3 General forced response So far, all of the driving forces have been sine or cosine excitations In this chapter we examine the response to any form of excitation such as Impulse Sums of sines and cosines Any integrable function © D, J, Inman 1/47 Mechanical Engineering at Virginia Tech

Mechanical Engineering at Virginia Tech Linear Superposition allows us to break up complicated forces into sums of simpler forces, compute the response and add to get the total solution © D, J, Inman 2/47 Mechanical Engineering at Virginia Tech

3.1 Impulse Response Function F(t) Impulse excitation t -e t +e t e is a small positive number Figure 3.1 © D, J, Inman 3/47 Mechanical Engineering at Virginia Tech

Mechanical Engineering at Virginia Tech From sophomore dynamics The impulse imparted to an object is equal to the change in the objects momentum F(t) i.e. area under pulse t -e t +e t © D, J, Inman 4/47 Mechanical Engineering at Virginia Tech

Mechanical Engineering at Virginia Tech We use the properties of impulse to define the impulse function: Dirac Delta function Equal impulses F(t) t © D, J, Inman 5/47 Mechanical Engineering at Virginia Tech

The effect of an impulse on a spring-mass-damper is related to its change in momentum. Just after impulse Just before impulse Thus the response to impulse with zero IC is equal to the free response with IC: x0=0 and v0 =FDt/m © D, J, Inman 6/47 Mechanical Engineering at Virginia Tech

Recall that the free response to just non zero initial conditions is: © D, J, Inman 7/47 Mechanical Engineering at Virginia Tech

So for an underdamped system the impulse response is (x0 = 0) k x(t) c 1 0.5 h(t) -0.5 -1 10 20 30 40 Time Response to an impulse at t = 0, and zero initial conditions © D, J, Inman 8/47 Mechanical Engineering at Virginia Tech

Mechanical Engineering at Virginia Tech The response to an impulse is thus defined in terms of the impulse response function, h(t). © D, J, Inman 9/47 Mechanical Engineering at Virginia Tech

Mechanical Engineering at Virginia Tech 1 t=0 h1 For example: If two pulses occur at two different times then their impulse responses will superimpose -1 10 20 30 40 1 t=10 h2 -1 10 20 30 40 1 h1+h2 -1 10 20 30 40 © D, J, Inman 10/47 Time Mechanical Engineering at Virginia Tech

Consider the undamped impulse response © D, J, Inman 11/47 Mechanical Engineering at Virginia Tech

Example 3.1.1Design a camera mount with a vibration constraint Consider example 2.1.3 of the security camera again only this time with an impulsive load © D, J, Inman 12/47 Mechanical Engineering at Virginia Tech

Mechanical Engineering at Virginia Tech Using the stiffness and mass parameters of Example 2.1.3, does the system stay with in vibration limits if hit by a 1 kg bird traveling at 72 kmh? The natural frequency of the camera system is From equations (3.7) and (3.8) with  = 0, the impulsive response is: The magnitude of the response due to the impulse is thus © D, J, Inman 13/47 Mechanical Engineering at Virginia Tech

Mechanical Engineering at Virginia Tech Next compute the momentum of the bird to complete the magnitude calculation: Next use this value in the expression for the maximum value: This max value exceeds the camera tolerance © D, J, Inman 14/47 Mechanical Engineering at Virginia Tech

Example 3.1.2: two impacts, zero initial conditions. (double hit) © D, J, Inman 15/47 Mechanical Engineering at Virginia Tech

Example 3.1.2 two impacts and initial conditions Note, no need to redo constants of integration for impulse excitation (others, yes) © D, J, Inman 16/47 Mechanical Engineering at Virginia Tech

Computation of the response to first impulse: © D, J, Inman 17/47 Mechanical Engineering at Virginia Tech

Total Response for 0< t < 4 © D, J, Inman 18/47 Mechanical Engineering at Virginia Tech

Next compute the response to the second impulse: Here the Heaviside step function is used to “turn on” the response to the impulse at t = 4 seconds. © D, J, Inman 19/47 Mechanical Engineering at Virginia Tech

To get the total response add the partial solutions: © D, J, Inman 20/47 Mechanical Engineering at Virginia Tech

3.2 Response to an Arbitrary Input The response to general force, F(t), can be viewed as a series of impulses of magnitude F(ti)Dt Response at time t due to the ith impulse zero IC xi(t) = [F(ti)Dt ].h(t-ti) for t>ti ti xi t F(t) Impulses F(ti) (3.12) t t1,t2 ,t3 ti © D, J, Inman 21/47 Mechanical Engineering at Virginia Tech

Properties of convolution integrals: It is symmetric meaning: © D, J, Inman 22/47 Mechanical Engineering at Virginia Tech

Mechanical Engineering at Virginia Tech The convolution integral, or Duhamel integral, for underdamped systems is: The response to any integrable force can be computed with either of these forms Which form to use depends on which is easiest to compute © D, J, Inman 23/47 Mechanical Engineering at Virginia Tech

Example 3.2.1: Step function input Figure 3.6 Step function To solve apply (3.13): © D, J, Inman 24/47 Mechanical Engineering at Virginia Tech

Mechanical Engineering at Virginia Tech © D, J, Inman 25/47 Mechanical Engineering at Virginia Tech

