# 2 Cumulative Effect of Tol

## Full text

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### 2 CUMULATIVE EFFECT OF TOLERANCES

All piece parts are eventually used for assembly. The probable lengths of the resulting combinations are the one to be concentrated upon. In large volume

production the problem is complicated by the fact the lengths of the assembly members are unknown to the designer. They may only know that the lengths of individual parts lie somewhere within the specified tolerances.

Variations in a dimension occur despite of keeping production conditions as constant as possible. The manner in which these variations are distributed within the tolerance therefore becomes important in determining the length of the resulting assembly. In addition, the drawings are done at one location, but the fabrication and assembly take place at far-distant places under conditions unknown to the designer. Nevertheless, it becomes necessary that the designer makes the best estimate possible so as to find out the probability of assembly variations be. The contacting surfaces should be flat and free of burrs and other imperfections. Thermal effects must also be taken into account.

Types of Tolerance Specification: 1. Unilateral tolerance 2. Bilateral tolerance

• Equal

• Unequal

Unspecified tolerances for the present consideration could be employed as ±0.3. The open tolerance for various dimensions are illustrated in the Table 2.2

For ease of calculation, bilateral tolerancing should be used. Unilateral tolerance can be changed temporarily changed to the bilateral form for the purpose of making calculations. Thus a dimension 4.000,+0.000,-0.008 mm would become

3.996±0.004mm. Tolerance will be represented as u and will be used to represent the permissible variation each way from the mean. The other nomenclatures foregoing u

are as follows:

A Area under distribution curve

k Class width

f(x) Function of x

N Number of pieces

u Engineering tolerance

un Natural tolerance or 3σ

ua Natural tolerance for assembly

x Dimension

x Weighted average of dimension x

z Number of standard deviations between specified dimension x and mean

xof normal curve

σ Standard deviation or root-mean-square error

σa Standard deviation for assembly

σt standard deviation for truncated normal curve

σ2 Variance, Square of standard deviation

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∞ Infinity

xa Mean value

### 2.1 Arithmetic or Sure-Fit Law

The length of a simple part is shown in Fig 1. A mean dimension x is given with

the tolerance ´u´ expressed bilaterally. The mean value xa for an assembly is the sum

of means of separate part as given by the Eq 2.1.

xa= x1 + x2 + …...= ∑x ………(2.1)

The simplest rule for estimating the tolerance ua for the assembly is merely to

add the tolerances ´u´ of the individual parts. The results are given in equation 2.2.

ua= u1 + u2 + …………..= ∑u ……….(2.2)

Fig 1 Length of simple part expressed with bilateral tolerance

Example1:

The lengths for the three parts that are placed in end-to-end assembly are shown in

Figure 2 determine the probable tolerance ua for the assembly by the arithmetic rule.

Fig 53 Piece parts

Solution:

By Eq. (2.2),

ua= u=0.050+0.075+0.150=0.275 mm

Equation (2.2) usually gives a value too large for ua. In fact, all the parts could

simultaneously be at their high limit or all simultaneously at their low limit. Although physically possible, such an event would be of very rare occurrence. In general, some parts will be longer than their mean and sum will be shorter, so that after assembly the

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The inverse problem sometime occurs. Mere assembly tolerance ua would be

specified with piece part tolerances u1, u2 …to be determined. The arithmetic rule may

impose a severe restriction of small tolerances on the individual parts.

### 2.1.1 Assembly When Some Parts Double Back on Others

Assemblies are sometimes arranged where some of the parts double back on the others. It can be easily shown that the mean of the resulting assembly is equal to the algebraic sum of the means of the component parts and the tolerance

### on the others. Assembly dimension is 35.56±0.011 .

on the assembly is merely the sum of the tolerances on the components all taken positively when the arithmetic rule is used.

Example 2:

Suppose that the three parts of Fig. 54(a) are arranged as in Fig.54(b) and (c). Find the dimensions of the resulting assembly.

