Engineering Vibrations 3rd Edition Inman.pdf

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494

Distributed-Parameter Systems

Chap. 6

Euler

B

ournoulll Beam Theory

. . d

beam

wi

th

the

transverse direction

of

vibration

Figure

6.10

tUustratesa

~ n u l e v e r e

) . .

they

direction]. T he beam is of rectangu

indicated (i.e.,

the

defiec_Mn ,_w \ ~ ~ ) m e s s h and

length[.

Also associated

with

Jar cross section

A x)

wtth ~ t d t h Jf El x) h e r e E is Young s elastic modulus

the beam is a [lexural b e n d m g ) st n ~ s s al area moment

of

nertia about the z

axis.

for

the

beam a_nd I

(

x ts t ~ e o t h s s ~ e ustains a

be

nding moment M

(

x,

I),

which is

Fr

om mcchamcs of matenals, e earns . ( ) b

d

h

b

m

deflection or bending dcformatwn, w x, t , y

relate to t e ea

a2w x, t) (6 85)

M x, t) EI x) ax2

.

. . . be derived from examining the force diagram of

A m o ~ e . l of_bendmg vtbr

auon

rnay as indicated in Figure

6 1

0.

Assuming the de

an infinttcstmal element

of

the

beha:

t th s

hear

deformation is much smaller than

formation

to be sma

ll enough sue a e

w x,r)

dx

A x)

l

Sec. 6.5 Bending Vibration of a Beam 95

w x,

1)

(i.e.,

so

that

the

si

de

s

of

the

element

dx

do

not bend), a summation

of

forces

in they direction yields

av x,

t )

)

a

2

w x, t)

V x,

t) + ax

dx

-

V x,

t) + f x, t) dx

=

pA x) dx

at

2

(6.86)

Here V

(

x, t is the shear force at the eft

end of

the element dx, V

(

x, t + V.r(x, t) d x

is the shear force at the right end of he element dx,f x, t is the total external force

applied

to

the element per unit length, and the term on the right side

of

the equality

is the inertial force

of

the element.The assumption

of

small shear deformation used

in the force balance

of

equati

on (6.86)

is t rue i /h,

10

and l/ h

y 10

(i.e., for

long slender beams or Eu ler-Bernoulli beams .

Next the moments acting on the el

ement

dx

ab out the z axis through point Q

are summed.T his yields

[

aM x,

t) J [

aY x, t)

J dx

M x,

t)

+

ax

dx -

M x, t)

+ V x

, t) + ax

dx dx +

[f x,

t)

dx]

T 0

(6.87)

Here the left-handside

of

the equation is zero since t s also assumed

that

the rotary

inertia of the eleme

nt

dx is negligible. Simplifying this expression yields

[

aM x,t)

+

V x t)]dx + [iJV x,t)

+ f x,t)] dx)

2

0

(6.88)

ax ax

2

Since dx is assumed

to

be very small, (dx )

2

is assumed to be almost zero, so that this

m

oment

expression yields

dx

is small, but not zero)

aM x,

t)

V x, t) = - ax

(6.89)

This states that the shear force is proportional to the spatial change in the bending

moment. Substitution

of

this expressionfor the shear force into equation

(6.86)

yields

a

2

a

2

w x, 1)

[

M x

, t) ]dx

+

f x,t)dx

=

pA x)dx

2

{6.90)

h

F urther sub

st

itution

of

equation

(6.85)

into

(6.90)

and dividing by dx yields

.

a

2

w x

1)

a

2

[ a

2

w x t ]

pA(x)

2

+ -

2

El

x)

2

= f x,

t)

at i lx

x

(6.91)

If

no external force is applied

so

thatf x, t)

=

0 and if

EI x)

and

A x)

are

assumed

to be constant, equation

(6.91)

simplifies

so

that free vibration is governed by

Pw x,

t) +

cz a4w x,

t

=

0 c = {Ei (6.92)

2

ax

4

Y A

Note

that

unJike

the

previous equations, the free vibration equation (6.92) contains

four

spat

ial derivat\ves and hence requires four (instead of two) boundary conditions



96


Chap. 6

in calcul

at

ing a solution.

