx0; tÞ

›x2 ¼ 0 ð4:135Þ

›

›x

›2vðx0; tÞ

›x2

􀁻 !

¼ 0 ð4:136Þ

Substitute Equation 4.126 into Equation 4.135 and Equation 4.136:

d2Y ðx0Þ

dx2 qðtÞ ¼ 0

d2Y ðx0Þ

dx2 qðtÞ ¼ 0

which are true for all qðtÞ: Hence, the following modal BCs result for a free end:

d2Y ðx0Þ

dx2 ¼ 0 ð4:137Þ

d2Y ðx0Þ

dx2 ¼ 0 ð4:138Þ

For a uniform beam, Equation 4.137 becomes

d3Y ðx0Þ

dx3 ¼ 0

Some common conditions and the corresponding modal BC equations for the bending vibration of a

beam are listed in Box 4.2.

4.5.4 Free Vibration of a Simply Supported Beam

To illustrate this approach, consider a uniform beam of length l that is pinned (simply supported) at both

ends. In this case, both displacement and the bending moment will be zero at each end. Accordingly, we

have the modal boundary conditions

Y ð0Þ ¼ 0 ¼ Y ðlÞ ð4:139Þ

d2Y ð0Þ

dx2 ¼ 0 ¼

d2Y ðlÞ

dx2 ð4:140Þ

where l ¼ length of the beam. Substitute Equation 4.134 into Equation 4.139:

C1 þ C3 ¼ 0 ð4:141Þ

C1 cosh ll þ C2 sinh ll þ C3 cos ll þ C4 sin ll ¼ 0 ð4:142Þ

To apply the bending moment BCs, first differentiate Equation 4.134 to obtain

dx ¼ lC1 sinh lx þ lC2 cosh lx 2 lC3 sin lx þ lC4 cos lx

d2Y

dx2 ¼ l2C1 cosh lx þ l2C2 sinh lx 2 l2C3 cos lx 2 l2C4 sin lx

and the substitute these into the bending moment BCs (Equation 4.140). We obtain

C1 2 C3 ¼ 0 ð4:143Þ

C1 cosh ll þ C2 sinh ll 2 C3 cos ll 2 C4 sin ll ¼ 0 ð4:144Þ

Distributed-Parameter Systems 4-31

as l – 0 in general, due to the oscillatory nature of most modes. Equation 4.141 and Equation 4.143 give

C1 ¼ 0 ¼ C3: Then,

Equation 4.142 becomes:

Equation 4.144 becomes: C2 sinh ll þ C4 sin ll ¼ 0

C2 sinh ll 2 C4 sin ll ¼ 0

Add

C2 sinh ll ¼ 0

However, sinh ll ¼ 0 if and only if l ¼ 0: This corresponds to zero-frequency conditions (no

oscillations), and is rejected as it is not true in general. Hence, we have C2 ¼ 0: Accordingly, we are left

Box 4.2

BOUNDARY CONDITIONS FOR TRANSVERSE

VIBRATION OF BEAMS

1. Simply supported (pinned) Deflection ¼ 0 ) Y ¼ 0

Bending moment ¼ 0 )

d2Y

dx2 ¼ 0

2. Clamped (fixed)

Deflection ¼ 0 ) Y ¼ 0

Slope ¼ 0 )

dx ¼ 0

3. Free Bending moment ¼ 0 )

d2Y

dx2 ¼ 0

Shear force ¼ 0 )

d2Y

dx2 ¼ 0

4. Sliding Slope ¼ 0 )

dx ¼ 0

Shear force ¼ 0 )

d2Y

dx2 ¼ 0

5. Dynamic (flexible, inertial, etc.) Transverse equation of motion (or force balance)

Substitute Equation 4.128, if needed

Rotatory equation of motion (or moment balance)

4-32 Vibration and Shock Handbook

with the remaining equation:

C4 sin ll ¼ 0 ð4:145Þ

However, if C4 ¼ 0 then Y ðxÞ ¼ 0; which corresponds to a stationary beam with no oscillations, and is

rejected as the trivial solution. Hence, the valid solution is given by sin ll ¼ 0 which gives the infinite set

of solutions:

lil ¼ ip for i ¼ 1; 2; 3; … ð4:146Þ

Note that we must have i . 0 because l has to be nonzero, thereby giving nonzero natural frequencies

according to Equation 4.131, as required for the given problem.

4.5.4.1 Normalization of Mode Shape Functions

For absorbing the yet unknown constant C4 into qðtÞ; we will normalize the mode shape functions. The

commonly used normalization condition is

ðl

Y 2

i dx ¼

2 ð4:147Þ

Hence,

2 ¼

ðl

4 sin2 ipx

dx ¼ C2

ðl

sin2 ipx

dx ¼

Note that we used cos 2u ¼ 1 2 2 sin2u prior to integration. Then, for normalized mode shape

functions, we have C4 ¼ 1: Hence, the normalized eigenfunctions (mode shape functions) for various

modes are given by

YiðxÞ ¼ sin

ipx

for i ¼ 1; 2; 3; … ð4:148Þ

Using the result (Equation 4.146) in Equation 4.131, the natural frequencies of the ith mode are

vi ¼

i2p2

ffiffiffiffiffi

for i ¼ 1; 2; 3; … ð4:149Þ

In this manner, we have obtained an infinite set of mode shape functions YiðxÞ for a simply supported

beam. Hence, according to the solution (Equation 4.126), we have a corresponding infinite set of

