Bureau of Engineering

Research

Robert t . Dinssn

and Cy rus- 0 . Varrn

DISCLAIMER

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FORCED VIBRATION OF CONTINUOUS ELASTIC MEMBERS

Rober t L. Dineen

and

Cyrus 0. Varan

Department o f C i v i l ~ n g i n e e r i n g and

Bureau of Engineer ing Research The U n i v e r s i t y of New Mexico

Albuquerque, New Mexico

I

F i n a l Report CE-41(72) SL-216-2

August 1972

N O T I C E I

'.

: I

This report was prepared as an account of work sponsored by the United States Government. Neither

. the United States nor the United States Atomic Energy Commission, nor any of their employees, nor any of their contractors, subcontractors, or their employees, makes any warranty, express o r implied, or assumes any legal liability GE responsibility for tho occuroop, o o m ~ pleteness o r usefulness of any information, apparatus, product o r process disclosed, or represents that its use would not infringe privately owned rights.

I . .

Robert L. Dineen'" and Cyrus 0. Varan (2)

SUMMARY

-The t r ans ien t and steady-state. response o f continuous l i n e a r

e.1asti c member t o per iod ic load ing i s invest igated. The continuous

elements are assumed t o have uni form proper t ies . A genera l ly app l i -

cable numeri ca l so l u t i on and an ana l y t i ca l so l u t i o n based on apply i ng

Laplace Transform t o reduce the p a r t i a l d i f f e r e n t i a l equation i n terms

o f t ime t o an ord inary d i f f e r e n t i a l equation are formulated.

(1 Research Assistant , C i v i l Engineering Department ,. Un i ve rs i t y o f : -.. ....- . -

. N e w Mexico, A1 buquerque, New Mexico 87106 :: ....;-- ,.. -2.:. -., .U .. ..,., . x:

( 2 ) ~ s s o c i a t e Professor, C i v i 1 Engineering ~ e ~ & t n j < i t , .uniVersi ty of New Mexico , A 1 buquerque , New Mexico 871 06 ',S','> "

. . . .. , . . ,, '.. , . . . . 5 % . * .

. .

. 1. INTRODUCTION . .. - . .. . - . - . . . .. . . . . -

I&tmdc loads may be c l a s s i f i e d i n t o two main groups:. periodic .i m d nonperiodic loads. For r e a l s t ructures-.with damping, t h e e f f e c t

of the periodic loading may be divi,ded i n t o two parts: the t rans ient . 3

motion and the steady-state motion. The transient motion r e s u l t s , . .

f m m the i n i t i a l conditions and the reaction of the system charac-

t e r i s t i c s t o these conditions a s well a s the external load. This

e f f e c t eventually diminishes if damping i s present. The steady-state

motion is the resul t ing motion once the t ransient motion i s damped

. out. Nonperiodic loads may be considered t o have two ~ a h s also: '. .

. . t h e first, which occurs during the time of the loading, i s considered

t o be the t rans ient motion, while'.the second, which occurs a f t e r the

end of t h e loading interval , i s the f ree vibration condition. For . .

systems with damping, the f r e e vibration motion i s eventually

suppressed and the system comes f i n a l l y t o res t .

In both of these load types, the t rans ient e f fec t s a r e of

considerable importance. For t h e nonperiodic loading, .they represent

the iiesponse of the 3yotem during Its i n i t i a l stage and determine the

i n i t i a l conditions f o r . t h e f r e e vibration phase. In the case of

periodic loading, the t rans ient e f fec ts i n m a n y instances resu l t i n

s t r e s ses and deflections i n the s tructure well above those corre-

sponding t o the steady-state motion. This is t rue even f o r damped . . . . 1.

systems, s ince damping has l i t t l e e f fec t during the i n i t i a l phase . . : . -

. . of t h e response.

, . . >

. . . .

Considerable effort has been devoted to determining

steady-state motion, due in large tothe fact that the time

variable.may.be eliminated in governing equation. Such a . .

simplification is not possible if transient effects are to

be determined. Hence less work has been done on this aspect

of the forced motion problem for many structures.

The purpose of this work is to derive a method for

determining the complete solution, i.e., the transient and .. . .

steady-state response. The system will be considered as

undamped and initially at rest, with the elements of the

structure taken as uniform.

. . Method of Soluti.on -

The structure will be analyzed using the discrete.

mechanics approach.' using this method, the system is con-'

sidered to be composed of uniform elements which are con-

nected at discrete points. By determining the behavior of

a typical element for arbitrary end moments and imposing the

required force and deformation requirements at the ends of

adjacent beams, a recurrence relation in the form of a dif-

ference equation is obtained. A closed' form solution of

this equation in terms of a finite series is used t o give a

discrete field solution for the moments in the system at the

support points.

The response of the typ ica l element i s complicated by t h e presence

of t h e time variable. For the cases of free-vibration and steady-state

forced motion, t h i s var iable may. be eliminated and t h e problem takes

'the form of a s t a t i c solution. The more general problem of t r ans ien t

motion cannot use t h i s technique. I n t h i s paper, two approaches a r e

investigated. The first u t i l i z e s a var ia t ion of the separation of

' variables technique' t o -obtain the response o f ' a typical: element. The

end r e s u l t i s an i n t e g r a l equation of t h e first kind. ' Solution of

t h i s expression'was done using a numerical method. This procedure

was applied only t o the continuous beam problem.

In t h e second method, which r e s u l t s i n a closed form solutiori

f o r both systems, t h e Laplace transform was used t o eliminate the

time variable. The solut ion of the difference equation was then

inverted t o obtain the moment expressions i n r e a l t i m e . -. - 7

The''most widely. used .procedure i n determining t h e response of . -, -

' . continuous' beams t o . time-dependent loads is the normal mode method. 1

'Tbe response of t h e beam, usually t h e deflection, i s expressed a s

an i n f i n i t e s e r i e s i n which t h e time and s p a t i a l functions a r e

separable. The s p a t i a l functions a r e t h e natural vibrat ion modes

of t h e beam. The method, then, requires p r io r knowledge of the

natural frequencies (eigenvalues) and t h e na tura l v ibra t ional modes

( e ~ ~ e n f u n c t i o n s ) of thc beam hefore the response can be determined.

The na tu ra l frequencies of beams a r e themselves important pr inc ipal ly

I Hurty, W. C. 'and Rubinstein; M. F., rentic ice-Hall, Englewood Cliffs, N. J. :

3

t o avoid resonance conditions. Therefore, a considerable amount of

e f f o r t has been expended by invest igators on the problem of deter-

mining t h e na tura l frequencies and mode shapes of continuous beams. . .

2 Ilsrnley was apparently the first t o solve f o r the na tu ra l

fmquencies of a continuous'beam. The beam had spans of a r b i t r a r y .

length but with equal f l e x u r a l s t i f fness . Using t h e general solut ion

f o r a beam, t h e force and compatibili ty conditions a t t h e ends of

each span were enforced a s well a s t h e boundary conditions f o r t h e

system. For a beam of "nv spans under a f r e e vibrat ion condition,

t h i s resul ted i n a s e t of (n - 1 ) homogeneous equations i n ( n - 1 )

unknowns. By s e t t i n g the determinate of these equations equal t o

zero , , the frequency equation f o r t h e system was obtained. The roots

of t h i s equation a r e then d i r e c t l y r e la ted t o the na tu ra l frequencies

of t h e system. This approach is g e n e r a w . referred t o as t h e

' "classical" method and is included i n several t e x t s on vibration. 3,4

Another approach t o t h i r problem was taken by s a i b e l O S He first ' . -

determined t h e mode shapes f o r t h e beam with the given boundary

Z Darnley, E. R., he Transverse Vibrations of Beams' and t h e ' Whirling of Shafts Supported a t Intermediate Points," P h i l o s o ~ h i c a l Magazine, Vol. 41 (192b), p. 81.

Timoshenko, S. and Young, D o H., v ibrat ion Problem i n - P l ~ i n e e r i n g (D. Van Nostrand, New York: 342-345.

. .

4 Volterra, E. and Zachmanoglou, E. C., Dynamics of Vibration (Charles E. Merri l l , Columbus: 1965)) pp. 324-327.

I1 ' Saibel, E., Vibration Frequencies of Continuous ~eams,"

Journal of the'Aeronautica1 Sciences, Vol. U., NO. 1, Jan. 194.4, . . p p . g 5 5 -

conditions a t i ts ends but with all i n t e r i o r supports removed. The

Lagrange equation of motion was then used with the potent ia l energy

term modified t o account f o r the 'effects of the in te rna l constraints.

I h e potent ia l energy due t o a constraint was e'kpressed a s the product

of an undetermined coeff icient and the expansion i n terms of the

beam mode shapes evaluated a t the constraint location. The minimi-

zation of t h e t o t a l potent ia l energy resulted i n the frequency

equation of r e s t r a in t ' i n the form of an i n f i n i t e ser ies . The

frequency values are. obtained approximately by truncating the se r i e s

and solving the resul tant equation.

For beams with spans of uniform f lexura l s t i f fness and spacing,

a f i n i t e difference f o m t i o n was employed by Miles6 t o determine

the natura l frequencies and mode shapes. A s i n the c lass ica l

method, t h e general solution of a s ingle span i s obtained first.

Since all spans a r e identical , t h i s equation along with the boundary - conditions a t the ends of t h e beam and the force and compat.ibility

requirements a t the i n t e r i o r supports a r e employed t o derive a

difference equation. A closed form solution i s then obtained f o r

t h i s difference equation i n the form of a f i n i t e ser ies . This

represents t h e frequency equation f o r the system, the m o t s of which

Miles, J . W., "vibrations of Beams on Many Supports, " J o u m l of the h ineerin Mechanics Division, ASCE, Paper No. 863, V- --+-&

' Jan. 195 , p. 3-1-863-9.

may be used t o ca lcu la te the frequencies of t h e system. Similar

7 approaches have been reported by Lin and by Wah and Calcote. 8

The e a r l i e s t known.investigation.of t h e dynamic response of a

continuous beam was by Ayre, Ford and ~ a c o b ~ e n . ' ' This study was an

extension of e a r l i e r work. 10yny12 on t h e behavior of s i n g l e span

milway bridges t o a constant force moving a t a f ixed ve loc i ty

across t h e s t ruc ture . The beam had two uniform spans with t h e ioad ..

, .

moving from l e f t t o r i g h t across both'spans. No damping was

considered. Af ter determining t h e na tu ra l frequencies and mode

shapes using the c l a s s i c a l technique, t he modes were separated i n t o

symmetric and antisymmetric par t s . The solut ion was obtainkd by

' Lin, Y. K., " ~ r e e Vibrations of Continuous Beams on E l a s t i c 11 Supports, In te rna t ional Journa l of' Mechanical Sciences, Vol. 4,

Sept.-Oct., 1962, pp. kg--

8 Wah, T. and Calcote, L. R., S t ruc tu ra l Analysis & F i n i t e Difference Calculus (van Nostrand Reinhold, New York: 1970- 53-55.

n Ayre, R. S . , Ford, George and Jacobsen, I. S .., Tmnsverse Vibration of a Two-span Beam Under t h e Action of a Moving Constant ~ o r c e , " Journal of A ~ p l i e d Mechanics, March 1950, pp. 1-12. -

. .

