263676512 Mechanical vibration solved examples PDF

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Mechanical Vibrations: 4600-431 March 1, 2010

Example Problems

Contents 1 Free Vibration of Single Degree-of-freedom Systems 1.1 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Unsolved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2 29

2 Forced Single Degree-of-freedom Systems 2.1 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Unsolved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

36 36 50

3 Frictionally Damped Systems 3.1 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Unsolved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

64 72

4 Multi Degree-of-freedom Systems 4.1 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Unsolved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

76 76 87

1

2

64

1

Free Vibration of Single Degree-of-freedom Systems

1.1

Solved Problems

Problem 1: For the mechanical system shown to the right, the uniform rigid bar has mass m and pinned at point O. For this system:

ˆ  k

a) find the equations of motion;

ı ˆ

b) Identify the damping ratio and natural frequency in terms of the parameters m, c, k, and ℓ.

m ℓ 2

y

c) For:

ℓ 2

θ

m = 1.50kg, c = 0.125N/(m/s),

ℓ = 45cm, O

k = 250N/m,

2m x

find the angular displacement of the bar θ(t) for the following initial conditions: θ(0) = 0,

c

k

˙ θ(0) = 10rad/s.

Assume that in the horizontal position the system is in static equilibrium and that all angles remain small. Solution: a) In addition to the coordinate θ identified in the original figure, we also define x and y as the displacment of the block and end of the bar respecively. The directions ˆı and ˆ are defined as shown in the figure. A free body diagram for this system is shown to the right. Note that the tension in the cable between the bar and the block is unknown and represented with T while the reaction force F R is included, although both its magnitude and direction are unspecified. In terms of the identifed coordinates, the angular acceleration of the bar αβ and the linear acceleration of the block aG are ˆ αβ = θ¨ k,

−k y ˆ

−T ˆ FR

T ˆ

−k x ˆ

−c x˙ ˆ

aG = x ¨ ˆ.

We can also relate the identified coordinates as x=

ℓ θ, 2

y = ℓ θ.

The equations of motion for this system can be obtained with linear momentum balance applied to the block and angular momentum balance aout O on the bar. These can be

2

written as

X

X

MO

  F = m aG −→ T − k x − c x˙ ˆ =   ℓ ˆ = = I O αβ −→ − T − k y ℓ k 2

2mx ¨ ˆ, m ℓ2 ¨ ˆ θ k. 3

Solving the first equation for T and substituting into the second equation yields − (2 m x ¨ + k x + c x) ˙

m ℓ2 ¨ ℓ −kyℓ = θ. 2 3

Using the coordinate relations we can obtain the equation of motion as 5 m ℓ2 c ℓ2 ˙ 5 k ℓ2 θ¨ + θ+ θ = 0. 4 6 4 b) In the above equation the equivalent mass, damping, and stiffness are meq =

5 m ℓ2 , 6

beq =

c ℓ2 , 4

keq =

5 k ℓ2 . 4

From these the damping ratio and natural frequency are √ c ℓ2 beq 3c 4 ζ= p = q = √ , 2 2 2 keq meq 2 50 k m 2 5 k4ℓ 5 m6 ℓ s s r 5 k ℓ2 3k keq 4 = ωn = = 5 m ℓ2 2m meq 6

c) Evaluating the damping ratio and natural frequency we find that for the given values of the parameters ωn = 15.8rad/s, ζ = 7.91 × 10−4 . Therefore the system is underdamped and the general solution can be written as   p   p  θ(t) = e−ζ ωn t a sin ωn 1 − ζ 2 t + b cos ωn 1 − ζ 2 t , where a and b are arbitrary constants used to fit the initial conditions. Evaluating θ(t) ˙ at t = 0 yields and θ(t) p ˙ θ(0) = −ζ ωn b + ωn 1 − ζ 2 a = 10rad/s, θ(0) = b = 0, so that the general solution becomes   θ(t) = 0.632 e−0.0125 t sin 15.8 t .

3

Problem 2: For the mechanical system shown to the right, the uniform rigid bar has mass m and pinned at point O. For this system:

ℓ 2

k

a) find the equations of motion;

θ

m b) Identify the damping ratio and natural frequency in terms of the parameters m, c, k, and ℓ.

z ℓ 2

m

ˆ

c) For:

x m = 2kg,

ℓ = 25cm,

c = 0.25N/(m/s),

k = 50N/m,

k

c

ˆı

find the angular displacement of the bar θ(t) for the following initial conditions: θ(0) = 0,

˙ θ(0) = 10rad/s.

d) for this motion, find the tension in the cable connecting the rod and the block as a function of time. Assume that the system is in static equilibrium at θ = 0, and that all angles remain small. Solution: a) We identify the coordinates x and z as shown above, which are related to the angular displacement θ as: ℓ ℓ x = θ, z = θ. 2 2 An appropriate free-body diagram is shown to the right. Applying linear momentum balance on the block yields X F = m aG , ¨ ˆ. (T − k x − c x) ˙ ˆ = m x

