机械工程导论第四章105-120页翻译课件

1、EXAMPLE 4.2The control lever rotates when the 10-lb and 25 lb forces act on it as show in Figure 4.8(a). Determine the magnitude and direction of the resultant by using the vector polygon approach.SOLUTIONThe two forces are combined in a sketch using the head-to-tail rule. In Figure 4.8(b), the 25-l

2、b force is sketched first and the 10=1b force is added at the angle of 500 from vertical. The resultant R extends from the tail of the 25-1b force vector (which is labeled as the start point) to the head of the 10-lb force vector (labeled as the endpoint). The three vectors form a side-angle-side tr

3、iangle. Applying the law of cosines from Appendix B, we solve for the side length R throughfrom which R33.29 lb. The angle at which R acts is similarly determined by applying the law of sinesThe resultant acts at the angle 11.13 shown in Figure 4.8(b).控制杆转动的时候，10磅和25磅的力作用于它如图4.8所示（一）。利用矢量多边形的方法确定的合力

4、的大小和方向。解决方案这两个力结合在一个素描用头去尾规则。图4.8（b），25磅的力画第一和10 = 1b力在垂直500增加。所得到的R的25-1b力向量的尾部延伸（这是标示为出发点）到10磅的力向量的头（标记为端点）。三个向量形成一个边角边三角形。应用从附录B余弦定理，我们解决的边长r通过从R33.29磅。在R的行为是通过施加正弦定理同样确定角在角11.13如图4.8所示的所得的行为（B）。4.3 MOMENT OF A FORCEWhen you are trying to loosen; a frozen bolt, the bolt is more easily turned when

5、 a wrench with a long handle is used. The tendency of a force.(in this case, applied to the handles end) to make an object rotate is called a moment. The magnitude of a moment depends both on the force that is applied and on the lever arm that separates the force from a pivot point. Perpendicular Le

6、ver ArmThe magnitude of a moment is found from its definition (4.8)where Mo is the moment of the force about point O , F is the magnitude of the force, and d is the perpendicular lever arm distance from the forces line of action to point O. The term torque can also be used to describe the effect of

7、a force acting over a lever arm, but mechanical engineers generally reserve torque to describe moments that cause rotation of a shaft in a motor, engine, or gearbox. We will discuss those applications in Chapter 7.4.3力矩当你想放松；冷冻螺栓，螺栓更容易转向时，用长柄扳手的使用。力的方向。（在这种情况下，应用于手柄的一端）使物体旋转称为矩。一个时刻的大小取决于上的力被施加在杠杆将力

8、从一个支点。垂直的杠杆臂一个时刻的大小被发现从它的定义 (4.8)在莫力的一点O，F是力的大小，和D是垂直的杠杆臂距离行动力的线点。长期的扭矩也可以用来描述一个力作用在一个杆臂效应，但机械工程师一般储备扭矩描述的时刻，在电机，轴发动机产生旋转，或变速箱。我们将讨论这些应用程序在7章。Based on Equation 4.8, the unit for Mo is the product of force and, distance. In the USCS ， the unit for a moment is in-lb or ft-lb. In the SI, the unit N-m i

9、s used, and various prefixes are applied when the numerical value is either very large or very small. For instance; 5000N-m5kN-m and 0.002N-m2mN-m. Conversion factors between the two systems of units are shown in Table 4:2, where you can see that 1 ft-lb1.356 N-m. Work and energy, which are other qu

10、antities that arise in mechanical engineering, also have units that are the product of force and distance. For instance, when working in the SI, a joule (J) is defined as one newton-meter. It is the amount of work performed by a 1 newton force that moves through a distance of 1 meter. However, the p

11、hysical quantities of work and energy are quite different from moments and torques, and in order to be clear when distinguishing them, the unit of N-m (and not J) should be used in the SI for moment and torque. The expression MoFd can best be understood by applying it to a specific structure. In Fig

12、ure 4.9(a), the force F is directed generally downward and to the right on the bracket. One might be interested in the moment of F about the base of the support post, which is labeled in the figure as point O. The structure could break at that location, and an engineer would examine the post to make

