tpages - USQ OpenDesk€¦ · 5 SYSTEM MODELING Preamble 5.1 Objectives 5.1 5.1 Stiffness elements...

MEC3403

Dynamics IIFaculty of Engineering and Surveying

S tudy book

Written by

SC FokBASc (Hons) Ottawa, PhD Monash,Faculty of Engineering and SurveyingUniversity of Southern Queensland

TABLE OFCONTENTS

PAGE

1 KINEMATICS

Preamble 1.1Objectives 1.11.1 Particle kinematics 1.11.2 Vector derivatives in rotation systems 1.51.3 Motion of a particle in a moving coordinate system 1.71.4 Rigid-body kinematics 1.91.5 Instantaneous center 1.131.6 Inertial frames, generalized coordinates and degrees of freedom 1.15Activity feedback 1.16

2 KINEMATICS ANALYSIS OF MECHANISMS

Preamble 2.1Objectives 2.12.1 Kinematic-chains 2.12.2 Kinematics analysis of Four-Bar Mechanisms 2.22.3 Equivalent linkages 2.92.4 Complex linkages 2.11Activity feedback 2.13

3 DYNAMICS ANALYSIS OF PLANAR MECHANISMS

Preamble 3.1Objectives 3.13.1 Principles of linear and angular momentums 3.13.2 Work and energy 3.43.3 Analysis of forces and torques in four-bar linkages 3.43.4 Balancing of rotating machinery 3.83.5 Balancing of reciprocating machines 3.10Activity feedback 3.13

4 RIGID BODY DYNAMICS

Preamble 4.1Objectives 4.14.1 Inertia elements 4.1Example 4.1 4.64.2 Euler’s equations of motion 4.11Example 4.2 4.134.3 Gyroscopic dynamics 4.17Example 4.3 4.18Activity feedback 4.21

5 SYSTEM MODELING

Preamble 5.1Objectives 5.15.1 Stiffness elements 5.15.2 Dissipation elements 5.55.3 Model construction 5.75.4 Force-balance and moment balance methods 5.8Example 5.1 5.95.5 Lagrange’s equations 5.12Example 5.2 5.14Activity feedback 5.18

6 VIBRATION OF SINGLE DEGREE OF FREEDOM SYSTEMS

Preamble 6.1Objectives 6.16.1 Natural frequency and damping ratio 6.26.2 Governing equations for different types of applied forces and their solutions

using Laplace transform 6.4Example 6.1 6.66.3 General solution of one degree of freedom systems 6.96.4 Free responses of one degree of freedom systems 6.126.5 Responses of one degree of freedom systems to single harmonic excitation 6.136.6 Examples of one degree of freedom systems to harmonic excitation 6.206.7 Responses of one degree of freedom systems to excitations with many

harmonic components 6.276.8 Response of one degree of freedom systems to impulse input 6.306.9 Response of one degree of freedom systems to step input 6.33Activity feedback 6.34

7 VIBRATION OF MULTI-DEGREE-OF-FREEDOM SYSTEMS

Preamble 7.1Objectives 7.17.1 General form of matrix equation for linear multi-degree of freedom systems 7.17.2 Free responses of un-damped systems: natural frequencies and mode

shapes 7.57.3 Properties of mode shapes 7.77.4 Characteristics of damped systems 7.97.5 Normal mode approach 7.117.6 State-space formulation 7.15Activity feedback 7.16

TutorialsTutorial 1a T.3Tutorial 1b T.4Tutorial 2a T.7Tutorial 2b T.9Tutorial 3a T.11Tutorial 3b T.12Tutorial 4a T.13Tutorial 4b T.15Tutorial 5a T.17Tutorial 5b T.19Tutorial 6a T.22Tutorial 6b T.23Tutorial 7a T.23Tutorial 7b T.23Tutorial 8a T.24Tutorial 8b T.24Tutorial 9a T.24Tutorial 9b T.24Tutorial 10a T.25Tutorial 10b T.25Tutorial 11a T.25Tutorial 11b T.25Tutorial 12a T.26Tutorial 12b T.26Tutorial 13a T.26

Selected Readings

Module 1 – Kinematics

Module 1

KINEMATICS 1

Module 1 – Kinematics 1.1

Preamble

This module reviews the basic principles of kinematics, with emphasis on the use of vectors to represent the position, velocity and acceleration. It is intended to give you an opportunity to revise the material learned in MEC2401 along with some essential mathematical skills (particularly in vector algebra such as cross product and the time derivative of vectors). You should try to understand these kinematics principles so that you can apply them for the analysis of 3D rigid bodies.

Objectives

On completion of this module you should be able to:

1. describe the linear position, velocity and acceleration of a particle in terms of vectors with respect to a fixed reference frame

2. differentiate a position (velocity) vector of a particle with respect to time to obtain the velocity (acceleration) vector in a rotating coordinate system

3. differentiate a position (velocity) vector of a particle with respect to time to obtain the velocity (acceleration) vector in a moving coordinate system

4. describe the translational motion of any point on a rotating rigid body in terms of the linear position, linear velocity and linear acceleration of another point on the body (such as the center of mass) and the absolute angular displacement, angular velocity and angular acceleration of the body

5. determine the instantaneous centers for bodies in planar motion

6. appreciate the concepts of inertial frame, generalized coordinates and degrees of freedom.

1.1 Particle kinematics

Reading

Text: Before you proceed, please read section 1.2.1 ‘Particle Kinematics’,pp. 4–5 of the textbook.

This section introduces the use of vectors to describe the linear position, linear velocity and linear acceleration of a particle in 3D space. To do this, we need a reference frame that is fixed in time. The reference frame can be represented using a set of unit vectors i, j, k corresponding

1.2 MEC3403 – Dynamics II

to the x, y and z-axis respectively (Unit vector means that the magnitude of the vector is 1 unit, i.e. ):

Given a vector , its magnitude is defined as

Example: For , the magnitude is

For convenience, the reference frame is normally a right-handed orthogonal system of coordinates (see figure 1.1). A right-handed orthogonal system of coordinates will satisfy the following relationships (note: i, j, and k are the unit vectors for the x, y, and z-axis respectively):

This simply means that the vectors i, j, and k are perpendicular to each other.

Figure 1.1: Right-handed orthogonal system of coordinates

i j k 1= = =

i1

0

0

= j0

1

0

= k0

0

1

=

a

ax

ay

az

= 2 2 2x y za a a a= + +

a1

2

3

= 2 2 21 2 3 14a = + + =

, ,i j k j k i k i j× = × = × =

0, 0, 0i j j k k i⋅ = ⋅ = ⋅ =


Note that the dot and cross products are defined as follow:

Given and , their dot product is a scalar:

where θ is the angle between the two vectors. Note that means that a and b are perpendicular (orthogonal). The cross product of a and b is a vector

The cross product of vectors a and b will provide a vector that is perpendicular to both a and b.

Activity 1.1

Objective: revision of algebra

Given and

• Find their dot product and the angle between them. Explain how you would use the dot product to determine if two vectors are perpendicular. What is the magnitude of vector a and how is the magnitude related to the dot product?

• Find their cross product and verify that the resulting vector is perpendicular to both a and b.

With a fixed reference frame defined, the absolute position of the particle at any time can be described by the vector r that starts from ‘O’ the origin of the reference frame and ends at the particle. A path can be used to describe the changes in position of the particle with time (see figure 1.2).

a

ax

ay

az

= b

bx

by

bz

=

cos( )x x y y z za b a b a b a b a b θ⋅ = + + =

a b⋅ 0=

dety z z y

x y z z x x z

x y z x y y x

i j k a b a b

a b a a a a b a b

b b b a b a b

− × = = − −

sin( )a b a b θ× =

1

3

5

a

=

2

4

6

b

=


Figure 1.2: Path of a particle

The absolute position of the particle along the path at any instance of time can be represented by

where , and are functions of time. For example

As the particle travels along the path, its velocity and acceleration can change. With the reference frame fixed in time, the absolute linear velocity of the particle is a vector defined as

i

j

k

At time t1 At time t2

Path of the particler

O

p

p

p

x

r y

z

=

xp yp zp

2

3

p

p

p

x t

r y t

z t

= =

p

pp

p

p

p

dx

dtx

dydrv y

dt dtz

dz

dt

= = =


Similarly, the absolute linear acceleration of the particle with respect to a fixed reference frame is also a vector defined as

For example, given ; i.e. ; ; and .

The absolute linear velocity and acceleration are respectively

and

Activity 1.2

Objective 1

Find the linear velocity and acceleration for the following particle motion

1.2 Vector derivatives in rotation systems

Reading

Selected reading 1.1: Before you proceed, please read Greenwood, ‘2-8 Vector derivatives in rotating systems’, pp. 47–9.

2

2

2

2

2

2

p

pp

p

p

p

d x

dtx

d ydva y

dt dtz

d z

dt

= = =

2

3

t

r t

t

=

px t= 2py t=

3pz t=

2

1

2

3

p

p

p

x

v y t

z t

= =

0

2

6

p

p

p

x

a y

z t

= =

2 3

32

Tt t

r t

=


The previous section assumed that the vector r is referenced to a frame a=i, j, k that is fixed in time. The differentiation of the vector r with respect to time based on the fixed reference frame is simply . This section deals with the situation where the vector r is referenced to a frame b = (e1, e2, e2) that is rotating at angular velocity ω relative to the reference frame a = (i, j, k) that is fixed in time (refer to figure 2-15 in the selected reading). In this case, the differentiation of the vector r is more complicated and is given by:

In other words, the absolute velocity v obtained from the differentiation of the vector r in a frame b rotating with absolute angular velocity vector ω relative to a frame a fixed in time is

= the rate of change of the vector r with respect to the rotating frame b. The component is the linear velocity component generated as a result of the rotating frame. Note the

similarity of the component with the linear velocity due to rotation in 2D. Figure 1.3 shows a 2D body rotating with angular velocity ω radians per second. The linear velocity v of point ‘P’ is given by ω multiplied by r:

where the direction of v is perpendicular to r.

Figure 1.3: Linear velocity due to rotation in 2D

In the 3D case, the linear velocity due to rotation corresponding to figure 1.3 has the form

Notice that the multiplication is replaced by the cross product. The other difference is that in the 3D case: v, ω and r are 3 by 1 vectors. The velocity v is perpendicular to both ω and r.

r·

( )r

drr r

dtω= + ×

( )rv r rω= + ×

r·( )rω r×

ω r×

( )v rω=

v

rω

P

( )v rω= ×


The direction of angular velocity vector ω is governed by the right-hand rule as shown in figure 1.4: the four fingers of the right-hand curl in the direction of positive rate of rotation and the thumb will point towards the positive direction for vector ω.

Figure 1.4: Right-hand rule for finding the direction of vector ω

Activity 1.3

Objective 2

The position of a particle r is described with respect to frame b= (e1, e2, e3), which is rotating at absolute angular velocity ω about fixed frame a= (x, y, z). Find the absolute linear velocity of the article at time t = 1 given

and

1.3 Motion of a particle in a moving coordinate system

Reading

Selected reading 1.2: Before you proceed, please read Greenwood, ‘2-9 Motion of a particle in a moving coordinate system’, pp. 49–51.

The previous section only deals with a coordinate system that is rotating with respect to a reference frame that is fixed in time. This section deals with the absolute velocity and acceleration of a particle in a moving coordinate system involving both translation and rotation motion relative to a reference frame that is fixed in time.

Direction of positive angular rotation

Positive direction of vector ω

2

2

t

r t

t

=

0

0

2

ω =


The results from the selected reading can be summarized as follow:

• If r is the position vector referenced to a moving coordinate system involving both translation and rotation motion relative to a reference frame that is fixed in time, the absolute velocity v of the particle is given by

= absolute linear velocity of the origin of the moving coordinate system due to translation with respect to the fixed reference frame, = the rate of change of the particle position with respect to the moving coordinate system, ω = absolute angular velocity vector of the moving coordinate system rotating with respect to the fixed reference frame, and r = the position vector of the particle relative to the origin of the moving coordinate system.

• The absolute acceleration a of the particle is

= absolute linear acceleration of the origin of the moving coordinate system due to translation with respect to the fixed reference frame, = the rate of change of the particle linear velocity with respect to the moving coordinate system, and = absolute angular acceleration vector of the moving coordinate system rotating with respect to the fixed reference frame.

Again, note the similarity of the definitions of some of the acceleration terms between 2D and 3D. Figure 1.5 shows a 2D body rotating with angular velocity ω rad/s and angular acceleration α rad/s2. The linear acceleration of point ‘P’ due to the angular acceleration has two components:

• A tangential component αt in the direction of motion given by α multiplied by r:

• A centripetal component αr towards the centre of rotation given by ω2 multiplied by r:

In the 3D case, the tangential component is and centripetal component is . The multiplication is replaced by the cross product and v, ω, α and r are 3 by 1 vectors. The direction of vector α also follows the right-hand rule of figure 1.4. The term with the form

is called the Coriolis acceleration. Note that this term is zero if

( )rv R r rω= + + ×

R·

r·( )r

( ) ( ) 2 ( )r ra R r r r rα ω ω ω= + + × + × × + ×

R( )rr

α ω=

( )ta rα=

2 ( )ra rω=

α r× ω ω× r( )×

2 ( )rrω × ( ) 0.rr =


Figure 1.5: Linear acceleration due to rotation in 2D

Activity 1.4

Objective B

Do problem 1.4 on page 16 of textbook.

1.4 Rigid-body kinematics

Reading

Text: Before you proceed, please read section 1.2.1 ‘Planar rigid-body kinematics’, pp. 5–10 of the textbook.

The previous sections deal with the position, velocity, and acceleration of a particle described with respect to fixed and moving coordinate systems. This section extends these concepts to rigid bodies. The main difference between a particle and a rigid body is that for a particle, the size and shape are not important. A particle is treated as a point mass in 3D space. A rigid body can be viewed as consisting of infinite number of particles. For a rigid body, the size and shape have to be considered (rigid means that the body cannot be deformed). It is important to consider the orientation of the rigid body because of its shape (the orientation of a particle is not an issue). Figure 1.6 illustrates this point.

Figure 1.6 shows that the absolute position of a rigid body can be described by the vector that starts from ‘O’ the origin of the reference frame (that is fixed in time) and ends at

‘G’, the center of mass of the body. The concept of the center of mass will be discussed later. In both figures 1.6(a) and (b), the center of mass ‘G’ of the rigid body with respect to ‘O’ (origin of frame x, y, z) has the same where

at

rω, α

P

ar

/G Or

/G Or


Figure 1.6: A rigid body with similar position but different orientations

In figures 1.6(a) and (b), the velocity and acceleration of ‘G’ the center of mass of the rigid body with respect to ‘O’ (i.e. origin of fixed frame (x, y, z)) are the same and are defined respectively as

and

In figures 1.6(a) and (b), the positions of ‘P’ are different. Obviously, position of the center of mass is not enough to describe the motion of point ‘P’ on the rigid body. The absolute positions of point ‘P’ relative to point ‘G’ in figures 1.6(a) and (b) are different due to the orientation of the body.

To define the orientation of the body, we will use a set of reference frame b represented by unit vectors e1, e2, and e3 (See figure 1.6 where e3 is not shown but is orthogonal to e1 and e2). Note that frame b = (e1, e2, e3) is fixed to the body and rotates with respect to framea = (i, j, k) as the rigid body orientation changes with time. In figure 1.6, for simplicity assume that the body is only rotating about the z-axis (i.e. with angular displacement ). Try to visualize this using figure 1.6. For general rigid body orientation, there can be angular displacements of , , and of frame b about the x, y, and z-axis of frame a respectively.

/G

G OG

G

x

r y

z

=

(a) (b)

i i

j j

k k

G G e1

e1

e2e2

O O

rG/O

P

P

rG/O

vG O⁄x· G

y·G

z·G

= aG O⁄x··G

y··Gz··G

=

θz

θx θy θz


Let the absolute angular velocity of the rigid body be denoted by a vector . Note that

all points on the rigid body rotate with the same angular velocity. Let the angular acceleration

of the rigid body be denoted by a vector . Note that all points on the rigid body will

have the same angular acceleration.

The next activity helps you in the revision on the use of vectors for the description of the position of a rigid body.

Activity 1.5

Objective 4

Find the vectors r1, r2 and r3 to describe the mechanism in figure 1.7 for . (O1B = 4 mm, BC = 18mm, O3C = 11mm, O1O3 = 10mm).

Figure 1.7: Mechanism for activity 1.5

In figure 1.6, the relative position ‘P’ on the rigid body with respect to ‘G’ (the center of mass) is defined as . Note that this vector is described in frame b, which is rotating at ω with respect to fixed frame a. Hence, the relative linear velocity of point ‘P’ with respect to ‘G’ in figure 1.6 is:

ωθ· x

θ· y

θ· z

=

αθ··x

θ·· y

θ·· z

=

θ1 120°=

Link 1

Link 2

Link 3

r1

r2

r3

r0

B

C

O1 O3

y

x

z

θ2

θ3

θ1

rP G⁄

vP G⁄ ω rP G⁄×=


ω = angular velocity of the rigid body. The above simply implies that the velocity of ‘P’ with respect to ‘G’ is due to the rotation at angular velocity ω. There is no translation of ‘P’ with respect to ‘G’ (i.e. distance is fixed and hence ; otherwise the body is not rigid). Note that different points on the rigid body will have different relative linear velocity although they share the same angular velocity ω.

The absolute position of ‘P’ in figure 1.6 can be described as

Note that is described in fixed frame a. The differentiation of the position vector would give the absolute velocity of point ‘P’:

The differentiation is based on the motion of a particle in a moving coordinate system with and (refer to previous section). The absolute acceleration of point ‘P’ is

obtained by differentiation of the velocity vector:

The differentiation is again based on the motion of a particle in a moving coordinate system with , and (refer to previous section).

Using the concept of , the rates of change of the unit vectors e1, e2, and e3 are:

To summarise, the velocity and acceleration of point ‘P’ in figure 1.6 are respectively:

rP G⁄ r·( )r 0=

rP O⁄ rG O⁄ rP G⁄+=

rG O⁄

vP O⁄ vG O⁄ vP G⁄+ vG O⁄ ω rP G⁄×+= =

r·( )r 0= R· vG O⁄=

aP O⁄ aG O⁄ aP G⁄+ aG O⁄ α rP G⁄×+= = ω ω× rP G⁄×+

r·( )r 0= r··( )r 0= R·· aG O⁄=

v ω r×=

de1

dt-------- ω e1×=

de2

dt-------- ω e2×=

de3

dt-------- ω e3×=

vP O⁄ vG O⁄ ω rP G⁄×+=

aP O⁄ aG O⁄ α rP G⁄ ω vP G⁄×+×+=


Figure 1.8: Vector representation of the absolute position of P

We do not necessary have to use points ‘P’ and ‘G’ in the above equations (see figure 1.8). The two equations are valid for any 2 points on the rotating rigid body (examine examples 1.1, 1.2 and 1.3 in the textbook). Note the similarity of the above two equations with that based on the motion of a particle in a moving coordinate system. As long as and , the equations are valid. Otherwise, the velocity and acceleration terms corresponding to and

have to be taken into consideration. You should realise that the goal of all the above mathematics is to obtain the absolute velocity (acceleration) of a point on the rigid body given the velocity (acceleration) of another point on the body along with the angular velocity (and angular acceleration) of the rotating body.

Activity 1.6

Objective 4


1.5 Instantaneous center

Reading

Selected reading 1.3: Before you proceed, please read Greenwood, ‘2-10 Plane motion’, pp. 52–55.

Any two bodies having motion relative to one another have an instantaneous center. An instantaneous center is an instantaneous point common to the two bodies that has the same velocity in each (i.e. the velocity at the instantaneous center is zero). This can also be viewed as a point on one body about which the other body is instantaneously rotating.

x

y

z

O

G

P

rG/O

rP/O

rP/G

‘P’ and ‘G’ are any 2 points on the rigid body

ω, α

r·( )r 0= r··( )r 0=r·( )r

r··( )r


The following are some commonly encountered instantaneous centers:

• Given a rigid body where the directions of the linear velocities of two points in the body are known (figure 1.9). The instantaneous center is the point of intersection of the lines perpendicular to the velocities. In figure 1.9, the velocities VA and VB are known at points ‘A’ and ‘B’ respectively. The dashed lines are perpendiculars to these velocities. The intersection of the dashed lines is the instantaneous center where

Figure 1.9: Instantaneous center for a body with known velocities at two points

• The instantaneous center for a body, which rolls without slipping on another body, is the point of contact between the two bodies.

Activity 1.7

Objective 5


VA ω CA( )× VB ω CB( )×= = =

A B VA

VB

Instantaneous center C

ω


1.6 Inertial frames, generalized coordinates and degrees of freedom

Reading

Text: Before you proceed, please read section 1.2.2 ‘Generalized coordinates and degrees of freedom’, pp. 10–12 of the textbook.

A frame of reference is not an inertial frame if a body not acted upon by outside influences can accelerate on its own accord. Measurements of time, space and mass made with respect to the inertial frame is absolute and interactions are instantaneous. These conditions are valid only if the velocities involved are much smaller than the speed of light. For most machines operating on the earth’s surface, these assumptions are valid.

The configuration of a given system can be expressed in terms of various sets of coordinates. For example, one can describe the position of a particle in 3D space using the Cartesian coordinates (x, y, z). Alternatively, one can also describe the same particle position using cylindrical coordinates (r, θ, z) or spherical coordinates (ρ, θ, φ). Generalized coordinates are any minimum sets of coordinates that can be used to describe a system.

The minimum number of independent coordinates needed to describe the motion of a system is called the degrees of freedom of a system.

Activity 1.8

Objective 6



Activity feedback

Activity 1.1

Given and ,

Two vectors are perpendicular if their dot product is zero.

Hence the cross product is perpendicular to a and b.

Activity 1.2

1

3

5

a

=

2

4

6

b

=

( ) 1(2) 3(4) 5(6) 44a b⋅ = + + =

2 2 2 2 2 21 3 5 35x y za a a a a a= ⋅ = + + = + + =

2 2 2 2 2 22 4 6 56x y zb b b b b b= ⋅ = + + = + + =

1 44cos

35 56θ − =

2

( ) (3(6) 5(4)) (5(2) 1(6)) (1(4) 3(2)) 4

2

a b i j k

− × = − + − + − = −

( ) 2 12 10 0

( ) 4 16 12 0

a b a

a b b

× ⋅ = − + − =× ⋅ = − + − =

2 3

32

Tt t

r t

=

222 3 2

1 132 2

T Tt t t

v t = =

20 2

2

T

a t =


Activity 1.3

. At time t = 1:

. At time t = 1:

Activity 1.4

From example 1.3 in text:

The angular velocity can be expressed as a vector (i.e. rotating in the z or

e3 axis). Using the concept of :

2

2

t

r t

t

=

1

2

1

r

=

( )1

2

2r

r

t

=

( )1

2

2r

r

=

( )rv r rω= + ×

1 1 1 0 1 3

2 2 2 0 2 4

2 1 2 2 1 2

v ω−

= + × = + × =

( ) ( )1 2p p p pv x y e y x eω ω= − + +

( ) ( )1 2p p p p

da x y e y x e

dtω ω = − + +

( ) ( ) ( ) ( )1 21 2p p p p p p p p p p

de dea x y y e x y y x x e y x

dt dtω α ω ω α ω= − − + − + + + + +

[ ]0 0Tω

i ie eω= ×

1 21 2 2 1

0 0 1 0 0 0

0 0 0 0 0 1

0 0

de dee e e e

dt dtω ω

ω ω ω ω

= × = × = = × = × = −

( ) ( )2 21 22 2p p p p p p p pa x y x y e y x y x eω ω α ω ω α= − − − + + − +


Activity 1.5

Notice that you are given the following: and , ,

and . The vector r1 is given by the following:

mm

To get the other vectors, we join B and O3 and use the cosine rule to find the length of l:

mm

But radians

radians

radians

mm

mm

θ1 120°= 1 4r = 2 18r =3 11r =

0 10r =

1 1

1 1 1

cos( ) 2

sin( ) 3.4641

0 0

r

r r

θθ

− = =

Link 1

Link 2

Link 3 r1

r2

r3

r0

BC

O1 O3

y

x

z

θ2

θ3

θ1

l

φ1

φ2

Hφ1+φ2

2 220 1 0 1 12 cos( ) 156 12.49l r r r r lθ= + − = → =

1 1 1 1sin( ) sin( ) 0.281l rφ θ φ= → =

2 222 3 3 2 22 cos( ) 1.7427r l r l r φ φ= + − → =

3 1 22 4.2595θ π φ φ= − − =

3 3

3 3 3

cos( ) 4.8136

sin( ) 9.8909

0 0

r

r r

θθ

− = = −

3 3 1 1sin( ) sin( ) 6.4268H r rθ θ= − =


radians

mm

Activity 1.6

We will assign the frame of reference attached to the body as shown

The position vector of the mass with respect to fixed point ‘O’ is

The absolute velocity of the slider is then given by

The absolute acceleration of the mass is

2 22

sin( ) 0.3651H

rθ θ= → =

2 2

2 2 2

cos( ) 16.8136

sin( ) 6.4268

0 0

r

r r

θθ

= =

O

e1e2

θ

r1

m

r r1e1=

( )1 1 1 1 1 1 1 1v re re re r eω= + = + ×

1 1 1 1 1 1 2

0 1

0 0

0

v re r re r eθθ

= + × = +

1 1 1 1 1 2 1 2 1 2a re re r e r e r eθ θ θ= + + + +

( ) ( )1 1 1 1 1 2 1 2 1 2a r e r e r e r e r eω θ θ θ ω= + × + + + ×

1 1 1 1 2 1 2 1

0 1 0 0

0 0 0 1

0 0

a re r r e r e rθ θ θθ θ

= + × + + + ×


Activity 1.7

The point of contact is an instantaneous center. From example 1.2 in the textbook:

Hence, the acceleration at the instantaneous center is not zero.

Activity 1.8

(a) There are three coordinates and one constraint (the length L is constant). Therefore the number of degrees of freedom is 2.

(b) Each particle needs 3 coordinates. For 3 particles, the total is 9 coordinates. There are 3 constraints (the three distances between the particles are fixed). Therefore, the number of degrees of freedom is 6.

(c) This is a rigid body. There are 6 degrees of freedom: 3 translation and 3 rotational.

( ) ( )1 1 1 2 1 2 1 2 1 1r e r e r e r e r eθ θ θ θ θ= + + + −

( ) ( )21 1 1 1 1 22a r r e r r eθ θ θ= − + +

( )/ / / /C O G O C G C Ga a r rω ω α= + × × + ×

/ /

0 0 0

, 0 , 0 , 0

0 0

C G G O

r

r r a

θω α

θ θ

= − = = = − −

/ 2

0

0

C Oa rθ =

Module 2 – Kinematics analysis of mechanisms

Module 2

KINEMATICS ANALYSIS OF MECHANISMS 2

Module 2 – Kinematics analysis of mechanisms 2.1

Preamble

This module applies the basic principles (discussed in module 1) to analyze the kinematics of practical mechanisms. Developing programs in MATLAB can facilitate the solving of some of these applications. You should try to understand how the basic kinematics principles are applied in the analysis of the mechanisms.

Objectives


1. define the following terms: kinematic-pair, open and closed kinematic-chains

2. analyse the kinematics of mechanisms such as four-bar linkage, slider-crank linkage, etc.

3. analyse direct-contact mechanisms using equivalent four-bar linkages

4. apply the concepts to analyse complex linkages

5. appreciate the use of computer programs in the kinematics analysis of mechanisms and develop simple Matlab programs to assist in the kinematics analysis of mechanisms.

2.1 Kinematic-chains

Reading

Selected reading 2.1: Before you proceed, please read Wilson & Sadler ‘1.4 Terminology and definitions’, pp. 7–14.

A kinematic-pair is the coupling or joint between 2 rigid bodies that constraints their relative motion. The kinematic-pair can be classified according to the contact between the jointed bodies:

• Lower kinematic-pairs: there is surface contact between the jointed bodies

• Higher kinematic-pairs: the contact is localized to lines, curves, or points

A kinematic-chain is a system of rigid bodies that are joined together by kinematic-joints to permit the bodies to move relative to one another. Kinematic-chains can be classified as:

• Open kinematic-chain: There are bodies in the chain with only one associated kinematic-joint


• Closed kinematic-chain: Each body in the chain has at least two associated kinematic-joints

A mechanism is a closed kinematic-chain with one of the bodies fixed (designated as the base). An industrial robot is normally an open kinematic-chain. In a structure, there can be no motion of the bodies relative to one another. A linkage is an assemblage of rigid bodies connected by kinematic-joints.

Activity 2.1

Objective 1

Explain the difference between open and closed kinematic-chains.

2.2 Kinematics analysis of Four-Bar Mechanisms

Reading

Selected reading 2.1: Before you proceed, please read Wilson & Sadler ‘3.4 Application of analytical vector and matrix methods to linkages’, pp. 178–82 and ‘4.2 Analysis of a four-bar linkage by analytical vector methods’, pp. 266–70.

θ1, ω1, α1

Link 1

Link 2

Link 3 r3B

C

O1 O2

θ2

θ3

D


Figure 2.1: A four-bar linkage

A four-bar linkage is used to describe the application of the basic concepts discussed in module 1. Figure 2.1 shows a four-bar linkage, which is also called the crank-and-rocker mechanism. Link 1, the crank drives the system and can rotate in a full circle. Link 3, the rocker, can only oscillate. Link 2 is called the coupler or connecting rod. The position problem generally involves finding the position vectors of the links when the dimensions of all members are given together with the crank angle θ1. This has been covered in module 1 where the positions of the links are expressed as vectors, i.e. r0, r1, r2, r3. With the known positions of the linkages, the kinematics problem generally involves finding the linear (angular) velocity and acceleration of any point in the mechanism given the absolute angular velocity and acceleration of link 1, i.e. ω1, and α1 are given (For most engineering applications, the mechanism is driven such that α1 = 0).

The approach to solve the kinematics problem involves finding the absolute angular velocities and accelerations of links 2 and 3, i.e. find ω2, ω3, α2 and α3. Using the positions, angular velocities and angular accelerations, we can then determine the linear velocity and acceleration of any point (example point ‘D’ in figure 2.1) in the mechanism.

To find the angular velocities of the coupler and the rocker, you must first write the constrained position equation (in terms of position vectors) for the mechanism. The four-bar linkage can be described by the position vector polygon:

The position vector polygon can then be differentiated with respect to time to get the velocity vector equation:

y

x

z

θ1, ω1, α1

r1

r2

r3

r0

rD/B

B

D

C

O1 O2

θ2

θ3

0 1 2 3 0r r r r+ + + =

0 1 2 3 0v v v v+ + + =


The velocity terms in the above equation can be found based on the concepts discussed in module 1. For example:

• as the points O1 and O2 are fixed (see figure 2.1)

• = velocity of point ‘B’ (Only rotation is involved and ω1 is the absolute

angular velocity of link 1)

• (Only rotation is involved and ω2 is the absolute angular velocity of link 2)


Note that ω1 is the driving velocity for link 1, the crank. The other angular velocities ω2 and ω3 are generated as a result of rotating the crank at ω1. All the links can only rotate about thez-axis. Hence, let

where only ω1z is known. The vectors r1, r2, and r3 can be found (refer to activity 1.5 in module 1). The position vectors are constrained in the x-y plane and have the following form (note the absence of the z-axis components):

The following velocities can be evaluated using the cross product:

The velocity vector equation can be written as

0

0

0

0

v

=

1 1 1v rω= ×

2 2 2v rω= ×

3 3 3v rω= ×

1 2 3

1 2 3

0 0 0

0 , 0 , 0

z z z

ω ω ωω ω ω

= = =

1 2 3

1 1 2 2 3 3, ,

0 0 0

x x x

y y y

r r r

r r r r r r

= = =

1 1 2 2 3 3

1 1 1 1 1 2 2 2 2 2 3 3 3 3 3, ,

0 0 0

z y z y z y

z x z x z x

r r r

v r r v r r v r r

ω ω ωω ω ω ω ω ω

− − − = × = = × = = × =

1 1 2 2 3 3

1 1 2 2 3 3

0

0

0 0 0 0

z y z y z y

z x z x z x

r r r

r r r

ω ω ωω ω ω− − − + + =


The above consists of two equations with two unknowns: ω2z and ω3z.