(1 lo Xxyloff, A. N., Uber die errunpenen Schwingungen von gleich- fozmigen e l a s t i s c h e n Staben," Mathematische Annalen, Vol. 61, 1905, p. 2u.

Inglis, C. E., A Mathematic+% Treatise- on Vibrations i n - Fbiluay Bridges (cambridge University Press, ~ a z r i c i g e : 1934) 1

Timoshenko, S. and Young, D. H., Vibration Problem i n Engineering ( D. Van Nostqnd, Princeton: m. 351-355 .-, ,

considering three time eras: the time t h e load is i n t h e first span,

t h e time t h e load is i n the second span and t h e time a f t e r it leaves

t h e beam. The generalized coordinate f o r each of these time periods

' and for t h i symmetric and antisymmetric modes a r e obtained from t h e .

so lu t ion of Lagrange's equation of motion. This solut ion was

generalized i n a succeeding piper13 i n which the load w a s assumed

t o vary sinusoidal ly with time.

Saibel and a l s o examined t h i s problem including t h e case

where t h e intermediate supports a r e e l a s t i c . .The 'solut ion was

obtained using t h e nonnal mode method with t h e shape function

corresponding t o the beam with the a'ctual boundary conditions but . .

with t h e i n t e r i o r constraints removed. The constraints ' en te r i n t o

t h e solut ion by modifying the po ten t i a l energy t e r n i n the formulation

. . . o f t h e Lagrange equation. The r e s u l t i s a nonhomogeneous d i f f e r e n t i a l

equation f o r t h e generalized coordinate. This equation i s solved i n

two parts : a homogeneous and a pa r t i cu la r solution. The f i n a l

result contains two a r b i t r a r y constants which a r e determined from

the i n i t i a l conditions of the problem.

11 l3 Pyre, R.. S. and Jacobsen, L. S., Transverse Vibration of a Two-span Beam Under the Action 0 f . a Moving Alternating ~ o r c e , " Journal of Applied Mechanics, Vol. 17, no. 3 (1950)~ pp.. 283-290. -

. . l4 Saibel, E. and Lee, 8. F. Z., " ~ i b r a t ions of a ~ o n t inuous Beam

Under a Constant Movfng Force, Journal of the Franklin Ins t i tu te , --.-

( ~ e c . 1952), pp. 499-516.

- . The response of a continuous beam t o t h e impact of a spherical

mass was solved by ~ o ~ ~ m a n n . ~ ~ The beam had three spans of equal .

length and cross sections. with , the mass impacting a t t h e center of

t h e beam. The frequencies and mode shapes were determined using

t h e c l a s s i c a l method and were separated i n t o symmetrical and a n t i -

symmetrical parts . The impact force between the sphere and t h e

beam was calculated ' using H e r t ~ , ~ s ; impact theory. Using Lagrange 's

equation, t h e governing equation was derived and s'olved. While

symmetry i n both t h e loading and the geometry was present, t h e method

is applicable f o r impact along any p a r t of the. beam. The author a l s o

s t a t e s t h a t the method is applicable f o r a beam with.any number of

spans. A considerable amount of work would no doubt be involved i n

solving a more general condition.

1 4 Saibel and ~ ' ~ ~ ~ o l o n i a l ~ used the same method previously out l ined . .

and extended it t o permit more general loads. These loads a r e taken

as sepal-able i n t h e ti,me and s p a t i a l variables. . .

Bradshaw and warburton17 d e a l t with t h e t rans ient response of a

uniform beam on r i g i d o r f l e x i b l e supports when subjected t o p e s c r i b e d

nd 45 Hoppmann, W. Ha 2 , Impact on a ' ~ u l t i s ~ a n Beam, It Journal of - Applied Mechanics ( k c . 1950), pp. 409-414. .

16 Saibe1.E. and D'Appolonia, E., 11 Forced Vibrations of Continuous

~eams," Transactions ASCE, Paper no.. 2526, Vol. 117 (1952), pp. 1075- logo* .

'I7 Bradshaw, A. and Warburton, G. B., " ~ r a n s i e n t Flesponse:of . '.

~eams," Journal Mechanical Ensineerinq Science, ~ o l . 9, no. 4, 1967, pp. 2 9 0 - r

support motion. A preliminary t o the so lu t ion was t h e determination

of , t h e na tu ra l frequencies and mode shapes ' f o r .the beam with cons t ra in t s . :I

The response so lu t ion was i n t h e f o m of t h e i n t e g r a l equation of motion

' i n which t h e motion due t o t h e supports is accounted f o r as an addi-

t i o n a l term. By expressing t h e influence functions f o r t h e supports.

and t h e .beam def lect ion as an i n f i n i t e s e r i e s expansion i n terms 'of

t h b normal mode shapes and applying the orthogonali ty re la t ionsh ip

between t h e modes, t h e i n t e g r a l equation is reduced t o an ordinary.

nonhomogeneous d i f f e r e n t i a l equation f o r t h e generalized coordinate.

The so lu t ion of t h i s equation was obtained using Duhamel's in tegra l .

Another va r i a t i on of t he normal-mode method was developed by

18 Pilkey who again first determined t h e na tu ra l frequencies and mode

shapes f o r t he beam with homogeneous in t e rna l - cons t ra in t s and boundary

conditions. The a c t u a l beam def lec t ion shears and moments were then

assumed t o be an i n f i n i t e s e r i e s expansion i n t e rns of these mode

shapes. By multiplying t h e o r i g i n a l governing equation f o r a beam @ - .

t h e ith mode shape and in t eg ra t ing over t h e length of t h e beam, an

i n t e g r a l equation i n terms of t h e a c t u a l def lec t ion and t h e above mode

shapes was obtatned. Using in t eg ra t ion by p a r t s and subs t i t u t ing t h e

expressions f o r shear and moments f o r t h e corresponding d i f f e r e n t i a l r

terms a t t h e cons t ra in t loca t ions , t he d i f f e r e n t i a l equation f o r t h e

Pilkey, W. D a y "~ormal- ode Solutions f o r t h e Qmamic Response of S t ruc tu ra l Members with Time-Dependent In-Span ~ o n d i t i o n s , " The Journa l -- of t h e Acoustical Society of h w i c a , Vol. 44, No. 6, lw. -

generalized coordinate was obtained. The r ight s ide of t h i s expression

contains a generalized force due t o t h e nonhorno~eneous span conditions. .!

From the solut ion of t h i s equation t h e f i n a l form of the generalized

.coordinate is determined.

Recently a closed form solut ion by GangaRao k d smith19 was

obtained f o r t h e more genexal problem of a . t o r s i o n l e s s grid. The system I

was assumed t o consist of beams of uniform length and s t i f f n e s s i n each

d i rec t ion (but not necessarily t h e same i n both direct ions) , simply

supported on a l l outer edges. Using the macro mechanics f l e x i b i l i t y

formulation, t h e governing d i f f e r e n t i a l equation f o r a typ ica l beam of

t h e g r i d was solved f o r the case of steady-state vibrat ion due t 0 . a

Mmc d e l t a loading. This resul t?was then converted i n t o t h e form of

8 f i n i t e s e r i e s i n which t h e loads represent the unknown in te lac t ion

.forces between beams of t h e g r i d a t t h e mesh point locations. By

sa t i s fy ing the force and deflect ion requirements a t t h e mesh points

of t h e g r i d an expression was obtained f o r t h e in terac t ion force.

. . Once the in terac t ion forces a r e known, t h e equation f o r t h e deflect ion

. ' at t h e mesh points may be determined. This appmach may be eas i ly . .

special ized for t h e continuom beam problem. However.it has two

I lmitat ions: it represents only a steady-state solution and the

external loads can be applied only a t evenly spaced d i sc re te points.

Therefore, t r ans ien t e f fec t s cannot be determined nor can d is t r ibuted

loads be t r e a t e d a s continuous. . . . ,

" GangaRao, Hota V. S . and Smith, J. C., "Dynamic Fie ld Analysis of Torsionless Grids," Joufnsl of the Enpineering Hechanics Division,. -- ASCE) . . PI3, .Paper No. &993, June i972, op. 079-693.

Duncan2' a l s o f o m d a t e d a s teady-s ta te solution of the continuous . .

I

beam problem by assuming simple harmonic forces and using t h e admit- I

tance method. Admittances a r e here defined a s " f l e x i b i l i t i e s measured

under t h e conditions of steady forced osc i l l a t ions of given frequency

and ' ident ica l phase. "21 The method resu l t s i n the dynamic equivalence ;

of t h e three moment equation. I

Anothei approach amendable t o t h e continuous beam problem is. t h a t

devised by Berger and shore2* f o r determining the dynamic response of

a la t t ice- type structure. They determined t h e response a t t h e ends

of an element of t h e s t ruc tu re due t o uni t s t ep loads a t those points

us ing . the governing p a r t i a l d i f f e r e n t i a l equations f o r transverse,

. . t o r s i o n a l and longitudinal motion. This resul ted i n a matrix equation . .

' re la t ing forces and displacements. Three addit ional matrix equations

. ' . were formed from the requirements of jo in t equilibrium and compati- '

. .

b i l i t y and t h e support condit ions. ~ h e s e matrix equations were then

- . reduced t o l i n e a r expressions using the Laplace transform. The

solut ions o f , these equations resul ted i n t h e transformed expressions

If 20 Duncan, W. J., Free and Forced Osci l lat ions of Continuous ~-

Beams: Treatment by t h e Admittance Method," Philosophical Magazine, Serial 7, Vol. 34, 1943, pp. 49-63. ,

. . I1 22 Berger, S. and Shore, Sidney, Dynamic Response of Lat t ice- . '

!Pype Structures, " hgineer inn Mechanics Division, ASCE EM2, Paper No. 3484, Apri l 1963, pp. 47-70.

-' .

for t h e end deformations and forces . A s e r i e s expansion of orthogonal.

functions was used t o obtain approximate inversions of these equations. I

TO insure uniform convergence of t h e s e r i e s with time, ex terna l

.damping and forcing function cutoff was employed. I I

As i n t h e case of t h e g r id solut ion, t h i s method may be used t o

so lve t h e continuous beam problem. Again, however, t he re a r e some,

stringent l imi ta t ions , s ince only point loads can be appl ied a t t h e

ends of t h e elements and t h e i n v e r s i 0 n . i ~ not exact. . .

. .

. . . NUMERICAL SOLUTION

3& General Comments and Assumptions - The procedure used i n determining t h e def lec t ion of a continuous

beam f o r time-dependent loads is t o first der ive t h e def lec t ion

equation f o r a t y p i c a l span, supported a t both ends, subjected t o

unknown end moments and given normal loads. Zly s a t i s f y i n g t h e

. f o r c e a n d compat ibi l i ty requirements between .ad Jacent spans, a

difference equation f o r t h e moments a t t h e support po in ts is obtained.

