Chapter W2 Mechanical Systems I - Elsevier.com

Lobontiu: System Dynamics for Engineering Students Website Chapter 2 1

Chapter W2 Mechanical Systems I

Introduction

This companion website chapter gives more details on a few topics that have been

introduced in the printed-book Chapter 2, namely:

• Equivalent stiffness for series and parallel spring connections

• Viscous damping coefficient for translatory and rotary motions

• Equivalent damping coefficients for linear-motion dampers connected in series

and in parallel

• Mechanical lever systems

• Gear shaft systems

• Pulley systems

• The principle of work and energy method applied to single-DOF conservative and

nonconservative systems

• Critically-damped and overdamped free response of single-DOF mechanical

systems

W2.1 Series and Parallel Combinations of Translatory Springs ( 2.1.2)

To vary the stiffness properties of a mechanical system, several springs can be

combined in series or in parallel, as sketched in Fig. W2.1 where it is assumed that the

serial and parallel translatory spring chains are clamped at one end.

Figure W2.1 Translatory spring combinations: (a) series; (b) parallel

(a) (b)

x1

k1

f k2

x2 x k1

f k2


For serial springs, the force is the same in each component, whereas the total deformation

is the sum of individual deformations; conversely, for parallel spring combinations, the

displacements are identical, whereas the sum of individual spring forces equals the

externally applied force.

The task is to obtain a system whose force-deformation equation is:

e eqf f k x= = (W2.1)

where keq = ks is the equivalent series stiffness and keq = kp is the equivalent parallel

stiffness. For the series connection, the total displacement x is expressed as:

1 21 2 1 2

1 1f fx x x fk k k k

= + = + = +

(W2.2)

Comparing Eqs. (W2.1) and (W2.2), it follows that the equivalent series stiffness is:

1 2

1 1 1

sk k k= + (W2.3)

When n springs instead of two are coupled in series, Eq. (W2.3) becomes

1 2

1 1 1 1...= + + +s nk k k k

(W2.4)

For the parallel connection sketched in Fig. W2.1(b), the total force can be expressed as:

( )1 2 1 2 1 2e ef f f k x k x k k x= + = + = + (W2.5)

Comparing Eqs. (W2.1) and (W2.5), it follows that:

1 2pk k k= + (W2.6)

For a parallel connection comprising n springs, the equivalent stiffness is

1 2 ...= + + +p nk k k k (W2.7)

Equations (W2.4) and (W2.7) are collected in Eqs. (2.18) in the printed book

W2.2 Translatory and Rotary Viscous Damping Coefficients ( 2.1.3)

W2.2.1 Translatory Viscous Damping Coefficient

Consider the translatory (linear-motion) system with viscous damping of Fig. W2.2. It

consists of a piston of diameter Di and length l inside a cylinder of inner diameter Do

filled with fluid of dynamic viscosity μ and mass density ρ.


Figure W2.2 Linear-motion damper

When the piston translates to the left with a velocity v, two categories of damping

forces act on it that are produced at the fluid-piston interface. One force is due to the

lateral friction between the cylindrical face and the surrounding fluid while the other

force is generated by the pressure difference between the front and rear cylinder planar

faces. Equation (2.26) in the printed book gives the damping force applied by a fluid to a

plate when the gap is g and the common (superposition) area is A as

,d lAf A v

gµτ= = (W2.8)

For this case the gap is

2

o iD Dg −= (W2.9)

while the contact area is approximated to

iA D lπ= (W2.10)

Equations (W2.9) and (W2.10) are substituted in Eq. (W2.8), which becomes:

,2 i

d lo i

D lf A vD Dπµτ= =−

(W2.11)

l

R (hydraulic resistance)

fixed cylinder

Di Do

piston

v

fluid


The frontal damping force can be found by first expressing the fluid mass flow rate

(more details on this topic are provided in Chapter 4) corresponding to the fluid motion to

the right through the fluid resistor of resistance R as:

mdq pdt R

∆= (W2.12)

where Δp is the pressure difference between the two fluid compartments lying in the front

and in the rear of the piston and qm is the mass flow rate. The quantity of fluid being

displaced within the cylinder is equal to the quantity of fluid moving backwards through

the resistance conduit. For an elementary distance Δx travelled horizontally by the piston

the corresponding mass flow rate is:

