1、n this work. The method consists of designing the systems based on standard circuits made for each change on the state of the actuators, these changes are called steps. The first part is to design those kinds of standard circuits for each step, the next task is to link the standard circuits and the
2、last part is to connect the control elements that receive signals from sensors, switches and the previous movements, and give the air or electricity to the supply lines of each step. In Figs. 1 and 2 the standard circuits are drawn for pneumatic and electro-pneumatic system 8. It is possible to see
3、the relations with the previous and the next steps. 3. The method applied inside the controller The result of the method presented before is a sequence of movements of the actuator that is well defined by steps. It means that each change on the position of the actuators is a new state of the system
4、and the transition between states is called step. The standard circuit described before helps the designer to define the states of the systems and to define the condition to each change between the states. In the end of the design, the system is defined by a sequence that never chances and states th
5、at have the inputs and the outputs well defined. The inputs are the condition for the transition and the outputs are the result of the transition. All the configuration of those steps stays inside of the microcontroller and is executed the same way it was designed. The sequences of strings are progr
6、ammed inside the controller with 5 bytes; each string has the configuration of one step of the process. There are two bytes for the inputs, one byte for the outputs and two more for the other configurations and auxiliary functions of the step. After programming, this sequence of strings is saved ins
7、ide of a non-volatile memory of the microcontroller, so they can be read and executed. The controller task is not to work in the same way as a conventional PLC, but the purpose of it is to be an example of a versatile controller that is design for an specific area. A conventional PLC process the con
8、trol of the system using a cycle where it makes an image of the inputs, execute all the conditions defined by the configuration programmed inside, and then update the state of the outputs. This controller works in a different way, where it read the configuration of the step, wait the condition of in
9、puts to be satisfied, then update the state or the outputs and after that jump to the next step and start the process again. It can generate some limitations, as the fact that this controller cannot execute, inside the program, movements that must be repeated for some time, but this problem can be s
10、olved with some external logic components. Another limitation is that the controller cannot be applied on systems that have no sequence. These limitations are a characteristic of the system that must be analyzed for each application.4. Characteristics of the controller The controller is based on the
