14.1 Introduction

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14.1.1 Basic Concepts

Reinforced concrete structures are capable of free and forced vibration when disturbed from their

equilibrium configuration (see Chapter 1 and Chapter 2). Free vibration takes place when a structure

vibrates under the action inherent to the structure itself without being impressed by external forces. The

structure under free vibration vibrates at its natural frequency, which is one of the dynamic properties of

the structure. Forced vibration takes place under the excitation of an external force and at the frequency

of the exciting force, which is independent of the natural frequency of the structure. When the frequency

of the exciting force coincides with the natural frequency of a structure, resonance occurs and

dangerously large amplitudes may result. Therefore, the calculation of natural frequency and the

examination of resonance are of practical importance. Also, in reality, the influence of viscous dampening

on a vibrating structure should be taken into account. In this section, the fundamental dynamic

properties are introduced.

14-1

© 2005 by Taylor & Francis Group, LLC

14.1.2 Natural Frequencies

An undamped single-degree-of-freedom (single-DoF) system (Figure 14.1) is used to explain natural

frequency. Figure 14.1(c) shows the free-body diagram with inclusion of the inertial force my€: This force

is equal to the mass, m; multiplied by the acceleration, y€; and should always be directed negatively with

respect to the corresponding coordinate; note that k is the spring constant of the system. The summation

of forces in the y-direction directly gives the following equation of motion:

my€ þ ky ¼ 0 ð14:1Þ

To solve Equation 14.1, we assume

y ¼ A cos wt ð14:2Þ

or

y ¼ B sin wt ð14:3Þ

where A and B are constants depending on the initiation of the motion and w is a quantity denoting

a physical characteristic of the system. The substitution of Equation 14.2 or Equation 14.3 into

Equation 14.1 gives

ð2mw2 þ kÞA cos wt ¼ 0 ð14:4Þ

If this equation is to be satisfied at any time, the factor in parentheses must be equal to zero:

2mw2 þ k ¼ 0 ð14:5Þ

The positive root of Equation 14.5

w ¼

ffiffiffiffi

k

m

s

rad=sec ð14:6Þ

is known as the circular or angular natural frequency of the system (Paz, 1997; Clough and Penzien,

1993).

Note: The Greek symbol v is commonly used rather than w to denote this natural frequency in the

literature, which is expressed in rad/sec and is called “angular” natural frequency.

The “cyclic” natural frequency f is usually expressed in hertz (Hz) or cycles per second (cps). The

quantity w differs from the natural frequency f only by the constant factor, 2p: The period, T; is usually

expressed in seconds per cycle (i.e., the value reciprocal to the natural frequency, f ). Therefore, it gives

f ¼

1

T ¼

w

2p ð14:7Þ

Example 14.1

Determine the natural frequency of the system shown in Figure 14.2, consisting of a weight of

W ¼ 230 kN attached to a horizontal reinforced concrete cantilever beam through the coil spring k2:

m

k

y mg mg

ky ky mÿ

N N

(a) (b) (c)

FIGURE 14.1 Free-body diagrams: (a) showing single-DoF; (b) showing only external forces; (c) showing external

and inertial forces.

14-2 Vibration and Shock Handbook

© 2005 by Taylor & Francis Group, LLC

The cantilever beam has a cross section of 305 mm

ðhÞ £ 152 mm ðbÞ; modulus of elasticity

E ¼ 30 kN/mm2, and a length L ¼ 3 m. The coil

spring has stiffness k2 ¼ 200 N/mm.

Solution

The deflection, D, at the free end of a cantilever

beam that is acted upon by a static force, P, at the

free end is given by

D ¼

PL3

3EI

The corresponding spring constant, k1, is then

k1 ¼

P

D ¼

3EI

L3

where I ¼ ð1=12Þbh3 (for rectangular section)

when the contribution of reinforcing bars is

neglected. The cantilever and the coil spring of

this system are connected as springs in series.

Consequently, the equivalent spring constant is

1

ke ¼

1

k1 þ

1

k2

Substituting corresponding numerical values, we obtain

I ¼

1

12 £ 152ð305Þ3 ¼ 3:59 £ 108 mm4

k1 ¼

3 £ 30;000 £ 3:59 £ 108

ð3000Þ3 ¼ 1197 N=mm

and

1

ke ¼

1

1197 þ

1

200

ke ¼ 171 N=mm

The natural frequency for this system is then given by Equation 14.6 as

v ¼

ffiffiffiffiffiffi

ke=m

p

ðm ¼ W =g and g ¼ 9:8 m=sec2Þ

v ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

171 £ 9:8=230 p

v ¼ 2:7 rad=sec

or, using Equation 14.7

f ¼ 0:43 Hz

14.1.3 Viscous Damping

Let us assume that we have modeled a structural system as a simple oscillator with viscous damping, as

shown in Figure 14.3. In this figure, m and k are the mass and spring constant of the oscillator,

respectively, and c is the viscous damping coefficient (or damping constant). The summation of forces in

(a) W = 230 kN

(b)

A

A

3000 mm

305.0 mm

276.0 mm

23.8 mm

2 - #3 (fy = 413.8 MPa)

3 - #6 (fy = 475.85 MPa)

152.4 mm

#2@89 mm

f'c = 39 N/mm2

A's = 142 mm2

As = 852 mm2

k2 = 200 N/mm

FIGURE 14.2 System for Example 14.1.