generalized coordinates qiðtÞ; i ¼ 1; 2; 3; …; which satisfy Equation 4.128. It follows that the overall

response of the beam is

vðx; tÞ ¼

YiðxÞqiðtÞ ð4:150Þ

4.5.4.2 Initial Conditions

We have yet to solve Equation 4.128 for determining qiðtÞ: To do so, we need to know the initial conditions

qið0Þ and q_ið0Þ: These are determined from the beam initial conditions of displacement and speed, which

have to be known:

vðx; 0Þ ¼ dðxÞ ð4:151Þ

›vðx; 0Þ

›t ¼ sðxÞ ð4:152Þ

Substitute Equation 4.150 into Equation 4.151 and Equation 4.152 to obtain

YiðxÞqið0Þ ¼ dðxÞ ð4:153Þ X

YiðxÞq_ið0Þ ¼ sðxÞ ð4:154Þ

Distributed-Parameter Systems 4-33

Multiply by YjðxÞ and integrate from x ¼ 0 to l; using the orthogonality property of

YiðxÞ ¼ sinðipx=lÞ; namely

ðl

sin

ipx

sin

jpx

dx ¼

0 for i – j

for i ¼ j

8><

ð4:155Þ

We obtain

qjð0Þ ¼

ðl

dðxÞYjðxÞdx ð4:156Þ

q_jð0Þ ¼

ðl

sðxÞYjðxÞdx ð4:157Þ

In this manner, qiðtÞ is completely determined for each vi by solving Equation 4.128, using the

initial conditions (Equation 4.156 and Equation 4.157). Hence, the complete solution (Equation

4.150) is determined for the free bending vibration of a simply supported beam.

4.5.5 Orthogonality of Mode Shapes

We have seen that the mode shapes of simply supported beams in bending vibrations are orthogonal

(see Equation 4.155). This property is not limited to simply supported beams but holds for most BCs, as

we will show now. First, from integration by parts, twice, we have

ðl

d4Yj

dx4 dx ¼ Yi

d3Yj

dx3

􀀈 􀀈 􀀈 􀀈 􀀈

ðl

dYi

d3Yj

dx3 dx ¼ Yi

d3Yj

dx3 2

dYi

d2Yj

dx2

" #l

0þ

ðl

d2Yi

dx2

d2Yj

dx2 dx ð4:158Þ

Now consider two separate modes, i and j; which have the modal equations

Mode i :

d4Yi

dx4 ¼ l4iYi ðaÞ

Mode j :

d4Yj

dx4 ¼ l4j

Yj ðbÞ

Multiply Equation (a) by Yj, multiply Equation (b) by Yi, integrate both with respect to x from 0 to l;

make use of Equation 4.158 and subtract the second result from the first. We obtain

ðl4i

2 l4j

ðl

YiYjdx ¼

ðl

d4Yi

dx4 2 Yi

d4Yj

dx4

􀁻 !

¼ Yj

d3Yi

dx3 2

dYj

d2Yi

dx2

" #l

2 Yi

d3Yj

dx3 2

dYi

d2Yj

dx2

" #l

ð4:159Þ

Clearly, the two RHS terms are zero for typical BCs, such as pinned, fixed, free, and sliding. Now, since

li – lj for i – j (unequal modes), we have

ðl

YiYj dx ¼ 0 for i – j lj for i ¼ j ð4:160Þ

Note that normalized mode shape functions may be used here to obtain the constant lj ¼ l=2:

4-34 Vibration and Shock Handbook

4.5.5.1 Case of Variable Cross Section

The orthogonality of mode shapes holds for nonuniform beams as well. Here, EI is not constant. We use

integration by parts:

ðl

dx2 EI

d2Yi

dx2 dx ¼ Yj

dx EI d2 Yi

dx2 |fflffl{zfflffl}

664

775

dYj

EI d2 Yi

dx2 |ffl{zffl}

ðl

d2Yj

dx2

d2Yi

dx2 dx ð4:161Þ

Again, the first two terms on the RHS are zero for typical BCs. Then, as before, we use the modal

equations (Equation 4.129) for two different modes, i and j:

dx2 EI

d2Yi

dx2 ¼ v2i

rAYi

dx2 EI

d2Yj

dx2 ¼ v2j

rAYj

Multiply the first equation by Yj; the second equation by Yi; subtract the second result from the first,

integrate the result form x ¼ 0 to l; and finally use Equation 4.161 to cancel the equal terms. We obtain

ðv2i

2 v2j

ðl

rAYiYj dx 2 Yj

d2Yi

2 Yi

d2Yj

" #l

dYj

d2Yi

dx2 2

dYi

d2Yj

dx2

" #l

0¼ 0

ð4:162Þ

Now, as before, for common BCs, the second and third boundary terms in Equation 4.162 will vanish.

Hence, after canceling the term v2i

2 v2j

; which is – 0 i – j; we obtain the orthogonality condition for

nonuniform beams as

ðl

rAYiYj dx ¼

0 for i – j

aj for i ¼ j

(

ð4:163Þ

The general steps for the modal analysis of a distributed-parameter vibrating system are summarized in

Box 4.3.

