Chapter 3 Kinematics and forces analysis of planar mechanisms (第3章 平面机构的运动分析和力分析) w14 w21 w23 v34 j M1 FR21 FR12 FR32 FR23 FR43 FR41 h 1 2 3 4 F

Tasks and methods of kinematic analysis of mechanisms(机构运动分析的目的和方法) Positions(轨迹)/Angular positions(角位移） Velocities(速度)/Angular velocities(角速度） Accelerations(加速度)/Angular Accelerations(角加速度）. Problems： It is a base for force analysis to understand the kinematics properties of existing mechanisms(了解现有机构的运动性能，为受力分析奠定基础). Purpose： Means： 1)Instant center (瞬心法)（v and ω）; 2)Vector equation Graphical Method(矢量方程图解法) 3)Analysis method(解析法)（上机计算）.

3.1 Instant center of velocity (速度瞬心) 3.1.1 Instant center of velocity(速度瞬心) The definition of an instant center of velocity is a coincident point, common to two links/bodies in plane motion, which point has the same absolute velocity and no relative veloctiy( 两个互作平行平面运动的构件上绝对速度相等、相对速度为零的瞬时重合点称为这两个构件的速度瞬心, 简称瞬心。瞬心用符号Pij表示).

2 绝对速度为零的 A 1 瞬时重合点。 vA1A2 B vB1B2 P12 绝对速度不为零 的瞬时重合点。 e.g.P12 is the instant center of velocity between link 1 and link 2. 1.Absolute instant center (绝对瞬心) P12 1 2 A B vA1A2 vB1B2 绝对速度为零的 瞬时重合点。 2.Relative instant center (相对瞬心) 绝对速度不为零 的瞬时重合点。

3.1.2 Number of instant centers of a mechanism (机构中瞬心的数目) Denoted as K： Note: The frame is included in the number N . 注意: N为包括机架在内的所有构件数。

3.1.3 Location of the instant center of two links connected by a kinematic pair(机构中瞬心位置的确定) 1.Observable instant center(直接构成运动副的两构件的瞬心) (1)Revolute pair(转动副连接) The center of the revolute pair is the instant center(铰链点即为瞬心). 1 2 P12

(2)Sliding pair(移动副连接) Infinity in either direction perpendicular to the guide-way(瞬心在垂直于导路无穷远处) v12 1 2

(3)Planar higher pairs(平面高副)(rolling and sliding pair) n t A 2 1 v12 Somewhere on the common normal n-n through the connecting point A(瞬心在接触点的公法线上；若为纯滚动，瞬心在接触点上) If v12=0 ， P12 is A point； If v12≠0 ,P12 lies on the common normal n-n.

2.Theorem of three centers Aronhold-Kennedy Theorem(三心定理) (unobervable instant centers) Theorem： Three indepentdent links in general plane motion have three instant centers and the three instant centers must lie on a straight line.

proof： vK2≠ vK3 Suppose P23 is at K point，as vK2 vK3 1 vK2 vK3 w2 w3 vK2≠ vK3 To obtain the same direction of two velocity, K point must lie on the line through P12 and P13 points. So，P12, P23 and P13 must lie on the same line.

Example1：For the four-bar mechanism shown as follow, locate all instant centers for the mechanism. Solution： P12、P23、P34、P14 (observable instant centers) P24 1 2 3 4 { P12、P23 P13 P14 P34 P12 P23 P14、P34 { P12、P14 P24 P23、P34 P13

Linear graph/Instant center polygon 框图法(瞬心多边形) The solution of each link instant centers represent by a polygon graph. A line between the dots representing the link pairs each time we find an instant center. The triangle composed by lines represents three instant centers and three centers are collinear. 1 2 3 4 P12 P24 P13 P14 P34 P23 Example：

Example 2：Finding all instant centers for a slider-crank mechanism。 Solution： 1 2 3 4 P12 P24 P13 P14 P34 P23 1 4 3 2

3.1.4 Velocity analysis with instant centers Example 1:w1 is given in the slider-crank mechanism, find the ratio of transmission i13 and w3. P12 P34 P14 P23 ∞ 1 2 3 4 w1 P13 ω3 Solution： 1 2 3 4 P13

Example2: w1 is given in the cam mechanism, find the velocity v2 of the follower 2. 1 2 3 w1 Solution: n P13 P23 ∞ P12 P23 ∞

Exercise: for the five-bar linkage ,suppose that the driving links 2 and 3 are rotating with constant angular velocity w2 =10rad/s (ccw) and w3 =5rad/s(cw), respectively. Determine the couplers w4 and w5 . 5 4 3 2 1

