Solution manual for dynamics of structures 4 e 4th edition 0132858037 by geraldine.roberson709

Solution Manual for Dynamics of Structures, 4/E 4th Edition :

0132858037

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CHAPTER 3

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Problem 3.1 From the given data, k = ω2 = (2π f )2 = (8π)2 = 64π2 (a) m n n k = (ω′)2 = (2π f ′)2 = (6π)2 = 36π2 m + Δm n n (b) Dividing Eq. (a) by Eq. (b) gives 1 + Δm m = 16 9 m = 9 Δm = 9 5 = 6.43 lbs g (c) 7 From Eq. (a), 7 g g k = 64π 2 m = 64π2 6.43 g = 10 52 lbs in (d)

Problem 3.2

At ω = ω n , from Eq. (3.2.15), 1

o = (ust )o 2ζ = 2 (a)

At ω = 0 1ω n, from Eq. (3.2.13),

uo ≈ (ust )o = 0 2

Substituting (ust )o = 0 2 in Eq. (a) gives

ζ = 0.05

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Assuming that damping is small enough to justify the approximation that the resonant frequency is ω n and the resonant amplitude of Rd is 1 2ζ, then the given data implies:

0.0576

of small damping implied in Eq. (a) is reasonable; otherwise we would have to use the exact resonant frequency

and exact resonant

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) n n n n ⎝

Problem 3.3

(uo )ω = ω = (u ) 1 st o 2ζ (a) (uo )ω = 12ω 1 = (ust )o 2 2 1 (ω ωn )2 + 2ζω ω n = (ust )o [1 1 (1 2)2 ]2 + [2ζ (1 2)]2 (b) Combining Eq. (a) and Eq. (b): 2 ⎛ 2 ⎞ 1 ⎜ (uo )ω=1 2ωn ⎟ 1 For (2ζ )2 ⎜ (uo 2 ⎟ ω=ωn ⎠ = ( 0.44)2 +(2.4ζ )2 (c) (uo )ω = 12ω = 1 (uo )ω = ω 4 Equation

64ζ2

5 76ζ2 ⇒ ζ

= ω n 1 ⎡ 2ζ2 2 ⎤

amplitude = (u st ) o ⎢ ⎣2ζ 1 2ζ .

= 0.1935 +

Assumption

Problem 3.4

(a) Machine running at 20 rpm

(d) Summarizing these results together with given data:

The isolator is effective at ωω n = 0.9; it reduces the deformation amplitude to 39% of the response without

isolators. At ωω = 0 1 or 3, the isolator has essentially

n no influence on reducing the deformation.

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ωω n (uo )ζ = 0 (uo )ζ=0.25 0.1 0.2 0.1997 0.9 1.042 0.4053 3.0 0.0248 0.0243 st o 2

ω = ω n 20 200 = 0.1 (u ) uo = (a) 1 (ω ω n ) or

0 2 = (ust )o ⇒ (u ) = 0 1980 in

1 (0 1)2 st o

For ζ = 0 25 and ωω n = 0 1, 1 uo = (ust )o 2 2 2 (b) 1 (ω ωn ) + 2ζωωn or uo = 0.1980 [1 1 (0 1)2]2 + [2 (0 25) (0 1)]2 = 0 1997 in

ω = ωn 180 200 = 0 9 From Eq. (a), (u ) 1 042 = st o ⇒ (ust )o = 0 1980 in 1 (0.9)2 For ζ = 0 25 and ωω n = 0 9, Eq. (b) reads 1 uo = 0.1980 2 2 1 = 0 4053 in (0 9)2 + 2 (0 25)(0 9)

(b) Machine running at 180 rpm

ω = ωn 600 = 3 200

Eq. (a),

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This publication

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in any form or by any

electronic, mechanical, photoco5pying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. (u ) 0.0248 = st o ⇒ (ust )o = 0.1980 in. 1 (3)2

0 25 and ωωn = 3, Eq. (b) reads 1 uo = 0.1980 [1 (3)2 ]2 + [2 (0.25)(3)]2 = 0.0243 in.

