achapt 4

# achapt 4 - MOTION IN A STABILITY REGION (PART I) 4 MOTION...

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Unformatted text preview: MOTION IN A STABILITY REGION (PART I) 4 MOTION IN A STABILITY REGION (PART I) When motion is confined to one independent degree-of-freedom, the linearized equation that governs the motion is of the form (4 – 1) f kx x c x m = + + & & & In this section, we analyze Eq. (4 – 1) in more detail than in Chapter 1. The system’s free motion ( f = 0) is analyzed first and then its forced motion. The analysis performed for the free motion is called a transient analysis . The transient analysis is independent of the non-homogeneous term f that appears on the right side of the differential equation. The non-homogeneous term is often a force but it can also arise as a result of prescribing the displacement at a point in the system. The right side of the differential equation is generally called the excitation . It is shown how the time dependence of an excitation affects a system’s time response. We start with the constant excitation. Static loads, weight forces, and prescribed displacements are the most frequent examples. The constant excitation change’s a MAE 461: DYNAMICS AND CONTROLS MOTION IN A STABILITY REGION (PART I) system’s equilibrium position. Next, we consider the harmonic excitation. Systems that contain unbalanced rotating elements, like a washing machine, milling machines, and rotating shafts, are systems that are acted on by harmonic excitations. The harmonic excitation causes a system to undergo harmonic motion. 1. Free Undamped Motion Figure 4 – 1: The mass-spring-damper system First consider the free undamped system (See Fig. 4 – 1). Letting f = 0 and c = 0 in Eq. (4 – 1) yields (4 – 2) = + kx x m & & Equation (4 – 2) is a homogeneous constant- coefficient linear differential equation. As with any constant-coefficient linear differential equation, the solution is a combination of complex exponential functions. We start by looking at the single complex exponential function (4 – 3) st e x = MAE 461: DYNAMICS AND CONTROLS MOTION IN A STABILITY REGION (PART I) where s is a complex number that needs to be determined. Substitute Eq. (4 – 3) and its second time derivative into Eq. (4 – 2) to get ) ( 2 = + st st ke e s m Dividing by st e (4 – 3) 2 = + k ms The values of s for which x = satisfies the differential equation are st e (4 – 4 a , b ) m k i s n n = ± = ω ω , where . 1 − = i The two solutions are (4 – 5) t i t i n n e x e x ω ω − = = 2 1 These two solutions may seem a bit odd; after all they’re complex. The complex solutions are actually just building blocks from which the real solution is constructed. Recall that and in Eq. (4 – 5) are complex harmonic functions of the form (See Fig. 4 – 2) t i n e ω t i n e ω − (4 – 6) t i t e t i t e n n t i n n t i n n ω ω ω ω ω ω sin cos sin cos − = + = − MAE 461: DYNAMICS AND CONTROLS MOTION IN A STABILITY REGION (PART I) Figue 4 – 2: e st for pure imaginary s The real solution is a linear combination of the two complex solutions. From Eq. (4 – 5) and Eq. (4 – complex solutions....
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## This note was uploaded on 09/17/2009 for the course MAE 469 taught by Professor Silverberg during the Fall '08 term at N.C. State.

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achapt 4 - MOTION IN A STABILITY REGION (PART I) 4 MOTION...

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