Seventh Edition
CHAPTER
1 9
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University
Mechanical Vibrations
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Vector Mechanics for Engineers: Dynamics Contents Introduction
Sample Problem 19.4
Free Vibrations Vibrations of Particles. Particles. Simple Harm
Forced Vibrations
Simple Pendulum (Approximate Solution)
Sample Problem 19.5
Simple Pendulum (Exact Solution)
Damped Free Vibrations
Sample Problem 19.1
Damped Forced Vibrations
Free Vibrations of Rigid Bodies
Electrical Analogues
Sample Problem 19.2 Sample Problem 19.3 Principle of Conservation of Energy
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Vector Mechanics for Engineers: Dynamics Introduction • Mechanical vibration is the motion of a particle or body which oscillates about a position of equilibrium. equilibrium. Most vibrations in machines and structures are undesirable due to increased stresses and energy losses. • Time interval required for a system to complete a full cycle of the motion is the period the period of of the vibration. • Number of cycles per unit time defines the frequency the frequency of the vibrations. • Maximum displacement displacement of the system from the equilibrium position is the amplitude of the vibration. • When the motion is maintained by the restoring forces only, the vibration is described as free as free vibration. vibration . When a periodic force is vibration . applied to the system, the motion is described as forced as forced vibration. • When the frictional dissipation of energy is neglected, the motion is said to be undamped . Actually, all vibrations are damped to damped to some degree.
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Vector Mechanics for Engineers: Dynamics Free Vibrations of Particles. Simple Harmonic Motion xm from its d istance x • If a particle is displaced through a distance equilibrium position and released with no velocity, the particle will undergo simple undergo simple harmonic motion, motion , ma = F = W − k ( δ st + x ) = − kx m x + kx = 0
• General solution is the sum of two particular solutions , two particular solutions, k k + x = C 1 sin t C t cos 2 m m
= C 1 sin ( ω n t ) + C 2 cos( ω n t ) a periodic function and ω n is the natural circular • x is a periodic frequency of the motion. • C 1 and C 2 are determined by the initial conditions:
x = C 1 sin ( ω n t ) + C 2 cos( ω n t ) v = x = C 1ω n cos( ω n t ) − C 2ω n sin ( ω n t )
= x0 C 1 = v0 ω n C 2
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Vector Mechanics for Engineers: Dynamics Free Vibrations of Particles. Simple Harmonic Motion v0
C 1
=
C 2
= x0
ω n
• Displacement is equivalent to the x the x component of the sum of two vectors C 1 + C 2 which rotate with constant angular velocity ω n . x = xm sin ( ω n t + φ )
xm
= ( v0
ω n ) 2 + x02
φ = tan −1 ( v0 x0ω n ) τ n
=
2π
ω n 1 f n = τ n
=
=
=
amplitude phase angle
period
ω n = = natural frequency 2π
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Vector Mechanics for Engineers: Dynamics Free Vibrations of Particles. Simple Harmonic Motion • Velocity-time and acceleration-time curves can be represented by sine curves of the same period as the displacement-time displacement-time curve but different phase angles.
x = xm sin ( ω n t + φ ) v = x
= xmω n cos( ω n t + φ ) = xmω n sin ( ω n t + φ + π 2) a = x
= − xmω n2 sin ( ω n t + φ ) = xmω n2 sin ( ω n t + φ + π )
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Vector Mechanics for Engineers: Dynamics Simple Pendulum (Approximate Solution) • Results obtained for the spring-mass system can be applied whenever the resultant force on a particle is proportional to the displacement and directed towards the equilibrium position. • Consider tangential components of acceleration and force for a simple pendulum, F t = mat : − W sin θ = ml θ
∑
+ g sin θ = 0 θ l for small angles,
+ g θ = 0 θ l θ = θ m sin ( ω n t + φ ) τ n
= 2π = 2π ω n
l g
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Vector Mechanics for Engineers: Dynamics Simple Pendulum (Exact Solution) g An exact solution for θ + sin θ = 0 l
leads to
τ n
=4
l g
π 2
∫
0
d φ 1 − sin 2 (θ m 2 ) sin 2 φ
which requires numerical solution.