13、 sure that it can support F. The moment is calculated based upon both the magnitude of F and the perpendicular offset distance d between the forces line of action and point O. In fact, F could be applied to the bracket at any point along its line of action, and the moment produced about O would rema

14、in unchanged because d would likewise not change. The direction of the moment is clockwise, because F tends to cause the post to rotate that way (even though the rigid mounting would prevent the post from actually moving in this case).基于方程4.8，Mo单位都是力量的产物，距离。在统一，一会儿单位在四磅或英尺-磅。，单位N-M使用，及各种前缀时应用数值是非常大或

15、非常小的。例如；5000n-m5kn-m和0.002n-m表2显示2mn-m.转换因子的单位之间的两个系统，在那里你可以看到1英尺磅1.356 N-m。工作和能源，这是其它数量在机械工程中出现的，也有是力和距离的乘积单位。例如，在SI工作时，焦耳（J）被定义为一个牛顿米。它是由一个1牛顿的力通过距离1米的工作量。然而，功和能的物理量的时刻和力矩不同，为了明确区分它们的时候，N-M单位（而不是J）应用于刻Si和扭矩。表达的莫FD可以应用于特定结构的理解。图4.9（a），力F大致向下和支架上的权利。一个可能感兴趣的一对支撑杆的基地，这是标记在图点。结构能在那个位置的休息，和一个工程师将审查后确保它

16、能支持这个时刻是基于F的大小和垂直偏移距离D动作点。事实上，力线之间的计算值，f可以应用于支架在任何点沿其作用线，产生O的时刻将保持不变，因为D同样会不会改变。目前的方向是顺时针，因为F往往会造成后旋转方式（尽管刚性安装防止后从在这种情况下，实际上运动）。In Figure 4.9(b) the direction of F has been changed. The forces line of action now passes directly through point O, and the offset distance becomes d0. No moment is produc

17、ed, and the force tends to pull the post directly out of its base. In short, the orientation of a force as well as its magnitude must be taken into account when calculating a moment.图4.9（b）F的方向已更改。行动力的线将直接通过点O，和偏移距离变成D0。不产生力矩，和力往往拉后直接从其基地。总之，力的方向以及它的大小必须考虑计算的时候。EXAMPLE 4.3The open-end wrench shown i

18、n Figure 4.10 tightens a hexagonal head nut and bolt. Calculate the moments produced by the 35-lb force about the center of the nut when the force is applied to the wrench in the two orientations shown. The overall length of the handle, which is inclined slightly upward, is in. long between centers

19、of the open and closed ends.SOLUTION(a) In Figure 4.10(a), the 35-lb force acts vertically downward. The perpendicular distance from the center of the nut to the forces line of action is d 6 in. The incline and length of the wrenchs handle are immaterial insofar as calculating d is concerned; the ha

20、ndles length is not necessarily the same as the perpendicular lever arm distance. The moment has magnitude(35 1b) (6 in.)210 in. lb17.5 ft.lbwhich is directed clockwise（CW）. (b) In Figure 4.10(b); the force has shifted to an inclined angle, and its line of action has changed so that d is measured to

21、 be. in., The moment is reduced to(35 lb) (5.375 in.)188 in.lb15.7ft.lbWe report the answer as Mo = 15.7 ft-lb (CW).实例4.3如图4.10所示的开口扳手紧固六角头螺栓和螺母。计算力矩的35磅力的螺母中心当外力施加到扳手在两个方向所示。手柄的长度，这是轻微的向上倾斜，在。长之间的开口端和封闭端中心。解决方案（一）在图4.10（a），35磅的力垂直向下作用。从螺母的中心作用的力线的垂直距离为d6。该扳手的手柄的倾斜和长度都无关紧要只要计算D有关；手柄的长度是不一定的垂直距离相同的杠杆

22、臂。目前有大小（35 1b）（6。210）。LB17.5 ft.lb这是导演的顺时针（CW）。（b）在图4.10（b）；力已经转移到一个倾斜角度，和它的作用线了，D测得。在那一刻，减少。（35磅）（5.375。188）LB15.7ft.lb。我们报告的回答为Mo = 15.7英尺磅（CW）。Moment ComponentsJust as we can break a force into rectangular components; it is sometimes useful to calculate a moment as the sum of its components. The