Solution:

In Fig. 54 (b) the parts have lengths that give the largest assembly dimension and in Fig.54 (c) the lengths give the smallest assembly dimension. It is seen that the assembly dimension, 35.56 mm, is equal to the algebraic sum of the means of the

parts. The tolerance ua of 0.011 mm. is the sum of the tolerances on the three parts

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Example 3:

Arrange the three parts of Fig.55 similarly to Fig. 54 Find the value of tolerance u2 if the

tolerance of the resulting opening by the arithmetic rule is to be +0.4572 mm. Make drawings to demonstrate the validity of the solution.

Solution:

The calculations are carried out in the figure. The maximum opening occurs when parts 1 and 2 have their maximum lengths and part 3 has its minimum length.

The figure indicates that u2 is equal to 0.010.

u

u

u

Fig. 55 Determination of component tolerance.

The minimum opening occurs when parts 1 and 2 have their minimum lengths and part

3 has its maximum length. The figure shows that u2 is equal to 0.254 mm.

### Distributed

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f(x) = 2 2 1 2 1       − − σµ

### π σ

x e

The variations in the length of the assembly dimension form a distribution that

has a variance σ 12 whose value is determined

by Eq. (2.3)

## )

### ( )

N x f x x11 2 1 1

The distribution for another lot to be used in the assembly will have a variance

σ 22 . The third lot will have the variance σ 32, and so on.

Let a part be selected at random from each lot and placed in end-to-end

assembly. Assembly lengths will vary depending on the lengths of the parts of which it is composed. A sufficient number of assemblies will form a distribution for the

assembly lengths and this distribution will have a variance σ a2.

σ a2 = σ 12 + σ 22+…=Σ σ 2……….(2.3)

In this equation σ 12 , σ 22 refer to the variances of the piece part distributions regardless

of their shape.

Variance σ a2 can also be calculated from the assembly histogram in the usual

way. However, variance σ a2 as calculated by Eq. (2.3) tells nothing about the shape of

the distribution curve for the assemblies. It can be demonstrated, however, that the shape of the distribution curve rapidly approaches the normal form as the number of parts in the assembly increases. This statement is valid regardless of the shape of the distribution curves for the errors of the individual parts. It can be shown, however, that if the piece-part error distribution does not depart too much from the normal form that only a few parts are needed to give the normal form to the assembly curve. Thus only the two parts are needed for truncated normal curves such as those in Fig.2.8(b), (c) and (d). However, for distribution curves such as Fig.2.8 (e) and (f), at least three parts

are needed.The natural tolerance un for the normal curve could be taken as three

standard deviations σ of the curve, or

un=3σ ………..……….(2.4)

Assume that the assembly has a sufficient number of parts that its distribution curve

will have the normal form. Its natural tolerance ua will then have the length

ua=3σ a ………..(2.5)

Example 4:

Let the tolerance in Fig 53 natural tolerances with normally distributed errors centered

on the mean of the dimension. Find the tolerance ua for the assembly.

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For part 1, 3 002 . 0 1 = σ For part 2, 0.001 3 003 . 0 2 = = σ For part 3, 0.002 3 006 . 0 3 = = σ By Eq 2.3 2 2 2 2 3 2 2 2 1 2 0.001 0.002 3 002 . 0 + +       = + + =σ σ σ σa 9 000049 . 0 = 3 007 . 0 = a σ By Eq 2.5 ua = 3σa = 0.007

It is noted that 0.007 is a marked reduction from 0.011 as found by arithmetic rule in Example 1. In fact, unless there is full assurance that the errors are normally

distributed and centered on the mean dimension with 3σ of the errors equal to the engineering tolerances u , there is a grave possibility that a solution will give too small a value for ua.

### 2.32.4 Assembly Length Determined By Tolerances

Let the error distributions for each of the parts of an assembly have the normal form centered on the mean of the dimension. Let Eq. (2.4) and (2.5) be substituted into Eq(2.3). ... 9 9 9 2 2 2 1 2 + + = u u ua or ua2 = u1 2 + u22 + …. = ∑ u2 ……….. (2.6)

It must be carefully noted that this equation is valid only when all piece part errors are distributed normally and centered on the mean of the dimension with 3σ equal to u. Example 5:

Let the errors for the parts of Fig. 53 be normally distributed and centered on the mean

of the dimension. Find the natural tolerance ua for the assembly by Eq (2.6).

