Grashof's criterion for Double Crank Four-Bar Mechanism is almost the same as Crank-Rocker Mechanism. They're both defined as S+L < P+Q. The difference is on the location of the shortest link (S). For double crank mechanism, S must be on the ground link (frame). But for the crank-rocker, S will be on the side link and it must be the input link.
We then move the shortest link from the frame to the side link on LH. With the help of SAM 7.0 Professional software (by Artas Engineering), it can display the paths of desired nodes as shown in the picture.
Grashof's Criterion: Crank-Rocker Mechanism using SAM 7.0 Professional Software
The pink curves represent paths of desired nodes. The input link which is now the shortest link is able to make a full revolution. It's called a crank. And the output (can be either L, P or Q) will only oscillate around its pivoting point. It's called a rocker. But if we change the driver to node 4 and let it drives from the link which is not the shortest link, it can't be reversed. The shortest link can't make a full revolution. What we have to do is to swap as shown in the following picture and it becomes a crank-rocker mechanism again which is now driving from the RH pivoting point.
Grashof's Criterion: Crank-Rocker Mechanism Driver on RH side
How can we see the velocity? From the path display, we can see how they move. But to see the velocity, we can choose to display the hodograph.
Path is the line that a moving point describes in the fixed reference system.
Velocity Hodograph is the locus of the arrowhead of the velocity vectors (rotated 90 degree) of a moving point.
The hodograph in SAM 7.0 is shown as follows.
Velocity Hodograph in SAM 7.0 The Ultimate Mechanism Designer
The lines on the outside of the path represent CCW direction of the node. And the lines at the inside of the path represent CW direction of the node. From the above picture, we can see that the input link of the crank-rocker mechanism rotates at a constant speed in CCW direction since the hodograph displays uniform lines at outside of the circular path. However, the output link oscillates back and forth with changes in velocity since there are lines of the hodograph both on the outside and inside of the its curve path. And we can quickly point out the position where the mechanism has highest velocity from longest line on the hodograph. SAM 7.0 is also able to plot various design parameters of the mechanism e.g. velocity, displacement, acceleration, length, etc. In the above picture, LH graph shows the velocity profile of the output link which can be traced manually. The following shows the hodograph of the double crank mechanism for a comparison.
Hodograph of Double Crank Mechanism
The following is the video showing how to use SAM 7.0 Software to simulate the double crank and crank-rocker mechanisms.
One of the most commonly used linkages is the four-bar linkage. It consists of 3 moving links and 1 ground link (also called a frame). There are 4 pin joints connected between those links. And from Gruebler's Equation, it's the mechanism with 1 degree of freedom (1DOF) which requires only 1 actuator to drive and control position of all linkages. The four-bar mechanism consists of the following components.
Four-bar mechanism components
The link that connects to the driver or power source is called the input link. The other link connected to the fixed pivot is called the output link. The remaining moving link connected between input and output links is called the coupler. It couples the motion of the input link to the output link.
There are different configurations of lengths of the four-bar mechanism and it results in different movements of the mechanism. Grashof's Criterion helps classifies into the following categories:
Double Crank --- also called a drag link mechanism
Crank-Rocker
Double Rocker
Change Point
Triple Rocker
In this post, we're going to explore the case of Double Crank which both input and output links are able to rotate through a full revolution. We use Autodesk ForceEffect app on Google chrome to illustrate and demonstrate how the mechanism moves. The links can be setup easily by just dragging. The exact length dimension can also be specified. Once we setup all required items i.e. links, pivots and drive, we can play and see the animation of the linkages with the path of desired tracing points.
Autodesk ForceEffect app on Google chrome
Grashof's criterion states that a four-bar mechanism has at least one revolving link if:
S + L ≤ P + Q
where:
S = length of the shortest link
L = length of the longest link
P = length of one of the intermediate length links
Q = length of the other intermediate length link
For the Double Crank category, the following criteria must be satisfied:
Double Crank:
S + L < P + Q
S is the length of the frame (ground link)
So we setup the links in ForceEffect app as follows.
Grashof's Criteria for Double Crank 4-bar Mechanism
The shortest link (S) is the frame which is 400 mm long. The length of the longest link (L) is 1300 mm. The remaining 2 intermediate links (P and Q) have length 700 mm and 1200 mm.
This satisfies Grashof's criteria since (S)400 + (L)1300 < (P)700 + (Q)1200. And this is how this mechanism moves.
Paths of double crank four-bar mechanism
Both input and output links can make a full revolution as desired. Let's find more details of other categories in later post.
The following video shows how to use ForceEffect app to simulate the motion of double crank four-bar linkage.
