Keywords

1 Introduction

Various planar mechanisms and manipulators are widely used in different practical applications [1,2,3,4,5,6,7,8,9]. Their synthesis and analysis are important tasks that allow improving different existing mechanisms, as well as creating new structures. It is well known that any mechanism is designed as a combination of one or several Assur groups with zero degree-of-freedom (DoF) and a driving link with mobility more than zero [10, 11]. A series connection of Assur groups into a kinematic chain of a mechanism does not change its overall mobility.

The simplest planar Assur group is a dyad, which includes two links with three one-DoF kinematic pairs. Totally five different types of dyads are known [12]. Various mechanisms can be created on their basis beginning with four-bar mechanisms [13, 14]. Four-bar Assur groups are the following by number of links after dyads. They are divided into III and IV class. Assur group of the III class is shown in Fig. 1a, it includes a three-paired link with three levers. Various planar mechanisms and manipulators are designed on the basis of this group [15,16,17,18]. Figure 1b shows a four-bar Assur group of the IV class, which includes a closed variable contour [19]. This group is used in designing of planar transport mechanisms, bolt insertion systems, mining machines and other technique.

Fig. 1
figure 1

Planar four-bar Assur groups: a of the III class; b of the IV class

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Links of both Assur groups shown in Fig. 1a, b are connected only by rotational kinematic pairs. However, rotational pairs can be replaced with prismatic pairs. In this case, the mobility of these groups is not changed, it remains equaled zero. At the same time, it becomes possible to design novel mechanical systems on the basis of these groups for performing various technological operations. In the presented study turn to a mechanism that includes a four-bar Assur group of the III class with rotational and prismatic kinematic pairs. Discuss the kinematic scheme of this mechanism and carry out its structural analysis in Sect. 2.

2 Structural Analysis

Figure 2a shows a six-bar quick-return mechanism that includes crank 1, slide block 2, swing lever 3, rocker arm 4, slider 5 and fixed link 6. The number of DoF of this mechanism can be calculated in accordance with the Chebyshev’s formula written as follows

Fig. 2
figure 2

a Kinematic scheme of quick-return mechanism; b Four-bar Assur group with three levers; c Dyads RRP; d Dyads RRR and RRP

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$$W_{3} = 3n - 2p_{5}$$
(1)

where W3—mobility, defining number of DoF of a planar kinematic chain with three imposed constraints, n—number of movable links of a mechanism, p5—numbers of one-DoF kinematic pairs [20, 21]. The quick-return mechanism shown in Fig. 2a includes n = 5 and p5 = 7. According to (1) its mobility equals one (W3 = 1). It means that to have predefined motions of slider 5, it needs to give an input motion to a single link. Such a motion is the rotational and applied to crank 1.

As crank 1 is the driving link in the mechanism, it is possible to separate out only one Assur group, which is a four-bar group of the III class with levers 2 (RP), 4 (RR) and 5 (PR) shown in Fig. 2b. If we chose another driving link, the class of Assur groups included in the mechanism changes. For example, with driving link 4 it is possible to separate out two dyads RRP (Fig. 2c), with driving link 5 we can separate out dyad RRR and RRP (Fig. 2d).

Structural analysis of the six-bar quick-return mechanism allows solving a further kinematic task. Its algorithm depends on the DoF of the investigated mechanism and included Assur groups.

3 Direct Kinematic Analysis

Turn to a kinematic analysis of the six-bar quick-return mechanism. We apply grapho-analytical method to determine directions and numerical values of velocities and accelerations of its links. Draw these diagrams for the position of the mechanism shown in Fig. 3a. Accept angular velocity of crank 1 is w1 = 60 min−1, the dimensions of the movable links are OA = 74 mm, BC = 104 mm, BD = 364 mm. Links 1 and 2 have identical velocity of point A (VA1 = VA2), which is calculated as VA = w1 · OA = 0.074 m/s. Draw vector pa = 74 mm at the velocity vector diagram shown in Fig. 3b. The scale of the velocity diagram is calculated as µV = VA/pa = 0.001 (m/s)/mm.

