ch.03 single dof systems - governing equations

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03. Single DOF Systems:

Governing Equations

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

Vibrations 3.01 Single DOF Systems: Governing Equations

§1.Chapter Objectives

• Obtain the governing equation of motion for single degree-of-

freedom (dof) translating and rotating systems by using force

balance and moment balance methods

• Obtain the governing equation of motion for single dof

translating and rotating systems by using Lagrange’s

equations

• Determine the equivalent mass, equivalent stiffness, and

equivalent damping of a single dof system

• Determine the natural frequency and damping factor of a

system



§2.Force-Balance and Moment-Balance Methods

1.Force Balance Method

Newtonian principle of linear momentum

𝐹 − 𝑝 = 0 (3.1a)

𝐹 : the net external force vector acting on the system

𝑝 : the absolute linear momentum of the considered system

For a system of constant mass 𝑚 whose center of mass is

moving with absolute acceleration 𝑎, the rate of change of

linear momentum 𝑝 = 𝑚 𝑎

𝐹 − 𝑚 𝑎 = 0 (3.1b)

−𝑚 𝑎 : inertial force

⟹The sum of the external forces and inertial forces acting on

the system is zero; that is, the system is in equilibrium

under the action of external and inertial forces




Vertical Vibrations of a Spring-Mass-Damper System

- Obtain an equation to describe the motions of the spring-mass-

damper system in the vertical

The position vector of

the mass from the fixed

point 𝑂 𝑟 = 𝑟 𝑗= (𝐿 + 𝛿𝑠𝑡 + 𝑥) 𝑗

Force balance along

the 𝑗 direction

𝑓 𝑡 𝑗 + 𝑚𝑔 𝑗 − 𝑘𝑥 + 𝑘𝛿𝑠𝑡 𝑗 − 𝑐𝑑𝑟

𝑑𝑡 𝑗 − 𝑚

𝑑2𝑟

𝑑𝑡2 𝑗 = 0




- Noting that 𝐿 and 𝛿𝑠𝑡 are constants, rearranging terms to get

the following scalar differential equation

𝑚𝑑2𝑥

𝑑𝑡2 + 𝑐𝑑𝑥

𝑑𝑡+ 𝑘 𝑥 + 𝛿𝑠𝑡 = 𝑓 𝑡 + 𝑚𝑔




Static Equilibrium Position

- The static-equilibrium position of a system is the position that

corresponds to the system’s rest state; that is, a position with

zero velocity and zero acceleration

- The static-equilibrium position is the solution of

𝑘 𝑥 + 𝛿𝑠𝑡 = 𝑚𝑔

- The static displacement

𝛿𝑠𝑡 =𝑚𝑔

𝑘⟹ 𝑥 = 0 is the static-equilibrium position of the system

- The spring has an unstretched length 𝐿, the static-equilibrium

position measured from the origin 𝑂 is given by

𝑥𝑠𝑡 = 𝑥𝑠𝑡 𝑗 = (𝐿 + 𝛿𝑠𝑡) 𝑗



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Equation of Motion for Oscillations about the Static-EquilibriumPosition

𝑚𝑑2𝑥


𝑑𝑡+ 𝑘 𝑥 + 𝛿𝑠𝑡 = 𝑓 𝑡 + 𝑚𝑔


𝑘

⟹ 𝑚𝑑2𝑥


𝑑𝑡+ 𝑘𝑥 = 𝑓 𝑡

Equation (3.8) is the governing equation of motion of a single

dof system for oscillations about the static-equilibrium position

• The left-hand side: the forces from the components that

comprise a single dof system

• The right-hand side: the external force acting on the mass


(3.8)



Horizontal Vibrations of a Spring-Mass-Damper System

Consider a mass moving in a direction normal

to the direction of gravity

• It is assumed that the mass moves without

friction

• The unstretched length of the spring is 𝐿, and

a fixed point 𝑂 is located at the unstretched

position of the spring

• The spring does not undergo any static

deflection and carrying out a force balance

along the 𝑖 direction

• The static-equilibrium position 𝑥 = 0 coincides with the

position corresponding to the unstretched spring




Force Transmitted to Fixed Surface

The total reaction force due to the spring and

the damper on the fixed surface is the sum of

the static and dynamic forces

𝐹𝑅 = 𝑘𝛿𝑠𝑡 + 𝑘𝑥 + 𝑐𝑑𝑥

𝑑𝑡

If considering only the dynamic part of the

reaction force-that is, only those forces created

by the motion 𝑥(𝑡) from its static equilibrium

position, then

𝐹𝑅𝑑 = 𝑘𝑥 + 𝑐𝑑𝑥

𝑑𝑡




- Ex.3.1 Wind-drivenOscillationsaboutaSystem’sStatic-EquilibriumPosition

The wind flow across civil structures typically generates a

excitation force 𝑓(𝑡) on the structure that consists of a steady-

state part and a fluctuating part

𝑓 𝑡 = 𝑓𝑠𝑠 + 𝑓𝑑(𝑡)

𝑓𝑠𝑠 : the time-independent steady-state force

𝑓𝑑(𝑡) : the fluctuating time-dependent portion of the force

A single dof model of the vibrating structure

𝑚𝑑2𝑥


𝑑𝑡+ 𝑘𝑥 = 𝑓𝑠𝑠 + 𝑓𝑑 𝑡 ⟹ 𝑥 𝑡 = 𝑥0 + 𝑥𝑑(𝑡)

𝑥0 : the static equilibrium position, 𝑥0 = 𝑓𝑠𝑠/𝑘

𝑥𝑑(𝑡) : motions about the static position

⟹ 𝑚𝑑2𝑥


𝑑𝑡+ 𝑘𝑥 = 𝑓𝑑 𝑡




- Ex.3.2 EardrumOscillations:NonlinearOscillatorandLinearizedSystems

Determine the static-equilibrium positions of this system and

illustrate how the governing nonlinear equation can be

linearized to study oscillations local to an equilibrium position

Solution

The governing nonlinear equation

𝑚𝑑2𝑥

𝑑𝑡2 + 𝑘𝑥 + 𝑘𝑥2 = 0

The restoring force of the eardrum has a component with a

quadratic nonlinearity

Static-Equilibrium Positions

Equilibrium positions 𝑥 = 𝑥0 are solutions of the algebraic equation

𝑘 𝑥0 + 𝑥02 = 0 ⟹ 𝑥0 = 0, 𝑥0 = −1


(𝑎)



