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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
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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).
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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).
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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。
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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。
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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
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(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
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(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.
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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.
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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.
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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
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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:
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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
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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
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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 ∞
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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
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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:
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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
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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 :
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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。
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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 )
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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.
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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= FQ=FR21 ∑MO= Md=Mr
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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
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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。
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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。
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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。
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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
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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
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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
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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
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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
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(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
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(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。
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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;
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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
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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)
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FR21 F M1 1 2 3 4 w21 w23 FR12 FR32 w14 FR41 q 所以自锁范围为
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