CHAPTER 5: TRANSIENT VIBRATIONS OF SDOF SYSTEMS Short Answer Problems 5.1 True: The convolution integral solves the differential equation for a SDOF system. 5.2 True: Alternate derivations of the convolution integral use variation of parameters (assuming a particular solution which is a linear combination of the homogenous solution with coefficients that vary with time and deriving integrals for the coefficients) or the Laplace transform method (applying the Laplace transform method to the differential equation, reducing the solution for the transform of the differential equation to the product of two transforms and using the convolution property to invert) 5.3 False: The effect of an impulse applied to a SDOF system is to cause a discrete change in velocity. 5.4 False: The Laplace transform method involves the initial conditions in the solution of the differential equation. 5.5 True: Numerical integration of the convolution integral can be obtained by interpolating the forcing function and exactly integrating the interpolation times . The method uses piecewise impulses, piecewise constants or piecewise linear functions to interpolate the forcing function. 5.6 True: Self-starting methods use initial conditions to start the integration process. 5.7 False: The transfer function for a SDOF system is the ratio of the Laplace transform of the output to the Laplace transform of the input. 5.8 False: The transfer function is the Laplace transform of the impulsive response of a system. 5.9 False: The maximum force transferred to the foundation from a machine mounted on an isolator due to an impulsive force is minimized by selecting the damping ratio of the system to be 0.25. 5.10 True: The transmitted force is flat for a damping ratio between 0.23-0.3. 5.11 The function impulsive response.
is the response of a system due to a unit impulse, or the system's
5.12 The principle of impulse and momentum is used in the derivation of the convolution integral. 369 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
5.13 The convolution integral represents the solution to the differential equation governing the motion of a SDOF system due to any type of excitation. represents the response of a system with an impulsive 5.14 1 1 response h(t) at a time of 1 second due to an applied force F(t). 5.15 A pulse of short duration is applied over a short enough time such that the shape of the pulse has little effect on the response of the system. Only the total impulse imparted to the system by the pulse has an effect. The system is then modeled as a system undergoing free vibrations with an initial displacement equal to zero and an initial velocity equal to I/m where I is the impulse imparted by the pulse. 5.16 The response spectrum of a pulse is a plot of time / 2 .
/ versus the nondimensional
5.17 The impulsive response of a system due to motion input does not exist because an impulsive motion input means an instantaneous change in motion input which is not possible. 5.18 Given: m = 2 kg, k = 1000 N/m, N· 6 m/s.
12 N·s. The velocity imparted to the system is
5.19 Given: system shown,
5.20 Given: system shown,
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Chapter 5: Transient Vibrations of SDOF Systems
5.21 Given: system shown,
0 The natural frequency of the system is
5.22 Given: m = 5 kg, k = 1000 N/m, N/
14.1 rad/s. The impulsive response of an undamped system is
sin
/
sin 14.1
0.141 sin 14.1 .
15 N · s, m = 0.5 kg, k = 200 N/m. The natural frequency of the system is
5.23 Given: N/ . N· .
.
/
20 rad/s. The response of the system is sin 20
sin
1.5 sin 20 .
5.24 (a)-(vi); (b)-(iii); (c)-(i); (d)-(i); (e)-(iii); (f)-(ii)
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Chapter 5: Transient Vibrations of SDOF Systems
Chapter Problems 5.1 A SDOF system with m = 20 kg, k = 10000 N/m and c = 540 N · s/m is at rest in equilibrium when a 50 N · s impulse is applied. Determine the response of the system. Given: m = 20 kg, k = 10000 N/m and c = 540 N · s/m, I = 50 N · s Find: Solution: The natural frequency and damping ratio of the system are 22.36
rad s
0.604
2
The system’s damped natural frequency is 17.83 rad/s
1
The response of an underdamped SDOF system to an impulse is sin 0.1402
.
50 N · s 20 kg 17.83 rad/s sin 17.83
.
.
sin 17.83
Problem 5.1 illustrates the response of a SDOF system subject to an impulse.
5.2 A SDOF system is with m = 10 kg, k = 40,000 N/m, and c = 300 N · s/m is at rest in equilibrium when a 80 N · s impulse is applied. This is followed by a 40 N · s impulse 0.02 sec later. Determine the response of the system. Given: m = 10 kg, k = 40,000 N/m and c = 300 N · s/m, 0.02 s
= 80 N · s,
= 40 N · s,
Find: Solution: The natural frequency and damping ratio of the system are 63.25
rad s
0.237
2 372
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Chapter 5: Transient Vibrations of SDOF Systems
The system’s damped natural frequency is 61.44 rad/s
1
The response of an underdamped SDOF system to the two impulses is obtained using the method of superposition as sin
sin 80 N · s
rad 10 kg 61.44 s 40 N · s rad 10 kg 61.44 s 0.1302
sin 61.44
.
.
.
.
sin 61.44 .
.
0.0651
sin 61.44
sin 61.44
0.02
0.02
0.02 0.02
Problem 5.2 illustrates application of multiple impulses to an underdamped SDOF system.
5.3 A SDOF system with m = 1.3 kg, k = 12,000 N/m, and c = 400 N · s/m is at rest in equilibrium when a 100 N · s impulse is applied. This is followed by a 150 N · s impulse 0.12 sec later. Determine the response of the system. Given: m = 1.3 kg, k = 12,000 N/m, and c = 400 N · s/m, 0.12 s
= 100 N · s,
= 150 N · s,
Find: Solution: The natural frequency and damping ratio of the system are 96.07
rad s
1.60
2
The response of an overdamped SDOF system to the two impulses is obtained using the method of superposition as 2
Noting that
1 2
1
1
120.2 rad/s
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Chapter 5: Transient Vibrations of SDOF Systems 10 N · s rad 2 120.2 s
.
.
.
150 N · s rad 2 120.2 s .
0.4161
.
.
.
.
.
.
0.6242
.
.
.
.
.
0.12
.
0.12
Problem 5.3 illustrates application of multiple impulses to an overdamped SDOF system.
5.4 Use the method of variation of parameters to obtain the general solution of Equation (5.1) &x& + 2ζω n x& + ω n2 x =
F (t ) m
(1)
and show that it can be written in the form of the convolution integral, Equation (5.25).
x(t ) =
1 mωd
t
∫ F (τ )e
−ζωnt
sin ωd (t − τ )dτ
(2)
0
Given: Equation (1) Show: Equation (2) using variation of parameters Solution: The homogeneous solution of eq. (1) is
x h (t ) = C1 e −ζω nt cos ω d t + C 2 e −ζω nt sin ω d t
Application of the method of variation of parameters involves assuming a particular solution of the form of the homogeneous solution, but with the constants replaced by unknown functions of time, x(t ) = C1 (t )e −ζω nt cos ω d t + C 2 (t )e −ζω nt sin ω d t
(3)
Differentiating eq. (3) with respect to time
(
)
x& (t ) = C1 − ζω n e −ζω nt cos ω d t − ω d e −ζω nt sin ω d t + C& 1e −ζω nt cos ω d t + C − ζω e −ζω nt sin ω t + ω e −ζω nt cos ω t + C& e −ζω nt sin ω t 2
(
n
d
d
d
)
2
d
The algebra is simplified by choosing
e −ζω nt cos ω d tC& 1 + e −ζω nt sin ω d tC& 2 = 0
(4)
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Chapter 5: Transient Vibrations of SDOF Systems
Then
(
&x&(t ) = C& 1 − ζω n e −ζω nt cos ω d t − ω d e −ζω nt sin ω d t
(
) )
+ C 1 ζ 2ω n2 e −ζω nt cos ω d t + 2ζω nω d e −ζω nt sin ω d t − ω d2 e −ζω nt cos ω d t + C& − ζe −ζω nt sin ω t + ω e −ζω nt cos ω t 2
(
d
d
d
)
(
+ C 2 ζ 2ω n2 e −ζω nt sin ω d t − 2ζω nω d e −ζω nt cos ω d t − ω d2 e −ζω nt sin ω d t
(5)
)
Substituting into eq.(1) and simplifying leads to
C&1e −ζωnt (− ζωn cos ωd t − ωd sin ωd t )
F (t ) + C& 2 e −ζωnt (− ζωn sin ωd t + ωd cos ωd t ) = m
(6)
Equation (6) is simplified by using eq. (4)
F (t ) − ωd C&1e −ζωnt sin ωd t + ωd C& 2 e −ζωnt cos ωd t = mωd
(7)
Equations (5) and (8) can be solved simultaneously yielding F (t ) ζωnt C&1 = e sin ω d t mω d
F (t ) ζωnt C& 2 = e cos ω d t mω d
F (τ ) ζωnτ C1 (t ) = − ∫ e sin ω dτ dτ mω d 0 t
t
C2 (t ) = ∫ 0
(8)
F (τ ) ζωnτ e cos ω dτ dτ mω d
Substitution of eq. (8) into eq. (3) leads to t
x(t ) = ∫
0
F (τ ) −ζωn (t −τ ) (− cos ωd t sin ωdτ + sin ωd t cos ωdτ ) dτ e mωd F (τ ) −ζωn (t −τ ) =∫ e sin ωd (t − τ ) dτ mωd 0 t
Problem 5.4 illustrates derivation of the convolution integral using the method of variation of parameters. 375 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
5.5 Use the convolution integral to determine the response of an underdamped SDOF system of mass m and natural frequency ωn when the excitation is the unit step function, u(t).
