1. Mechanical Vibrations
Section 9
Mechanical Vibrations
Repetitive motion relative to an equilibrium position.
Evident in virtually all machines.
Vibrational loads can exceed static loads by orders of magnitude.

2. Undamped, Free Vibration
One-degree of motion
SF = 0:
-kx - ma = 0
mass, m
stiffness, k Fi
Fs = kx
FBD
of non-equilibrium forces a

3. Motion Equation Equilibrium: kx + ma = 0 Rewrite as
Define: (circular frequency)
Then:

4. Motion Equation Second order, differential equation has the solution
x = xm sin(pt+f)
where:

5. Motion Curve xm t
Displacement
x = xm sin(pt+f)
Period, t
Frequency, f

6. Differentiating the Solution:
Position
x = xm sin(pt+f)
Velocity
v = xm p cos(pt+f)
Acceleration
a = -xm p2 sin(pt+f)

7. Problem 9-1:
A 100 lb electronics case is supported by the spring suspension as shown. The case is able to move along two linear guides. The spring has a constant of 75 lb/in. The case is displaced 0.5 in then released. Determine the resulting frequency of vibrations. Also determine the maximum velocity and acceleration of the case. Neglect the effects of friction.

8. Problem 9-10:
The bent link shown has negligible mass and supports a 4 kg collar at its end. Determine the frequency of vibration if the collar is displaced a small amount and released.
250 mm
100 mm
300 N/m
4 kg

9. Equivalent Springs Springs in parallel kequiv = k1 + k2 + ...
Springs in series k2 k1 k2 k1

10. Problem 9-3:
A 40 lb machine base is supported by the double spring suspension as shown. The machine base is able to move along two linear guides. Both springs have a constant of 15 lb/in. The machine base is displaced 1.25 in then released. Determine the resulting frequency of vibrations. Also determine the maximum velocity and acceleration of the base.
Neglect the effects of friction
machine base

11. Undamped, Forced Vibration
One-degree of motion
SF = 0:
-kx - ma = F0 sin wt
mass, m
stiffness, k
F0sin wt
FBD
of non-equilibrium forces Fi
Fs = kx a
F0sin wt

12. Motion Equation Equilibrium: kx + ma = F0sin wt Rewrite as
Define: (circular frequency)
Then:

13. Motion Equation Second order, differential equation has the solution
Free vibration Forced vibration

14. Motion Curve In time, the free vibration will dampen out.
Forced
Full solution
Displacement
Time
In time, the free vibration will dampen out.
Forced vibration term is called the steady state solution.

15. Steady State Solution The steady state vibration is:
Differentiating the solution:

16. Problem 9-13
The electric motor has a mass of 50 kg and is supported by four springs, each having a stiffness of 100 N/m. The motor turns a 7 kg disk, which is mounted eccentrically, 20 mm from the disks center. Determine the speed of the motor at which resonance occurs. Assume the motor only vibrates in the vertical direction

17. Problem 9-14
Determine the amplitude of the steady-state vibration of the motor described in the previous problem, when it is running at
1200 rpm.

18. Frequency Response Plot

19. Displacement Excitation
Some machines have a periodic displacement of the support.
Simply replace F0 with kd0
mass, m
stiffness, k
d0sin wt

20. Problem 9-24:
The 18 lb instrument shown is used for on site measurements, and is carried in a truck. It is centered uniformly on a platform, which is isolated from the truck by four springs, each having a stiffness of 13 lb/in. Determine the frequency of the vibrations of the truck body, which will cause resonance to occur. The platform is only able to vibrate in the vertical direction

21. Problem 9-25:
Determine the amplitude of the steady-state vibration of the instrument described in the previous problem, when the truck floor is vibrating at 7 Hz with an amplitude of 2 in.

22. Viscous Damping
Many cases, damping is attributed to resistance created by a substance, such as oil, air or water.
This type of viscous force is proportional to the speed of the rigid body.
Damping
Force c 1
Relative Velocity

23. Damped, Forced Vibration
One-degree of motion
SF = 0:
-kx - cv - ma = F0 sin wt
F0sin wt
mass, m
stiffness, k
damping, c
FBD
of non-equilibrium forces Fi
Fs = kx a
F0sin wt
Fs = cv

24. Motion Equation Equilibrium: kx +cv+ ma = F0sin wt Rewrite as
Define: (circular frequency)
Then:

25. Motion Equation Second order, differential equation has the solution
Free vibration Forced vibration

26. Damping Ratio Critical damping ratio Damped natural frequency
if c > cc system does not oscillate
Damped natural frequency

27. Motion Curve Again, the free vibration will dampen out.
Forced
Full solution
Displacement
Time
Again, the free vibration will dampen out.
Forced vibration term is called the steady state solution.

28. Steady State Solution The steady state vibration is:
Differentiating the solution:

29. Problem 9-28:
The electric motor has a mass of 40 kg and is supported by four springs, each having a stiffness of 200 N/m. The motor turns a 4 kg disk, which is mounted eccentrically, 60 mm from the disks center. Determine the speed of the motor at which resonance occurs. The damping factor c/cc = Assume the motor only vibrates in the vertical direction

30. Problem 9-29:
Determine the amplitude of the steady-state vibration of the motor described in the previous problem, when it is running at
100 rpm.

31. Frequency Response