τ n
=
2 K
π
2π
l
g
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Vector Mechanics for Engineers: Dynamics Sample Problem 19.1 SOLUTION: • For each spring arrangement, determine the spring constant for a single equivalent spring. • Apply the approximate relations for the harmonic motion of a spring-mass system. A 50-kg block moves between vertical guides as shown. The block is pulled pulled 40mm down from its equilibrium position and released. For each spring arrangement, determine a) the period of the vibration, b) the maximum velocity of the block, and c) the maximum acceleration of the block.
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Vector Mechanics for Engineers: Dynamics Sample Problem 19.1 k 1 = 4 kN m
k 2
= 6 kN
m
SOLUTION: • Springs in parallel: - determine the spring constant for equivalent spring - apply the approximate relations for the harmonic motion of a spring-mass system
P = k 1δ + k 2δ k =
P
δ
= k 1 + k 2
= 10 kN
4
m = 10 N m
k
ω n
=
τ n
=
vm
= x m ω n = ( 0.040 m) (14.14 rad s )
m
=
104 N/m 20 kg
= 14.14 rad
2π
ω n
s
τ n
= 0.444 s
vm
= 0.566 m s
am
= 8.00 m
am = x m an2
= ( 0.040 m)() (14.14 rad s ) 2
s2
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Vector Mechanics for Engineers: Dynamics Sample Problem 19.1 k 1 = 4 kN m
k 2
= 6 kN
m
• Springs in series: - determine the spring constant for equivalent spring - apply the approximate relations for the harmonic motion of a spring-mass system
P = k 1δ + k 2δ k =
P
δ
= k 1 + k 2
= 10 kN
m = 10 4 N m
k
=
τ n
=
vm
= x m ω n = ( 0.040 m ) ( 6.93 rad s )
m
=
2400N/m
ω n
20 kg
= 6.93 rad s
2π
ω n
τ n
= 0.907 s
vm
= 0.277 m s
am
= 1.920 m
am = x m an2
= ( 0.040 m ) ( 6.93 rad s ) 2
s2
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Vector Mechanics for Engineers: Dynamics Free Vibrations of Rigid Bodies • If an equation of motion takes the form x + ω n2 x = 0
or θ + ω n2θ = 0
the corresponding motion may be considered as simple harmonic motion. • Analysis objective is to determine ω n. • Consider the oscillations of a square plate − W ( b sin θ ) = ( mbθ ) + I θ
1 m ( 2b ) but I = 12
2
+ ( 2b ) 2 = 23 mb 2 ,
θ + 3 g sin θ ≅ θ + 3 g θ = 0 5b
then ω n
=
5b
3 g , τ n 5b
=
2π
ω n
= 2π
5b 3 g
• For an equivalent simple pendulum, l = 5b 3
W = mg
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Vector Mechanics for Engineers: Dynamics Sample Problem 19.2 SOLUTION:
k
• From the kinematics of the system, s ystem, relate the linear displacement and acceleration to the rotation of the cylinder. • Based on a free-body-diagram equation for the equivalence of the external and effective forces, write the equation of motion.
W is suspended A cylinder of weight W is as shown. Determine the period and natural frequency of vibrations of the cylinder.
• Substitute the kinematic relations to arrive at an equation involving only the angular displacement and acceleration.
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Vector Mechanics for Engineers: Dynamics Sample Problem 19.2 SOLUTION: • From the kinematics of the system, relate the linear displacement and acceleration to the rotation of the cylinder. x = r θ δ = 2 x = 2r θ
α = θ
a
= r α = r θ
a
= r θ
• Based on a free-body-diagram equation for the equivalence equivale nce of the external and effective forces, write the equation of motion. M A = ( M A ) eff : Wr − T 2 ( 2r ) = ma r + I α
∑
but
T 2
∑
= T 0 + k δ = 12 W + k ( 2r θ )
• Substitute the kinematic relations to arrive at an equation involving only the angular displacement and acceleration.
