M17_CHOPRA_Dinamica_Systems With Distributed Mass and Elasticity

17 Systems with Distributed Mass and Elasticity

PREVIEW So far in this book we have focused on discretized systems, typically with lumped masses; such a system is an assemblage of rigid elements having mass (e.g., the floor diaphragms of a multistory building) and massless elements that are flexible (e.g., the beams and columns of a buil building) ding).. A major part of this book is dev devoted oted to lumped-mass lumped-mass discretized discretized systems, for two reasons. First First,, such systems can effectively effectively idealize idealize many classes of struct structures, ures, especially multistory buildings. Second, effective methods that are ideal for computer implementation are available to solve the system of ordinary differential equations governing the motion of such systems. How Howev ever, er, a lumpe lumped-mass d-mass idealization idealization,, altho although ugh applicable, applicable, is not a natural approach for certain types of structures, such as a bridge (Fig. 2.1.2e), a chimney (Fig. 2.1.2f), an arch dam (Fig. 1.10.2), or a nuclear containment structure (Fig. 1.10.1). In this chapter we formulate the structural dynamics problem for one-dimensional systems with distributed mass, such as a beam or a tower, and solutions are presented for simple systems (e.g., a uniform beam and a uniform tower). tower). The solutions solutions presented for these simple cases provide insight into the dynamics of distri distribute buted-mass d-mass systems that have an infinite number of DOFs and how they differ from lumped-mass systems with a finite number of DOFs. The chapter ends with a discussion of why this infinite-DOF approach is not feasible for practical systems, pointing to the need for discretized methods for distribute distri buted-mass d-mass systems. systems. The results presented presented for the simpl simplee system systemss provi provide de the exac exactt solution against which results from discretized methods can be compared (Chapter 18). 697

698

Systems Syst ems with Dist Distribut ributed ed Mass and Elas Elastici ticity ty

Chap.. 17 Chap

17.1 EQU EQUA ATION OF UNDA UNDAMPED MPED MOTION: APPLIED FORCES In this section we develop the equation governing the transverse vibration of a straight beam without without damping subjected subjected to exte external rnal force. Figur Figuree 17.1.1a shows such a beam I (( x ) and mass m ( x x )) per unit length, both of which may vary with with flexural rigidity E I position x . The exter external nal forces forces p ( x , t t )), which may vary with position and time, cause x ,, t t )) (Fig motion of the beam described by the transverse displacement u ( x (Fig.. 17.1.1b). 17.1.1b). The equation of motion to be developed will be valid for support conditions other than the simple supports shown and for beams with intermediate supports. p( x,t x,t )

p dx V

), EI ( x m( x x ), x ) x

•

M +

M

L

•

+ V +

(a) f I = m dx

u( x,t x,t )

u

∂2u ∂t 2

∂ M dx ∂ x

∂V dx ∂ x

dx

x

(b)

(c)

distributed mass mass and elasticity: (a) beam and applied force; Figure 17.1.1 System with distributed (b) displacement; (c) forces on element.

The system has an infinite number of DOFs because its mass is distributed. Therefore, we consider a differential element of the beam, isolated by two adjoining sections. The forces on the element are shown in Fig. 17.1.1c, where an inertia force has been in x ,, t ) is the transverse shear cluded following D’Alembert’s principle (Section 1.5.2); V ( x t )) is the bending moment. Equilibrium of forces in the y -direction gives force and M ( x , t 2

∂ V

∂ u p−m = ∂ x ∂ t 2

(17.1.1)

Without the inertia force this equation is the familiar relation between the shear force in a beam and exte external rnal transverse transverse force. The inertia inertia force modifies the external external force in recognition of the dynamics of the problem. If the inertial moment associated with angular acceleration of the element is neglected, rotational equilibrium of the element gives the standard relation

== ∂∂M x

V

(17.1.2)

We now use Eqs. (17.1.1) and (17.1.2) to write the equation governing the trans x ,, t t )). Wit verse displacement u ( x With h shear deformation deformation neglected, neglected, the momen moment–cur t–curvatu vature re

Sec.. 17.2 Sec 17.2

699

Equati Equ ation on of Und Undamp amped ed Mot Motion ion:: Support Support Exc Excita itatio tion n

relation is M

∂ 2u

= E I ∂ x

2

(17.1.3)

Substituting Eqs. (17.1.3) and (17.1.2) into Eq. (17.1.1) gives m ( x x ))

∂ 2u ∂ t 2

∂2

+ ∂ x

2



E I I (( x )

∂ 2u ∂ x 2

=

p( x x ,, t t ))

(17.1.4)

This is the partial differential equation governing the motion u u(( x , t t )) of the beam subjected x ,, t ). To obtain a unique solution to this equation, we must to external dynamic forces p( x specify two boundary conditions at each end of the beam and the initial displacement u ( x x ,, 0) and initial velocity u ( x x ,, 0).

˙

17.2 EQUA EQUATION TION OF UNDAMPED MOTION: SUPPORT EXCITATION EXCITATION Consider two simpl Consider simplee cases: a canti cantilev lever er beam subjec subjected ted to horiz horizontal ontal base motion (Fig. 17.2.1a) or a beam with multiple supports subjected to identical motion in the vertical direction (Fig. 17.2.1b). The total displacement of the beam is u t ( x , t t ))

=u

t )) g (t

+ u( x , t t ))

(17.2.1)

where the bea where beam m dis displa placem cement ent u ( x , t measured ed rel relati ative ve to the sup suppor portt mot motion ion u g (t t )), measur t )), results from the deformations of the beam. •

u( x,t x,t ) u( x,t x,t ) L

ug(t )

x x •

ug(t )

(b)

(a) Cantilever beam subjected subjected to base excitation; excitation; (b) continu continuous ous beam Figure 17.2.1 (a) Cantilever subjected to identical motion at all supports.

For these simple cases of a beam excited by support motion the derivation of the equation of motion is only slightly different than that for applied forces. Recognizing that the inertia forces are now related to the total accelerations and that external forces p ( x x ,, t t )) do not exist, Eq. (17.1.1) becomes

700

Systems Syst ems with Dist Distribut ributed ed Mass and Elas Elastici ticity ty 2 t

∂ V

2

2

d u ∂ u ∂ u m m m = − = − − d t ∂ x ∂ t ∂ t 2

2

g

Chap.. 17 Chap

2

(17.2.2)

wherein Eq. (17.2.1) has been used to obtain the second half of the equation. Substituting Eqs. (17.1.3) and (17.1.2) into Eq. (17.2.2) gives m ( x x ))

∂ 2u ∂ t 2

∂2

+ ∂ x

2



E I I (( x x ))

∂ 2u ∂ x 2

=−

m ( x x ))u g (t )

¨

(17.2.3)

By comparing Eqs. (17.2.3) and (17.1.4), it is clear that the deformation response u ( x x ,, t t )) of the beam to support acceleration u g (t t )) will be identical to the response of the system m ( x x ))u g (t ). The support excitation can with stationary supports due to external forces therefore be replaced by effective forces (Fig. 17.2.2):

¨

=−

peff ( x x ,, t t ))

¨

= −m ( x x ))u¨

t )) g (t

u¨g

(17.2.4) u¨g

u¨g

m( x x )

=

x

x

peff ( x,t x,t ) = −m( x x )u¨ ¨ g(t )

=

peff ( x,t x,t ) = −m( x x )u¨ g(t)

u¨ g(t )

Stationary base

Stationary supports

(a)

(b) Figure 17.2.2

Effective forces peff ( x , t t )).

This formulation can be generalized to include the possibility of different motions of the various supports supports of a struct structure. ure. Such multiple multiple suppor supportt exci excitatio tation n may exist in sev several eral practical situations (Section 9.7), but is not included in this chapter because it is usually not possible to analyze such practical problems as infinite-DOF systems. They are usually discretized by the finite element method (Chapter 18) and analyzed by extensions of the procedures of Section 13.5.