Fig. 3
figure 3

a Chosen position of the quick-return mechanism; b Velocity vector diagram; c Accelerator vector diagram

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As the mechanism includes the three-bar Assur group shown in Fig. 2b, it needs to apply the method of Assur points to solve kinematic task. It is possible to find three Assur points in the mechanism. The first point is S1, which lies at the intersection of the perpendiculars to links 3 and 6 drawn through points A and D (Fig. 2a). The second point is S2, which lies at the intersection of the perpendicular to link 6 drawn through point D and an extension of link 4. The third point is S3, which lies at the intersection of the perpendicular to link 3 drawn through point A and an extension of link 4. We will use point S1. Turn to its velocity calculation. Velocity of point S1 is determined from the following system of equations.

$$\left\{ {\begin{array}{*{20}l} {\overline{V}_{{S_{1} }} = \overline{V}_{A} + \overline{{V_{{A_{3} A}} + V_{{S_{1} A_{3} }} }} ,\;\;{\text{where}}\;\;\overline{{V_{{A_{3} A}} + V_{{S_{1} A_{3} }} }} \bot S_{1} A;} \hfill \\ {\overline{V}_{{S_{1} }} = \overline{V}_{{D_{6} }} + \overline{{V_{{D_{5} D_{6} }} + V_{{S_{1} D_{5} }} }} ,\;\;{\text{where}}\;\;V_{{D_{6} }} = 0,\text{ }\overline{{V_{{D_{5} D_{6} }} + V_{{S_{1} D_{5} }} }} \bot S_{1} D.} \hfill \\ \end{array} } \right.$$
(2)

Then we can define velocities of points B, A3 and D5 through equation systems (3–5).

$$\left\{ {\begin{array}{*{20}l} {\overline{V}_{B} = \overline{V}_{{S_{1} }} + \overline{V}_{{BS_{1} }} ,\;\;{\text{where}}\;\;\overline{V}_{{BS_{1} }} \bot BS_{1} ;} \hfill \\ {\overline{V}_{B} = \overline{V}_{C} + \overline{V}_{BC} ,\;\;where\;\;V_{C} = 0,\overline{V}_{BC} \bot BC.} \hfill \\ \end{array} } \right.$$
(3)
$$\left\{ \begin{aligned} \overline{V}_{{A_{3} }} = \overline{V}_{B} + \overline{V}_{{A_{3} B}} ,\;\;{\text{where}}\;\;\overline{V}_{{A_{3} B}} \bot AB; \hfill \\ \overline{V}_{{A_{3} }} = \overline{V}_{A} + \overline{V}_{{A_{3} A}} ,\;\;{\text{where}}\;\;\overline{V}_{{A_{3} A}} ||AB. \hfill \\ \end{aligned} \right.$$
(4)
$$\left\{ {\begin{array}{*{20}l} {\overline{V}_{{D_{5} }} = \overline{V}_{B} + \overline{V}_{{D_{5} B}} ,\;\;{\text{where}}\;\;\overline{V}_{{D_{5} B}} \bot BD;} \hfill \\ {\overline{V}_{{D_{5} }} = \overline{V}_{{D_{6} }} + \overline{V}_{{D_{5} D_{6} }} ,\;\;{\text{where}}\;\;V_{{D_{6} }} = 0,\overline{V}_{{D_{5} D_{6} }} ||D_{5} D_{6} .} \hfill \\ \end{array} } \right.$$
(5)

Velocities of points D3 and D5 equal to each other (VD3 = VD5), because links 3 and 5 connect to each other by the rotational kinematic pair. Numerical values of velocities VS1, VB, VA3 and VD5 are determined by multiplication of vector lengths ps1, pb, pa3 and pd5 from the velocity diagram by its scale µV. Finally velocities equal: VS1 = 0.075 m/s, VB = 0.097 m/s, VA3 = 0.112 m/s, VD5 = 0.149 m/s.