Solution :

By Eq.(2.6),

ua2 = u1 2 + u22 + u32 = 0.050 2 + 0.075 2 + 0.150 2 = 0.031

ua = 0.176 mm

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σ = 0.0434 σ

σ σ σ

Care must be exercised in distinguishing between the engineering tolerance u as

placed on the drawing by the designer and the natural tolerance un as determined by

the accuracy of the production equipment.

### Tolerances

Sometimes the piece-part errors overrun the given tolerances as shown in Fig.2.5. In large-volume production the parts may not be individually checked, If the entire production is assembled , the variation in the length of the assembly will thereby be increased.

For the normal distribution the proportion of parts that overrun the tolerances can be represented by the shaded areas in the tails of the curve. These areas can be found by the use of the Table 1.1, as previously shown. Coordinate z is required , which represents the number of standard deviations σ in the space between some chosen

value of error x and mean x of the normal curve. Coordinate z is given by the

following Equation:

σ

x x

z = − ……….(2.7)

When x is taken at the edge of the engineering tolerance, x- x becomes equal to u and

Eq.(2.7) becomes

σ u

z = ………..(2.8)

The σ in these equations refers to the σ of the full normal curve. When applied to the normal curve for an assembly, Eq.(2.7) becomes

a a a x x z σ − = ……….(2.9) where

σa isthe standard deviation of the assembly curve.

Example 6:

In Fig. 56 suppose that the parts have normally distributed errors, but 4% of them are smaller than the specified tolerance and 4% are larger. If all of them are assembled ,

find the natural tolerance ua for the resulting assembly.

Solution:

Since the entire production is assembled, the standard deviation to be used are those of the full normal curve. By Table1.1, for A=0.04 and z= -1.75, using Eq.(2-7) for the part on left,

044 . 0 75 . 1 0762 . 0 1 1 1 1 = − − = = − = z u z x x σ

Eq (2.7) for the part on right,

070 . 0 75 . 1 1219 . 0 21 2 2 2 = − = = − = z u z x x σ σa 2 = ∑ σ2 = 5 * 0.0442 + 2 * 0.0702 = 0.019

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σa = 0.140 mm

ua = 3 σa = 3 * 0.140 = 0.420 mm.

Fig 56 Assembly curve for piece parts that overrun their tolerance

### 2.32.6 Truncated Normal Curve

A symmetrically truncated normal curve results when equal amounts of the curve are cut off at each and. Sometimes the distribution curve for a set of errors resembles a

truncated curve. For making calculations, its standard deviation σ t will be required. It

is obvious that σ t for the truncated portion is smaller thanσ for the full normal curve

before truncation.

σ σ

σ σ

Fig 57 Truncated Normal Curve. For symmetrical truncation:

2

t

### σ

z is a positive value

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Values of the ratio

### σ

t can be found from the graphs of Fig.59. such values depend z

or u/σ as determined by Eq

σµ

=

z .The empirical equations shown on the figure can be

used if desired. The curves of Fig 57 do not apply to nonsymmetrical truncation.

The lower curve in Fig 57 is convenient , as it enables one to pass directly from z or u/σ to u/σ t.

Example 7:

Consider the curve of Fig 59, where u is 0.102 and σ is 0.064. Find the standard

deviation σ t for the retained centralportion of the curve.

σ σ

σ

Fig 58. Areas under the Normal curve when spread of production exceeds the engineering tolerance.

Solution:

Let x be taken at the edge of the tolerance zone. Then ,

σµ = z = 0.102/0.064=1.59 By Fig 59,

### σ

t = 0.776 σ t = 0.776 σ = 0.776*0.064=0.050 mm.

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s = u u u u u σ σ σ σ σ σ σ σ σ σ σ σ σ = σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ = 0.880 σ σ σ σ σ σ σ

σ

σ

σ

### σ

σ

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Example 8:

A normal curve is given a symmetrical truncation equal to σ of the full normal curve.

Find the value of σ t for the truncated curve.

Solution:

For σ truncation, u=2σ or σ =u/2.