Linearization

Equilibrium positions 𝑥 = 𝑥0 are solutions of the algebraic equation

𝑘 𝑥0 + 𝑥02 = 0 ⟹ 𝑥0 = 0, 𝑥0 = −1

Subtitute 𝑥 𝑡 = 𝑥0 + 𝑥(𝑡) into (a) with note that

𝑥2 𝑡 = 𝑥0 + 𝑥 𝑡2

≈ 𝑥02 + 2𝑥0 𝑥 𝑡 + ⋯

𝑑2𝑥

𝑑𝑡2 =𝑑2 𝑥0 + 𝑥 𝑡

𝑑𝑡2 =𝑑2 𝑥

𝑑𝑡2

⟹ 𝑚𝑑2 𝑥

𝑑𝑡2 + 𝑘 𝑥0 + 𝑥(𝑡) + 𝑘 𝑥02 + 2𝑥0 𝑥 𝑡 = 0

𝑥0 = 0 ⟹ 𝑚𝑑2 𝑥

𝑑𝑡2 + 𝑘 𝑥(𝑡) = 0

𝑥0 = −1 ⟹ 𝑚𝑑2 𝑥

𝑑𝑡2 − 𝑘 𝑥(𝑡) = 0

⟹ the equations have different stiffness terms



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2. Moment-Balance Methods

For single dof systems that undergo rotational motion, the

moment balance method is useful in deriving the governing

equation

The angular momentum about the center of mass of the disc

𝐻 = 𝐽𝐺 𝜃𝑘

⟹ 𝑀 = 𝐽𝐺 𝜃𝑘




The governing equation of motion

𝑀 𝑡 𝑘 − 𝑘𝑡 𝜃𝑘 − 𝑐𝑡

𝑑𝜃

𝑑𝑡𝑘 − 𝐽𝐺

𝑑2𝜃

𝑑𝑡2 = 0

⟹ 𝐽𝐺𝑑2𝜃

𝑑𝑡2 + 𝑐𝑡

𝑑𝜃

𝑑𝑡+ 𝑘𝑡𝜃 = 𝑀 𝑡




All linear single dof vibratory systems are governed by a linear

second-order ordinary differential equation with an inertia term,

a stiffness term, a damping term, and a term related to the

external forcing imposed on the system

• Translational motion

𝑚𝑑2𝑥



• Rotational motion

𝐽𝐺𝑑2𝜃

𝑑𝑡2 + 𝑐𝑡

𝑑𝜃

𝑑𝑡+ 𝑘𝑡𝜃 = 𝑀 𝑡




Ex.3.3 Hand Biomechanics

The moment balance about

point 𝑂

𝑀 − 𝐽0 𝜃𝑘 = 0

𝐽0: the rotary inertia of the

forearm and the object

held in the hand

The net moment 𝑀 acting

about the point 𝑂 due to gravity loading and the forces due to

the biceps and triceps

𝑀 = −𝑀𝑔𝑙𝑐𝑜𝑠𝜃𝑘 − 𝑚𝑔𝑙

2𝑐𝑜𝑠𝜃𝑘 + 𝐹𝑏𝑎𝑘 − 𝐹𝑡𝑎𝑘

⟹ −𝑀𝑔𝑙𝑐𝑜𝑠𝜃𝑘 − 𝑚𝑔𝑙

2𝑐𝑜𝑠𝜃𝑘 + 𝐹𝑏𝑎𝑘 − 𝐽0 𝜃𝑘 = 0




−𝑀𝑔𝑙𝑐𝑜𝑠𝜃𝑘 − 𝑚𝑔𝑙

2𝑐𝑜𝑠𝜃𝑘 + 𝐹𝑏𝑎𝑘 − 𝐽0 𝜃𝑘 = 0

Note that: 𝐹𝑏 = −𝑘𝑏𝜃, 𝐹𝑡 = 𝐾𝑡𝑣 = 𝐾𝑡𝑎 𝜃, 𝐹0 = 𝑚𝑙2/3 + 𝑀𝑙2

⟹ 𝑀 +𝑚

3𝑙2 𝜃 + 𝐾𝑡𝑎

2 𝜃 + 𝑘𝑏𝑎𝜃 + 𝑀 +𝑚

2𝑔𝑙𝑐𝑜𝑠𝜃 = 0

Static-Equilibrium Position

The equilibrium position 𝜃 = 𝜃0 is a solution of the

transcendental equation

𝑘𝑏𝑎𝜃0 + 𝑀 +𝑚

2𝑔𝑙𝑐𝑜𝑠𝜃0 = 0




Linear System Governing “Small” Oscillations about the Static-

Equilibrium Position

Consider oscillations about the static-equilibrium position and

expand the angular variable 𝜃 𝑡 = 𝜃0 + 𝜃 𝑡 with note that

𝑐𝑜𝑠𝜃 = cos 𝜃0 + 𝜃 ≈ 𝑐𝑜𝑠𝜃0 − 𝜃𝑠𝑖𝑛𝜃0 + ⋯

𝑑𝜃(𝑡)

𝑑𝑡=

𝑑(𝜃0 + 𝜃)

𝑑𝑡= 𝜃(𝑡)

𝑑2𝜃(𝑡)

𝑑𝑡2 =𝑑2(𝜃0 + 𝜃)

𝑑𝑡2 = 𝜃(𝑡)

⟹ 𝑀 +𝑚

3𝑙2 𝜃 + 𝐾𝑡𝑎

2 𝜃 + 𝑘𝑒 𝜃 = 0

where

𝑘𝑒 = 𝑘𝑏𝑎 − 𝑀 +𝑚

2𝑔𝑙𝑠𝑖𝑛𝜃0



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§3.Natural Frequency and Damping Factor

1.Natural Frequency

Translation Vibrations: Natural Frequency

𝜔𝑛 = 2𝜋𝑓𝑛 =𝑘

𝑚(𝑟𝑎𝑑/𝑠)

𝑘 : the stiffness of the system, 𝑁/𝑚

𝑚 : the system mass, 𝑘𝑔

𝑓𝑛 : the natural frequency, 𝐻𝑧

For the mass-damper-spring system

𝜔𝑛 = 2𝜋𝑓𝑛 =𝑔

𝛿𝑠𝑡(𝑟𝑎𝑑/𝑠)

𝛿𝑠𝑡: the static deflection of the system, 𝑚


(3.15)

(3.14)



Rotational Vibrations: Natural Frequency

𝜔𝑛 = 2𝜋𝑓𝑛 =𝑘𝑡

𝐽(𝑟𝑎𝑑/𝑠)

𝑘𝑡 : the torsion stiffness of the system, 𝑁𝑚/𝑟𝑎𝑑

𝐽 : the system mass, 𝑘𝑔𝑚/𝑠2

𝑓𝑛 : the natural frequency, 𝐻𝑧

Period of Undamped Free Oscillations

For an unforced and undamped system, the period of free

oscillation of the system is given by

𝑇 =1

𝑓𝑛=

2𝜋

𝜔𝑛


(3.16)

(3.17)



𝛿𝑠𝑡(𝑟𝑎𝑑/𝑠) (3.15)


Ex.3.4 Natural Frequency from Static Deflection of a Machine System

The static deflections of a machinery are found to be 0.1, 1,

10(𝑚𝑚). Determine the natural frequency for vertical vibrations

Solution

𝑓𝑛1 =1

2𝜋

𝑔

𝛿𝑠𝑡1=

1

2𝜋

9.81

0.1 × 10−3 = 49.85𝐻𝑧

𝑓𝑛2 =1

2𝜋

𝑔

𝛿𝑠𝑡2=

1

2𝜋

9.81

1 × 10−3 = 15.76𝐻𝑧

𝑓𝑛3 =1

2𝜋

𝑔

𝛿𝑠𝑡3=

1

2𝜋

9.81

10 × 10−3 = 4.98𝐻𝑧




- Ex.3.5 Static Deflection and Natural Frequency of the Tibia

Bone in a Human Leg

Consider a person of 100𝑘𝑔 mass standing upright. The tibia

has a length of 40𝑐𝑚, and it is modeled as a hollow tube with an

inner diameter of 2.4𝑐𝑚 and an outer diameter of 3.4𝑐𝑚. The

Young’s modulus of elasticity of the bone material is 2 ×1010𝑁/𝑚2. Determine the static deflection in the tibia bone and

an estimate of the natural frequency of axial vibrations

Solution

Assume that both legs support the weight of the person

equally, so that the weight supported by the tibia

𝑚𝑔 = 100/2 × 9.81 = 490.5𝑁




𝛿𝑠𝑡(𝑟𝑎𝑑/𝑠) (3.15)


The stiffness of the tibia

𝑘 =𝐴𝐸

𝐿=

1 × 1010 ×𝜋4

3.4 × 10−2 2 − 2.4 × 10−2 2

40 × 10−2

= 22.78 × 106𝑁/𝑚2

The static deflection


𝑘=

490.5

22.78 × 106 = 21.53 × 10−6𝑚

The natural frequency

𝑓𝑛 =1

2𝜋

𝑔

𝛿𝑠𝑡=

1

2𝜋

9.81

21.53 × 10−6 = 107.4𝐻𝑧




Ex.3.6 System with A Constant Natural Frequency

Examine how the spring-mounting system can be designed and

discuss a realization of this spring in practice

Solution

In order to realize the desired objective of constant natural

frequency regardless of the system weight, we need a

nonlinear spring whose equivalent spring constant is given by

𝑘 = 𝐴𝑊

𝐴: a constant, 𝑊 = 𝑚𝑔: the weight, 𝑔: the gravitational constant


𝑓𝑛 =1

2𝜋

𝑘

𝑚=

1

2𝜋

𝑘𝑔

𝑊=

1

2𝜋𝐴𝑔𝐻𝑧

⟹ 𝑓𝑛 is constant irrespective of the weight of the mass



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Nonlinear Spring Mounting

When the side walls of a rubber cylindrical tube are

compressed into the nonlinear region, the equivalent spring

stiffness of this system approximates the characteristic given

by 𝑘 = 𝐴𝑊

For illustrative purposes, consider a spring that has the

general force-displacement relationship

𝐹 𝑥 = 𝑎𝑥

𝑏

𝑐

𝑎, 𝑏: scale factors, 𝑐: shape factor


𝑥0 = 𝑏𝑊

𝑎

1/𝑐




For “small” amplitude vibrations about 𝑥0, the linear equivalent

stiffness of this spring is determined

𝑘𝑒𝑞 = 𝑑𝐹(𝑥)