11、 MICROCHIP microcontroller PIC16F877 6,7 with 40 pins, and it has all the resources needed for this project .It has enough pins for all the components, serial communication implemented in circuit, EEPROM memory to save all the configuration of the system and the sequence of steps. For the execution
12、of the main program, it offers complete resources as timers and interruptions. The list of resources of the controller was created to explore all the capacity of the microcontroller to make it as complete as possible. During the step, the program chooses how to use the resources reading the configur
13、ation string of the step. This string has two bytes for digital inputs, one used as a mask and the other one used as a value expected. One byte is used to configure the outputs value. One bytes more is used for the internal timer , the analog input or time-out. The EEPROM memory inside is 256 bytes
14、length that is enough to save the string of the steps, with this characteristic it is possible to save between 48 steps （Table 1）.The controller (Fig.3) has also a display and some buttons that are used with an interactive menu to program the sequence of steps and other configurations.4.1. Interacti
15、on components For the real application the controller must have some elements to interact with the final user and to offer a complete monitoring of the system resources that are available to the designer while creating the logic control of the pneumatic system (Fig.3):Interactive mode of work; funct
16、ion available on the main program for didactic purposes, the user gives the signal to execute the step.LCD display, which shows the status of the system, values of inputs, outputs, timer and statistics of the sequence execution.Beep to give important alerts, stop, start and emergency. Leds to show p
17、ower on and others to show the state of inputs and outputs. 4.2. Security To make the final application works property, a correct configuration to execute the steps in the right way is needed, but more then that it must offer solutions in case of bad functioning or problems in the execution of the s
18、equence. The controller offers the possibility to configure two internal virtual circuits that work in parallel to the principal. These two circuits can be used as emergency or reset buttons and can return the system to a certain state at any time 2. There are two inputs that work with interruption
19、to get an immediate access to these functions. It is possible to configure the position, the buttons and the value of time-out of the system. 4.3. User interface The sequence of strings can be programmed using the interface elements of the controller. A Computer interface can also be used to generat
20、e the user program easily. With a good documentation the final user can use the interface to configure the strings of bytes that define the steps of the sequence. But it is possible to create a program with visual resources that works as a translator to the user, it changes his work to the values th
