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Vibrations 2.01 Modeling of Vibratory Systems Vibrations 2.02 Modeling of Vibratory Systems

Chapter Objectives

• Compute the mass moment of inertia of rotational systems

• Determine the stiffness of various linear and nonlinear elastic

components in translation and torsion and the equivalent

stiffness when many individual linear components are

02. Modeling of Vibratory combined

System

• Determine the stiffness of fluid, gas, and pendulum elements

• Determine the potential energy of stiffness elements

• Determine the damping for systems that have different sources

of dissipation: viscosity, dry friction, fluid, and material

• Construct models of vibratory systems

HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Vibrations 2.03 Modeling of Vibratory Systems Vibrations 2.04 Modeling of Vibratory Systems

§1. Introduction §2.Inertia Elements

- Translational motion

- Three elements that comprise a vibrating system - Rotational motion

Slender bar

• Inertia elements: stores and releases kinetic energy

Circular disk

• Stiffness elements: stores and releases potential energy = 1 2

• Dissipation elements: express energy loss in a system 12

- Components comprising a vibrating mechanical system = 1 2

• Translational motion • Rotational motion 2

Mass, ( ) Mass moment of inertia, ( 2)

Sphere = 2 2

Stiffness, ( / ) Stiffness, ( / )

Damping, ( / ) Damping, ( / ) 5

External force, ( ) External moment, ( )

Circular cylinder = = 1 (3 2 + ℎ2)

12

= 1 2

2

- Parallel-axes theorem

= + 2 (2.1)

: distance from the center of gravity to point

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Vibrations 2.05 Modeling of Vibratory Systems Vibrations 2.06 Modeling of Vibratory Systems

§2.Inertia Elements §2.Inertia Elements

- For a mass translating with a - To arrive at Eq. (2.3) in a different manner, let us consider the

work-energy theorem

velocity of magnitude ሶ in the −

plane under the driving force Ԧ Assume that the mass is translated from an

initial rest state, where the velocity is zero at

time 0, to the final state at time . Then, the

• The equation governing the motion of the mass work done under the action of a force Ԧ

Ԧ = ( ሶ Ԧ ) = න Ԧ . Ԧ = න ሷ Ԧ . Ԧ = න ሷ

when and Ԧ are independent of time

= 0 = 0 = 1 0

2

න ሷ ሶ න ሶ ሶ ሶ 2

= ሷ (2.2) 0 0

(2.3)

• The kinetic energy, , of mass The kinetic energy

1 1 ቚ − ቚ = = 1 ሶ 2 (2.5)

= 2 ሶ Ԧ ∙ ሶ Ԧ 2 2

= ሶ 2 = = 0

HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien = 1 ሶ Ԧ ∙ ሶ Ԧ = 1 ሶ 2 (2.3)

22 Nguyen Tan Tien

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Vibrations 2.07 Modeling of Vibratory Systems Vibrations 2.08 Modeling of Vibratory Systems

§2.Inertia Elements §2.Inertia Elements

- For a rigid body undergoing only - Ex.2.1 Determination of mass moments of inertia

rotation in the plane with an angular Illustrate how the mass moments of inertia of several different

speed ሶ rigid body distributions are determined

Solution

• Uniform Disk

• The equation governing the rotation of the mass of inertia The mass moment of inertia about the point , which

= ሷ (2.6) 1 is located a distance from point

2

: the moment acting about the center of mass or a = 2 = + 2 = 1 2 + 2 = 3 2

fixed point along the direction normal to the plane of 2 2

motion • Uniform Bar

: the associated mass moment of inertia The mass moment of inertia about the point

• The kinetic energy of the system 2 1 1 1

2 12 4 3

1 = + = 2 + 2 = 2

2

= ሶ 2 = 1 2

12

HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Vibrations 2.09 Modeling of Vibratory Systems Vibrations 2.10 Modeling of Vibratory Systems

§2.Inertia Elements §2.Inertia Elements

- Ex.2.2 Slider mechanism: system with varying inertia property 2 = 2 + 2 − 2

A slider of mass slides along a uniform bar of 2 = ( /2)2+ 2 −

mass with a pivot at point . Another bar, which 2 = ( /2)2+ 2 − ( − )

is pivoted at point ′, has a portion of length that

has a mass and another portion of length that The rotary inertia of this system

has a mass . Determine the rotary inertia of = + ( ) + ( ) + ( )

this system and show its dependence on the

where

Solution angular displacement coordinate

= 1 2

From geometry, , , can be described in terms of 3

2 = 2 + 2 − 2 ( ) = 2( )