Given: m, ωn, ζ < 1, F(t) = F0u(t) Find: x(t) Solution: Using the convolution integral t
x(t ) = ∫ F (τ ) h(t − τ ) dτ 0
1
= ~ mω d
(1)
t
∫ F u(τ )e
−ζω n (t −τ )
0
sin ω d (t − τ ) dτ
0
The integral is easiest evaluated by letting
u = t −τ
(2)
Changing the variable of integration from τ to u leads to
F x(t ) = − ~ 0 mω d
0
∫
e −ζω nu sin ω d u du
(3)
t
The integral in eq. (3) can be evaluated by referring to a table of integrals or integration by parts twice. Either method leads to u =0
F ⎡ω − e −ζω nu (ζ sin ω d t + ω d cos ω d t ) ⎤ x(t ) = ~ 0 ⎢ d ⎥ mω d ⎣ ω n2 ⎦ u =t F 1 − ζ 2 − e −ζω nt ζ sin ω d t + 1 − ζ 2 cos ω d t = ~ 0 mω nω d
[
)]
(
Problem 5.5 illustrates the application of the convolution integral to determine the response of an underdamped one-degree-of-freedom system to the unit step function.
5.6 Let g(t) be the response of an underdamped system to a unit step function and h(t) the response of an underdamped system to the unit impulse function. Show
h(t ) =
dg dt
(1)
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Chapter 5: Transient Vibrations of SDOF Systems
Given: h(t), g(t) Show: eq.(1) Solution: The response of an underdamped system to a unit impulse is 1 h(t ) = ~ e −ζω nt sin ω d t mω d
(2)
From problem 5.5 the response of an underdamped system to the unit step function is F g (t ) = ~ 0 mω n ω d
[ 1−ζ
2
)]
(
− e −ζω nt ζ sin ω d t + 1 − ζ 2 cos ω d t
(3)
Differentiating eq.(3) with respect to t, using the product rule
[
)
(
F dg =− ~ 0 − ζω n e −ζω nt ζ sin ω d t + 1 − ζ 2 cos ω d t dt mω n ω d
(
+ e −ζω nt ζω d cos ω d t − ω d 1 − ζ 2 sin ω d t
[(
(
)
)]
)
]
F dg = ~ 0 e −ζω nt ω n 1 − ζ 2 + ζ 2 sin ω d t + ζ 1 − ζ 2 − ζ 1 − ζ 2 cos ω d t dt mω nω d
Hence F dg = ~ 0 e −ζω nt sin ω d t dt mω d
= h(t )
Problem 5.6 illustrates that the response of a system due to a unit impulse function equals the derivative of the response of the system due to the unit step function.
5.7 Use the convolution integral and the notation and results of Chapter Problem 5.6 to derive the following alternative expression for the response of a system subject to an excitation, F(t): t
x(t ) = F (0 ) g (t ) + ∫
0
dF (τ ) g (t − τ ) dτ dτ
(1)
Given: x(t) Find: Show eq. (1) 377 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
Solution: The response of a one-degree-of-freedom system subject to the excitation F(t) is determined using the convolution integral as t
x(t ) = ∫ F (τ ) h(t − τ ) dτ
(2)
0
The general integration by parts formula is
∫ u dv = uv − ∫ v du
Let u = F (τ ) dv = h (t − τ ) dτ
The results of problem 5.6 show that
h(t ) =
dg dt
Hence dF dτ dτ v = − g (t − τ ) du =
Application of integration by parts to eq. (2) results in
x(t ) = [− F (τ ) g (t − τ )]
τ =t τ =0
t
dF g (t − τ )dτ dτ
+∫ 0
or t
x(t ) = F (0 ) g (t ) − F (t ) g (0 ) + ∫ 0
dF g (t − τ )dτ dτ
But g(0) = 0, hence eq.(1) is obtained Problem 5.7 illustrates an alternate form of the convolution integral.
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Chapter 5: Transient Vibrations of SDOF Systems
5.8 A SDOF undamped system is initially at rest in equilibrium and subject to a force F(t) = F0te-t/2. Use the convolution integral to determine the response of the system.
Given: m, ω, F(t) Find: x(t) Solution: The convolution integral yields the response of an undamped system as t
1 F (τ )sin ω n (t − τ ) dτ x(t ) = mω n ∫0
For the excitation given τ
t
− 1 2 F τ e sin ω n (t − τ ) dτ x(t ) = 0 mω n ∫0
Evaluation of the integral leads to t t ⎧ − − 2 2 ⎪ ω te 1 ω ne F − x(t ) = − 0 ⎨− n 1 2 ⎛ 2 1 ⎞2 mω n ⎪ ω 2 + n ⎜ω n + ⎟ 4 ⎩ 4⎠ ⎝
⎫ ⎪ ⎡⎛ 1 ⎤⎪ 1 1 2⎞ − ω n ⎟ sin ω nt + ω n cos ω nt ⎥ ⎬ + 2 ⎢⎜ 2 ⎠ ⎦⎪ ⎛ 2 1 ⎞ ⎣⎝ 4 ⎜ω n + ⎟ ⎪⎭ 4⎠ ⎝
Problem 5.8 illustrates use of the convolution integral to determine the transient response of an undamped one-degree-of-freedom system.
5.9 The mass of Figure P5.9 has a velocity v when it engages the spring-dashpot system. Let x(t) be the displacement of the mass from the position where the mechanism is engaged. Use the convolution integral to determine x(t). Assume the system is underdamped.
Given: m, v, k, c, θ Find: x(t)
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Chapter 5: Transient Vibrations of SDOF Systems
Solution: Since x(t) is measured from when the mechanism is engaged, the force in the spring in the initial position is zero. Thus, consider free-body diagrams of the system at an arbitrary time. mg
=
:
Kx
mx
. Cx
N EXTERNAL FORCES
EFFECTIVE FORCES
Summing forces in the direction along the surface
(∑ F )
ext
= (∑ F )eff
leads to mg sin θ − kx − c x& = m &x& m&x& + cx& + kx = mg sin θ
&x& + 2ζω n x& + ω n2 x = mg sin θ
where
ωn =
k c , ζ = m 2 mk
The initial conditions for the motion are
x(0 ) = 0, x& (0 ) = v
The convolution integral with non-zero initial conditions, is used with
F (t ) = mg sin θ
leading to x(t ) =
v
ωd
e
−ζω n t
sin ω d t +
1 mω d
t
∫ mg sin θ e
−ζω nτ
sin ω d (t − τ ) dτ
0
⎛ ⎞⎤ g sin θ ⎡ ζ −ζω n t ⎜ ⎟⎥ ⎢ = + − − e −ζω nt sin ω d t + 1 e sin ω t cos ω t d d 2 ⎜ ⎟⎥ ωd ω n2 ⎢ 1−ζ ⎝ ⎠⎦ ⎣
v
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Chapter 5: Transient Vibrations of SDOF Systems
Problem 5.9 illustrates (a) that when the generalized coordinate is measured from a position other than the system’s equilibrium position, nonhomogeneous terms occur in the governing equation and, (b) the use of the convolution integral with non-zero initial conditions.
5.10 Use the convolution integral to determine the response of the system of Figure P5.10.
Given: System shown Find: x(t) Solution: Free body diagrams of the system at an arbitrary instant are shown below. = R
Mo e-t/5 2K (
m
L θ 6
Lθ ) 3
EXTERNAL FORCES
EFFECTIVE FORCES
1 mL2 θ 12
:
Lθ ) 3
L .2 θ 6
:
K(
m
Summing moments about the point of support
(∑ M )
= (∑ M 0 )eff .
0 ext .
−k
L L L L 1 L L θ − 2 k θ + M 0 e −t / 5 = mL2θ&& + m θ&& 3 3 3 3 12 6 6 t 2 2 − L L m θ&& + k = M 0e 5 9 3
The system is undamped with a natural frequency of
ωn =
3k m
The convolution integral is used to write the solution as t
τ
− 9M θ (t ) = 2 0 ∫ e 5 sin ω n (t − τ ) mL ω n 0
Integration and application of limits results in
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Chapter 5: Transient Vibrations of SDOF Systems
θ (t ) =
t − ⎛ ⎞ 1 9M 0 1 5 ⎜ e + sin ω nt − ω n cos ω nt ⎟⎟ ω n 2 ⎜ 5 mL ω n ω 2 + 1 ⎝ ⎠ n 25
Problem 5.10 illustrates application of the convolution integral to determine the response of a one-degree-of-freedom system.
5.11 Use the convolution integral to determine the response of an underdamped SDOF system of natural frequency ωn and damping ratio ζ when subject to a harmonic excitation F(t) = F0 sin ωt.
Given: F(t) = F0 sin ωt Find: x(t) Solution: The convolution integral for the response of the system is
x(t ) =
1 mω d
t
∫F
0
sin(ωτ )e −ζω n (t −τ ) sin[ω d (t − τ )] dτ
0
The integral is evaluated to yield x(t ) = −
[
F0 − 2ζωω n cos(ωt ) + (ω n2 − ω 2 ) sin ωt mD
]
Fω 0 −ζω nt e − 2ζωω d cos(ω d t ) + (ω n2 (1 − 2ζ 2 ) − ω 2 ) sin(ωt ) mω d D
[
]
where D = (ω n2 + ω 2 + 2ωω d )(ω n2 + ω 2 − 2ωω d )
Problem 5.11 illustrates the use of the convolution integral to determine the forced response of a damped system.
5.12 A machine tool with a mass of 30 kg is mounted on an undamped foundation of stiffness 1500 N/m. During operation, it is subject to the machining forces shown in Figure P5.12. Use the principle of superposition and the convolution integral to determine the response of the system to each force.