3.3 Forces analysis of mechanisms (机构的受力分析) 3.3.1 Goals and contents of forces analysis considering influences of friction Goals： Analyze the effects of friction in kinematic pairs in mechanisms and reduce the disadvantages. 1）Introduction of friction in several kinematic pairs； 2） Force analysis considering influences of friction ； 3）Mechanical efficiency and self-locking. Contents：

F21 3.3.2 Force analysis considering influences of friction 1.Friction in kinematic pairs （1）Friction in sliding pairs 1）determination of friction force in sliding pairs 1 2 Ft FN21 FR21 F Fn b F21 j v12 a FR21—composite force(合外力) FN21—normal reaction force(法向反力)； F21—friction(摩擦力)。 direction：direction of F21 is opposite to that of v12. magnitude  ： F21 is proportional to FN21 with friction coefficient f being a certain value. F21=fFN21 F21

2）Determination of reaction force in kinematic pairs FN21 and F21 are the forces link 2 acting on link 1 and the reaction force FR21 is the composition of them. 1 2 Ft FN21 FR21 F Fn b F21 j v12 a suppose included angle between FR21 and FN21is j ∴j =arctanf ，friction angle。 direction：angle between FR21and v12 is (90°+j ) FR21 magnitude   ：

Under different situations, reaction force should be as follows: j FR21 v12 2 F 1 a b d c (a) 当外力F 的作用线位于接触表面ab 之内时 构件1与构件2仅一面受力，FR21 如图a 所示。 2 1 j c d b a F FR21 v12 (b) 当外力F的作用线位于接触表面ab 之外时构件1除了移动之外，还要发生倾 转， FR21如图b 所示。 j 2 1 c d b a h v12 FR21 F (c) 当外力F的作用线平行移动轴线并 距移动轴线h 时，构件1除了移动之外， 还要发生倾转， FR21如图c 所示。

FQ + F + FR21 = 0 F = FQ tan(a +j ) FR21 = FQ /cos(a +j ) Example: Link 1 moves at a constant speed on a slant of link 2，loading is FQ , friction coefficient is f , driving force is F (horizontal )and a is given。Determine FR21 =? F=? when link 1 move along a slant up and down at a constant speed. FQ 1 2 a F a j 1 (a+j) FQ F v12 FR21 Solution：1) Force analysis on link 1 when it move up at a constant speed. FQ + F + FR21 = 0 F = FQ tan(a +j ) FR21 = FQ /cos(a +j )

1 a (a-j) FQ F v12 j FR21 2) Force analysis on link 1 when it move down at a constant speed. FQ + F + FR21 = 0 F = FQ tan(a -j ) FR21 = FQ /cos(a -j ) Conclusion：the different direction of FR21 is due to the different direction of relative velocity.

2.Friction in revolute pairs （radial shaft friction径向轴颈摩擦） 1 2 FQ FN21 (a) A 1)Fig.a，当构件 1与构件2 没有相对转动时，其接触点在A 点，此时FN21= -FQ ，此二力共 线，等值反向。 2)Fig.b，在Md的驱动下使轴颈 匀速转动，由于摩擦力的存在，构件 1在构件2的AB弧段向上爬升，直至B 点达到平衡。 r FQ FN21 F21 FR21 Md O B 1 2 (b) 平衡条件为 ∑Fy=0 FQ=FR21 ∑MO=0 Md=Mr

Mr=F21r =FR21r fv(当量摩擦系数)通过理论推导，有 跑合轴颈：fv=1.27f 非跑合轴颈：fv=1.57f

1)当e<r 时，FQ′作用在摩擦圆内, 此时Md<Mr,机构若原来运动则减 速直至静止,若原来静止则自锁。 r = fv r 为定值。 r FQ FR21 Md O 式中r 称为摩擦圆半径。 r FQ FR21 r FQ FR21 根据力学观点，若Md与FQ的总合 力为FQ′，其距离为e， 1)当e<r 时，FQ′作用在摩擦圆内, 此时Md<Mr,机构若原来运动则减 速直至静止,若原来静止则自锁。 r FR21 O B e FQ′ r FR21 O B e FQ′ r FQ FR21 Md O B r FR21 O B e FQ′ 2)当e= r 时，FQ′切于摩擦圆，此 时Md=Mr,构件1达到惯性平衡。 3)当e>r 时，FQ′作用在摩擦圆外， 此时Md>Mr,构件1作加速转动。 在考虑摩擦受力分析时，常将FQ′ 用FRij表示。

转动副中总反力的确定： 1)FRAB作用线切于摩擦圆； 2)FRAB对轴心取矩的方向与wBA转 向相反； FRAB 唯一确切位置。 整体平衡条件包括：若该构件为二力构件，明 确其受拉还是受压；若该构件是三力构件，此 三力必汇交于一点；若该构件受一个力矩和两 个力作用，此时该二力必构成一力偶与力矩达 到平衡；等等。

