NOMBRE Y FIRMA DEL PRESIDENTE SELLO DE LA ENTIDAD

It has been shown in the preceding sections that the equation of motion of a simple spring-mass system with damping can be expressed as

m ¨v(t) + c ˙v(t) + k v(t) = p(t) (2-19)

in which v(t) represents the dynamic response (that is, the displacement from the static-equilibrium position) and p(t) represents the effective load acting on the system, either applied directly or resulting from support motions.

The solution of Eq. (2-19) will be obtained by considering first the homogeneous form with the right side set equal to zero, i.e.,

m ¨v(t) + c ˙v(t) + k v(t) = 0 (2-20)

Motions taking place with no applied force are called free vibrations, and it is the free-vibration response of the system which now will be examined.

The free-vibration response that is obtained as the solution of Eq. (2-20) may be expressed in the following form:

v(t) = G exp(st) (2-21)

where G is an arbitrary complex constant and exp(st) ≡ est_{denotes the exponential}

ANALYSIS OF FREE VIBRATIONS 21 in expressing dynamic loadings and responses; therefore it is useful now to briefly review the complex number concept.

Considering first the complex constant G, this may be represented as a vector plotted in the complex plane as shown in Fig. 2-4. This sketch demonstrates that the vector may be expressed in terms of its real and imaginary Cartesian components:

G = GR+ i GI (2-22a)

or alternatively that it may be expressed in polar coordinates using its absolute value G(the length of the vector) and its angle θ, measured counterclockwise from the real axis:

G = G exp(i θ) (2-22b)

In addition, from the trigonometric relations shown in the sketch, it is clear that Eq. (2-22a) also may be written

G = G cos θ + i G sin θ (2-22c)

Using this expression and noting that cos θ = sin θ + π 2

and sin θ = − cos θ +

π 2

it is easy to show that multiplying a vector by i has the effect of rotating it counterclockwise in the complex plane through an angle of π

2 radians or 90 degrees.

Similarly it may be seen that multiplying by −i rotates the vector 90◦ _clockwise.

Now equating Eq. (2-22c) to Eq. (2-22b), and also noting that a negative imaginary component would be associated with a negative vector angle, leads to Euler’s pair of equations that serve to transform from trigonometric to exponential functions:

exp(iθ) = cos θ + i sin θ exp(−iθ) = cos θ − i sin θ

)

(2-23a) Furthermore, Eqs. (2-23a) may be solved simultaneously to obtain the inverse form of Euler’s equations: cos θ = 1 2  exp(iθ) + exp(−iθ) sin θ = −2i  exp(iθ) − exp(−iθ) ) (2-23b) Im Re G θ FIGURE 2-4

Complex constant representation in complex plane.

 G G_R=Gcos θ i G_I= iGsin θ G = G_R+ i G_I or G =G exp(i θ)

To derive a free-vibration response expression, Eq. (2-21) is substituted into Eq. (2-20), leading to

(m s2+ c s + k) G exp(st) = 0

and after dividing by mG exp(st) and introducing the notation

ω2≡ _mk (2-24)

this expression becomes

s2+ c ms + ω

2_{= 0 (2-25)}

The two values of s that satisfy this quadratic expression depend on the value of c relative to the values of k and m; thus the type of motion given by Eq. (2-21) depends on the amount of damping in the system.

Considering now the undamped system for which c = 0, it is evident that the two values of s given by solving Eq.(2-25) are

s1,2= ± i ω (2-26)

Thus the total response includes two terms of the form of Eq. (2-21), as follows:

v(t) = G1 exp(iωt) + G2 exp(−iωt) (2-27)

in which the two exponential terms result from the two values of s, and the complex constants G1and G2represent the (as yet) arbitrary amplitudes of the corresponding

vibration terms.