Given: m = 30 kg, k = 1500 N/m, F(t) 382 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
Find: x(t) Solution: The natural frequency of the system is
ωn =
k = m
N m = 7.07 rad 30 kg sec
1500
The graphical breakdown of F(t) into functions whose response is presented in Table 5.1 is shown below. 3000N
3000N
.5
.5
3000N
+
.5
2.0
2
The mathematical form of F(t) is
F (t ) = 6000t [u (t ) − u (t − 0.5 )]+ (− 2000t + 4000 )[u (t − 0.5 ) − u (t − 2.)] N
The response of the system is given by
x(t ) = x1 (t ) − x2 (t ) + x3 (t ) − x4 (t )
where each xi(t), i = 1, 2, 3, 4 is determined using Table 5.1 as shown below. x1(t): Ramp function, A = 6000 N/sec, B = 0, t0 = 0 N sec ⎡t − 1 sin 7.07 t ⎤ u (t ) x1 (t ) = ⎥⎦ N ⎢⎣ 7.07 1500 m = 4 [t − 0.141sin 7.07 t ]u (t ) m 6000
x2(t): Ramp function, A = 6000 N/sec, B=0, t0 = 0.5 sec N sec [t − 0.5 cos 7.07(t − 0.5) − 0.141sin 7.07(t − 0.5)]u (t − 0.5) x2 (t ) = N 1500 m = 4 [t − 0.5 cos(7.07t − 3.54) − 0.141sin (7.07t − 3.54)]u (t − 0.5) m 6000
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Chapter 5: Transient Vibrations of SDOF Systems
x3(t): Ramp function, A = -2000 N/sec, B = 4000 N, t0 = 0.5 sec N sec [t − 2. − (0.5 − 2 ) cos 7.07(t − 0.5) x3 (t ) = N 1500 m − 0.141sin 7.07(t − 0.5)]u (t − 0.5) = −1.33 [t − 2. + 1.5 cos(7.07t − 3.54 ) − 0.141sin (7.07t − 3.54 )]u (t − 0.5) m − 2000
x4(t): Ramp function, A = -2000 N/sec, B = 4000 N, t0 = 2.0 sec N sec [t − 2. − 0.141sin 7.07(t − 2.)]u (t − 2.) x4 (t ) = N 1500 m = −1.33 [t − 2. − 0.141sin (7.07t − 14.14 )]u (t − 2.) − 2000
Problem 5.12 illustrates (a) graphical breakdown of an excitation whose form changes at discrete times and (b) use of superposition and Table 5.1 to determine the response of an undamped system.
5.13 A machine tool with a mass of 30 kg is mounted on an undamped foundation of stiffness 1500 N/m. During operation, it is subject to the machining forces shown in Figure P5.13. Use the principle of superposition and the convolution integral to determine the response of the system to each force.
Given: m = 30 kg, k = 1500 N/m, F(t) Find: x(t) Solution: The natural frequency of the system is
ωn =
k = m
N m = 7.07 rad 30 kg sec
1500
The graphical breakdown of the excitation into functions whose responses are available in Table 4.1 is shown below.
-
2
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Chapter 5: Transient Vibrations of SDOF Systems
The mathematical from of the excitation is
F (t ) = 1000 sin πt [u (t ) − u (t − 2)] N
or since sinπt is periodic of period 2,
F (t ) = 1000 sin πtu (t ) − 1000 sin π (t − 2)u (t − 2) N
The response of the system is
x(t ) = x1 (t ) − x2 (t )
where x1(t) and x2(t) are obtained using Table 5.1 x1(t): Sinusoidal excitation, A = 1000 N, ω = π rad/sec, t0 = 0 Note that ω/ωn = 0.444, then 1000 N ⎡ 1 (sin πt − sin 7.07t )− 1 (sin πt + sin 7.07t )⎤⎥ ⎢ N 0.444 + 1 ⎛ ⎞ 0.444 − 1 ⎦ 2⎜1500 ⎟ ⎣ m⎠ ⎝ 1 = (− 1.799 sin πt + 1.7999 sin 7.07t − 0.693 sin πt − 0.693 sin 7.07 t ) 3 1 = (− 2.49 sin πt + 1.097 sin 7.07 t ) m 3
x1 (t ) =
x2(t): Sinusoidal excitation, A = 1000 N, ω = π rad/sec, t0 = 2 sec 1000 N {− 1.799[sin π (t − 2) − sin 7.07(t − 2)] N⎞ ⎛ 2⎜1500 ⎟ m⎠ ⎝ − 0.693[sin π (t − 2 ) + sin 7.07(t − 2 )] }u (t − 2 )
x2 (t ) =
=
1 [− 2.49 sin πt + 1.097 sin (7.07 t − 14.14)]u (t − 2) m 3
Problem 5.13 illustrates (a) the graphical breakdown of an excitation whose form changes with time and (b) application of superposition and Table 5.1 to determine the response of an undamped system to an excitation whose form changes at discrete times.
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Chapter 5: Transient Vibrations of SDOF Systems
5.14 A machine tool with a mass of 30 kg is mounted on an undamped foundation of stiffness 1500 N/m. During operation, it is subject to the machining forces shown in Figure P5.14. Use the principle of superposition and the convolution integral to determine the response of the system to each force.
Given: m = 30 kg, k = 1500 N/m, F(t) as shown Find: x(t) Solution: The system’s natural frequency is
ωn =
k = m
N m = 7.07 rad 30 kg sec
1500
The graphical breakdown of the solution is as shown below 500
500
-
1 500
+ 1
1.5
1.5
The mathematical form of the excitation is
F (t ) = 500[u (t ) − u (t − 1)] + (− 1000t + 1500 )[u (t − 1) − u (t − 1.5 )] The system response is given by
x(t ) = x1 (t ) − x2 (t ) + x3 (t ) − x4 (t )
where x1, x2, x3, and x4 are determined from Table 5.1 as follows x1: Step function, A = 500 N, t0 = 0
x1 =
500 N 1 ( 1 − cos 7.07 t ) u (t ) = (1 − cos 7.07 t ) u (t ) m N 3 1500 m 386
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Chapter 5: Transient Vibrations of SDOF Systems
x2: Step function, A = 500 N, t0 = 1 sec x2 (t ) =
500 N 1 [ 1 − cos 7.07(t − 1)]u (t − 1) = [1 − cos(7.07t − 7.07 )]u (t − 1) m N 3 1500 m
x3: Ramp function, A = -1000 N/sec, B = 1500 N, t0 = 1 sec N sec ⎡t − 1.5 + 0.5 cos 7.07(t − 1) − 1 sin 7.70(t − 1)⎤u (t − 1) x3 (t ) = − ⎥⎦ N ⎢⎣ 7.07 1500 m 2 = [t − 1.5 + 0.5 cos(7.07t − 7.07 ) − 0.141sin (7.07t − 7.07 )]u (t − 1) m 3 1000
x4: Ramp function, A = -1000 N/sec, B = 1500 N, t0 = 1.5 sec N sec ⎡t − 1.5 − 1 sin 7.07(t − 1.5)⎤u (t − 1.5) x4 (t ) = ⎥⎦ N ⎢⎣ 7.07 1500 m 2 = − [t − 1.5 − 0.141sin (7.07t − 10.61)]u (t − 1.5) m 3 − 1000
Problem 5.14 illustrates (a) the graphical breakdown of an excitation whose form changes with time, and (b) the use of Table 5.1 to and the superposition principle to determine the response of an undamped system.
5.15 A machine tool with a mass of 30 kg is mounted on an undamped foundation of stiffness 1500 N/m. During operation, it is subject to the machining forces shown in Figure P5.15. Use the principle of superposition and the convolution integral to determine the response of the system to each force.
Given: m = 30 kg, k = 1500 N/m, F(t) Find: x(t) Solution: The natural frequency of the system is
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Chapter 5: Transient Vibrations of SDOF Systems
ωn =
k = m
N m = 7.07 rad sec 30 kg
1500
The graphical breakdown of F(t) into functions whose response are available from Table 5.1 is shown below.
