ChapterBilingual学习教案

1、会计学1ChapterBilingual第一页，共42页。2022-6-252第1页/共42页第二页，共42页。Chapter 3 Formulation and Dynamic Behavior of Translational Mechanical Systems线性机械系统的公式和动态(dngti)性能Syllabus3.1 Introduction3.2 Variables3.3 Element Laws元件定律元件定律(dngl)3.4 Interconnection Laws连接定律连接定律(dngl)3.5 Obtaining the System Model第2页/共42页第三

2、页，共42页。2022-6-254第3页/共42页第四页，共42页。2022-6-255pIn order to analyze, synthesize and design a dynamic system, an accurate mathematical model must be determined in advance事先事先(shxin).pMathematical models describe描述描述 the motion patterns 运动模式运动模式of physical systems.pMathematical models quantitatively量化地量化

3、地 reveal揭示揭示 the relationship between systems parameters参数参数 and performance功能功能, and the dynamic behavior of the system as well.pThe derivation of the model 模型的推导模型的推导is based on the fact that the dynamic system can be completely described完全描述完全描述 by known differential equations or experimental tes

4、t data.pThe ability to analyze the system and determine its performance depends on how well the characteristics can be expressed mathematically.3.1.1 Concepts of Mathematical Models第4页/共42页第五页，共42页。2022-6-2561.Static models: reveal the system motion patterns 系统运动系统运动(yndng)模式模式independent of time.2.

5、Dynamic models: reveal the system motion patterns dependent of time, including:3.External models: they only reveal relationship between the input and the corresponding output of a control system, e. g. differential equation, transfer function, etc.4.Internal models: they reveal relationship among th

6、e input, output and the internal variables, e. g. state space expression, etc.3.1.2 Types of Mathematical Models第5页/共42页第六页，共42页。2022-6-257第6页/共42页第七页，共42页。2022-6-258The symbols for the basic variables used to describe the dynamic behavior of translational mechanical systems are:p x - Displacement 位

7、移位移(wiy)in meter ( m )p v - Velocity速度速度 in meter per second ( m/s )p f - Force力力 in Newton ( N )p a - Acceleration加速度加速度 in meter per second square ( m/s2 )第7页/共42页第八页，共42页。2022-6-259第8页/共42页第九页，共42页。2022-6-2510第9页/共42页第十页，共42页。2022-6-2511第10页/共42页第十一页，共42页。pPhysical devices are represented by one

8、or more idealized elements that obey laws involving the variables associated with the elements. pSome degree of approximation is required in selecting the elements to represent a device, and the behavior of the combined elements may not correspond exactly to the behavior of the device. pThe major el

9、ements that we include in translational systems are mass 质量质量, friction(damping)摩擦摩擦(mc)（阻尼）（阻尼）, stiffness(spring)刚性（弹簧）刚性（弹簧）. 2022-6-2512第11页/共42页第十二页，共42页。2022-6-2513Newtons First LawWhen viewed in an inertial reference frame, an object either remains at rest or continues to move at a constant v

10、elocity, unless acted upon by an external force. Newtons Second LawThe vector sum of the external forces 外力外力(wil)F on an object is equal to the mass m of that object multiplied by the acceleration vector a of the object: F = ma.Newtons Third LawWhen one body exerts a force on a second body, the sec

11、ond body simultaneously exerts a force equal in magnitude and opposite in direction 大小相等方向大小相等方向(fngxing)相反相反on the first body.Review第12页/共42页第十三页，共42页。2022-6-25142.)()()(F(t)(B(t)(BF(t)()()(F(t)dttaKdttvKtKxdttaxBtvtxmtvmtma第13页/共42页第十四页，共42页。3.3.1 Mass质量质量(zhling)pA mass M, which has units of kilo

12、grams (kg), subjected to a force f. pNewtons second law states that the sum of the forces acting on a body is equal to the time rate of the change of the momentum动量动量(dngling):()dMvfdt which, for a constant mass定常质量定常质量(zhling), can be written as:dvMfdt (3.4)2022-6-2515第14页/共42页第十五页，共42页。Mass M22( )

13、( )( )MddftMv tMx tdtdt 2022-6-2516fM(t)(3.5)第15页/共42页第十六页，共42页。3.3.2 Friction A mass sliding on an oil film油膜油膜that has laminar flow层流层流, as depicted描绘描绘(miohu) in Fig. 3.3(a), is subject to viscous friction粘性摩擦粘性摩擦and obeys the linear relationship: fB vpThe friction coefficient B摩擦系数摩擦系数 has units

14、 单位单位(dnwi)of Newton-second per meter (Ns/m).pB is proportional成正比成正比 to the contact area and to the viscosity粘性粘性 of the oil, and inversely proportional 成反比成反比to the thickness of the film. pv = v2-v1Figure 3.3 (a) Friction described by (3.6) with v = v2-v12022-6-2517(3.6)第16页/共42页第十七页，共42页。Viscous