We can solve for the unknowns ω2z and ω3z to get:

Once ω2z and ω3z are found, we can use them to find the linear velocity of any point in the linkage. For example, to find the linear velocity at point ‘D’, we again start with the position vector equation:

We can differentiate this to get

(This is because O1 and O2 are both part of link 4, the ground link that is stationary). The velocity terms in the above equation can be found based on the concepts discussed in module 1:

• = velocity of point ‘B’ (Only rotation is involved and ω1 is the absolute

angular velocity of link 1)

• (Revised module 1 again if you have difficulties understanding this)

Hence

After obtaining the angular velocities of the various links, the next step is to get the angular acceleration by differentiating the velocity vector equation:

1 1 2 2 3 3 0z y z y z yr r rω ω ω+ + =

1 1 2 2 3 3 0z x z x z xr r rω ω ω+ + =

( )1 1 3 3 1

22 3 3 2

z y x y x

zy x y x

r r r r

r r r r

ωω

− −=

−

( )1 2 1 1 2

32 3 3 2

z y x y x

zy x y x

r r r r

r r r r

ωω

− −=

−

1/ /1

D O D Br r r= +

1 2/ //1

D O D OD B Dv v v v v= + = =

1 1 1v rω= ×

/ /2

D B D Bv rω= ×

/1 1 2

D D Bv r rω ω= × + ×

0 1 2 3 0a a a a+ + + =


Again we use the concepts discussed in module 1 to evaluate these terms:

• as the points O1 and O2 are fixed (see figure 2.1)

• = acceleration of point ‘B’ (Only rotation is involved and α1 is the absolute angular acceleration of link 1)

• (Only rotation is involved and α2 is the absolute angular

acceleration of link 2)

• (Only rotation is involved and α3 is the absolute angular

acceleration of link 3)

Note that α1 is the driving acceleration for link 1. The other angular accelerations α2 and α3 are generated as a result of rotating link 1 at α1. All the links can only rotate about the z-axis. Hence, let

where only a1z is known. The following accelerations can be evaluated using the cross product:

0

0

0

0

a

=

1 1 1 1 1 1a r rα ω ω= × + × ×

2 2 2 2 2 2a r rα ω ω= × + × ×

3 3 3 3 3 3a r rα ω ω= × + × ×

1 2 3

1 2 3

0 0 0

0 , 0 , 0

z z z

α α αα α α

= = =

21 1 1 1

21 1 1 1 1 1 1 1 1 1

0 0

z y z x

z x z y

r r

a r r r r

α ωα ω ω α ω

− − = × + × × = + −

22 2 2 2

22 2 2 2 2 2 2 2 2 2

0 0

z y z x

z x z y

r r

a r r r r

α ωα ω ω α ω

− − = × + × × = + −

23 3 3 3

23 3 3 3 3 3 3 3 3 3

0 0

z y z x

z x z y

r r

a r r r r

α ωα ω ω α ω

− − = × + × × = + −


The acceleration vector equation can be written as

The above consists of two equations with two unknowns: α2z and α3z.

We can solve for the unknowns α2z and α3z to get:

Once the angular accelerations are found, we can use them to find the linear acceleration of any point in the linkage. For example, to find the linear acceleration at point ‘D’ in figure 2.1, we again start with the position vector equation:

We can differentiate this to get

Differentiating this again gives

(This is because O1 and O2 are both part of link 4, the ground link that is stationary). The acceleration terms in the above equation can be found based on the concepts discussed in module 1:

• = acceleration of point ‘B’ (Only rotation is involved and α1 is

the absolute angular acceleration of link 1)

2 2 21 1 1 1 2 2 2 2 3 3 3 3

2 2 21 1 1 1 2 2 2 2 3 3 3 3

0

0

0 0 0 0 0 0 0

z y z x z y z x z y z x

z x z y z x z y z x z y

r r r r r r

r r r r r r

α ω α ω α ωα ω α ω α ω

− − − − − − + − + + − + + − =

2 2 21 1 1 1 2 2 2 2 3 3 3 3 0z y z x z y z x z y z xr r r r r rα ω α ω α ω+ + + + + =

2 2 21 1 1 1 2 2 2 2 3 3 3 3 0z x z y z x z y z x z yr r r r r rα ω α ω α ω− + − + − =

( ) ( )2 2 2 2 2 23 1 1 1 1 2 2 3 3 3 1 1 1 1 2 2 3 3

22 3 3 2

x z y z x z x z x y z x z y z y z y

zy x y x

r r r r r r r r r r

r r r r

α ω ω ω α ω ω ωα

− − − − − − + + +=

−

( ) ( )2 2 2 2 2 22 1 1 1 1 2 2 3 3 2 1 1 1 1 2 2 3 3

32 3 3 2

x z y z x z x z x y z x z y z y z y

zy x y x

r r r r r r r r r r

r r r r

α ω ω ω α ω ω ωα

− − − − − + − + + +=

−

1/ /1

D O D Br r r= +

1 2/ //1

D O D OD B Dv v v v v= + = =

1 2/ //1

D O D OD B Da a a a a= + = =

1 1 1 1 1 1a r rα ω ω= × + × ×


• (Revised module 1 again if you have difficulties

understanding this)

Hence

The above is the kinematics analysis of the four-bar linkage. Don’t try to memorise the equations as the mechanism can change. Try to understand the approach and the utilization of the concepts in module 1 in the kinematics analysis. The concepts can be applied to other closed kinematic-chains to give different equations. The concepts can also be extended from planar to spatial mechanisms.

Notice that for a simple mechanism, it is tedious to perform manual calculations for more than two different configurations. You should also go through the approach and examine the recursive nature of the vector equations and the ease of developing programs in Matlab to assist in the analysis (The recursive nature of the vector equations can simplify the program). Matlab has a built-in function for cross product. To find in Matlab, use the function c = cross (a,b).

Activity 2.2

Objectives 2 and 5

Figure 2.2: Slider-crank mechanism for activity 2.2

A slider-crank mechanism is shown in figure 2.2. The crank (link 1) is rotating at ω1 rad/s and α1 rad/s2.

• The position vectors to describe the mechanism were found in tutorial 1b. Differentiate the position vector equation to get the velocity vector equation in terms of ω1, ω2, and vc.

/ / /2 2 2

D B D B D Ba r rα ω ω= × + × ×

/ /1 1 1 1 1 2 2 2

D D B D Ba r r r rα ω ω α ω ω= × + × × + × + × ×

c a b×=

O1

Link 1 Link 2

Link 3

B

C

θ1

y

x

z

φ

D


• Describe the position and velocity vectors in terms of their x, y, and z components. Evaluate the cross products in the velocity vector equation to get the equations for finding ω2, and vc.

• Differentiate the velocity vector equation to get the acceleration vector equation in terms of ω1, ω2, α1, α2, vc and ac.

• Describe the acceleration vectors in terms of their elements of their x, y, and z components. Evaluate the cross products in the acceleration vector equation to get the equations for finding α2, and ac.

• Develop a Matlab program to find vc and ac.

• Explain how you would determine the velocity and acceleration of point ‘D’ (the midpoint of BC)?

Note that O1C is horizontal. The position vectors description was determined in tutorial 1b.

2.3 Equivalent linkages

Reading

Selected reading 2.4: Before you proceed, please read Martin ‘7-5 Equivalent linkages’, pp. 128–32.

This section deals with the kinematics analysis of direct contact mechanisms using equivalent four-bar linkages. For example, the equivalent linkage concept can be used to analyze the cam and follower mechanism shown in figure 2.3(a). The cam, rotating about O1, is the driving link, which imparts motion to the follower. The profile of the cam will determine the desired motion of the follower. Points ‘C’ and ‘D’ are the centers of curvatures. A center of curvature is not the same as the center of rotation.

For acceleration analysis purposes, the mechanism of figure 2.3(a) is equivalent to the four-bar linkage shown in figure 2.3(b). The position vectors associated with the links in the equivalent four-bar linkages are r0 = O2O1, r1 = O1C, r2 = CD, and r3 = DO2. The analysis of the acceleration in a four-bar linkage has been covered in the previous section. Note that if the cam surface consists of segments of circular arcs, then a different four-bar linkage is needed to represent each section of the surface and there can be instantaneous changes in angular acceleration of the follower arm.


(a)

(b)

Figure 2.3: Equivalent four-bar linkage for cam and follower

CLink 1 (cam)

Link 2 (follower)

O1

O2

D

O2

C

D

O1

r0

r1

r2

r3


Activity 2.3

Objective 3

Sketch the equivalent linkage for the mechanism shown in figure 2.4.


2.4 Complex linkages

Reading

Selected reading 2.5: Before you proceed, please read Martin ‘Complex linkages’, pp. 189–90.

The above concepts can be extended to complex linkages. In most cases, the mechanism can be broken down into simpler basic linkages that can easily be analyzed. However, in some situations, the solution may not be so simple. The approach in such situations would then involve trial and error. The use of computer programs in these situations could be indispensable.

O

A

B


Activity 2.4

Objective 4

Explain how you would attempt to analyze the mechanism shown in figure 2.5. Note that link 1 is the driver and link 6 is constrained to slide horizontally. Link 4 is the ground (fixed link).


θ1, ω1, α1

Link 1

Link 2

Link 3

B

C

O1 O2

y

x

zθ2

θ3

Link 5

Link 6 D


Activity feedback

Activity 2.1

The bodies in an open kinematic-chain can have only one associated kinematic-joint. In a closed kinematic-chain, each body has at least two associated kinematic-joints

Activity 2.2

Figure 2.2 can be represented by the following position vector polygon:

Since we want the position of point ‘C’ with respect to point ‘O1’, the position vector equation is

The position vector polygon can be differentiated with respect to time to get the velocity vector equation:

The velocity terms in the above equation are defined as follow:

• as the point C has rectilinear motion along the x-axis (i.e. O1C is not fixed)

• = velocity of point ‘B’ (Only rotation is involved and ω1 is the absolute angular velocity of link 1)


Therefore:

r1r2

r0

O1 C

B

0 1 2r r r= +

0 1 2v v v= +

0 0 cxv r v= =

1 1 1v rω= ×

2 2 2v rω= ×

0 1 1 2 2v r rω ω= × + ×


Links 1 and 2 can only rotate about the z-axis. The slider Velocity v0 only has a component along the x-axis. Hence

For the planar mechanism, the position vectors have the following form (note the absence of the z-axis components):

The position vectors are used in the evaluation of the cross products in the velocity vector equation to give

The above consists of two equations with two unknowns: ω2z and vcx.

We can solve for the unknowns ω2z and vcx to get:

The velocity vector equation is differentiated to get the acceleration vector equation:

1 2 0

1 2

0 0

0 , 0 , 0

0

cx

z z

v

vω ωω ω

= = =

1 2

1 1 2 2 0, , 0

0 0 0

x x cx

y y

r r r

r r r r r

= = =

1 1 2 2

1 1 2 2 0

0 0 0

z y z y cx

z x z x

r r v

r r

ω ωω ω− − + =

1 1 2 2z y z y cxr r vω ω− − =

1 1 2 2 0z x z xr rω ω+ =

1 12

2

z xz

x

r

r

ωω −=

( )1 1 2 2 1

2

z y x y x

cxx

r r r rv

r

ω− −=

0 1 2a a a= +


where a0 = ac. Note that

The acceleration vector equation now as the form

Hence

Once the angular accelerations are found, we can use them to find the linear acceleration at any point in the linkage. For example, to find the linear acceleration at point ‘D’, we again start with the position vector equation:

We can differentiate this to get the velocity vector equation

where

Differentiating the velocity vector equation gives

where

1 2 0

1 2

0 0

0 , 0 , 0

0

cx

z z

a

aα αα α

= = =

2 21 1 1 1 2 2 2 2

2 21 1 1 1 2 2 2 2 0

0 0 0 0 0

z y z x z y z x cx

z x z y z x z y

r r r r a

r r r r

α ω α ωα ω α ω

− − − − + − + + − =

2 21 1 2 2 1 1

22

z y z y z xz

x

r r r

r

ω ω αα

+ −=

2 21 1 1 1 2 2 2 2( )cx z y z x z y z xa r r r rα ω α ω= − + + +

1/ /1

D O D Br r r= +

1/ /1

D O D B Dv v v v= + =

/ /2

D B D Bv rω= ×

1/ /1

D O D B Da a a a= + =

/ / /2 2 2

D B D B D Ba r rα ω ω= × + × ×


The following Matlab program can be used to find velocity and acceleration at the slider.

Activity 2.3

The sketch of the equivalent linkage is given below. Link 3 is called the slider. Point ‘C’ is the center of curvature of the cam (i.e. for surface at the point of contact ‘A’ in figure 2.4). OC is link 1, the driving link. Link 4 is the ground (fixed link).

% ENG 3403 Activity 2.2 % Sample solution %clear clcR = input('Input length R=O1B in m :'); L = input('Input length L=BC in m :'); th1_deg = input('Input theta_1 in degrees :'); %th1 = th1_deg*pi/180; % find angle phi n = L/R; phi =asin(sin(th1)/n); th2 = 2*pi - phi; % find vectors r1 and r2 r1 = [R*cos(th1); R*sin(th1); 0]; r2 = [L*cos(th2); L*sin(th2); 0]; w1z = input('Input w1 in rad/s :'); al1z =input('Input alpha1 in rad/s^2 :'); w1 = [0; 0; w1z]; al1 = [0; 0; al1z]; w2z = -w1z*r1(1)/r2(1); vcx = -w1z*(r1(2)*r2(1) - r2(2)*r1(1))/r2(1) al2z = (r1(2)*w1z^2 + r2(2)*w2z^2 - al1z*r1(1))/r2(1); acx = -(al1z*r1(2) + r1(1)*w1z^2 + al2z*r2(2) + r2(1)*w2z^2)

O

A

B

C

Link 2

Link 2

Link 1

Link 3

Link 4

Parallel to edge of follower


Activity 2.4

The complex mechanism can be separated into two simpler linkages as shown below. The analysis will first determine the angular velocities and angular accelerations of the four-bar linkage. These angular velocities and accelerations are then used in the analysis of the crank-slider mechanism, which now has O2C as the driver. Note that point ‘D’ can only slide along the horizontal (it can only have a linear velocity).

θ1, ω1, α1

Link 1

Link 2

Link 3

B

O1 O2

C

C

Link 5

Link 6 D

Link 3

O2


Module 3 – Dynamics analysis of planar mechanisms

Module 3

DYNAMICS ANALYSIS OF PLANAR MECHANISMS 3

Module 3 – Dynamics analysis of planar mechanisms 3.1

Preamble

This module reviews some basic principles of dynamics, with emphasis on the use of vectors to represent the force, linear momentum, moment, and angular momentum. The concepts reviewed included the principle of linear momentum, the principle of angular momentum, work and energy. Following the revision, the module focuses on the applications of these basic principles in the dynamics analysis of practical mechanisms such as the four-bar linkages and slider-crank mechanisms. The static and dynamic balancing of rotating machines and the balancing of reciprocating machines are also discussed. Developing programs in MATLAB can facilitate the solving of some of these applications. You should try to understand how the basic dynamics principles are applied in the analysis rather than just memorizing the equations.

Objectives


1. demonstrate an understanding of the principle of linear momentum

2. demonstrate an understanding of the principle of angular momentum

3. demonstrate an understanding of the concepts of work and energy

4. analyse the dynamics of mechanisms such as four-bar linkage, slider-crank linkage, etc.

5. analyse the static and dynamic balancing of rotating machinery

6. analyse the balancing of reciprocating machines

7. develop simple Matlab programs to assist in the dynamic analysis of mechanisms.

3.1 Principles of linear and angular momentums

Reading

Text: Before you proceed, please read section 1.2.3 ‘Particle and rigid-body dynamics’, pp. 12–15 of the textbook.

This section covers two important principles: the principle of linear momentum and the principle of angular momentum.


The linear momentum is defined as the product of the mass and linear velocity, i.e. mv (For a system of many particles or a rigid body, the velocity is normally taken at the center of mass of the system or body). Note that linear momentum is a vector. The principle of linear momentum states that

F = total external force acting on the system/body, m = total mass of the system/body, v and a are respectively the linear velocity and linear acceleration normally taken at the center of mass of the system/body. Without external forces, the linear momentum is conserved.

The principle of linear momentum is also called Newton’s second law of motion. This law is used to derive the equations of motion for systems of particles or rigid-bodies.

Activity 3.1

Objective 1

Consider a particle of mass m attached to the midpoint of a weightless rod of length l (figure 3.1). The mass cannot slide along the rod. The ends of the rod are constrained to move along the x and y-axis. A uniform gravitational field acts on the mass. Determine the equation of motion for the mass using the principle of linear momentum.

Figure 3.1: System for activity 3.1

The angular momentum is defined as . The angular momentum is normally taken about a fixed point (or the center of mass for a system/body). Note that r is the position vector from that fixed point to the line of action of the linear momentum vector. The angular momentum H is a vector. For a system of particles or a rigid body moving in the x-y plane (e.g. rotating at ωz about the z-axis), the angular momentum is related to the system/body mass moment of inertia Jz along the z-axis: i.e. , (where Jz is the mass moment of inertia taken about the fixed point (or the center of mass of the system/body) along the z-axis.

dF mv ma

dt= =

m x

y

θ

l/2

l/2

( )H r mv= ×

H Jzωz=


The principle of angular momentum states that

M = total moment of the external forces acting about the fixed point (or the center of mass for a system/body), where the angular momentum was taken.

Note that for a system of particles or a rigid body moving in the x-y plane (e.g. rotating at ωz about the z-axis), the principle of angular momentum can be written as

where Mz = total moment of external forces about the z-axis taken with respect to a fixed point (or the center of mass); αz = angular acceleration of the system/body about the z-axis.

Without external moments, the linear momentum is conserved. The principle of angular momentum is used to derive the equations of motion for a system of particles or a rigid-body.

Activity 3.2

Objective 2

Derive the equation of motion for the pendulum of length ‘L’ and mass ‘m’ shown in figure 3.2 using the principle of angular momentum.

Figure 3.2: Pendulum for activity 3.2

dHM

dt=

z z z z zM J Jω α= =

y

x

zm

O

θ

L


3.2 Work and energy

Reading

Text: Before you proceed, please read section 1.2.4 ‘Work and energy’,pp. 15–16 of the textbook.

This section reviews two important concepts: work and energy. Energy is a scalar quantity (not a vector). There are two basic forms of energies: kinetic energy and potential energy. The section only deals with kinetic energy. The kinetic energy for a rigid body can generally be separated into two translational and rotational kinetic energies.

The work-energy theorem is important for the derivation of the equations of motion for a system of particles or a rigid body.

Activity 3.3

Objective 3


3.3 Analysis of forces and torques in four-bar linkages

In this section, a four-bar linkage is used to illustrate the application of the basic dynamics concepts discussed in the previous sections. Figure 3.3 shows a four-bar linkage, the kinematics analysis of which was discussed in module 2. To refresh your memory, link 1 (the crank) drives link 3 (the rocker). In the kinematics analysis, we have found ω2, ω3, α2 and α3 using known values of θ1, r0, r1, r2, and r3, along with the driver angular velocity ω1 and angular acceleration α1 (For most engineering applications, the mechanism is driven such that

). This section will continue with these known data to determine the instantaneous torque T required to drive link 1 to produce the specified motion. This is a dynamics problem and the mass, center of mass (CG), and moment of inertia for each linkage element must first be determined. (Please read the following selected reading to find out more on the calculation of the mass and mass moment of inertia for linkage elements)

Reading

Selected reading 3.1: Before you proceed, please read Wilson & Sadler ‘Calculation of mass and mass moment of inertia for linkage elements’,pp. 689–92.

α1 0=


It is assumed that the following had been found for each link:

• m1 = Mass of link 1, = position vector of CG1 of link 1 from O1, IG1 = Mass moment of inertia of link 1 about the axis of rotation through CG1

• m2 = Mass of link 2, = position vector of CG2 of link 2 from B, IG2 = Mass moment of inertia of link 2 about the axis of rotation through CG2

• m3 = Mass of link 3, = position vector of CG3 of link 3 from O3, IG3 = Mass moment of inertia of link 3 about the axis of rotation through CG3

(a)

(b)

Figure 3.3: Forces and torques analysis in four-bar linkage

The approach to analyse the dynamics starts by finding the accelerations at CG1, CG2 and CG3. The determination of the acceleration was covered in module 2. The accelerations of CG1, CG2, CG3 can be found respectively using:

1 1/CG Or

2 /CG Br

3 3/CG Or

θ1, ω1, α1

Link 1

Link 2

Link 3

r1

r2

r3

r0

B

C

O1 O3

y

x

z

ω2, α2CG2

CG1

CG3

ω3, α3

C1r1

r2

r3

B

C

O1 O3

C2

CG2

CG1

CG3

C3

TF1

F2

F3


Using Newton’s 2nd law of motion, F = ma, the force Fi1 at CG1, the force Fi2 at CG2 and the force Fi3 at CG3 due to the linear accelerations can be found respectively using:

From the concepts learned in the previous sections, the angular momentum for link 1 about

CG1 is defined as or

Using the principle of angular momentum , the moment for link 1 about CG1 due

to the angular acceleration is

Similarly, the moment for link 2 about CG2 is and the moment for link 3 about

CG3 is .

Note that Fi1, Fi2, Fi3, Ci1, Ci2, Ci3, are internal forces and torques. Using d’Alembert’s principle, the reverse-effective or inertial forces and torques are , ,

, , , . These inertial forces and torques together with the external forces and torques must maintain static equilibrium. These inertia forces and torques are shown together with the input torque T in figure 3.3b. The input torque T to link 1 can now be determined based on static equilibrium analysis.

Reading

Selected reading 3.2: Before you proceed, please read Wilson & Sadler ‘Analytical solution for the four-bar linkage’, pp. 642–5.

1 1 1 1 1/ /1 1 1

CG CG O CG Oa r rα ω ω= × + × ×

2 2 2/ /1 1 1 1 1 2 2 2

CG CG B CG Ba r r r rα ω ω α ω ω= × + × × + × + × ×

3 3 3 3 3/ /3 3 3

CG CG O CG Oa r rα ω ω= × + × ×

11 1

CGiF m a=

22 2

CGiF m a=

33 3

CGiF m a=

1 1 1G GH I ω=

11 1

GG

dHI

dtα=

dHM

dt=

1 1 1i GC I α=

2 2 2i GC I α=

3 3 3i GC I α=

F1 Fi1–= F2 Fi2–=F3 Fi3–= C1 Ci1–= C2 Ci2–= C3 Ci3–=


The section on ‘Analytical solution for the four-bar linkage’ introduces an approach to solve for the input torque T once F1, F2, F3, C1, C2, and C3 are found. The approach involves drawing free body diagram for each link and then analyzing the forces and moments to maintain equilibrium (Revised the free body diagrams covered in MEC2401 if you have problems understanding the approach). The approach also documents the procedure for finding the unknown bearing forces at the joints.

The above is the dynamics analysis of the four-bar linkage. Try to understand the approach and the utilization of the concepts in the dynamics analysis. The concepts can be applied to analyse other mechanisms.

Notice that for a simple mechanism like the four-bar linkage, it is tedious to perform manual calculation for more than two different configurations. You should go through the approach and try to develop programs in Matlab to assist in the analysis.

Activity 3.4

Objective 4

Perform a dynamic-force analysis of the slider-crank mechanism shown in figure 3.4. Derive the equations for the instantaneous input torque T at link 1 and the pin forces at ‘O1’, ‘B’ and ‘C’ in terms of the inertial forces (F1, F2) at (CG1, CG2) and inertial torques (C1, C2) at link 1 and link 2 (the inertial forces and torques are shown in figure 3.4). Hint: use free-body diagrams for the links and determine the required forces and torques based on static equilibrium. Explain how you would get the inertial forces based on the kinematics analysis of the slider-crank mechanism.

Figure 3.4: Slider-crank mechanism for activity 3.4

O1

Link 1

Link 2

Link 3

B

C

θ1

y

x

z

φF1 F2

C1 C2

T


3.4 Balancing of rotating machinery

Reading

Selected reading 3.3: Before you proceed, please read ‘10-7 Balancing of rigid bodies’, pp. 730–9.

This section deals with the problem of unbalanced mass in machines. The unbalanced dynamic forces will result in vibration, a topic that will be covered later. The section covers two procedures for machine balancing: static balance and dynamic balance.

Static balancing involves mounting the shaft horizontally on knife-edge bearings. Mass is either added or removed from the shaft in such a way that the combined center of mass is coincident with the bearing centerline and the resultant shaking force is zero. The mass added is represented by mc (the mass removed is represented by –mc). The position of mc can be defined using two coordinates: rc and θc. For a system with n masses, i.e. m1, m2, ..., mn,

Be careful when you find the inverse tangent using calculators (Most calculators will give the

answer in the range . You will have to correct the answer from the calculator based on the quadrant in which the angle lies. Revised your trigonometry if you have problems). Notice that there can be many possible solutions as mc and rc are coupled (you have to specify one to find the other). The above equations are valid if the balancing involves removing mass –mc at –rc.

Activity 3.5

Objective 5

The front and side views of a rigid rotor are shown in figure 3.5. This mechanism is to be statically balanced by the addition of a fourth mass at a 178 mm radius. Determine the mass to be added and the angular position of the balancing mass. (r1 = 385 mm, r2 = 255 mm, r3 = 305 mm).

1 1

1

sin( )tan

cos( )

n

i i iic n

i i ii

m r

m r

θθ

θ− =

=

− = −

∑∑

( ) ( )2 2

1 1cos( ) sin( )

n n

c c i i i i i ii im r m r m rθ θ

= == +∑ ∑

90 90cθ− ≤ ≤


Figure 3.5: Static balancing of rotor

Dynamic balancing involves rotating the shaft. Masses are either added or removed so that the bearings do not experience any forces and moments. For dynamic balancing using two masses mc1 and mc2 (located at rc1, θc1 and rc2, θc2 respectively) on a system with n unbalanced masses, i.e. m1, m2, ..., mn,

(You must specify sc2, the position along the shaft where mass mc2 is added). Notice that there can be many possible solutions as mc2 and rc2 are coupled (you have to specify one to find the other).

Notice that there can be many possible solutions as mc1 and rc1 are coupled (you have to specify one to find the other).

(side view) (front view)

m2=2.27 kg

120°105°

105°

m1=1.81 kg

m3=0.91 kg

r1

r2

r3

shaft

supportsupport

1 12

1

sin( )tan

cos( )

n

i i i iic n

i i i ii

m s r

m s r

θθ

θ− =

=

− = −

∑∑

( ) ( )2 2

2 2 1 12

1cos( ) sin( )

n n

c c i i i i i i i ii ic

m r m s r m s rs

θ θ= =

= +∑ ∑

( )( )

2 2 2111

2 2 21

sin( ) sin( )tan

cos( ) cos( )

n

i i i c c ci

c n

i i i c c ci

m r m r

m r m r

θ θθ

θ θ=−

=

− − = − −

∑∑

( ) ( )2 2

1 1 2 2 2 2 2 21 1cos( ) cos( ) sin( ) sin( )

n n

c c c c c i i i c c c i i ii im r m r m r m r m rθ θ θ θ

= == + + +∑ ∑


Activity 3.6

Objective 5

Figure 3.6: Dynamic balancing of rotor

The front and side views of a rigid rotor are shown in figure 3.6. This mechanism is to be dynamically balanced by the addition of two masses in the planes ‘P’ and ‘Q’. Determine the corrections needed. (m1 = 4 kg, m2 = 2 kg, r1 = 2 m,r1 = 2.5 m).

3.5 Balancing of reciprocating machines

Reading

Selected reading 3.4: Before you proceed, please read Martin ‘20-3 Slider-crank mechanism’, pp. 410–13.

(side view) (front view)

90°

m2=2 kg

m1=4 kg r2

r1

P Q

6 m 6 m 6 m

support support

shaft


Figure 3.7: A typical reciprocating machine: a slider-crank mechanism

The slider-crank mechanism shown in figure 3.7 is a typical reciprocating machine. This section uses the slider-crank mechanism to illustrate the balancing of reciprocating machines. The analysis in the above selected reading approximates the forces at the pin on the slider by

• Relacing the mass m1 (at CG1) of link 1 with an equivalent mass m1B located at point B.

• Separating the mass m2 of the connecting rod (link 2) into two masses: m2B located at ‘B’ and m2C located at ‘C’ (refer to selected reading).

The final results show that the forces at the pin on the slider is in the direction of sliding

where m3 is the mass of the slider and R, L, θ1, ω1 are as defined in figure 3.7. Note that the force is also called the shaking force and it contains two components: a primary part and a secondary part. The analysis of the shaking force is only an approximation. The exact solution is the mass of the slider multiplied by its acceleration.

Reading

Selected reading 3.5: Before you proceed, please read Wilson & Sadler ‘Balancing of reciprocating machines’, pp. 740–3.

The material covered in the above reading again stated that the shaking force consists of primary and secondary parts (Although the symbols used are different, the equation for the shaking force is the same as that discussed above. Please compare them and note that

for constant angular velocity). As the slider crank mechanism is often found in engine, the use of counterweight to reduce the shaking force is discussed. The counterweight mc is normally mounted on the crank at a distance rc and at an angle θ1+180° (refer to figure 10.24 in the selected reading). With the counterweight, the total shaking force is now given by

Note: O1B=R and BC=L

O1

Link 1 Link 2

Link 3

B

C

θ1, ω1

y

x

z

2 22 3 1 1 2 3 1 1( ) cos( ) ( ) cos(2 )p C C

Rf m m R m m R

Lω θ ω θ= + + +

ω1t θ1=


Note that with the addition of the counterweight, the shaking force in the x-component is reduced but another force is generated in the direction of the y-axis.

