Hello All,

We are looking to understand the Spring Connection with and without damping in a nonlinear dynamic study. We are using 2012 Sp4 or Sp5...the final Sp. Anyway, I am now looking at a simple model to understand the Spring Connection. The model consists of a thick plate 12 in x 12 in x 2in with a smaller 2 in^(cubed) block positioned 2 inches above the plate and centered. The larger plate is fixed and the smaller block is constrained only to allow motion along the gravity vector. Each part has a reference point in the center of the opposing faces. The Spring Connection is applied between the two points thus aligning the spring with the gravity vector. Values of the spring are chosen such that when gravity is applied, as a constant function in time, the resulting displacement of the smaller block should be 0.25 inch (compression) or there about with no pre-load. However, the results suggest the displacement is 0.5 inch! Here is the linear model I assume for the spring, Axial Stiffness is 9.017 lbf/in. The steel small block weight is 2.254 lbf. Axial Spring Force = Axial Spring Stiffness times the Axial Displacement. That is, F = ky. If I am not mistaken, F/k, for the values above, is more equal to 0.25 inch than the reported results. See images for setup and the above stated response.

With a pre-load, the resulting displacement is zero or there about. The pre-load is set to be equal to that of the weight of the small block. And thus the results make more sense. See the next set of images...

A delta force is then considered and applied to the smaller block with a direction the same as the gravity vector. The force magnitude is chosen such that the initial force plus the delta force will cause a 0.25 inch displacement. That is Fo + dF = k(yo + dy). I should also let you know the force is constant through time. So, Fo = Mass * Gravity or ma and dF = the additional applied force chosen to be 2.254 lbf. The spring is still defined as k = 9.017 lbf/in with a pre-load of 2.254 lbf. Now, Fo + dF = 4.508 lbf and dividing the total force by 9.017 lbf/in suggest a displacement of 0.499 inch. And, that all makes sense yet there is a pre-load defined for the spring accounting for half of the force and thus should account for half of the displacement. Please help me sort out my math if I am in error. See the results...

So, in the end I have some questions:

- Why is the time history of the displacement results sinusoidal?
- Yes, the averages give the expected linear result but why if the input forces as a function of time are constant?

- The Damping value (not discussed above) appears to have no effect when the forces are not constant in time. Why?
- The connector details are reporting the spring units are not honored. Is this a bug affecting the functional results?

The fundamental reason why we need to use springs is to emulate the effect of compliant, damped suspension. We are not interested in the suspension response so modeling the nonlinear material data is not a reasonable work around. The spring connections are described in the SW help as the preferred method of choice for this type of simulation.

Any help resolving the above issues is greatly appreciated and of course, time is of the essence.

Thanks again,

-Stewart

P.S. Wesley posted a similar question in the past but there appears to be no responses...

*"Wesley Morgan wrote:*

* *

*When I create a point to point spring link with some value of damping, the damping value resets to zero when I go back to "edit definition." This happens for non-linear dynamic studies. I'm guessing that damping is never applied in the first place, anything I test does not appear to be damped. Any help or commentary would be appreciated. I'm using SolidWorks Simulation Premium 2011. "*

The answer to this issue is that my observations are correct...springs do not work as intended in non-linear simulation. SolidWorks has created SPR 710092 to address this issue. The solidWorks developers agree this behavior is incorrect and will be addressed.