 # Two Discussions on Cantilevered Snap Fit and Calculating Max Shear Stress Q: I designed a cantilevered snap fit that extended as a continuation of a wall, but some of the prototypes cracked at the snap finger barb. Usually, the highest stresses are at the base of the cantilever. So why did the snap finger crack where it did? A: The snap finger would likely have cracked at its base if it were extending from a flat surface rather than as a continuation of a wall.  In this case, because it was extending from a wall there was no stress concentration at the base,…

Q: I designed a cantilevered snap fit that extended as a continuation of a wall, but some of the prototypes cracked at the snap finger barb. Usually, the highest stresses are at the base of the cantilever. So why did the snap finger crack where it did?

A: The snap finger would likely have cracked at its base if it were extending from a flat surface rather than as a continuation of a wall.  In this case, because it was extending from a wall there was no stress concentration at the base, and the largest stress concentration occurred at the barb on the snap finger.

There are several things that can be done to redesign the snap fit. First, add as much radius to the barb notch as possible so as to reduce the stress concentration. Also, make the snap finger longer so that less strain builds up in the snap finger during deformation at assembly. If you can, provide a mechanical stop behind the snap finger so that it cannot be deformed beyond its yield point. Make sure to review the notch sensitivity of the plastic to see if there are alternate materials that are less prone to failure in this application.

Q: I calculated the max shear stress in a rod and then checked it with an FEA simulation. It was a straight .125″ diameter rod subjected to 35 in-lb of pure torque. The standard torsion equation I used was: Max Shear Stress = 16 * Applied Torque / (pi * rod diameter^3). The value I got from the torsion calculation was 91 KSI but the analysis indicated a max stress of 158 KSI. The analysis was just a simple rod fixed on one end and with the torque applied close to the opposite end. The 158 KSI value was from the outer surface in the middle, away from either end. Somehow I must have set up the analysis wrong. Any ideas on how to do it correctly? The next step is to add a flat to the shaft and I want to make sure the analysis is set up correctly before including features that cannot be checked with simple calculations.

A: Your FEA analysis is probably displaying results as von Mises equivalent stress. In the von Mises failure theory (a.k.a. the Distortion Energy Theory) the tensile, compressive and shear stresses acting at a point are run through a series of equations to combine and simplify them into one equivalent stress that can be compared to the published tensile strength data. This is done because typically the material’s tensile strength that is found in manufacturer’s literature is based on tensile tests done with all of the force applied along one axis, such as in an Instron tensile testing machine. In real life situations, there could be forces acting in two or even three directions (axes), plus torsional shear stresses.

The Distortion Energy Theory draws a distinction between those forces that tend to change a part’s volume and those forces that tend to change, or distort, its shape. The basic premise of the theory is that the material will break when the distortion stresses exceed the yield strength of the material. Shear stress has more of a tendency to cause the part to distort than tensile stress, so this is taken into account in the equivalent stress equations. The equations for the equivalent stress can be found in good machine design texts. When only shear stress is involved, the equation simplifies to: equivalent stress = 1.73 * shear stress.

In your case, the equivalent stress was: 1.73 * 91 KSI = 157.6 KSI. 