Reinforced Concrete Structures 14-3

© 2005 by Taylor & Francis Group, LLC

the y-direction gives the differential equation of

motion.

my€ þ cy_ þ ky ¼ 0 ð14:8Þ

To solve Equation 14.8, we try y ¼ C ept :

Substituting this function into Equation 14.8

results in the equation

mCp2 ept þ cCp ept þ kC ept ¼ 0 ð14:9Þ

After cancellation of the common factors,

Equation 14.9 reduces to an equation called the

characteristic equation for the system, given by

mp2 þ cp þ k ¼ 0 ð14:10Þ

The roots of this quadratic equation are

p1; p2 ¼ 2

c

2m

^

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

c

2m

􀀏 􀀐2

2

k

m

s

ð14:11Þ

For a system oscillating with critical damping, the expression under the radical in Equation 14.11 is

equal to zero, that is

ccr

2m

􀀏 􀀐2

2

k

m ¼ 0 ð14:12Þ

ccr ¼ 2

ffiffiffiffi

km p ð14:13Þ

where ccr designates the critical damping value (Clough and Penzien, 1993; Paz, 1997).

In general, the damping of the system is expressed as

c ¼ j ccr ð14:14Þ

where j is the damping ratio of the system.

Note: Often the symbol z is used instead of j to denote the damping ratio.

The dampening value depends on structural materials. According to Dowrick (1987), the values for

concrete structures are listed in Table 14.1. It can be seen from this table that for reinforced and

prestressed concrete the values are 5 and 2%, respectively.

Example 14.2

A reinforced concrete beam consists of a weight of 5.0 kN and a stiffness of k ¼ 4:0 kN/cm. Find

1. the undamped natural frequency

2. the damping coefficient

mÿ

ky

y

y

k

c

m

(a) (b)

cy·

FIGURE 14.3 (a) Viscous damped oscillator; (b) freebody

diagram.

TABLE 14.1 Typical Damping Ratios for Concrete Structures

Type of Construction Damping j; % of Critical

Concrete frame, with all walls of flexible construction 5

Concrete frame, with stiff cladding and all internal walls flexible 7

Concrete frame, with concrete or masonry shear walls 10

Concrete and/or masonry shear wall buildings 10

Prestressed concrete 2

Notes: (1) The term frame indicates beam and column bending structures as distinct from shear structures. (2) The term

concrete includes both reinforced and prestressed concrete in buildings. For isolated prestressed concrete members such as in

bridge decks, damping values less than 5% may be appropriate, e.g., a value of 1 to 2% may apply if the structure remains

substantially uncracked.

14-4 Vibration and Shock Handbook

© 2005 by Taylor & Francis Group, LLC

Solution

1. w ¼

ffiffiffiffiffi

k=m p ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ð4:0 kN=cm £ 980 cm=sec2Þ=5:0 kN

p

¼ 28:0 rad=sec:

2. c ¼ jccr ¼ 0:05 £ 2 £

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

pð5:0 £ 4:0Þ=980 ¼ 14 N sec=cm:

14.1.4 Damped Harmonic Excitation

Now consider the case of one-DoF system in Figure 14.4, vibration due to an external load of a sine

function under the influence of viscous damping. The differential equation of motion is obtained by

equating to zero the sum of the forces in the free-body diagram of Figure 14.4(b). Hence,

my€ þ cy_ þ ky ¼ F0 sin w􀀊 t ð14:15Þ

The total response is then obtained by summing the complementary solution and the particular solution

yðtÞ ¼ e2jwt ðA cos wDt þ B sin wDtÞ þ

ffiyffiffistffiffisffiiffinffiðffi4ffiffiffitffiffi2ffiffiffiuffiÞffiffiffiffi

ð1 2 r2Þ2 þ ð2rjÞ2

p ð14:16Þ

where r is the frequency ratio that is equal to

forced vibration frequency, 4; divided by natural

frequency, w:

Note: A further discussion of this topic is found

in Chapter 2.

By examining the transient component of the

response, it may be seen that the presence of

the exponential factor, e2jwt, will cause this

component to vanish, leaving only the steadystate

motion, Y ; which is given by the second term

of Equation 14.16.

The ratio of the steady-state amplitude of Y to

the static deflection yst; defined above, is known as

the dynamic magnification factor, D:

D ¼

Y

yst ¼

1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ð1 2 r2Þ2 þ ð2rjÞ2

p ð14:17Þ

D varies with the frequency ratio, r; and the

damping ratio, j: Equation 14.17 is plotted in

Figure 14.5 (Paz, 1997; Clough and Penzien,

1993). It can be seen from this figure that

0:8 # r # 1:2 is in the resonance zone. Therefore,

for design, r . 1:2 or r , 0:8 is required to

avoid resonance.

x = 0.7

x = 0.4

0.2

0.25

0.15

0.125

x=0

0 1

4

2

x=1

1

2

3

4

2 3

Dynamic magnification factor D

Frequency ratio γ = ω /ω

FIGURE 14.5 Dynamic magnification factor as a

function of the frequency ratio for various amounts of

damping.

k

y

(a) (b)

c

ky mÿ

Fo sin ωt cy· Fo sin ωt

FIGURE 14.4 (a) Damped oscillator harmonically excited; (b) free-body diagram.

Reinforced Concrete Structures 14-5

© 2005 by Taylor & Francis Group, LLC

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