Ezy Mechanic

Grashof's Criterion: Crank-Rocker Four-Bar Mechanism

2015-08-08T15:13:00.001+07:00

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.

Grashof's Criterion: Four-bar Mechanism Double Crank

2015-08-05T23:28:00.002+07:00

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
Autodesk ForceEffect

Gruebler's Equation for calculating Degrees of Freedom of the Mechanism

2015-07-29T20:34:00.001+07:00

Mechanism with 1 degree of freedom

To design the mechanism, the first thing we should check is the number of degrees of freedom (DOF) of the mechanism. The degree of freedom is the number of inputs required to control the position of all links of the mechanism. It's usually the number of actuators needed to operate the mechanism.

We can use Gruebler's equation to calculate the number of degrees of freedom of the mechanism as follows.

Gruebler's Equation

where:

F = number of degrees of freedom

n = total number of links in the mechanism

L = total number of lower pairs (1 DOF such as pins and sliding joints)

H = total number of higher pairs (2 DOF such as cam and gear joints)

4-bar linkage

In the machine, we often require one degree of freedom which we can position all linkages with only 1 actuator. The four-bar linkage as shown in the picture is the example of the mechanism with 1 DOF. It has 4 links (3 bars with 1 ground link) and 4 revolute joints which the degree of freedom (F) can be calculated as follows.

n = 4 --- 4 links
L = 4 --- 4 revoulte joints
H = 0 --- no higher pairs
F = 3(4-1) - 2(4) - 0 = 1

Another example of 1 DOF mechanism is the slider-crank mechanism where it has the following number of links and joints.

Slider-Crank Mechanism

n = 4 --- 2 links + 1 ground link + 1 slider
L = 4 --- 3 pins + 1 slider
H = 0 --- no higher pairs
F = 3(4-1) - 2(4) - 0 = 1

The gears mechanism also has 1 DOF since it has the following values.

Gears Mechanism

n = 3 --- 2 gears + 1 ground link
L = 2 --- 2 revolute joints
H = 1 --- 1 gear joint
F = 3(3-1) - 2(2) - 1 = 1

These 3 examples have 1 degree of freedom which requires only 1 actuator to move the mechanism. It could be a motor, an air cylinder, etc.

If we change the slider joint of the slider-crank mechanism to the fixed pin joint, the mechanism will be locked since it has 0 DOF which is considered as a structure. The calculation using Gruebler's equation is as follows.

Structure: Degree of freedom = 0

n = 3 --- 2 links + 1 ground link
L = 3 --- 3 pin joints
H = 0 --- no cam or gear joints
F = 3(3-1) - 2(3) - 0 = 0 --- F=0 then it can't move.

If one of the pin joint of the 4-bar linkage changes to the slider joint, it will increase both the number of links and number of lower pairs. This makes the mechanism unconstrained because it has 2 DOF and required 2 actuators to control the position of the mechanism.

n = 5 --- 3 links + 1 slider + 1 ground link
L = 5 --- 4 pin joints + 1 slider joint
H = 0 --- no higher pairs
F = 3(5-1) - 2(5) - 0 = 2 --- F > 1, the mechanism is unconstrained.

References:

Machines & Mechanisms 3rd Edition, David H. Myszka
4 Basic Kinematics of Constrained Rigid Bodies
Linkages sketch and animation using SAM 7.0 Mechanism Design Software

Stiffness of a lever with eccentric loading - Part 2

2015-07-29T00:14:00.000+07:00

Let's continue from the previous post. We clearly see that the small post must be shortened because it the weakest point and creates torsional load to the lever arm. We're going to try redesigning with a bent lever arm so that the loading point stays in the middle of the center line of the arm.

Lever bending to eliminate torsion

The eccentric distance "r₂" is shorter than previous design (r₁). It's now the eccentric dimension with respect to the bent portion of the lever. Torsion still exists at end portion which has shorter length. This gives much less torsion compared with the earlier design. It also eliminate the torsion from the longer part of the lever since the loading point stays at the center line of the lever.

Two lever designs at the same loading point

From the overlay, we can see that the pull rod connecting point is still at the same location. The lever arm still connect to the same hub. But we can reduce the length of the small post which is the pin for the pull rod connection. The longer portion of the lever is now under bending only. Most torsion has been eliminated. It exists at the small portion as explained.