σµ = z = σσ 2 =2 By Fig.2.8

### σ

t = 0.88 σ t = 0.88 σ = 0.88 * 2 u = 0.44 u

This result can also be obtained from the lower curve in Fig.4. For z = 2, t u σ =2.27 σ t = 27 . 2 u = 0.44u

Normal curves with truncations of 0.5σ , σ ,1.5σ and 2σ are shown in Fig.59.

Thus if σ t is equal to 0.44 u, the normal curve has been given a symmetrical

truncation equal to σ . Such a rule of thumb applies strictly only to the normal curve

but may be approximately true when the frequency is highest near the mean and becomes less on either side.

### 2.32.7 Assemblies Made from Truncated Normal Curve

Fig 59 Standard deviations for various distributions of errors. Ratio

### σ

t Obtained.

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Fig 60 Individual parts and their assemblies

Let the piece-part as shown in Fig.60 errors be approximated by a truncated normal curve. Such distributions can occur when normal curve distributions overrun the specified tolerance and the parts are sorted by go/no go gauges and only the acceptable parts are assembled.

In working examples it will be necessary to find σt, the standard deviation of the

acceptable central portion of the error distribution curves for the piece parts. These are then substituted into Eq.(2.3).

2 2 2 1 2 t t a σ σ σ = + 88 . 0 1 2 1 =

### σ

t (for 1σ truncation) σ t1=0.44u1 σ t2=0.44u2 2 2 2 1 t t a σ σ σ = + = 0.44 2 2 2 1 u u + or 2 2 2 1 32 . 1 u u ua = + Example 9:

Suppose that only the parts falling within the engineering tolerances in Fig 56 are

assembled. Find the tolerance ua for the assembly.

Solution :

By Table 1.1 , for A= 0.04 and z = 1.75 , by Fig 55,

### σ

t = 0.82

σt = 0.82 σ

From Example 6,

σ1 = 0.043, σ2 = 0.070 for full normal curves.

Then

σt1 = 0.82 * 0.043 , σt2 = 0.82 * 0.070

By Eq.(2.3),

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= 0.013 mm σa = 0.113 mm

The assembly distribution can be assumed to be normal , so that ua = 3 σa = 3 * 0.113 = 0.339 mm.

Example 10:

Let the error distribution in Fig 54 be truncated normal curves of the form shown in

Fig 59 c. Find the natural tolerance ua of the resulting assembly distribution.

Solution. In Fig 59(c) , σt = 0.44 u. For part 1, σt1 = 0.44 * 0.051 =0.022 For part 2, σt2 = 0.44 * 0.076 =0.034 For part 3, σt3 = 0.44 * 0.152 =0.067 By Eq.(2.3), σa 2 = ∑ σt2 = 0.0222+0.0342+0.0672 σ a = 0.078 ua = 3 σa = 3 * 0.078 = 0.235 mm.

This result is intermediate between that of Example 1 using the arithmetic law and

Example 4 when the errors were assumed to be normally distributed with 3σequal to u.

Example 11:

Let the errors in part1,Fig 54 have a distribution like that of Fig.59 (c) .For part 2 the distribution is that of Fig.59(d), and for part 3, the errors have the distribution of

Fig. 59(e).Find the natural tolerance ua for end- to -end assembly.

Solution:

By Fig. 59,

For part 1, σt1 = 0.440u = 0.440* 0.051 =0.022mm

For part 2, σt2 = 0.496u =0.496* 0.076 =0.038mm

For part 3, σt3 = 0.540u =0.540* 0.152 =0.082mm

By Eq(2.3)

2

2

### )

2 2 = 0.0224 + 0.0378 + 0.0823 a σ = 0.009 mm a =0.093 σ

The assembly curve can be assumed to have the normal form. By Eq(2.5)

ua=3σa = 3*0.093=0.288mm.

This example indicates that when the piece part distributions resemble normal curves with the greater amount of truncation, the excepted tolerance on the assembly is increased.

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The difference between the expected tolerance for an assembly when calculated by the arithmetic method and then by statistical method becomes greater as the number of parts in the assembly is increased.

