Mechanical vibration

1. Isolators = Springs + Dampers
Undamped Spring Mount
Pneumatic Rubber Mount
Damped Spring Mount
A press mounted on four
pneumatic rubber mounts
VIBRATION ISOLATION
Vibration isolation means to design isolators i.e., springs
and/or dampers in such a way that an object/equipment can
be protected from the harmful effects of vibrations (large
amplitudes and transmission forces)

2. Harmonic Base Excitation
• The vibrations are being transmitted to the structure through the excitation of the base
or support.
• Ex: vehicle vibrations due to rough road, earthquake, sensors attachment, camera
attached to a fighter jet or a vehicle & a machine placed on a floor that is vibrating.
• Actually, displacement and force both will be transmitted through base excitations.
• This problem is also known as vibration isolation (except the sensor applications).
ẍ
net effect
EOM: (∑F = mẍ)
- c(ẋ-ẏ) – k(x-y) = mẍ
Note we must select the signs
of x and y such that the
coefficients of m c and k in the
EOM remains +ve wrt the
response variable (i.e. x)
or
We know that: y = Ysinωt & ẏ = Yωcosωt
(1)
Put in (1)
mẍ + cẋ+ kx = Asin(ωt+α)
or
Coz RHS of (2) represents sum of two harmonic motions
(2)
ẍ

3. Recall that resultant/sum of two harmonic motions is given by:
Resultant =

4. mẍ + cẋ+ kx = Asin(ωt+α)
For the sum of two harmonic motions A and α are as follows:
α = tan-1 (cω/k)
&
Note that the unit of A is Newton (N)
Therefore,
mẍ + cẋ+ kx = sin(ωt+α) (3)
So we can say that Eq (3) again becomes EOM of damped forced vibration.
The particular solution of (3) is:
xp (t) = X sin(ωt + α – φ)
&
or,
here,
Fo=

5. • It is again important to see that how much displacement is transferred to the structure
due to the excitation of the base and what measures could be taken to reduce /isolate it.
Fo=
Finally,
Td = X/Y is the displacement transmissibility.
Td is the ratio of the amplitude of the mass to that of the base.
We have,
Put in above Eq.
(X/Y in terms of ζ & r)

6. • We can control Td through ζ & r.
• Td = X/Y increases as r tends to 1.
• Td tends to infinity when ζ = 0
(undamped) @ r = 1.
• Td reaches its maximum only for
0 < ζ < 1 at r = rm =
(dTd/dr = 0)
• For any value of ζ Td begins to decrease
as r exceeds 1, it becomes unity when
r = √2 and further continuously decreases
if r > √2. Hence, for Td < 1 the operating
region should be r > 2. Note that in this
region even ξ needs not to be large.
• Td is always unity when r = √2 for any
value of ζ.
• If r < √2 then smaller damping ratio
leads to larger value of Td (not good) but
if r > √2 smaller damping ratio leads to
smaller values of Td (good).

7. Force Transmissibility
• The problem of base excitation also generates the problem of force transmitted to the mass
through the isolators connected (i.e. due to springs and dampers attached).
• In fact any vibrating structure is capable of transmitting force to its base/foundation via isolator
connections.
We know that for forced damped vibrations the steady response is given as follows:
Differentiate twice to get ẍ = -Xω2sin(ωt-φ) and put in (1).
F(t) = mXω2sin(ωt-φ)
Where,
F(t) = FT sin(ωt-φ)
(1)
F(t) = k(x-y) + c(ẋ-ẏ) = -mẍ
FT = mXω2 (2)
Therefore,
(Max amplitude of transmitted force)
- c(ẋ-ẏ) – k(x-y) = mẍ
EOM:
or,
• Note that the transmitted force is also
harmonic & it is in phase with the
displacement of the mass.

8. Recall eq (2): FT = mXω2
For forced damped vibrations X is:
and for the base excitation we have:
Fo=
Put above values in eq (2):
mω2
FT =
After taking k common from the numerator and denominator and substituting
m/k = 1/ωn
2 , cω/k = 2ζr and r = ω/ωn we can write the above equation as follows:
The ratio FT/k Y is known as the force transmissibility and is due to the motion of the
base.

9. • Unlike the displacement
transmissibility (X/Y) the force
transmissibility (FT/k Y) does not
necessarily decreases if r > √2.
• For critically damped systems
FT/k Y continuously increases with
r. (drive slow on rough tracks!)
• FT/k Y reaches infinity near
resonance if ζ is small.
• if r > √2 then ζ must remain small
(ζ ≤ 0.2) to reduce the force
transmissibility.
• FT/k Y reaches 1 if ζ reaches 0.
• The best way to reduce force
transmissibility FT/k Y
irrespective of ζ is to keep r low
(i.e. keep r near 0)
• During base excitation we need to
have a compromise b/w the force
transmissibility & displacement
transmissibility

10. In a 1DOF underdamped forced vibration system the maximum vibration amplitude is measured
to be 2 mm. If the frequency ratio is to be 0.9 find the stiffness of the isolator mechanism in
terms of the excitation force.
Let the amplitude of force excitation = Fo
Than Xmax = 2 mm = 0.002 m.
We know that for damped force vibration:
M
However @ M = Mmax
r = (1-2ζ2)1/2
=
1 − 𝑟𝑟2
2
ζ =
1−0.92
2
ζ = 0.308
Therefore from (1):
While:
Mmax = Xmax/δst = Xmax/Fo/ k = Xmax k/Fo (1)
𝑘𝑘 =
𝐹𝐹𝑜𝑜
2𝑋𝑋𝑚𝑚𝑚𝑚𝑚𝑚ζ 1 − ζ2
𝑘𝑘 =
𝐹𝐹𝑜𝑜
2 × 0.002 × 0.3 1 − 0.32
k = 1.1447×10-3 Fo ANS