4.5.6 Forced Bending Vibration

The equation of motion is

›2

›x2 EI

›2v

›x2 þ rA

›2v

›t2 ¼ f ðx; tÞ ð4:164Þ

Assume a separable, forced response:

vðx; tÞ ¼

q􀀊iðtÞYiðxÞ ð4:165Þ

where q􀀊iðtÞ are the generalized coordinates in the forced case. Substitute Equation 4.165 into Equation

4.164 in the beam equation:

q􀀊iðtÞ

dx2 EI

d2YiðxÞ

dx2 þ rA

€􀀊

qiðtÞYiðxÞ ¼ f ðx; tÞ

Distributed-Parameter Systems 4-35

The first term on the LHS, on using the mode shape equation 4.129, becomes 􀀊qiðtÞrAv2i

YiðxÞ: Multiply

the result by YjðxÞ and integrate with respect to x½0; l􀀉: We obtain

v2j

q􀀊jðtÞ

ðl

rAY 2

j ðxÞdx þ €

􀀊

qjðtÞ

ðl

rAY 2

j ðxÞdx ¼

ðl

YjðxÞf ðx; tÞdx

|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl}

fj ðtÞ

Each of the two integrals on the LHS evaluates aj according to Equation 4.163. Hence,

€􀀊

qjðtÞ þ v2j

q􀀊jðtÞ ¼

fjðtÞ for j ¼ 1; 2; 3; … ð4:166Þ

Box 4.3

MODAL ANALYSIS OF CONTINUOUS SYSTEMS

Equation of free (unforced) motion:

Lðx; tÞvðx; tÞ ¼ 0 ðiÞ

where

vðx; tÞ ¼ system response

Lðx; tÞ ¼ partial differential operator in space ðxÞ and time ðtÞ

Model solution:

Assume a separable solution

vðx; tÞ ¼ Y ðxÞqðtÞ ðiiÞ

because a modal response is separable in time and space.

Note:

Y ðxÞ ¼ mode shape

qðtÞ ¼ generalized coordinate for free response

Note: For two- and three-dimensional space systems, time and space will still be separable for a

modal response. However, the space function itself may not be separable along each coordinate

direction.

Steps:

1. Substitute Equation (ii) in Equation (i) and separate the space function (of x ) and the time

function (of t ), each of which should be equal to the same constant.

2. Solve the resulting ODE for Y ðxÞ using system boundary conditions. We obtain an infinite

set of mode shapes YiðxÞ; up to one unknown (removed by normalization), and natural

frequencies vi:

3. Solve the ODE for qðtÞ using system initial conditions to determine qiðtÞ for mode i:

(The orthogonality of YiðxÞ will be needed to establish the initial conditions for qiðtÞ.)

4. Overall response

vðx; tÞ ¼

YiðxÞqiðtÞ

4-36 Vibration and Shock Handbook

We can then solve this equation to determine the generalized coordinates q􀀊jðtÞ; using the knowledge of

the forcing function fjðtÞ and the initial conditions q􀀊jð0Þ and ðdq􀀊j=dtÞð0Þ: Specifically, if the initial

displacement and speed of the beam are given by Equation 4.151 and Equation 4.152, respectively, by

following the procedure that was adopted to obtain the results (Equation 4.156 and Equation 4.157), we

determine

q􀀊jð0Þ ¼

ðl

rAdðxÞYjðxÞdx ð4:167Þ

dq􀀊jð0Þ

dt ¼

ðl

rAsðxÞYjðxÞdx ð4:168Þ

Finally, we obtain the overall response of the forced system as

vðx; tÞ ¼

q􀀊jðtÞYjðxÞ ð4:169Þ

The main steps in the forced response analysis are summarized in Box 4.4.

Box 4.4

FORCED RESPONSE OF CONTINUOUS

SYSTEMS

Equation of forced motion:

Lðx; tÞvðx; tÞ ¼ L1ðx; tÞf ðx; tÞ ðiÞ

where

vðx; tÞ ¼ forced response of the system

f ðx; tÞ ¼ distributed force per unit space

L and L1 are partial differential operators in space and time

Steps:

1. Substitute the modal expansion

vðx; tÞ ¼

YiðxÞq􀀊iðtÞ ðiiÞ

in Equation (i), where

YiðxÞ ¼ mode shapes

q􀀊iðtÞ ¼ generalized coordinates for forced motion

2. Multiply by YjðxÞ and integrate with respect to space ðxÞ using orthogonality

ðl

mðxÞYiðxÞYjðxÞdx ¼0 for i – j ðiiiÞ

Note: Additional boundary terms are present in Equation (iii) when there are lumped elements

at the system boundary.

3. Determine the initial conditions for q􀀊iðtÞ: We require Equation (iii) for this.

4. Solve the ODE for q􀀊iðtÞ using the initial conditions.

5. Substitute the results into Equation (ii).

Distributed-Parameter Systems 4-37

Example 4.6

A pump is mounted at the midspan of a simply supported thin beam of uniform cross section and length

l: The pump rotation generates a transverse force f0 cos vt, as schematically shown in Figure 4.12.

Initially, the system starts from rest, from the static equilibrium position of the beam, such that

vðx; tÞ ¼0 and

›vðx; tÞ

›t ¼ 0 at t ¼ 0

It is necessary to obtain the transverse response vðx; tÞ of the beam in the form of a modal summation

during operation of the pump.

First, determine qjðtÞ in terms of f0; v; vj and the beam parameters r; A; and l; assuming that the beam

is completely undamped. Are all modes of the beam excited by the pump? If the beam is lightly damped,

what would be its steady-state response?