2

4i

mDdq dxπ ρ= (W2.13)

By combining Eqs. (W2.12) and (W2.13), and by also taking into account that dx/dt = v,

it follows that:

2

4m idq Dp R R v

dtπ ρ∆ = = (W2.14)

The frontal damping force can now be calculated as area times pressure difference:

2 2 4

, 4 16i i

d fD Df p Rvπ π ρ= ∆ = (W2.15)

The total damping force is found by adding up the two components of Eqs. (W2.11) and

(W2.15):

2 4

, ,2

16i i

d d l d f to i

D l Df f f v Rv c vD Dπµ π ρ= + = + =−

(W2.16)

where the translatory-motion damping coefficient is:

2 4 32 216 16

i i it i

o i o i

D l D Dlc R D RD D D Dπµ π πµρ π ρ

= + = + − −

(W2.17)

as given in Eq. (2.28) in the printed book.

W2.2.2 Rotary Viscous Damping Coefficient

Consider the piston-cylinder system of Fig. W2.2 is studied when the piston rotates

with an angular velocity ω inside the fluid-filled cylinder; it is also assumed the cylinder


does not translate inside the cylinder and therefore, does not change its axial position, as

indicated in Fig. W2.3. There are also no losses due to fluid-piston frontal interaction.

Figure W2.3 Rotary-motion damper

In this situation, since the linear tangential velocity of a point on the cylinder’s

surface is angular velocity times radius, the damping force which is produced at the

rotary piston-fluid interface is:

2

id

DAfgµ ω= × × (W2.18)

The interface area A of Eq. (W2.10) and g of Eq. (W2.9) are substituted in Eq. (W2.18),

which results in

22

2i i i

do i o i

D l D D lfD D D Dπµ πµω ω= × × = ×− −

(W2.19)

As a consequence, the damping torque is computed as:

( )

2 3

2 2 2i i i i

d do i o i

D D D l D lm fD D D Dπµ πµω ω= × = × × = ×

− − (W2.20)

According to its definition of Eq. (2.22), the rotary damping torque is:

d rm c ω= (W2.21)

Comparing Eqs. (W2.20) and (W2.21), it follows that the viscous damping coefficient in

rotary motion is:

( )

3

2i

ro i

D lcD Dπµ

=−

(W2.22)

fixed cylinder

rotary piston

fluid

l

ω

Di Do


as indicated in Eq. (2.28) in the printed book.

W2.3 Series and Parallel Combinations of Viscous Dampers ( 2.1.3)

Translatory viscous dampers can be combined in series, in parallel or in series-

parallel to change the damping coefficient provided by single dampers. Figure W2.4

shows two linear-motion dampers defined by the coefficients c1 and c2, which are

connected in series – Fig. W2.4(a) and in parallel – Fig. W2.4(b).

Figure W2.4 Translatory damper combinations: (a) series; (b) parallel

The objective of combining dampers in either series or parallel is to obtain an

equivalent system with a single damper whose force-velocity equation is:

d eqf f c x= = (W2.23)

where ceq = cs is the equivalent series damping coefficient and ceq = cp is the equivalent

parallel damping coefficient. For the series connection, the total velocity at the chain's

end in Fig. W2.4(a) can be expressed as the sum of individual dampers velocities taking

into account that the force is identical on each component:

1 21 2 1 2

1 1f fx x x fc c c c

= + = + = +

(W2.24)

Comparison of Eqs. (W2.23) and (W2.24) yields the equivalent series damping

coefficient:

1 2

1 1 1

sc c c= + (W2.25)

When n linear-motion dampers are connected in series, the equivalent damping

coefficient becomes

(a) (b)

c1 c2

x1 x2

f f

x c1

c2


1 2

1 1 1 1...= + + +s nc c c c

(W2.26)

For the parallel connection of the 2 dampers shown in Fig. W2.4(b), the total force

considers that individual dampers have identical velocities and the total damping force is

the sum of individual damping forces:

( )1 2 1 2 1 2d df f f c x c x c c x= + = + = + (W2.27)

Comparing Eqs. (W2.23) and (W2.27), it follows that the parallel-connection equivalent

damping coefficient is:

1 2pc c c= + (W2.28) When n linear-motion dampers are coupled in parallel, the equivalent damping

coefficient becomes:

1 2 ...= + + +p nc c c c (W2.29)

Equations (W2.26) and (W2.29) are collected in Eq. (2.31) in the printed book.

