Mechanical Vibrations

BMM3653

Mohd Hafizi Bin Zohari, Ph.D.

 Lesson Outcome

Students will be able to:

1. Identify the different types of element of

vibration
2. Identify the damping elements in a system
3. Analyze a system with spring elements
 ELEMENTARY PARTS OF
VIBRATING SYSTEMS
A general vibratory system includes means:
(1)For storing potential energy (elasticity)
Spring element

(2) For storing kinetic energy ( mass, inertia)

Mass element

(3) By which energy is lost (damping)

Damping element
Free Vibration:
A system is left to vibrate on its own after an initial disturbance
and no external force acts on the system. E.g. simple pendulum.

Forced Vibration:
A system that is subjected to a repeating external force. E.g.
oscillation arises from diesel engines.
Resonance occurs when the frequency of the external force
coincides with one of the natural frequencies of the system.
Un-damped Vibration:
When no energy is lost or dissipated in friction or other
resistance during oscillations

Damped Vibration:
When any energy is lost or dissipated in friction or other
resistance during oscillations
Nonlinear Vibration:
If any of the components behave nonlinearly.

Linear Vibration:
When all basic components of a vibratory system, i.e. the
spring, the mass and the damper behave linearly.
Deterministic Vibration:
If the value or magnitude of the excitation (force or
motion) acting on a vibratory system is known at any
given time.

Nondeterministic or random Vibration:

When the value of the excitation at a given time
cannot be predicted.
 System of 1 Physical 2 3 Solution to 4
Governing
Interest Model Equations GE

1. Use to construct model

2. Construct FBD, employ F = ma and M p = I p a

3. Solve analytically, numerically, computationally etc.
4. Interpret results
5. Refine as necessary.
Step 1 : Mathematical Modeling
• What do we include in our model?
• For example considering an automotive on uneven road, do we model the
vehicle as a single, two or multi degree of freedom?
• How do we model the road?
• The suspension system?
• The tires? etc.

Step 2: Derivation of Governing Equations

• Given the above modeling assumptions,
• What physical laws do we use to obtain a mathematical model?
Step 3: Solution of the Governing Equations
– Can we solve the mathematical model in Closed form?
– Approximately?
– Numerically?
Step 4: Interpretation of Results
– What information does the analysis of our model provide?
– Is it consistent we experimental results?
– Can we this information to predict future behavior or
design a better system?
Example of the modeling of
a forging hammer:
Develop a sequence of three mathematical models of the
system for investigating vibration in the vertical direction.
Consider the elasticity of the tires, elasticity and damping
of the struts (in the vertical direction), masses of the wheels,
and elasticity, damping, and mass of the rider.
• Start with the simplest model and refine it gradually.
• Then the equivalent values of the mass, stiffness, and
damping of the system are used.

Equivalent mass (meq) includes the

mass of the wheels, vehicle body and
the rider

Equivalent damping constant (ceq)

includes the damping of the struts
and the rider

Equivalent stiffness (keq) includes

the stiffness of the tires, struts, and
rider

SDOF model of the

motorcycle with a rider
Previous model can be refined by representing the
masses of wheels, elasticity of tires, and
elasticity and damping of the struts separately

Mass of the vehicle body (mv)

Mass of the rider (mr)

Mass of the wheel (mw)

The refined previous model can be obtained
• Damping is very important for an accurate prediction of vibration
response of the system

• Damper is assumed to have neither mass nor elasticity

• Damping force exists only if there is relative velocity between two

ends of the damper.

• Damping element is an element that resists relative velocity across it.

• Dashpot or damper is the basis of a shock absorber.

 Symbol of damper

dx
Damping force, F0 = cv = cx = c
dt
Damping constant or damping coefficient
SHEAR STRESS ( ) developed in the fluid layer at a distance y from
the fixed plate is: du
 =
dy
where du/dy = v/h is the velocity gradient.

SHEAR OR RESISTING FORCE (F) developed at the bottom surface of

the moving plate is: Av
F = A =  = cv
h
Where A is the surface area of the moving plate and
A is the damping constant
c=
h
If a damper is nonlinear, the operating velocity
(v*) and the equivalent damping constant is:

dF
c=
dv v*
Structural damping (material or solid)
• Due to the internal molecular friction of the material of the structure
• Due to the connections inherent in a structural system

Viscous damping
• Occurs in a system vibrating in a fluid.
• Very important
• Damping in our study normally considered to be viscous type unless
otherwise specified. dx
Damping force, F0 = cv = cx = c
dt
Coulomb damping (Dry friction)
• Occurs when the motion of the body is on a dry surface
• Negative damping
Damping force, F0 = N
Spring Elements (Spring)

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 Spring Elements

❑ Work done (U) in deforming a spring or the strain (potential)

energy is given by: 1 2
U= kx
2
❑ When an incremental force ΔF is added to F:
F + F = F ( x* + x)
dF
= F (x ) +
*
(x)
dx x*