23、moment is determined as the sum of portions that are associated with the two components of the force, rather than the full resultant value of the force. A motivation for calculating, the moment in this manner is, that it is often easier to find the lever arms for individual components than for the r

24、esultant force. When applying this technique, it is necessary to use a sign convention and keep track of whether the contribution made by each force component is clockwise or counterclockwise: To illustrate this method, we calculate the moments about point O of the forces shown in Figure 4.11. We fi

25、rst choose the following sign convention: A moment that is directed clockwise is positive, and a counterclockwise moment is negative. The sign convention is just a bookkeeping tool for combining the various clockwise and counterclockwise moment components. Any contribution to the moment about O that

26、 acts clockwise is given a positive sign; and any contribution in the other direction is negative. This choice of positive and negative directions is arbitrary; and we could just as easily have selected the counterclockwise direction as being positive. However, once the sign convention is chosen, we

27、 stick with it and apply it consistently.力矩分量正如我们可以打破力为矩形组件；它计算一个时刻作为它的部分的和有时是有益的。现在是确定的部分，随着力的两个组成部分之和，而不是力的全合成值。据计算，这样的时刻，它往往比合力更容易找到个别组件的杠杆臂。应用这种技术时，有必要使用的符号和追踪是否各分力的贡献是顺时针或逆时针：为了说明这种方法，我们计算矩点如图4.11所示的力量。我们首先选择以下签署公约：那就是向顺时针和逆时针方向为正，负弯矩。在签署公约只是一种记帐相结合的不同的顺时针和逆时针的力矩分量的工具。任何贡献的时刻O行为给出一个积极的迹象，顺时针方向；

28、而在其它方向的任何贡献都是阴性。这种选择的正面和负面的方向是任意的；我们还可以轻松地选择逆时针方向为正。然而，一旦签署公约的选择，我们坚持，始终如一地应用它。In Figure 4.11(a). the force is broken down into the components Fx and Fy. Rather than determine the distance from point O to the line of action of F, which ought involve a geometrical construction that we want to avoid,

29、we instead calculate the individual lever arm distances for Fx and Fy, which are more straightforward. Keeping track of the sign convention, the moment about O becomes . Each contribution to Mo is positive because Fx and Fy each tend to cause clockwise rotation. Their effects combine constructively.

30、 The orientation of F has been changed in Figure 4.11(b). While the component Fx continues to exert a positive moment, Fy now tends to cause counterclockwise rotation about O. It therefore makes a negative contribution, and the net moment becomes .Here the two components combine in a deconstructive

31、manner. For the particular orientation in which, the two terms precisely cancel. The moment in that case is zero because the line of action for F passes directly through O, as in Figure 4.9(b). In the general case of the moment components method, we write (4.9)and the positive and negative signs are

32、 selected depending on whether the component tends to cause clockwise or counterclockwise rotation:图4.11（一）。力的分解成部件FX和FY。而不是确定从O点到行动F线的距离，应涉及的几何结构，我们希望避免的，我们不是计算个人的杠杆臂的距离为外汇和FY，这是更简单。保持轨道的符号约定，O瞬间变得。每个贡献莫正因为FX，FY每个往往导致顺时针旋转。结合建设性的影响。F的方向已经在图4.11（b）的变化。当组件FX继续发挥积极的时刻，比起现在往往会造成逆时针旋转大约O。因此负贡献，和净力矩而成。这里

33、的两个组件结合以解构的方式。在这特别的方向，这两个术语精确的抵消。在这种情况下的力矩为零因为F作用线直接穿过O，如图4.9（b）。在那一刻的部件的方法，一般情况下，我们写的4.9。和积极的和消极的迹象是根据组件是否会导致顺时针或逆时针转动选：Regardless of which method you use to calculate a moment, when reporting an answer you should state (1) the numerical magnitude of the moment, (2) the units, and (3) the direction