15、friction also may be used to model a dashpot阻尼器阻尼器, such as the shock absorbers 吸振器吸振器on an automobile. As indicated in Fig. 3.4(a), a piston moves through an oil-filled cylinder装满油的装满油的圆柱圆柱(yunzh), and there are small holes in the face of piston through which the oil passes as the parts零件零件 move re

16、lative to each other. The symbol符号符号 often used for a dashpot is shown in Fig. 3.4(b). Figure 3.4 (a) A dashpot阻尼器阻尼器. (b) Its representation表示表示(biosh).2022-6-2518(a)(b)第17页/共42页第十八页，共42页。Damping B 1212( )( )( )( )( )( )( )Bdd x tftBx tx tBdtdtB v tv tB v t 2022-6-2519(3.7)第18页/共42页第十九页，共42页。3.3.3

17、Stiffness刚刚性性(n xn) The most common stiffness element刚性刚性(n xn)元件元件 is the spring.Figure 3.6 (a)d0: the length of the spring when no force is applied;x: the elongation变形变形(bin xng) caused by the force f.d(t) = d0 + x: the total length at any instantFigure 3.6 (b) The stiffness property refers to the

18、 algebraic relationship between x and f For a linear spring, the curve in Figure 3.6(b) is a straight line and f = Kx, where K is a constant with units of Newtons per meter (N/m).Figure 3.6 Characteristic of a general spring2022-6-2520(a)(b)第19页/共42页第二十页，共42页。Spring k1212( )( )( )( ) ( )( )( )kftk x

19、 tx tk x tkv tv t dtkv t dt2022-6-2521(3.8)第20页/共42页第二十一页，共42页。2022-6-2522第21页/共42页第二十二页，共42页。3.4.1 DAlemberts Law达朗贝尔定理达朗贝尔定理(dngl)DAlemberts law is just a restatement of Newtons Second Law governing the rate of the change of momentum动量动量(dngling). For a constant mass, we can write:ext()iidvfMdt (3

20、.9)(fext)i: all the external forces acting on the body;Forces and velocity: vector quantities, can be treated as scalars (the motion is constrained to be in a fixed direction). Rewriting (3.9) as:ext()0iidvfMdt (3.10)2022-6-2523第22页/共42页第二十三页，共42页。ext()0iidvfMdt Suggests from Eq.(3.10): the mass can

21、 be considered to be in equilibrium. That is the sum of the forces = 0 (Mdv/dt as an additional force).Mdv/dt: the Fictitious force虚力虚力, the Inertial force内力内力(nil).Including Inertial force along with the external forces, rewrite (3.10) as :(3.10)0iif (3.11)2022-6-2524fi: all the forces acting on th

22、e body, including the Inertial force.第23页/共42页第二十四页，共42页。3.4.2 The Law of Reaction Forces相互作用力定律相互作用力定律(dngl)Newtons Third Law states( regarding reaction forces): Accompanying any force of one element on another, there is a reaction force on the first element of equal magnitude and opposite directio

23、n.fk : exerted by mass on the right end of spring, with positive sense defined to be to right.fk : acts on mass a reaction force of equal magnitude with its positive sense to left.fk : the fixed surface exerts a force on the spring with the positive sense to the left.fk : the spring exerts an force

24、with the equal magnitude and opposite sense on the surface.2022-6-2525Figure 3.8 Example of reaction forces.第24页/共42页第二十五页，共42页。2022-6-2526第25页/共42页第二十六页，共42页。2022-6-2527第26页/共42页第二十七页，共42页。Some notes:pThe system model must obey both the element laws(displacement x, velocity v, and acceleration a=dv

25、/dt) and the interconnection laws.pWrite all the element laws in terms of displacement x and its derivatives导导数数(do sh) (or in terms of x, v, dx/dt and dv/dt). pChoose the assumed positive directions for a, v, and x to be the same, so it will not be necessary to indicate all three positive direction

26、s on the diagram. pThroughout this PPT, dots over the variables are used to denote derivatives with respect to time. For example,22 and dxd yxydtdt2022-6-2528第27页/共42页第二十八页，共42页。3.5.1 Free-Body DiagramsEXAMPLE 3.1Draw the free-body diagram and apply DAlemberts law to write a differetial equation for

27、 the system shown in Fig. 3.9(a). The mass is assumed to move horizontally on frictionless bearings, and the spring and dashpot are linear. fa(t) is the applied force, also is the input variable. x(t) is the displacement of the mass M, also is the output variable.(v(t) is just an intermediate variab

28、le中间中间(zhngjin)变量变量 and should not exist in the final differential equation)Fig. 3.9 (a) Translational system 线性系统线性系统2022-6-2529第28页/共42页第二十九页，共42页。ANALYSISThe horizontal forces水平水平(shupng)力力, which are included in the free-body diagram, are:p fK, the force exerted by the springp fB, the force exer