The so lu t ion of t h i s d i f fe rence equation gives t h e expressions f o r t h e

end moments on each span. The def lec t ion of any span may then be

ca lcu la ted using t h e de f l ec t ion equation f o r the t y p i c a l span with . .

. . .

t h e appropr ia te end moments and normal loads. . .

Tn t h i s chapter, t h e de f l ec t ion of a t y p i c a l span i s determined

.using a modified separa t ion of var iab les method i n which t h e time

variable is reta ined. The r e s u l t i n g expressions f o r moments and

de f l ec t ions a r e i n t h e form of i n t e g r a l equations, which are solved

by approximate methods. A two-span beam problem is 'used t o i l l u s t r a t e

t h e method.

The following assumptions were made i n der iving thqgovern ing

equations :

1. The Bernoulli-Euler beam theory i s used (no account i s taken . .

. . . of t h e e f f e c t s of shear and ro t a to ry i n e r t i a ) .

. 2. A l l spans of t h e continuous beam are of equal l ength and

. . s t i f f n e s s . - . .

3. The beam behaves i n a l i n e a r e l a s t i c manner and experiences

only small deflections.

4. The external normal load ac t ing on t h e beam i s separable

in its time and s p a t i a l variables.

3.2 Derivation of the Governing In tegra l Equation -- A continuous beam of N spans. is shown i n Figure 4.1, i n which

the span and t h e support numbers a r e indicated. The coordinate

system adopted i s a l s o shown i n t h i s diagram. The slope-deflection

' . s ign convention w i l l be used f o r the moments throughout t h i s and the

following chapters; i.e., clockwise moments a r e positive.

3.2.1 Response of a Typical Span -- The deflect ion f o r t h e span shown i n Figure 4.2 w i l l be deter-

. . .mined using the procedure of Mindlin and ~oodman.' This method

. . employs a modified separation o f 'va r i ab les technique ,in which t h e

deflect ion i s considered t o be t h e sum of two parts:

(%) . a "s ta t ic" solut ion which s a t i s f i e s the homogeneous

B e r n o u l l i - N e r equation with t h e i n e r t i a term suppressed

and with t h e ' a c t u a l time-dependent boundary conditions,

(b) an " iner t ia" solut ion which s a t i s f i e s the B e r n o u l l i - N e r . .

equation with homogeneous boundary conditions.

n Mindlin, R. D. and ~oodman, L.. E . , Beam Vibrations with ,Time-Dependent Boundary conditions, " ~ o u & a l of Amlied Mechanics - (~ecember 1950), pp. 377-380.

l " ~ t a t i c " Solution

' F r o m Reference 1 t h e form of the "stat ic" . solut ion is, [

L

vhere f ( t ) a r e the time-dependent boundary conditions, two f o r i

each end. of t h e beam, and g (x) a r e the s p a t t a l functions corre- i

sponding t o each o f t h e functions, f i ( t ) . The general expression

for g (x) corresponds t o t h e polynomial solut ion f o r beams loaded i

s t a t i c a l l y ; i.e,,

The time functions, f ( t ) f o r . t h e span of Figure 4,1 are: i

and -

. . . . -

, where S ' = beam s t i f f n e s s (modulus of e l a s t i c i t y x moment of i n e r t i a ) ..

The functions ~ ( t ) and M'(t) a r e t h e unknown mdments a t t h e ends . . .

x = 0 and x = L of t h e span. The sign' on ~ ' ( t ) has been changed t o

conform t o t h e slope-deflection convention.

The constants i n Equation (3-2) a r e detelrnined using t h e boundary

conditions given by Equations '(3-3). Since the functions f i ( t ) and . .

f ( t ) are both zem, only g2(x) and gl(x) must be determined.. The 3 constants for these two functions may be found using the following

relationships: 2

Using the first s e t of conditions above ( x ) takes the form, , g2

The second s e t o f conditions gives, . ,

Substituting Equations (3-3), (3-5) and (3-6) into (3-1); the

n staticn solution is, . , . .

3.2.1.2 "Inert ia" Solution

Again from Reference 1, t h e form of t h e "inert ia" solut ion is,

where X (x) = jth orthogonal mode shape of a typ ica l span with 3 homogeneous boundary conditions,

' . and ~ ~ ( t ) = a ' t i m e function corresponding t o t h e jth mode.

The homogeneous boundary conditions a r e f ( t ) = 0 ( ~ ~ u a t i o n s (3-3) i

. set equal t o zero). The mode shape i n t h i s instance would correspond

t o a simply-supported beam of length L and s t i f fness , S = 331. -The

mode shape is, therefore, .

The.time function is determined f r o m the expression, 3

Tj( t ) = AJ cos ~ ' t + B s i n 3 3

. . The f irs t two terms depend on t h e i n i t i a l conditions w h i l e ' t h e - l a s t

term is a function of the "s ta t ic" deflection, the mode shape and

the external loading. w is t h e c i r c u l a r frequency of t h e jthmode 3 of vibration. For a simply-supported beam i ts value is,

where, S = beam stiff'ness

p = mass density of the beam material

A = cros's-section area of the beam L.

and, L = beam length.

coefficients of the first two terms of Equation (3-10) a r e . '

given by the expressions,

where j w = w(x,0) = i n i t i a l displacement of the beam 0

1

and, . w = w(x,0) = i n i t i a l velocity of the.beam, 0

Since it has been assumed tha t t h e beam is i n i t i a l l y a t rest ,

Fmm Equations (3-3)) the functions f i ( t ) = f ( t ) = 0. Also note 3

. . that the remaining two functions, f 2 ( t ) and f4( t ) , are second . . . .

. . derivatives 'with respect t o x of the deflection. Therefore,

- a2v(o,t) f2(0) - 4

The same reasoning holds f o r . fb ( t ) . Subst i tut ion of these r e s u l t s

= 0 , since. W(X,O) = o t = O

.' a2;(o, t j . , and, $0) =

4

i n t o Equations (3-12) and (3-13) gives,

= o , s ince w ( x , ~ ) = o , ( 3-15b) t = O . .

Gquation (3-10) reduces, then, t o

. ' -

The function i n the integrand, P ( t ) , is given by il

where, q(x) p ( t ) = t h e external load on the span,,

and

The constants Q 3' =i3

and G* a r e coeff ic ients of t h e generalized i d

Iv Fourier expansion of the functions q(x), gi(x) and gi (x ) about

the orthogonal mode shapes, X (x); i.e., 3

From the general expression' for gi(x) given i n Equation (3-2)

Therefore,

Equation (3-18) reduces, then, t o

,The time function is, therefore,

.The final form of the "inertian part of the Solution i s obtained by

, . aubitituting the above results into (3-8)) alongwith the values of

GZj a d G43 as given i n Appendix A.

. . I . s i n m ( t - 7) d? .

3

3.2.1.3 Deflection Equation f o r the nth Soan ---L

The def lec t ion equation f o r the nth span is obtained by.adding . .

the "s ta t ic" and t h e "inert ia" sclutions. ,

3.2.2 Force and Compatibility Conditions' - -- . .

For continui ty t o e x i s t i n the beam, the ro ta t ions of adjacent

. spans must be equal a t a l l times. Consider the nth support of the

beam. Continuity requires t h a t the ro ta t ion of the l e f t end of

the n+lst span, 0 ( t ) , must be equal t o the ro ta t ion of the r i g h t n+l

end of the nth span, 0 ' ( t ) , where the subscripts denote t h e span n .

number. Using Boole's difference operator, E, the continui ty equation

may be wri t ten as,

where ( t ) = ~ 0 ~ ( t ) .

The ro ta t ions a t t h e ends of the nth span may be determined by

d i f fe ren t i a t ing the deflect ion equatiqn, (3-269, with respect t o . .

7

x and evaluating the r e s u l t a t t h e ends, x = 0. The general

, expression f o r rotat ion is,

e T + - 1 s i n UI j . . ' ( t - T) d~ . With

and

. t h e continui ty condition reduces to ,

sin w ( t - T) d? . . 3

The relat ionship between the moments a t the ends of adjacent ' ,

spans is obtained from the force condition a t the nth support.

Figure 4.3 is a f r e e body diagram of t h i s joint. Since moments .are

the only forces of concern here, the diagram represents a moment

equilibrium condition.

Therefore,

I

,I or,.

q t , + Mn+l ( t ) = 0

Substltuting t h i s relationship in to Equation (3-29) results in,

e sin u, ( t - 7) dT , 3 . .

where A% = E - 2 + E - ~ , is the cent& difference operator. .

Let

where ~ ~ ( t ) now represents the moment a t the nth support. Also l e t

the load'tem f o r th'e nth support be,

" 6a t p ~ t ) = z 5 [E - (-1)JI ,J ~ ~ ~ p ~ ( ~ i s i n w J ( t - T) dr .

3=1 (3-32)

Fmm .these relations, t h e final form of the difference equation is,

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Q)

AV + 6 M ( t ) .= P ( t ) + I n I n n

3.2m3 Solution of the Difference Eauation -- : Assume t h a t the t i m e and t h e d i sc re te s p a t i a l variables a r e

2 separable and nay be expressed a s a f i n i t e s e r i e s expansion; i.e*,

. .

m-1 ~ , ( t ) = C % ( t ) s i n n ,

i=l +

where n is the d i sc re te var iable and N is the t o t a l number of spans.

Also assume t h a t the external load t e r n may be represented i n t h e

same manner s o tha t ,

. R-1 p,(t) = C Fi(t) s i n n

i=l

For t h e general case where a value f o r moment and load e x i s t s a t

every disc re te point, t h e coeff ic ients . of Equations (3-34) and . .

- (3-35) .are given, respectively, as ,

N-1 i n q ( t ) = - M ( t ) s i n rj-- I , * 5-1 5 .

N-1 i n P i ( t ) = , C ~ ~ ( t ) s i n 5 . . . .

5=1

. - Applying the d i f f e r e n t i a l and difference operations indicated

in, Equation (3-33) o n t h e assumed solution, (3-34), t h e following .

re la t ions a r e obtained: , .

C Wah, Thein and Calcote, Lee R., S t ruc tura l Difference Calculus ( ~ e w York: Van Nostrand

N-1 N-1 .. i n G .n ( t ) = -$ (C q ( t ) s i n n = x % ( t ) s i n n ,

at i=1 in i=l

N-1 .. N-1 in .. AV M ( t ) = A V ~ C fii(t) s i n n = - i n n n 4 C s i n fii(t) s i n n ,

i=1 i=1

N-1 , N-1 i n AV M ( t ) = bVn x fii(t) s i n $! n = - 4 C sin2 9 R ( t ) s i n - n ..

n n i=1 i=1 2 N i N

Subst i tut ion of t h e above expressions i n t o the difference equation

I-1 14-1 C ( 6 - 4 s in2 2) fii(t) s i n n = Ci(t) s i n $ n + i=1 i=1

2 in i n 1211 - ( -~ . ) j ] + b(-l)J s i n f i i ( ~ ) s i n n 0

. .

sin U (t - 7) dT . . 3

. - Since both s ides of t h i s equation represent an expansion i n the same

f i n i t e ser ies , t h e coeff ic ients may be equated.. The equation, then,

.. reduces t o

+ 4(- l ) j s i n Ri(7) s i n ru (t - T) a? . . , J; . ;

. .

