97407357-Vibration-of-Continuous-Systems-By-singiresu-S-Rao.pdf

Vibration of Continuous Systents

Singiresu S. RaoProfessor and Chairman Department of Mechanical and Aerospace Engineering University of Miami Coral Gables, Florida-

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81CEHTllHNIAI.

~lll 8 0 7 z

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: G?WILEY : z z~ 2007;

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JOHN WILEY & SONS. INC.

This book is printed on acid-free paper.@> Copyright

2007 by John Wiley & Sons, Inc. All rights reserved.

Published by John Wiley & Sons, Inc., Hoboken, New Jersey Published simultaneously in Canada. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400, fax 978-646-8600, or on the web at www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 'Ill River Street, Hoboken, NJ 07030, 201-748-6011, fax 201-748-6008, or online at http://www.wiley.comlgo/permissions. Limit of LiabilitylDisclaimer of Warranty: While the publisher and author have used their best efforts in preparing this book, they make no representations or warranties with respect to the accuracy or completeness of the contents of this book and specifically disclaim any implied warranties of merchantability or fitness for a particular purpose. No warranty may be created or extended by sales representatives or written sales materials. The advice and strategies contained herein may not be suitable for your situation. You should consult with a professional where appropriate. Neither the publisher nor author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. For general information on our other products and services, or technical support, please contact our Customer Care Department within the United States at 800-762-2974, outside the United States at 317-572-3993 or fax 317-572-4002. Wiley. also publishes its books in a variety of electronic formats. Some content that appears in print may not be available in electronic books. For more information about Wiley products, visit our Web site at http://www.wiley.com. Library of Congress Cataloging-in-Publication Data:

Rao, S. S. Vibration of Continuous Systems / Singiresu S. Rao.

p.

cm.

Includes index. ISBN-13 978-0-471-77171-5 (cloth) ISBN-IO 0-471-77171-6 (cloth)' I. Vibration-Textbooks. 2. Structural dynamics-Textbooks. 1. Title. TA355.R378 2007 624.1'71-dc22 2006008775 Printed in the United States of America 10987654321

/~~(J 0~ i~jiI " . ,J __

-:. ./.r; -;.0,

.I

...,

\~

~'.,

; W~~;;t"'\':"'t~~~'1l~~'IJ!";'l'_,...,.".,'_-~,,_;"fJ"'~":";'r:_"""'l'~\~~'

Vibration of Discrete Systems: Brief Review2.1 VmRATION OF A SINGLE-DEGREE-OF -FREEDOM SYSTEMThe number of degrees of freedom of a vibrating system is defined by the minimum number of displacement components required to describe the configuration of the system during vibration. Each system shown in Fig. 2.1 denotes a single-degree-of-freedom system. The essential features of a vibrating system include (1) a mass m, producing an inertia force: mx; (2) a spring of stiffness k, producing a resisting force: kx; and (3) a damping mechanism that dissipates the energy. If the equivalent viscous damping coefficient is denoted as c, the damping force produced is eX.

2.1.1 Free Vibration In the absence of damping, the equation of motion of a single-degree-of-freedom system is given by

mx +kx

= f(t)

(2.1)

where f(t) is the force acting on the mass and x(t) is the displacement of the mass m. The free v~bration of the system, in the absence of the forcing function f(t), is governed by the equation

mx + kxThe solution of Eq. (2.2) can be expressed asx(t) = xocoswnt

=0

(2.2)

Xo +W

. smwnt

(2.3)

n

where

Wn

is the natural frequency of the system, given by(2.4)

Xo

x(t = 0) is the initial displacement and xo dx(t of the system. Equation (2.3) can also be expressed as x(t) = A cos(wnt - lfJ)

=

=

= O)/dt

is the initial velocity(2.5)

33

34

Vibration of Discrete Systems: Brief Review

c

9(t)

(a)

(b)

(c)

~

x(t)

f(t)

k

yet) (d) (e)

Figure 2.1

Single-degree-of-freedom systems.

orx(t)

= A sin(wnt

+ 4>0)

(2.6)

where

(2.7)

4> = tan--A,.

] XoXOWn

(2.8) (2.9)

o=an

t

-]

XOWn

-.-

XO

in Fig. 2.2.

The free vibration response of the system indicated by Eq. (2.5) is shown graphically The equation of motion for the vibration of a viscously damped system is given

by

mx + ex +kx = f(t)

(2.] 0)

2.1x(t)I

Vibration of a Single-Dcgree-of-Freedom'Syslci\I-35

I

/..-

Slope

= Xo~

/

. 1): x(t) where

+ wnxo)

t] e-wnt

(2.18)

= Cje(-s+~)Wnt

+ C2e(-s-~)Wnt + ~)(~ ~)

(2.19)

Cj =

XOWn (~

+ Xo-

2wn~ C2 =The motions indicated Fig. 2.3.-XOWn

(2.20)

xo(2.21) in

2wn~

by Eqs. (2.16), (2.18), and (2.19) are shown graphically

2.1.2

Forced Vibration under Harmonic ForceFor an undamped system subjected to the harmonic force f(t) of motion is

= fo cos wt, the equation(2.22)

mx

+ kx

= fo cos wt

where fo is the magnitude and w is the frequency of the applied force. The steady-state solution or the particular integral of Eq. (2.22) is given by xp(t) = X coswt where (2.23)

X

=

fok - mw2

=

OS! 1 - (w/wn)2 response and

(2.24)

denotes the maximum amplitude

of the steady-state fo OS! =k

(2.25)

indicates the static deflection of the mass under the force fo. The ratio

-=OS!

X

1 1 - (w/wn)2 (2.26)

represents the ratio of the dynamic to static amplitude of motion and is called the amplification factor, magnification factor, or amplitude ratio. The. variation of the amplitude

2.1x(t)

Vibration of a Single-Degree-ot"-Freedom'System ~;-37

, "" ,

"

(a)x(t)

_!Xo

--0

i

,..... /,,--

---

Cze(-l;-,[fCi}w,.l

,/

I I I I

Cz(b) x(t)

(c)

Figure 2.3 Damped free vibration response: (a) underdamped vibration (~ < 1); (b) over damped vibration (~ > 1); (c) critically damped vibration (~ = I).

38

Vibration of Discrete Systems: Brief ReviewXlo.t

3

o

2

3

4

-1

-2

-3 Figure 2.4 Magnification factor of an undamped system.

ratio with frequency ratio is shown in Fig. 2.4. The total solution of Eq.(2.22), including the homogeneous solution and the particular integral, is given by x(t)

=

(xo -

fo

" - mw

2)

coswnt

+ XoWn

sinwnt

+

fo k - mw

2 coswt

(2.27)

At resonance,

wjwn = 1, and the solution given by Eq. (2.27) can be expressed as(2.28 )

This solution can be seen to increase indefinitely, with time as shown in Fig. 2.5. When a viscously damped system is subjected to the harmonic force f (t) fo cos wt, the equation of motion becomes

=

mx + ex + kx = fo cos wtThe particular solution of Eq. (2.29) can be expressed xp(t) as

(2.29)

= X cos(wt - f/

(2.30)

--

,

~,"~'"

. .Y~

..

""~H"',n,-t",,,,'1,~;''':''''/''I'T'''1~''-1'>.~~ ~.

'~''.C:::;":l-l!:'l'l'1!."_:;t,.,t; .... T'''''0'~''

.:. .rd""'~.-~"""""-'~'''''"''''''"-'-

",.=

-'o\:l'~'

,,.-I;,~,~.~r~.

"-~"'~'r-''''~~'~~--~~'-'~

"".'~.~""f"" ...

~,,,.~~~

",..

r.~1

2.1

Vibration of a Single-Degree-of-Freedom System ';;39

--- --- --"o

--- --- --Figure 2.5 Response when r = wjwn = 1 (effects of Xo and :co not considered).

where X is the amplitude and 0, the force acts in the negative direction. As can be seen from Eqs. (3.1) and (3.3), Newton's second law of motion and D' Alembert's principle are equivalent. However, Newton's second law 6f motion is more commonly used in deriving the equations of motion of vibrating bodies and systems. The equations of motion of the axial vibration of a bar, transverse vibration of a thin beam, and the transverse vibration of a thin plate are derived using the equilibrium approach in the following sections.

=

3.4

EQUATION OF MOTION OF A BAR IN AXIAL VIBRATIONConsider an elastic bar of length 1 with varying cross-sectional area A(x), as shown in Fig. 3.1. The axial forces acting on the cross sections of a small element of the bar of length dx are given by P and P + dP withP = (J' A = EA-

au

ax

(3.4)

where

(1

is the axial stress, E is Young's modulus, u is the axial displacement, and

au/ax is the axial strain. If f(x, t) denotes the external force per unit length, theresulting force acting on the bar element in the x direction is (P +dP) - P

+f

dx

= dP + f dx

The application of Newton's second law of motion gives mass x acceleration = resultant force or pAdx-2

a2u

at

= dP

+f

dx

(3.5)

70

Derivation of Equations: Equilibrium Approach

z

t

I IaI

C I I

".""

o

"------~-_.~._I II

fix,

"

t)

I I

I I

x, u

II

II I

b

I I~

x

.1(a)

Equilibrium position

r-

dx

-I I-_c'c - P+dP

Displaced position

P

---.b

d-------"d'

I(b)

u+du

~I

"Figure 3.1

Longitudinal vibration of a bar.

where p is the mass density of the bar. By using the relationdP = (apjax)dx and Eq. (3.4), the equation of motion for the forced longitudinal vibration of a nonunifonn bar, Eq. (3.5), can be expressed as a [ au(x, t)] -a EA(x) x ax

+ I(x,

t) = P(x)A(x)-,

i'llat-

(x, t)

(3.6)

For a unifonn bar, Eq. (3.6) reduces to EA ax2 (x, t)

a2u

+ I(x,

t) = pA 8t2 (x,

a2u

1)

(3.7)

.-o;',...".--~~..,.0t>O t>O

(3.22) (3.23) (3.24) (3.25)

awa

ax,(x = 0, t) = 0, w(x=l,t)=O,

2w -2

ax

(x =l,t)

=0,

t>O

Other possible boundary conditions

of the beam are given in Chapter 11.