The

presence of lhe two time derivatives again requires that

two initial conditions,

one

for the displacementand one for the velocity, be specified.

The boundary conditions

required to

solve

th

e spatial equation in a

separation-of-variables sol ution of

equation

(6.92) are ob tained by examining the

deflection

w(x, t) , the slope of

the deflection

iJw(x,

t)/ilx,

the bending

moment

EJiPw(x, t)

jax

2

, and

the sh

ear force

a[EliJ

2

w(x,

t) jax']fa

x

at

each end

of

the

beam.

A co

mmon configuration is

clamped-free

or

cantilevered

as illustra ted in Fig

ure 6.10. In addition to a boundary being clamped or free,

the

end

of

a beam could

be resting on a support restrained from bending or deflecting.The

sit

uation is called

simply

supported or

pinned.

A

sliding

bou ndary is one in which displacement

is

allowed but rotat

ion

is not.

The shear

load at a sliding boundary is zero.

f a beam in transverse vibration is free at one end, the deflectiOJl

and

slope at

that end arc unrestricted,

but

the bending moment and shear force must vanish:

c

2

w

bending moment =

l

2

= 0

ax

c

[

i w]

hear

force = -

El -

= 0

ax ax

2

6.93)

If, on the other hand , the

end

of a beam is clamped (or fixed), the bending moment

and shear force are unrestricted,but the deflection and slope must vanish at that end:

deflection = w = 0

aw

slope

= = 0

ax

6.94)

At

a simply supported or

pinned end,

the slope

and

shear force are unrestricted and

the deflection

and

bending

moment

must vanish:

deflection = w = 0

aw

bending mom en t =

l -

2

= 0

iJx

6.95)

At a sliding end, the slope or rotation is zero and no shear force is allowed. On the

other hand, the deflection and bending moment arc unrestricted. Hence,

at

a sliding

boundary,

aw

slope= - = 0

ax

shear force = _ ___ ET = 0

x x

6.96)

Other boundary conditions

are

possible by connecting the ends

of

a beam to a vari

ety of devices such as lumped masses, springs, and so on.These boundar y conditions

can be determined by force and moment balances.

In

addition to sa tisfying four boundary conditions, the solution of equation

(6.92) for free vibration can be calcula ted only if two initial conditions (in time) are

Sec.

6.5

Bending Vibration

o

a Beam

97

specified. As in the case of the rod tr d b . . .

fied initial deflection and velocity ~ o ~ : ~ :

ar

, these InJtJal conditionsare speci-

w(x, 0) = Wo(x) and w,(x, 0)

=

ti.lo x)

assuming that

t

== 0 is

the

initial time N

t

h .

no

motion will r

es

ult. ·

0

e

t

at

Wo a

nd

10

o cannot

both

be zero, or

The

so

lution of equation

(6.92) subject

to four bounda r d . . .

tial conditions proceeds following tl h Y JtJonsand two mt-

exac

y t e same st

eps

used

10

previo t A

separation-of-variables solution oftheform

w(x

t) =

X(

·)T(

) . .

us s e ~ t?n

s.

stitutcd into the equation ofmotion e . x t tsassumed.This tssub-

quation

(6.92),

to yteld (after

rea

rrangement)

2

X' '(x) 1 (1)

c = - - - -

w 2

X(x)

T(t) -

(6.97)

where the partial derivatives have been re 1 d . I . .

(note: X' ' = d4X/dx4 z I 2 pace

i l l

t

otal denvauve

s as before

made b d . -

d T dt ).

