CHAPTER 3: TOLERANCE ALLOCATION
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2D Analyzer: CHAPTER 3: TOLERANCE ALLOCATION | Manuals |
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.
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.
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.
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".
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
The clearance tolerance is obtained by summing the component tolerances by root sum squares:
TSUM = R(TA2 + TB2 + TC2 + TD2 + TE2 + TF2 + TG2)
= R(.00152 + .0082 + .00252 + .0022 + .0062 + .0022 + .00252)
= .01108 (too small)
Solving for the proportionality factor with weight factors:
Ti' = PWiTi
TASM = R(TA2+ P2WB2TB2+ TC2+ P2WD2TD2+ P2WE2TE2+ P2WF2TF2+ TG2)
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
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