𝑑𝑥𝑥=𝑥0

=𝑎𝑐

𝑏

𝑥𝑜

𝑏

𝑐−1

=𝑎𝑐

𝑏

𝑊

𝑏

𝑐−1𝑐

The natural frequency of this system

𝑓𝑛 =1

2𝜋

𝑘𝑒𝑞

𝑊/𝑔

=1

2𝜋

𝑔𝑐

𝑏

𝑊

𝑎

−1/𝑐

=1

2𝜋

𝑔𝑐

𝑏

𝑊

𝑎

−1/2𝑐

𝐻𝑧




Representative Spring Data

Consider the representative data of a

nonlinear spring shown in the figure

Using lsqcurvefit in Matlab to identify

𝑎 = 2500𝑁, 𝑏 = 0.011𝑚, 𝑐 = 2.77

⟹ 𝑓𝑛 =1

2𝜋

𝑔𝑐

𝑏

𝑊

𝑎

−1/2𝑐

= 32.4747𝑊−1/3.54𝐻𝑧

Plot 𝑓𝑛(𝑊)




Representative Spring Data

From the figure of 𝑓𝑛(𝑊)

• over a sizable portion of the load

range, the natural frequency of the

system varies within the range of 8.8%

• The natural frequency of a system with

a linear spring whose static

displacement ranges from 12 ÷ 5𝑚𝑚varies approximately from 4.5 ÷ 7.0𝐻𝑧or approximately 22% about a

frequency of 5.8𝐻𝑧

1

2𝜋

9.8

0.012≈ 4.5𝐻𝑧,

1

2𝜋

9.8

0.005≈ 7𝐻𝑧

of 5.8 Hz




2.Damping Factor

Translation Vibrations: Damping Factor

For translating single dof systems, the damping factor or

damping ratio 𝜉 is defined as

𝜉 =𝑐

2𝑚𝜔𝑛=

𝑐

2 𝑘𝑚=

𝑐𝜔𝑛

2𝑘

𝑐: the system damping coefficient, 𝑁𝑠/𝑚

𝑘: the system stiffness, 𝑁/𝑚

𝑚: the system mass, 𝑘𝑔

Critical Damping, Underdamping, and Overdamping

Defining the critical damping 𝑐𝑐

𝑐𝑐 = 2𝑚𝜔𝑛 = 2 𝑘𝑚, 𝜉 = 𝑐/𝑐𝑐 (3.19)

0 < 𝜉 < 1: underdamped,𝜉 > 1: overdamped,𝜉 = 1: criticallydamped


(3.18)



Rotational Vibrations: Damping Factor

For rotating single dof systems, the damping factor or damping

ratio 𝜉 is defined as

𝜉 =𝑐𝑡

2𝐽𝜔𝑛=

𝑐𝑡

2 𝑘𝑡𝐽

𝑐𝑡: the system damping coefficient, 𝑁𝑚𝑠/𝑟𝑎𝑑

𝑘𝑡: the system stiffness, 𝑁𝑚/𝑟𝑎𝑑

𝐽: the system moment of inertia, 𝑘𝑔𝑚2



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Governing Equation of Motion in Terms of Natural Frequency

and Damping Factor

Rewriting the equation of motion

𝑑2𝑥

𝑑𝑡2 + 2𝜉𝜔𝑛

𝑑𝑥

𝑑𝑡+ 𝜔𝑛

2𝑥 =𝑓(𝑡)

𝑚If we introduce the dimensionless time 𝜏 = 𝜔𝑛𝑡 , then the

equation can be written

𝑑2𝑥

𝑑𝜏2 + 2𝜉𝑑𝑥

𝑑𝜏+ 𝑥 =

𝑓(𝜏)

𝑘




- Ex.3.7 Effect of Mass on the Damping Factor

A system is initially designed to be critically damped - that is,

with a damping factor of 𝜉 = 1. Due to a design change, the

mass of the system is increased 20% - that is, from 𝑚 to 1.2𝑚.

Will the system still be critically damped if the stiffness and the

damping coefficient of the system are kept the same?

Solution

The damping factor of the system after the design change

𝜉𝑛𝑒𝑤 =𝑐

2 𝑘(1.2𝑚)= 0.91

𝑐

2 𝑘𝑚= 0.91

𝑐

𝑐𝑐= 0.91

⟹ The system with the increased mass is no longer critically

damped; rather, it is now underdamped




- Ex.3.8 Effects of System Parameters on the Damping Ratio

An engineer finds that a single dof system with mass 𝑚 ,

damping 𝑐, and spring constant 𝑘 has too much static deflection

𝛿𝑠𝑡. The engineer would like to decrease 𝛿𝑠𝑡 by a factor of 2,

while keeping the damping ratio constant. Determine the

different options

Solution

The problem involves vertical vibrations


𝑘

2𝜉 =𝑐

𝑚

𝛿𝑠𝑡

𝑔= 𝑐

𝛿𝑠𝑡

𝑔𝑚2 =1

𝑚

𝑐2𝛿𝑠𝑡

𝑔

⟹ there are three ways that one can achieve the goal




First choice

Let 𝑐 remain constant, reduce 𝛿𝑠𝑡 by one-half


𝑘

𝛿𝑠𝑡′ =

𝛿𝑠𝑡

2=

𝑚𝑔

2𝑘=

𝑚′𝑔

𝑘′Comparing (a) and (b)

𝑚′𝑔

𝑘′=

𝑚𝑔

2𝑘=

𝑚/ 2 𝑔

𝑘 2⟹ 𝑚 → 𝑚′ =

𝑚

2, 𝑘 → 𝑘′ = 𝑘 2

Check the damping ratio

2𝜉′ = 𝑐𝛿′

𝑠𝑡

𝑔𝑚′2 = 𝑐𝛿𝑠𝑡

2𝑔 𝑚/ 22 = 𝑐

𝛿𝑠𝑡

𝑔𝑚2 = 2𝜉


Before (a)

After (b)



Second choice

Let 𝑚 remain constant, reduce 𝛿𝑠𝑡 by one-half

2𝜉 = 𝑐𝛿𝑠𝑡

𝑔𝑚2 =1

𝑚

𝑐2𝛿𝑠𝑡

𝑔

2𝜉′ =1

𝑚

𝑐′2𝛿𝑠𝑡′

𝑔=

1

𝑚

𝑐′2𝛿𝑠𝑡

2𝑔

Comparing (c) and (d)

𝑐′2

2= 𝑐2 ⟹ 𝑐 → 𝑐′ = 𝑐 2


𝛿𝑠𝑡′ =

𝑚𝑔

𝑘′=

𝛿𝑠𝑡

2=

𝑚𝑔

2𝑘⟹ 𝑘 → 𝑘′ = 2𝑘


Before (c)

After (d)



Third choice

Let 𝑘 remain constant, reduce 𝛿𝑠𝑡 by one-half


𝑘

𝛿𝑠𝑡′ =

𝛿𝑠𝑡

2=

𝑚𝑔

2𝑘=

𝑚′𝑔

𝑘Comparing (e) and (f)

𝑚′ =𝑚

2⟹ 𝑚 → 𝑚′ =

𝑚

2The constant damping ratio

2𝜉′ = 𝑐′𝛿′

𝑠𝑡

𝑔𝑚′2 = 𝑐′𝛿𝑠𝑡

2𝑔 𝑚/2 2 = 𝑐′2𝛿𝑠𝑡

𝑔𝑚2 = 𝑐𝛿𝑠𝑡

𝑔𝑚2 = 2𝜉

⟹ 𝑐 → 𝑐′ = 𝑐 2


Before (e)

After (f)


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𝐹 𝑥 = 𝜇𝑚𝑔𝑠𝑔𝑛( 𝑥) (2.52)

𝑚𝑑2𝑥


𝑑𝑡+ 𝑘𝑥 = 𝑓 𝑡 (3.8)

§4.Governing Equations for Different Type of Damping

The governing equations of motion for systems with different

types of damping are obtained by replacing the term

corresponding to the force due to viscous damping with the force

due to either the fluid, structural, or dry friction type damping

Coulomb or Dry Friction Damping

Using Eq. (2.52) and Eq. (3.8), the governing equation of motion

takes the form

𝑚𝑑2𝑥


𝑑𝑡+ 𝜇𝑚𝑔𝑠𝑔𝑛( 𝑥) = 𝑓(𝑡)

which is a nonlinear equation because the damping

characteristic is piecewise linear



𝑛𝑜𝑛𝑙𝑖𝑛𝑒𝑎𝑟 𝑑𝑟𝑦 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒

𝐹 𝑥 = 𝑐𝑑 𝑥2𝑠𝑔𝑛 𝑥 = 𝑐𝑑| 𝑥| 𝑥 (2.54)

𝐹 = 𝑘𝜋𝛽ℎ𝑠𝑔𝑛 𝑥 |𝑥| (2.57)