We now establish the relation between these constants by expressing each of them in terms of its real and imaginary components:

G1= G1R+ i G1I ; G2= G2R+ i G2I

and by transforming the exponential terms to trigonometric form using Eqs. (2-23a), so that Eq. (2-27) becomes

v(t) = G1R+ i G1I
cos ωt + i sin ωt
+ G2R+ i G2I
cos ωt − i sin ωt

or after simplifying

v(t) = (G1R+ G2R) cos ωt − (G1I− G2I) sin ωt

+ ih(G1I + G2I) cos ωt + (G1R− G2R) sin ωt

i

ANALYSIS OF FREE VIBRATIONS 23 However, this free-vibration response must be real, so the imaginary term (shown in square brackets) must be zero for all values of t, and this condition requires that

G1I = −G2I ≡ GI G1R= G2R≡ GR

From this it is seen that G1and G2are a complex conjugate pair:

G1= GR+ i GI G2= GR− i GI

and with these Eq. (2-27) becomes finally

v(t) = (GR+ i GI) exp(iωt) + (GR− i GI) exp(−iωt) (2-29)

The response given by the first term of Eq. (2-29) is depicted in Fig. 2-5 as a vector representing the complex constant G1rotating in the counterclockwise direction with

the angular velocity ω; also shown are its real and imaginary constants. It will be noted that the resultant response vector (GR+ i GI) exp(iωt)leads vector GRexp(iωt)by

the phase angle θ; moreover it is evident that the response also can be expressed in terms of the absolute value, G, and the combined angle (ωt + θ). Examination of the second term of Eq. (2-29) shows that the response associated with it is entirely equivalent to that shown in Fig. 2-5 except that the resultant vector G exp[−i(ωt+θ)] is rotating in the clockwise direction and the phase angle by which it leads the component GRexp(−iωt) also is in the clockwise direction.

The two counter-rotating vectors G exp[i(ωt + θ)] and G exp[−i(ωt + θ)] that represent the total free-vibration response given by Eq. (2-29) are shown in Fig. 2-6;

Im

Re

θ

−θ

FIGURE 2-6

Total free-vibration response.

(GR+ i GI) exp (iωt) =G exp[i( ω t + θ)] (GR− i GI) exp (− iω t) =G exp[− i(ω t + θ)] G_Rexp(iω t) GRexp(− iω t) 2G cos (ω t + θ) (ω t + θ) − (ω t + θ) Im Re θ =phase angle FIGURE 2-5

Portrayal of first term of Eq. (2-29).

ωt ωt i GIexp(iωt) G_Rexp(iω t) (GR+ i GI) exp(iωt) =Gexp[i ( ω t + θ)] whereG = G R2+ GI 2

it is evident here that the imaginary components of the two vectors cancel each other leaving only the real vibratory motion

v(t) = 2 G cos(ωt + θ) (2-30)

An alternative for this real motion expression may be derived by applying the Euler transformation Eq. (2-23a) to Eq. (2-29), with the result

v(t) = A cos ωt + B sin ωt (2-31)

in which A = 2GR and B = −2GI. The values of these two constants may be

determined from the initial conditions, that is, the displacement v(0) and velocity ˙v(0)at time t = 0 when the free vibration was set in motion. Substituting these into Eq. (2-31) and its first time derivative, respectively, it is easy to show that

v(0) = A = 2GR

˙v(0)

ω = B = −2GI (2-32)

Thus Eq. (2-31) becomes

v(t) = v(0) cos ωt + ˙v(0)

ω sin ωt (2-33)

This solution represents a simple harmonic motion (SHM) and is portrayed graphically in Fig. 2-7. The quantity ω, which we have identified previously as the angular velocity (measured in radians per unit of time) of the vectors rotating in the complex plane, also is known as the circular frequency. The cyclic frequency, usually referred to as the frequency of motion, is given by

f = ω 2π (2-34) Its reciprocal 1 f = 2π ω = T (2-35) t FIGURE 2-7

Undamped free-vibration response.

v(t) T = 2 (t + ) − . v(0) v(0)

ANALYSIS OF FREE VIBRATIONS 25 is the time required to complete one cycle and is called the period of the motion. Usually for structural and mechanical systems the period T is measured in seconds and the frequency is measured in cycles per second, commonly referred to as Hertz (Hz).

