Ch 04 spring and damper elements in mechanical systems

of Technology, Mechanical Engineering Department Nguyen Tan Tien System Dynamics 4.01 Spring & Damper Elements in Mechanical Systems §1.Spring Elements 1.Force-Deflection Relations Linea

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4 Spring and Damper

Elements in Mechanical

Systems

HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien

System Dynamics 4.01 Spring & Damper Elements in Mechanical Systems

§1.Spring Elements 1.Force-Deflection Relations

Linear force deflection model

𝑓 = 𝑘𝑥

𝑓: the restoring force, 𝑁 𝑥: the compression or extension distance, 𝑚

𝑘: the spring constant, or stiffness, 𝑁/𝑚

For the helical coil spring

𝑘 = 𝐺𝑑 4 64𝑛𝑅3

𝑑: the wire diameter, 𝑚 𝑅: the radius of the coil, 𝑚 𝑛: the number of coils 𝐺: the shear modulus of elasticity, 𝑃𝑎 HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien System Dynamics 4.02 Spring & Damper Elements in Mechanical Systems

§1.Spring Elements

2.Tensile test of a rod

§1.Spring Elements 3.Analytical Determination of the Spring Constant

Derive the spring constant expression for a cylindrical rod subjected to an axial force𝑓 Solution

The force deflection relation of the rod

𝑥 = 𝐿

𝐸𝐴𝑓 =

4𝐿 𝜋𝐸𝐷2𝑓 𝑥: axial deformation, 𝑚 𝐿: length of the rod, 𝑚 𝑓: applied axial force, 𝑁 𝐸: modulus of elasticity, 𝑃𝑎 𝐴: cross-sectional area, 𝑚2 𝐷: rod diameter, 𝑚 The spring constant𝑘

𝑘 ≡𝑓

𝑥=

𝐸𝐴

𝐿 =

𝜋𝐸𝐷2 4𝐿 , 𝑁/𝑚 HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien System Dynamics 4.04 Spring & Damper Elements in Mechanical Systems

§1.Spring Elements

- Example 4.1.2 Spring Constant of a Fixed-End Beam

Derive the spring constant expression of a fixed-end beam Solution

The force-deflection relation of a fixed-end beam

𝑥 = 𝐿

3 192𝐸𝐼𝐴𝑓 =

𝐿3 16𝐸𝑤ℎ3𝑓 𝑥: deflection, 𝑚 𝐿: length of the beam, 𝑚

𝑓: applied force, 𝑁 𝐸: modulus of elasticity, 𝑃𝑎

𝐼𝐴: area moment of inertia,𝐼𝐴= 𝑤ℎ3/12, 𝑚2

The spring constant𝑘

𝑘 ≡𝑓

𝑥=

16𝐸𝑤ℎ3

𝐿3 , 𝑁/𝑚 HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien

§1.Spring Elements

- Spring constants of common elements

HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien System Dynamics 4.06 Spring & Damper Elements in Mechanical Systems

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§1.Spring Elements

- Spring constants of common elements

§1.Spring Elements

- Torsional spring constants of common elements

§1.Spring Elements

4.Series and parallel spring elements

- Series spring elements

1

𝑘𝑒= 𝑖=1

𝑛 1 𝑘𝑖

- Parallel spring elements

𝑘𝑒= 𝑖=1

𝑛 𝑘𝑖

§1.Spring Elements

- Example 4.1.3 Spring Constant of a Stepped Shaft

Determine the expression for the equivalent torsional spring constant for the stepped shaft

Solution Torque applied to the shaft

𝑇 = 𝑘𝑇1𝜃1= 𝑘𝑇2𝜃2 The equivalent spring constant 1

𝑘𝑇𝑒=

1

𝑘𝑇1+

1

𝑘𝑇2, 𝑘𝑇𝑖=𝐺𝜋𝐷𝑖

4 32𝐿𝑖, 𝑖 = 1,2

⟹𝑘𝑇𝑒= 𝑘𝑇1𝑘𝑇2

𝑘𝑇1+ 𝑘𝑇2

§1.Spring Elements

- Example 4.1.4 Spring Constant of a Lever System

Consider a horizontal force𝑓 acting on a lever that is attached to two springs

Assume that the resulting motion is small

Determine the expression for the equivalent spring constant that relates the applied force𝑓 to the resulting displacement 𝑥 Solution