Wr − ( 12 W + 2kr θ )( 2r )
= m( r θ ) r + 12 mr 2θ
+ 8 k θ = 0 θ 3m ω n
=
8k 3m
τ n
=
2π
ω n
= 2π
3m 8k
f n
=
ω n 2π
=
1
8k
2π 3m
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Vector Mechanics for Engineers: Dynamics Sample Problem 19.3 SOLUTION: • Using the free-body-diagram equation for the equivalence of the external and effective moments, write the equation of motion for the disk/gear and wire.
W = 20 lb
τ n
= 1.13s
τ n
= 1.93 s
• With the natural frequency and moment of inertia for the disk known, known , calculate the torsional spring constant.
The disk and gear undergo torsional vibration with the periods shown. • With natural frequency and spring Assume that the moment exerted by the constant known, calculate the moment of wire is proportional to the twist angle. inertia for the gear. Determine a) the wire torsional spring • Apply the relations for simple harmonic constant, b) the centroidal moment of motion to calculate the maximum gear inertia of the gear, and c) the maximum velocity. angular velocity of the gear if rotated through 90o and released.
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Vector Mechanics for Engineers: Dynamics Sample Problem 19.3 SOLUTION: • Using the free-body-diagram equation for the equivalence of the external and effective moments, write the equation of motion for the disk/gear and wire.
∑ M O = ∑ ( M O ) eff :
+ K θ = − I θ + K θ = 0 θ I
W = 20 lb
τ n
= 1.13s
τ n
= 1.93 s ω n
=
K I
τ n
=
2π
ω n
= 2π
I K
• With the natural frequency and moment of inertia for the disk known, calculate the torsional t orsional spring constant. 2
I = 12 mr
1.13 = 2π
1 20
=
2 32.2
0.138
K
2
8 = 0.138 lb ⋅ ft ⋅ s 2 12
K = 4.27 lb ⋅ ft rad
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Vector Mechanics for Engineers: Dynamics Sample Problem 19.3 • With natural frequency and spring constant known, calculate the moment of inertia for the gear. 1.93 = 2π
ω n
=
= 1.13s
K I
τ n
τ n
=
= 1.93 s
2π
ω n
= 2π
K = 4.27 lb ⋅ ft rad
I = 0.403 lb ⋅ ft ⋅ s 2
4.27
• Apply the relations for simple harmonic motion to to calculate the maximum gear velocity.
W = 20 lb
τ n
I
θ = θ m sin ω nt
I K
ω = θ mω n sin ω nt
θ m
= 90° = 1.571 rad
ω m
2π 2π ( ) 1 . 571 rad = θ m = 1 . 93 s τ n ω m
ω m
= θ mω n
= 5.11rad s
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Vector Mechanics for Engineers: Dynamics Principle of Conservation of Energy • Resultant force on a mass in simple harmonic motion is conservative conservative - total energy energy is conserved. 1 m x 2 + 1 kx 2 = constant T + V = constant 2
2
x 2 + ω n2 x 2 = • Consider simple harmonic motion of the square plate,
= Wb(1 − cosθ ) = Wb[2 sin 2 (θ m 2 ) ] ≅ 12 Wbθ m2
T 1
=0
T 2
= 12 mvm2 + 12 I ω m2
V 1
V 2
=0
ω n
=
= 12 m( bθ m ) 2 + 12 ( 23 mb 2 )ω m2
= 12 ( 53 mb 2 )θ m2
T 1 + V 1
= T 2 + V 2 2 = 12 ( 53 mb 2 )θ m2 ω n2 + 0 0 + 12 Wbθ m
3 g 5b
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Vector Mechanics for Engineers: Dynamics Sample Problem 19.4 SOLUTION: • Apply the principle of conservation of energy between the positions of maximum and minimum potential energy. • Solve the energy equation for the natural frequency of the oscillations. o scillations.
