Download - Ch.02 Modeling of Vibratory Systems

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02. Modeling of Vibratory

System

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

Vibrations 2.01 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

combined

• 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



§1.Introduction

- Three elements that comprise a vibrating system

• Inertia elements: stores and releases kinetic energy

• Stiffness elements: stores and releases potential energy

• Dissipation elements: express energy loss in a system

- Components comprising a vibrating mechanical system

• Translational motion • Rotational motion

Mass, 𝑚 (𝑘𝑔) Mass moment of inertia, 𝐽 (𝑘𝑔𝑚2)

Stiffness, 𝑘 (𝑁/𝑚) Stiffness, 𝑘𝑡 (𝑁𝑚/𝑟𝑎𝑑)

Damping, 𝑐 (𝑁𝑠/𝑚) Damping, 𝑐𝑡 (𝑁𝑚𝑠/𝑟𝑎𝑑)

External force, 𝐹 (𝑁) External moment, 𝑀 (𝑁𝑚)



§2.Inertia Elements

- Translational motion 𝑚

- Rotational motion

Slender bar 𝐽𝐺 =1

12𝑚𝐿2

Circular disk 𝐽𝐺 =1

2𝑚𝑅2

Sphere 𝐽𝐺 =2

5𝑚𝑅2

Circular cylinder 𝐽𝑥 = 𝐽𝑦 =1

12𝑚(3𝑅2 + ℎ2)

𝐽𝑧 =1

2𝑚𝑅2

𝐽𝑂 = 𝐽𝐺 + 𝑚𝑑2, 𝑑: distance from the center of gravity to point 𝐺


- Parallel-axes theorem



- For a mass 𝑚 translating with a

velocity of magnitude 𝑥 in the 𝑋−𝑌

plane under the driving force 𝐹

• The equation governing the motion of the mass 𝑚

𝐹 𝑖 =𝑑

𝑑𝑡(𝑚 𝑥 𝑖)

when 𝑚 and 𝑖 are independent of time

𝐹 = 𝑚 𝑥 (2.2)

• The kinetic energy, 𝑇, of mass 𝑚

𝑇 =1

2𝑚 𝑥 𝑖 ∙ 𝑥 𝑖 =

1

2𝑚 𝑥2




- For a rigid body undergoing only

rotation in the plane with an angular

speed 𝜃

• The equation governing the rotation of the mass of inertia

𝑀 = 𝐽 𝜃 (2.6)

𝑀 : the moment acting about the center of mass 𝐺 or a

fixed point 𝑂 along the direction normal to the plane of

motion

𝐽 : the associated mass moment of inertia

• The kinetic energy of the system

𝑇 =1

2𝐽 𝜃2



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- Ex.2.1 Determination of mass moments of inertia

Illustrate how the mass moments of inertia of several different

rigid body distributions are determined

Solution

• Uniform Disk

The mass moment of inertia about the point 𝑂 ,

which is located a distance 𝑅 from point 𝐺

𝐽𝑂 = 𝐽𝐺 + 𝑚𝑅2 =1

2𝑚𝑅2 + 𝑚𝑅2 =

3

2𝑚𝑅2

• Uniform Bar

The mass moment of inertia about the point 𝑂

𝐽𝑂 = 𝐽𝐺 + 𝑚𝐿

2

2

=1

12𝑚𝐿2 +

1

4𝑚𝐿2 =

1

3𝑚𝐿2


𝐽𝐺 =1

2𝑚𝑅2

𝐽𝐺 =1

12𝑚𝐿2



- Ex.2.2 Slider mechanism: system with varying inertia property

A slider of mass 𝑚𝑠 slides along a uniform bar of

mass 𝑚𝑙 with a pivot at point 𝑂. Another bar,

which is pivoted at point 𝑂′, has a portion of length

𝑏 that has a mass 𝑚𝑏 and another portion of

length 𝑒 that has a mass 𝑚𝑒. Determine the rotary

inertia 𝐽𝑂 of this system and show its dependence

on the angular displacement coordinate 𝜑

From geometry, 𝑟, 𝑎𝑏, 𝑎𝑒 can be described in terms of 𝜑

𝑟2 𝜑 = 𝑎2 + 𝑏2 − 2𝑎𝑏𝑐𝑜𝑠𝜑

𝑎𝑏2 𝜑 = (𝑏/2)2+𝑎2 − 𝑎𝑏𝑐𝑜𝑠𝜑

𝑎𝑒2 𝜑 = (𝑒/2)2+𝑎2 − 𝑎𝑒𝑐𝑜𝑠(𝜋 − 𝜑)