W2.4 Lever Systems ( 2.2.2)

Equations for calculating the mass and stiffness corresponding to point masses and springs

that are relocated on a lever are derived here. Also discussed in this section is the topic of serially

coupling several levers.

W2.4.1 Relocating Mass and Stiffness on Levers

Not always masses and/or springs are connected at the same point on a lever; in cases

where masses and/or springs are attached at various positions, it is useful to transfer these

lumped inertia and stiffness properties from their original positions to a mobile end of the

lever. The mass mA for instance is placed at point A on the lever of Fig. W2.5; it is of

interest to determine the mass mB, which is located at the free end B, and which is

dynamically equivalent to the original mass mA. The equivalence between the levers of

Fig. W2.5 requires that the two systems have identical displacement fields (the same

displacement zB for instance) and the same kinetic energy at all times. Considering the

rod is massless, the kinetic energy of the original lever system is:


( )212

= A AT m z (W2.30)

Taking into account Eq. (2.47) in the printed book, which is:

B B

A A

z lz l

= (W2.31)

the following relationship can be written between the velocities at A and B:

=

AA B

B

lz zl

(W2.32)

Figure W2.5 Equivalent lever-mass systems with: (a) original mass position; (b) displaced mass position

Substitution of Eq. (W2.32) into Eq. (W2.30) produces:

( )2

212

=

AA B

B

lT m zl

(W2.33)

On the other hand, the kinetic energy of the equivalent lever system with a mass mB

located at point B (and which is equal to the kinetic energy of the original system, as

mentioned) is:

( )212

= B BT m z (W2.34)

Equations (W2.33) and (W2.34) indicate that the equivalent mass is:

2

=

AB A

B

lm ml

(W2.35)

which is Eq. (2.50) in the printed book.

Stiffness properties can similarly be transferred between various locations on a lever

system, as indicated in Fig. W2.6.

A

lA

lB

zB O

B

zA mA

lB

zB O

B

mB

(a) (b)


Figure W2.6 Equivalent lever-spring systems with: (a) original spring position; (b) displaced spring position

The original and equivalent lever-spring systems of Fig. W2.6 need to possess the same

potential energy and their displacement fields need to be identical. Taking into account

Eq. (W2.31), the elastic potential energy of the original system is:

( ) ( )2

2 21 12 2

= =

A

A A A BB

lU k z k zl

(W2.36)

whereas the elastic potential energy of the equivalent system is:

( )212

= B BU k z (W2.37)

Comparison of Eqs. (W2.36) and (W2.37) results in:

2

=

AB A

B

lk kl

(W2.38)


W2.4.2 Serially-Coupled Levers

Levers can be coupled serially, as sketched in Fig. W2.7, to increase displacement

amplification and force reduction. The output displacement from the first stage, zB, is the

input to the second stage, whose output displacement is zD (displacements are not shown

in Fig. W2.7). The following stage amplifications (a1 and a2) can be calculated:

1

2

= = = =

B B

A A

D D

C C

z l az lz l az l

(W2.39)

A

lA

lB

zB O

B

zA

kA

B lB

zB O

kB

(a) (b)


Figure W2.7 Two-stage lever system

The push-rod BC is assumed rigid and, for small displacements, it moves vertically such

that zB = zC. It then follows that:

1 2

= × =

D B D

A A C

z z z a az z z

(W2.40)

which indicates that the motion at D is a1 times a2 larger than the motion at A.

Considering that the work done by fA is equal to the work done by the force fD, it follows

that:

1 2

1= =D A

A D

f zf z a a

(W2.41)

Equation (W2.41) indicates that the force fD is a1 times a2 smaller than the force at A, fA

in the quasi-static case.

