1-DOF Forced Harmonic Vibration. MCE371: Vibrations. Prof. Richter. Department of Mechanical Engineering. Handout 8 Fall 2011

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1 MCE371: Vibrations Prof. Richter Department of Mechanical Engineering Handout 8 Fall 2011

2 Harmonic Forcing Functions Transient vs. Steady Vibration Follow Palm, Sect. 4.1, 4.9 and 4.10 Harmonic forcing functions have the general form F(t) = F o sin(wt) Note that w is independent of w n This type of forcing appears in rotation-induced vibrations: Machine tool chatter Motor mounts Gas turbine engine shafts Forcing functions which are not harmonic (or even cyclic) can be decomposed into a weighted sum of harmonic functions via Fourier analysis.

3 Transient vs. Steady Vibration Harmonic Forcing Solution via Laplace Note: This deviates from the textbook, but the results will be the same Take the 1-DOF equation of motion with harmonic forcing: mẍ +cẋ +k = F o sin(wt) Take Laplace transform with initial conditions: m(s 2 w X sx(0) ẋ(0))+c(sx x(0))+kx = F o s 2 +w 2 Solve for X(s): X(s) = F o w (ms +c)x(0)+mẋ(0) (s 2 +w 2 )(ms 2 + +cs +k) ms 2 +cs +k The first term (red) is called the steady-state response and is independent of the initial conditions. The second term is the transient response, and depends on initial conditions.

4 Steady vs. Transient Response Transient vs. Steady Vibration The transient response is what we would obtain if there was no forcing (F o = 0). We have see that the solution to this case has the form (see Handout 7) x(t) = Be ζwnt sin(w d t +φ) where B and φ are determined by x(0) and ẋ(0). In damped vibrations, ζw n > 0 and the solution is a decaying oscillation. This means that the transient solution practically disappears after some time, and only the steady oscillation remains. The steady response is what we would obtain if the initial conditions were zero.

5 Steady-State Response Transient vs. Steady Vibration The steady-state response is found by inverting the second term of X(s) on page 3 of this handout. For c > 0: where x(t) = X(w)sin(wt +φ(w) X(w) = F o (k mw2 ) 2 +(cw) 2 φ(w) = tan 1 cw mw 2 k X(w) and φ(w) are called the magnitude and phase of the steady solution. The ratio X(w)/F o is called the magnitude ratio. The above formulas match Palm, Eqs , and

6 Example 1-DOF Forced Harmonic Vibration Transient vs. Steady Vibration Use ode45 or Simulink to plot the transient, steady and total solutions for m = 1, c = 0.1, k = 10, w = 0.1 and F o = 2.

7 Magnitude Ratio and Phase Formulas via Transfer Function Define the transfer function as the Laplace ratio of output to input (can be displacement to forcing, damper force to forcing, etc.) with zero initial conditions. For displacement output and forcing input: For 1-DOF vibration we get T(s) = X(s) F(s) X(s) F(s) = 1 ms 2 +cs +k If the output of interest is the force due to the damper and spring, we have f t = cẋ +kx, so F t (s) = csx(s)+kx(s). Substituting F t (s)/(cs +k) F(s) = 1 ms 2 +cs +k F t (s) F(s) = cs +k ms 2 +cs +k Compare this with Palm, Eq (no need to use the e st stuff).

8 The s = iw substitution For any transfer function T(s) (vibrations-related or not) with harmonic input and decaying transients the magnitude ratio and phase are given by: M(w) = output amplitude input amplitude = T(iw) φ(w) = T(iw) where stands for angle (result in radians). This formula is a big deal and it will haunt you through the linear controls course. Learn how to use it!

9 Example 1-DOF Forced Harmonic Vibration Predict the displacement amplitude at steady-state and the phase shift between input and output for the previous example. The main skill to be learned here is predicting the output amplitude when the forcing frequency and amplitude are known. Think about the applications.

10 Dimensionless Magnitude/Phase Charts Define the frequency ratio as r = w w n (forcing frequency to natural frequency). Then the magnitude ratio and the phase can be worked out as a function of ζ and r: M = x ss = 1 1 F o k (1 r2 ) 2 +(2ζr) 2 φ = tan 1 2ζr r 2 1 where x ss indicates the amplitude of the displacement at steady-state. Note that when r = 0 (zero frequency, or constant forcing F o ), M = 1/k, so x ss = F o /k (the static deflection, δ st ). Proper use of dimensionless vibration charts is usually part of engineering licensing exams ( PE exam ).

11 Example 1-DOF Forced Harmonic Vibration Confirm the results of the earlier example using the dimensionless charts/formulas (p.214):

12 Logarithmic Plots and the Decibel The magnitude ratio can become really large for small ζ and r close to 1. It s difficult to plot the M-chart in a linear scale. Define the decibel unit as m = 20logM (base 10 log used). To go back from db to linear units: M = 10 m/20 Once in db, magnitude ratios are plotted on a linear scale (log operation has been built into the definition of m, you don t want to use the log two times). It s convenient to use a logarithmic scale for frequency (or freq. ratio), since M changes very slowly with r for small r and very fast for large r.

13 The Bode Plot The Bode plot (Henrik Bode, ) shows m in db on a linear scale against a logarithmic frequency scale. It also shows the phase in degrees on a linear scale against a logarithmic frequency scale. In Matlab: Create a transfer function object by entering the coefficients of s in descending order: >> T=tf(100,[1 2 25]) (T(s) = 100/(s 2 +2s +25)) Use the bode command with defaults: >> bode(t) Exercise: Find the transfer function for the earlier example and check that the Bode plot gives the same results.

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