Upper Saddle River,

reserved.

is protected by

should

obtained

or transmission

means,

For

Problem

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3.5

= 1200 lbs, E = 30 × 106 psi, I = 10 in 4 , L = 8 ft; ζ = 1% ⎛300⎞ po = 60 lbs; ω = ⎜ ⎟ 2π = 10πrads sec

of two beams: ⎝ 60 ⎠ k = 2 ⎜ 48 ⎟ = 32,552 lbs in. ⎛ EI ⎞ ⎝ L3 ⎠ Natural frequency: k k 32,552 ωn = = = = 102 3 rads sec m w g 1200 386 Steady state response: 1 Rd = 2 2 1 (ω ωn )2 + 2ζω ω n where ωωn = 10π102.3 = 1 Rd = 0.3071. Therefore, = 1 104 [1 0 0943]2 + [2 × 0 01 × 0 3071]2 Displacement: p uo = (ust )o Rd = o Rd k = 60 32,552 × 1.104 = 2.035 × 10 3 in Acceleration amplitude: 2 2 3 u && o = ω uo = (10π) 2.035 × 10 = 2 009 in sec 2 = 0 0052g

Given: w

Stiffness

Problem 3.6

In Eq. (3.2.1) replacing the applied force by po cosωt and dividing by m we get

u && + 2ζω nu & + ω 2 u = po cosωt m (a)

(a) The particular solution is of the form:

u p (t) = C sinωt + D cosω t (b)

Differentiating once and then twice gives

u & p (t) = Cω cosωt Dω sinωt (c)

u && p (t) = Cω 2 sinωt Dω 2 cosωt (d)

Substituting Eqs. (b)-(d) in Eq. (a) and collecting terms: 2 2 (ω n ω ) C 2ζω nω D

sinωt p + 2ζω n ω C + (ω n ω ) D cosωt = o cosωt m

Equating coefficients of sin ωt and of cosωt on the two sides of the equation:

Solving Eqs. (e) and (f) for C and D gives

(ωn ω ) + (2ζωnω)

(g)

Substituting Eqs. (g) and (h) in Eq. (b) gives u (t) = po 1 (ωωn ) cos ωt + 2ζωω n sin ωt k 1 (ω ω )2 2 + 2ζω ω 2 n n

(b) Maximum deformation is uo =

Substituting for C and D gives C2 + D2

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n n 2

C = D =

2 2

(ω n ω ) C (2ζωnω) D = 0 p (e)

ζωnω) C + (ω

ω2 ) D = o m (f)

2 2 2 2

p ω2 ω2 o n 2 2 2 2

ζωnω

(h)

(ωn ω ) + (2ζωnω)

u = po 1

This is same as Eq (3 2 11) for the amplitude of deformation due to sinusoidal force.

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o k 2 2 1 (ω ωn )2 + 2ζω ω n

Problem 3.7

(a) The displacement amplitude in given by Eq. (3.2.11):

where β = ωω n Resonance occurs at β when uo is maximum, i.e., duo respect to β gives

= 0. Differentiating Eq. (a) with

(b) Substituting Eq. (b) in Eq. (a) gives the resonant amplitude:

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2 2 2 1 2 uo

ust

o (1 β ) + (2ζβ

(a)

= (

)

1 2 2 2 3 2 (1 β ) 2 + (2ζβ) × 2(1 β 2 ) ( 2β) + 2 (2ζβ) 2ζ = 0 or 4 (1 or β 2) β + 8ζ 2β = 0 1 β 2 = 2ζ 2 ⇒ β = 1 2ζ 2 (b) Resonant frequency: ω r = ω n 1 2ζ 2

uo = (ust )o (2ζ2 )2 1 + 4ζ2 (1 2ζ2 ) or uo = (ust )o 2ζ 1 1 ζ 2 (c)

Problem 3.8

(a) From Eq. (3.2.19) the acceleration amplitude is

where β = ωω n Resonance occurs at β where u && o is maximum, i.e., du && o dβ = 0 Differentiating Eq. (a) with respect to β and setting the result equal to zero gives