3.6 EQUATION OF MOTION OF A PLATE IN TRANSVERSE VIBRATIONThe following assumptions are made in deriving the differential a transversely vibrating plate: equation of motion of

1. The thickness h of the plate is small compared to its other dimensions.2. The middle plane of the plate does not undergo in-plane deformation middle plane is a neutral surface). 3. The transverse deflection w is small compared (i.e., the

to the thickness of the plate.

4. The influence of transverse shear deformation is neglected (i.e., straight lines normal to the middle surface before deformation remain straight and normal after deformation). S. The effect of rotary inertia is neglected. The plate is referred to a system of orthogonal coordinates xyz. The middle plane of the plate is assumed to coincide with the xy plane before deformation, and the deflection of the middle surface is defined by w(x, y, t), as shown in Fig. 3.3(a).

74

Derivation of Equations: Equilibrium Approach

~I

/// / / / / / / / / / /

O'yx

(a)

._.-._~

x

dX

/yj/ dQ" d Q ,,+--" y

Q,. ~"')dy

;/

"

dy

/.Figure 3.3 a plate.(0)

dx

(b)

Stresses in a plate; (b) forces and induced moment resultants in an element of

3.6

Equation of Motion of a Plate in Transverse Vibrauon ''"'''75

3.6.1 State of Stress For thin plates subjected to bending forces (i.e., transverse loads and bending moments), the direct stress in the z direction (0",,) is usually neglected. Thus, the nonzero stress components are O"xx. O"yy. (J'xy, O"yl.' and O"Xl.' As we are considering flexural (bending) deformations only. there will be no resulting force in the x and y directions; that is.

h~ 1-h/2

(J'xx dz

= 0,

1h~-h/2

(J'yy dz

=0

(3.26)

It can be noted that in beams, which can be considered as one-dimensional analogs of plates, the shear stress (J'xy will not be present. As in beam theory, the stresses O"xx (and O"yy) and (J'Xl. (and O"Yl.) are assumed to vary linearly and parabolically, respectively, over the thickness of the plate, as indicated in Fig. 3.3(a). The shear stress O"xy is assumed to vary linearly over the thickness of the plate, as shown in Fig. 3.3(a). The stresses O"xx, O"yy. O"xy, (J'Yl.' and O"xz are used in defining the following force and moment resultants per unit length:Mx =

My

Mxy

1 1 =1= =

h/2

O"xxZ dz

-h/2h/2

(jyyZ dzh/2

-h/2,

O"xyzdz

-h/2h/2

= Myx

since

(3.27)

Qx

1

O"xz dzh/2

-h/2

Qy

=

1

O"Yl.dz

-h/2

These force and moment resultants are shown in Fig. 3.3(b). 3.6.2 Dynamic Equilibrium Equations By considering an element of the plate, the differential equation of motion in terms of force and moment resultants can be derived. For this we consider the bending moments and shear forces to be functions of x, y. and t. so that if M x acts on one side of the element, Mx + dMx = Mx + (aMx/ax)dx acts on the opposite side. The resulting equations of motion can be written as follows. Dynamic equilibrium of forces in the z direction:( Qx

+ aa:x

dX)

dy

+ ( Qy + aa~y

dY)

dx

+ f dxdyzdirection

- Qx dy - Qydx

= mass of element x acceleration in the=phdxdYat2

a

2w

76

Derivation of Equations: Equilibrium Approach or

-ox + -oy

oQx

oQy +f(x,y,t)

= phot2

02W (3.28)

where f(x,y,t) is the intensity of the external distributed the material of the plate. Equilibrium of moments about the x axis: ( Qy+ oQ) o/dy (OM) My+ - Mydx By neglecting terms involving products ten as

load and p is the density of

dxdy=

o/dY - Mxydy

dx+

( Mxy+ dy

x) OM oxYdx

dy

- f dxdy-

2of small quantities, this equation can be writ-

Qy = oMy oy Equilibrium of moments about the y axis:

+ oMxyox

(3.29)

( Qx

oQx) dx + a;-

dydx

= (OMx) Mx +-Mxdy

a;- dx- Myxdx

dy

OMyx) dy + ( Myx + -ay-

dx

- f dxdYT

dx

or(3.30)

3.6.3

Strain-Displacement RelationsTo derive the strain-displacement relations, consider the bending deformation of a small element (by neglecting shear deformation), as shown in Fig. 3.4. In the edge view of the element (in the xz plane), PQRS is the undeformed position and P' Q' R' S' is the deformed position of the element. Due to the assumption that "normals to the middle plane of the undeformed plate remain straight and normal to the middle plane after deformation," line AS will become A' B' after deformation. Thus, points such as K will have in-plane displacements u and v (parallel to the x and y axes), due to rotation of the normal AB about the y and x axes, respectively. The in-plane displacements of K can be expressed as (Fig. 3.4b and c) u =

.

-7_

ow ow ay(3.31)

~ ax

v= -z-

-

3.6

Equation of Motion of a Plate in Transverse-V1brat-ion 77

___~x

t z(a)

J ~-_.P Aw(x,y)

Q

'--114- 1w(x,y)

CAD

Figure 3.4yz plane.

(a)

Edge view of a plate; (b) deformation in the

xz

plane;

(c)

deformation in the

The linear strain-displacement relations are given by

auexx = ax'eyy

av= oy'exy

= oy + ox

ou

ov

(3.32)

where exx and eyy are normal strains parallel to the x and y axes, respectively, and is the shear strain in the ry plane. Equations (3.31) and (3.32) yield

exy

exx = eyy

au ox = av ay

(_zaw) ox = ~ (_zow) oy oy=~

ox

ox2 = _zo2w oy2=

= _zo2w

(3.33)

ex =y

au + ov ay ox

=~

ay

(_zaw) + ~ (_zaw) ax ox oy

-2z o2w oxoy

Equations (3.31) show that the transverse displacement w(x, y, t) completely describes the deformation state of the plate.

78

Derivation of Equations: Equilibrium Approach

3.6.4

Moment-Displacement

Relations

We assume the plate to be in a state of plane stress. Thus, the stress-strain relations can be expressed as (1xx = -1E

-vE

z xx

+ -l-yy -v Z + -l--v ZexxvE

vE

(1yy = -l--Zeyy -v (1xy = Gexy

(3.34)

where E is Young's modulus, G is the shear modulus, and v is Poisson's ratio. By substituting Eq. (3.33) into Eq. (3.34) and the resulting stress into the first three equations of (3.27), we obtain, after integration, azw Mx = -D ( --z ax aZw aZw) + v-z ay aZw) (3.35) aZw

My = -D ( --z + v-ay ax 2 Mxy

= Myx = -(1 Eh3

v)D--

axay

where D, the flexural rigidity of the plate, is given by

D=

12(1 - vZ)

(3.36)

The flexural rigidity D is analogous to the flexural stiffness of a beam (EI). In fact, D = EI for a plate of unit width when v is taken as zero. The use of Eqs. (3.35) in Eqs. (3.29) and (3.30) lead to the relations aZw) Qx=-D- a (aZw -+ax axz ay2 a (aZw aZw) Q =-D-+_ Y ay axz ayZ 3.6.5 Equation of Motion in Terms of Displacement By substituting Eqs. (3.35) and (3.37) into Eqs. (3.28)-(3.30), we notice that moment equilibrium equations (3.29) and (3.30) are satisfied automatically, and Eq. (3.28) gives the desired equation of motion as a4w D ( ax4 If (3.37)

+ 2 axZayZ + ay4 + ph

a4w

a4w)

aZw at2 as

= f(x,a2w at2

y, t)

(3.38)

f (x. y.

t)

= 0, we obtain the free vibration equationD(

4 a w a4w a4w) -+2--+-4 2 ax ax ay2 ay4

+ph-=O

(3.39)

3.6

Equation of Motion of a Plate in Transverse Y;bration

79

Equations (3.38) and (3.39) can be written in a more general form: DV'4w

+ ph-2 at

a2w

=

f

(3.40) (3.41 )

DV'4w + ph-2whereV'4

a2w at4

=0

= '\12V'2, the

biharmonic operator, is given by'\14

a a a =+ 2 ax2ay2 + ax4 ay4

4

4

(3.42)

i.~f:

in Cartesian coordinates.

3.6.6

Initial and Boundary ConditionsAs the equation of motion, Eq. (3.38) or (3.39), involves fourth-order partial derivatives with respect to x and y, and second-order partial derivatives with respect to t, we need to specify four conditions in terms of each of x and y (Le., two conditions for any edge) and two conditions in terms of t (usually, in the form of initial conditions) to find a unique solution of the problem. If the displacement and velocity of the plate at t = 0 are specified as wo(x, y) and wo(x, y), the initial conditions can be expressed as'w(x, aw(x,

y, 0) = wo(x, y) y, 0) = wo(x, y)

(3.43) (3.44)

at

The general boundary conditions that are applicable for any type of geometry of the plate can be stated as follows. Let n and s denote the coordinates in the directions normal and tangential to the boundary. At a fixed edge, the deflection and the slope along the normal direction must be zero:

w=o aw =0 an w=oMn =0 where the expression by [5] for Mn in terms of normal and tangential coordinates

(3.45) (3.46)

For a simply supported edge, the deflection and the bending moment acting on the edge about the s direction must be zero; that is, (3.47) (3.48) is given

Mn = -D ['\1

2

W -

(1 - v) (~

~:

+ ~:~)]

(3.49)

where R denotes the radius of curvature of the edge. For example, if the edge with y = b = constant of a rectangular plate is simply supported, Eqs. (3.47) and (3.48)

80

Derivation of Eq.uations: Equilibrium Approach become w(x, b) = 0, My

0:::::x ::::: ao2W ( oy2

(3.50)

= -D

+ v ox2

02W)

(x, b)

= 0,

(3.51)

where the dimensions of the plate are assumed to be a and b parallel to the x and y axes, respectively. The other possible boundary conditions of the plate are discussed in Chapter 14.

