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2D Analyzer:CHAPTER 3: TOLERANCE ALLOCATION | Manuals |

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In the previous chapter it was shown that analytical modeling of assemblies
provides a quantitative basis for the evaluation of design variations and
specification of tolerances. An important distinction in tolerance specification
is that engineers are more commonly faced with the problem of tolerance
*allocation* rather than tolerance *analysis*.

The difference between these two problems is illustrated in Figure 3.1. In tolerance analysis the component tolerances are all known or specified and the resulting assembly variation is calculated. In tolerance allocation, on the other hand, the assembly tolerance is known from design requirements, whereas the component tolerances to meet these requirements are unknown. The available assembly tolerance must be distributed or allocated among the components in some rational way. The influence of the tolerance accumulation model and the allocation rule chosen by the designer on the resulting tolerance allocation will be demonstrated.

**Figure 3.1: Tolerance Analysis vs.
Tolerance Allocation. **

Another difference in the two problems is the *yield or acceptance
fraction* of the assembly process. The assembly yield is the quality
level. It is the percent of assemblies which meet the engineering tolerance
requirements. It may be expressed as the fraction of acceptable assemblies
or as the fraction of rejects (DPU). The rejects may also be expressed in
parts-per-million (PPM), that is, the rejects per million assemblies .

In tolerance analysis the assembly yield is unknown. It is calculated by summing the component tolerances to determine the assembly variation, then applying the upper and lower Specification limits to the calculated assembly distribution. In tolerance allocation, on the other hand, the assembly yield is specified as a design requirement. The component tolerances must then be set to assure that the resulting assembly yield meets the specification.

The rational allocation of component tolerances requires the establishment of some rule for distributing the assembly tolerance among the components. Sections 3.1-3.6 discuss the rules available in TI/TOL.

3.1 Allocation By Proportional Scaling

The designer begins by assigning reasonable component tolerances based on process or design guidelines. The component tolerances are summed to see if they meet the specified assembly tolerance. If not, the component tolerances are scaled by a constant proportionality factor. In this way the relative magnitudes of the component tolerances are preserved.

This algorithm is demonstrated graphically in Figure 3.2** **for an
assembly tolerance Tasm, which is the sum of two component tolerances, T1
and T2. The straight line labeled as the **Worst Case Limit** is the
locus of all possible combinations of T1 and T2 which, added linearly, equal
Tasm. The parabola labeled **Statistical Limit** is the locus of root
sum squares of T1 and T2 which equal Tasm.

**Figure 3.2: Graphical interpretation
of tolerance allocation by proportional scaling. **

Suppose the designer chooses initial values for T1 and T2 based on typical
process tolerances for the two component parts. This combination is the
point labeled **Original Tolerances** in the figure. By drawing a line
from the origin through this point, then extending it until it intersects
the Worst Case Limit, the largest possible tolerances for T1 and T2 are
obtained, which satisfy the worst case condition and which still have the
original ratio of T1 to T2. Extending this line further, until it intersects
the Statistical Limit curve, new values for T1 and T2 are obtained which
satisfy the assembly tolerance limit by root sum squares.

Although Figure 3.2 illustrates 1-D tolerance accumulation models, the algorithm may be applied to 2-D or 3-D tolerance accumulation by pre-multiplying each component tolerance by its tolerance sensitivity.

The Proportional Scaling option does not exist explicitly in the TI/TOL 2D Analyzer. However, by leaving the weight factors at their default settings of 1.0, the Weight Factors allocation option (discussed in section 3.3) becomes equivalent to proportional scaling.

3.2 1-D Example: Tolerance Allocation by Proportional Scaling

The following example is based on the shaft and housing assembly shown in Figure 3.3.

**Figure 3.3. Shaft and housing assembly.
**

Dimensions **A**, **C** and **G** represent dimensions on vendor-supplied
parts, and already have tolerances associated with them. Initial tolerances
for parts **B**, **D**, **E**, and **F** are selected from tolerance
guidelines such as those illustrated in Figure 3.4[1]. Tolerances are chosen
from the middle of the range for each part size. This table is available in
the TI/TOL 2-D Analyzer's on-line Tolerance Reference Handbook.

**Figure 3.4. Tolerance range of machining
processes [1]. **

The retaining ring (**A**) and the two bearings (**C** and **G**)
supporting the shaft are vendor-supplied, hence their tolerances are fixed
and must not be altered by the allocation process. The critical clearance
is the shaft end-play, which is determined by tolerance accumulation in
the assembly. The vector diagram overlaid on the figure is the assembly
loop that models the end-play. The average clearance is the vector sum of
the average part dimensions in the loop:

**Initial Tolerance Specifications**

Required Clearance = .020+/-.015

Average Clearance = -A + B - C + D - E + F - G

= -.0505 + 8.000 - .5093 + .400 - 7.711 + .400 - .5093

= .020

__Dimension A B C D E F G__

Nominals .0505 8.000 .5093 .400 7.711 .400 .5093

Tolerances(+/-)

Design .008 .002 .006 .002

Fixed .0015 .0025 .0025

The clearance tolerance is obtained by summing the component tolerances by worst case:

TSUM = + TA + TB + TC + TD + TE + TF + TG

= + .0015 + .008 + .0025 + .002 + .006 + .002 + .0025

= .0245 (too large)

Now, solving for the proportionality factor:

TASM = .015 = .0015 +.0025 +.0025 + P (.008 + .002 + .006 + .002)

P = .47222

Note that the three fixed tolerances were subtracted from the assembly tolerance before computing the scale factor. Thus only the four design tolerances are re-allocated:

TB = .47222 (.008) = .00378 TE = .47222 (.006) = .00283

TD = .47222 (.002) = .00094 TF = .47222 (.002) = .00094

Each of the design tolerances has been scaled down to meet assembly requirements as shown in Figure 3.5. This procedure could also be followed assuming a statistical sum for the assembly tolerance, in which case the tolerances would be scaled up. Results are summarized in Table 3.1.