Activity 3.7

Objectives 6 and 7

Approximate the shaking force for the slider-crank mechanism given in question 2 of tutorial 3b (Tutorial questions and solutions are available on WebCT) when α1 = 0 rad/s2, ω1 = 10 rad/s and θ1 = 70°. Obtain the exact shaking force using the results in question 2 of tutorial 3b. Compare the approximated shaking force with the exact shaking force using a simple Matlab program and discuss on the accuracy of

2 2 22 3 1 1 2 3 1 1 1 1

21 1

( ) cos( ) ( ) cos(2 ) cos( )

sin( )

0

C C c c

p c c

Rm m R m m R m r

Lf m r

ω θ ω θ ω θ

ω θ

+ + + −

= −

2 22 3 1 1 2 3 1 1( ) cos( ) ( ) cos(2 )p C C

Rf m m R m m R

Lω θ ω θ= + + +


Activity feedback

Activity 3.1

Using Newton’s second law of motion along the x-axis:

Hence

Using Newton’s second law of motion along the y-axis:

Hence

Activity 3.2

Using principle of angular momentum (taking moments about fixed point ‘O’):

0xF mx= =

2sin( ) cos( ) sin( ) cos( ) 02 2 2 2

l l l lx x xθ θ θ θ θ θ θ= → = → = − + =

2 cos( )

sin( )

θθ θθ

=

yF mg my= − =

2cos( ) sin( ) cos( ) sin( )2 2 2 2

l l l ly y yθ θ θ θ θ θ θ= → = − → = − −

2cos( ) sin( )2 2

l lmg m mθ θ θ θ− = − −

cos( )cos( ) sin( )

2 sin( ) 2

l lg

θθ θ θ θθ

= +

2 2sin( ) cos ( ) sin ( )2 2 2

l l lg θ θ θ θ θ θ= + =

2sin( )

g

lθ θ=

2sin( )OM J mgL mLθ θ θ= ⇔ − =

sin( ) 0g

Lθ θ+ =


Activity 3.3

The rotating disc has rotational kinetic energy about the point ‘O’. The mass

m can move along the x and y axes and to find the its translational kinetic energy, we need to get the velocity along the x and y axes with respect to point ‘O’:

Hence, the total kinetic energy of the system is Ke = Ke1 + Ke2:

21 0

1

2Ke J θ=

( )2 22

1

2Ke m x y= +

cos( ) cos( ) sin( ) ( )sin( )x R r x R rθ θ ϕ θ θ θ ϕ θ ϕ= + + → = − − + +

sin( ) sin( ) cos( ) ( ) cos( )y R r y R rθ θ ϕ θ θ θ ϕ θ ϕ= + + → = + + +

2 2 2 2 2 2 2( ) sin ( ) ( ) sin ( ) 2 ( )sin( )sin( )x R r Rrθ θ θ ϕ θ ϕ θ θ ϕ θ θ ϕ= + + + + + +

2 2 2 2 2 2 2( ) cos ( ) ( ) cos ( ) 2 ( ) cos( )cos( )y R r Rrθ θ θ ϕ θ ϕ θ θ ϕ θ θ ϕ= + + + + + +

2 2 2 2 2 2( ) ( ) 2 ( ) cos( )x y R r Rrθ θ ϕ θ θ ϕ ϕ+ = + + + +

2 2 2 2 20

1 1( ( ) 2 ( ) cos( ))

2 2Ke J m R r Rrθ θ θ ϕ θ θ ϕ ϕ= + + + + +

θ

ϕ

Ox

y m

R

r


Activity 3.4

The free body diagrams are as follow:

For link 3, using Newton’s law:

Sum of vertical forces equal zero:

For link 2: note that

and

For static equilibrium, sum of the forces must be zero, i.e.

, where

, where

Link 1

Link 2

Link 3

B

C

θ1

φ

F1

F2

C1

C2

T

B

C

F01y

F01x

F21y

F21x

F12x

F12y

F32y

F23y

F32x

F23x

F03y

lG

l

L1

LG1

y

x

z

O1

323 3

CGxF m a=

03 23y yF F= −

22

22 2 2 2

0 0

CGx x

CGy i y

F a

F F F m a

= = − = −

2 2 2GC I α= −

12 32 2 0x x xF F F+ + = 22 2

CGx xF m a= −

12 32 2 0y y yF F F+ + = 22 2

CGy yF m a= −


We take the moment of the force about point ‘B’:

Similly for link 1, note that

and

For static equilibrium, sum of the forces must be zero, i.e.

, where

, where

We take the moment of the force about point ‘O1’:

Note that once aCG3 is found, we will know . We can then find ;

follow by (using the moment equation of link 2); follow by ; follow by F01x, F01y and finally T (using the moment equation for link 1). A simple Matlab program can be easily developed based on the above sequence of determination.

The inertial forces and torques can be obtained from the linear and angular accelerations as shown in the equations above.

(Note: In the free body diagrams, we have assumed the forces and torques to act in certain directions. If you use the above the equations and find that the forces or torques are negative, it simply means that the assumed directions of these forces and torques in the free body diagrams are incorrect (they should point in the opposite directions).

Activity 3.5

Given 3 masses: m1 = 1.81 kg, r1 = 385 mm, θ1 = 120°; m2 = 2.27 kg, r2 = 255 mm, θ2 = 225°; m3 = 0.91 kg, r3 = 305 mm, θ3 = 330°;

Therefore for rc = 178 mm, mc = 2.92 kg.

2 32 32 2 2( cos( )) ( sin( )) ( cos( )) ( sin( )) 0y x y G x GC F l F l F l F lφ φ φ φ+ + + + =

11

11 1 1 1

0 0

CGx x

CGy i y

F a

F F F m a

= = − = −

1 1 1GC I α= −

01 21 1 0x x xF F F+ + = 11 1

CGx xF m a= −

01 21 1 0y y yF F F+ + = 11 1

CGy yF m a= −

1 21 1 1 21 1 1 1 1 1 1 1 1( cos( )) ( sin( )) ( sin( )) ( cos( )) 0y x x G y GC F L F L F L F L Tθ θ θ θ+ − − + + =

23 32x xF F= − 12 21x xF F= −

32 23y yF F= − 12 21y yF F= −

1 1

1

sin( )tan 6.11

cos( )

n

i i iic n

i i ii

m r

m r

θθ

θ− =

=

− = = − −

∑∑

( ) ( )2 2

1 1cos( ) sin( ) 520.33

n n

c c i i i i i ii im r m r m rθ θ

= == + =∑ ∑


Activity 3.6

Given 2 unbalanced masses: m1 = 4 kg, r1 = 2 m, θ1 = 0°; s1 = 6 m, m2 = 2 kg, r2 = 2.5 m,s2 = 12 m, θ2 = 90°; and sc2 = 18 m:

For rc2 = 1 m, mc2 = 4.2687 kg.

For rc1 = 1 m, mc1 = 5.5877 kg.

Activity 3.7

Given rad/s2, rad/s and . From question 2 of tutorial 3b,

kg, kg, R = 2 m and L = 3.76 m. Assume that the center of mass for link 2 is

located in the middle of link 2. Therefore, kg. The approximated shaking force is

given by

N

The acceleration of the slider is 20.9 m/s2. The exact shaking force is . The error is 16.5%.

1 12

1

sin( )tan 128.66

cos( )

n

i i i iic n

i i i ii

m s r

m s r

θθ

θ− =

=

− = = − −

∑∑

( ) ( )2 2

2 2 1 12

1cos( ) sin( ) 4.2687

n n

c c i i i i i i i ii ic

m r m s r m s rs

θ θ= =

= + =∑ ∑

( )( )

2 2 2111

2 2 21

sin( ) sin( )tan 162.65

cos( ) cos( )

n

i i i c c ci

c n

i i i c c ci

m r m r

m r m r

θ θθ

θ θ=−

=

− − = = − − −

∑∑

( ) ( )2 2

1 1 2 2 2 2 2 21 1cos( ) cos( ) sin( ) sin( )

n n

c c c c c i i i c c c i i ii im r m r m r m r m rθ θ θ θ

= == + + +∑ ∑

1 1 5.5877c cm r =

α1 0= ω1 10= θ1 70°=

m2 2= m3 3=

m2c 1=

2 22 3 1 1 2 3 1 1( ) cos( ) ( ) cos(2 ) 52.36p C C

Rf m m R m m R

Lω θ ω θ= + + + =

m3aCG3 62.7 N=


The following Matlab file was created to perform the above analysis.

% ENG 3403 Activity 3.7 % Sample solution %clearclcR = 2; L = 3.76; th1_deg = 70; % in degrees w1z = 10; % in rad/s alp1z = 0; % in rad/s^2 th1 = th1_deg*pi/180; % in radians %% Link 1 %sth1=sin(th1); cth1=cos(th1); s2th1=sin(2*th1); c2th1=cos(2*th1); r1=[R*cth1; R*sth1; 0]; %% connecting rod %phi=asin(R*sth1/L); th2 = 2*pi-phi; sth2 = sin(th2); cth2 = cos(th2); r2 = [L*cth2; L*sth2; 0]; w2z = -r1(1)*w1z/r2(1); alp2z = (w1z*w1z*r1(2) + w2z*w2z*r2(2) - alp1z*r1(1))/r2(1); %% Slider %vcx = w1z*(r1(2)*r2(1) - r2(2)*r1(1))/r2(1); acx = alp1z*r1(2) + w1z*w1z*r1(1) + alp2z*r2(2) + w2z*w2z*r2(1) %% Find the force at the slider %m2 = 2; m2c = m2/2; m3 = 3; M = m2c + m3; %fp_true = m3*acx % exact force at slider fp_appro = M*R*w1z*w1z*(cth1 +R*c2th1/L) % approximate force accuracy = abs((fp_appro - fp_true)/fp_true)*100 % in percentage

Module 4 – Rigid body dynamics

Module 4

RIGID BODY DYNAMICS 4

Module 4 – Rigid body dynamics 4.1

Preamble

The dynamics of planar mechanisms such as the four-bar and slider-crank linkages were covered in module 3. This module extends the concepts learned to general 3D rigid body motion. The module covers the mass moment of inertia for general rotation in 3D space, Euler’s equations of motion and gyroscopic dynamics.

Objectives


1. determine the mass moment of inertia for a planar mechanism

2. determine the mass moment of inertia for a 3D rigid body

3. demonstrate an understanding of Euler’s equations of motion and apply them to analyse the dynamics of general 3D rigid body motion

4. demonstrate an understanding of gyroscopic dynamics and apply them to analyse the dynamics of 3D rigid body motion.

4.1 Inertia elements

Reading

Text: Before you proceed, please read section 2.2 ‘Inertia elements’, pp. 22–6 of the textbook.

This section reviews the basic concept of mass and mass moment of inertia. Forces and translational motion are characterized with respect to the mass of the body. Moments and rotational motion are characterized with respect to the mass moment of inertia of the body. The selected reading explains the approach to find the mass moment of inertia (including the use of the parallel axis theorem); the relationship between moment, inertia and angular acceleration; and the relationship between the kinetic energy of rotational motion with the mass moment of inertia. The selected reading only covers these principles with respect to a body undergoing planar motion (i.e. in the x-y plane). The important points are:

• The mass moment of inertia is usually defined about its center of mass or a fixed point ‘O’ with respect to a reference frame


• For planar rotation (i.e. in the x-y plane) with inertia JG and J0 defined with respect to the z-axis about points ‘G’ and ‘O’ respectively (figure 4.1), the parallel axis theorem states that where m = mass and d = distance between ‘G’ and ‘O’.

• There is an analogy between mass m and mass moment of inertia J. The principle of linear

momentum can be viewed as while the principle of angular momentum can

be viewed as . Similarly, the translational kinetic energy is while the

rotational kinetic energy is . Notice that the equations for translational motion are

applicable for rotational motion if we interchange the mass moment of inertia J for mass m, moment/torque M for force F, and angular displacement θ for linear displacement x.

Figure 4.1: Parallel axis theorem for planar motion

Activity 4.1

Objective 1


You should try to understand the above principles, as these concepts will now be extended from planar to general 3D motion. First, the concept of mass moment of inertia and the parallel axis theorem will be extended to general 3D space.

Reading

Selected reading 4.1: Before you proceed, please read ‘Appendix B – Mass moments of inertia’, pp. 665–71, 682–7, 713–16.

J0 JG md2+=

F mx=∑

M Jθ=∑ 21

2mx

21

2Jθ

x

y

z

G

xO

y

d


The complete mass moment of inertia for a solid body about a point ‘O’ with respect to a set of 3 right-handed reference coordinates is a 3 × 3 matrix of the form

If we choose the point ‘O’ to be the center of mass ‘G’ and the 3 referenced coordinates to be the principal axes of the body, the mass moment of inertia of symmetrical bodies can be simplified to

For symmetrical bodies, the 3 principal axes are the axes of symmetry. For most rigid bodies found in engineering, you do not have to determine the mass moment of inertia through integration (The mass moment of inertia of common solids are listed in table D/4 (Meriam & Kraige, pp. 713–16).

For a displacement between points ‘G’ and ‘O’ in 3D space, the parallel axis

theorem is . Note that the referenced

frames used for JG and J0 must be parallel.

Assume we are given the following:

• The mass moment of inertia JO (i.e. taken about point ‘O’) with respect to the x, y, and z-axis

• An axis ‘OM’ that is represented by unit vector λ, i.e. with respect to the x, y,

and z-axis

The mass moment of inertia about point ‘O’ along the axis ‘OM’ is

xx xy xz

o yx yy yz

zx zy zz

I I I

J I I I

I I I

− − = − − − −

0 0

0 0

0 0

xx

G yy

zz

I

J I

I

=

x

y

z

d

d d

d

=

2 2

2 20

2 2

( )

( )

( )

y z x y x z

G x y x z y z

x z y z x y

d d d d d d

J J m d d d d d d

d d d d d d

+ − − = + − + − − − +

l

m

n

λ =

2 2 2( ) ( ) ( ) 2 ( ) 2 ( ) 2 ( )OM xx yy zz xy xz yzJ I l I m I n I lm I nl I mn= + + − − −


The above can also be expressed as

Note that , , , and hence the above again reduces to

Given the mass moment of inertia JO (i.e. taken about point ‘O’) with respect to a referenced frame a, the above can be extended to determine the mass moment of inertia about the same point ‘O’ but with respect to another referenced frame b. This situation is depicted in figure 4.2 where there are two referenced frames: let reference frame a be the set of coordinates defined by vectors (x, y, z) and reference frame b be the set of coordinates defined by vectors (e1, e2, e3). Both frames share the same origin ‘O’. However, frames a and b are not parallel, i.e.

Figure 4.2: Two non-parallel right-handed referenced frames with same origin

Let (J0)A be the mass moment of inertia about ‘O’ with respect to frame a. Let (J0)B be the mass moment of inertia about ‘O’ with respect to frame b. If (J0)B is known, then (J0)A can be found using

[ ]

[ ]

2 2 2

xx xy xz

OM yx yy yz

zx zy zz

xx xy xz

yx yy yz

zx zy zz

xx xy xz yx yy yz zx zy zz

I I I l

J l m n I I I m

I I I n

I l I m I n

l m n I l I m I n

I l I m I n

I l I ml I nl I lm I m I nm I nl I mn I n

− − = − − − − − − = − + − − − −

= − − − + − − − +

Ixy Iyx= Ixz Izx= Izy Iyz=

2 2 2( ) ( ) ( ) 2 ( ) 2 ( ) 2 ( )OM xx yy zz xy xz yzJ I l I m I n I lm I nl I mn= + + − − −

1 1 1 1( ) ( ) ( )x y ze e i e j e k= + +

2 2 2 2( ) ( ) ( )x y ze e i e j e k= + +

3 3 3 3( ) ( ) ( )x y ze e i e j e k= + +

xy

ze3

e1

e2O


where

These concepts on the mass moment of inertia will be illustrated using the following example:

0 0( ) ( )A B TJ R J R=

1 1 1

2 2 2

3 3 3

x y zT

x y z

x y z

e e e

R e e e

e e e

=


Example 4.1

Consider the thin uniform rod (of mass m and length l with radius ) shown in figure 4.3. The mass moment of inertia of the rod about its center of mass ‘G’ with respect to its principal axes (1, 2, 3) can be found from the given tables in the selected reading:

Hence

Figure 4.3: Uniform rod for example 4.1

Now consider the mass moment of inertia J0 of the rod about ‘O’ with respect to the frame

. Note that the (x, y, z) axes are parallel with the (1, 2, 3) axes respectively.

Point ‘A’, the end of the rod, is at a distance from ‘O’. In this case, we can use the parallel

r 0≈

211 22 33

10

12I I ml I= = =

2

112

22

33

10 0

120 01

0 0 0 012

0 0 0 0 0

G

mlI

J I ml

I

= =

x

y

z

O

A

a

b

3

1

2

G

2

l

a x y z, ,( )=

ab0


axis theorem to find J0 (the distance from point ‘O’ to ‘G’ with respect to frame a is

. Hence

J0 is the mass moment of inertia about ‘O’ with respect to frame .

In the above analysis, we have defined the principal axes such that they are parallel to the x, y, and z-axis. Now, consider the same thin uniform rod but with the principal axes as defined in figure 4.4. In this case, the principal axes are not parallel to the x, y, and z-axis. The mass moment of inertia of the rod about its center of mass ‘G’ with respect to its principal axes(1, 2, 3) in figure 4.4 is

With frame b = (1, 2, 3) axes, the mass moment of inertia about ‘G’ with respect to frame b is:

2

x

y

z

d a

d d b

d l

= =

( )

( )

22

22

0

2 2

2 2

2 20

2 2

( )4 2

( )4 2

2 2

1

3 2

1

3 2

2 2

G

l alb ab

l blJ J m ab a

al bla b

almb ml mab m

blJ mab ma ml m

al blm m m a b

+ − −

= + − + −

− − +

+ − − = − + −

− − +

a x y z, ,( )=

211 22 33

10

12I I ml I= = =

( )11

222

332

0 0 00 0

10 0 0 0

120 0 1

0 012

B

G

I

J I ml

Iml

= =


Figure 4.4: A change in the definition of the principal axes for example 4.1

If we apply the parallel axis theorem now to find mass moment of inertia about ‘O’, the mass moment of inertia determined will be with respect to the frame (1’, 2’, 3’), which is parallel to frame b = (1, 2, 3) as shown in figure 4.5. Notice that the distance between ‘O’ and ‘G’ defined with respect to frame (1’, 2’, 3’) is

x

y

z

O

A

a

b

1

2

3

G

2

l

2x

y

z

ld

d d a

d b

= =

( ) ( )

( )

( )

( )

( )

( )

2 2

22

0

22

2 2

2 20

2 2

2 2

2 4

2 4

2 2

1

2 3

1

2 3

B B

G

B

al bla b

al lJ J m b ab

bl lab a

al blm a b m m

alJ m mb ml mab

blm mab ma ml

+ − −

= + − + −

− − +

+ − − = − + −

− − +


where (J0)B indicates that the mass moment of inertia is still with respect to frame

and not (x, y, z).

Figure 4.5: Consequence of applying the parallel axis theorem to figure 4.4

To find (J0)A with respect to frame a= (x, y, z) from (J0)B, we must use the relationship

From figures 4.4 and 4.5, we can see that the unit vector (1’) is defined with respect to the

(x, y, z) axes as . Similarly the unit vectors (2’) and (3’) are respectively

and (i.e. with respect to the (x, y, z) axes). We can use these vectors to determine R:

and

b 1′ 2′ 3′, ,( ) 1 2 3, ,( )= =

2’

3’

1’

O

A

a

b

1

2

3

G

2

l

0 0( ) ( )A B TJ R J R=

1

0

0

1

e

=

2

1

0

0

e

=

3

0

1

0

e

=

0 0 1

1 0 0

0 1 0

TR

=

0 1 0

0 0 1

1 0 0

R

=


Hence

This is the same answer as obtained using the frame assignment in figure 4.3.

( )

2 2

2 20 0

2 2

1

3 21

( ) ( )3 2

2 2

A B T

almb ml mab m

blJ R J R mab ma ml m

al blm m m a b

+ − − = = − + − − − +


Activity 4.2

Objective 2

Find the mass moment of inertia of the cube (of dimensions a × a × a) about ‘O’ in figure 4.6. The cut-off portion has dimensions (a/2) × (a/2) × (a/2). Hint, the mass moment of inertia is additive, i.e. the mass moment of inertia of a composite body can be found by summing the mass moment of inertias of its components, all taken about the same point and with respect to the same set of referenced axes.

Figure 4.6: Cube for activity 4.2

4.2 Euler’s equations of motion

Reading

Selected reading 4.2: Before you proceed, please read Torby ‘Rigid body dynamics’, pp. 205–10.

It is convenient to think of Euler’s equations of motion in terms of two coordinate systems, one of which is a fixed frame a = (x, y, z) and the other of which is a moving frameb = (e1, e2, e3) that rotates with the body. Note that it is also convenient to choose (e1, e2, e3) as the principal axes with origin located at the center of mass ‘G’ so that the mass moment of inertia matrix will simplify to

x

y

z

O

11

22

33

0 0

0 0

0 0G

I

J I

I

=


Let the angular velocity of the body with respect to the fixed frame a be ω, and let the angular velocity of the moving coordinate system (e1, e2, e3) with respect to (x, y, z) be Ω. If the moving system of coordinates (e1, e2, e3) is fixed with respect to the body, then

ω = Ω . For the coordinate system (e1, e2, e3) corresponding to the principal axes, the

Euler’s equations of motion can be written as

• M1 is the net moment of external forces about the center of mass (or a stationary point) with respect to the principal axis e1



Euler’s equations of motion are commonly used for the dynamics analysis of rigid bodies undergoing general 3D motion. The application of Euler’s equations of motion will be illustrated in the following example:

1

2

3

ωωω

=

1 11 1 33 22 2 3( )M I I Iω ω ω= + −

2 22 2 11 33 3 1( )M I I Iω ω ω= + −

3 33 3 22 11 1 2( )M I I Iω ω ω= + −


Example 4.2

Figure 4.7: Unbalanced shaft for example 4.2

A rotor is imperfectly mounted on a shaft as shown in figure 4.7. The shaft is rotating at a

constant angular velocity of . Find the moment induced on the shaft (assume the

rotor is a thin disk of mass m and radius r).

Let frame a = (x, y, z) and frame b = (1, 2, 3). Frame b is obtained by rotating frame a by an angle of α about the y-axis. Frame b is also the set of principal axes for the rotor.

The angular velocity of the disk with respect to frame a is

The angular velocity with respect to frame b is

1

x

z

3zω α

G

0

0

z

ωω

=

0

0

zω

Ω =

1

2

3

sin( )

0

cos( )

z

z

ω ω αω ω

ω ω α

− = =


The mass moment of inertia of the rotor with respect to the principal axes of frame b about the center of mass ‘G’ is

We can now apply Euler’s equations of motion taking moments of external forces about the center of mass ‘G’ with respect to frame b, i.e.

For the unbalanced shaft, there is no external force. Assume that (i.e. the rotor is spinning at constant angular velocity). Therefore M1 = 0 and M3 = 0 (This is because ω2 = 0).

Notice that M2 is also acting in the direction of the y-axis (which is parallel to the 2-axis). This is the moment acting on the disk. The moment acting on the shaft will be negative of M2.

We have evaluated the above using the moment about ‘G’ with respect to frame b. We should get the same answer if we evaluate it using moment about ‘G’ with respect to frame a. In this case, the angular velocity with respect to frame a is

The mass moment of inertia of the rotor with respect to frame a can be obtained from the mass moment of inertia with respect to the principal axes of frame b about the center of mass ‘G’ using

2

112

22

332

10 0

40 01

0 0 0 04

0 0 10 0

2

G

mrI

J I mr

Imr

= =

1 11 1 33 22 2 3( )M I I Iω ω ω= + −

2 22 2 11 33 3 1( )M I I Iω ω ω= + −

3 33 3 22 11 1 2( )M I I Iω ω ω= + −

1 2 3 0ω ω ω= = =

( ) 2 22 11 33 3 1

1sin( )cos( )

4 zM I I mrω ω ω α α= − =

0

0x

y

z zω

Ω Ω = Ω = Ω

( ) ( )A TG GJ R J R=


Let (e1, e2, e3) be the unit vectors for frame b:

We can now apply the full Euler’s equations of motion taking moments of external forces about the center of mass ‘G’ with respect to frame a, (i.e. equations (7.15), 7.16) and (7.17)

in Torby 1994.) With :

Notice that you get the same answer (i.e. ) as the y-axis and the ‘2’-axis are parallel.

cos( ) 0 sin( )

0 1 0

sin( ) 0 cos( )

TR

α α

α α

− =

2

2

2

10 0

4cos( ) 0 sin( ) cos( ) 0 sin( )1

( ) 0 1 0 0 0 0 1 04

sin( ) 0 cos( ) sin( ) 0 cos( )10 0

2

AG

mr

J mr

mr

α α α α

α α α α

− = −

2

2

2

1 sin ( ) 0 sin( ) cos( )1

( ) 0 1 04

sin( ) cos( ) 0 1 cos ( )

xx xy xzA

G yx yy yz

zx zy zz

I I I

J mr I I I

I I I

α α α

α α α

+ = = +

0x y x y zΩ = Ω = Ω = Ω = Ω =

2 0x yz zM I= − Ω =

2 2 21sin( )cos( )

4y zx z zM I mr α α ω= Ω =

0zM =

2yM M=


Activity 4.3

Objective 3

A thin circular disk of mass m and radius r rolls with a constant angular velocity

about its own axis of symmetry as shown in figure 4.8. The center of mass of the disk traces a circular path of radius R about the vertical axis and the angle θ is observed to remain constant during the motion. Use the principle of linear momentum and Euler’s equations to determine the forces, toques and the

relationship between and θ. (Note that axes (1, 2, 3) are the principal axes of

the disk and frame (x, y, z) is the fixed frame. Assume no slip condition and is

constant).

Figure 4.8: Unconstrained rolling disk for activity 4.3

ψ

φ

φ


4.3 Gyroscopic dynamics

Reading

Selected reading 4.3: Before you proceed, please read ‘Applied dynamics’, pp. 195–8.

In the example given in the selected reading, the body is spinning with angular velocity about one of its principal axis. At the same time, the principal frame is also rotating with angular velocity , which is also called the rate of precession, about the fixed frame. Under these rotations, a gyroscopic moment is generated that is perpendicular to both the above axes of spins.

The gyroscopic dynamics can be expressed as

where ω is the total angular velocity of the rotating body; and Ω is the rate of precession. The moment M and angular momentum H should be taken about the center of mass or a stationary point. The application of the gyroscopic dynamics will be illustrated in the following example:

φ

ψ

( )GM H J ω= Ω× = Ω×


Example 4.3

Analyze activity 4.3 using the gyroscopic dynamics equation.

In figure 4.8, the body is spinning with angular velocity about the principal 3-axis e3. The

rate of precession of frame b with respect to frame a is about the z-axis. We shall take

moments about the center of mass with respect to frame b. The mass moment of inertia of the thin disk about ‘G’, the center of mass, with respect to (e1, e2, e3) is:

Notice that as the disk rolls an angle of ψ, the distance traveled is . If there is no slip, we can view this distance traveled with respect to frame a = (x, y, z) as

(note the negative sign is due to the direction of rotation of φ). Hence

for no slip:

The total angular velocity of the body with respect to frame b = (e1, e2, e3) = (1, 2, 3) under the no slip assumption is

ψ

φ

2

2

2

10 0

41

0 04

10 0

2

G

mr

J mr

mr

=

s rψ=

( )cos( )s R r θ φ= − +

( ) ( )cos( ) cos( )r R r r R rψ θ φ ψ θ φ= − + → = − +

( )cos( )R r

r

θψ φ

+= −

( )

1

2

3

0 0 0

sin( ) sin( ) sin( )

cos( ) cos( )cos( )

RR rrr

ωω ω φ θ φ θ φ θ

ω ψ φ θ θ φφ θ

= = = = + + − −


The rate precession with respect to frame b = (e1, e2, e3) is given by

Using the equation for the gyroscopic dynamics, the moment of the forces about ‘G’ with respect to frame b is now written as

Note that FAy is the force due to centripetal acceleration and . The force FAz is due

to the weight of the disk and . The moment of external forces taken about e1 is

The above equation can be reduced to

The answer is similar to that derived earlier using Euler’s equations (refer to solution for activity 4.3).

0

sin( )

cos( )

φ θφ θ

Ω =

12

2

32

00 0

1sin( ) ( ) sin( ) sin( )

4cos( ) cos( )

2

G G

M

M M J mr

M Rmr

r

φ θ ω φ θ φ θφ θ φ θ

φ

= = × = ×

−

2AyF Rφ=

AzF mg= −

2 2 2 21

1 1sin( ) cos( ) sin( ) sin( ) cos( )

2 4Ay Az

RM F r F r mr mr

rθ θ φ θ φ θ θ= − = − −

2 2 2 2 21 1sin( ) cos( ) sin( ) sin( )cos( )

2 4

RmR r mgr mr mr

rφ θ θ φ θ φ θ θ− = − −

2 2 2

2

1 1sin( ) cos( ) sin( ) sin( )cos( )

2 41 1

sin( ) cos( ) cos( )2 4

R g R r

R R r g

φ θ θ φ θ φ θ θ

φ θ θ θ

− = − −

+ + =

2 6 cos( )sin( ) cos( )

4

R rg

θφ θ θ+ =

( )4 cos( )

sin( ) 6 cos( )

g

R r

θφθ θ

=+


Activity 4.4

Objective 4

Figure 4.9: Top for activity 4.4

Analyze the dynamics of the spinning top in figure 4.9 and determine the relationship between ψ and φ (Note that the distance ‘OG’ = L (‘G’ is the center of mass of the body). The figure also shows the side view of the body in the y-z plane. Assume a symmetrical body such that

and that the angular velocities and are constants.11 22I I= φ ψ


Activity feedback

Activity 4.1

Given Jd = rotary inertia of disc about ‘O’; JG = rotary inertia of crank about ‘G’; mG = mass of the crank; and mp = mass of the slider.

The mass moment of inertia of the slider about point ‘O’ is where rp is the distance from ‘O’ to the slider.

The mass moment of inertia of the crank about ‘O’ is obtained using the parallel axis theorem:

where rG is the distance from ‘O’ to ‘G’.

Furthermore, note that

Hence

The total rotary inertia about ‘O’ is

Using the above relationships, the total rotary inertia can be expressed as a function of θ.

2p p pJ m r=

2 2 2( cos( ) cos( ))pr r l dθ γ= + +

2/ 0G G G GJ J m r= +

2 2 2( cos( ) cos( )) ( sin( ) sin( ))Gr r a r aθ γ θ γ= + + −

sin( ) sin( )r d lθ γ= +

1 sin( )sin

r d

l

θγ − − =

/d G O pJ J J J= + +


Activity 4.2

We first determine the mass moment of inertia of the whole cube without removing the cut-out portion:

From tables, the mass moment of inertia of the big cube (of mass ‘M’) with respect to frame b = (1, 2, 3) taken about ‘G’ is:

x

y

z

subtract

G

3

1

2a

a

a

x

y

z

O

2

21

2

10 0

61

0 06

10 0

6

G

Ma

J Ma

Ma

=


Distance ‘OG’ with respect to (x, y, z) is

We can use the parallel axis theorem to get the mass moment of inertia with respect to frame a = (x, y, z) about ‘O’ since both frames a and b are parallel, i.e.

where M = mass of the cube.

Similarly, we can find the mass moment of inertia for the cut-off portion shown below with respect to its center of mass ‘G2’ and frame b = (1, 2, 3)

Distance ‘OG2’ with respect to (x, y, z) is

2

2

2

a

d a

a

=

2 2 2

2 2 2 21 1

2 2 2

2 1 11 1 1

3 4 42 4 41 1 1 1 2 1

4 2 4 4 3 41 1 1 1 1 24 4 2 4 4 3

O G

a a a

J J M a a a Ma

a a a

− −− − = + − − = − − − − − −

G2

3

1

2

(a/2)×(a/2)×(a/2)

x

y

z

O

2

22

2

10 0

241

0 024

10 0

24

G

ma

J ma

ma

=

3 4

4

3 4

a

d a

a

=


We can again use the parallel axis theorem to get the mass moment of inertia with respect to frame a = (x, y, z) about ‘O’ since both frames a and b are parallel, i.e.

where m = mass of the cut-out portion. Note that

The total mass moment of inertia of the body with respect to frame a = (x, y, z) about ‘O’ is given by

Activity 4.3

Let frame a = (x, y, z) and frame b = (1, 2, 3). Notice that as the disk rolls an angle of ψ, the distance traveled is . If there is no slip, we can view this distance traveled with

respect to frame a as (note the negative sign is due to the direction

of rotation of φ). Hence for no slip:

The free body diagram of the unconstrained rolling disk is given in the figure below. The

angular velocity of the disk expressed in frame b is . This can be viewed as a

vector pointing in the direction of the positive e3 or 3-axis. This vector can be resolved into 2

components: a component along the z-axis and a component along the y-axis. Hence, the angular velocity of the disk expressed in frame a is

2 2 2

2 2 2 22 2

2 2 2

10 3 9 2 3 9

16 16 16 3 16 163 18 3 3 7 3

16 16 16 16 6 169 3 10 9 3 2

16 16 16 16 16 3

O G

a a a

J J m a a a ma

a a a

− − − − = + − − = − − − − − −

8

Mm =

1 2O O OJ J J= −

s rψ=

( )cos( )s R r θ φ= − +

( ) ( )cos( ) cos( )r R r r R rψ θ φ ψ θ φ= − + → = − +

( )cos( )R r

r

θψ φ

+= −

0

0bψψ

=

cos( )ψ θ sin( )ψ θ−

0

sin( )

cos( )aψ ψ θ

ψ θ

= −


Taking into consideration , the total angular velocity of the body expressed in frame a is

Similarly, the angular velocity can be viewed as a vector pointing in the direction of the

positive z-axis. This vector can be resolved into 2 components: a component along

the e2 or 2-axis and a component along the e3 or 3-axis. Hence, the total angular velocity expressed in frame b is

We will analyze the body using Euler’s equations by taking moments with respect to the center of mass G with respect to frame b.

