A technique to predict the effects of tolerance stack-up on assembly forces and stresses combines a rigid body assembly model with a finite-element analysis.

**KARL MERKLEY**

KEN CHASE

Brigham Young University

Provo, Utah

KEN CHASE

Brigham Young University

Provo, Utah

**EDITED BY MARTHA K. RAYMOND**

Tolerance variation in assemblies results from three sources: size, form, and kinematics. Size variation comes from inconsistent dimensional length. Form variation is due to geometric differences such as flatness or cylindricity. Kinematic variation is produced by small adjustments between mating parts in response to dimensional and geometric irregularities. During assembly, tolerances in each part can accumulate to create large residual stack-ups, inducing poor product performance and high cost.

Flexible assemblies containing nonrigid parts, produce an additional complexity to tolerance prediction. These assemblies, constructed of thin sheet metal, fiberglass, plastics, or rubber, can deform substantially from their geometrical shape by the nature of their weight or flexibility or by assembly forces applied during the manufacturing process.

Several tolerance models account for the stack-up problem in assemblies with rigid components and point to a variety of solutions. One method, however, has been applied early in the design process and calculates stresses in rigid assemblies consisting of flexible parts.

**BUILDING TOLERANCE MODELS**

One tolerance model developed at Brigham Young University, based on a Direct Linearization Method, uses a linear approximation of nonlinear assembly functions. Even simple assemblies, such as a pin inserted in a hole, are described by complex linear equations. DLM has less than 1% error in predicting the effects of tolerance stack-up for a wide range of assemblies. Another advantage of the method is that it frames the tolerance analysis problem in familiar terms such as datums, vectors, and kinematic joints. For a complete rigid body tolerance analysis, the DLM modeler is available in TI/TOL 3D+ from Texas Instruments. This package is integrated with the parametric capabilities of Pro/Engineer, a solid-modeling system from Parametric Technologies Corp.

The software also handles functions such as estimating the number of rejects in a set of assemblies and indicating critical component dimensions. It also indicates areas prone to gaps or interferences and gives a statistical variation for these values.

Though these tools are useful for dealing with purely geometric quantities, they don’t consider inherent flexibility of parts, a concern when using thin-shell components. A slightly warped airplane skin, for example, can be riveted in place, creating residual stresses. These stresses, especially in composite parts, may lead to failure. A technique called linear contact analysis determines the probability of failure due to specified tolerance variations.

This method calculates contact stresses in parts such as press-fit cylinders using finite-element methods, and requires contact over entire mating surfaces. The points where the mating surfaces contact are described as the nodes in a model. The technique includes several assumptions. First, small geometric variations in a part create insignificant changes in the part stiffness. In addition, friction between the parts is negligible and the material in the assembly behaves linearly.

A demonstration of the technique starts by bringing two parts into contact, for example bolting two pieces of sheet metal together, to create interference or close an assembly gap, *δ _{0}*. The gap is the sum of the displacements of the individual parts to their final equilibrium point such that:

*δ _{0} = δ_{a} + δ_{b}*

where *δ _{a}* and

*δ*are the deformations required to close the gap.

_{b}Applying Hooke’s law to each part:

Closing the gap brings the forces in each spring to equilibrium such that:

*F _{a} = F_{b} = F_{eq}*

By combining equations, δ* _{0}* is defined as:

Because the gap is the known quantity, its value is substituted to solve for Because the gap is the known quantity, its value is substituted to solve for * F _{eq}*:

In stress analysis, displacement carries the most significant. Substituting force equations into displacement equations, calculates displacements of individual components as described by:

The displacements of *δ _{a}* and

*δ*depend on the stiffness of the mating parts.

_{b}To design bolted or bonded joints in assemblies, extend the results to a matrix formulation describing the closure deformations such as:

where {*δ _{a}*}, {

*δ*}, and {

_{b}*δ*} are displacement vectors and Ka and Kb are stiffness matrices. Solving three matrix inversions, however, is numerically tricky, even without iterations or convergence difficulties. A simpler method condenses these matrices so that the values of the elements are numerically reduced to terms called super-elements. Dealing with smaller matrices is simpler, and this process makes the stiffness matrices the same size so that they can be added or multiplied.