Likewise, angular momentum balance on the bar provides X M O = I O αβ ,   m ℓ2 ¨ ˆ ℓ ˆ ℓ k = θ k. −T − k z 2 2 12

k z ˆ −T ˆ T ˆ FR

−k x ˆ

−c x˙ ˆ

Combining these equations and eliminating the tension, the equation of motion can be written as 7m ¨ θ + c θ˙ + 2 k θ = 0. 6

4

b) For the above equation the equivalent mass, damping, and stiffness are meq =

7m , 6

beq = b,

keq = 2 k,

and the natural frequency and damping ratio are s r √ beq 12 k keq 3b √ = , ζ= p ωn = . = meq 7m 2 keq meq 28k m Problem 3: The block shown to the right rests on a frictionless surface. Find the response of the system if the block is displaced from its static equilibrium position 15cm to the right and released from rest. m = 4.0kg, k1 = 1.5N/m,

ˆ x ˆı

k1

k2 m

b = 0.25N/(m/s),

b

k2 = 0.50N/(m/s).

Solution: An appropriate free-body diagram is shown to the right. Notice that the two springs are effectively in parallel, as the displacement across each spring is identical. Linear momentum balance on this block provides X F = m aG , (−k1 x − k2 x − b x) ˙ ˆı

−k1 x ıˆ

−b x˙ ˆı

= mx ¨ ıˆ,

or, writing this in standard form mx ¨ + b x˙ + (k1 + k2 ) x = 0. Further, the system is released from rest so that the initial conditions are x(0) = x0 = 15cm,

5

x˙ (0) = 0cm/s.

−k2 x ˆı

Problem 4: For the system shown to the right, the disk of mass m rolls without slip and x measures the displacement of the disk from the unstretched position of the spring. The surface is inclined at an angle of φ with respect to vertical.

x

k z

a) find the equations of motion. Do not neglect gravity;

c

b) if the system is underdamped, what is the frequency of the free vibrations of this system in terms of the parameters k, c, and m;

φ

θ

(m, r)

ˆ

C

eˆ2

c) for what value of the damping constant c is the system critically damped;

ˆı

d) what is the static equilibrium displacement of the disk?

eˆ1

Solution: a) In addition to x, the displacement of the center of the disk, we identify the coordinates z and θ, the displacement across the spring and the rotation of the disk respectively. These additional coordinates are related to x as z = 2 x,

x = −r θ.

An appropriate free-body diagram is shown to the right. We note that (ˆı, ˆ) are related to the directions (ˆ e1 , ˆe2 ) as ıˆ = cos φ e ˆ 1 + sin φ e ˆ2 , ˆ2 . ˆ = − sin φ ˆe1 + cos φ e

−k z eˆ1 −m g ˆ

−c x˙ ˆe1

fr e ˆ1 N ˆe2

The moment produced by gravity about point C is M gravity

= rGC × (−m g ˆ),

= (r ˆe2 ) × (−m g ) ˆ = −m g r sin φ ˆk. Angular momentum balance about the contact point C yields X M C = I C αD ,     3 m r2 ¨ ˆ ˆ (2 r) k z + r c x˙ − m g r sin φ k = θ k. 2 Eliminating the coordinates z and θ, we can write the equation of motion in terms of x as 3m x ¨ + c x˙ + 4 k x = m g sin φ. 2

6

Since the gravitational force has been included in the development of this equation of motion, the coordinates are measured with respect to the unstretched position of the spring. b) Assuming the system is underdamped, the frequency of the free vibrations is ωd = p ωn 1 − ζ 2 , where s r c beq 8k keq = √ , , ζ= p = ωn = meq 3m 2 keq meq 2 6km so that ωd =

r

8k 3m

r

1−

c2 , 24 k m

c) The system is critically damped when ζ = 1, which corresponds to a damping coefficient of √ ccr = 2 6 k m. d) The system is stationary in static equilibrium, so that x ≡ x0 = constant—both x˙ and x ¨ vanish, and the equation of motion reduces to 4 k x0 = m g sin φ. Solving for x0 , the equilibrium displacement is x0 =

m g sin φ . 4k

Problem 5: In the figure shown to the right, in the absence of gravity the springs are unstretched in the equilibrium position.

z2 k2 r2

a) Determine the deflection of each spring from its unstretched length when the system shown is in equilibrium.

θ ˆ

b) If the system is released from the unstretched position of the springs, what is the maximum angular velocity of the disk during the resulting motion?

r1

I

ˆı

z1

m x k1

Solution: 7

a) We define the coordinates x, θ , z1 , and z2 as shown in the figure, which are related as x = −r1 θ,

z1 = r1 θ,

−k2 z2 ıˆ

z2 = −r2 θ.