Finite element analysis - displacement of the lever with bent end

Shortening the small post and keeping the loading point on the center line of the lever can reduce displacement from 2.4 mm to 1.6 mm (33% reduction). Small twisting is present at the end but there is no twisting over the long portion of the lever.

von Mises stress of the bent lever

The von Mises stress reduces about 77%. This improvement can be used in most cases even when the lever shaft is short and the pull rod location is beyond the lever hub location and it's stiffer than the big lever with small post. The following picture is the example of bent levers on the machine.

Example of a bent lever on the machine

Stiffness of a lever with eccentric loading - Part 1

2015-07-28T22:33:00.000+07:00

A lever is a mechanical part that has arms and a fixed pivot used to transmit torque. The cross section of the arm is usually a rectangular shape which makes it stiff against bending. However, its stiffness may become lower if the loading point is not on the right place. We're going to see the effect of torsion on the lever with the improvement idea.

The following picture is the components of a general cam and lever system. The lever has the hub in the middle. It's where bearings are placed inside and it's the fixed pivoting point on a lever shaft (not shown). There are 2 arms to the left and the right. They're both welded to the hub and rotate together. The right arm is connected to a cam follower that rolls over cam surface. When the cam rotates the lever will swing up and down since there will be spring forces pushing the cam follower to keep contact with the cam surface all the time. The left arm has another point to connect to other machine parts, which usually is a pull rod. It will rotate at the same angle (degrees) with the right arm, but the distance (mm) may be different depending on a lever ratio.

Components of cam & lever system

The pull rod connecting point on the left in this example is not good since it has a long distance from the arm and the arm will not only be subjected to bending but also torsion. The following is the lever in the machine.

Lever with eccentric pull rod connecting point

The eccentric distance "r" usually comes from space limitation on the lever shaft. There may be some other machine parts blocking the lever and it can't move any further. The hub stays at the same location and the lever is just straight from the hub. Because of this, there will be a gap between the lever and the pull rod. That's why the designer add that small connection post at the end.

Load from the pull rod

The small post will be subjected to bending and the lever arm will also subjected to bending as well as torsion. If the distance "r₁" is much less then it could improve the stiffness. Here is the finite element analysis result of this lever.

Finite element analysis - displacement of the lever under eccentric load

We can see from the finite element analysis result that the lever bends down due to the load and it also twists to the right because of the eccentric load. However, the small post that connects to the pull rod reduces the overall stiffness since it also bends down.

von Mises stress of the lever under torsional load

The highest von Mises stress is at the weakest point on the small post as shown in the picture.

To reduce the overall displacement, the overall stiffness will improve. We can change some designs to eliminate the torsion away from the lever arm and reduce the length of the post. Let's see how it's done in the next post.

Stiffness of a welded straight square tube

2015-07-27T14:14:00.003+07:00

To modify or improve process of the existing machine in a production line may have some constraints regarding available spaces. There may be other moving objects or fixed parts which obstruct mounting of newly design parts. The part that comes later may be more complex since it has a limited space for mounting. The green support, design C from earlier post, is the example.

Part with other object occupied spaces

This is the side view of the above picture. The moving object occupies the space above the horizontal square tube and the design is in the L shape as shown below in order to avoid interference.

Side view of design C

This green support takes both loads in x and z directions. The load in z will try to bend the part and the load in x will try to twist the part. The dimension L is quite long from the center of the horizontal tube and the torque which is product of F_x and L will easily twist the horizontal tube. Since L is long, only a few twisting degrees of the tube will result in having larger displacement at the end (at the top flange where the loads applied).

Loads on welded part (design C)

If there is no obstruction above the horizontal tube, we can then change the design not to form the L shape in order to avoid twisting of the tube as shown below.

Straight tube design when space is available

The force in z is still trying to bend the straight tube up. However, the force in x will not try to twist the tube anymore. It's trying to bend the tube to the left instead. And with this design we can reduce the mass from 0.92 kg of design C to 0.71 kg which is about 23% reduction. And here is the finite element analysis result using the same fixations and loads.

Finite element analysis result: displacement of straight tube design

Not only mass reduction, but also the stiffness increases. The displacement reduces from 9.7 mm of design C to 6.98 mm which is about 28% reduction. This is because we eliminate the torque from the tube. But of course, this is when there is available spaces above the horizontal tube of design C.