17.3 NA NATURAL TURAL VIBRATION VIBRATION FREQUENCIES AND MODES For the case of free vibration, Eqs. (17.1.4) and (17.2.3) become m ( x )

∂ 2u ∂ t 2

∂2

+ ∂ x

2



E I I (( x )

∂ 2u ∂ x 2

=

0

(17.3.1)

Sec.. 17.3 Sec 17.3

701

Natura Nat urall Vib Vibrat ration ion Freq requen uencie cies s and Mod Modes es

We attempt a solution of the form u ( x x ,, t )

= φ( x x ))q (t )

(17.3.2)

Then ∂ 2u ∂ t 2

∂ 2u

= φ( x x ))q¨ (t t ))

∂ x 2

= φ ( x x ))q (t t ))

(17.3.3)

t )) where overdots denote a time derivative and primes denote an x derivative; thus q (t 2 2  2 2 and φ ( x ) Substitutin tuting g Eq. (17.3 (17.3.3) .3) in Eq. (17.3.1) (17.3.1) dq /d t , q (t ) d q /dt , and φ d φ/d x . Substi leads to

¨ =

˙ ≡

=

¨ + q (t ) E I I (( x x ))φ( x x ))  = 0



m ( x x ))φ( x )q (t t ))



which, when divided by m m(( x x ))φ( x )q (t t )), becomes

−q¨ (t ) = [ E I I (( x x ))φ( x x ))]

(17.3.4)

m ( x x ))φ( x )

q (t t ))

The expression on the left is a function of t t only and the one on the right depends only on x . For Eq. (17.3.4) to be valid for all values of x and t , the two expressions must therefore be constant, say ω 2 . Thus the parti partial al differential differential equation equation (17.3 (17.3.1) .1) becomes two ordina ordinary ry differential equations, one governing the time function q (t t )) and the other governing the x )): spatial function φ function φ( x 2

q

¨ + ω q = 0  E I I (( x x ))φ ( x ) − ω m ( x x ))φ( x x )) = 0



(17.3.5)

2



(17.3.6)

Equation (17.3.5) has the same form as the equation governing free vibration of an SDF I (( x x )) and system with natural frequency ω . For any given stiffness and mass functions, E I m ( x x )), respectively, there is an infinite set of frequencies ω and associated modes φ( x ) that satisfy the eigenvalue problem defined by Eq. (17.3.6) and the support (boundary) conditions for the beam. Forr th Fo thee spe speci cial al ca case se of a un unif ifor orm m be beam am,, E I and m ( x ) m , and and Eq. (17 (17.3. .3.6) 6) I (( x ) E I and becomes

=

E I φIV ( x x ))

2

− ω m φ( x ) = 0

or

φIV ( x )

=

4

− β φ( x x )) = 0

(17.3.7)

where β4

2

= ω E mI

(17.3.8)

The general solution of Eq. (17.3.7) is (see Derivation 17.1) φ( x )

= C sin β x + + C cos β x + + C sinh β x + + C cosh β x 1

2

3

4

(17.3.9)

This solution contains four unknown constants, C 1 , C 2 , C 3 , and C 4 , and the eigenvalue parameter β . Application of the four boundary conditions for a single-span beam, two at each end of the beam, will provide a solution for β and hence for the natural frequency ω [from Eq. (17.3.8)] and for three constants in terms of the fourth, resulting in the natural mode of

702


Chap.. 17 Chap

Eq. (17.3.9). This procedure is illustrated next by two examples: a simply supported beam and a cantilever beam. Results are also available for other boundary conditions but are not included in this book.

17.3.1 Uniform Simply Simply Supported Supported Beam The natural frequencies and modes of vibration of a uniform beam simply supported at both bot h ends are det determ ermine ined d next. At x 0 and x L , the displacement and bending moment are zero. Thus, using Eqs. (17.3.2), (17.1.3), and (17.3.9) at x 0 gives

=

=

= =

= 0 ⇒ φ(0) = 0 ⇒ C + C = 0 M(0, t t )) = 0 ⇒ E I φ (0) = 0 ⇒ β (−C + C ) = 0 These two equations give C = C = 0 and the general solution reduces to φ( x x )) = C sin β x + + C sinh β x = L , Then at x = u ( L , t t )) = 0 ⇒ φ( L L)) = 0 ⇒ C sin β L + C sinh β L = 0 M( L , t t )) = 0 ⇒ E I φ ( L L)) = 0 ⇒ β (−C sin β L + C sinh β L L)) = 0 u (0, t t ))

2

4

2

2

2

4

(17.3.10a) (17.3.10b)

4

1

3

1

3

2

1

3

(17.3.11)

(17.3.12a) (17.3. (17 .3.12b 12b))

Adding these two equations after dropping β 2 from the second equation gives C 3 sinh β L

= 0

Since sinh β L cannot be zero (otherwise, ω will be zero, a trivial solution implying no vibration at all), so C 3 must be zero. This leads to the frequency equation: C 1 sin β L

= 0

(17.3.13)

= 0, which gives φ( x ) = 0, a trivial solution.

This equation can be satisfied by selecting C 1 Therefore, sin β L must be zero, from which β L

= nπ

= 1, 2, 3, . . .

n

(17.3.14)

Equation (17.3.8) then gives the natural vibration frequencies: ωn

=

n2π 2 L 2



E I

= 1, 2, 3, . . .

n

m

(17.3.15)

The natural vibration mode corresponding to ω n is obtained by substituting Eq. (17.3.14) in Eq. (17.3.11) with C 3 0 as determined earlier:

=

φn ( x )

= C sin n Lπ x 1

(17.3.16)

Sec.. 17.3 Sec 17.3

703


φ m, EI

•

x

L

•

/ x ) = sin(π x L) φ1( x

ω 1 =

π2

EI m

L2

/ x ) = sin(2π x L) φ2( x

ω 2 =

4π2 L2

EI m

/ x ) = sin(3π x L) φ3( x

ω 3 =

9π2 EI m L2 Figure 17.3.1 Natural vibration modes and frequencies of uniform simply supported beams.

• • •

The value of C 1 will make the maximum value of φ of φn ( x ) equal to unity. C 1 is arbitrary; C 1 These natural modes are shown in Fig. 17.3.1. For a simply supported uniform beam, we have determined an infinite series of modes each with its vibration frequency. Equations (17.3.15) and (17.3.16) and Fig. 17.3.1 π 2 ( E I I / /m L 4 )1/2 . tell us that the first mode is a half sine wave and that its frequency ω 1 The second mode is a complete sine wave with frequency ω2 4ω1 ; the third is one and a half sine waves with frequency ω3 9w1 ; and so on.

=

=

=

=

17.3.2 Uniform Cantilever Cantilever Beam Beam In this section the natural vibration frequencies and modes of a uniform cantilever beam are determined. determined. At the clamped end, x 0, the displacement displacement and slope are zero. zero. Thus Eq. (17.3.9) gives

=

= 0 ⇒ φ(0) = 0 ⇒ C + C = 0 ⇒ C = −C (17.3.17a) u  (0, t t )) = 0 ⇒ φ (0) = 0 ⇒ β( β(C C + C ) = 0 ⇒ C = −C (17.3.17b) At the free end, x = L , of the cantilever the bending moment and shear are both zero. u (0, t t ))

2

4

1

4

3

Thus, from Eqs. (17.3.9) and after using Eq. (17.3.17), we obtain

2

3

1

704


Chap.. 17 Chap

= 0 ⇒ E I φ ( L L)) = 0 ⇒ C (sin β L + sinh β L L)) + C (cos β L + cosh β L L)) = 0  L)) = 0 V ( L , t ) = 0 ⇒ E I φ ( L ⇒ C (cos β L + cosh β L L)) + C (− sin β L + sinh β L L)) = 0

M( L , t )

1

2

1

2

(17.3.18a)

(17.3. (17 .3.18b 18b))

Rewriting Eqs. (17.3.18a) and (17.3.18b) in matrix form yields



sin β L cos β L

+ sinh β L cos β L + cosh β L + cosh β L − sin β L + sinh β L

  =   C 1 C 2

0 0

(17.3.19)

Equation (17.3.19) can be satisfied by selecting both C 1 and C 2 equal to zero, but this C 1 and C 2 to be would woul d give a trivial solution solution of no vibra vibration tion at all. For either either or both of C nonzero, the coefficient matrix in Eq. (17.3.19) must be singular (i.e., its determinant must be zero). This leads to the frequency equation: 1 cos β L cosh β L 0 (17.3.20)

+

=

No simple solution is available for β β L L , so Eq. (17.3.20) is solved numerically to obtain βn L

= 1.8751, 4.6941, 7.8548, and 10.996 (17.3.21) for n = 1, 2, 3, and 4. For n > 4, β L  (2n − 1)π/ 2. Equat Equation ion (17.3.8) (17.3.8) then giv gives es the n

first four natural frequencies: ω1

=

3.516 L 2



E I

ω2

m

=

22.03 L 2



E I m

ω3

=

61.70 L 2



E I

ω4

=

m



E I

120.9 L 2

m

(17.3.22) β n L , the natural vibration mode is Corresponding to each value of β

φn ( x x ))

= C

1



cosh βn L



β L − cos β x −− sinh β L ++ cos − sin β x x )) cosh β x − (sinh β x − sin β L n

n

n

n

n

n

n

(17.3.23) x •

x ) φ3( x

x ) φ2( x L

x ) φ1( x

m, EI

•

x ) φ4( x

• • •

φ

ω 1 =

3.516 EI m L2

Figure 17.3.2

ω 2 =

22.03 EI m L2

ω 3 =

61.70 EI m L2

ω 4 =

120.9 EI m L2

Natural vibration modes and frequencies of uniform cantilever beams.