In particular, what is the steady-state response of the beam at the pump location? Sketch its amplitude

as a function of the excitation frequency v:

Solution

Using the Dirac delta function dðxÞ; we can express the equation of forced motion of the beam as

›4v

›x4 þ rA

›2v

›t2 ¼ f0 cos vtd x 2

􀀏 􀀐

ð4:170Þ

Substitute vðx; tÞ ¼

i YiðxÞqiðtÞ; where normalized mode shapes for the simply supported beam are

YiðxÞ ¼ sin ipx=l, multiply by YjðxÞ, integrate over x ¼ ½0; l􀀉 and use the orthogonality of mode shapes.

We obtain

􀀏 􀀐4 l

qjðtÞ þ rA

q€j ¼ f0 cos vt sin

Note:

ðaþ

pðxÞdðx 2 aÞdx ¼ pðaÞ

Hence, from Equation 4.149, we obtain

q€j þv2j

qj ¼ aj cos vt ð4:171Þ

where

aj ¼

2f0

rAl

sin j

The given initial conditions are satisfied if and only if qjð0Þ ¼ 0 and q_jð0Þ ¼ 0 for all j:

l/2 x

Pump

f0 cos wt

FIGURE 4.12 A pump mounted on a simply supported beam.

4-38 Vibration and Shock Handbook

Hence, the complete solution is

qjðtÞ ¼

ðv2j

2 v2Þ ½cos vt 2 cos vjt􀀉 ð4:172Þ

It follows that the total response is

vðx; tÞ ¼

2f0

rAl

X sin jp=2

ðv2j

2 v2Þ

sin j

l ½cos vt 2 cos vjt􀀉 ð4:173Þ

Clearly, sinðjp=2Þ ¼ 0 for even values of j: We see that even modes of the beam are not excited by the

pump. This is to be expected because, for even modes, the midspan is a node point and has no motion.

If the beam is lightly damped, its natural response terms ðcos vjtÞ will decay to zero with time. Hence, the

steady-state response will be

vssðx; tÞ ¼

2f0

rAl

cos vt

X sin jp=2 sin jpx=l

ðv2j

2 v2Þ ð4:174Þ

At the pump location ðx ¼ l=2Þ; the steady-state response is

vssðl=2; tÞ ¼

2f0

rAl

cos vt

X sin2jp=2

ðv2j

2 v2Þ ¼

2f0

rAl

cos vt

ðv21

2 v2Þ þ

ðv23

2 v2Þ þ

ðv25

2 v2Þ þ · · ·

" #

Note again that only the odd modes contribute to the response. Furthermore, for a simply supported

beam

vj ¼ j2v1 for j ¼ 1; 2; 3; …

where

v1 ¼

􀀏 􀀐2

ffiffiffiffiffi

Nondimensionalize the midspan response at steady state as

v􀀊 ¼

rAlv21

2f0

vssðl=2; tÞ ¼

ð14 2v􀀊2Þ þ

ð34 2v􀀊2Þ þ

ð54 2v􀀊2Þ þ · · ·

􀀒 􀀓

cos vt

Its amplitude is

v0ðv􀀊Þ ¼

ð14 2v􀀊2Þ þ

ð34 2v􀀊2Þ þ

ð54 2v􀀊2Þ þ · · ·

􀀈 􀀈 􀀈 􀀈

The characteristic of the amplitude as a function of the nondimensional excitation frequency is sketched

in Figure 4.13.

Example 4.7

Perform a modal analysis to determine natural frequencies and mode shapes of transverse vibration of a

thin cantilever (i.e., a beam with “fixed – free” or “clamped – free” end conditions). The coordinate system

and the beam parameters are as shown in Figure 4.14.

Solution

As usual, the mode shapes are given by

Y ðxÞ ¼ C1 cosh lx þ C2 sinh lx þ C3 cos lx þ C4 sin lx ð4:134Þ

Distributed-Parameter Systems 4-39

Its first three derivatives are

dY ðxÞ

dx ¼ C1l sinh x þ C2l cosh lx 2 C3l sin lx þ C4l cos lx ð4:175Þ

d2Y ðxÞ

dx2 ¼ C1l2 cosh lx þ C2l2 sinh lx 2 C3l2 cos lx 2 C4l2 sin lx ð4:176Þ

d3Y ðxÞ

dx3 ¼ C1l3 sinh lx þ C2l3 cosh lx þ C3l3 sin lx 2 C4l3 cos lx ð4:177Þ

The BCs of the beam are

At x ¼ 0 : vð0; tÞ ¼0 and

›vð0; tÞ

›x ¼ 0

At x ¼ l : EI

›2vðl; tÞ

›x2 ¼0 and EI

›3vðl; tÞ

›x3 ¼ 0

The corresponding modal BCs are

Y ð0Þ ¼ 0;

dY ð0Þ

dx ¼ 0;

d2Y ðlÞ

dx2 ¼ 0;

d3Y ðlÞ

dx3 ¼ 0

ð4:178Þ

Substitute Equation 4.134, Equation 4.175 to

Equation 4.177 into Equation 4.178. We obtain

C1 þ C3 ¼ 0 ðiÞ

C2 þ C4 ¼ 0 ðiiÞ

Steady-State

Response

Amplitude

(Nondimensional)

0.1 1 10 25 49 100

Excitation Frequency w/w1

0.5

1.0

1.5

2.0

0.0

FIGURE 4.13 Amplitude of the steady response at the pump as a function of excitation frequency.

0 x

E, I, r, A l

y, v(x,t)

FIGURE 4.14 A cantilever in bending vibration.