Example 12:

Same as Example 11 except there are three pieces for each of the parts of Fig 54. Find the excepted tolerance for the end-end assembly.

Solution: From Example 11, σ t1 = 0.022 σ t2 = 0.038 σ t3 = 0.082 By Eq(2.3), σ a 2 = Σ σ t 2 = 3*0.0222 + 3*0.0382 +3*0.0822 = 0.026 σ a = 0.162 mm ua = 0.485 mm

By the arithmetic rule,

ua = 3(0.051 + 0.076 + 0.152) = 0.838 mm

Hence the variation between the statistical method and the arithmetic rule becomes greater when the number of parts is increased and the arithmetic rule may indicate too large a tolerance for the assembly.

The inverse problem that of finding the piece-part tolerances for a specified assembly tolerance, can be handled by the foregoing equations. The following example will aid in visualizing the large difference between arithmetic tolerances and those determined by probability methods.

Example 13:

It is desired that an end-to-end assembly of four equal parts have a tolerance ua equal

to 0.203mm . Find the piece-part tolerances that should be specified by the arithmetic rule, the normal law with a truncation of one standard deviation.

Solution:

For the arithmetic rule, by Eq.(2.1)

ua = Σ u = 4u = 0.203 mm

u = 1/4*0.203 = 0.051 mm For the normal distribution , by Eq.(2.7)

ua 2 = Σ un 2 = 4un 2 =0.0232 = 0.041

un 2 =1/4*0.041= 0.010

un = 0.102 mm

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σ a 2 = Σ σ 2 = 4*σ t 2 σ a= 2σ t = 2* 0.44u = 0.88u ua = 3*σ a = 3*0.88u = 0.203 mm 0.203 u = --- = 0.077 mm 3*0.88

The results in Example 13 indicate that the arithmetic rule demands the smallest tolerances for the piece parts. The tolerances required by assuming normal

distributions may be too liberal. The truncated normal assumptions, for

σ t = 0.44u ,gives results about halfway between.

In an assembly, the dimensions that are more expensive to maintain can be given larger tolerances, while the dimensions that are inexpensive can be given smaller tolerances. The following example illustrates how this can be done using estimates by the designer.

Example 14:

Suppose in an assembly of three parts that ua must be maintained at 0.229 .Let u1 be

the least expensive tolerance and let the designer feel that u2 can be 1.5 u1 and that u3

can be 2 u1.Ifthe piece part errors are assumed to have a truncated normal distribution

with

σ t=0.44u,find suitable values for the tolerances.

Solution: σ t=0.44u σ a= ua/3 = 0.229/3 = 0.076 For part1, σ t1 = 0.44u1 For part1, σ t2 = 0.44u2 = 0.44 * 1.5 u1 = 0.66 u1 For part3 σ t3 = 0.44u3= 0.44 * 2 u1 = 0.88 u1 σ a 2 = Σ σ t 2 = u1 2 (0.442+0.662+0.882)=1.404 u1 2 σ a = 1.184 u1 = 1/3* 0.229=0.076 mm u1 = 0.076/1.184 = 0.064 mm u2 = 1.5 u1 = 0.097 mm u3 = 2u1 = 0.129 mm Check: σ t1 = 0.44u1 = 0.028 σ t1 2 = 0.0008 σ t2 = 0.44u2 = 0.043 σ t2 2 = 0.0018 σ t3 = 0.44u3 = 0.057 σ t3 2 = 0.0032 --- σ a 2 = Σ σ t 2 = 0.0058

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σ a = 0.076 ua = 3σ a = 0.229

Example 15:

Seven similar parts with normal distributions for the errors are placed in end-to-end assembly. The normal curves overrun the engineering tolerance u of 0.043 .Parts are

gauged and only good parts are assembled. The Natural tolerance ua of the assembly

must be 0.152mm.Find the proportion of oversize and undersize parts that are discarded.