φ

0

sin( )

cos( )

ψ θψ θ φ

Ω = − +

φsin( )φ θ

cos( )φ θ

1

2

3

0

sin( )

cos( )

ωω ω φ θ

ω ψ φ θ

= = +


Free body diagram of example 4.2

We can eliminate ψ in the above equation using the no slip condition, i.e.

Let (i, j, k) be the unit vectors of frame a = (x, y, z) and (e1, e2, e3) be the unit vectors of frame b. In terms of the unit vectors:

The angular acceleration of the body with respect to frame b is

Note that and e2 and e3 are rotating with respect to fixed frame a at and

respectively. From module 1:

( )

1

2

3

0 0 0

sin( ) sin( ) sin( )

cos( ) cos( )cos( )

RR rrr

ωω ω φ θ φ θ φ θ

ω ψ φ θ θ φφ θ

= = = = + + − −

( ) ( )2 3sin( )R

e er

ω φ θ φ= −

( ) ( ) ( ) ( )2 2 3 3sin( ) cos( ) sin( )d R R

e e e edt r r

ω φ θ φ θ θ φ θ φ φ= + + − −

0θ ψ φ= = = 2eω

3eω


The respective angular velocities and of e2 and e3 are expressed in frame b. Using the above relationships, we get

Notice that there are no angular accelerations in the directions of e2 and e3. The angular acceleration is in the direction of e1, which is parallel to the x-axis, i.e.

The mass moment of inertia for a thin disk about the center of mass with respect to the principal axis defined by frame b = (e1, e2, e3) can be obtained from tables:

Notice that point ‘O’ is an instantaneous center (instantaneous stationary point). For convenience, we will place the origin of frame a = (x, y, z) axes at this stationary point ‘O’. The position of ‘G’, the center of mass of the disk, from ‘O’ with respect to frame a is given by

2

3

22 2

33 3

0 0 cos( )

sin( ) 1 0

cos( ) 0 0

0 0 sin( )

sin( ) 0 0

cos( ) 1 0

e

e

dee e

dt

dee e

dt

φ θω φ θ

φ θ

φ θω φ θ

φ θ

− = = × = × = = = × = × =

2eω3eω

2 22

1

2

3

cos( )sin( ) sin( )sin( )cos( )

0 0 0

0 0 0

R R r

r rd

dt

θφ θ φ θω φ θ θω ω

ω

+ − − − = = + =

( ) 2cos( )sin( )

0

0

R r

r

θφ θ

ω

+ −

= Ω =

2

112

22

332

10 0

40 01

0 0 0 04

0 0 10 0

2

G

mrI

J I mr

Imr

= =


The acceleration of the center of mass ‘G’ with respect to ‘O’ expressed in frame a is given by

Basically, the above equation indicates that the acceleration due to rotation has two components: a tangential component and a normal centripetal component (refer to module 1).

or

The above result is obvious, as the center of mass will only have a centripetal acceleration in

the direction of the y-axis due to the constant rotation velocity , i.e. .

We can now use the free body diagram to analyze the forces and moments. First consider the forces with respect to the (x, y, z) axes. The total external forces acting on the body is given by

where and . Using the principle of linear

momentum:

The above vector equation indicates that .

The moment of the external forces should be taken about the center of mass (or stationary point ‘O’). In the above Euler’s equations, we are taking the moments about the center of mass ‘G’ with respect to frame b. Application of the Euler’s equations of motion leads to:

/

0

tan( )

G Or R

R θ

= −

/ / /G O G O G Or r r= Ω× + Ω×Ω×

2

/

cos( )sin( ) 0 0 0 0

0 sin( ) sin( )

0 tan( ) cos( ) cos( ) tan( )

G O

R r

rr R R

R R

θ φ θ

ψ θ ψ θθ φ ψ θ φ ψ θ θ

+ − = × − + − × − × − + +

/ 2

0

0

G Or Rφ =

/ 2

0

0

CG Oa Rφ =

φ 2AyF mRφ=

ext A cF F F= +Ax

A Ay

Az

F

F F

F

=

0

0cF

mg

= −

/ 2

0 0

0

0

Ax

CG Oext Ay

Az

F

F ma F m R

F mg

φ = ⇒ + = −

20, ,Ax Ay AzF F mR F mgφ= = =


Similarly,

and

From

we get the following relationship between and θ:

1 11 1 33 22 2 3sin( ) cos( ) ( )Ay AzM F r F r I I Iθ θ ω ω ω= − = + −

( )2 2 2 2 2cos( )1 1 1sin( ) cos( ) sin( ) sin( )

4 2 4

R r RmR r mgr mr mr mr

r r

θφ θ θ φ θ φ θ φ

+ − = − + − −

( )2 2 21 1sin( ) cos( ) cos( ) sin( ) sin( )

4 4mR r mgr mr R r mr Rφ θ θ θ φ θ φ θ − = − + −

( )( )2 22 2 11 33 3 1 0 0M I I Iω ω ω= + − ⇒ =

( )( )3 33 3 22 11 1 2 0 0M I I Iω ω ω= + − ⇒ =

( )2 2 21 1sin( ) cos( ) cos( ) sin( ) sin( )

4 4mR r mgr mr R r mr Rφ θ θ θ φ θ φ θ − = − + −

φ

2 2 21 2sin( ) cos( ) cos( ) sin( ) sin( )

4 4R g r Rφ θ θ θ φ θ φ θ− = − −

2 sin( )(6 cos( )) 4 cos( )R r gφ θ θ θ+ =

4 cos( )

sin( )(6 cos( ))

g

R r

θφθ θ

=+


Activity 4.4

The free body diagram of the spinning body is given below:

The mass moment of inertia of the top about ‘G’, the center of mass, with respect to the principal axes of frame b = (1, 2, 3) is:

Assuming a symmetry top, I11 = I22.

Let frame a = (x, y, z). The total angular velocity of the body with respect to frame b is

The rate precession with respect to frame b is given by

11

22

33

0 0

0 0

0 0G

I

J I

I

=

1

2

3

0

sin( )

cos( )

ωω ω φ θ

ω ψ φ θ

= = +

0

sin( )

cos( )

φ θφ θ

Ω =


Using the equation for the gyroscopic dynamics, the moment of the forces about ‘G’ with respect to frame b is now written as

To determine M1 above, we need to use the principle of linear momentum, i.e. from the free body diagram:

Note that the only acceleration presence is the centripetal acceleration due to (i.e. assuming constant angular velocities). Evaluating the above vector equation leads to

The moment M1 about e1 or the ‘1’-axis can now be evaluated:

Hence

1

2 22

3 33 33

0 0 0

sin( ) ( ) sin( ) sin( )

cos( ) cos( ) cos( )G G

M

M M J I

M I I

φ θ ω φ θ φ θφ θ φ θ ψ φ θ

= = × = × +

/ 2

0 0

0 sin( )

0

Ox

CG Oext Oy

Oz

F

F ma F m L

F mg

θ φ = ⇒ + = −

φ

20, sin( ) ,Ox Oy OzF F mL F mgθ φ= = =

2 21 cos( ) sin( ) sin( )cos( ) sin( )Oy OzM F L F L mL mgLθ θ φ θ θ θ= + = +

2 2 2 233 33 22sin( )cos( ) sin( ) sin( ) sin( ) cos( ) sin( ) cos( )mL mgL I I Iφ θ θ θ ψφ θ φ θ θ φ θ θ+ = + −

( ) ( )2 233 33 22cos( ) cos( )mL g L I I Iφ θ ψφ φ θ+ = + −

( ) ( )2 233 22

33

cos( ) cos( )mL g L I I

I

φ θ φ θψ

φ+ − −

=


Module 5 – System modeling

Module 5

SYSTEM MODELING 5

Module 5 – System modeling 5.1

Preamble

This module covers the mathematical modeling of systems for vibration analysis. The stiffness and dissipation elements are introduced along with the approaches to derive the equations governing the motion of the systems. You should try to focus on the model development process so that you can apply the concept to develop mathematical models for engineering systems.

Objectives


1. demonstrate an understanding of the behaviour of stiffness elements

2. demonstrate an understanding of the behaviour of dissipation elements

3. appreciate the concept of using inertia, stiffness, and dissipation elements to construct a system model

4. apply the force-balance and moment-balance methods to derive the equations of motion for a system

5. apply Lagrange’s equations to derive the system equations of motion.

5.1 Stiffness elements

Reading

Text: Before you proceed, please read sections 2.3.1 and 2.3.2, pp. 26–37 of the textbook.

The stiffness elements (commonly called springs) can store potential energy. For the spring (figure 5.1) that is extended from 0 to ∆x by an external force F, the potential energy V is the work needed to un-deform the spring to its original shape:

In general, a linear spring can be modelled as F = k∆x or T = k∆θ, depending whether it is a translational or torsional element (F = force, T = torque, x = linear displacement, θ = angular displacement, and k = spring constant). Note that the potential energy of a torsional spring is

0

x

V Fdx= ∫


Figure 5.2 shows a typical translational spring. Notice that point ‘A’ and point ‘B’ can move independently of each other. Let the displacement at point ‘A’ be x1 and displacement at point ‘B’ be x2. Frames of reference are attached at points ‘A’ and ‘B’ to indicate the direction of positive displacements for x1 and x2.

Figure 5.1: Simple stiffness element

Figure 5.2: Translation spring

Consider the following two cases when the translation spring is in tension (figure 5.3a) and compression (figure 5.3b):

0

V Tdθ

θ= ∫

F

k

Initial un-deformed length without force F

Extension ∆x due to force F

Equilibrium position x=0

kA B

x x

y y

x1 x2


Figure 5.3: Translational springs in tension and compression

Since force is a vector, the reference frame in figure 5.3 is also applicable to the force F. For the spring in tension, the force and the potential energy are respectively governed by:

Notice that in tension, the force at point ‘A’ is negative (this corresponds to –x1) while the force at point ‘B’ is positive (this corresponds to x2). If point ‘A’ is fixed (i.e. x1 = 0), thenF = kx2.

For the spring in compression, the force and the potential energy are respectively governed by:

Notice that in compression, the force at point ‘A’ is positive (this corresponds to x1) while the force at point ‘B’ is negative (this corresponds to –x2). If point ‘A’ is fixed (i.e. x1 = 0), thenF = –kx2.

Figure 5.4 shows a typical torsional spring in tension (figure 5.4a) and compression (figure 5.4b). Notice that point ‘A’ and point ‘B’ can rotate independently of each other. Let the angular displacement at point ‘A’ be θ1 and angular displacement at point ‘B’ be θ2. Frames of reference are attached at points ‘A’ and ‘B’ to indicate the direction of positive angular displacements of θ1 and θ2.

Figure 5.4: Torsional springs in tension and compression

(a) Tension (b) Compression

k k

F

A B

x1 x2

F F

A B

x1 x2

F

( ) ( ) ( )2 2

2 1 2 1 1 2

1 1

2 2F k x x V k x x k x x= − = − = −

( ) ( ) ( )2 2

1 2 1 2 2 1

1 1

2 2F k x x V k x x k x x= − = − = −

(a) Tension (b) Compression

A B

Tk

x x

y y

T

θ1 θ2

Tk

x x

y y

T

θ1 θ2

BA


Since torque is a vector, the reference frame in figure 5.4 is also applicable to the torque T. For the torsional spring in tension, the torque and the potential energy are respectively governed by:

Notice that in tension, the torque at point ‘A’ is negative (this corresponds to –θ1) while the torque at point ‘B’ is positive (this corresponds to θ2). If point ‘A’ is fixed (i.e. θ1 = 0), thenT = kθ2.

For the torsional spring in compression, the torque and the potential energy are respectively governed by:

Notice that in compression, the torque at point ‘A’ is positive (this corresponds to θ1) while the torque at point ‘B’ is negative (this corresponds to –θ2). If point ‘A’ is fixed (i.e. θ1 = 0), thenT = –kθ2.

Figure 5.5 shows ‘n’ springs with different spring constants connected in parallel. These ‘n’ springs in parallel can be combined and the result is equivalent to one spring with an equivalent spring constant of keq, i.e.

Figure 5.5: Combining springs in parallel

Figure 5.6 shows ‘n’ springs with different spring constants connected in series. These ‘n’ springs in series can be combined and the result is equivalent to one spring with an equivalent spring constant of keq, i.e.

( ) ( ) ( )2 2

2 1 2 1 1 2

1 1

2 2T k V k kθ θ θ θ θ θ= − = − = −

( ) ( ) ( )2 2

1 2 1 2 2 1

1 1

2 2T k V k kθ θ θ θ θ θ= − = − = −

1 21

n

eq n ii

k k k k k=

= + + + =∑

k1

…

k2 kn

keq

11 2

1 1 1 1 1n

ieq n ik k k k k=

= + + + =∑


Figure 5.6: Combining springs in series

The above combinations of springs in parallel and series are applicable to both translational and torsional springs.

Table 2.3 in the textbook shows some elastic members that can be modelled as springs.

Activity 5.1

Objective 1


5.2 Dissipation elements

Reading

Text: Before you proceed, please read section 2.4.1, pp. 45–8 of the textbook.

These elements convert motion to energy that is not recoverable. For this course, we are mainly concerned with viscous damping. In general, a viscous damper can be modelled as

or depending whether it is a translational or torsional element

(F = force, T = torque, x = linear displacement, θ = angular displacement, and c = damping coefficient). The energy dissipated by elements with linear damping characteristics is

governed either by or .

k1

…

k2 kn

keq

F c x= ∆ T c θ= ∆

21

2D c x= ∆ 21

2D c θ= ∆


Figure 5.7: Translational dampers in tension and compression

Figure 5.7 shows a typical translational viscous damper in tension (figure 5.7a) and compression (figure 5.7b). Notice that point ‘A’ and point ‘B’ can move independently of each other. Let the velocity at point ‘A’ be and velocity at point ‘B’ be . Frames of reference are attached at points ‘A’ and ‘B’ to indicate the direction of positive directions for velocities

and .

Since force is a vector, the reference frame in figure 5.7 is also applicable to the force F. For the damper in tension, the force and the dissipated energy are respectively governed by:

Notice that in tension, the force at point ‘A’ is negative (this corresponds to – ) while the

force at point ‘B’ is positive (this corresponds to ). If point ‘A’ is fixed (i.e. = 0), then

F = c .

For the damper in compression, the force and the energy dissipated are respectively governed by:

Notice that in compression, the force at point ‘A’ is positive (this corresponds to ) while the

force at point ‘B’ is negative (this corresponds to – ). If point ‘A’ is fixed (i.e. = 0), then

F = –c .

The concept can be extended to torsional dampers.

Similar to springs, ‘n’ viscous dampers connected in parallel can be combined and the result is equivalent to one damper with an equivalent damper coefficient of ceq, i.e.

(a) (b)

c

F

A B

x1 x2

F F

A B

x1 x2

F

c

1x2x

1x 2x

( ) ( ) ( )2 2

2 1 2 1 1 2

1 1

2 2F c x x D c x x c x x= − = − = −

1x

2x 1x

2x

( ) ( ) ( )2 2

1 2 1 2 2 1

1 1

2 2F c x x D c x x c x x= − = − = −

1x

2x 1x

2x


Similarly, ‘n’ viscous dampers connected in series can be combined and the result is equivalent to one damper with an equivalent damper coefficient of ceq, i.e.

Activity 5.2

Objective 2


5.3 Model construction

Reading

Text: Before you proceed, please read section 2.5, pp. 50–5 of the textbook.

One of the basic ideas behind engineering analysis is to develop a mathematical model of the physical system and then uses the mathematical model to predict the system behaviours under different conditions. The examples given in the selected reading show that many systems can be modelled using basic inertia, stiffness and damping elements. The approach to use these elements to construct a system model is covered in the next section.

Activity 5.3

Objective 3


1 21

n

eq n ii

c c c c c=

= + + + =∑

11 2

1 1 1 1 1n

ieq n ic c c c c=

= + + + =∑


5.4 Force-balance and moment balance methods

Reading

Text: Before you proceed, please read section 3.2, pp. 64–73 and section 7.2.1, pp. 310–20 of the textbook (Ignore linearization on pp. 317).

The force-balance method uses Newton’s 2nd law F = ma to derive the equation of motion. The forces in springs and dampers were covered in the previous sections. Notice that the forces of the spring and damper elements should be taken about the static equilibrium position, as we are mainly concerned with the dynamic forces.

The moment-balance method uses the principle of linear momentum:

M is the total moment due to the forces acting on the system. This moment should be taken about a stationary point or at the center of mass of the system whichever is more convenient.

The procedure for model construction using these methods is as follow:

• Separate the system into inertia, stiffness and damping elements.

• Assign reference frames to all appropriate parts of these elements.

• Put in the external forces and torques.

• Assume the internal forces and torques. Put these into the free body diagram.

• Balance the forces and moments to get the set of governing equations.

• Combine and reduce these equations to get the equations of motion for the system.

The following example will illustrate the concept. (Note: linearization is not included in this course).

M Jθ=


Example 5.1

Derive the equation of motion for the system shown in figure 5.8.

Figure 5.8: System for example 5.1

We are concerned mainly with the displacement x about the static equilibrium position with mass m (At this static equilibrium position: kx0 = mg. We are not concerned with x0). The free body diagram of the system showing the separated elements is given in figure 5.9.

In figures 5.9a and 5.9b, the reference coordinate for positive displacement (i.e. x) has been defined (the reference frame should be the same in both figures). There are no external forces and torques for this example. The internal forces are F2 and F1. Figure 5.9a assumes that the spring is in tension. Figure 5.9b assumes that the spring is in compression. We will show that both assumptions in figures 5.9a and 5.9b will lead to the same system model as long as the force assignments after that are consistent, i.e. in figure 5.9a, if the spring is in tension, the force F1 on the cable attached to the disk must be pointing down (i.e. equal and opposite to F1 on the spring side.). Similarly, for the spring in compression, the force F1 on the cable attached to the disk must be pointing up in figure 5.9b. Notice also that if F2 (on the mass m) is pointing up, then the same force F2 on the cable attached to the disk must be pointing down and vice versa.

m

k

R

Note: Jo=J

Static equilibrium position with mass m

Static equilibrium position without mass m

x

O

x0


Figure 5.9: Free body diagram for example 5.1

We will start with figure 5.9a. Assume the mass m has displaced a distance x about the static equilibrium position. Note that x = Rθ (where θ is the rotation of the disk about point ‘O’). Using Newton’s 2nd law on the mass:

(Note the – F2 as the reference x is positive in the downwards direction). For the spring:

Taking moments about point ‘O’ (Note θ is positive in the counter-clockwise direction following the definition of positive direction for the reference x-axis):

Substitute the other equations into the moment equation, we can combine and reduce the result to:

We will now analyse the case given in figure 5.9b. Using Newton’s 2nd law on the mass:

(a) (b)

k

m

R

x

F2 F1

F1 x

Oθ

k

m

R

x

F1

F1 x

Oθ

F2

F2

F2

J0 J0

2F mx mRθ− = =

1F kx kRθ= =

2 1( ) OF F R J θ− =

( )2 2 0OJ mR kRθ θ+ + =


(Note that F2 is now positive as the reference x is positive in the downwards direction). For the spring (now in compression):

Taking moments about point ‘O’ (Note θ is positive in the counter-clockwise direction following the definition of positive direction for the reference x-axis):

Substitute the other equations into the moment equation, we can combine and reduce the result to:

The model is the same with that obtained using the assumptions in figure 5.9(a). Note that if you have assumed the direction of the internal force/torque but after substituting numerical values, you found the answer for the force/torque to be negative, this simply implies that your assumed direction for that force/torque is wrong. Note that the reference axis is fixed.

2F mx mRθ= =

1F kx kRθ= − = −

1 2( ) OF F R J θ− =

( )2 2 0OJ mR kRθ θ+ + =


Activity 5.4

Objective 4


Derive the equation of motion for the system shown in figure 5.10.

5.5 Lagrange’s equations

Reading

Text: Before you proceed, please read section 3.6, pp. 87–106 and section 7.2.3, pp. 323–336 of the textbook (Ignore linearization).

Lagrange’s equations offer another approach to construct the models for systems. Unlike the force-balance and moment-balance methods, Lagrange’s equations are based on principle of energy and virtual work.

The previous section shows that the equations for translational and rotational systems are similar in form except for the coordinates (i.e. x is used for linear displacement while θ is used for angular displacement). Lagrange’s method resolves this issue using generalized coordinates qi, i = 1, 2, ..., N for a system with N-degrees of freedom (i.e. N separate masses).

m

k

R

Note: Jo=J

Static equilibrium position with mass m

x

O

cF


The procedure for using Lagrange’s equations to construct the system model is as follow:

• Find the total kinetic energy T of the inertia elements and then determine

and

• Find the total potential energy V of the stiffness elements and then determine

• Find the total dissipative energy D of the dampers and then determine

• Convert the applied forces Fl and torques Ml to generalised forces Qj, where

Note that rl is the linear displacement in the direction of force Fl due to the application of Fl. Similarly ωl is the angular velocity in the direction of moment Ml due to the application of Ml.

• Substitute the above into the following Lagrange’s equations to get the system model:

j

T

q

∂∂ j

T

q

∂∂

j

V

q

∂∂

j

D

q

∂∂

l lj l l

l lj j

rQ F M

q q

ω∂ ∂= ⋅ + ⋅∂ ∂∑ ∑

, 1, 2, ,jj j j j

d T T D VQ j N

dt q q q q

∂ ∂ ∂ ∂− + + = = ∂ ∂ ∂ ∂


Example 5.2

Model the system given in figure 5.11 using Lagrange’s equations and generalised coordinates q1 = x1 and q2 = x2.

Figure 5.11:System for example 5.2

Find the total kinetic energy T of masses m1 and m2:

Note that q1 = x1, q2 = x2.

and

and

Find the total potential energy V of springs with constants k1, k2 and k3:

and

m1

FA

rA

x1k1

c1

m2

FB

rB

x2k2

c2

k3

c3

2 21 1 2 2

1 1

2 2T m x m x= +

1 1

0T T

q x

∂ ∂= =∂ ∂ 2 2

0T T

q x

∂ ∂= =∂ ∂

1 11 1

T Tm x

q x

∂ ∂= =∂ ∂ 2 2

2 2

T Tm x

q x

∂ ∂= =∂ ∂

( )22 21 1 2 2 1 3 2

1 1 1

2 2 2V k x k x x k x= + − +

1 1 2 2 11 1

( )V V

k x k x xq x

∂ ∂= = − −∂ ∂ 3 2 2 2 1

2 2

( )V V

k x k x xq x

∂ ∂= = + −∂ ∂


Find the total dissipative energy D of the dampers:

and

Convert the forces FA and FB into generalised forces. Note that rA is the displacement in the direction of FA caused by FA and rA = x1. Similarly, rB is the displacement in the direction of FB caused by FB and rB = x2. There is no external applied toque. Hence

This is because and as x1 and x2 are independent.

Similarly and :

Substituting the above into the Lagrange’s equations to get the following two equations:

and

For the first equation:

For the second equation:

( )1 2

22 21 2 2 1 3

1 1 1

2 2 2D c x c x x c x= + − +

( )1 1 2 2 11 1

D Dc x c x x

q x

∂ ∂= = − −∂ ∂ ( )3 2 2 2 1

2 2

D Dc x c x x

q x

∂ ∂= = + −∂ ∂

1 21

1 1 1 1 1

l A Bl A B A B A

l

r r r x xQ F F F F F F

q q q x x

∂ ∂ ∂ ∂ ∂= ⋅ = + = + =∂ ∂ ∂ ∂ ∂∑

1

1

1x

x

∂ =∂

2

1

0x

x

∂ =∂

2

2

1x

x

∂ =∂

1

2

0x

x

∂ =∂

1 22

2 2 2 2 2

l A Bl A B A B B

l

r r r x xQ F F F F F F

q q q x x

∂ ∂ ∂ ∂ ∂= ⋅ = + = + =∂ ∂ ∂ ∂ ∂∑

11 1 1 1

d T T D VQ

dt q q q q

∂ ∂ ∂ ∂− + + = ∂ ∂ ∂ ∂ 2

2 2 2 2

d T T D VQ

dt q q q q

∂ ∂ ∂ ∂− + + = ∂ ∂ ∂ ∂

( ) ( ) ( )

( ) ( )

11 1 1 1

1 1 1 1 2 2 1 1 1 2 2 1

1 1 1 2 1 2 2 1 2 1 2 2

0 A

A

d T T D VQ

dt q q q q

dm x c x c x x k x k x x F

dtm x c c x c x k k x k x F

∂ ∂ ∂ ∂− + + = ∂ ∂ ∂ ∂

− + − − + − − =

+ + − + + − =


The two governing equations are:

( ) ( ) ( )

( ) ( )

22 2 2 2

2 2 2 2 1 3 2 2 2 1 3 2

2 2 2 3 2 2 1 2 3 2 2 1

0 B

B

d T T D VQ

dt q q q q

dm x c x x c x k x x k x F

dtm x c c x c x k k x k x F

∂ ∂ ∂ ∂− + + = ∂ ∂ ∂ ∂

− + − + + − + − =

+ + − + + − =

( ) ( )1 1 1 2 1 2 2 1 2 1 2 2 Am x c c x c x k k x k x F+ + − + + − =

( ) ( )2 2 2 3 2 2 1 2 3 2 2 1 Bm x c c x c x k k x k x F+ + − + + − =


Activity 5.5

Objective 5

Derive the equation of motion for the system given in activity 5.4 using Lagrange’s method.


Activity feedback

Activity 5.1

The two cantilever beams act as two springs in parallel. Using entry 4 in Table 2.3 (pp. 31 of the textbook), the spring constant for each of the cantilever beam is

The combined spring constants of these two springs in parallel is

The beam pinned by the two cantilever beams has a spring constant given by entry 5 in Table 2.3 (pp. 31 of the textbook):

The spring k3 is in series with kT, hence the equivalent spring constant of the system is

Activity 5.2

Dampers c1 and c2 are connected in parallel. Their combined damping constant is

Damper c3 is connected in parallel to cT (notice that all the dampers are joined to a fixed wall). Hence the equivalent damping constant is

1 11 2 3

1

3E Ik k

L= =

1 11 2 3

1

6T

E Ik k k

L= + =

3 2 2 3

3 48

2 2

EIL EIk

LL L= =

3

1 1 1

eq Tk k k= +

1

3

1 1eq

T

kk k

−

= +

1 2Tc c c= +

3 1 2 3eq Tc c c c c c= + = + +


Activity 5.3

Springs with constants k1 and k2 are connected in parallel. Hence the equivalent spring constant is

Similarly, the dampers are also connected in parallel and the equivalent damping constant is

The system reduces to a system with one mass, one equivalent spring, and one equivalent damper. One end of the damper (and spring) is connected to the fixed wall and the other to the mass. Using the force balance method:

Activity 5.4

The free body diagram with the external force F and the assumed internal forces are given in figure above. Note that .

Using the force balance approach:

1 2eqk k k= +

1 2eqc c c= +

0eq eqmx c x k x+ + =

m

k

R

Note: Jo=J

x

O

cF

θ

x

F2 F1

F3 F4

x Rθ=

2F F mx− =


Using the moment balance approach:

Note that

We can combined and simplify the above equations:

The equation of motion for the system is simplified to

Activity 5.5

The system is of one degree of freedom and we will use the generalised coordinate .

The total kinetic energy T of the inertia elements (both translational and rotational) is

The total potential energy V of the stiffness elements is

1 3 4F F F kx cx= + = +

2 1( ) OF F R J θ− =

x R x R x Rθ θ θ= ⇒ = ⇒ =

( )

( )

O

O

F mx cx kx R J

F mR cR kR R J

θθ θ θ θ

− − − =

− − − =

2 2 2( )OJ mR cR kR FRθ θ θ+ + + =

1q θ=

( )2 2 2 2 2 21 1 1 1 1( )

2 2 2 2 2O O OT mx J m R J J mRθ θ θ θ= + = + = +

2

1

( )O

T dTJ mR

q dθ

θ∂ = = +∂

1

0T dT

q dθ∂ = =∂

2 2 21 1

2 2V kx kR θ= =

2

i

V dVkR

q dθ

θ∂ = =∂


The total dissipative energy D of the dampers is

Convert the applied forces F to generalised forces Q, (there is no external torque)

Using the Lagrange’s equations:

The equation of motion is similar to that found in activity 5.4.

2 2 21 1

2 2D cx cR θ= =

2

i

D dDcR

q dθ

θ∂ = =∂

( )11

l Aj l A

l j

r r dx dQ F Q F F F R FR

q q d dθ

θ θ∂ ∂= ⋅ ⇒ = = = =∂ ∂∑

1jj j j j

d T T D V d T T D VQ Q

dt q q q q dt θ θ θ θ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ − + + = ⇒ − + + = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

( )( ) ( )2 2 20O

dJ mR cR kR FR

dtθ θ θ+ − + + =

( )2 2 2OJ mR cR kR FRθ θ θ+ + + =


Module 6 – Vibration of single degree of freedom systems

Module 6

VIBRATION OF SINGLE DEGREE OF FREEDOM SYSTEMS 6

Module 6 – Vibration of single degree of freedom systems 6.1

Preamble

This module covers the vibration analysis of systems with one degree of freedom. The concepts of natural frequency and damping ratio along with the use of Laplace transform to solve the mathematical models are first introduced. This is followed by a discussion of the general response characteristics, which are divided into un-damped, under-damped, critically damped, and over-damped cases. Both the responses to initial conditions and to forcing functions such as harmonic and impulse excitations are presented. You should try to focus on the analysis process and understand the general response characteristics so that you can apply the concept to analyze the vibration of single degree of freedom engineering systems.

Objectives


1. demonstrate an understanding of the concept of natural frequency and damping ratio associated with single degree of freedom systems and their characteristics in vibration analysis

2. appreciate the form of governing equations for single degree of freedom systems, the concept of transfer function and the use of Laplace transform to solve the equation of motion for single degree of freedom systems

3. demonstrate an understanding of the general form of solution for single degree of freedom systems

4. determine the free responses of un-damped, under-damped, critically damped and over-damped single degree of freedom systems

5. appreciate the frequency response characteristics of single degree of freedom systems to harmonic excitation and the concept of resonance in vibration analysis

6. apply the frequency response concepts to analyze the vibration of single degree of freedom systems, including the accelerometer, systems with unbalanced rotating mass, systems with base excitation, and vibration isolation

7. appreciate the frequency response characteristics of single degree of freedom systems to multiple harmonic excitations and the use of Fourier series to transform input excitations to multiple harmonic components for vibration analysis

8. appreciate the response characteristics of single degree of freedom systems to impulse excitations and apply the concept for vibration analysis

9. appreciate the response characteristics of single degree of freedom under-damped systems to step excitations and apply the concept for vibration analysis.