𝑚𝑑2𝑥


𝑑𝑡+ 𝑘𝑥 = 𝑓 𝑡 (3.8)

§4.Governing Equations for Different Type of Damping

Fluid Damping


𝑚𝑑2𝑥

𝑑𝑡2 + 𝑐𝑑| 𝑥| 𝑥 + 𝑘𝑥 = 𝑓(𝑡)

which is a nonlinear equation due to the nature of the damping

Structural Damping


𝑚𝑑2𝑥

𝑑𝑡2 + 𝑘𝜋𝛽ℎ𝑠𝑔𝑛 𝑥 |𝑥| + 𝑘𝑥 = 𝑓(𝑡)



𝑛𝑜𝑛𝑙𝑖𝑛𝑒𝑎𝑟 𝑓𝑙𝑢𝑖𝑑 𝑑𝑎𝑚𝑝𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒

§5.Governing Equations for Different Type of Applied Forces

1.System with Base excitation

- The base-excitation model is a prototype that is useful for studying

• buildings subjected to earthquakes

• packaging during transportation

• vehicle response, and

• designing accelerometers

- The physical system of interest is represented by a single dof

system whose base is subjected to a displacement

disturbance 𝑦(𝑡), and an equation governing the motion of

this system is sought to determine the response of the

system 𝑥(𝑡)




- A prototype of a single dof system subjected to a base excitation

• The vehicle provides the base excitation 𝑦(𝑡) to the

instrumentation package modeled as a single dof

• The displacement response 𝑥(𝑡) is measured from the

system’s static-equilibrium position

Assume that no external force is applied directly to the mass;

that is, 𝑓 𝑡 = 0




- The following governing equation of motion

𝑚𝑑2𝑥


𝑑𝑡+ 𝑘𝑥 = 𝑐

𝑑𝑦

𝑑𝑡+ 𝑘𝑦

⟹ 𝑚𝑑2𝑥


𝑑𝑥

𝑑𝑡+ 𝜔𝑛

2𝑥 = 2𝜉𝜔𝑛

𝑑𝑦

𝑑𝑡+ 𝜔𝑛

2𝑦

𝑦(𝑡) and 𝑥(𝑡) are measured from a fixed point 𝑂 located in an

inertial reference frame and a fixed point located at the

system’s static equilibrium position, respectively




- If the relative displacement is desired, the governing equation

of motion

𝑚𝑑2𝑧

𝑑𝑡2 + 𝑐𝑑𝑧

𝑑𝑡+ 𝑘𝑧 = −𝑚

𝑑2𝑦

𝑑𝑡2

with 𝑧 𝑡 ≡ 𝑥 𝑡 − 𝑦(𝑡)

⟹𝑑2𝑧


𝑑𝑧

𝑑𝑡+ 𝜔𝑛

2𝑧 = −𝑑2𝑦

𝑑𝑡2



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2.System with Unbalanced Rotating Mass

- Assume that the unbalance generates a force that acts on the

system’s mass. This force, in turn, is transmitted through the

spring and damper to the fixed base

- The unbalance is modeled as a mass 𝑚0 that rotates with an

angular speed 𝜔, and this mass is located a fixed distance 𝑒from the center of rotation




- From the free-body diagram (FBD) of the unbalanced mass 𝑚0

𝑁𝑥 = −𝑚0( 𝑥 − 𝜖𝜔2𝑠𝑖𝑛𝜔𝑡)

𝑁𝑦 = 𝑚0𝜖𝜔2𝑐𝑜𝑠𝜔𝑡

- From the FBD of mas 𝑀

𝑀𝑑2𝑥


𝑑𝑡+ 𝑘𝑥 = 𝑁𝑥

⟹ (𝑀 + 𝑚0)𝑑2𝑥


𝑑𝑡+ 𝑘𝑥 = 𝑚0𝜖𝜔

2𝑠𝑖𝑛𝜔𝑡

⟹𝑑2𝑥


𝑑𝑥

𝑑𝑡+ 𝜔𝑛

2𝑥 =𝐹(𝜔)

𝑚𝑠𝑖𝑛𝜔𝑡

where 𝑚 = 𝑀 + 𝑚0, 𝜔𝑛 = 𝑘/𝑚, 𝐹 𝜔 = 𝑚0𝜖𝜔2

- The static displacement of the spring

𝛿𝑠𝑡 =𝑀 + 𝑚0 𝑔

𝑘=

𝑚𝑔

𝑘




3.System with Added Mass Due to a Fluid

- The equation of motion of the system

𝑚𝑑2𝑥

𝑑𝑡2 + 𝑘𝑥 = 𝑓 𝑡 + 𝑓1(𝑡)

𝑥(𝑡) : measured from the unstretched position of the spring

𝑓(𝑡) : the externally applied force

𝑓1(𝑡) : the force exerted by the fluid on the mass due to the

motion of the mass




- The force generated by the fluid on the rigid body

𝑓1 𝑡 = −𝐾0𝑀𝑑2𝑥

𝑑𝑡2 − 𝐶𝑓

𝑑𝑥

𝑑𝑡

𝑀 : the mass of the fluid displaced by the body

𝐾0 : an added mass coefficient

𝐶𝑓 : a positive fluid damping coefficient

- The governing equation of motion

𝑚 + 𝐾0𝑀𝑑2𝑥

𝑑𝑡2 + 𝐶𝑓

𝑑𝑥


𝐾0𝑀 : the added mass due to the fluid



§6.Lagrange’s Equations

Consider a system with 𝑁 degrees of freedom that is described

by a set of 𝑁 generalized coordinates 𝑞𝑖 , 𝑖 = 1,2,…𝑁. In terms

of the chosen generalized coordinates, Lagrange’s equations

have the form

𝑑

𝑑𝑡

𝜕𝑇

𝜕 𝑞𝑗−

𝜕𝑇

𝜕𝑞𝑗+

𝜕𝐷

𝜕 𝑞𝑗+

𝜕𝑉

𝜕𝑞𝑗= 𝑄𝑗 , 𝑗 = 1,2,… , 𝑁

𝑞𝑗 : generalized coordinate

𝑞𝑗 : generalized velocity

𝑇 : the kinetic energy of the system

𝑉 : the potential energy of the system

𝐷 : the Rayleigh dissipation function

𝑄𝑗 : the generalized force that appears in the 𝑗𝑡ℎ equation


(3.41)



The generalized force 𝑄𝑗 that appears in the 𝑗𝑡ℎ equation

𝑄𝑗 =

𝑙

𝐹𝑙

𝜕 𝑟𝑙𝜕𝑞𝑗

+

𝑙

𝑀𝑙

𝜕𝜔𝑙

𝜕 𝑞𝑗

𝐹𝑙, 𝑀𝑙 : the vector representations of the externally

applied forces and moments on the 𝑙𝑡ℎ body

𝑟𝑙 : the position vector to the location where the force

𝐹𝑙 is applied

𝜔𝑙 : the 𝑙𝑡ℎ body’s angular velocity about the axis

along which the considered moment is applied



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Linear Vibratory Systems

For vibratory systems with linear characteristics

𝑇 =1

2

𝑗=1

𝑁

𝑛=1

𝑁

𝑚𝑗𝑛 𝑞𝑗 𝑞𝑛

𝑉 =1

2

𝑗=1

𝑁

𝑛=1

𝑁

𝑘𝑗𝑛𝑞𝑗𝑞𝑛

𝐷 =1

2

𝑗=1

𝑁

𝑛=1

𝑁

𝑐𝑗𝑛 𝑞𝑗 𝑞𝑛

𝑚𝑗𝑛 : the inertia coefficients

𝑘𝑗𝑛 : the stiffness coefficients

𝑐𝑗𝑛 : the damping coefficients




Single Degree-Of-Freedom

The case of a single degree-of-freedom system, 𝑁 = 1, the

Lagrange’s equation

𝑑

𝑑𝑡

𝜕𝑇

𝜕 𝑞1−

𝜕𝑇

𝜕𝑞1+

𝜕𝐷

𝜕 𝑞1+

𝜕𝑉

𝜕𝑞1= 𝑄1

where the generalized force is obtained from

𝑄1 =

𝑙

𝐹𝑙

𝜕 𝑟𝑙𝜕𝑞1

+

𝑙

𝑀𝑙

𝜕𝜔𝑙

𝜕 𝑞1


(3.44)