Solution:

By Eq.(2.5)

σ a = ua/3=0.152/3=0.051

For the assembly, by Eq.(2.3)

σ a 2 = Σ σ 2 =7σ t 2 OR σ t 2 = σ a 2 /7

Then

σ t = σ a / 7 = 0.051/2.646=0.019 For piece parts

With only u and σ t are known ,let the following proportion be formed:

u/σ t = 0.043/0.019 = 2.250

By the lower curve, Fig 4, for u/σ t of 2.250 , then

Z = 1.97 By Table 2.1, 97 . 1 − ∞ − A = 0.024 or 2.44%

Thus 2.44% of the parts will be undersize and 2.44% will be lower size.

### :

1. Find the resulting dimension in both cases.

1. 2.

Fig 2.10 Piece parts

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σ σ σ σ

Fig 61 Piece parts

The resulting dimension x, will have a tolerance of addition of individual tolerances. (i.e.) parts: U1+u1,U2+u2,U3+u3,…..,Un+un

Case 1:

By Sure Fit Law: Assembly: Ua = U1±U2 = 30+40 = 70 mm ua = u1+u2 = 0.3+0.4 = 0.7 mm X = 70±0.7 mm Case 2:

By Sure Fit Law: Assembly: Ua = U1±U2 = 70-30 = 40 mm ua = u1+u2 = 0.3+0.5 = 0.8 mm X = 70±0.8 mm 2. Find the resultant dimension

Fig 62 Normal Distribution-Piece parts By Normal law: µa = µ1 + µ2 = 30 + 40 = 70 mm σ a = 22 2 1 σ σ + ua = u12+u22 =

2

### )

2 4 . 0 3 . 0 + = 0.5 Resulting Dimension is 70±0.5mm

3. Three parts are to be placed in end-to-end assembly. The tolerances are 0.2286 mm, 0.3048 mm and 0.9144 mm. Form an estimate for the variations in the lengths of the assemblies by using the arithmetic and normal laws.

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Solution:

Arithmetic or Sure-fit law:

ua = 0.229 + 0.305 + 0.914

= 1.448 mm

Normal law:

ua =

0.229

2 +

0.305

2 +

0.914

### )

2

= 0.991 mm

4. Find the resultant dimension X in Fig 63

Fig 63 Dimension of Parts Solution:

Parts: U1±u1 ; U2±u2

Assembly:Ua±ua

Sure Fit Law : ua=u1+u2

Normal Law : ua= u12+u22

1σ truncated normal law ua= 1.32* 22

2 1 u

u + Mean dimension of 30.2 and 29.9 = 30.05±0.15 Mean dimension of 20.7 and 19.9 = 20.3±0.4 The resulting length X will be :

1.By sure fit law Ua=U1+U2 = 50.35 X=50.35±0.55

ua=u1+u2 = 0.55

2.By Normal law Ua=U1+U2 = 50.35

ua= u12 +u22 =

0.15

### ) ( )

2 + 0.4 2 =0.427 X= 50.35±0.427

3.By 1σ truncated Normal law

Ua=U1+U2 = 50.35

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=1.32

2

### )

2 3 / 4 . 0 * 2 3 / 15 . 0 * 2 + X= 50.35±0.376 =0.376

### 5.

Analyze the interference of the interference fit 20H7/r6 by the above methods. If the

piece part

errors are having 1σ truncated normal distribution, find the percentage of

assemblies that have

more than 20 microns interference.

### 20.028

Fig.64 Normal Distribution

σ a 2 2 2 1 t t σ σ + = = 0.44 u12+u22 = 0.44 (0.007 )2 +(0.004)2 = 0.004 Area above 20=0.5+ 24 20 A Solution :

By sure fit law:

Mean (interference) =µ1−µ2 or u1−u2

= 27.5µ

Tolerance on mean interference= u1+u2

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i.e., Mean interference = 27.5µ Minimum interference = 7µ Maximum interference = 48µ By Normal law: Mean interference=u1−u2=27.5µ

Tolerance on mean interference= 2

2 2 1 u u + = (0.011)2 +(0.010 )2 = 0.015 = 14.5µ Minimum interference=27.5-14.5=13 µ Maximum interference=27.5+14.5=42µ