Reference:
Machines & Mechanisms 3rd edition, David H. Myszka
As we can see from previous post that design C is still the best choice since it's light-weight and relatively rigid compared to other designs. Design E is the most stiff design but it's also the most heavy design. Let's continue with the remaining 2 designs with 4 ribs at the mounting base to see whether they're better than design C or not.
Finite Element Analysis result: Displacement of Design G
The weight of this design is in the same level as design D, but it's more rigid. When compared to design A, its weight is 123% and the displacement is 72.8%. Then we increase lengths of those 4 ribs in order to reduce the displacement. This is design H which its mass increases to 133% of design A. Here is the FEA result.
Finite Element Analysis result: Displacement of Design H
The displacement is 9.72 mm which is 65.8% of design A. It's comparable to design C, but heavier. So design C is still the best choice. It's simple, not difficult to make, light-weight and stiff. The following is the summary of displacement and mass of all designs. The value in percentages are compared to design A.
Summary of displacement and mass of different welding designs
The red, yellow and green are used to express what is desirable and what isn't. Red is undesirable and green is desirable.
We hope the experiment about the stiffness using finite element analysis in these 4 posts may provide you the idea of improvement. Sometimes, we add extra materials (more weight, more cost) but gain very little. Sometimes, only minor changes change significantly improve the design with acceptable weight and cost.
The following is the video showing more details of this experiment.
Design C is good when we can't have any parts above the horizontal tube. It may interfere with other parts, for instance. However, if there is no space limitation above the horizontal tube, we could greatly improve the stiffness of the structure and reduce the weight. Let's find out in the next post.
From [Stiffness comparison of welded parts - Part 1], we have 3 different designs of supports subjected to the same loads which we're going to compare their stiffness. Let's start with the Finite Element Analysis (FEA) model of design A.
Finite element analysis (FEA) model of Design A
The mounting plate on the right is constrained so that there is no displacement in x, y and z directions. The flange on the left is subjected to face loads in x and z directions which means the support is under bending and torsion. After solving the equations, we get the post processor result as follows.
FEA result: Displacement of Design A -- typical welding connection
The max displacement magnitude is 14.77 mm and displacement in x, y, and z are as follows.
Max displacement in x = -12.2 mm
Max displacement in y = 4.4 mm
Max displacement in z = -7.5 mm
The same conditions applied on design B and C. And these are the FEA results.
FEA result: Displacement of Design B -- additional square tube
The max displacement magnitude is 10.77 mm (27% decreased) and displacement in x, y, and z are as follows.
Max displacement in x = -10.1 mm (17% decreased)
Max displacement in y = 3.2 mm (27% decreased)
Max displacement in z = -2.4 mm (68% decreased)
The displacement in z direction reduces 68% since the additional square tube helps support the bending. The third tube also helps reduce torsion which result in reduction of displacement in x direction by 17%.
FEA result: Displacement of Design C -- reinforced plate inserted between square tubes
The max displacement magnitude is 9.7 mm (34% decreased) and displacement in x, y, and z are as follows.
Max displacement in x = -8 mm (34% decreased compared to design A)
Max displacement in y = 3.1 mm (29% decreased compared to design A)
Max displacement in z = -4.8 mm (36% decreased compared to design A)
The displacement in z direction reduces 36% which is less than design B. However, when subjected to torsion, the displacement in x direction is much less than design B. This design is only adding a small piece of steel plate.
Summary:
Displacement Magnitude:
Design A = 14.77 mm (100%)
Design B = 10.77 mm (-27%)
Design C = 9.7 mm (-34%)
Mass:
Design A = 0.9 kg (100%)
Design B = 1.3 kg (+44%)
Design C = 0.92 kg (+2%)
So design C could improve stiffness with small increment of mass and it isn't difficult to do. There may be other better designs compared to design C, but this is one of the improvements that we can easily gain by minor changes to design.
The following video shows the 3D model in Unigraphics and finite element analysis model with results in LISA finite element analysis software.
A differential screw is a mechanism that provides very fine motions of machine parts. There are several forms of its configuration.
The picture shows one common form of the differential screws. There are 3 main components as follows:
Spindle (differential screw) -- The spindle has two different threads on the same axis.
Base -- It's the base of the whole mechanism which has one threaded hole.
Nut -- It has one threaded hole with sliding joint. This is the end mover where we will get fine motion. This part may be connected to other machine components to provide precise motion.
How it works
Different threads on the same spindle / distance between marks
The spindle has two different thread sizes. In this example, the larger one (thread A) has M12 coarse thread which has a pitch of 1.75 mm. Another thread (B) is M10 coarse thread which has a pitch of 1.5 mm.