For small angles𝜃, the upper spring deflection is 𝑥

and the deflection of the lower spring is𝑥/2

For static equilibrium

𝑓𝐿 = 𝑘𝑥𝐿 − 𝑘𝑥

2

𝐿

2= 0 ⟹ 𝑓 = 𝑘 𝑥 +

𝑥

4 =

5

4𝑘𝑥 The equivalent spring constant𝑘𝑒= 5𝑘/4

§1.Spring Elements 5.Nonlinear spring elements

- In some case, the use of a nonlinear model is unavoidable

This is the case when a system is designed to utilize two or more spring elements to achieve a spring constant that varies with the applied load Even if each spring element is linear, the combined system will be nonlinear

- Example 4.1.5 Deflection of a Nonlinear System

Obtain the deflection of the system model shown in figure as

a function of the weight𝑊 Assume that each spring exerts a

force that is proportional to its compression

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§1.Spring Elements

Solution

At equilibrium points: the weight force must balance the

spring forces

𝑊 = 𝑘1𝑥 + 2𝑘2𝑥 − 𝑑 𝑥 ≥ 𝑑⟹

𝑥 =𝑊

𝑥 =𝑊 + 2𝑘2𝑑

𝑘1+ 2𝑘2 𝑥 ≥ 𝑑

These relations can be used to generate the plot of𝑥 vs 𝑊

§1.Spring Elements

- Most spring elements display nonlinear behavior if the deflection is large enough

- Force-deflection relations for three types of spring elements

• A hardening spring element

• A softening spring element

• The linear spring element: straight line

Hardening spring element Softening spring element

§2.Modeling of Mass-Spring Systems

Point-mass assumption: (1) the object is a rigid body and (2)

neglect the force distribution within the object

1.Real versus ideal spring elements

- All real spring elements have mass and are not rigid bodies

• Decide whether and how the system can be modeled as

one or more rigid bodies

- If the system consists of an object attached to a spring

• Neglect the spring mass relative to the mass of the object

• Take the mass center of the system to be located at the

mass center of the object

- An ideal spring element is massless A real spring element

can be represented by an ideal element either by

• neglecting its mass, or

• including it in another mass in the system

§2.Modeling of Mass-Spring Systems 2.Effect of spring free length and object geometry

Consider a cube: mass𝑚, side length 2𝑎; and the spring with

free length𝐿; the horizontal surface is frictionless

• The mass is in equilibrium position 𝐺 when the spring is at its free length𝐿

• The mass displaced a distance 𝑥 from the equilibrium position, the spring has been stretched a distance𝑥 from its free length, and thus its force is𝑘𝑥

• Equation of motion from free body diagram

𝑚 𝑥 = −𝑘𝑥

• Note: 𝐿 and 𝑎 does not appear in the equation of motion⟹ we can represent the object as a point mass, as shown

3.Effect of gravity

Consider the object slides on an inclined

frictionless surface

• At equilibrium position, the spring stretches

a distance𝛿𝑠𝑡(static spring deflection)

𝑚𝑔𝑠𝑖𝑛𝜙 − 𝑘𝛿𝑠𝑡= 0

• The spring has been stretched a distance

𝑥 + 𝛿𝑠𝑡 from its free length, and thus its force is𝑘(𝑥 + 𝛿𝑠𝑡)

• From the free body diagram

𝑚 𝑥 = −𝑘 𝑥 + 𝛿𝑠𝑡 + 𝑚𝑔𝑠𝑖𝑛𝜙

= −𝑘𝑥 + (𝑚𝑔𝑠𝑖𝑛𝜙 − 𝑘𝛿𝑠𝑡)

• because 𝑚𝑔𝑠𝑖𝑛𝜙 = 𝑘𝛿𝑠𝑡⟹𝑚 𝑥 = −𝑘𝑥

Force due to gravity is canceled out of the equation of motion

§2.Modeling of Mass-Spring Systems 4.Choosing the equilibrium position as coordinate reference

- Static deflection in a mass-spring system

- Advantages of choosing the equilibrium position as the coordinate origin

• Do not need to specify the geometric dimensions of the mass

• Simplify the equation of motion by eliminating static forces HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien System Dynamics 4.18 Spring & Damper Elements in Mechanical Systems

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- Example 4.2.1 Equation of Motion of a Two-Mass System

Derive the equations of motion of the two-mass system

shown in the figure

Solution Choose the coordinates𝑥1 and𝑥2 as the displacements of the masses from their equilibrium positions