1000
1000
1000
+ .1
.1
1000
.1
1000
-
+ -
.3
.3
.5
-1000
+
0.5 0.6
-
-1000
.6
The mathematical form of F(t) is
F (t ) = 10000 t [u (t ) − u (t − 0.1)] + 1000 [u (t − 0.1) − u (t − 0.3 )] + (− 10000 t + 40000 )[u (t − 0.3 ) − u (t − 0.5 )] + (10000 t − 6000 )[u (t − 0.5 ) − u (t − 0.6 )]
Superposition is used to write
x(t ) = x1 (t ) − x2 (t ) + x3 (t ) − x4 (t ) + x5 (t ) − x6 (t ) + x7 (t ) − x8 (t )
where xi(t), i = 1,…,8 are determined from Table 5.1 as shown below x1(t): Ramp function, A = 10000 N/sec, B = 0, t0 = 0 10000 N ⎛ 1 ⎞ sin 7.07t ⎟ u (t ) ⎜t − N ⎠ 1500 ⎝ 7.07 m = 6.67 (t − 0.141sin 7.07t )u (t ) m
x1 (t ) =
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Chapter 5: Transient Vibrations of SDOF Systems
x2(t): Ramp function, A = 10000 N/sec, B = 0, t0 = 0.1 sec 10000 N [t − 0.1cos 7.07(t − 0.1) − 0.141sin 7.07(t − 0.1)]u (t − 0.1) N 1500 m = 6.67[t − 0.1 cos(7.07t − 0.707 ) − 0.141sin (7.07t − 0.707 )]u (t − 0.1) m
x2 (t ) =
x3(t): Step function, A = 1000 N, t0 = 0.1 sec 1000 N [1 − cos 7.07(t − 0.1)]u (t − 0.1) N 1500 m = 0.667[1 − cos(7.07t − 0.707 )]u (t − 0.1) m
x3 (t ) =
x4(t): Step function, A = 1000 N, t0 = 0.3 sec 1000 N [1 − cos 7.07(t − 0.3)]u (t − 0.3) N 1500 m = 0.667[1 − cos(7.07t − 2.12 )]u (t − 2.12 ) m
x4 (t ) =
x5(t): Ramp function, A = -10000 N/sec, B = 4000 N, t0 = 0.3 sec
x5 (t ) = −6.67[t − 0.4 + 0.1cos(7.07t − 2.12) − 0.141sin(7.07t − 2.12)]u (t − 0.3) m
x6(t): Ramp function, A = -10000 N/sec, B = 4000 N, t0 = 0.5 sec
x6 (t ) = −6.67[t − 0.4 − 0.1cos(7.07t − 3.54) − 0.141sin(7.07t − 3.54)]u (t − 0.5) m
x7(t): Ramp function, A = 10000 N/sec, B = -6000 N, t0 = 0.5 sec
x7 (t ) = 6.67[t − 0.6 + 0.1cos(7.07t − 3.54) − 0.141sin(7.07t − 3.54)]u (t − 0.5) m
x8(t): Ramp function, A = 10000 N/sec, B = -6000 N, t0 = 0.6 sec
x8 (t ) = 6.67[t − 0.6 − 0.141sin(7.07t − 4.24)]u (t − 0.6) m
Problem 5.15 illustrates (a) graphical breakdown of an excitation that changes form at discrete times to functions whose responses can be determined from Table 5.1 and (b) use of superposition and Table 5.1 to determine the response of an undamped system due to an excitation that changes form at discrete times.
389 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
5.16 A machine tool with a mass of 30 kg is mounted on an undamped foundation of stiffness 1500 N/m. During operation, it is subject to the machining forces shown in Figure P5.16. Use the principle of superposition and the convolution integral to determine the response of the system to each force.
Given: F(t), m = 30 kg, k = 1500 N/m, Find: x(t) Solution: the natural frequency of the system is
ωn =
rad k 30 kg = = 7.07 N sec m 1500 m
The graphical breakdown of F(t) is shown below.
-
.5
-1000t u(t-.5)
-1000t
(1000t-1000) u(t-1.5) 500 u(t-1.5)
+
-
.5
+ 1.5
1.5
(1000t-1000) u(t-.5)
The mathematical form of F(t) is
F (t ) = −1000 t [u (t ) − u (t − 0.5 )] + (1000 t − 1000 )[u (t − 0.5 ) − u (t − 1.5 )] + 500 u (t − 1.5 )
The response of the system is
x(t ) = x1 (t ) − x2 (t ) + x3 (t ) − x4 (t ) + x5 (t )
where x1 , x2 , x3 , x4 , and x5 are determined using Table 5.1 as show below x1(t): Ramp function, A = -1000 N/sec, B = 0, t0 = 0 390 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
N sec (t − 0.141sin 7.07t )u (t ) x1 (t ) = − N 1500 m 1000
x2: Ramp function, A = -1000 N/sec, B = 0, t0 = 0.5 sec
x2 (t ) = −0.667[t − 0.5 cos(7.07t − 3.54 ) − 0.141sin(7.07t − 3.54 )]u (t − 0.5 )
x3: Ramp function, A= 1000 N/sec, B=-1000 N, t0=0.5 sec
x3 (t ) = −0.667[t − 1 + 0.5 cos(7.07t − 3.54 ) − 0.141sin(7.07t − 3.54 )]u (t − 0.5 )
x4: Ramp function, A = 1000 N/sec, B = -1000 N, t0 = 1.5 sec
x4 (t ) = 0.667[t − 1 − 0.5 cos(7.07t − 10.61) − 0.141sin(7.07t − 10.61)]u (t − 1.5 )
x5: Step function, A = 500 N, t0 = 1.5 sec
x5 (t ) = 0.333[1 − cos(7.07t − 10.61)]u (t − 1.5 )
Problem 5.16 illustrates (a) graphical breakdown of an excitation whose form changes with time and (b) the use of Table 5.1 and superposition to determine the response of a onedegree-of-freedom system due to an excitation whose form changes at discrete values of time.
5.17 A machine tool with a mass of 30 kg is mounted on an undamped foundation of stiffness 1500 N/m. During operation, it is subject to the machining forces shown in Figure P5.17. Use the principle of superposition and the convolution integral to determine the response of the system to each force.
Given: F(t), m = 30 kg, k = 1500 N/m Find: x(t) Solution: the natural frequency of the system is
ωn =
rad k 30 kg = = 7.07 N sec m 1500 m
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Chapter 5: Transient Vibrations of SDOF Systems
The graphical breakdown of F(t) is shown below.
300tu (t)
300tu (t-2)
-
600e
+
2
-.2(t-2)
2
The mathematical form of F(t) is
F (t ) = 300t [u (t ) − u (t − 2 )] + 600e −0.2(t −2 )u(t − 2 )
The response of the system is
x(t ) = x1 (t ) − x2 (t ) + x3 (t )
where x1, x2, and x3 are determined using Table 5.1 as shown below x1: Ramp function, A = 300 N/sec, B = 0, t0 = 0 N sec (t − 0.141sin 7.07t )u (t ) x1 (t ) = − N 1500 m 300
x2: Ramp function, A = 300 N/sec, B = 0, t0 = 2 sec
x2 (t ) = 0.2[t − 2 cos(7.07 t − 14.14 ) − 0.141sin(7.07t − 14.14 )]u (t − 2 )
x3: Exponential function, A = 600 N, α = 0.2 sec-1, t0 = 2 sec
[
0 .2 sin (7.07 t − 14.14 ) − 7.07 1 cos(7.07 t − 14.14 )] u (t − 2 ) 2 ⎛ 0 .2 ⎞ 1+ ⎜ ⎟ ⎝ 7.07 ⎠ = 0.3997 e −0.2 (1− 2 ) + 0.0283 sin (7.07 t − 14.14 ) − cos(7.07 t − 14.14 ) u (t − 2 ) x3 (t ) = 0.4 e −0.2 (t − 2 ) +
[
]
Problem 5.17 illustrates (a) graphical breakdown of an excitation whose form changes with time and (b) the use of Table 5.1 and superposition to determine the response of a onedegree-of-freedom system due to an excitation whose form changes at discrete values of time. 392 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
5.18 A machine tool with a mass of 30 kg is mounted on an undamped foundation of stiffness 1500 N/m. During operation, it is subject to the machining forces shown in Figure P5.18. Use the principle of superposition and the convolution integral to determine the response of the system to each force.
Given: F(t), m = 30 kg, k = 1500 N/m Find: x(t) Solution: the natural frequency of the system is
ωn =
k 30 kg rad = = 7.07 N m sec 1500 m
The graphical breakdown of F(t) is shown below. 100tu (t-1) 100tu (t)
-
100 u (t-1)
+ 1
1
100 u (t-4)
-
450 δ (t-6)
+ 4
6
The mathematical form of F(t) is
F (t ) = 100t [u (t ) − u (t − 1)] + 100[u (t − 1) − u (t − 4 )] + 450δ (t − 6 ) The response of the system is
x(t ) = x1 (t ) − x2 (t ) + x3 (t ) − x4 (t ) + x5 (t )
where x1, x2, x3, x4, and x5 are determined using Table 5.1 as show below x1: Ramp function, A = 100 N/sec, B = 0, t0 = 0
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Chapter 5: Transient Vibrations of SDOF Systems
N sec (t − 0.141sin 7.07t )u (t ) x1 (t ) = M 1500 m 100
x2: Ramp function, A = 100 N/sec, B = 0, t0 = 1 sec
x2 (t ) = 0.0333[t − cos(7.07 t − 7.07 ) − 0.141sin (7.07 t − 7.07 )]u (t − 1)
x3: Step function, A = 100 N, t0 = 1 sec
x3 (t ) = 0.0333[1 − cos(7.07t − 7.07 )]u(t − 1)
x4: Step function, A = 100 N, t0 = 4 sec
x4 (t ) = 0.0333[1 − cos(7.07t − 28.28 )]u(t − 4 )
x5: Impulse function, A = 450 N-sec, t0 = 6 sec
x5 (t ) =
(450 N − sec)⎛⎜ 7.07 rad ⎞⎟
sec ⎠ ⎝ sin (7.07t − 42.42 )u (t − 6 ) N 1500 m = 2.12 sin (7.07t − 42.42 )u (t − 6 )
Problem 5.18 illustrates (a) graphical breakdown of an excitation whose form changes with time and (b) the use of Table 5.1 and superposition to determine the response of a onedegree-of-freedom system due to an excitation whose form changes at discrete values of time.
5.19 The force applied to the 120 kg anvil of a forge hammer during operation is approximated as a rectangular pulse of magnitude 2000 N for a duration of 0.3 s. The anvil is mounted on a foundation of stiffness 2000 N/m and damping ratio 0.4. What is the maximum displacement of the anvil?