W2.5 Gear Shaft Transmissions ( 2.2.2)

Rotary motion needs to be transmitted from one shaft to another in many mechanical

engineering applications and one modality to realize this is through gear trains. Part of a

geared wheel is shown in Fig. W2.8 (a). The median circle is the pitch circle and the

distance measured on this circle between two consecutive teeth is known as the circular

pitch and is denoted by p. The radius of the pitch circle is R. Two gears that mesh (or

engage) are schematically shown in Fig. W2.8 (b) where the corresponding pitch circles

are tangent.

lC

A

lA

lB

O1

B

fA

fD

lD

C

D O2


Figure W2.8 Toothed gears: (a) Main parameters of a toothed gear pair; (b) Meshing gears

The length of the pitch circle can be computed as:

2= =l R pNπ (W2.42)

where N represents the number of teeth. It is necessary that two engaging toothed gears

have the same circular pitch, and therefore, by using Eq. (W2.42) for the meshing gears 1

and 2 of Fig. W2.8 (b), it follows that:

1 1

2 2

=R NR N

(W2.43)

Since there is no slippage between the two pitch circles (the actual gear engaging is

equivalent to a pure rolling of the two pitch circles), the following is true:

1 1 2 2=R Rθ θ (W2.44)

which leads to:

1 2

2 1

=RR

θθ

(W2.45)

The same non slippage condition makes that the two engaging gears have the same

velocity at the contact point, namely:

1 1 2 2= =v R Rω ω (W2.46)

and this equation, combined with Eq. (W2.45), results in:

1 2 2

2 1 1

= =

RR

ω θω θ

(W2.47)

mt1, ω1

v

mt2, ω2

R1

R2

N1

N2

R1

R2

p

pitch circles

(b) (a)


Equation (W2.47) could have been obtained by taking the time derivative of Eq.

(W2.45). By applying the time derivative twice to Eq. (W2.45) yields:

1 2

2 1

=

RR

θθ

(W2.48)

Under ideal conditions, the power is fully transmitted from one gear to the other in

meshing. Assuming the power torque on gear 1 is mt1 and that a torque mt2 is transmitted

to gear 2 through meshing, the power equality requires that:

1 1 2 2=

t tm mθ θ (W2.49)

which results in:

22

1 1

=

t

t

mm

θθ

(W2.50)

Equations (W2.43), (W2.45), (W2.47), (W2.48) and (W2.50) can be combined into a

single equation, namely:

11 1 2 2 2

2 2 1 1 1 2

= = = = =

t

t

mN RN R m

θ θ θθ θ θ

(W2.51)


W2.6 Pulley Systems

Pulleys are utilized to change the direction and characteristics (displacement, velocity

and acceleration) of translatory motion by utilizing their rotation capability. Quasi-

statically, pulleys are similar to levers, as they can amplify (or reduce) the mechanical

displacement. Figure W2.9 for instance shows two discs or radii RA and RB, which are

assembled rigidly on the same shaft. One wire is fixed over the circumference of each

cylinder and two forces, fA and fB pull the wires in opposite directions.


Figure W2.9 Two-disc pulley

Considering the discs rotate by a small angle θ in the direction of zA, the two

displacements at the discs peripheries are:

=

=A A

B B

z Rz R

θθ

(W2.52)

which indicates that:

= =B B

A A

z R az R

(W2.53)

where a denotes the displacement amplification, very much similar to the function

realized by levers. Since the quasistatic works done by the two forces are equal, it follows

that:

1= =B A

A B

f zf z a

(W2.54)

which is, again, identical to the mechanical lever.

The similitude between levers and pulleys in the quasistatic domain is also valid for a

multi-stage (specifically, a two-stage) pulley system, as the one sketched in Fig. W2.10.

The following individual displacement amplification ratios can be formulated:

1

2

= = = =

B B

A A

D D

C C

z R az Rz R az R

(W2.55)

zA

fA

RA

RB

fB

zB

A B


Figure W2.10 Two-stage disc pulley system

By also taking into account that the periphery displacements along the wire that connects

the RC and RB discs are identical, it follows that:

1 2= × =D B D

A A C

z z z a az z z

(W2.56)

and therefore the displacement at the periphery of the disc RD is a1 times a2 larger than

the displacement at the periphery of disc RA. By equating the works done by the two

forces, fA and fD, it follows that:

1 2

1=D Af f

a a (W2.57)

which indicates that the force fD is a1 times a2 smaller than the force fA.