(b) Dividing both the numerator and denominator of Eq (a) by

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⎜ ω 2 ⎧ 2 o ⎟ ⎜ ⎟

u && o = po 1

m 2ζ 1 ζ2 po u & = R 2 = po ⎛ ω ⎞ o m a Rd m ⎝ n ⎠ p β= 2 = o (a) m (1 β2 )2 1 2 + (2ζβ)

1 (1 β 2)2 +(2ζβ )2 ⎨2β ⎩ [(1 β 2)2 + (2ζβ )2 ]1 2 ⎜β ⎟ [(1 β 2)2 + (2ζβ )2] 1 2[ 4β (1 β 2) + 8ζ 2β] =0 ⎛ 2 ⎞ ⎜ ⎟ ⎬ ⎝ ⎠ ⎪ ⎭ Multiplying the numerator by [(1 and dividing it by β

β 2)2 + (2ζβ)2 ] 1 2 2 (1 β 2)2 + 4ζ 2β 2 β= 2 2 4 (1 β 2) + 8ζ 2 = 0 or 1 2β 2 + β 4 + 4ζ 2β 2 + β 2 β 4 2ζ 2β 2 = 0 or 1 = β 2 (1 2ζ 2) ⇒β = 1 (b) 1 2ζ2 Resonant frequency: ω r = ωn (c) 1 2ζ 2

gives

β 2

u & o p 1 = 1 2 (d)

gives:

Substituting Eq. (b) in Eq. (d) gives the resonant amplitude:

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⎝ 2 ⎠ m ⎡⎛ 1 ⎞2 4ζ 2 ⎤ ⎢⎜ 1⎟ + ⎥ ⎢⎜β 2 ⎟ ⎣ β 2 ⎥ ⎦

p 1 u && o = o m ( 2ζ2 )2 + 4ζ2 (1 1 2 2ζ )

(a) From Eq (3.2.17), the velocity amplitude is

0. Differentiating Eq. (a) with respect to β and setting the derivative equal to zero gives

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0 0 ⎛ω ⎞ p u &0 = p Rv = ⎜ ω ⎟ d km km ⎝ n ⎠ = p0 β km [(1 β2)2 + (2ζβ)2]1/ 2 (a) where β=ω/ω n Resonance occurs

u & is maximum, i.e., du &

/ dβ =

[(1 β 2 )2 +(2ζβ)2] 1/2 β[(1 β ) +(2ζβ) ] [2(1 β )( 2β)+2(2ζβ)2ζ ]=0 2 Multiplying the numerator by [(1 β 2 )2 +(2ζβ)2 ]3/2 gives [(1 β2)2 +(2ζ β)2] β[2(1 β2)( 2β)+2(2ζβ)(2ζ )]=0 2 or 1 2β 2 +β 4 +4ζ 2β 2 β ( 2β+2β 3 +4ζ 2β) = 0 or β 4 =1⇒ β=1 (b) Resonant frequency: ω r =ω n (b) Substituding Eq. (b) into Eq. (a)

where

gives

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Problem 3.10

We assume that the structure has no mass other than the roof mass.

From Eq. (3.3.4),

At ω = ω n , Ra = 1 2ζ and Eq (a) gives

ζ = 1 me e ω2 (b)

2u && o m

Given:

me = 2 × 50 g = 100 lbs g = 0.1kips g

m = 500 kips g

e = 12 in

ω n = 2π fn = 2π × 4 = 8π 2

u && o = 0.02g = 7.72 in. sec

Substituting the above data in Eq (b) gives

ζ = 1 01 (12) (8π)2 2 (7.72) 500 = 0.0982 = 9.82%

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m ω

m e ω2 Fω I 2

&& o = e n Ra

n (a)

Problem 3.11

The given data is plotted in the form of the frequency response curve shown in the accompanying figure:

(a) Natural frequency

The frequency response curve peaks at fn = 1 487 Hz

Assuming small damping, this value is the natural frequency of the system

(b) Damping ratio

The acceleration at the peak is rpeak = 8.6 ×10 3 g. 3 Draw a horizontal line at rpeak ÷ 2 = obtain fa and fb in Hz: 6 08 × 10 g to fa = 1 473 Hz fb = 1 507 Hz

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Then, ζ = fb fa = 2 fn 1 507 1 473 2 (1 487)

= 1.14%

= 0.0114

Problem 3.12

(a) Transmissibility is given by Eq. (3.5.3) with ζ = 0. Thus the force transmitted is

The negative value is invalid. Therefore,

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2 st

1 ( fT)o = po 1 (ω ω n ) where ω n = g δst and ω 1 = 2π f .

