howkers
Mechanical
- Nov 4, 2003
- 29
I was asked to check the torsional stress in the wire of a spring and plugged the numbers into the standard equation from the text books:
shear stress = (8 x load x mean diameter x w-factor)/(pi x wire-diameter cubed)
I was surprised to get a result which exceeded the shear ultimate stress for the solid length and the estimated shear yield at the working length. I've done the calc in both metric and imperial units.
These springs have been in service for a number of years and we have had no experience of either breakage or permanent set. Can anyone explain this?
shear stress = (8 x load x mean diameter x w-factor)/(pi x wire-diameter cubed)
I was surprised to get a result which exceeded the shear ultimate stress for the solid length and the estimated shear yield at the working length. I've done the calc in both metric and imperial units.
These springs have been in service for a number of years and we have had no experience of either breakage or permanent set. Can anyone explain this?