We will apply this torque reduction design to other applications in later posts. Please stay tuned!

Stiffness comparison of welded parts - Part 4

2015-07-25T01:18:00.000+07:00

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.

Stiffness comparison of welded parts - Part 3

2015-07-25T00:32:00.001+07:00

There are a lot of designs of welded parts which are commonly seen. The design A as shown in previous post are one of the examples. However, there are some more examples which may seem to be much better than design C. Let's have a look at these 5 more designs and see how stiff they are. Normally, to make the welded parts more rigid, we have to add some more materials. This usually increases the weight (mass) of the part which, in most cases, is undesirable. Therefore, we want to get the design that provides as low as possible displacement with relatively small increases of the weight. These are 5 new designs to consider.

Five more designs of welded parts for stiffness comparison

Let's recall the designs from previous post (design A, B and C) in order to compare with these new designs.

Welded parts of previous analysis

Design C is the best solution compared with design A and B as shown in earlier post. However, we may think that design A is not so bad, probably we can add 2 more plates on both sides of the square tubes which will be stronger for torsional load. Because of this, we get design D which we have more masses due to those 2 plates. The mass of design D is 1.12 kg which is 124% of the original mass (design A). What about the displacement?

Finite Element Analysis result: Displacement of Design D

The displacement of this design is in the same level as design A. Design A has displacement of 14.77 mm and this design has displacement of 14.49 mm. The bad thing is that it's heavier by 24% but still has the same displacement. We can see from the finite element analysis result that there is small displacement on the horizontal tube up to half of its length which is good. But larger displacement starts at the end of the tube. Length of the rib is not long enough to prevent the horizontal tube from twisting.

Then, we have a new improvement by making those 2 side ribs larger as shown in design E. We expect to see much reduction of displacement because the ribs are all over the length of the tube.

Here is the FEA result.

Finite Element Analysis result: Displacement of Design E

The displacement is much less as expected. It has only 8.17 mm which is only 55.3% of the original design. We can reduce the displacement almost by half since those ribs make it strong against torsional load. However, we have to sacrifice much more weight. Total mass becomes 202% of the original design. So it's rigid, but heavy. Probably, we can reduce some weights but still get good stiffness.

We then have Design F which both side ribs are cut to reduce weight from 1.82 kg to 1.44 kg (160% of original mass). It's still quite heavy, but let's see the displacement.

Finite Element Analysis result: Displacement of Design F

This design is still not good since it's heavier than design C and has more displacement than design C. So among these 3 new designs, design E is the most rigid design but it's too heavy.

Let's continue in the next post for the remaining 2 designs.

Stiffness comparison of welded parts - Part 2

2015-07-16T12:50:00.000+07:00

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.

Stiffness comparison of welded parts - Part 1

2015-07-15T12:47:00.000+07:00

In the machine, we usually make parts to support other machine components from standard steel profiles because they are relatively low cost. In this post, we're going to see how different constructions affect the stiffness of the welded parts. We have 3 different constructions of supports as shown in the picture.

Supports with different constructions

The constructions are slightly different.

A -- typical welding (mass = 0.9 kg)
B -- typical welding with extra tube (mass = 1.3 kg)
C -- square tubes welding with thin plate insertion between tubes (mass = 0.92 kg)

All 3 constructions are subjected to the same forces and constraints as shown in the picture. Main parts are square tubes size 25 mm x 25 mm with thickness of 2 mm. The flange on the right is the mounting plate. It is constrained not to move in x, y and z directions. And we assume this is the only mounting plate allowed to use due to the space limitation in the machine.

Welded square tubes subjected to face loads

There is another flange on the left is subjected to face loads in both x and z directions. So we can expect the displacement of the parts to the left (-x direction) and up (+y direction) directions when loaded.

Let's start with design A -- typical welding with 45 deg angle connection. There is only small area on the perimeter for welding.

Design A: Cut-away picture of typical welded square tubes with 45 deg. connection

Design B is the extended version of design A. An additional square tube is welded to withstand more forces but we get more weight and required more space. Assume there is nothing above the L-shape welded part and we can put additional square tube as shown in the picture.