4-40 Vibration and Shock Handbook

C1 cosh ll þ C2 sinh ll 2 C3 cos ll 2 C4 sin ll ¼ 0 ðiiiÞ

C1 sinh ll þ C2 cosh ll þ C3 sin ll 2 C4 cos ll ¼ 0 ðivÞ

Eliminate C1 and C2 in Equation (iii) and Equation (iv) by substituting Equation (i) and Equation (ii).

We obtain

½cosh ll þ cos ll􀀉C3 þ ½sinh ll þ sin ll􀀉C4 ¼ 0 ðvÞ

½sinh ll 2 sin ll􀀉C3 þ ½cosh ll þ cos ll􀀉C4 ¼ 0 ðviÞ

or, in the vector-matrix form

cosh ll þ cos ll sinh ll þ sin ll

sinh ll 2 sin ll cosh ll þ cos ll

" #

ð4:179Þ

The trivial solution of Equation 4.179 is C3 ¼ 0 ¼ C4: Then, from Equation (i) and Equation (ii), we also

have C1 ¼ 0 ¼ C2: This solution corresponds to Y ðxÞ ¼ 0 and is not acceptable in general for a vibrating

system. Hence, the matrix in Equation 4.179 must be noninvertible (i.e., singular). Hence, the

determinant of the matrix must vanish (see Appendix 3A); thus

ðcosh ll þ cos llÞ2 2 ðsinh ll þ sin llÞðsinh ll 2 sin llÞ ¼ 0

cosh2 ll þ 2 cosh ll cos ll þ cos2ll 2 sinh2ll þ sin2ll ¼ 0

However, it is well known that

cosh2ll 2 sinh2ll ¼1 and cos2ll þ sin2ll ¼ 1

Hence,

cos ll cosh ll ¼ 21 ð4:180Þ

This equation has an infinite number of solutions li for i ¼ 1; 2; 3; … giving an infinite number of

natural frequencies:

vi ¼ l2i

ffiffiffiffiffi

ð4:181Þ

The corresponding mode shapes are given by Equation 4.134 subject to Equation (i), Equation (ii), and

Equation (v) or Equation (vi). This gives YiðxÞ ¼ C3ðcos lix 2 cosh lixÞ þ C4ðsin lix 2 sinh lixÞ with

C3 ¼ 2 ½sinh lil þ sin lil􀀉

½cosh lil þ cos lil􀀉

It follows that

YiðxÞ ¼ C4½sin lix 2 sinh lix􀀉 þ C4

sinh lil þ sin lil

cosh lil þ cos lil

􀀒 􀀓

½2cos lix þ cosh lix􀀉

The unknown multiplier C4 simply scales the mode shape and is absorbed into the generalized

coordinate qiðtÞ as usual. In fact, this is a process of normalization of mode shapes, where C4 ¼ 1 is used.

So, we have the normalized mode shapes:

YiðxÞ ¼ a sin lix þ b sinh lix þ ai½c cos lix þ d cosh lix􀀉 ð4:182Þ

with

a ¼ 1; b ¼ 21; c ¼ 21; d ¼1 and ai ¼

sinh lil þ sin lil

cosh lil þ cos lil ð4:183Þ

The first three roots of Equation 4.180 are

l1l ¼ 1:875104; l2l ¼ 4:694091; l3l ¼ 7:854757

Distributed-Parameter Systems 4-41

The corresponding three mode shapes are sketched in Figure 4.15. Note, in particular, that the node

points are not physically fixed but remain stationary during a modal motion. This completes the

solution.

The modal information corresponds to the infinite set of natural frequencies that are obtained by

solving Equation 4.180 subject to Equation 4.181, and the mode shapes are given by Equation 4.182

subject to Equation 4.183. Modal information corresponding to other common BCs may also be put in

this form. Table 4.3 summarizes such data. Table 4.4 provides numerical values corresponding to this

modal information for the first three modes.

Location Along Beam x/l

Mode

Shape

Yi(x)

FIGURE 4.15 First three modes of a cantilever (fixed – free beam) in transverse vibration.

TABLE 4.3 Modal Information for Bending Vibration of Beams

End

Conditions

Natural Frequencies

vi ¼ l2i

ffiffiffiffiffiffiffiffi

pEI=rA

where li are roots of

Mode Shapes

Yi ðxÞ ¼ a sin li x þ b sinh li þ ai ½c cos li x þ d cosh li x􀀉

a b c d ai

Pinned – pinned

sin li l ¼ 0

i ¼ 1; 2; 3; …

1 0 0 0 0

Fixed – fixed

cosh li l cos li l ¼ 1

i ¼ 1; 2; 3; …

1 2 1 21 1

sinh li l 2 sin li l

cosh li l 2 cos li l

Free – free

cosh li l cos li l ¼ 1

i ¼ 0; 1; 2; …

1 1 2 1 21 Same

Fixed – pinned

tanh li l ¼ tan li l

i ¼ 1; 2; 3; …

1 2 1 21 1 Same

Fixed – free cosh li l cos li l ¼ 21 1 2 1 21 1

sinh li l þ sin li l

cosh li l þ cos li l

Fixed – sliding tanh li l ¼ 2tan li l 1 2 1 21 1

cosh li l þ cos li l

sin li l 2 sin li l

Pinned – free tanh li l ¼ tan li l 1 ai 0 0

sin li l

sinh li l

4-42 Vibration and Shock Handbook

4.5.7 Bending Vibration of Beams with Axial Loads

In practice, beam-type members that undergo flexural (transverse) vibrations may carry axial forces.