𝑎𝑏 : the distance from the midpoint of bar of mass 𝑚𝑒 to 𝑂

𝑎𝑒 : the distance from the midpoint of bar of mass 𝑚𝑏 to 𝑂


Solution



𝑟2 𝜑 = 𝑎2 + 𝑏2 − 2𝑎𝑏𝑐𝑜𝑠𝜑

𝑎𝑏2 𝜑 = (𝑏/2)2+𝑎2 − 𝑎𝑏𝑐𝑜𝑠𝜑

𝑎𝑒2 𝜑 = (𝑒/2)2+𝑎2 − 𝑎𝑒𝑐𝑜𝑠(𝜋 − 𝜑)

The rotary inertia 𝐽𝑂 of this system

𝐽𝑂 = 𝐽𝑚𝑙+ 𝐽𝑚𝑠

(𝜑) + 𝐽𝑚𝑏(𝜑) + 𝐽𝑚𝑒

(𝜑)

where

𝐽𝑚𝑙=

1

3𝑚𝑙𝑗

2

𝐽𝑚𝑠(𝜑) = 𝑚𝑠𝑟

2(𝜑)

𝐽𝑚𝑏𝜑 = 𝑚𝑏

𝑏2

12+ 𝑚𝑏𝑎𝑏

2 = 𝑚𝑏

𝑏2

3+ 𝑎2 − 𝑎𝑏𝑐𝑜𝑠𝜑

𝐽𝑚𝑒𝜑 = 𝑚𝑒

𝑒2

12+ 𝑚𝑒𝑎𝑒

2 = 𝑚𝑒

𝑒2

3+ 𝑎2 − 𝑎𝑒𝑐𝑜𝑠(𝜋 − 𝜑)



§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




- Some representative types of stiffness elements that are

commercially available along with their typical application


Building or highway

base isolation for

lateral motion using

cylindrical rubber

bearings

Wire rope isolators to isolate vertical motions of machinery

Steel cable

springs

used in a

chimney

tuned

mass

damper to

suppress

lateral

motions

Air springs used in

suspension systems to

isolate vertical motions

Typical steel

coil springs

for isolation

of vertical

motions



- The stiffness elements store and release the potential energy

of a system

- Consider a spring under acting force of magnitude 𝐹 is

directed along the direction of the unit vector 𝑗

𝐹𝑠 = −𝐹 𝑗

𝐹𝑠 tries to restore the stiffness element to its undeformed

configuration, it is referred to as a restoring force


Stiffness element with a force acting on it Free-body diagram


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- As the stiffness element is deformed, energy is stored in this

element, and as the stiffness element is undeformed, energy

is released

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

the stiffness element from the deformed position to the

undeformed position; that is, the work needed to undeform

the element to its original shape

𝑉 𝑥 = 𝑑𝑒𝑓𝑜𝑟𝑚𝑒𝑑 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛

𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑜𝑟 𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛

𝐹𝑠𝑑𝑥

=

𝑥

0

𝐹𝑠𝑑𝑥 =

𝑥

0

−𝐹 𝑗 ∙ 𝑑𝑥 𝑗 =

𝑥

0

𝐹𝑑𝑥




2.Linear Springs

- Translation Spring

• Deformation

𝐹 𝑥 = 𝑘𝑥 (2.9)

𝐹 : the applied force

𝑘 : the spring constant

𝑥 : the spring deflection

• The potential energy 𝑉 stored in the spring

𝑉 𝑥 =

0

𝑥

𝐹 𝑥 𝑑𝑥 =

0

𝑥

𝑘𝑥𝑑𝑥 = 𝑘

0

𝑥

𝑥𝑑𝑥 =1

2𝑘𝑥2


(2.10)



- Torsion Spring

• Deformation

𝜏 𝜃 = 𝑘𝑡𝜃 (2.11)

𝜏 : the applied moment

𝑘𝑡 : the spring constant

𝜃 : the spring deflection

• The potential energy 𝑉 stored in the spring

𝑉 𝜃 =

0

𝜃

𝜏 𝜃 𝑑𝜃 =

0

𝜃

𝑘𝑡𝜃𝑑𝜃 =1

2𝑘𝑡𝜃

2


(2.12)



