Chapter- 4(2)

Dr. Millerjothi, BITS Pilani, Dubai Campus

Transient vibration of single dof systems, Laplace and finite difference methods

Chapter 4


Transient Vibration Excited by a suddenly applied non-periodic excitation F(t) ⇒ transient response Steady-state oscillations are generally not produced.IMPULSE EXCITATION

A force of very large magnitude that acts for a very short time but with a time integral that is finite.

Transient Vibration & IMPULSE EXCITATION


Figure shows an impulsive force of magnitude with a time duration of εAs ε approaches zero, such forces tend to become infinite; however, the impulse is considered to be finiteWhen is unity, such a force in the limiting case ε 0 is called the unit impulse, or the delta function


A delta function at t = is (t - ) and has the following properties:

If (t - ) is multiplied by any time function f(t), the product will be zero everywhere except at t = and its time integral will be


Because Fdt = mdv, the impulse acting on the mass will result in its velocity equal to without an appreciable change in its displacement. For free vibration, Hence, the response of a spring-mass system initially at rest and

excited by an impulse is

Where is the response to a unit impulse.


When damping is present, with

Or

In either the damped or undamped case, the equation for the impulsive response can be expressed in the form


Equation for the response of the system excited by an arbitrary force f(t)The arbitrary force to be a series of impulses as shown in Figure

ARBITRARY EXCITATION


One of the impulses at time t = its strength isand its contribution to the response at time t, is dependent upon the elapsed time t - , or where where , where h(t – ) is the response to a unit impulse started at t = .Because the system is linear, the principle of superposition holds. Thus the response to the arbitrary excitation f(t), is

This integral is called the convolution integral and is sometimes referred to as the superposition integral.


In terms of the relative displacement z = x – y,

For an undamped system initially at rest, the solution for the relative displacement becomes

Base excitation:


Complete solution: both transient and forced vibration Ex 4.3.1 Formulate the Laplace transform solution of a viscously damped spring-mass system with initial Conditions x(0) and

LAPLACE TRANSFORM

)0(x


The response x(t) is found from the inverse of the following equations, the first term represents the forced vibration and the second term represents the transient solution due to the initial conditions.

For the more general case,

where A(s) and B(s) are polynomials


Its reciprocal


The question of how far a body can be dropped without incurring damage is of frequent interest. Such considerations are of paramount importance in the landing of airplanes or the cushioning of packaged articles. In this example, we discuss some of the elementary aspects of this problem by idealizing the mecha

nical system in terms of linear spring-mass components.

- initial conditions

[Example 4.3.2] (Drop Test)


Consider the time response of the undamped spring-mass system to three different excitations. The time response must be

considered in two parts, t < t1 and t > t1.Rise time

- The sum of two ramp functions

PULSE EXCITATION AND RISE TIME 13 4.4 PULSE EXCITATION AND RISE TI

ME Consider the time response of the undamped spring-mass system to three different excitations. The time response must be considered in two parts, 1tt and 1tt. Rise time


For the first ramp function, the terms of the convolution integral are

and the response becomes


For the second ramp function starting at t1, the solution can be written as

By superposition, the response for t1 > t, becomes


Sum of two step functions The response of the step function

The peak response is equal to 2.0 at t = τ/2The response to the second step function started at t = t1 is

The response in the second interval t > t1, becomes

Rectangular pulse


The excitation is

and the differential equation of motion is

The general solution isWhere p = π/ t1

With the initial conditions

Half-sine pulse


The previous solution reduces to

To determine the solution for t > t1, we can use the previous equation but with t replaced by t - t1.

For t - t1, the excitation force is zero and we can obtain the solution as a free vibration with t’ = t - t1.


The initial values can be obtained from previous equation which becomes


• When the time duration t1 for a pulse excitation is smallcompared to the natural period τ of the spring-mass oscillator, theexcitation is called a shock.• Shock vibration tests for certification of satisfactory design• The maximum peak response is a measure of the severity of

the shock.

SHOCK RESPONSE SPECTRUM


Plot of the maximum peak response of the single DOF oscillator as a function of the natural period of the oscillator.

The maximum of the peaks, often labeled maximax, represents only a single point on the time response curve.

It does not uniquely define the shock input because it is possible for two different shock pulses to have the same maximum peak response.

The peak response for the response spectrum plot is

Shock response spectrum (SRS):


In the case where the shock is due to the sudden motion of the support point, f(t), is replaced by the acceleration of thesupport point,

The maximum value of x(t) or z(t) is plotted as a function of t1/τwhere τ is the natural period of the oscillation and t1 is the pulseduration time.


SRS for a time response to the rectangular pulse:For t1/τ = 1/8, the peak response is at tm =

0.32τ.Thus, we have one point, 0.8 on the SRS plot.


If we change the pulse duration time to t1/τ = 0.4, the peak response is at time tm = 0.45τ.This then gives us a second point on the SRS plot, etc.


The dashed line curves are called the residual spectrum, and theupper curve, which is equal to 2.0 for t1/τ > 0.5, represents the

envelope of all peaks.


The following figures show the SRS for the half-sine and the triangle pulse.


For shock isolation, the maximum peak response or the transmissibility must be less than unity.For the rectangular pulse,

and

Vibration isolation is then possible for

and

and the natural period of the isolated system must be greater than six times the pulse time.

SHOCK ISOLATION


Consider a more general pulse bounded by a rectangular pulse,such as shown in Fig. 4.6.1.The impulse of these force pulses is clearly less than that of the

rectangular pulse. It is reasonable to assume that the maximum peak response of

the rectangular pulse must be the upper bound to that of the pulse of general shape.


In Runge-Kutta method, the second order differential equation is first reduced to two first order equation.

Consider the differential equation for the single DOF system


x and y can be expressed in terms of Taylors series. Let the time increment h = Δt.

Replacing the first order derivative by average slope and ignoring higher orders,


Using Simpson rule, the average slope in the interval h becomes

In 4th order Runge-Kutta method, the center term is split into 2 terms and 4 values of are computed for each point i as follows


Solve numerically the differential equation

With initial conditions and forcing function shown in Fig. 4.7.2.

Example 4.7.1


Solution


Although the Runge-Kutta method does not require the evaluation of derivatives beyond the first, its higher accuracy is achieved by four evaluations of the first derivatives to obtain agreement with the Taylor series solution through terms of order h4.

Thus this method is more versatile and can be used for a single or more variables.

For 2 variables, we can let and write 2 first-order equations as

or

Chapter- 4(2)

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