Sec.. 17.3 Sec 17.3

705


where C 1 is an arbitrary constant. To arrive at Eq. (17.3.23), C 2 is expressed in terms of C C 1 from Eq. (17.3.18a) and substituted in the general solution, Eq. (17.3.9), and Eq. (17.3.17) is used. The first four natural vibration modes are shown in Fig. 17.3.2. Derivation 17.1 The solution of the fourth-order ordinary differential equation (17.3.7) is of the form φ( x )

= Aeax

(a)

where A is an arbit arbitrary rary constant. constant. Substi Substitutin tuting g for φ( x ) and its fourth derivative in Eq. (a) yields the characteristic equation a4

− β 4 = 0 or (a2 − β 2)()(aa 2 + β 2 ) = 0 which gives a = ±β and a = ±i β . Thus the general solution of Eq. (17.3.7) is x )) = A 1 ei β x + A 2 e−i β x + A 3 eβ x + A 4 e−β x φ( x

(b)

(c)

Equation (c) can be rewritten as Eq. (17.3.9) because e±β x

= cosh β x ± ± sinh β x

e±i β x

= cos β x ± ± i sin β x

(d)

17.3.3 Shear Deformation Deformation and Rotational Rotational Inertia In the preceding derivation of the equation of motion for the transverse vibration of a beam, the inertial moment associated with rotation of the beam sections was ignored in Eq. (17.1.2), and only the deflection associated with bending stress in the beam was included clude d in Eq. (17.1.3), (17.1.3), thus ignoring the deflection deflection due to shear stress stress in the beam. The analysis of beam vibration, including both the effects of rotational inertia and shear deformation, is called the Timoshenko beam theory . The following equation governs such free vibration of a uniform beam with m ( x ) m and E I I (( x ) E I :

=

=

m

∂ 2u ∂ t 2

∂ 4u

2

+ E I ∂ x − mr 4

E

∂ 4u

κG

∂ x 2 ∂ t 2

+  1

== √

m 2 r 2 ∂ 4 u

+ κ G A ∂ t = 0 4

(17.3.24)

I / I / A is the radius of gyration of the beam cross where G is the modulus of rigidity, r section, A is th thee ar area ea of cr cross oss se sect ctio ion, n, and and κ is a con consta stant nt tha thatt dep depend endss on the cro cross-s ss-sect ection ional al shape and accounts for the nonuniform distribution of shear stress across the section. The constant κ is derived for various cross-sectional shapes in textbooks on solid mechanics 9 (e.g., κ is 56 for rectangular cross section and 10 for circular cross section). Consider a beam with both ends simply supported. Assuming a solution of the form sin (n π x / L u ( x x ,, t ) C sin L)) sin ωn t , which satisfies the necessary end conditions, the frequency equation is obtained. Denoting a natural frequency of the beam by ω n if shear and

=

706


Chap.. 17 Chap

rotational inertia effects are included, by ω n if these effects are neglected [Eq. (17.3.15)], ωn /ωn , this frequency equation can be written as and defining n

=

(1

−

2n )

−

2n

   + +   n π r

2

1

L

E

4n

κG

n π r L

4

E κG

=0

(17.3.25)

r / n π r r / L))4 term may be dropped and Eq. (17.3.25) If it is assumed that n r / L << 1, the ( (n / L reduces to ωn

= ω

n

1

 + 1

r / L))2 (1 (n π r / L

(17.3.26)

E /κ G) + E /κG

which implies that ωn < ωn . The correction correction due to rotational rotational inertia inertia is represented represented by 2 the term ( (n n π r r / / L L)) in the denominator, whereas the shear deformation correction appears 2 r / / L L)) ( E /κ /κ G ). Thu E /κ /κG G times as (n π r Thuss the correct correction ion term for shear deforma deformatio tion n is E larger larg er than the rotat rotational ional inertia inertia correction correction term. For steel beams of rectangular rectangular cross section, E approximately 3.2. Values of n E /κ /κG G is approximately ωn /ωn are plotted in Fig. 17.3.3 /κ G ; using the solution of Eq. (17.3.25), a quadratic equation in  2n , for three values of E /κ these results are valid for all natural frequencies because n is included in the abscissa scale.. Simi scale Similar lar results are presented presented in Fig. 17.3.4 17.3.4 for the first five natural frequencie frequenciess E /κ /κ G of a beam with E 3.2. Also included included is the appro approximat ximatee value of the frequ frequency ency

=

=

1.0

0.8

n

ω

0.6

/

n ′ ω

2

0.4 4

E/ κ κG = 1

3

0.2

0.0 0.0

0.2

0.4

0.6

0.8

nr / L Figure 17.3.3 Infl Influen uence ce of she shear ar def deform ormati ation on and rot rotati ationalinerti onalinertiaa on nat natura urall fre freque quencie nciess of simply supported beams.

1.0

Sec. Se c. 17 17.4 .4

707

Moda Mo dall Ort Ortho hogo gona nali lity ty

1.0 n = 1

0.8

n

ω

n = 2

0.6

/

′ n ω

n = 3

0.4

n = 4 n = 5

Exact Approximate

0.2

0.0

0.0

0.05

0.10 r/L

Influen uence ce of she shear ar def deform ormati ation on and rot rotati ationalinerti onalinertiaa on nat natura urall fre freque quencie nciess Figure 17.3.4 Infl of simply supported beams.

nr / / L < 0.2 the error in the approximate equation is less given by Eq. (17.3.26). given (17.3.26). When nr than 5%. Shear deformation and rotational inertia have the effect of lowering the natural frequencies, quenc ies, as show shown n in Figs. 17.3.3 17.3.3 and 17.3. 17.3.4. 4. For a fixed value of the slenderness slenderness ratio L/ L /r of the beam, the frequency reduction due to shear deformation and rotational inertia /κ G value and with mode number. The latter observation implies that increases with the E /κ while the corrections corrections due to shear deformation deformation and rotat rotational ional inertia inertia may be unimp unimportant ortant for the fundamental natural frequency, they could be significant for the higher frequencies. E /κG G and mode number, the frequency reduction increases with r r / / L , For a fixed value of E /κ implying its significance for less slender or stubby beams. From the results presented one can estimate estimate whether these corrections corrections need to be inclu included ded in a particular particular problem. For earthquake response analysis of many practical structures, these corrections are not significant, but it is important to realize that these corrections do exist. If significant, they can be included in the finite element formulation for practical structures which are not amenable to solution as infinite-DOF systems.

17.4 MODA MODAL L ORTHOGONALITY ORTHOGONALITY In this section we derive the orthogonality properties of natural vibration modes of systems with distributed mass and elasticity. For convenience, the derivation is restricted to singlespan beams with hinged, clamped, or free ends and without any lumped mass at the ends, although the final result applies in general.

708


Chap.. 17 Chap

The starting point for this derivation is Eq. (17.3.6), which governs the natural frequencies and modes; for mode r , E I I (( x x ))φr  ( x x ))





ωr 2 m ( x x ))φr ( x x ))

 =  = = 

(17.4.1)

Multiplying both sides by φ by φn ( x ) and integrating from 0 to L gives L



φn ( x )

0

L

 E I I (( x x ))φ ( x x )) d x



r

ωr 2

m ( x x ))φn ( x x ))φr ( x x )) d x

(17.4.2)

0

The left side of this equation is integrated by parts; applying this procedure twice leads to L



φn ( x x ))

0

E I I (( x )φ ( x x ))



r





L

 = =

φn ( x x ))[ E I I (( x x ))φ ( x )]

d x

r

 −

L



φ ( x x ))[ E I I (( x )φ ( x )] n

r

0

0

L

 + 0

E I I (( x x ))φn ( x x ))φr  ( x ) d x

(17.4.3)

It is ea easy sy to se seee th that at th thee qu quan anti titi ties es en encl clos osed ed in are zer zero o at x 0 an and d L if th thee en ends ds of the the φ beam are free, simply supported, or clamped. This is true at a clamped end because 0 and φ 0, at a simply supported end because φ 0 and the bending moment is zero (i.e., E I φ 0), and at a free end because the bending moment is zero (i.e., E I φ 0) and the   shear force is zero [i.e., ( 0]. With the quantities in set to zero, Eq. (17.4.3) ( E E I φ ) substituted in Eq. (17.4.2) gives

{···} =

= =

= =

=

{···}

L

 0

=

=

L

E I I (( x x ))φ ( x x ))φ ( x ) d x = = ω2 n

r

r



m ( x )φn ( x )φr ( x ) d x

(17.4.4)

0

Similarly, starting with Eq. (17.3.6) written for mode n , multiplying both sides by φr ( x x )), integrating from 0 to L , and using integration by parts twice leads to L

 0

L

= ω2 E I I (( x )φ ( x )φ ( x x )) d x = n

r

n



m ( x x ))φn ( x x ))φr ( x x )) d x

(17.4.5)

0

Subtracting Eq. (17.4.4) from Eq. (17.4.5) gives L

(ωn2

−

Therefore, if ω ω n

ωr 2 )



m ( x x ))φn ( x x ))φr ( x ) d x

= 0 =

0

= ω , r

L



m ( x x ))φn ( x x ))φr ( x x )) d x

= 0 =

0

(17.4.6a)

and this substituted in Eq. (17.4.2) leads to L

 0



φn ( x x )) E I I (( x )φr  ( x ) d x





= 0 =

(17.4.6b)

Equations (17.4.6a) and (17.4.6b) are the orthogonality relations for the natural vibration

Sec.. 17.5 Sec 17.5

709

Modal Mod al Ana Analys lysis is of Fo Force rced d Dyn Dynami amic c Res Respon ponse se

modes. If a system has repeated repeated frequencies, frequencies, modes φn ( x x )) still exist such that any two r , satisfy the orthogonality relations even if ω ω n ωr . modes, n

=

=

17.5 MODA MODAL L ANALYSIS ANALYSIS OF FORCED DYNAMIC RESPONSE We now return to the partial differential equation (17.1.4), which is to be solved for a given x ,, t ). Assuming that the associated eigenvalue problem of Eq. (17.3.6) has applied force p( x been solved for the natural frequencies and modes, the displacement due to each mode is given by Eq. (17.3.2) and the total displacement by