21、at the controller understands. To implement the communication between the computer interface and the controller a simple protocol with check sum and number of bytes is the minimum requirements to guarantee the integrity of the data. 4.4. Firmware The main loop works by reading the strings of the ste
22、ps from the EEPROM memory that has all the information about the steps. In each step, the status of the system is saved on the memory and it is shown on the display too. Depending of the user configuration, it can use the interruption to work with the emergency circuit or time-out to keep the system
23、 safety. In Fig.4,a block diagram of micro controller main program is presented.5. Example of electro-pneumatic system The system is not a representation of a specific machine, but it is made with some common movements and components found in a real one. The system is composed of four actuators. The
24、 actuators A, B and C are double acting and D-single acting. Actuator A advances and stays in specified position till the end of the cycle, it could work fixing an object to the next action for example (Fig. 5) , it is the first step. When A reaches the end position, actuator C starts his work toget
25、her with B, making as many cycles as possible during the advancing of B. It depends on how fast actuator B is advancing; the speed is regulated by a flowing control valve. It was the second step. B and C are examples of actuators working together, while B pushes an object slowly, C repeats its work
26、for some time.When B reaches the final position, C stops immediately its cycle and comes back to the initial position. The actuator D is a single acting one with spring return and works together with the back of C, it is the third step. D works making very fast forward and backward movement, just on
27、e time. Its backward movement is the fourth step. D could be a tool to make a hole on the object. When D reaches the initial position, A and B return too,. 二次函数精讲基础题型一认识二次函数1、y=mxm2+3m+2是二次函数，则m的值为（ ）A、0，-3B、0，3C、0D、-32、关于二次函数y=ax2+b，命题正确的是（ ）A、若a0,则y随x增大而增大B、x0时y随x增大而增大。C、若x0时，y随x增大而增大D、若a0则y有最大值。二
28、简单作图1在一个坐标系内做出，你发现了什么结论2同样的在同一个坐标系内做出，的图像，你又发现了什么结论，并且与上一题的图像比较的话，你又有什么样新的发现3 已知抛物线，五点法作图。2、已知y=ax2+bx+c中a0，c0 ，0,0 B.a0, 0 C.a0, 0 D.a0, 0,b0时,它的图象经过()A.一、二、三象限B.一、二、四象限 C一、三、四象限D.一、二、三、四象限12已知二次函数yax2bxc(a0)的图象如图所示，给出以下结论：Oa0.该函数的图象关于直线对称. 当时，函数y的值都等于0.其中正确结论的个数是（ ） A3 B2 C1 D0四，二次函数的性质：顶点，与X轴的焦点，
29、对称轴，最值问题1抛物线y=4x2-11x-3与y轴的交点坐标是_2抛物线y= -6x2-x+2与x轴的交点的坐标是_抛物线y=(x-1)2+2的对称轴是直线_顶点坐标为_3、 方程ax2+bx+c=0的两根为-3，1则抛物线y=ax2+bx+c的对称轴是直线_。4、 函数y=-x2+4x+1图象顶点坐标是（ ）A、（2，3）B、（-2，3）C、（2，1）D、（2，5）5、 抛物线的顶点坐标是（ ）A（2，3） B（2，3） C（2，3） D（2，3）6、 二次函数的图象的顶点坐标是（）ABCD7、 抛物线的顶点坐标为（A）（-2，7） （B）（-2，-25） （C）（2，7） （D）（2，-
30、9）8、 向上发射一枚炮弹，经x秒后的高度为y公尺，且时间与高度关系为y=ax2+bx。若此炮弹在第7秒与第14秒时的高度相等，则再下列哪一个时间的高度是最高的？ (A) 第8秒 (B) 第10秒 (C) 第12秒 (D) 第15秒 。9、 二次函数的最小值是（ ） A2 B1 C3 D 10、 已知二次函数 ， 为常数，当y达到最小值时，x的值为 （ ）（A） （B） （C） （D）11、 如图，直角坐标系中，两条抛物线有相同的对称轴，下列关系不正确的是（ ）ABCD12、 7当 x=4时，函数的最小值为8，抛物线过点（6，0）求：（1） 顶点坐标和对称轴；（2）函数的表达式；（3）x取什么
