NOT COMPLETE. θ 4 B 2 = O 2 O 4 = A 2 = A 1 B 1 O 2 KINEMATIC SYNTHESIS

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ME 35 NOT COMPLETE Design Design a crank-rocker four-bar (Grashof) where the input link rotates completely and the output link (the follower) rocks back and forth with a prescribed angle The design

1 ME 35 NOT COMPLETE Design Design a crank-rocker four-bar (Grashof) where the input link rotates completely and the output link (the follower) rocks back and forth with a prescribed angle The design requires equal time forward and back for the rocker assuming a constant speed motor turning the crank s shown in the diagram, link is the crank and link 4 is the rocker with a prescribed angle Since there are no other requirements, there are infinite numbers of designs that would fulfill the stated requirements The design process is summarized in the following steps O Select a length L 4 for the rocker (follower) Place the pin joint at a convenient location on the ground Construct the rocker at its limit positions, based on the given angle This process gives us the position of the pin joint at its two limit positions, and L 4 L 4 Draw an axis through L 3 Place the pin joint O at a convenient location on this axis (infinite choices) This established the ground link and the length L = O L 4 Measure the distance This is twice the length of the crank, L 5 Draw a circle with its center at O and the radius L O 6 Position the pin joint at two positions, and, at the intersections of the axis and the circle This establishes the length of the coupler L 3 = = 7 Test for Grashof s condition If non-grashof, repeat the process from step 3 by locating O at a different location on the axis O PE Nikravesh

2 ME 35 This is the designed four-bar O Example Design an equal time forward and back four-bar mechanism to operate the windshield wiper of the rear window of an automobile The dimensions are given in the figure The range of motion of the rocker (blade) is 05 o The designed four-bar must fit within a compartment shown as a shaded area 6 blade Time Ratio In Design- it is required that the rocker to have equal time forward and back assuming a constant speed motor as the input We achieved this requirement by assuring the crank to rotate 80 o when the rocker moved forward, and 80 o when the rocker moved backward For such mechanism, the time ratio is one-to-one; ie, T R = : In another design, it may be required for the output link to move slow in one direction and fast in the other direction; for example, the time ratio could be T R = :5 (slow:fast) This is normally called a quick-return mechanism To design such a mechanism, we must assure that the crank circle is split into two portions, α and β, such that α / β = T R Since α + β = 360 o, we can determine the two angles: α = T α R 360 o, β = 360 o (a) + T R + T R The difference between any of these two angles from 80 o is O denoted as δ : δ = 80 o α = β 80 o (b) β Design This is similar to Design, but here the time ratio is not one-to-one; eg, T R = : PE Nikravesh

3 ME 35 (Same as step in Design ) Select a length L 4 for the rocker (follower) Place the pin joint at a convenient location Construct the rocker at its limit positions, knowing the angle This process gives us the position of the pin joint at its two limit positions, and δ Compute α, β, and δ 3 Draw two axes from and such that the angle O between them is δ For this purpose, you may draw an axis through first (infinite choices), and draw a line making an angle δ with the first axis Then from draw an axis parallel to the line The intersection δ of the two axes is O This establishes the ground link and the length L = O 4 With its center at O, draw a circle arc from until it intersects the O axis at Measure the distance between and This is twice the length of the crank, L L 5 Draw a circle with radius L, centered at O 6 Position the pin joint at two positions, and, at the intersections of the two axes and the circle This establishes the length of the coupler L 3 = = 7 Test for Grashof s condition If non-grashof, repeat the process from step 3 by drawing the axis through in a different orientation Note that: O = L 3 + L, O = L 3 L Therefore, L O δ L = O O, L 3 = O + O δ This is the designed four-bar O O Note: This method is adequate for time ratios up to :5 For larger time ratios we need to design a six-bar or a more complex mechanism PE Nikravesh 3

4 ME 35 Example Design 3 This problem requires designing a crank-rocker four-bar, where an extension to the rocker link should find two limiting positions, C D and C D, as shown This design can easily be turned into Design or Design depending on the required time-ratio D C D C Draw the perpendicular bi-sector to C C Draw the perpendicular bi-sector to D D 3 The intersection of the two perpendicular bi-sectors is the location of the pin joint D C D 4 Construct the rocker link at its two limits O C D and C O C D 5 Select a proper value for L 4 and place the pin joint at its two limits and 6 Continue with either Design- or Design- procedure D D C C L 4 L 4 Example 3 Design 4 In some four-bar mechanisms, the output link could be the coupler, not the follower In this design problem two positions of the coupler links are provided These are not necessarily the limiting positions We assume that and are two positions of the pin joint, and and are two positions of the pin joint ; ie, L 3 is known PE Nikravesh 4