∞

 =

u ( x , t t ))

φr ( x )qr (t t ))

(17.5.1)

r 1

=

Thus the respon response se u ( x x ,, t ) has been expressed as the superposition of the contributions of the individual modes; the r th term in the series of Eq. (17.5.1) is the contribution of the r th mode to the response. We will see next that Eq. (17.1.4) can be transformed to an infinite set of ordinary t )) as the unknown. Subdifferential equations, each of which has one modal coordinate qn (t stituting Eq. (17.5.1) in Eq. (17.1.4) gives

∞



m ( x )φr ( x )qr (t t ))

¨

r 1

=

∞

 + 



E I I (( x )φr  ( x x )) qr (t t ))



r 1

=

= p( x , t t ))

x )), integrate it over the length of the beam, and interNow we multiply each term by φ n ( x change the order of integration and summation to get

∞

L

 ¨  qr (t t ))

m ( x )φn ( x x ))φr ( x x )) d x

0

r 1

=

∞

L

  + + qr (t t ))

0

r 1

=



I (( x x ))φr  ( x x )) d x φn ( x ) E I



  =

L

p(( x , t p t ))φn ( x x )) d x

0

By virtue of the orthogonality properties of modes given by Eq. (17.4.6), all terms in each of the summations on the left side vanish except the one term for which r n , leaving L

qn (t )

¨



L

2

m ( x x )) [φn ( x x ))] d x

0

+ q (t t )) + n

This equation can be rewritten as



φn ( x x ))

0

M n qn (t t ))

¨

= =   = = L

 E I I (( x x ))φ ( x x )) d x



n

+ K q (t t )) = P (t t )) n n

p(( x , t p t ))φn ( x x )) d x

0

n

(17.5.2)

where L

M n

 =

L

2

m ( x x )) [φn ( x x ))] d x

K n

0

 = 0

L

Pn (t )

 = 0



φn ( x x )) E I I (( x )φn ( x x )) d x

p(( x , t p t ))φn ( x x )) d x





(17.5.3)

710


Chap.. 17 Chap

If each end of the beam is free, hinged, or clamped, Eq. (17.4.3) and subsequent discussion gives an alternative equation for K n : L

K n

 = 0

2

E I I (( x x )) φn ( x )

 

d x

(17.5.4)

The generalized mass M n and generalized stiffness K n for the n th mode are related: K n

2 n

= ω M

n

(17.5.5)

This relation can be derived by writing Eq. (17.4.1) for the n th mode, multiplying both side si dess by φn ( x ), int integ egrat rating ing ov over er 0 to L , and and uti utiliz lizing ing the defi definit nition ionss of M term rm M n and K n . The te Pn (t t )) in Eq generalized for force ce for Eq.. (1 (17. 7.5. 5.2) 2) is ca call lled ed th thee generalized for th thee n th mo mode. Equat )), and the generalized properties M n , tion (17.5.2) governs the n th modal coordinate coordinate q n (t mode φ n ( x ). K n , and Pn (t t )) depend only on the n th mode φ Thus we have an infinite number of equations like Eq. (17.5.2), one for each mode. t )) has been transThe partial differential equation (17.1.4) in the unknown function u ( x , t t )). formed to an infinite set of ordinary differential equations (17.5.2) in unknowns qn (t Recall that the same equations (12.3.3), N in in number, were obtained for N -DOF systems. For applied dynamic forces defined by p( x x ,, t ), the motion u ( x , t t )) of the system can t )). The equation be determined by solving the modal equations for q n (t equation for each each mode is independent of the equations for all other modes and can therefore be solved separately. Furt Fu rthe herm rmor ore, e, ea each ch mo moda dall eq equa uati tion on is of th thee sa same me fo form rm as th thee eq equa uati tion on of mo moti tion on fo forr an SD SDF F system. Thus the results obtained in Chapters 3 and 4 for the response of SDF systems to various dynamic forces—harmonic force, impulsive force, etc.—can be adapted to obtain t )) for the modal equations. solutions q n (t t )) have been determined, the contribution of the n th mode to the disOnce the q n (t placement u u(( x x ,, t t )) is given by u n ( x x ,, t t ))

= φ ( x )q (t t )) n

n

(17.5.6)

The total displacement is the combination of the contributions of all the modes: u ( x , t t ))

∞

 =

u n ( x , t t ))

n 1

=

∞

 =

φn ( x )qn (t t ))

(17.5.7)

n 1

=

The bending moment and shear force at any section along the length of the beam are related to the displacements u u(( x ) as follows: M( x )

= E I I (( x )u ( x x ))

V ( x x ))

E I I (( x )u  ( x x ))



=





(17.5.8)

x )) replaced by u n ( x , t t )), These static relationships apply at each instant of time with u ( x which is given by Eq. (17.5.6). Thus the contribution of the n th mode to the internal forces is given by Mn ( x x ,, t )

= E I I (( x )φ ( x )q (t t )) n

n

V n ( x , t t ))

=



E I I (( x )φn ( x x )) qn (t t ))





(17.5.9)

Sec.. 17.5 Sec 17.5

711


Combining the contributions of all modes gives the total internal forces: M( x , t t ))

∞

 =

Mn ( x , t t ))

n 1

V ( x x ,, t t ))

∞

E I I (( x )φn ( x )qn (t t ))

(17.5.10a)

(17.5.10b)

n 1

=

 =

∞

 = =

V n ( x x ,, t )

n 1

∞

 = 



n 1

=



E I I (( x x ))φn ( x x )) qn (t t ))

=

Example 17.1 Derive mathe Derive mathematic matical al expre expressions ssions for the dynami dynamicc respon response—dis se—displacem placement ent and bendin bending g moments—of a uniform simply supported beam to a step-function force po at distance ξ from the left end (Fig. E17.1). Specialize the results for the force applied at midspan.

p po po t

0

m, EI

ξ •

L

•

(a)

(b) Figure E17.1

Solution 1. Determine the natural vibration frequencies and modes. ωn

=

n2π 2 L 2



E I m

φn ( x )

= sin n Lπ x

(a)

Substituting φ φ n ( x ) in Eq. (17.5.3a) gives M n , which is 2. Set up the modal equations. Substituting 2 substituted in Eq. (17.5.5) together with ω n to get K n : 4

4

(b) = m2L K n = n 2π L E3 I t )) = po δ( x − x − Substituting p ( x , t − ξ ), where δ( − ξ ) is the Dirac delta function centered centered at ξ , δ( x δ( x M n

in Eq. (17.5.3c) gives

Pn (t )

= po φn (ξ )

(c)

Then the n th modal equation is M n qn (t t ))

¨

+ K n qn (t t )) = po φn (ξ )

(d)

712


Chap.. 17 Chap

3. Solve the modal equations. Equation (4.3.2) describes the response of an SDF system t )) to q n (t t )) to a step force. We will adapt this solution to Eq. (d) by changi changing ng the notation u (t and noting that ( u st )o po φn (ξ)/ K n . Thus (u (ξ)/K

=

3

= po φK n (ξ ) (1 − cos ωn t t )) = 2π p4o E LI φnn(ξ4 ) (1 − cos ωn t t ))

qn (t )

(e)

n

Substituting Eq. (e) in Eq. (17.5.7) and noting that φn ( x ) is known from Eq. (a), we obtain the u(( x x ,, t t )). displacement response u 4. Specialize for ξ latter in Eq. Eq. (17.5 (17.5.7) .7) ξ L /2. Substituting ξ L /2 in Eq. (e) and the latter gives

= =

= =

∞

2 po L 3

u ( x , t t ))

π 4 E I

where

L/ φn ( L /2)

 =   L

φn

n 1

=

=

2

n4

0 1 1

(1

− cos ωn t t )) sin n Lπ x

(f)

= 2, 4, 6, . . . = 1, 5, 9, . . . = 3, 7, 11, . . .

n n n

−

(g)

from Eq. (a) and Fig. 17.3.1. Substituting Eq. (g) in Eq. (f) gives u ( x , t t ))

=

2 po L 3 π 4 E I

− 1

cos ω1 t 1

sin

π x L

3π x ω3 t − 1 − cos sin L 81

1

cos ω5 t 5π x 1 − cos ω7 t 7π x sin sin + − 625 − +··· 2401 L L



(h)

The displacement at midspan is u

 = L

2

, t

2 po L 3 π 4 E I

− 1

cos ω1 t 1



1 − cos ω5 t 1 − cos ω7 t ω3 t + − cos + + +··· 81 625 2401 1

(i)

The coefficients 1, 81, 625, 2401, and so on, in the denominator suggest that the first-mode contribution is dominant and that the series converges rapidly. The bending moments are obtained by substituting Eq. (h) in Eq. (17.5.9a): t )) M( x , t

=−

2 po L π2

− 1

cos ω1 t 1

sin

π x L

3π x ω3 t − 1 − cos sin L 9



5π x 1 − cos ω7 t 7π x ω5 t + − cos sin − sin +··· L L 25 49 1

(j)