Using integration by parts twice on the integral expression in (3-39)

gives the relationship,

Equation (3-39) becomes, then,

Collecting like terms, the governing integral equation for'the . .

coefficients of Equation (3-34) is,

w . aifii(t) '= ~ ~ ( t ) - 9 1211 - (-l)j] + 4(-I.)' sin

j=l L

.at ,(TI sin w (t - T) i r 3

where, .

Wow examine a Expanding-gives, 1.-

3 . .. These s e r i e s have the values, .. .

and,

Therefore, 1

Equation (3-39) reduces, then, t o

QD '

12a 1 2[1 - ( - l ) ~ ] + 4(-1) s i n fii(7) s i n U) ( t - T) d~ = Pi(t) . 3=1 L 3

Let - 3 2 2 - */2[1 - ( -1 ) j ] + 4(-1) s i n 2N $13

. .

. The f inal fonn of Equation (3-43) is,

. This equation has t h e form of an inhomogeneous Volterra equation

of t h e first kind.

'

Numerical Solution of the Integral Equation -- The i n t e g r a l equation derived i n the prevlous sect ion was solved

using a numerical method i n which the in tegra l i s approximated by t h e

sum of a f i n i t e ser ies . This sum is arranged so t h a t the moment a t

Weast, R. C., ed., Handbook of Tables for Vathematics --8. '4th Edition levelan and: The ~hernicd-~ubber Co., 1970 , p

the time tn i s determined from t h e moments calculated a t the previous

times, t , t , , t . The procedure is an extension ofFleming

and Romualdi ' s method developed f o r discretd systems. 4

Rearranging Equation (3-45),

Approximate the in tegra l by a f i n i t e sum of N equal time increments,

TI. Using the trapezoidal ru le and a time t = N 0 TI, (*3-46) takes

t h e form,

Q)

+ .... +(%[(Ii - I)TI] Bid s i n UJ [ ( N - A + l ) T l ] + j=1 3

OD

+ %(B 0 TI) . pi s i n . w [ ( N - N)TI] = F i ( ~ . TI) . . (3-47) 3 I 3-1

bsuming t h a t the system is i n i t i a l l y a t rest, ~ ~ ( 0 ) = 0. Therefore',

0 = 0 Making t h i s subst i tut ion, t h e last expression may be'

rewri t ten as,

Q) 8-2 TI fii [ (N - 11~11 C pij s i n w [ (N - N + TI] + TI c fii(k . TI)

j=l j k=l

4 It Fleming, J. Fa and Romualdi, J. P., A General Pmcedure For Calc,ulating Dynanic Response ilue To Impulsive Loads," Journal of the -- Franklin' Lnst i tute ( ~ o l . 2.75, n.2, Feb. 1963), pp. 107-120.

Now rearrange t h i s equation t o obtain the expression f o r t h e moment

a t time, t = ( N - TI. . .

".he f i n a l form of t h i s equation' f o r computational use is obtained by

truncating t h e i n f i n i t e s e r i e s a t some value J = J.

J

5 where = c pij s i n w [(N - ~ I T I ] . j=l 3

The value of the moment a t time t = TI may be found from

- Equation (3-50) by s e t t i n g N = 2. Therefore,

Qr incementing N by TI, successive values of Ri(t) may be determined.

The accuracy of the results using t h i s method is dependent on

the maximum number of frequencies used (i.e,, the value of M) and

the ~ i z e of t h e time increment, TI. . .

3.3.1 Example Problem - Consider t h e two-span beam o f s ec t ion 4.5.2.3. using t h e same

span parameters and loading. Then,

~ ( x , t ) = q(x) o p ( t ) = 50 s i n 500t' . To determine. the.moments i n t he beam a t t h e support location,

t h e values of $ and 9 ( t ) must be calculated f i r s t . From 1 3 1

. Equation (3-44) with i = 1 and N = 2, ,

Expanding t h e expressions f o r t h e moment and load coe f f i c i en t s

gives

and F l ( t ) = pl( t )

F ina l ly t h e load t e r n may be calculated from Equation (3-32). With

i l ( t ) = 2 $$I- - ( - l ) ~ ] Jt ,A(,) s i n ur ( t - T) d~ '. j=1 3

s i n c e t h e loading is the same f o r both spans. From Equation (3-20),

where t h e mode shape function, X (x) , i s given by (3-g),.

Theref ore,

Then,

The f i n a l form of t h e load t e r n using t h e first th ree terms of t h e

s e r i e s is ,

Fl(t) = 3.5186 s i n 500t - 1.88016 s i n 925t +

- 118~68 x s i n 8315t - 8.5422 i s i n 23140t . Using these values of $ and Fl(t), Equation (3-50) was

programmed t o obtain t h e moments a t d i s c r e t e times.- A time increment

o f secs. and the first seven frequencies gave t h e r e s u l t s shown

i n Figure 3.1. The exact so lu t ion obtained by the Laplace transform

method is included f o r comparison.

The def lec t ions of t h e second span were obtained using t h e above

. . values and numerically i n t eg ra t ing Equation (3-26), i n which t h e

second derfvat ives of t h e end moment were eliminated using in tegra t ion

by p a r t s twice. The r e s u l t s a r e shown i n Figure 3.2 f o r one m i l l i -

second time i n t e r v a l s a f t e r t h e load i s applied.

3.3.2 Discussion of t he Results -- The e f f e c t s on t h e moaents of varying t h e t i m e and t runca t ing

t h e in f in i te aeries were examined. Considering a t o t a l time .equal

~ ( x , t ) = 50 s i n .500t 1

, Support P t . Numb e r

+-$' 0 1 2

Time (milliseconds)

N g u r e 3.1 Moment i n t h e Two-span Continuous Beam,at t h e -- --- Center Suppo,-t

t o t h e first ha l f cycle of t h e loading function, t h e f irst th ree

frequencies were used, and the time increment was progressively

reduced u n t i l convergence .occurred. This procedure was then

repeated f o r t h e first f ive , seven and nine frequencies, succes-

sively. The r e s u l t s a r e shown i n Figure 3.3.

The number of frequencies used ( o r a l t e rna t e ly , t h e point a t

which i n f i n i t e s e r i e s i s t runcated) a f f e c t s t h e r e s u l t s i n t h r e e .

ways. M r s t , incieasing the number of frequencies tends t o decrease

t h e peak moment. The peak value f o r t h r e e frequencies was nearly

9 percent l a r g e r than t h e exact value, while f o r nine frequencies

the difference was about 3 percent. The second e f f e c t i s a not ice-

able phase s h i f t i n t ime, .with t h e numerical r e s u l t s s h i f t i n g c lo se r

t o t h e exact values. Final ly , t h e r e s u l t s near t h e end of t h e t o t a l

time period tend t o move c lose r t o t h e exact curve.

The s i z e of t h e time increment appears t o be c r i t i c a l up t o a

c e r t a i n point . Since t h e functions involved a r e sinusoidal, t h e

' e f f e c t of numerically i n t eg ra t ing a s ine function with d i f f e r e n t

t i m e increments was checked. From t h i s work, it was found t h a t qu i t e

. accurate i n t eg l a t ions were possible i f t h e ha l f period of t h e funct ion

w e r e divided i n t o s i x time increments. For example using t h e first

* five frequencies with time increments of 2.268 x secs., which

corresponds t o one s i x t h of t h e ha l f period of t he f i f t h frequency,

and 1.0004 x sccs., t h e orily change i n t h e moment was about a

one percent change i n t he peak load.

3 Frequencies ----- 5 Frequencies --- .7 Frequencies --- 9 Frequencies

2 3 4

Time (milliseconds)

Figure 3.3 The Effect of Truncating the Infinite Series in - - - the ~ u m e r i c a Solution -

The s i z e of t h e time increment becomes l e s s important a s t h e

n b b e r of . frequencies increases r e f l ec t ing t h e progressively smaller

contributions of the higher frequencies t o the response. A check

using nine frequencies showed no perceivable difference i n t h e p lo t ted

moment values when the time increment was decreased from 2.000 x 10 -5 - 6 secs. t o 6.9906 x 10 secs.

For t h i s pa r t i cu la r problem, the most accurate peak value of the

moment appears t o be about 3 percent Larger than the exact value. The

numerical solut ion appears a l s o t o be s l i g h t l y out of phase with t h e

exact solution. The. numerical solut ion f o r t h i s pa r t i cu la r problem,

then, generally gives values f o r . t h e moments t h a t a r e l a rge r than the

a c t u a l values. Its use f o r t r ans i en t loa,ding should be qui te s a t i s -

factory.

he results f o r the def lect ion show the e f fec t s of the l a r g e r

moment values. The maximum deflect ion a t th ree milliseconds i s about

12 percent l e s s than the exact value. This l a rge r e r r o r probably

r e f l e c t s an accumulating e r r o r i n the numerical method.

4. CONTINUOUS BEAMS: ANALYTIC SOLUTION

4.1 General Comments and Assum~tions - - ' The problem of determining the r e s p o n s e . o f - a continuous beam t o

an ex te rna l dynamic load w i l l be ' so lved based on t h e in t e r ac t ion of

t h e ind iv idua l spans. An a r b i t r a r y i n t e r i o r span i s considered which

is subjected t o an a r b i t r a r y ex t e rna l load and unknown end moments.

A di f fe rence equation f o r t h e moments a t t h e i n t e r i o r supports is

then derived by s a t i s f y i n g the f o r c e and compat ibi l i ty conditions

between adjacent spans of t h e o r i g i n a l continuous beam. The solu-

t i o n of t h i s d i f fe rence equation gives a f i e l d so lu t ion f o r t h e moment

a t any i n t e r i o r support a s a funct ion of time. Once t h e end moments

are known, t h e def lec t ion of any'span may be obtained based on contin-

uum mechanics.

The following assumptions w i l l be adopted i n der iving t h e gov-

e rn ing equations of motion:

1. he Bernoulli-Euler beam theory w i l l be used (no account f

w i l l be taken of t h e e f f e c t s of shear o r ro ta tory i n e r t i a ) .

2. A l l spans w i l l be of equal l ength and s t i f f n e s s .

3. The beam behaves i n a l i n e a r e l a s t i c manner and experiences

only small def lec t ions .

4. The ex te rna l load is separable i n t h e time and s p a t i a l

var iab les .