3.7 ADDITIONAL CONTRIBUTIONSIn the equilibrium approach, the principles of equilibrium of forces and moments are used by considering an element of the physical system. This gives the analyst a physical feel of the problem. Hence, the approach has been used historically by many authors to derive equations of motion. For example, Love [6] considered the free-body diagram of a curved rod to derive coupled equations of motion for the vibration of a curved rod or beam. Timoshenko and Woinowsky-Krieger derived equations of motion for the vibration of plates and cylindrical shells [7]. Static equilibrium equations of symmetrically loaded shells of revolution have been derived using the equilibrium approach, and the resulting equations have subsequently been specialized for spherical, conical, circular cylindrical, toroidal, and ellipsoidal shells by Ugural [3] for determining the membrane stresses. The approach was also used to derive equilibrium equations ofaxisymmetrically loaded circular cylindrical and general shells of revolution by including the bending behavior. In the equilibrium approach, the boundary conditions are developed by considering the physics of the problem. Although the equilibrium and variational approaches can give the same equations of motion, the variational methods have the advantage of yielding the exact form of the boundary conditions automatically. Historically, the deyelopment of plate theory, in terms of the correct forms of the governing equation and the boundary conditions, has been associated with the energy (or variational) approach. Several investigators, including Bernoulli, Germain, Lagrange, Poisson, and Navier, haye attempted to present a satisfactory theory of plates but did not succeed completely. Later, Kirchhoff [8] derived the correct governing equations for plates using minimization of the (potential) energy and pointed out that there exist only two boundary conditions on a plate edge. Subsequently, Lord Kelvin and Tait [9] gave physical insight to the boundary conditions given by Kirchhoff by converting twisting moments along the edge of the plate into shearing forces. Thus, the edges are subject to only two forces: shear and moment.

REFERENCESL F. P. Beer, E. R. Johnston, Jr., and 1. T. DeWolf, Mechanics of Materials, 3rd ed., McGrawHill, New York, 2002. 2. S. S. Rao, Mechanical Vibrations, 4th ed., Prentice Hall, Upper Saddle River, NJ, 2004. 3. A. C. Ugural. Stresses in Plates and Shells, McGraw-Hill, New York, 1981.

--

ProoJems ,814. S. S. Rao, The Finite Element Method in Engineering, 3rd ed., Butterworth-Heinemann. Boston, 1999. 5. E. Ventsel and T. Krauthammer. Thin Plates and Shells: Theory, Analysis. and Applications, Marcel Dekker, New York, 2001. 6. A. E. H. Love, A Treatise on the Mathematical Theory of Elasticity. 4th ed, Dover, York, 1944., .'~x

New

7. S. P. Timoshenko New York, 1959.

and S. Woinowsky-Krieger,

Theory of Plates and Shells, McGraw-Hill.Scheibe, JourPress.

8. G. R. Kirchhoff, Uber das Gleichgewicht

und die Bewegung einer elastishen

nalfuer die Reine und Angewandte Mathematik, Vol. 40, pp. 51-88, 1850. 9. Lord Kelvin and P. G. Tait, Treatise on Natural Philosophy, Vol. I, ClarendonOxford, 1883. 10. W. Riigge,

Stresses in Shells. 2nd ed., Springer-Verlag,

New York, 1973.

PROBLEMS3.2 Consider a prismatic bar with one end (at x 0) 3.1 The system shown in Fig. 3.5 consists of a cylinder connected to a spring of stiffness Ko and the other of mass Mo and radius R that rolls without slipping on a end (at x I) attached to a mass Mo as shown in horizontal surface. The cylinder is connected to a viscous Fig. 3.6. The bar has a length of I, cross-sectional area A, damper of damping constant c and a spring of stiffness k. mass density p, and modulus of elasticity E. Derive the A uniform bar of length I and mass M is pin-connected equation of motion for the axial vibration of the bar and to the center of the cylinder and is subjected to a force F at the other end. Derive the equations of motion of '. the boundary conditions using the equilibrium approach. the two-degree-of-freedom system using the equilibrium approach.

=

=

x

.1

c-.':1{

\

"

Figure 3.5

82

Derivation of Equations: Equilibrium Approach

M . rx~~

u(x)

Figure 3.6

(a)

w(l, t)

V(l,i) ..

1

;

J

)(M(l, t)V(l, t)(b)

Figure 3.7

3.3 A beam resting on an elastic foundation and subjected to a distributed transverse force f (x, t) is shown in Fig. 3.7(a). One end of the beam (at x = 0) is simply supported and the other end (at x = I) carries a mass Mo. The'free-body diagram of the end mass Mo is shown in Fig. 3.7(b). (a) Derive the equation of motion of the beam using the equilibrium approach. (b) Find the boundary conditions of the beam.

3.4 Consider a differential element of a membrane under uniform tension T in a polar coordinate system as shown in Fig. 3.8. Derive the equation of motion for the transverse vibration of a circular membrane of radius R using the equilibrium approach. Assume that the membrane has a mass of m per unit area. 3.5 Consider a differential element of a circular plate subjected to the transverse distributed force f (r, (), t) as shown in Fig. 3.9. Noting that Qr and Mrr vanish

~.problems83y

//

//

x

Figure 3.8

x

d8y

--dr dr

aM,

(b)

Figure 3.9

84

Derivation of Equations: Equilibrium Approach resisting force offered by the foundation to a transverse deflection of the plate w is given by Jew per unit area. The plate is subjected to a transverse force f (x, y, t) per unit area. Derive a differential equation of motion governing the transverse vibration of the plate using the equilibrium approach.

due to the symmetry, derive an equation of motion for the transverse vibration of a circular plate using the equilibrium approach. 3.6 Consider a rectangular plate resting on an elastic foundation with a foundation modulus k so that the

,

I

4Derivation of Equations: Variational Approach4.1 INTRODUCTIONAs stated earlier, vibration problems can be formulated using an equilibrium, a variational, or an integral equation approach. The variational approach is considered in this chapter. In the variational approach, the conditions of extremization of a functional are used to derive the equations of motion. The variational methods offer the following advantages: 1. Forces that do no work, such as forces of constraint on masses, need not be considered. 2. Accelerations of masses need not be considered; only velocities are needed. 3. Mathematical operations are to be performed on scalars, not on vectors, in deriving the equations of motion. Since the variational methods make use of the principles of calculus of variations, the basic concepts of calculus of variations are presented. However, a brief review of the calculus of a single variable is given first to indicate the similarity of the concepts.

4.2 CALCULUS OF A SINGLE VARIABLETo understand the principles of calculus of variations, we start with the extremization of a function of a single variable from elementary calculus [2]. For this, consider a continuous and differentiable function of one variable, defined in the interval (X"X2), with extreme points at a, b, and c as shown in Fig. 4.1. In this figure the point x = a denotes a local minimum with f(a) ::: f(x) for all x in the neighborhood of a. Similarly, the point x = b represents a local maximum with f(b) ::::f(x) for all x in the neighborhood of b. The point x = c indicates a stationary or inflection point with f(c) ::: f(x) on one side and f(c) ::::f(x) on the other side of the neighborhood of c. To establish the conditions of extreme values of the function f(x), consider a Taylor series expansion of the function about an extreme point such as x a:

=

f(x)

= f(a)

+ df -

dx

I (x a

a)

1df + -2

2

2! dx

I (x a

3

a) 2

1d + --

f

3! dx3

I (x a

a) 3

+ ...(4.1)

8S

f~

."

(

.

f'

II

88

Derivation of Equations: Variational Approach

,I

/1I

.

j= 4>(x) + EI'/(X)

,/ " II",'

.. ,

\

.. ....

--

-- -- ---- -- --"4>= 4>(X)

I I I I Ix

Figure 4.2

Exact and trial solutions.

It can be observed that the necessary condition for the extremum of

I is that(4.14)

dldeE=O

Using differentiation of an integral! and noting that both and we obtain /E

cPx are functions of e,aj)

=

dl de

1 =Xl

x2 j (a a a ae

+ ax

aj al/lx) ae

dx = xI

1

x2 j (a a'1

+ ax'1x

dx

(4.15)

. When e is set equal to zero, (, l/lx) are replaced by (l/l,l/lx) and Eq. (4.15) reduces to

/E(O)

dl = -(0) = de

1 (ax2

XI

j -'1 al/l

a) + -T/xj al/lx f(x,E)dE dXIdE

dx

=0

(4.16)

1=I(E)=then

1=

X2(t)

(a)

XI(t)

ItIfXI

d1 dX2 == f(x2.E)dE dE

f(xl,E)-

1-'2\f) of + .. -dxX,(f)

aE

(b)

and

X2

are constants, Eq. (b) reduces to It

=

dl dE

1XI

x2

afaf:

dx

(c)

'~

4.4

Variation 0per.aror

.89

Integrating the second term of the integral in Eq. (4.16) by parts, we obtainX

Ie(O)

=

af I 2 -'I at/>x.q

+

l

x2

XI

f [a - - -d( at/> dx

af --

)] T/dx at/>x

=0

(4.17)

In view ofEq. (4.9), Eq ..(4.17) givesIe(O) =

l

x2

Xl

f [a _ ~ at/> dx

(:!L)]at/>x

T/dx = 0

(4.18)

Since Eq. (4.18) must hold for all

'I. we have

f) d at/>- dx . at/>x

af

(a

=0

(4.19)

This equation. known as the Euler-Lagrange equation, is, in general, a second-order differential equation. The solution of Eq. (4.19) gives the function t/>(x) that makes the integral I stationary.