34、. You can indicate the direction by using notation provided that you have also shown the sign convention on your drawing. The notation CW or CCW is also useful for denoting whether the moment acts clockwise or counterclockwise.不管你使用计算力矩法，当报告答案你应该状态（1）时刻的数值的大小，（2）的单位，和（3）的方向。你可以用表示只要你也显示在你的绘图符号规则指明方向

35、。符号正转或反转的也有用是否顺时针或逆时针的力矩作用。EXAMPLE 4.4Determine the moment about the center of the nut as the 250-N force is applied to the handle of the adjustable wrench in Figure 4.12. Use (a) the perpendicular lever arm method, and (b) the moment components method.SOLUTION(a) As shown in Figure 4.13(a), we deno

36、te the center of the nut as point A, and the point of application of the force as point B. Using the dimensions given, distance AB is calculated as Although this is the distance from A to the location at which the force is applied, it is not the perpendicular lever arm distance d. For this reason, w

37、e need to calculate the length of segment AC. Because the applied force is tilted 35 from vertical, a line perpendicular to the force is oriented 35 from horizontal. As depicted in Figure 4.13(a); line AB lies at the angle tan-1(75/200)20.6 below horizontal, and so it is offset by from AC. The corre

38、ct lever arm distance therefore becomes .The wrenchs moment is . directed clockwise (CW).(b) In Figure 4.13(b), the 250-N force is broken into its components. The horizontal portion is (250N)sin35143N, and the vertical component is (250N)cos35205N. Those components are oriented leftward and downward

39、, respectively, and they each exert clockwise moment contributions about A. In Figure 4.13(b), we have shown our positive sign convention for moments as being clockwise. By summing the moment of each component, we haveBecause the net result is positive, the moment is directed clockwise.实例4.4确定力矩螺母的中

40、心为250-n施加力的可调扳手，图4.12中的处理。使用（一）垂直的杠杆臂的方法，和（b）一组分的方法。解决方案（一）如图4.13所示（一），我们表示螺母的中心点，和力为B点使用的尺寸应用的角度，距离AB计算这虽然是从一个位置处的力施加的距离，它不是垂直的杠杆臂的距离D。为此，我们需要计算的实际长度段因作用力是倾斜35垂直，垂直的力线是面向35水平。如图4.13（a）；AB线位于角tan-1（75 / 200）20.6水平以下，从而抵消从AC。正确的杠杆臂的距离因此成为扳手的时刻。向顺时针（CW）。（b）在图4.13（b），该250-n力分解成它的组成。水平部分（250N）sin35143N，

41、和垂直分量（250N）cos35205n。这些组件是面向向左、向下，分别，和他们每个人产生顺时针弯矩贡献大约在图4.13（b），我们已经表明我们积极的符号的规定时刻为顺时针方向。通过每个组件的时刻，我们有因为最终的结果是积极的，此刻正方向顺时针。4.4 EQUILIBRIUM OF FORCES AND MOMENTSWith groundwork for the properties of forces and moments now in place, we next turn to the task of calculating (unknown) forces that act on

42、structures and machines in response to other (known) forces that are present. This process involves applying the principle of static equilibrium to systems that are either stationary or moving at constant velocity. In either case, no acceleration is present, and in accordance with the laws of motion

43、, the resultant force on the system is zero.Particles and Rigid BodiesA mechanical system can include either la single object (for instance, an engine piston) or multiple objects that are connected together (the entire engine). When the physical dimensions of the object are unimportant in calculatin

44、g forces, the object is called a particle. This concept idealizes an object as being concentrated at a single point, rather than distributed over, an extended area or volume. For the purposes of problem solving, a particle can therefore be treated as having negligible dimensions. On the other hand,

45、if the length, width, and breadth of an object are important for the problem at hand, it is called a rigid body. When looking at the motion of the Space Shuttle as it orbits the Earth, for instance, the spacecraft is regarded as a particle because its dimensions are small compared to the size of the

46、 orbit. However, when the Shuttle is landing and engineers are interested in its aerodynamics and flight characteristics, the vehicle would instead be analyzed as a rigid body. Figure.4.14 illustrates the distinction between forces that are applied to a particle and to a rigid body; you can see how