- Combinations of Linear Springs

• Parallel Springs

Translation springs

Total force

𝐹 𝑥 =𝐹1 𝑥 +𝐹1 𝑥 =𝑘1𝑥+𝑘2𝑥= 𝑘1 +𝑘2 𝑥 =𝑘𝑒𝑥

Equivalent spring

𝑘𝑒 = 𝑘1 + 𝑘2

Torsion springs

Total moment

𝜏 𝜃 = 𝜏1 𝜃 + 𝜏2 𝜃

= 𝑘𝑡1𝜃 + 𝑘𝑡2

𝜃 = 𝑘𝑡1+ 𝑘𝑡2

𝜃 = 𝑘𝑡𝑒𝜃

Equivalent spring

𝑘𝑡𝑒= 𝑘𝑡1

+ 𝑘𝑡2




• Series Springs

Translation springs

𝑥 = 𝑥1 + 𝑥2 =𝐹

𝑘1+

𝐹

𝑘2=

1

𝑘1+

1

𝑘2𝐹 =

𝐹

𝑘𝑒

Equivalent spring

𝑘𝑒 =1

𝑘1+

1

𝑘2

−1

=𝑘1𝑘2

𝑘1 + 𝑘2

Torsion springs

𝜃 = 𝜃1 + 𝜃2 =𝜏

𝑘𝑡1

+𝜏

𝑘𝑡2

=1

𝑘𝑡1

+1

𝑘𝑡2

𝜏 =𝜏

𝑘𝑡𝑒

Equivalent spring

𝑘𝑡𝑒=

1

𝑘𝑡1

+1

𝑘𝑡2

−1

=𝑘𝑡1

𝑘𝑡2

𝑘𝑡1+ 𝑘𝑡2


Displacement

Displacement



- Spring Constants for Some Common Elastic Elements

1. Axially loaded rod or cable

𝑘 =𝐴𝐸

𝐿

𝐴 : cross-sectional area, 𝑚2

𝐸 : Young’s modulus of elasticity, 𝑁/𝑚2

𝐿 : length of the rod, 𝑚

2. Axially loaded tapered rod

𝑘 =𝜋𝐸𝑑1𝑑2

4𝐿


𝑑1 : rod diameter, 𝑚

𝑑2 : rod diameter, 𝑚




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3. Hollow circular rod in torsion

𝑘𝑡 =𝐺𝐼

𝐿

𝐺 : shear modulus of elasticity, 𝑁/𝑚2

𝐼 : the torsion constant (polar moment of inertia), 𝑚4

For the concentric circular tubes,


𝑑𝑜 : outside rod diameter, 𝑚

𝑑𝑖 : inside rod diameter, 𝑚

4. Cantilever beam

𝑘 =3𝐸𝐼

𝑎3, 0 < 𝑎 ≤ 𝐿


𝐼 : the area moment of inertia about the bending axis, 𝑚4

𝑎 : position of applied force, 𝑚

𝐿 : length of the beam, 𝑚


𝐼 =𝜋(𝑑𝑜

4 − 𝑑𝑖4)

32



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

𝑘 =3𝐸𝐼(𝑎 + 𝑏)

𝑎2𝑏2



𝑎,𝑏: position of applied force, 𝑚

6. Clamped-clamped beam (Fixed-fixed beam)

𝑘 =3𝐸𝐼(𝑎 + 𝑏)3

𝑎3𝑏3



𝑎,𝑏: position of applied force, 𝑚




7. Two circular rods in torsion

𝑘𝑡𝑒= 𝑘𝑡1

+ 𝑘𝑡2, 𝑘𝑡𝑖

=𝐺𝑖𝐼𝑖𝐿𝑖

𝐺𝑖 : modulus of elasticity, 𝑁/𝑚2

𝐼𝑖 : the torsion constant (polar moment of inertia), 𝑚4

𝐿𝑖 : position of applied force, 𝑚

8. Two circular rods in torsion

𝑘𝑡𝑒=

1

𝑘𝑡1

+1

𝑘𝑡2

−1

, 𝑘𝑡𝑖=

𝐺𝑖𝐼𝑖𝐿𝑖

𝐺𝑖 : modulus of elasticity, 𝑁/𝑚2

𝐼𝑖 : the torsion constant (polar moment of inertia), 𝑚4

𝐿𝑖 : position of applied force, 𝑚




9. Coil springs

𝑘 =𝐺𝑑4

8𝑛𝐷3

𝐺 : modulus of elasticity, 𝑁/𝑚2

𝑑 : wire diameter, 𝑚

𝑛 : number of active coil

𝐷 : mean coil diameter, 𝑚

10. Clamped rectangular plate, constant thickness, force at center

𝑘 =𝐸ℎ3

12𝛼𝑎2(1 − 𝜈2)


ℎ : thickness of plate, 𝑚

𝛼 : coefficient

𝑎 : width of the plate, 𝑚

𝜈 : poison ratio


𝑏/𝑎 𝛼1.0 0.005601.2 0.006471.4 0.006911.6 0.007121.8 0.007202.0 0.00722



- Force-displacement relationships may also be used to

determine parameters such as 𝑘 that characterize a stiffness

element




- Ex.2.3 Equivalent stiffness of a beam-spring combination

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

• The force is applied to the free end of the spring

𝑘𝑒 =1

𝑘𝑏𝑒𝑎𝑚+

1

𝑘1

−1

𝑘𝑏𝑒𝑎𝑚 =3𝐸𝐼

𝐿3

• The force is applied simultaneously to the free end of the

cantilever beam

𝑘𝑒 = 𝑘𝑏𝑒𝑎𝑚 + 𝑘1


⟹ parrallel

⟹ series


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- Ex.2.4 Equivalent stiffness of a cantilever beam with a