FromNewton’s law, equation of motion

𝑚1 𝑥1= −𝑘1𝑥1− 𝑥2 + 𝑓

𝑚2 𝑥2= −𝑘1𝑥2− 𝑥1 − 𝑘2𝑥2 or

𝑚1 𝑥1+ 𝑘1𝑥1− 𝑥2 = 𝑓 𝑚2 𝑥2+ 𝑘1𝑥2− 𝑥1 + 𝑘2𝑥2= 0

These equations can be derived directly from original figure

§2.Modeling of Mass-Spring Systems 5.Solving the Equation of Motion

- The general equation of motion for mass-spring systems

𝑚 𝑥 + 𝑘𝑥 = 𝑓

- The solution in the form

𝑥 𝑡 = 𝑥(0)

𝜔𝑛 𝑠𝑖𝑛𝜔𝑛𝑡 + 𝑥 0 𝑐𝑜𝑠𝜔𝑛𝑡

𝜔𝑛= 𝑚𝑘: natural frequency or

𝑥 𝑡 = 𝐴𝑠𝑖𝑛 𝜔𝑛𝑡 + 𝜙 𝑠𝑖𝑛𝜙 =𝑥(0)

𝐴 , 𝐴 = [𝑥(0)]2+

𝑥(0)

𝜔𝑛 2

The vertical motion of the mass 𝑚 attached to the beam can be modeled as

a mass supported by a spring Assume that the beam mass is negligible compared to𝑚 so that the beam can be modeled as an ideal spring Determine the system’s natural frequency of oscillation Solution

The spring constant of fixed-end beam𝑘 = 16𝐸𝑤ℎ3/𝐿3

The mass𝑚 on the beam will oscillate with a frequency of

𝜔𝑛= 𝑘

𝑚=

16𝐸𝑤ℎ3

𝑚𝐿3 HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien

- Example 4.2.3 A Torsional Spring System

Consider a torsional system An inertia𝐼 is attached to a rod, whose torsional spring constant is𝑘𝑇 The angle of twist is𝜃 The inertia of the rod can be negligible so that the rod can be modeled as an ideal torsional spring Obtain the equation of motion in terms

of𝜃 and determine the natural frequency Solution

The rod is modeled as an ideal torsional spring, this system is conceptually identical to that shown in the figure

From the free body diagram: 𝐼 𝜃 = −𝑘𝑇𝜃 and the natural frequency: 𝜔𝑛= 𝑘𝑇/𝐼 HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien System Dynamics 4.22 Spring & Damper Elements in Mechanical Systems

6.Displace inputs and spring elements

- Consider the mass-spring system and its free body diagram

Motion equation

𝑚 𝑥(𝑡) + 𝑘𝑥(𝑡) = 𝑓(𝑡) 𝑓(𝑡): external force input

- Given the displacement 𝑦(𝑡) of the left-hand end

of the spring Motion equation

𝑚 𝑥 𝑡 + 𝑘1𝑥 𝑡 − 𝑦 𝑡 + 𝑘2𝑥(𝑡) = 0 𝑦(𝑡): external displacement input

- External displacement input can be generated from a cam

follower system

§2.Modeling of Mass-Spring Systems 7.Simple harmonic motion

- Consider thesimple harmonic motion

𝑚 𝑥 + 𝑘𝑥 = 0 the acceleration 𝑥(𝑡) = −𝑘𝑥(𝑡)/𝑚 = −𝜔𝑛2𝑥(𝑡), is proportional

to the displacement𝑥(𝑡) but opposite in sign

- The solution

𝑥 𝑡 = 𝐴𝑠𝑖𝑛 𝜔𝑛𝑡 + 𝜙

𝑥 𝑡 = 𝐴𝜔𝑛𝑐𝑜𝑠 𝜔𝑛𝑡 + 𝜙

= 𝐴𝜔𝑛𝑠𝑖𝑛 𝜔𝑛𝑡 + 𝜙 +𝜋

2

𝑥 𝑡 = −𝐴𝜔𝑛2𝑠𝑖𝑛 𝜔𝑛𝑡 + 𝜙

= 𝐴𝜔𝑛𝑠𝑖𝑛 𝜔𝑛𝑡 + 𝜙 + 𝜋

- Figure for the case of

𝑥 0 = 1, 𝑥 0 = 0, 𝜔𝑛= 2 HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien System Dynamics 4.24 Spring & Damper Elements in Mechanical Systems

