Modeling of a Mechanical System

Transcription

1 Chapter 3 Modeling of a Mechanical System 3.1 Units Currently, there are two systems of units: One is the international system (SI) metric system, the other is the British engineering system (BES) the English system. We prefer to use the SI system in this course, buy may use either of them whenever necessary. Also, within each system, we define certain base units. In SI system, they are meter, ilogram, and second. In BES, they are foot, pound, and second. Almost all the other units used in mechanical system can be derived from these base units, and are categized as derived units. In this course, the following basic and derived units are used me often. Quantity Length Mass Time ce Energy Power SI m g s N=g-m/s 2 J=N-m W=N-m/s BES ft slug s lb ft-lb hp 3.2 Mechanical Elements Inertia Elements The inertia elements include masses f translation and moments of inertia f rotation. The mass is usually denoted by m with unit as g slug. The moment of inertia often represented as J with unit as g m Spring Elements A linear spring is a mechanical element that can be defmed by an eternal fce such that the defmation is directly proptional to the fce tque applied to it. a translational spring, the relation between the acting fce and the net displacement is = = ( 1 2 ) (3.2.1) where is a proptionality constant called a spring constant, the unit of is N/m lb/in. a rotational motion with a tsional spring, the relation between the acting tque T and the net angular displacement θ is T = θ = (θ 1 θ 2 ) (3.2.2) where is also a proptionality constant called a tsional spring constant. Note: When a linear spring is overstretched over certain point, it will become nonlinear. In this course, we will assume the springs are always wing within its linear limit. urther, although all practical springs have inertia and damping, we assume that the effect of them are negligibly small. Therefe, all the springs in this course will be ideal springs with neither mass n damping and will obey the linear fce-displacement law linear tque-angular displacement law. 21

2 CHAPTER 3. MODELING O A MECHANICAL SYSTEM 22 Eercise Given two springs with spring constant 1 and 2, obtain the equivalent spring constant eq f the two springs connected in (1) parallel (2) serial Solution: (1) The two springs have same displacement, therefe, = = eq, eq = (2) The fces on each spring are same,. However, their displacements are different. Let them be 1 and 2. Then, 1 1 = 2 2 = 1 = 1 2 = 2 Since the total movement is = 1 + 2, and we have = eq, then we can obtain eq = eq = = Damping Elements When the viscosity drag is not negligible in a system, we often model them with the damping fce. The damping fce always depends on the relative velocity between the fluid and the surface. In this class, we will only deal with the linear damping fce, which is a linear function of the relative velocity. In engineering, we often encounter three inds of damping elements: a dashpot, a sliding viscous friction a journal bearing. The fmer two can be classified as translational damper. Given the relative velocity, the drag fce generated is f = cẋ = c(ẋ 1 ẋ 2 ) where c is a proptionality constant called the damping coefficient. bearing is a special ind of rotational damper, with the drag tque: T = c ω = c(ω 1 ω 2 ) On the other hand, the journal where ω = θ is the relative angular velocity. Lie spring elements, we assume the damping elements are only wing in their linear range in this course. Also, we consider all the damping elements are idea, that is, with no inertia and spring effects. When analyze a system, we often use these symbols to denote different damping elements. Note that the direction of the fce is always opposite to the direction of the relative motion. Eercise Given two dampers with damping constant c 1 and c 2, obtain the equivalent damping constant c eq f the two dampers connected in (1) parallel (2) serial.:

3 CHAPTER 3. MODELING O A MECHANICAL SYSTEM 23 c1 c2 1 2 c1 c2 Solution: Using the analysis analogous to spring, ecept that with c instead of, ẋ instead of, we have (1) in parallel c eq = c 1 + c 2 (2) in serial c eq = c 1c 2 c 1 + c 2 Eercise How about putting two masses in parallel and in series? 3.3 Modeling the Simple Translational System Newton s laws: 1. A particle iginally at rest, moving in a straight line with a constant velocity, will remain that way as long as it is not acted upon by an eternal fce. 2. The time rate of change of momentum equals the eternal fce. f = d (mv) dt 3. Every action is opposed by an equal reaction. = m dv dt Eample 1. A simple hizontal spring-mass system on a frictionless surface. = ma (3.3.1) m. m rom the free body diagram (.B.D.) of the mass, we can see that the only fce acting on the mass is the spring fce provided that there is no eternal fce. Using Newton s laws, we can easily obtain the dynamic equation: mẍ = mẍ + = 0 Eample 2. A simple vertical spring-mass system. Now, there is one me eternal fce acting on the mass: the gravity. Therefe, the equation becomes: mÿ = y + mg However, if we inspect the system me closely, we can see that the gravitational fce is always being opposed statically by the equilibrium spring deflection δ, mg = δ. If we measure the displacement from the equilibrium position, that is, = y δ, y = + δ, then the dynamic equation can be simplified to mẍ = ( + δ) + mg, mẍ + = 0.