transverse end load

Acantilever beam:𝐸 = 72×109𝑁/𝑚2, 𝑎 = 750𝑚𝑚, 𝑑𝑖 = 110𝑚𝑚,

𝑑𝑜 = 120𝑚𝑚. Determine the equivalent stiffness of this beam

Solution

The area moment of inertia 𝐼 about the bending axis

𝐼 =𝜋

32𝑑𝑜

4 − 𝑑𝑖4

=𝜋

32120 × 10−3 4 − 110 × 10−3 4

= 5.98 × 10−6𝑚4

The equivalent stiffness of the cantilever beam

𝑘 =3𝐸𝐼

𝐿3 =3 × 72 × 109 × 5.98 × 10−6

750 × 10−3 3 = 3.06 × 106𝑁/𝑚




- Ex.2.5 Equivalent stiffness of a beam with a fixed end and a

translating support at the other end

Consider a uniform beam of length 𝐿 with flexural rigidity 𝐸𝐼. When the

beam is subjected to a transverse loading 𝐹at the translating support end, determine

the equivalent stiffness of this beam

Solution

By observation

⟺




To this end, we use Case 6 of Table 2.3 and set 𝑎 = 𝑏 = 𝐿 and

obtain

𝑘𝑓𝑖𝑥𝑒𝑑 = 3𝐸𝐼 𝑎 + 𝑏 3

𝑎3𝑏3

𝑎=𝑏=𝐿

=3𝐸𝐼 𝐿 + 𝐿 3

𝐿3𝐿3 =24𝐸𝐼

𝐿3

Recognizing that the equivalent stiffness of a fixed-fixed beam

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

stiffness of a parallel spring combination of two end loaded

beams, we obtain that

𝑘𝑒 =1

2𝑘𝑓𝑖𝑥𝑒𝑑 =

1

2

24𝐸𝐼

𝐿3 =12𝐸𝐼

𝐿3




- Ex.2.6 Equivalent stiffness of a micro-electromechanical

system (MEMS) fixed-fixed flexure

A micro-electromechanical sensor

system (MEMS) consisting of four

flexures. Each of these flexures is

fixed at one end and connected to a

mass at the other end. The length of each flexure is𝐿 =100𝜇𝑚, the thickness of each flexure is ℎ = 2𝜇𝑚, and the

width of each flexure is 𝑏 = 2𝜇𝑚. A transverse loading acts on

the mass along the 𝑍-direction, which is normal to the 𝑋 − 𝑌plane. Each flexure is fabricated from a poly-silicon material,

which has a Young’s modulus of elasticity 𝐸 = 150𝐺𝑃𝑎

Determine the equivalent stiffness of the system




Solution

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

at one end and free to translate only

at the other end, similar to the

system in Ex.2.5

The equivalent stiffness of each

flexure is given by

𝑘𝑓𝑙𝑒𝑥𝑢𝑟𝑒 =12𝐸𝐼

𝐿3 , 𝐼 =𝑏ℎ3

12The equivalent stiffness of the system

𝑘𝑒 = 4 × 𝑘𝑓𝑙𝑒𝑥𝑢𝑟𝑒 = 4 ×12𝐸 ×

𝑏ℎ3

12𝐿3 = 4

𝐸𝑏ℎ3

𝐿3

= 4150 × 109 × 2 × 10−6 × 2 × 10−6

100 × 10−6 = 9.6𝑁/𝑚




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

Determine of the equivalence spring constant for

the parallel springs subjected to unequal forces

Solution

From similar triangles

𝑥 = 𝑥2 +𝑏

𝑎 + 𝑏𝑥1 − 𝑥2 =

𝑏

𝑎 + 𝑏𝑥1 +

𝑎

𝑎 + 𝑏𝑥2

Consider the bar

𝐹 = 𝐹1 + 𝐹2

𝑏𝐹2 = 𝑎𝐹1

Therefore

𝑥1 =𝐹1

𝑘1=

𝑏𝐹

𝑘1 𝑎 + 𝑏, 𝑥2 =

𝐹2

𝑘2=

𝑎𝐹

𝑘2 𝑎 + 𝑏


⟹ 𝐹1 =𝑏𝐹

𝑎 + 𝑏, 𝐹2 =

𝑎𝐹

𝑎 + 𝑏


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⟹ 𝑥 =𝑏

𝑎 + 𝑏𝑥1 +

𝑎

𝑎 + 𝑏𝑥2

=𝑏

𝑎 + 𝑏

𝑏𝐹

𝑘1 𝑎 + 𝑏+

𝑎

𝑎 + 𝑏

𝑎𝐹

𝑘2 𝑎 + 𝑏

=𝐹

𝑎 + 𝑏 2

𝑘1𝑎2 + 𝑘2𝑏

2

𝑘1𝑘2

For the equivalent system

𝐹 = 𝑘𝑒𝑥

⟹ 𝑘𝑒 =𝐹

𝑥The equivalence spring constant for the parallel

springs subjected to unequal forces

𝑘𝑒 =𝑘1𝑘2 𝑎 + 𝑏 2

𝑘1𝑎2 + 𝑘2𝑏

2




3.Nonlinear Springs

- Nonlinear stiffness elements appear in many applications,

including leaf springs in vehicle suspensions and uniaxial

micro-electromechanical devices in the presence of

electrostatic actuation

- The spring force 𝐹(𝑥)