_{0}Consider a linear assembly of springs with subassemblies A and B. This is a one-dimensional series spring problem that easily solves without software. The spring lengths are subject to random manufacturing variations. Each spring in subassembly A has an unloaded length of *L*⁄4±0.01 and each spring in subassembly B has an unloaded length of *L*⁄3±0.01. Assume each tolerance *T* represents a 3ο variation in length. Add the component tolerances by a root-sum-squares technique to obtain a statistical estimate of each subassembly’s variation. The statistical assembly variation due to the accumulated component tolerances is:

The gap variation between the free length of the two spring subassemblies is defined by the root-sum-squares of the variation of each subassembly so that:

If the nominal lengths of A and B are equal, gap variation is the maximum probable difference between the ends of the springs due to manufacturing errors. Springs with unequal nominal lengths require additional calculations based on an average gap.

For example, two sets of springs might be assembled by stretching one and compressing the other until they are the same length. This deformation results in a spring force that depends on gap size. The equilibrium position of the combined assembly, must be a function of spring stiffness. When the springs act as simple truss elements, the stiffness matrices for each assembly are:

The condensed super-element matrices are described by:

These results may be substituted into the closure equations:

Because subassembly A has more components and a larger variation than subassembly B the resulting deformations represent a statistical variation in the springs’ displacement in each subassembly.

The assembly force required to close the gap is proportional to the statistical deformation of the springs. The nominal force in each spring subassembly is zero when the ideal length of each subset is equal to *L*. However, accounting for the 3δ variation in the lengths induces an equilibrating force of:

in each subassembly.

This analysis can be applied to assemblies consisting of flexible parts, such as blocks press-fit inside a base. MSC/Patran and MSC/Nastran from MacNeal-Schwendler Corp. calculates solutions. MSC/Patran defines a finite-element model and super-elements. The super-element stiffness matrices are assembled using MSC/Nastran and matrix manipulations are performed with Matlab from The MathWorks.

As a second example, consider three blocks that fit inside a base. By describing the parts as rigid bodies, the solution uses assumptions for a one-dimensional tolerance stack-up problem. However, when considering deformations, the problem becomes two dimensional.

The assembly’s nominal dimensions are such that the blocks are press-fit into the base. The nominal interference is 0.025 in. with a tolerance no more than ±0.025 in. Summing the nominal and variable interferences gives a range of 0 to 0.05 in. The parts have a yield strength of 70 ksi and are required to be assembled without yielding. The problem is to determine the number of rejected parts.

First, constrain the FE model in the *X* and *Y *directions at the midpoint of the base. The rest of the base is on rollers, and rollers are also between the blocks and base to allow kinematic adjustments. Assume that the friction is very high between the mating surfaces on the left and right edges so that the nodes on the vertical edges will move together. The blocks are thick relative to their length and width, so the problem is modeled with plane strain finite elements.

In the model, the blocks fit exactly inside the base. Nominal dimensions are used to create the finite-element model since small variations in geometry don’t change part stiffness.

Once the displacements along the boundaries are known, the interior stresses and displacements can be calculated for each part.

The pressure from the interference fit causes the base to expand horizontally with the greatest deflection in the side rails. The blocks contract horizontally with the greatest contraction near the base.

However, the peak stress is 102 ksi, which exceeds a 70-ksi design limit. In fact, the material exceeds the plastic limit and can never reach 102 ksi. Because the analysis describes deformations in the linear range, a simple scale factor of 70/102 applied to the maximum interference reduces maximum stress. All stresses and displacements throughout the assembly are reduced by the same factor.

Assuming interference has a normal distribution and the stress/strain relationship is linear, the simple scale factor may be applied to the statistical stress area.

As the interference varies from 0 to 0.05 in., maximum Von Mises stress in the assembly varies from 0 to 102 ksi. The mean Von Mises stress is 51 ksi and corresponds to the mean interference of 0.025 in. Above 70 ksi, the distribution for the maximum stress is unknown from the linear analysis due to discontinuity at the yield limit.

The following example shows how the model predicts failure rate of assemblies. When the stress distribution about the mean has a 3s variation, calculate the number of standard deviations corresponding to 70 ksi from:

where z is the z-score of a normalized distribution.

For normal distributions, this value corresponds to 86.86%, which means that 13.14% of all assemblies will exceed the design limits and be rejected due to assigned tolerances.

Though material properties extend into the plastic region, the model uses linear results to predict the number of parts exceeding the yield strength. These design constraints are typical for nonlinear analysis in assemblies where parts such as bolts are stressed almost to the yielding point or where composite members are functional while the interlaminar shear does not induce delamination.

For future applications, this technique is being verified on real world problems with multiple elements and two and three-dimensi onal tolerance variations.