6.1 Natural frequency and damping ratio

Reading


The equation of motion of a translational linear mass-spring-damper system shown in figure 6.1 is a typical single degree of freedom system. Under an external force f(t), the system can be modelled as a second order differential equation of the form:

The corresponding equation for a rotational system under external torque M is

The natural frequency of the translational system expressed in radians per second is defined as

rad/s

For the rotational system, it is

rad/s

The symbol for the natural frequency, expressed in Hertz, is fn (Hz) where

Figure 6.1: A mass-spring-damper system

( )mx cx kx f t+ + =

J c k Mθ θ θ+ + =

n

k

mω =

n

k

Jω =

2n nfω π=

m

f(t)

x

k

c


The free response of a system occurs when there is no external force or torque. The free

response is a result of initial conditions x(0) and . The system without damper, i.e. c = 0, is called un-damped. The period T of free oscillation of an un-damped one degree of freedom system is associated with the natural frequency:

The unit for the period T is seconds. (Note: non-linear spring is not covered in this course).

The damping ratio for a translational system is defined as

The damping ratio for a rotational system is defined as

The damping ratio is related to friction and is normally greater than zero.

If then c = 2mωn. This is called critical damping.

If , the system is referred to as under-damped.

If , the system is referred to as over-damped.

If , the system is referred to un-damped.

The general second order differential equations for the translational and rotational systems can be written in terms of the natural frequency and damping ratio as

and

This course only deals with linear viscous damping.

Activity 6.1

Objective 1


(0)x

1 2

n n

Tf

πω

= =

ζ

2 22n

n

cc c

m kkm

ωζω

= = =

ζ

2 22n

n

cc c

J kkJ

ωζω

= = =

ζ 1=

ζ 1<

ζ 1>

ζ 0=

2 ( )2 n n

f tx x x

mζω ω+ + = 22 n n

M

Jθ ζω θ ω θ+ + =


6.2 Governing equations for different types of applied forces and their solutions using Laplace transform

Reading

Text: Before you proceed, please read sections 3.5.1 and 3.5.2, pp. 83–6 of the textbook.

Notice that the equation of motion of a single degree of freedom system with base excitation or unbalanced rotating mass can also be expressed as a second order differential equation of the form:

For the case of base excitation, we can put q = x(t) – y(t) (i.e. the relative displacement) and

(see figure 3.6 in selected reading). For the case of the unbalanced rotating

mass, q = x and (see figure 3.7 in selected reading).

Given the mathematical equation for the system together with the known values for the mass m, damping ratio , natural frequency ωn, the applied force f(t), and the initial conditions q(0)

and , we need to solve the above second order differential equation to get the response q as a function of time. We can make use of Laplace transform to solve the differential equation.

Reading

Text: Before you proceed, please read section 4.1, pp. 115–16 and Appendix A, pp. 557–63 of the textbook.

Given , the procedure to apply Laplace transform to solve the

differential equation is as follow:

1. Covert the differential equation to the Laplace domain using the ‘s’ operator, i.e. change

x(t), , to X(s) and f(t) to F(s) using the following:

2 ( )2 n n

f tq q q

mζω ω+ + =

( ) ( )f t m y t= −2( ) sin( )of t m tεω ω=

ζ

(0)q

2 ( )2 n n

f tx x x

mζω ω+ + =

( )x t ( )x t

[ ]2 2 1( ) (0) (0) 2 ( ) (0) ( ) ( )n ns X s x sx sX s x X s F s

mζω ω− − + − + =


Note that for zero initial conditions, i.e. x(0) = 0 and , the transfer function G(s)

is defined as

2. Find the Laplace transform F(s) of f(t) using Table A, pp. 559–60 in the textbook.

3. Rearrange the equation to express X(s) as a function of s and use partial fractions to simplify the expression to forms available in Table A, pp. 559–60 in the textbook.

4. Use Table A, pp. 559–60 in the textbook to convert X(s) back to get the response x(t), i.e. response x as a function of time.

The following example will illustrate the approach.

(0) 0x =

( )2 2

( ) 1( )

( ) 2 n n

X sG s

F s m s sζω ω= =

+ +


Example 6.1

Get the response for the system shown in figure 6.1 with mass m = 1 kg, damping ratio , natural frequency ωn = 3 rad/s, the applied force f(t) = u(t) (a unit step input at t0=0),

and the initial conditions are x(0) = 0 and . Is the response similar if the damping

ratio is changed to ?

With ωn = 3 rad/s, m = 1 kg, and , the equation of motion for a unit step input is

With zero initial conditions, the Laplace transform of the equation is given by

Using entry 6 in Table A (pp 559 of textbook): . Therefore the equation reduces to

We do not have to use partial fractions to simplify the expression as we can use entry 15 in Table A (p. 560 of the textbook) to get x(t), the inverse Laplace transformation of X(s) for the under-damped system:

where rad/s and rad or 60°.

To get the response for ωn = 3 rad/s, m = 1 kg, and , for a unit step input, we have


ζ 0.5=

(0) 0x =ζ 1=

ζ 0.5=

2 ( )2 3 9 ( )n n

f tx x x x x x u t

mζω ω+ + = ⇒ + + =

2 ( ) 3 ( ) 9 ( ) ( )s X s sX s X s U s+ + =

1( )U s

s=

( )2 1( ) 3 9X s s s

s+ + =

2 2

1 1 1 9( )

( 3 9) 9 ( 3 9)X s

s s s s s s= =

+ + + +

( ) 1.51( ) 1 sin 0.1111 0.1283 sin(2.5981 1.0472)

9nt tn

dd

x t e t e tζωω ω ϕω

− − = − + = − +

21 2.5981d nω ω ζ= − = 1cos ( ) 1.0472ϕ ζ−= =

ζ 1=

2 ( )2 6 9 ( )n n

f tx x x x x x u t

mζω ω+ + = ⇒ + + =

2 ( ) 6 ( ) 9 ( ) ( )s X s sX s X s U s+ + =


Using entry 6 in Table A (p. 559 of textbook): . Therefore the equation reduces to

Using partial fractions to simplify the expression:

where . When s = 0, we get ; When s = –3, we get

; Comparing the s2 terms: A + B = 0 or B = –A. Hence

We can use entry 7 in Table A (p. 559 of the textbook) to get the inverse Laplace transformation for the first and second terms. To get the inverse Laplace transformation for the third term, we have to combine the concepts in entries 7 and 22 in Table A (pp. 559–60 of the textbook), i.e.:

The response for the critically damped case is

Notice that the forms of the response for the under-damped and critically damped cases are different. The plots of responses for the displacement x versus time t for the under-damped and critically damped cases are given figure 6.2. The two responses are different. Each response consists of two parts: the transient response and the steady state response. The transient responses are dependent on the damping ratio. For the under-damped case, there are oscillations. The steady states for the under-damped and critically damped cases are similar.

1( )U s

s=

( ) ( )2

2

1 1 1( ) 6 9 ( )

( 3)( 3)6 9X s s s X s

s s s ss s s+ + = ⇒ = =

+ ++ +

2 2

1 1( )

( 3) ( 3) ( 3)

A B CX s

s s s s s= = + +

+ + +

2( 3) ( 3) 1A s Bs s Cs+ + + + = 1

9A =

1

3C = −

2

1 1 1 1 1 1( )

9 9 ( 3) 3 ( 3)X s

s s s = − − + +

11 1( ) ( ) , ( 1, 2,3, )

( ) ( 1)!n at

nG s g t t e n

s a n− −= ⇔ = =

+ −

3 3( ) 0.1111 0.1111 0.3333t tx t e te− −= − −


Figure 6.2: Step Responses for example 6.1

Activity 6.2

Objective 2

Get the response for the system given in example 6.1 when the damping ratio . What is the response if the system becomes un-damped, i.e. ?

(Assume zero initial conditions)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.02

0.04

0.06

0.08

0.1

0.12

0.14

Time (s)

Dis

plac

emen

t (x

)

Under-damped Critically damped

ζ 1.5= ζ 0=


6.3 General solution of one degree of freedom systems

Reading


In the previous section, we have discussed the use of Laplace transformation to solve the second order differential equation. The Laplace transform of

is

We can observe the following from the Laplace transformation:

1. With no forcing function (i.e. f(t) = 0 or F(s) = 0), the solution to the differential equation is dependent on the initial conditions x(0) and .

2. With zero initial conditions, the solution of the differential equation is dependent on the forcing function f(t) or F(s)

For convenience, we can view the general solution to the differential equation as consisting of two parts: the response to the initial conditions and the response to the forcing function.

In the previous section (refer to example 6.1 and activity 6.2), we have seen that the response characteristics for un-damped, under-damped, critically damped and over-damped systems are

different. We will separate the response to the initial conditions x(0) = X0 and from

the response to the forcing function for each of these cases.

2 ( )2 n n

f tx x x

mζω ω+ + =

[ ]2 2 1( ) (0) (0) 2 ( ) (0) ( ) ( )n ns X s x sx sX s x X s F s

mζω ω− − + − + =

(0)x

0(0)x V=


General solution for the Un-damped case

and

The forcing function response will depend on the nature of the applied force. Without the applied force, the initial condition response will oscillate sinusoidally about the equilibrium position x = 0 with amplitude A0.

General solution for the Under-damped case

and .

is called the damped natural frequency. The forcing function response will

depend on the nature of the applied force. Without the applied force, the initial condition response will oscillate with frequency ωd and the amplitude A0 of the oscillation will decay exponentially to the equilibrium position x = 0.

0

0

1( ) sin( ) sin( ) ( )

t

n d nn

x t A t f t dm

ω ϕ ω η η ηω

= + + −∫

2

2 00 0

n

VA X

ω

= +

1 0

0

tan nd

X

V

ωϕ − =

0

0

( ) sin( ) sin( [ ]) ( )n

n n

ttt

d d dd

ex t A e t e t f d

m

ζωζω ζω ηω ϕ ω η η η

ω

−−= + + −∫

2

2 0 00 0

n

d

V XA X

ζωω

+= +

1 0

0 0

tan dd

n

X

V X

ωϕζω

− = +

21d nω ω ζ= −


General solution for the Critically damped case

The forcing function response will depend on the nature of the applied force. Without the applied force, the initial condition response will not oscillate and will decay exponentially to the equilibrium position x = 0.

General solution for the Over-damped case

. The forcing function response will depend on the nature of the applied

force. Without the applied force, the initial condition response will not oscillate and will decay with time to the equilibrium position x = 0. The hyperbolic sine and cosine are defined as follow:

[ ]0 0 0

0

1( ) ( )n n n

tt t

nx t X e V X te e f t dm

ω ω ω ηω η η η− − −= + + + −∫

' '0 00 '

''

0

( ) cosh( ) sinh( )

1sinh( ) ( )

n n

n

t tnd d

d

t

dd

V Xx t X e t e t

e f t dm

ζω ζω

ζω η

ζωω ωω

ω η η ηω

− −

−

+= +

+ −∫

' 2 1d nω ω ζ= −

( )1sinh( )

2z zz e e−= −

( )1cosh( )

2z zz e e−= +


Activity 6.3

Objective 3

Get the general responses for the system given in example 6.1 subjected to the applied force F = u(t) for damping ratios , , and .

The initial conditions are x(0) = 0.7 and .

6.4 Free responses of one degree of freedom systems

Reading


The previous section deals with the general responses (i.e. both initial condition response and forcing function response) of un-damped, under-damped, critically damped and over-damped single degree of freedom systems. This section will concentrate on the free responses of these systems (i.e. there is no applied force or f(t) = 0).

The free response is dependent on the initial conditions x(0) = X0 and . The

equations to the free responses of un-damped, under-damped, critically damped and over-damped single degree of freedom systems are given in the previous section. When x(0) = 0 but

, we have initial velocity response. The selected reading shows that the initial velocity

response for critically damped system reaches its equilibrium position in the shortest possible time.

The free response of the under-damped system warrants further study. It is oscillatory with amplitude decaying exponentially to 0.

Reading


Given an under-damped system with unknown damping ratio, the characteristics of the logarithmic decrement of the displacement response can be used to determine the damping ratio.

ζ 0= ζ 0.5= ζ 1= ζ 1.5=

(0) 0.3x =

0(0)x V=

V0 0≠


Activity 6.4

Objective 4


6.5 Responses of one degree of freedom systems to single harmonic excitation

Reading


The previous section deals with free responses under the initial conditions. This section assumes the initial conditions are zero and focuses on the response of the under-damped single degree of freedom to single harmonic excitation, i.e. given

We need to find the response x(t) to harmonic input excitations such as f(t) = F0sin(ωt) (sine function) or f(t) = F0cos(ωt) (cosine function). Do not confuse the natural frequency ωn with the forcing function frequency ω. Both have the same units of rad/s. The natural frequency ωn for a system is fixed by the spring constant and mass. However, the input frequency ω can change as you can apply the forcing function at any frequency.

The frequency ratio is defined as .

The general response to harmonic excitation for both the sine and cosine forcing functions contains two parts: the transient portion and the steady state portion.

The transient and steady state responses for the sine and cosine functions are tabulated in Table 6.1.

2 ( )2 n n

f tx x x

mζω ω+ + =

n

ωω

Ω =

( ) ( ) ( )trans ssx t x t x t= +


Table 6.1: Transient and steady state responses for sine and cosine excitations

As , and only the steady state response remains. Notice that for a sine input

forcing function f(t) = F0sin(ωt) (i.e. with amplitude F0 at frequency ω), the steady state response xsss is also a sine function oscillating at frequency ω but with a phase shift of θ(Ω)

and amplitude of Furthermore H(Ω) is not constant but changes with ω.

The same characteristics exist for the cosine input (i.e. f(t) = F0cos(ωt)). As ,

and only the steady state response remains. Notice that for a cosine input forcing

function f(t) = F0cos(ωt) (i.e. with amplitude F0 at frequency ω), the steady state response xcss

Input Response

f(t) = F0sin(ωt) Transient response

Steady state response

f(t) = F0cos(ωt) Transient response

Steady state response

( )20

2

( )( ) sin 1 ( )

1

nt

strans n st

F H ex t t

k

ζω

ω ζ θζ

−Ω Ω= − + Ω−

( )0( ) ( )sin ( )sss

Fx t H t

kω θ= Ω − Ω

( )20

2

( )( ) cos 1 ( )

1

nt

ctrans n ct

F H ex t t

k

ζω

ω ζ θζ

−Ω= − − Ω−

( )0( ) ( ) cos ( )css

Fx t H t

kω θ= Ω − Ω

( ) ( )2 22

1( )

1 2H

ζΩ =

− Ω + Ω

12

2( ) tan

1

ζθ − Ω Ω = − Ω

21

2 2

2 1( ) tan

2 (1 )st

ζ ζθζ

− − Ω = − − Ω

21

2 2

(1 )( ) tan

( 1) 1ct

ζθζ

− − + Ω Ω = Ω − −

t ∞→ xstrans 0→

0 ( ).F

Hk

Ω

t ∞→xctrans 0→


is also a cosine function oscillating at frequency ω but with a phase shift of θ(Ω) and

amplitude of Furthermore H(Ω) is not constant but changes with ω.

Activity 6.5

Objective 5

Find and plot the time response for the system governed by the following equation:

To analyze the transient response, let td be the time and Nd be the number of periods that it

takes for to decay to d. Then

and for , .

Activity 6.6

Objective 5

A single degree of freedom system with natural frequency of 10 rad/s and damping ratio is excited by a harmonic forcing function oscillating at 2 rad/s. How long does it takes for the transient response to decay to within 5% of the steady state amplitude?

We will now focus on the steady state response characteristics of one degree of freedom systems to harmonic excitation.

Reading


0 ( ).F

Hk

Ω

( ) 0.2586 ( ) 0.2929 ( ) 0.75sin(3 )x t x t x t t+ + =

0

( )

( )

x t k

F H Ω

211lnd

n

dt

ζζω

−≥ −

Ω

21ln

2d

dN

ζπζ

−Ω≥ − Ω

ζ 1< d < 1<

ζ 0.2=


Given

The steady state response (regardless of the damping ratio) is

where

Remember that the steady state response will oscillate at ω, the same frequency as the forcing function. If the input is a sine forcing function, the steady state displacement is also a sine function. If the input is a cosine function, the steady state displacement is also a cosine function. There can be a phase shift θ(Ω) between the input excitation and the output response.

The response amplitude is dependent on ω. We can differentiate the steady state

displacement to get the steady state velocity and acceleration. The concept is equally applicable to rotating system subjected to sinusoidal input moment M = M0sin(ωt). For rotating system, we have to replace force with torque (or moment) and x with θ in all the above equations.

Reading


The force response to a sine excitation for an un-damped single degree of freedom system

(with zero initial conditions) for is

Notice that the equation for H(Ω) has the following characteristics: for (un-damped case), when Ω = 1, . This means that the spring in un-damped system will break if

2 0 sin( )2 n n

F tx x x

m

ωζω ω+ + =

( )0( ) ( )sin ( )sss

Fx t H t

kω θ= Ω − Ω

( ) ( )2 22

1( )

1 2H

ζΩ =

− Ω + Ω

12

2( ) tan

1

ζθ − Ω Ω = − Ω

0 ( )F

Hk

Ω

nω ω≠

[ ]02

( ) sin( ) sin( )(1 ) n

Fx t t t

kω ω= − Ω

− Ω

ζ 0=H Ω( ) ∞=


, i.e. the input excitation frequency ω = ωn. This phenomenon is commonly

called resonance.

Activity 6.7

Objective 5


We will now examine in detail the characteristics of H(Ω) and θ(Ω) as Ω changes.

Reading

Text: Before you proceed, please read section 5.2.4, pp. 187–91, section 5.3.1, pp. 192–3, and section 5.3.4, pp. 201–2 of the textbook.

The following equations for H(Ω) and θ(Ω) are valid for un-damped, under-damped, critically damped and over-damped systems:

Figure 6.3: Amplitude and phase versus frequency plots

1n

ωω

Ω = =

( ) ( )2 22

1( )

1 2H

ζΩ =

− Ω + Ω

12

2( ) tan

1

ζθ − Ω Ω = − Ω


Figure 6.3 shows the plots of H(Ω) versus Ω and θ(Ω) versus Ω as the damping ratio increases from . Figure 6.3a is called the amplitude versus frequency plot and figure 6.3b is called the phase shift versus frequency plot. For all values of damping ratio , the following are true:

• At small values of Ω, i.e. , and rad.

• At (i.e. ω = ωn), and rad.

• As Ω increases from 1, i.e. Ω >> 1, and rad.

Note: and ωn is fixed. Therefore implies and implies

. The amplitude of the displacement to the harmonic excitation is given by .

For small Ω (or small ω), the amplitude can be approximated as as . This is

called the stiffness-dominated region, where the phase shift is rad.

Figure 6.3a shows that for , H(Ω) increases from 1 as ω increases from 0. Hence the magnitude of the displacement response is magnified. H(Ω) continues to increase as ω

increases until it reach a maximum at rad. (for small , ). After

that, H(Ω) decreases as ω increases. The region around (i.e. Ω = 1) is called the

damping dominated region, where the phase shift is rad.

For , figure 6.3a shows that H(Ω) decreases from 1 as ω increases from 0.

For Ω >> 1, . The displacement amplitude can be approximated as

. This is called the inertia-dominated region, where the phase

shift is rad.

The transfer function was defined in section 6.2 as

ζ 0=ζ

Ω 0→ H Ω( ) 1→ θ Ω( ) 0→

Ω 1=1

( )2

Hζ

Ω = ( )2

πθ Ω =

2

1( )H Ω ≈

Ωθ Ω( ) π→

n

ωω

Ω = Ω 0→ ω 0→ Ω ∞→

ω ∞→ 0 ( )F

Hk

Ω

0F

kH Ω( ) 1≈

θ Ω( ) 0≈

ζ 0.707<

21 2r nω ω ζ= − ζ ωr ωn≈

ω ωn≈

( )2

πθ Ω =

ζ 0.707≥

2

2 2

1( ) nH

ωω

Ω ≈ =Ω

20 0 0

2 2( ) nF F F

Hk k m

ωω ω

Ω = =

θ Ω( ) π≈

( ) 22 2

( ) 1 1( )

( ) 21 2

n n

n n

X sG s

F s m s s s sk

ζω ωζ

ω ω

= = = + + + +


The frequency response function is obtained from the transfer function by substituting s = jω,

where (the imaginary number). Hence

Remember that a complex number c = a + bj has a real part (i.e. a) and an imaginary part (i.e.

b). The magnitude of the complex number is . The angle θ for the complex

number is given by . The same concept applied to G(jω) gives the following:

We called G(jΩ), which has magnitude and phase – the frequency

response function.

Activity 6.8

Objective 5

Plot H(Ω) versus Ω, magnitude versus Ω, and phase versus Ω for the frequency response function of the system given in activity 6.7. (The range for Ω is 0 to 10).

1j = −

2 2

1 1( ) ( )

1 221

n n

G j G jk j

k j

ωζω ζω

ω ω

= ⇒ Ω = − Ω + Ω − +

2 2c a b= +

1tanb

aθ − =

( )( ) ( )2 22

1 1( )

1 2G j H

kk ζΩ = = Ω

− Ω + Ω

12

2( ( )) tan ( )

1G j

ζ θ− Ω ∠ Ω = − = − Ω − Ω

1( )H

kΩ 1

2

2tan

1

ζ− Ω − Ω


6.6 Examples of one degree of freedom systems to harmonic excitation

This section will focus on some applications of single degree of freedom systems subjected to harmonic excitation. We shall first examine systems with rotating unbalanced mass.

Reading


The general equation for systems with rotating unbalanced mass has the form:

where and . Let . Define the dimensionless time as

. The governing equation can be expressed in terms of dimensionless time as

The steady state displacement response has the form:

where

The steady state velocity and acceleration responses can be obtained by differentiating x(τ) with respect to time. Please note the following:

and

22 0 0( ) 2 ( ) ( ) sin( ) sin( )n n

eq eq

F mx t x t x t t t

m m

εωζω ω ω ω+ + = =

20 0F m εω= 0eqm M m= + o

eq

mM

mεε=

ntτ ω=

22

22 sin( )

d x dxx M

d d εζ ττ τ

+ + = Ω Ω

( ) ( )2( ) ( )sin ( ) ( )sin ( )ubx M H M Hε ετ τ θ τ θ= Ω Ω Ω − Ω = Ω Ω − Ω

( ) ( )

22

2 22

( ) ( )1 2

ubH Hζ

ΩΩ = Ω Ω =− Ω + Ω

12

2( ) tan

1

ζθ − Ω Ω = − Ω

( ) ( ),x t x τ=n

dx dx

dt dω

τ=

2 22

2 2n

d x d x

dt dω

τ=


We can analyse the characteristics of Hub(Ω) and θ(Ω) based on the above equations. The characteristics are summarized in Table 6.2.

Table 6.2: Characteristics for systems with rotating unbalanced mass

Note that , i.e. the magnitude of the force is . For

Ω >> 1, the magnitude of the response displacement is (the negative

sign is a result that the response and excitation is out of phase by π radians).

Activity 6.9

Objective 6

A machine shown in figure 6.4 has mass M = 10 kg and an unbalanced rotating mass m = 1 kg at ε = 0.5m. The spring and damper have values of k = 20 N/m and c = 5 Ns/m respectively. Find the steady state displacement and the force transmitted to the floor if the unbalanced mass was found to be rotating atω = 2 rad/s.

Conditions Characteristics Displacement and velocity

rad

rad

Ω 0→ 2( )ubH Ω → Ω

( ) 0θ Ω →

( )2( ) sinx M ετ τ≈ Ω Ω

( )3( ) cosv M ετ τ≈ Ω Ω

Ω 1= 1( )

2ubHζ

Ω =

( )2

πθ Ω =

( ) sin2 2

Mx ε πτ τ

ζ ≈ −

( ) cos2 2

Mv ε πτ τ

ζ ≈ −

Ω >>1 ( ) 1ubH Ω →

( )θ πΩ =( )( ) sinx M ετ τ π≈ Ω −

( )( ) cosv M ετ τ π≈ Ω Ω −

2( ) sin( )f M ετ τ= Ω Ω 2( )f M ετ = Ω

2

( )( )

fx M ε

ττ −≈ =Ω


Figure 6.4: Machine for activity 6.9

Next, we will look at systems with base excitation.

Reading


The governing equation for systems with base excitation can be expressed in terms of dimensionless time as

For

Note that the sine and cosine functions with the same frequency can be combined, i.e.

M

k c

m

ε

ω

x

2

22 2 ( ) ( )

d x dx dyx y f

d d dζ ζ τ τ

τ τ τ+ + = + =

0( ) sin( )y yτ τ= Ω

0 0( ) 2 ( ) 2 cos( ) sin( )dy

f y y yd

τ ζ τ ζ τ ττ

= + = Ω Ω + Ω

sin( ) cos( ) sin( ) cos( )A t B t C t C tω ω ω ϕ ω φ+ = + = −


where

Therefore

where

The system equation is

The steady state displacement response is

This can be put into the form:

where

The steady state velocity and acceleration responses can be obtained by differentiating x(τ) with respect to time. Please note the following:

and

2 2 1 1, tan , tanB A

C A BA B

ϕ φ− − = + = =

( ) ( )2 220 0 0( ) 2 sin( ) 1 2 sin( )f y y yτ ζ τ ϕ ζ τ ϕ= + Ω Ω + = + Ω Ω +

( )1tan 2ϕ ζ−= Ω

22

022 1 (2 ) sin( )

d x dxx y

d dζ ζ τ ϕ

τ τ+ + = + Ω Ω +

20( ) 1 (2 ) ( )sin( ( ))x y Hτ ζ τ ϕ θ= + Ω Ω Ω + − Ω

0( ) ( )sin( ( ))mbx y Hτ τ ψ= Ω Ω − Ω

( ) ( )

2

2 22

1 (2 )( )

1 2mbH

ζ

ζ

+ ΩΩ =

− Ω + Ω

( )3

1

2 2

2( ) tan

1 4 1

ζψζ

− Ω Ω = + Ω −

( ) ( ),x t x τ=n

dx dx

dt dω

τ=

2 22

2 2n

d x d x

dt dω

τ=


We can analyse the characteristics of and ψ(Ω) based on the above equations. The

characteristics are summarized in Table 6.3:

Table 6.3: Characteristics for systems with base excitation

Activity 6.10

Objective 6


Conditions Characteristics Displacement and velocity

rad

rad

( )mbH Ω

Ω 0→ ( ) 1mbH Ω →

( ) 0ψ Ω →

( )0( ) sin ( )x y yτ τ τ≈ Ω =

( )0( ) cosv yτ τ≈ Ω Ω

Ω 1= 21 (2 )(1)

2mbHζ

ζ+

=

1 1(1) tan

2ψ

ζ− =

( )0( ) (1)sin (1)mbx y Hτ τ ψ≈ −

( )0( ) (1)cos (1)mbv y Hτ τ ψ≈ −

Ω >>1 2( )mbH

ζΩ →Ω

11 2

2( ) tan

4 1

ζψζ

− ΩΩ = −

( )01

2( ) sin ( )

yx

ζτ τ ψ≈ Ω − ΩΩ

( )0 1( ) 2 cos ( )v yτ ζ τ ψ≈ Ω − Ω

m

k c

x

y


For the system given in figure 6.5, derive the transmissibility TR for the displacement, which is defined as the ratio of the maximum amplitude of the mass displacement to the maximum amplitude of the input displacement. The input displacement is given as

Next, we will look at how the vibration characteristics can be used for acceleration measurements.

Reading


The accelerometer is designed based on the principles of vibration using the concept of base excitation. The governing equation for such a system in terms of relative displacement and

(dimensionless time ) is

where and . Hence is the

maximum acceleration of the base. Generally, the accelerometer uses the relative displacement z to determine the base acceleration . The base displacement is obtained by solving the

governing equation:

Therefore

Note that from the above definitions

The maximum acceleration of the base can be found from

0( ) sin( )y t y tω=

ntτ ω=

2 2

02 22 sin( )

d z dz d yz a

d d dζ τ

τ τ τ+ + = − = Ω

0( ) sin( )y yτ τ= Ω 20( ) sin( )y yτ τ= − Ω Ω 2

max 0( )y yτ = Ω

y

0( ) ( )sin( ( ))z a Hτ τ θ= Ω Ω − Ω

max0 ( )

za

H=

Ω

20 0a y= Ω

22 2 2 max

max max 0 0( ) ( )( )

nn n

zy t y y a

H

ωω τ ω ω= = = =

Ω


For small , and we can use the maximum relative displacement to

approximate the maximum acceleration, i.e. .

For small damping ratio, i.e. and

Note that if the damping ratio is small, H(Ω) will increase from 1 as Ω increases from 0 (this means that for small Ω, H(Ω)>1. The accelerometer is normally design so that

At , (where ωa = 2πfa is the frequency in rad/s such that H(Ωa) = 1+d), we

have

Activity 6.11

Objective 6

An accelerometer has a natural frequency of 5 Hz. Approximate the input frequency range that can be measured with a 1% error.

Next, we will look at vibration isolation.

Reading


The transmissibility ratio TR is often used to measure the vibration isolation. The TR for systems with unbalanced rotating mass (activity 6.9) and for systems with base excitations (activity 6.10) are similar, i.e.

Ω 0→ H Ω( ) 1→2

max max ny z ω≈

0ζ →

( ) ( )22 22

1 1( )

11 2H

ζΩ = ≈

− Ω− Ω + Ω

2

1( ) 1

1H dΩ ≈ ≤ +

− Ω

a aa

n n

f

f

ωω

Ω = =

11

1a dΩ ≈ −

+


and

To minimize the vibration in both cases, we need TR << 1. This means that Hmb<<1. For system with low damping ratio (i.e. ), we get

for and

for

The reduction in transmissibility is

Activity 6.12

Objective 6

A mass-spring-damper system with mass 15 kg has a rotating unbalanced force. If the damping ratio is 0.05, find the spring constant such that only 10% of the unbalanced force is transmitted when the fan is rotating at 50 rad/s.

6.7 Responses of one degree of freedom systems to excitations with many harmonic components

The previous section deals with single periodic excitations with only one input frequency ω. This section extends the concept to input with multiple periodic excitations where there are many frequencies ωk, k = 1, 2, ..., n.

Reading

Before you proceed, please read section 5.9, pp. 232–46, and Appendix B, p. 564 of the textbook.

0

( )Tmb

FTR H

F= = Ω max

0

( )mb

xTR H

y= = Ω

ζ 0→

( ) ( )

2

22 22

1 (2 ) 1( )

11 2mbTR H

ζ

ζ

+ Ω= Ω = →

− Ω− Ω + ΩΩ 1≤

( ) ( )

2

22 22

1 (2 ) 1( )

11 2mbTR H

ζ

ζ

+ Ω= Ω = →

Ω −− Ω + ΩΩ 1≥

1R TR= −


The main concept dealing with multiple excitations is the principle of superposition. Let

• x1(t) be the response to input f1(t); and

• x2(t) be the response to input f2(t)

For a linear system (i.e. a system that can be represented by a linear differential equation) excited by multiple inputs of the form:

where A and B are constants, the principle of superposition states that the total response x(t) will be the combined responses to the individual inputs, i.e.