( fT)o = po 4π2 2 (a) 1 f g δ st (b) For ( fT)o 0 1 = = 0 1 po and f =

(a)

1 ⇒ 1 (4π 2 386)(20)2 δ 4π 2 2 1 (20) 386 δst = ±10 ⇒ 4π 2 2(20) δst = 9 or 11 386

Therefore

20, Eq.

gives

δst =

(4π 2

(20)2 =

386)

0 269 in

Problem 3.13

The equation governing the deformation u(t) in the suspension system is

mu&& + cu & + ku = mu&& g (t) (a)

where ug (t) = ugo sinωt and ω = 2π v L

Substituting in Eq. (a) gives

mu&& + cu & + ku = mω 2 ugo sinωt (b)

The amplitude of deformation is

uo = (ust )o Rd

= 2 u = Rd k

The amplitude of the spring force is

Numerical calculations:

k = 800 lbs in.; m = 4000 386 lb-sec

ugo = 3 in.

ω n = 8 768 rads sec; ω = 3 686 rads sec

ωω n = 0.420

Thus Eq. (d) gives

Substituting ωn and Rd in Eq. (c) gives

fSo = 4000 386 (3.686)2 3(1124) = 474 8 lbs

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2 fSo

where Rd = kuo = mω ugoRd 1 2 2 (c) (d) 1 (ω ωn )2 + 2ζω ω n

2 ]2 1

= [1 (0 42)

+ [2 (0.4)(0.42)]

= 1.124

Spring force: f S = ku

The resonant frequency for f S is the same as that for uo , which is the resonsnat frequency for Ra (from section 3.2.5):

Resonance occurs at this forcing frequency, which implies a speed of

10.655g100 =1696 ft /sec =116 mph

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o d 2 2

ug btg= ugo sinωt ⇒ u && g btg= ω ugo sinω t peff btg= mu && g btg= mω ugo sinωt ≡ po sinωt mω ugo Fω I u0 =bust g R = 2 2 Rd = ugo Rd = ugo Ra k ω n

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Problem 3.14

res = ωn 1 2ζ 2 =

/ sec 1 2(0.4)2

8786 =10655 rads

= ωL = b

2π 2π

Problem 3.15

Given: w = 2000 lbs, f = 1500 cpm = 25 Hz, and TR = 0 10

For an undamped system, TR = 1

= 0.1

For TR < 1, ωω n > 2 and the equation becomes

11 6 kips in

(47 31) 2000 386

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1 (ω ωn )

1 (ω ω )2

⇒ 1 (ω ω )2 =

⇒ ωω =

n n ω = ω = 2πf = 2π

= n 3 32 3.32 47.31rads

3.32 2 2 2 k = ωnm = ωn w g =

= 0.1

3 32

(25)

sec

Problem 3.16

The excitation is

The equation of motion is mu&& where

+ cu & + ku = peff (t) (b) peff (t) = m u && g (t) = ω m ugo sinωt (c)

The deformation response is given by Eq. (3.2.10) with po replaced by ω2 m ugo :

2 m u

(t) = go R sin (ωt φ) (d)

where Rd and φ are defined by Eqs. (3.2.11) and (3.2.12), respectively.

The total displacement is

Substituting Eqs. (a), (d), (3.2.11) and (3.2.12) gives

Equation (e) is of the form u t (t) = C sin

hence uo = C

+ D . Substituting for C and D gives

Equation (f) is the same as Eq. (3.6.5).

