Knowledge Answer

Why did a retaining ring fail by 'Shear' instead of 'Groove Deformation' in a tool steel assembly?

2026-06-16 FAQ

In assemblies where the housing/shaft is made of hardened tool steel ($HRC > 50$), the groove will not yield or 'roll'. Therefore, the full axial load is transmitted as a pure shear stress to the ring. The ring fails when the shear stress $\tau = \frac{P}{\pi \cdot D \cdot t}$ exceeds the shear strength $\tau_{ult}$ of the ring material. This is a sudden,...

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In assemblies where the housing/shaft is made of hardened tool steel ($HRC > 50$), the groove will not yield or 'roll'. Therefore, the full axial load is transmitted as a pure shear stress to the ring. The ring fails when the shear stress $\tau = \frac{P}{\pi \cdot D \cdot t}$ exceeds the shear strength $\tau_{ult}$ of the ring material. This is a sudden, brittle-looking failure. Troubleshooting involves verifying the load $P$; if the load was within limits, the ring material may have had 'inclusion' defects or was 'over-tempered' (too brittle). For these rare cases, increasing the ring thickness $t$ is the only mechanical solution.

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