𝐹 𝑥 = 𝑘𝑥𝑙𝑖𝑛𝑒𝑎𝑟 𝑠𝑝𝑟𝑖𝑛𝑔 𝑒𝑙𝑒𝑚𝑒𝑛𝑡

+ 𝛼𝑘𝑥3

𝑛𝑜𝑛𝑙𝑖𝑛𝑒𝑎𝑟 𝑠𝑝𝑟𝑖𝑛𝑔 𝑒𝑙𝑒𝑚𝑒𝑛𝑡

(2.23)

𝛼 : the stiffness coefficient of the nonlinear term

𝛼 > 0 hardening spring 𝛼 < 0 softening spring

𝑘 : the linear spring constant

- The potential energy 𝑉

𝑉 𝑥 = 0

𝑥

𝐹 𝑥 𝑑𝑥 =1

2𝑘𝑥2 +

1

2𝛼𝑘𝑥4


(2.24)


𝐹 𝑥 = 𝑘𝑥 + 𝛼𝑘𝑥3 (2.23)


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

graph is no longer a straight line. The slope of this graph at a

location 𝑥 = 𝑥𝑙 is given by

𝑑𝐹

𝑑𝑥𝑥=𝑥𝑙

= 𝑘 + 3𝛼𝑘𝑥2

𝑥=𝑥𝑙

= 𝑘 + 3𝛼𝑘𝑥𝑙2

⟹ in the vicinity of displacements in a neighborhood of 𝑥 =𝑥𝑙, the cubic nonlinear stiffness element may be replaced by

a linear stiffness element with a stiffness constant (2.25)

- The constant of proportionality 𝛼𝑘 for the nonlinear cubic

spring is determined experimentally


(2.25)