2 = ( /2)2+ 2 −

2 = ( /2)2+ 2 − ( − ) = 2 + 2 = 2 + 2 −

12 3

: the distance from the midpoint of bar of mass to

= 2 + 2 = 2 + 2 − ( − )

: the distance from the midpoint of bar of mass to 12 3

HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Vibrations 2.11 Modeling of Vibratory Systems Vibrations 2.12 Modeling of Vibratory Systems

Exercises Exercises

- E.2.1 Examine Eqs. (2.1) and (2.5) and verify that the units - E.2.2 Consider the slider mechanism and show that the rotary

(dimensions) of the different terms in the respective equations inertia 0 about the pivot point ′ , is also a

are consistent function of the angular displacement

Solution Solution

where,

= + 2 (2.1)

(2.5)

ȁ = − ȁ = 0 = = 1 ሶ 2

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Vibrations 2.13 Modeling of Vibratory Systems Vibrations 2.14 Modeling of Vibratory Systems

Exercises Exercises

- Ex.2.3 Determine the rotary inertia of this system about the

point and express it as a function of the angular displacement

The disc has a rotary inertia

about the point . The crank

has a mass and rotary

inertia about the point at

the center of mass of the crank.

The mass of the slider is

Solution

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Vibrations 2.15 Modeling of Vibratory Systems Vibrations 2.16 Modeling of Vibratory Systems

Exercises §3.Stiffness Elements

1.Introduction

- Stiffness elements are manufactured from different materials

and they have many different shapes

- Application

• to minimize vibration transmission from machinery to the

supporting structure

• to isolate a building from earthquakes

• to absorb energy from systems subjected to impacts

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Vibrations 2.17 Modeling of Vibratory Systems Vibrations 2.18 Modeling of Vibratory Systems

§3.Stiffness Elements §3.Stiffness Elements

- Some representative types of stiffness elements that are - The stiffness elements store and release the potential energy

of a system

commercially available along with their typical application

- Consider a spring under acting force of magnitude Ԧ is

Building or highway directed along the direction of the unit vector Ԧ

base isolation for

lateral motion using

cylindrical rubber

bearings

Steel cable Wire rope isolators to isolate vertical Ԧ = − Ԧ

springs motions of machinery

used in a

chimney Air springs used in Typical steel Stiffness element with a force acting on it Free-body diagram

tuned suspension systems to coil springs

mass isolate vertical motions for isolation Ԧ tries to restore the stiffness element to its undeformed

damper to of vertical configuration, it is referred to as a restoring force

suppress motions

lateral

motions

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Vibrations 2.19 Modeling of Vibratory Systems Vibrations 2.20 Modeling of Vibratory Systems

§3.Stiffness Elements §3.Stiffness Elements

- As the stiffness element is deformed, energy is stored in this 2.Linear Springs

element, and as the stiffness element is undeformed, energy

is released - Translation Spring

- The potential energy is defined as the work done to take the • Deformation

stiffness element from the deformed position to the

undeformed position; that is, the work needed to undeform the = (2.9)

element to its original shape

: the applied force

: the spring constant

: the spring deflection

= න Ԧ • The potential energy stored in the spring

1

2

00 0 = න = න = න = 2 (2.10)

= න Ԧ = න − Ԧ ∙ Ԧ = න 0 00

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Vibrations 2.21 Modeling of Vibratory Systems Vibrations 2.22 Modeling of Vibratory Systems

§3.Stiffness Elements §3.Stiffness Elements

- Torsion Spring - Combinations of Linear Springs

• Parallel Springs

• Deformation

= (2.11) Translation springs

: the applied moment

Total force

: the spring constant = 1 + 2 = 1 + 2 = 1 + 2 =

: the spring deflection

Equivalent spring

• The potential energy stored in the spring = 1 + 2

1 Torsion springs

2

= න = න = 2 (2.12) Total moment

= 1 + 2

00 = 1 + 2 = 1 + 2 =

Equivalent spring

= 1 + 2

HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Vibrations 2.23 Modeling of Vibratory Systems Vibrations 2.24 Modeling of Vibratory Systems