This result is useful for the analysis of multiple harmonic excitations. For a single harmonic excitation of the form:

The steady state response is given by

Using the principle of superposition, the same system excited by multiple harmonic excitations of the form

will have a combined steady state response of the form:

where

The above section deals with the approach to handle multiple harmonic excitations. The next issue deals with the representation of the input excitation as a combination of harmonic components. This is accomplished using Fourier series. Any periodic function f(t) with period

1 2( ) ( ) ( )f t Af t Bf t= +

1 2( ) ( ) ( )x t Ax t Bx t= +

2 0( ) 2 ( ) ( ) sin( )n n

fx t x t x t t

mζω ω ω+ + =

0

1( ) ( )sin( ( ))x t f H t

kω θ= Ω − Ω

2

1

1( ) 2 ( ) ( ) sin( )

N

n n i i ii

x t x t x t f tm

ζω ω ω φ=

+ + = +∑

1

1( ) ( )sin( ( ))

N

i i i i ii

x t f H tk

ω φ θ=

= Ω + − Ω∑

ii

n

ωω

Ω =


(where ω0 is called the fundamental frequency) can be approximated as a Fourier

series

where

Previously, we have combined sine and cosine functions with the same frequency, i.e.

where

and

We can also extend the concept to combine the terms in the Fourier series as follow:

where

and

0

2T

πω

=

[ ]00 0

1

( ) cos( ) sin( )2 n n

n

af t a n t b n tω ω

∞

=

≈ + +∑

0

0

2( ) cos( ) 0,1, 2,

T

na f t n t dt nT

ω= =∫

0

0

2( )sin( ) 1, 2,

T

nb f t n t dt nT

ω= =∫

cos( ) sin( ) sin( )A t B t C tω ω ω ψ+ = +

2 2C A B= +1tan

A

Bψ − =

( )0 0 01 1

cos( ) sin( ) sin( )n n n nn n

a n t b n t c n tω ω ω ψ∞ ∞

= =

+ = +∑ ∑

2 2n n nc a b= +

1tan nn

n

a

bψ −

=


Activity 6.13

Objective 7


Determine the steady state response of the mass-spring-damper system subjected to the saw-tooth excitation shown in figure 6.6.

6.8 Response of one degree of freedom systems to impulse input

The previous section deals with harmonic excitations. This section will focus on response of under-damped systems to impulse excitation.

Reading

Before you proceed, please read section 6.2, pp. 262–5 of the textbook.

The governing equation of an under-damped system excited by an impulse is given by

f(t)m

xk

c

time

f(t)

0 τ-τ

0( )

0

AtA t

f tAt

t

ττ

ττ

+ − ≤ ≤= ≤ ≤

0( ) ( )f t f tδ=


The response of an under-damped system to the impulse input is

The form of the response is similar to the initial condition response of an under-damped

system with and : i.e.

In other words, we can treat the impulse response as the initial velocity response if we let

and . The impulse responses for the un-damped, under-damped, critically

damped and over-damped cases are tabulated in Table 6.4 together with their corresponding initial velocity responses.

Table 6.4: Initial velocity and impulse responses

Damping Condition Initial Velocity Response Impulse Response

Undamped

Under-damped

Critically damped

Over damped

2 0 ( )2 n n

f tx x x

m

δζω ω+ + =

0( ) sin( )ntd

d

fx t e t

mζω ω

ω−=

0(0) 0x X= = 0(0)x V=

0( ) sin( )ntd

d

Vx t e tζω ω

ω−=

0 0X = 00

fV

m=

ζ 0= 0( ) sin( )nn

Vx t tω

ω= 0( ) sin( )n

n

fx t t

mω

ω=

0 ζ 1< <0( ) sin( )nt

dd

Vx t e tζω ω

ω−= 0( ) sin( )nt

dd

fx t e t

mζω ω

ω−=

ζ 1=0( ) ntx t V te ω−= 0( ) ntf

x t tem

ω−=

ζ 1>20

2( ) sinh( 1)

1

nt

n

n

V ex t t

ζω

ω ζω ζ

−

= −−

20

2( ) sinh( 1)

1

nt

n

n

f ex t t

m

ζω

ω ζω ζ

−

= −−


The maximum of the impulse response for the under-damped case occurs at

where .

The maximum transmitted force by the impulse for the under-damped case through the spring

and damper to a fixed boundary is where

.

Activity 6.14

Objective 8

An impact hammer is used to apply an unit impulse δ(t) to the system shown in figure 6.7 (m = 2 kg, k = 8 N/m, c = 2 Ns/m). Find the system response. What is the impulse response if the damper is removed? (Assume zero initial conditions)


maxx

21m nt

ϕτ ωζ

= =−

21 1

tanζϕ

ζ− − =

( ) / tane ϕ ψ ϕ− −

21

2

2 1tan

1 2

ζ ζψζ

− − = −

x

k c

m


6.9 Response of one degree of freedom systems to step input

The previous section deals with the impulse response of a single degree of freedom system. This section deals with the response of single degree of freedom systems to a step input.

Reading


The focus of the section is on the characteristics of the step response for under-damped systems. The transient response for under-damped systems is oscillatory. The key parameters to characterise the response are rise time tr, maximum percentage overshoot Po, and settling time ts. These parameters can be approximated using the following relationships:

%

where the settling time is defined as the time for the step response to decay to ±d of the steady state value.

Activity 6.15

Objective 9

What is the maximum percentage overshoot for a mass-spring-damper system subjected to a step input if the damping ratio ? (Assume zero initial conditions)

21tan( )d rt

ζωζ

−=

−

21100OP e

ζπζ

− − =

1ln

100n s

dtω

ζ = −

ζ 0.36=


Activity feedback

Activity 6.1

From activity 5.3 in module 5, the system was shown to reduce to a mass m, equivalent spring constant , and equivalent damping coefficient . The single degree of freedom system can be represented by the equation:

Hence, the natural frequency is rad/s. The damping ratio is given by

Activity 6.2

To get the response for ωn = 3 rad/s, m = 1 kg, and , for a unit step input, we have




where . When s = 0, we

get ; When s = –7.8541, we get ; When s = –1.1459, we get

; Hence

keq k1 k2+= ceq c1 c2+=

0eq eqmx c x k x+ + =

1 2eqn

k k k

m mω += =

1 2

2 2eq

n n

c c c

m mζ

ω ω+= =

ζ 1.5=

2 ( )2 9 9 ( )n n

f tx x x x x x u t

mζω ω+ + = ⇒ + + =

2 ( ) 9 ( ) 9 ( ) ( )s X s sX s X s U s+ + =

1( )U s

s=

( ) ( )2

2

1 1 1( ) 9 9 ( )

( 7.8541)( 1.1459)9 9X s s s X s

s s s ss s s+ + = ⇒ = =

+ ++ +

1 1( )

( 7.8541)( 1.1459) ( 7.8541) ( 1.1459)

A B CX s

s s s s s s= = + +

+ + + +

( 7.8541)( 1.1459) ( 1.1459) ( 7.8541) 1A s s Bs s Cs s+ + + + + + =

0.1111A = 0.019B =0.1301C = −


We can use entry 7 in Table A (p. 559 of the textbook) to get the inverse Laplace transformation for the first, second, and third terms. The response for the over-damped case is

For :




where . When s = 0, we get ; Comparing coefficient

for s2: ; Comparing coefficient for s: ; Hence

We can use entry 7 in Table A (p. 559 of the textbook) to get the inverse Laplace transformation for the first term and entry 18 to get the inverse Laplace transformation for the second term. The response for the un-damped case is

The responses for the over-damped and un-damped systems are shown in the following figure. Notice that the un-damped response oscillates forever. The over-damped transient response does not oscillate. Compare these with the responses given in example 6.1.

1 0.019 0.1301( ) 0.1111

( 7.8541) ( 1.1459)X s

s s s = + − + +

7.8541 1.1459( ) 0.1111 0.019 0.1301t tx t e e− −= + −

ζ 0=

2 ( )2 9 ( )n n

f tx x x x x u t

mζω ω+ + = ⇒ + =

2 ( ) 9 ( ) ( )s X s X s U s+ =

1( )U s

s=

( ) ( )2

2

1 1( ) 9 ( )

9X s s X s

s s s+ = ⇒ =

+

2 2

1 1( )

( 9) ( 9)

A Bs CX s

s s s s

+= = ++ +

2( 9) ( ) 1A s s Bs C+ + + = 0.1111A =

0.1111B = − 0C =

2 2

1 1 1( ) 0.1111 0.1111

( 9) ( 9)

sX s

s s s s= = −

+ +

( ) 0.1111 0.1111cos(3 )x t t= −


Activity 6.3

The given parameters and conditions are ωn = 3 rad/s, m = 1 kg, x(0) = X0 = 0.7, and

subjected to a unit step input.

The step responses for zero initial conditions were found in example 6.1 and activity 6.2 for , , and . To get the total responses, we need to get the

responses to the initial conditions when , , and .

For , the initial condition response is

where and


0(0) 0.3x V= =

ζ 0= ζ 0.5= ζ 1= ζ 1.5=ζ 0= ζ 0.5= ζ 1= ζ 1.5=

ζ 0=

0( ) sin( ) 0.7071sin(3 1.4289)n dx t A t tω ϕ= + = +

2

2 00 0 0.7071

n

VA X

ω

= + =

1 0

0

tan 1.4289nd

X

V

ωϕ − = =

ζ 0.5=

1.50( ) sin( ) 0.8718 sin(2.5981 0.9322)nt t

d dx t A e t e tζω ω ϕ− −= + = +


where and

rad and rad/s



where rad/s.

The responses are summarised in the above table. For each damping condition, the total response is given by

Damping Response

Response to Initial conditions

Step response


Step response


Step response


Step response

2

2 0 00 0 0.8718n

d

V XA X

ζωω

+= + =

1 0

0 0

tan 0.9322dd

n

X

V X

ωϕζω

− = = +

21 2.5981d nω ω ζ= − =

ζ 1=

[ ] 3 30 0 0( ) 0.7 2.4n nt t t t

nx t X e V X te e teω ωω− − − −= + + = +

ζ 1.5=

' '0 00 '

4.5 4.5

( ) cosh( ) sinh( )

0.7 cosh(3.3541 ) 1.0286 sinh(3.3541 )

n nt tnd d

d

t t

V Xx t X e t e t

e t e t

ζω ζωζωω ωω

− −

−

+= +

= +

' 2 1 3.3541d nω ω ζ= − =

ζ 0= ( ) 0.7071sin(3 1.4289)Ax t t= +

( ) 0.1111 0.1111cos(3 )Bx t t= −

ζ 0.5= 1.5( ) 0.8718 sin(2.5981 0.9322)tAx t e t−= +

1.5( ) 0.1111 0.1283 sin(2.5981 1.0472)tBx t e t−= − +

ζ 1= 3 3( ) 0.7 2.4t tAx t e te− −= +

3 3( ) 0.1111 0.1111 0.3333t tBx t e te− −= − −

ζ 1.5= 4.5 4.5( ) 0.7 cosh(3.3541 ) 1.0286 sinh(3.3541 )t tAx t e t e t−= +

7.8541 1.1459( ) 0.1111 0.019 0.1301t tBx t e e− −= + −

( ) ( ) ( )A Bx t x t x t= +


Activity 6.4

Figure E4.5 shows that the free response is oscillatory about the equilibrium position. The system is under-damped. The period of oscillation can be estimated from the graph to be about

0.6 s. The damped natural frequency is therefore rad/s. The damping

factor can be estimated from the logarithmic decrement. At time t0 = 0.25 s, response amplitude is x0 = θ0= 3.9 radians. One cycle later (i.e. p = 1) at time t1 = 0.75 s, x1 = θ1= 1 radian. Therefore using equation (4.38) in the textbook:

The damping ratio can be found using equation (4.40) in the textbook:

The natural frequency is rad/s.

Activity 6.5

Given

Hence rad/s and . Let mass

m = 1, then k = 0.2929 N/m.

This is an under-damped system. The excitation is where

.

The response consists of 2 parts. The transient response is

210.47

0.6d

πω ≈ =

0

1

1 1 3.9ln ln 1.36

1 1

x

p xδ

= = =

2 2 2

10.212

421

δζδ ππ

δ

= = =+ +

210.71

1d

n

ωωζ

= =−

( ) 0.2586 ( ) 0.2929 ( ) 0.75sin(3 )x t x t x t t+ + =

2 0.2929 0.5412n nω ω= ⇒ = 2 0.2586 0.2389nζω ζ= ⇒ =

0 sin( ) 0.75sin(3 )F t tω =

3 5.5432n

ωωω

= ⇒ Ω = =

( )20

2

( )( ) sin 1 ( )

1

nt

strans n st

F H ex t t

k

ζω

ω ζ θζ

−Ω Ω= − + Ω−

( )0.1293( ) 0.4898 sin 0.5255 0.0155tstransx t e t−= +


where

rad.

The steady state response is

where

rad.

The total response is

The response is plotted for time 0 to 50 seconds.

( ) ( )2 22

1( ) 0.0335

1 2H

ζΩ = =

− Ω + Ω

21

2 2

2 1( ) tan 0.0155

2 (1 )st

ζ ζθζ

− − Ω = = − − Ω

( )0( ) ( ) sin ( )sss

Fx t H t

kω θ= Ω − Ω

( )( ) 0.0858sin 3 3.0527sssx t t= −

12

2( ) tan 3.0527

1

ζθ − Ω Ω = = − Ω

( )0.1293( ) 0.4898 sin 0.5255 0.0155 0.0858sin(3 3.0527)tx t e t t−= + + −


Activity 6.6

Given d = 5%=0.05, ωn = 10 rad/s, and ω = 2 rad/s. Hence

s.

Therefore it takes about 0.7034 s for the response to reach within 5% of the steady state value. Note that the non-dimensional time is

Activity 6.7

Given m = 150 kg, k = 30 × 103 N/m and c = 1500 Ns/m. Hence

rad/s and

ζ 0.2= 0.2n

ωω

Ω = =

211 1 0.05 1 0.04ln ln 0.7034

2 0.2dn

dt

ζζω

− −≥ − = − = Ω

7.034d n dtτ ω= =

14.14n

k

mω = = 0.354

2 n

c

mζ

ω= =


The forcing function has amplitude F0 = 70 N and excitation frequency ω = 2π(3) rad/s.

Hence

rad.

The steady state displacement will have an amplitude = m.

Activity 6.8

The frequency response function G(jΩ) has magnitude and phase

From activity 6.7: k = 30 × 103 N/m and ; and 0 < Ω < 10.

1.333n

ωω

Ω = =

( ) ( )2 22

1( 1.333) 0.818

1 2H

ζΩ = = =

− Ω + Ω

12

2( 1.333) tan 0.882

1

ζθ − Ω Ω = = = − − Ω

0 ( ) 0.0019F

Hk

Ω =

1( )H

kΩ

12

2tan

1

ζ− Ω − − Ω

ζ 0.354=

( ) ( )2 22

( ) 1

1 2

H

k k ζ

Ω =− Ω + Ω


You can use Matlab or a spreadsheet to generate the following plots:

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Ω

H( Ω

)

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6x 10

-5

Ω

Mag

nitu

de H

( Ω)/

k


Notice the shape of the H(Ω) versus Ω plot agrees with that for an under-damped system. The increase of H(Ω) from 1 (around Ω = 1) is called dynamic amplification. Dynamic amplification will not exist for . The phase shift is –90° at Ω = 1.

Activity 6.9

Given mass M = 10 kg, unbalanced rotating mass m = 1 kg, Meq = M + m = 11 kg. At ε = 0.5m

The spring and damper have values of k = 20 N/m and c = 5 Ns/m respectively. Hence

rad/s and

At ω = 2 rad/s, . Hence

0 1 2 3 4 5 6 7 8 9 10-180

-160

-140

-120

-100

-80

-60

-40

-20

0

Ω

Pha

se θ

(Ω

)

ζ 0.707≥

0.0455eq

mM

Mεε= =

1.3484neq

k

Mω = = 0.1685

2 eq

c

kMζ = =

1.4832n

ωω

Ω = =

( ) ( )

2

2 22

( ) 1.69231 2

ubHζ

ΩΩ = =− Ω + Ω


rad.

The spring and damper separate the mass from the floor. The force transmitted to the floor will be the combined force from the spring and damper:

Notice that the sine and cosine terms have the same frequency and phase shift. Hence the sine and cosine terms can be combined, i.e.

where

The maximum transmitted force is 1.7201 N. The transmissibility TR is defined as the ratio of the maximum transmitted force to the maximum impressed force. The maximum impressed force is F0 = moew2 =2 N (see equation 5.63a in textbook). The transmissibility for the system is hence

12

2( ) tan 2.7468

1

ζθ − Ω Ω = = − Ω

( )( ) ( )sin ( ) 0.0769sin(1.4832 2.7468)ubx M Hετ τ θ τ= Ω Ω − Ω = −

( )( ) ( ) cos ( ) 0.1141cos(1.4832 2.7468)ubv M Hετ τ θ τ= Ω Ω Ω − Ω = −

( ) ( )( ) ( ) 2T

dx t dxF kx t c kx t k

dt d

τζτ

= + = +

0.0769 sin(1.4832 2.7468) 0.1141(2 )cos(1.4832 2.7468)TF k kτ ζ τ= − + −

1.5385sin(1.4832 2.7468) 0.7692cos(1.4832 2.7468)TF τ τ= − + −

sin( ) cos( ) sin( )A t B t C tω φ ω φ ω φ ϕ+ + + = + +

2 2 1, tanB

C A BA

ϕ − = + =

( ) ( )2 21.5385 0.7692 sin(1.4832 2.7468 ) 1.7201sin(1.4832 2.7468 0.3551)TF τ ϕ τ= + − + = − +

20

1.72010.86

2TF

TRm εω

= = =


Activity 6.10

The input displacement is given by:

The maximum input displacement is y0.

The governing equation for systems with base excitation is

The steady state displacement response is

The maximum output displacement is

Hence, the transmissibility ratio TR is:

Activity 6.11

The approximated maximum acceleration is

The actual theoretical acceleration is

The error generated is

For an error of 1% or less:

0 0( ) sin( ) ( ) sin( )y t y t y yω τ τ= ⇒ = Ω

2

22 2 ( )

d x dx dyx y f

d d dζ ζ τ

τ τ τ+ + = + =

20( ) 1 (2 ) ( )sin( ( ))x y Hτ ζ τ ϕ θ= + Ω Ω Ω + − Ω

20 1 (2 ) ( )y Hζ+ Ω Ω

2 20

2 2 20

1 (2 ) ( ) 1 (2 )

(1 ) (2 )

y HTR

y

ζ ζζ

+ Ω Ω + Ω= =

− Ω + Ω

2max max ny z ω≈

22 max

max 0 ( )n

n

zy a

H

ωω= =

Ω

22 max

max

2

max

( )( ) 1

( )

nn

n

zz

He H

z

H

ωω

ω

−Ω= = Ω −

Ω

( ) 1 0.01

( ) 1.01

H

H

Ω − ≤Ω ≤


Hence d = 0.01 and

Given fn = 5 Hz;

Hz

Activity 6.12

TR = (maximum force transmitted)/(maximum input force)

TR = (10% of maximum input force)/(maximum input force)

Hence TR = 10% = 0.1. For damping ratio = 0.05:

Simplify to get

or (frequency ratio cannot be negative)

rad/s

Given m = 15 kg; hence the spring constant is N/m.

Alternatively, for small damping ratio, we can approximate

N/m

The error of the approximation is

11

1a

an

f

f dΩ = ≈ −

+

11 0.4975

1a a n nf f fd

= Ω ≈ − =+

2

2 2 2

1 (2*0.05* )0.1

(1 ) (2*0.05* )TR

+ Ω= =

− Ω + Ω

4 20.001 0.0299 0.99 0Ω − Ω − =

2 11.5566Ω = 2 8.5666Ω = −

503.3995 14.7081

3.3995n

ωωΩ = ⇒ = = =Ω

2 3245nk mω= =

2

10.1 3.3166

1TR = ≈ ⇒ Ω =

Ω −

25015.0756 3409

3.3166n nk mωω ω= = = ⇒ = =Ω

3409 32455%

3245e

−= =


Activity 6.13

The saw-tooth function can be approximated as a Fourier series

where

The steady state response to is

The steady state response to

( )00 0

1

( ) cos( sin( )2 n n

n

af t a n t b n tω ω

∞

=

≈ + +∑

0

2πωτ

=

2

0

02

2 2 2( )cos( ) ( ) cos 0,1,2,

T

na f t n t dt f t n t dt nT

τ

τ

πωτ τ

−

= = = ∫ ∫

2

0

02

2 2 2( )sin( ) ( )sin 1,2,

T

nb f t n t dt f t n t dt nT

τ

τ

πωτ τ

−

= = = ∫ ∫

20

0

2 0

2 At Ata A dt dt A

τ

ττ τ τ−

= + + =

∫ ∫

20

2 0

2 2 2cos cos 0 1,2,n

At n At na A t dt t dt n

τ

τ

π πτ τ τ τ τ−

= + + = =

∫ ∫

20

2 0

2 2 2sin sin 1,2,n

At n At n Ab A t dt t dt n

n

τ

τ

π πτ τ τ τ τ π−

= + + = − =

∫ ∫

2 4 6( ) sin sin sin

2 2 3

A A A Af t t t t

π π ππ τ π τ π τ

≈ − − − −

1( )2

Af t = 1 2

Ax

k=

2

2 4 6( ) sin sin sin

2 3

A A Af t t t t

π π ππ τ π τ π τ

= − − − −


is

where

, , , ...

Note

Using the principle of superposition, the total steady state response to the saw-tooth input is

Note that the process of finding the Fourier series is very tedious. In general, for common periodic functions, we do not need to resolve to first principles to get the Fourier series. Appendix B on page 564 of the textbook has a list of Fourier series for some common periodic functions (note that the Fourier series are normalised to unit magnitude). The Fourier series derived in the activity can be obtained from entry ‘c’ in Table B (after multiplying by its actual magnitude ‘A’).

2 1 1 2 2

3 3

2 4( ) ( )sin ( ) ( )sin ( )

2

6( )sin ( )

3

A Ax t H t H t

k k

AH t

k

π πθ θπ τ π τ

π θπ τ

= − Ω − Ω − Ω − Ω

− Ω − Ω −

1

2

n

π τω

Ω =2

4

n

π τω

Ω = 3

6

n

π τω

Ω =

n

k

mω =

( ) ( )2 22

1( )

1 2k

k k

Hζ

Ω =− Ω + Ω

12

2( ) tan

1k

kk

ζθ − ΩΩ = − Ω

1 2 1 1 2 2

3 3

2 4( ) ( ) ( ) ( )sin ( ) ( )sin ( )

2 2

6( )sin ( )

3

A A Ax t x t x t H t H t

k k k

AH t

k

π πθ θπ τ π τ

π θπ τ

= + = − Ω − Ω − Ω − Ω

− Ω − Ω −


Activity 6.14

The system has a natural frequency rad/s and a damping ratio

. The system is under-damped. The response of the under-damped system

to the impulse input is

Notice that f0 = 1 for a unit impulse and the response will be 0 in the steady state (as ).

If the damper is removed, then and the impulse response is given by

The impulse response without damper will oscillate forever.

Activity 6.15

Given , i.e. an under-damped system. The percentage overshoot is given by

Alternatively, we can use rad.

2n

k

mω = =

0.252 n

c

mζ

ω= =

0.25(2) 20

2

1( ) sin( ) sin(2 1 0.25 )

2(2 1 0.25 )nt t

dd

fx t e t e t

mζω ω

ω− −= = −

−

0.5( ) 0.26 sin(1.94 )tx t e t−=

t ∞→

ζ 0=

0 1( ) sin( ) sin(2 ) 0.25sin(2 )

2(2)nn

fx t t t t

mω

ω= = =

ζ 0.36=

2 2

0.36

1 1 0.36100 100 29.75%OP e e

ζπ πζ

− − − − = = =

2sin( ) 1 1.2025ϕ ζ ϕ= − ⇒ =

tan( )100 29.75%OP eπ

ϕ − = =


Module 7 – Vibration of multi-degree-of-freedom systems

Module 7

VIBRATION OF MULTI-DEGREE-OF-FREEDOM SYSTEMS 7

Module 7 – Vibration of multi-degree-of-freedom systems 7.1

Preamble

This module covers the vibration analysis of systems with multi-degree-of-freedom. The general form of the matrix equation for multi-degree-of-freedom systems is introduced together with the concepts of natural frequencies and mode shapes for un-damped free vibration. The properties of mode shapes and the characteristics of the responses of damped systems are discussed. The module also presents the approach to solve the matrix equation using the normal mode method. The approach is used to analyze the harmonic responses of multi-degree-of-freedom systems. These concepts are then applied to reduce vibrations through isolation and absorbers. The module ends with a discussion on the use of the state space approach for vibration analysis. You should try to focus on the analysis process and understand the general characteristics so that you can apply the concept to analyze the vibration of multi-degree-of-freedom engineering systems.

Objectives


1. demonstrate an understanding of the matrix equation for the vibration analysis of multi-degree-of-freedom systems

2. demonstrate an understanding of the concept of natural frequencies and mode shapes of un-damped multi-degree-of-freedom systems

3. appreciate the properties of the mode shapes and their applications in vibration analysis

4. demonstrate an understanding of the characteristics of damped vibrations in multi-degree-of-freedom systems

5. demonstrate an understanding of the normal mode approach and apply the concept to find the responses of multi-degree-of-freedom systems

6. appreciate the use of the state space approach to analyse the vibration of multi-degree-of-freedom systems.

7.1 General form of matrix equation for linear multi-degree of freedom systems

Reading



The governing equations of motion for a linear ‘n’ degrees of freedom system can be represented by a matrix equation of the form:

is the displacement vector of generalised coordinates q1, ..., qn

is the velocity vector and is the acceleration vector.

[M] is called the inertia matrix. [M] is a matrix:

[C] is called the damping matrix. [C] is a matrix of the form

The system is dynamically uncoupled if [M] and [C] are diagonal matrices with off-diagonal elements equal to zeros.

[G] is called the gyroscopic matrix. [G] is a skew symmetrical matrix of the form

[ ] [ ][ ][ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]M q C G q K H q Q+ + + + =

1

[ ]

n

q

q

q

=

n 1×( )

1

[ ]

n

q

q

q

=

n 1×( )1

[ ]

n

q

q

q

=

n 1×( )

n n×( )

11 12 1

21 22 2

1 2

[ ]

n

n

n n nn

m m m

m m mM

m m m

=

n n×( )

11 12 1

21 22 2

1 2

[ ]

n

n

n n nn

c c c

c c cC

c c c

=

n n×( )

11 12 1

21 22 2

1 2

[ ]

n

n

n n nn

g g g

g g gG

g g g

=


[K] is called the stiffness matrix. [K] is a matrix of the form

The system is statically uncoupled if [K] is a diagonal matrix with off-diagonal elements equal to zero.

[H] is called the circulatory matrix. [H] is a skew symmetrical matrix of the form

[Q] is the force vector:

For free response, there is no external forces and [Q] = [0].

The matrix equation for the multi-degree-of-freedom system can be derived using force-balance, moment-balance, and Lagrange’s method.

n n×( )

11 12 1

21 22 2

1 2

[ ]

n

n

n n nn

k k k

k k kK

k k k

=

n n×( )

11 12 1

21 22 2

1 2

[ ]

n

n

n n nn

h h h

h h hH

h h h

=

n 1×( )

1

2[ ]

n

Q

QQ

Q

=


Activity 7.1

Objective 1


Consider the system shown in figure 7.1. The system is constrained to motion in the x-y plane. The mass moment of inertia about the z-axis through the center of mass ‘G’ is IG. Determine the matrix equation representing the system motion using generalised coordinates xA and θ, assuming all motions are taken from the static equilibrium position. What happen to the [M]

and [C] matrices if point ‘A’ = ‘G’ and ? What happen to the [K] matrix if

?

k1

x

l3 l4

l1 l2

l

GA

c2k2

GA

xA

θxG

x2

x1

y

1 1 2 2c l c l=

1 3 2 4k l k l=


7.2 Free responses of un-damped systems: natural frequencies and mode shapes

This section concentrates on the approach to find the natural frequencies of multi-degree-of-freedom systems and the concept of mode shapes.

Reading


Gyroscopic forces and circulatory forces mainly occur in rotating systems such as shaft. This section focuses on the free response of an un-damped systems without these forces, i.e.[G] = [0] and [H] = [0]. For convenience, we replace the vector [q] with [x]. For an un-damped system, [C] = [0]. For free response [Q] = [0]. Hence the matrix equation for the free response of an ‘n’ degree of freedom un-damped system reduces to

We can compare the above matrix equation with the free response of a single degree of freedom un-damped system, which has the governing equation of the form

From the differential equation of the single degree of freedom system, we have defined the

natural frequency as rad/s. The matrix equation basically consists of ‘n’

differential equations. We will now discuss the determination of the ‘n’ natural frequencies of the system represented by the matrix equation.

The single and multi-degree-of-freedom systems have the following differences:

• The matrix equation consists of ‘n’ differential equations. As such, there should be ‘n’ natural frequencies (unlike the single degree-of-freedom system, which can have only one natural frequency)

• For a single differential equation, we can find the natural frequency by dividing the spring constant k by mass m (as these are scalars). In the matrix equation, we cannot divide the

[K] matrix by a [M] matrix. This means that we will have to use matrix algebra to find the ‘n’ natural frequencies in the matrix equation. This is the eigenvalue problem described in the textbook.

For real and symmetrical matrices [M] and [K], the eigenvalues and associated eigenvectors are real. The procedure to find the natural frequencies given the matrix equation is as follow:

1. Solve the eigenvalue problem associated with the matrix equation, i.e. find (the eigenvalues), using

[ ][ ] [ ][ ] [0]M x K x+ =

0mx kx+ =

n

k

mω =

n n×( ) n n×( )

λ2


2. For an ‘n’ degree of freedom system, there will be ‘n’ solutions for , i.e. .

Associate with each eigenvalue is a natural frequency, i.e. the kth natural frequency is

. An ‘n’ degree of freedom system will have ‘n’ natural frequencies, i.e. ω1,

ω2, ..., ωn, which can be arranged such that .

Activity 7.2

Objective 2

Determine the natural frequencies for the system given in activity 7.1 withm = 2 kg, IG = 1.32 kgm2, c1 = c2 =0, k1 = 2 N/m, k2 = 1.2 N/m, l1 = l3 = 1 m and l2 = l4 = 1.1 m (i.e. point ‘A’ = ‘G’)

Associated with each natural frequency ωj is an eigenvector vector [X]j, also called the mode shape. For convenience, the mode shapes are gathered into a modal matrix. After the natural frequencies have been found, the procedure to find the mode shapes and modal matrix is as follow:

1. For each natural frequency ωj, j = 1, 2, ..., n find the eigenvector (mode shape or modal vector) [X]j using the following matrix equation:

or

2. To get the normal mode, normalise the first element in the eigenvector to 1 as shown below:

Alternatively, you can also normalise any element in the eigenvectors to 1 (Be consistence. If you normalise the 3rd element of one eigenvector to one, make sure you repeat the process and normalise the third elements of all the other eigenvectors to one). Another option is to normalise the magnitude of the eigenvector to one.

3. Place the normalised mode shapes into a modal matrix [Φ] as follow:

2det [ ] [ ] 0M Kλ + =

λ2 2 2 21 2, , nλ λ λ

2k kω λ= −

1 2 nω ω ω≤ ≤ ≤

2[ ] [ ] [ ] [0]j jM K Xλ + = 2[ ] [ ] [ ] [0]i jK M Xω − =

1

2 2 11

1

1

[ ]

j

j j j

j j

nj nj j

X

X X XX X

X X X

= =

[ ]1 2[ ] [ ] [ ] [ ]nX X XΦ =


Please revise the above concepts such as determinant, eigenvalues, and eigenvectors. For a quick revision in matrix algebra, please read section 7.8 (pp. 370–379) and section 7.10 (pp. 386–391) in Kreyszig E., ‘Advanced Engineering Mathematics’, John Wiley & Sons, 1993.