2 Response of a - m i c a 1 S ~ a n - -- The continuous beam cons is t ing of N equal ' spans with a d i s t r i b -

u ted normal loading, ~ ( x , t ) , i s shown i n Figure 4.1. Consider a

typ ica l span of t h i s system as i l l u s t r a t ed i n Figure 4.2. This span

is subjected t o a given normal load, ~ ( x , t ) , and end moments,

~ ( t ) and M' ( t ) , due t o deformations.

!the governing equation of motion fo r a beam using the Bernoulli-

N e r theory is,

where w(x,t) = span deflection . .

.k2 = S / ~ A

S = beam s t i f fness (i.e., modulus of

e l a s t i c i t y x moment of ine r t i a )

p = mass density of the beam material

A = cross-sectional area of the beam

. and ~ ( r , t ) = the external load on the span.

The nments a t the ends of the beam follow the slope-deflection sign

. . convention ( i .e., clockwise moments are positive).

. The boundary conditions fo r the beam are:

w(0,t) = 0 , (4-2a)

and

The sign change i n Equation (4-3b) is i n accord with the slope- . .

deflection convention. I. . , : .

41

Support No. n - 1 n

1-I . ,

w

Figure 4.2 Free Sody Magram of t h e nth Span

Figure 4.3 Free Body Diagram of t he nth Support -

The i n i t i a l conditions f o r a beam or ig inal ly a t r e s t .are:

w(x,o) = 0 , ( 4-4a i

and

To obtain a solut ion f o r the governing equation the method of

Carslaw and 3aeger1 will be used. The p a r t i a l d i f f e r e n t i a l equation

(4-1) is reduced t o an ordinary d i f f e r e n t i a l equat ion by using a

Laplace transform on the time variable. This equation becomes, then,

where ;(x,P) = =et w(x;t) 9

and t (x,p) = $ ~ ( x , t ) . Using Equations (4-ha), and (4-kb), the above expression becomes,

2 ' - - E the r e s u l t i s t h e ordinary d i f f e r e n t i a l equation, Let t ing q - k2 '

2 ' . The solut ion of t h i s equation i s , : - w = A sinh qx + B cosh qx + C s i n qx + D cos qx '

L Carslaw, H. S. and Jaeger, J. C., ,hers,tional Methods in &plied Mathemtics ( ~ e w York: Dover Publications, 19- x5

The coeff ic ients i n (4-7) a r e determined from the boundaly

conditions. The transformed boundary conditions, from Equations (4-2)

aqd (4-3)' are:

. . gt w(o, t ) '= G ( o , ~ ) = 0 , (4-8)

From Equations' (4-8) and (4-g),

p a2w(x,t)

ax 2

Using Equation (4-10) we find,

A sinh qL + B cosh qL + C s i n qL + D cos qL +

. .

- d2; - - dx2 x=L

+ ' 3 f L y [ r i n h q ( ~ - 5 ) + 2q- 0

- - f i t (p) S

x=L

Subst i tut ing f o r B and D, t he values given i n (4-12) and (4-13), t h i c

equation becomes,

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R i.:

A sinh qL + C s i n qL = (cosh qL - .cos ,qL) + ;: 2q S 'i ,I

Finally, using Equations (4-11), (4-12) and ( 4-13), 1

fi 1 A sinh q~ - C sin q~ = + (cosh qL + cos q ~ ) + 2 q2s- 2Sq . - .

+ s i n q ( ~ - g ) ] d5 . (4-15)

Adding (4-14) and (4-15) gives,

fil(p) + hi(p) cosh qL + A = . .

2 q 2 ~ s inh qL

1 - f L ~ ( Z J P ) s inh q ( ~ - 5) d5 (4-16) .

Zq36 sinh qL 0

Subtracting ( 4-15) f mm (4-14) gives,

+ G(p) cos aL +

C = . -.

2;'s s i n q~ ' I

1 '

L +

3 f P(b,p) 61x1 q ( ~ - 6 ) dS .. 2q S s i n qL 0

( 4-17)

The final form o f t h e transformed deflect ion ' equation is,

- fi' (s in qL sinh qx - sinh qL s i n qx) + # ' -

2 q 2 ~ s i n qL sinh qL .

fi + 2 [ s in qL cosh qL s'inh qx +

2q S s i n qL sinh qL

- cos qL sinh qL s i n qx + s i n qL ~ i n h , ' ~ ~ ( c o s qx - cosh qx)] +

+ 1 3

P(5,p) [s inh q~ s i n q ( ~ - 5 ) s i n qx + 2q S s i n qL sinh qL 0

1 - s i n qL sinh q ( ~ - 5) s inh qx d5 +

+ - - L I X ? ( 5 , p ) [ s i b q(x - 5) - s i n q ( x - 511 d~ i (4-18) zq3s o

Given t h e end moments, the deflect ion equation i n r e a l time is obtained

by invert ing (4-18); i.e.,

. . - 4.3 Formulation -- of the Difference Equation f o r Moments a t the Supoorts -- To determine the unknown moments a t the i n t e r i o r supports of the

. . continuaus beam, the force and compatibili ty requirements must be

s a t i s f i e d between each adjacent span. Figure 4.3 is a f r e e body

diagram of the nth support. Since ro ta tory i n e r t i a e f fec t s a r e

neglected, t h i s diagram represents a s t a t i c equilibrium condition.

Using subscripts t o denote the span number, the moment equation is,

Taking: the Laplace transform of t h i s equation gives,

E again denotes t h e Boolean difference operator such that,

~ o m p a t i b i l i t y of deformstion a t t h e nth support is expressed by

after transforming.

The expression f o r t h e transformed slope a t the ends of a span

may be obtained by d i f fe ren t i a t ing Equation (4-18) with respect t o

t h x.. For t h e n span,

d;, - a 1 [ s i n qL cosh qx - sinh qL cos qx] + dx

I . . . + % [ i i n q~ cosh q~ cosh qx - cos q~ i i n h q~ cos qx +

. .

- s i n qL s inh qL ( s i n qx + sinh qx)] +

+ 1 q [J' %({,PI sib qi. s i n q ( ~ - 5) cos qx +

- s i n qL sinh q ( ~ - 5) cosh qx .dS + ( s i n qL sinh q ~ ) c 1

Subst i tut ion of (4-24) i n t o (4-23) with the. appropriate subscripts

gives,

pn - f i ~ + ~ ) ( s in , qL - s inh q ~ ) +

+ (ii - f i n + ( s i n q~ cosh q~ - cos q~ s i ~ q ~ )

i

= [ L ( , l ( g , ~ ) q [ s i n h qL s i n q ( ~ - 5) - s i n qL si* q ( ~ - 5 ) I +

' + %(?,p) s inh qL s i n q5 - s i n qL s i n q5 de . I 11 ( 4-25

Using r e l a t i o n (4-21), t h e above equation may be wr i t t en as t h e

d i f fe rence equation,

n (4-2&)

where AVn = t h e c e n t r a l d i f fe rence operator

kl = s i n h qL - s i n qL . (4-26b)

k2 = sin qL cosh qL - cos qL s inh qL ( 4 - 2 6 ~ )

en = f lL ( PI 1 s inb qL s i n q ( ~ - 5) +

- s i n q~ s i h q ( ~ - S ) ] +

+ a(5,~) [ s inh qL s i n qS - s i n qL slnh qS

(4-26a)

Le t t i ng f i = fin , n . .

where 2 now denotes t h e moment i n t h e continuous beam a t t h e n t h n

support, Equation (4-26a) becomes,

mls i s t h e governing d i f fe rence equation f o r moments a t t h e support

l o a t ions.

4.4 Solution of the Difference Equation f o r a Beam with Simply- - -- ---- . Supported Ends i -

Assume t h a t the time and the d iscre te variables i n the

expressions f o r t h e moments are ' separable and may be expressed a s

a f i n i t e ~ e r i e s . ~ Therefore, l e t . .

N-1 i n ~ , ( t ) = C Gi(t) s i n n ,

i=1

.where n i s ' t h e d i sc re te variable and N is t h e total 'number of spans.

Assume a l s o t h a t the external load term may be represented i n a

s imi lar manner as,

N o 1 ~ ~ ~ ( t ) = C qi(t) s i n n .

i=1

Using inverse sum f ormulas , t h e coeff ic ients of Equations (4-28)

and (4-29) a r e given, respectively, by

N-1 in ~ ~ ( t ) = - C M ( t ) s i n j , j=1 3

and * N-1

= - C i n ~ ~ ( t ) s i n j . j=1

&plying t h e Iapbace transformation to (4-28) tiud (4-29) t o eliminate . .

t h e time variable r e su l t s i n

N-1 %(PI = C 5i(p) s i n n ,

i=1

'

Wah, l h e i n and Calcote, Lee R., S t ruc tura l Anal sis BJ Fin i t e Difference Calculus ( ~ e v York: Van Nostrand Reinho-), p. 22.

Subs t i t u t ing (4-32) and (4-33) i n t o t h e governing d i f fe rence

equation and performing t h e ind ica ted operat ions gives,

Matching t h e s e r i e s coe f f i c i en t s -we f ind ,

The moment coe f f i c i en t s a r e obtained by inver t ing Equation (4-35);

. ,

T ~ P ) -1 - 2 , i r r 0 '

2(kl + k2) - 4kl s i n - ( b-36) 2N

- PS-1 , . '.u,(P)

2 in s i n in n

i=1 2(k1 + k2) - 4k1 s i n - N ( 4-37)

2N

4.5 Applications

To i l l u s t r a t e the so lu t ion procedure and t o verif 'y . the r e s u l t s

obtained us ing . the d i sc re te approach, two problems.are examined i n

t h e following sections. The first is a two-span continuous beam with

a uniform load over both spans. The load is assumed t o vary sinus-

o ida l ly with time. The symmetry of t h e geometry and t h e loading

about t h e center support make t h i s example equivalent t o a propped

cantilever'beam. Therefore, two solu t ions of the propped canti lever ,

by Laplace transform and t h e normal mode methods, were used t o check

the resu l t s of t h e d i sc re te method.

The second problem, a three-span beam with a normal load i n t h e

t h i r d span t h a t var ies l i n e a r l y with distance, was chosen a s an

example of a. more application. The first and second spans

i r - t h i s case have deflect ions t h a t a re dependent only on the support

a support moment. Again, the load is assumed t o vary sinusoidal ly

with time.

In both examples the response of t h e system i n terms of the

moments a t t h e oupport locaLions and the deflect ions of the spans

is obtained.

The solu t ion procedure consists of two parts: . f irs t determining . .

t h e transformed moments a t t h e support locat ions and second solving

f o r t h e deflect ion expressions. In t h e following examples, t h e

transformed expressions f o r t h e support moments were obtained and

then inverted. Using t h e transform moments then, t h e transformed ,

deflect ion equations were derived, and inverted. If one i s .

in te res ted only i n the deflect ion of t h e beam, inversion of t h e

'moment expressions need not be done.

To determine t h e expressions f o r the support moments the

following procedure was used:

1. Calculate En(P) f o r each i n t e r i o r support using

Equation (4-26d).

