[Psychology] Mechanical Assemblies Phần 3 doc

96 4 CONSTRAINT IN ASSEMBLY

TABLE 4-11. Constraint Analysis of Pin-Hole and

Pin-Slot Joint

FIGURE 4-32. Illustrating the Equivalence of Three Dif￾ferent Positions for the Z Force in WRU in Table 4-11. The

arrows for FZ-\ and MX represent the original entries in the

first row of WR and are equivalent to Fz2 without MX and

Fza also without MX￾measured about the global Y axis. Note that there is no cor￾responding problem regarding pin axis parallelism about

the global X axis because the right pin can rock in the slot

about its x axis.

4.F.4.b. Remarks

A constraint analysis of a complex assembly may result

in numerous reported overconstraints. Several reasons are

possible:

• The engineer made an error.

• The engineer intended that those features be over￾constrained.

• The overconstraints are there in a mathematical sense

but not in a practical sense.

Let us consider these cases one at a time.

The engineer made an error. In this case, the error

may be immediately obvious to the engineer, who can

correct it. In a complex assembly, however, this may

not be easy, especially in the presence of mathematical

overconstraints.

The engineer intended that those features be overcon￾strained. The engineer does not need to take any action in

this case.

The overconstraints are mathematical but not practi￾cal. This is the most interesting situation. It arises in par￾ticular in the case of two-sided constraints. These, in turn,

can arise in several ways:

1. The engineer can create a new two-sided feature by

intersecting elementary surface contacts. For exam￾ple, as we saw in Section 4.F.2.C, a pin-slot feature

can be created by intersecting two plane surfaces

with a cylinder. The result will contain a two-sided

constraint, which will generate an overconstraint

report in the analysis. Possibly, the report will be

cluttered with overconstraint reports from such fea￾tures. We avoided this clutter in the examples above

by creating a toolkit feature comprising the allowed

motions of a pin-slot. The way we described the

allowed motions in the twist matrix suppressed the

overconstraint that we know is really inside it. This

permitted us to focus on achieving desired con￾straints and detecting errors.

The engineer may relieve an overconstraint

within a feature with two-sided constraint by pro￾viding a small amount of clearance in the final de￾sign if the resulting location uncertainty, backlash,

vibration, or other consequences are tolerable. If the

consequences are intolerable, the engineer may pro￾vide for a small amount of interference, as long as

the resulting compressive stress is tolerable. In the

first case, any resulting location uncertainty must be

included in the tolerance analysis, while in the other

case the resulting stress must be investigated to en￾sure that it does not cause damage, cracks, fatigue,

and so on.

2. The engineer can combine two elementary features

that collectively create two-sided or other means of

overconstraint. This occurs, for example, in the pin￾hole plus pin-slot. The engineer may relieve such

overconstraints by providing a little clearance. All

the cautions listed above for single features apply

here.

»TU = [Tl;T2]

TU =

0012- 2 0

0016- 2 0

000 0 1 0

100 0 0- 6

»WR = recip(TU)

WR =

00160 0

00001 0

4.F. DESIGN AND ANALYSIS OF ASSEMBLY FEATURES USING SCREW THEORY 97

Alternatively, he can construct a complex new

feature containing the geometries of several ele￾mentary features, but defined so that interfeature

overconstraints among them are suppressed by def￾inition. This approach should be avoided for the

following reason: It is relatively easy to control the

dimensions and variations of a single feature, which

usually involves creating surfaces that are near each

other relative to the size of the feature. Providing the

necessary small clearance to avoid overconstraint

without compromising accuracy is also relatively

easy. However, controlling interfeature dimensions

and variations when these features are far from each

other relative to the size of each individual feature

is much more difficult and prone to errors that can

cause overconstraint. It is better to confront this pos￾sibility by using simple individual features rather

than defining a new complex feature that suppresses

the overconstraint and pretending it will not happen.

Defining the overconstraint away will simply keep

the engineer from finding constraint errors or non￾robust aspects of the design.

It is also up to the engineer to decide which toolkit fea￾tures best represent the problem at hand, or to design an

appropriate feature instead. The overconstraints discussed

in the previous subsection will not arise if toolkit features

9, 18, and 19 are used instead of features 2 and 4. One of

the thought questions at the end of the chapter asks you to

investigate this alternate formulation.

Constraint analysis is useful for finding constraint mis￾takes. Choosing features like 18 or 19 that optimistically

assume away some overconstraint opportunities may re￾sult in an optimistic constraint analysis that fails to identify

a mistake. A possible design technique is to use the most

internally constrained toolkit features like 2 and 4 first,

examine the resulting report, and separate the intended

constraints from the extraneous ones and the mistakes.