Given: m = 120 kg, F = 2000 N, t0 = 0.3 s, k = 2000 N/m, ζ = 0.4 Find: xmax Solution: The natural frequency of the anvil is
ωn =
k = 4.08 rad/s m 394
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Chapter 5: Transient Vibrations of SDOF Systems
The damped natural frequency is
ω d = ω n 1 − ζ 2 = 3.74 rad/s The convolution integral is used to determine the response of the system for t < t0 as
x(t ) =
1 mω d
t
∫F e
−ζω ( t −τ )
0
sin(ω d (t − τ )) dτ
0
which is evaluated to
[
x(t ) = F0 5.88 × 10 −2 − 5.88 × 10 −3 e −4t cos(3.74t ) − 6.29 × 10 −3 sin(3.74)
]
A plot of the above expression reveals that x(t) does not reach a maximum in the interval from t = 0 to t = 0.3 The appropriate expression for x(t) for t > 0.3 s is t 1 0 −ζω ( t −τ ) x(t ) = F0 e sin(ω d (t − τ )) dτ ∫ mω d 0 which is evaluated to
x(t ) = 0.396e −4t +1.2 cos(3.74t ) + 1.115e −4t +1.2 sin(3.74t ) − 1.177e −4t cos(3.74t ) − 0.126e −4t sin(3.74t ) The maximum value of x(t) is evaluated as xmax = 0.909 m. Note: MATLAB was used to symbolically integrate the convolution integral and to determine the maximum of x(t). Problem 5.19 illustrates the use of the convolution integral to determine the maximum response of a system.
5.20 A one-story frame structure houses a chemical laboratory. Figure P5.20 shows the results of a model test to predict the transient force to which the structure would be subject if an explosion would occur. The equivalent mass of the structure is 2000 kg and its equivalent stiffness is 5 × 106 N/m. Approximate the maximum displacement of the structure due to this blast.
Given: m = 2000 kg, k = 5 × 106 N/m, F(t) Find: xmax 395 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
Solution: The blast force is approximated by the force shown. The natural frequency of the structure is
ωn =
k = m
N m = 50 rad 2000 kg sec
5 × 10 6
The force for t < 0.2 sec is approximated by a ramp,
F (t ) = 2.5 × 104 t N, 0 < t < 0.2 sec
The response during this time is determined using Table 5.1 for a ramp function with A = 2.5 × 104 N/sec, B = 0, t0 = 0,
x(t ) =
2.5 × 10 4 N (t − 0.02 sin 50t ), 0 < t < 0.2 sec 6 N 5 × 10 sec
The extrema occur during this time interval at values of t such that
3 π 1 − cosω nt = 0, ω nt = 0, π ,7 ,... 2 2
It is also possible for the absolute maximum during the interval to occur at t = 0.2 sec. The values are checked and it is found that a maximum displacement of 1.055 × 10-3 m occurs at t = 0.2 sec. During the time interval between 0.2 sec and 1.0 sec, the excitation is approximated as a constant force of 5000 N. The response during this period is obtained using the principle of superposition and Table 5.1 The response is the response of the ramp function previously obtained minus the response due to the same ramp function but starting at 0.2 sec plus the response due to a step excitation starting at 0.2 sec. Using Table 5.1 Ramp function, A = 2.5 × 104 N/sec, B = 0, t0 = 0.2 sec
xb (t ) = 0.005[t − 0.2 cos 50(t − 0.2 ) − 0.02 sin 50(t − 0.2 )]
Step function, A = 5000 N, t0 = 0.2 sec
xc (t ) = 0.001[1 − cos 50(t − 0.2 )]
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Chapter 5: Transient Vibrations of SDOF Systems
Combining the excitations x(t ) = X a (t ) − X b (t ) + X c (t ) = 0.001 − 1x10 −4 sin 50t + 1 × 10 −4 sin 50(t − 0.2)
The times at which the maxima occur are obtained by taking the derivative and setting it to zero. This leads to the equation
cos 50t = cos 50(t − 0.2 ) whose solutions are
ω nt = tan −1 (3.380 )
This leads to a maximum of xmax = 1.16 × 10 −3 m
Problem 5.20 illustrates (a) the approximation of an excitation by a combination of functions whose responses are catalogued, (b) The use of Table 5.1, and (c) determination of the maximum response.
5.21 A 20 kg radio set is mounted in a ship on an undamped foundation of stiffness 1000 N/m. The ship is loosely tied to a dock. During a storm, the ship experiences the displacement of Figure P5.21. Determine the maximum acceleration of the radio.
Given: m = 20 kg, k = 1000 N/m Find: amax Solution: The natural frequency of the radio is
ωn =
k = 7.07 rad/s m
The radio is subject to the base motion of the ship. The mathematical representation of the base motion is
y (t ) = 0.15{1.67t [u (t ) − u (t − 0.6)] + (7 − 10t )[u (t − 0.6) − u (t − 0.7)]}
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Chapter 5: Transient Vibrations of SDOF Systems
The convolution integral solution is used
x(t ) = mω
t
2 n
1
∫ y(τ ) mω 0
sin[ω n (t − τ )] dτ n
Since the system is undamped, Table 5.1and the principle of linear superposition can be used to determine the system response,
x(t ) = x a (t ) − xb (t ) + xc (t ) − x d (t ) where each x(t) is the response due to a ramp function. To this end, note that k = mω n2 = 1000 and 0.15k = 150. The responses to each of the ramp inputs are determined using Table 5.1 as follows xa(t): Ramp , A = 150(1.67) = 250, B = 0, t0 = 0
⎛ ⎞ 1 x a (t ) = 0.25⎜⎜ t − sin ω n t ⎟⎟u (t ) ⎝ ωn ⎠ xb(t): Ramp, A = 250, B = 0, t0 = 0.6
⎡ ⎤ 1 xb (t ) = 0.25⎢t − 0.6 cos ω n (t − 0.6) − sin ω n (t − 0.6)⎥u (t − 0.6) ωn ⎣ ⎦ xc(t): Ramp, A = 150(-10) = -1500, B = 150(7) = 1050, t0 = 0.6
⎡ ⎤ 1 xc (t ) = −1.5⎢t − 0.7 + 0.1 cos ω n (t − 0.6) − sin ω n (t − 0.6)⎥u (t − 0.6) ωn ⎣ ⎦ xd(t): Ramp, A = -1500, B = 1050, t0 = 0.7
⎡ ⎤ 1 x d (t ) = −1.5⎢t − 0.7 − sin ω n (t − 0.7)⎥u (t − 0.7) ωn ⎣ ⎦ Note that the differential equation leads to
&x& + ω n2 x = ω n2 y &x& = ω n2 ( y − x)
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Chapter 5: Transient Vibrations of SDOF Systems
To this end the acceleration is determined as 0 s < t < 0.6 s
⎡ 0.25 ⎤ &x& = ω n2 ⎢ sin ω n t ⎥ ⎣ ωn ⎦ 0.6 s < t < 0.7 s
⎡ 1 ⎤ &x& = ω n2 ⎢ (0.25 sin ω n t − 1.25 sin ω n (t − 0.6) )⎥ ⎣ω n ⎦ t > 0.7 s
⎡ 1 ⎤ &x& = ω n2 ⎢ (0.25 sin ω n t − 1.25 sin ω n (t − 0.6) + 1.5 sin ω n (t − 0.7) )⎥ ⎣ω n ⎦ The maximum acceleration is 0.1722 m/s2. Problem 5.21 illustrates the maximum acceleration from a shock type input.
5.22 A personal computer of mass m is packed inside a box such that the stiffness and damping coefficient of the packing material are k and c, respectively. The package is accidentally dropped from a height h and lands on a hard surface without rebound. Set up the convolution integral whose evaluation leads to displacement of the computer relative to the package.