Angular velocities of slide block 2 and swing lever 3 equal to each other (w2 = w3), because links 2 and 3 connect to each other by prismatic kinematic pair, their numerical values are w2 = w3 = 0.285 s−1. Angular velocity of rocker arm 4 equals 0.929 (w4 = 0.929 s−1). Angular velocity of slider 5 equals zero (w5 = 0).

Turn to the calculation of accelerations. Accept that crank 1 rotates with constant angular velocity (w1 = const). Then acceleration of point A equals aA = w 21  · OA = 0.074 m/s2. Draw vector πa, which describes acceleration of point A at the accelerator diagram shown in Fig. 3c. The scale of the accelerator diagram is calculated as µa = aA/πa = 0.001 (m/s2)/mm. Acceleration of point S1 is calculated from the following equation system

$$\left\{ {\begin{array}{*{20}l} {\overline{a}_{{S_{1} }} = \overline{a}_{{A_{3} }} + \overline{a}_{{S_{1} A_{3} }}^{n} + \overline{a}_{{S_{1} A_{3} }}^{\tau } = \overline{a}_{A} + \overline{{a_{{A_{3} A}}^{k} + a_{{S_{1} A_{3} }}^{n} }} + \overline{{a_{{A_{3} A}}^{r} + a_{{S_{1} A_{3} }}^{\tau } }} ;} \hfill \\ {\overline{a}_{{S_{1} }} = \overline{a}_{{D_{5} }} + \overline{a}_{{S_{1} D_{5} }}^{n} + \overline{a}_{{S_{1} D_{5} }}^{\tau } = \overline{a}_{{D_{6} }} + \overline{a}_{{S_{1} D_{5} }}^{n} + \overline{{a_{{D_{5} D_{6} }}^{r} + a_{{S_{1} D_{5} }}^{\tau } }} .} \hfill \\ \end{array} } \right.$$
(6)

Accelerations of points B, A3 and D5 can be found from equation systems (7–9).

$$\left\{ {\begin{array}{*{20}l} {\overline{a}_{B} = \overline{a}_{{S_{1} }} + \overline{a}_{{BS_{1} }}^{n} + \overline{a}_{{BS_{1} }}^{\tau } ,\;\;{\text{where}}\;\;\overline{a}_{{BS_{1} }}^{n} ||BS_{1} ,\overline{a}_{{BS_{1} }}^{\tau } \bot BS_{1} ;} \hfill \\ {\overline{a}_{B} = \overline{a}_{C} + \overline{a}_{BC}^{n} + \overline{a}_{BC}^{\tau } ,\;\;{\text{where}}\;\;a_{C} = 0,\overline{a}_{BC}^{n} ||BC, \overline{a}_{BC}^{\tau } \bot BC.} \hfill \\ \end{array} } \right.$$
(7)
$$\left\{ \begin{aligned} \overline{a}_{{A_{3} }} = \overline{a}_{B} + \overline{a}_{{A_{3} B}}^{n} + \overline{a}_{{A_{3} B}}^{\tau } ,\;\;{\text{where}}\;\;\overline{a}_{{A_{3} B}}^{n} ||AB,\overline{a}_{{A_{3} B}}^{\tau } \bot AB; \hfill \\ \overline{a}_{{A_{3} }} = \overline{a}_{A} + \overline{a}_{{A_{3} A}}^{k} + \overline{a}_{{A_{3} A}}^{r} ,\;\;{\text{where}}\;\;\text{ }\overline{a}_{{A_{3} A}}^{k} \bot AB,\overline{a}_{{A_{3} B}}^{r} ||AB. \hfill \\ \end{aligned} \right.$$
(8)
$$\left\{ {\begin{array}{*{20}l} {\overline{a}_{{D_{5} }} = \overline{a}_{B} + \overline{a}_{{D_{5} B}}^{n} + \overline{a}_{{D_{5} B}}^{\tau } ,\;\;{\text{where}}\;\;\overline{a}_{{D_{5} B}}^{n} ||DB, \overline{a}_{{D_{5} B}}^{\tau } \bot DB;} \hfill \\ {\overline{a}_{{D_{5} }} = \overline{a}_{{D_{6} }} + \overline{a}_{{D_{5} D_{6} }}^{k} + \overline{a}_{{D_{5} D_{6} }}^{r} ,\;\;{\text{where}}\;\;a_{{D_{6} }} = 0,a_{{D_{5} D_{6} }}^{k} = 0,\overline{a}_{{D_{5} D_{6} }}^{r} ||D_{5} D_{6} .} \hfill \\ \end{array} } \right.$$
(9)