Pitches (leads) of the differential screw
The pitch (or lead) of a screw is distance the screw advances when it turns one revolution. Therefore, when the handle turns one revolution, thread A rotates one revolution and moves in a distance equal to the pitch of thread A (1.75 mm). Since thread B is on the same spindle, it also moves together with thread A (1.75 mm) and also rotates one revolution. However, thread B connects to the nut which is unable to rotate. So, the nut retracts a distance equal to the pitch of thread B which is 1.5 mm. Hence, the motion of the nut is the advance distance of thread A minus the retracted distance. It is the difference between the pitch of threads. This is why it is called the differential screw.
Differential screw displacement formula
where:
ΔSnut = travelling distance of the nut (mm)
LA = pitch of thread A (mm)
LB = pitch of thread B (mm)
Δθscrew = number of turns of the screw (rev)
From this example, we have ΔSnut = 1.75 - 1.5 = 0.25 mm. That means the nut travels 0.25 mm per each turn of the spindle. As shown in the above picture, the distance between 2 marks is 2.5 mm. Then we need to turn the spindle 10 revolutions so that the nut will travel 2.5 mm.
As we can see from the formula, if we need the nut to move 0.1 mm per one turn of the spindle, we need to select the different screw threads. Since we know that the standard metric coarse threads have the following values:
M5, pitch = 0.8 mm
M4, pitch = 0.7 mm
The difference between pitches is 0.8 - 0.7 = 0.1 mm which is as per the requirement. So, thread A will be M5 and thread B is M4 and we will get 0.1 mm per turn.
Watch the following video to see how it moves. We use Unigraphics NX4 motion simulation to show all motions.
Reference:
Machines & Mechanisms Third Edition by David H. Myszka
A cam-driven mechanism is commonly used in most production machines since all motions and timings can be controlled. Not only the displacement is controlled, but also the velocity and acceleration as well as jerk can be controlled. Cam-driven mechanism allows overlapping motion between machine parts since the positions of all relevant parts can be determined from the timing diagram which is desirable for high speed application.
A simple cam-driven mechanism consist of the following parts as shown in the kinematic diagram:
Cam: for motion generation (displacement, velocity, acceleration and timing).
Cam follower: rolling part mounted on a lever.
Spring (not shown): to keep contact between cam surface and cam follower.
Lever: to transfer continuous cam rotation to swinging motion.
Pull rod: to transfer the motion from the lever to the slider.
Slider: end equipment (processing equipment)
In this post, we're going to focus on the pull rod (also known as push rod or tie rod) which is one of the common parts for most machines. The pull rod allows position adjustment of its connected parts since its length can be adjusted. Normally, the pull rod consists of the following parts:
Pull rod
Rod end bearing RH thread
Rod end bearing LH thread
Nut RH thread
Nut LH thread
Pull rod with both female rod end bearings
The pull rod usually made of a hexagonal post. The mechanic can use a wrench on the hexagonal part to tighten or loosen the pull rod from the rod end bearings. Rod end bearings must have RH thread on one side and LH thread on the other side otherwise the distance between the rod ends will remain the same.
Example of pull rods on the machine
Adjustment of the pull rod length usually happens when both sides of the rod end bearings are already connected to other parts in the machine (in this example, it is connected to the lever and the slider already). To adjust the length, no need to disconnect the rod end bearing, first we have to loosen both RH and LH nuts so that the pull rod can be turned. Then turn the pull rod in either direction and its length will change. By doing this, we can then adjust the position of the connected parts which, in the case, is the slider. After the slider is at the desired position, tighten both nuts.
The male rod end bearings version is also available. We can use the same nuts, but the hexagonal post will have threaded holes instead (see the following picture).
Pull rod with both male rod end bearings
The increment of the pull rod length (distance between both rod end bearings) is determined by the pitch of the thread on the rod end bearings. For this example, the M10 thread has a pitch of 1.5 mm. One turn of the pull rod will change the distance of each rod end bearing by 1.5 mm. Therefore, the increment is 2 times the pitch (2 x 1.5 = 3 mm/turn).
The pull rod length is increased or decreased according to the following directions.
Pull rod length extension and retraction according to the turning direction
The following is the animated picture showing how the slider position can be adjusted by turning the pull rod.
Animated picture of pull rod length adjustment
Watch the following video for how the cam driven-mechanism works and where the pull rod is used in the system. The simulation uses NX4 motion simulation module.
To configure a rotary indexer, there are several parameters to decide such as number of stops (S), indexing angle (bm), maximum output torque, etc. In this post, we will show one more parameter which may affect the design if misunderstood. It's a number of dwell. In most cases, we use 1-dwell (single dwell) for the application. The indexer performs single index within the designated indexing angle and wait (dwell) until cycle complete.
However, for some indexer models, there will be the option to select 2-dwell cam type (double dwell, double indexes). Or there may be only 2-dwell version especially for the models which have large number of stops and long indexing angle e.g. S=20, bm = 210 deg.