The motion represented by Eq. (2-33) and depicted by Fig. 2-7 also may be interpreted in terms of a pair of vectors, v(0) and ˙v(0)

ω rotating counter-clockwise in

the complex plane with angular velocity ω, as shown in Fig. 2-8. Using previously stated relations among the free-vibration constants and the initial conditions, it may be seen that Fig. 2-8 is equivalent to Fig. 2-5, but with double amplitude and with a negative phase angle to correspond with positive initial conditions. Accordingly, the amplitude ρ = 2G, and as shown by Eq. (2-30) the free vibration may be expressed as

v(t) = ρ cos(ωt + θ) (2-36)

in which the amplitude is given by ρ = r  v(0)2+h ˙v(0) ω i2 (2-37) and the phase angle by

θ = tan−1  − ˙v(0) ω v(0)  (2-38) 2-6 DAMPED FREE VIBRATIONS

If damping is present in the system, the solution of Eq. (2-25) which defines the response is s1,2 = − c 2m± r _c 2m 2 − ω2 _(2-39)

Three types of motion are represented by this expression, according to whether the quantity under the square-root sign is positive, negative, or zero. It is convenient to discuss first the case when the radical term vanishes, which is called the critically- damped condition.

Im

ρ

Re

FIGURE 2-8

Rotating vector representation of undamped free vibration. v(0) −θ ω t ω t (ω t +θ) . v(0) _ω

Critically-Damped Systems

If the radical term in Eq. (2-39) is set equal to zero, it is evident that c/2m = ω; thus, the critical value of the damping coefficient, cc, is

cc= 2 m ω (2-40)

Then both values of s given by Eq. (2-39) are the same, i.e., s1= s2= −

cc

2m= −ω (2-41)

The solution of Eq. (2-20) in this special case must now be of the form

v(t) = (G1+ G2t) exp(−ωt) (2-42)

in which the second term must contain t since the two roots of Eq. (2-25) are identical. Because the exponential term exp(−ωt) is a real function, the constants G1and G2

must also be real.

Using the initial conditions v(0) and ˙v(0), these constants can be evaluated leading to

v(t) =v(0) (1 − ωt) + ˙v(0) t exp(−ωt) (2-43)

which is portrayed graphically in Fig. 2-9 for positive values of v(0) and ˙v(0). Note that this free response of a critically-damped system does not include oscillation about the zero-deflection position; instead it simply returns to zero asymptotically in accordance with the exponential term of Eq. (2-43). However, a single zero-displacement crossing would occur if the signs of the initial velocity and displacement were different from each other. A very useful definition of the critically-damped condition described above is that it represents the smallest amount of damping for which no oscillation occurs in the free-vibration response.

v(t)

t

FIGURE 2-9

Free-vibration response with critical damping. v(0)

. v(0)

ANALYSIS OF FREE VIBRATIONS 27

Undercritically-Damped Systems

If damping is less than critical, that is, if c < cc(i.e., c < 2mω), it is apparent

that the quantity under the radical sign in Eq. (2-39) is negative. To evaluate the free-vibration response in this case, it is convenient to express damping in terms of a damping ratio ξ which is the ratio of the given damping to the critical value;

ξ ≡ c cc

= c

2mω (2-44)

Introducing Eq. (2-44) into Eq. (2-39) leads to

s1,2= −ξω ± i ωD (2-45)

where

ωD≡ ω

p

1 − ξ2 _(2-46)

is the free-vibration frequency of the damped system. Making use of Eq. (2-21) and the two values of s given by Eq. (2-45), the free-vibration response becomes

v(t) =G1exp(iωDt) + G2 exp(−iωDt)



exp(−ξωt) (2-47)

in which the constants G1and G2must be complex conjugate pairs for the response

v(t)to be real, i.e., G1 = GR+ i GI and G2= GR− i GI similar to the undamped

case shown by Eq. (2-27).

The response given by Eq. (2-47) can be represented by vectors in the complex plane similar to those shown in Fig. 2-6 for the undamped case; the only difference is that the damped circular frequency ωDmust be substituted for the undamped circular

frequency ω and the magnitudes of the vectors must be forced to decay exponentially with time in accordance with the term outside of the brackets, exp(−ξωt).