Notice that because of these coordinate definitions, a rotation with positive θ gives rise to a negative value in both x and z2 . Likewise, we see that x = −z1 . Using the free-body diagram shown to the right, linear momentum balance on the block provides X F = m aG ,

FR k1 z1 ˆ

−T ˆ T ˆ

¨ ˆ, (T − m g) ˆ = m x

−m g ˆ

while angular momentum balance on the disk yields X

MO

(T r1 − k1 z1 r1 + k2 z2 r2 ) ˆk

= I O αD , ˆ = I θ¨ k.

Eliminating the unknown tension T from these equations and using the coordinate relations, the equation of motion becomes     I + m r21 θ¨ + k1 r12 + k2 r22 θ = m g r1 . The equilibrium rotation of the disk thus is found to be m g r1 θeq = . k1 r21 + k2 r22 With this, the equilibrium deflection of each spring is found to be m g r21 , k1 r21 + k2 r22 m g r1 r2 =− . k1 r12 + k2 r22

z1,eq = r1 θeq = z2,eq = −r2 θeq

b) The general free response of the disk can be expressed as θ(t) = θeq + A sin(ωn t) + B cos(ωn t), where θeq is given above, A and B are arbitrary constants, and s k1 r12 + k2 r22 . ωn = I + m r12 The system is released with the initial conditions: θ(0) = 0, 8

˙ = 0, θ(0)

so that solving for the arbitrary constants A = 0,

B = −θeq .

Therefore the solution is m g r1 θ(t) = θeq (1 − cos(ωn t)) = k1 r21 + k2 r22

1 − cos

s

k1 r12 + k2 r22 t I + m r12

!!

.

The angular velocity of the disk becomes   ˙ = θeq ωn sin(ωn t), θ(t) which has amplitude Ω = θeq ωn = p

m g r1 (k1 r21 + k2 r22 ) (I + m r12)

Problem 6: For the mechanical system shown to the right, the uniform rigid bar has mass m and pinned at point O. For this system:

k

a) find the equations of motion;

z2

b) what value of the damping constant c gives rise to a critically damped system?

m

ℓ 2

k

θ

m z1 ℓ 2

ˆ

c ˆı

Solution: a) In addition to θ, we define two additional coordinates, z1 and z2 , to measure the deflection at the left and right ends of the bar. These coordinates are related as: ℓ z1 = θ, 2

−k z2 ˆ

ℓ z2 = θ = z1 . 2

A free-body diagram for this system is shown to the right. Applying angular momentum balance on the bar eliminates the appearance of the reaction force and leads to: X M G = I G θ¨ ˆk, ℓ  m ℓ2 ¨ ˆ kˆ = θ k. T − k z1 − cz˙1 2 12 9

−T ˆ T ˆ

k z1 ˆ G

FR c z˙1 ˆ

Likewise, application of linear momentum balance on the block yields: X F = m aG ,   − k z2 − T ˆ = m¨ z2 ˆ Eliminating the unknown tension T and solving for z1 in terms of z2 , the equation of motion becomes: m ℓ2 ¨ c ℓ2 ˙ k ℓ2 θ+ θ+ θ = 0. 3 4 2 b) A critically damped system occurs when ζ = 1. For this system: √ 3c ζ =√ . 32 k m Solving for ccr yields: ccr =

r

Problem 7: Find the response of the system shown to the right if the block is pulled down by 15cm and released form rest. m = 2.0kg, k1 = 0.5N/m,

32 k m . 3

ˆ

k1 b

b = 0.5N/(m/s), k2 = 0.25N/(m/s).

ıˆ

k2

x m

Solution: For this system, the two springs in series may be replaced by an equivalent spring, with constant: k1 k2 1 = keq = 1 . 1 k + k2 1 + k2 k1 Therefore, the free-body diagram is shown to the right. Applying linear momentum balance to the block yields: X F = m aG ,   ¨ ˆ, keq x + b x˙ ˆ = −m x

keq x ˆ

b x˙ ˆ

which can finally be written as:

mx ¨ + b x˙ + keq x = 0. With the numerical values given above, this becomes:     1 N 3 1N (2kg) x ¨+ m, x = 0, x(0) = x˙ + 6m 2 m/s 20 10

x˙ (0) = 0m/s.

With this, the damping ratio and natural frequency are: r √ 1 3 −1 ωn = s , ζ= 4 . 12 Therefore, the system is underdamped and the general response can be written as:   x(t) = e−ζ ωn t A cos(ωd t) + B sin(ωd t) . Using the initial conditions to solve for A and B, we find: ! r √ 3 3 −t/8 13 √ t + x(t) = cos e cos 13 20 8 3

Problem 8: In the figure shown to the right, in the absence of gravity the springs are unstretched in the equilibrium position.

