2 Replies Latest reply on Oct 27, 2011 8:23 AM by Jason Richardson

    Simulation parameters for shear pin failure?

    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.

       

      Thanks,

       

      Jason

        • Re: Simulation parameters for shear pin failure?
          John Sanchez

          So is it accurate to say that the pin will fail in the areas of the reduced cross section (The 5mm grooves you describe)? Maybe elaborate on the loading scenario a bit for us.

           

          It is pretty typical (for designing purposes) to assume that a cross section in shear will fail at about 70% of its UTS in tension. Now this is a pretty broad assumption considering the array of material properties that can affect these types of failures but it's a place to start.

           

          Could you approach this problem with hand calcs first?

            • Re: Simulation parameters for shear pin failure?
              Jason Richardson

              I've been able to establish a shear factor of 0.758 from a previous destructive test having used 0.7 initially. This was done using hand calcs.

               

              The shear pin indeed did fracture at its area of smallest cross section. The shaft is part of a weak-link in a chain, with the two shear areas being pulled apart in equal and opposing directions.

               

              Having played around quite abit, it seems the 2nd Principle Stress approaches the UTS at the failure load found using hand calculations. Does this have any significant and how would I have known this was the failure criteria?