Following the same procedure used in arriving at Eq. (2-31), Eq. (2-47) also can be expressed in the equivalent trigonometric form

v(t) =A cos ωDt + B sin ωDt



exp(−ξωt) (2-48)

where A = 2GRand B = −2GI. Using the initial conditions v(0) and ˙v(0), constants

Aand B can be evaluated leading to v(t) =  v(0) cos ωDt +  ˙v(0) + v(0)ξω ωD  sin ωDt  exp(−ξωt) (2-49)

Alternatively, this response can be written in the form

in which ρ =  v(0)2₊ ˙v(0) + v(0)ξω ωD 21/2 (2-51) θ = − tan−1  ˙v(0) + v(0)ξω ωD v(0)  (2-52) Note that for low damping values which are typical of most practical structures, ξ < 20%, the frequency ratio ωD/ωas given by Eq. (2-46) is nearly equal to unity.

The relation between damping ratio and frequency ratio may be depicted graphically as a circle of unit radius as shown in Fig. 2-10.

A plot of the response of an undercritically-damped system subjected to an initial displacement v(0) but starting with zero velocity is shown in Fig. 2-11. It is of interest to note that the underdamped system oscillates about the neutral position, with a constant circular frequency ωD. The rotating-vector representation of Eq. (2-47) is

equivalent to Fig. 2-6 except that ω is replaced by ωD and the lengths of the vectors

diminish exponentially as the response damps out.

0 1

1 circle

FIGURE 2-10

Relationship between frequency ratio and damping ratio.

t

FIGURE 2-11

Free-vibration response of undercritically-damped system. D v(t) e− t v(0) v2 v1 v0  Dv(0)  D 2  D 4  D . v(0) = 0 3  D

ANALYSIS OF FREE VIBRATIONS 29 The true damping characteristics of typical structural systems are very complex and difficult to define. However, it is common practice to express the damping of such real systems in terms of equivalent viscous-damping ratios ξ which show similar decay rates under free-vibration conditions. Therefore, let us now relate more fully the viscous-damping ratio ξ to the free-vibration response shown in Fig. 2-11.

Consider any two successive positive peaks such as vn and vn+1which occur

at times n 2π ωD

and (n + 1)2π

ωD, respectively. Using Eq. (2-50), the ratio of these two

successive values is given by

vn/vn+1= exp(2πξω/ωD) (2-53)

Taking the natural logarithm (ln) of both sides of this equation and substituting ωD =

ωp1 − ξ2_{, one obtains the so-called logarithmic decrement of damping, δ, defined}

by

δ ≡ ln_vvn

n+1

= p2πξ

1 − ξ2 (2-54)

For low values of damping, Eq. (2-54) can be approximated by

δ= 2πξ. (2-55)

where the symbol .= represents “approximately equal,” thus, vn vn+1 = exp(δ) . = exp(2πξ) = 1 + 2πξ +(2πξ) 2 2! + · · · (2-56)

Sufficient accuracy is obtained by retaining only the first two terms in the Taylor’s series expansion on the right hand side, in which case

ξ=. vn− vn+1

2π vn+1 (2-57)

To illustrate the accuracy of Eq. (2-57), the ratio of the exact value of ξ as given by Eq. (2-54) to the approximate value as given by Eq. (2-57) is plotted against the approximate value in Fig. 2-12. This graph permits one to correct the damping ratio obtained by the approximate method.

1.00 0.75 0.50_{0 0.05 0.10 0.15 0.20} (approx.) (exact) _(approx.) FIGURE 2-12

Damping-ratio correction factor to be applied to result obtained from Eq. (2-57).

FIGURE 2-13

Damping ratio vs. number of cycles required to reduce peak amplitude by 50 percent. Damping ratio N o. cy cl es to re du ce pe ak am pl itu de by 50 % 6 5 4 3 2 1 0 0 0.05 0.10 0.15 0.20

For lightly damped systems, greater accuracy in evaluating the damping ratio can be obtained by considering response peaks which are several cycles apart, say m cycles; then

ln vn vn+m

= p2mπξ

1 − ξ2 (2-58)

which can be simplified for low damping to an approximate relation equivalent to Eq. (2-57):

ξ=. vn− vn+m 2 m π vn+m

(2-59) When damped free vibrations are observed experimentally, a convenient method for estimating the damping ratio is to count the number of cycles required to give a 50 percent reduction in amplitude. The relationship to be used in this case is presented graphically in Fig. 2-13. As a quick rule of thumb, it is convenient to remember that for percentages of critical damping equal to 10, 5, and 2.5, the corresponding amplitudes are reduced by 50 percent in approximately one, two, and four cycles, respectively.