Next, eliminate the mistakes. Finally, replace the inter￾nally constrained features with similar ones that have a

little internal clearance or use some of the less internally

constrained features like 18 and 19 and judge whether the

intended final constraint arrangement has been achieved.

4.F.5. Graphical Technique for Conducting

Twist Matrix Analyses

A simple graphical technique can be used to help keep

track of which twist matrices should be collected into

unions and which should be intersected ([Shukla and

Whitney]). The technique is presented here in a series of

increasingly complex examples.

4.F.5.a. A Single Feature with a Single Twist Matrix

Consider the single feature illustrated in Figure 4-33.

To set up the graphical technique, we make a graph

that represents parts and the features that join them. A

simple graph of this type is shown in Figure 4-34. Then

we trace a path or paths in the graph from the moving

part to the fixed part, passing through the necessary fea￾tures and other parts on the way. In this case, part A is the

moving part while part B is the fixed part. The diagram is

shown in Figure 4-35.

The procedure is:

• Identify every path from the moving part to the fixed

part.

• For each path, construct the twist matrix for the mov￾ing part for each feature on the path, using the same

reference coordinate frame (such as one attached to

the fixed part), and form the union of all these twist

matrices.

FIGURE 4-33. Single Feature to

Illustrate Graphical Technique.

FIGURE 4-34. Definition of Terms

for Graph Representation of an

Assembly.

FIGURE 4-35. Diagram for Analyzing the

Feature Situation in Figure 4-33. This

case is trivial because there is only one

path from the moving part (A) to the fixed

part (B).

98 4 CONSTRAINT IN ASSEMBLY

Form the intersection of all the twist unions using

the procedure in 4.E.2.d. A nonempty TR represents

underconstraint in the assembly.

In this case, there is only one path and on this path there

is only one feature, so the procedure is trivial.

4.R5.b. Two Parts Joined by Two Features

Next, consider the feature in Figure 4-31. The analysis di￾agram is shown in Figure 4-36. In this case, there are two

paths from the moving part (A) to the fixed part (B). On

each path we find one feature. Fl is chosen arbitrarily to

be the pin-hole while F2 is chosen to be the pin-slot. The

motion analysis matrix 77? is obtained by intersecting the

twist matrices corresponding to Fl and F2, as illustrated

in Table 4-10.

4.F.5.C. An Assembly with a Moving Part

Finally, consider the 4-bar linkage shown, together with

its analysis diagram, in Figure 4-37. The problem is to

determine the degrees of freedom of link L3, considered

to be the moving part, with respect to LI, considered to

be the fixed part.

FIGURE 4-36. Diagram for Two Parts Joined by Two Fea￾tures. In this case there are two paths.

FIGURE 4-37. Four-Bar Linkage and Its Analysis Dia￾gram. The problem is to determine the degrees of freedom

of link L3 with respect to link L1. We have two paths with

two features on each path. The left path connects L3 to L1

via R2, L2, and R1. The right path connects L3 to L1 via R3,

l_4, and R4. R1 through R4 are rotary joints, each consisting

of a pin-hole feature. Links L1 through L4 are of equal length

and all lie nominally in the X-Y plane. The Z axis points out

of the paper.

The motion analysis goes as follows:

For the left path, we find that there are two features,

R2, and Rl, between L3 and LI. We need to find

how those features generate motion for L3. First, we

erase the right path and all its features. Then we form

a twist allowing R2 to move L3 while Rl is frozen,

and then we form a twist that allows Rl to move L3

while R2 is frozen. Each of these twists is calculated

using the same fixed reference associated with LI,

say one centered on Rl. We then form the union of

these two twists to get a representation of the left path.

For the right path, the process is similar, except that

we erase the left path and consider R3 and R4, and

we again use a coordinate reference centered on Rl.

Finally, we intersect the left path union and the right

path union to find the net motion allowed to L3.

The whole process is shown in Table 4-12.

We have shown that this simple technique permits anal￾ysis of single joints made of several features as well as

analysis of several parts connected by several joints. If

these joints are made of several features, then the user

should analyze each joint separately, finding the net twist

allowed by all its constituent features by intersecting their

individual twists, and then combine the joints using the

method shown here.

This method can be used on any assembly or linkage as

long as it does not contain cross-coupling. We saw in Fig￾ure 4-6 a mechanism that has cross-coupling. The method

above will not be able to find the motions of the top hori￾zontal link if the bottom horizontal link is fixed. However,

the motion of the top link can be found if the left or right

vertical link is considered fixed, and the answer can be

rewritten to conform to the situation where the bottom

link is fixed.