Design B: Typical welded square tubes with extra support

Design C is the modified version of design A where we insert a thin steel plate (thickness = 2 mm) between both square tubes.

Design C: Cut-away picture of welded parts with a reinforced steel plate

Let's see how we setup the finite element model with constraints and loads in the next post.

Moment of inertia

2015-07-14T22:12:00.001+07:00

The Moment of Inertia or Mass Moment of Inertia is the measure of a body's resistance to change in it's rotational speed. The moment of inertia must be specified with respect to a chosen rotational axis.
The moment of inertia depends on the body's mass distribution and the rotational axis chosen. The larger moment of inertia requiring more torque to change the body's rotational speed.

A point mass
The moment of inertia is the mass times the radius from the rotational axis squared.

Moment of Inertia of Point Mass

A collection of point mass
The moment of inertia is just the sum of the point mass moment of inertia.

Moment of Inertia of Collection of Point Mass

A continuous mass distributions
The moment of inertia require an infinite sum of all the point mass moment of inertia which make up the whole part. This can be calculated by an integration over the whole mass.

Moment of Inertia of Continuous Mass Distributions

A common shape
For a common shape, we usually calculate the moment of inertia about the certain point and use it to apply for another rotation axis by use the parallel axis theorem.

Moment of Inertia of Common Shape Body

Parallel Axis Theorem
The moment of inertia about any rotation axis which parallel to a certain axis at center of mass (I_parallel) is the moment of inertia about the center of mass (I_cm) plus the product of mass times the distance between the center of mass and the rotation axis squared.

Parallel Axis Theorem

Parallel Axis Theorem can also apply to any rotation axis which parallel to any certain axis at a point that already know the standard moment of inertia (I_xx , I_yy or I_zz).

Parallel Axis Theorem & Radius of Gyration

Radius of gyration [K]
It is the distance from a rotation axis to a certain point which the mass of a body may be assumed to be concentrated and at which the moment of inertia will be equal to the moment of inertia of the actual mass about the rotation axis.

Differential screw for fine adjustments of precision equipment

2015-07-10T23:51:00.002+07:00

Differential screw components

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:

ΔS_nut = travelling distance of the nut (mm)
L_A = pitch of thread A (mm)
L_B = pitch of thread B (mm)
Δθ_screw = number of turns of the screw (rev)

From this example, we have ΔS_nut = 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 pull rod for position adjustment of a cam-driven mechanism

2015-07-08T17:34:00.000+07:00

Kinematic diagram of a cam-driven mechanism

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.

Example of rotary indexer sizing calculation for table plate drive application (3/3)

2015-07-08T10:21:00.005+07:00

Step 4 : Rotary Indexer Model Selection

Minimum Follower Wheel Pitch Diameter (Credit : Sankyo)

Recommended Size of Rotary Indexer can be estimated. By calculate Radius of Gyration divide by Follower Wheel Pitch Radius, this value should be less than 5 (this value may vary depend on difference manufacturer)

Recommended Follower Wheel Pitch Radius of Rotary Indexer

Minimum Follower Wheel Pitch Radius, [PR_min] = 427/5 = 85.424 ≈ 85.4 mm
Then, Minimum Follower Wheel Pitch Diameter, [PD_min] ≈ 85.4 x 2 ≈ 171 mm

Recommended Size of Rotary Indexer also can be estimated by another method. By calculate Table Diameter divide by center distance between input & output shaft, this value should be less than 7 (this value may vary depend on difference manufacturer)

Distance Between Input & Output Shaft

Recommended Center Distance between Input & Output Shaft of Rotary Indexer

Then, Minimum Distance Between Input & Output Shaft, [CD_min] = 1000/7 = 142.857 ≈ 143 mm
Total Output Torque for Indexer Selection, [T_{t select}] = T_t x SF = 307.271 x 1.2 = 368.726 ≈ 369 N.m @ Input Shaft(camshaft) Speed 60 rpm

Thus, Select Indexer that Dynamic Rated Output Torque [T_op] is more than 369 N.m at input shaft speed more than 60 rpm, 8 stops, Indexing Angle 270 degree, MS motion curve, minimum center distance between input & output shaft 143 mm and minimum follower wheel pitch diameter 171 mm

Step 5 : Gear & Motor Selection
Find the maximum torque on gear output shaft at working shaft speed