§3.Stiffness Elements §3.Stiffness Elements

• Series Springs - Spring Constants for Some Common Elastic Elements

Translation springs

Displacement 1 1 1.Axially loaded rod or cable

= 1 + 2 = 1 + 2 = 1 + 2 =

Equivalent spring =

: cross-sectional area, 2

11 −1 1 2 : Young’s modulus of elasticity, / 2

1 + 2 1 + 2

= = : length of the rod,

Torsion springs 2.Axially loaded tapered rod

Displacement 11 = 1 2

= 1 + 2 = 1 + 2 = 1 + 2 = 4

: Young’s modulus of elasticity, / 2

Equivalent spring 1 : rod diameter,

2 : rod diameter,

11 −1 1 2 : length of the rod,

1 + 2 1 + 2

= =

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Vibrations 2.25 Modeling of Vibratory Systems Vibrations 2.26 Modeling of Vibratory Systems

§3.Stiffness Elements §3.Stiffness Elements

3.Hollow circular rod in torsion

5.Pinned-pinned beam (Hinged, simply supported)

= 3 ( + )

: shear modulus of elasticity, / 2 = 2 2

: Young’s modulus of elasticity, / 2

: the torsion constant (polar moment of inertia), 4 : the area moment of inertia about the bending axis, 4

For the concentric circular tubes, ( 4 − 4 ) , : position of applied force,

32

=

: length of the rod,

: outside rod diameter, 6.Clamped-clamped beam (Fixed-fixed beam)

3 ( + )3

: inside rod diameter,

= 3 3

4.Cantilever beam : Young’s modulus of elasticity, / 2

3 0 < ≤ : the area moment of inertia about the bending axis, 4

= 3 ,

, : position of applied force,

: Young’s modulus of elasticity, / 2

: the area moment of inertia about the bending axis, 4

: position of applied force,

: length of the beam,

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Vibrations 2.27 Modeling of Vibratory Systems Vibrations 2.28 Modeling of Vibratory Systems

§3.Stiffness Elements §3.Stiffness Elements

7.Two circular rods in torsion 9.Coil springs

4

= 1 + 2, =

= 8 3

: modulus of elasticity, / 2

: modulus of elasticity, / 2

: wire diameter,

: the torsion constant (polar moment of inertia), 4 : number of active coil

: position of applied force, : mean coil diameter,

8.Two circular rods in torsion 10. Clamped rectangular plate, constant thickness, force at center

1 1 −1 ℎ3 /

= 1 + 2 , = = 12 2(1 − 2) 1.0 0.00560

: Young’s modulus of elasticity, / 2 1.2 0.00647

: modulus of elasticity, / 2 0.00691

ℎ : thickness of plate, 1.4 0.00712

: the torsion constant (polar moment of inertia), 4 0.00720

: coefficient 1.6 0.00722

: position of applied force, : width of the plate, 1.8

2.0

: poison ratio

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Vibrations 2.29 Modeling of Vibratory Systems Vibrations 2.30 Modeling of Vibratory Systems

§3.Stiffness Elements §3.Stiffness Elements

11. Clamped circular plate, constant thickness, force at center - Force-displacement relationships may also be used to

4.189 ℎ3 determine parameters such as that characterize a stiffness

element

= 2(1 − 2)

: Young’s modulus of elasticity, / 2

ℎ : thickness of plate,

: width of the plate,

: poison ratio

12. Cantilever plate, constant thickness, force at center of free edge

0.496 ℎ3

= 2(1 − 2)

: Young’s modulus of elasticity, / 2

ℎ : thickness of plate,

: width of the plate, ≫

: length of the plate,

: poison ratio

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Vibrations 2.31 Modeling of Vibratory Systems Vibrations 2.32 Modeling of Vibratory Systems

§3.Stiffness Elements §3.Stiffness Elements

- Ex.2.3 Equivalent stiffness of a beam-spring combination - Ex.2.4 Equivalent stiffness of a cantilever beam with a

transverse end load

Consider a cantilever beam that has a spring attached at its free end

Acantilever beam: = 72 × 109 / 2, = 750 , = 110 ,

• The force is applied to the free end of the spring ⟹ parrallel = 120 . Determine the equivalent stiffness of this beam