Linear Single Degree-Of-Freedom Systems

The expressions for the system kinetic energy, the system

potential energy, and the system dissipation function reduce to

𝑇 =1

2

𝑗=1

1

𝑛=1

1

𝑚𝑗𝑛 𝑞𝑗 𝑞𝑛 =1

2𝑚11 𝑞1

2 ≡1

2𝑚𝑒 𝑞1

2

𝑉 =1

2

𝑗=1

1

𝑛=1

1

𝑘𝑗𝑛𝑞𝑗𝑞𝑛 =1

2𝑘11𝑞1

2 ≡1

2𝑘𝑒𝑞1

2

𝐷 =1

2

𝑗=1

1

𝑛=1

1

𝑐𝑗𝑛 𝑞𝑗 𝑞𝑛 =1

2𝑐11 𝑞1

2 ≡1

2𝑐𝑒 𝑞1

2

𝑚𝑒, 𝑘𝑒, 𝑐𝑒 : the equivalent mass, stiffness, and damping

From Lagrange’s equation

𝑚𝑒 𝑞1 + 𝑐𝑒 𝑞1 + 𝑘𝑒𝑞1 = 𝑄1



(3.46)


To obtain the governing equation of motion of a linear vibrating

system with viscous damping

• Obtains expressions for the system kinetic energy 𝑇 ,

system potential energy 𝑉, and system dissipation function 𝐷

• Identify the equivalent mass 𝑚𝑒, equivalent stiffness 𝑘𝑒,

and equivalent damping 𝑐𝑒

• Determine the generalized force

• Apply the governing equation

𝑚𝑒 𝑞1 + 𝑐𝑒 𝑞1 + 𝑘𝑒𝑞1 = 𝑄1

• Determine the system natural frequency

𝜔𝑛 =𝑘𝑒

𝑚𝑒, 𝜉 =

𝑐𝑒

2𝑚𝑒𝜔𝑛=

𝑐𝑒

2 𝑘𝑒𝑚𝑒




- Ex.3.9 Motion of A Linear Single Degree-Of-Freedom System

Obtain the governing equation for the mass-damper-spring

system

Solution

Identify the following

𝑞1 = 𝑥, 𝐹𝑙 = 𝑓(𝑡) 𝑗, 𝑟𝑙 = 𝑥 𝑗, 𝑀𝑙 = 0

Determine the generalized force

𝑄1 =

𝑙

𝐹𝑙


+ 0 = 𝑓 𝑡 𝑗𝜕𝑥 𝑗

𝜕𝑥= 𝑓(𝑡)

The system kinetic energy 𝑇, system potential energy 𝑉, and

system dissipation function 𝐷

𝑇 =1

2𝑚 𝑥2, 𝑉 =

1

2𝑘𝑥2, 𝐷 =

1

2𝑐 𝑥2




Identify the following

𝑞1 = 𝑥, 𝐹𝑙 = 𝑓(𝑡) 𝑗, 𝑟𝑙 = 𝑥 𝑗, 𝑀𝑙 = 0

Determine the generalized force

𝑄1 =

𝑙

𝐹𝑙


+ 0 = 𝑓 𝑡 𝑗𝜕𝑥 𝑗

𝜕𝑥= 𝑓(𝑡)

The system kinetic energy 𝑇, system potential energy

𝑉, and system dissipation function 𝐷

𝑇 =1

2𝑚 𝑥2, 𝑉 =

1

2𝑘𝑥2, 𝐷 =

1

2𝑐 𝑥2

⟹ 𝑚𝑒 = 𝑚, 𝑘𝑒 = 𝑘, 𝑐𝑒 = 𝑐

The governing equation

𝑚𝑑2𝑥

𝑑𝑡2 + 𝑐𝑑𝑦

𝑑𝑡+ 𝑘𝑥 = 𝑓(𝑡)



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- Ex.3.10 Motion of A System that Translates and Rotates

Obtain the governing equation of motion for “small” oscillations

about the upright position

Solution

Choose the generalized coordinate

𝑞1 = 𝜃, 𝐹𝑙 = 0, 𝑀𝑙 = 𝑀 𝑡 𝑘, 𝜔𝑙 = 𝜃𝑘

The generalized force

𝑄1 =

𝑙

𝑀𝑙 ∙𝜕𝜔𝑙

𝜕 𝑞1= 𝑀 𝑡 𝑘 ∙

𝜕 𝜃𝑘

𝜕 𝜃= 𝑀(𝑡)



𝐽𝐺 =1

2𝑚𝑟2


The potential energy

𝑉 =1

2𝑘𝑥2 =

1

2𝑘(𝑟𝜃)2=

1

2𝑘𝑟2𝜃2

⟹ the equivalent stiffness

The kinetic energy of the system

𝑇 =1

2𝑚 𝑥2 +

1

2𝐽𝐺 𝜃2

⟹ 𝑇 =1

2𝑚𝑟2 + 𝐽𝐺 𝜃2 =

1

2

3

2𝑚𝑟2 𝜃2

⟹ the equivalent mass of the system



𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦

𝑡𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑖𝑜𝑛𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦

𝑘𝑒 = 𝑘𝑟2

𝑚𝑒 =3

2𝑚𝑟2


The dissipation function

𝐷 =1

2𝑐 𝑥2 =

1

2𝑐(𝑟 𝜃)2=

1

2(𝑐𝑟2) 𝜃2

⟹ the equivalent damping coefficient

𝑐𝑒 = 𝑐𝑟2

The governing equation of motion3

2𝑚𝑟2 𝜃 + 𝑐𝑟2 𝜃 + 𝑘𝑟2𝜃 = 𝑀(𝑡)

Natural frequency and damping factor

𝜔𝑛 =𝑘𝑒

𝑚𝑒=

𝑘𝑟2

3𝑚𝑟2/2=

2𝑘

3𝑚

𝜉 =𝑐𝑒

2𝑚𝑒𝜔𝑛=

𝑐𝑟2

2(3𝑚𝑟2/2) 2𝑘/3𝑚=

6

6𝑘𝑚




- Ex.3.11 Inverted Pendulum

Obtain the governing equation of motion for “small” oscillations

about the upright position

Solution

The total rotary inertia of the system

𝐽𝑂 = 𝐽𝑂1+ 𝐽𝑂2

𝐽𝑂1: mass momentof inertia of 𝑚1 about point𝑂

𝐽𝑂2: massmomentof inertiaof thebaraboutpoint𝑂

𝐽𝑂1=

2

5𝑚1𝑟

2 + 𝑚1𝐿12

𝐽𝑂2=

1

12𝑚2𝐿2

2 + 𝑚2

𝐿2

2

2

=1

3𝑚2𝐿2

2




Choosing 𝑞1 = 𝜃 as the generalized coordinate, the system

kinetic energy takes the form

𝑇 =1

2𝐽𝑂 𝜃2 =

1

2𝐽𝑂1

+ 𝐽𝑂2 𝜃2

=1

2

2

5𝑚1𝑟

2 + 𝑚1𝐿12 +

1

3𝑚2𝐿2

2 𝜃2

For small 𝜃 ⟹ 𝑥1 ≈ 𝐿1𝜃

The system potential energy

𝑉 =1

2𝑘𝑥1

2 −1

2𝑚1𝑔𝐿1𝜃

2 −1

2𝑚2𝑔

𝐿2

2𝜃2

=1

2𝑘𝐿1

2 − 𝑚1𝑔𝐿1 − 𝑚2𝑔𝐿2

2𝜃2

𝐷 =1

2𝑐 𝑥1

2 =1

2𝑐𝐿1

2 𝜃2





The equivalent inertia, the equivalent stiffness, and the

equivalent damping properties of the system

𝑇 =1

2

2

5𝑚1𝑟

2 +𝑚1𝐿12 +

1

3𝑚2𝐿2

2 𝜃2

𝑉 =1

2𝑘𝐿1

2 − 𝑚1𝑔𝐿1 − 𝑚2𝑔𝐿2

2𝜃2

𝐷 =1

2𝑐 𝑥1

2 =1

2𝑐𝐿1

2 𝜃2

The governing equation of motion 𝑚𝑒 𝜃 + 𝑐𝑒

𝜃 + 𝑘𝑒𝜃 = 0

Natural frequency

𝜔𝑛 =𝑘𝑒

𝑚𝑒=

𝑘𝐿12 − 𝑚1𝑔𝐿1 − 𝑚2𝑔𝐿2/2

𝐽𝑂1+ 𝐽𝑂2

𝑘𝑒 can be negative, which affects the stability of the system



⟹𝑚𝑒 =2

5𝑚1𝑟

2 +𝑚1𝐿12 +

1

3𝑚2𝐿2

2

⟹𝑘𝑒 =𝑘𝐿12 −𝑚1𝑔𝐿1 −𝑚2𝑔

𝐿2

2

⟹ 𝑐𝑒 = 𝑐𝐿12

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• Natural Frequency of Pendulum System