In this case, gear is directly connect to the rotary indexer. Then the maximum torque on gear output shaft is equal to the camshaft torque of rotary indexer
The camshaft torque of rotary indexer[T_c] is determined by following formula:

Rotary Indexer Cam Shaft Torque

Internal Indexer Inertia Torque, [T_oi] = 2.943 N.m (from rotary indexer manufacturer data)
Camshaft Friction Torque, [T_x] = 16.677 N.m (from rotary indexer manufacturer data)
Maximum Camshaft Torque Coefficient, [Q_m] = 0.987 (This is standard value, find the Qm for Modified Sine Cam Curve from Cam Curve Characteristic Table)
Then, Camshaft Torque, [T_c] = 87.552 N.m @ camshaft speed 60 rpm (This is the torque at input shaft of rotary indexer which is equal to torque at output shaft of gear)

Next, calculate the equivalent camshaft torque for gear selection [T_{ce select}] by considering of operating condition & safety factor

Calculated Operation Factor, [f] = 1.8 (Calculate from type of load(steady or shock load), operating hours per day, frequency of starts/stops, ambient temperature, type of lubrication or others factor. Please check gear manufacturer information. In this case use 1.8)
Then, T_{ce select}= T_ce x f x SF = 87.552 x 1.8 x 1.2 = 189.113 N.m @ Output Shaft Speed 60 rpm

Calculate Gear Input Shaft Torque

Gear Ratio, [i_g] = 10.33 (Check from gear manufacturer info.)
Gear Input Shaft Speed, [N_g] = N_rpmx i_g= 60 x 10.33 = 619.8 ≈ 620 rpm
Gear Running Efficiency, [Eff_g] = 92% (Check from gear manufacturer info.)
Gear Input Shaft Friction Torque, [T_xg] = 0.9 N.m (Check from gear manufacturer info.)
The Gear Input Shaft Torque[T_g] is determined by following formula:

Gear Input Shaft Torque

Gear Input Shaft Torque, [T_g] = 10.113 N.m @ Input Shaft Speed 620 rpm

Calculate Motor Torque

Motor Revolution per Minute, [N_motor] = 1730 rpm (from motor manufacturer data)
Gear Ratio Required, [i_req] = N_motor / N_rpm = 1730 / 60 = 28.833
Pulley Speed Ratio, [i_pulley] = I_req / I_g= 28.833 / 10.33 = 2.791 ≈ 2.8 (←if very close to 1, motor can be mounted directly to gear, then set Eff_pulley = 100% & T_xp = 0

Table Drive Application with Pulley (Credit : Sankyo)

Pulley Running Efficiency, [Eff_pulley] = 90% (estimated)
Pulley Shaft Friction Torque, [T_xp] = 2 N.m (estimated)
The Motor Torque [T_motor] is determined by following formula:

Motor Torque

Then, Motor Torque, [T_motor] = 10.003/(2.791 x 0.9) + 2 = 6.026 N.m @ Speed 1730 rpm

Calculate Motor Power

The Peak Motor Power [P_{motor peak}] is determined by following formula:

Then, The Peak Motor Power [P_{motor peak}] = 2 x 3.1416 x 1730 x 6.026 / 60 = 1091.617 Watt ≈ 1.092 Kw

Calculate Motor Power for Motor Selection

Power for Motor Selection, [P_{motor sel}] = P_{motor peak} x SF = 1.092 x 1.2 = 1.310 Kw

Then, select gear that maximum continuous output torque is more than 189 N.m at output shaft speed more than 60 rpm, gear ratio 10.33
Motor power 1.31 Kw at speed 1730 rpm
Ratio of pulley between gear and motor 2.8

Example of rotary indexer sizing calculation for table plate drive application (2/3)

2015-07-08T10:18:00.002+07:00

Calculations

Rotary Indexer with Dial Plate Driving Application Torque Diagram

Step 1 : Moment of Inertia Calculations

How to calculate the Moment of Inertia?
Dial Plate Inertia, [I_d] = 1/2 x 43.040 x ((1000/1000)/2)^2 = 5.380 Kg.m²

Dial Plate Inertia

Stations Inertia, [I_st] = 160 x ((450/1000)/2)^2 = 32.400 Kg.m²

Stations Inertia

Additional Part Inertia, [I_{add part}] = 8 x ((300/1000)/2)^2 = 0.720 Kg.m²

Additional Parts Inertia

If there are others group of additional parts, do the same method.
Total Moment of Inertia, [I_total or ∑MK²] = 5.380 + 32.400 + 0.720 = 38.500 Kg.m²

Total Inertia

GD² = 38.500 x 4 x 9.81 = 1510.739 N.m²

GD Square

Total Moving Mass, [∑M] = 43.040 + 120 + 8 = 211.040 Kg.