Here

the

chOice of separation constant

w2

is

c o m e ~ f ; ~ ~ 1

~ ~ : ; : ~ ~ : ~ : ~ i : : : systems of Section 6.4,

that the

natural

r e q ~ e d c y

(6.98)

which is the rightside

of

equation

6.97).This

temporal equation has asolution

of

he form

T(t)

=

Asinwc

+

Bcoswr

(6.99)

~ ~ ~ ~ t i ~ ~ : ~ ~ ~ s : ~ : ~ : c ~ ~ d b ~ ~ v ~ t ~ e ~ : ~ a s p l ~ a ~ r r n i n e d by the specified initial

Th t I

u

wn.

e spa

Ja

equation comes from rearranging equation

(6.97),

which yields

X (x) - ( ~ c )X(x)

=

0

(6.100)

By

defining [recall

equation

(6.92)]

(6.101)

and ~ s u r n i n g a solution to eq uation (6.100) of the form Ae<T.r the eneral solution

equatiOn (6.100) can be calculated to be

of the

form (see ~ b l e ;

6.42)

of

X(x)

= al sin j3x + a2cos{3x + a3sinhl3x + a4cosh{3x (6.102)

H.ere the value for 3 and thr

ee of

the four constants

of inte

rati

: ~ ~ ~ ~ ~ t c ~ ~ n ~ d

from

the

four boundary conditions.The f ~ u r t : ~ ~ ~ t : ~ t a b ~ : : : :

termined f ; ~ m ~ h : ~ ~ ~ ~ ~ n t s

~ . a ? d

B the e ~ p o r a l equation, which arc then de-

rocedure f r .

J

con JtJons. be followmg example illustrates the solution

at

end and mpl supported

at

other end.



P

o a

f1xed

one

si y the

498


Chap. 6

Example 6.5-l

Calculate

the

natural frequencies and mode

s

hapes

for the

tran

sve rse

vibration of

a

beam of

length

I that

is fixed at one

end and pinned at

the other end.

Solution The boundaryconditions in this case are given by equation (6.94) at the fixed

end and equation (6.95) at the pinned end. Substitution of the general solution gi ven by

equation (6.102)

into

equation (6.94) at x

=

0 yields

X O) = 0

=>

a

2

+ a

4

=

0 (a)

X'(O) = 0

=>

(3(a

1

+ a

3

) =0 (b)

Similarly, at x = I the boundary conditions result in

X f) =

0 =>

a

1

sin 13/ -t- a

2

cos [3/

+ a

3

sinh 13/

+ a4

cosh

3

= 0 (c)

EIX (l)

= 0

=>

f3

2

(- a

1

sinf3/-

a

2

cosf31 + a

3

sinbf3/ + a

4

coshf3/) = 0 (d)

These

four boundary conditions thus yield

four

equations [(a) through (ci)J

in the

four un

known coefficients a

1

, a

2

, a

3

, and a

4

•These can be written as the siuglc vector equation

0

13

sinhl3/

13

2

sinh 3

1 ] a•]

[OJ

1

2 0

cosh f3/ a

3

= 0

f3

2

cosh f3/ a. 0

Recall

from

Chapter

4 t

hat this vector equation

can

have

a

nonzero solution

for the

vector a = [a

1

a

2

a

3

a

4

f only if the

determinant

of the coefficient matrix vanishes

(i.e., if the coefficient matrix is singular). Furthermore, recall that sin

ce

the coefficient

matrix is singular,

not

all

of the elements of the vector

a can be calculated.

Selling the detem1inant above

equal

to

zero

yields the

characteri

stic equation

tan 31 = tanh

f31

This equality is satisfied for an infinite number

of

choices for 13 denoted 13,.The solution

can be visualized by

plotting

both tan

3

and tanh

3

versus

3 on

the same plot. This

is

similar to

the solution technique

used in

Example

6.4.]

and

illustrated in Figure

6.9.