Fig 65 (a)Normal Distribution Fig 65 (b) 1σ Truncated Normal Distribution

By truncated Normal law: Refer 65 (b) µa =µ1−µ2=27.5 microns. σ a 2 2 2 1 t t σ σ + = = 0.44 2 2 2 1 u u + = 0.44 (0.0070 )2+(0.0043 )2 = 0.00361 = 3.61µ µ σ µ σ µ σ µ σ 67 . 16 61 . 3 * 3 5 . 27 3 ; 007 . 0 010 . 0 * 3 / 2 2 33 . 38 61 . 3 * 3 5 . 27 3 ; 007 . 0 011 . 0 * 3 / 2 2 2 1 = − = − = = = = + = + = = = a a a a u u

Therefore Maximum Interference = 38.33µ

Minimum Interference = 16.67µ

7.The clearance permitted between stator and rotor in a motor assembly is 0.120 to 0.280.assuming hole basis system, determine the design sizes for both the stator

bore

and rotor to meet the above functional requirements. The nominal size of the rotor may

be taken as φ 200.Distribute the machining tolerance equally between the stator and

rotor. Solution:

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Maximum clearance=0.280mm Minimum clearance=0.120mm Mean clearance = (280-120)/2+0.120 200µ 6σ a=0.280-0.120 =160µ σ a=160/6=26.6µ =0.003mm Hole based system is followed.

Therefore,

Fundamental deviation of hole is zero i.e., Min hole=basic size of hole

Fig 66 Stator and Rotor in a Motor Assembly

σ a= 22 2 1 σ

σ +

If we distribute the machining tolerances equally between stator and rotor u1 =u2 2 1 σ σ = Therefore σ a= 2σ σ =σ a/ 2 =0.027/ 2=0.019 and 3σ = 0.056 6σ = 0.113mm

8. Two parts have normally distributed errors centered on their tolerances u of 0.076 mm, 0.122mm as produced, 5% of the parts are undersize and 5% are oversize. Find the natural tolerance for an end-to-end assembly of the two parts if the entire production is assembled.

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A = 0.05 Z = -1.64 Left

## )

z x x − = 1 1

### σ

= -0.076/-1.64 = 0.047 mm

## )

z x x − = 2 2

### σ

= -0.122/-1.64 = 0.074 mm σ a2 = Σ σ 2 = 2 * (0.047)2 + 2 * (0.074)2 = 0.015 mm σ a = 0.124 mm ua = 3σ a = 3* 0.124 = 0.372 mm

9.The assembly tolerance ua for the three parts must be held to 0.508 mm.

the designer decided that the tolerance on the second part should be twice as great as for the first part, and for the third part should be three

times as great as for the first. Distributions are such that σ t = 0.440u .

Find the tolerance that should be specified for each part. Solution: ua = 0.508 mm u2 = 2 u1 u3 = 3 u1 σ t = 0.44u σ t1 =

a

### u

= 3 508 . 0 = 0.169 mm σ t2 = 0.44 u2 = 0.88 u1 σ t3 = 0.44 u3 = 1.32u1 σ a2 = σ t2 =u12 (0.442 +0.882 +1.322) = 2.711 u12 σ a = 0.169 mm

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u1 = 0.103 mm

u2 = 0.206 mm

u3 = 0.309 mm

10.Three parts in an end-to-end assembly have a natural tolerance of ua for the

assembly of 0.419 mm. The entire production has to be assembled. If piece part errors are normally distributed, find the proportion of production that fulfills the engineering tolerance of u = 0.152 mm. Solution: ua =0.419 mm u = 0.152 mm σ a = ua/3 = 0.419/3 = 0.139 mm σ a2= Σ σ t2= 3 σ t2 σ t2 = 3 2 a

### σ

σ t = 0.081 mm t u σ = 0.152/0.081 = 1.889 From Table 2.1 z = 1.33 For z = 1.33 z A−∞ = 0.092 = 91.8%

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Fig.67 Drill Bush

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Solution:

Drill bush to datum[A] gives:

From bush I.D(G1) to O.D(H1) = 0.0±0.01(concentricity)

From H1 to E1 = 0.0±0.00(assume perfect fit)

From E1 to E2 = 38.0±0.01

From E2 to H2 = 0.0±0.00(assume perfect fit)

