Contact Convergence -- Solver Window

cwharpe

I've been reviewing Youtube's Mecway Channel and trying to recreate examples. Newbie Focus of study lately is Non-Linear Analysis. I've been trying to find more detailed explanations for the Analysis Settings buttons, and also for the Solver Window convergence graphs. Seems there often needs a bit of adjusting to the time step increments, choice for quasi-static run, etc. to arrive at a solution. But what determines a "successful" solution?

Judging from many of the examples this forum has offered, "successful" could be witnessed by the decaying convergence plots seen in the Solver Window. Many of your examples graph a deep dive as the Displacement Correction, Residual Force, no. of Contacts decidedly diminish over time.

I'm trying to figure out whether it's ever normal? for these plots to jig-jag and bounce all over the place (see attached pic), and what the corrective action should be. If I have overlooked prior threads to this subject, my apologies, but please post.

Victor

That looks normal. At each time step, it's solving a new static problem and convergence has to begin again, so the graphs jump back up to high values. Even with quasi-static turned off, there are still time steps quietly happening in CCX so it may still do this.

There's also a misleading feature of the graph in that very small values (at or near 0) of the displacement and force curves are excluded because they won't fit on the logarithmic axis. So the final iteration of a time step might not appear when it converges very well.

cwharpe

Thanks for the reassurance.

Regarding initial Contact Stiffness (C.S.) between two dissimilar materials, say E1=30e6 psi, E2=10e6 psi. Let E1 material be 2" tall, 3" wide, 4" long, and E2 material be 1" tall, 2" wide, 3" long. Using your formula of Contact Stiffness = (10*Young's Modulus)/thickness...

Which Young's Modulus is appropriate to apply, E1 or E2? And which thickness is appropriate (assuming in the normal direction of the contact plane)??

So, if E1, then C.S.=30e7/2" = 15e7 lb/in^3. Similarly, if E2: C.S. = 10e7/1" lb/in^3 ??

Lastly, I note several C.S.'s for pin/lug problems used the radius of the pin. I imagine a flat bar of nom. thickness curved around on itself ??

Victor

In your example, either value is OK because they're so similar. Anything within the same order of magnitude is basically equivalent. Same goes for radius vs diameter - it doesn't matter.

If the materials were really different (eg. rubber gasket on steel casting), probably use the smaller thickness and/or the higher Young's modulus so that the artificial contact stiffness is greater than the real stiffness of both parts.

The aim is to get the penetration, which is error, low enough that it's acceptable. You're making a trade-off between low penetration (high contact stiffness) and fast or successful convergence of the solver (low contact stiffness).

cwharpe

Thanks! This helps me begin at least within the ballpark of the successful solution.