4 Replies Latest reply on Oct 24, 2011 4:59 AM by Jason Richardson

    Finding the load limit for a component?

    Jason Richardson

      I'm having difficulties finding the force which is required to make a shear pin break in a weak-link chain.

       

      I have the simulation all up and running and the von mises FOS is coming out at 0.3 when I know the modelled component has been tested to break at this load.

       

      So I guess the question is, how do I know what load will rip the shear pin appart? There's some pretty large deformation shown. Do any points with a FOS below 1 result in failure?

       

      Appologies for the vagueness of the question. It's my first week of work and the first time I've used solidworks in a few years.

        • Re: Finding the load limit for a component?
          Ryan Werner

          Hi Jason,

           

          What is the method of failure for your pin?  Is using the von Mises FOS the correct one?  If the pin is likely to fail in shear than perhaps the Shear FOS will give you more accurate results.  Just a thought.

           

          Ryan W.

          • Re: Finding the load limit for a component?
            David Maxham

            Just a guess, but maybe your constraints need to be re-checked, especially between the shear pin and the links.

            • Re: Finding the load limit for a component?
              Stuart Moore

              Why would you use simulation for this? all you need is the cross sectional area of the pin and the shear strength of the material and it's a simple calculation.  It would be quicker than setting-up a simulation.

              • Re: Finding the load limit for a component?
                Jason Richardson

                I understand the hand calculations are easy enough, and were certainly accurate when a destructive test of said shear pin was conducted. That doesn't take away from the fact that I'd like to obtain similar results from the solidworks simulation, so I can be sure of results from it in future - Particularly considering the cost of each destructive test being approx. £3000.

                 

                I've provided a copy of the part I'm trying to test for. The two sections penultimate to the ends are the areas where the force can be considered to be applied - in opposing directions (although not significant for modelling purposes). Both of these areas also receive approximately half of the applied load. The rest of the part can be considered fixed.

                 

                The material used is 817M40'T' and has been tested to have a yield of 683MPa and a UTS of 971MPa.

                 

                Hand calculations and the destructive test indicate that failure will occur at around 200,000Kg, with 100,000Kg being applied to each of the described faces.

                 

                I would be very appreciative of any indicaters of where I'm going wrong.

                 

                Thanks in advance,

                 

                Jason

                 

                Edit - It appears the 2nd principle stress gives the results expected. With it reaching the UTS at around 201,000Kg total load. I'm a little confused as to why it was the 2nd principle stress which gives the correct results. It also doesn't seem possible to obtain any FOS plots for the P2 Stress. Any help with regards to that, or the reasoning why I should have known to chose the 2nd principle stress for attaining the maximum load the pin could support would be helpful.