4.4

VARIATION OPERATOREquation (4.17) can also be deriyed using a variation operator 0, defined as ot/>= (x) - t/>(x) (4.20)

where t/>(x) is the true function of x that extremizes I, and (x) is another function of x which is infinitesimally different from t/>(x) at every point x in the interval XI < X < X2 The variation of a function t/> (x) denotes an infinitesimal change in the function at a given value of x. The change is virtual and arbitrary. The variation differs from the usual differentiation, which denotes a measure of the change in a function (such as t/ resulting from a specified change in an independent variable (such as x). In view of Eq. (4.8), Eq. (4.20) can be represented as ot/>(x) = (x) - t/>(x)

= ST/(x)

(4.21)

where the parameter S tends to zero. The variation operator has the following important properties, which are useful in the extremization of the functional I. 1. Since the variation operator is defined to cause an infinitesimal change in the functiont/>for a fixed value of x, we have

ox

=0

(4.22)

and hence the independent variable x will not participate in the variation process. 2. The variation operator is commutative with respect to the operation of differentiation. For this, consider the derivative of a variation: d -ot/> dx

=

d -ST/(x) dx

dT/(x) = s-dx

(4.23)

90

Derivation of Equations: Variational Approach

Next, consider the operation of the variation of a derivation, o (d>/ dx). Using the definition of Eq. (4.21),d> o dx

=

d d> dx - dx

= dx

d

(> -

=

d dx ETJ(X)

= E~

dTJ(x)

(4.24)

Thus, Egs. (4.23) and (4.24) indicate that the operations of differentiation and variation are commutative:d -o> dx

= 0dx

d>

(4.25)

3. The variation operator is commutative with respect to the operation of integration. For this, consider the variation of an integral:2 o X >(x)dx= 1Xl = x2 1x2 (x)dxxl

1xl-

>(x)dx

2 X [(x) 1xl

- >(x)]dx

= 1X2xl

o>(x)dx

(4.26)

Equation (4.26) establishes that the operations of integration and variation are commutative:o

1Xl

X2 >(x)dx

=

lx2xl

o>(x)dx

(4:27)

For the extremization of the functional I of Eq. (4.7), we follow the procedure used for the extremization of a function of a single variable and define the functional I to be stationary if the first variation is zero: OJ = 0 (4.28)

Using Eg. (4.7) and the commutative property of Eg. (4.27), Eg. (4.28) can be written asOJ

=

1Xl

X2

of dx = 0 > (x):

(4.29)

where the variation ofof = f(x,

f is caused by the varying function

, >x)- f(x, >, >X)= f(x, > + cTJ, >x+ ETJx)- f(x, >, >x)

(4.30)

The expansion of the function f (x, > + ETJ,>x+ r/x) about (x, >, >x) givesf(x, > + ETJ,>x+ ETJx) = f(x, af af >, >X)+ a>ETJ+ a>xETJx+ ...E

(4.31 )

Since that

E

is assumed to be small, we neglect terms of higher order inaf af) ( -TJ+-TJx a> a>x

in Eg. (4.31), so

Of=E

...... ~~~.....,,-~.T"'I '. "~~'""'.,;:i8Uj dV dt

aUj aUj d

V

(4.160)

v is the kinetic energy of the vibrating body. Thus, Eq. (4.156) can be expressed as

/12

8(rr - T) dt =

/'2 IIIv

+

/12 II52

ct>j8ujd5 dt

(4.161)

where rr denotes the total strain energy of the solid body: rr =

IIIv

rro dV

(4.162)

If the external forces acting on the body are such that the sum of the integrals on the right-hand side of Eq. (4.161) denotes the variation of a single function W (known as the potential energy of loading), we have

IIIv

cf>i8uid~

+-..

II52

ct>j8ujd5 = -8W

(4.163)

Then Eq. (4.161) can be expressed as8[12 LdtI]

=[12I]

(rr _ T

+ W)dt = 0

(4.164)

where L=rr-T+W(4.165)

is called the Lagrangian function and Eq. (4.164) is known as Hamilton's principle. Note that a negative sign is included, as indicated in Eq. (4.163), for the potential energy of loading (W). Hamilton's principle can be stated in words as follows: The time integral of the Lagrangian function between the initial time t1 and the final time t2 is an extremum for the actual displacements (motion) with respect to all admissible virtual displacements that vanish throughout the entire time interval: first, at all points of the body at the instants tl and t2, and second, over the surface 51, where the displacements are prescribed. Hamilton's principle can be interpreted in another way by considering the displacements Uj(xj, X2, X3, t), i = 1, 2, 3, to constitute a dynanlic path in space. Then Hamilton's principle states: Among all admissible dynamic paths that satisfy the prescribed geometric boundary conditions on 51 at all times and the prescribed conditions at two arbitrary instants of time t1 and t2 at every point of the body, the actual dynamic path (solution) makes the Lagrangian function an extremum.

"r"""'-'-~-~.'--"-''''''.---, ..;~!t:...""~.~' _""",~;_~,."".,.""N."1';,~""_'_"'-:."r.k_,

.. 7., ~"_':"""'I""'~~l __ .~-,:r.:>~-,,c."-"".-"

I

.."'!'_-'"

'o::,.,.....~... ""-~".~""'~

. ~~III':~--,..,,\"".'"

, ..-.,-.-. .. ,~,;"..., .. ~

.~~~_~~~~J'i:1""f"7~~~.~~'~"\f"l--.\'.'.

__

.."..,:r'." .,~'"

4.11

Applications of Hamilton's PtincrpleUS

4.114.11.1

APPLICATIONS OF HAMILTON'S PRINCIPLEEquation of Motion for Torsional Vibration of a Shaft (Free Vibration) Strain EnergyTo derive a general expression for the strain energy of a shaft, consider the shaft to be of variable cross section under a torsional load as shown in Fig. 4.5. If O(x, t) denotes the angular displacement of the cross section at x, the angular displacement of the cross section at x + dx can be denoted as B(x, t) + [oO(x, t)jox]dx, due to the distributed torsional load m,(x, t). The shear strain at a radial distance r is given by y = r(aOjox). The corresponding shear stress can be represented as r Gy Gr(aO jox), where G is the shear modulus. The strain energy density Jro can be represented as Jro ~ry iGr2 (aOjax)2. The total strain energy of the shaft can be determined as

=

=

=

=

Jr =

fffv

L

JrodV = i

ffA

lGr

2

(:~y

L

dAdx

= li

GJ

(:~ydx

(4.166)

where V is the volume, L is the length, A is the cross-sectional moment of inertia (for a uniform circular shaft) of the shaft.

area, and J = I p polar

Kinetic Energyexpressed as

The kinetic energy

of a shaft with variable cross section can be

T

='2 10

1

{L lo(x) (ao(xat' t)2

dx

(4.167)

where lo(x) = plp(x) is the mass moment of inertia per unit length of the shaft and p is the density. By using Eqs. (4.166) and (4.167), Hamilton's principle can be used to obtain

;; 1'I

'2

(T -Jr)dt

=;; {L

10

[1102

{L 10 (00)2

at

dx

_1 102'2

{L GJ (aO)2 dX] dt = 0

ax

(4.168) By carrying out the variation operation, the various terms in Eq. (4.168) can be rewritten, noting that;; and ajat as well as ;; and ajax are commutative, as

;;

1 [IlL'2

'I

2

0

10 (aO)2] dx at

dt = -

11'I 0

L

2 a 0 10-280dxdt at

(4.169)

aI

C

,_,_,l,-!,-.-,_._. J~,_.xI I I I I

x

1Figure 4.5 Torsional vibration of a shaft.

116

Derivation of Equations: Variational Approach

assuming that

e is prescribed

at

t1 and t2 so that oe = 0 at t1 and t2. Similarly,=

o

1 [IlL12II

2

0

GJ (ae)2] dx dt ax

1',)2 [ GJ-oe ae ax2 at

IL 0

lL -ox 0 (af)) GJax011

OfJdx] dt(4.170)

Thus, Eq. (4.168) becomes

ae 1(2) dt = 0 oedx - GJ-oe (4.171) ax Assuming that 8e = 0 atx = 0 and x = L, and 8e is arbitrary in 0 < x < L, Eq. (4.171)/110

ae) 1 jl [O -ax ( GJ-ax2L

a e]2 10-

requires that

a ( GJ oe) ox ax -

10

2e a at2

= 0,

O F=Psine+Psin> ~ P(tane =P (E5.1.2)

+ tan(E5.1.3)

LlJ~(L - ~)

Equation (E5.1.3) can be solved for lJ, which upon substitution in Eq. (E5.1.3) results inw(x) = -g(x,~)

F P

(E5.1.4)

where g (x , ~) is the impulse response junction, also known as Green's junction, given by x(L - ~)

Lg(x,~)=

'(E5.1.5)'

{

HL-x) L

If the external load applied to the string is distributed with a magnitude of f(~) unit length, the transverse displacement of the string can be expressed asw(x) =

per

~lL

g(x, Of(~)d~

(E5.1.6)

If the displacement variation w(x) is specified, Eq. (E5.1.6) becomes an integral equation of the first kind for the unknown force distribution f(x). For free vibration, the force per unit length, due to inertia, is given byf(x, t) = -p(x)

a2w(x,t) 2 at

(E5.1.7)

"1

_=-< .....-r-~ ..... "'.,.,....-""_"': ......,....".,...... '""""_~=~

.=.,.,."...oJ~_...,~-:'~~'l

I

.. ~~.""".,.

, ~.,.."t"::L =L

.J>::or

sinhLor

=0n = 1,2, ...

(E5.2.9)

(E5.2.1O)

130

Derivation of Equations: Integral Equation Approach Equations (E5.2.5) and (E5.2.1O) lead to

X(x) = asm-where a is a constant.