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a f a f 2 | V 2 2 2 = |

ug (t) = ugo sinωt or u && g (t) =

ugo sinωt (a)

k d

u t (t) = u(t) +

(

aω ωf2 u u t (t)= u go sinω n t + 1 aω ω n f n go 2 2 2 + 2ζω ω n × n 1 aω ω n f sinω n t 2ζ ω ω n cosω ts 1 aω ω n f u go 2 2 2 + 2ζω ω n × { 1 aω ω n f + 4ζ aω ω n f sinω n t 2ζaω ω n f cosω t} 2 2 2 3

t 2 2

U1/2 t | 1+b2ζωωn g

uo = ugo S T 1 bω ω n g2 2 2 + 2ζωω n W (f)

ωt (e) + D cos

t ;

Problem 3.17

The accelerometer properties are fn = 50 cps and ζ = 0 7; and the excitation is

u && g (t) = 0 1g sin (2π f t); f = 10, 20 and 40 Hz (a)

Compare the excitation with the measured relative displacement

( 1 ω2 ) R u && g (t φ ω)

(b)

(c)

Table

P3.17

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n d n

Rd =

2 2 1

ωωn )2

ζω ω n and ⎡ 2ζ ωω ⎤ φ = tan 1 ⎢ n ⎥ (d) 1

u(t) = where ωω n )2

(

+ 2

(

t) = Rd u && g (t φ ω)

Because the instrument has been calibrated at low excitation frequencies, after separating the instrument constant 1 ω2 , the recorded acceleration is

(

(e)

For a given f (or ω), ωω n is computed and substituted in Eqs. (c) and (d) to calculate Rd and φ With Rd and φ known for a given excitation frequency, the recorded acceleration is given by Eq. (e).

f (Hz) ωω n Rd φ ω, sec % error in Rd 10 0.2 1 0.0045 0 20 0.4 0.991 0.0047 0.9 40 0.8 0.850 0.0050 15

The computed amplitude Rd and time lag φ ω agrees with Fig. 3.7.3. The difference between Rd and 1 is the error in measured acceleration (Table P3.17).

We want to bound Rd as follows:

The relevant condition is Rd ≤ 1 01 because we are interested in a continuous range of frequencies over which the error is less than 1%. Therefore, impose

This publication

the

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River,

07458. n d g n 2 Problem 3.18 From Eq. (3.7.1), ω2 u(t) = R u && (t φ ω) f ≤ 4 86 Hz = Rd u && go sin (2π f t and therefore, 2 φ) ω nuo = R u&& d = 1 2 2 (a) go 1 (ω ωn )2 + 2ζω ω n Equation (a) is plotted for ζ = 0 6, the accelerometer damping ratio: 1 5 1 01 ω 2 uo & u&go 1 0 5 0 194 0 723 0 0 0 5 1 1 5 2 Error < 1% f fn or ωωn

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0 99 ≤ Rd ≤ 1 01 (b)

1 = 1 01 (c) 1 (ω ω )2 2 + 2ζω ω 2 Defining β n n ≡ ωωn , Eq. (c) can be rewritten as (1 β 2 )2 + (2ζβ)2 = ⎜ 1 ⎟ ⇒ ⎛ ⎞ β4 0 56β2

Choose β = 0 194 frequency range: (see figure) which gives the desired

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+ 1 = ⎝ 1.01⎠ 0.980 3 ⇒ β4 0.56β2 + 0 0197 = 0 ⇒ β

2 = 0 0377, 0 5223 ⇒ β = 0 194,

723

f ≤

⇒

0 194 fn = 0 194 (25)

We want to bound Rd as follows 0.99

The relevant condition is 0 99 ≤ Rd because Rd is always smaller than 1.01 in this case. Therefore, impose

which gives the desired frequency range:

in a

2 2 2

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Problem 3.19

From Eq. (3.7.1), f ≤ 0.405 fn = 0 405(50) ⇒ ω n u(t) = Rd u&& g (t φ ω) f ≤ 20 25 Hz = Rd u && go sin (2π f t and therefore, 2 φ) ω nuo = R u&& d = 1 2 2 (a) go 1 (ωωn )2 + 2ζω ω n Equation (a) is plotted for ζ = 0 7, the accelerometer damping ratio: 1 5 ω n uo &u&go 1 0 5 0 99 0 405 0 0 0 5 1 1 5 2 Error < 1% f f n or ωω n