Accelerations of points D3 and D5 equal to each other (aD3 = aD5), because links 3 and 5 connect to each other by the rotational kinematic pair. Figure 3c provides accelerator vector diagram from which numerical values of accelerations aS1, aB, aA3 and aD5 are determined by multiplication of the vector lengths πs1, πb, πa3 and πd5 by diagram scale µa. Finally accelerations equal: aS1 = 0.052 m/s2, aB = 0.093 m/s2, aA3 = 0.036 m/s2, aD5 = 0.039 m/s2.

Angular accelerations of slide block 2 and swing lever 3 equal to each other (ε2 = ε3), because links 2 and 3 connect to each other by the prismatic kinematic pair, their numerical values are ε2 = ε3 = 0.332 s−2. Angular acceleration of rocker arm 4 equals 0.878 (ε4 = 0.878 s−2). Angular acceleration of slider 5 equals zero (ε5 = 0).

Note that points b, a3 and d5 (d3) are placed on one line at the velocity and acceleration diagrams. Such a placement is proved by Fig. 3a where points B, A and D are placed on one line (line BD). When driving link is 4 or 5, the kinematic analysis of quick-return mechanism is simplified as it is not necessary to find Assur points. In these cases the linkage includes only pair of dyads shown in Fig. 2c, d.

4 Simulation Development

Numerical analysis of the quick-return mechanism has been carried out for the position measurements of distinguished points of movable links and tracing their trajectories. Figure 4 presents 3D model of the quick-return mechanism with red-colored trajectories of points A3, B and D3(D5). These trajectories can be varied by using data from the kinematic analysis and changing of link lengths of the mechanism.

Fig. 4
figure 4

Three-dimensional model of the quick-return mechanism with motion trajectories of its distinguished points (color figure online)

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Table 1 provides coordinate variation of distinguished points A1(A2), A3, B and D3(D5) of the quick-return mechanism depending on the tilt angle of crank 1. The initial position of crank 1 is when it lies in the negative part of axis X, so the Y-coordinate equals zero. Next positions are set after each 60°. The data from Table 1 allows solving inverse kinematic problem, i.e. define parameters of the quick-return mechanism by desired law motion.

Table 1 Coordinate variation of the distinguished points of links depending on a tilt angle of crank 1
Full size table

5 Conclusions

The presented study provides results of structural and kinematic analysis of the planar quick-return mechanism based on the four-bar Assur group of the III class. The structural analysis of the mechanism has allowed defining various Assur groups when different driving links were accepted. The kinematics analysis has been solved by grapho-analytical method for the case when a four-bar Assur group was included in the mechanism. CAD-supported 3D modeling has allowed detecting motion trajectories and numerical values of coordinates of all movable links. The data obtained from the structural analysis can be applied for synthesis of new planing mechanisms, quick-return systems and other types of planar linkages. Results of kinematic calculations can be used in kinetostatic and dynamic analysis of the quick-return mechanism.