2-dwell indexer will perform differently from 1-dwell indexer though they're both having the same S and bm. The 1-dwell indexer will index only one once per each full revolution of the input cam shaft. But the 2-dwell indexer will index 2 times and also stop 2 times per each revolution of the input cam shaft as can be seen in the following displacement diagram.
Displacement diagram of both 1-dwell indexer and 2-dwell indexer
The 2-dwell indexer divides displacement into 2 halves. The first half take half indexing angle (bm/2) to rotate the output shaft to next station. The dwell angle is also divided by 2. It has 2 times displacement compared with the displacement of the 1-dwell cam as can be seen in the red line on the above chart.
In this post, we use indexers with following parameters for comparison.
They both have ...
Number of stops, S = 12 stops. So the displacement (hm) becomes hm = 360/12 = 30 deg.
Indexing angle, bm = 210 deg.
Same input shaft speed (w)
1-dwell and 2-dwell rotary indexers with S = 12 and bm = 210 deg.
According to the explanation of the 2-dwell cam motion, the second indexer will perform 2 indexes and 2 dwells per 1 turn of the input shaft. The actual indexing angle will become bm/2 = 210/2 = 105 deg. Therefore, the 2-dwell indexer takes 105 degrees to complete the first index with output shaft displacement (hm) of 30 deg. Then it waits (dwell) until the input shaft angle reaches 180 deg. and it restart the next indexing from 180 deg. to 180 + 105 = 285 deg. After that, it waits until the input shaft complete its turn. Then the next cycle starts...
Here is the animation of how both indexers move.
How 1-dwell indexer and 2-dwell indexer move
Watch the following video for the animation made with Unigraphics NX4 motion simulation.
In post [Indexing Cam Angles Comparison (1/2)], we explained how to calculate the displacement of indexer (hm) from number of stops (S) and the meaning of the indexing angle (bm). In this post, we will show how the rotary indexers with different indexing angles move.
We have 2 indexers. Both of them have S = 4, but the first indexer has bm = 120 deg. and the second indexer has bm = 240 deg.
Indexers with bm=120 deg. and bm=240 deg.
The input shafts of both indexers rotate continuously at the same speed. The first indexer completes its indexing in 120 deg. while the second indexer takes longer time and completes its indexing in 240 deg. We can clearly see from the animation that the longer the indexing time, the smoother indexing motion (lower acceleration).
The indexer has many displacement profiles e.g. modified sine, modified trapezoidal, etc. But for the explanation, we will use cycloidal motion profile.
The cycloidal displacement profile can be expressed as ...
... (eq. 1)
where:
h = displacement (deg.)
q = angle of the input shaft (deg.)
hm = displacement of the turret plate (deg.) bm = indexing angle (deg.)
Displacement diagram of 2 indexers with different indexing angles (bm)
We can see from the displacement diagram that both curves have smooth continuous displacement. There is not much displacement at approximate 10% and 90% of the indexing angle.
For bm = 120 deg.: 10% = 12 deg.
Very small displacement within first 12 deg. and from 108 deg. to 120 deg.
For bm = 240 deg.: 10% = 24 deg.
Very small displacement within first 24 deg. and from 216 deg. to 240 deg.
This fact can be used for timing diagram design later.
Maximum velocity can be calculated from ...
... (eq. 2)
where:
vmax = maximum velocity (deg./s)
hm = displacement of the turret plate (deg.)
w = angular velocity of the input shaft (rad/s)
tm = indexing time (s) -- indexing time is proportional to indexing angle (bm)
Velocity diagram of 2 indexers with different indexing angles (bm)
Both indexers have same number of stops (same displacement hm) and same input velocity (w). So the relationship between velocities can be expressed as
... (eq. 3)
Since the indexing time of the second indexer is 2 times the first one. The velocity of the second indexer then becomes half of the first one as can be seen in eq. 3
The maximum tangential acceleration of cycloidal profile when the input shaft rotates at a constant speed (constant w) can be expressed as ...
... (eq. 4)
where:
amax = maximum tangential acceleration (deg./s2)
hm = displacement of the turret plate (deg.)
w = angular velocity of the input shaft (rad/s)
tm = indexing time (s) -- indexing time is proportional to indexing angle (bm)
Both indexers have the same number of stops (same displacement hm) and same input velocity (w). So the relation between accelerations can be expressed as
... (eq. 5)
Since the indexing time of the second indexer is 2 times the first one. The acceleration of the second indexer then becomes 1/4 of the first one as can be seen in eq. 5 -- This is quite interesting. By selecting longer indexing angle (indexing time), we can reduce its acceleration by square of indexing time ratio.
Acceleration diagram of 2 indexers with different indexing angles (bm)
In this example, the second indexer runs at only 25% tangential acceleration of the first indexer by selecting 2 times indexing angle. However, there will be less time during dwell period for other machine units to work with.
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