2. Take t h e Laplace transform of Equation (4-31) and expand

. it t o derive. an expression f o r (P) . i

3. U s e Equation (4-35) t o obtain the corresponding expression

4. Subst i tu te t h e r e s u l t s f r o m Step 3 i n t o the expanded form

of Equation (4-32) t o ge t the equations f o r the transformed

moments a t the supports.

5. &vert the resu l t s of Step 4 t o obtain the ac tua l moments.

The deflect ion of each span may be found once the expressions

for the transformed support moments are available (3tsp 4 above). .

The transformed deflect ion f o r each span may then be determined using

Equation ( 4-18) with t h e appropriate t ransf ormed end moments and span

loading. Again the f i n a l s t e p is the inversion of t h e resul t ing

equation t o real time.

Since . t he spans a r e uniform, it i s only necessary t o determine

t h e response of a simply-supported span t o each of t hese end moments

and to each of t h e ex t e rna l loads. The r e s u l t a n t response f o r any

span, then, i s t h e superposi t ion of t h e ex te rna l load response f o r

that span p lus t h e responses due t o t h e appropr ia te end moments.

4.5.2 Two-Span Continuous Beam - Consider t h e two-span continuous beam shown i n Figure b.4. The

load i s uniform and of t h e same magnitude i n both spans. The v a r i a t i o n

of t h e load with time is taken t o be s inuso ida l with a maximum value,

Fo, and a frequency, w.

4.5.2.1 Moment a t t h e Center S u p ~ o r t -- Since t h e r e is only one i n t e r i o r support, only one moment

. . - expression needs t o be determined and consequently only one load tern

L- needs t o be evaluated. From Equation (4-26d) and not ing t h a t t h e loads

In both spans a r e equal, . . . .

2FOw ( s i n qL + s inh qL - - s i n qL cosh qL - s i n h qL cos q ~ )

where *ow P +l!J

Transforming Equation ( 4-31) and expanding gives,

' , Subst i tut ing t h i s value i n t o (4-35) and noting t h a t N = 2,

The value of k2 is given by Equation (4-26~) .

From Equation (4-32),

%(PI =

~ ~ w ( s i n qL + sinh qL - s i n qL cosh qL - sinh qL cos q ~ ) %(PI = (4-40)

( s i n qL cosh qL 7 cos qL sinh q ~ )

The inversion of Equation (4-40) may be done using the Inversion

4 formula,

where = sum of t h e residues,

The moment expression becomes, then,

a e ~t F ~ W -.- r --. .- .

$(x2 + $)(sin C O L ~ YL - cos /IL si* w) ( s in pL + s inh pL - s i n w' cosh pL - sinh pL cos pL) .dA .

- !bornson, 1 . To, h a a c e Transfornation ( ~ e v Jersey: renti ice- Hall, 1960),, p. 142,

I

Noting t h a t no pole e x i s t s a t $ = 0, the f i r s t s e t of poles

corresponds t o the roots of the 'second fac to r i n t h e denominator

of Equation (4-kl),

~ h e s e roots are,

? . = * i w ,

The residue f o r these poles i s obtained by evaluating

Xt e ~,w(s in p~ + sinh p~ - s i n p . ~ cash p~ - sinh cos p ~ ) V

5 = 2 p ( 2 ~ ) ( s i n uL cosh - cos pL sinh pL)

. . a t t h e above values.

. . The residue obtained is,

k ~ ~ ( s i n cp + sinh m - s i n m cosh m - sinh m cos a) s i n w t

9 = w(sin cg cosh cp - cos cp sinh cp) (4-43)'

where g = .-fie The second s e t of poles corresponds t o the roots of the t h i r d

f a c t o r of t h e denominator i n (4-41) . .

s i n pL cosh pL - cos.pL s i n . DL = 0

which may be rewritten a s

taa pL = tanh yC . ( 4-44 )

Let t h e roots of t h i s transcendental equation be designated as a me

Then there a l s o e x i s t imaginary roots, i a s ince m '

tan iam = i tanh a m

tanh ism = i tan am . Then, tan lam = t@ ism

reduces t o (4-44). The poles i n t h i s case are, therefore,

with

The residues a r e obtained from,

eXtkFgdSin + sinh a - sin a C O S ~ LIT' - sinh u~ cos' LJ~) ' % = . . ifiUL(X2 + 9) s i n p~ s i ~

. .

The residue f o r t h e second s e t of poles is,

i n am + sinh am - cos a sinh am m

- s i n a cosh a m . m

m

!he final expression f o r the moment a t t h e i n t e r i o r support point

is the sum of the residues. Therefore,.

kF0(sin rn + sinh m - sin cp cosh rp - sinh ip cos cp) ~ ( t ) = w(sin cp cosh cp - cos cp sinh cp)

e sin w t +

+ sinh am m

- cos a ' sinh a - sin a cosh a ) m m m * m = l a (w2 - a4k2/~4) sin a sinh am

m m m

. . 4.5.2.2 Deflection Equation ,

Having the expression for f$(p), the transformed deflection of

either span may be found using Equation (4-18). Considering the

second span and noting that

and, therefore,

Equation ( 4-18) becomes,

G(x,P) = 2 [sin q~ cosh q~ sinh qx + 2q S sin qL sinh qL

- cos qL sinh qL sin qx + sin qL sinh qL (cos qx - cosh qx)] +

+ 1 3 JL i(s,P)jsinh qL sin q ( ~ - 5 ) sin qx +

2q S sin qL sinh qL 0

. . For a s inuso ida l forcing function,

Subs t i t u t ing t h i s expression i n t o (4-47) and in t eg ra t ing gives

t h e f i n a l form of t h e transformed de f l ec t ion equation,

G(x,P) = 2q4~(p2 . + ur2)(cos q~ s inh q~ - s i n q~ cosh ;L)

0 {(cosh qL + cos qL - 2 c o s qL cosh q ~ ) s inh qx +

+ ( 2 cos qL cosh qL - cos qL - cosh qL) s i n qx:+

9 ( s i n qL + s inh qL - 2 s i n qL cosh qL) cos qx +

+ (2 cos qL s inh qL'- s i n qL - s inh q ~ ) cosh qx +

- 2(cos qL s inh qL - s i n qL cosh q ~ ) (4-48)

Following t h e same procedure used t o i n v e r t q(p), t h e def lec t ion

: is; . .

.,k2 4 x , t ) = 2

(0 c ( s i n x si* H x ) + 2w ~ ( c o s rp s inh co - s i n cp cosh Q)

+ c cos I. x + c cosh 2 x - c, 1 s i n w t + 2 3

a m i n x + m = l sa;(ui? - k2a4/L4) s i n a s inh am

m ' m

a a a ' 2

- ~ i n h $ x) + c6 cos - m x + c . C O S ~ - m L 7 L

The expressions f o r t h e coe f f i c i en t s ci are given i n Appendix B.

4.5.2.3 Numerical Example

Numerical values f o r the resu l t s derived i n t h e two preceding

sekt ions were determined f o r a beam with spans having t h e following

parameters :

I = 229 in. 4

A = 10.3 in. 2

L = 100 in.

The maximum value of the uniform load was taken a s

Fo = SO lbs./in.

with a forcing frequency of

.The &ts of t h e transcendental ,equation, (4-44), were solved

separately and a r e l i s t e d i n Table 4.1.

. . . Table 4.1

Roots, of Equation (4-44)

. . . Index, Root,

The resul tant moment i n t h e beam a t the center support is shown

In Figure 4.5 a s a function of time. Also included i s a p lo t of the

moment due t o the steady-state condition, along with the value of

t h e moment due t o a uniform s t a t i c load,

Figure 4.6 i s a p lo t of t h e second span'deflect ion a t three

one-millisecond time in tervals a f t e r applying the load, with t h e

, deflect ion a t threemil l i seconds being very near t h e maximum value.

, ' ' The s t a t i c deflect ion of t h e beam has a l s o been included f o r

comparison purposes. ' A typ ica l deflection-time relat ionship f o r a

point on the beam i s i l l u s t r a t e d by Figure 4.7 f o r x = 60 in. . $

Both t h e moment at the center support and t h e deflect ion of

t h e span were determined using the first seven terms i n the i n f i n i t e

s e r i e s p a r t of t h e solutions.

4.5.2.4 Discussion of the Results -- - . The expressions f o r the moment a t t h e center support and t h e

Span deflect ion a r e exact closed form solutions, since the inversions

of the transformed quant i t ies were exact. This was ve r i f i ed by

applying the n o d mode and t h e Laplace transform methods t o the

equivalent propped cant i lever beam problem, The process of

inversion a l s o yielded some of t h e na tura l frequencies of t h e two-

span beam..

Tke tmnscendental equation given by (4-44) corresponds t o t h e

. . f reguency equation f o r a propped, cant i lever beam, the roots of which

I .

. . )Harris, Cyri l M. and ,Crede, ~ h a r l e s E., Shock and Vibration --

Handbook, vole 1 ( ~ e w York: M c G m - H i l l , 1961), p. 7-14. . .

Resultant Moment

Figure 4.5 Resultant and Steady-State Moments f o r t h e Two-Span Continuous Beam - -- - at the Center Support Point -- -

F = 50 s i n 500 t

Span No,

. . Distance x, in.

o 10 20 40 50 60 100

Figure 4.6 Span Deflection f o r t h e Two-Span Continuous Beam a t -- -- One Millisecond Time In t e rva l s a f t e r t h e Loading i s - - -- - A ~ ~ l i e d

Figure 4.7 Time Variation of the Deflection at x . = 60 in. - -- - -

may be used t o determine t h e values of the na tura l frequencies of

t h e beam. Due t o t h e symmetry i n geometry and..loading of t h e t l

two-span beam about its center support, it becomes equivalent t o '

'.two back-to-back propped can t i l eve r beams, and s o the poles of

t h e inversion i n t e g r a l correspond t o t h e roots of the frequency

equation. However, t h i s symmetry condition has a l s o suppressed

any odd mode response, therefore not all of t h e na tu ra l frequencies

of t h e beam a r e obtained. The inversion, then, has t h e e f f e c t of

choosing discriminately only those values which a r e needed.

Equations (4-46) and ( 4-49) represent t h e complete so lu t ions

.for t h e undamped problem. In both cases, t h e so lu t ion cons is t s of

two pa r t s . The f i r s t pa r t , t h e s teady-state solution, is given by

the first term i n t h e equation and represents the long-time response

o f ' a damped system t o a periodic loading. The second par t , t h e

t r a n s i e n t solut ion, i s given by t h e second term i n t h e equation.

It r e f l e c t s t h e e f f e c t of t h e i n i t i a l conditions and t h e na tu ra l

frequencies of the system on its response. Since al l r e a l systems

have some damping, t h i s e f f e c t would eventually be suppressed leaving

only the .s teady-s ta te response. Damping, however, has l i t t l e e f f e c t

on t h e response durlng its i n i t i a l stages, so t h a t t r a n s i e n t e f f e c t s

are of importance even i n systems with damping. Therefore, t h e

solutions derived herein f o r an undamped beam nonetheless y i e l d

r e s u l t s appl icablc t o a damped beam.

