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