Given: m, k, c, h Find: zmax Solution: The velocity of the package is
v(t ) = − gt [1 − u (t − t0 )], t0 =
2h g
The velocity of the package acts as a base excitation to the computer. The displacement of the computer relative to the package is
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Chapter 5: Transient Vibrations of SDOF Systems
z (t ) = −
=
g
e
1−ζ 2
1−ζ
1−ζ
−ζω n t
2
∫ v(τ )e
−ζω n (t −τ )
ζω ( ∫ τ [1 − u (τ − t )]e −
0
2
sin ω d (t − τ − χ )dτ
0
t
g
=
t
1
n
t −τ )
sin ω d (t − τ − χ )dτ
0
t ⎧⎪ t ζω τ ⎫⎪ n sin ω d (t − τ − χ )dτ − u (t − t 0 )∫ τe ζω nτ sin ω d (t − τ − χ )dτ ⎬ ⎨∫ τe ⎪⎩ 0 ⎪⎭ t0
Note that ζω τ ∫ τe sin ω d (t − τ − χ )dτ = − n
+
eζω nτ
ω n2
[(2ζ
[
]
τeζω τ ζ sin ω d (t − τ − χ ) − 1 − ζ 2 cos(t − τ − χ ) ωn n
)
]
− 1 sin ω d (t − τ − χ ) − 2ζ 1 − ζ cos ω d (t − τ − χ )
2
2
Thus the system response is
+
e −ζω nτ
+
ω
2 n
g
[(2ζ
1
ω n2
[(2ζ
e −ζω n (t −t0 )
ω
2 n
)
− 1 sin ω d χ + 2ζ 1 − ζ 2 cos ω d χ
]
(
g
2
[(2ζ − 1)sin ω χ + 2ζ 1 − ζ cos ω χ ] t [ζ sin ω (t − t − χ ) − 1 − ζ cos ω (t − t ω 2
)
2
d
d
2
0
d
0
n
2
)
ω n2
[(2ζ
]
⎫ − 1 sin ω d (t − χ ) − 2ζ 1 − ζ 2 cos ω d (t − χ ) ⎬ ⎭
1
− e −ζω n (t −t0 ) +
2
2
)
⎧ t ζω n sin ω d χ + 1 − ζ 2 cos ω d χ ⎨− ω 1−ζ ⎩ n
− u (t − t0 ) +
(
⎧ t − ζ sin ω d χ − 1 − ζ 2 cos ω d χ ⎨ 2 1 − ζ ⎩ω n
x(t ) =
d
0
]
− χ)
]
⎫ − 1 sin ω d (t − t0 − χ ) − 2ζ 1 − ζ 2 cos ω d (t − t0 − χ ) ⎬ ⎭
)
The maximum response will probably occur for t > t0. However the algebra is too complicated to obtain an analytical solution. A numerical solution must be used. Problem 5.22 illustrates the use of the convolution integral to determine the time dependent displacement of a one-degree-of-freedom system with viscous damping. 400 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
5.23 Use the Laplace transform method to determine the response of a system at rest in equilibrium when subject to
F (t ) = F0 cosωt[1 − u(t − t0 )]
for (a) ζ = 0, (b) 0 < ζ < 1, (c) ζ = 1, (d) ζ > 1. Given: F(t), ζ , ωn Find: x(t) Solution: The Laplace transform of a system at rest in equilibrium when subject to an excitation F(t) is X (s ) =
1 F (s ) 2 m s + 2ζω n s + ω n2
where, for this problem
F (s ) = F0
[
]
s 1 − e −st0 2 s +ω 2
Hence X (s ) =
(a)
)[
]
F0 s 1 − e − st0 2 2 2 m s + ω s + 2ζω n s + ω n2
(
)(
0.The system is undamped. Hence 1
A partial fraction decomposition yields
Setting up equations to solve for A, B, C and D lead to 0 0 401 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
0
The solutions of the above equations are ,
0,
,
0
which leads to
Then 1
Using the second shifting theorem the inversion of this transform is cos
cos
cos
cos
(b) ζ < 1, The quadratic term in the denominator has no real roots. Thus it is left as a quadratic factor and partial faction decomposition yields G (s ) =
s s + ω + s + 2ζω n s + ω n2
(
2
2
) (
2
(
)
)
⎡ ωn2 − ω 2 s + 2ζω nω 2 = ⎢ 2 s 2 +ω 2 ωn2 − ω 2 + 4ζ 2ω 2ωn2 ⎣
(
)
1
−
(ω
)
− ω 2 s + 2ζω n3 ⎤ ⎥ s + 2ζω n s + ωn2 ⎦ 2 n 2
Note that
s 2 + 2ζω n s + ω n2 = (s + ζω n ) + ω d2 2
Hence after some algebra g (t ) = L−1 {G (s )} =
(
)
− ω −ω e 2 n
2
(ω
−ζωnt
2 n
−ω
)
2 2
cos ωd t −
1 + 4ζ ω ω 2
ζ 1− ζ 2
2
2 n
(ω
2
[(ω
2 n
)
− ω 2 cos ωt + 2ζωω n sin ωt
)
+ω e 2 n
−ζω nt
⎤ sin ω d t ⎥ ⎥⎦
Note that 402 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
[
X (s ) =
]
F0 G(s ) 1 − e −st0 m
Thus using linearity and the second shifting theorem
x(t ) =
F0 [g (t ) − g (t − t0 )u(t − t0 )] m
(c) ζ = 1, The denominator has equal roots. Then As + B C D + + 2 2 s +ω s + ω n (s + ω n )2
G (s ) =
where A= B=
ω n2 − ω 2 ω n2 (ω n2 + ω 2 ) + ω 2 (ω 2 + ωω n )
1 ω nω 2 + ω 3 + 2ω 2ω n 2 ω n2 ω 2 + ω n2 + ω 2 ω 2 + ωω n
(
)
(
)
C = −A
ω n (2ω 2 + ω n2 + ωω n ) D= 2 2 ω n (ω n + ω 2 ) + ω 2 (ω 2 + ωω n ) then
g (t ) = A cosωt +
B
ω
sin ωt + Ce−ω nt + Dte−ω nt
and
x(t ) =
F0 [g (t ) − g (t − t0 )u(t − t0 )] m
(d) ζ > 1,The denominator has two real roots
( = ω (− ζ +
) − 1)
s1 = ω n − ζ − ζ 2 − 1
s2
n
ζ2
In this case 403 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
G (s ) =
As + B C D + + 2 2 s +ω s − s1 s − s2
where A,B,C, and D solve A+C + D = 0 2ζω n A + B − s2C − s1 D = 0
ω A + 2ζω n B + ω 2C + ω 2 D = 1 2 n
A + ω n2 B − ω 2 s2C − ω 2 s1 D = 0 Then
g (t ) = A cosωt +
B
ω
sin ωt + Ce s1t + Des2t
Problem 5.23 illustrates the use of the Laplace transform method to determine the transient response of (a) an undamped system (b) an underdamped system, (c) a critically damped system, and (d) an overdamped system.
5.24 Use the Laplace transform method to determine the response of an undamped SDOF system initially at rest in equilibrium when subject to a symmetric triangular pulse of magnitude F0 and total duration t0.
Fo
Given: m, k, F(t), x(0) = 0, x(0) = 0 t o/2
Find: x(t)
to
Solution: The graphical breakdown of the triangular pulse is shown below (2Fo - 2Fo t ) u (t -t o/2) to
2Fo tu (t) to
-
2Fo tu (t -t o/2) to
+
to/2
-
to/2
to
to
(2Fo - 2Fo t ) u (t -t o ) to
The mathematical form of the excitation is 404 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
F (t ) =
⎞⎡ ⎛ t ⎞ ⎤ 2 F0 t ⎡ ⎛ t ⎞⎤ ⎛ − 2 F0 u (t ) − u⎜ t − 0 ⎟⎥ + ⎜⎜ t + 2 F0 ⎟⎟ ⎢u⎜ t − 0 ⎟ − u (t − t0 )⎥ ⎢ t0 ⎣ 2 ⎠ ⎦ ⎝ t0 2⎠ ⎝ ⎦ ⎠⎣ ⎝ 2F 4F = 0 tu (t ) − 0 t0 t0
⎛ t0 ⎞ ⎛ t0 ⎞ 2 F0 (t − t0 )u (t − t0 ) ⎜ t − ⎟u⎜ t − ⎟ + 2⎠ ⎝ 2 ⎠ t0 ⎝
The Laplace transform of F(t) is obtained using the second shifting theorem and transform pair 2 of Table B.1, F (s ) =
2 F0 t0 s 2
t −s 0 ⎡ ⎤ 2 − + e −st0 ⎥ 1 2 e ⎢ ⎣ ⎦
The transform of the system response is X (s ) =
F (s ) m s 2 + ω n2
(
)
Partial fraction decomposition yields
1 1 ⎛1 1 = 2 ⎜⎜ 2 − 2 2 s s + ωn ωn ⎝ s s + ω n2
2
(
2
)
(
)
⎞ ⎟⎟ ⎠
Hence
X (s ) =
t −s 0 ⎞ 2 F0 ⎛ 1 1 ⎞⎛ 2 ⎜ ⎜ ⎟ − − + e −st0 ⎟⎟ 1 2 e 2 ⎜ 2 2 2 ⎟⎜ mt0ω n ⎝ s s + ω n ⎠⎝ ⎠
Application of the first shifting theorem and transform pairs 2 and 4 of table B1 are used to invert the transform and obtain the system response x(t ) =
2 F0 mt0ω n2
⎧⎛ ⎞ 1 sin ω n t ⎟⎟ u (t ) ⎨⎜⎜ t − ⎠ ⎩⎝ ω n
⎡ t 1 ⎛ t ⎞⎤ ⎛ t ⎞ − 2 ⎢t − 0 − sin ω n ⎜ t − 0 ⎟⎥ u ⎜ t − 0 ⎟ ⎝ 2 ⎠⎦ ⎝ 2 ⎠ ⎣ 2 ωn ⎫ ⎡ ⎤ 1 sin ω n (t − t0 )⎥u (t − t0 )⎬ + ⎢t − t0 − ωn ⎣ ⎦ ⎭
Problem 5.24 illustrates application of the Laplace transform to determine the response of an undamped one-degree-of-freedom system subject to an excitation whose form changes with time.
405 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
5.25 Use the Laplace transform method to determine the response of an underdamped SDOF system to a rectangular pulse of magnitude F0 and time t0.
Given: F(t) = F0[u(t) – u(t – t0)] Find: x(t) using Laplace transform method Solution: The Laplace transform of F(t) is obtained using the second shifting theorem F ( s) =
(
F0 1 − e − st0 s
)
Assume the initial conditions are both zero then
X ( s) =
F0 / m(1 − e − st0 ) s( s 2 + 2ζω n s + ωn2 )
Partial fraction decomposition leads to
X ( s) =
F0 mωn2
⎡1 ⎤ s + 2ζωn (1 − e −st0 ) ⎢ − 2 2⎥ s s + 2 s + ζω ω n n ⎦ ⎣
For an unerdamped system the quadratic denominator can be rewritten as ( s + ζω n ) 2 + ω d2 . Then application f both the First Shifting Theorem and the Second Shifting Theorem leads to ⎤ ζω n F0 ⎡ 1 − e −ζω nt (cos ω d t + sin ω d t ⎥u (t ) − 2 ⎢ ωd mω n ⎣ ⎦ ⎤ ζω n F0 ⎡ 1 − e −ζω n (t −t0 ) (cos ω d (t − t 0 ) + sin ω d (t − t 0 )⎥u (t − t 0 ) 2 ⎢ ωd mω n ⎣ ⎦
x(t ) =
Problem 5.25 illustrates the application of the Laplace transform method to derive the response of an underdamped one-degree-of-freedom system to a rectangular pulse. The solution requires application of the shifting theorems.