Activity 7.3

Objective 2

Determine the mode shapes and modal matrix for the system given in activity 7.2.

7.3 Properties of mode shapes

The previous section concerns the determination of the natural frequencies, mode shapes and modal matrix. This section discusses the properties of the mode shapes with respect to the mass and stiffness matrices.

Reading


The mode shapes or modal vectors are said to be orthogonal with respect to the mass and stiffness matrices as they satisfied the following conditions:

if

and

if

If , for j = 1, 2, ..., n, then

and are respectively called the modal mass and modal stiffness of the j mode.

The above concept can be extended from modal vector to the modal matrix, i.e.

[ ] [ ][ ] 0T

j kX M X = j k≠

[ ] [ ][ ] 0T

j kX K X = j k≠

j k=

[ ] [ ][ ] ˆT

jjj jX M X M=

[ ] [ ][ ] ˆT

jjj jX K X K=

ˆjjM ˆ

jjK


The orthogonality properties mean that any vibratory motion of the system can be viewed as a weighted sum of the oscillations in the individual modes.

Activity 7.4

Objective 3

Show the mode shapes of the system given in activity 7.3 are orthogonal with respect to the mass and stiffness matrices. Find the modal mass and modal stiffness of the two modes for the given system using the modal vectors. Re-calculate the modal mass and modal stiffness for the two modes using the modal matrix. Are the modal mass and modal stiffness for the two modes similar in the two approaches?

The modal vector can be normalised with respect to the modal mass to form the mass normalised mode, i.e. for j = 1, 2, ..., n,

The mass normalised modes can be combined to form the mass normalised modal

matrix :

If we used the mass normalised modal matrix, then

and

[ ] [ ][ ] [ ]11

22

ˆ 0 0

ˆ0 0

ˆ0 0

T

D

nn

M

MM M

M

Φ Φ = =

[ ] [ ][ ] [ ]11

22

ˆ 0 0

ˆ0 0

ˆ0 0

T

D

nn

K

KK K

K

Φ Φ = =

[ ] [ ]1

ˆM j j

jj

X XM

=

[ ]M jX

[ ]MΦ

[ ]1 2[ ] [ ] [ ] [ ]M M M M nX X XΦ =

[ ] [ ][ ] [ ]T

M MM IΦ Φ = [ ] [ ][ ] 2T

M M DK ω Φ Φ =


where [I] is the identity matrix and is the diagonal matrix, whose diagonal elements are

squares of the natural frequencies (with off-diagonal terms equal to zeros).

Activity 7.5

Objective 3

Find the mass normalised modal vectors and mass normalised modal matrix for

the system given in activity 7.4. Verify that and

.

7.4 Characteristics of damped systems

The previous sections deal with an un-damped system. This section extends the concept of natural frequencies to damped systems.

Reading


The section only deals with damped systems without gyroscopic and circulatory forces, i.e. [G] = [H] = [0] and [Q] = [0]. Hence the matrix equation for the free response of an ‘n’ degree of freedom damped system reduces to

The eigenvalues for the system can be obtained from the roots of the characteristics equation:

The characteristics discussed are restricted to the case of proportional damping, i.e.

where α and β are real-valued constants. The eigenvalue problem for this class of damped systems has the following characteristic equations:

2Dω

[ ] [ ][ ] [ ]T

M MM IΦ Φ =

[ ] [ ][ ] 2T

M M DK ω Φ Φ =

[ ][ ] [ ][ ] [ ][ ] [0]M x C x K x+ + =

2det [ ] [ ] [ ] 0M C Kλ λ + + =

[ ] [ ] [ ]C M Kα β= +

( )[ ] [ ]det 1 ( ) 0K Mβλ λ λ α + + + =


Solving the characteristics equation will give a total of ‘n’ eigenvalues. Let ζk be the modal

damping factor associated with the kth mode. If , then the eigenvalue will be a

complex-valued quantity of the form:

where ωk is the natural frequency of the un-damped system and is the damped natural frequency of the kth mode.

Activity 7.6

Determine if the given system has proportional damping and find the natural frequencies, modal damping factors and damped natural frequencies for the two modes:

The above activity indicates that the natural frequencies of the damped system are the same as the corresponding un-damped system (i.e. obtained by ignoring the damping). Let [Φ] = modal matrix obtained using the eigenvectors in the corresponding un-damped case. For proportional damping

There are some exceptions to the above diagonalization of the damping matrix as discussed in the textbook.

Activity 7.7

Objective 4

For the system discussed in activity 7.6, verify that .

0 1kζ≤ <

1,2dk k k dkjλ ζ ω ω= − ±

21dk k kω ω ζ= −

1 1 1

2 2 2

2 0 0.8 0.2 2 1 0

0 1 0.2 0.4 1 1 0

x x x

x x x

− − + + = − −

[ ] [ ][ ] [ ]1 1 11

2 2 22

ˆ2 0 0

ˆ0 2 0

ˆ0 0 2

T

D

n n nn

M

MC C

M

ζ ωζ ω

ζ ω

Φ Φ = =

[ ] [ ][ ] [ ]T

DC CΦ Φ =


7.5 Normal mode approach

This section discusses the normal mode approach, which is suitable for finding the responses for systems with proportionally damping without gyroscopic and circulatory forces.

Reading


Given

The procedure to find the response based on the normal model approach is as follow:

1. Find the natural frequencies, mode shapes, and modal matrix using

2. Find using

3. Find using

4. Find using

[ ][ ] [ ][ ] [ ][ ] [ ]M x C x K x F+ + =

[ ]Φ

[ ][ ] [ ][ ] [ ]0M x K x+ =

[ ]DM

[ ] [ ] [ ][ ]11

22

ˆ 0 0

ˆ0 0

ˆ0 0

T

D

nn

M

MM M

M

= Φ Φ =

[ ] 1

DM−

[ ]

11

1

22

10 0

ˆ

10 0

ˆ

10 0

ˆ

D

nn

M

MM

M

−

=

[ ]DC

[ ] [ ] [ ][ ]T

DC C= Φ Φ


5. Find using

(Note that if you have values for ζj and ωj for j = 1, 2, ..., n, you can skip some steps and form the above matrix very easily without any matrix multiplication)

6. Find

(Note that if you have the natural frequencies, you can skip some steps and form the above matrix very easily without any matrix multiplication)

7. Find using

8. Formulate the uncoupled set of equations

This is equivalent to ‘n’ equations of the form:

The change in coordinates from [x] to [η] is based on the following transformation

With the above, we now have de-coupled the multi-degree-of-freedom systems to ‘n’ single degree-of-freedom systems. We can use the concepts learned in module 6 to solve each of

these equations for . Remember that we are given the initial conditions for and

. We will need to transform these initial conditions to and in order to

solve the above de-coupled equations. This is accomplished using

( )2D

ζω

( ) [ ] [ ]1 1

1 2 2

2 0 0

0 2 02

0 0 2

D DD

n n

M C

ζ ωζ ω

ζω

ζ ω

−

= =

2Dω

[ ] [ ][ ]

21

22 2

2

0 0

0 0

0 0

T

D M M

n

K

ωω

ω

ω

= Φ Φ =

[ ]Q

[ ] [ ] [ ] [ ]1 T

DQ M F−= Φ

[ ] ( ) [ ] [ ] [ ] [ ][ ] [ ][ ] [ ][ ] [ ]22 DDQ M x C x K x Fη ζω η ω η + + = ⇔ + + =

2( ) 2 ( ) ( ) ( ) 1,2, ,j j j j j j jt t t Q t j nη ζ ω η ω η+ + = =

[ ] [ ][ ]x η= Φ

( )j tη [ ](0)x

[ ](0)x [ ](0)η [ ](0)η


9. After we have obtained all the solutions for for j = 1, 2, ..., n, we need to convert it back to [x(t)] using

where

Activity 7.8

Objective 5

Determine the steady state responses of the following system (assume zero initial conditions)

The above deals with the normal mode approach to find the forced responses of multi-degree-of-freedom systems. We will now examine the responses due only to initial conditions, i.e.[F] = [0]

Reading


The free (initial conditions) responses for multi-degree-of-freedom system are summarized in table 7.1 (note that we are mainly concern here with the un-damped and under-damped cases as these will cause oscillations or vibrations)

[ ] [ ][ ]1(0) (0)

ˆT

j jjj

X M xM

η =

[ ] [ ][ ]1(0) (0)

ˆT

j jjj

X M xM

η =

( )j tη

[ ] [ ][ ] [ ]1

( ) ( ) ( )n

jjj

x t t X tη η=

= Φ =∑

[ ]1

2

( )

( )( )

( )n

t

tt

t

ηη

η

η

=

1 1 1

2 2 2

2 0 0.8 0.2 2 1 sin(3 )

0 1 0.2 0.4 1 1 2sin(3 )

x x x t

x x x t

− − + + = − −


Table 7.1: Initial condition responses

Activity 7.9

Objective 5

Determine the initial condition responses of the following system

with

and

The above deals with the initial conditions responses. We will now examine the response of multi-degree-of-freedom to harmonic excitation with zero initial conditions.

Damping Condition Initial Condition Response

Un-damped

and

Under-damped

and

[ ] [ ]1

( ) sin( )n

j j jjj

x t X A tω ϕ=

= +∑

2

2 (0)(0) j

j jj

Aη

ηω

= +

1 (0)tan

(0)j j

jj

ω ηϕ

η−

=

0jζ =

[ ] [ ]1

( ) sin( )j j

nt

j dj djjj

x t X A e tζ ω ω ϕ−

=

= +∑2

2 (0) (0)(0) j j j j

j jdj

Aη ζ ω η

ηω

+= +

1 (0)tan

(0) (0)dj j

djj j j j

ω ηϕ

η ζ ω η−

= +

0 1jζ< <

1 1 1

2 2 2

2 0 0.8 0.2 2 1 0

0 1 0.2 0.4 1 1 0

x x x

x x x

− − + + = − −

[ ] 0.5(0)

0.5x

=

[ ] 0(0)

0x

=


Reading


The response to the harmonic excitation of a proportional damped two degree-of-freedom system can be determined using the normal mode approach discussed earlier.

Activity 7.10

Objective 5

Determine the responses of the following system (assume zero initial conditions)

7.6 State-space formulation

Reading


The state space formulation is an alternative approach to determine the responses of a multi-degree-of-freedom system. In state space form, the matrix equation can be easily solved using numerical methods (Matlab has functions to solve state space equation). Note that the state variables of the system are not unique. The textbook introduces one way of forming the state space equation.

Activity 7.11

Objective 6

Get the state space equations and use Matlab to determine the responses for the systems discussed in activities 7.8, 7.9, and 7.10.

1 1

2 2

2 0 2 1 1sin(3 )

0 1 1 1 2

x xt

x x

− + = −


Activity feedback

Activity 7.1

The figure below shows the displacement of the system.

We first consider the left end of the beam. Note that

For small θ, . Therefore . Note that .

Similarly, we can consider the right end of the beam using the following figure:

Note that . Similarly, we can determine that .

We will now consider the forces and moments acting on the system using the following free body diagram:

l3

A

xAθh1

x1

1

3

sin( )h

lθ =

1

3

sin( )h

lθ θ≈ =

1 3h l θ≈ 1 1 3A Ax x h x l θ= − = −

l4

A

xAθ

h2

x2

2 2 4A Ax x h x l θ= + = +1 3( )G Ax x l l θ= + −


Using :

Taking moments about point ‘A’: :

k1

l4

l1 l2

l

GA

c1 k2 c2

F1 F2 F3 F4

F1+F2 F3+F4

Gxm

l3

y

x

G GF ma mx= =∑

1 2 3 4 GF F F F mx+ + + =

1 1 1 1 2 2 2 2 Gk x c x k x c x mx− − − − =

1 3 1 3 2 4 2 4 1 3( ) ( ) ( ) ( ) ( ( ) )A A A A Ak x l c x l k x l c x l m x l lθ θ θ θ θ− − − − − + − + = + −

0A GM I θ− =∑

1 3 2 3 3 4 4 4 1 3( ) 0G GF l F l F l F l mx l l I θ+ − − − − − =

1 3 3 1 3 3 2 4 4 2 4 4

1 3 1 3

( ) ( ) ( ) ( )

( ( ) )( ) 0

A A A A

A G

k x l l c x l l k x l l c x l l

m x l l l l I

θ θ θ θ

θ θ

− + − − + − +

− + − − − =


The two equations can be combined in matrix form:

, ,

, and [G] = [H] = [0]

If point ‘A’ = ‘G’, then and . In addition if , then the matrix equation

reduces to

Notice that the [M] and [C] are diagonal matrices and the off diagonal terms are 0. This condition is called dynamically uncoupled.

If , then the matrix equation becomes

Notice that the [K] is a diagonal matrix and the off diagonal terms are 0. This condition is called statically uncoupled.

Activity 7.2

With m = 2 kg, IG = 1.32 kgm2, c1 = c2 = 0, k1 =2 N/m, k2 =1.2 N/m, I1 = I3 = 1 m andI2 = I4 = 1.1 m, the matrix equation becomes

1 3 1 2 1 3 2 42 2 2

1 3 1 3 1 3 2 4 1 3 2 4

1 2 1 3 2 42 2

3 3 2 4 1 3 2 4

( ) ( ) ( )

( ) ( ( ) ) ( ) ( )

( ) ( ) 0

( ) ( ) 0

A A

G

A

m m l l c c c l c lx x

m l l I m l l c l c l c l c l

k k k l k l x

k l k l k l k l

θ θ

θ

− + − + + − + − − + +

+ − + + = − + +

1 32

1 3 1 3

( )[ ]

( ) ( ( ) )G

m m l lM

m l l I m l l

− = − + −

1 2 1 3 2 42 2

1 3 2 4 1 3 2 4

( ) ( )[ ]

( ) ( )

c c c l c lC

c l c l c l c l

+ − + = − + +

1 2 1 3 2 42 2

3 3 2 4 1 3 2 4

( ) ( )[ ]

( ) ( )

k k k l k lK

k l k l k l k l

+ − + = − + +

3 1l l= 4 2l l= 1 1 2 2c l c l=

1 22 2

1 3 2 4

1 2 1 3 2 42 2

3 3 2 4 1 3 2 4

0 ( ) 0

0 ( ) 0 ( )

( ) ( ) 0

( ) ( ) 0

A A

G

A

m c cx x

I c l c l

k k k l k l x

k l k l k l k l

θ θ

θ

+ + +

+ − + + = − + +

1 3 2 4k l k l=

1 3 1 2 1 3 2 42 2 2

1 3 1 3 1 3 2 4 1 3 2 4

1 22 2

1 3 2 4

( ) ( ) ( )

( ) ( ( ) ) ( ) ( )

( ) 0 0

0 ( ) 0

A A

G

A

m m l l c c c l c lx x

m l l I m l l c l c l c l c l

k k x

k l k l

θ θ

θ

− + − + + − + − − + +

+ + = +

2 0 3.2 0.68 0

0 1.32 0.68 3.452 0G Gx x

θ θ−

+ = −


The eigenvalue problem associated with the matrix equation can be used to find :

The 2 solutions for are and .

Associate with each solution is a natural frequency, i.e. and

The 1st natural frequency is ω1 = 1.204 rad/s and the 2nd natural frequency is ω2 = 1.663 rad/s.

Activity 7.3

The two natural frequencies found in activity 7.2 are ω1 = 1.204 rad/s and ω2 =1.663 rad/s.

For the first natural frequency ω1, the eigenvector [X]1 can be found using:

For the second natural frequency ω2, the eigenvector [X]2 can be found using:

2λ

22

2

2 3.2 0.68det [ ] [ ] det 0

0.68 1.32 3.452M K

λλλ

+ − + = = − +

( )( )2 22 3.2 1.32 3.452 0.4624 0λ λ+ + − =

4 22.64 11.128 10.584 0λ λ+ + =

2λ 21 1.4497λ = − 2

2 2.7654λ = −

21 1 1.4497 1.204ω λ= − = =

22 2 2.7654 1.663ω λ= − = =

1121 1

21

2 0 3.2 0.68 0[ ] [ ] [ ] 1.4497

0 1.32 0.68 3.452 0

XM K X

Xλ

− + = − + = −

[ ]

11

21

11111

11

0.3006 0.68 0

0.68 1.5384 0

1

0.442 0.442

X

X

XX X

X

− = −

= =

[ ] [ ] [ ] 1222 2

22

2 0 3.2 0.68 02.7654

0 1.32 0.68 3.452 0

XM K X

Xλ

− + = − + = −

12

22

2.3308 0.68 0

0.68 0.1983 0

X

X

− − = − −

[ ] 12122

12

1

3.4278 3.4278

XX X

X

= = − −


Notice that the first and second eigenvectors are both normalised such that their first elements have the same value 1. The normalised eigenvectors are

and

The normalised mode shapes are placed into a modal matrix [Φ] as follow:

The sketches of the mode shapes for [X]1 and [X]2 are as follow:

[ ]1

1

0.442X

=

[ ]2

1

3.4278X

= −

[ ]1 2

1 1[ ] [ ] [ ]

0.442 3.4278X X

Φ = = −

1 2

1

0.442

DOF

Mode 1

1 2

1

-3.4278

DOF

Mode 2

-1

-2

-3

-4


The significant of the mode shapes can be interpreted by considering the system depicted by the following diagram:

The displacement of any point ‘P’ on the beam can be obtained with respect to point ‘G’. We define positive y if point ‘P’ is on the right of ‘G’. The displacement of point ‘P’ is

. In the diagram, there is a pivot point ‘O’, which does not have any

displacement. Note that

where is position from ‘G’ to ‘O’ ( is positive if O is on the right of G). For small θ,

and

Notice that in activity 7.2, the generalised coordinates are xG and θ. The mode shapes obtained concerns the relative displacement of the generalised coordinates, i.e. for the first natural frequency ω1 = 1.204 rad/s or the first mode:

We can use the above mode shape to obtain the position of the pivot point for the first mode:

This means that the pivot point is 2.2624 m to the left of ‘G’ for the first mode.

yP

G

xPθ

xGx2

x1

O

P

p G Px x y θ= +

0

tan( ) Gx

yθ =

−

0y 0y

tan( )θ θ≈

Go

xy

θ= −

[ ]11

1

0.442Gx

Xθ

= ≡

12..2624

0.442G

O

xy

θ= − = − = −


Similarly, for the second mode:

The pivot point for the second mode is

This means that the pivot point is 0.2917 m to the right of ‘G’ for the second mode.

The above indicates that the pivot point can be determined from the mode shape. The pivot point is a stationary point. If a sensor is placed at a pivot point, it will not be able to measure the vibration at that mode. Hence, once we obtain the pivot point for a particular mode, we should not place the sensor at that location if we want to measure the vibration at that mode.

G

θxG

x2

x1

O

2.2624m

[ ]22

1

3.4278Gx

Xθ

= ≡ −

10.2917

3.4278G

O

xy

θ= − = =

G

θ

xG

x2

x1

O

0.2917 m


Activity 7.4

From activity 7.3, the mass matrix, stiffness matrix, normalized modal vectors for the two modes, and the modal matrix are:

, , , and

The orthogonal properties for the modal vectors with respect to the mass matrix are shown as follow:

The following shows that the modal masses determined using the modal vectors and the modal matrix are similar:

The orthogonal properties for the modal vectors with respect to the stiffness matrix are shown as follow:

[ ] 2 0

0 1.32M

=

[ ] 3.2 0.68

0.68 3.452K

− = −

[ ]1

1

0.442X

=

[ ]2

1

3.4278X

= −

[ ]1 2

1 1[ ] [ ] [ ]

0.442 3.4278X X

Φ = = −

[ ] [ ][ ] [ ]1 2

2 0 11 0.442 0

0 1.32 3.4278T

X M X

= = −

[ ] [ ][ ] [ ]2 1

2 0 11 3.4278 0

0 1.32 0.442T

X M X

= − =

[ ] [ ][ ] [ ] 111 1

2 0 1 ˆ1 0.442 2.25790 1.32 0.442

TX M X M

= = =

[ ] [ ][ ] [ ] 222 2

2 0 1 ˆ1 3.4278 17.50940 1.32 3.4278

TX M X M

= − = = −

[ ] [ ][ ] [ ]1 0.442 2 0 1 1 2.2579 0

1 3.4278 0 1.32 0.442 3.4278 0 17.5094T

DM M

Φ Φ = = = − −

[ ] [ ][ ] [ ]1 2

3.2 0.68 11 0.442 0

0.68 3.452 3.4278T

X K X−

= = − −

[ ] [ ][ ] [ ]2 1

3.2 0.68 11 3.4278 0

0.68 3.452 0.442T

X K X−

= − = −


The following shows that the modal stiffness determined using the modal vectors and the modal matrix are similar:

The two natural frequencies for the system are ω1 = 1.204 rad/s and ω2 = 1.663 rad/s. Notice that the natural frequencies can be obtained from

Note the similarity of the above equation with for single degree-of-freedom systems.

Activity 7.5

From activity 7.4, we have the following data:

, , , ,

, ;

The mass normalised modal vectors are:

[ ] [ ][ ] [ ] 111 1

3.2 0.68 1 ˆ1 0.442 3.27330.68 3.452 0.442

TX K X K

− = = = −

[ ] [ ][ ] [ ] 222 2

3.2 0.68 1 ˆ1 3.4278 48.42120.68 3.452 3.4278

TX K X K

− = − = = − −

[ ] [ ][ ] [ ]1 0.442 3.2 0.68 1 1 3.2733 0

1 3.4278 0.68 3.452 0.442 3.4278 0 48.4212T

DK K−

Φ Φ = = = − − −

[ ] [ ]1 2

1 122

2.2579 0 3.2733 0 1.4497 0 0

0 17.5094 0 48.4212 0 2.7654 0D DM Kω

ω

−−

= = =

2 k

mω =

[ ] 2 0

0 1.32M

=

[ ] 3.2 0.68

0.68 3.452K

− = − 11

ˆ 2.2579M = 22ˆ 17.5094M =

[ ]1

1

0.442X

=

[ ]2

1

3.4278X

= −

[ ] [ ]1 1

11

1 0.66551 10.442 0.2942ˆ 2.2579

MX XM

= = =

[ ] [ ]2 2

22

1 0.2391 1

3.42782 0.8192ˆ 17.5094MX X

M

= = = − −


The mass normalised modal matrix is

Using the mass normalized modal matrix:

Activity 7.6

We first must ensure that the system has proportional damping. If the system has proportional damping:

Equating the elements in the above matrix equation:

and

Since α and β are real-valued constants, the system has proportional damping.

Next we get the eigenvalues of the system using the characteristic equation:

[ ]1 2

0.6655 0.239[ ] [ ] [ ]

0.2942 0.8192M M MX X

Φ = = −

[ ] [ ][ ] [ ]0.6655 0.239 2 0 0.6655 0.239 1 0

0.2942 0.8192 0 1.32 0.2942 0.8192 0 1

TT

M MM I

Φ Φ = = = − −

[ ] [ ][ ] 0.6655 0.239 3.2 0.68 0.6655 0.239

0.2942 0.8192 0.68 3.452 0.2942 0.8192

TT

M MK−

Φ Φ = − − −

[ ] [ ][ ]2

2122

1.4497 0 0

0 2.7654 0

T

M M DKω ω

ω

Φ Φ = = =

[ ] [ ] [ ]0.8 0.2 2 0 2 1 2 2

0.2 0.4 0 1 1 1C M K

α β βα β α β

β α β− − + −

= = + = + = − − − +

2 2 0.8 0.4α β α β+ = ⇒ + =

0.2 0.2β β− = − ⇒ = 0.2α =

2 2 2 0 0.8 0.2 2 1det [ ] [ ] [ ] det 0

0 1 0.2 0.4 1 1M C Kλ λ λ λ

− − + + = + + = − −

2

2

(2 0.8 2) ( 0.2 1)det 0

( 0.2 1) ( 0.4 1)

λ λ λλ λ λ

+ + − −= − − + +

2 2(2 0.8 2)( 0.4 1) ( 0.2 1)( 0.2 1) 0λ λ λ λ λ λ+ + + + − − − − − =

4 3 22 1.6 4.28 1.2 1 0λ λ λ λ+ + + + =


The four roots of the characteristic equation can be found using the Matlab ‘roots’ command:

As the system has proportional damping: where

For k = 1:

Hence using the real part of , we get rad/s. The damped

natural frequency for this mode is rad/s.

For k = 2:

Hence using the real part of , we get rad/s. The

damped natural frequency for this mode is rad/s.

Note that the two natural frequencies can also be obtained using the corresponding un-damped system, i.e. by ignoring the damping. In this case, the natural frequencies are associated with the characteristic equation:

1,21 0.1293 0.5255d jλ = − ±

1,22 0.2707 1.2782d jλ = − ±

1,2dk k k dkjλ ζ ω ω= − ± 21dk k kω ω ζ= −

211

11 1 1

10.52550.2389

0.1293d

ζω ζζ ω ζ

−= = ⇒ =

1,21dλ1 1 10.1293 0.5412ζ ω ω= ⇒ =

21 1 11 0.5255dω ω ζ= − =

222

22 2 2

11.27820.2072

0.2707d

ζω ζζ ω ζ

−= = ⇒ =

1,22dλ 2 2 20.2707 1.3066ζ ω ω= ⇒ =

22 2 21 1.2782dω ω ζ= − =

22 2

2

2 0 2 1 (2 2) 1det [ ] [ ] det det 0

0 1 1 1 1 ( 1)M K

λλ λλ

− + − + = + = = − − +

2 2(2 2)( 1) 1 0λ λ+ + − =

4 22 4 1 0λ λ+ + =

2 4 8 11

4 2λ − ± = = − ±


The natural frequencies are

or rad/s.

or rad/s.

These natural frequencies are the same as that obtained earlier. In general, you can get the natural frequencies of the system by ignoring the damping.

Activity 7.7

For the system discussed in activity 7.6, the two natural frequencies are at rad/s

and rad/s.

For the first natural frequency ω1, the eigenvector [X]1 associated with the un-damped case is:

For the second natural frequency ω2, the eigenvector [X]2 associated with the un-damped case is:

1

11 0.2929 0.5412

2ω = − = = 1 0.5412ω =

2

11 1.7071 1.3066

2ω = + = = 2 1.3066ω =

1 0.5412ω =

2 1.3066ω =

112 21 1 1

21

2 0 2 1 0[ ] [ ] [ ]

0 1 1 1 0

XM K X

Xλ ω

− + = − + = −

[ ]

11

21

11111

11

1.4142 1 0

1 0.7071 0

1

1.4142 1.4142

X

X

XX X

X

− = −

= =

[ ] [ ][ ] [ ] 111 1

2 0 1 ˆ1 1.4142 40 1 1.4142

TX M X M

= = =

122 22 2 2

22

2 0 2 1 0[ ] [ ] [ ]

0 1 1 1 0

XM K X

Xλ ω

− + = − + = −

[ ]

12

22

12122

12

1.4142 1 0

1 0.7071 0

1

1.4142 1.4142

X

X

XX X

X

− − = − −

= = − −


We have to verify that

From activity 7.6, and rad/s. We found . Hence

From activity 7.6, and rad/s. We found . Hence

This agrees with the above [CD] obtained using .