Integrating (use a table, code or calculator) yields the solution: Fig 3.7 © D, J, Inman 26/47 Mechanical Engineering at Virginia Tech

Mechanical Engineering at Virginia Tech Example: undamped oscillator under IC and constant force For an undamped system: F0 t1 t2 The homogeneous solution is x(t) m Good until the applied force acts at t1, then: k © D, J, Inman 27/47 Mechanical Engineering at Virginia Tech

Next compute the solution between t1 and t2 © D, J, Inman 28/47 Mechanical Engineering at Virginia Tech

Now compute the solution for time greater than t2 © D, J, Inman 29/47 Mechanical Engineering at Virginia Tech

Total solution is superposition: 0.3 0.2 Displacement x(t) 0.1 -0.1 2 4 6 8 10 © D, J, Inman 30/47 Time (s) Mechanical Engineering at Virginia Tech

Example 3.2.3: Static versus dynamic load This has max value of , twice the static load © D, J, Inman 31/47 Mechanical Engineering at Virginia Tech

Numerical simulation and plotting At the end of this chapter, numerical simulation is used to solve the problems of this section. Numerical simulation is often easier then computing these integrals It is wise to check the two approaches against each other by plotting the analytical solution and numerical solution on the same graph © D, J, Inman 32/47 Mechanical Engineering at Virginia Tech

3.3 Response to an Arbitrary Periodic Input We have solutions to sine and cosine inputs. What about periodic but non-harmonic inputs? We know that periodic functions can be represented by a series of sines and cosines (Fourier) Response is superposition of as many RHS terms as you think are necessary to represent the forcing function accurately 2 T 1.5 1 0.5 Displacement x(t) -0.5 -1 -1.5 -2 2 4 6 Time (s) Figure 3.11 © D, J, Inman 33/47 Mechanical Engineering at Virginia Tech

Recall the Fourier Series Definition: © D, J, Inman 34/47 Mechanical Engineering at Virginia Tech

The terms of the Fourier series satisfy orthogonality conditions: © D, J, Inman 35/47 Mechanical Engineering at Virginia Tech

Fourier Series Example F(t) Step 1: find the F.S. and determine how many terms you need F0 t1 t2=T © D, J, Inman 36/47 Mechanical Engineering at Virginia Tech

Fourier Series Example 1.2 1 F(t) 2 coefficients 0.8 10 coefficients 100 coefficients 0.6 Force F(t) 0.4 0.2 -0.2 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Time (s) © D, J, Inman 37/47 Mechanical Engineering at Virginia Tech

Having obtained the FS of input The next step is to find responses to each term of the FS And then, just add them up! Danger!!: Resonance occurs whenever a multiple of excitation frequency equals the natural frequency. You may excite at 100rad/s and observe resonance while natural frequency is 500rad/s!! © D, J, Inman 38/47 Mechanical Engineering at Virginia Tech

Solution as a series of sines and cosines to The solution can be written as a summation Solutions calculated from equations of motion (see section Example 3.3.2) © D, J, Inman 39/47 Mechanical Engineering at Virginia Tech

Mechanical Engineering at Virginia Tech 3.4 Transform Methods An alternative to solving the previous problems, similar to section 2.3 © D, J, Inman 40/47 Mechanical Engineering at Virginia Tech

Mechanical Engineering at Virginia Tech Laplace Transform Laplace transformation Laplace transforms are very useful because they change differential equations into simple algebraic equations Examples of Laplace transforms (see page 216) in book) f(t) F(s) Step function, u(t) e-at sin(w t ) 1/s 1/(s+a) w / ( s2 + w 2) © D, J, Inman 41/47 Mechanical Engineering at Virginia Tech

Mechanical Engineering at Virginia Tech Laplace Transform Example: Laplace transform of a step function u(t) u(t) t Example: Laplace transform of e-at e-at t © D, J, Inman 42/47 Mechanical Engineering at Virginia Tech

Laplace Transforms of Derivatives Laplace transform of the derivative of a function Integration by parts gives, © D, J, Inman 43/47 Mechanical Engineering at Virginia Tech

Laplace Transform Procedures Laplace transform of the integral of a function Steps in using the Laplace transformation to solve DE’s Find differential equations Find Laplace transform of equations Rearrange equations in terms of variable of interest Convert back into time domain to find resulting response (inverse transform using tables) © D, J, Inman 44/47 Mechanical Engineering at Virginia Tech

Laplace Transform Shift Property Note these shift properties in t and s spaces... thus © D, J, Inman 45/47 Mechanical Engineering at Virginia Tech

Mechanical Engineering at Virginia Tech Example 3.4.3: compute the forced response of a spring mass system to a step input using LT Compare this to the solution given in (3.18) © D, J, Inman 46/47 Mechanical Engineering at Virginia Tech

Mechanical Engineering at Virginia Tech Fourier Transform From Fourier series of non-periodic functions Allow period to go to infinity Similar to Laplace Transform Useful for random inputs Corresponding inverse transform Fourier transform of the unit impulse response is the frequency response function 1 2 3 4 -0.1 -0.05 0.05 0.1 Time (s) h(t) w n =2 and M=1 Fourier Transform 1 2 3 4 5 -20 -10 10 20 Frequency (Hz) Normalized H(w) (dB) © D, J, Inman 47/47 Mechanical Engineering at Virginia Tech