Structural Dynamics - ESA 322 Lecture 2

Structural Dynamics - ESA 322

Lecture 2 19/2/2016

Today’s lecture overview • • • • •

Simple harmonic motion Vibration analysis procedure Vibration responses SDOF – Free vibration SDOF – Undamped forced vibration

The Big Picture (Part 1) Week

Week 1

Syllabus

Introductory class Introduction to vibration

Week 2

SDOF Free Undamped Vibration SDOF Force Undamped Vibration

Week 3

SDOF Free Damped Vibration SDOF Force Damped Vibration

Week 4

SDOF Base Excitation SDOF Rotating Unbalanced Mass

Week 5

Numerical Methods 2DOF Free Vibration

Week 6

2DOF Forced Vibration Dynamic Vibration Absorber

Week 7

Test 1

Notes SHM

System identification

Plot responses

Simple Harmonic Motion • A type of periodic motion – where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. – that undergoes sinusoidal oscillation where the acceleration is proportional to the displacement and directed toward the mean position. • Most physical systems exhibit this type of motion during oscillation • The expression for SHM is given by

x(t) = Asin(wt + f )

Simple Harmonic Motion x(t) = Asin(wt + f ) Phase angle amplitude

Oscillation frequency in rad/s

Example of simple harmonic motion response

Simple Harmonic Motion • Phase angle

Simple Harmonic Motion

Vibration analysis procedure 1. Convert the physical system to simplified model. 2. Draw free body diagram (FBD) of the simplified model. 3. Determine the equation of motion of the system from FBD – – –

Newton's 2nd Law, Lagrange's equation D Alembert Principle

4. Solve the equation of motion to obtain the response. – –

Numerically integrate EOM (simulation) Analytical method

5. Interpretation of the result for the physical system.

Vibration analysis procedure • From physical structure to simplified model

FBD

Solve

EOM

Example of vibration responses

Amplitude (y axis)

Time response or Time history

Fourier transform/ Inverse Fourier transform

Spectral response or frequency response

Fourier Transform • The Fourier transform is used to perform – the mapping between a signal as a function of time, and the constituent amplitudes as a function of frequency. – In other words, a signal can be viewed in the time domain or in the frequency domain. The frequency domain representation is also called the spectrum of the signal. – For certain signals, this can be performed analytically with calculus. – For arbitrary signals, the signal must first be digitized, and a Discrete Fourier Transform (DFT) performed. The standard numerical algorithm used for the DFT is called the Fast Fourier Transform (FFT) or Discrete FFT (DFFT). – Due to limitations inherent in digitization and numerical algorithms, the FFT will result in an approximation to the spectrum.

Example of vibration responses • Example of real vibration measurement result

Time domain

Frequency domain

SDOF Free Vibration • Let’s consider a simple harmonic oscillator as shown in figure 1 x(t) k

k - spring stiffness m – mass x(t) – degree of freedom

m Figure 1

Assumptions 1) Stiffness is linear 2) No friction exist 3) Spring has no mass 4) Oscillates with SHM

SDOF Free Vibration • A spring is a type of mechanical link, which in most applications and in this course is assumed to have negligible mass and damping.

SDOF Free Vibration • Free body diagram x(t)

m

Fs • From Newton’s 2nd law

 F  ma Fs  mx

positive Fs – Elastic force

d 2x where a  dt 2 a  x

…………………………………………………………………………(1)

SDOF Free Vibration • From Hooke’s law

Fs  kx

(2)

• Substitute equation 2 into 1, we get • Rearrange

 kx  mx mx  kx  0

• Divide by m

k x  x  0 m

(3) Equation of motion for free vibration

SDOF Free Vibration k x  x  0 m

(4)

• This is 2nd order ODE. To solve it, we must use trial functions. We want to solve this as a function of time. • Lets us assume that the solution for the equation is

x(t )  A cos(nt   )

Substitute (5) into (4)

(5)

SDOF Free Vibration • Before we can substitute we must differentiate (5) with time twice,

Therefore,

d x(t )  x  n A sin(nt   ) dt d2 2   x ( t )  x    n A cos( n t   ) 2 dt

k   A cos(nt   )  A cos(nt   )  0 m 2 n

(6)