47、a force imbalance could cause the rigid body to rotate.A particle is in equilibrium if the forces acting on it balance with zero resultant. Because forces combine as vectors, the resultant must be zero in two perpendicular directions, which we label x and y: and (4.10)For a rigid body to be in equil

48、ibrium, it is necessary that (1) the resultant of all forces is zero ， and (2) the net moment is also zero. When those conditions are met, there is no tendency for the object either to move (in response to the forces) or to rotate (in response to the moments). The requirements for equilibrium of a r

49、igid body involve Equation 4.10 and (4.11)The notation Moi is used, to denote the ith moment that is applied about point O. In summary, the two relationships in Equation 4.10 specify that no net force acts in4.4力平衡和力矩随着基础性质的力和力矩在现在的位置，我们接下来的计算任务（未知），在响应于其他结构和机械行为的力量（已知）的力量，是目前。这个过程包括应用静力平衡，以恒定的速度是固定

50、的或移动的系统原理。在任何情况下，没有加速度的存在，并根据运动规律，对系统的合力为零。质点和刚体一个机械系统可以包括La单一的对象（例如，发动机活塞）或多个对象连接在一起（整个发动机）。当物体的物理尺寸是不重要的计算力，对象是所谓的粒子。这一概念将一个对象作为被集中在一个点，而不是分布在一个扩展的面积或体积。为解决问题的目的，因此，一个粒子可以被视为具有可忽略的尺寸。另一方面，如果长度，宽度，和一个对象的宽度是重要的在手的问题，它被称为一个刚体。当看到航天飞机的运动，它绕地球运行，例如，飞船被视为颗粒由于其尺寸相对于轨道的大小是小的。然而，当飞机着陆和工程师在其空气动力学和飞行特性的兴趣，车辆

51、将不是被分析为一个刚体。figure.4.14说明应用到粒子和刚体的力之间的区别；你可以看到一个力不平衡可能导致刚体转动。A particle is in equilibrium if the forces acting on it balance with zero resultant. Because forces combine as vectors, the resultant must be zero in two perpendicular directions, which we label x and y: and (4.10)For a rigid body to be in

52、 equilibrium, it is necessary that (1) the resultant of all forces is zero ， and (2) the net moment is also zero. When those conditions are met, there is no tendency for the object either to move (in response to the forces) or to rotate (in response to the moments). The requirements for equilibrium

53、of a rigid body involve Equation 4.10 and (4.11)The notation Moi is used, to denote the ith moment that is applied about point O. In summary, the two relationships in Equation 4.10 specify that no net force acts in一个粒子在平衡如果力作用在它上面，零产生的平衡。因为力结合为载体，最终必须为零的两个垂直方向上，这是我们的标签x和y：4.10。对于刚体处于平衡状态，它是必要的，（1）所有

54、的合力为零，和（2）的净磁矩为零。当这些条件得到满足，没有对对象移动的趋势（反应部队）或转动（反应时间）。刚体的平衡要求，涉及方程4.104.11。我是用符号，表示第i的时刻，是关于点。总之，在方程4.10的两个关系指定没有净力作用either the x or y directions, and Equation 4.11 states that no net moment tends to cause rotation.Sign conventions are a bookkeeping method to distinguish forces acting in opposing dir

55、ections and moments that are oriented clockwise or counterclockwise. The summations in the equilibrium equation are conducted overall forces and moments present, whether or not their directions and magnitudes are known in advance. Forces that are unknown at the start of the problem are always includ

56、ed in the summation algebraic variables are assigned to them, and the equilibrium equations are then applied to determine numerical values. Mathematically speaking, the equilibrium equations for a rigid body comprise a system of three linear equations involving the unknown forces and moments. One im

57、plication of this characteristic is that it is possible to determine at most three unknown, quantities when Equations 4.10-4.11 are applied to a single rigid body. By contrast; when applying the equilibrium requirements to a particle, the moment equation is not used. Therefore, only two independent

58、equations are produced, and only two unknowns can be determined. It is not possible to obtain more independent equations of equilibrium by resolving moments about an alternative point or by summing forces in different directions. The additional equations will still be valid, but they will simply be combinations of the other (already derived) ones. As such, they will provide no new information.