≤ R

≤ 1 01 (b)

1 = 0 99 (c) 1 (ω ω )2 2 + 2ζω ω 2 Defining β ≡ n n ωωn , Eq. (c) can be rewritten as (1 β 2 )2 + (2ζ β)2 = ⎜ 1 ⎟ ⇒ ⎛ ⎞

0.99

β 4 004β 2 + 1 = 10203 ⇒

β 4 0.04β 2 0.0203 = 0 ⇒

β 2 = 0.1239, 01639

Take only the positive value:

β 2 = 01639 ⇒ β = 0405

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⎝

⎠

Problem 3.20

From Eq. (3.7.3),

For maximum accuracy, uo ugo = 1; this condition after using Eqs. (3.2.20) and (3.2.11) for Ra gives

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uo = Ra ugo (a)

(ωω n )2 = 1 (b) 2 2 1 (ωωn )2 + 2ζω ω n Squaring, simplifying and defining β = ωω n : (1 β 2)2 + (2ζβ)2 = β 4 ⇒ 1 2β 2 + 4ζ 2β 2 = 0 ⇒ ζ 2 = 1 1 2 4β 2 For ωω n >> 1 (β >> 1), ζ 2 = 1 ⇒ ζ = 0 707 2

Problem 3.21

From Eq. (3.7.3),

Equation (a) is plotted for ζ = 0 6, the instrument damping ratio:

We want to bound Ra as follows (see figure) Ra ≤ 1 01 (b)

Therefore imposing Ra = 1 01 which, after defining

, gives

Squaring this equation gives

1.383, 5149

Choose β = 5.149 frequency range: (see figure) which gives the desired

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⎜ ⎠

u (ω ω )2 o = Ra = n (a) ugo

ω ω

2ζω ω 2 n n

1 (

)2 2 +

1 5 1 Ra 0 5 1 01 1 383 5 149 Error < 1% 0 0 1 2 3 4 5 6 f fn or ωω n

β ≡ ωωn

β 2

101 (1 β 2)2 + (2ζβ)2

2 ⎛ β 2 ⎞ (1 β 2)2 + (2ζβ)2 =⎜ ⎝ ⎟ ⇒ 1.01⎟ 1 2β 2 + β 4 + 144β 2 = 0.9803β 4 ⇒ 0.0197β 4 0.56β 2 + 1 = 0 ⇒ β

2 = 1.914, 26.51 ⇒ β =

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Problem 3.22

From Eq. (3.7.3),

Equation (a) is plotted for ζ = 0 7, the instrument damping ratio:

Since Ra is always smaller than 1.01, we want to bound Ra as follows

Therefore imposing Ra = 0.99 (b) which, after defining

Squaring this equation gives

Take only the positive value:

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u (ω ω )2 o = Ra = n (a) ugo 1 (ω ω )2 2 + 2ζω ω 2 n n

1 5 1 0 99 0 5 Error < 1% 2 470 0 0 1 2 3 4 5 6 f f n or ω ωn

a ≥ 0.99

β ≡ ωω

β 2

0.99 (1 β 2)2

(2ζβ)2

n , gives

2 ⎛ ⎞ (1 β 2 )2 + (2ζβ)2 = ⎜β ⎟ ⇒ ⎜ 0.99⎟ 1 2β 2 + β 4 ⎝ + 196β 2 ⎠ = 10203β 4 ⇒ 0.0203

0.04β 2 1 = 0 ⇒ β 2

= 8073, 6.102

β2 = 6.102 ⇒ β = 2 470

which gives the desired frequency range:

f ≥ 2 470 fn = 2.470 (0.5) ⇒

f ≥ 1.235 Hz

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Problem 3.23

From Eq. (3.8.1),

ED where

= 2πζ (ω ω n ) ku2 (a)

u = po R (b)

o k d

In Eq (a) substituting Eq (b) and Eq (3 2 11) for Rd gives

π p 2 2ζωω

o 2

E = o n k 1 (ω ω )2 + 2ζωω 2 n n

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Problem 3.24

From Eq. (3.8.9) the loss factor is

(a)

where ED =πcω u 2 and ESo = k u 2 2. Substituting these in Eq. (a) gives

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o o

1 ED 2π ESo

ξ =

ξ = c ω k (b)