The numerical r e s u l t s f o r the moment a t the support show the

significance of the t r ans ien t e f f e c t over t h e steady-state solution.

For t h i s p a r t i c u l a r case, there i s more than a 305 increase i n the

.moment over the steady-state value. The dynamic amplification f a c t o r

(the r a t i o of the maximum dynamic moment t o the s t a t i c moment) i s

about 1.5. The same, of course, i s t r u e f o r t h e deflect ions.

Ngures 4.5 and 4.7 'both show t h e e f f e c t of the i n i t i a l conditions,

with t h e values of t h e functions and t h e i r f i rs t derivat ives being

The numerical r e s u l t s were obtained using the first seven terms.

of t h e t r ans ien t solution. For t h e s e t of parameters used i n t h i s

example, no difference occurred i n t h e f i f t h s ign i f i can t d i g i t f o r

deflect ions when only three terms of t h e s e r i e s were used. Reducing

t h e number of terms from seven t o three i n evaluating t h e moments

resul ted i n a change of t h e fourth s ign i f i can t figure. Very accurate

results f o r t h i s example, then, were obtained using only th ree terms

of t h e ser ies .

. . THIS PAGE

WAS INTENTIONALLY

LEFTBLANK

u. Three-Span Continuous Beam . . - An example of a more general application' of the d i sc re te method

is t h e three-span beam shown i n Figure 4.8. The load i n t h i s case

- occurs only i n the t h i r d span and i s a l i n e a r function of the distance

from support Number 2. The t h i r d span,. then, w i l l have a response

tha t i s dependent on t h e externa l load and t h e support moment. The

, response of t h e first and second spans w i l l be functions only of

t h e support moments.

The response of the beam w i l l be obtained by f i r s t determining

t h e moments a t t h e supports. Once these moments a r e known, t h e beam

ma$ be divided i n t o spans and each span response determined using

. , t h e appropriate end moments and norms1 loads.

. . . . .

4.5.3.1 Sup~ort Moments . -

The load i n t h e t h i r d span may be expressed as,

. . ~ ( x , t ) = Fox s i n w t

where x is measured from support Number 2. . . . .

Its Laplace transform with respect t o t i h e is,

For 8 three-span beam ( N = 3), t h e transformed coe f f i c i en t s of ' t he .

f i n i t e s e r i e s ' f o r t h e load terms become,

~ ( x , t ) = Fox sin wt

\

Support No. 0 . . 1 ' 2 3

I . . Note: spans have the same E I

Figure 4.8 Three-Span Continuous - Beam

and :,(PI = 0*57735[4(P) - g2(P)]

Since only the t h i r d span is loaded, there i s only one load term,

6 (p), t o be calculated. Therefore, 2

Fo" ,(p) =*- ( q ~ s i n qL + qL sinh qL - 2 s i n qL sinh q ~ ) .

With q ( p ) = 0, - q2(p) = -G1(p)

where

0.57735 F0w : 1 ~ = 3 2 1

( q ~ s i n qL + qL sinh qL - 2 s i n qL sinh q ~ ) . . ( + w

Then, from Equation (.4-35),

Expansion of Equation ( 4-32) gives the following expressions

, for the transformed moments a t t h e supports:

- F ~ W ( ~ L s i n qL + qL sinh qL - 2 s i n qL sinh q ~ ) kl %(PI = ,

q3(p2 + ui?)(2k2 + k1)(2k2 - kl)

and

~ F ~ W ( ~ L s i n qL + qL sinh qL - 2 s i n qL sinh 6 ~ ) k2 $(PI =

q3 (p2 + u2) (a2 + kl) (a2 - kl) .

Using residue theory, these values of the transformed moments may be

inverted t o r e a l time.

Considering t h e moment a t support Number 1 first,

. (pL s i n pL + pL sinh pL - 2 s i n uL sinh G) k1 d l

The poles of t h e integrand are, . .

X a * i w ,

and tlse NULS of the factors ,

and

As i n t h e case o f t h e previous example, there is no pole a t X = 0.

The l a s t two equations i n expanded form a r e

sin.. q ~ ( 1 - 2 cos q ~ ) = s i n q ~ ( 1 - 2 cosh q ~ ) ( 4-52 i

and s i n q ~ ( 1 + 2 cosh q l ) = sinh qL(1 + 2 cos q l ) . (4-53)

Let t h e r e a l roots of these two transcendental equations be am and

b , respectively. Then a corresponding s e t of imaginary roots, 3

and i b a l s o exists . This may be ve r i f i ed by subs t i tu t ion i n t o .I '

' , t h e above expression.

For t h e f i r s t s e t of poles, t h e residue i s obtained by evaluating

t h e following expression a t X = 2 i w with = irp,rp where cp = L fi ,

(@ s i n pL + pL sinh pL - 2 s i n 'pL sinh UL) . The r e s u l t is,

3 - FoL (rp s i n rp + cp sinh cq - 2 s i n cp s inh ~ ) ( s i n h m - s i n m)

% = 3 s i n w t . cp [k(sin cp cosh rp - cos rp s inh cp12 - (s inh r p - s i n cp12 1

. .

h e second residue is detelmined from the expression,

A t . .

e F~I&(UL s i n UL + UL sinh pL - 2 s i n J. sinh @) % = 2 2 14 (X + w2)( 4 s i n p~ sinh p~ + cosh UL - cos p ~ ) ,'

. . . 2 2 Evaluation a t pL = f a wi th corresponding values of X = kam/l , m

- 2 k U 0 w ( a m s i n am + am sinh am - 2 s i n am sinh am a 2 : - -

a2[s2 - , 2 a 2 ~ 4 ) ( b s i n a si* a + cosh a - cos a ) . m m m . m m

.i

Finally, the last residue is determined from

- eXt~ouk(w s i n pL + pL sinh pL 2 s i n pL sinh UL) R = iu2(~2 + u2)(4 s i n pL sinh pL - cosh + cos pL) .

!

Evaluating t h i s expression at uL = f b with the corresponding values 3

2 2 of = j kb /L , gives, 3

+ b sinh b - 2 s i n b sinh b 3

b2

s i J b 3 sinhjb 3 - cosh b 3 + cos~! 3 ) sb ' L

The equation f o r the moment a t support Number 1 is the sum'of

. . the residues f o r each of t h e poles. In the case of t h e second and

third s e t of poles, the transcendental equations have an i n f i n i t e

number of roots s o t h a t the residues i n these cases a r e i n f i n i t e

series. The f inal form of the moment is therefore,

3 - FOL (sinh cp - s i n cp) . .~

--

5 .2 . v 4(sin tp cash cp - ~ O S rp sinh cp) - (s inh p - s i n cp) ] . .

, . (9, s i n rp + tp sinh cp - 2 s i n p s inh tp) s i n w t +

2 am

( 2 w 0 w ) s i n k - t + L~

2kLFow e

b2 3 (w2 - k 2b4 /~4 ) 3 (4 s i n b s i n h b - cosh b + cos b . .

3 3 3

b2 s i n b + sinh b - 2 s i n b s inh bj) s i n k

3 3 3 A t . L~

Examining Equations (4-49) and (4-50), we note t h a t the

denominators a r e the same f o r both transformed moment re la t ions and

that

The moment a t t h e second support may then be obtained by multiplying

(4-57) by t h e above f a c t o r evaluated a t t h e appropriate pole.

.Theref ore, . .

3 2FOL (O s i n rp + rp sinh 9 - 2 s i n cp sinh cp)

% = 3[ 2 cp 4(sin cp cosh rp - cos q s inh cp) - (sinh cp - s i n cp) 2

( s i n rp cosh cp - cos cp s inh cp) s i n w t +

kLFOw . .

e . .

m = l a:(ur2 - a s inh a + cosh a m m m - .OS

2

(am s i n a + am s inh am am m - 2 s i n am sinh a s i n k - t +

m) b2

. .

: / ~ ~ ) ( 4 s i n b si J

4.5.3.2 Span ~ e f l e c t i o n s

Using t h e expressions f o r the transformed moments a t the support

loca t ions and Equation (4-18), the equation f o r t h e t ransfomed

deflect ion of each of the three spans may be obtained. Inversion of

these r e s u l t s gives the ac tua l deflect ion relationships. -

Equation (4-18) represents t h e transformed response of a simply-

supported span which i s subjected t o end moments and a normal load. . .

The response of the continuous system is therefore 'determined span- . .

by-span ra the r than a s an ent i ty . Since spans adjacent a t a support

, point have the same end moments (except f o r a change i n s ign) , t h e

calculat ions may be simplified.by first determining the response of

a simple span t o each of t h e end moments and t o each of the normal

. loads separately and then using t h e superposition pr inc ip le t o obtain

the t o t a l response of each span. . . . -

To determine t h e response f o r ~ ( t ) consider t h i s moment t o be

ac t ing a t the r igh t end of the f i rs t span. The transformed deflect ion

expression is, f mm Equation (4-l8).,

. . %(P) ( s i n qL sinh ax - sinh qL s i n ax) + % P ) = 7 sin qL s i n h q11 ( 4-59)

2q S

me t ransfomed response due t o % act ing on the righ* end of the

second span, may be obtained by replacing q ( p ) above by $(p).

merefore,

. (sin q~ si**~ ax - st* s in t q x ~ G,(x,P) = - s i n qL sinh qL 0 (4-60) 2q2s

Only the t h i r d span i s loaded i n t h i s part icula ' r problem, and i t s

deflect ion due t o t h e l i n e a r l y varying normal load i s given by

- w = 1

3 /I :rw [si* qL s i n q ( ~ - 5) s i n qx + 2q S s i n qL sinh qL 0 p + w

- s i n q~ sinh q ( ~ - S) sinh qx] dS +

x FoSw +'/ p2 [ s i ~ q(x - I) - s i n q(x - 511 d s .

2 q 3 ~ o + ' w

Integration of t h i s equation gives,

- w = L 2q5s(p2 + w2) s i n q~ sinh q~ [ q ~ ( s i n h qL sin qx + s i n qL sinh qx) +

- 2qx s i n qL sinh qL . I Subst i tut ing t h e expressions f o r q ( p ) and $(p) derived i n

the preceding sect ion i n t o the first two deflect ion re la t ions , r e s u l t s

:. s sin qL + sinh q ~ ) - 2 s i n qL sinh qL G,(x,P) =

2q5s(p2 + w2)(2k2 + k1)(2k2 - kl) s i n qL sinh qL a

a ( s i n qL sinh qx - sinh qL s i n qx) , and

. . sin qL + sinh q ~ ) - 2 in qL sinh q G,(x,P) = 2 a

2q5s(p2 + w )(a2 + k1)(2k2 kl) s i n qL sinh qL

( s i n qL sinh qx - sinh qL s i n qx) , respectively, where

and = s i n qL cosh qL - cos qL sinh qL . k2

The inversion in tegra l s are, respectively,

X t 1 i - e sin UL sinh ux - sinh UL s i n UX)

w,(x, t) = - 2ni - ~U'S (1' + w2)(2k2 + k,)(2k2 - k$ s i n uL sinh uL

I [ u ~ ( s i n UL + sinh 111) - 2 s i n UL sinh UL d l , (4-62) .