= + 1

3 Solution

= 3

The area moment of inertia about the bending axis

= 32 4 − 4

• The force is applied simultaneously to the free end of the 120 × 10−3 4 − 110 × 10−3 4

cantilever beam ⟹ series = 32

1 1 −1 = 5.98 × 10−6 4

= + 1

The equivalent stiffness of the cantilever beam

3 3 × 72 × 109 × 5.98 × 10−6 = 3.06 × 106 /

= 3 = 750 × 10−3 3

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Vibrations 2.33 Modeling of Vibratory Systems Vibrations 2.34 Modeling of Vibratory Systems

§3.Stiffness Elements §3.Stiffness Elements

- Ex.2.5 Equivalent stiffness of a beam with a fixed end and a To this end, we use Case 6 of Table 2.3 and set = = and

translating support at the other end

obtain

Consider a uniform beam of length with flexural rigidity . When the 3 + 3 3 + 3 24

beam is subjected to a transverse loading = 3 3 อ = 3 3 = 3

at the translating support end, determine

the equivalent stiffness of this beam = =

Recognizing that the equivalent stiffness of a fixed-fixed beam

Solution of length 2 loaded at its middle is equal to the total equivalent

By observation stiffness of a parallel spring combination of two end loaded

beams, we obtain that

1 1 24 12

= 2 = 2 3 = 3

⟺ Nguyen Tan Tien HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Vibrations 2.35 Modeling of Vibratory Systems Vibrations 2.36 Modeling of Vibratory Systems

§3.Stiffness Elements §3.Stiffness Elements

- Ex.2.6 Equivalent stiffness of a micro-electromechanical Solution

system (MEMS) fixed-fixed flexure

Each of the four flexures can be treated as a beam that is fixed

A micro-electromechanical sensor at one end and free to translate only

at the other end, similar to the system

system (MEMS) consisting of four in Ex.2.5

flexures. Each of these flexures is The equivalent stiffness of each

fixed at one end and connected to a flexure is given by

mass at the other end. The length of each flexure is = 100 ,

the thickness of each flexure is ℎ = 2 , and the width of each 12 ℎ3

flexure is = 2 . A transverse loading acts on the mass = 3 , = 12

along the -direction, which is normal to the − plane. Each

The equivalent stiffness of the system

flexure is fabricated from a poly-silicon material, which has a

Young’s modulus of elasticity = 150 = 4 × = 4 × 12 × ℎ3 = ℎ3

3 12 4 3

Determine the equivalent stiffness of the system 150 × 109 × 2 × 10−6 × 2 × 10−6

= 4 100 × 10−6 = 9.6 /

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Vibrations 2.37 Modeling of Vibratory Systems Vibrations 2.38 Modeling of Vibratory Systems

§3.Stiffness Elements §3.Stiffness Elements

- Ex.2.7 Equivalent stiffness of springs in parallel: removal of a restriction

⟹ = + 1 + + 2

Determine of the equivalence spring constant for

the parallel springs subjected to unequal forces = + 1 + + + 2 +

Solution 1 2 + 2 2

+ 2 1 2

From similar triangles =

For the equivalent system

= 2 + + 1 − 2 = + 1 + + 2

=

Consider the bar

= 1 + 2 ⟹ =

2 = 1 ቋ ⟹ 1 = + , 2 = +

The equivalence spring constant for the parallel

Therefore

springs subjected to unequal forces

1 2

1 = 1 = 1 + , 2 = 2 = 2 + = 1 2 + 2

1 2 + 2 2

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Vibrations 2.39 Modeling of Vibratory Systems Vibrations 2.40 Modeling of Vibratory Systems

§3.Stiffness Elements §3.Stiffness Elements

3.Nonlinear Springs - For a nonlinear stiffness element described by Eq. (2.23), the

- Nonlinear stiffness elements appear in many applications, graph is no longer a straight line. The slope of this graph at a

including leaf springs in vehicle suspensions and uniaxial

micro-electromechanical devices in the presence of location = is given by

electrostatic actuation

+ 3 2 ቚ = + 3 2 (2.25)

ቤ =

=

=

- The spring force ( )

= ด + 3 (2.23) ⟹ in the vicinity of displacements in a neighborhood of =

, the cubic nonlinear stiffness element may be replaced by a

linear stiffness element with a stiffness constant (2.25)