1 d 2F
+ (x) 2 + ...
2! dx 2 x*

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 • Longitudinal motion
EA • A is the cross-sectional area (m2)
k=
• E is the elastic modulus (Pa=N/m2)
• l is the length (m)
• k is the stiffness (N/m)

x(t)
• Jp is the polar moment of inertia
of the rod
GJ p • J is the mass moment of inertia
Jp
k= •
of the disk
G is the shear modulus, l is the
length
0

J  (t)
 Spring Elements

❑Combination of Springs:
1) Springs in parallel – if we have n spring constants k1,
k2, …, kn in parallel, then the equivalent spring constant
keq is:

keq = k1 + k 2 + ... + k n

34
 Spring Elements

❑Combination of Springs:
2) Springs in series – if we
have n spring constants k1,
k2, …, kn in series, then the
equivalent spring constant
keq is:

1 1 1 1
= + + ... +
keq k1 k 2 kn

35
 Example:
Torsional Spring Constant of a Propeller Shaft

Determine the torsional spring constant of

the speed propeller shaft shown in Fig 1.25:

Fig 1.25

36
 Solution
We need to consider the segments 1-2 and
2-3 of the shaft as springs in combination.
From Fig. 1.25, the torque induced at any
cross section of the shaft (such as AA or BB)
can be seen to be equal to the torque
applied at the propeller, T. Hence, the
elasticity (springs) corresponding to the two
segments 1-2 and 2-3 are to be considered
as series springs. The spring constants of
segments 1-2 and 2-3 of the shaft (kt12 and
kt23) are given by

37
 Solution

GJ12 G ( D124 − d124 ) (80 109 ) (0.34 − 0.2 4 )

kt12 = = =
l12 32l12 32(2)
= 25.5255 106 N - m/rad

GJ 23 G ( D234 − d 234 ) (80 109 ) (0.254 − 0.154 )

kt23 = = =
l23 32l23 32(3)
= 8.9012 106 N - m/rad

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 Solution

Since the springs are in series, gives

kt12 kt23 (25.5255 10 )(8.9012 10 )

6 6
kteq = =
kt12 + kt23 (25.5255 10 + 8.9012 10 )
6 6

= 6.5997 10 N - m/rad

6

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 Example:
Equivalent k of a Crane

The boom AB of crane is a

uniform steel bar of length 10 m
and x-section area of 2,500 mm2.
A weight W is suspended while
the crane is stationary. Steel
cable CDEBF has x-sectional
area of 100 mm2. Neglect effect
of cable CDEB, find equivalent
spring constant of system in the
vertical direction.

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 Solution

A vertical displacement x of point B will cause the

spring k2 (boom) to deform by x2 = x cos 45º and the
spring k1 (cable) to deform by an amount x1 = x cos
(90º – θ). Length of cable FB, l1 is as shown.

l12 = 32 + 10 2 − 2(3)(10) cos135 = 151.426

 l1 = 12.3055 m

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✓The angle 𝜃 satisfies the relation:
2
l1 + 3 − 2(l1)(3) cos  = 10
2 2

cos  = 0.8184,  = 35.0736

✓The total potential energy (U):

1 1
U = k1 ( x cos 45) + k 2 [ x cos(90 −  )]2
2

2 2
A1E1 (100  10−6 )(207  109 )
k1 = = = 1.6822  106 N/m
l1 12.0355

A2 E2 (2500  10−6 )(207  109 )

k2 = = = 5.1750  107 N/m
l2 10 42
✓Potential Energy of the equivalent spring is:
1
U eq = keq x 2
2
✓By setting U = Ueq, hence:

keq = 26.4304  106 N/m

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A, E

A = Area of Cross Section

AE
K= E= Modulus of Elasticity
L L = Length of the bar
• E= Modulus of Elasticity
• I = Area Moment of Inertia relative to bending axis
• Kb = Spring stiffness of beam

3EI
F= 3 
L
3EI
Kb = 3
=
FL 3
for a cntilever beam
L
3EI
Similarly obtain Kb for other boundary
conditions from the deflection equations
 F
L

Keq
K X

3EI F
X
K eq = K + 3
L
Equivalent
System
Series or Parallel? Case 1
F1 = K eq X
K 2 K3
K eq = K1 +
K 2 + K3

Case 2
F2 = K eq X
K 2 K1
K eq = K 3 +
K 2 + K1
Stiffness Elements: Beams in Bending

Fixed-Fixed Simply supported

L L

FL 3 FL3
F = K max ,  max = F = K max ,  max =
192 EI 48EI
192 EI 48EI
Kb = Kb = 3
L3 L
 Stiffness Element: Torsion

M= Moment d 4
Kt = GJ
L ,J =
M = K t ( L −  R ) = K t 
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For a round shaft of diameter, d
M = Kt Nm
J is the polar area moment , units =
rad
Thank You