Figure W2.11 Equivalence between pulley and gear connections

Since pulleys are normally located on shafts, the inertia and stiffness of a given

pulley-shaft systems can be transferred to the connecting pulley-shaft system exactly the

D

zD

RC

RD

fD zA

fA

RA

RB

zC

B ZB

A

C

Pulley system Gear system


same as for gear trains, as long as the belt connecting the two adjacent pulleys functions

without slippage – Fig. W2.11 is an illustration of this similitude.

W2.7 The Principle of Work and Energy (the Energy Method)

Energy equations are given in the printed-book Chapter 2 for the mass (inertia),

springs and dampers. While being systematically introduced and applied in Chapter 3, the

energy method can also be utilized to model basic mechanical systems, such as the

single-DOF ones of Chapter 2. Presented here is the principle of work and energy (or

simply, the energy method) as applied to conservative and nonconservative systems.

W2.7.1 Conservative Systems

For a mechanical system with no external forcing acting on it and no losses (such as

those owing to damping), the total energy E (which comprises kinetic T, elastic potential

Ue and gravitational potential Ug energy fractions) is constant and the corresponding

systems are known as conservative systems. As a consequence, the time derivative of the

total energy is zero for a conservative system, namely:

( ) 0e gdE d T U Udt dt

= + + = (W2.58)

The generic Eq. (W2.58) can be used to derive the mathematical model of conservative

systems, as an alternative to Newton’s second law of motion.

Example W2.1 Derive the mathematical model and determine the natural frequency corresponding to the free

vibrations of the torsional dynamic vibration absorber sketched in Fig W2.12. Assume the rod

connecting the two masses is rigid and massless, and that the displacements are small, such that

the four springs’ directions remain unchanged. Also assume the motion takes place in a horizontal

plane such that gravitational effects can be neglected. Known are: m = 1kg, k = 100 N/m, l1 =

0.25 m, and l2 = 0.5 m.


Figure W2.12 Torsional dynamic vibration absorber

Solution: The total energy of this system is formed of kinetic and elastic potential energy contributions,

namely:

2 2 2 21 12 42 2 2e

kE T U mx y mx ky = + = + × = +

(W2.59)

because there are two identical masses and four identical springs. For small motions:

2

1

x ly l=

=

θθ

(W2.60)

Substituting Eqs. (W2.60) in Eq. (W2.59) changes the total energy to:

2 2 2 22 1E ml kl= +θ θ (W2.61)

The system is conservative and therefore, the time derivative of E is zero:

( )2 22 12 0dE ml kl

dt= + = θ θ θ (W2.62)

Equation (W2.62) is satisfied at all times only when the parenthesis is zero (because the rod

angular velocity cannot be zero always):

2 22 1 0ml kl+ =θ θ (W2.63)

Equation (W2.63) is the mathematical model of the torsional dynamic vibration absorber; its

natural frequency is

2

1 122 2

nkl l kml l m

= =ω (W2.64)

With the numerical parameters of this example, the natural frequency of Eq. (W2.65) is ωn= 5

rad/s.

θ

l1 l1

l2 l2

k/2

k/2

k/2

k/2

m

x

m

x

y

y


Example W2.2 Consider the mechanical system of Fig. W2.13 comprising a wheel of mass m1 and a spring

of stiffness k1. The wheel undergoes pure rolling on a fixed horizontal surface during the system’s

free vibrations. Determine the conditions k1 and m1 need to satisfy such that the natural frequency

of this system be identical to the one of the system studied in Example W2.1 and sketched in Fig.

W2.12.