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§3.Energy Methods

- The force exerted by a spring is a conservative force

- For a linear spring, resisting force is given by𝑓 = −𝑘𝑥 and

thus the potential energy of a linear spring is given by

𝑉 𝑥 =1

2𝑘𝑥2

𝑘: spring constant, 𝑁/𝑚

𝑥: the deflection from the free length of the spring, 𝑚

- For a torsional spring, resisting moment is given by𝑀 = 𝑘𝑇𝜃

The work done by this moment and stored as potential energy

in the spring is

𝑉 𝜃 =

0

𝜃 𝑀𝑑𝜃 = 0

𝜃 𝑘𝑇𝜃 𝑑𝜃 =1

2 𝑇𝜃 2

𝑘𝑇: spring constant,𝑁/𝑚

𝜃: the twist angle, 𝑟𝑎𝑑

§3.Energy Methods

- The conservation of energy principle

𝑇 + 𝑉 = 𝑇0+ 𝑉0= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

or Δ𝑇 + Δ𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑇: the system’s kinetic energy 𝑉: the system’s potential energy

- Consider the horizontal spring-mass system The system has kinetic and elastic potential energy

1

2𝑚 𝑥2+

1

2𝑘𝑥2=

1

2𝑚 𝑥0+1

2𝑚𝑥0= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 or

1

2𝑚 𝑥

2− 𝑥0 +1

2𝑘 𝑥

2− 𝑥0 = 0

§3.Energy Methods

- Consider the vertical spring-mass system The system has

kinetic, elastic potential, gravitational energy

- For this system, we must include the effect of

gravity, and thus the potential energy is the sum

of the spring’s potential energy 𝑉𝑠 and the

gravitational potential energy𝑉𝑔, which

𝑇 + 𝑉𝑔+ 𝑉𝑠= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

or Δ𝑇 + Δ𝑉 = Δ𝑇 + Δ𝑉𝑔+ Δ𝑉𝑠= 0

- The spring is at its free length when𝑦 = 0, so we can write

1

2𝑚 𝑦

2− 𝑚𝑔𝑦 +1

2𝑘𝑦

2= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

§3.Energy Methods

A spring system is used to isolate the foundation from the force of a falling object

Suppose the weight𝑊 is dropped from a height

ℎ above the platform Determine the maximum spring compression and the maximum force transmitted to the foundation The given values are𝑘1= 104𝑁/𝑚,𝑘2= 1.5 × 104𝑁/𝑚,𝑑 = 0.1𝑚, and ℎ = 0.5𝑚 Consider two cases: (a)𝑊 = 64𝑁 and (b) 𝑊 = 256𝑁 Solution

The velocity of the weight is zero initially and also when the maximum compression is attained,𝑇 = 0, and we have

Δ𝑇 + Δ𝑉 = Δ𝑇 + Δ𝑉𝑠+ Δ𝑉𝑔= 0

⟹Δ𝑉𝑠+ Δ𝑉𝑔= 0

§3.Energy Methods

The weight is dropped from a heightℎ above the platform and

if choosing the gravitational potential energy

reference at that height, then the maximum

spring compression𝑥 can be found by

1

2 1 𝑥

2− 0 + [0 − 𝑊 ℎ + 𝑥) = 0, 𝑥 < 𝑑

1

2 1𝑥

2− 0 +1

2 2𝑘2 𝑥 − 𝑑

2− 0 + [0 − 𝑊 ℎ + 𝑥 ], 𝑥 ≥ 𝑑

or

1

2 1𝑥

2− 𝑊ℎ − 𝑊𝑥 = 0, 𝑥 < 𝑑

𝑘1+ 2𝑘2𝑥2− 2𝑊 + 4𝑘2𝑑 𝑥 + 2𝑘2𝑑2− 2𝑊ℎ = 0, 𝑥 ≥ 𝑑

§3.Energy Methods

With the given values

104𝑥2− 2𝑊𝑥 − 𝑊 = 0, 𝑥 < 0.1 (1)

4 × 104𝑥2− 2𝑊 + 6000 𝑥 + 300 − 𝑊 = 0, (2)

𝑥 ≥ 0.1

a.𝑊 = 64𝑁

(1): 104𝑥2− 128𝑥 − 64 = 0

⟹ 0 < 𝑥 = 0.0867 < 0.1

⟹ 𝑥𝑚𝑎𝑥= 0.0867, 𝐹𝑚𝑎𝑥= 𝑘1𝑥𝑚𝑎𝑥= 104× 0.0867 = 867𝑁

b.𝑊 = 256𝑁

(1): 104𝑥2− 512𝑥 − 256 = 0 ⟹ 0 > 𝑥 = 0.1876 > 0.1 Because𝑥 > 0.1, using equation (2)