Now locate the pivot point 𝑂 on the top, the

equivalent stiffness of this system

𝑘𝑒 = 𝑘𝐿12 + 𝑚1𝑔𝐿1 + 𝑚2𝑔

𝐿2

2and the natural frequency of this system

𝜔𝑛 =𝑘𝑒

𝑚𝑒=

𝑘𝐿12 + 𝑚1𝑔𝐿1 + 𝑚2𝑔𝐿2/2

𝐽𝑂1+ 𝐽𝑂2

If 𝑚2 ≪ 𝑚1, 𝑟 ≪ 𝐿1, and 𝑘 = 0, then

𝜔𝑛 =𝑚1𝑔𝐿1 1 + 𝑚2𝐿2/𝑚1𝐿1

𝑚1𝐿12 1 + 2𝑟2/5𝐿1

2 →𝑔

𝐿

→ the natural frequency of a pendulum composed of a rigid,

weightless rod carrying a mass a distance 𝐿1 from its pivot




- Ex.3.12 Motion of A Disk Segment

Derive the governing equation of motion of a disk segment

Solution

The position vector from the fixed point 𝑂 to the

center of mass 𝐺

𝑟 = −𝑅𝜃 + 𝑏𝑠𝑖𝑛𝜃 𝑖 + (𝑅 − 𝑏𝑐𝑜𝑠𝜃) 𝑗

The absolute velocity of the center of mass 𝑟 = − 𝑅 − 𝑏𝑐𝑜𝑠𝜃 𝜃 𝑖 + 𝑏𝑠𝑖𝑛𝜃 𝜃 𝑗

Selecting the generalized coordinate 𝑞1 = 𝜃 ,

the system kinetic energy

𝑇 =1

2𝐽𝐺 𝜃2 +

1

2𝑚 𝑟 ∙ 𝑟

⟹ 𝑇 =1

2𝐽𝐺 𝜃2 +

1

2𝑚 𝑅2 + 𝑏2 − 2𝑏𝑅𝑐𝑜𝑠𝜃 𝜃2



Taylor series expansion

𝑐𝑜𝑠𝜃 = 𝑐𝑜𝑠 𝜃0 + 𝜃 ≈ 𝑐𝑜𝑠𝜃0 − 𝜃𝑠𝑖𝑛𝜃0 −1

2 𝜃2𝑐𝑜𝑠𝜃0 + ⋯

𝑠𝑖𝑛𝜃 = 𝑠𝑖𝑛 𝜃0 + 𝜃 ≈ 𝑠𝑖𝑛𝜃0 − 𝜃𝑐𝑜𝑠𝜃0 −1

2 𝜃2𝑠𝑖𝑛𝜃0 + ⋯


Choosing the fixed ground as the datum, the system potential

energy

𝑉 = 𝑚𝑔 𝑅 − 𝑏𝑐𝑜𝑠𝜃

Small Oscillations about the Upright Position

Express the angular displacement as

𝜃(𝑡) = 𝜃0 + 𝜃(𝑡)

Since 𝜃0 = 0 , and small 𝜃 , using 𝑠𝑖𝑛𝜃 ≈ 𝜃 ,

𝑐𝑜𝑠𝜃 ≈ 1 −1

2 𝜃2, rewrite the energy functions

𝑇 ≈1

2𝐽𝐺 + 𝑚 𝑅 − 𝑏 2 𝜃2, 𝑉 ≈ 𝑚𝑔 𝑅 − 𝑏 +

1

2𝑚𝑔𝑏 𝜃2




The equivalent inertia of the system

𝑇 ≈1

2𝐽𝐺 + 𝑚 𝑅 − 𝑏 2 𝜃2

The potential energy is not in standard form because of the

constant term 𝑚𝑔 𝑅 − 𝑏

𝑉 ≈ 𝑚𝑔 𝑅 − 𝑏 +1

2𝑚𝑔𝑏 𝜃2

However, since the datum for the potential energy is not

unique, we can shift the datum for the potential energy from

the fixed ground to a distance (𝑅 − 𝑏) above the ground

𝑉 =1

2𝑚𝑔𝑏 𝜃2

Then, the equivalent stiffness can be defined

𝑘𝑒 = 𝑚𝑔𝑏



⟹ 𝑚𝑒 = 𝐽𝐺 + 𝑚 𝑅 − 𝑏 2



𝐽𝐺 + 𝑚 𝑅 − 𝑏 2 𝜃 + 𝑚𝑔𝑏 𝜃 = 0

Natural Frequency

𝜔𝑛 =𝑘𝑒

𝑚𝑒

=𝑚𝑔𝑏

𝐽𝐺 + 𝑚 𝑅 − 𝑏 2

=𝑔

𝐽𝐺 + 𝑚 𝑅 − 𝑏 2

𝑚𝑏




- Ex.3.13 Translating System with a Pre-tensioned/compressedSpring

Derive the governing equation of motion for vertical

translations 𝑥 of the mass about the static

equilibrium position of the system

Solution

The equation of motion will be derived for

“small” amplitude vertical oscillations; that is,

𝑥/𝐿 ≪ 1

The horizontal spring is pretensioned with a tension, which is

produced by an initial extension of the spring by an amount 𝛿0

𝑇1 = 𝑘1𝛿0

The kinetic energy of the system

𝑇 =1

2𝑚 𝑥2



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

2𝑛(𝑛 − 1)𝑥2 + ⋯


The potential energy of the system

𝑉 =1

2𝑘1 𝛿0 + ∆𝐿 2

𝑓𝑜𝑟 𝑠𝑝𝑟𝑖𝑛𝑔 𝑘1

+1

2𝑘2𝑥

2

𝑓𝑜𝑟 𝑠𝑝𝑟𝑖𝑛𝑔 𝑘2

∆𝐿 : the change in the length of the spring with

stiffness 𝑘1 due to the motion 𝑥 of the mass

∆𝐿 = 𝐿2 + 𝑥2 − 𝐿 = 𝐿 1 + (𝑥/𝐿)2− 𝐿

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+⋯

⟹ ∆𝐿 ≈ 𝐿 1+(𝑥/𝐿)2/2 − 𝐿 = 𝐿(𝑥/𝐿)2/2

⟹ 𝑉 =1

2𝑘1 𝛿0 +

𝐿

2

𝑥

𝐿

2 2

+1

2𝑘2𝑥

2



𝑑

𝑑𝑡

𝜕𝑇

𝜕 𝑞1−

𝜕𝑇

𝜕𝑞1+

𝜕𝐷

𝜕 𝑞1+

𝜕𝑉

𝜕𝑞1= 𝑄1 (3.44)