4.F.6. Graphical Technique for Conducting

Constraint Analyses22

Systematic constraint analysis begins the same way

that motion analysis does, by drawing the graph and

enumerating the paths. However, constraint analysis

is considerably more tedious because the intersection

method has to be applied to all combinations of features

2 See [Shukla and Whitney 2001b].

4.F. DESIGN AND ANALYSIS OF ASSEMBLY FEATURES USING SCREW THEORY 99

TABLE 4-12. Motion Analysis of Four-Bar Linkage

Motion analysis:

The first step is to analyze the left path to find L3's motions as if it is connected to the fixed link only by L2. This is done by considering the

motion that each remaining feature could give L3 while considering other remaining motions frozen:

Rotate L3 about R2 in Rl coordinates with Rl frozen:

»tl = [001100 ]

tl =

00110 0

Rotate L3 about Rl in Rl coordinates with R2 frozen:

»t2 =[001000 ]

t2 =

00100 0

Form the union of these to get the possible motions of L3 provided by the left path:

»tlp = [tl;t2]

tip =

00110 0

00100 0

Now analyze the right path to find the motions of L3 as if it is connected to the fixed link only by L4. Again, this is done by considering the

motion that each remaining feature could give L3 while considering the other remaining motions frozen:

Rotate L3 about R3:

»t3 =[0011-10 ]

t3 =

0011-1 0

Rotate L3 about R4:

»t4 =[0010-10 ]

t4 =

0010-1 0

Form the union of these to get the motions of L3 provided by the right path:

»trp = [t3;t4 ]

trp =

0011-1 0

0010-1 0

Form the intersection of tip and trp to see what the net allowed motion is

»wlp = recip(tlp)

wlp =

0 1.000 0 000 0

0 0 1.000 0 0 0 0

0 0 0 1.000 0 0 0

000 0 1.0000 -0.000 0

»wrp = recip(trp)

wrp=

0 1.0000 00 0 1.0000

0 0 1.000 0 0 0 0

0 0 0 1.0000 0 0

000 0 1.000 0 0.000 0

(continued)

100 4 CONSTRAINT IN ASSEMBLY

TABLE 4-12. (Continued)

»WU = [wlp;wrp]

WU =

0 1.000 0 000 0

0 0 1.0000 00 0

0 0 0 1.0000 0 0

0 0 0 0 1.0000 -0.000 0

0 1.0000 00 0 1.0000

0 0 1.0000 00 0

0 0 0 1.0000 0 0

0 0 0 0 1.0000 0.000 0

»TR = recip(WU)

TR = -0.0000 -0.000 0 -0.0000 1.0000 0.000 0 0.000 0

This says that L3 is permitted to move in the X direction in Rl coordinates. This is the answer we expect based on intuition.

The reciprocal of TR, WU, shows what forces and moments can be resisted by the linkage. These are Fy, Fz

, Mx, My, and Mz. But we

do not know if any of these is overconstrained. This question is resolved in Table 4-13.

if we want to identify every feature that contributes to

overconstraint.

In brief, the process is as follows:

• Choose any path and check to see if its twists, formed

into a union, overconstrain the parts.

• Choose a second path and intersect its twist union

with the first path's twist union to find a combined

feature that allows only those motions common to

those paths. Check to see if this combination over￾constrains the parts.

• Continue in this way, adding one path's twist union

at a time until all have finally been combined and

checked for overconstraint.

To see how this works, consider a part A joined to an￾other part B by four features Fl, F2, F3, and F4, as shown

in Figure 4-38.

On each path there is one twist, so we have four twists,

one for each path: T\, T^, ?3, and T4. By using the rela￾tionship Wi = recip(Tf), we first find the corresponding

FIGURE 4-38. Path Diagram

for Two Parts Joined by Four

Features.

wrenches: W\, W2, W3, and W4. We then systematically

form (the order is arbitrary) T\2,T\23, and ^234, and

Wi2, Wi23, and W\234, as follows:

Tiz = 0(7,, r2) = recip(U(W}

, W2))

= net motions allowed by path 1 and path 2

^123 = n(7i, T2, 7s) = ncip(\J(Wi, W2, Wj))

= net motions allowed by path 1, path 2,

= and path 3

r,234 = n(r,, r2, r3, r4) = rmp(u(w,, w2, w3, w4))

= net motions allowed by all four paths

w}2 = n(Wi, w2) = mcip(u(Ti, r2))

= overconstraint provided by path 1 and path 2

Wm = n(W12, W3) = recip(U(Tl2, r3))

= overconstraint provided by path 1, path 2,

and path 3

W,234 = n(W123, W4) = recip((J(Tm, T4))

= overconstraint provided by all four paths

Note that

W1234 / H(Wi, W2, W3, W4) = mcip(U(Ti, T2, T3, T4))

= the force(s) or moment(s) that can be supported

by all four paths at once