(2): 4 × 104𝑥2− 6512𝑥 + 44 = 0 ⟹ 0 > 𝑥 = 0.1557 > 0.1

⟹ 𝑥𝑚𝑎𝑥= 0.1557, 𝐹𝑚𝑎𝑥= 𝑘1𝑥𝑚𝑎𝑥+ 2𝑘2(𝑥𝑚𝑎𝑥− 0.1) = 3240𝑁 HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien System Dynamics 4.30 Spring & Damper Elements in Mechanical Systems

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§3.Energy Methods

1.Obtaining Motion Equation

- In mass-spring systems with negligible friction and damping,

we can often use the principle of conservation of energy to

obtain the equation of motion and, for simple harmonic

motion, to determine the frequency of vibration without

obtaining the equation of motion

- Example 4.3.2 Equation of Motion of a Mass-Spring System

Use the energy method to derive the equation of

motion of the mass𝑚 attached to a spring and

moving in the vertical direction

§3.Energy Methods

Solution Take the reference at 𝑥 = 0 , the total potential energy of the system (𝑘𝛿𝑠𝑡= 𝑚𝑔)

𝑉 = 𝑉𝑠+ 𝑉𝑔=1

2𝑘(𝑥 + 𝛿𝑠𝑡)2−𝑚𝑔𝑥

=1

2𝑘𝑥2+ 𝑘𝛿𝑠𝑡𝑥 +1

2𝑘𝛿𝑠𝑡2− 𝑚𝑔𝑥 =1

2𝑘𝑥2+

1

2𝑘𝛿𝑠𝑡2 The total energy of the system

𝑇 + 𝑉 =1

2𝑚 𝑥

2+1

2𝑘𝑥

2+1

2𝑘𝛿𝑠𝑡

2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

⟹𝑑 𝑇 + 𝑉

𝑑𝑡 = 0 ⟹ 𝑚 𝑥 𝑥 + 𝑘𝑥 𝑥 = 0 The equation of motion: 𝑚 𝑥 + 𝑘𝑥 = 0

§3.Energy Methods

2.Reileigh’s Method

- Based on the principle of conservation of energy

- In simple harmonic motion

• at the equilibrium position 𝑥 = 0

𝑇 = 𝑇𝑚𝑎𝑥

𝑉 = 𝑉𝑚𝑖𝑛

• when 𝑥 = 𝑥𝑚𝑎𝑥

𝑉 = 𝑉𝑚𝑎𝑥

𝑇 = 𝑇𝑚𝑖𝑛= 0

- From conservation of energy

𝑇𝑚𝑎𝑥+ 𝑉𝑚𝑖𝑛= 𝑇𝑚𝑖𝑛+ 𝑉𝑚𝑎𝑥= 0 + 𝑉𝑚𝑎𝑥

⟹𝑇𝑚𝑎𝑥= 𝑉𝑚𝑎𝑥− 𝑉𝑚𝑖𝑛

§3.Energy Methods

- Example for a mass-spring system

𝑇 = 𝑚 𝑥2/2

𝑉 = 𝑘 𝑥 + 𝛿𝑠𝑡2/2 − 𝑚𝑔𝑥 From conservation of energy𝑇𝑚𝑎𝑥= 𝑉𝑚𝑎𝑥− 𝑉𝑚𝑖𝑛

⟹1

2𝑚 𝑥𝑚𝑎𝑥

2 =1

2𝑘(𝑥𝑚𝑎𝑥+ 𝛿𝑠𝑡)

2−𝑚𝑔𝑥𝑚𝑎𝑥−1

2𝑘𝛿𝑠𝑡 2 with𝑘𝛿𝑠𝑡= 𝑚𝑔

⟹1

2𝑚 𝑥𝑚𝑎𝑥2 =1

2𝑘𝑥𝑚𝑎𝑥2

In simple harmonic motion

| 𝑥𝑚𝑎𝑥| = 𝜔𝑛|𝑥𝑚𝑎𝑥|

⟹1

2𝑚(𝜔𝑛|𝑥𝑚𝑎𝑥|)

2=1

2𝑘𝑥𝑚𝑎𝑥 2

⟹𝜔𝑛= 𝑘/𝑚

§3.Energy Methods

- Example 4.3.3 Natural Frequency of a Suspension System

Consider the suspension of one front wheel of a car:𝐿1= 0.4𝑚, 𝐿2= 0.6𝑚, spring constant𝑘 = 3.6 × 104𝑁/𝑚, and the car weight associated with that wheel is3500𝑁

Determine thesuspension’s natural frequency for vertical motion

Solution

Frame moving: 𝐴𝑓= 𝐶𝐶′

Spring deflection:𝐴𝑠= 𝐷𝐸 − 𝐷′𝐸′ ≈ 𝐸𝐸′

From∆𝐹𝐶𝐶′, approximately:

𝐴𝑠≈𝐿2− 𝐿1

𝐿2 𝐴𝑓=1

3𝐴𝑓 HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien

§3.Energy Methods

The change in potential energy (𝑘𝛿𝑠𝑡= 𝑚𝑔)

𝑉𝑚𝑎𝑥− 𝑉𝑚𝑖𝑛=12𝑘 𝐴𝑠+ 𝛿𝑠𝑡2− 𝑚𝑔𝐴𝑠−12𝑘𝛿𝑠𝑡2

=12𝑘𝐴𝑠=12𝑘 13𝐴𝑓2 The amplitude of the velocity of the mass in simple harmonic motion is𝜔𝑛𝐴𝑓, and thus the maximum kinetic energy 𝑇𝑚𝑎𝑥=12𝑚(𝜔𝑛𝐴𝑓) 2

FromRayleigh’s method, 𝑇𝑚𝑎𝑥= 𝑉𝑚𝑎𝑥− 𝑉𝑚𝑖𝑛, we obtain 1

2𝑚(𝜔𝑛𝐴𝑓)2=12𝑘 13𝐴𝑓2

⟹ 𝜔𝑛=1 3

𝑘

𝑚=

1 3

3.6 × 104 3500 9.8

= 3.345𝑟𝑎𝑑/𝑠

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§3.Energy Methods

3.Equivalent Mass of Elastic Elements

- In an elastic element is represented as in the figure, we

assume the mass of the element

• is negligible compared to the rest of the system’s

mass, or

• has been included in the mass attached to the

element

- In the second case, this included mass is called the

equivalent mass of the element and to be computed by using

kinetic energy equivalence

- Example 4.3.4 Equivalent Mass of a Rod

The rod shown in the figure acts like a spring when

an axially applied force stretches or compresses

the rod Determine the equivalent mass of the rod

§3.Energy Methods

Solution Mass of an infinitesimal element 𝑑𝑚𝑟= 𝜌𝑑𝑦 Kinetic energy of the element 𝑑𝐾𝐸 = 𝑑𝑚𝑟 𝑦2/2 Kinetic energy of the entire rod

𝐾𝐸 =1

2 0

𝐿

𝑦2𝑑𝑚𝑟=1

2 0

𝐿

𝑦2𝜌𝑑𝑦 Assume that the velocity 𝑦 of the element is linearly proportional to its distance from the support, then 𝑦 = 𝑥𝑦/𝐿

𝐾𝐸 =1

2 0

𝐿

𝑥𝑦 𝐿

2 𝜌𝑑𝑦 =1 2

𝜌 𝑥2

𝐿2 0

𝐿

𝑦2𝑑𝑦 =1 2

𝜌 𝑥2

𝐿2

𝑦3 3 0 𝐿

⟹ 𝐾𝐸 =1 2

𝜌 𝑥2

𝐿2

𝐿3

3=

1 2

𝜌𝐿

3 𝑥

2=1 2

𝑚𝑟

3 𝑥

2⟹𝑚𝑒=𝑚𝑟

3

Then 𝑚 = 𝑚𝑐+ 𝑚𝑒= 𝑚𝑐+ 𝑚𝑟/3 HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien System Dynamics 4.38 Spring & Damper Elements in Mechanical Systems

§3.Energy Methods

Equivalent masses of common elements–translational systems

§3.Energy Methods

Equivalent masses of common elements–translational systems

§3.Energy Methods

Equivalent inertias of common elements–rotational systems

§3.Energy Methods

Equivalent inertias of common elements–rotational systems

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§3.Energy Methods

- Example 4.3.5 Equivalent Mass of a Fixed-End Beam

A motor mounted on a beam with two fixed-end supports An imbalance in the motor’s rotating mass will produce a vertical force 𝑓 that oscillates at the same frequency as the motor’s rotational speed The

resulting beam motion can be excessive if the frequency is

near the natural frequency of the beam, and excessive beam

motion can eventually cause beam failure

Determine the natural frequency of the beam-motor system

§3.Energy Methods

Solution

Treating this system as if it were a single mass located at the middle of the beam results in the equivalent system, where 𝑥 is the displacement of the motor from its equilibrium position Equivalent mass of the system