Chose the generalize coordinate 𝑞1 = 𝑥

𝑇 =1

2𝑚 𝑥2

⟹𝑑

𝑑𝑡

𝜕𝑇

𝜕 𝑥=

𝑑

𝑑𝑡𝑚 𝑥 = 𝑚 𝑥

𝜕𝑇

𝜕𝑥= 0,

𝜕𝐷

𝜕 𝑥= 0, 𝑄 = 0

𝑉 =1

2𝑘1 𝛿0 +

𝐿

2

𝑥

𝐿

2 2

+1

2𝑘2𝑥

2

⟹𝜕𝑉

𝜕𝑥=𝑘1 𝛿0 +

𝐿

2

𝑥

𝐿

2 2𝑥

𝐿+𝑘2𝑥= 𝑘1 +

𝑘1𝛿0

𝐿𝑥+

𝑘1

2

𝑥3

𝐿2≈ 𝑘2 +

𝑇1𝐿

𝑥





𝑚 𝑥 + 𝑘2 +𝑇1

𝐿𝑥 = 0


𝜔𝑛 = 𝑘𝑒/𝑚𝑒 = 𝑘2 + 𝑇1/𝐿 /𝑚

If the spring of constant 𝑘1 is compressed instead of being in

tension, then we can replace 𝑇1 by −𝑇1 , and the natural

frequency

𝜔𝑛 = 𝑘𝑒/𝑚𝑒 = 𝑘2 − 𝑇1/𝐿 /𝑚

The natural frequency 𝜔𝑛 can be made very low by adjusting

the compression of the spring with stiffness 𝑘1. At the same

time, the spring with stiffness 𝑘2 can be made stiff enough so

that the static displacement of the system is not excessive




- Ex.3.14 Equation of Motion for a Disk with An Extended Mass

Determine the governing equation of motion

and the natural frequency for the system

Solution

The velocity of 𝑚

𝑣𝑚 =𝑑 𝑟𝑚𝑑𝑡

=𝑑

𝑑𝑡𝑥 + 𝐿𝑠𝑖𝑛𝜃 𝑖 + 𝐿 − 𝐿𝑐𝑜𝑠𝜃 𝑗

= −𝑅 𝜃 + 𝐿 𝜃𝑐𝑜𝑠𝜃 𝑖 + 𝐿 𝜃𝑠𝑖𝑛𝜃 𝑗




The kinetic energy of the system 𝑇 = 𝑇𝑑 + 𝑇𝑝

𝑇𝑑 =1

2𝑚𝑑 𝑥2 +

1

2𝐽𝐺 𝜃2

=1

2𝑚𝑑𝑅2 𝜃2 +

1

2𝐽𝐺 𝜃2

𝑇𝑝 =1

2𝑚𝑣𝑚

2

=1

2𝑚 −𝑅 𝜃 + 𝐿 𝜃𝑐𝑜𝑠𝜃 𝑖 + 𝐿 𝜃𝑠𝑖𝑛𝜃 𝑗

2

=1

2𝑚(𝑅2 + 𝐿2 − 2𝐿𝑅𝑐𝑜𝑠𝜃) 𝜃2

≈1

2𝑚 𝐿 − 𝑅 2 𝜃2

⟹ 𝑇 = 𝑇𝑑 + 𝑇𝑝 =1

2𝑚 𝐿 − 𝑅 2 + 𝑚𝑑𝑅2 + 𝐽𝐺 𝜃2 ≡

1

2𝑚𝑒

𝜃2




The potential energy of the system

𝑉 =1

2𝑘𝑥2 + 𝑚𝑔(𝐿 − 𝐿𝑐𝑜𝑠𝜃)

=1

2𝑘𝑅2𝜃2 + 𝑚𝑔𝐿 1 − 𝑐𝑜𝑠𝜃

⟹ 𝑉 =1

2𝑘𝑅2𝜃2 +

1

2𝑚𝑔𝐿𝜃2 𝑐𝑜𝑠𝜃 ≈ 1 −

𝜃2

2

=1

2𝑘𝑅2 + 𝑚𝑔𝐿 𝜃2 ≡

1

2𝑘𝑒𝜃

2


𝐷 =1

2𝑐 𝑥2 =

1

2𝑐𝑅2 𝜃2 ≡

1

2𝑐𝑒

𝜃2

𝑚𝑒 𝜃 + 𝑐𝑒

𝜃 + 𝑘𝑒𝜃 = 0,𝜔𝑛 =𝑘𝑒

𝑚𝑒=

𝑘𝑅2 + 𝑚𝑔𝐿

𝑚(𝐿 − 𝑅)2+𝑚𝑑𝑅2 + 𝐽𝐺




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- Ex.3.15 Micro-Electromechanical System

Determine the governing equation of motion and the natural

frequency for the micro-electromechanical system

Solution

The potential energy

𝑉 =1

2𝑘𝑡𝜑

2 +1

2𝑘 𝑥0 𝑡 −𝑥1

2 +1

4𝑚2𝑔(𝐿2 −𝐿1)𝜑

2

=1

2𝑘𝑡𝜑

2 +1

2𝑘 𝑥0 𝑡 −𝐿2𝜑

2 +1

4𝑚2𝑔(𝐿2 −𝐿1)𝜑

2

The kinetic energy

𝑇 =1

2𝐽0 𝜑2 +

1

2𝑚1 𝑥1

2 =1

2𝐽0 +𝑚1𝐿2

2 𝜑2 ≡1

2𝑚𝑒 𝜑2

Dissipation function


𝐷 =1

2𝑐 𝑥2

2 =1

2𝑐𝐿1

2 𝜑2 ≡1

2𝑐𝑒 𝜑2



𝑉 =1

2𝑘𝑡𝜑

2 +1

2𝑘 𝑥0 𝑡 − 𝐿2𝜑

2 +1

4𝑚2𝑔(𝐿2 − 𝐿1)𝜑

2

The potential energy is not in the standard form ⟹ the

governing equation must be derived from Lagrange’s equation𝜕𝑉

𝜕𝜑= 𝑘𝑡 + 𝑘𝐿2

2 +1

2𝑚2𝑔(𝐿2 − 𝐿1) 𝜑 − 𝑘𝐿2𝑥0 𝑡

= 𝑘𝑒𝜑 − 𝑘𝐿2𝑥0 𝑡


𝑚𝑒 𝜑 + 𝑐𝑒 𝜑 + 𝑘𝑒𝜑 = 𝑘𝐿2𝑥0(𝑡)


𝜔𝑛 =𝑘𝑒

𝑚𝑒=

𝑘𝑡 + 𝑘𝐿22 + 𝑚2𝑔(𝐿2 − 𝐿1)/2

𝐽0 + 𝑚1𝐿22




Ex.3.16 Slider Mechanism

Obtain the equation of motion of the slider mechanism

Solution

The geometric constraints on the motion

𝑟2 𝜑 = 𝑎2 + 𝑏2 − 2𝑎𝑏𝑐𝑜𝑠𝜑 (a)

⟹ 𝑟 𝜑 =𝑎𝑏

𝑟(𝜑) 𝜑𝑠𝑖𝑛𝜑

𝑟 𝜑 𝑠𝑖𝑛𝛽 = 𝑏𝑠𝑖𝑛𝜑 (b)

𝑎 = 𝑟 𝜑 𝑐𝑜𝑠𝛽 + 𝑏𝑐𝑜𝑠𝜑 (c)

⟹ 𝑟 𝜑 𝑐𝑜𝑠𝛽 − 𝑟 𝜑 𝛽𝑠𝑖𝑛𝛽 − 𝑏 𝜑𝑠𝑖𝑛𝜑 = 0

⟹ 𝛽 = 𝑟 𝜑 𝑐𝑜𝑠𝛽 − 𝑏 𝜑𝑠𝑖𝑛𝜑

𝑟(𝜑)𝑠𝑖𝑛𝛽=

𝜑

𝑟2(𝜑)𝑎𝑏𝑐𝑜𝑠𝜑 − 𝑏2




System Kinetic Energy

𝑇 =1

2𝐽𝑚𝑏 +𝐽𝑚𝑒 𝜑2 +

1

2𝐽𝑚𝑙

𝛽2 +1

2𝑚𝑠 𝑟2 +

1

2𝑚𝑠𝑟

2 𝛽2

𝐽𝑚𝑏 =1

3𝑚𝑏𝑏

2

𝐽𝑚𝑒 =1

3𝑚𝑒𝑒

2

𝐽𝑚𝑙 =1

3𝑚𝑙𝑙

2

𝑚 𝜑 ≡ 𝐽𝑚𝑏 + 𝐽𝑚𝑒 + 𝐽𝑚𝑙 + 𝑚𝑠𝑟2

𝑎𝑏𝑐𝑜𝑠𝜑 − 𝑏2

𝑟2(𝜑)

2

+ 𝑚𝑠

𝑎𝑏𝑠𝑖𝑛𝜑

𝑟(𝜑)

2

⟹ 𝑇 =1

2𝑚(𝜑) 𝜑2


where


𝑑

𝑑𝑡

𝜕𝑇

𝜕 𝑞1−

𝜕𝑇

𝜕𝑞1+

𝜕𝐷

𝜕 𝑞1+

𝜕𝑉

𝜕𝑞1= 𝑄1 (3.44)


System Kinetic Energy

𝑇 =1

2𝑚(𝜑) 𝜑2

System Potential Energy

𝑉 =1

2𝑘𝑟2(𝜑) +

1

2𝑘𝑑 𝑑 𝑡 − 𝑒𝜑 2

Equation of motion

𝑑

𝑑𝑡

𝜕𝑇

𝜕 𝜑=

𝑑

𝑑𝑡𝑚(𝜑) 𝜑 = 𝑚(𝜑) 𝜑,

𝜕𝑇

𝜕𝜑= 𝑚′ 𝜑 𝜑2

𝜕𝑉

𝜕𝜑= 𝑘𝑟 𝜑 𝑟′ 𝜑 + 𝑘𝑑𝑒2𝜑 − 𝑘𝑑𝑒2𝑑(𝑡)

⟹ 𝑚 𝜑 𝜑 +1

2𝑚′ 𝜑 𝜑2 + 𝑘𝑟 𝜑 𝑟′ 𝜑 + 𝑘𝑑𝑒2𝜑 = 𝑘𝑑𝑒2𝑑(𝑡)