X t v+i= e 2 ~ , u k ~ ( s i n UL sinh p~ - sinh UL s i n UX)

- i w *'s(x2 + w2)(2k2 + kl) (2k2 - kl) s i n UL sinh uL '

sin uL + sinh UL) - 2 s i n uL sinh ul d~ , I ( 4-63) and

X t e Few 2u4s(12 + u2) s i n p~ sinh a

[ ~ ( s i n h s i n ux + s i n @ sinh px) - 2x s i n @ sinh uL d l . I The'residues f o r the deflect ions due t o t h e end moment a r e

- obtained f o r t h e poles corresponding t o the roots of

and s i n uL sinh pL = 0 . The poles corresponding t o t h e first 'three expressions were determined

. i n the previous sect ion f o r t h e inversion of the moment expressions.

The last expression has poles a t L.

. , UL = nn

arid UL = inn .

The form o f t h e residue f o r the f i r s t s e t o i p o l e s f o r wl is,

, . - e X t sin UL sinh ux - sinh UL s i n UX)

5 = i k5sX(2k2 + kl)(2k2 - kl ) s i n uL sfnh uL

[ a ( s i n a + sinh u ~ ) - 2 s i n UL sinh UL . For the second s e t of poles the residue is,

I e'tk~ow[U~(sin VL + sinh UL) - 2 s i n UL sinh UL]

R2 L

2 p 4 i s ~ ( 1 2 + w2)(cosh UL - cos (II + 4 s i n sinh

j s i n uL sinh ux - sinh uL s i n w) s i n uL sinh pL

The residue f o r the t h i r d s e t is,

. . X t - e mow uL(sin UL + sinh a) - 2 s i n UL sinh - R3 - a4i~~(12 I ;,,cos fi - cos, + , s i n si, J,)

. ( s i n uL sinh ux - sinh uL s i n ux) s i n uL sinh pL . .

And f i n a l l y f o r the last s e t ,

- e'tkFgw[u~(sin uL + sinh &) - 2 s i n sinh ull R4 =

p4i~~(12 + w2)(sin pL cosh uL + cos uL sinh uL)

(s inh uL - s i n uL)(sin uL sinh ux - sinh uL s i n ux)

b ( s i n . u ~ cosh s - cos UL sinh U L ) ~ - (sib UL - s i n 2 . ]

The residues f o r w (x , t ) have the same denominators a s those 2

of w (x , t ) with t h e only changes occurring i n the numerators. 1

For v2(x,t) the re a r e only two s e t s of poles involved

. corresponding t o t h e l a s t . two fac to r s i n the denominator of (4-64). I *

.. . . ..- . The residue forms are,

eXt~6wk(sinh~~ sin ux + sin p~ sinh ux) - 2x sin UL iinh % L = 4u4sX sin pL sinh uI.

I .I

and

eAbourrl(sinh UL sin ux + sin UL sinh UX) - 2~ sin a sinh ~ LI % =

u3is~(h2 + w2)(sin uL cosh uL + cos UL si* u ~ ) .

, The final ebuations for the deflections are

5 m - FOL ~~(tq)~~(cp)(sin cp sinh x - sinh m sin x) wl(x,t) = sin urt +

2 . a m sin k - t L~

a m '(sin am si* x - sinh a sin $ x ) + m

3 FO&L Jl b b2 sin k +- t

j-1 b4s(w2 - k2a L J:

b . (sin bJ sinh $ x - .inh II sin 2 :) + 3

3 2FOt&L sin x 2 2

- 3 3 2 n n

2 sin k - L n=l 3n n ~ ( w - k n4n4/~4)(-1)n L~

5 m cP FOL Jl(q)k2(sin m sinh x - sinh cu sin - x) L

w,(x,t) = sin urt + 05s~3(m)~4(rol . ,

a rn a s i n am sinh x - sinh o. s i n -

m L

3 b2 3 s i n k - t 0

L2

b (sin b ' sinh $ x - sinh b s i n 2 x) +

.3 3

3 2F0&L s i n x 2 2 n n s i n k - t j ( 4-66)

, + - k2n\4/~4) L~

and

s i n m sinh f x + L sinh r~ s i n f x - 2x s i n p sinh Q

2rp4s s i n m sinh rp e

. sin w t +

3 nn 2 2 2kF0& s i n i;-- x

- 3 3 n n

2 s i n k - t . . . $

(4-67) n=l n n s (w2 - k n4n4/~4) cos nn ,

. . . .

The expressions f o r the coeff ic ients J~(Y) a r e l i s t e d i n Appendix B.

The deflect ion f o r each of the spans may. be obtained f r o m the

previous deflect ion re la t ions using the pr inc ip le of superposition

along with t h e appropriate change i n the s p a t i a l variable. These

deflect ions are

./ . ws3 ( x , t ) = - w*(L. - x , t ) + w&x,t)

f o r t h e first, second and t h i r d ' spans, respect ively.

40503.3 Numerical Example

I Take t h e p rope r t i e s of t h e spans f o r t h i s example t o be t h e

same a s those of t h e previous two span beam problem. For . the

ex t e rna l load, assume

Fo = 5 lbs./in.

and t h e frequency of t h e forc ing funct ion t o be

The first f i v e roo ts of t h e two t ranscendental equations, (4-52)

. . and ( 4 0 5 3 ) ~ a r e given i n Tables 4.2 and 4.3, respect ively. . N g u r e 4.9 i l l u s t r a t e s t h e va r i a t i on of t h e moments i n t h e beam

a t t h e i n t e r i o r support points . The s teady-s ta te 'values are a l s o

included a s wel l as t h e s t a t i c moments corresponding t o t he maximum

loading condition.

The corresaonding def leot iono for the.beam a t one niillisecond

t i m e i n t e r v a l s a f t e r t h e load is appl ied a r e shown i n Figure 4.10.

The s t a t i c de f l ec t ion corresponding t o t h e maximum loading condi t ion

has a l s o been included f o r comparison.

Table 4.2

Firs t Five Roots of Equation (4-52)

Index, Root,

a = u*L m

. Table 4,3

First Five Roots of Equation (4-53)

Index, R o o t ,

4

I .

Steady-State .

Resultant Mcrnent -

Figure 4.9 S teaciy -State and Resultant Moments i n t h e Three-Span Continuous Beam - -- - a t Support Number 1 7 -

Resultant Moment

Steady-State Moment

Time (sees.) -

'

Figure 4.10 Steady-State and Resultant Moments i n t h e Three-span Continuous - -- Beam a t Support Number 2

' -- -

o 20 40 60 80 o 20 40 60 80 0 2 0 ~ 6 0 8 0 ~ 0 o

Span Distance ( i n . )

F igure 4.11 Def lec t ions -- of t h e !Three-span Continuous Beam -

Discussion of t he Results -- .;I

As i n t h e two-span beam example, t h e poles used i n t h e invers ion

i n t e g r a l a r e r e l a t e d t o t h e n a t u r a l frequencies of t h e beam. The

.po les f o r t h e invers ion of t h e def lec t ions , Equations (4-62) and

(4-36) apparent ly r e s u l t i n a complete s e t of frequencies f o r t h e

system. The frequencies were ca lcu la ted using these poles and ordered

t o ob ta in t h e first f i v e frequencies. Table 4.4 l is ts these fre-

quencies along with t h e corresponding values calculated.from

6 information i n t h e l i t e r a t u r e .

The poles f o r i nve r t i ng t h e moments i n t h e beam a t t h e i n t e r i o r

support po in t s do not, however, provide a l l t h e beam frequencies.

The expressions f o r t h e transformed def lec t ions , Equations (4-62),

(4-63) and (4-64), have an add i t i ona l set of poles no t found i n t h e

transformed moment expression, Equation (4-51). This set of poles

corresponds t o those modes composed of h a l f s i n e shapes. The support

loca t ions , then, f o r these de f l ec t ion configurat ions occur a t

i n f l e c t i o n poin ts where t h e moments vanish. Hence t h e frequency

equation f o r these modes does no t appear i n t h e invers ion expression.

Both of t h e moment diagrams show a subs t an t i a l d i f fe rence between

t h e r e s u l t a n t so lu t ion and t h e s teady-s ta te solution. A t support

Number 1, t h e r e su l t an t moment i s over twice the s teady-s ta te value.

Table 4.4

. Natuml Frequencies f o r a Uniform Continuous Beam of Three Equal Spans

Natural Frequericies ( rad./sec . ) I From Laplace . . From Ref. 5

Invers ion Poles

A t support Number 2, there i s a 20$ increase, Another in te res t ing

result is t h a t the dynamic load f a c t o r i s not the same f o r both of

t h e support moments. For the first. support it is nearly 3, while

for t h e second it is 1.45,

The moment diagram of Figure 4.9 a l s o indicates a wave e f fec t

i n t h e beam by the time delay a t the beginning of the plot . A

s ign i f i can t time in te rva l passes before the e f fec t of the load i n -

the t h i r d span reaches t h e first support point. This same e f f e c t

is i l l u s t r a t e d i n the deflect ion diagrams of Figure 4.11 by the

progressive shape changes i n the beam with increasing time. The

wave act ion i s even more evident i n Figure 4,12 where deflect ions

i n the first span a t x = 70 , in . and x = 30 in . are p lo t ted against

time. Since t h e disturbance is travel ing from r igh t t o l e f t , it

reaches x = 70 in . first passing on then t o x = 30 in.

Good convergence was achieved i n t h i s problem using f ive . te rms

of t h e i n f i n i t e ser ies . The l a r g e s t difference between.the t h i r d

and t h e f i f t h term of t h e s e r i e s occurred i n the fourth s igni f icant

figure f o r the moments and the deflections.

APPENDIX - A:

Values of G2j and Gbj i n Equation (3-24)' ./

6" g 2 ( x ) x j ( ~ ) - = J L 1 - - k + L x 2 jn 3 2 - & x3) s i n , x

L L ,J B ~ ( x ) x ~ ( x ) - = 6 (- x + x3) s i n L x

- -- ~3 . L3 3 COS jn = - ( - 1 1 ~

(,In> ( j d 3

. <

Appendix - B: ; Values .of the coefficients J ~ ( ~ ) in ~~uations (4-65) and

(4-661). '

J1(y) = y(sin y +*sinh y) - 2 sin y sinh y J2(y) = sinh y - sin y J3(y) = h(sin y cosh y - cos y sinh y)2 - (sinh y - sin y)2 J4(y) = sin y sinh y

J~(Y) = cosh y - cos y + 4 sin y sinh y . .

J&Y) = cos y - cosh y + 4 sin y sinh y