2.考虑摩擦的机构受力分析 步骤如下： 1)明确机构中驱动力（矩）和阻力（矩）， 搞清运动趋势。 2)画受力图。必须从受力最少构件入手，然 后分别画出其他构件受力图。 3)求未知量。从有已知力（矩）的构件入手， 根据平衡条件求出未知力（矩），然后以 此构件为基准，由近及远分析其它构件， 直至求出未知量。

Example1: As shown in figure, four-bar linkage, positions, sizes, fv ,r, driving force F and resistant torque M3 of the mechanism are given, find the position and direction of reaction forces in each kinematic pair by the effect of F. Link 2: Solution： r = fv r FR12 + FR32 = 0 w 21 w 23 FR12 FR21 FR23 FR32 1 2 3 4 F M3 Link 1: FR43 h F + FR21 + FR41 = 0 FR41 Link 3: w 14 w 34 FR23 + FR43 = 0 M3 = FR23 h

FR12 + FR32 = 0 FQ + FR23 + FR43 = 0 FR21 + FR41 = 0 M1 = FR21 h Example2: As shown in figure, guide-bar mechanism, positions, sizes, fv ,r, driving torque M1 of the mechanism are given. FQ is resistance, find the resistance FQ =? M1 FQ 3 4 1 2 FR43 FR23 FR21 Solution： r = fv r，j = arctan f w14 FR12 + FR32 = 0 Link 2： FR41 h Link 3： FQ + FR23 + FR43 = 0 3 2 FR12 FR32 w21 v23 Link 1： FR21 + FR41 = 0 M1 = FR21 h 受力面 w34 Choose mN，FR21=M1/h ∴FQ=∣FQ∣mN FR43 FQ FR23 FR41 FR21 FR12 FR32

v21 F FQ w23 Link 1： F + FR21 + FR31 = 0 Link 2： FQ + FR12 + FR32 = 0 Example3:Positions, sizes, fv ,r, driving force F and resistance FQ are given. Find the reaction line of total reaction force. 1 2 3 F FQ Solution： Link 1： F + FR21 + FR31 = 0 Link 2： FQ + FR12 + FR32 = 0 FR32 v21 FR21 FR12 j FR31 j w23 v13

Example4: Positions,sizes, friction angle j between the sliding pair and friction circles are given in the following clamping mechanism， Please draw all reaction force lines of kinematic pair under the driving force F and the resistance force FQ. F Solution ： FR41 Link2： FR12 + FR32 = 0 1 j w14 FR12 FR32 w21 FR21 Link 1： F + FR21 + FR41 = 0 2 j FR43 Link 3： FQ + FR23 + FR43 = 0 3 w23 v34 4 FR23 FQ

3.Mechanical efficiency and self-locking(机械效率和自锁) (1)机械效率 机械对能量的有效利用程度，用h 表示。 式中：Wd ——输入功； Wr —— 输出功； Wf —— 损耗功。 因为Wf总是存在的，所以h＜1。 希望Wf↓、h↑的措施有： 1)简化机械系统，减少运动副。 2)减少摩擦，合理选材。 h的表达形式： (a)h用功率表示

(b)h用力的比值表示 i.有摩擦时 F为驱动力, FQ为阻力 传动系统 F v FQ vQ ii.无摩擦时 F0为理想驱动力,FQ为阻力

(2)机械自锁 自锁： 运动副的自锁条件： (c)h用力矩的比值表示 无论驱动力多大，在该驱动力单独作用下 都不能使机械运动，这种现象称为自锁。 运动副的自锁条件： r FQ (FR21) FR21 e O 1 2 b j v12 F FR21 1 2 b≤j自锁 e≤r自锁 当取“=”时，机构需原来静止。

机械的自锁条件：h≤0 机械的自锁用效率确定： 当Wf＞＞Wd时，h≤0。 h=0时,说明机械处于自锁的临界状态；

为驱动力，M1 为阻力矩，摩擦系数f 、当量摩 擦系数fv、轴颈r均已知，求机构的自锁范围。 h FR41 FR21 w21 w23 F M1 1 2 3 4 FR12 FR32 w14 解： r = fv r，j = arctan f 构件2： FR12 + FR32 = 0 构件1： FR21 + FR41 = 0 M1 = FR21 h

F M1 1 2 3 4 w21 w23 FR12 FR32 w14 h FR41 FR21 FR21 F M1 1 2 3 4 w21 w23 FR12 FR32 w14 FR41 1) FR21 F M1 1 2 3 4 w21 w23 FR12 FR32 w14 FR41 2) FR21 F M1 1 2 3 4 w21 w23 FR12 FR32 w14 FR41 3)

FR21 F M1 1 2 3 4 w21 w23 FR12 FR32 w14 FR41 q 所以自锁范围为