KNJ4383 Mechanical Vibrations

Introduction to course content

Aim of the course

To provides knowledge, understanding and synthesis in

vibration monitoring.

Course Outcomes

By the end of the course, students should be able to : • Apply vibration analysis to solve dynamic

problems • Develop vibration solutions to solve

complex engineering problems • Demonstrate proper techniques in

measuring vibration • Explain noise, the measurement and the

practical applications

Topics • Vibration - Introduction

Vibration theory, Free Vibration, Free body diagram of the system

• Single-Degree of Freedom (SDOF)

Transient and stead state response. Frequency response functions. Mass, stiffness and damping controlled regions. Practical applications.

• Vibration of Multi-Degree of Freedom (MDOF) systems

Lagrange’s equation of motion. Free and forced response using direct solutions of equation of motion. Forced Response by modal analysis. Introduction to orthogonality and generalised coordinates. Transmissibility.

• Vibration

Measurement transducers. Experimental frequency analysis. Experimental modal analysis.

• Free and forced vibration of continuous elastic systems

• Rayleigh’s method for the fundamental natural frequency

• Mechanical Vibration - Noise

Course Assessment

20%

30%

50%

Percentage %

Tests

Assignment/Quiz/Project/Tutorials/PBL

Final Exam

Reference books

1. S.S. Rao, Mechanical Vibrations, Prentice Hall, 2005.

2. B.H. Tongue, Principle of Vibrations, Oxford Press, 2002.

3. W.T. Thomson, Theory of Vibration with Applications, Chapman & Hall, 1993.

4. Balachandran, EB. Magrab, Vibrations, Thomson, 2009.

Importance of Study Vibration • Human activities (hear because of eardrums vibrate, see

because light waves undergo vibration, breathe associated with the vibration of lungs, speak due to oscillatory motion of larynges & tongues).

• In turbines, vibration cause mechanical failures. • Natural frequency of vibration of a machine coincides

with the frequency of the external excitation – Resonance.

• Besides detrimental effects, vibration can be utilise profitably – in vibratory conveyors/ hoppers, washing machine, electric toothbrushes, dentist’s drills, clocks, massage units, pile driving, vibratory testing of materials, electronic circuits to filter out the unwanted frequencies.

Introduction to Vibration • Engineering systems possessing mass and elasticity are

capable of relative motion. If the motion of such systems repeats itself after a given interval times, the motion is know as vibration (oscillation).

• In general, vibration is a form of wasted energy & undesirable in many cases, e.g. in machinery – noise generated, breaks down parts, transmission of unwanted forces & movements to closely-by objects.

• Generally, a vibratory system includes means for storing potential energy (spring/elasticity), kinetic energy (mass/inertia) & by which energy is gradually lost (damper).

Introduction to Vibration • To vibrate, a system must have two key attributes: Mass &

Elasticity. • Real system will possess some damping – acts to dissipate

energy. • The minimum number of independent coordinates required

to determine completely the positions of all parts of a system at any instant of time defines the number of degrees of freedom of the system.

Single degree of freedom systems

Introduction to Vibration Example of simple pendulum with m (mass), θ (angular displacement)

At 1 – Kinetic energy = 0

At 2 – Potential energy = mgl(1-cos θ)

Gravitational force (mg) induces a torque

mgl sinθ about the point O, the pendulum

swing to the left from 1. At 2, all PE convert to KE.

Then continue to swing to 3.

Pendulum keep repeating and will have oscillatory

motion.

The magnitude of oscillatory (θ) gradually decreases and the pendulum ultimately stops due to resistance (damping) offered by air (dissipation of vibration).

Classification of Vibration

Classified in several ways:

• Free vibration & forced vibration

• Undamped & damped vibration

• Linear & nonlinear vibration

• Deterministic & random vibration

Free Vibration

• Free vibration - the periodic motion observed as the system is displaced from its static equilibrium position.

• The forces acting are the spring force, the friction force, and the weight of the mass.

• The vibration will diminish with time due to the presence of friction.

• The free vibration is sometimes called transient.

Vibration Analysis Procedure • A vibratory system is a dynamic system for which

the variables such as the excitations (inputs) and responses (outputs) are time dependent. The response of a vibrating system generally depends on the initial conditions as well as the external excitations.

• Often the overall behaviour of the system can be determined by considering even a simple model of the complex physical system. Thus the analysis of a vibrating system usually involves mathematical modeling, derivation of the governing equations, solution of the equations, and interpretation of the results.

Example 1 : Mathematical Model of a motorcycle Figure below shows a motorcycle with a rider. Develop a sequence of three mathematical models of the system for investigating vibration in the vertical direction. Consider the elasticity of the tires, elasticity and damping of the struts (in the vertical direction), masses of the wheels, and elasticity, damping and mass of the rider.

Motorcycle with a rider – a physical system and mathematical model

Solution: We start with the simplest model and refine it gradually. When the equivalent values of the mass, stiffness, and damping of the system are used, we obtain a single-degree-of-freedom model of the motorcycle with a rider as indicated in Fig(b). In this model, the equivalent stiffness(keq) includes the stiffnesses of the tires, struts, and rider. The equivalent damping constant (ceq)includes the damping of the struts and the rider. The equivalent mass includes the masses of the wheels, vehicle body, and the rider. This model can be refined by representing the masses of wheels, elasticity of the tires, and elasticity and damping of the struts separately, as shown in Fig(c). In this model, the mass of the vehicle body (mv) and the mass of the rider(mr) are shown as a single mass, mv + mr When the elasticity (as spring constant kr) and damping (as damping constant Cr)of the rider are considered, the refined model shown in Fig(d) can be obtained. Note that the models shown in Figs(b) to (d) are not unique. For example, by combining the spring constants of both tires, the masses of both wheels, and the spring and damping constants of both struts as single quantities, the model shown in Fig(e) can be obtained instead of Fig(c).