The bending moment at midspan is M

 =−  − L

2

, t

2 po L π2

1

cos ω1 t 1

1 − cos ω5 t 1 − cos ω7 t ω3 t + − cos + + +··· 9 25 49 1



(k)

This series with n 2 in the denominator converges slowly compared to Eq. (i) with n 4 in the denominator. This difference implies that higher modes contribute more significantly to forces

Sec.. 17.5 Sec 17.5

713


than to displacements, a result consistent with the conclusions of Chapters 12 and 13 for discretized systems. Example 17.2 A simply supported bridge with a single span of length L has a deck of unifo uniform rm cross section with mass m per feet length and flexural rigidity E I . A single single wheel load load po travels across the bridge at uniform velocity of v , as shown in Fig. E17.2a. Neglecting damping, determine an equation for the deflection at midspan as a function of time. 2 1 = =

n

Pn po

n

3 =

n

po v t

m, EI

x

L / v

L

•

•

-po

(a) 0

50

(b)

100

150

200 Location of load, ft

) 7 . 4 5 4 2 /

0

0

1

2

2.5 Time t , sec

1 mode

0.5

0

p (

1

) 2 /

1.5

÷

L ( u

(c)

10 modes

Figure E17.2

The properties of a prestressed concrete box girder elevated freeway connector are 200 ft, m 11 kips/g per foot, I 700 ft4 , and E 576,000 kips/f kips/ftt2 . If v L v 55 mph, determine an equation for the deflection at midspan as a function of time. Also determine the maximum value of deflection over time.

=

=

=

=

=

Solution We assume that the mass of the wheel load is small compared to the bridge mass, and it can be neglected. 1. Determine the natural vibration frequencies and modes. ωn

=

n2π 2 L 2



E I m

φn ( x )

= sin n Lπ x

(a)

2. Determine the generalized mass and stiffness. Substituting φ Substituting φ n ( x ) in Eq. (17.5.3a) gives M n , which is substituted in Eq. (17.5.5) together with ω n2 to get K n : M n

= m2L

4

K n

4

= n 2π L E3 I

(b)

714


Chap.. 17 Chap

3. Determine the generalized force. A load po traveling with a velocity v takes time t d L /v to cross the bridg bridge. e. At any time t the posit position ion is as shown in Fig. E17.2a. E17.2a. Thus the d moving load can be described mathematically as

=

≤ t ≤ ≤ t d d ≥≥ t d d x − where δ( − v t t )) is the Dirac delta function centered at x = δ( x p(( x p x ,, t )

=



p o δ( x δ( x 0

− vt t )) −

0 t

(c) Substi stitut tuting ing Eq. (c) (c) in v t . Sub

Eq. (17.5.3c) gives

Pn (t t ))

  =  =  =

L L 0

− vt t ))φn ( x )d x −

po δ( x δ( x

0

≤ t ≤ ≤ t d d ≥≥ t d d

0 t

0 t

p o φn (v t )) (vt 0

≤ t ≤ ≤ t d d ≥ t d d ≥ 0 ≤ t ≤ p o sin (n π t t / ≤ t d d /t d d ) 0 t ≥ ≥ t d d

(d)

n half-cycles This generalized force is shown in Fig. E17.2b; for the n th mode, it consists of n of the sine function. 4. Set up modal equations. M n qn (t t ))

¨

(e)

+ K n qn (t t )) = Pn (t t ))

t )) are given by Eqs. (b) and (d). Pn (t t )) represents n half-cycles of where M n , K n , and Pn (t a sine function. To solve these modal equations, equations, we first determine determine the response of an SDF system to such an excitation. t )) po sin ωt . The equation of motion 5. Response of SDF system to n half-cycles of p (t is

=

≤≤ t d d (f) ≥ t d d ≥ During t ≤ ≤ t d d , the force is the same as the harmonic force p(t t )) = po sin ωt considered considered earlier mu

¨ + ku

 =

p o sin(n π t t / /t d d ) 0

t t

with frequency:

ω

= nt π = n Lπ v

(g)

d d

The response is given by Eq. (3.1.6b), which is repeated here for convenience, with ( u st )o (u po /k : u (t t )) (u st )o

1

= 1 − (ω/ω )2 n



sin ωt

ω

− ω −

n



sin ωn t

t

≤ t d d ≤

≡

(h)

t d After the force ends (i.e., t d ) the system vibrates freely with its motion described by Eq. (4.7.3 (4.7.3). ). The displac displaceme ement nt u (t d d ) and velocity u (t d d ) at the end of the excitation are

≥

˙

Sec.. 17.5 Sec 17.5

715


determined from Eq. (h): u (t d d ) (u st )o u (t d d )

ω/ωn = 1 −−(ω/ω sin ωn t d d )2

ω = 1 − (ω/ω (−1)n − cos ωn t d d )2

˙

(u st )o



n

Substituting Eqs. (i) in Eq. (4.7.3) gives u (t t )) (u st )o



n = 1 −ω/ω − t d d ) − sin ωn t (−1)n sin ωn (t − 2 (ω/ω ) n

(i1)

n





(i2)

≥ t d d ≥

(j)

t

6. Solve modal equations. The solution of Eq. (f) is given by Eqs. (h) and (j). We will u((t t )) to qn (t t )) and noting adapt this solution to the modal equations (e) by changing the notation u that

= Lv

t d d

≡ pk o = PK no = m2L po 2

(u st )o

ωn

n

where ω n and ω are given by Eqs. (a) and (g), respectively. The results are qn (t t ))

2 po

1

= m L 2 − L))2 ωn (n π v/ v/ L



sin

n π v t L



nπ v

− ω L sin ωn t n

t

≤ L /v ≤

(k)

t ≥ = 2m pLo 2 − n1 L)2 ωnπ Lv (−1)n sin ωn (t − − L L/v) ≥ L /v (l) /v) − sin ωn t ωn ( π v/ v/ L ) n L or T n  This solution is valid provided that ω n  = nπ v/ = 2 L L//nv. Note that when specialized v/ L for n = 1, Eqs. (k) and (l) reduce to Eqs. (g) and (h) of Example 8.4.



qn (t t ))



7. Determine the total response. The displacement response of the beam is given by Eq. (17.5.7):

∞

 =

u ( x , t t ))

(m)

t )) φn ( x )qn (t

n 1

=

x )) is given by Eq. (a) and q n (t ) by Eqs. (k) and (l). where φ where φn ( x 8. Determine deflection at midspan. Substituting x L /2 in Eq. (m) gives

= =

∞

 =

u ( L L/ t )) /2, t

L/ /2)qn (t ) φn ( L

(n)

n 1

=

where φn

 = L

2

0 1 1

−

n n n

= 2, 4, 6, . . . = 1, 5, 9, . . . = 3, 7, 11, . . .

(o)

Equation (o) indicates that the even-numbered modes, the antisymmetric modes, do not contribute to the midspan deflection.

716


Chap.. 17 Chap

9. Numerical results. For the given prestressed concrete structure and vehicle speed:

= 3211.2 = 0.3416 kip-sec2 /ft2 E I = = 576,000 × 700 = 4.032 × 108 kip-ft2 4.032 × 108 π2 = 8.477 rad/sec ω = m

1

2002



0.3416

T 1

= 0.74 sec ωn = n 2 ω1 v = 55 mph = 80.67 ft/sec πv = 1.267 L = Lv = 80200 = 2.479 sec .67 Because the duration of the excitation t d d = L /v is greater than T n /2 for all n , the maximum t d d

response occurs while the moving response moving load is on the bridge span. This phase of the response is given by Eqs. (k) and (n):

u ( L L/ t )) /2, t

2 po

= (0.3416)200 po

= 2454.7

∞

 n 1

=

∞

 n 1

=

L/ /2) φn ( L (8.477n 2 )2

− (1.267n)2

L/ /2) φn ( L n 4 (1

− 0.02234/n 2)





sin1.267nt

1.267n

2



− 8.477n2 sin8.477n t −

sin1.267nt

− −

0.1495 n

2



sin8.477n t

(p)

Equation (p) is valid for 0 2.479 sec. The values of midspan deflection deflection calculated calculated t from Eq. (p) at many values of t are shown in Fig. E17.2c; the maximum deflection u o 1.1869 po /2454.7 is attained when the moving load is near midspan. midspan. This deflection deflection is only 3 17% larger than the static deflection ( po L /48 E I ) due to a stationary load po at midspan, implying that the dynamic effects of a moving load are small. Also shown is the result considering only the contribution of the first vibration mode [i.e., the n 1 term in Eq. (p)]. (Recall that this was the result obtained in Example 8.4.) It is clear that the response contributions of higher modes are negligible.