: the stiffness coefficient of the nonlinear term - The constant of proportionality for the nonlinear cubic

spring is determined experimentally

> 0 hardening spring < 0 softening spring

: the linear spring constant

- The potential energy

= 1 2 + 1 4 (2.24) = + 3

2 2

= න Nguyen Tan Tien HCM City Univ. of Technology, Faculty of Mechanical Engineering (2.23)

0 Nguyen Tan Tien

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Vibrations 2.41 Modeling of Vibratory Systems Vibrations 2.42 Modeling of Vibratory Systems

§3.Stiffness Elements §3.Stiffness Elements

- Experimentally obtained data used to determine the nonlinear - Ex.2.8 Nonlinear stiffness due to geometry

spring constant

a. Nonlinear stiffness due to geometry

• The initial tension force

0 = ቚ = 0

=0

• The force in the spring

= 0 + 2 + 2 −

• The force in the -direction is obtained

= = = 0 2 + 2 −

2 + 2 + 2 + 2

2 + 2

⟹ the spring force opposing the motion is a nonlinear function

of the displacement . Hence, this vibratory model of the

system will have nonlinear stiffness

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Vibrations 2.43 Modeling of Vibratory Systems Vibrations 2.44 Modeling of Vibratory Systems

§3.Stiffness Elements §3.Stiffness Elements

Cubic Springs and Linear Springs 3

= 0 + 2

Assume that ȁ / ȁ ≪ 1, using binomial expansion

2 2 1/2 1 2 1 4 When the nonlinear term is negligible

1 + = 1 + = 1 + 2 + 8 + ⋯

/ 1 + / 2 − 1 = 0 = 0

⟹ = 0 + and the spring constant is proportional to the initial tension in

1 + / 2 the spring

1 + / 2 b.Nonlinear spring composed of a set of linear springs

/ 1 + 1 2 Another example of a nonlinear spring is one that is piecewise

2+ 2 linear as shown in figure

−1

= 0 1 1 2

2 2

1 + 1 +

3

= 0 + 2

Binomial expansion 1 + = 1 + + 1 ( − 1) 2 +⋯

2

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Vibrations 2.45 Modeling of Vibratory Systems Vibrations 2.46 Modeling of Vibratory Systems

§3.Stiffness Elements §3.Stiffness Elements

4.Other Forms of Potential Energy Elements • The equivalent spring constant of this fluid system

Consider other stiffness elements in which there is a = = 2 0 (2.27)

mechanism for storing and releasing potential energy. The

source of the restoring force is a fluid element or a gravitational

loading • The potential energy

- Fluid Element = 1 2 = 0 2

2

• The magnitude of the total force of the Alternatively, the potential energy can also be obtained

displaced fluid acting on the rest of the fluid

directly from the work done

Manometer = 2 0

: the mass density of the fluid, / 3

: gravitational constant, / 2

0 : the manometer cross-sectional area, 2 = න

: the fluid displacement, 0

= 2 0 න

0

= 0 2

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Vibrations 2.47 Modeling of Vibratory Systems Vibrations 2.48 Modeling of Vibratory Systems

§3.Stiffness Elements §3.Stiffness Elements

- Compressed Gas • The magnitude of the force on the piston

• When the piston moves by an amount along = 0

= 0 0 −

the axis of the piston, 0 decreases to a volume = 0 0 0− 1 − / 0 −

⟹ = 0 0 1 − / 0 −

= 0 − 0 = 0 0 1 − / 0 (2.33)

Gas compression ⟹ = 0 1 − / 0 The Eq. (2.33) describes a nonlinear force

with a piston 0: the piston cross-sectional area, 2

versus displacement relationship

• The equation of state for the gas At the vicinity of = , the stiffness of an equivalent linear

stiffness element

= 0 0 = 0 = ⟹ = 0 −

: gas pressure, / 2 : gas volume, 3 = ቤ = 0 0 1 − / 0 − −1

/ 0 0

: the ratio of specific heats of the gas, when compressed =

- slowly, the compression is isothermal, = 1 For ≪ 1, = 0 0

0

- rapidly, the compression is adiabatic, = / = 1.4

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Vibrations 2.49 Modeling of Vibratory Systems Vibrations 2.50 Modeling of Vibratory Systems