The

first

Cive

so

lutions

are

f31/

=

3.926602

f34/ 13.351768

[3

2

1

=

7.068583

13sl = 16.49336143

f3

J/

= 10.210176

For the rest of the mode

s (i.e., for values

of the

index

n > 5),

the soluti

ons

to

the

char

acterist ic

equation

are well

approximated

by

(41'1

+

1)11

[3 .1

= 4

With these values of he weighted frequencies 13,./, he individual

modes of

vibration can

be calculated. Solving the preceding matrix equation

for

the individual coefficients a;

yields

a

1

= - a

3

,

a

2

= -a

4

, a

nd

(sinhf3,.1-

sinl3,./)a

3

+

(coshf3

11

/ -

cosl3,./)a

4

==

0

Sec. 6.5

Bending Vibration of a

eam

499

Thus

a

3

= _ cosh 3 ,.1 - cos (3,./

sinh

3

,1

- sin

13,1 a

4

for each n. Tbe fourth coefficient a cannot b d . .

cause the

coefficient matrix

is

sm·

u l ~

( h e. et

crmmed

by th1s

set of

equations,

be-

. . . ar Ot erw1se eacha would b Tb.i .

mg

coefflctcnt becomes

the

arb tr . d ' , e

zero

. s

rcmam

-

1

ary

magmtu e of

the

· f · .

depends on 11, denote it by

a )

S b

li .

e•gen uncuon. As this constant

X x)

for

the spatial solution ;i;icts r ~ ~ ~ ~ h o f t t ~ s _values a in the exp r

ession

have the form a e eJgenfuncltons or

mode

shapes

X

·)- coshl3,1-cosa l

,

P • •

sinh 13 1 - sin

13,1 (smh

[3,x

-

sm

[3.,x

-

coshf3,.x +

cos

3 ~ ~ x ]

II

=

J , 2, 3, ...

The first

three

mode shapes are plotted in Figure 6 11

for(''

) -

J

a d 1 2

2 11

- n II = , ,3.

Mode3

x/1

Figure 6. U Plot of the first ihree mode sha f

beam

of

Example 6

5

1

b . .

1

. pes

0

th

e clampcd- piJtncd

· · •ar llran Ynormalized

to

unity.

These mode

shapes can

be shown to be orthogonal,

so that

i'x,(; ')X.,(x)dx

=

0

for n * m

(sec

Exercise6.45).

As

in

Exa

1

6

. .

tial

con dit ions can be used to calcul t tbmp e

.3.2,

this orthogonality, along with the ini-

h

. ' a e

econstantsA andB

·

th · .

t e displacement

1

0 e scnes

so

lut1on

for

tu(x,t) = f A, sin w,t +B., cosw t)X

x )

n 1 n n

0

Table 6.6 summarizes a number of dif£ b . .

slender beam.

The

slender beam m d l . ~ n t

o ~ d a r y

conftgurations for the

·' o e gtven m equatton (6.91)

is

often referred to



NDARY CONFIGURATIONS OF A SLENDER

T

LE

6.6 SAMPLE

OF

VARIOUS

:uLENGTH

I ILLUSTRATING WEIGHTED NATURAL

BEAM IN TRANSVERSE VIBRATION

FREQUENCIES AND MODE SHAPES•

Co

nfig

ur

ati

on

t=x

I

___ ]

Free free

a x

~ l

Clamped- free

Clamped-pinned

Clarnpcd-;;liding

r· f

ClaJRped-<:larnped

Weighted frequen_ci':

a ~ d

charactensuc

eq

uauon

0 (rigid-body mode)

4.73004074

7.85320462

10.9956078

14.1371655

17.2787597

(

211

+ l )n

for 11

>

5

2

cos 3

cosh -

1

1.87510407

4.69409113

7.85475744

l0.99554073

14.13716839

(211- 1)T1 s

for

11

>

2

cos 3 cosh

3.92660231

7.06858275

10.210176t2

13.35176878

16.49336143

(4n

+

1)11'

for 11 >

S

4

ta

n 13/ - tanh 13/

Mode

shape

cosh 13.rt .._ cos \3,.,.

