Fluid_Monkey
Mechanical
- Oct 28, 2020
- 5
Hello All,
I wanted to know how to work out the Permissible Shear Stress of Mild Steel having Yield Stress (YS) of 230 MPa and Ultimate Tensile Stress (UTS) of 410 MPa. Do you simply do 0.577xYS or do I need to consider a factory of Safety over and above this? 0.577 is 1/√3 and taken from Von Mises Criteria. Some websites seem to suggest that you need to take 0.577xUTS and then divide this by a reasonable Factor of Safety. Any help would be very much appreciated. I have attached the mechanical properties table for the material in question which is a low carbon mild steel.
I wanted to know how to work out the Permissible Shear Stress of Mild Steel having Yield Stress (YS) of 230 MPa and Ultimate Tensile Stress (UTS) of 410 MPa. Do you simply do 0.577xYS or do I need to consider a factory of Safety over and above this? 0.577 is 1/√3 and taken from Von Mises Criteria. Some websites seem to suggest that you need to take 0.577xUTS and then divide this by a reasonable Factor of Safety. Any help would be very much appreciated. I have attached the mechanical properties table for the material in question which is a low carbon mild steel.