29、ted by the dashpot.p fI, the inertial force / the fictitious force.p fa(t), the applied force.2022-6-2530Fig. 3.9 (a) Translational system.(b) Free-body diagram.(c) Free-body diagram including element laws.p fK = Kxp fB = Bvp fI =Mdv/dt p fa(t), be given第29页/共42页第三十页，共42页。Fig. 3.9 (a) Translational

30、system.(b) Free-body diagram.(c) Free-body diagram including element laws.DAlemberts law can now be applied to the free-body diagram in Fig. 3.9(c), pay attention to the assumed arrow directions. If forces acting to the right are regarded as positive, the law yields:( )()0af tMvBvKx Replacing v and

31、by and by , respectively分别分别(fnbi), and rearranging the terms, we can rewrite this equation as:( )aMxBxKxf t2022-6-2531v SOLUTION第30页/共42页第三十一页，共42页。EXAMPLE 3.2Draw the free-body diagrams for the two-mass system shown in Fig. 3.10(a) and use DAlemberts law to write the two differential equations tha

32、t describe mass 1 and mass 2, respectively. x1(t) and x2(t) are the displacements of the mass 1 and mass 2, respectively. fa (t) is the applied force. K1, K2 and B are coefficients for spring 1, spring 2 and dashpot, respectively.Fig. 3.10 (a) Translational system.2022-6-2532第31页/共42页第三十二页，共42页。SOLU

33、TIONBecause there are two masses that can move with different unknown velocities, a separate free-body diagram is drawn for each one:Fig. 3.10 (b) Free-body diagram for mass 1.Fig. 3.10 (c) Free-body diagram for mass 2.212211111()()0B xxKxxM xK x2221221( )()()0af tM xB xxKxx2022-6-2533第32页/共42页第三十三页

34、，共42页。EXAMPLE 3.3Fig. 3.11 (a) Translational system with vertical motion.SOLUTIONFig. 3.11 (b) Free-body diagram.Draw free-body diagram, including the gravity重力重力(zhngl), and find the differential equation describing the motion of the mass shown in Fig. 3.11(a). x is the displacement of the mass. fa

35、(t) is the applied force. K and B are coefficients for spring and dashpot, respectively.2022-6-2534Assume that x is the displacement from the position corresponding to a spring that is neither stretched nor compressed. The gravity on the mass is Mg. By summing the forces on the free-body diagram, we

36、 obtain:( )aMxBxKxf tMg第33页/共42页第三十四页，共42页。Two springs or dashpots are said to be in parallel if the first end of each is attached to the same body and if the remaining ends are also attached to a common body.3.5.2 Parallel Combinations并联并联(bnglin)EXAMPLE 3.4The system shown in Fig. 3.12(a) includes

37、 two linear springs between wall and mass M.Write the differential equation describing the motion of the mass. (Assume that the springs have the same unstretched length相同相同(xin tn)未拉伸长度未拉伸长度/自然长度自然长度)Find the spring constant Keq for a single spring that could replace K1 and K2. (Assume that the spri

38、ngs have the same unstretched length)Fig. 3.12(a) Translational system2022-6-2535第34页/共42页第三十五页，共42页。SOLUTION 1.If the unstretched lengths of the two springs are identical相等相等(xingdng)的的, then they will have the same elongation, denoted by x, when the mass is in motion. Summing the forces gives：Fig.

39、 3.12(b) Free-body diagrams when the springs have the same unstretched lengths.12()( )aMxBxKKxf tFig. 3.12(a) Translational system.2022-6-2536第35页/共42页第三十六页，共42页。Fig. 3.12(a) Translational system.Fig. 3.12(c) Free-body diagram when the combination of K1 and K2 is replaced by a single equivalent spri

40、ng.If the combination of K1 and K2 is replaced by a single equivalent spring, then the system reduces to that shown in Fig. 3.12(d):( )eqaMxBxK xf tComparing the above two equations reveals that：12( )()( )eqaaMxBxK xf tMxBxKKxf t 12eqKKKSOLUTION 2.2022-6-2537第36页/共42页第三十七页，共42页。SUMMARYTwo parallel s

41、prings or dashpots have their respective ends joined，as shown in Fig. 3.13. From the last example, we see that for the parallel combination of two springs,12eqKKKSimilarly, it can be shown that for two dashpots in parallel, as in part (b) of Fig. 3.13,12eqBBBFigure 3.13 Parallel combinations. (a) Keq = K1+K2 (b) Beq= B1+B22022-6-2538第37页/共42页第三十八页，共42页。3.5.3 Series Combinations串联串联(chunlin)Two springs or dashpots are said to be in series if they are joined at only one end of each element and if there is no other element connected to their common junction.EXAMPLE 3.