.

nrrx

L

(E5.2.11)

5.4 GENERAL FORMULATION OF THE EIGENVALUE PROBLEM5.4.1 One-Dimensional SystemsFor a one-dimensional continuousw(x,

system, the displacement t)

w (x, t) can"be expressed as(5.29)

=

l

L

a(x, ~)f(~,

t) d~

where a(x,~) is the flexibility influence function that satisfies the boundary conditions of the system and f (~, t) is the distributed load at point ~ at time t. For a system undergoing free vibration, the load represents the inertia force, so that f() x, t = -m (x )a2w(x,

at2

t) (5.30)

where m (x) is the mass per unit length. Assuming during free vibration,

a harmonic motion, of frequency w

w(x, t)= W(x)coswtEq. (5.30) can be expressed as

(5.31)

f(x, t)Substituting

= w m(x)W(x)2

cos wt

(5.32)

Eqs. (5.31) and (5.32) into Eq. (5.29) results in

W(x) =

w21L a(x, ~)m(~)W(~)d~

(5.33)

It can be seen that Eq. (5.33) is a homogeneous integral equation of the second kind and represents the eigenvalue problem of the system in integral form. Consider a membrane of area A whose equilibrium shape lies in the xy plane. Let the membrane be fixed at its boundary, S, and subjected to a uniform tension P (force per unit length). Let the transverse displacement of point Q (x, y) due to the transverse load f (~, 1]) d ~ d 1] applied at the point R (~, 1]) be w( Q). By considering the equilibrium of a small element of area d.x dy of the membrane, the differential equation can be derived as

Example 5.3 Free Transverse Vwration of a Membrane

(E5.3.1)

5.4

General Formulation of the Eigenvalue Problem

. 131

The Green's function of the membrane, K(x, y;~, 17), is given by [4]K(Q, R)

= K(x,

y;~,

17)

= log -r

1

- h(Q, R)

(E5.3.2)

where r denotes the distance between two points Q and R in the domain of the membrane:(E5.3.3)

and h (Q, R) is a hannonic function whose values on the boundary of the membrane, S, are the same as those of log(l / r) so that K (Q, R) will be zero on S. For example, if the membrane is circular with center at (0,0) and radius a, the variation of the function K(Q, R) will be as shown in Fig. 5.2. Since the membrane is fixed along its boundary S, the transverse displacement of point Q can be expressed asw(Q) = _1_ 2rr P

IfA

K(Q, R)f(R)dA

(E5.3.4)

K(Q. R)

-'Boundary, S

Figure 5.2

Variation of the Green's function for a circular membrane. (From Ref. [4]).

132

Derivation of Equations: Integral Equation Approach

From this static relation, the free vibration relation can be obtained by substituting _p(R)[02w(R)/ot2] for f(R) in Eq. (E5.3.4) so thatw(Q) = --2 1 7r

P

JfA

K(Q, R)P(R)-2 02w (R)dA

ot

(E5.3.5)

Assuming harmonic motion with frequency w, we havew(Q)

= W(Q)eiwt

(E5.3.6)

where W(Q) denotes the amplitude of vibration at point Q. Substitution ofEq. (E5.3.6) in Eq. (5.3.5) yields the relationW(Q) = ~

27rP

JfA

K(Q, R)p(R)W(R)dA

(E5.3.7)

5.4.2 General Continuous SystemsThe general form of Eq. (5.33), valid for any continuous system, can be expressed asW(x) =

Ai

g(x, ~)m(~)W(~)dV(~)

(5.34)

where W(x) and W(O denote the displacements at points x and ~, respectively. Depending on the dimensionality of the problem, points x and ~ may be defined by one, two, or three spatial coordinates. The general flexibility influence function g(x, ~), also known as the Green'sfunction, is symmetric in x and~, [i.e., g(x,~) = g(~, x)] for a self-adjoint problem. Note that the kernel, g(x, ~)m(~), in Eq. (5.34) is not symmetric unless m(~) is a constant. However, the kernel can be made symmetric by noting the fact that m(~) > 0 and introducing the function (x):(x)

= Jm(x)W(x)

(5.35)

By multiplying both sides of Eg. (5.34) by' .Jm(x) and using Eg. (5.35), we obtain(x) = A

i

K(x, ~)(~) dV(~)

(5.36)

where the kernel (5.37) can be seen to be symmetric. An advantage of the transformation above is that a symmetric kernel usually possesses an infinite number of eigenvalues, A, for which Eq. (5.36) will have nonzero solutions. On the other hand, a nonsymmetric kernel mayor may not have eigenvalues [1]. For any specific eigenvalue Ai, Eq. (5.36) has a nontrivial solution i(X), which is related to Wi(x) by Eq. (5.35). The function Wi(x) represents the eigenfunction corresponding to the eigenvalue Ai of the system.

.

)0\."'.-

...

---~5.5 Solution of Integral Eqtiations

~

.. , ............... ',...

-'133

5.4.3

Orthogonality of EigenfunctionsIt can be shown that the eigenfunctions 4>i (x) are orthogonal in the usual sense. while the functions Wi(x) are orthogonal with respect to the functions m(x). For this. consider Eq. (5.36), corresponding to two distinct eigenvalues Ai and A j:

i(X)

= Ai

j(x) Multiply obtain Eq. (5.38) by j(x),

= Aj

Iv Iv

K(x, ~)i(~)d V(~) K(x, ~)j(~) dV(~) over the domain

(5.38) (5.39)

integrate

V, and use Eq. (5.39) to

(5.40)

which yields

(5.41)

Since Ai and Aj are distinct, Ai :/; Aj, Eq. (5.41) leads to the orthogonality

relation

(5.42)When Eq. (5.35) is used in Eq. (5.42), eigenfunctions Wi (x) as we obtain the orthogonality relation for the

for Ai :/; A j for Ai Aj

=

(5.43)

5.5

SOLUTION OF INTEGRAL EQUATIONSSeveral methods, both exact and approximate methods, can be used to find the solutions of integral equations [1,4-6]. The method of undetermined coefficients and the Rayleigh-Ritz, Galerkin, collocation, and numerical integration methods are considered in this section.

134

Derivation of Equations: Integral Equation Approach

5.5.1

Method of Undetermined Coefficients In this method the unknown function is assumed to be in the form of a power series of a finite number of terms. The assumed function is then substituted into the integral equation and the regular part is integrated. This results in a set of simultaneous equations in terms of the unknown coefficients. Solution of these simultaneous equations yields the solution of the integral equation. Example 5.4 Find the solution of the integral equation

21)SOLUTION

(1- ~+xO(~)d~

= -x

+1

(E5.4.I)

Assume the solution of (x) in a power series of two terms as (E5.4.2)

where c) and C2 are constants to be determined. Substitute Eq. (E5.4.2) into Eq. (E5.4.1) and carry out the integration to obtain

21)

(1 - ~ + x~)(c)

+ C2~)

d~ = -x

+1

(E5.4.3)

Upon integration, Eq. (E5.4.3) becomes

(C) + ~C2)

+x

(C) + ~C2)= 1 = -1

= -x

+1

(E5.4.4)

Equating similar terms on both sides of Eq. (E5.4.4), we obtainC)

C)

+ ~C2 + ~C2C2

(E5.4.5)

Equations (E5.4.5) yield Eq. (E5.4.1) is given by

C)

= -3

and (x)

= 3.

Thus, the solution of the integral

= - 3 + 3x

(E5.4.6)

5.5.2

Iterative Method An iterative method similar to the matrix iteration method for the solution of a matrix eigenvalue problem can be used for the solution of the integral Eq. (5.34). The iteration method assumes that the eigenvalues are distinct and well separated such that A) < A2 < A3 .... In addition, the iteration method is based on the expansion theorem related to the eigenfunctions Wj(x). Similar to the expansion theorem of the matrix eigenvalue problem, the expansion theorem related to the integral forrimlation of the eigenvalue problem can be stated as00

W(x)

= LCiWi(X)j=)

(5.44)

~I

5.5

Solution of Integral Equations

135

where the coefficients

Cj

are determined asCj

=

Iv

m(x)W(x)Wj(x)dV(x)

(5.45)

Equation (5.44) indicates that any function W(x) that satisfies the boundary conditions of the system can be represented as a linear combination of the eigenfunctions Wi (x) of the system. First Eigenfunction The iteration method starts with the selection of a trial function wil) (x) as an approximation to the first eigenfunction or mode shape, WI (x). Substituting Wil)(x) for W(x) on the right-hand side of Eq. (5.34) and evaluating the integral, the next (improved) approximation to the eigenfunction WI (x) can be obtained: (5.46) Using Eq. (5.44), Eq. (5.46) can be expressed asW?)(x)

= =

Lj=100

00

Cj

f

g(x, ~)m(~)Wj(~)dV(~)

v

"Cjj=1

Wi (x) A'I

(5.47)

The definition of the eigenvalue problem, Eq. (5.34), yieldsWj(x)

= Aj

Iv

g(x, ~)m(~)Wj(~) dV(~)

(5.48)

Using W}2)(x) as the trial function on the right-hand side of Eq. (5.48), we obtain the new approximation, W?)(x), as

00

_

"

Cj

Wj(x)2

-~ i=1

(5.49)

A;

The continuation of the process leads to00

(n)

WI

(x) = ~j=1

"

Cj

Wj(x)n-I'

n = 2,3, ...