Total Moving Mass

Radius of Gyration, [K] = (I_total/∑M)^1/2 = 427.118 mm

Radius of Gyration

Step 2 : Maximum Angular Acceleration Calculation

Cam Curve Characteristic Value

From the table, find the Dimensionless Maximum Angular Acceleration for Modified Sine Cam Curve, [A_m] = 5.53
Then, the Maximum Angular Acceleration, [a_m] = 0.785 / 0.75^2 x 5.53 = 7.721 rad/sec²

Maximum Angular Acceleration

^{Step 3 : Required Torque Calculations}

Torque on Rotary Indexer

Inertia Torque at Index Shaft, [T_i] = 38.500 x 7.721 = 297.271 N.m

External Friction Torque to Indexer, [T_f] = 10 N.m (Such as outboard support bearings, assume to 10 N.m)
Work Torque, [T_w] = 0 N.m (Indexer is doing work such as lifting parts, assume to 0)
Total Output Torque, [T_t] = 297.271 + 10 + 0 = 307.271 N.m

Total Torque at Output Shaft

Example of rotary indexer sizing calculation for table plate drive application (1/3)

2015-07-03T16:53:00.000+07:00

Indexing System Information

Physical Properties Info.

Rotary Indexer with Dial Table Driving Application

Dial Table Info.

Diameter, [D_dial] = 1000 mm
Thickness, [Y] = 20 mm
Material = Aluminum (Density 0.00274 g/mm³)

Mass of dial table

Then Table Mass, [M_dial] = 43.040 Kg.

Station Info.

Number of Stations, [N_stations] = 8 stations
Mass per Station, [M_station] = 20 Kg/staion
Then Total Station Mass, [SM_stations] = 8 x 20 = 160 Kg
Rotating Radius to Station Center, [R_st] = 450 mm (If the station is large or complex shape, a radius of gyration of each station should be calculated.)

Additional Part Info. (The same part at the same rotation radius)

Number of parts, [N_{add part}] = 4 parts
Part Mass, [M_{add part}] = 2 Kg/part
Total Additional Part Mass, [SM_addpart] = 4 x 2 = 8 Kg
Rotating Radius to Part Center, [R_{add part}] = 300 mm (If the part is large or complex shape, a radius of gyration of each part should be calculated.)
If there are others group of additional parts, do the same method.

Movement Info.

Rotary Indexer Movement Term

Number of Stops, [S] = 8 stops

Displacement per 1 index

Dial Plate Displacement per Index, [h_m] = 2 x 3.1416 / 8 = 0.785 rad
Indexer Input Shaft Cycle, [T] = 1 sec (Assume to 1 second, this can be adjusted later)

Indexing Rate

Indexing Rate, [N_rpm] = 60/1 = 60 rpm
Indexing Period, [B_m] = 270 deg (Assume to 270 degree, should be taken as long as possible for smoother movement. Please check to the manufacturer information)

Indexing Time

Indexing Time, [t_m] = 270/360 x 60/60 = 0.750 sec

Dwell Time

Dwell Time, [t_dwell] = 1 - 0.750 = 0.250 sec
Required Dwell Time, [t_{dwell required}] = 10 sec (This is actual required dwell time according to the working process.)
In case of t_{dwell required} > t_dwell we can use clutch and brake to extend the dwell period, but if t_{dwell required} < t_dwellthat mean we can reduce t_dwell to be the same as t_{dwell required}
Reducing t_{dwell required} can be consider in 2 cases. Reducing indexer input shaft cycle [T] or Increasing indexing period [B_m].
This case, t_{dwell required} > t_dwell