≤ ≤

=

=

=

17.6 EARTHQUAKE RESPONSE HISTORY HISTORY ANALYSIS ANALYSIS t )) of the supports, the equation of As shown earlier, when the excitation is acceleration u g (t t )) except that this force is replaced motion for a beam is the same as for applied force p ( x , t by peff ( x , t t )) given by Eq. (17.2.4). Thus the modal analysis procedure of Section 17.5 can readily be extended to the earthquake problem. From Eq. (17.2.4) the effective earthquake forces are

¨

peff ( x x ,, t t ))

= −m ( x x ))u¨

t )) g (t

(17.6.1)

Sec.. 17.6 Sec 17.6

717

Earthqu Eart hquak ake e Res Respon ponse se His History tory Ana Analys lysis is

The spatial distribution of these forces is defined by m ( x ). Thi Thiss for force ce distrib distributi ution on can be expanded as a summation of inertia force distributions sn ( x ) associated with natural vibration modes (see Section 12.8): m ( x x ))

∞

 =

sr ( x )

r 1

∞

 =

r m ( x )φr ( x x ))

(17.6.2)

r 1

=

=

Premultiplying both sides by φ by φ n ( x x )), integrating over the length of the beam, and utilizing modal orthogonality with respect to the mass distribution, Eq. (17.4.6a), leads to n

L

L hn

L hn

where

= M

n

 =

m ( x )φn ( x ) d x

(17.6.3)

0

The contribution of the n th mode to m ( x x )) is sn ( x x ))

=  m ( x )φ ( x ) n

n

(17.6.4)

Observe that these modal expansion equations for a distributed-mass system are similar to the corresponding equations (13.2.2) and (13.2.4) for lumped-mass systems. For uniform cantilever towers with mass m per unit length the modal expansion of Eq. (17.6.2) (17.6.2) is as sho shown wn in Fig Fig.. 17.6.1. 17.6.1. The functi functions ons sn ( x ) were ev evaluat aluated ed from Eq. (17.6.4) using Eqs. (17.6.3) and (17.5.3a) and the φ n ( x ) given by Eqs. (17.3.23) and (17.3.21). 1.566m

=

-m

0 m s( x x )

-0.868m

0.509m

+

-m

Figure 17.6.1

0 m s1(x)

+

-m

0 m s2(x)

-0.364m

+

-m

0 m s 3(x)

+

-m

• • •

0 m s 4(x)

Modal expansion of effective effective earthquake forces for uniform towers. towers.

x ,, t t )) in Retu Re turn rnin ing g no now w to th thee mo moda dall an anal alys ysis is pr proc oced edur uree of Se Sect ctio ion n 17 17.5 .5,, p( x x ,, t t )) given by Eq. (17.6.1), to obtain Eq. (17.5.3c) is replaced by peff ( x Pn (t t ))

h n

= − L u¨

t )) g (t

(17.6.5)

Substituting Substituti ng Eq. (17.6.5) in Eq. (17.5.2), dividing by M n , and using Eqs. (17.5.5) and (17.6.2a) gives the modal equations of an undamped tower subjected to earthquake excitation: qn ωn2 qn n u g (t t )) (17.6.6)

¨ +

= − ¨

718


Chap.. 17 Chap

For classically damped systems, Eq. (17.6.6) becomes 2 n n

¨ + 2ζ ω q˙ + ω q = − u¨ (t t ))

qn

n

n n

n

g

(17.6.7)

where ζ n is the damping ratio for the n th mode. mode. Thi Thiss is the same same as Eq. (13.1. (13.1.7) 7) deh rived earlier for N -DOF systems, except that L n and M n that enter into  n are now given by Eqs. (17.6.3) and (17.5.3), respectiv respectively ely.. As shown in Section 13.1.3, the solution of Eq. (17.6.7) is qn (t )

=  D (t ) n

n

(17.6.8)

where Dn (t t )) is the deformation response of the n th-mode SDF system. This is an SDF system with vibration properties—natural frequency ω n and damping ratio ζ n —of the n th mode of the distributed-mass system. Thus q n (t t )) is readily available once the SDF system response respon se has been determined determined by the metho methods ds of Chapte Chapterr 6. The contribution contribution of the n th t )). Substimode to the earthquake response of the tower can be expressed in terms of Dn (t tuting Eq. (17.6.8) in Eqs. (17.5.6) and (17.5.9) gives the displacements, bending moments, and shear forces due to the n th mode: u n ( x , t t )) Mn ( x x ,, t t ))

=  E I I (( x x ))φ ( x x )) D (t t )) n

n

n

=  φ ( x x )) D (t t ))  V ( x , t t )) =  E I I (( x )φ ( x ) D n

n

(17.6.9)

n

n



n



n

t )) n (t

(17.6.10)

Alternati Altern ative vely ly,, as in Cha Chapte pterr 13 for an N -DO -DOF F sys system tem,, the int intern ernal al for forces ces can be det deter er-mined from the equivalent static forces associated with displacements u n ( x , t t )) computed from dynamic analysis. analysis. To derive an equation for these forces, we introduce introduce a famil familiar iar equation from elementary beam theory relating deflections u ( x x )) to applied forces f ( x ( x )). For a uniform beam E I u IV ( x x ))

=

f ( x ( x ))

(17.6.11)

d 4 u /d x 4 . The more general where E I is the flexural rigidity and u IV general version version of this equation applicable to nonuniform beams with flexural rigidity E I I (( x x )) is

=



E I I (( x )u  ( x x ))



 =

f ( x ( x )

(17.6.12)

u(( x x )) by the time-varying displacements u n ( x , t t )) from Eq. (17.6.9) gives Replacing u f n ( x x ,, t )

=

n



E I I (( x x ))φn ( x x )) Dn (t t ))





(17.6.13)

which, by using Eq. (17.4.1) rewritten for the n th mode, becomes f n ( x , t t ))

= s ( x x )) A (t ) n

n

(17.6.14)

x )) is given by Eq. (17.6.4), and An (t t )), the pseudo-acceleration response of the where s n ( x n th-mode SDF system, is given by Eq. (13.1.12), which is repeated: An (t t ))

2 n

= ω D (t t )) n

(17.6.15)

Observe the similarity between Eqs. (17.6.14) and (13.2.7) for a lumped-mass system. At

Sec.. 17.6 Sec 17.6

719

Earthqu Eart hquak ake e Res Respon ponse se His History tory Ana Analys lysis is

any time instant the contribution r n (t t )) of the n th mode to any response quantity r r ((t )— deflection, shear force, or bending moment at any location of the beam—is determined by static analysis of the beam subjected to external forces f n ( x , t t )), and can be expressed as r n (t )

st t )) n An (t

= r

(17.6.16)

The modal static response r nst is determined determined by stati staticc analysis of the tow tower er due to exte external rnal forces sn (ξ ) (Fig. 17.6.2). 17.6.2). As shown in Fig. 17.6.1, these forces forces due to the fundamental fundamental mode all act in the same direction, but for the second and higher modes they will change direction as one moves up the tower. •

sn(ξ) = Γ nm(ξ)φn(ξ) L

ξ x Figure 17.6.2 Static problem to be solved to determine modal static responses.

•

The modal static responses are presented in Table 17.6.1 for five response quantities: x )) at location x , the bending moment M( x x )) at location x , the base shear the shear V ( x V b V (0), the base moment M b M(0), and deflection u ( x x )). The first four equations come from static analysis of the system in Fig. 17.6.2. The result for u ( x x )) is obtained by comparing compa ring Eqs. (17.6.9) and (17.6.16) (17.6.16) and using Eq. (17.6.15). (17.6.15). Par Parts ts of the equations equations for st st V bn and Mbn are obtained by substituting Eq. (17.6.4) for sn (ξ ), using Eq. (17.6.3) for L nh , and defining

=

=

L

θ

L h ∗ = n

M ∗ = n L h n

n

n

L nh

TABLE 17.6.1

M( x ) V b Mb

u ( x )

x m ( x x ))φn ( x x )) d x

(17.6.17)

0

MODAL STA STATIC RESPONSES Modal Static Response, r nst

Response, r V ( x x ))

 =

L θ n

= x L L sn (ξ ) d ξ ξ L L ( x x ) = (ξ − − x )sn (ξ ) d ξ ξ Mst n ( x st = 0 L L sn (ξ ) d ξ ξ = = n L hn = M n∗ V bn Mst = 0 L L ξ ξ ssn (ξ) d ξ ξ = = n L θ n = h ∗n M n∗ bn u st ( x x ) = (n /ωn2 )φn ( x ) n ( V nst ( ( x x )

   

Observe the similarity between the equations in Table 17.6.1 and those for a lumped-mass system in Table 13.2.1. The approach symbolized by Eq. (17.6.16) to determine shear and

720


Chap.. 17 Chap

moment is preferable over Eq. (17.6.10) because it avoids computation of the second and third derivatives of the mode shapes; obviously, both methods will give identical results. The base shear V bn (t t )) and base moment Mbn (t t )) due to the n th mode are obtained st by specializing Eq. (17.6.16) for V b and Mb and substituting for V bn and Mst bn from Table 17.6.1: V bn (t t ))

= M ∗ A (t t ))

Mbn (t )

n

n

= h∗ V n

bn (t )

(17.6.18)

Because Eq. (17.6.18) is identical to Eqs. (13.2.12b) and (13.2.15b) for lumped-mass systems, following Section 13.2.5, M n∗ and h ∗n may be interpreted as the effective modal mass and effective modal height for the n th mode. Obser Observe ve that Eq. (17.6.17a (17.6.17a and b) is identical to Eq. (13.2.9a) for lumped-mass systems; the definitions of M n , L nh , and L θ n differ, however, between distributed-mass and lumped-mass systems. The sum of the effective modal masses over all modes is equal to the total mass of the tower:

∞

L

 =  M ∗ n

m ( x x )) d x

(17.6.19)

