Mechanical Vibrations

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

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. Displacement x = xm sin(pt+f) xm t Motion Curve • 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. 4 kg 100 mm 250 mm 300 N/m 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.

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

10. machine base 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

11. F0sin wt stiffness, k mass, m FBD of non-equilibrium forces a Fi Fs = kx Undamped, Forced Vibration One-degree of motion SF = 0: -kx - ma = F0 sin wt 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. Full solution Forced Displacement Time Free Motion Curve • 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. d0sin wt stiffness, k mass, m Displacement Excitation • Some machines have a periodic displacement of the support. • Simply replace F0 with kd0

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

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 • if c > cc system does not oscillate • Damped natural frequency

27. Forced Full solution Displacement Time Free Motion Curve • 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 = 0.15. 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