Example E2-1. A one-story building is idealized as a rigid girder sup- ported by weightless columns, as shown in Fig. E2-1. In order to evaluate the dynamic properties of this structure, a free-vibration test is made, in which the roof system (rigid girder) is displaced laterally by a hydraulic jack and then suddenly released. During the jacking operation, it is observed that a force of 20 kips [9, 072 kg]is required to displace the girder 0.20 in [0.508 cm]. After the instantaneous release of this initial displacement, the maximum displace- ment on the first return swing is only 0.16 in [0.406 cm] and the period of this displacement cycle is T = 1.40 sec.

From these data, the following dynamic behavioral properties are deter- mined:

ANALYSIS OF FREE VIBRATIONS 31 k 2 k 2 c Weight W = mg v p = jacking force FIGURE E2-1

Vibration test of a simple building.

(1) Effective weight of the girder: T =2π ω = 2π s W g k = 1.40 sec Hence W = 1.40 2π 2 g k = 0.0496 20 0.2386 = 1, 920 kips [870.9 × 10 3_kg]

where the acceleration of gravity is taken to be g = 386 in/sec2

(2) Undamped frequency of vibration: f = 1 T = 1 1.40= 0.714 Hz ω = 2πf = 4.48 rad/sec (3) Damping properties: Logarithmic decrement: δ = ln0.20 0.16 = 0.223 Damping ratio: ξ =. δ 2π = 3.55% Damping coefficient: c = ξ cc= ξ 2mω = 0.0355 2(1, 920) 386 4.48

= 1.584 kips · sec/in [282.9 kg · sec/cm]

Damped frequency: ωD= ω

p

1 − ξ2_{= ω(0.999)}1/2_{= ω}.

(4) Amplitude after six cycles: v6=  v1 v0 6 v0=  4 5 6 (0.20) = 0.0524 in [0.1331 cm]

Overcritically-Damped Systems

Although it is very unusual under normal conditions to have overcritically- damped structural systems, they do sometimes occur as mechanical systems; therefore, it is useful to carry out the response analysis of an overcritically-damped system to make this presentation complete. In this case having ξ ≡ c/cc > 1, it is convenient to

write Eq. (2-39) in the form

s1,2 = −ξω ± ω p ξ2_{− 1 = −ξω ± ˆω (2-60)} in which ˆ ω ≡ ωpξ2_{− 1 (2-61)}

Substituting the two values of s given by Eq. (2-60) into Eq. (2-21) and simplifying leads eventually to

v(t) = [A sinh ˆωt + B cosh ˆωt] exp(−ξωt) (2-62)

in which the real constants A and B can be evaluated using the initial conditions v(0)and ˙v(0). It is easily shown from the form of Eq. (2-62) that the response of an overcritically-damped system is similar to the motion of a critically-damped system as shown in Fig. 2-9; however, the asymptotic return to the zero-displacement position is slower depending upon the amount of damping.

PROBLEMS

2-1. The weight W of the building of Fig. E2-1 is 200 kips and the building is set into free vibration by releasing it (at time t = 0) from a displacement of 1.20 in. If the maximum displacement on the return swing is 0.86 in at time t = 0.64 sec, determine:

(a) the lateral spring stiffness k (b) the damping ratio ξ (c) the damping coefficient c

2-2. Assume that the mass and stiffness of the structure of Fig. 2-1a are as follows: m = 2 kips · sec2_{/in, k = 40 kips/in. If the system is set into free vibration}

with the initial conditions v(0) = 0.7 in and ˙v(0) = 5.6 in/sec, determine the displacement and velocity at t = 1.0 sec, assuming:

(a) c = 0 (undamped system) (b) c = 2.8 kips · sec/in

2-3. Assume that the mass and stiffness of the system of Fig. 2-1a are m = 5 kips · sec2_{/inand k = 20 kips/in, and that it is undamped. If the initial displacement}

is v(0) = 1.8 in, and the displacement at t = 1.2 sec is also 1.8 in, determine: (a) the displacement at t = 2.4 sec