Based on equivalent viscous damping, the displacement amplitude is given by

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o T 2 g

Problem 3.25

uo = {1 [(4 π) (F p )]2 }1 2 (a) (ust ) o From the given data 1 (ω ω n ) F po ω ω n = 50 = 100 = Tn = T 1 2 0 25 = 0.3 0 833 Substituting these data in Eq. (a) gives uo = (u st ) o Now, {1 [(4 π) (1 2)]2 }1 2 1 (0 833)2 = 2 52 (u where st )o = po k po = 100 kips 2 ⎛ π⎞ w k = ω 2 m = ⎜ 2 ⎟ = (8π)2 500 n ⎜ ⎟ ⎝ n ⎠ 386 = 818kips in. Thus (ust )o = 100 818 = 0.1222 in. and uo = 2

= 0

1222)

308

Because of the j4 in the denominator of the series, two

terms are enough to obtain reasonable convergence of the

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to:

o T 0 0 0 u T T 0 0T | Problem 3.26 where (ust )o = po k and β j = jω0 ω n . Equation (f) is (a) p(t) is an even function: indeterminate when β j = 1; corresponding values of T0 are T n , 3T n , 5T n , etc. p(t) = ⎛ 2 ⎞ p o ⎜1 t ⎟ 0 ≤ t ≤ T0 2 (a) (c) For T T = 2, β = jω ω = j T T = j 2 and ⎜ ⎟ 0 n j 0 n n 0 ⎝ T0 ⎠ T 2 T 2 Eq. (f) becomes ∞ 1 2 ⎛ 2 ⎞ u(t) 1 16 1 2πjt a = p(t)dt = p ⎜1 t ⎟ dt = + 2 ∑ 2 2 cos ( ) 0 T ∫ T ∫ o ⎜ T ⎟ (u ) 2 π j (4 j ) T 0 p = 2 T T0 2 z 0 0 ⎝ cos ω 0 ⎠ (b) st o j =1,3,5,L 0 1 two terms 2 a j = T0 2pbtg b j 0tgdt 0.8 0.6 three terms 0 T0 2 u (t ) = 4 p T0 z 2F 2 t I cos b jω tgdt ( ) st o 0.4 0 H G1 0 0 0 0 = p0 R ω b jω tg T 2 z t b jω tgdt U 0.2 0 4 1 0 2 T0 2 T0 | T j 0 8 0 z sin 0 0 F 2π I cos 0 W 0 0.2 0.4 0.6 0.8 1 t T T = p0 2 0 t jt dt 2 cos T 0 0 0 4 p = L Mt F 2πj I t T0 2 T L F 2πj IO t T0 2

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SN M sin J KQ P + 0 McosG JP V

π j

(b) The steady-state response of an undamped system is obtained by substituting Eqs. (b), (c) and

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N 0 b g = t Q 2 ∞ 2 ∞ T | T T 2π j M N H T KP Q W | series solution. 0 0 0 0 0 L OT0 2 2 p0 M F2π j I P = cos 2 j2 T π 0 0 2 p = cos π j 1 π 2 j2 ⎧ 4 po ∴a j ⎪ ⎨ π j 2 ⎪ j = 1,3,5,L (c) ⎩ 0 j = 2,4,6,L bn = 0 because p(t)is an even function (d) Thus the Fourier series representation of p(t) is p(t) = po + 4 po ∑ 1 cos ( jω0t) (e) 2 2 2 π j =1,3,5,L j

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u(t) = 1 + 4 ∑ 1 2 2 cos ( jω0t) (ust )o 2 π j =1,3,5,L j (1 β j ) (f)

(d) in Eq (3.13.6)

obtain