0

n 1

=

and the sum of the first moments about the base of the effective modal masses M n∗ located at heights h ∗n is equal to the first moment of the distributed mass about the base:

∞

 n 1

=

L

h ∗ M ∗ = n

n



xm x m ( x ) d x

(17.6.20)

0

These relations can be proven in the same manner as the analogous equations (13.2.14) and (13.2.17) (13.2.17) for a lumpe lumped-mass d-mass system. system. In particular particular,, Eq. (17.6.19) (17.6.19) can be proven by integrati inte grating ng Eq. (17.6. (17.6.2) 2) over the height of the tower and using Eq. (17.6 (17.6.3b). .3b). Simil Similarly arly,, Eq. (17.6.20) can be derived from the modal expansion of forces x m ( x ). The effective modal masses M n∗ and effective modal heights h ∗n for a uniform cantilever tower are shown in Fig. 17.6.3; note that h ∗n are plotted without their algebraic signs. M n∗ and h ∗n were determined from Eq. (17.6.17) using the known modes (Fig. 17.3.2). Ob M n∗ for the first four modes gives 90% of the total mass of the tower. serve that the sum of M •

0.613mL

L

=

L 6 2 7 . 0

m( x x ) = m •

0.188mL

L 9 0 2 . 0

¨ ug(t)

Mode Figure 17.6.3

1

2

L0.065mL 7 2 1 . 0

3

L

0.033mL

0 9 0 . 0

••• 4

Effective modal modal masses and effective effective modal heights.

¨ ug(t)

Sec.. 17.7 Sec 17.7

721

Earthqu Eart hquak ake e Res Respons ponse e Spe Spectr ctrum um Ana Analys lysis is

Combining the response contributions of all the modes the earthquake response of the system: r ((t r t ))

∞

 =

r n (t t ))

n 1

∞

 =

r nst A n (t t ))

(17.6.21)

n 1

=

=

where Eq. (17.6.16) has been used for r n (t t )). This n th-mode contribution to the response t )), the pseudocan be determined from the modal static response (Table 17.6.1) and An (t accele acc elerat ration ion res respon ponse se of the n thth-mod modee SD SDF F sys system tem,, jus justt as for N -DO -DOF F syst systems ems (Fi (Fig. g. 13. 13.1.1 1.1). ).

17.7 EARTHQUAKE RESPONSE SPECTRUM ANALYSIS ANALYSIS The peak response of a distributed-mass system, such as a cantilever tower, can be estimated from the earthquake response (or design) spectrum by procedures analogous to those developed in Chapter 13, Part B for lumped-mass systems. The exact peak value of the n th-mode response r n (t t )) is r no no

st n An

= r

(17.7.1)

where An A(T n , ζ n ) is the ordinate of the pseudo-acceleration spectrum corresponding to natural period T n and damping ratio ζ n . Alternativ Alternatively, ely, r no no may be viewed as the result of static analysis of the tower subjected to external forces

≡

f no no ( x )

= s ( x ) A n

n

(17.7.2)

x ,, t ) defined in Eq. (17.6.14). which are the peak values of the equivalent static forces f n ( x r ((t t )) can be estimated by combining the modal The peak value r o of the total response r peaks r no no according to one of the modal combination rules presented in Section 13.7.2. Because the natural frequencies of transverse vibration of a beam are well separated, the SRSS combination rule is satisfactory. Thus 1/2

  

r o

∞

2 r no

(17.7.3)

n 1

=

Example 17.3 A reinforced-concrete chimney, 600 ft high, has a uniform hollow circular cross section with outsidee diameter 50 ft and wall thickness outsid thickness 2 ft 6 in. (Fig. E17.3a). E17.3a). For purposes purposes of earth earthquake quake analysis, the chimney is assumed clamped at the base and its mass and flexural stiffness are computed comput ed from the gross area of the concrete (neglecting (neglecting the reinforcing reinforcing steel). The elastic modulus for concrete E c 3600 ksi, and its unit weight is 150 lb/ft 3 . Modal damping damping ratios are estimated estimated as 5%. Deter Determine mine the displacements, displacements, shear forces, and bending moments due to an earthquake characterized by the design spectrum of Fig. 6.9.5 scaled to a peak ground acceleration of 0.25g. Neglect shear deformations and rotational inertia.

=

722


Chap.. 17 Chap

g 1 0 6 . 0 =

0.75

(a)

2

A

•

•

(b)

•

•

•

600′

g 8 7 6 . g 0 8 7 = 6 . 4 3 0 A A =

50′

g /

0.50 g 4 2 1 . 0 =

A

•

0.25

1

A

•

•

2′ - 6″

••

0

0 T 4 T 3 T 2

1

2

T 1

3

4

T n, sec Figure E17.3a, b

Solution 1. Determine the chimney properties.

m

2 2 = π [25 − (3222..25) ]0.15 = 1.738 kip-sec2 /ft2

= (3600 × 144) π4 =

E I

254



− (22.5)4 = 5.469 × 1010 kip-ft2



Equati ation on (17 (17.3. .3.22) 22) gi gives ves the nat nat-2. Determine the natural vibration periods and modes. Equ ural frequencies of vibration, and the corresponding periods, in seconds, are T 1 3.626, T 2 0.5787, T 3 0.2067, T 4 0.1055, and so on. The natural natural modes, given given by Eq. (17.3.23) with β n L defined by Eq. (17.3.21), were evaluated numerically for many values of x and are shown in Fig. 17.3.2, normalized to unit value at the top. 3. Compute the modal properties. With the mode shapes known, the properties M n , h θ L n , L n , M n∗ , and h ∗n were obtained by numerically evaluating their respective integrals, and are presented in Table E17.3. ordinates. The design spectrum of Fig. 6.9.5 scaled to a 4. Read the design spectrum ordinates. peak acceleration of 0.25g is shown in Fig. E17.3b, wherein the pseudo-acceleration ordinates corresponding to the first four periods are noted: An /g 0.124, 0.678, 0.678, and 0.601. 5. Compute the displacements. The peak displacements u no ( x ) due to the n th mode x )) given in Taare given by Eq. (17.7.1), where the modal static response r nst becomes u st n ( x 2 ble 17.6.1. Substi Substitutin tuting g known values of  1, 2,  n , φ n ( x ), ω n , and An leads to u no ( x ), n 3, and 4, show shown n in Fig. E17.3c. E17.3c. At each location x these peak modal displacements are combined according to the SRSS rule, Eq. (17.7.3), to obtain an estimate of the total displacements u o ( x ), which are also shown. Observe that the total displacements are due primarily to the first mode.

=

=

=

=

=

=

Sec.. 17.7 Sec 17.7

25.0

-25

0 u1o

-1.93

25 -25

10.86

-40

723

Earthqu Eart hquak ake e Res Respons ponse e Spe Spectr ctrum um Ana Analys lysis is

0 f 1o

0 u2o

-32.9

40 -40

0 f 2o

0.144

25 -25

0 u3o

-0.024

25 -25

0 u4o

25.1

25 -25

0 25 Total uo

(c) Lateral displacements, in. 19.29 -12.24

40 -40

0 f 3o

40 -40

0 f 4o

40

(d) Equivalent static forces, kips

-6

0

6 -6

V 1o

0

6 -6

V 2o

0

6 -6

V 3o

0

6 -6

0 6 Total V o

1.5 -1.5

0 1.5 Total Mo

V 4o

(e) Shears, 103 kips

-1.5

0 M1o

1.5 -1.5

0 M 2o

1.5 -1.5

0

1.5 -1.5

M3o

0 M4o

(f) Bending moments, 10 6 kip-ft Figure E17.3c–f

724

Systems Syst ems with Dist Distribut ributed ed Mass and Elas Elastici ticity ty TABLE E17.3

Chap.. 17 Chap

MODAL PROPERTIES

Mode

M n /m L

1 2 3 4

0.2500 0.2500 0.2500 0.2498

L hn /m L

0.3915

n

1.5660

2 L θ n /m L

0.2844

−0.2170 −0.8679 −0.0454 0.1272 0.5089 0.0162 −0.0909 −0.3637 −0.0082

Determine ine the modal expansion expansion of m ( x ), Eq. (17.6.2 x )) known from (17.6.2). ). Wi With th φn ( x 6. Determ Fig. 17.3.2 and  n from Table E17.3, the functions s n ( x ) are deter determined mined from Eq. (17.6. (17.6.4). 4). Actually, these were presented in Fig. 17.6.1. Computee the equival equivalent ent static forces forces for the nth mode. These forces f no 7. Comput no ( x ) are x )) of Fig. 17.6.1 and the A n values of Fig. E17.3b. determined from Eq. (17.7.2) using the s n ( x The results for the first four modes are shown in Fig. E17.3d. 8. Compute the shears and bending moments. For each mode the peak values of shears and ben bendin ding g mom moment entss at loc locati ation on x are com comput puted ed by sta static tic ana analys lysis is of the chi chimne mney y sub subjec jected ted to forces f no . The Th e resu re sult ltin ing g shear she ars s and an d bend be ndin ing g mome mo ment nts s due du e to the th e first fir st four fo ur mode mo des s are ar e show sh own n x ) x ( ) no in Fig. E17.3e E17.3e and f. At each section section x , these modal responses are combined by the SRSS rule [Eq. (17.7.3)] (17.7.3)] to obtain an estim estimate ate of the total forces, forces, which are also shown. Obser Observe ve that the first two modes contribute significantly to the total response, with the second-mode contribution more significant for the shears than for moments. interest est to compare the results above 9. Compare with Rayleigh’s method. It is of inter considering conside ring response in four modes with the approx approximate imate solution solution using Rayleigh’s Rayleigh’s metho method d (Example 8.3). The approximate analysis predicts the displacements reasonably well but not the bending moments or shears. There are two reasons for the larg larger er errors in forces: (a) The approximate results differ from the exact response due to the first mode because the assumed shape function in Rayleigh’s method is an approximation to this mode; this discrepancy introduces trodu ces larger errors errors in forces than in displa displacement cements. s. (b) The second and higher modes, whose response contributions to forces are more significant than they are for displacements, are neglected in Rayleigh’s method.

