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Example Problems	AutoCAD Verification: STACK BLOCKS
Home : Example Problems : AutoCad - Verification - Stack Blocks

Figure 3.1: Schematic of the stack blocks and corresponding 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 dependent 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 use the Six-Sigma analysis model with weight-factor allocation to adjust the non-fixed component tolerances until the +/-3[[sigma]] calculated assembly variation corresponds to the specified assembly limits.

3.4 Part Names and DRFs

Figure 3.2: Diagram showing the location of the part DRFs.

Remarks>> Datum reference frames (DRFs) should correspond to locations on the parts to which the component dimensions are referenced.

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 3.4 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>> In order to generate loops identical to the ones shown, first manually create loop 3. The remaining two loops can then be created using the automatic loop generation option.

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. If CATS does not automatically make those vectors equivalent, it must be done manually or the analysis results will be invalid.

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 5habe a single geometric tolerance applied because only the flatness of the block affects the assembly variations.

3.8 Sensitivity Matrices

Constraint Sensitivities

A Matrix

	B_C	F	G	I	J	K
X1	1.2635	0.0000	0.0000	0.0000	0.0000	0.0000
Y1	-.96467	0.0000	0.0000	0.0000	0.0000	0.0000
1	0.0000	0.0000	0.0000	0.0000	0.0000	0.0000
X2	0.0000	1.0000	0.0000	-.26347	0.0000	0.0000
Y2	0.0000	0.0000	1.0000	.96467	0.0000	0.0000
2	0.0000	0.0000	0.0000	0.0000	0.0000	0.0000
X3	0.0000	-1.0000	0.0000	0.0000	1.0000	0.0000
Y3	0.0000	0.0000	-1.0000	0.0000	0.0000	1.0000
3	0.0000	0.0000	0.0000	0.0000	0.0000	0.0000

B Matrix

	A	D	E	H	L	q1	q2	q3	q4
X1	0.0000	-.96467	0.0000	0.0000	0.0000	-18.718	10.048	0.0000	0.0000
Y1	1.0000	-.26347	-1.0000	0.0000	0.0000	6.6200	0.0000	0.0000	0.0000
1	0.0000	0.0000	0.0000	0.0000	0.0000	-1.0000	1.0000	0.0000	0.0000
X2	0.0000	0.0000	0.0000	-.96467	0.0000	0.0000	10.048	4.0600	0.0000
Y2	0.0000	0.0000	-1.0000	-.26347	0.0000	0.0000	0.0000	-3.9050	0.0000
2	0.0000	0.0000	0.0000	0.0000	0.0000	0.0000	1.0000	1.0000	0.0000
X3	0.0000	0.0000	0.0000	.96467	-.96467	0.0000	0.0000	-4.0600	10.675
Y3	0.0000	0.0000	0.0000	.26347	-.26347	0.0000	0.0000	3.9050	-28.125
3	0.0000	0.0000	0.0000	0.0000	0.0000	0.0000	0.0000	-1.0000	1.0000

F Matrix

	1	2	3	4	5	6	7
X1	1.0000	1.0000	.26347	.26347	-1.0000	0.0000	0.0000
Y1	0.0000	0.0000	-.96467	-.96467	0.0000	0.0000	0.0000
1	0.0000	0.0000	0.0000	0.0000	0.0000	0.0000	0.0000
X2	0.0000	0.0000	0.0000	0.0000	-1.0000	-.26347	0.0000
Y2	0.0000	0.0000	0.0000	0.0000	0.0000	.96467	0.0000
2	0.0000	0.0000	0.0000	0.0000	0.0000	0.0000	0.0000
X3	0.0000	0.0000	0.0000	0.0000	0.0000	.26347	-.26347
Y3	0.0000	0.0000	0.0000	0.0000	0.0000	-.96467	.96467
3	0.0000	0.0000	0.0000	0.0000	0.0000	0.0000	0.0000

Tolerance Sensitivities

-B^-1A Matrix

-B^-1F Matrix

	1	2	3	4	5	6	7
A	.27312	.27312	1.0366	1.0366	-3.89E-16	.76903	.26759
D	1.0366	1.0366	.27312	.27312	-1.0366	-.35799	.35799
E	0.0000	0.0000	0.0000	0.0000	.27312	1.1270	-.09040
H	0.0000	0.0000	0.0000	0.0000	-1.0366	-.02590	-.24722
L	0.0000	0.0000	0.0000	0.0000	-1.0366	-.02590	-.24722
q1	0.0000	0.0000	0.0000	0.0000	1.49E-17	.03983	-.03983
q2	0.0000	0.0000	0.0000	0.0000	1.49E-17	.03983	-.03983
q3	0.0000	0.0000	0.0000	0.0000	-1.49E-17	-.03983	.03983
q4	0.0000	0.0000	0.0000	0.0000	-1.49E-17	-.03983	.03983

3.9 Predicted Assembly Variation

Table 3.4: Independent Variable Tolerances and Control Factors

Table 3.5: Kinematic Assembly Variables (No Geometric Tolerances)

Table 3.6: Geometric Tolerances

Feat.	Joint	Part Name	Feature Type	Tolerance Band	Char. Length
1	1	Ground	Flatness	.080 mm	N/A
2	1	Cylinder	Circularity	.020 mm	N/A
3	2	Cylinder	Circularity	.020 mm	N/A
4	2	Block	Flatness	.050 mm	N/A
5	3	Ground	Flatness	.080 mm	N/A
6	4	Block	Flatness	.050 mm	N/A
7	5	Block	Flatness	.050 mm	N/A

Table 3.7: Kinematic Assembly Variables (Geometric Tolerances Included)

Remarks>> As is apparent from the geometry of the stack blocks, all angular variations are identical.

Table 3.8: Six-Sigma Percent Contributions To A (No Geometric Tolerances)

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

Table 3.9: Six-Sigma Percent Rejects

Spec. Name	Spec. Type	Nominal Dimension	(+/-) Computed Variation	With Geometric Tolerances		Without Geometric Tolerances
	Dep. Length	18.718	.40018	Z	Rej.	Z	Rej.
ZASM = 3.000		USL 19.018	Upper Tail	2.25	12256	2.26	11962
(Rejects in PPM)		LSL 18.418		-2.25	12256	-2.26	11962

Remarks>>The Z-scores are significantly below the specified 3[[sigma]] limits. The component tolerances must be tightened to meet the specifications.

3.10 Tolerance And Nominal 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 3.10: Six-Sigma Weight Factor Tolerance Allocation (No Geometric Tolerances).

Assembly Specs.	Nom.	USL	LSL	+/- ZASM
Dep. Length (mm)	18.7182	19.0182	18.4182	3.000

Nominal Allocation

Nominal Allocation is used to estimate the dimension nominals that will center the assembly variable or match the upper or lower assembly variation with the upper or lower specification limit. The larger the weight factor assigned to a dimension, the more that nominal may be adjusted.

Table 3.11: Six-Sigma Nominal Allocation (No Geometric Tolerances).

Lower Specification Limit Justified.

Assembly Specs.	Nom.	USL	LSL	+/- ZASM
Dep. Length (mm)	18.7182	19.0182	18.4182	3.000