𝑚𝑒= 𝑚𝑐+ 0.38𝑚𝑑 Equivalent spring constant of the beam

𝑘 =4𝐸𝑤ℎ 3

𝐿3 The system model 𝑚𝑒 𝑥 + 𝑘𝑥 = 𝑓 The natural frequency 𝜔𝑛= 𝑘

𝑚𝑒= 16𝐸𝑤ℎ3/𝐿3 𝑚𝑐+0.38𝑚𝑑 HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien System Dynamics 4.44 Spring & Damper Elements in Mechanical Systems

§3.Energy Methods

- Example 4.3.6 Torsional Vibration with Fixed Ends

An inertia𝐼1rigidly connected to two shafts, each with inertia𝐼2 The other ends of the shafts are rigidly attached to the supports

The applied torque is𝑇1

a Derive the equation of motion

b The cylinder𝐼1 is a cylinder𝜙100𝑚𝑚 × 75𝑚𝑚; the shafts

𝐼2are cylinders𝜙50𝑚𝑚 × 150𝑚𝑚 The three cylinders are

made of steel with a shear modulus𝐺 = 75𝐺𝑃𝑎 and a

density𝜌 = 7850𝑘𝑔/𝑚3

Calculate thesystem’s natural frequency

§3.Energy Methods

Solution

a Equation of motion The equivalent inertia

𝐼𝑒= 𝐼1+ 2 1

3𝐼2 The spring constant

𝑘 =𝐺𝜋𝐷 4 32𝐿 Equation of motion is derived from the free body diagram

𝐼𝑒 𝜃 = 𝑇1− 𝑘𝜃 − 𝑘𝜃 = 𝑇1− 2𝑘𝜃

§3.Energy Methods

b.System’s natural frequency

The value of𝑘

𝑘 =𝐺𝜋𝐷 4 32𝐿 =

75 × 103× 𝜋 × 504

32 × 150

= 3.078 × 105𝑁𝑚𝑚/𝑟𝑎𝑑 The moment of inertia of a cylinder of

𝐼 =12𝑚 𝐷2 2=12𝜋𝜌𝐿 𝐷2 4

𝐼1=1

2𝜋 7.85 × 10−675 504= 5.78 × 10−3𝑘𝑔𝑚2

𝐼2=1

2𝜋 7.85 × 10−6150 254= 0.73 × 10−3𝑘𝑔𝑚2

⟹ 𝐼𝑒= 5.78 × 10−3+ 2 1

30.73 × 10−3 = 6.67 × 10−3𝑘𝑔𝑚2 System’s natural frequency: 𝜔𝑛= 𝑘/𝐼𝑒= 6793𝑟𝑎𝑑/𝑠

§4.Damping Elements

- A spring element exerts a reaction force in response to a

displacement, either compression or extension, of the element

- A damping element is an element that resists relative velocity

across it

- A common example of a damping

element, or damper, is a cylinder

containing a fluid and a piston with one or more holes

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1.A Door Closer

An example from everyday life of a device that contains a

damping element as well as a spring element is the door closer

§4.Damping Elements 2.Shock Absorbers

- A typical shock absorber is very complex but the basic principle of its operation is the damper concept

- The damping resistance can be designed to be dependent

on the sign of the relative velocity

- If the two spring constants are different

or if the two valves have different shapes, then the flow resistance will be dependent on the direction of motion Oleo strut: an aircraft shock absorber

3.Ideal Dampers

- Real damping elements have mass, such as the masses of

the piston and the cylinder in a shock absorber

- If the system consists of an object attached to a damper, a

simplifying assumption is to neglect the damper mass

relative to the mass of the object and take the mass center of

the system to be located at the mass center of the object

- If the piston mass and cylinder mass are substantial, the

damper must be modeled as two masses, one for the piston

and one for the cylinder

- An ideal damping element is one that is massless

§4.Damping Elements 4.Damper Presentations

- The linear model for the damping force

𝑓 = 𝑐𝑣

𝑐: the damping coefficient, 𝑁𝑠/𝑚 𝑣: relative velocity, 𝑚/𝑠

- The damping coefficient of a piston-type damper with a single hole

𝑐 = 8𝜋𝜇𝐿 𝐷

𝑑

2

− 1 2

𝜇: the viscosity of the fluid, 𝑁𝑠/𝑚2 𝐿: the length of the hole through the piston, 𝑚 𝑑: the diameter of the hole, 𝑚

