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Example Problems	PRO-E Verification: CHAPTER 3: STACK BLOCKS
Home : Example Problems : Pro-E 2D - Verification - Stack Blocks

Figure 3.1: Schematic of the stack blocks with dimension variables.

3.0 Problem Description

The stack blocks problem is an imaginary assembly used to teach tolerance analysis techniques. It consists of three parts: a base (ground), a sliding block, and a cylinder. The block slides on the base until it contacts the left side wall. The cylinder also contacts the left side wall.

Table 3.1: Manufactured Variables (Independent).

3.1 Design Requirements

Table 3.2: Assembly Variables and Specification Limits.

Remarks>> Three closed loops are needed to solve for all nine kinematic variables.

3.2 Modeling Considerations

• Cylinder radii B and C are not independent dimensions. They must be equivalenced for analysis.

3.3 Design Goal

The goal for this problem is to allocate the non-fixed component tolerances so the +/-3[[sigma]] assembly variation of A corresponds to the specified assembly limits.

3.4 Part DRFs And Feature Datums

Figure 3.2: Diagram showing the location of the part DRFs and feature datums.

Remarks>> Datum reference frames (DRFs) should correspond to locations on the parts to which the component dimensions are referenced. Feature datums are used to define the paths from the joint to the part DRFs. They allow the joint path vectors to correspond to dimensioned lengths.

3.5 Kinematic Joints

Five joints are required to model the stack blocks.

Figure 3.3: Kinematic joint diagram.

Remarks>> Cylindrical slider and edge slider joints both have a rotational and a translational degree of freedom, making ten total dependent variables. However, since joint 1 and joint 2 share the same cylinder center, the rotational degrees of freedom are reduced by one, leaving nine total degrees of freedom. This allows us to solve for this assembly's dependent variables with only three loops.

Table 3.3: Kinematic Joints of the Stack blocks.

3.6 Network Diagram, Vector Loops, and Design Specifications

The network diagram in figure 2.4 shows that three loops are necessary to describe the stack blocks assembly. A design specification has been applied to the dependent length A (vertical distance from the Ground DRF to the cylinder center).

Figure 3.4: Network diagram and loop diagram for the stack blocks assembly.

Remarks>> The autoloop generator will not create loops like those shown above. TI TOL allows the user to manually define the desired loop paths. However, as long as a valid combination of loops is chosen, the only difference between the results will be due to round-off error. The autoloop routine minimizes round-off error by choosing the shortest possible loop combination. In order to generate loops identical to the ones shown, first manually create loop 3. The remaining two loops can then be created using autoloop.

Loop 1 contains a pair of redundant vectors. Redundant vectors occur because each loop that passes though a part must pass through that part's DRF. The path from joint 2 to the block DRF ends with a vector between a feature datum and the block DRF. The path from joint 3 to the block DRF is a vector between joint 3 and the block DRF. These are actually the same vector, but with different endpoints. TI TOL does not automatically recognize that these vectors are equivalent, so the user must do it manually or the analysis results will be invalid. The vectors appear on top of each other, so it's necessary to use the 'query select' or 'menu select' option to find them.

Don't forget to equivalence the cylinder radii as well.

3.7 Geometric Tolerances

Seven geometric tolerances have been applied to the stack blocks assembly.

Figure 3.5: Geometric tolerance diagram.

Remarks>> In general, each joint will have one or two geometric tolerances applied to it. Joint 1 has two geometric tolerances because the roundness of the cylinder and the flatness of the ground both affect the assembly. Joints 4 and 5 have a single geometric tolerance applied because only the flatness of the block affects the assembly variations.

3.8 Sensitivity Matrices

Table 3.4: -B^-1A Matrix

Table 3.5: -B^-1F Matrix

	1	2	3	4	5	6	7
A	0.27312	0.27312	1.03663	1.03663	-3.89E-16	0.76903	0.26759
D	1.03663	1.03663	0.27312	0.27312	-1.03663	-0.35799	0.35799
E	0.00000	0.00000	0.00000	0.00000	0.27312	1.12702	-0.09040
H	0.00000	0.00000	0.00000	0.00000	-1.03663	-0.02590	-0.24722
L	0.00000	0.00000	0.00000	0.00000	-1.03663	-0.02590	-0.24722
q1	0.00000	0.00000	0.00000	0.00000	1.49E-17	0.03983	-0.03983
q2	0.00000	0.00000	0.00000	0.00000	1.49E-17	0.03983	-0.03983
q3	0.00000	0.00000	0.00000	0.00000	1.49E-17	0.03983	-0.03983
q4	0.00000	0.00000	0.00000	0.00000	-1.49E-17	-0.03983	0.03983

3.9 Predicted Assembly Variation

Table 3.6: Independent Variable Tolerances and Control Factors

Table 3.7: Kinematic Assembly Variables (Geometric Tolerances Not Applied)

Table 3.8: Normalized Sensitivities To A (Geometric Tolerances Not Applied)

Table 3.9: SSC Percent Contributions To A (Geometric Tolerances Not Applied)

Remarks>> The roller radius is the dominant cause of variation in A.

Table 3.10: SSC Percent Rejects (Geometric Tolerances Not Applied)

Table 3.11: Geometric Tolerances

Name	Part Name	Type	Joint	Tolerance Band	Char. Length
1	Ground	Flatness	1	0.08000	N/A
2	Cylinder	Circularity	1	0.02000	N/A
3	Cylinder	Circularity	2	0.02000	N/A
4	Block	Flatness	2	0.05000	N/A
5	Ground	Flatness	3	0.08000	N/A
6	Block	Flatness	4	0.05000	N/A
7	Block	Flatness	5	0.05000	N/A

Table 3.12: Kinematic Assembly Variables (Geometric Tolerances Applied)

Table 3.13: SSC Percent Rejects (Geometric Tolerances Applied)

Remarks>> Applying geometric tolerances to the tolerance model does not significantly increase the number of rejects. For allocation purposes, the geometric tolerance variations will not be included.

3.10 Tolerance Allocation

Weight Factor Tolerance Allocation

Weight Factor Tolerance Allocation adjusts dimension tolerances according to the user-assigned weight factors. The weight factors that are large compared to the others receive a greater portion of the unused variation when there is a positive variance pool (assembly variation is smaller than the specified assembly limits) and are reduced the least when there is a negative variance pool (assembly variation exceeds the specified assembly limits). The user should assign larger weight factors to the tolerances he wants to become (or remain) as large as possible.

Table 3.14: SSC Weight Factor Tolerance Allocation

(Geometric Tolerances Not Applied).

Table 3.15: WC Weight Factor Tolerance Allocation

(Geometric Tolerances Not Applied).

Table 3.16: SSA Weight Factor Tolerance Allocation

(Geometric Tolerances Not Applied).