§3.Stiffness Elements = 1 §3.Stiffness Elements

≠ 1

• For arbitrary / 0, the potential energy - Pendulum System

• Pendulum systems: bar with uniformly

distributed mass

= න

0 At , the vertical distance through which the

center of gravity of the bar moves up from the

= 0 0 න 1 − / 0 −

reference position

0

− 0 0 0 1 − / 0

⟹ = ቐ 0 0 0 1 − / 0 − − 1 Pendulum systems: bar = 2 − 2 = 2 1 −

− 1 with uniformly distributed

mass The increase in the potential energy

= න = න =

00

or in

1

= 2 (1 − )

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Vibrations 2.51 Modeling of Vibratory Systems Vibrations 2.52 Modeling of Vibratory Systems

§3.Stiffness Elements §3.Stiffness Elements

1 • Pendulum systems: mass on a weightless rod

= 2 1 −

1 The increase in the potential energy

2 2

≈ 2 ≈ 1 1 2 = 1 2

2 2

1

= 2 2 where the equivalent spring constant

where the equivalent spring constant = 1

Pendulum systems: bar If the weightless bar is replaced by one that

with uniformly distributed = 2

Pendulum systems: mass has a uniformly distributed mass , then the

mass

on a weightless rod total potential energy of the bar and the mass

≈ 1 2 + 1 1 2 = 1 2 = 1 2

4 2 2 2 + 1 2

Taylor series approximation = 1 − 2 + ⋯ where the equivalent spring constant

2

= 2 + 1

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Vibrations 2.53 Modeling of Vibratory Systems Vibrations 2.54 Modeling of Vibratory Systems

§3.Stiffness Elements §3.Stiffness Elements

• Pendulum systems: inverted mass on a - Ex.2.9 Equivalent stiffness due to gravity loading

weightless rod

For “small” rotations about the upright position θ = 0,

The decrease in the potential energy the potential energy

≈ − 1 1 2 = 1 1 2 − 1 2 2

2 2 2

1

=2 1 − 2 2

Pendulum systems: There is a gain or loss in potential energy depending

inverted mass on a

on whether 1 > 2 or vice versa

weightless rod • When the bar has a uniformly distributed mass

1 = 1/ , 2 = 2/ , = 1 + 2, = 1/2, = 2/2

= 21 − 22 2 = 1 1 + 2 2 = 1 2

4( 1 + 2) 2 2 2

where the equivalent stiffness = 1 + 2 /2

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Vibrations 2.55 Modeling of Vibratory Systems Vibrations 2.56 Modeling of Vibratory Systems

Exercises Exercises

- E.2.4 Find the equivalent length of a spring of constant - E.2.5 Find the equivalence stiffness for the spring combinations

cross section of diameter 2 that has the same spring constant

as the tapered spring rod. Both springs have the same Young’s Solution

modulus

Solution

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Vibrations 2.57 Modeling of Vibratory Systems Vibrations 2.58 Modeling of Vibratory Systems

Exercises Exercises 2ℎ

≈ 2 1 − 1 − /2 2

- E.2.6 Consider the mechanical spring system shown in the

figure. Assume that the bars are rigid and

determine the equivalent spring constant ,

which we can use in the relation =

Solution

Binomial expansion 1 + = 1 + + 1 ( − 1) 2 + ⋯

2

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Vibrations 2.59 Modeling of Vibratory Systems Vibrations 2.60 Modeling of Vibratory Systems

Exercises Exercises

- E.2.7 Consider the three beams connected as shown in the - E.2.8 For the weightless pulley system shown in the figure,

figure. The beam that is attached to the determine the equivalent spring constant.

ends of the two cantilever beams is Recall that when the center of the pulley moves

pinned so that its ends can rotate by an amount , the rope moves an amount 2

unimpeded. Determine the system’s

equivalent spring constant for the Solution

transverse loading shown

Solution

with Nguyen Tan Tien HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

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Vibrations 2.61 Modeling of Vibratory Systems Vibrations 2.62 Modeling of Vibratory Systems

Exercises Exercises

- E.2.9 Determine the equivalent stiffness for each of the - E.2.10 For the two cantilever beams whose free ends are

systems shown in the figure. Each system consists of three

linear springs with stiffness 1, 2, and 3 connected to springs shown in the figure,

Solution give the expressions for the spring

constants 1 and 2 and determine the

equivalent spring constant for the

system

Solution

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Vibrations 2.63 Modeling of Vibratory Systems Vibrations 2.64 Modeling of Vibratory Systems