(5.50)

Ai

Since the eigenvalues are assumed to satisfy the relation A\ < )..2" " the first term on the right-hand side of Eq. (5.50) becomes large compared to the other terms and as

136

Derivation of Equations: Integral Equation Approach n ~00,

Eq. (5.50) yields lim W(n-I\x)n-+oo I

= CI WI (x)An-ZI

(5.51)

lim W(n)(x) =n-+oo I

CI

WI (x)An-I I

(5.52) AI as

Equations (5.51) and (5.52) yield the converged

eigenvalue

A] = hmn-+oo

.

wt-I)(x) wt)(x) (5.53)

and the converged eigenvector

can be taken as W,(x)

= n_oo lim

wt)(x)

(5.54)

Higher Eigenfunctions

To determine the second eigenfunction, the trial function Wil) (x) used must be made completely free of the first eigenfunction, W, (x). For this we use any arbitrary trial function Wil) (x) to generate wil) (x) as (5.55)

where is a constant that can be determined eigenfunctions:

a,

from the orthogonality

condition of the

Ivor

m(x)Wi')

(x)W, (x) dV(x)

= Iv m(x)Wi')

(x)W, (x) dV(x) - a]

Iv

m(x)[W] (x)f

dV(x)

=0

(5.56)

m(x)Wil)(x) w] (x) dV(x) a, = Iv Iv m(x)[W, (x)]2 dV(x) When

(5.57)

w] (x)

is normalized

according to Eg. (5.43),

IvEq. (5.57) becomes a, =

m(x)[WI (x)f

dV(x)

= I

(5.58)

Iv

m(x)Wi')(x)W,(x)dV(x)

(5.59)

Once Gj is determined, we substitute Eg. (5.55) for W(x) on the right-hand side of Eq. (5.34), evaluate the integral, and denote the result as WiZ)(x), the next (improved) approximation to the true eigenfunction Wz(x): (5.60)

5.5 For the next iteration,

Solution of hllegral Equations

1J7

we generate W?) (x) that is free of WI (x) as (5.61)

where

02

can be found using an equation .similar to Eq.(5.59) as (5.62)

when WI (x) is normalized according to Eq. (5.43). When the iterative process is continued, we obtain, result as)..2 =. w(n-l)

as n -+

00,

the converged

limn~oo

(x)_

_2

wt) (x)

(5.63) (5.64)

W2(X) = limn~oo

win>(x)

To find the third eigenfunction of the system, we start with any arbitrary trial function WJI)(x) and generate the function WJI) (x) that is completely free of the first and second eigenfunctions WI (x) and W2(X) as (5.65)

where WI (x) to find to find

the constants al and a2 can be found by making Wjl)(x) orthogonal to both and W2(X). The procedure used in finding the second eigenfunction can be used the converged solution for )..3 and W3(X). In fact, a similar process can be used all other higher eigenvalues and eigenfunctions.

Example 5.5 Find the first eigenvalue and the corresponding

eigenfunction of a tightly stretched string under tension using the iterative method with the trial functionw(l)( ) _ I X -

x(L - x)2

SOLUTION Let the mass of the string be m per unit length and the tension in the string be P. The Green's function or the flexibility influence function, g(x, ~), can be derived by applying a unit load at point ~ and finding the resulting deflection at point x as shown in Fig. 5.3. For vertical force equilibrium, we have

a a P-+P--=l~ L-~

(ES.5.1)

138

Derivation of Equations: Integral Equation Approach

(a)

r----~;---i-L-;xI II

I .-.-.-.- .. -.-.-.-.-+-.-.-.-. Y

)r-L-X~ ---)

a

(b)

Figure 5.3

which yields a= Thus, the Green's function is given by axg(x,~) ~(L - ~) (E5.5.2)

PL

={

a(L _ x) L -~ '

T'

~>x(E5.5.3)

~i(X)l7ii=1

(5.91)

where 1/; are coefficients or generalized coordinates to be determined. When Eq. (5.91) is substituted into Eq. (5.90), the equality will not hold; hence an error function or residual c (x) can be defined as (5.92)

By substituting

Eq. (5.91) into (5.90), the error function can be expressed as c(X) = w(x) - >..

Iv

g(x, ~)m(~)w(~)

dV(~) (5.93)

=

Li=1

n

1/iUi(X) -

5:.

Li=1

n

1/i

1v

g(x, ~)m(;)Ui(~)

dV(;)

To determine the coefficients T/k, the error function is set equal to zero at n distinct points. By setting the error, Eq. (5.93), equal to zero at the points xk(k = 1,2, ... , n), we obtain k = 1, 2, ... , n Equations (5.93) and (5.94) lead to the eigenvalue problem (5.94)

Li=1

n

(mki - hki)r/i

= 0,

k = 1, 2, ... , n

(5.95)

which can be expressed

in matrix form as [m]ij = 5:.[k]ij (5.96)

where the elements

of the matrices [m] and [k] are given bymki

=

Ui(Xk)

(5.97) dV(;) (5.98)

kki = Iv g(Xk, ;)m(;)ui(;)

It is to be noted that the matrices [m] and [k] are, in general, not symmetric. The solution of the eigenvalue problem with nonsymmetric matrices [m] and [k] is more complex than the one with symmetric matrices [3].

146

Derivation of Equations: Integral Equation Approach Numerical Integration Method

5.5.6

In the numerical integration method, the regular part of the integral equation is decomposed into the form of a sum, and the equation is then reduced to a set of simultaneous linear equations with the values of the unknown function at some points in the domain of integration treated as the unknown quantities. The procedure is illustrated through the following example.

Example 5.7

Find the solution of the integral equation

(x)

+

1o

1

(1 + x;)(;)

d; = f(x)

== x2

-

23 -x 24

4 +_ 3

(E5.7.1)

numerically and compare the result with the exact solution (x) = x2-

2x

+1

(E5.7.2)

SOLUTION We use the Gauss integration method for the numerical solution of Eg. (E5.7.1). In Gauss integration, the integral is evaluated by using the formula1

{get)-I

dt =

L wig(ti)i=1

n

(5.7.3)

where n is called the number of Gauss points, Wi are called weights, and ti are the specified values of t in the range of integration. For any specified n, the values of Wi and ti are chosen so that the formula will be exact for polynomials up to and including degree 2n - 1. Since the range of integration in Eq. (E5.7.3) for x is -1 to + 1, the formula can be made applicable to a general range of integration using a transformation of the variable. Thus, an integral of the form f(x) dx can be evaluated, using the Gauss integration method, as

f:

lis used so that

b

b f(x) dx = ; a

L w;f(x;)n i=1

(E5.7.4)

a

where the coordinate transformation x=(b - a)t 2

+a +b(E5.7.5)

Xi

=

(b - a)ti

+a +b2 (E5.7.6)

.......-

'

'

5,6

Recent Contributions

147

Using n = 4, the corresponding values of Wi and ti are given by (2]WI

=

W4

= 0.347854845147454

(E5.7.7)

W2 = W3 = 0.652145154862546tl t2

= -0.861136311594053 = -0.339981043584856 (E5.7.8)

The values of the variable

Xi

given by Eq. (E5.7.6) for a XI = 0.06943184 X2 = 0.33000946 X3 = 0.66999054 X4 = 0.93056816

= 0 and b = 1 are(E5.7.9)

Treating the values of

act> dy

act> dz

act>.

act>

=0 .

(10.158)

using Eqs. (10.156) and (10.157). Equation (10.158) indicates that the stress function ct> is a constant on the boundary of the cross section of the rod. Since the magnitude

n.(

t

A.

/

1~-dy (a) (b)

I

Figure 10.8

Boundary condition on the stresses.

10.9 Torsional RigidiD' of Noncircular Shafts

305

of this constant does not affect the stress, which contains only derivatives of , we choose. for convenience, (10.159) to be the boundary condition. Next we derive a relation between the unknown angle ,B(angle of twist per unit length) and the torque (Mt) acting on the rod. For this. consider the cross section of the twisted rod as shown in Fig. 10.9. The moment about the x axis of all the forces acting on a small elemental area dA located at the point (y,z) is given by (10.160) The resulting moment can be found by integrating the expression in Eq. (10.160) over the entire area of cross section of the bar as (10.161)

Each term under the integral sign in Eq. (10.161) can be integrated by parts to obtain (see Fig. 10.9):

=

I (yl~;-l~dz z

dY)

=-

I 1~2dz

dy

=-

IIA

dA

(10.162)

t

I

O'xy

LI I

y

IdA

L. . .

(y,z)

z

.L--

Figure 10.9

Forces acting on the cross section of a rod under torsion.

306

Torsional Vibration of Shafts

z

tt

+b

b

--1-

~a-+-a~Figure 10.10 Elliptic cross section of a rod.

since

= 0 at the points

PI and P2. Similarly,

IfA

a -zdA az

If =I=A

a -zdydz az

=

I ldy

p4

P3

a -zdz az

dy (z

I~: -l:4IIA

4>dZ) = -

IIA

4>dA

(10.163)

Thus, the torque on the cross section (M,) is given by .. M,._.= 2 dA (10.164)

The function 4>satisfies the linear differential (Poisson) equation given by Eq. (10.153) and depends linearly on GfJ, so that Eq. (10.164) produces an equation of the form M, = GJfJ = CfJ, where J is called the torsional constant (J is the polar moment of inertia of the cross section for a circular section) and C is called the torsional rigidity. Thus, Eq. (10.164) can be used to find the torsional rigidity (C). Note There are very few cross-sectional shapes for which Eq. (10.164) can be evaluated in closed form to find an exact solution of the torsion problem. The following example indicates the procedure of finding an exact closed-form solution for the torsion problem for an elliptic cross section. ExampLe 10.6 (Fig. 10.10). SOLUTION Find the torsional rigidity of a rod with an elliptic cross section The equation of the boundary of the ellipse can be expressed as fey, Noting that\72

z)

= 1-

v2~2 -

Z2

b2

=0(y, z) as

(E1O.6.l)

f is a constant, we take the stress function(y, z) = c (1 - ~: - ~~)

(E1O.6.2)

10.9 Torsional Rigidity of Noncircular Shafts where c is a constant. Using Eq. (ElO.6.2) in Eq. (l0.153), V'2 If we choose c = a2 we obtain

307

a + a = -2c =ay2 az2

2

2

( 1

-2 a

) + -1 2

b

= -2GfJ

(ElO.6.3)

GfJa2b2

+ b2

(EIO.6.4) but also the

the function satisfies not only the differential equation, Eq. (l0.153), boundary condition, Eq. (l0.159). The stresses O"xyand (1xz become

(1xy = O"xz=The torque (Mt) can be obtained as

az

a

=

2GfJa'1 (a2 + b2) z

(E10.6.5)