Motor stop time

Input Shaft Stop Time, [t_stop] = 9.750 sec

Machine cycle time

Machine Cycle Time, [Tmachine] = T + t_stop = 1 + 9.750 = 10.750 sec
Machine Speed, [Speedmachine] = 3600/10.750 x 8/8 = 334.884 UPH

Machine Speed

Displacement VS Indexing Angle

Cam Curve : Modified Sine (assumed, please check from rotary indexer manufacturer data)
Safety Factor, [SF] = 1.2
Operating Condition : 24 Hrs running, Steady load, Starts/stops 2 times per hour, Ambient temperature 30 C, Without cooling fan

Rotary indexer sizing calculation for table plate drive application

2015-07-03T16:34:00.000+07:00

In this post we will explain how to calculate the size of rotary indexer, gear & motor for dial plate table application.

Rotary Indexer with Table Plate Driving Application

There are 2 things we need to know for the calculation

Physical properties info.
Movement info.

Physical properties information of movement part is data we need to use for calculate the total moment of inertia of the moving system. It consists of dial plate diameter, thickness, density for calculate mass, the number of stations, station mass, station radius to rotation center, etc.

Movement information is data we need for find the maximum angular acceleration of the moving system. It consists of number of stops, RPM of rotary indexer input shaft, indexing angle, dwell time, type of cam curve, etc.

After we know the value of the total moment of inertia and the maximum angular acceleration of the moving system, we can easily calculate the required torque for driving the system. Then we can select the rotary indexer that match the required torque.

For selection of gear, we can use this required torque at output shaft of indexer to calculate the required torque of input shaft of the indexer by using maximum camshaft torque coefficient, internal indexer inertia torque, camshaft torque from indexer manufacturer data, number of stops and indexing angle. When we know the torque required at indexer camshaft(which is output shaft torque of gear), we can select the gear.

For the motor selection, we have calculate the gear input shaft torque by converting gear output shaft torque with gear ratio. Then use this torque for calculate the motor power required.

Example of Rotary Indexer Sizing Calculation for Table Plate Drive Application

Indexers with single dwell (1-dwell) VS double dwell (2-dwell)

2015-07-01T19:45:00.000+07:00

To configure a rotary indexer, there are several parameters to decide such as number of stops (S), indexing angle (b_m), 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, b_m = 210 deg.

2-dwell indexer will perform differently from 1-dwell indexer though they're both having the same S and b_m. 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 (b_m/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 (h_m) becomes h_m = 360/12 = 30 deg.
Indexing angle, b_m = 210 deg.
Same input shaft speed (w)

1-dwell and 2-dwell rotary indexers with S = 12 and b_m = 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 b_m/2 = 210/2 = 105 deg. Therefore, the 2-dwell indexer takes 105 degrees to complete the first index with output shaft displacement (h_m) 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.

Indexing Cam Angles Comparison (2/2)

2015-06-29T23:50:00.000+07:00

In post [Indexing Cam Angles Comparison (1/2)], we explained how to calculate the displacement of indexer (h_m) from number of stops (S) and the meaning of the indexing angle (b_m). 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 b_m = 120 deg. and the second indexer has b_m = 240 deg.

Indexers with b_m=120 deg. and b_m=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.)
h_m = displacement of the turret plate (deg.)
b_m = indexing angle (deg.)

Displacement diagram of 2 indexers with different indexing angles (b_m)

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 b_m = 120 deg.: 10% = 12 deg.

Very small displacement within first 12 deg. and from 108 deg. to 120 deg.

For b_m = 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:
v_max = maximum velocity (deg./s)
h_m = displacement of the turret plate (deg.)

w = angular velocity of the input shaft (rad/s)
t_m = indexing time (s) -- indexing time is proportional to indexing angle (b_m)

Velocity diagram of 2 indexers with different indexing angles (b_m)

Both indexers have same number of stops (same displacement h_m) 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:
a_max = maximum tangential acceleration (deg./s²)
h_m = displacement of the turret plate (deg.)

w = angular velocity of the input shaft (rad/s)
t_m = indexing time (s) -- indexing time is proportional to indexing angle (b_m)

Both indexers have the same number of stops (same displacement h_m) 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 (b_m)

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.