- Experimentally obtained data used to determine the

nonlinear spring constant 𝛼𝑘




- Ex.2.8 Nonlinear stiffness due to geometry

a. Nonlinear stiffness due to geometry

• The initial tension force

𝑇0 = 𝐹𝑠𝛾=0

= 𝑘𝛿0

• The force in the spring

𝐹𝑠 𝑥 = 𝑘𝛿0 + 𝑘 𝐿2 + 𝑥2 − 𝐿

• The force in the 𝑥-direction is obtained

𝐹𝑥 𝑥 = 𝐹𝑠𝑠𝑖𝑛𝛾 =𝐹𝑠𝑥

𝐿2 + 𝑥2=

𝑥𝑘𝛿0

𝐿2 + 𝑥2+

𝑘𝑥 𝐿2 + 𝑥2 − 𝐿

𝐿2 + 𝑥2

⟹ the spring force opposing the motion is a nonlinear function

of the displacement 𝑥 . Hence, this vibratory model of the

system will have nonlinear stiffness



Binomial expansion 1 + 𝑥 𝑛 = 1 + 𝑛𝑥 +1

2𝑛(𝑛 − 1)𝑥2 + ⋯


Cubic Springs and Linear Springs

Assume that |𝑥/𝐿| ≪ 1, using binomial expansion

1 +𝑥

𝐿

2

= 1 +𝑥

𝐿

2 1/2

= 1 +1

2

𝑥

𝐿

2

+1

8

𝑥

𝐿

4

+ ⋯

⟹ 𝐹𝑥 𝑥 = 𝑘𝛿0

𝑥/𝐿

1 + 𝑥/𝐿 2+

𝑘𝑥 1 + 𝑥/𝐿 2 − 1

1 + 𝑥/𝐿 2

= 𝑘𝛿0

𝑥/𝐿

1 +12

𝑥𝐿

2 +𝑘𝑥 1 +

12

𝑥𝐿

2− 1

1 +12

𝑥𝐿

2

= 𝑘𝛿0

𝑥

𝐿+

𝑘

2𝐿

𝑥

𝐿

3



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𝐹𝑥 𝑥 = 𝑘𝛿0

𝑥

𝐿+

𝑘

2𝐿

𝑥

𝐿

3

When the nonlinear term is negligible

𝐹𝑥 𝑥 = 𝑘𝛿0

𝑥

𝐿= 𝑇0

𝑥

𝐿and the spring constant is proportional to the initial tension in

the spring

b. Nonlinear spring composed of a set of linear springs

Another example of a nonlinear spring is one that is piecewise

linear as shown in figure




4.Other Forms of Potential Energy Elements

Consider other stiffness elements in which there is a

mechanism for storing and releasing potential energy. The

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

loading

- Fluid Element

• The magnitude of the total force of the

displaced fluid acting on the rest of the fluid

𝐹𝑚 𝑥 = 2𝜌𝑔𝐴0𝑥

𝜌 : the mass density of the fluid, 𝑘𝑔/𝑚3

𝑔 : gravitational constant, 𝑚/𝑠2

𝐴0 : the manometer cross-sectional area, 𝑚2

𝑥 : the fluid displacement, 𝑚


Manometer



• The equivalent spring constant of this fluid system

𝑘𝑒 =𝑑𝐹𝑚

𝑑𝑥= 2𝜌𝑔𝐴0

• The potential energy

𝑉 𝑥 =1

2𝑘𝑒𝑥

2 = 𝜌𝑔𝐴0𝑥2

Alternatively, the potential energy can also be obtained

directly from the work done

𝑉 𝑥 = 0

𝑥

𝐹𝑚 𝑥 𝑑𝑥

= 2𝜌𝑔𝐴0 0

𝑥

𝑥𝑑𝑥

= 𝜌𝑔𝐴0𝑥2




- Compressed Gas

• When the piston moves by an amount 𝑥 along

the axis of the piston, 𝑉0 decreases to a

volume 𝑉𝑐

𝑉𝑐 𝑥 = 𝑉0 − 𝐴0𝑥 = 𝐴0𝐿0 1 − 𝑥/𝐿0

⟹ 𝑉𝑐 𝑥 = 𝑉0 1 − 𝑥/𝐿0

𝐴0: the piston cross-sectional area, 𝑚2

• The equation of state for the gas

𝑃𝑉𝑐𝑛 = 𝑃0𝑉0

𝑛 = 𝑐0 = 𝑐𝑜𝑛𝑠𝑡 ⟹ 𝑃 = 𝑐0𝑉𝑐−𝑛

𝑃 : gas pressure, 𝑁/𝑚2 𝑉 : gas volume, 𝑚3

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

- slowly, the compression is isothermal, 𝑛 = 1

- rapidly, the compression is adiabatic, 𝑛 = 𝑐𝑝/𝑐𝑣 = 1.4


Gas compression

with a piston



• The magnitude of the force on the piston

𝐹 𝑥 = 𝐴0𝑃

= 𝐴0𝑐0𝑉𝑐−𝑛

= 𝐴0𝑐0𝑉0−𝑛 1 − 𝑥/𝐿0

−𝑛

⟹ 𝐹 𝑥 = 𝐴0𝑃0 1 − 𝑥/𝐿0−𝑛 (2.33)

The Eq. (2.33) describes a nonlinear force

versus displacement relationship

At the vicinity of 𝑥 = 𝑥𝑙, the stiffness of an equivalent linear

stiffness element

𝑘𝑒 = 𝑑𝐹

𝑑𝑥𝑥=𝑥𝑙

=𝑛𝐴0𝑃0

𝐿01 − 𝑥𝑙/𝐿0

−𝑛−1

For 𝑥𝑙/𝐿0 ≪ 1,


𝑘𝑒 =𝑛𝐴0𝑃0

𝐿0



• For arbitrary 𝑥/𝐿0, the potential energy

𝑉 𝑥 = 0

𝑥

𝐹 𝑥 𝑑𝑥

= 𝐴0𝑃0 0

𝑥

1 − 𝑥/𝐿0−𝑛𝑑𝑥

⟹ 𝑉 𝑥 = −𝐴0𝑃0𝐿0𝑙𝑛 1 − 𝑥/𝐿0 𝑛 = 1

𝐴0𝑃0𝐿0

𝑛 − 11 − 𝑥/𝐿0

−𝑛 − 1 𝑛 ≠ 1



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8


- Pendulum System

• Pendulum systems: bar with uniformly

distributed mass

At 𝜃, the vertical distance through which the

center of gravity of the bar moves up from the

reference position

𝑥 =𝐿

2−

𝐿

2𝑐𝑜𝑠𝜃 =

𝐿

21 − 𝑐𝑜𝑠𝜃

The increase in the potential energy

𝑉 𝑥 = 0

𝑥

𝐹 𝑥 𝑑𝑥 = 0

𝑥

𝑚𝑔𝑑𝑥 = 𝑚𝑔𝑥

or in 𝜃

𝑉 𝜃 =1

2𝑚𝑔𝐿(1 − 𝑐𝑜𝑠𝜃)