- Ex.3.17 Oscillations of A Crankshaft

Obtain the equation of motion of the crankshaft

Solution

• Kinematics

The position vector of the

slider mass 𝑚𝑝

𝑟𝑃 = 𝑟𝑐𝑜𝑠𝜃 + 𝑙𝑐𝑜𝑠𝛾 𝑖 + 𝑑 𝑗

The position vector of the center of mass 𝐺 of the crank

𝑟𝐺 = 𝑟𝑐𝑜𝑠𝜃 + 𝑎𝑐𝑜𝑠𝛾 𝑖 + 𝑟𝑠𝑖𝑛𝜃 + 𝑎𝑠𝑖𝑛𝛾 𝑗

From geometry

𝑟𝑠𝑖𝑛𝜃 = 𝑑 + 𝑙𝑠𝑖𝑛𝛾

The slider velocity

𝑣𝑃 = 𝑟𝑃 = −𝑟 𝜃𝑠𝑖𝑛𝜃 − 𝑙 𝛾𝑠𝑖𝑛𝛾 𝑖 = −𝑟 𝜃 𝑠𝑖𝑛𝜃 +𝑡𝑎𝑛𝛾𝑐𝑜𝑠𝜃 𝑖



⟹ 𝑟 𝜃𝑐𝑜𝑠𝜃 = 𝑙 𝛾𝑐𝑜𝑠𝛾 ⟹ 𝛾 =𝑟

𝑙

𝑐𝑜𝑠𝜃

𝑐𝑜𝑠𝛾 𝜃

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The velocity of the center of mass 𝐺 of the crank

𝑣𝐺 = −𝑟 𝜃𝑠𝑖𝑛𝜃 − 𝑎 𝛾𝑠𝑖𝑛𝛾 𝑖 + 𝑟 𝜃𝑐𝑜𝑠𝜃 − 𝑎 𝛾𝑐𝑜𝑠𝛾 𝑗

⟹ 𝑣𝐺 = − 𝑠𝑖𝑛𝜃 +𝑎

𝑙𝑡𝑎𝑛𝛾𝑐𝑜𝑠𝜃 𝑟 𝜃 𝑖 +

𝑏

𝑙𝑐𝑜𝑠𝜃 𝑟 𝜃 𝑗

• System Kinetic Energy

The total kinetic energy of the system

𝑇 =1

2𝐽𝑑 𝜃2 +

1

2𝑚𝐺𝑣𝐺

2 +1

2𝐽𝐺 𝛾2 +

1

2𝑚𝑃𝑣𝑃

2 ≡1

2𝐽(𝜃) 𝜃2

𝐽 𝜃 = 𝐽𝑑 + 𝑟2𝑚𝐺 𝑠𝑖𝑛𝜃 +𝑎

𝑙𝑡𝑎𝑛𝛾𝑐𝑜𝑠𝜃

2

+𝑏

𝑙𝑐𝑜𝑠𝜃

2

+𝐽𝐺𝑟

𝑙

𝑐𝑜𝑠𝜃

𝑐𝑜𝑠𝛾

2

+ 𝑟2𝑚𝑃 𝑠𝑖𝑛𝜃 + 𝑡𝑎𝑛𝛾𝑐𝑜𝑠𝜃 2



where,

𝛾 = 𝑠𝑖𝑛−1𝑟

𝑙𝑠𝑖𝑛𝜃 −

𝑑

𝑙


• Equation of Motion

The governing equation of motion has the form

𝑑

𝑑𝑡

𝜕𝑇

𝜕 𝜃−

𝜕𝑇

𝜕𝜃= −𝑀(𝑡)

After performing the differentiation operations

𝐽 𝜃 𝜃 +1

2

𝜕𝐽(𝜃)

𝜕𝜃 𝜃2 = −𝑀(𝑡)

The angle 𝜃 can be expressed

𝜃 𝑡 = 𝜔𝑡 + 𝜙(𝑡)

Then

𝐽 𝜃 𝜙 +1

2

𝜕𝐽(𝜃)

𝜕𝜃𝜔 + 𝜙

2= −𝑀(𝑡)




- Ex.3.18 Vibration of A Centrifugal Governor

Derive the equation of motion of

governor by usingLagrange’sequation

Solution

The velocity vector relative to point

𝑂 of the left hand mass

𝑉𝑚 = −𝐿 𝜑𝑐𝑜𝑠𝜑 𝑖 + 𝐿 𝜑𝑠𝑖𝑛𝜑 𝑗

+(𝑟 + 𝐿𝑠𝑖𝑛𝜑)𝜔𝑘

The kinetic energy

𝑇 𝜑, 𝜑 = 21

2𝑚 𝑉𝑚𝑉𝑚

= 𝑚 −𝐿 𝜑𝑐𝑜𝑠𝜑 2 + 𝐿 𝜑𝑠𝑖𝑛𝜑 2 + 𝑟 + 𝐿𝑠𝑖𝑛𝜑 𝜔 2

= 𝑚𝜔2 𝑟 + 𝐿𝑠𝑖𝑛𝜑 2 + 𝑚 𝜑2𝐿2




The potential energy with respect to the static equilibrium position

𝑉 𝜑 =1

2𝑘 2𝐿 1−𝑐𝑜𝑠𝜑 2 −2𝑚𝑔𝐿𝑐𝑜𝑠𝜑

Using equation (3.44) with

𝑞1 = 𝜑

𝐷 = 0

𝑄1 = 0

and performing the required

operations, to obtain the following

governing equation

𝑚𝐿2 𝜑 − 𝑚𝑟𝐿𝜔2𝑐𝑜𝑠𝜑 − 𝑚𝜔2 + 2𝑘 𝐿2𝑠𝑖𝑛𝜑𝑐𝑜𝑠𝜑

+𝐿 𝑚𝑔 + 2𝑘𝐿 𝑠𝑖𝑛𝜑 = 0



𝑑

𝑑𝑡

𝜕𝑇

𝜕 𝑞1−

𝜕𝑇

𝜕𝑞1+

𝜕𝐷

𝜕 𝑞1+

𝜕𝑉

𝜕𝑞1= 𝑄1 (3.44)


𝑚𝐿2 𝜑 − 𝑚𝑟𝐿𝜔2𝑐𝑜𝑠𝜑 − 𝑚𝜔2 + 2𝑘 𝐿2𝑠𝑖𝑛𝜑𝑐𝑜𝑠𝜑

+𝐿 𝑚𝑔 + 2𝑘𝐿 𝑠𝑖𝑛𝜑 = 0

Introducing the quantities

𝛾 ≡𝑟

𝐿, 𝜔𝑝

2 ≡𝑔

𝐿, 𝜔𝑛

2 ≡2𝑘

𝑚Rewrite the equation

𝜑 − 𝛾𝜔2𝑐𝑜𝑠𝜑

− 𝜔2 + 𝜔𝑛2 𝑠𝑖𝑛𝜑𝑐𝑜𝑠𝜑

+ 𝜔𝑝2 + 𝜔𝑛

2 𝑠𝑖𝑛𝜑 = 0

Assume that the oscillation 𝜑 about 𝜑 = 0 are small (𝑐𝑜𝑠𝜑 ≈ 1,𝑠𝑖𝑛𝜑 ≈ 𝜑) to get the final equation

𝜑 + 𝜔𝑝2 − 𝜔2 𝜑 = 𝛾𝜔2




- Ex.3.19 Oscillations of A Rotating System

Determine the change in the equilibrium position of

the wheel and the natural frequency of the system

about this equilibrium position

Solution

The spring force = the centrifugal force

𝑘𝛿 = 𝑚(𝑅 + 𝛿)Ω2 ⟹ 𝛿 =𝑅

𝜔1𝑛2

Ω2 − 1

, 𝜔1𝑛2 =

𝑘

𝑚

For small angles of rotation, the kinetic energy

𝑇 =1

2

1

2𝑚𝑟2

𝑥

𝑟

2

+1

2𝑚 𝑥2 =

1

2

3

2𝑚 𝑥2

The potential energy for oscillations about the equilibrium

position



2/7/2014

15


The potential energy for oscillations about the equilibrium position

𝑉 =1

2𝑘𝑥2

The Lagrange equation for this undamped system

𝑑

𝑑𝑡

𝜕𝑇

𝜕 𝑥−

𝜕𝑇

𝜕𝑥+

𝜕𝐷

𝜕 𝑥+

𝜕𝑉

𝜕𝑥= 𝑄𝑥 = 𝑚𝑥Ω2

where the centrifugal force 𝑚𝑥Ω2 is treated as an external force


3

2𝑚 𝑥 + 𝑘 − 𝑚Ω2 𝑥 = 0


𝜔𝑛 =𝑘

𝑚=

2

3𝜔1𝑛

2 − Ω2



Excercises



ch.03 single dof systems - governing equations

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