Activity 7.8

In activities 7.6 and 7.7, the following were found:

• , rad/s and

• , rad/s and

and

[ ] [ ][ ] [ ] 222 2

2 0 1 ˆ1 1.4142 40 1 1.4142

TX M X M

= − = = −

[ ]1 2

1 1[ ] [ ] [ ]

1.4142 1.4142X X

Φ = = −

[ ] [ ][ ] 1 1 0.8 0.2 1 1 1.0343 0

1.4142 1.4142 0.2 0.4 1.4142 1.4142 0 2.1657

TT

C−

Φ Φ = = − − −

[ ] [ ][ ] [ ] 1 1 11

2 2 22

ˆ2 0

ˆ0 2

T

D

MC C

M

ζ ωζ ω

Φ Φ = =

1 0.2389ζ = 1 0.5412ω =11

ˆ 4M =

1 1 11ˆ2 1.0343Mζ ω =

2 0.2072ζ = 2 1.3066ω =22

ˆ 4M =

2 2 22ˆ2 2.1657Mζ ω =

[ ] [ ][ ]TCΦ Φ

1 0.2389ζ = 1 0.5412ω = 11ˆ 4M =

2 0.2072ζ = 2 1.3066ω = 22ˆ 4M =

[ ]1

1

1.4142X

=

[ ]2

1

1.4142X

= −

[ ]1 2

1 1[ ] [ ] [ ]

1.4142 1.4142X X

Φ = = −


The two un-coupled equations are:

For zero initial conditions, i.e. and , we have and

. The steady state solutions to the above harmonic excitations are

and

[ ] [ ] 111

22

ˆ 0 4 0 0.25 0

ˆ 0 4 0 0.250D D

MM M

M

− = = ⇒ =

( ) 1 1

2 2

2 0 0.2586 02

0 2 0 0.5415D

ζ ωζω

ζ ω

= =

22 1

32

0.2929 00

0 1.70720D

ωωω

= =

[ ] [ ] [ ] [ ]1 0.25 0 1 1 1sin(3 )

0 0.25 1.4142 1.4142 2T

DQ M F t−

= Φ = −

0.75sin(3 )

0.3535Q t

= −

1 1 1( ) 0.2586 ( ) 0.2929 ( ) 0.75sin(3 )t t t tη η η+ + =

2 2 2( ) 0.5415 ( ) 1.7072 ( ) 0.3535sin(3 )t t t tη η η+ + = −

[ ] [ ](0) 0x = [ ] [ ](0) 0x = [ ] [ ](0) 0η =

[ ] [ ](0) 0η =

1 1 1

0.75( ) ( )sin(3 ( ))

0.2929t H tη θ= Ω − Ω

2 2 2

0.3535( ) ( )sin(3 ( ))

1.7072t H tη θ= − Ω − Ω

1

35.5432

0.5412Ω = = 2

32.296

1.3066Ω = =

( ) ( )1 2 22

1 1 1

1( ) 0.0335

1 2H

ζΩ = =

− Ω + Ω


and

Activity 7.9

In activity 7.8, the same system has been un-coupled to give the following equations:

In activities 7.6 and 7.7, the following were found:

• , rad/s, rad/s and

• , rad/s, rad/s and

and

1

1 1 11 2

2( ) tan 3.0527

1

ζθ − ΩΩ = = − Ω

( ) ( )2 2 22

2 2 2

1( ) 0.2285

1 2H

ζΩ = =

− Ω + Ω

1 2 22 2

2

2( ) tan 2.9224

1

ζθ − ΩΩ = = − Ω

1( ) 0.0858sin(3 3.0527)t tη = − 2 ( ) 0.0473sin(3 2.9224)t tη = − −

1 1

2 2

( ) ( ) 1 1 0.0858sin(3 3.0527)[ ]

( ) ( ) 1.4142 1.4142 0.0473sin(3 2.9224)

x t t t

x t t t

ηη

− = Φ = − − −

1( ) 0.0858sin(3 3.0527) 0.0473sin(3 2.9224)x t t t= − − −

2 ( ) 0.1213sin(3 3.0527) 0.0669sin(3 2.9224)x t t t= − + −

1 1 1( ) 0.2586 ( ) 0.2929 ( ) 0t t tη η η+ + =

2 2 2( ) 0.5415 ( ) 1.7072 ( ) 0t t tη η η+ + =

1 0.2389ζ = 1 0.5412ω = 21 1 11 0.5255dω ω ζ= − =

11ˆ 4M =

2 0.2072ζ = 2 1.3066ω = 22 2 21 1.2782dω ω ζ= − = 22

ˆ 4M =

[ ]1

1

1.4142X

=

[ ]2

1

1.4142X

= −

[ ] [ ][ ] [ ]1 111

2 0 0.51 1(0) (0) 1 1.4142 0.4268

ˆ 0 1 0.54T

X M xM

η = = =


Note that .

and radians

and radians

Activity 7.10

The system is similar to that of activity 7.6 except that there is no damping. The natural

frequencies were found in activity 7.8 as rad/s and rad/s. Based

on the information in activity 7.8, the de-coupled equations are:

For and , the responses are

[ ] [ ][ ] [ ]2 222

2 0 0.51 1(0) (0) 1 1.4142 0.0732

ˆ 0 1 0.54T

X M xM

η = = − =

[ ] [ ](0) 0η =

[ ] [ ]1

( ) sin( )j j

nt

j dj djjj

x t X A e tζ ω ω ϕ−

=

= +∑

2

2 1 1 11 1

1

(0)(0) 0.4395

d

Aζ ωηη

ω

= + =

1 11

1 1

tan 1.3296dd

ωϕζ ω

− = =

2

2 2 2 22 2

2

(0)(0) 0.0748

d

Aζ ω ηη

ω

= + =

1 22

2 2

tan 1.3621dd

ωϕζ ω

− = =

[ ] 0.1293

0.2707

1( ) 0.4395 sin(0.5255 1.3296)

1.4142

10.0748 sin(1.2782 1.3621)

1.4142

t

t

x t e t

e t

−

−

= +

+ + −

1 0.5412ω = 2 1.3066ω =

1 1( ) 0.2929 ( ) 0.75sin(3 )t t tη η+ =

2 2( ) 1.7072 ( ) 0.3535sin(3 )t t tη η+ = −

1ω ω≠2ω ω≠

[ ]02

( ) sin( ) sin( )(1 )i

i i ii

Ft t t

kη ω ω= − Ω

− Ω


Note that

and

Activity 7.11

Given

Let

and

11

5.5432ωω

Ω = = 22

2.296ωω

Ω = =

[ ] [ ]1

1 12

0.75 1( ) sin(3 ) sin(0.5412 ) 0.0861 sin(3 ) 5.5432sin(0.5412 )

0.2929 1t t t t tη

= − Ω = − − − Ω

[ ] [ ]2 222

0.3535 1( ) sin(3 ) sin(1.3066 ) 0.0485 sin(3 ) 2.296sin(1.3066 )

1.7072 1t t t t tη

= − − Ω = − − Ω

1 1 1

2 2 2

( ) ( ) ( )1 1[ ]

( ) ( ) ( )1.4142 1.4142

x t t t

x t t t

η ηη η

= Φ = −

[ ][ ]

[ ][ ]

1

2

0.0861 sin(3 ) 5.5432sin(0.5412 ) 0.0485 sin(3 ) 2.296sin(1.3066 )( )

0.1218 sin(3 ) 5.5432sin(1.3066 ) 0.0685 sin(3 ) 2.296sin(1.3066 )( )

t t t tx t

t t t tx t

− − − = + − − − −

1 1 1

2 2 2

2 0 0.8 0.2 2 1 sin(3 )

0 1 0.2 0.4 1 1 2sin(3 )

x x x t

x x x t

− − + + = − −

[ ]1 1

2 2

1 1

2 2

x x

x xY Y

x x

x x

= ⇒ =

[ ] [ ] 12 0 0.5 0

0 1 0 1M M

− = ⇒ =

[ ] [ ]1 1 0.5

1 1M K

− − = −

[ ] [ ]1 0.4 0.1

0.2 0.4M C

− − = −


The state space equation is:

For activity 7.8,

The responses for the state-space equation were obtained using Matlab ODE45 function. First, create a Matlab function file act711.m as follow:

Then use the following main program to get the responses (note that the equations obtained in activity 7.8 were also plotted to compare the responses from the 2 methods)

[ ][ ] [ ]

[ ] [ ] [ ] [ ]1 1

0 0 1 00 0 0 0 1

1 0.5 0.4 0.1

1 1 0.2 0.4

IA

M K M C− −

= = − −− − − −

[ ][ ]

[ ] 1

0 00 0 0

0.5 0

0 1

BM

−

= =

[ ][ ] [ ][ ]Y A Y B F = +

[ ] 1sin(3 )

2F t

=

function ydot=act711(t, y) ydot = [0 0 1 0; 0 0 0 1; -1 0.5 -0.4 0.1; 1 -1 0.2 -0.4]*y

+ [0 0; 0 0; 0.5 0; 0 1]*[1; 2]*sin(3*t);

% Main Matlab program %[t, y]=ode45('act711', [0, 60], [0;0;0;0]); n=length(t);x1=0.0858*sin(3*t-3.0527*ones(n,1))-0.0473*sin(3*t-2.9224*ones(n,1));plot(t, y(:,1), t, x1); xlabel(' Time (s) '); ylabel(' x_1 '); legend(' State-space method ', ' Steady-state in Activity 7.8 '); pausex2=0.1213*sin(3*t-3.0527*ones(n,1))+0.0669*sin(3*t-2.29224*ones(n,1));plot(t, y(:,2), t, x2); xlabel(' Time (s) '); ylabel(' x_2 '); legend(' State-space method ', ' Steady-state in Activity 7.8 ');


The plots are as followed:


For activity 7.9, there is no forcing function and the initial conditions are

and

The responses for the state-space equation were obtained using Matlab ODE45 function. The previous Matlab function file is modified to act711a.m as follow:

The main program is also modified to get the responses (note that the equations obtained in activity 7.9 were also plotted to compare the responses from the 2 methods)

[ ] 0.5(0)

0.5x

=

[ ] 0(0)

0x

=

function ydot=act711a(t, y) ydot = [0 0 1 0; 0 0 0 1; -1 0.5 -0.4 0.1; 1 -1 0.2 -0.4]*y

+ [0 0; 0 0; 0.5 0; 0 1]*[0; 0];

% Main Matlab program %[t, y]=ode45('act711a', [0, 60], [0.5;0.5;0;0]); n=length(t);x1=0.4395*exp(-0.1293*t).*sin(0.5255*t+1.3296*ones(n,1))+0.0748*exp(-0.2707*t).*sin(1.2782*t+1.3621*ones(n,1));plot(t, y(:,1), t, x1, '+'); xlabel(' Time (s) '); ylabel(' x_1 '); legend(' State-space method ', ' Initial conditions response in Activity 7.9 '); pausex2=1.41412*0.4395*exp(-0.129*t).*sin(0.5255*t+1.3296*ones(n,1))-1.4142*0.0748*exp(-0.2707*t).*sin(1.2782*t+1.3621*ones(n,1));plot(t, y(:,2), t, x2, '+' ); xlabel(' Time (s) '); ylabel(' x_2 '); legend(' State-space method ', ' Initial conditions response in Activity 7.9 ');


The plots are as followed:


For activity 7.10, the forcing function is same as in activity 7.8 but now the system has no damping. Assume zero initial conditions.

The responses for the state-space equation were obtained using Matlab ODE45 function. The previous Matlab function file is modified to act711b.m as follow:

The main program is also modified to get the responses (note that the equations obtained in activity 7.10 were also plotted to compare the responses from the 2 methods)

function ydot=act711b(t, y) ydot = [0 0 1 0; 0 0 0 1; -1 0.5 0 0; 1 -1 0 0]*y+ [0 0; 0 0; 0.5 0; 0 1]*[1; 2]*sin(3*t);

% Main Matlab program %[t, y]=ode45('act711b', [0, 60], [0;0;0;0]); n=length(t);dum1=(sin(3*t)-5.5432*sin(0.5412*t)); dum2=(sin(3*t)-2.296*sin(1.3066*t)); x1=-0.0861*dum1 + 0.0485*dum2; plot(t, y(:,1), t, x1 ); xlabel(' Time (s) '); ylabel(' x_1 '); legend(' State-space method ', ' Response in Activity 7.10 '); pausex2=-0.1218*dum1 - 0.0685*dum2; plot(t, y(:,2), t, x2 ); xlabel(' Time (s) '); ylabel(' x_2 '); legend(' State-space method ', ' Response in Activity 7.10 ');


The plots are as followed (the differences are due to round-off errors):

Tutorials T.1

Tutorials

Tutorials

Module 1 Kinematics: Tutorial 1aTutorial 1b

Module 2 Kinematics analysis of mechanisms: Tutorial 2aTutorial 2b

Module 3 Dynamics analysis of planar mechanisms Tutorial 3aTutorial 3bTutorial 4aTutorial 4b

Module 4 Rigid body dynamics Tutorial 5aTutorial 5bTutorial 6a

Module 5 System modeling Tutorial 6bTutorial 7aTutorial 7b

Module 6 Vibration of single degree of freedom systems Tutorial 8aTutorial 8bTutorial 9aTutorial 9bTutorial 10aTutorial 10bTutorial 11a

Module 7 Vibration of multi-degree-of-freedom systems Tutorial 11bTutorial 12aTutorial 12bTutorial 13a

T.2 MEC3403 – Dynamics II

Tutorials T.3

Tutorial 1a

1. Find the eigenvaules for

2. Given

• Find

• Show that

• Show that

3. Given

Show that

4. Find the linear velocity and acceleration for the following particle motion described relative to a frame fixed in time by

5. Do problem 1.2 on page 16 of the textbook.

1 2 3

0 4 2

0 0 7

A

= −

2 (2 3)

, 1

(2 1)

t t

a t b

t t

− = − = + −

( ), (3 )d d

a adt dt

( ) ( ) ( )d d d

a b a b a bdt dt dt

× = × + ×

( ) ( ) ( )d d d

a b a b a bdt dt dt

⋅ = ⋅ + ⋅

2 2 1 0

, 0 3

1 2 0 3

t t

b t A t

t

= =

( )d d d

Ab A b A bdt dt dt

= +

2cos( )

2sin( )

2

t

r t

t

=


6. Given that the vector r is in a moving coordinate system and its velocity is given by

Show that the differentiation of the velocity with respect to time will result in

Tutorial 1b

1. The slide-crank mechanism shown in figure 1 is a special case of the four-bar linkage. Link 1, the crank, is the driver. Link 2 is the connecting rod. Link 3 is called the slider. Point ‘C’ on link 3 can only have rectilinear motion. This mechanism is widely used in engines. Find the vectors r0, r1 and r2 to describe the slider-crank mechanism shown in figure 1(a) for θ1 = 70° (O1B = 2 m, BC = 3.76 m, O1C is horizontal). The position vectors description is given in figure 1(b).

Figure 1: A slider-crank mechanism

2. Let n be the number of links in a mechanism. The number of instantaneous centers is

( )rv R r rω= + + ×

( ) ( ) 2 ( )r ra R r r r rα ω ω ω= + + × + × × + ×

( 1)

2

n nN

−=

Tutorials T.5

Figure 2: Mechanism for question 2

Kennedy’s theorem states that any three bodies having relative motion to another have three instantaneous centers that lie on a straight line. Use the above information to sketch the instantaneous centers for the four-bar mechanism given in figure 2.

3. Find the degrees of freedom for the following spatial linkages

• This open-loop kinematic chain includes cylindrical, prismatic, and spherical pairs

• This is a schematic representation of a piece of construction machinery. It has two hydraulic cylinders (links 1 and 2, and links 6 and 7), which may be considered cylindrical pairs. The other pairs are revolute joints, including a double revolute joint where links 2, 3 and 8 meet. Find the number of degrees of freedom if this linkage is treated as a spatial linkage

Link 1

Link 2

Link 3 B

C

O1 O2




Tutorials T.7

Tutorial 2a

1. The four-bar linkage is shown in figure 1. It is driven by the crank (link 1), which is rotating at 94.2 rad/sec counter clock-wise (assume the angular velocity is constant). Find the following when θ1 = 120°:

• The angular velocities of links 2 and 3• The angular accelerations of links 2 and 3• The linear velocity and linear acceleration at point ‘D’• Write a program in Matlab to compute the above velocities and accelerations

(Lengths O1B = 4 mm, BC = 18mm, O3C = 11mm, O1O3 = 10mm)

Figure 1: Four-bar linkage for question 1

2. The slider-crank mechanism in figure 2 is driven by the crank (link 1), which is rotating at

an angular velocity of rad/sec and an angular acceleration of rad/s2.

Figure 2: Slider-crank mechanism for question 2

θ1

Link 1

Link 2

Link 3

r1

r2

r3

r0

BD

C

O1 O3

y

xz

θ2

θ3

7mm

3mm

1 10ω = 1 40α =

O1

Link 1 Link 2

Link 3

B

C

θ1

y

xz

φ


Find the following:

• The angular velocity and angular acceleration of link 2 when θ1 = 70°.

• The linear velocity and linear acceleration of the slider (i.e. at point ‘C’) when θ1 = 70°.

• Find the linear velocity and linear acceleration of the center of mass of the connecting rod (assumed the center of mass to be located at the midpoint of link 2) when θ1 = 70°.

• Write a program in Matlab to compute the above velocities and accelerations

Note: O1B = 2 m, BC = 3.76 m, and O1C is horizontal. The position vectors description was determined in tutorial 1b.

3. We can obtain a general equation for the slider velocity in question 2. Let lengths O1B = R and BC = L. Show that the slider speed is given by

4. The slider-crank mechanism shown in figure 3 has an offset. Given link O1B is rotating at constant velocity. Explain the procedure for finding the angular velocity/acceleration of link 2 and the linear velocity/acceleration of the slider. (Let lengths O1B = R and BC = L).


( )1

1 1 2 21

cos( )sin( ) 1

1 sin ( )

Rv R

L R L

θω θθ

= + −

O1

Link 1

Link 2

Link 3

B

C

θ1

y

xz

φ

Offset ε

δ

r1

r2

Tutorials T.9

Tutorial 2b

1. In figure 1, bodies 2 and 3 are in sliding contact. The radius of curvature of link 2 is at C2 and the radius of curvature of link 3 is at C3. Sketch the equivalent four-bar linkage for this mechanism.

Figure 1: Sliding contact mechanism for question 1

2. Explain how you would attempt to analyze the mechanism shown in figure 2. Note that link 1 is the driver and point ‘D’ on link 5 is constrained to slide horizontally.


O1 O2

C2

C3

2 3

P3P2


3. The four-bar linkage in figure 3 is driven by the crank (link 1) rotating at ω1 = 95 rad/sec counter clock-wise and input acceleration α1 = 0. Find the angular velocities and acceleration of links 2 and 3. Find the linear velocity and acceleration of the centres of mass for links 2 and 3. Note that θ1 = 135°, O1B = 30 mm, BC = 100mm, O3C = 50mm, O1O3 = 70mm. The centres of mass for links 2 (i.e. CG2) and link 3 (i.e. CG3) are assumed to be located at the midpoints of the respective links.


4. Figure 4 shows a four-bar linkage. Using Kennedy’s theorem (refer to tutorial 1b), determine the instantaneous centers for the mechanism. Use the instantaneous centers to derive the relationship between (the linear velocity at point ‘B’) and (the linear velocity at point ‘C’).


θ1Link 1

Link 2

Link 3

B

C

O1, CG1 O3

y

xz

CG2

CG3

vB vC

ω1

Link 1

Link 2

Link 3 B

C

O1, O2

y

xz

vB

vC

ω3

Tutorials T.11

Tutorial 3a

1. Do problem 1.7 on page 17 of textbook.






Tutorial 3b

1. The four-bar linkage in figure 1 is driven by the crank (link 1) rotating at ω1 = 95 rad/sec counter clock-wise and input acceleration α1 = 0. Find the torque T required to drive the crank under the following conditions: θ1 = 135°, O1B = 30 mm, BC = 100mm,O3C = 50mm, O1O3 = 70mm. Except for link 1 where CG1 = O1, the center of mass for links 2 and 3 are assumed to be located at the midpoints of the respective links. The link parameters are m1 = 0.2 kg, IG1 = 20 kg.mm2, m2 = 0.2 kg, IG2 = 400 kg.mm2, m3 = 0.3 kg, IG3 = 100 kg.mm2.

Figure 1: Four-bar linkage for question 1

2. Perform a dynamic-force analysis for the slider-crank mechanism given in question 2 of tutorial 2a for θ1 = 70°, ω1 = 10 rad/s and α1 = 40 rad/s2. The centers of mass for link 1 and link 2 are assumed to be located at the midpoints of the respective links. The link parameters are m1 = 1 kg, IG1 = 0.01 kg.m2, m2 = 2 kg, IG2 = 0.02 kg.m2, m3 = 3 kg. Find the following:

• The instantaneous torque required at link 1 for the position shown, i.e. θ1 = 70°.• The bearing forces for the pins at ‘O1’, ‘B’ and ‘C’ when θ1 = 70°.• Write a program in Matlab so that you can vary the crank angle θ1 from 0° to 360° in

5° increments and plot the instantaneous torque versus θ1.

Note: O1B = 2 m, BC = 3.76 m, O1C is horizontal. The position vectors description was determined in tutorial 1b.



θ1Link 1

Link 2

Link 3 B

C

O1, CG1 O3

y

xz

θ2

φ3

CG2

CG3

Tutorials T.13

Tutorial 4a

1. For the slider-crank mechanism given in figure 1, find the reactions at each pin joint and the external torque to be applied to the crank (O1A = 3 m; AB = 12 m; rad/s;

rad/s2; ; ; ; ;

the centre of mass of link 1 is at O1; the centre of mass of link 2 is 4.5 m away from point ‘A’).


2. The four-bar linkage of figure 2 has a constant crank velocity rad/s counter clockwise. Link 2 has a mass of 0.3 kg and a moment of inertia about the center of mass CG2 of 1000 kg.mm2. Link 3 has a mass of 0.2 kg and a moment of inertia about its center of mass CG3 of 150 kg.mm2. Find the crank torque required for dynamic equilibrium. (O1O3 = 900 mm; O1B = 400 mm; BC = 600 mm; CG2B = 400 mm; CG2C = 300 mm; O3C = 500 mm; O3CG3 = 250 mm; assume the centre of mass of link 1 is at O1).


ω1 210=

α1 0= m1 1 kg= m2 3.4 kg= m3 2.85 kg= IG1 IG2 0.1 kg.mm2= =

ω1 60=

θ1=60°

Link 1

Link 2

Link 3

CG2

B

C

O1 O3

y

xz

θ2

θ3

CG3


3. The four-bar linkage of figure 3 has a constant input angular velocity rad/s counter clockwise. Link 2 has a mass of 1 kg and a moment of inertia about the center of mass CG2 of 700 kg.mm2. Link 3 has a mass of 0.6 kg and a moment of inertia about its center of mass CG3 of 500 kg.mm2. Find the crank torque required for dynamic equilibrium. (O1O3 = 200 mm; O1B = 80 mm; BC = 160 mm; CG2B = 100 mm;O3C = 100 mm; O3CG3 = 50 mm; assume the centre of mass of link 1 is at O1).


ω1 60=

θ1=45°

Link 1

Link 2

Link 3

CG2

B

C

O1 O3

y

xz

θ2

θ3

CG3

Tutorials T.15

Tutorial 4b

1. Figure 1 shows the centers of mass of the three blades of a fan. Determine the location and the counterweight that will statically balance the rotor. Will the corrections found above changed as the fan speed varied? (m1 = 10 kg, m2 = 10 kg, m3 = 10 kg, r1 = 3 mm,r2 = 6 mm, r3 = 2 mm).

Figure 1: Static balancing of rotor for question 1

2. Determine the corrections needed in planes ‘P’ and ‘Q’ to dynamically balance the rotor shown in figure 2. Are the corrections dependent on the angular speed of the shaft?(r1 = 5.25 mm, r2 = 4 mm)

Figure 2: Dynamic balancing of rotor for question 2

3. Determine the corrections needed in planes ‘P’ and ‘Q’ to dynamically balance the rotor shown in figure 3. Note that m1 = 2 kg, m2 = 3 kg, m3 = 2 kg, r1 = 3 m, r2 = 3 m,

r2 = 2 m, , , and .

120°120°

120°

m1

m2

m3

r1

r2

r3

90°

m2=6 kg

m1=4 kg

r2

r1

P Q

4 m 9 m

7 m

120°

1 60θ = 2 150θ = 3 270θ =


Figure 3: Rotor for question 3

4. Discuss the approximate reduction of the shaking force for the slider-crank mechanism

given in question 2 of tutorial 3b (when rad/s and rad/s2) with the

addition of a counterweight mc mounted on the crank at a distance rc = 2 m and at an angle θ1 + 180°. Investigate the approximate shaking forces before and after the addition of the counterweight for the following cases: mc = 1 kg; mc = 5 kg and mc = 10 kg. Write a simple

Matlab program to plot the shaking forces for to 360° in increments of 10°.

m2m1

m3 P

Q

2 m 1 m

m3

1 m 1 m 1 m

m1 m2

z

y

1 10ω = 1 0α =

1 0θ =

Tutorials T.17

Tutorial 5a

1. Do problem 2.4 on page 56 of textbook

2. Find the mass moment of inertia of the body given in figure 1 about point ‘G’ with respect to frame b = (1, 2, 3). Assume that the body is made up of two components: a solid circular cylinder and a hollow cylinder. (Note that the 2-axis and the x-axis are parallel)

Figure 1: Composite body for problem 2

3. Find the mass moment of inertia of the body given in figure 1 about point ‘O’ with respect to frame a = (x, y, z). Assume the mass of the solid cylinder is equalled to the mass of the hollow cylinder (each is 1 kg). The length of the hollow cylinder is 1m and the length of the solid cylinder is 2m. The outer radius of the hollow cylinder is 2m and the inner radius is 1m.

4. The mass moment of inertia for a discrete system of ‘n’ particles shown in figure 2 about ‘O’ with respect to the frame a= (x, y, z) is given by


where

The position of particle ‘p’ is defined as . Find the mass moment of inertia of

the system with two masses: m1 located at and m2 located at with respect to

the origin of frame a.

xx xy xz

yx yy yz

zx zy zz

I I I

J I I I

I I I

=

2 2

1

( )m

xx p p pp

I y z m=

= +∑

2 2

1

( )m

yy p p pp

I x z m=

= +∑

2 2

1

( )m

zz p p pp

I x y m=

= +∑

1

( )m

xy p p p yxp

I x y m I=

= − =∑

1

( )m

xz p p p zxp

I x z m I=

= − =∑

1

( )m

yz p p p zyp

I y z m I=

= − =∑

p

p p

p

x

R y

z

=

1

0

0

0.45

1

0.8

Tutorials T.19

Figure 2: System of particles for problem 4

Tutorial 5b

1. An idealised crankshaft is shown in figure 1. Assume rod AB is mass is negligible andm = 1 kg. Determine the dynamic forces on the bearings at points ‘A’ and ‘B’ under the following two conditions (Note that the bearing reactions do not include the static or stationary components due to gravity force acting on the masses):

(a) The shaft is rotating with a constant velocity rad/s

(b) The shaft is started up from rest with angular acceleration rad/s2 using a start

up torque . Find Tz.

mp

Frame a

O

x

y

z

PR

0

0

z

ωω

=

0

0

z

αα

=

0

0

z

T

T

=


Figure 1: Crankshaft of problem 1

2. A thin uniform rectangular plate of mass m shown in figure 2 can rotate freely about thez-axis. Bearings at points ‘A’ and ‘B’ support the shaft. The plate is released from rest from the horizontal position shown. Find the reactions at ‘A’ and ‘B’ under the following two conditions:

(a) At the instance just after its release (plate is in the x-z plane)

(b) At the instance when the plate reaches the vertical position (plate is in the y-z plane). (Hint: use the conservation of energy to find .

Note: frame a = (x, y, z) and frame b = (1, 2, 3)

Figure 2: Plate for problem 2

ωz

Tutorials T.21

3. For the wheel with radius ‘r’ and centre of mass ‘G’ shown in figure 3, find the forces acting at points ‘A’ and ‘O’ (Assume all the forces in the direction of the y-axis are

provided at point ‘O’, i.e. ). Note that the angular velocities and are

constants.

Figure 3: Constrained rolling disk for problem 3

4. A thin disk of mass m and radius r rotates with constant angular spin about a pin at ‘O’

as shown in figure 4. The mechanism is spinning about the z-axis at constant angular

velocity . Find the total moment acting on the pin at ‘O’.

Figure 4: Rotating disk for problem 4

0AyF = ψ φ

ψ

φ


Tutorial 6a

1. Figure 1 shows a fan, which is balanced when the rotor is not rotating. When the rotor rotates, the mass m has to be moved from point ‘B’ to point ‘C’ to prevent the fan from

rotating about the x-axis. Let be the distance ‘BC’. Note that the angular velocities

and are constants. Find the relationship for in terms of and using the

gyroscopic dynamics equation.

Figure 1: Rotating fan for problem 1

2. Solve problem 3 of tutorial 5a using the gyroscopic dynamics equation.

3. Solve problem 4 of tutorial 5a using the gyroscopic dynamics equation.

BCx

ψ φ BCx φ ψ

3

21

z

y

x

G

φ

B

C

ψ

m

G2

Tutorials T.23

Tutorial 6b






Tutorial 7a

1. Do problem 3.10 on page 108 of the textbook using force-balance and moment-balance methods.





Tutorial 7b


2. Do problem 3.13 on page 108 of the textbook using Lagrange’s equations.





Tutorial 8a

1. Do problem 3.2 on page 107 of the textbook (Note: the area moment of inertia of an annular cross-section (with r0 = external radius and ri = internal radius) is




Tutorial 8b





Tutorial 9a






Tutorial 9b




4. A spring-mass-damper system is subjected to a harmonic force. The amplitude at resonance is found to be 20 mm. If excited at the frequency , the amplitude is only 10 mm. What is the damping ratio?

Iπ4--- ro

4 ri4–( )=

ω 0.75ωn=

Tutorials T.25

Tutorial 10a






Tutorial 10b




Tutorial 11a

1. Do problem 6.3 on page 305 of the textbook. Assume the input excitation frequency is rad/s.


3. Consider a mass-spring damper system with mass m = 1 kg, damper constant c = 6 Ns/m, and spring constant k = 25 N/m. Estimate the rise time, maximum overshoot and the settling time (to reach 98% of the steady state value) when the system is subjected to a unit step input.

Tutorial 11b


2. Do problem 7.21 on page 376 of the textbook (Hint: Use Matlab ‘eig’ function).

3. Do problem 7.31 on page 377 of the textbook (Hint: Use Matlab ‘eig’ function).

4. Do problem 7.41 on page 379 of the textbook. (Note: Nm/rad; and N/m)

ω 1.4=

kt 600=k 1000=


Tutorial 12a






Tutorial 12b



3. Do problem 8.10 on page 470 of the textbook (Do part (a) only)

4. Do problem 8.11 on page 470 of the textbook. (Find only the steady-state solution)

Tutorial 13a


2. Derive the equation for the system represented by figure E8.26 on page 473 of the textbook. Given m1 = 80 kg; m2 = 1100 kg; k1 = 300 kN/m; k2 = 30 kN/m; and

c2 = 5000 N/(m/s). Find the initial responses of m1 and m2 using Matlab if m,

m/s, and .

3. Figure 1a shows that mass m1 is having forced vibrations. To reduce the amplitude of the vibration, an auxiliary spring-mass is added to mass m1 as shown in figure 1b. Let the steady state displacements of x1 and x2 in figure 2 be represented respectively as

and

Specify suitable values for k2 and m2 for the auxiliary spring-mass vibration absorber system so that the amplitude A is minimized.

1(0) 0.2x =

1(0) 0.2x = 2 2(0) (0) 0x x= =

1 sin( )x A tω= 2 sin( )x B tω=

Tutorials T.27

Figure 1: Vibration absorber system


Reading 1.1

Greenwood, DT 1965, ‘2-8 Vector derivatives in rotating systems’, Principles of dynamics, Prentice Hall, Engelwood Cliffs, NJ, pp. 47–9.

Read

ing 1

Read

ing 1

Read

ing 1

Reading 1.2

Greenwood, DT 1965, ‘2-9 Motion of a particle in a moving coordinate system, Principles of dynamics, Prentice Hall, Engelwood Cliffs, NJ, pp. 49–51.

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ing 1.

2

Read

ing 1.

2

Read

ing 1.

2

Reading 1.3

Greenwood, DT 1965, ‘2-10 Plane motion’, Principles of dynamics, Prentice Hall, Engelwood Cliffs, NJ, pp. 52–5.

Read

ing 1.

3

Read

ing 1.

3

Read

ing 1.

3

Read

ing 2.

1

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ing 2.

1

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ing 2.

1

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ing 2.

1

Read

ing 2.

1

Reading 2.2

Wilson, CE & Sadler, JP 2003, ‘3.4 Application of analytical vector and matrix methods to linkages’, Kinematics and dynamics of machinery, Pearson Education, Upper Saddle River, NJ, pp. 178–82.

Read

ing 2.

2

Read

ing 2.

2

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ing 2.

2

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ing 2.

2

Reading 2.3

Wilson, CE & Sadler, JP 2003, ‘4.2 Analysis of a four-bar linkage by analytical vector methods’, Kinematics and dynamics of machinery, Pearson Education, Upper Saddle River, NJ, pp. 266–70.

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ing 2.

3

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ing 2.

3

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ing 2.

3

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ing 2.

3

Reading 2.4

Martin, GH 1982, ‘7-5 Equivalent linkages’, Kinematics and dynamics of machines, McGraw Hill, NY, pp. 128–32.

Read

ing 2.

4

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ing 2.

4

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ing 2.

4

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ing 2.

4

Reading 2.5

Martin, GH 1982, ‘Complex linkages’, Kinematics and dynamics of machines, McGraw Hill, NY, pp. 189–90.

Read

ing 2.

5

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ing 2.

5

Reading 3.1

Wilson, CE & Sadler, JP 2003, ‘Calculation of mass and mass moment of inertia for linkage elements’, Kinematics and dynamics of machinery, Pearson Education, Upper Saddle River, NJ, pp. 689–92.

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ing 3.

1

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ing 3.

1

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ing 3.

1

Reading 3.2

Wilson, CE & Sadler, JP 2003, ‘Analytical solution for four-bar linkage’, Kinematics and dynamics of machinery, Pearson Education, Upper Saddle River, NJ, pp. 642–5.

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ing 3.

2

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ing 3.

2

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ing 3.

2

Reading 3.3

Wilson, CE & Sadler, JP 2003, ‘10.7 Balancing of rigid bodies’, Kinematics and dynamics of machinery, Pearson Education, Upper Saddle River, NJ, pp. 730–9.

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ing 3.

3

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ing 3.

3

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ing 3.

3

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ing 3.

3

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ing 3.

3

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ing 3.

3

Reading 3.4

Martin, GH 1982, ‘20-3 Slider-crank mechanism’, Kinematics and dynamics of machines, McGraw Hill, NY, pp. 410–13.

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ing 3.

4

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ing 3.

4

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ing 3.

4

Reading 3.5

Wilson, CE & Sadler, JP 2003, ‘10.8 Balancing reciprocating machines’, Kinematics and dynamics of machinery, Pearson Education, Upper Saddle River, NJ, pp. 740–3.

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ing 3.

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Reading 4.1

Meriam, JL & Kraige, LG 1998, ‘Appendix B-Mass moments of inertia’, Engineering mechanics: dynamics, Wiley, NY, pp. 665–71, 682–7, 713–16.

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Reading 4.2

Torby BJ 1984, ‘Rigid body dynamics’, Advanced dynamics for engineers, Holt Reinhart Winston, NY, pp. 205–10.

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Reading 4.3

Moon FC 1998, ‘Gyroscopic dynamics’, Applied dynamics, Wiley, NY, pp. 195–8.

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