Pendulum systems: bar

with uniformly distributed

mass


Taylor series approximation 𝑐𝑜𝑠𝜃 = 1 −𝜃2

2+ ⋯


𝑉 𝜃 =1

2𝑚𝑔𝐿 1 − 𝑐𝑜𝑠𝜃

≈1

2

𝑚𝑔𝐿

2𝜃2

=1

2𝑘𝑒𝜃

2

where the equivalent spring constant

𝑘𝑒 =𝑚𝑔𝐿

2


Pendulum systems: bar

with uniformly distributed

mass



• Pendulum systems: mass on a weightless rod

The increase in the potential energy

𝑉 𝜃 ≈1

2𝑚1𝑔𝐿𝜃2 =

1

2𝑘𝑒𝜃

2


𝑘𝑒 = 𝑚1𝑔𝐿

If the weightless bar is replaced by one that

has a uniformly distributed mass 𝑚, then the

total potential energy of the bar and the mass

𝑉 𝜃 ≈1

4𝑚𝑔𝐿𝜃2 +

1

2𝑚1𝑔𝐿𝜃2 =

1

2

𝑚

2+ 𝑚1 𝑔𝐿𝜃2 =

1

2𝑘𝑒𝜃

2


𝑘𝑒 =𝑚

2+ 𝑚1 𝑔𝐿


Pendulum systems: mass

on a weightless rod



• Pendulum systems: inverted mass on a

weightless rod

The decrease in the potential energy

𝑉 𝜃 ≈ −1

2𝑚1𝑔𝐿𝜃2


Pendulum systems:

inverted mass on a

weightless rod



- Ex.2.9 Equivalent stiffness due to gravity loading

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

the potential energy

𝑉 𝜃 =1

2𝑚1𝑔𝑏𝜃2 −

1

2𝑚2𝑔𝑎𝜃2

=1

2𝑚1𝑏 − 𝑚2𝑎 𝑔𝜃2

There is a gain or loss in potential energy depending

on whether 𝑚1𝑏 > 𝑚2𝑎 or vice versa

• When the bar has a uniformly distributed mass 𝑚

𝑚1 = 𝑚𝐿1/𝐿, 𝑚2 = 𝑚𝐿2/𝐿, 𝐿 = 𝐿1 + 𝐿2, 𝑏 = 𝐿1/2, 𝑐 = 𝐿2/2

𝑉 𝜃 =𝐿12 − 𝐿2

2

4(𝐿1 + 𝐿2)𝑚𝑔𝜃2 =

1

2

𝐿1 + 𝐿2

2𝑚𝑔𝜃2 =

1

2𝑘𝑒𝜃

2

where the equivalent stiffness 𝑘𝑒 = 𝐿1 + 𝐿2 𝑚𝑔/2



§4.Dissipation Elements

- Damping elements are assumed to have neither inertia nor the

means to store or release potential energy

- The mechanical motion imparted to these elements is

converted to heat or sound and, hence, they are called non-

conservative or dissipative because this energy is not

recoverable by the mechanical system

- There are four common types of damping mechanisms used to

model vibratory systems

• Viscous damping

• Coulomb or dry friction damping

• Material or solid or hysteretic damping

• Fluid damping

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

function of velocity



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1.Viscous damping

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

a cylinder, the damping force generated is proportional to the

relative velocity between the two boundaries confining the fluid

- A common representation of a viscous damper is a cylinder

with a piston head

- Depending on the damper construction and the velocity

range, the magnitude of the damper force 𝐹( 𝑥) is a nonlinear

function of velocity or can be approximated as a linear

function of velocity




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

𝐹 𝑥 = 𝑐 𝑥 (2.46)

𝑐 : the damping coefficient, 𝑁/(𝑚/𝑠)

Viscous damping of the form given by Eq. (2.46) is also

called slow-fluid damping

- In the case of a nonlinear viscous damper described by a

function 𝐹 𝑥 , the equivalent linear viscous damping around

an operating speed 𝑥 = 𝑥𝑙 is determined as follows

𝑐𝑒 = 𝑑𝐹 𝑥

𝑑 𝑥 𝑥= 𝑥𝑙

- Linear viscous damping elements can be combined in the

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

proportional to velocity instead of displacement


(2.47)



• Energy Dissipation

The energy dissipated by a linear viscous damper

𝐸𝑑 = 𝐹𝑑𝑥 = 𝐹 𝑥𝑑𝑡 = 𝑐 𝑥2𝑑𝑡 = 𝑐 𝑥2𝑑𝑡

• Parallel-Plate Damper

An example of a viscous damper is shown in the figure

The shear force acting on the

bottom plate

𝐹 𝑥 =𝜇 𝑥

ℎ𝐴 =

𝜇𝐴

ℎ 𝑥

The damping coefficient 𝑐 for the parallel-plate construction

𝑐 =𝜇𝐴

ℎ


(2.48)

(2.49)