5.26 Use the Laplace transform method to derive the response of a SDOF system initially at rest in equilibrium when subject to a harmonic force F0 sin ωt, when (a) ω ≠ ωn and (b) ω = ωn.
Given: ω, ωn , F0 Find: x(t) when (a) ω ≠ ωn, (b) ω = ωn 406 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
Solution: The transform of the response of an undamped one-degree-of-freedom system is attained using eq. (5. 25) with the initial conditions taken to be zero as 1 F (s ) X (s ) = ~ 2 m s + ω n2
(a) For F(t) = F0sinωt, ω ≠ ωn,
F0ω s + ω2
F (s ) =
2
Hence Fω 1 X (s ) = 0~ 2 2 m s + ω s 2 + ω n2
(
)(
)
Partial fraction decomposition is used leading to
Fω X (s ) = ~ 20 2 m ω − ωn
(
)
⎛ 1 1 ⎞ ⎜⎜ 2 ⎟ − 2 2 s + ωn2 ⎟⎠ ⎝ s +ω
Linearity and Table B.1 is used to invert the transform and give
Fω x(t ) = ~ 20 2 m ω − ωn
(
)
⎞ ⎛1 1 ⎜⎜ sin ωt − sin ω nt ⎟⎟ ωn ⎠ ⎝ω
(b) When ω = ωn, F (s ) =
F0ω n s + ω n2 2
and
X (s ) =
F0ωn 1 ~ m s 2 + ωn2
(
)
2
A table of transforms more extensive than those listed in Table B.1 or use of properties not listed in Table B.2 used in conjunction with transforms listed in Table B.1 are required to invert the preceding transform. The result is
x(t ) =
F0 (sin ω nt − ω nt cos ω nt ) ~ 2 mω n2
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Chapter 5: Transient Vibrations of SDOF Systems
Problem 5.26 illustrates the use of the Laplace transform method to derive the response of a one-degree-of-freedom undamped system.
5.27 Determine the transfer function for the relative displacement of a SDOF system with
where Z(s) is the Laplace transform of the relative
base motion defined as
displacement and Y(s) is the Laplace transform of the motion of the base. Given: y(t) Determine: Solution: The differential equation governing the relative displacement is 2
Taking the Laplace transform of the equation assuming all initial conditions are zero leads to 2
Solving for Z(s) leads to
2
Problem 5.27 illustrates the transfer function of a system with base motion.
5.28 Determine the transfer function for the force transmitted to the foundation for a SDOF
where
system. The transfer function is defined as
is the Laplace
transform of the transmitted force and F(s) is the Laplace transform of the applied force. Given: F(s) Find: Solution: The force transmitted to the foundation with a machine mounted on a spring and viscous damper is
408 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
Taking the Laplace transform of this equation leads to
The transform of the system output is related to the transform of the system input by
Substituting for X(s) in the equation for
leads to
Solving for the transfer function yields
which can also be written as 2
2
Problem 5.28 illustrates the transfer function for the transmitted force.
5.29 Use the transfer function to determine the response of an SDOF system excited by motion of its base with m = 3 kg and k = 18,000 N/m where the base motion is shown in Figure P5.29.
Given: m = 3 kg, k = 18,000 N/m Find: Solution: The equation for the displacement of the undamped system due to motion of its base is
The transfer function is obtained as
409 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
The natural frequency of the system is 18000 N/m 3 kg
77.5 rad/s
Thus 6000 6000
The input motion is given by 0.05
0.2 0.05
0.02
0.02
0.05
0.05
0.01
0.01
0.2 0.2
2
0.5
0.025
2
0.05
0.005
0.05
0.5
0.5
0.5
0.5
The second shifting theorem is employed to take the Laplace transform of y(t) resulting in 0.05
0.01
0.05
.
.
Substituting Y(s) into the transfer function yields 6000 6000
0.05
.
0.01
0.05
.
A partial fraction decomposition leads to 6000 6000
1
1 6000
Thus 1
Noting that
1 6000
0.05
and
0.01
.
.
sin 77.5
0.05
.
0.0129 sin 77.5 and using
the second shifting theorem the inverse of X(s) is 0.05
0.0129 sin 77.5 0.01 0.2 0.05 0.5 0.0129 sin 77.5
0.0129 sin 77.5 0.5 0.5
0.2
0.2
Problem 5.29 illustrates the application of the Laplace transform method using transfer functions.
410 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
5.30 Use the transfer function to determine the response of a SDOF system with m =1 kg, k = 100 N/m, and c = 6 N · s/m when the system is subject to motion of its base shown in Figure P5.30.
Given: m = 1 kg, k = 100 N/m, c = 6 N · s/m Find: x(t) Solution: The natural frequency for the system is 10 rad/s and its damping ratio is 0.3. The transfer function for the system is 6 6
100 100
The motion of the base is given by 0.1
1
0.1
0.1
0.1
1
1 1
The Laplace transform of the base motion is 0.1
0.1
Then using the transfer function 6
100 6 100
0.1
0.1
6
100 6 100
0.1
0.1
A partial fraction decomposition leads to 6
1
100 6 100
1 6
100
Hence 1
1 6
100
0.1
0.1
It is noted that 1 6
100
1 3
91
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Chapter 5: Transient Vibrations of SDOF Systems
and 1 6
1 100
sin √91
√91
0.105
sin 9.54
The second shifting theorem is used to invert the transform yielding 0.1
0.105
sin 9.54
0.1
1
0.105
sin 9.54
1
1
Problem 5.30 illustrates application of the transfer function to determine the response of a damped SDOF system due to base motion.
5.31 Repeat Chapter Problem 5.30 if the system parameters are m = 1 kg, k = 200 N/m, and c = 30 N · s/m.
Given: m = 1 kg, k = 200 N/m, c = 30 N · s/m Find: x(t) Solution: The natural frequency for the system is 14.14rad/s and its damping ratio is 1.061 The transfer function for the system is 30 200 30 200
The motion of the base is given by 0.1
1
0.1
0.1
0.1
1
1 1
The Laplace transform of the base motion is 0.1
0.1
Then using the transfer function 30 200 30 200
0.1
0.1
30
200 30 200
0.1
0.1
A partial fraction decomposition leads to 30
200 30 200
1
1 30
200
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Chapter 5: Transient Vibrations of SDOF Systems
Hence 1
1 30
200
0.1
0.1
It is noted that 1 30
1 10
200 1 30
200
20 0.1
0.1 10
0.1 20
0.1
The second shifting theorem is used to invert the transform yielding 0.1
0.1
0.1
0.1
1
0.1
0.1
1
Problem 5.31 illustrates application of the transfer function to determine the response of a damped SDOF system due to base motion.
5.32 For the system of Figure P5.32(a), complete the following. (a) Determine its transfer function defined as . (b) Use the transfer function to find the response of the
system due to y(t) as shown in Figure P5.32(b). Use m = 1 kg, k = 100 N/m, and c = 30 N · s/m. Given: (b) m = 1 kg, k = 100 N/m and c = 30 N · s/m, F(t) as shown Find:
,
Solution: (a) The differential equation governing the motion of the block is 3
Taking the Laplace transform of both sides of the equation setting all initial condition to zero leads to 3
Solving for the transfer function leads to 3
413 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
(b) Substituting numbers into the transfer function leads to 30 100 30 300
The base motion is given by 0.001 0.05
0.05 0.02
0.001
0.02
0.05
0.05
0.05
The Laplace transform of the base motion is 0.02
0.02
.
which when substituted into the transfer function yields 0.02 30 30
100 300
1
.
A partial fraction decomposition leads to 0.02 30 30
100 300
0.02
1
1 30
300
0.02
1 15
1
75
The inverse Laplace transform of X(s) is x(t), 0.02
1 √75 0.02
sin √75 0.05
1
.
√75
sin √75
0.05
0.05
Problem 5.32 illustrates the use of the transfer function is solving base motion problems.
414 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
5.33 For the system of Figure P5.33(a), complete the following.
(a) Determine its transfer function defined as
where
is the Laplace
transform of the angular displacement of the bar. (b) Use the transfer function to determine Given:
N
1000 ,
Find:
N
2000 ,
due to y(t), as showing in Figure P5.33(b). 12 kg,
20 cm,
80 cm,
,
Solution: FBDs of the system at an arbitrary instant are shown. Summing moments about ∑ the point of support using ∑ leads to 1000
0.2
0.2
2000 0.8
0.8
12 0.3
0.3
1 12 1 12
which becomes upon simplification 2.08
155.2
200
Taking the Laplace transform of both sides of the above equation leads to 200 2.08 155.2
(b) The excitation is given by 0.1 0.1
0.1 0.2
0.002
0.1
0.1
0.1
0.1
0.1
2
0.2 0.2
The Laplace transform of y(t) is 0.1
1
2
.
.
415 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
Substitution into the transfer function yields 20 2.08
2
.
1 74.61
1
1
155.2
.
A partial fraction decomposition leads to 20 2.08 74.61
1
2
.
.
Inversion leads to 0.129
0.116 sin 8.67 2 0.1 0.116 sin 8.67 0.2 0.116 sin 8.67 0.2 0.2
0.1
0.1
Problem 5.33 illustrates application of the transfer function to base motion problems.