Indexing cam angles comparison (1/2)

2015-06-29T22:49:00.000+07:00

In post [Rotary Indexer For Indexing Motion], we explained the meaning of indexing with the example of displacement and velocity diagram to give an idea how the indexing mechanism works. We also showed the location of input and output shafts of the rotary indexer. In this post, we're going to see how both shafts work and the meaning of indexing angle.

Geared motor and Turret plate connected to the rotary indexer

Basic equipment setup of indexing system is as shown in the picture. The input shaft of the rotary indexer is connected to the geared motor or other drive components e.g. pulley, gear, etc. In this example, we use a "hollow shaft" geared motor which allows indexer's input shaft to insert directly into the hollow gear shaft without using any shaft couplings. But they must be carefully align in order to avoid misalignment between shafts.

At the output side of the rotary indexer, the output shaft or flange can be mounted to a turret plate. There will be tools mounted on the turret plate at the same P.C.D. (pitch circle diameter). The tool at each station is usually work-piece holder. When the turret plate indexes, it will transfer the work-piece from one station to another station in order to complete all required processes at different stations.

Indexing motion from the rotary indexer doesn't require stopping of the motor since it has internal construction with cam and rollers that generate indexing motion at the output shaft while the input shaft runs continuously. As a machine designer, to select the right indexer for the application, the first thing to do is to select number of stations. Usually we will provide additional spared positions for future improvement i.e. if required number of stations is 6, we may select 8 stations instead.

Number of stations is usually "number of stops (S)" on the indexer. In the following picture, the indexer has number of stops = 4. In each turn of the input shaft (or geared motor's shaft), the output shaft (turret plate) moves 1/4 turn. Therefore, the displacement of the turret plate (h_m) in degree can be calculated using:

h_m = 360/S

where:
h_m = displacement of the turret plate (deg.)
S = number of stops

Number of stops (S), indexing angle (b_m) and displacement (h_m)

Hence, the displacement of the turret (h_m) is 360/4 = 90 deg. The displacement is also an important factor of the acceleration which we we explore more in later posts.

The indexing angle (b_m) is the total angle at the input shaft to rotate the output shaft (turret) to another station. If the rotary indexer has S=4 and b_m = 120 deg., it means that when the input shaft turns 120 deg., the turret will completely turn from one station to another with the displacement of 90 deg.

See how different indexing angles move in the next post.

Rotary Indexer for Indexing Motion

2015-06-22T23:31:00.000+07:00

Rotary Indexer

Indexing is the process of intermittent motion where machine starts and stops at precise locations in specified intervals. Indexing motion is required in most machines for many applications since the stopping period (dwell period) of indexing allows other units in the machine do their jobs. If machine runs continuously without indexing, it may be much more difficult and complex for other units to follow the moving object compared with object that stops.

Indexing Motion

The easiest way to provide the intermittent motion is to start and stop the motor. Probably, a servo motor or a stepping motor can do that with precise intervals and locations. However, when required a high degree of rigidity, a positive motion mechanism to ensure no backlash, the cam-driven indexer is most suitable for the application. The indexer is available from many manufacturers which we can choose from product catalogs according to desired load, indexing interval, indexing profile, number of stops, etc.

Ideal Displacement and Velocity Diagram of a rotary indexer

The above chart shows 2 cycles of displacement and velocity of the rotary indexer. In each cycle, the rotary indexer rotates (indexing) then stops. The indexing time is the time of indexing and the dwell time is the time when the indexer has no movement (zero velocity). The following picture shows input and output shafts of the indexer.

The input shaft is normally connected to a geared drive (motor) that rotates at a constant speed. The construction inside the indexer (roller gear, cam, conjugate plate cams, etc.) makes the output shaft rotates and stops as shown in the earlier displacement diagram. However, in some applications, it may require longer dwell time for other machine units to do their jobs. For this case, the input shaft (motor) will rotate and stop and wait to extend the dwell time until other machine units have completed their processes then it rotates again. This is also called "cycle-on-demand" application.

Input and output shafts of a rotary indexer

To select the right rotary indexer for your application, there are several parameters to take into account such as number of stops, indexing angle, load (inertia), torque, etc. We will explore more details about those parameters. At the end we will make an excel calculation sheet to calculate values of necessary parameters for selection the right rotary indexer from the manufacturers.

Next: [Indexing Cam Angles Comparison (1/2)]

Reference:

3D CAD model (SANDEX right angle indexer: ED11)