!! √ 13 √ t . 8 3

x2 k2

a) Determine the deflection of each spring from its unstretched length when the system shown is in equilibrium.

r2 θ r1

b) If the system is released from the unstretched position of the springs, what is the maximum angular velocity of the disk during the resulting motion? ˆ ˆı

m x1 k1

Solution: a) We define θ, x1 , and x2 as indicated in the above figure. In particular, x1 and x2 represent the displacement in their respective springs as measured from their unstretched position. These coordinates are related through the following transformations: x1 = −r1 θ,

11

x2 = −r2 θ.

An appropriate free-body diagram for this system is shown to the right. Notice that the gravitational force must be included to determine the equilibrium deflection in the system. To eliminate the reaction force on the disk, angular momentum balance is applied about the center, yielding: X ˆ M G = I G θ¨k,   ˆ = I G θ¨ ˆk. T r1 + k2 r2 x2 k

Also, applying linear momentum balance to the block yields: X F = m aG ,   ¨1 ˆ. T − k1 x1 − m g ˆ = m x

−k2 x2 ˆı

G FR −T ˆ T ˆ −m g ˆ −k1 x1 ˆ

Finally, eliminating the unknown tension from these equations and using the above coordinate transformations, this single-degree-of-freedom system can be modeled with the equation:     I G + m r12 ¨θ + k1 r21 + k2 r22 θ = (m g r1 ). This equation of motion determines the equilibrium position θeq (with ¨θeq = 0) to be: m g r1 . θeq = k1 r21 + k2 r22 Therefore, the equilibrium displacements in each spring are: x1,eq = −r1 θeq =

m g r12 , k1 r21 + k2 r22

x2,eq = −r2 θeq =

m g r1 r2 . k1 r21 + k2 r22

b) Define new coordinates z1 and z2 , which measure the displacement in springs 1 and 2 with respect to the static equilibrium position, that is: z1 = x1 − x1,eq ,

z2 = x2 − x2,eq .

Likewise, let φ represent the angular displacement of the disk from the static equilibrium position: φ = θ − θeq . Therefore, the potential energy of this system can be written as: V

= =

1 1 k1 z 21 + k2 z22, 2 2  1 k1 r12 + k2 r22 φ2 . 2

Also, the kinetic energy becomes: T

= =

1 G ˙2 1 I φ + m z˙12, 2 2  1 G I + m r21 φ˙2 . 2 12

If the system is released from rest at the unstretched position of the springs, then: φ(0) = −θeq = −

m g r1 , k1 r21 + k2 r22

φ˙ (0) = 0.

At this initial state, the potential and kinetic energies become: T0 = 0,

V0 =

(m g r1 )2 . 2(k1 r12 + k2 r22 )

Because this system is conservative, the total energy, E = T + V remains constant. Therefore, when the kinetic energy is maximal, the potential energy is minimal, that is:  1 G V1 = 0, T1 = I + m r12 φ˙ 2max . 2 Finally, conservation of energy implies that V0 = T1 , and solving for φ˙ max we find that: s (m g r1 )2 ˙φmax = . G (I + m r12 )(k1 r21 + k2 r22 )

Problem 9: For the system shown to the right, the disk of mass m rolls without slip and x measures the displacement of the disk from the unstretched position of the spring.

ˆ

x z

3k

ˆı θ

a) find the equations of motion; k

b) if the system is underdamped, what is the frequency of the free vibrations of this system in terms of the parameters k, c, and m;

m

c

Solution: a) We define the three coordinates as shown as the figure, related as: x = −r θ,

z = −2 r θ,

A free-body diagram for this system is shown to the right. Notice that the force in the upper spring depends on z, rather than x, while the friction force has an unknown magnitude f . Because the disk is assumed to roll without slip, we are unable to specify the value of f , but instead can relate the displacement and rotation of the disk through the coordinate relations above.

z = 2 x.

−3k z ˆı G −k x ˆı

C

−c x˙ ˆı

f ıˆ

The equations of motions can be developed directly with angular momentum balance about the contact point, so that: X M C = I C θ¨ ˆk,   2 ˆ = 3 m r θ¨ k. ˆ (3k z) 2r + (k x) r + (c x) ˙ r k 2 13

Finally, writing this equation in terms of a single coordinate, we obtain:     3 m r2 ¨ θ + (c r2 ) θ˙ + 13 k r2 θ = 0. 2 b) For an underdamped response, the frequency of oscillation is ωd = ωn this system, we find that: r c 26 k , ζ= √ ωn = , 3m 78 k m so that: ωd =

r

p

1 − ζ 2 . With

2 c2 26 k − . 3m 9 m2

Problem 10: In the system shown to the right, the pulley has mass m and radius r, so that the moment 2 of inertia about the mass center is IG = mr2 .

r

a) Find the governing equations of motion;