4 CHAPTER 3. MODELING O A MECHANICAL SYSTEM 24 c y m Eample 3. A single-mass model of a car s suspension. When we want to mae a preliminary eamination of a car s suspension, we always model it as a mass-spring-damper system. Here, we model it with a single mass. The elasticity of both the tire and the suspension spring can be modeled with a spring with spring constant. The shoc absber is modeled as a translational damper with constant c. If we assume the weight of the car is evenly distributed to the four wheels, and the mass of the wheel, tire and ale are negligible, the mass m of the model is one quarter of the total mass ecept the moving part. We want to find out the motion of the system. c m1 rom the free body diagram, we have mẍ = c(ẏ ẋ) (y ) mẍ + cẋ + = cẏ + y Eample 4. A two-mass model of a suspension. Here we model the system with me compleity. We consider the suspension model with the wheel-tire-ale assembly. As the previous one, the mass m 1 is one-fourth mass of the car body, and m 2 is the mass of the wheel-tire-ale assembly. c m1 1 m 2 2 rom the free body diagram, the equation f m 1 is m 1 ẍ 1 = c 1 (ẋ 2 ẋ 1 ) + 1 ( 2 1 ) while the equation f m 2 is m 1 ẍ 1 + c 1 ẋ c 1 ẋ = 0 m 2 ẍ 2 = c 1 (ẋ 2 ẋ 1 ) 1 ( 2 1 ) + 2 (y 2 )

5 CHAPTER 3. MODELING O A MECHANICAL SYSTEM 25 m 2 ẍ 2 + c 1 ẋ 2 + ( ) 2 c 1 ẋ = 2 y 3.4 Modeling the simple rotational system Rotation about a fied ais: M = dh dt = I dω (3.4.1) dt where M is the eternal applied moments tque, H is the angular momentum, I is the body s moment of inertia about the ais, and ω is the vect angular velocity. Eample 5. A model with angular displacement input. Given a shaft suppted by a bearing with damping, we want to find out the dynamic equation of the system. The input is the rotation θ i. Because of the elasticity of the shaft, the angular displacement of the shaft is different, and denoted by θ. The elasticity of the shaft is modeled by a rotational spring, while the damping constant is c. rom the free body diagram, we have I θ = (θ i θ) c θ I θ + c θ + θ = θ W, Energy, and Power The w done in a mechanical system is the dot product of fce and distance ( tque and angular displacement), when they are epressed in vect fm. Let fce be and distance, then the w W done by the fce along the displacement is W = (3.5.1) Note that the w is a scalar. It can be positive negative, but comes with no direction. The unit of w is joule (J) in SI system, 1J = 1N m Energy is the ability capacity to do w. It can be as many fms, such as chemical energy, electrical energy, mechanical energy. In mechanical system, the energy always eists as the potential energy inetic energy, both. The potential energy of a mass (m) in mechanical system always dues to the gravitational field. It is defined as U = ˆ h 0 mgd = mgh where h is the altitude and g is the local gravity acceleration.

6 CHAPTER 3. MODELING O A MECHANICAL SYSTEM 26 The potential energy of a spring is defined as U = U = ˆ 0 ˆ θ 0 d = θdθ = 1 2 θ2 (Translational) (Rotational) Eercise: What are the changes of the potential energy f a mass at different positions, and a spring with different stretch compression? Only inertia elements in mechanical systems can ste inetic energy. a mass m in pure translation, it is T = 1 2 mẋ2 where ẋ is the velocity of the mass. a moment of inertia J in pure rotation, it is T = 1 2 J θ 2 where θ is the angular velocity of the moment of inertia. Eercise: How to calculate the change of inetic energy with different velocity angular velocity? Power is the time rate of doing w. Note that there is no time issue when we consider w and energy, while power is the w per unit time. Conservation of energy: When there is no friction element (such as a damper) in a mechanical system, the total energy will be constant, that is T 1 + U 1 = T 2 + U 2 where the subscript 1, 2 denote two distinct instances. Eercise: Given the spring-mass system shown by eample 1, the motion of the mass can be solved as (t) = 0 cos( m t) provided that the initial condition is (0) = 0 and ẋ 0 = 0. Verify that the total energy is constant. Advanced Topic: When there is no energy enter leave the mechanical system, we often use the Euler-Lagrange Equation to find out the dynamic equation. That is, define the Lagrangian L = T U, the generalized variables q, the motion of the system can be described by d dt ( L q ) L q = 0 (3.5.2) To probe further, you are encouraged to read related boos tae Advanced Dynamics and other similar courses.