31、值时，y随x的增大而增大；x取什么值时，y随x增大而减五平移问题1、在平面直角坐标系中，将二次函数的图象向上平移2个单位，所得图象的解析式为A B C D2、将抛物线向下平移1个单位，得到的抛物线是（）ABCD3、将函数的图象向右平移a个单位，得到函数的图象，则a的值为A1B2C3 D4 4、把抛物线向左平移1个单位，然后向上平移3个单位，则平移后抛物线的解析式为ABCD5、把二次函数的图象向左平移2个单位，再向上平移1个单位，所得到的图象对应的二次函数关系式是（ ） （A） （B） （C） （D）六二次函数的应用1某涵洞是抛物线型，它的截面如图l上52，得水面宽AB=16m，涵洞顶点O到水面
32、的距离为 24m，在图中直角坐标系中，涵洞所在抛物线的函数关系式是_ 2是一个横断面为抛物线形状的拱桥，当水面在l时，拱顶（拱桥洞的最高点）离水面2m，水面宽4m。（2）建立平面直角坐标系，则抛物线的关系式是（）A B C DOxyABC3如图，某隧道口的横截面是抛物线形，已知路宽AB为6米，最高点离地面的距离OC为5米以最高点O为坐标原点，抛物线的对称轴为y轴，1米为数轴的单位长度，建立平面直角坐标系，求（1）以这一部分抛物线为图象的函数解析式，并写出x的取值范围；（2） 有一辆宽2.8米，高1米的农用货车（货物最高处与地面AB的距离）能否通过此隧道？4有一座抛物线形拱桥，在正常水位时水面A
33、 B的宽为20m，如果水位上升3米时，水面CD的宽为10m（1）建立如图1256所示直角坐标系，求此抛物线的解析式；（2）现有一辆载有救援物质的货车从甲地出发需经过此桥开往乙地，已知甲地距此桥 280km（桥长忽略不计）货车正以 40kmh的速度开往乙地，当行驶1小时，忽然接到通知；前方连降暴雨，造成水位以每小时025m的速度持续上涨（货车接到通知时水位在CD处，当水位到达最高点O时，禁止车辆通行）试问：如果货车按原来速度行驶，能否安全通过此桥？若能，请说明理由，若不能，要使货车安全通过此桥，速度应超过每小时多少千米？ 5已知如图 1253，ABC的面积为2400cm2，底边BC长为多80cm
34、，若点D在BC边上，E在AC边上，F在AB边上，且四边形BDEF为平行四边形，设BD=xcm，SBDEF=y cm2 求：（1）y与x的函数关系式； （2）自变量 x的取值范围； （3）当x取何值时，y有最大值？最大值是多少？6某商店将进货每个10元的商品，按每个18元售.机械设计课程设计姓名： 班级： 学号： 指导教师： 成 绩： 日期：2011 年 6 月 目 录1. 设计目的22. 设计方案33. 电机选择54. 装置运动动力参数计算75.带传动设计 96.齿轮设计187.轴类零件设计288.轴承的寿命计算319.键连接的校核3210.润滑及密封类型选择 3311.减速器附件设计 33
35、12.心得体会 3413.参考文献 351. 设计目的 机械设计课程是培养学生具有机械设计能力的技术基础课。课程设计则是机械设计课程的实践性教学环节，同时也是高等工科院校大多数专业学生第一次全面的设计能力训练，其目的是： （1）通过课程设计实践，树立正确的设计思想，增强创新意识，培养综合运用机械设计课程和其他先修课程的理论与实际知识去分析和解决机械设计问题的能力。 （2）学习机械设计的一般方法，掌握机械设计的一般规律。 （3）通过制定设计方案，合理选择传动机构和零件类型，正确计算零件工作能力，确定尺寸和掌握机械零件，以较全面的考虑制造工艺，使用和维护要求，之后进行结构设计，达到了解和掌握机械零
36、件，机械传动装置或简单机械的设计过程和方法。 （4）学习进行机械设计基础技能的训练，例如：计算，绘图，查阅设计资料和手册，运用标准和规范等。2. 设计方案及要求 据所给题目：设计一带式输送机的传动装置（两级展开式圆柱直齿轮减速器）方案图如下：1输送带2电动机3V带传动4减速器5联轴器 技术与条件说明：1）传动装置的使用寿命预定为 8年每年按350天计算， 每天16小时计算；2）工作情况：单向运输，载荷平稳，室内工作，有粉尘，环境温度不超过35度；3）电动机的电源为三相交流电，电压为380/220伏；4）运动要求：输送带运动速度误差不超过；滚筒传动效率0.96；5）检修周期：半年小修，两年中修，
37、四年大修。设计要求 1）减速器装配图1张； 2）零件图2张（低速级齿轮，低速级轴）； 3）设计计算说明书一份，按指导老师的要求书写 4）相关参数：F=8KN，V=0.6，D=400mm。3. 电机选择3.1 电动机类型的选择 按工作要求和工作条件选用Y系列鼠笼三相异步电动机。其结构为全封闭自扇冷式结构，电压为380V。3.2 选择电动机的容量工作机有效功率P=,根据任务书所给数据F=8KN，V=0.6。则有：P=4.8KW从电动机到工作机输送带之间的总效率为 =式中，分别为V带传动效率, 滚动轴承效率，齿轮传动效率，联轴器效率，卷筒效率。据机械设计手册知=0.96，=0.99，=0.97，=0
38、.99，=0.99，则有： =0.96 =0.85所以电动机所需的工作功率为： P=5.88KW 取P=6.0KW3.3 确定电动机的转速按推荐的两级同轴式圆柱斜齿轮减速器传动比I=840和带的传动比I=24，则系统的传动比范围应为：I=I=（840）（24）=16200工作机卷筒的转速为 n= 所以电动机转速的可选范围为 n=I=（16200）28.7 =（4595740）符合这一范围的同步转速有750r/min,1000r/min和1500r/min三种，由于本次课程设计要求的电机同步转速是1000r/min。查询机械设计手册（软件版）【常有电动机】-【三相异步电动机】-【三相异步电动机的
39、选型】-【Y系列（IP44）三相异步电动机技术条件】-【电动机的机座号与转速对应关系】确定电机的型号为Y160M-6.其满载转速为970r/min,额定功率为7.5KW。4. 装置运动动力参数计算4.1 传动装置总传动比和分配各级传动比1）传动装置总传动比 I=2）分配到各级传动比 因为I=已知带传动比的合理范围为24。故取V带的传动比则I分配减速器传动比，参考机械设计指导书图12分配齿轮传动比得高速级传动比，低速级传动比为4.2 传动装置的运动和动力参数计算电动机轴：转速：n=970输入功率：P=P=6.0KW输出转矩：T=9.55=9.55 =5.9N轴（高速轴）转速：n=输入功率：P=P输入转矩T=9.55轴（中间轴）转速：n=输入功率：P=P =5.5KW输入转矩：T=9.55 轴（低速轴）转速：n=输入功率：PP =5.28KW输入转矩：TN 卷筒轴：转速：n输入功率：P=P =5.28 =5.17KW输入转矩： N各轴运动和动力参数表4.1轴 号功率（KW）转矩（N）转速（）电机轴65.99701轴5.761.34402轴5.55.693.63轴5.281.7628.6卷同轴5.171.7328.6图4-15.带传动设计5.1 确定