SDOF Free Vibration • Simplifying, it becomes k 2  n A cos(nt   )  A cos(nt   )  0 m k 2  n   0 m

k n  m

(7)

SDOF Free Vibration • Substitute (7) into (5)

 k  x(t )  A cos t     m  • A can be found from initial conditions i.e. x(0) - Initial displacement x (0) - Initial velocity

Free Vibration - SDOF • A natural frequency is the frequency at which the structure would oscillate if it were disturbed from its rest position and then allowed to vibrate freely.

k n (rad / s)  m • There is no external force applied to the mass during oscillation; hence the motion resulting from an initial disturbance will be free vibration. • There is no element that causes dissipation of energy during the motion of the mass, the amplitude of motion remains constant with time.

SDOF Forced Vibration • Let’s consider a simple harmonic oscillator as shown in figure 2 with external force acting on the system

x(t) k

m Figure 2

k – spring stiffness m – mass F(t) – External force x(t) – degree of freedom

F(t)

Assumptions 1) Stiffness is linear 2) No friction exist 3) Spring has no mass 4) Oscillates with shm

SDOF Forced Vibration • Draw FBD,

x(t)

m

Fs

F(t)

• Apply Newton’s 2nd law

 F  mx

• Rearrange

 kx  F (t )  mx mx  kx  F (t )

Equation of motion for forced vibration (8)

SDOF Forced Vibration F (t )  Fo cost  • If • Substitute (9) into (8) mx  kx  Fo cost  • Please keep in mind that  is not equal to n

 Is the forcing frequency applied by the external force n Is the natural frequency of the system

(9)

(10)

SDOF Forced Vibration • In forced vibration, when external force is applied to the system with frequency not equal to the system natural frequency, the system will want to oscillate at n . Instead, it will be forced to follow the forcing frequency . • Equation (10) is a non homogenous ODE. This means that it has two solutions; – Steady state (particular solution) – Transient (homogenous solution)

• The total response is the superposition of the transient response and steady state response. Steady state response will dominate after sometime.

SDOF Forced Vibration • Transient and steady state response

SDOF Forced Vibration • Assume the trial function to be in the form of (11) x(t )  A cost  • Differentiate x(t) x  A sin t  x   2 cost 

• Substitute into m  2 A cost   kA cost   Fo cost 

(12)

(13)

SDOF Forced Vibration • Simplifying

k   mA cost   F cost  2

o

Fo k 2 A     m  m Fo m A k 2 m Fo k A 2 1      n 

Cancel out cos(wt)

Simplify into

SDOF Forced Vibration • Equation (14) shows that Fo k A 2 1      n  If If If

  

0 then A = Fo ∞ then A = 0

k

n then A = ∞

which is the static deflection

(14)

SDOF Forced Vibration

• Left figure shows that the displacement and excitation force are in phase when the frequency ratio is less than 1, if it is greater than 1 they have opposite sign, i.e. they are 180 degrees out of phase.

SDOF Forced Vibration • Resonance response

SDOF Forced Vibration • When the forcing frequency is close to but not at the natural frequency, a beating phenomenon occurs. This appears as a low frequency impressed over the frequency of the system. Sometimes the two sinusoids add to each other, and at other times they cancel each other out, resulting in a beating phenomenon.

Exercise i.

What effect does a decrease in mass have on the natural frequency of a system? ii. What effect does a decrease in stiffness have on the natural frequency of a system? iii. Why does the amplitude of free undamped vibration does not diminish after initial disturbance? iv. Why is it important to know the natural frequency of a system?

Exercise v. A harmonic oscillation test gave the natural frequency of a water tower to be 0.41 Hz. Given that the mass of the tank is 150 tonnes, what deflection will result if a 50 kN horizontal load is applied? You may neglect the mass of the tower. vi. Describe in brief what is Fourier transform

Exercise vii. An SDOF system with m = 2 kg, k = 35 N/m) is given an initial displacement of 10 mm. vii. Find the natural frequency; viii. Period of vibration; ix. Amplitude of vibration; • An undamped system vibrates with a frequency of 10 Hz and amplitude 1 mm. Calculate the maximum amplitude of the system's velocity and acceleration.

Past year Exam Question