Exercises Exercises

- E.2.11 For the system of translation and torsion springs shown in - E.2.15 Find the torsion-spring constant of the stepped shaft

the figure, determine the equivalent spring shown in the figure, where each shaft has the

constant for torsional oscillations. The disc has a

radius , and the translation springs are same shear modulus

tangential to the disc at the point of attachment

Determine the equivalent spring length of a shaft of constant

Solution

diameter and length that has the same spring constant as

the torsion spring shown in the figure

Solution

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Vibrations 2.65 Modeling of Vibratory Systems Vibrations 2.66 Modeling of Vibratory Systems

Exercises Exercises

- E.2.16 Consider the two nonlinear springs in parallel that are

shown in the figure. The force-displacement

relations for each spring are, respectively

= + 3, = 1,2

a. Determine the equivalent spring constant

b. If = 1000 , 1 = 2 = 50,000 / , = 2 −2,

determine

Solution

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Exercises Exercises

- E.2.17 Consider the two nonlinear springs in series, the force-

displacement relations for each spring are, respectively,

= + 3, = 1,2

a. Determine the equivalent spring constant

b. If = 1000 , 1 = 50,000 / , 2 = 25,000 / , =

2 −2, determine

Solution

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Vibrations 2.69 Modeling of Vibratory Systems Vibrations 2.70 Modeling of Vibratory Systems

Exercises Exercises

- E.2.20 Consider “small” amplitude angular oscillations of the

pendulum. Considering the gravitational loading and the

torsion spring at the pivot point, determine the

expression for the system’s equivalent spring constant.

The masses are held with rigid, weightless rods for the loading

Solution

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Vibrations 2.71 Modeling of Vibratory Systems Vibrations 2.72 Modeling of Vibratory Systems

§4.Dissipation Elements §4.Dissipation Elements

- Damping elements are assumed to have neither inertia nor the 1.Viscous damping

means to store or release potential energy

- When a viscous fluid flows through a slot or around a piston in a

- The mechanical motion imparted to these elements is cylinder, the damping force generated is proportional to the

converted to heat or sound and, hence, they are called non- relative velocity between the two boundaries confining the fluid

conservative or dissipative because this energy is not

recoverable by the mechanical system

- There are four common types of damping mechanisms used to - A common representation of a viscous damper is a cylinder

model vibratory systems with a piston head

• Viscous damping

• Coulomb or dry friction damping - Depending on the damper construction and the velocity range,

• Material or solid or hysteretic damping the magnitude of the damper force ( ሶ ) is a nonlinear

• Fluid damping function of velocity or can be approximated as a linear

function of velocity

In all these cases, the damping force is expressed as a function

of velocity

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§4.Dissipation Elements §4.Dissipation Elements

• Energy Dissipation

- In the linear case, the damper force is expressed as

ሶ = ሶ (2.46) The energy dissipated by a linear viscous damper

: the damping coefficient, /( / ) = න = න ሶ = න ሶ 2 = න ሶ 2 (2.48)

Viscous damping of the form given by Eq. (2.46) is also called • Parallel-Plate Damper

slow-fluid damping

- In the case of a nonlinear viscous damper described by a An example of a viscous damper is shown in the figure

function ሶ , the equivalent linear viscous damping around The shear force acting on the

bottom plate

an operating speed ሶ = ሶ is determined as follows

ሶ

= ሶ ቤ (2.47) ሶ = ℎ = ℎ ሶ (2.49)

ሶ ሶ = ሶ

- Linear viscous damping elements can be combined in the The damping coefficient for the parallel-plate construction

same way that linear springs are, except that the forces are

= ℎ

proportional to velocity instead of displacement (2.50)

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Vibrations 2.75 Modeling of Vibratory Systems Vibrations 2.75 Modeling of Vibratory Systems

§4.Dissipation Elements §4.Dissipation Elements

- Ex.2.10 Design of a parallel-plate damper - Ex.2.11 Equivalent damping coefficient and

equivalent stiffness of a vibratory system

A parallel-plate damper with a top plate of dimensions 100

× 100 is to be pulled across an oil layer of thickness 0.2 , Consider the vibratory system in which the motion of mass is

restrained by a set of linear springs and linear viscous

which is confined between the moving plate and a fixed plate. dampers. Determine and