5.34 During its normal operation, a 144-kg machine tool is subject to a 15,000 N · s impulse. Design an efficient isolator such that the maximum force transmitted through the isolator is 2500 N and the maximum displacement is minimized.
Given: I = 25000 N · s, m = 144 kg, FT,max. = 2500 N Find: k and c such that xmax. is a minimum Solution: For a given FT,max., xmax. is minimized by selecting ζ such that S(ζ) is a minimum. This requires setting 0.4
The natural frequency is calculated from .
.
0.4
where from impulse-momentum 15000 N · sec
and from eq. (5.110) 0.4
0.88
Thus
416 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems 2500 N 15000 N · sec 0.88
0.189
rad sec
The isolator stiffness is 144 kg
0.189
rad sec
1.14
N m
The isolator damping coefficient is 2
2 0.4 144 kg
0.189
rad sec
43.55
N · sec m
Problem 5.34 illustrates design of an isolator to protect a foundation from impulsive loading with minimum displacement.
5.35 A 110 kg pump is mounted on an isolator of stiffness 4 × 105 N/m and a damping ratio of 0.15. The pump is given a sudden velocity of 30 m/s. What is the maximum force transmitted through the isolator and what is the maximum displacement of the pump?
Given: m = 110 kg, k = 4 × 105 N/m, ζ = 0.15, v = 30 m/sec Find: FT,max , xmax. Solution: The natural frequency of the system is 4
N m 110 kg 10
60.3
rad sec
The maximum transmitted force is calculated from .
0.15
where Q(0.15) = 0.844 as calculated using eq.(5.110). Thus .
0.844 110 kg
30
m sec
60.3
rad sec
1.68
10 N
The maximum displacement is obtained from .
1 2
,
0.15
where S(0.15) = 1.823 is obtained from eq.(5.112). Then 417 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems 1.823 110 kg .
2 1.68
m sec N 10 m 30
0.537 m
Problem 5.35 illustrates the use of Q(ζ) and S(ζ).
5.36 During operation, a 50-kg machine tool is subject to the short-duration pulse of Figure P5.36. Design an isolator that minimizes the maximum displacement and reduces the maximum transmitted force to 5000 N. What is the maximum displacement of the machine tool when this isolator is used?
Given: m = 50 kg, Fmax =5000 Find: isolator design, xmax Solution: The total impulse applied to the machine due to the impulsive excitation is t0
I = ∫ F (t ) dt 0
1 ⎡ ⎤ I = (30,000 N) ⎢(0.005 s) + (0.005 s)⎥ 2 ⎣ ⎦ I = 225 N ⋅ s The initial velocity imparted to the machine due to the impulse is v=
I 225 N ⋅ s = = 4.5 m/s m 50 kg
The maximum displacement is minimized by designing an isolator of damping ratio ζ = 0.4. Then if the maximum transmitted force is limited to 5,000 N, the maximum displacement is
x max
1 1 (50 kg)(4.5 m/s) 2 mv 2 = 2 S ( 0 .4 ) = 2 1.04 = 0.105 m 5000 N Fmax
The natural frequency of the isolator is calculated by using Eq.(5.110)
ωn =
Fmax 5000 N = = 25.3 rad/s mvQ (0.4) (50 kg)(4.5 m/s)(0.88)
The maximum isolator stiffness is 418 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
k = mωn2 = (50 kg)(25.3 rad/s) 2 = 3.19 × 10 4 N/m Problem 5.36 illustrates design of an isolator to protect against a short duration pulse.
5.37 Repeat Chapter Problem 5.36 for the short-duration pulse of Figure P5.37.
Given: m = 50 kg, Fmax = 5000 N Find: isolator design, xmax Solution: The total impulse applied to the machine due to the impulsive excitation is t0
I = ∫ F (t ) dt 0
1 I = (20,000 N)(0.01 s) 2 I = 100 N ⋅ s The initial velocity imparted to the machine due to the impulse is v=
I 100 N ⋅ s = = 2 m/s m 50 kg
The maximum displacement is minimized by designing an isolator of damping ratio ζ = 0.4. Then if the maximum transmitted force is limited to 5,000 N, the maximum displacement is
x max
1 1 (50 kg)(2 m/s) 2 mv 2 = 2 S (0.4) = 2 1.04 = 0.0208 m 5000 N Fmax
The natural frequency of the isolator is calculated by using Eq.(5.110)
ωn =
Fmax 5000 N = = 56.8 rad/s mvQ (0.4) (50 kg)(2 m/s)(0.88)
The maximum isolator stiffness is k = mωn2 = (50 kg)(56.8 rad/s) 2 = 1.61 × 105 N/m Problem 5.37 illustrates design of an isolator to protect against a short duration pulse.
419 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
5.38 A ship is moored at a dock in rough seas and frequently impacts the dock. The maximum velocity change caused by the impact is 15 m/s. Design an isolator to protect a sensitive 80-kg navigational control system such that its maximum acceleration is 30 m/s .
Given: m = 80 kg, ∆
15 m/s,
30 m/s
Find: k,c Solution: Since minimizing the transmitted accelerations is the only consideration, the damping ratio is chosen as 0.25 such that is a minimum with a value of Q(0.25) = 0.81. Then the maximum transmitted acceleration is 0.25
0.81
which leads to 30 m/s 0.81 15 m/s
0.81
2.47 rad/s
The maximum stiffness is calculated as 80 kg 2.47 rad/s
487.7 N/m
The isolator damping coefficient is 2
2 0.25 80 kg
2.47
rad s
98.8 N · s/m
Problem 5.38 illustrates shock isolation theory applied to a transmitted acceleration problem.
5.39 A one-story frame structure with an equivalent mass 12,000 kg and stiffness 1.8 × 10 N/m is subject to a blast whose force is given in Figure P5.39. What is the maximum deflection of the structure?
Given: 35000 N,
12,000 kg, 0.6 s
1.8
10
N
,
Find: Solution: Calculations show that
420 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
2
2
0.6 s 1.8x 10 N/m 2π 12000 kg
1.17
From the displacement spectrum for an undamped system subject to a triangular pulse 1.5
The maximum displacement is calculated as 1.5 35000 N 1.8 10 N/m
2.92 cm
Problem 5.39 illustrates use of a displacement spectrum to determine the maximum displacement of a structure.
5.40 A 20 kg machine is on a foundation that is subject to an acceleration that is modeled as a versed sine pulse of magnitude of 20 m/s and duration of 0.4 s. Design an undamped isolator such that the maximum acceleration felt by the machine is 15 m/s . What is the maximum displacement of the machine tool relative to its foundation when the isolator is used?
Given: m = 20 kg,
20 m/s ,
= 15 m/s ,
0.4 s,
0
Find: k, Solution: The ratio of the acceleration of the machine to the acceleration of its foundation is 0.75
The force spectrum for a versed sine pulse is used to determine that for the acceleration ratio to be less than 0.76, 2
0.29
0.29 2π 0.4 s
4.5 rad/s
rad s
400 N/m
The isolator stiffness is given by 20 kg
4.5
The maximum relative displacement is determined using the displacement spectrum. However since the isolator is undamped the displacement spectrum is the same as the force spectrum. For transmitted acceleration problems the vertical scale of the displacement spectrum becomes 421 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems 0.75 20 m/s 4.5 rad/s
0.75
0.74 m
Problem 5.40 illustrates the use of force and displacement spectra to design an isolator to protect a machine from a foundation acceleration.
5.41 During operation, a 100 kg machine tool is exposed to a force that is modeled as a sinusoidal pulse of magnitude of 3100 N and duration of 0.05 s. Design an isolator with a damping ratio of 0.1 such that the maximum force transmitted through the isolator is 2000 N and the maximum displacement of the machine tool is 3 cm. 3100 N,
Given: m = 100 kg,
0.05 s,
2000 N,
= 3 cm,
0.1
Find: k, c Solution: The maximum value of pulse yields
N
is π .
0.25 or
0.645. The force spectrum for a sinusoidal
N
31.4 rad/s. The displacement spectrum for a
.
sinusoidal pulse corresponding to a value of 0.25 on the horizontal scale yields 0.67 0.21 m which is less than the 3 cm required. The stiffness and damping coefficient of such an isolator are 100 kg 31.4 rad/s 2
2 0.1 100 kg
31.4
98600 N/m
rad s
528.0 N · s/m
Problem 5.41 illustrates use of the force and displacement spectra for sinusoidal pulse excitation.
5.42 During operation a 80 kg machine is subject to a triangular pulse with a magnitude of 30,000 N and duration of 0.15 s. What is the range of undamped isolator stiffness such that the maximum transmitted force is 15,000 N and the maximum displacement is 5 cm?
Given: m = 80 kg,
= 30,000 N,
0.15 s,
5 cm,
15,000N
Find: range of k Solution: Limiting the maximum force to 15000 N implies that
0.5 which, from the
force spectrum for an undamped system subject to a triangular pulse, leads to which implies .
0.3
0.05 s. 422
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5: Transient Vibrations of SDOF Systems
This leads to 2π 0.05 s
125.7 rad/s
which leads to 80 kg
125.7
rad s
1.26
10 N/m
This is an upper bound on the stiffness. Checking the maximum displacement with this stiffness 0.5 30000 N 1.26 10 N/m
0.5
The maximum value of For this value of k,
is 1.6. With this value 1.89
1.1
1.19 cm .
N .
3.3
4.8
10 N/m.
10 . The iteration can
continue, but this is close enough. Thus 3.3
10
1.26
10 N/m
Problem 5.42 illustrates the use of the force and displacement spectra to determine a lower bound and an upper bound on the stiffness.
423 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.