(2.50)


𝑐 =𝜇𝐴

ℎ(2.50)


- Ex.2.10 Design of a parallel-plate damper

A parallel-plate damper with a top plate of dimensions100𝑚𝑚× 100𝑚𝑚 is to be pulled across an oil layer of thickness

0.2𝑚𝑚, which is confined between the moving plate and a

fixed plate. We are given that this oil is SAE30 oil, which has a

viscosity of 345𝑚𝑃𝑎𝑠 (345 × 103𝑁𝑠/𝑚2)

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

𝑐 =𝜇𝐴

ℎ=

345×10−3 100×10−3 ×100×10−3

0.2 × 10−3 = 17.25𝑁𝑠/𝑚




- Ex.2.11 Equivalent damping coefficient and

equivalent stiffness of a vibratory system

Consider the vibratory system in which the motion of mass 𝑚is restrained by a set of linear springs and linear viscous

dampers. Determine 𝑘𝑒 and 𝑐𝑒

Solution

The equivalence system

𝑘𝑒 = 𝑘1 +𝑘2𝑘3

𝑘2 + 𝑘3, 𝑐𝑒 = 𝑐1 + 𝑐2




𝑑 𝑥 𝑥= 𝑥𝑙

(2.47)


- Ex.2.12 Equivalent linear damping coefficientof a nonlinear damper

It has been experimentally determined that the damper force-

velocity relationship is given by the function

𝐹 𝑥 = 4𝑁𝑠/𝑚 𝑥 + (0.3𝑁𝑠3/𝑚) 𝑥3

Determine the equivalent linear damping coefficient around an

operating speed of 3𝑚/𝑠

Solution

Using the Eq. (2.47)


𝑑 𝑥 𝑥=3𝑚/𝑠

= 4 + 0.9 𝑥2

𝑥=3𝑚/𝑠= 4 + 0.9 × 32

⟹ 𝑐𝑒 = 12.1𝑁𝑠/𝑚



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2.Other Forms of Dissipation

- Coulomb Damping or Dry Friction

This type of damping is due to the force caused by friction

between two solid surfaces

The friction force acting on the system

𝐹 𝑥 = 𝜇𝑁𝑠𝑔𝑛( 𝑥) (2.51)

𝜇 : the kinetic coefficient of friction

𝑁 : the force compressing the surfaces, 𝑁

𝑠𝑔𝑛: the signum function, 𝑠𝑔𝑛 𝑥 = +1 𝑥 > 0−1 𝑥 < 0

0 𝑥 = 0




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

𝐹 𝑥 = 𝜇𝑚𝑔𝑠𝑔𝑛( 𝑥) (2.52)

The energy dissipated in this case

(2.53)

- Fluid Damping (Velocity-Squared Damping)

This type of damping is associated with a system whose mass is

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

𝐹 𝑥 = 𝑐𝑑 𝑥2𝑠𝑔𝑛 𝑥 = 𝑐𝑑| 𝑥| 𝑥 (2.54)

𝑐𝑑 : friction coefficient, 𝑐𝑑 = 𝐶𝜌𝐴/2

𝐶 : drag coefficient

𝜌 : the mass density of the fluid

𝐴 : the projected area of the mass in a direction normal to 𝑥

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


𝐸𝑑 = 𝐹𝑑𝑥 = 𝐹 𝑥𝑑𝑡 = 𝜇𝑚𝑔 𝑠𝑔𝑛 𝑥 𝑥𝑑𝑡



The energy dissipated

𝐸𝑑 = 𝐹𝑑𝑥 = 𝐹 𝑥𝑑𝑡 = 𝑐𝑑 𝑠𝑔𝑛( 𝑥) 𝑥3𝑑𝑡

- Structural or Solid or Hysteretic Damping

This type of damping describes the losses in materials due to

internal friction. The damping force is a function of

displacement and velocity and is of the form

𝐹 = 𝑘𝜋𝛽ℎ𝑠𝑔𝑛 𝑥 |𝑥| (2.57)

𝛽ℎ : an empirically determined constant

The energy dissipated

𝐸𝑑 = 𝐹𝑑𝑥 = 𝐹 𝑥𝑑𝑡 = 𝑘𝜋𝛽ℎ 𝑠𝑔𝑛( 𝑥)|𝑥|𝑑𝑡


(2.54)

(2.58)


§5.Model Construction

1.Introduction

- In this section, four examples are provided to illustrate how

the previously described inertia, stiffness, and damping

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




2.A Micro-electromechanical System

Micro-electromechanical accelerometer and a vibratory model

of this sensor




3.The Human Body

Human body and a vibratory model



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11


4.A Ski

Cross-country ski, which is a physical system with distributed

stiffness and inertia properties, and its vibratory model




5.Cutting Process

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

system



§6.Design for Vibration



Exercises