**Figure 3.5. Tolerance allocation by
proportional scaling. **

3.3 Allocation by Weight Factors

A more versatile method of assigning tolerances is by means of weight factors. Using this algorithm, the designer assigns weight factors to each tolerance in the chain and the system distributes a corresponding fraction of the tolerance pool to each component. A larger weight factor for a given component means a larger fraction of the tolerance pool will be allocated to it. In this way, more tolerance can be given to those dimensions which are the more costly or difficult to hold, thus making the design more robust.

Figure 3.6** **illustrates this algorithm graphically for a two component
assembly. The original values for component tolerances T1 and T2 are selected
from process considerations and are represented as a point in the figure,
as before. The tolerances are scaled, similar to proportional scaling, only
the scale factor is weighted for each component tolerance so the greater
scale factors yield the least reduction in tolerance.

**Figure 3.6. Graphical interpretation
of tolerance allocation by weight factors. **

Again, although the figure illustrates 1-D tolerance accumulation models, the algorithm may be applied equally well to 2- D and 3-D stacks. It may also be applied to worst case, statistical, or six sigma tolerance sums.

Note that any components which are vendor-supplied or subject to other design considerations can be excluded from the allocation process by declaring them to be "fixed".

3.4 1-D Example: Tolerance Allocation by Weight Factors

The shaft and housing assembly of section 3.2 will be revisited, using weight factors to allocate tolerances. The three tolerances A, C and G are fixed. Design tolerances B, D, E and F are assigned weight factors of 10, 20, 10 and 20, respectively. Weights were determined on the basis of machining difficulty.

The clearance tolerance is obtained by summing the component tolerances by worst limits:

TSUM = + TA + TB + TC + TD + TE + TF + TG

= + .0015 + .008 + .0025 + .002 + .006 + .002 + .0025

= .0245 (too large)

Solving for the proportionality factor with weight factors:

Ti' = PWiTi

TASM = .015 = .0015 +.0025 +.0025 + P [(10)(.008) + (20)(.002) + (10)(.006) + (20)(.002)]

P = 2.31818

Note that only the four design tolerances are re-allocated:

TB = 2.31818(10/60)(.008) = .00309 TE = 2.31818(10/60)(.006) = .00232

TD = 2.31818(20/60)(.002) = .00155 TF = 2.31818(20/60)(.002) = .00155

3.5 1-D Example: Statistical (RSS) Allocation by Weight Factors

The clearance tolerance is obtained by summing the component tolerances by root sum squares:

TSUM = R(TA

^{2}+ TB^{2}+ TC^{2}+ TD^{2}+ TE^{2}+ TF^{2}+ TG^{2})= R(.0015

^{2}+ .008^{2}+ .0025^{2}+ .002^{2}+ .006^{2}+ .002^{2}+ .0025^{2})= .01108 (too small)

Solving for the proportionality factor with weight factors:

Ti' = PWiTi

TASM

^{ = }R(TA^{2}+ P^{2}WB^{2}TB^{2}+ TC^{2}+ P^{2}WD^{2}TD^{2}+ P^{2}WE^{2}TE^{2}+ P^{2}WF^{2}TF^{2}+ TG^{2})P = 7.57238

Note that the three fixed tolerances are not re-allocated:

TB = (7.57238)(10/60)(.008) = .01010 TE = (7.57238)(10/60)(.006) = .00757

TD = (7.57238)(20/60)(.002) = .00505 TF = (7.57238)(20/60)(.002) = .00505

All of the preceding examples of allocation are compared in Table 3.1. A graphical comparison is shown in Figure 3.7.

**Table 3.1 Comparison of Allocation
Methods **

Proportional |
Weight Factors |
|||||

Original Dim |
Worst Case | Stat ±3 | Weight Factor | Worst Case | Stat ±3 | |

A | .0015* | .0015 | .0015 | 0 | .0015 | .0015 |

B | .008 | .00378 | .01116 | 10 | .00309 | .0101 |

C | .0025* | .0025 | .0025 | 0 | .0025 | .0025 |

D | .002 | .00094 | .00279 | 20 | .00155 | .00505 |

E | .006 | .00283 | .00837 | 10 | .00232 | .00757 |

F | .002 | .00094 | .00279 | 20 | .00155 | .00505 |

G | .0025* | .0025 | .0025 | 0 | .0025 | .0025 |

Assembly Tolerance | .0150 | .0150 | .0150 | .0150 | ||

Scale Factor (P) | .47222 | 1.39526 | 2.31818 | 7.57238 |

*Fixed tolerances

**Figure 3.7. Tolerance allocation by
proportional scaling and weight factors **

3.5 Tolerance Allocation Using the SSA Model

- When used for allocation purposes, the Six-Sigma Assembly Drift model does
not operate on the same assumptions as the RSS and SSC models. The RSS and
SSC models widen or tighten the component tolerances until the user-specified
quality level (acceptance fraction) is achieved. The SSA model is primarily
intended for analysis. Since it is based directly on the RSS analysis model,
when used for allocation purposes, it yields the same combination of component
tolerances as the RSS model would. The calculated reject fraction shown for
SSA will be much higher, though.

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