From bush O.D (H2) to I.D(G2) = 0.0±0.01(concentricity)

---Net Dimension = 38.0±0.03(acceptable)

The desired location of hole F2 as measured from datum [A] is 23.5±0.1 mm and tracing the link from the drill bush(G2) to datum[A] gives:

From bush I.D(G2) to O.D(H2) = 0.0±0.01(concentricity)

From H2 to E2 = 0.0±0.00(assume perfect fit)

From bush at hole E2 to column D1 = -38.0±0.01–21.0±0.01+20.0±0.01

= -39.0 ±0.03

From column D1 to dowel pin = -10.0±0.00

From dowel pin B to datum [A] = -12.5±0.05+85.0±0.05

= 72.5±0.10

---Net Dimension 23.5±0.14(violated)

Considering tolerance stacks in the vertical orientation(downwards as positive),it can be seen that the desired location of hole F2 as measured from datum [B] is 23.5±0.1mm and tracing the link from the drill bush G to datum [B] gives:

From bush I.D(G) to O.D(H) = 0.0±0.01(concentricity)

From H to E = 0.0±0.00(assume perfect fit)

From bush at hole E to column D = -45.0±0.01 From column D1 to dowel pin B = 58.0±0.00

From dowel pin B to datum [B] = 10.5±0.2-47.0±0.05 = -36.5±0.25

---Net Dimension = -23.5±0.27(violated)

Problems:

### 1.

Five similar parts are placed in end-to-end assembly. Parts overrun the engineering

tolerance of 0.127 mm, with 2% of the parts being oversize and 2% undersize. Only

parts within tolerance are assembled. Find the natural tolerance ua for the assembly.

[ua =0.371mm]

### 2.

The engineering tolerances for the two parts in end-to-end assembly are u1=0.076

mm and u2=0.102 mm. Drawings are marked “95% of parts to fulfill given

tolerances”. The entire production is to be assembled. Assume that the errors are normally distributed and centered on the tolerances. Find the natural tolerance for

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the resulting assembly.

[ua = 0.193 mm]

### 3.

The natural tolerance ua for an assembly is to be 0.305. There are four parts with

similar tolerances. Drawings are marked “98% of parts to fulfill given tolerances”. Assume that the part errors are normally distributed and centered on the

tolerances. Find the tolerance u that should appear on the drawing. The entire production is to be assembled.

[u = 0.117 mm]

### 4.

Seven parts in end-to-end assembly are to have a natural tolerance ua for the

assembly of 0.154 mm. Part errors are normally distributed but overrun the tolerances, with 0.62% of them too small and 0.62% too large. Only the good

parts are assembled. Find the engineering tolerance u that should appear on

the drawing.

[u = 0.051 mm]

### 5.

Seven parts in end-to-end assembly have truncated normal curves for the error

distributions, with 1% of the parts discarded as too large and 1% discarded as too small. Engineering tolerance u = 0.051 mm. Only the good parts are assembled.

Find the natural tolerance ua for the assembly.

[ua = 0.163 mm]

### 6.

Six parts in an end-to-end assembly are dimensioned u = 0.048 mm. Errors have

truncated normal curve distributions. The assembly tolerance ua = 0.152 mm. Find

the percent of over length and under length parts that are discarded.

[1.79%] 7.A case study illustrating cumulative effect of tolerances

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Table 2.1 Proportion of total area between -

and z

### 0.0985

Z z A−∞ Z Az∞ Z Az∞ Z Az

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### 0.0559

Table 2.1 (continued)

### 0.0019

Z Az ∞ − Z Az∞ Z Az∞ Z Az

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### 1

Table 2.2 Free Dimensions Tolerance

OVER UPTO TOLERANCE

0 3 0.1 3 10 0.15 10 18 0.20 18 30 0.25 30 50 0.30 50 80 0.35 80 120 0.40 120 250 0.50 250 315 0.60 315 500 0.70 References:

Merhyle Spotts,F.,”Dimensioning and Tolerancing for Quantity Production”,Prentice-Hall,Inc.,Englewood Cliffs,1983.

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