17.8 DIFFICUL DIFFICULTY TY IN ANAL ANALYZING YZING PRACTICAL SYSTEMS It is evident that the dynamic response of systems with distributed mass and elasticity can be determined by the modal analysis procedure once the natural vibration frequencies and modes of the system have been determined. Both examples solved in Section 17.3 involved uniform beams, and we found the natural frequencies and modes analytically, although the frequency equation for the cantilever had to be solved numerically. This classical approach is rarely feasible if the flexural rigidity E I or mass m vary along the length of the beam, several intermediate supports are involved, or the system is an assemblage of several members with distributed mass. In this section we identify some of the difficulties in obtaining analytical solutions for the above-mentioned systems. Consider a single-span beam with mass m ( x x )) and flexural stiffness E I I (( x x )). To determine the natural frequencies and modes, we need to solve Eq. (17.3.6), which can be

Sec.. 17.8 Sec 17.8

725

Difficu Dif ficulty lty in Ana Analyz lyzing ing Pra Practi ctical cal Sys System tems s

rewritten as E I I (( x x ))φIV ( x x ))

2

+ 2 E I  ( x x ))φ( x ) + E I  ( x x ))φ ( x ) − ω m ( x x ))φ( x x )) = 0

(17.8.1)

I (( x x )), E I  ( x ), E I  ( x x )), and m m(( x ) of this fourth-order differential Because the coefficients E I equation vary with x , an analytical solution is rarely feasible for ω 2 and and φ φ ( x x )). Therefore, I (( x ) and it is not practical to use the classical approach for practical problems in which E I m ( x x )) may be complicated functions. In finding the natural frequencies and modes of a beam on multiple supports, the uniform segment between each pair of supports is considered as a separate beam with its origin origi n at the left end of the segment. segment. Equat Equation ion (17.3.9) (17.3.9) applies to each segment, segment, there is one such equation for each segment, and the necessary boundary conditions are:

1. At each end of the beam the usual boundary conditions are applicable, depending on the type of support. 2. At each intermediate support the deflection is zero, and since the beam is continuous, the slope and the moment just to the left and to the right of the support are the same. This process quickly becomes unmanageable because of the four constants in Eq. (17.3.9), which must be evaluated evaluated in each segm segment. ent. An analytical analytical solution is rarely feasible feasible for 2 I (( x ) vary within each ω and φ( x ), especially if the span lengths vary and m ( x ) and E I segment, as would often be the case for a multispan bridge. Consider the two-member frame shown in Fig. 17.8.1. Each member is axially rigid and an d ha hass un unif ifor orm m pr prop oper erti ties es—fl —flex exur ural al ri rigi gidi dity ty an and d ma mass— ss—as as in indi dica cate ted; d; ho howe weve verr, they may differ differ from one member to the other. other. Each member member is consid considered ered as a separ separate ate beam with its origin at one end. Equat Equation ion (17.3.9) applies applies to each uniform member, member, there is one such equation for each member, and the necessary end and joint conditions are: 1. At the supports of the frame the usual boundary conditions are applicable, depending on the type of support, resulting in four equations for the frame of Fig. 17.8.1: φ(1) ( L 1 )

=0

φ(1) ( L 1 )

φ(2) (0)

=0

=0

x 1) φ(1)( x L1

•

•

x 1 •

EI 1, m1 L2

)

2

x

( )

x 2

EI 2, m2

2 (

•

φ

Figure 17.8.1

φ(2) (0)

=0

726


Chap.. 17 Chap

members should be compatible; this 2. At the joint the end displacements of the joining members condition for axially rigid members gives φ(1) (0)

=0

φ(2) ( L 2 )

=0

3. At the joint the end slopes of the joining members should be compatible; thus φ(1) (0)

= φ

(2) ( L 2 )

4. At the joint the bending moments should be in equilibrium; thus E I 1 φ(1) (0)

+ E I φ 2

(2) ( L 2 )

=0

A simple two-member frame requires setting up these eight conditions and the evaluation of eight constants. The process becomes unmanageable for a frame with many members. It sho should uld no now w be ev evide ident nt tha thatt the cla classi ssical cal pro proced cedure ure to det determ ermine ine the nat natura urall fre freque quenncies and modes of a distri distribute buted-mas d-masss system with infinite number of DOF DOF,, is not feasible for practical practical struc structures. tures. Such problems problems can be analy analyzed zed by discre discretizi tizing ng them as systems with a finite number of DOFs, as discussed in the next chapter.

F U R T H ER

R E A D IN G

Clough, R. W., and Penzien, J., Dynamics of Structures, McGraw-Hill, New York, 1993, Chapters 17–19. Humar, J. L., Dynamics of Structures, 2nd ed., A. A. Balkema Publishers, Lisse, The Netherlands, 2002, Chapters 14–17. Stokey, W. F., “Vibration of Systems Having Distributed Mass and Elasticity,” Chapter 8 in Shock and Vibration Handbook (ed. (ed. C. M. Harris), McGraw-Hill, New York, 1988. Timoshenko, S., Young, D. H., and Weaver, W., Jr., Vibration Problems in Engineering, Wiley, New York, 1974.

PROBLEMS 17.1

Find the first three three natural vibrati vibration on frequencies frequencies and modes of a uniform beam beam clamped at both ends. Sketch the modes. Comment on how these frequencies compare with those of a simply supported beam.

17.2

Find the first three three natural vibrati vibration on frequencies frequencies and modes of a uniform beam beam clamped at one end and simply supported at the other. Sketch the modes.

17.3

Find the first five five natural natural vibration vibration frequencie frequenciess and modes of a uniform beam beam free at both ends. Sketch the modes. Comment on how these frequencies compare with those for a beam clamped at both ends. ( Hint : The first two modes are rigid-body modes.)

Chap Ch ap.. 17 17.4

727

Prob Pr oble lems ms

A weigh weightt W is suspended from the midspan of a simply supported beam as shown in Fig.. P17 Fig P17.4. .4. If the wire by whi which ch the wei weight ght is susp suspende ended d sudd suddenl enly y sna snaps, ps, describe describe the subsequent subsequ ent vibration vibration of the beam. Specia Specialize lize the general result to obtain the deflection deflection at midspan. Neglect damping.

L

•

•

m, EI

Wire W

17.5

Figure P17.4

Derive mathematical expressions for the displacement displacement response of a simply supported uniform beam to the force distribution distribution shown in Fig. P17.5 P17.5;; the time variation variation of the force is a x ,, t t )) in terms of the natural vibration modes of step function. Express the displacements u ( x the beam. Identify the modes that do not contribute to the response. Specialize the general result to obtain the deflection at midspan. Neglect damping.

p(t )

m, EI L

• 17.6

•

Figure P17.5

Derive mathematical expressions for the displacement displacement response of a simply supported uniform beam to the force distribution distribution shown in Fig. P17.6 P17.6;; the time variation variation of the force is a x ,, t t )) in terms of the natural vibration modes of step function. Express the displacements u ( x the beam. Identify the modes that do not contribute to the response. Specialize the general result to obtain the deflection at quarter span. Neglect damping.

p(t )

m, EI p(t )

•

L

•

Figure P17.6

17.7

Solve Problem 8.25 considering all natural vibration modes of the bridge.

17.8

Solve Problem 8.26 considering all natural vibration modes of the bridge.

17.9

Prove that Eq. (17.6.19) is valid for a vertical cantilever cantilever beam.

17.10

Prove that Eq. (17.6.20) (17.6.20) is valid for a vertical cantilever cantilever beam.

728 17.11


Chap.. 17 Chap

A free-standing intake–outlet tower tower 200 ft high has a uniform hollow hollow circular cross section with outside outside diameter 25 ft and wall thickness 1 ft 3 in. Assume that the tower tower is clamped at the base and that its mass and flexural stiffness are computed from the gross area of the concrete concre te (neglecting (neglecting reinforcing reinforcing steel). The elastic modulus for concrete concrete is 3600 ksi, and 3 its unit weight is 150 lb/ft . Modal damping ratios are estimated as 5%. Determine the top displacement, base shear, and base overturning moment due to an earthquake characterized by the design spectrum of Fig. 6.9.5 scaled to the peak ground acceleration of 13 g. Neglect shear deformations and rotational inertia.

M17_CHOPRA_Dinamica_Systems With Distributed Mass and Elasticity

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