-ay

a

2GfJb2 = (a2 + b2) Y

(ElO.6.6)

MtNoting that

=

IIA

(O"xzY- O"xyz)dA

= a;?b2

IIA

Clb

2

+z

2 2 a ) dA

(ElO.6.7)

II idA=11 II = IIA A

ldydz 2 z dzdy

= Iz = ~rrba3

(E10.6.8)

z2dA

= Iy = ~rrab3

(E10.6.9)

A

A

Eq. (E1O.6.7) yields (ElO.6.1O) When Mt is expressed as

MtEq. (ElO.6.10)

= GJfJJ=---

= CfJ

(E10.6.1l)

gives the torsional constant J as

rra3b3

a2

+ b2

(E10.6.12)

and the torsional rigidity C as

CThe rate of twist can be expressed

=

rra3b3 Mt a2 + b2 G =

7i

(E10.6.13)

in terms of the torque as

fJ =

C=

Mt

Mt (a2

+ b2)

Grra3b3

(ElO.6.14)

f"

308

Torsional Vibration of Shafts

10.10 PRANDTL'S MEMBRANE ANALOGYPrandtl observed that the differential equation for the stress function, Eq. (10.153), is of the same form as the equation that describes the deflection of a membrane or soap film under transverse pressure [see Eq. (13.1) without the right-hand-side inertia term]. This analogy between the torsion and membrane problems has been used in determining the torsional rigidity of rods with noncircular cross sections experimentally [3, 4]. An actual experiment with a soap bubble would consist of an airtight box with a hole cut on one side (Fig. 10.11). The shape of the hole is the same as the cross section of the rod in torsion. First, a soap film is created over the hole. Then air under pressure (p) is pumped into the box. This causes the soap film to deflect transversely as shown in Fig. 10.11. If P denotes the uniform tension in the soap film, the small transyerse deflection of the soap film (w) is governed by the equation [see Eq. (13.1) without the right-band-side inertia term] (10.165) or (10.166) in the hole region (cross section) and

w=O

(10.167)

on the boundary of the hole (cross section). Note that the differential equation and the boundary condition, Eqs. (10.166) and (10.167), are of precisely the same form as for the stress function 4>,namely, Eqs. (10.153) and (10.159):V 4> = -2Gf32

(10.168)

in the interior, and (10.169) on the boundary. Thus, the soap bubble represents the surface of the stress function with

wpiP

4>

= 2Gf3Deflected soap film

(10.170)

Hole in box with shape similar to the cross section of the rod in torsion

Air pressure

Figure 10.11

Soap film for the membrane analogy.

l;,,~"~~,,~~ "M'ii'''~'k~~~~~~11'''"'ii{~il~':z.~~~Aii~~~i:.\o~-..i~~~~~;~~~~,Q:~l.'iWilS!i~.i;;.K~.;.id~~~~~,:>,

10.10

Prandtl's Membrane Anal~gy

309

or = Cw

(10.171)

where

9 denotes

a proportionality constant: C

= 2Gf3PP

(10.172)

-

The analogous quantities in the two cases are given in Table 10.2. The membrane analogy provides more than an experimental technique for the solution of torsion problem. It also serves as the basis for obtaining approximate analytical solutions for rods with narrow cross sections and open thin-walled cross sections. Table 10.3 gives the values of the maximum shear stress and the angle of twist per unit length for some commonly encountered cross-sectional shapes of rods. Example 10.7 Determine the torsional rigidity of a rod with a rectangular cross section as shown in Fig. 10.12.SOLUTION We seek a solution of the membrane equation, (10.166), and use it for the stress function, Eq. (10.171). The governing equation for the deflection of a membrane is

-a :::y ::: a, -b ::: z :::b

(E1O.7.1)

Table 10.2

Prandtl's Membrane Analogy Torsion problem

Soap bubble (membrane) problemW 1

P

P

G 2{3

- az' ay

aw aw

2 (volume under bubble)

z

-- ...M,

--xM,

Figure 10.12

Rod with a rectangular cross section .

I .,

310

Torsional Vibration of Shafts

Table 10.3 Cross section

Torsional Properties of Shafts with Various Cross Sections Angle of twist per unit length, () Maximum shear stress, Tmax

1. Solid circular shaft

2. Thick-walled tube

3. Thin-walled tube

'-.1 Ofm"" RO RI

2T 7r R3

T G

= torque = shear modulus2T 7rG(R6 - Ri)

2TRo 7r(Rri - Ri)

4. Solid elliptic shaft

.j2(a2 + b2)T 47rGa2b2t

T

27rabt

(continued

on next page)

10.10

Prandtl"s Membrane Analogy

311

Table 10.3

(continued) Angle of twist per unit length, f) Maximum shear stress, !max

Cross section 6. Solid square shaft

7.092T

(];;4

4.808T -a-3-

7. Solid rectangular shaftu b

T otGab3 ot 0.141 0.229 0.263 0.291 0.312 0.333 f3 0.208 0.246 0.267 0.292 0.312 0.333 1.0 2.0 3.0 5.0 10.000

8. Hollow rectangular shaft T 2abtl T 2abt2

T .itl b

(at2 + btl)T 2Gtlt2a2b2

!maxl

= =

!max 2

9. Solid equilateral triangular shaft

26T

Ga4

-;;3

13T

10. Thin-walled tube

SA

= =

circumference of the centerline of the tube (midwall perimeters) area enclosed by the midwall perimeters

T 2.4t

312

Torsional Vibration of Shafts and the boundary conditionswe-a, w(y,

arez) -b)

= w(a,= w(y,

= 0, b) = 0,z)

-b

:s z

:S b

(EIO.7.2) (ElO.7.3) at y

-a :S y :S athe boundary conditions

The deflection shape, w(y, z), that satisfies Eq. (EIO.7.2), can be expressed as00

= =fa,

w(y, z)

=

'"' ~

ai cos irry 2a Zi(Z)

(EIO.7.4) Substituting

i=I,3,5.... where ai is a constant and Zj (z) is a function Eq. (E1O.7.4) into Eq. (ElO.7.I), we obtain of z to be determined.

(ElO.7.5) When the relation

tj=I.3.5 .... is introduced on the right-handdz2

~(_1)(i-I)/2cos irr

irry

2a

=1

(ElO.7.6)

side of Eq. (ElO.7.5), the equation yields2

d2 Z 2 --j - ~Zi The solution of this second-orderZi(Z)

4a2 ...

=-

4 P -_(_1)(i-1)/2P irraj

(ElO.7.7) as

differential sinh - irrz 2a

equation can be expressed

= Aj cosh -

irrz 2a

+ Bj

+.

I6pa2

(_1)(1-1)/2

.

Pi2rr2ai

(ElO.7.8)

where Ai and Bj are constants to be determined from the boundary conditions at = =fb. Using the condition Zj(z = -b) = 0 in Eq. (EIO.7.8) yields Bj = 0, and the condition Zj (z = b) = 0 leads toZ

A - -------I -

I6pa2

1cosh(irrb/2a)

Pi2rr2aj

(ElO.7.9) membrane can be

Thus, the function expressed asZ(z)

Zj (z) and the deflection2 I6pa Pi2rr2aj I6pa2 Prr2

of the rectangular

=

(_1)(i-1)/2

[1-

cosh(irrz/2a)]cosh(irrb/2a)

(ElO.7.1O)

w(y, z

)

=

1=1.3.5 ....

~ . ~

1i2 (-1)

(i-I)/2 [

coshUrrZ/2a)] 1 - cosh(irrb/2a)

irr)' cos 2a (ElO.7.JJ)

,.l,~~Lt-'~~1o.."'--'~~$O~~~jl~U"~~~L:~~~~~~~~~~\o;,~.i&~~~...b.>~l~~~;..

10.11

Recent Contributions

313

Equations (ElO.7.11), (10.171), and (10.172) yield the stress function (y, z) as "'( y ) _ 32GfJa2 ~ 1 - ---::-L-- -(2 , rr i=I.3.5.... i2Z

1)(i-l)/2 [

cosh(irrZ/2a)] 1 - -----:---:-cosh(irrb/2a)

irry cos 2a

(E1O.7.12)

The torque on the rod, Mt, can be determined as [see Eq. (10.164)] Mt = 2

ffA

d A

64GfJa2ja jb ~ 1 ( 1 (i-l)/2 [ cosh(irrZ/2a)] = -~- 2 L-- -2 - ) 1- ---'--rr -a -b i=I.3.5.... i cosh(irrb/2a)

irry cos-dydz 2a (ElO.7.!3)

= 32GfJ(2a)\2b)rr4 Using the identity

~ -'4 L-- I i=I.3.5....

_

64GfJ(2a)4 rr5

~ tanh _irr_b L-- i5 2a i=I.3.5....

.!..

(E1O.7.14)

Eq. (EI0.7.l3) can be rewritten as .. 1 3 ( 192a ~ 1 irrb) Mt = 3GfJ(2a) (2b) 1 - rr5b i=~ .... i5 tanh 2a The torsional rigidity of the rectangular cross section (C) can be found as C = Mt = kG(2a)3(2b) (ElO.7.l5)

fJ

(EI0.7.16)

where 1 ( 1 - -192 a k =3 rr5 b

L

oo

i=I.3.5....

. -1 tanhlrr5i

a

b)

(E1O.7.17)

For any given rectangular cross section, the ratio b/a is known and hence the series in Eq. (ElO.7.1?) can be evaluated to find the value of k to any desired accuracy. The values of k for a range of b/a are given in Table 10.4.

10.11

RECENT CONTRIBUTIONSThe torsional vibration of tapered rods with rectangular cross section, pre-twisted uniform rods, and pre-twisted tapered rods is presented by Rao [4]. In addition, several refined theories of torsional vibration of rods are also presented in Ref. [4].