Mechanical System Components

• A mechanical system comprises inertia components, stiffness components, and damping components. The inertia components have kinetic energy when the system is in motion. The kinetic energy of a rigid body undergoing planar motion is

T = 𝟏

𝟐 𝒎𝒗𝟐 +

𝟏

𝟐 𝑰𝒘𝟐

where v is the velocity of the body’s mass center, w is its angular velocity about an axis perpendicular to the plane of motion, m is the body’s mass, and I is its mass moment of inertia about an axis parallel to the axis of rotation through the mass center. • A linear stiffness component (a linear spring) has a force displacement relation of the

form

F = kx where F is applied force and x is the component’s change in length from its unstretched length. The stiffness k has dimensions of force per length. • A dashpot is a mechanical device that adds viscous damping to a mechanical system.

A linear viscous damping component has a force-velocity relation of the form

F = cv where c is the damping coefficient of dimensions mass per time.

Equivalent Systems Analysis • All linear 1-degree-of-freedom systems with viscous

damping can be modeled by the simple mass-spring-dashpot system of Fig. Let x be the chosen generalized coordinate. The kinetic energy of a linear system can be written in the form

T = 𝟏

𝟐 𝒎𝒆𝒒𝒙

𝟐

• The potential energy of a linear system can be written in the form

V= 𝟏

𝟐 𝒌𝒆𝒒𝒙

𝟐

• The work done by the viscous damping force in a linear system between two arbitrary locations x1 and x2 can be written as

W = - 𝒄𝒆𝒒𝒙𝟐𝒙𝟏

𝒙 𝒅𝒙

Torsional Systems • When an angular coordinate is used as a generalized coordinate for a linear

system, the system can be modeled by the equivalent torsional system of Figure. The moment applied to a linear torsional spring is proportional to its angular rotation while the moment applied to a linear torsional viscous damper is proportional to its angular velocity. The equivalent system coefficients for a torsional system are determined by calculating the total kinetic energy, potential energy, and work done by viscous damping forces for the original system in terms of the chosen generalized coordinate and setting them equal to

T =

𝟏

𝟐 𝑰𝒆𝒒ө

𝟐

V = 𝟏

𝟐 𝒌𝒕𝒆𝒒ө

𝟐

W = - 𝒄𝒆𝒒𝒙𝟐𝒙𝟏

ө 𝒅ө

Static Equilibrium Position • Systems, such as the one in Figure, have elastic

elements that are subject to force when the system is in equilibrium.

• The resulting deflection in the elastic element is called its static deflection, usually denoted by, ∆𝑠𝑡.

• The static deflection of an elastic element in a linear system has no effect on the system’s equivalent stiffness.

Example 2: Spring constant of a Rod Find the equivalent spring constant of a uniform rod of length l, cross-sectional area A, and Young’s modulus E subjected to an axial tensile (or compression) force F as shown in Figure (a) below.

Solution: The elongation (or shortening) δ of the rod under the axial tensile (or compressive) force F can be expressed as

δ = δ𝒍𝒍 = Ɛ 𝒍 =

𝝈

𝑬 𝒍 =

𝑭𝒍

𝑨𝑬

Where Ɛ = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ

𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ =

δ𝒍 is the strain and 𝝈 =

𝑓𝑜𝑟𝑐𝑒

𝑎𝑟𝑒𝑎=

𝐹

𝐴 is the stress induced in the rod.

Using the definition of the spring constant k, we obtain from Equation

k = 𝑓𝑜𝑟𝑐𝑒 𝑎𝑝𝑝𝑙𝑖𝑒𝑑

𝑟𝑒𝑠𝑢𝑙𝑡𝑖𝑛𝑔 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 =

𝐹

𝜎=

𝐴𝐸

𝑙

The significance of the equivalent spring constant of the rod is shown in Fig(b).

Example 3: Spring constant if a cantilever beam

Find the equivalent spring constant of a cantilever beam subjected to a concentrated load F at its end as shown in Figure below.

Solution: We assume, for simplicity, that the self weight (or mass) of the beam is negligible and the concentrated load F is due to the weight of a point mass (W = mg). From strength of materials, we know that the end deflection of the beam due to a concentrated load F = W is given by

𝝈 = 𝑾𝒍𝟑

𝟑𝑬𝑰

where E is the Young’s modulus and I is the moment of inertia of the cross section of the beam about the bending or a-axis (i.e., axis perpendicular to the page). Hence the spring constant of the beam is Figure (b):

𝒌 = 𝑾

𝜹 =

𝟑𝑬𝑰

𝒍𝟑

Example 4: Equivalent k of a Suspension System

Figure shows the suspension system of a freight truck with a parallel-spring arrangement. Find the equivalent spring constant of the suspension if each of the three helical springs is made of steel with a shear modulus G = 80 X 109 N/m2 and has five effective turns, mean coil diameter D = 20 cm., and wire diameter d = 2 cm.

Solution: The stiffness of each helical spring is given by

𝒌 = 𝑮𝒅𝟒

𝟖𝒏𝑫𝟑

Since the three springs are identical and parallel, the equivalent spring constant of the suspension system is given by

𝒌𝒆𝒒 = 3k = 3(40,000) = 120,000 N/m

Helical Spring under axial load (d = wire diameter, D = mean coil diameter,

n = number of active turns); 𝒌 = 𝑮𝒅𝟒

𝟖𝒏𝑫𝟑

The End