𝐷: the diameter of the piston, 𝑚

For two holes, multiply the result by 2

4.Damper Presentations

- Symbols for translational and torsional damper elements

- The linear model of a torsional damper

𝑇 = 𝑐𝑇𝜔

𝑐𝑇: the torsional damping coefficient,𝑁𝑚𝑠/𝑟𝑎𝑑

𝜔: the angular velocity, 𝑟𝑎𝑑/𝑠

𝑇: the torque, 𝑁𝑚

§4.Damping Elements 5.Modeling Mass-Damper Systems

- Example 4.5.1 Damped Motion on an Inclined Surface

Derive and solve the equation of motion of the block sliding on an inclined, lubricated surface

Assume that the damping force is linear For this application the damping coefficient𝑐 depends on the contact area of the block, the viscosity of the lubricating fluid, and the thickness of the fluid layer Solution

The equation of motion

𝑚 𝑣 = 𝑚𝑔𝑠𝑖𝑛𝜃 − 𝑐𝑣

⟹𝑣 𝑡 = 𝑣 0 −𝑚𝑔𝑠𝑖𝑛𝜃

−𝑐𝑚𝑡 +𝑚𝑔𝑠𝑖𝑛𝜃 𝑐

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- Example 4.5.2 A Wheel-Axle System with Bearing Damping

A wheel-axle system in which the axle is supported by two sets of bearings that produce damping Each bearing set has a torsional damping coefficient 𝑐𝑇 The torque𝑇 is supplied by a motor The force

𝐹 is the friction force due to the road surface Derive the equation of motion Solution

The damping torque from each bearing 𝑐𝑇𝜔

Inertia of the wheel and shafts 𝐼 = 𝐼𝑤+ 2𝐼𝑠

𝐼 𝜔 = 𝑇 − 𝑅𝐹 − 2𝑐𝑇𝜔

- In a journal bearing, the axle passes through an opening in a support A commonly used formula for the damping coefficient of such a bearing isPetrov’s law

𝑐𝑇=𝜋𝐷

3𝐿𝜇 4𝜖

𝐿: the length of the opening 𝐷: the axle diameter 𝜖: the thickness of the lubricating layer (the radial clearance between the axle and the support)

- Example 4.4.3 A Generic Mass-Spring-Damper System

The figure represents a generic

mass-spring-damper system with an external force𝑓 Derive

its equation of motion and determine its

characteristic equation

Solution

𝑚 𝑥 = −𝑐 𝑥 − 𝑘 𝑥 + 𝛿𝑠𝑡 + 𝑚𝑔 + 𝑓 with𝑘𝛿𝑠𝑡= 𝑚𝑔

𝑚 𝑥 = −𝑐 𝑥 − 𝑘𝑥 + 𝑓

⟹𝑚 𝑥 + 𝑐 𝑥 + 𝑘𝑥 = 𝑓

For the system shown in the figure, the mass is

𝑚 = 1 and the spring constant is 𝑘 = 16 Investigate the free response as we increase the damping for the four cases: 𝑐 = 0,4,8,10

Use the initial conditions𝑥(0) = 1 and 𝑥(0) = 0

Solution The characteristic equation

𝑚 𝑥 + 𝑐 𝑥 + 𝑘𝑥 = 𝑓 ⟹𝑠2+ 𝑐𝑠 + 16 = 0 The roots

𝑠 =−𝑐 ± 𝑐

2− 64 2 HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien System Dynamics 4.58 Spring & Damper Elements in Mechanical Systems

The solution

𝑐 = 0: 𝑥 𝑡 = 𝑐𝑜𝑠4𝑡

𝑐 = 4: 𝑥 𝑡 = 1.155𝑒−2𝑡sin( 12𝑡 + 1.047)

𝑐 = 8: 𝑥 𝑡 = (1 + 4𝑡)𝑒−4𝑡

𝑐 = 10: 𝑥 𝑡 =4

3𝑒−2𝑡−1

3𝑒−8𝑡

Derive the equations of motion of the two-mass system shown in the figure

Solution For the mass𝑚1 𝑚1 𝑥1+ 𝑐1 𝑥1+ 𝑐2 𝑥1− 𝑥2 + 𝑘1𝑥1

+𝑘2 𝑥1− 𝑥2 = 0 For the mass𝑚2

𝑚2 𝑥2+ 𝑐2 𝑥2− 𝑥1 + 𝑘2 𝑥2− 𝑥1 = 𝑓 or

𝑚1 𝑥1+ (𝑐1+𝑐2) 𝑥1− 𝑐2 𝑥2+ 𝑘1+ 𝑘2𝑥1− 𝑘2𝑥2= 0 𝑚2 𝑥2+ 𝑐2 𝑥2− 𝑐2 𝑥1+ 𝑘2𝑥2− 𝑘2𝑥1= 𝑓

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