Determine the period of small s mall oscillations of a cylinder which rolls without slipping inside a curved surface.
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Vector Mechanics for Engineers: Dynamics Sample Problem 19.4 SOLUTION: • Apply the principle of conservation of energy between the positions of maximum and minimum potential energy.
T 1 + V 1
= T 2 + V 2
T 1
=0
T 2
= 12 mvm2 + 12 I ω m2
V 1
= Wh = W ( R − r ) (1 − cosθ ) ≅ W ( R − r ) (θ m2 2) V 2 2
R − r 2 = 12 m( R − r )θ m2 + 12 ( 12 mr 2 ) θ m r = 34 m( R − r ) 2θ m2
=0
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Vector Mechanics for Engineers: Dynamics Sample Problem 19.4 • Solve the energy equation for the natural frequency of the oscillations.
T 1
=0
T 2
= 34 m( R − r ) 2θ m2
T 1 + V 1
θ m2 0 + W ( R − r ) 2 θ m2 ( mg ) ( R − r ) 2
ω n2
=2
g
3 R − r
V 1
≅ W ( R − r ) θ m2
V 2
=0
2
= T 2 + V 2 = 34 m( R − r ) 2θ m2 + 0
= 34 m( R − r ) 2 (θ mω n ) 2m
τ n
=
2π
ω n
= 2π
3 R 2
− r
g
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Vector Mechanics for Engineers: Dynamics Forced Vibrations Forced vibrations - Occur when a system is subjected to a periodic force or a periodic displacement of a support.
ω f
=
forced frequency
∑ F = ma : P m sin ω f t + W − k ( δ st + x ) m x + kx = P m sin ω f t
= mx
W − k δ st + x − δ m sin ω f t m x + kx = k δ m sin ω f t
= m x
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Vector Mechanics for Engineers: Dynamics Forced Vibrations x = xcomplement ary + x particular
= [ C 1 sin ω n t + C 2 cos ω n t ] + xm sin ω f t Substituting particular solution into governing equation,
− mω f 2 xm sin ω f t + kxm sin ω f t = P m sin ω f t xm
=
P m 2 k − mω f
=
P m k 1 − (ω f ω n )
2
=
δ m 1 − (ω f ω n )
2
m x + kx = P m sin ω f t
m x + kx = k δ m sin ω f t
At ω f = ω n, forcing input is in resonance with the system.
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Vector Mechanics for Engineers: Dynamics Sample Problem 19.5 SOLUTION: • The resonant frequency is equal to the natural frequency of the system. • Evaluate the magnitude of the periodic force due to the motor unbalance. Determine the vibration amplitude from the frequency ratio at 1200 rpm. rp m. A motor weighing 350 lb is supported by four springs, each having a constant 750 lb/in. The unbalance of the motor is equivalent to a weight of 1 oz located 6 in. from the axis of rotation. Determine a) speed in rpm at which resonance will occur, and b) amplitude of the vibration at 1200 rpm.
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Vector Mechanics for Engineers: Dynamics Sample Problem 19.5 SOLUTION: • The resonant frequency is equal to the natural frequency of the system.
m=
350 32.2
= 10.87 lb ⋅ s 2
k = 4( 750 ) W = 350 lb
ft
= 3000 lb in
= 36,000 lb ft
k = 4(350 lb/in)
ω n
=
k
=
36,000
10.87 m = 57.5 rad/s = 549 rpm Resonance speed = 549 rpm
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Vector Mechanics for Engineers: Dynamics Sample Problem 19.5 • Evaluate the magnitude of the periodic force due to the motor unbalance. unbalance. Determine the vibration amplitude from the frequency ratio at 1200 rpm.