Figure W2.13 Mechanical system with rolling wheel and spring

Solution:

The mechanical system is conservative and its total energy comprises the kinetic energy of

the wheel and the elastic potential energy stored in the spring. For the system moving to the left,

as shown in Fig. W2.13, the total energy is expressed as:

2 21

1 12 2e I CE T U J k x= + = +θ (W2.65)

The kinetic energy contribution is rotational because the wheel undergoes a pure rotation with

respect to the instant center I (the point of contact between wheel and the horizontal surface). The

wheel mass moment of inertia with respect to I is calculated by using the parallel axis theorem as

2 2 2 21 1 1 1

1 32 2I CJ J m R m R m R m R= + = + = (W2.66)

where JC = 1/2mR2 is the wheel mass moment of inertia with respect to its center C. The

horizontal motion coordinate xC of the center C is calculated considering the instantaneous

rotation of this point with respect to I as

Cx R= θ (W2.67)

Substituting JI and xC of Eqs. (W2.66) and (W2.67) into Eq. (W2.65) results in

2 2 2 2 2 2 21 1 1 1

1 3 1 3 12 2 2 4 2eE T U m R k R m k R = + = + = +

θ θ θ θ (W2.68)

The time derivative of the total energy is zero and therefore, the following equation results from

Eq. (W2.68) as

C

x

m1

I

k1

R θ


21 1

32

dE m k Rdt

= +

θ θ θ (W2.69)

Equation (W2.69) is valid at all times when

1 13 02

m k+ =θ θ (W2.70)

which yields the natural frequency

1

1

23n

km

=ω (W2.71)

This natural frequency is identical to the one of the vibration absorber – given in Eq. (W2.64) for:

1 1

2 1

23

l kkl m m

= (W2.72)

which yields

2

1 12

1 2

32

k l km l m

= × × (W2.73)

For the particular parameter values of Example W2.1, it follows from Eq. (W2.73) that k1/m1 =

37.5 rad/sec2. When a mass m1 = 1 kg is used for instance, the resulting stiffness is k1 = 37.5 N/m.

W2.7.2 Nonconservative Systems

A mechanical system which receives and/or loses energy is a nonconservative system.

For such a system, forces (or moments) other the inertial, spring or gravity act on the

respective system and which are known as nonconservative. As known from dynamics,

the principle of work and energy applies in these situations, according to which the

variation in the total energy (kinetic and potential) between two system states (positions)

is equal to the work done by the nonconservative forces between the same positions.

Mathematically, this is formulated as

12 1 2E W →∆ = (W2.74)

Applying the time derivative to Eq. (W2.74) results in

12 1 2d dE Wdt dt →∆ = (W2.75)


Equation (W2.75) can be utilized as an alternative to Newton’s second law of motion to

derive the mathematical model of nonconservative mechanical systems.

Example W2.3 Use the energy method to derive the mathematical model of the mass-spring system (spring of

stiffness is k) shown in Fig. W2.14 by studying the downward motion of the body of mass m. Consider

there is friction between the body and the incline, the coefficient of kinematic friction being μ.

Figure W2.14 Mass-spring single DOF mechanical system: (a) schematic representation; (b) two positions on

the incline

Solution:

Figure W2.14 (b) shows the original position of the body when the spring is undeformed, and

also indicates a generic position, underneath the original one, where the spring has a deformation

x and the body is situated at a distance h underneath the datum line, which passes through the

original position. The total energy variation of this mechanical system between the two points

shown in Fig. W2.14 (b) is:

( ) 2 21 1 sin2 2e gE T U U mx kx mgx α∆ = ∆ + + = + − (W2.76)

where the minus sign appears in front of the gravitational potential energy because the body is

underneath the datum line in its generic position; it was also considered that h = xsinα. The work

done by the friction force between the motion origin and the current point of abscissa x is

negative and is expressed as:

cosfW f x mgxµ α= − = − (W2.77)

According to the principle of work and energy

( ) ( )sin cosx mx kx mg x mgα µ α+ − = − (W2.78)

x

k

m

α

x

datum line

h

original position

generic (displaced) position

(a) (b)


which results in

sin cosmx mg mg kxα µ α= − − (W2.79)

Newton’s second law of motion is applied to the body of mass m, which is

t f emx f f f= − − (W2.80)

where ft (the tangential component of the gravity force), ff (the friction force), and fe (the spring

elastic force) are:

sincos

t

f

e

f mgf mgf kx

αµ α

= = =

(W2.81)

Substituting the forces of Eqs. (W2.81) in Eq. (W2.80) yields Eq. (W2.79), which was the

mathematical model for this mechanical system derived by the principle of work and energy.