We are given that this oil is SAE30 oil, which has a viscosity of

345 (345 × 103 / 2)

Determine the viscous damping coefficient of this system

Solution

To this end, using Eq. (2.50) and substitute the given values

into this expression and find that Solution

345 × 10−3 100 × 10−3 × 100 × 10−3 The equivalence system

= ℎ = = 17.25 /

0.2 × 10−3

2 3

= 1 + 2 + 3 , = 1 + 2

= (2.50) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

ℎ Nguyen Tan Tien

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Vibrations 2.76 Modeling of Vibratory Systems Vibrations 2.77 Modeling of Vibratory Systems

§4.Dissipation Elements §4.Dissipation Elements

- Ex.2.12 Equivalent linear damping coefficient of a nonlinear damper 2.Other Forms of Dissipation

It has been experimentally determined that the damper force- - Coulomb Damping or Dry Friction

velocity relationship is given by the function This type of damping is due to the force caused by friction

ሶ = 4 / ሶ + (0.3 3/ ) ሶ 3 between two solid surfaces

Determine the equivalent linear damping coefficient around an

operating speed of 3 /

Solution

Using the Eq. (2.47) The friction force acting on the system

= ሶ ቤ= 4 + 0.9 ሶ 2 ቚ = 4 + 0.9 × 32 ሶ = ( ሶ ) (2.51)

ሶ ሶ =3 / ሶ =3 / : the kinetic coefficient of friction

⟹ = 12.1 / : the force compressing the surfaces,

= ሶ ቚ (2.47) +1 ሶ > 0

ሶ : the signum function, ሶ = ቐ−1 ሶ < 0

ሶ = ሶ Nguyen Tan Tien

0 ሶ = 0

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§4.Dissipation Elements §4.Dissipation Elements

The energy dissipated

If the normal force is due to the system weight, then =

ሶ = ( ሶ ) (2.52) = න = න ሶ = න ( ሶ ) ሶ 3 (2.54)

The energy dissipated in this case

= න = න ሶ = න ሶ ሶ (2.53) - Structural or Solid or Hysteretic Damping

- Fluid Damping (Velocity-Squared Damping) This type of damping describes the losses in materials due to

internal friction. The damping force is a function of

This type of damping is associated with a system whose mass is displacement and velocity and is of the form

vibrating in a fluid medium. The magnitude of the damping force

ሶ = ሶ 2 ሶ = ȁ ሶ ȁ ሶ (2.54) = ℎ ሶ ȁ ȁ (2.57)

ℎ : an empirically determined constant

: friction coefficient, = /2 The energy dissipated

: drag coefficient

: the mass density of the fluid = න = න ሶ = ℎ න ( ሶ )ȁ ȁ (2.58)

: the projected area of the mass in a direction normal to ሶ

Fluid damping of (2.54) is often referred to as fast-fluid damping

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Vibrations 2.81 Modeling of Vibratory Systems Vibrations 2.82 Modeling of Vibratory Systems

Exercises Exercises

- E.2.21 Determine the equivalent damping of the system - E.2.22 Determine the equivalent damping of the system

Solution Solution

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Vibrations 2.80 Modeling of Vibratory Systems Vibrations 2.81 Modeling of Vibratory Systems

§5.Model Construction §5.Model Construction

1.Introduction 2.A Micro-electromechanical System

- In this section, four examples are provided to illustrate how Micro-electromechanical accelerometer and a vibratory model

the previously described inertia, stiffness, and damping of this sensor

elements are used to construct system models

- Modeling is an art, and often experience serves as a guide in

model construction

- In this section, the examples are drawn from different areas,

and are presented in a progressive order proceeding from the

use of discrete inertia, stiffness, and damping elements in a

model to distributed elements, and finally, to a combination of

distributed and discrete elements

- As discussed in the subsequent chapters, the mass, stiffness,

and damping of a system appear as parameters in the

governing equations of the system

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§5.Model Construction §5.Model Construction

3.The Human Body 4.A Ski

Human body and a vibratory model Cross-country ski, which is a physical system with distributed

stiffness and inertia properties, and its vibratory model

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Vibrations 2.84 Modeling of Vibratory Systems Vibrations 2.85 Modeling of Vibratory Systems

§5.Model Construction §6.Design for Vibration

5.Cutting Process

Work-piece-tool turning system and vibratory model of this

system

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