LECTURE 3: CHAPTER 2

1. LEARNING OBJECTIVES

a. Understanding types of motions

b. Describing these motions by applying kinematic principles.

c. Describing degrees of freedom (DOF) of physical systems.

d. Applying kinetic principles using Work-Energy method, and Equation of

Motion method.

e. Deriving SDOF systems with spring elements.

2. MODELLING

A process of developing equations or sets of equations to describe motions of

physical systems.

3. TYPES OF MOTIONS

Dynamics – involves study of motion or response of bodies due to action of forces

Can be categorized into:

a. Kinematics – study of motion of bodies which does not involve the study / analysis of the forces causing the motion.

b. Kinetics - the analysis of motion / response of bodies and the forces acting on it.

Focus on rigid body (dimension not negligible, thus capable of rotating about its own axis) motions.

Rigid Body Motions – 2D (Plane Motion)

Translation and rotation of bodies in 2D.

a. Rectilinear translation

b. Curvilinear translation

c. Fixed axis rotation

d. General plane motion

KMEM 3211 – MECHANICAL VIBRATION

Examples of Plane Motions

Choosing a suitable coordinate system

Coordinate – an independent quantity to define positions in space.

3D coordinate systems 2D coordinate systems

a. Cartesian (x, y, z)

b. Cylindrical (r, , L)

c. Spherical (r, , )

a. Cartesian (x, y)

b. Polar (r, )

Relationship between Cartesian and polar coordinates in 2D

2

P

y

x

r

θ

y

x

22 yxr

x

y1tan

KMEM 3211 – MECHANICAL VIBRATION

Constraint – restrict motions of bodies. Provide less Degrees of Freedom (DOF) for bodies in motion. Less Equation of Motion (EOM) to solve for the response.

Generalized Coordinate – provide complete information about motion of bodies including the constraints.

Generalized Coordinate and DOFs

3

KMEM 3211 – MECHANICAL VIBRATION

4. PLANE MOTION OF RIGID BODY

Newton’s Law of Motions

1st Law: A body at rest or in motion (constant velocity) will remain unchange until there exist unbalance force acting on it.

2nd Law: The sum of all external forces is equal to the rate of change of momentum.

The net force vector and acceleration vector act on the same line of action.

The quantity of F and a, becomes scalar if the mass is constrained to move in one direction.

3rd Law: Action and reaction between interacting bodies are equal in magnitude, collinear and acts in the opposite direction.

Solving plane motion problem

4

dt

vmdF

am

dt

vdmF .

KMEM 3211 – MECHANICAL VIBRATION

Construct Free Body Diagram (FBD) and include all the forces and reaction forces, dimensions

Mass moment of inertia

About a reference axis is defined as

where r is the distance from the reference axis to the element mass, dm.

Review Moment of Inertia calculation (page 69 & 70)

Study Example 2.2-1 page 71.

Read through section on General Plane Motion pg 72-73.

5

KMEM 3211 – MECHANICAL VIBRATION

6

KMEM 3211 – MECHANICAL VIBRATION

Read through Section on Curvilinear Translation page 73 & 74 (Instantaneous Centre of Rotation)

5. WORK-ENERGY METHOD

a. Parameter required: force, displacement, velocity only.

b. Integrate EOM over displacement

c. Using principle of Energy Conservation

“The sum of potential energy (PE) and kinetic energy (KE) of a body remains constant at each instant of time, throughout the motion”

d. Work = force x displacement

where, T = kinetic energy (KE) ; V = Potential Energy (PE) subscript 1 and 2 are the values at time t1 and t2.

The total kinetic and potential energy under Conservation of Energy:

7

KMEM 3211 – MECHANICAL VIBRATION

Consider a Single Degree Of Freedom (SDOF) Spring-Mass System

Energy is conserved, then:

We know that energy is a scalar quantity

When energy is conserved, we have;

Substituting T, VSPRING and VGRAVITY into (1), yield;

8

Assume periodic motion.

Conditions:

a. When t1 passes equilibrium position, V1 = 0, since h = 0

b. As t2 reaches maximum displacement, velocity = 0, thus T2 = 0

m

kΔ

mg

KMEM 3211 – MECHANICAL VIBRATION

Differentiate with respect to t;

From static FBD, forces from kΔ = mg at static equilibrium condition. Therefore, they cancel out, leaving with;

For vibrating system,

But,

Therefore, Eq (2) is the Differential Equation of Motion describing the mass-spring system.

*Note: Study section 2.3 and all the examples.

6. EQUIVALENT MASS AND INERTIA

a. Use in Work-Energy method

b. Simplify description of system whose motions are coupled. (i.e. rotation coupled with translation). Example: Cart-wheel on page 87-90.

7. EQUIVALENT MASS OF SPRING ELEMENTS

a. Generally, spring mass is negligible relative to the system’s mass.

b. When the mass is not negligible i.e. cantilever, it can be calculated using the Work-Energy method. Study example 2.5-3 and go through Table 2.5-1.

8. SYSTEM WITH SPRING ELEMENTS

a. Ideal case: massless spring (relative to a body).

b. Real case: spring with specified mass (normally it will be stated in the problem), use equivalent mass method.

c. Lumped mass (concentrated mass) for homogeneous mass (i.e. uniform mass distribution)

9