Repetitive motion relative to an equilibrium position.

Evident in virtually all machines.

Vibrational loads can exceed static loads by orders of magnitude.

Mechanical VibrationsSection 9

Undamped, Free Vibration

One-degree of motion

F = 0:

-kx - ma = 0

mass, mstiffness, k

FiFs = kx

FBDof non-equilibrium forces

a

Motion Equation

Equilibrium: kx + ma = 0

Rewrite as

Define: (circular frequency)

Then:

02

2

kxdt

xdm

m

kp

022

2

xpdt

xd

Motion Equation

Second order, differential equation has the solution

x = xm sin(pt+)

where:

0

01tanv

px

20

2

0 xp

vxm

Motion Curve

Period,

Frequency, f

Time

Volts

xm

Displacementx = xm sin(pt+)

p

2

m

kpf

2

1

2

Differentiating the Solution:

Position

x = xm sin(pt+)

Velocity

v = xm p cos(pt+)

Acceleration

a = -xm p2 sin(pt+)

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.

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

Equivalent Springs

Springs in parallel

kequiv = k1 + k2 + ...

Springs in series

...111

21

kkkequiv

k2

k1

k2k1

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

Undamped, Forced Vibration

One-degree of motion

F = 0:

-kx - ma = F0 sin t

mass, mstiffness, k F0sin t

FBDof non-equilibrium forces

Fi

Fs = kx

a

F0sin t

Motion Equation

Equilibrium: kx + ma = F0sin t

Rewrite as

Define: (circular frequency)

Then:

tFkxdt

xdm sin02

2

m

kp

tFxpdt

xd sin02

2

2

Motion Equation

Second order, differential equation has the solution

Free vibration Forced vibration

tp

kFptxx m

sin

)/(1

/sin

20

Motion Curve

In time, the free vibration will dampen out. Forced vibration term is called the steady state

solution.

Free

Forced Full solutionDisplacement

Time

Steady State Solution

The steady state vibration is:

Differentiating the solution:

tp

kFxss

sin

)/(1

/2

0

tp

kFvss

cos)/(1

/2

0

tp

kFass

sin)/(1

/2

02

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

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.

Frequency Response Plot

Displacement Excitation

Some machines have a periodic displacement of the support.

Simply replace F0 with k0

mass, mstiffness, k 0sin t

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

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.

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.

DampingForce

Relative Velocity

c1

Damped, Forced Vibration

One-degree of motion

F = 0:

-kx - cv - ma = F0 sin t

F0sin tmass, m

stiffness, k

damping, c

FBDof non-equilibrium forces

Fi

Fs = kx

a

F0sin t

Fs = cv

Motion Equation

Equilibrium: kx +cv+ ma = F0sin t

Rewrite as

Define: (circular frequency)

Then:

tFkxdt

dxc

dt

xdm sin02

2

m

kp

tFxpdt

dxp

dt

xd sin02

2

2

Motion Equation

Second order, differential equation has the solution

Free vibration Forced vibration

)'sin()/)(/(2)/(1

/)sin(

222

0)2/(

tpccp

kFtpexx

c

dtmc

m

Damping Ratio

Critical damping ratio

– if c > cc system does not oscillate

Damped natural frequency

mpm

kmcc 22

22

12

cd c

cp

m

c

m

kp

Again, the free vibration will dampen out. Forced vibration term is called the steady state

solution.

Motion Curve

Free

Forced Full solutionDisplacement

Time

Steady State Solution

The steady state vibration is:

Differentiating the solution:

)'sin(

)/)(/(2)/(1

/222

0

tpccp

kFx

c

ss

t

pccp

kFv

c

ss

cos

)/)(/(2)/(1

/222

0

t

pccp

kFa

c

ss

sin

)/)(/(2)/(1

/222

02

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

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.

Frequency Response