= ω = 1200 rpm = 125.7 rad/s 1 lb 1 0.001941 lb ⋅ s 2 = m = (1 oz ) 16 oz 32.2 ft s 2
ω f
W = 350 lb
P m
= man = mr ω 2 = ( 0.001941) (126 )(125.7 ) 2 = 15.33 lb
xm
=
k = 4(350 lb/in)
ω n
= 57.5 rad/s
P m k
1 − (ω f ω n )
2
=
ft
15.33 3000 1 − (125.7 57.5)
2
= −0.001352 in xm = 0.001352 in. (out of phase)
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Vector Mechanics for Engineers: Dynamics Damped Free Vibrations • All vibrations are damped to some degree by forces due to dry friction, friction, fluid friction, friction , or internal friction. friction. • With viscous damping due damping due to fluid friction,
∑ F = ma :
W − k ( δ st + x ) − c x = m x m x + c x + kx = 0
• Substituting x Substituting x = eλ t t and dividing through by eλ t t equation , yields the characteristic equation, 2
mλ
+ cλ + k = 0
2
c k ± λ = − − 2m 2m m c
• Define the critical damping coefficient such that 2
cc − k = 0 2m m
cc
= 2m
k m
= 2mω n
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Vector Mechanics for Engineers: Dynamics Damped Free Vibrations • Characteristic equation, 2
mλ
+ cλ + k = 0
cc
2mω n
=
=
2
c k ± λ = − − 2m 2m m c
critical damping coefficient
• Heavy damping : c > cc x = C 1e λ 1t + C 2 e λ 2t
- nega negati tive ve root rootss - nonv nonvib ibra rato tory ry mot motio ion n
• Critical damping : c = cc x = ( C 1 + C 2t ) e
ω nt
−
- double root oots - nonv nonvib ibra rato tory ry mot motio ion n
• Light damping : c < cc x = e −
( c 2 m ) t
( C 1 sin ω d t + C 2 cos ω d t ) 2
ω d = ω n
c = damped frequency 1 − cc
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Vector Mechanics for Engineers: Dynamics Damped Forced Vibrations
+ c x + kx = P m sin ω f t m x xm P m k
=
tan φ =
xm
δ
=
x = xcomplement ary
+ x particular
1
[1 − (ω f
2( c cc ) (ω f ω n ) 1 − (ω f ω n )
2
ω n )
] + [2( c cc ) (ω f
2 2
ω n ) ] 2
=
magnification factor
= phase difference between forcing and steady state response
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Vector Mechanics for Engineers: Dynamics Electrical Analogues • Consider an electrical circuit consisting of an inductor, resistor and capacitor with a source of alternating voltage E m sin ω f t − L Lq + Rq +
1
C
di dt
q
− Ri − = 0 C
q = E m sin ω f t
• Oscillations of the electrical system are analogous to damped forced vibrations of a mechanical system.
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Vector Mechanics for Engineers: Dynamics Electrical Analogues • The analogy between electrical and mechanical systems also applies to transient as well as steadystate oscillations. • With a charge q = q0 on the capacitor, closing the switch is analogous to releasing the mass of the mechanical system with no initial velocity at x at x = x0. • If the circuit includes a battery with constant voltage E , closing the switch is analogous to suddenly applying a force of constant magnitude P magnitude P to to the mass of the mechanical system.
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Vector Mechanics for Engineers: Dynamics Electrical Analogues • The electrical system analogy provides a means of experimentally determining determining the characteristics of a given mechanical system. • For the mechanical system, m1 x1 + c1 x1 + c2 ( x1 − x2 ) + k 1 x1 + k 2 ( x1 − x2 )
m2 x2
+ c2 ( x2 − x1 ) + k 2 ( x2 − x1 ) = P m sin ω f t
• For the electrical system, q q − q2 L1q1 + R1 ( q1 − q2 ) + 1 + 1 C 1 C 2
L2 q2 + R2 ( q2
=0
− q1 ) +
q2
− q1
C 2
=0
= E m sin ω f t
• The governing equations are are equivalent. The characteristics characteristics of the vibrations of the mechanical system may may be inferred from the oscillations of the electrical system.