AnsweredAssumed Answered

Simulation parameters for shear pin failure?

Question asked by Jason Richardson on Oct 27, 2011
Latest reply on Oct 27, 2011 by Jason Richardson

I am really having difficulties finding the value for which a shear pin will fracture and fail. I've no doubt that it's down to my own inexperience with solidworks simulations and possibly a gap in knowledge with regards to shear stress and failure criterion.


I have attached the part I want to analise and would appreciate anyone explaining their method to how they would go about obtaining accurate results.


The pin is made from 817M40'T' material which has been tested to have a UTS of 971MPa and a Yield of 683MPa.


Equal force can be considered to act on each of the shear faces which have 5mm grooves machined into the sides of them. The rest of the part can be considered fixed.


Possible areas I'm going wrong:

1. Imputting incorrect material properties - For tensile strength I am inputting the UTS, unsure about that.

2. Not all material values are known - In particular the shear modulus, I'm using 80GPa which I believe is roughly correct (Although I'm unsure if this is necessary for the simulation?) For Poissons ratio I'm using 0.3, and a Elastic modulus of 205GPa - Which again I believe correct but am unsure of their necessity in calculating failure force.

3. Not having correct failure criterion.


Any help from someone with a little more experience would be greatly appreciated and I would like to appologise in advance for my gaps in knowledge.