3 Replies Latest reply on Jul 6, 2017 10:18 PM by David Hooper

    Simulation edge weld results vs. Throat Shear Method

    Casey Kluesener

      Hi all,

      This is longer than it needs to be, so bear with me

       

      I am trying to model some welds and wanted to do a check of the SW edge weld connector results with the throat shear method before I continued on.

      I used the throat shear method outlined in this presentation: http://www.eng.uwo.ca/designcentre/FEA%20resources/116_Welds.pdf

      Here is the help page describing the formulas Solidworks uses: 2015 SOLIDWORKS Help - Weld Size Calculations (American Standard)

       

      I performed a simulation of two flat plates welded together with a double sided fillet weld. I did one simulation where the flat plate was joined using the double sided fillet weld connector, and another simply bonding the two plates and using splitlines to align the nodes. My fillet weld connector is a custom steel with a shear allowable of 40000 psi and a safety factor of 1. Both models were shell meshed with draft quality, and one to one node correspondence at the joint. Equal loading conditions and fixtures were applied to both models.

      With the throat shear method, I am getting a maximum weld size of about 1.60e-5 in, and using the Edge Weld connector I get a maximum weld size of about 1.65e-5 in. So this looks good!

       

      Now the problem is...

      When I do the throat shear method, I am basically extracting the loads at each node, dividing that by the "nodal length" (the total length of element acting on the node) to get the load per unit length. When I divide by the nodal length, I get a force per unit length unit (for Normal, Shear, and Bending), which is dependent obviously on element size.

      The total normal load going through the weld joint is 1.25 lbs

      The throat shear method, in the normal direction, gives me an average force per unit length of .25 lbf/in. The weld joint is 5 in long, so 5 in x .25 lbf/in is 1.25. So that makes sense.

      The solidworks weld, in the normal direction, gives me an average force per unit length of 11 lbf/in. 11 lbf/in x 5 in = 55 lbf, which to me doesn't make any sense.

       

      So, how exactly is the solidworks weld loading calculated? It must be correct, because I am getting the same weld size for both methods, just different loading.

       

      Attached are screenshots of my results