Examples are structural members such as columns, struts, and towers. Generally, a tension will increase

the natural frequencies of bending and compression will decrease them. Hence, one way to avoid the

excitation of a particular natural frequency (and mode) of bending vibration is to use a suitable tension

or compression in the axial direction.

The equation of motion for the transverse motion of a thin beam subjected to an axial tension P may

be easily derived by following the procedure that led to Equation 4.122. For simplicity, assume a thin

beam subjected to a constant tensile force P: A small element dx of the beam is shown in Figure 4.16. The

vertical component (in the positive y direction) of the axial force is

2P sin u þ P sinðu þ duÞ ø 2Pu þ Pðu þ duÞ ¼ P du

because the slope u ¼ ›v=›x is small. Also, the change in slope is

du ¼

›2v

›x2 dx

It is clear that the previous equation of transverse dynamics of element dx has to be modified simply by

adding the term Pð›2v=›x2Þdx to the f ðx; tÞdx side of the equation. Then, the resulting equation of

transverse vibration will be

›2v

›t2 þ

›2

›x2 EI

›2v

›x2

􀁻 !

2 P

›2v

›x2 ¼ f ðx; tÞ ð4:184Þ

TABLE 4.4 Roots of the Frequency Equation for

Bending Vibration of Beams

End Conditions First Three Roots li l

Pinned – pinned p

Fixed – fixed 4.730041

7.853205

10.995608

Free – free 0

4.730041

7.853205

10.995608

Fixed – pinned 3.926602

7.068583

10.210176

Fixed – free 1.875104

4.694091

7.854757

Fixed – sliding 2.365020

5.497804

8.639380

Pinned – free 0

3.926602

7.068583

10.210176

Distributed-Parameter Systems 4-43

For modal analysis of a uniform beam, then, we use the equation of free motion

›2v

›t2 þ EI

›4v

›x4 2 P

›2v

›x2 ¼ 0 ð4:185Þ

With a separable (modal) solution of the form

vðx; tÞ ¼ Y ðxÞqðtÞ ð4:186Þ

we have

q€ðtÞ

qðtÞ ¼ 2

EI ðd4Y =dx4Þ 2 P ðd2Y =dx2Þ

rAY ¼ 2v2 ð4:187Þ

which gives, as before, the time response equation of the generalized coordinates

q€ðtÞ þv2qðtÞ ¼ 0 ð4:188Þ

and the mode shape equation

d4Y

dx4 2 P

d2Y

dx2 2 rAv2Y ¼ 0 ð4:189Þ

Note that the mode shape equation is still fourth order, but is different. The analysis, however, may be

done as before by using four BCs at the two ends of the beam to determine the natural frequencies (an

infinite set) and the corresponding normalized mode shapes.

4.5.8 Bending Vibration of Thick Beams

In our derivation of the governing equation for the lateral vibration of thin beams (known as the

Bernoulli – Euler beam equation), we neglected the following effects in particular:

1. Deformation and associated lateral motion due to shear stresses

2. Moment of inertia of beam elements in rotatory motion

Note, however, that we did use the fact that shear forces ðQÞ are present in a beam cross section, even

though the resulting deformations were not taken into account. Also, in writing the equation for rotational

motion of a beam element dx; we simply summed the moments to zero, without including the inertia

moment. These assumptions are valid for a beam whose cross-sectional dimensions are small compared

with its length. However, for a thick beam, the effect of shear deformation and rotatory inertia must be

included in deriving the governing equation. The resulting equation is known as the Timoshenko beam

equation. Important steps in the derivation of the equation of motion for the forced transverse vibration of

beams, including the effects of shear deformation and rotatory inertia, are given now.

y, v(x,t)

q+dq

FIGURE 4.16 Beam element in transverse vibration and subjected to axial tension.

4-44 Vibration and Shock Handbook

Consider a small element dx of a beam. Figure 4.17(a) illustrates the contribution of the bending of an

element and the deformation due to transverse shear stresses towards the total slope of the beam neutral

axis. Let

u ¼ angle of rotation of the beam element due to bending

f ¼ increase in slope of the element due to shear deformation in transverse shear (this is equal to

shear strain)

Then, the total slope of the beam element is

›v

›x ¼ u þ f ð4:190Þ

Here, v and x take the usual meanings, as for a thin beam.

Figure 4.17(b) shows an element dx of the beam with the forces, moments, and the linear and angular

accelerations marked. With the sign convention shown in Figure 4.17(b), the linear shear-stress – shearstrain

relation can be stated as

Q ¼ 2kGAf ð4:191Þ

where the Timoshenko shear coefficient

k ¼

Average shear stress on beam cross section at x

Shear stress at the neutral axis at x

and

G ¼ shear modulus

The equation for translatory motion is

f ðx; tÞdx þ Q 2 Q þ

›Q

›x

􀀏 􀀐

¼ ðrA dxÞ

›2v

›t2

Hence,

f ðx; tÞ 2

›Q

›x ¼ rA

›2v

›t2 ð4:192Þ

∂ t2

∂2 θ

∂x

∂ v

y, v(x,t)

x+dx

(a) (b)

x Bent Only

Bent and

Sheared

x x+dx

f(x,t)

M δ x

∂x

∂M

Q δ x

∂ x

∂Q +

∂t2

∂ 2v

q +f =

FIGURE 4.17 A Timoshenko beam element: (a) combined effect of bending and shear; (b) dynamic effects

including rotatory inertia.