W2.8 Free Damped Response – Critical Damping and Overdamping ( 2.2.2)

The free underdamped response of single DOF-systems has been studied in Chapter 2

of the printed book. Briefly discussed are here the underdamped response (0 < ξ < 1), the

critically-damped response (ξ = 1) and the overdamped response (ξ > 1). The differential

equation governing the free damped response of a single-DOF system is

0mx cx kx+ + = (W2.82)

which can be written in terms of the natural frequency ωn and the damping ratio ξ as

22 0n nx x xξω ω+ + = (W2.83)

Searching for solutions of the type x = eλt further transforms Eq. (W2.83) into

( )2 22 0tn n eλλ ξω λ ω+ + = (W2.84)

and therefore to

2 22 0n nλ ξω λ ω+ + = (W2.85)

which is the characteristic equation.

Underdamping refers to situations where 0 < ξ < 1 and the eigenvalues (the roots of

Eq. (W2.85)) in this case are complex. As shown in Appendix A, the general solution to

the homogeneous Eq. (W2.85) contains terms of the form:

( ) ( )cos ; sint te t e tσ σω ω (W2.86)


where σ is the real part of the complex root and ω is the imaginary part. The roots of Eq.

(W2.85) can be written in the form:

21,2 1= − ± −n n jλ ξω ξ ω (W2.87)

so that:

2; 1= − = −n nσ ξω ω ξ ω (W2.88)

and the general solution is of the form:

( ) ( )

( ) ( )1 2

2 21 2

( ) cos sin

cos 1 sin 1−

= +

= − + − n

t t

n n

x t c e t c e t

e c t c t

σ σ

ξω

ω ω

ξ ω ξ ω (W2.89)

An alternative form to Eq. (W2.89) is:

( )2( ) sin 1−= − +ntnx t Xe tξω ξ ω ϕ (W2.90)

For critical damping, the damping ratio is equal to 1, and therefore, the two roots of

the characteristic Eq. (W2.85) are identical, namely:

1 2 nλ λ λ ω= = = − (W2.91)

As explained in Appendix A, the general solution in this case is of the form:

( ) ( )1 2 1 2( ) nttx t c c t e c c t e ωλ −= + = + (W2.92)

Known initial displacement x0 and velocity v0 lead to the following constants:

1 0

2 0 0

= = + n

c xc v xξω

(W2.93)

and therefore Eq. (W2.92) becomes:

( )0 0 0( ) −= + + nt

nx t x v x t e ωξω (W2.94)

It is probably the right place to elaborate a bit on the damping ratio, which is actually

defined as the ratio of an actual damping coefficient c and the critical damping coefficient

ccr, namely:

=cr

cc

ξ (W2.95)

As mentioned here, the two eigenvalues are identical in critical damping, and one can

analyze the characteristic equation, which results from Eq. (W2.82):

2 0+ + =m c kλ λ (W2.96)


In order for Eq. (W2.96) to have identical roots, the discriminant needs to be zero, which

means that the critical damping coefficient is:

2 4 0 or 2− = =cr crc mk c mk (W2.97)

and, therefore:

n

cr

mc mkc mk mk

ξ ω ξ ξ= = = (W2.98)

In overdamping, the damping ratio is larger than 1 and therefore, the roots of the

characteristic equation λ1 and λ2 are real. As shown in Appendix A, the general solution

of the homogeneous Eq. (W2.83) is of the form:

( ) ( )2 2

1 21 1

1 2 1 2( )− + − − − −

= + = +n nt tt tx t c e c e c e c eξ ξ ω ξ ξ ωλ λ (W2.99)

Known initial displacement x0 and velocity v0 lead to the following constants:

( )

( )

20 0

1 2

20 0

2 2

1

2 1

1

2 1

n

n

n

n

v xc

v xc

ξ ξ ω

ξ ω

ξ ξ ω

ξ ω

+ − − = − − + + − = −

(W2.100)

Chapter W2 Mechanical Systems I - Elsevier.com

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Transcript of Chapter W2 Mechanical Systems I - Elsevier.com