- a

11

(sinh 13,,. +sin 13,rr b

cosh 13.X cos 13,,x

11

(si

nh sin

3 , . ~

COW

13,rY

- COS

13,rY

11

(sinh

13,,x

-

sin 13 .x)

2.36502037

5.49780392

8.63937983

11.78097245

14.92256510

(411- 1)11

for >

5

4

tan

3

+ tanh 13/ 0

4.73004074

7.85320462

\0.9956079

14.1371

.655

17.2787597

cosh

j3

11

x cos 13.,r

-u

11

(sinh

13,»

- sin13.rr

2n + l)'TI'

for >5

2

cos

13/

cosh 131

1

sin 3 0

TX

s n

1

-

a .

0.9825

1.0008

0.9999

t.OOOO

0.9999

1 for 11

>

5

0.7341

1.0

185

0.9992

1.0000

1.0000

1 for

11 >

5

l.OOOB

1

[or 1

> l

0.9825

1

for

11 > 1

0.982502

1.00078

0.999966

t.OOOO

t.OOOO

1

for

11

> 5

none

Pinned-pinned .

. .,

1

are related

10

the nat 11

ral frequencaes

by

ere ed

r ~ e s P•

The

of

1

th

111

Sec . 6.5

Bending Vibration of a Be

am

5 1

as the

Euler Bemoufli or

BernouUi-E ul

er

beam equation.The assump tions used in

formulating this model

ar

e that the beam be

• Uniform along its span,

or

length,

and slender

• Composed of a linear,homogeneous, isotropic elastic materialwithout axial loads

• Such that plane sections remain plane

• Such that

the

plane of symmetry of

the

beam is also the plane of vibration so

that rotation and translation are decoupled

• Such that rotary inertia and shear

def

ormation can be

neg

lected

T he key to solvi ng for the time response of distributed-parameter systems is

th e orthogonality o( the mode shapes. Note from Table

6.

7 that the mode shapes are

quite complicated in many configurations. This does not mean that or thogonali ty

is

.necessarily violated, just that evaluating the integrals in the modal ana lysis proce

dure becomes more difficult.

TABLE 6.7 EQUATIONS FOR THE MODE SHAPE

COEFFICIENTS

FOR USE

IN

TABLE 6.4

8

Boundary condi

tion

Frec-[rec

Clamped-free

Clamped-pinned

Clamped-sliding

Clamped-damped

Formula for

ern

c o s h ~ , . / - cos ..

1

a.,

= sinh 13.,/

- sin f3.J

sinh 3.,/ - sin13.,/

a

=

cosh

13.,/ + cos

13.1

cosh

13.,/

-

cos ..1

a = sinh

13,./

- sin

13,./

sinh

..

1 - s i n ~ . , /

a

. = cos

h

3.,/

+ cos

13,,1

Same as [ree- free

•These coefficients

arc

used in

the

calculations for

the

mode

shapes, as

illustrated

in

Example

6.5.1.

Timoshenko Be m Theory

The model of .the tra nsverse vibrat ion of the beam presented in

eq

uation (6 .91 )

ignores the e ffects of sh

ea

r deformation a nd rotary inertia. Beams modeled includ

ing the effects of rotary inertia and shear deformation are called Timoshenko beams.

These

effects arc considered next. As mentioned previously, it is safe to ignore the

shear deformation as long as

th

e hz and

h,

. illustrated in Figure 6.10

are

sma ll com

pared with the length of the beam. As the beam becomes shorter, the efiect of shear

deformation becomes evident. Thi s is illustrated in Figure 6.12, which is a repeat of

the e lement dx

of

Figure 6.10 with shear deformation included.

Referring to the figure, the line OA is a line through the center of the element



•H the weight natural d . E rnple 6

5

1 values

a

for e. - aly£/fpA.asuse xa · · ·

equation (6. 101) or ' - P • formulas given in . able 6.5.

mode

shapes

are

computed from

the t

1

d

X

_

A(l

_ i/ 2)·

th

e first is

bTh

ere

arc two fccc- free m o d ~ shapes: Xo

=

cons an an

o - . '

translational,the second rota t•onal.

dx

perpendicular to

the

face at

the

ri

ght

side.

Th

e line

OB

,

on

the o

ther

h

and,

is the









































































































































































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