Distributed-Parameter Systems 4-45

The equation for the rotatory motion of the element, taking into account the rotatory inertia, is

M þ

›M

›x

dx 2 M 2 Q dx ¼ ðIr dxÞ

›2u

›t2

which becomes

›M

›x

2 Q ¼ Ir

›2u

›t2 ð4:193Þ

From the elementary theory of bending, as before

M ¼ EI

›u

›x ð4:194Þ

The relationship between the shear modulus and the modulus of elasticity is known to be

E ¼ 2ð1 þ nÞG ð4:195Þ

where n ¼ Poisson’s ratio. This relation may be substituted if desired.

Manipulation of these equations yields

›4v

›x4 þ rA

›2v

›t2 2 rI 1 þ

􀀏 􀀐

›4v

›x2 ›t2 þ

r2I

›4v

›t4

¼ f ðx; tÞ 2

kGA

›2f ðx; tÞ

›x2 þ

kGA

›2f ðx; tÞ

›t2 ð4:196Þ

This is the Timoshenko beam equation for forced transverse motion. Note that this equation is fourth

order in time, whereas the thin beam equation is second order in time. The modal analysis may proceed as

before, by using the free ðf ¼ 0Þ equation and a separable solution. However, the resulting differential

equation for the generalized coordinates will be fourth order in time and, as a result, additional natural

frequency bands will be created. The reason is the independent presence of shear and bending motions. The

differential equation of mode shapes will be fourth order in x and the solution procedure will be as before,

through the use of four BCs at the two ends of the beam.

4.5.9 Use of the Energy Approach

So far, we have used only the direct, Newtonian approach in deriving the governing equations for

continuous members in vibration. Of course, the same results may be obtained by using the

Lagrangian (energy) approach. The general approach here is to first express the Lagrangian L of the

system as

L ¼ T p 2V ð4:197Þ

where

Tp ¼ total kinetic co-energy (equal to kinetic energy T for typical systems)

V ¼ total potential energy

Then, for a virtual increment (variation) of the system through incrementing the system variable, the

following condition will hold:

ðt2

t1 ½dL þ dW 􀀉dt ¼ 0 ð4:198Þ

where dL is the increment in the Lagrangian and dW is the work done by the external forces on the system

due to the increment. Finally, using the arbitrariness of the variation, the equation of motion, along with

the BCs, can be obtained. This approach is illustrated now.

4-46 Vibration and Shock Handbook

First, consider the free motion. We can easily extend the result to the case of forced motion. Kinetic

energy is given by

T ¼

ðl

rA dx

›v

›t

􀀏 􀀐2

ðl

›v

›t

􀀏 􀀐2

dx ð4:199Þ

Potential energy due to bending results from work needed to bend the beam (angle denoted by u ¼ ›v=›x

as usual) under bending moment M; thus

V ¼

ðEnd 2

End 1

M df ¼

ðl

›2v

›x2

􀁻 !2

dx ð4:200Þ

Before proceeding further, note the following steps of variation and integration by parts with respect to t:

ðt2

›v

›t

􀀏 􀀐2

dt ¼

ðt2

›v

›t

›v

›t

􀀏 􀀐

dt ¼

ðt2

›v

›t

› dv

›t

dt ¼

›v

›t

􀀒 􀀓t2

ðt2

›2v

›t2 dv dt ð4:201Þ

Note that we interchanged the operations d and ›=›t prior to integrating by parts. Also, by convention,

we assume that no variations are performed at the starting and ending times (t1 and t2) of integration.

Hence,

dvðt1Þ ¼0 and dvðt2Þ ¼ 0 ð4:202Þ

Similarly, variation and integration by parts with respect to x are done as follows:

ðl

›2v

›x2

􀁻 !2

dx ¼

ðl

›2v

›x2 d

›2v

›x2

􀁻 !

dx ¼

ðl

›2v

›x2

›

›x

›v

›x

􀀏 􀀐

¼ EI

›2v

›x2 d

›v

›x

" 􀀏 􀀐#l

ðl

›

›x

›2v

›x2 d

›v

›x

􀀏 􀀐

dx ¼ EI

›2v

›x2 d

›v

›x

" 􀀏 􀀐#l

ðl

›

›x

›2v

›x2

› dv

›x

¼ EI

›2v

›x2 d

›v

›x

" 􀀏 􀀐#l

›

›x

›2v

›x2 dv

" #l

0þ

ðl

›2

›x2 EI

›2v

›x2 dv dx ð4:201aÞ

Now, for the case of free vibration ðdW ¼ 0Þ, substitute Equation 4.201 and Equation 4.201a in dT; and

dV of Equation 4.199 and Equation 4.200, to obtain

ðt2

dL dt ¼0 ¼ 2

ðt2

ðl

dx rA

›2v

›t2 þ

›2

›x2 EI

›2v

›x2

" #

dv þ

ðl

dxrA

›v

›t

􀀒 􀀓t2

ðt2

dt EI

›2v

›x2 d

›v

›x

" 􀀏 􀀐#l

0þ

ðt2

›

›x

›2v

›x2 dV

" #l

ð4:203Þ

Since Equation 4.203 holds for all arbitrary variations dvðtÞ; its coefficient should vanish. Hence,

›2v

›t2 þ

›2

›x2 EI

›2v

›x2 ¼ 0 ð4:204Þ

which is the same Bernoulli – Euler beam equation for free motion as we had derived before.