5 ME 35 Draw the perpendicular bi-sector to Place the pin joint O at a convenient location on this line (infinite choices) This establishes the length L 3 Draw the perpendicular bi-sector to 4 Place the pin joint at a convenient location on this line (infinite choices) This establishes the length L 4 and L If it is required for the four-bar to be Grashof, we must check for that If the designed mechanism is not Grashof, we repeat steps and 4 by placing O and at different locations on their corresponding bi-sectors We repeat this process until we find a Grashof four-bar If (a) the problem statement did not require a Grashof four-bar, or (b) we cannot find a Grashof four-bar, or (c) and should be the limiting positions of the coupler, we should do the followings: () we must assure that the designed four-bar can move from position to position and back continuously (without a need to disassemble/re-assemble the mechanism); and () design a dyad to drive this four-bar (dyads are discussed later in this chapter) Example 4 O O Design 5 variation of Design 4 could be that three orientations of the coupler link are provided Draw the perpendicular bi-sectors to and 3 The intersection of these two bi-sectors is the pin joint O This establishes the length L 3 Draw the perpendicular bi-sector to and 3 The intersection of these two bi-sectors is the pin joint This establishes the lengths L 4 and L 3 We observe that the design is unique Most likely, the design is non-grashof s long as the designed four-bar can move between the three configurations continuously, we can add a dyad to drive this four-bar 3 3 O PE Nikravesh 5

6 ME 35 Example 5 O Design 6 nother variation of Design-4 or Design- 5 is that two or three positions of an extension of the coupler link are provided We are given the freedom to shape the coupler link as desired ssume the extension is given at three EF positions as shown E F E F E 3 We can decide on the shape of the coupler link and construct it at one of the positions This step establishes the location of the pin joints and at that configuration, and the length of the coupler L 3 = Construct the same shape for the coupler at the other configurations This step establishes the location of the joints and at the other configurations E F F 3 The design can be continued as either Design-4 or Design-5 F E 3 Example 6 3 PE Nikravesh 6

7 ME 35 Dyad Most six-bar mechanisms are D combinations of two four-bars or a four-bar and a slider-crank Most often (3) (5) one of the four-bars is the driving C (6) mechanism; ie, it is a Grashof fourbar with a rotational motor (input) Dyad (4) Original four-bar () O The second mechanism, whether a O 6 four-bar or a slider-crank, contains the output link s an example, consider the six-bar mechanism shown containing two four-bars in series The original four-bar CDO 6 was designed first to perform certain task Link 6 (or link 5) is the output link This four-bar may or may not be Grashof Regardless of that, the four-bar O has been added to drive the original four-bar This second four-bar is called a dyad The output link of the dyad becomes the input link of the original mechanism Design 7 four-bar has been designed in such a way that its follower link can rock back and forth by an angle of θ 6 The input link of this four-bar negotiates in a range of Design a dyad to drive this four-bar with a given time ratio (T R could be : or :) (4) (5) θ 6 O 6 (6) Select an appropriate position for the pin joint on the input link of the original four-bar This point could be along the axis of the link, could coincide with pin joint C, or the link could be made triangular; ie, C,, and form a triangle Construct link 4 in its two limiting positions in order to determine the locations of and C D θ6 O 6 C D 3 Follow the procedure of Design- or Design-, depending on the time ratio, to complete the dyad In this example, we have assumed T R = : O The resulting six-bar is depicted in the form of an animation PE Nikravesh 7

8 ME 35 Example 7 PE Nikravesh 8

MENG 372 Chapter 3 Graphical Linkage Synthesis. All figures taken from Design of Machinery, 3 rd ed. Robert Norton 2003

MENG 372 Chapter 3 Graphical Linkage Synthesis All figures taken from Design of Machinery, 3 rd ed. Robert Norton 2003 1 Introduction Synthesis: to design or create a mechanism to give a certain motion