314

Torsional Vibration of Shafts Table 10.4 Values of k in Eq. (E1O.7.16) bla 1.0 1.5 2.0 2.5 3.0 k 0.141 0.196 0.229 0.249 0.263 bla 4.0 5.0 6.0 10.000

k 0.281 0.291 0.299 0.312 0.333

Torsional Vibration of Bars The torsional vibration of beams with a rectangular cross section is presented by Vet [8). An overview of the vibration problems associated with turbomachinery is given by Vance [9]. The free vibration coupling of bending and torsion of a uniform spinning beam was studied by Filipich and Rosales [16]. The exact solution was presented and a numerical example was presented to point out the influence of whole coupling. Torsional Vibration of Thin- Walled Beams The torsional vibration of beams of thinwalled open section has been studied by Gere [11). The behavior of torsion of bars with warping restraint is studied using Hamilton's principle by Lo and Goulard [12]. Vibration of a Cracked Rotor The coupling between longitudinal, lateral, and torsional vibrations of a cracked rotor was studied by Darpe et al. [13]. In this work, the stiffness matrix of a Timoshenko beam element was modified to account for the effect of a crack and all six degrees of freedom per node were considered. Torsional Vibration Control The torsional vibration control of a shaft through active constrained layer damping treatments has been studied by Shen et al. [14). The equation of motion of the arrangement, consisting of piezoelectric and viscoelastic layers, is derived and its stability and controllability are discussed. Torsional Vibration of Machinery Drives The startup torque in an electrical induction motor can create problems when the motor is connected to mechanical loads such as fans and pumps through shafts. The interrelationship between the electric motor and the mechanical system, which is effectively a multimass oscillatory system, has been examined by Ran et al. [15).

REFERENCES1. W. F. Riley, L. D. Sturges, and D. H. Morris, Mechanics of Materials, 5th ed., Wiley, New York, 1999. 2. S. Timoshenko, D. H. Young, and W. Weaver, Jr., Vibration Problems in Engineering, 4th ed., Wiley, New York, 1974. 3. W. B. Bickford, Advanced Mechanics of Materials, Addison-Wesley, Reading, MA, 1998. 4. J. S. Rao, Advanced Theory of Vibration, Wiley, New York, 1992.

Problems ..315 5. S. K. Clark, Dynamics of Continuous Elements, Prentice-Hall, Englewood Cliffs, NJ, 1972. 6. S. S. Rao, Mechanical Vibrations, 4th ed., Prentice Hall, Upper Saddle River, NJ, 2004. 7. J. H. Faupel and F. E. Fisher, Engineering Design, 2nd ed., Wiley-Interscience. New York, 1981. 8. M. Vet, Torsional vibration of beams having rectangular cross sections, Journal of tire Acoustical Society of America, Vol. 34, p. 1570, 1962. 9. J. M. Vance, Rotordynamics of Turbomachinery, Wiley, New York, 1988. 10. A. E. H. Love, A Treatise on the Mathematical Theory of Elasticity, 4th Ed., Dover. New York, 1944. 11. J. M. Gere, Torsional vibrations of beams of thin walled open section, Journal of Applied Mechanics, Vol. 21, p. 381, 1954. 12. H. Lo and M. Gou1ard, Torsion with warping restraint from Hamilton's principle, Proceedings of the 2nd Midwestern Conference on Solid Mechanics, 1955, p. 68. 13. A. K. Darpe, K. Gupta, and A. Chawla, Coupled bending, longitudinal and torsional vibration of a cracked rotor, Journal of Sound and Vibration, Vol. 269, No. 1-2, pp. 33-60, 2004. 14. I. Y. Shen, W. Guo, and Y. C. Pao, Torsional vibration control of a shaft through active constrained layer damping treatments, Journal of Vibration and Acoustics, Vol. 119, No.4, pp. 504-511, 1997. 15. L. Ran, R. Yacamini, and K. S. Smith, Torsional vibrations in electrical induction motor drives during start up, Journal of Vibration and Acoustics, Vol. 118, No.2, pp. 242-251, 1996. 16. C. P. Filipich and M. B. Rosales, Free flexural-torsional vibrations of a uniform spinning beam, Journal of Sound and Vibration, Vol. 141, No.3, pp. 375-387, 1990. 17. S. P. Timoshenko, Theory of bending, torsion and buckling of thin-walled member of open cross-section, Journal of the Franklin Institute, Vol. 239, pp. 201, 249, and 343, 1945.

PROBLEMS10.1 A shaft with a uniform circular cross section of diameter d and length 1 carries a heavy disk of mass moment of inertia h at the center. Find the first three natural frequencies and the corresponding modes of the shaft in torsional vibration. Assume that the shaft is fixed at both the ends. 10.2 A shaft with a uniform circular cross section of diameter d and length 1 carries a heavy disk of mass moment of inertia h at the center. If both ends of the shaft are fixed, determine the free vibration response of the system when the disk is given an initial angular displacement of eo and a zero initial angular velocity. 10.3 A uniform shaft supported at x 0 and rotating at an angular velocity n is suddenly stopped at the end x = O. If the end x = 1 is free and the cross section of the shaft is tubular with inner and outer radii rj and ro, respectively, find the subsequent time variation of the angular displacement of the shaft. 10.4 A uniform shaft of length 1 is fixed at x 0 and free at x = 1. Find the forced vibration response of the shaft if a torque Mt (t) = Mto cos nt is applied at the free end. Assume the initial conditions of the shaft to be zero. 10.5 Find the first three natural frequencies of torsional vibration of a shaft of length 1 m and diameter 20 rom for the following end conditions: (a) Both ends are fixed. (b) One end is fixed and the other end is free. (c) Both ends are free. Material of the shaft: steel with p = 7800 kglm3 and G = 0.8 X 10" N/m2.

=

=

316

Torsional Vibration of Shafts (b) Cross section: hollow with inner diameter 80 rom and outer diameter 120 mm; material: aluminum with p 2700 kglm3 and G 0.26 x IOJ] N/m2.

10.6 Solve Problem 10.5 by assuming the material of the shaft to be aluminum with p = 2700 kg/m3 and G 0.26 X 1011 N/m2

=

=

=

Consider two shafts each of length 1 with thinwalled tubular sections, one in the form of a circle and the other in the form of a square, as shown in Fig. 10.13.10.7

oFigure 10.13

10.10 Find the first three natural frequencies of torsional vibration of a shaft fixed at x 0 and a disk of mass moment of inertia II = 20 kgm2/rad attached at x = 1. Shaft: uniform circular cross section of diameter 20 rom and length 1 m; material of shaft: steel with p = 7800 kg/m3 and G = 0.8 X 1011 N/m2.

=

10.11 Find the fundamental natural frequency of torsional vibration of the shaft described in Problem 10.10 using a single-degree-of-freedom model.

10.13 Find the free torsional vibration response of a uniform shaft of length 1 subjected to an initial angular displacement O(x, 0) 0 and an initial angular velocity 10.8 Solve Problem 10.7 by assuming the tube wall thickness and the interior cavity areas of the tubes to be O(x,O) = Vo8(x -1) using modal analysis. Assume the the same. '. .shaft to be fixed at x = 0 and free at x = 1.

Assuming the same wall thickness of the tubes and the same total area of the region occupied by the material (material area), compare the fundamental natural frequencies of torsional vibration of the shafts.

10.12 Find the free torsional vibration response of a uniform shaft of length 1 subjected to an initial angular displacement O(x, 0) f)o(x / 1) and an initial angular velocity O(x, 0) 0 using modal analysis. Assume the shaft to be fixed at x 0 and free at x 1.

=

= =

=

=

10.9 Determine the velocity of propagation of torsional waves in the drive shaft of an automobile for the following data:

10.14 Derive the frequency equation for the torsional vibration of a uniform shaft with a torsional spring of stiffness kt attached to each end. 10.15 Find the steady-state response of a shaft fixed at both ends when subjected to a torque Mt (x, t) = Mto sin Ot at x 1/4 using modal analysis.

(a) Cross section: circular with diameter 100 rom; material: steel with p 7800 kg/m3 and G 0.8 X 1011N/m2

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=

=

,

"""'"""'~'~~~-"""'~~-"'.ul.""~"'-'~ori"""_-: ~t:J

.. .. -'-'

"-"

~I~

-3 .....:c, t:J

~ ~ '-' M ~ M31 >-: . ~ ~ \'5'-'

.-.

..

II

~ ~ '\l;:>.

:>

M

M "'< ~31'" :>

~

..

83I

.-. .. .. ciM ~ 31 ~ ~ '-' M

II

:>I~

.... ~

.g :.ac: ou

c:

....;

323

II

II0

II

,....,

,...., N '-: ...: ..

(EI5.6.6) (EI5.6.7) (EI5.6.8)

+ Zke8 E4>8 = ES8 + Zk4>8E88 = E28

where k4>4>,k88, and k4>8 are given by Eqs. (15.107)-(15.109): 1 au 1 a2w k4>4> = R2 a - R2 a 2k 88 -

(EI5.6.9)aw a

1 av R2 sin a e 1au a () -

1 R2 sin2

a2w a e2

+ Ji2" u

cot

-

Ji2"

cos

(EI5.6.1O)

k4>8 = R2 sin

1 a2w R2 sin a a eaw

+ R2

1 av a cotR2 sin ae

R2 sin -8 ae

. 1

a2wcos

+.R2 sin2 a () - -v+ R2

cot

aw

(EI5.6.1l)

Note that u, v, and w denote the components of displacement along , directions, respectively, in the mid plane of the shell.

e,

and z

15.4 STRESS-STRAIN RELATIONSIn a three-dimensional isotropic body, such as a thin shell, the stresses are related to the strains by Hooke's law asElI

1 =E

[all

-

v (0'22V (all

+ 0'33) ] + 0'33) ] + 0'22) ]

(15.110) (15.111) (15.112) (15.113) (15.114) (15.115)

E22 = E [0'22 E33 = - [0'33 E EI2=~0'12

1

1

V (all

E13=

E23 = G 0'13

G 0'23

-

G

l5.5

Force and Moment Resultants