CHAPTER

3

RESPONSE

TO HARMONIC

LOADING

3-1 UNDAMPED SYSTEM Complementary Solution

Assume the system of Fig. 2-1 is subjected to a harmonically varying load p(t) of sine-wave form having an amplitude p_oand circular frequency ω as shown by the equation of motion

m ¨v(t) + c ˙v(t) + k v(t) = p_o sin ωt (3-1)

Before considering this viscously damped case, it is instructive to examine the behavior of an undamped system as controlled by

m ¨v(t) + k v(t) = p_o sin ωt (3-2)

which has a complementary solution of the free-vibration form of Eq. (2-31)

vc(t) = A cos ωt + B sin ωt (3-3)

Particular Solution

The general solution must also include the particular solution which depends upon the form of dynamic loading. In this case of harmonic loading, it is reasonable to assume that the corresponding motion is harmonic and in phase with the loading; thus, the particular solution is

vp(t) = C sin ωt (3-4)

in which the amplitude C is to be evaluated. Substituting Eq. (3-4) into Eq. (3-2) gives

−m ω2C sin ωt + k C sin ωt = p_osin ωt (3-5)

Dividing through by sin ωt (which is nonzero in general) and by k and noting that k/m = ω2_{, one obtains after some rearrangement}

C =po k h ₁ 1 − β2 i (3-6) in which β is defined as the ratio of the applied loading frequency to the natural free-vibration frequency, i.e.,

β ≡ ω / ω (3-7)

General Solution

The general solution of Eq. (3-2) is now obtained by combining the comple- mentary and particular solutions and making use of Eq. (3-6); thus, one obtains

v(t) = vc(t) + vp(t) = A cos ωt + B sin ωt + p_o k h ₁ 1 − β2 i sin ωt (3-8)

In this equation, the values of A and B depend on the conditions with which the response was initiated. For the system starting from rest, i.e., v(0) = ˙v(0) = 0, it is easily shown that

A = 0 B = −poβ k h ₁ 1 − β2 i (3-9) in which case the response of Eq. (3-8) becomes

v(t) =po k h ₁ 1 − β2 i (sin ωt − β sin ωt) (3-10)

where p_o/ k = vst is the displacement which would be produced by the load po

applied statically and 1/(1 − β2_{)is the magnification factor (MF) representing the}

amplification effect of the harmonically applied loading. In this equation, sin ωt represents the response component at the frequency of the applied loading; it is called the steady-state response and is directly related to the loading. Also β sin ωt is the response component at the natural vibration frequency and is the free-vibration effect controlled by the initial conditions. Since in a practical case, damping will cause the last term to vanish eventually, it is termed the transient response. For this hypothetical undamped system, however, this term will not damp out but will continue indefinitely.

RESPONSE TO HARMONIC LOADING 35 Response Ratio — A convenient measure of the influence of dynamic loading is provided by the ratio of the dynamic displacement response to the displacement produced by static application of load p_o, i.e.,

R(t) ≡ v(t) vst =

v(t)

p_o/k (3-11)

From Eq. (3-10) it is evident that the response ratio resulting from the sine-wave loading of an undamped system starting from rest is

R(t) =h 1

1 − β2

i

(sin ωt − β sin ωt) (3-12)

It is informative to examine this response behavior in more detail by reference to Fig. 3-1. Figure 3-1a represents the steady-state component of response while Fig. 3- 1b represents the so-called transient response. In this example, it is assumed that β = 2/3, that is, the applied loading frequency is two-thirds of the free-vibration frequency. The total response R(t), i.e., the sum of both types of response, is shown in Fig. 3-1c. Two points are of interest: (1) the tendency for the two components

+

=

FIGURE 3-1

Response ratio produced by sine wave excitation starting from at-rest initial conditions: (a) steady state; (b) transient; (c) total R(t).

T_p= 2 T = 2 Frequency ratio = 2₃ t t t MF ×MF (a) (b) (c) R_p(t) R_s(t) R(t)

to get in phase and then out of phase again, causing a “beating” effect in the total response; and (2) the zero slope of total response at time t = 0, showing that the initial velocity of the transient response is just sufficient to cancel the initial velocity of the

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