The second integral term on the RHS of Equation 4.203 has no consequence. We conventionally pick

dvðt1Þ ¼ 0 and dvðt2Þ ¼ 0 at the time points t1 and t2:

The third integral term on the RHS of Equation 4.203 gives some BCs. Specifically, if the slope BC

›v=›x is zero (i.e., fixed end), then the corresponding bending moment at the end is arbitrary, as

expected. However, if the slope at the boundary is arbitrary, then the bending moment EIð›2v=›x2Þ at the

end should be zero (i.e., pinned or free end).

The last integral term on the RHS of Equation 4.203 gives some other BCs. Specifically, if

the displacement BC v is zero (i.e., pinned or fixed end) then the corresponding shear force at the

Distributed-Parameter Systems 4-47

end is arbitrary. However, if the displacement at the boundary is arbitrary, then the shear force

ð›=›xÞEIð›2v=›x2Þ at that end should be zero (i.e., free or sliding end).

Next, consider a forced beam with force per unit length given by f ðx; tÞ: Then, the work done by

f ðx; tÞdx in a small element dx of the beam, when moved through a displacement of dv; is

f ðx; tÞdx dv ð4:205Þ

Then, by combining Equation 4.205 with Equation 4.203, for arbitrary variation dv; we obtain the forced

vibration equation

›2v

›t2 þ

›2

›x2 EI

›2v

›x2 ¼ f ðx; tÞ ð4:206Þ

Note that external forces and moments applied at the ends of the beam can be incorporated into the BCs

in the same manner.

4.5.10 Orthogonality with Inertial Boundary Conditions

It can be verified that the conventional orthogonality condition (Equation 4.163) holds for beams in

transverse vibration, under common noninertial BCs. When an inertia element (rectilinear or rotatory) is

present at an end of the beam, this condition is violated. A modified and more general orthogonality

condition can be derived for application to beams with inertial boundary conditions.

To illustrate the procedure, consider a beam with a mass m attached at the end x ¼ l; as shown in

Figure 4.18(a). A free-body diagram giving the sign convention for shear force Q acting on m is shown in

Figure 4.18(b).

The BCs at x ¼ l are:

1. Bending moment vanishes, because there is no rotatory inertia at the end that is free. Hence,

›2vðl; tÞ

›x2 ¼ 0 ð4:207Þ

EI(x), ρA(x) l

(a) y, v(x,t)

Point

Mass

(b)

m v(l,t)

Q = EI

∂x2

∂2v (l,t)

∂x

∂

FIGURE 4.18 (a) A beam with an end mass in transverse vibration; (b) free-body diagram showing the shear force

acting on the end mass.

4-48 Vibration and Shock Handbook

2. Equation of rectilinear motion of the end mass

›2vðl; tÞ

›t2 ¼

›

›x

›2vðl; tÞ

›x2 ð4:208Þ

In the usual manner, by substituting vðx; tÞ ¼ YiðxÞqiðtÞ for mode i; along with €qiðtÞ ¼ 2v2i

qiðtÞ; into

Equation 4.207 and Equation 4.208, with the understanding that the result should hold for any qiðtÞ; we

obtain the corresponding modal boundary conditions:

d2YiðlÞ

dx2 ¼ 0 ð4:209Þ

d2YiðlÞ

dx2 þ v2i

mYiðlÞ ¼ 0 ð4:210Þ

for mode i:

Now, we return to Equation 4.162:

ðv2i

2 v2j

ðl

rAYiYj dx 2 Yj

d2Yi

dx2 2 Yi

d2Yj

dx2

" #l

dYj

d2Yi

dx2 2

dYi

d2Yj

dx2

" #l

0¼ 0

ð4:162Þ

The second and the third terms of Equation 4.162 will vanish at x ¼ 0 for noninertial BCs, as usual. At

x ¼ l; the third term will vanish in view of Equation 4.209. So, we are left with the second term at x ¼ l:

Substitute Equation 4.210:

d2Yi

dx2 2 Yi

d2Yj

dx2

􀁻 !

l¼ 2ðv2i

2 v2j

ÞmYiðlÞYjðlÞ ð4:211Þ

Substitute Equation 4.211 into Equation 4.162 and cancel v2i

2 v2j

– 0 for i – j: We obtain

ðl

rAYiYjdx þ mYiðlÞYjðlÞ ¼

0 for i – j

aj for i ¼ j

(

ð4:212Þ

This is the modified and more general orthogonality property. If the mass is at x ¼ 0; the direction of Q

that acts on m will reverse and, hence, the second term in Equation 4.212 will become 2mYið0ÞYjð0Þ:

4.5.10.1 Rotatory Inertia

If there is a free rotatory inertia at x ¼ l; without an associated rectilinear inertia, then the shear force will

vanish, giving

d2YiðlÞ

dx2 ¼ 0 ð4:213Þ

The equation of rotational motion of J will give

d2YiðlÞ

dx2 2 v2i

dYiðlÞ

dx ¼ 0 ð4:214Þ

Here, the second term in Equation 4.162 will vanish in view of Equation 4.213. Then, by substituting

Equation 4.214 into the third term of Equation 4.162, we obtain the modified orthogonality relation

ðl

rAYiYj dx þ J

dYiðlÞ

dYjðlÞ

dx ¼

0 for i – j

aj for i ¼ j

(

ð4:215Þ

Distributed-Parameter Systems 4-49

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